MAth 150B exam 3

Theorem 11.2: Estimate of the remainder

Le tn be a fixed positive integer. Suppose there exists a number m such that |f(ⁿ+1)(c)|≤m, for all c between a and x inclusive. The remainder in the nth order taylor polynomial for f centered at a satisfies |Rₙ(x)|=|f(x)-pₙ(x)|≤M|x-a|^n+1/(n+1)!

Areas of regions in polar coordinates

Let R be the region bounded by the graphs of r=f(θ) and r=g(θ), between θ=α and θ=β, where f and g are continuous and and f(θ)≥g(θ)≥0 on [α,β]. The area of r is ∫(β:α)1/2f(θ)^2-g(θ)^2 dθ.

Taylor polynomials

Let f be a function with f',f'', ..., and fⁿdefined at a. The nth order Taylor polynomial for f with its center at a, denoted fₙ, has the property that matches f in slope, value, and all derivatives up to the nth derivative at a , that is, pₙ(a)=f(a),pₙ'(a)=f'(a),...,and pₙⁿ(a)=fⁿ9a). The nth order Tsylor polynomial centered at a is pₙ(x)=f(a)+f'(a)(x-a)+f''(a)/2!(x-a0^2+...+fⁿ(a)/n!(x-a)ⁿ. More compactly, pₙ(x)=Σ(n:k=0)ck(x-a)^k, where the coeffiecints are ck=f^k(a)/k!, for k=0,1,2,...,n.

slopes on a tangent line

Let f be dufferentiable functiob a θ₀. the slope of the line tangent to the curve r=f(θ) at the point f(θ₀,θ₀) is dy/dx=f'(θ₀)sinθ₀+f'(θ₀)cosθ₀), provided the denominator is nonzeroat the point .At anglescθ₀ for whichf(θ₀)=0,f'(θ₀)≠0 and cosθ₀≠0, the tangent line is θ=θ₀, with slope tan θ₀.

Arclenght of a polar curve

Let f have a continuous derivaative on hte intervel [α,β]. The arc length of the polar curve r=f(θ) on [α,β] is L=∫(β:α) √(f(θ))^2+f'(θ)^2dθ.

Taylor's theorem (remainder theorem)

Let f have continuous derivatives up to f(ⁿ+1) on an open interval I conataining a. For all x in I, f9x)=pₙ(x)+Rₙ(x),\ where pₙ is the nth order Taylor polynomial for f centered at and the remainder is Rₙ9x)=f(ⁿ+1)(c)/(n+1)! (x-a0^n+1, for some point between x and a.

remainder in a Taylor polynomial

Let pₙ be the Taylor polynomial order n for f. The remainder in using pₙ to approximate f at the point is Rₙ(x)=f(x)-pₙ(x)

maclaurin convergence for cosx

Show that the maclaurin series 1-x^2/2!+x^4/4!-x^6/6!+...=Σ(∞:K=0)(-1)^KX^2K/(2K)! converges to f(x)=cosx, for -∞<x<∞. Solution: Th show that the power of the series conerges to f, we must show that lim n→∞|Rₙ(x)|=0, for -∞<x<∞. Accoding to Taylor's theorem, with a=0, Rₙ(x)=f^n+1(c)/(n+1)! x^n+1, where c is between 0 and x. Notice that f^n+1(c)=±sinc or f^n+1=±cosc. In all cases, |f^n+1(c)|≤1. Therefore, the absolute value of the remainder term is bounded as |Rₙ(x)|=|f^n+1(c)/(n+1)! x^n+1|≤(x|^n+1/(n+1)!. Holding x fixed and using lim n→∞x^n/n!=0, we see that lim n→∞ Rₙ(x)=0 for all x. Therefore, the given power series converges to f(x)=cosx, for all x; that is, cosx =Σ(∞:k=0)(-1)^k/(2k)!x^2k. The convergnece of the Taylor series is illustrated.

Remainder in the maclaurin sereis for e^x

Show that the maclaurin series for f(x0=e^x converges to f(x) for -∞<x<∞. Solution: As shown in example 4, the Maclaurin sereis for f(x)=e^x is Σ(∞:k=0)x^k/k!=1+x+x^2/2!+...+x^n/n!+.., which converges for -∞<x<∞ In example 7, of section 11.1, it was shown that the remainder of Rₙ(x)=e^c/(n+1)1 x(n+1, where c is between 0 and x. Notice hat the intermediate point c aries wiht n, but it is always between 0 and x. Therefore, e^c is between e^0 and e^x; In fact, e^c=e^|x|, for all n. It follows that |Rₙ(x)|≤e^|x/(n+1)! |x|^n+1. Holding x fixed, we have lim n→∞|Rₙ(x)|=lim n→∞ e^|x\/(n+1)! |x|^n+1. =e^|x|lim n→∞ ||x|^n+1/(n+1) |x|^n+1=e^|x| lim n→∞ |x|^n+1/(n+1)!=0, we used the fact that lim n→∞ x^n/n!=0, for -∞<x<∞ (section 10.2). Because lim n→∞ |rₙ(x)|=0, it follows that for all real numbers x, the Taylor series converges for e^x, or e^x=Σ(∞:k=0) x^k/k!=1+x+x2/2!+...+ x^n/n!+.... The convergence of the Taylor series is illustrated in figure 11.20, where Taylor poynomials of increasing degree are graphed together with e^x.

TAylor/maclaurin series for a function

Suppose the function f has derivatives of all orders on an interval centered at the point a. Teh Taylor serie for f centered a is f(a)+f'(a)(x-a)+f''(a)/2!*x-a)^2+f'''(a)/3!(x-a)^3+...=Σ(∞:k=0)f^k(a)/K!(x-a)^k A Taylor series centered at 0 is called a maclaurin series.

Differentiating and integrating power series

Suppose the power series ∑ck(x-a)^k converges for |x-a|<R and defines a function for that interval: 1. Then f is differentiable for |x-a|<R and is found by differentiating the power sereis for f term by term; that is, f'(x)=∑kck(x-a)^k, for |x-a|<R. 2. the indefinite integral of f is found by integrating the power series for f term by term; that is, ∫f(x)dx=∑ck(x-a0^(k=1)/k+1 +c, for |x-a|<R, where c is an arbitrary constant.

combining power series

Suppose the power series ∑ckx^k an d∑dkx^k converge to f(x0 and g(x), respectively on an intervl I. 1. sum and difference: The power series ∑(ck+dk)x^k converges to f(x) and g(x) on I. 2. Multiplication by a power: Suppose M is an integer such that k+m≥0, for all terms of the power series x^m ∑ckx^k=∑ckx^(k+m). This series converges to x^mf(x) for all x≠0 in I. when x=0, the series converges to lim k→∞ x^m f(x). 3. Compositon: If h(x)=bx^m, where m is a positive integer and b is a nonzero real number, the power series ∑ck(h(x))^k converges to the composite function f(h(x)) for all x such that h(x) is in I.

Symetry in polar equatios

Symetry about the x-axis occurs if the point (r,θ) is on the graph whenever (r,-θ) is on the graph. Symetry about the y-axis occurs if the point (r,θ) is on the graph whenever (r,pi-θ)-(-r,-θ) is on the graph. Symetry about the origin occurs if (r,θ) is on the graph whenever (-r,θ)=(r,θ+pi) is on the graph.

matching polar and cartesian graphs

The butterfly graph is described by the equation r=e^sinθ-2cos4θ, for 0≤θ≤2pi, which is plotted in cartesian and polar coordiantes. Follow the cartesian graphs through tthe points A, B, C,...,N,O and mark the crresponding poiints on the polar curve. Solution: point A has the cartesian coordinates (θ=0,r=-1). Teh corresponding mark i the polar plot with polar coordinates (-1,0) is marked A. Point b on the cartesian plot is on the θ- axis; therefore, r=0. zaThe corresponding plot on the polar plot is the origin. Teh same arguement used to locate B applies to F,H,J,L and N, all of which appear in the origin in the polar plot. In general , the local and endpoint maxima and minima in the cartesian graph (a,C,D,E,G,I,K,M, and O) correspond to the extreme points oof the loops and are marked accordingly.

onuations of a line

The equations x=x₀+at, y=y₀+bt, for -∞<t<∞, where x₀,y₀, a and b are constants, with a≠0 describes a line with slope b/a passing through te point (x₀,y₀). If a≠0 and b≠0, the line is vertical.

absolute value of the reaminder

The error made in approximating f(x) by pₙ(x). Equivalently, we have f(x)=pₙ(x)+Rₙ(x), which says that f consists of two componets: the polynomial approximation and the associated remainder.

mystery series

The power series Σ(∞:k=0)(-1)^kk'4^k appeared in the opening section of 11.2. Determinine the interval of convergence of the power series and find the function it represents on this interval. Solution: Applying the ratio test, we determine that it converges when |x^2/4^k|, which implies that |x|<2. A quickm check of the endpoints of the origional series confirms that it diverges at x=±2. Therefore, the iterval of convergence is |x|<2. To find the function represented by the series, we apply several maneuvers until we obtain a geommwtric series . First note that Σ(∞:k=0)(-1)^kx^k/4&kx^2k=Σ(∞:k=0)k(-1/4)^kx^2k. The series ion the right is a geometric series because of teh resence of teh facor k. The key is to relaize that k could appear this way through differentiation; specifically, something like d/dx(x^2k)=2kx^2k-1. To achieve terms of this form, we write Σ(∞:k=0)(-1)^kk/4^kx^2k=Σ(∞:k=0)(-/4)^kx^2k=1/2Σ(∞:k=0)2k(-1/4)^2kx^2k=x/2Σ(Σ(∞:k=0)2k(-1/4)x^2k-1. Now we identify the derivative of another series: Σ(∞:k=0)(-1)^kx^2k/4^kx^2k=x/2Σ(∞:k=1)(-1/4)^kx^2k/2k-1=x/2Σ(∞:k=1)((-1/4)^kd/dx(x^2k)=x/2d/dx(-x^2/40^k). This last series is a geometric series with ratio r=-x^2 and first term -x^2/4; thereofore, its value is -x^2/4/1+(x^2-4), ;provided |x^2/4|<1, or|x|<2. We now have Σ(∞:k=1)(-1)^k k/4^k x^2k=x/2 d/dx( Σ(∞:k=1)(-x^2/4)^k)=x/2 d/dx(-x^2/4=k(x^2/4))=x/2d/dx(-x^2/4+x^2)=-4x^2/(4+x^2)^2. therefore, the function represented by the power series on (-2,2) has been has been uncovered. it s f(x)=4x^2/(4+x^2)^2. Notic that f is defined for -∞<x<∞, but its power series converges for f only at (-2,2).

Approximationg a real number using Taylor polynomials

Use polynomials of order n=0.,1,2, and 3 to approximate √(18).]Solution: letting f(x)=√x, we choose the center at a=16 because it is near 18 nd f and its derivatives are easy to evaluate at 16. The Taylor polynomials have the for pₙ(x)=f(16)+f'(16)(x-16)+f''(16)/2!(x-16)^2+...+fⁿ(16)(x-16)^n. We now evaluate the required derivatives : f(x)=√(x)=f(16)=4 f'(x)/1/2√(x)=f'(16)=1/8 f''(x)=-1/4x^-3/2=f''(16)=-1/156, and f³(x)=3/8x^-5/2=f³(160=3/8192 therefore, the polynomial p₃ ( which includes p₀,p₁ and p₂) is p₃(x)=4+1/8(x-16)-1/512(x-16)²+1/16382(x-16)^3. Teh Taylor polynomials (Figuren11.9) give better approximations to f as the orer of teh approximation increases. Letting x=18, we obtain the approximations to √(18) and the associated absolute errors shown in table 11.2 (A calculator is used for the value of√(18).) As expected, the errors decrease as n increases. Based on these calcualtions, a reasonable approximation of √(18) is 4.24 Table 11.2 n:approximation pₙ(18):absolute error|√(18)-pₙ(18) 0:4:2.4*10^-1 1:4.25:7.4*10^-3 2:4.24.188:4.5*10^-4 3:4.242676:3.5*10^-5

Ex5: plotting polar graphs

Use the cartesian to polar emthod to graph the polar equation r=1+sinθ. Solution: Viewing r and θ as cartesian coordinates, the graph of r=1+ sinθ on ht interval [0,21] is a standard sine curve with an amplitude 1 shifted up 1 unit. Notice that the graph begins with r=1 at θ=0, increases to r=2 at θ=pi/2, decreases to r=0 and θ=3pi/2, which indicates an intersection with an origin on the polar graph and increases to r=1 at θ=2pi. The second row of figure 12.27 shows the final polar curve (a cardioid) as it is transfered from the cartesian curve.

radius and interval of convergence using the ration test

Use the ratio test to find the radus and interval of convergence of Σ(∞:k=1) (x-20^k/√(k). r=lim k→∞ \(x-20^k+1/√(k+1)|/| (x-20^k/√(k)|=(x-2) lim k→∞√(k/k+1)=|x-2|√(lim k→∞ k/k+1)=|x-2|; the series converges absolutely ( and therefore, converges) for all x such that that r<1, which implies that |x-2|<1, or 1<x<3; we seee that the radius of convergence is r=1 on the intervals -∞<x<1.and 3<x<∞. We have r1, so the series diverges. We now test the endpoints. Subsitiuting x=1 gives the series Σ(∞:k=0)(x-2)^k/√(k)=Σ(∞+k=1)(-1)^k/√(k). This series converges by the Alternating series test (the terms of the series decrease in magnitude and aproach 0 as k→∞). Substituting x=3 gives the series Σ(∞:k=1)(x-2)^k/√(k)=Σ(∞:k=1)1/√(k); which is a divergent p-series. We conclude that the interval of convergence is x≤3.

Taylor polynomials for e^x

a, Find the Taylor polynomials of order n=90,1,2 and 3 for f(x)=e^x centered at 0. b. Use the polynomials in part a to approximate e^.1 and e^-.25. Find absolute errors, |f(x)-pₙ(x)| in the approximations. Use calculator values for the exact values of f. solution:a. Recall that the coefficients for the Taylor polynomials centered at 0 are ck=f^k(0)/k!, for k=0,1,2,...,n. With f(x)=e^x, we have f^k(x)=e^x, f^k(0)=1 and ck=1/k!, for k=0,1,2,3,... . The first four polynomials are p₀(x)=f(0)=1 p₁(x)=f(0)+f'(0)x=1+x p₂(x)=f(0)+f'(0)+f''(0)/2!x²=1+x+x²/2 p₃(x)=f(0)+f'(0)+f''(0)/2!x²+f'''(0)/3!x³=1+x+x²/2+x³/6 Notice that each successive polynomial provides a better fit to f(x)=e^x near o (figure 11.8). Continuing the patter in these polynomials, the nth order Taylor polynomial for e^x centered at 0 is pₙ(x)=1+x+x²/2!+x³/3!+...+xⁿ/n!=Σ(n:k=0)x^k/k! b. We evaluate pₙ(.1)and pₙ(.25) for n=0,1,2 1nd 3, and compare these values to the calulator values of e^.1=1.1051709 and e^-.25=.71880078. The results are shown in table 11.1. Obsereve that the errors in the pproximations decrease as n increases. In additioon, the errors in approxin=mationg e^.1 are smaller in magnitude than the errors in approxin=mating e^-.25 because x=.1 is closer to the center of the polynomial than x=-.25. reasonable approximations based on these calculations are e^.1≈1.105 ande^-.25≈.78. Table 11.1 n:Approximation pₙ(.1):Absolute error|e^.1-pₙ(.1)|:approximation pₙ(-/25):Absolute error|(e^-.25-pₙ(-.25) 0:1:1.1*10^-1:1:2.2*10^-1 1:1.1:5.2*10^-3:.75:2.9*10^-2 2:1.105:1.7*10^-4:.78125:2.4*10^-3 3:1.105167:4.3*10^-6:.718646:1.5*1-^-4.

arc lenght of polar curves

a. Find the arclenght of the spiral r=f(θ)=θ, for 0≤θ≤∨2pi. b. Find the arclength of the cardioid r=1+cosθ. Solution: a. L=∫(2pi:0)√(0)^2+1dθ=(θ/2√√(0^2+11/2ln(θ+√(0^2+1)))|(2pi:θ)=pi√(4pi^2+1)+1/2ln(2pi+√(4pi^2+1))=21.26 b. the cardioid id symetric aboutbhte y-axis and its upper half is generated for 0≤θ≤pi. The length of the full curve is twice the length of its upper half: L=2∫(pi:0)√((1+cosθ)^2(-sinθ)^2) dθ=2∫(pi:0)√(2+2cosV)dV=2∫(pi:0)√(4cos^2θ)dθ=4∫(pi:0)cos(v/2)=8sin(θ/2)|(pi:0)=8.

Ex6: Evaluating infinite series

a. Use the maclaurin series for f(x=tan^-1xto evaluate 1-1/3+1+...=Σ(∞:k=0)(-1)^k/2k+1 b. Let f(x)=e^x-1/x for x≠0 and f(0)=1. Use the maclaurin series for f to evalutat f'(1) and Σ(∞:k=0)k/(k+1)!'solution: a. From table 11.5, we se that for |x|≤1, tan^-1x=x-x^3/3+x^5/5-..._(-1)^kx(2k+1/2k+1+... =Σ(∞:k=0)(-1)^kx^2k+1/2k+1. Substituting x=1, we have tan^-11=1-(1)^3/3+(1)^5/5-...=Σ(∞:k=0)(-1)^kx^2k+1/2k+1.Because tan^-11=Π/4, the value of the series is Π/4. b. Using the maclaurin series for e^x, the series for f(x)=(e^x-10/x=1/x(1+x+x^2/2!+x^3/3!+...)-1)=1+xx/2!+x^2/3!+...=Σ(∞:k=0)x^k-1/k!, which converges for -∞<x<∞. By the quortient rule, f'(x0=xe^e-(e^x-1)/x^2. Differentiating the series for f terme by term (theorem 11.5), we find that f'(x)=d/dx(1+x/2!+x^2/3!+x^3/4!+...)=1/2!+2x/3!+3x^2/4++...)=Σ(∞:k=0)kx(k-1)/(k+1)!. We now have two expressions for f' that are ecaluated atx=-1 to shw that f'(1)=Σ(∞:k=0)k/(k+1)!

IS3: PArametic equations of lines

a. consider the parametric equations x=-2+3y, y=4-6y, for -∞<t<∞ whhich describe a line. Find the slope intedrcept for =m of the line. b. Find two pairs of parametric equations for the lne with slope 1/3 passes through the point (1,3). cnd the parametric equatioins of the line segment starting at p(4,7) and ending at (2,-3). S: to eliminate teh parameter, fis=rst solve the x- equation for f to find that t=x^2/3. Replacing t in the y- equation yields y=4-6(x+2/3)=4-2x-4=2y. THe line y=2x passes tr=hrought the origin with slope=-2. b We use the general paramt=etric quations of a line given in the summary box. Because the slope of th line is 1/3, we choose a=3 and b=1. Letting x₀=2 and y₀=1. Parametric equations for the line are x=2+3t, y=1+y, for -∞<t<∞. The line passes throught (2,1) when t=0 and rises to teh rigth as t increses. Notice that other choices for a and b could also work. For example, a=6 and b=2, the equations are x=2-6t and y=1-2t, for -∞<t<∞.. These equations describe the same line, but now a s t increases, the line is generated in the opposite direction. the slope of this line is x-(-3)/4-2=5. However, notice that when the line segment is traversedfrom p=0, both x and y are decreasing to accoutnt for the distinction in which the line is generated, we let a=-1 and b=-5. Because p(4,7) is the starting segment, we choose x₀4 and y₀=7. The resulting equations are x=4-t , y=7-5t. Notice that t=0 corresponds to the starting function. Because the qequations describe line segments, the interval for f must be restricted. What value of t corresponds to the endpoint of the line segment Q(2,-3). Seting x=4-t-2, we find that t=2. As a check, we set y= 7-5t=-3, which also implies that t=2. (If these clculations do not give the same value of t, it probablly means that the slope was not completed correctly). Therfore, the equations for the line segmetn are x=4-3t, y=5-7t, fot 0≤t≤2.

Ex: converting coordinates

a. express th point with polar coordiantex P(2,3pi/4) in cartesian coordinates. b. Express the point with cartesian cooridnates q(1,-1) in polar coordinates. a. the polar point (2, pi/4) has cartesian coordinates x=rcosθ=2cos3pi/4=-√2 and y=rsinθ=3pi/4=√2. R has cartesian coordiantes (-√2,√√2). b. It is bes tot locate this point first to be sure that the angle is chosen correctly. th point (1,-1) is in the fourth quadrant at a distance r=√((1)^1+(-1)^1)=√2 from the origin. the coordinate θ satisfies tanθ=y/x.=-1/1=-1. The angle in the foutrth quadrant with tanθ=-2 is θ=pi/4 or7pi/4. Therfore, two (of infinitely many) polar representations of θ are (2,-pi/4) and (√2,pi/4).

Radius and interval of convergence

a.Σ(∞:k=0)x^k/k! b.Σ(∞:k=0)(-1)^k(x-2)^k/4^k c. Σ(∞:k=1)k!x^k Soultion: a. The center of the power series is 0 and the terms of the series are x^k/k! Due to the presence of the factor k!, we test the series for absolute convergence using the ratiio test r=lim k→∞|x^(k+1)/(k+1)!/x^k/k!|=lim k→∞ |x|^k+1/|x|^k*K!/(k+1)!=\x| lim k→∞1/k+1=0 Notice that in taking the limit as k→∞, x is held fixed. Because r=0 for all real numbers x, the series converges absolutely for all x, which implies that the series converges for all x. Therefore, the interval of convergence is (-∞,∞) and the rdius of convergence is R=∞. b. we test fr absolute convergence using the root test: p=lim k→∞ k√((-1)^k(x-2)^k/^k)=|x-2)/4 In this case, p depe nds on the value of x. For absolute convergence, x must satisfy p=|x-2|/4<1, which implies that |x-2|<4.Using standard tachniques for solving inequaities, the solution set is -4<x<-2<4, or -2<x<6. We conclude that the series converges absolutely (an therefore, converges) on (-2,6), which means that the radius is R=4. When -∞<x<-2 or 6<x<∞, we have p>1, so the series dierges on these intervals (the terms do not approach 0 as k→∞ and the divergence test applies). The root test does not give informatiion about convergence at the endpoint x=-2 and x=6., because at these points, the root test results in p1. To test for convergence at the endpoints, we substitute each endpoint into the series and carry out seprate tests. At x=-2, the power series becomes Σ(∞:k=0)(-1)^k(x-2)^k/4^k=Σ(∞:k=0)4^k/4^k=Σ(∞:k=0)1. The series clearly diverges at the left endpoint. At x=6, the power series is Σ(∞:k=0)(-1)^k(x-2)^k/4k=Σ(∞:k=0)(-1)^k4^k/4^k=Σ(∞:k=0)(-1)^k. This series also diverges at the right endpoint. therefore, the interval of convergence is (-2,6), excluding the endpoints. c. In this case, the ratio test is preferable: r=lim k→∞((k+1)!x^(k+1)/k!x^k|=|x|limk→∞(k+1)!/k!=|x|lim k→∞(k+1)=∞ We see that r>1 for all x≠0, so the series diverges on (-∞,0) and (0,∞). The only way to satisfy |r|<1 is to take x=0, in which case, the power series has a value of 0. The inteval of convergence of the power series consists of the single point x=0 and the rsadius of convergence is R=0.

Power series

an infinite series in the form of Σ(∞:k=0) Ckx^k=c₀+c₁X+c₂x^2+...+cₙx^n+1+..., or more generally,Σ(∞:k=0) ck(x-a)^k=c₀+c₁(x-a)+...+cₙ(x-a)^n+CN+1(X-A)^n+1+..., where the center of the series a and the coeffieic=ents ck are constants. This type of series is called a power series because it consists of powers of x or (x-a).

Example 1 b&c: linear and quadratic approximations for lnx

b. find the quadraic appproximation to f(x)=lnx at x=1. c. use these approximations to estimate ln 1.05. b. We first compute f''(x)=-1/x^2 and f''(x)=-1. Building on the linear approximation found in part a, the quadratic approximation is p₂(x)=x-1+1/2 f''(1)(x-1)^2=x-1-1/2(x-1)^2=.04875 Because p₂ mathches f in value, slope and concavity at x=1, it provides a better approximation to f near x=1. c. To approxin=mate ln 1.0, we substitute x=1.05 into each polynomial approximation: p₁(1.05)=1.05-1 and p₂(1.050=1.05-1-1/291.05-1)^3=.04875 the value of ln 1.05 is given by a calcualtor, rounded to five decimal places is .049=879, showing imporvement in quadratic over linear approximation.

IS: paramteric ircle

confirm that the parametric equations x=4cos^2πt,y=4sin^2πt, for 0≤t≤1 descrive a circle of r=4, centered at the oorigin. b. Suppose a tutrtle walks with a constaant speed in the counterclockwise derection on the circular path form part (a). Starting form the point (4,0), the turtle comletes 1 lap in 30 minutes. Find a parametric description of the path of theturtle at any time, t≥0, where t is measured in minutes.' Solution: for each value of t in table 12.2, the corresponding ordered pairs(x,y) are recorded. Plotting these points as t increases for t=0 to t=1 results i a graph that appears to be a circle of r=4.; it is generalized with positive orientation in the clockwise direction, beginning and endinga t (4,0)Letting t increase beyiond t=1 would simply retrace the same curve. To identify the curve onclusively, the arameter t is eliminated by observing that x^2+y^2=((4cos2πt)^2+(4sin2t)^2)= 16(cos^2πt)+(sin^2πt)=16. We have confirmed that the graph of the parametric equations of the circle is x^2+y^2=16. b. duplicating the calcualtions at the end of part a, with any nonzero real number b, it can be shown that the parametric equations x=4cosbt, y-=4sinat also descirbes a circle of readius 4. When b>0, the graph is generated in the counterclov=ckwise direction. Th angular frequency b must be chosed=n so that as t varies from 0 to 30, the product bt varies fron 0 to 2π. Specifically, when t=30, we must have 30t=2π, or t=π/5! rad/min. Therfore, the equation for the turtle's motion are x=4cosπt/15, y=4sinπt/15, for 0≤t≤30.≤

convergence of Taylor series

et f have derivatives of all orers on an open interval I conataining a. The Taylor series for f centered at a converges to f, for all x in I , if and only if lim k→∞ Rₙ(x)=0, for all x in I where Rₙ(x)=f^(n+1)(c)/(n+1)!(x-a)^(n+1) in the remainder at x, with c between x and a.

Ex6: slopes of tangent lines

find d/dx for the following curves. Integrate the result and etermine the point if any, at which the curve has a horizontal or vertical line. a. x=f(t)=t, y=g(t)=2√t, fort≤0. b. x=f*t)=4cost,y=g(t)=16sint, for 0≤t≤2π. solution: we find that f'(t)=1 and g'(t)=6)=1/√t. therfore, d/dxg(t)/f'(t)=t/1/√x/1/√x. provided t≠0. Notice that dy/dx≠0 for t>0, so the curve has no horizontal tangent lines. On the other hand, as t→0+, we see that dy/dx→∞. therfore, the curve has a vertical line a tthe point (0,0). to eliminate t from the parametric equations, we substiture x into the x-equation to find that y==2√x=3. Because f(x)≥0, the curve is ≥the upper half of a parabola. Slopes of tanf=gent lines at other points on the curve are found by substituting the corresponding values of t. For ec=xample, the point (4,4) corresponds to t=4 and the slope of the tangent line at that point is 1/√4=1/2. b. these parametric equations describe an elipse with a major axis of length 2=32 on the x-axis nd a minor axis of length 8 o the y-axis. In this case, f'(t0=-4sintand g'(t)=16cost. therfore, dy/dx=g'(t)/f'(t)=16cost/-4sint=-4cott.At t=0, and t=π, cott is undefined. Notice that lim t→∞dx/dt=lim t→0+(-4cot c) =-∞ and lim t→0-(-4cott)=∞. Consequently, a vertical tangent line occurs at the pppoitn corresponding to t=0, which is (4,0). A similar argurment shows that a vertical line occurs at the point corresponding to t=π, which is(-4,0). At t=π/2, and t=3π/2, ot t=0 and the curve has a horizontal tangent line at the corresponding points (0,±1). Slopes of tangent lines at otehr points on the curve may be found. For example the point (2√2,8√2) correspodns to t=π4. the slope of the tangent lien is -4cot π/4=-4.

binomial coefficients

for real numbers p and integers k≥1, (p:k)=p(p-1)(p-2)...(p-k+1)?/k!,(p:0)=1

parametrc parabola

graph and analyze the parametric equations. x=f(t=2t, y=g(t)=1/2t^-2-4, for 0≤t≤∞. Solution: pltting points often often helps in visualiszing a parametric curve. table 12.2 shows tha values f x and y corresponding to several values on the interval [0,8]. By plotting the (x,y) pairs in table 12.1 znd connecting them with a smooth curve, we obtain the graph as sown in figure 12.2. As t increases from its innital value of t=0 to its final value at t=8, the curve is generated form the initial point (0,-4) to the fianl point (16,28). Notice that the value of the parameter is the direction in which the curve is generated; In this case, it unfolds upwarscan to the right, as shown by the arrows on the curve. Sometimes, it is possible to eliminate the parameter form a set of parametric equations and obtain a description of the curve in terms of x and y. In this case, from the x-cardion, we have t=x/2, which may be substituted in the equation to give y=1/2t-4=1/2(x/2)-4=x^2/8-4. Expressed in this form, we identify the graphs as part of a arabola. Be cause t lies in th interval 0≤t≤16. therefore, the parametric equations generate the segment of teh parabola for 0≤x≤16.

points in polar coordinates

graph the following points in polar coordinates:Q(1,5π/4), R(-1,7π/4) and S(2,-3π/2). Give two alternattive representations for eacj point. solution: the point Q(1,5π/4) is one unit form the prigin on a line OQ that makes and angle of 5π/4 with the positive x-axis. Subtracting 2π from the angle, the point q can be represented as (1,=3π/2). Subtracting π form the angle and negating theradial coordinate implies that θ also has the coordinates (-1,π/4). To locate teh point R(-1,7π/4) in the 4th quadrant, it is esiest to find the point R*(1,7π/4) in the 4th quadrant. Then R(-1,7π/4) is the reflection of the point throught the origin. Other representations of R include (-1,pi/4) and (1/3pi/4)>The point S(2,-3pi/2is two units from the origin founded by roatating clockwise through an angle of 3pi/2. The point S can also be represented as (2,pi/2) or (-2,-pi/2).

anipulating maclaurin series

let f(x)=e^x a. Find the maclaurin series fro f. b. Find its interval of convergence c. Use the maclaurin series for the functions x^4e^x,e^-2x and e^-x^2. solution:a. The coefficients of the Taylor polynomials for f(x)=e^x centered at 0 are ck=1/k! (examplle 3, section 11.1). They are also coefficients of the maclaurin series. Therefore, the maclaurin series for e^x is 1+x/1!+x^2/2!+...+x^n/n!+...=Σ(∞:k=0)x^k/k! b. Bby the Ratio Test, r=lim k→∞(x^k+1/(k+1)!/x^k/k!|=lim k→∞(x/k+1|=0. Because r<1 for all x, the interval of convergence is -∞∞<x<∞. c. As stated in theorem 11.4, power series may be added, multiplied by powers of x, or composed with functions on their intervals of convergnece. Therefore, the maclaurin series for x^4e^x is x^4Σ(∞:k=0)x^k/k!=Σ(∞:k=0) x^k+4/k!=x^4+x^5/1!+x^6/2!+...+x6k+4/k!+.... Similarly, e^-2x is the composition of f(-2x). Replacing x with -2x in the maclaurin series for f, the the series is a representation for e^-2x is Σ(∞:k=0)(-2x)^k/k!=Σ(∞:k=0)-1)^k(2x0^k k!=1-2x+2x^2-4/3x^3+.... The maclaurin series for e^-x^2 is obtained by replacing y with -x^2 in the power series for f. The resulting series is Σ(∞:k=0)(-x^2)^k/k!=Σ(∞:k=0)x^2)^k/k!=Σ(-1)^kx^2k!=1-x^2=x^4/2!-x^6/3!+.... Because the interval of convergence of f(x)=e^x is -∞<x<∞,∞ the mnipulations used to obtain the series for x^4e^x,e^-2x, or e^-x^2 do not cahnge the interval of convergence. If in doubt about the interval of convergence of a new series, apply the ratio test.

derivative for parametric curves

let f(x0=f(t) and y=g(t), where f and g are differentiable on [ab,]. then the slope of the tangent line to the curve at the point corresponsing to t is dy/dx=dy/dx/dx/dt=f'(t)/g'(t0, provided f'(t)≠0.

parpmetric curve

parametric equtaions have the form x=f(t),y=g(t), where f and g are given functions and the parameterct typically vries over a specified interval a≤tb. Teh parametric ccurve described by these equations consists of points in the plane (x,y)=(f(t),g(t)), for a≤t≤b.

circumference of a circle

prove tha tthe circumference of a circle radiius a>0 is 2πa. solution: a circle radius is desribed by x=f(t)=acost,y=g(t)=asint, for 0≤t≤2π. NOtice that f'(t)=-asint and fg'(t)=acost. The circumference is L=∫(2π:0)√(f'(t)^2+f'(t)^2) dt=∫(2π:0)√(-asin^2t)+(acost)^2dt=a∫(2π:0)√((sin^2t)+(cos^2t)) dt=a∫(2π:0)1 dt=2πa.

idea of a remainder

remainders are associated with various Taylor polynaomials for f(x)=e^x centered at 0 fr fixed order n, the remainders tend to increase in magnitude as x moves farther and farther from the center of the polynomials. And for fixed x, remainders decrease in magitude with increasing n. The remainder for a Taylor polynomial may be written quite concisely, which enables us to estimate the remainders.

parametric equations

specify x and x interms of a third variable, t, which often represents time.

positive orientation

the direction in which a parametric curve is generated as the parameter increasex is acalled the positive orientation of the curve (as indicated b arrows on the curve).

circles in polar coordinates

the equation r=a describes a circle of radius a, centered at (0,00. The equation r=2acosθ+2bsinθ describes a circle of radius√a^2+b^2, centered at (a,b).

polar axis

the postiitve axis in the polar coordinate ssytem, with polar coordinates tha thave the same form (r,θ). the radial coordinate descirbes the singular or direted distance form the origin to b. the angular cooridinate θ, describes an angle whose initial side is the postiive x-axo=is and whose terminal side is lies on the ray passing through the origin and p.

Determining the radius and interval of convergence of Σck(x-a)^k

1. We use the ratio test or the roote test to find teh interval (a-r,a+r) on wjicj the series converge absolutely; the radius of convergence for the series is r. 2. We use the radius of convergence to find the interval of convergence. *If R=∞, the interval of convergence is (-∞,∞). * If R=0, the interval of convergence is the single point x=a. *If 0<R<∞, the interval of convergence consists of the interval (a-r, a+R) and possibly one or both of its endpoints. Determining whether the series Σck(x-a)^k at the endpoint x=a-R nd x=a+R amounts to anaylzing the series ∑ck(-)^k and ∑ckR^k.

table 11.5

1/1-x=1+x+x^2+...+x^k+..=Σ(∞:k=0)x^k, for |x|<1. 1/1+x=1-x+x^2-...+(-1)^k+...=Σ(∞:k=0)x^k/K!, or |x|<∞. sinx= x-x^3/3!+x^5/5!-...+(-1)^kx^2k+1/(2k+1)!+...=Σ(∞:=0)(-1)^kx^2k+1/(2k+1)!, for |x|<∞. cosx= 1-x^2/2!+x^4/4!-...+(-1)^kx^2k/(2k)!+...=Σ(∞:k=0)(-1)^kx^k/k, for |x|<∞ ln(1+x)=x+X^2/2+x^3/3+...+(-1)^k+1x^k/k=Σ(∞:k=0)(-1)^k+1x^k/k, for -1<x≤1. -ln(1-x)=x+x^2/2/+x^3/3+...+x^k/k+...=Σ(∞:k=0)x^kk/, for -1≤x≤1. tan^-1x=x-x^3/3+x^5/5-...+(-1)^kx^2k+1/2k+1+...=Σ(∞:k=0)(-1)^kx^2k+1/2k+1, for |x|≤1. sinh x=x+x^3/3!+x^5/5!-...+x^2k+1/(2k+1)!+...=Σ(∞:k=0)x^2k+1/(2k+1)!, for |x|≤∞. cosh x=1+x^2/2+x^4/4!+...+x^2k/(2k)!+...=Σ(∞:k=0) x^2k/(2k)!, for |x|<∞. (1+x)^p=Σ(∞:k=0)(p:k)x^k, for |x|<1 and (p:k) p(p-1)(p-2)...(p-k+1)/k!(p:0)=1.

converting coordinates

A point with polar coordinates (r,θ) has cartesian coordinates (x,y) , where x=rcosθ and y=rsinθ. A point wiht cartesian coordinates (x,y) has polar coordiantes (r,v), where r^2=x^2+y^2 and tanθ=y/x.

power series (section 11.2)

A power series has the general form Σ(∞:k=0) ck(x-a)^k centered at a converges in one of three ways: 1. The series converges absolutely for all x. It follows, by theorem 10.14, that the seris converges for all x, in which case, the interval of convergence is (-∞,∞) and the radius of convergence is r=∞. 2. The real number R>0 such that the series converges absolutely (and therefore, converges) for|x-a|<R and diverges for|x-a|>R, in which casse, the readius of convergence is R. 3. The series diverges only at a, in which case the radius of convergencee is R=0.

approximate a defintie integral

Approximatecthe vlaue of the integral ∫(1:0)e^-x²dx, with an error no greater than 5*10^-11. Solution: ∫e^-x² cannot be expressed in terms of familiar functions. the strategy is to write the macluarin sreies for e^-x² and integrate it term by term. recall tha the integration of power sero=ies is valid within its interval of convergence. Beginning with the macluarin series, e^x=+x+x²/2!+x³3/3!+...+xⁿ/n!+..., which converges for -∞<x<∞. We repllace x with -x^2 to obtain e^-x²=1-x62+x^4/4!-x^6/6!+...+(-1)^nx^2n/n!+..., which also converges -∞<x<∞.By the fundamentl theorem of calculus, ∫(1:0)e^-x²dx=(x-x^3/3!+x^5/5!-x^7/7!+...+(-1)^nx^2n+1/(2n_10!+.... Because the definite integral is expressed as an alternating sreis, the magnitudecofo teh remaining truncating sreies after n is lass than the magnitude of the first neglected term, which is |(-1)^n+1/(2n+30(n+1)|. By trial and erroe, we find that the magnitude of this term is less than 5*10^-4). the sum of the terms of the series up to n=5 gives the approximation ∫(1:0)e^-x²dx=1-1/3+1/5*2!-1/7*3!+1/9*4!=1/11*5@=.747

Estimating a remainder

COnsider again example 4, in which we approximated √(18) using the Taylor polynomial p₃(x)=4+1/3(x-16)-1/512(x-16)^2+1/16,384(x-16)^3. Find an upper bound on the magnitude of the remainder when using p₃(x) to approximate √(18). solution: IN example 4, we computed the error in the approximation knowing the exact value of √(18) (obtained with a calcultor). In the more realistic case in which we do not know the exact value, theorem 11.2 allows us to estimate ermainders (or errors). Applying this theorem with n=3, a=16, and x=18, we find that the remainder in approximating √(18) by p₃(18) satisfies the bound |R₃(18)|≤M (18-16)^4'4!=2/3M, where M is a number that satisfies |f⁴(c)|≤M, for all c between 16 and 18 inclusive. In this particular problem, we find that f⁴(c)=-15/16 c^-7/2, so m must be chosen (as small as possible) such that |f⁴(c)|=15/16c^-7/2=15/16c^7/2≤M, for 16≤c≤18. You can verify that 15/16c^7/2 is a decreasing function on c[16,18] and has a maximum value of approximately 5.7*10^-5 at c=16. Therefore, a bound o n the remainder is |R₃(18)|≤2/3m≈2/3-5.7*10^-5≈3.8&10^-5. Notice that the error computed in example 4(Table 11.2) is 3.5*10^-5, which is less than the bound on the remainder (as it should be).

working with binomial series

COnsider the functions f(x)=^3√(1+x) and g(x)=^3√(1+x), where c>0 is a constant. . Find the first four terms of the binomial series for f centered at 0. b. Use part (a) to find the first four terms of the binomial series g(x0 centered at 0. c. Use part (b) to apporximate ^3√(23),^3√(24),...,^3√(31). Assume the series converges for g on its interval of convergence. solution:a. Because f(x)=(1+x)^2/3, we find the bnomial coefficients with p=1/3. c₀=(1/3:0)=1 c₁=(1/3:1)=1/3/1!=1/3 c₂=(1/3:2)=(1/3)(1/3-1)/2!=-1/9 c₃=(1/3:3)=(1/3)(1/3-2)/3!=5/3! The first four terms of the binomial series are 1+1/3x-1/9x^2+5/8!x^3. b. To avoid deriving a new series for g(x)=^3√(1+x), a few steps of a;gebra allow us to use part (a). Note that g(x)=^3√(1+x)=^3√c(1+x/c^3√c/^3(1+xc)=^3√c f(x/c). In other words, g can be expressed in terms of f, for whcih we already have a binomial series. The binomial series for g is obtained by substituting x/c into the binomial series for f and by multipying by ^3√(c):g(x)=^3√(c)(1+1/3(x/c))-1/9(x/c)^2+5/8(x/c)^3 ...). It can be shown that the series for f n part (a) for |x|≤1. Therefore, the series for f(x/c) converges to f(x/cc), provided |x/c|≤1, or equivalently, for |x|≤c. c. The series f part (b)may be truncated after four terms to approximate cube roots. For example note that ^3√(29)=^3√(27+2), so we find that ^3√(29)≈^3√(27)(1+1/3(2/27)-1/9(2/270^2+5/81(2/27)^3≈2.0723; the same method is used to approximate the cube roots of 23,24,...,30,31 (table 11.4). The absolute erroe is the difference between the approximation and the value given by a calculator. Notice that the errors increase as we move away from 27.

Arclength for curves defined by parametric equations

Cionsider teh curve described by the parametric equations x=f(t), y=g(t), where f' and g' arecontinuous and the curves is traversed once for a≤t≤b. the arclenght of the curve bertween (f(a),g(a)) and (f(b),g(b)) is l=∫(b:a)√(f'(t)^2+g"(t)^2)dt.

Areas of polar regions

Consider the circle r=1 and the cardioid r=1+cosθ. a. Find the area of the region inside the circle and inside the cardioid. b. Find hte area of the region outside the circle and outside the cardioid.solution:a. The points of intersection can be found by solving 1+cosθ=1 or cosθ=0. the solutions are θ=±pi/2. The regiion inside the circle and inside the cardioid consists of two subregions * a semicircle with radius 1 in the first and fourth quadrants bounded by hte circle r=1 and * two crescent- shaped regions in the second and third quadrants bounded by the cardioid r=+cosθ and the y- axis. The area of the semicircle is pi/2. to find the area of the upper crescent- shaped region in the second quadrant, notice that it is bounded by r=1+cosθ, as θ varies from pi/2 to pi. thereofre, its area is ∫(pi:pi/2)1/2(1+cosθ)^2dθ=∫(pi:pi/2)1/2(1+2cosθ+cos2θ)dθ=1/2∫(pi:pi/2)(1+2cosθ=+1_2cos2θ/2)dθ=1/2(θ+sinθ+sin^2θ/4)|(pi:pi/2)=3i/8=1. The are of the entire region (two crescents and a semicircle) is 2(3pi/8-1)+pi/2=5pi/4-2 b. Tyhe region inside the circle and the outside cardioid is bounded by the outer curve r=1 and the inner curve r=1+cos on the interval [pi/2,3pi/2]. Using symetry about the x-axis, the area is 2∫(pi:pi/2)1/2(1^2(1-cos)^2)d=∫(pi:pi/2)(-2cosθ-cos^2θdθ=2-pi/4. Note tha the region in parts a and b make up the interior of a circle of radius 1; indeed they have a sum of pi.

plotting complete curves

Consider the closed curve described by r=cos(2θ/5). Give an interval that generates the entire cuve and graph the curve. Solution: Recall that cost has a period of 2pi. Therefore, cos 2θ/5 completes one cycle when 2θ/5 varies from 0 to 2pi or when θ varies from 0 to 5pi. Therefore, it is tempting to conclude that the complete curve r=cos(2θ/5) is generated as θ varies from 0 to 5pi. You can check tha the point corresponding to θ=0 is not the point corresponding to θ=5pi, which means that the curve does not close itself over the interval[0,5pi]. to graph the complete curve r=cos(2θ.\\/5), we must find an interval [o,p], where t is an integer multiple of 5pi(so that f(0)=f(p)) and on an integer multiple of 2pi (so that the points (0,f(0)) and (p,f(p)) are the same). The smallest number satisfyinig these conditions is 10pi. Graphing r=cos (2θ/5) over the interval [0,10pi] produces the complete curve.

Ex 5: binomial series

Consider the function f(x)=√(1+x) a. Find the first four terms centered at 0. b. Approximate √(1.15) to three decimal places. Assume the series for f converges to f on its interval of convergence [-1,1]. a. We use the formula for the for the binomial coefficients with p=1//2 to compute the first foru coefficients: c₀=1 c₁=(1/2:1)=1/2/1!=1/2 c₂=(1/2:2)=1/2(-1/2)/2!=-1/8, and c₃= (1/2:3)=1/2(-1/2)(-3/2)/!=1/16 the leading terms of the binomial series are 1+1/2x=1/3x^2+1/16x^3 b. Truncating the binomial used to approxo=imate f(.15)=√(1.15). With x=.15, we find the polynomial approximations shown. Four terms of the power series (n=3) give √(1.15=1.072. ecause the approximatiions with n=2 and n=3 agree to three decimal places, when redi=uced, the approximation 1.072 is accurate to three decimal places.

Ex: Differentiating and integrating power series

Consider the geometric series f(x)=1/1-x=∑(∞:k=0)x^k=1+x^2+x^3+..., for |x|<1. a. Differentiate this series term by term to find the power series for f' and identify the function it represents. b. Integrate this series term by term and identfy the function it represents. Soluion: a. We know that f'(x)=(1-x)^-2. Differentiating the series, we find that if f'(x)=d/dx1+x+x^2+x^3+...)=1+2x+3x^2+...=∑(∞:k=0)(k+1)x^k; Threfore, on the interval |x|<1, f'(x)=(1-x)^-2=∑(∞:k=0)(k+1)x^k. Theorem 11.5 makes no claim about convergence of the differentiated series to f' at endpoints. In this case, substituting x=±1 for the power series for f' reveals that the series diverges at both endpoints. b. Integrating f and integrating the power series term by term, we have ∫dx/1-x=∫(1+x+x^2+x^3+...)dx, which implies that -ln|1-x|=x+x^2/2+x^3/3+x^4/4+...+c, where c is an arbitrary constant . Notice that the left side is 0 when x=0. The right side is 0 when x=0, provided we choose c=0. Because |x|<1, the absolute value sign on the left side may be multiplying both sides by -1, we have a series ln(1-x)=-x=x^2/2-x^3/3-x^4/4-...=∑(∞:k=0) x^k/k. It is interesting to test the endpoints of the intervvaal |x|<1, when x=1, the series is, (a multiple of) the divergent harmonic seies and whenx=-1, the series is thw convergent alternating series (section 10.6). So thE intrval of convergence is -1≤x≤1. Although we know the series converges at x=1, theorem 11.5 guarantees conhnverge to ln(1-x) only at the interior points. We cannot use theorem 11.5 to claim that the series converges to ln2 at x=-1.In fact, it does, as shown in section 11.3.

polar to carsian coordinates

Conver the polar equation r=10 cosθ to t=4sinθ to cartesian coordinates and describe the corresponding graph. Solution: Multiplyng both sides of the equation by r produces the equation r=10cosθ+24sinθ. Using the conversion relations r^2=x^2+y^2=10x-24y=0. Convertinghte square in both equations gives the equation x^2-10x+25=225+y^2-24y+144-144=0 oor (x-5)^2+(y-12)^2=169. We recognize (x-5)^2+(y-12)^2 =169 as the equation of a circle of radius 3 centered at (5,12).

power series for derivatives

Differentiate the maclauri series f(x)=sinx to verify that d/dx (sinx)=cosx sinx=x-x3/3!+x^5/5!-x^7/71+. and sinx converges for -∞<x<∞. By theorem 11.5, the differeentiated series also converges for -∞<x<∞ and it converges for f'(x). Differenetiting, we have d/dx(x-x^3/3!+x^5/5!-x^7/7!...)=1-x^2/2!+x^4/4!-x^6/6!+...=cosx. The differentiated series is the maclaurin series for cosx, confirming that f(x)=cosx.

A limit by Taylor series

Evaluate lim x→∞ x²_2cosx-2 Solution: Because the limit has the form indeterminate form 0/0, L,Hopital's rule can be used, which requires applications of the rule. Alternatively, because the limit involves values near 0, we substitute the maclaurin series for f. Recalling that cosx=1-x²/2 +x⁴/24+x⁶/720+..., we have lim x→0 x²+2cosx-2/3x^4=lim x→0 x²+2(1-x²/2+x⁴/4-x⁶/720+...)-2/3x⁴=lim x→∞ x²+(2-x²+x⁴/12-x⁶/360)+...)-2/3x⁴=lim x→0 x⁴/12-x⁶/360+.../3x⁴ = lim x→0(1/36 -x²/1080+...0=1/36.

Example 2: A limit by Taylor series

Evaluate lim x→∞(6x^5sin1/x-6x^4+x^2). solution: A Taylor sereies maay be centered at any finite point in the function, but we don't have the tools needed to expand a function about x=∞. Using a technique introduced earlier, we replace x with 1+t an dnote that as t→∞,t→0+. the new limit becomes: lim x→∞(6sint-6t+t^3/t^5): the limit now has the form 0/0. We now expand sint in a taylor series centered at 0. because sint=t-t^3/6+t^5/120+t^7/5040+..., the value of the origional limit is lim x→0+(6sint-6t+t^3/t^5)=limt→0+6(t-t^3/6+t^5,120-t^7/5040+...-6t+t^3/t65)=lim t→∞(t^5/120-t^7/840+.../t^5)=limt→0+(1/20-t^2/840+...)=1/20

Ex6: computing areas

Example 5 discussed the points of intersection of the curves r=3cosθ (a circle) and r=1+cosθ (a cardioid)Use those results to compute the area of the following non- overlaping regions a. region A b. region B c. region C a. It is evident tha tregion A is bounded on hte inside by hte cardioid and on the outside by the circle between the points Q(θ=0) and P(θ=pi/3). therefore, the area of region A is 1/2 ∫((pi/3:0)(3+cos2θ-2cos2θ)dθ=1/2(3θ+2sin2V-2sinθ)|(3pi/2:0)=pi/2. b. Examining regionB, otice that ray frawn form the borigin enters the region immediately. there is no inner boundary, and the uter boundary is r=cosθ on 0≤θ≤pi/3 and r=3cosθ on pi/3≤θ≤pi/2 (recall from exMPLE 5 THAT θ=pi/2 is the angle at which the circle intersects the origin). Therefore, we slice the region into two parts at θ=pi/3 and write two integrals for its are: area of region B=1/2∫(pi/3:0)(1+cosθ)^2dθ +1/2∫(pi/2:pi3)(3cosθ)^2dθ. While these integrals may be evaluated directly, it's easier to notice that area of region B = area of semicircle OPQ-area of regio A. Because r=3cosθ is a semicirle wth raius 3pi/2, we have area of region B= 1/2 pi(3/2)^2-pi/2=5pi/8. c. It's easy to incorrectly identify the inner boundary of region C as the circle and the outer bboundary as the cardioid. While these identifications are true, whe n pi/3≤θ≤pi/2 (notice again th radial lines in figure 12.42), there is only one boundary curve (the cardioid) when pi/2≤θpi. We conclude that the area of region C is 1/2∫i/2:pi/3)((1+cosθ)^2-(3cosθ)^2dθ+1/2∫(pi:pi/2)(1+cosθ)^2dθ=pi/8

center other that 0

FInd the first four nonzero terms of the Taylor series for f(x)=^3√(x) centered at 8. Solution: Notice that f has derivatives of all orders at x=8. The Taylor series at 8 has the form Σ(∞:k=0)ck(x-8)^k, where ck=f^k(8)/k!. Next, we evaluate the derivatives: f(x)=x^1/3→f(8)=2 f'(x)=1/3x^-2/3→f'(8)=1/12 f''(x)=-2/9x^-5/3→f''(8)=-1/144, and f³(x)=10/27→f³(8)=5/3456 We now assemble the power series 1+1/12(x-8)+1/2!(-1/144)(x-8)^2+1/3!(5/3456)(x-8)^3+..2+1/12(x-8)-1/288(x-8)^2+5/20-736(x-8)^3

Estimating the remainder for cosx

Find a bound for the magnitude of the remainder for the Taylor polynomials f(x)=cosx centered at 0. Solution: According to theorem 11.1 with a=0, we have Rₙ(x)=f((n+1)/(n+1)!x^n+1, where c is a point between 0 and x. Notice that f^n+1 (c)=±sinc, or f^n+1)(c)=±cosc, dependig on the value of n. IN all cases, |f^(n+1)(c)|≤1. Therefore, we take M=1 in theorem 11.2 and the absolute value of the remainder can be boundeed as |Rₙ(x)|=|f^(n+1)(c)/(n+1)x^n+1|≤|x|^n+1/(n+1)!. For exmplle, if we apporxinate cos(.1) using the Taylor polynomial p(10), the remainder satisfies |R10(.1)|≤(.1)^11/11!≈2.4*10^-9

Estimating the remaimder for e^x

Find a bound on the remainder in approximating e^.45 using thee Taylor polynomial of order n=6 for f(x)=e^x centered at 0. Sollution: Using therem 11.2, a bound on the remainder is given by |Rₙ(x)|≤M |x-a|^(n+1)/(n+1)@, where M i chosen such that | f^(n+1)(c)|≤M for all c between a nd x inclusive. Notice that f(x)=e^x iimplies that f'(x)=e^x, for k=0,1,2,.... .In this particular problem, we have n=6, a=0, and x=.45, so the bound on the remainder takes the form |R(6)(.45)|≤M |1.45-0|^7/7!≈ 7.4 *10^7 m, where m is chosen such that |f^7(c)|=e^c≤M, for all c in the interval [0,0.45]. Because e^c is an increasing function of c, its maximum value on the interval [0,,46] occurs at c=.45 and is e^.46. However, e^.45 cannot be evaluated exactly (it is the number we are approximating), so we must find a number M such that e^.45≤M. Here is one of many ways to obtain a bound : we observe that e^.45<e^1/2<4^1/2=2 and take m=2(figure 11.12). therefore, bound on the remainder is |R(6)(.45)|≤7.4*10_-1M=1.5*10^-6. Using the Taylor polynomial derived in example 3, with n=6, the resulting approximation to e^.45 is p(.45) =Σ(6:k=0) .45^k/k!≈1.5683114: It has an error that does not exceed 1.5*10^-6.

parametric equations of curves

Find a parametric representationn for the following curves. A. The segment of the parabola y=9-x^2, for -1≤x2.\b. The complete curve =(y-5)^2+√(x) c. The piecewise linear path connecting p(-2,0)to Q(0,3) to R(4,0), where the parameter varies over te=he interval 0≤x≤2. Solutuion:A. the simplest way to represent a curve y=f(x) parametrically is to let x=t and y= f(t) where t is the parameter. We must then find the appropirate interval for the parameter. Using theis spporach, the curve y=9-x^2 has the parametric representation x=t, y=9-t^2, for -1≤t≤3. This representation is not nique. For example, you can also verify that the parametric equations x=1-t, y=90(1-t^2)^2, for -2≤t≤2 also do the job althought hte equations trace the parabola from right to keft, while the origional wquataions trace the curve form left to right. b. In this case, it is easier to let y=t. Then a parametric description of a curve is y= (x-5)^2+√(1)=y+t. Notice that t can take values on the interval [0,8). As t→∞, we see that x→∞ and y→∞. c. the path consists of two line segments that can be paramtertized separately in the form x=x₀+at and y=y₀+bt. the line segment PQ originates at P(-2,0) and unfolds in the positive direction with slope=3/2. It can be represented as x=-2+2x,y=2+t or 0≤t≤1. Finding the parametric equations requires for QR some ingenuity. We want the line segment to originate at QR(0,3) when t=1 and eand at r(4,0) when t=2. Observe that when t=1,x=0 and when t=2, x=4. Substituting these pairs of values into the general equations x₀+a=0 and x₀+2a=4,. SOlving for x₀ and a, we find that xt=4 and a=4. Applying a similar procedure to the general y- equation, y=y₀+a,the relevant equations are y₀+b=3, y₀+2b-0. Solving for y₀ and b, we find that y₀=6 nad b =-3. Putting it all together, the equation for the line segent QR is x=-4+9t, y=6-3t, for 1≤t≤2. You cna verify that the points Q(0,3) and R(4,)0 correspond to t-1 and t-3, respectively. Furthermore, teh slope of the line is b/a=-3/4, wjch is correct.

A differentiable equation

Find a power series solution of the differential equation y'(t)=t+2, subject to the iitial condition y(0)=5. Identify the function represented by the power series. Solution: Because the initial condition is given at t=0, assure the solution has a taylor series entered at a+0 of the form y(t)=Σ(∞:k=0)ckt^k, where the coefficitnts ck must be determined . Recall that the coefficients of the Taylor series are given by cky^k(0)/k!, for k= 0,1,2,.... If we can determine y^k(0), for k=0,1,2,..., the coefficients of the series are also determined.. Suvbstituting the initial condition t=0 and y=6 into the power series y(t)=c₀+c₁t+c₂t^2+..., we find that t=c₀+c₁(00+c₂(0)^2+.... It follows that =6. to determine y'(0), we sustitute t=0 into the differentiable equation.:Th result is y'(0)=y(0)=6+2=8. therefore, C₁=y'(0)/1!=8. the remaining derivatives are obtained by successively differeentiating the differentaial equation and substituting t=0. We find that y''(0)=y'(0)=8, y^3(0)=y^2(0)=8, and in generral ,y^k(0)=8, for kk=0,1,2,.... Therefore, ck-y^k(0)/k!/8/k, for k=0,1,2,... and the Taylor series for the solution is y*t)=c₀+c₁t+c₂t^2+...=6+8/1!t+8/2!t^2+8/3!t^3+.... to identify the unction represented by the series, we write y(t)=-2+8+8/1!t+8/2!t^2+...=(-2+8)(1+t+t^2/2!+t^3/3+...). The power series that appears is the Taylro sreis for s=e^t. Therefore, teh solution is x=-2+e^t.

choosing a different center

Find the Taylor series for f(x0=1/1-x centered at 5, and then determine its radius and interval of convergence. solution: Using the pattern discovered above in example1b, we find that the kth derivative of f is f^k(x)=k!/(1-x)^(k+1). therefore, the Taylor coefficients are ck=fk(5)/k!=k^4/k!(1-5)^(k+1)=(-1)^k+1/4^k=1, fpr k=0,1,2,... and the series centered at 5 for f is Σ(∞:k=0)(-1)^k+1(x-5)^k/4^k+1=-1/4+x^5/4^2=(x-50^2/4^3+(x-50^3/4^4 .... To determine the interval of convergence for this series, we apply the ratio test: r=lim k→∞|(-1)^(k+2)(x-5)^k+1/4^k+2/(-1)^k=1(x-50^k/4^k+1|=lim k→∞|x-5_/4=|x-5|/4 The serie sconverges absolutely when r=|x-5 |/4<11, or when 1<x<9, so the radius of convergence is 4. The series diverges at x= and x=9, so we conclude that the interval of convergence is (1,9). the graphs of f and the 7th- order Taylor polynomial p₂ centered at 5 are shown in figure 11.19, along with the graph of p₇ centered at 0 found in example 1. Notice that in both cases, the interval of convergence extends to the vertical asymptote (x=1) of f(x)=1/1-x. In fact, it can be shown that the Taylor series for f, with any choice of a=center at x≠1, hs an interval of convergence about a that extends to x=1. Remarkably, Taylor series for f "know" the location of the discontinnuity of f.

Area of a polar region

Find the area of the four- leaf rose. r=f(θ)=2cosθ. solution: The graph of the rose appears to be symetric about the x- and y- axes; in fact, these sumetries can be proved. Appealing to this symetry, we find the area of one- half a leaf and the multiply the reult by 8 to obbtain the area of the full leaf. The upper hal of the rightmost leaf is generated as θ, increases from θ=0 (when r=2) to θ=pi/4(when r=θ). therfore, teh area of the entire rose is 8∫(pi/4:0)1/2f(θ)^2=4∫(pi/4:0)(2cos2θ)^2dθ=16∫(pi/4)1+cos2θ/2=(8θ+2sin4θ)|(pi/4:0)=(2pi+0)-(0+0)=2pi.

maclaurin series and convergence

Find the maTaylor series centered at 0) for the following functoins, and then find the interval of convergence. a. f(x)=cosx b. f(x)=1/1-x Solution:aThe procedure for finding coefficients of a Taylor series is the same as for Taylor polym=nomials: most of the work is computing the derivative of f. a. the Maclaurin series has the form Σ(∞:k=0)ckx^k, where ck=f^k(0)/k!, for k=1,2,.... We evaluate derivatives of f(x)=cosx at x=0 f(x)=cos x→f(0)=1 f'(x)=-sinx→f'(0)=0 f''(x)=-cosx→f''(0)=-1 f'''(x)=sinx→f'''(0)=0 f''''(x)=cosx→f''''(0)=1 Because the odd-order derivatives are zero, ck=f^k(0)/k!=0 when k is odd. Using the ven-order derivatives, we have c₀=f(0)=1 c₂=f''(0)/2!=1/2 c₄=f^4(0)/4!=1/4! c₆=f^6(0)/6!=1/6! and in general, c₂K=(-1)^k/c₂k!. Therefore, the maclaurin series for f is 1-x^2/2!+x^4/4!-x^6/6!=Σ(∞:k=0)(-1)^kx^k/(2k)!x^k; Notice that this series conntains all the Taylor polynomials . In this case, it consists of only even powers of x, reflecting that cosx is and even functioni. For what values of x does the series converge? A discussed in sectio 11.2, we apply the ratio Test: r=lim k→∞ |(-1)^(k+1)x^2(k+1)/(2(k+1))!/(-1)^kx^2k/(2k)!|=lim k→∞|x^2/(2k+2)(2k+1)|=0; IN this case, r<1 for all x, so the Maclaurin series converges for all x: We conclude that the interval of convergence is -∞<x<∞. b. We proceed in a similar way with f(x)=1/1-x by evaluating the derivatives of f at 0: f(x)=1/1-x→f(0)=1,f'(x)=1/(1-x)^2→f'(0)=1,f''(x)=2/(1-x)^3→f''(0)=2!, f³(x)=3!,f⁴(x)=4*3*2/(1-x)^5→f^4(0)=4!, and in general, f^4(0)=k!. Therefore, the MAclaurin series coefficients are ck=f^k(0)/k!=k!/k!=1, for k=0,1,2,.... The series for f centered at 0 is 1_x^2+x^3...=Σ(∞:k=0)x^k. this power series is familiar: the maclaurin series for f(x)=1/1-x is a geometric series. We could apply the ratio test, but we have already demonstrated that the series converges for |x|<1.

Ex5: points of intersection

Find the poiints of intersection of the circle r=3cosθ and the cardioid r=1+cosθ. Solution: the fac that a point has multiple representations in polar cooordinates may lead to subtle difficulties in finding points. We first proceed algebraically. Equating the two expressions for r and solving for θ, we have 3cosθ=1+cosθ or cosθ=1/2, which has roots θ=±pi/3. therefore, two intersection points are (3/2, pi/3) and (3/2,-5/3). Without examining grahs of curves, we might be tempte to stop here. Yet the figure shows anothe intersection point at th origin O that has not yet been detected. To find this intersection point, we must investigate the way i which the two curves are generated. As θ increases fro ) to 2pi, the cardioid generated is counterlockwise, beginning at (2,0). the cardioid passes through 0 when θ=pi. As θ increases from 0 to Pi, teh cicle is generated counterclockwise, beginning at (3,0). The circle passes through 0 when θ=pi/2. therefore, the intersection on the point 0 is (0,pi) on the cardioid (and these points do not satidfy the point of the equation on the circle(=), where o is (0,2pi) on the circle (and these points do not satisfy the point on hte cardioid). There is no "fool proof" rule for detecting such hidden intersection points. Care must be used.

vertical and horizontal tangent lines

Find the points on the interval -pi≤θ≤≤pi at which the cardioid r=f(θ)=1-cosθ has a vertical or horizontal line. Solution: Applying theorem 12.2, we find that dy/dx=f'(θ)sinθ+f(θ)cosθ/f'(θ)cos(θ)-f(θ)sin(θ)=sinθsinθ+(1-cosθ)cosθ/sinθcosθ-(1-cosθ)sinθ=(2cos^2θ-cosV-1)/sinθ-cosθ-1=(2cosV+1)(cosθ-1)sinθ(2cosθ-1). THe points with a horizontal tangent line satisfy dx/dy=0 and occur where the numerator is zero and the denominator is nonzero. the numerator is zero when θ=±2pi/3. Because the denominator is not zero when θ=±2pi/3, horizontal tangent lines occur at θ=±2pi/3. Vertical tangent lines occur where the numerator of dy/dx is nonzero and the denominator is zero. the denominator is zero whenθ= 0,±pi and ±pi/3 and the numerator is not zero when θ=±pi and ±pi/3. Thereofe, the vertical tangent line occurs at θ=±pi and ±pi/3. The point (0,0) on the curve must be anchored carefully because both the numerator and the numerator and the denominator of dy/dx =0 at θ=0. Notice that f(θ)=1-cosθ, where f(0) and f'(0)=0. therefore, dy/dx may be computed as a limit using L'Hopital's Rule. As θ→0+, we find that dy/dx=limθ→o+(-(2cosθ+1)(cosθ-1)/sinθ(2cosθ-1))=lim θ→0+4cosθsinθ-sinV/-2sin^2θ+2cos^2θ-cosθ=0/1=0 A siialr calculation using L'Hopital's rule shows that as θ→o-,dy/dx→0. Therefore, the curve has a slope of 0 at (0,0).

functions to power series

Find the power series representations centered at 0 for the following functions and give their intervals of convergence. a. tan^-1x b.ln(1+x/1-x) Solution: In both cases, we work with known power series an we use differentiation, integration, and other combinations. a. the key is to recall that ∫dx/1+x^2=tan^-1x+c and that by exaple 3c, 1/1+x^2=1-x^2+x^4-..., provided |x|<1. WE now integrate both sides of the last expression: ∫dx/1+x^2=∫(1=x^2+x^4)dx, which implies that tan^-1x=x-x^3/3+x^5/5-...+c. Subsituting x=0 and noting that tan^-10=0, the two sides of the eqution agree provided we choose c=0. Tehrefreo, tan_-1x=x-x^3/3+x^5/5-...=∑(∞:k=0)(-1)^kx^2k+1/2k+1. By theorem 11.5, this power series converges÷ to tan^-1x for |x|<1. testing the endpoints separateyl, we find that it also converges at x=±1.Therefore, the interval of convergence is [-1,1]. b. We have already seen (Example 4) that ln(1-x)=-x-x^2/2-x^3/3-.... replacing x with -x (property 3 of theorem 11.4), we have ln(1-(-x)=ln(1+x)=x-x^2/2+x^3/3-... . Subtracting these two power series gives ln(1+x/1-x)=ln(1+x)=ln(1=x)=(x-x^2/2+x^3/3-...)-(-x-x^2/2-x^3/3-...), for |x|<1=2(x+x^3/3+x^5/4+...)=2∑(∞:k=0)x^2k1/2k+1

Ex1: slopes on a curve

Find the slopesof the lines tangent to the circle r=f(θ)=θ. Solution:In this case, f(θ) is constant (independent of θ).Therefore, f(θ)≠0 and the slope formula becomes dy/dx=f'(θ)(csinx)+f'(θ)cosθ)/f'(θ)(cosθ)-f'(θ)sinθ=-cosθ/sinθ=cotθ. We can check a few points to see that this result makes sense with θ=0 and θ=pi. the slopedy/dx=-cotθ is undefined, which implies the tangent lines are vertical at these points. With θ=pi/2 and θ=3pi/2, the slope is zero. With θ=3pi/4 and θ=7pi/4, the slope is 1 and with θ=pi/4 and θ=5pi/4, the slope is -1. At all points, p(r,θ) on the circle, teh slope of the line 0+ from the origin p is tanθ, which is the negative reciprocal of -cotθ. therfore, Op is peprpendicular to the tangent line at all points on hte circle.

binomial series

For real numbers p≠0, the Taylor seriess for f(x)=(1+x)^p centered at 0 is the binomial series Σ(∞:k=0)(p:k)x^k=1+Σ(∞:k=0) p(p-1)(p-2)...(p-1+k)/k!x^k =1+px+p(p-1)/2!x^2+p(p-1)(p-2)/3!x^3+.... The series converges for |x|<1 (and possibly at the endpoints, depending on p). If p is a mnonegative integer, the series terminates and results in a polynomialof degree p.

useful taylor series

For the Taylor series to be useful, we need to now two things: * The values of x for which the Taylor series converges, and * The values of x for which the Taylor series for F equals f. The second question is subtle and is postponed for a few pages. For now, we find the Taylor series for f centered at a point, but we refrain from saying f(x) equals the power series.

Example 3: Combining power series

Given the geometric series 1/1-x=∑(∞:k=0)x^k=1+x+x^2+x^3+..., for |x|<1, find the powwer and interval of convergence for the following functions. a. x⁵/1-x b. 1/1-2x c/1/1+x^2 a. x^5/1-x= x⁵(1+x+x²+...)=x⁵+x⁶+x⁷+...=∑(∞k=0)x^(k+5); this geometric series has a ratio r=x when |r}=|x|<1. The interval of convergence is |x}<1. b. We substitute 2x for x in the power series 1/1-x:1/1-2x=1+(2x)+(2x)²+...=∑(∞:k=0)(2x)^k; this geometric series has a ratio r=2 and converges provided |r\=|2X|<1 OR |X|<1/2. tHE INTERVAL OF convergence is |x|<1/2. c. We substitute -x^2 for x in the power series for 1/1-x: 1/1+x^2=1+(-x^2)+(-x^2)^2+...=1-x^2+x^4=Σ(∞:k=0)(-10^kx^2k: This geometric series has a ratio of r=-x^2 and converges provided |r|=|-x^2|=|x^2|<1 or |x|<1.

Ex7: plotting polar graphs

Graph hte polar equation r^= 9cosθ. Use a graphing utility to check your work. Solution : the graph of this equation has symetry about the origin (because of r^2) and about the y-axis (because of θ). These two symetries imply symetry about the x-axis. A preliminary step is required before using the cartesian to polar method for graphing the curve. solvin the gven equation, we find that r=±3√cosθ. Notice that cosθ<0, for pi/2<θ<3i/2; so the curve does not exist on that interval. Therefore, we plot the curve on the intervals 0≤θ≤pi/2 and 3pi/2≤θ2pi. (The interval [-pi/2,pi/2] woul aslo work). BOth the positive and negative values of r are include in the cartesian graph. Now we are ready to transfer the points from the final cartsian graph to the final polar graph. The resulting curve is called a lemniscate.

cartesian to polar method for graphing r=f(θ)

Graph r=f(θ) as if r were cartesian coordinates on with θ on the horic=zona=tal axis and r on the vertical axis. Be sure to choose an interval for θ for which the entire polar curve is produced. 2. Use cartesian graph that you created in step 1 s a guide to sketch the points (r,θ) on the final polar curve.

plotting a curve

Graph the polar equation f(θ)=1+sinb. Solution: the domin of f consists of all real values of θ; however, the complete curve is generated by letting θ vary over any inteval of length 2pi. The resulting curve, called a cardioid, is symetric about the y-axis.

Ex6: plotting the grpahs

Grpah the polar equation r=2sin2θ. Solution: the cartesian graph of r=3sin2θ on the interval[0,2pi] has an amplitude of 3 and a period of pi. the θintercepts occur at θ-0, pi/2,3pi/2 and 2pi, which correspond to the intersections within the origin on the polar graph. Furthermore, arches of the cartesian coordinates between θ intercepts correspond to loops in hte polar curve.The graph is symetric about the x-axis, the y-axis and the origin. It is instructive to see how these symetries are justified. To prove symetry about the y-axis, notice that (r,θ)on hte graph→r=3sinθ→r=-3sin(-θ)→-r=3sin2(-θ)→(-r,-θ) on hte graph. We see that if (r,θ) is on hte graph, then (-r,-θ) is also on hte graph, which implies symetry about the y-axis. Similarly, to prove symetry about the oprgin, notice that (r,θ) on the graph →r=3sin2θr=3sin(2θ+2pi)=(sin2(θ+pi))→(r;θ+pi) on the graph,, We have shown that if (r+θ) is on the graph, then (r,θ+pi) is also on the grap, which implies symetry about the origin. Symetry about the y-axis and the origin imply symetry only about the x-axis. Had we proved these symetries in advance, we could have graphed the curve only if i the first quadrant, reflectors about the x- and y- axes would produce the full curve.

Identify the sreies

Identifyb the function given by the power series Σ(∞:k=0)(x-2x)^k/k! and give its interval of convergence. solution: the maclaurin series for teh expont]etial functin e^x=Σ(∞:k=0)x^k/k! converges for -∞<x<∞. Replacing x with 1-2x produces that given series Σ(∞:k=0)(1-2x)^k/k!his replacenmetn is allowed because 1-2x is within the interval of convergence of the series for e^x, that is, -∞<1-2x<∞ for all x. Therefore, the givn series represents e^1-2x, for -∞<x<∞.

Working with the remainder

In Exampple 4b of section 11.2, we show that the nth order taylor polynomial for f(x)=1n(1-x) centered at 0 is pₙ(x)=-Σ(n:k=1)x^k/k!=-x-x²/2-x³/3-...xⁿ/n. a. Find a bound on the error in approximating ln(1-x)by p₃(x) for values of x in the inerval [-1/2,1/2] b. How many terms of the taylor polynomial are needed to approximate the values of f(x)=ln(1-x) with an erro less than 10^-3 on hte intercval [-1/2,1/2]? Soution:a. The remainder for the Taylor polynomial p₃ is R₃(x)=f⁴(c)/4!x⁴, where c is between 0 and x. Computing four derivatives of f, we find that f⁴(x)=-6/(1-x)^4. On the interval [-1/2,1/2], the [-1/2,1/2], the maximum magnitude of this derivative occurs at x=1/2 (because the denominator is smallest at x=1/2) and 6/(1/2)⁴=96. Similarly, the factor x⁴has its amximum magnitude at x=±1/2, and it is (1/2)⁴=1/16. Therefore, |R₃(x)|≤96/4!(1/16)=.25 on the interval [-1/2,1/2]. The error in approximating f(x)by p₃(x), for -1/2≤x≤1/2, does not excee .25. b. for any positive integer n, the remainder Rₙ(x0=f^(n+1)/(n+1)!x^(n+1). Differentiating f several times that f^(n+1)(x)=n!/(1-x)^(n+1). On the interval [-1/2,1/2], the maximum magntude of the remainder is |Rₙ(x)|=1/(n+1)!*}f^(n+1)(c)|*|x|^n+1≤1/(n+1)!*n!2^(n+1)*1/2^(n+1)=1/n+1 To ensure that the error is less than 10^-3 on the entire interval [-1/2,1/2], n must satisfy |Rₙ|≤1/n+1≤10^-3 or n>99. The eror is likely to be significantly less than 10^-3 if x is near 0.