Power series

cos(3x^2) - 1 / x^3

(1 - 1/2(3x^2)^2 + 1/24(3x^2)^4 + ... ) - 1 / (x^2) = (-1/2(9x^4) + 1/24(81 x^8) .....)/x^3 = -9/2x + 27/8 x^5 - .... beginning of a series representation for the function

Find a power series representation for (x+3)(1/(1 −x)).

(1+x+x^2+x^3+...)(x+3) (x+3)/(1-x) = (x+3)+x(x-3) + x^2(x+3) + .... -this is not a power series-- caution!

Find a power series representation for x^2/(1 −x).

(x^2)(1/(1-x)) = (1+x+x^2+x^3+...)x^2 x^2/1-x = x^2 + x^3 + x^4 + x^5 + ...

power series examples inf,n=0 ∑n(x^n)

0 + 1x + 2x^2 + 3x^3 + 4x^4 + ... plug x=1/2 ∑n/n^2 ratio test?

power series examples inf,n=0 ∑x^n

1 + x + x^2 + x^3 + x^4 + ....

use substitutions to find power series representations for: 1/(1+2x^3)

1+(-2x^3) + (-2x^3)^2 + (-2x^3)^3 = 1-2x^3+4x^6-8x^9+... = inf,n=0∑(-2)^n(x^2n)

power series examples inf,n=0 ∑ 1/n!(x^n)

1+x+1/2x^2 + 1/6x^3 + 1/24x^4 + .... plug x=1: inf,n=0 ∑ 1/n! = 1+1+1/2+1/6+1/24 +... plug x=2: inf,n=0 ∑2^n / n! = 1+2+1/2(4) + 1/6(8) + 1/24(16) + ... not that if you plug in a value of x, this becomes a regular series

question to ask

1+x+x^2+x^3+x^4+x^5 +... questions we normally ask about a series: - does it converge? - if it converges, can we find its sum? questions we ask about this family of series: - which values of x make I a convergent series? - is the sum a nice function of x?

Use a combination of substitution and integration/differentiationto represent the following functions as power series: 1/1+x^2

1+x^2 + x^4 + x^6 + x^8... arctan(x) = x- 1/3(x^3) + 1/5(x^5) - 1/7(x^7) + ... -1 < x < 1

2 ways to find a PS representation

1. use the Taylor Series formula 2. manipulate one you already know (1/1-x ?)

use substitutions to find power series representations for: 1/(2-x)

1/2(1-1/2x) = 1/2(1/(1-(1/2)x)) t=1/2x = 1/2(1+1/2x + 1/4x^2 + 1/8x^3 + ...) = 1/2 inf,n=0 ∑(1/2^n)(x^n) = 1/2+1/4x + 1/8x^2 + 1/16x^3 + ... = inf,n=0 ∑(1/(2^(n+1)))(x^n)

use substitutions to find power series representations for: 1/(4+3x^2)

1/4(1+3/4 x^2) = 1/4 (1+ 3/4 x^2)t = -3/4 x^2

writing up your logic

A Ratio Test argument requires much less logic than some other tests. Just: - Say that you are applying the Ratio Test - Compute the limit - Write a concluding sentence. Caution: You may not be able to use L'Hopital's Rule to compute these limits! They may not be limits of nice functions! Use techniques like in previous examples where we "divide by the biggest thing". Also, remember how to cancel in factorials.

what is a power series?

A power series is a series of the form inf,n=0 ∑ Cnx^n = C0 + C1x + C2x^2 + C3x^3 + C4x^4 + ... - this x is a variable - the Cn's are constant, called the coefficients - usually we have a formula for Cn in terms of n. (on previous slide, Cn=1) here are two other power series: inf,n=0 ∑nx^n = 0+1x+2x^2+3x^3 + 4x^4 + ... inf, n=0 ∑ 1/n! x^n = 1+x+1/2x^2 + 1/6x^3 + 1/24 x^4 + ...

example compute the degree - 2 (or more?) Taylor polynomial of f(x) = 1/√(1-x^2)

Application: special relativity. The kinetic energy of an object moving at speed v is k=mc^2 (1/√(1-x^2) - 1) where x= v/c f(x) = 1/√(1-x^2) = (1-x^2)^-1/2 f'(x) = -1/2((1-x^2)^-3/2 (-2x) = x(1-x^2)^-3/2 f''(x) = (1-x^2)^-3/2 + x(-3/2)(1-x^2)^-3/2 (-2x) f(0) = 1, f'(0) = 0 f''(0) = 1+0 1/√(1-x^2)~ 1+0x+1/2! x^2 ~ 1+ 1/2x^2 K=MC^2(1+1/2x^2 - 1) = MC^2 (1/2x^2) = MC^2 (1/2)(v^2/c^2) = 1/2MV^2

ex: ∑n!(x^n) {first}, ∑1/n! (x^n) {second}, ∑x^n (third)

C0 + C1(x-a) + C2(x-a)^2 + ... plug in x=a C0 + 0 + 0 + 0 +... must converge 1. R=0. In the center of number line is a. At the center a, converges. To the left of a, diverges. To the right of a, diverges. <--------a-------> 2. R=inf. In the center of number line is a. converges at a and to the right and left of a. <------a------> 3. R=number. In the center of number line is a. from a to a+r and a-r, it is convergent. from a+r, a-r to infinity and -infinity it is divergent <----a-r----a----a+r----->

story 1: 1/1-x and 1/1+x^2

Consider the following Taylor Series; - 1/1-x = 1+x+x^2+x^3+x^4+... - 1/1+x^2 = 1-x^2+x^4-x^6+x^8+... what are the radii of converge of these series and why? 1/1-x = f(x) = 1+x+x^2+x^3+x^4+... geoseries, common ratio x converges when -1 < x < 1 f(-1) = 1/2 f(1) is undefined <-----(-1)----0----(1) ------> -1 PS rep stops converging here 1 f(x) undefined here -1 to 1 converging f(-1) is defined but PS rep can't converge beyond -1 because f(1) is undefined 1/1-t = 1+ t+t^2+... t=-x^2 1/1+x^2 = 1-x^2+x^4-x^6+... geo series, common ratio -x^2 converges when -1<-x^2<1 1>x^2>-1 x^2 > -1 is redundant sqrt(x^2) < sqrt(1) |x| < 1 -1 < x < 1 <----(-1)----0----(1)----> converge from -1 to 1 f(x) = 1/1+x^2 f(-1) = 1/1+1= 1/2 f(1) = 1/2 both are defined where is f undefined? 1+x^2 = 0 x^2 = -1 x= +/- sqrt(-1) = +/- i represent complex numbers on a "number plane" *not the xy-plane i -1+i l 1+i l -1 ---------- 1 l l -1-i -i 1-i "interval of convergence" of 1-x^2 + x^4 - x^6 + ... = 1/1+x^2

Theorem (The Root Test)

Do exactly the same thing as the Ratio Test, but instead of |aₙ + 1/ aₙ|, use ₙ√|aₙ|. Usually, anything the Root Test can do, the Ratio Test can do better. Weird expressions like n^n are the exception.

theorem on power series representations

If f has a power series representation centered at a, then the coefficients are given by the formula Cn=f^(n) (a) /n! That is, if f has a power series representation, then it is f(x) = f(a)+ f'(a)/1(x-a) + f''(a)/2! (x-a)^2 + f'''(a)/3! (x-a)^3 + ....

Taylor polynomials

If the Taylor series for f(x) converges to f(x), then the partial sums should approach f(x) these partial sums are called Taylor polynomials, and we're saying they should be good approximations to f(x). That's useful! notation: Tn(x) is the nth degree Taylor polynomial of f centered at a. (sometimes also written Ta n f(x) if a and f are not clear from context) T1(x) is the same as the linear approximation to f at x=a. T2(x) is the best quadratic approximation to f at x=a! These approximations get better and better- but more complicated! - as you increase the degree. f(x)=cos(x) illustrates this. f(x) = cos(x), a=0 T0(x) = 1 T1(x) = 1 T2(x) = 1- 1/2x^2 (T3(x) = 1- 1/2x^2) T4(x) = 1 - 1/2x^2 + 1/24 x^4 T6(x) = 1 - 1/2x^2 - 1/720x^6

Application 3: differential equations (simple?)

In physics, analyzing the behavior of a pendulum requires solving the differential equation y ′′ = C sin(y ) for a term-74constant C. This is basically impossible, but if you approximate sin(y)≈y , it's much easier. y''= Csin(y) :( y''=Cy :)

Less Nice Functions

Less nice functions may not have a pattern in their derivatives. You can still find the beginning of their Taylor series: - Take the first few derivatives of f (x). - Evaluate those derivatives at x = a. - Plug the results into the formula for the Taylor series coefficients. Try this with f (x ) = sec(x ) and f (x ) = √x + 4.

Differentiating and Integrating: Example

Let f (x) = 1/1−x. f (x ) = 1 + x + x^2 + x^3 + x^4 + ···=∞,n=0∑x^n Then f′(x) = 1/(1−x)^2, so 1/(1-x)^2 = 0+1+2x+3x^2+4x^3+...= inf,n=1 ∑n(x^n-1). this gives a power series representation for 1/(1-x)^2

Basic Example Given a function f (x), can we find a power series representation for it? That is, can we find a power series so that ∑Cn(x^n) = f(x) ?

Let f(x) = 1/(1-x). Then we already know that f(x) = 1+x+x^2 + x^3 + x^4 + ... (note that this is only valid if -1 < x < 1. That's okay.) this is a power series representation for f(x) = 1/(1-x) 1+x+x^2+x^3+.... sums to 1/(1-x) 1/1-x is represented as a power series by 1+x+x^2+x^3+....

Other Center Points

Recall the general construction of the Taylor series: f(x) = f(a) + f'(a)/1 (x-a) + f''(a)/2! (x-a)^2 + f'''(a)/3! (x-a)^3 +... use this to find the first few terms of: - The Taylor series for sin(x ) centered at x = π - The Taylor series for ln(x ) centered at x = 1 - The Taylor series for 1/x centered at x = 2

manipulating power series

Some things you can do to change a power series: - Multiply by constants - Multiply by powers of x - Substitute into it - Differentiate it (!) - Integrate it (!)

differentiating and integrating

Suppose you have a power series for f. Then you can get a power series for f′ by differentiating each term of the series individually. You can also get a power series for ∫ f(x) dx by integrating each term individually. These power series have the same radius of convergence as the one you started with.

Differentiating and Integrating: Technical Details

Suppose ∑Cn(x −a)^n has radius of convergence R (R >0), and f(x ) =∞,n=0∑Cn(x −a)^n. Then f is differentiable on the interval (a −R ,a + R), and f′(x) =∞,n=0∑d/dx Cn(x−a)^n=∞,n=1∑nCn(x −a)^n−1. Also, on the interval (a −R ,a + R), ∫f (x) dx =∞, n=0∑∫Cn(x −a)^n dx = C +∞, n=0 ∑[C^n/(n + 1)] (x −a)^n+1.

theorem (the ratio test)

Suppose ∑aₙ is a series. If lim n-> infinity |aₙ+1 / aₙ| = L and L<1, then ∑aₙ converges absolutely. (and therefore, it converges) If lim n->infinity |aₙ+1 / aₙ| > 1, or infinity, then ∑aₙ diverges. If lim n-> infinity |aₙ + 1 / aₙ| = 1, or the limit does not exist, then the Ratio Test is inconclusive and does not tell us anything about ∑aₙ

convergence of Taylor Series

The Taylor Series for e^x, sin(x), and cos(x) converge for all x. But how do we know their sums are equal to the original functions? What is if e^x, sin(x), and cos(x) don't have power series representations?

The Binomial Series

The Taylor series for (1 + x)^k is called the binomial series: (1 + x)^k= 1 + kx + k(k −1)/2!(x^2) + k(k −1)(k −2)/3!(x^3)+ k(k −1)(k −2)(k −3)/4!(x^4) + ··· (where k is any real number) This converges for −1 < x < 1, but it's hard to prove.

Methods for Finding Representations

There are basically two ways to represent f (x) as a power series: - Today: Start with 1/1 −x = 1 + x + x^2 + x^3 + x^4 + ··· and manipulate both sides until you get f(x) on the left. - Friday/next week: use Taylor series

What is∫ 1/(1 −x^5) dx ?

This antiderivative is awful to compute. But we can find a series representation for the answer as follows: ∫1/(1 −x^5) dx =∫(∞,n=0∑(∫x^5)^n)dx = inf,n=0 ∑(∫x^5n dx) = C + inf,n=0 ∑ 1/(5n+1) (x^5n+1)

Taylor Series from Geometric Series

We already know the power series representation for 1/1 −x : 1/1 −x = 1 + x + x^2 + x^3 + ···=∞,n=0∑x^n we can use this to find two other useful Taylor series: - ln(1+x) 1/1+x = 1-x+x^2-x^3+... = inf,n=0 ∑(-x)^n = inf,n=0∑(-1)^n x^n ln(1+x) = x-1/2x^2 + 1/3x^3 - 1/4x^4 + ... = inf,n=0∑(-1)^n+1 / n x^n -arctan(x) x -> (-x^2) 1/(1+x^2) = 1-x^2 + x^4 - x^6 + ... = inf,n=0 ∑(-1)^n(x^2n) arctan(x) = x- 1/3x^3 + 1/5x^5 - 1/7x^7 +... = inf,n=0 ∑(-1)^n / 2n+1 (x^2n+1) each of these converges when -1<x<1

Application 1: Cute Sums

You can evaluate the following sums by recognizing them as Taylorseries of a function: inf,n=0 ∑1/n! = 1 + 1 + 1/2! + 1/3! + 1/4! inf,n=0 ∑(−1)^n/2n + 1 = 1 − 1/3 + 1/5 − 1/7 + 1/9 −··· This is usually not an effective way of evaluating sums, unless you get very lucky. You can use it to approximate e (very effectively) or π (very badly).

Manipulating Taylor Series

You can manipulate Taylor series by substituting in, differentiating,integrating, etc. Examples: - Find the Taylor series for e^x^2 centered at x = 0. - What happens if you differentiate the Taylor series for e^x?What about sin(x) and cos(x)? - Find a series representation for an antiderivative of e^x^2. - Find the first few terms of the Taylor series for e^xsin(x). In the last example, you can also multiply Taylor series, but youhave to be very careful and not expect a nice formula for thecoefficients.

intro to power series

a family of series consider the following expression: 1+x+x^2+x^3+x^4+x^5 if you plug in a value of x, you get a series. Different values of x give different series. This kind of thing is called a power series. x=2: 1+2+4+8+16+... Geo, r=2, diverges converges if -1<x<1 x=1/2: 1+1/2+1/4+1/8+1/16 geo, r=1/2, converges sum= 1/(1-1/2) = 2 x=-1/3 1-1/3+1/9-1/27+1/81 ... geo, r=-1/3, converges sum=1/(1-(-1/3)) = 3/4

power series representations

a power series representation for a function f centered at a is a power series. inf,n=0 ∑Cn(x-a)^n = C0+C1(x-a)+C2(x-a)^2 + C3(x-a)^3 +... such that the sum of the series is equal to f(x) for all x in the interval of convergence of the series. (most commonly a=0, but not always) goal: given an arbitrary function, find a power series representation for it

center points

a series of the form inf,n=0 ∑Cn(x-a)^n = C0 + C1(x-a) + C2(x-a)^2 + C3(x-a)^3 + ... is called a power series centered at a. (If a=0, we get the simpler power series in previous examples) ex: inf,n=0∑(x-a)^n = 1+(x-a) + (x-a)^2 + (x-a)^3 + ... is a geometric series with common ratio (x-a). It converges if a-1 < x < a+1, and diverges otherwise.

example: for what values of x does inf,n=0 ∑ n(x^n)

apply the ratio test before plugging in for x! test all x at once! aₙ+1/aₙ = [((n+1)x^n+1) / (nx^n)] = [(n+1)/n]x lim n->inf (n+1 / n)x = x(lim n->inf (n+1)/n) = x(1) RT: If -1<x<1, converges (abs) if x<-1 or x>1, div if x=1 or x=-1, don't know

Use the Root Test on ∞, n=1 ∑1/n^n. Is it better than the Ratio Test?

aₙ + 1/ aₙ = [1/(n+1)^n+1]/[1/n^n] = n^n / (n+1)^n+1 = n^n/[(n+1)^n(n+1)] = (n/n+1)^n 1/n+1 lim n->inf (n/n+1)^n is 1^inf root test: √|aₙ| = ₙ√1/n^n = 1/ₙ√n^n = 1/n lim n->inf ₙ√|aₙ| = lim n-> inf 1/n = 0, converges

use the ratio test to determine the convergence of the following series: inf, n=0 ∑1/n^3

aₙ = 1/n^3 aₙ+1 = 1/(n+1)^3 aₙ+1/aₙ = [(1/(n+1)^3)/ (1/n^3)] = n^3/(n+1)^3 lim n-> inf |aₙ+1/aₙ| = lim n->inf((n/n+1)(1/n / 1/n))^3 = lim n-> inf (1/(1+ 1/n))^3 = (1/1+0)^3 = 1 RT inconclusive

use the ratio test to determine the convergence of the following series: inf, n=0 ∑n^5 / 5^n

aₙ = n^5/5^n aₙ+1 = (n+1)^5/5^(n+1) aₙ+1/aₙ = ((n+1)^5/5^(n+1)) / (n^5/5^n)= [(n+1)^5(5^n) / (5^n+1)(n^5)] = 1/5((n+1)^5/n^5) = (n+1/n)^5 (1/5) = (1+1/n)^5(1/5) then lim n->inf |aₙ+1/aₙ| = lim n-> inf (1+ 1/n)^5(1/5) = (1+0)^5 (1/5) = 1/5 therefore, by the Ratio Test, inf,n=0∑n^5/5^n converges

use the ratio test to find the radius of convergence of the series inf,n=0∑3^n(x-2)^n

aₙ+1/aₙ = 3^n+1(x-2)^n+1 / 3^n(x-2)^n = 3(x-2) lim n->inf |aₙ+1/aₙ| = |3(x-2)| <1 (to converge) -1 < 3(x-2) < 1 -1/8 < x-2 < 1/3 5/3 < x < 7/3 numberline <----5/3---2---7/3----> center is 2 from 5/3 to 2 to 7/3, converges from -inf to 5/3, diverge from 7/3 to inf, diverge R=1/3 IOC (5/3, 7/3)

ex: find the radius and interval of convergence of each of the following: inf,n=0 ∑(2x+1)^n / 3^n

aₙ+1/aₙ = [((2x+1)^(n+1) / 3^(n+1))/((2x+1)^n / 3^n)] = 2x+1 / 3 lim n->inf |aₙ+1/aₙ| = lim n->inf |2x+1/ 3| = |2x+1/ 3| < 1 -1 < 2x+1/ 3 < 1 number line <----(-2)----(-1/2)----(1)----> -3< 2x+1 < 3 -1< 2x < 2 -2< x < 1 RoC= 3/2 x=-2 ∑(-3)^n/3^n = ∑(-1)^n div RoC= 3/2 IoC= (-2, 1)

for which values of x does inf,n=0∑ 1/n(x^n) converge?

aₙ= 1/n(x^n) aₙ+1 = 1/(n+1)(x^n+1) aₙ+1/aₙ = [(1/(n+1)(x^n+1))/(1/n(x^n))] =n/n+1 (x) lim n->inf |aₙ+1/aₙ| = lim n->inf n/n+1 |x| = 1-|x| = |x| |x| < 1: converges (-1<x<1) |x| > 1: diverges (x<-1 or x>1) |x|=1: ??? (x=-1 or x=1) number line: <----(-1)-----0-----(1)----> radius of convergence: 1 x=1 inf,n=1∑1/n(1^n) = inf,n=1∑(1/n) = diverges x=-1 inf,n=1∑1/n(-1)^n -- converges (AST) interval of convergence: [-1, 1)

use the ratio test to determine the convergence of the following series: inf, n=0 ∑ (10^n/n!)

aₙ= 10^n/n! aₙ+1= 10^n+1/(n+1)! aₙ+1/aₙ= 10^n+1/(n+1)!/ 10^n/n! = [(10^n+1)(n!) / (10^n)(n+1)!] = 10/n+1 lim n->inf |aₙ+1 / aₙ| = lim n->inf 10/n+1 = 0 series converge

infinity,n=0 ∑n!x^n

aₙ=n!x^n aₙ+1=(n+1)!x^n+1 aₙ+1/aₙ = (n+1)!x^n+1 / n!(x^n) = (n+1)x lim n->inf |(n+1)(x)| = inf except when x=0 RoC=0 IoC= {0} ∑Cn(x-a)^n (2x+1)^n = (2(x+1/2))^n = 2^n(x+1/2)^n center: x=-1/2

story 3: e^x, sin(x), and cos(x)

consider the following three Taylor series: - e^x = 1+x+ 1/2x^2 + 1/6x^3 + 1/24x^4 + 1/120 x^5 +... - sin(x) = x-1/6x^3 + 1/120x^5 + ... - cos(x) = 1-1/2x^2 + 1/24x^4 +... these look almost similar, but not quite. Why? e^x = 1+x+ 1/2x^2 + 1/6x^3 + 1/24x^4 + ... e^ix = 1+(ix) + 1/2(ix)^2 + 1/6(ix)^3 + .... 1/24 (ix)^4 (i)^2 = -1 (i)^3 = (i)(i)^2 = (i)(-1) = (-i) (i)^4 = (i)(i)^3 = (-i)(i) = (-1)(i)^2 = (-1)(-1) = 1 (i)^5 = (i)(i)^4 =(i) (i)^6 = -1 (i)^7 = -(i) (i)^8 = 1 = 1+ix -1/2x^2 - i1/6x^3 + 1/24x^4 + i1/120x^5 = 1-1/2x^2+1/24x^4 - ... + (i)(x-1/6x^3 + 1/120x^5 -....) = cos(x) + isin(x) e^ix = cos(x) isin(x) Let x=pi e^ipi = cos(pi) + isin(pi) / 0 e^ipi = -1

e^x = inf,n=0∑ 1/n!(x^n)

e= inf,n=0∑ 1/n! e^x ~ 1+x+1/2x^2 + 1/6x^3 + 1/24x4 e~ 1+1+1/2+1/6+1/24 = 2+ 17/24

manipulating the Taylor series: Find a series representation for an antiderivative of e^x^2.

e^x = 1+x+1/2x^2 + 1/6x^3 + ... = inf,n=0∑1/n!(x^n) e^x^2 = 1+(x^2) + 1/2(x^2)^2 + 1/6(x^2)^3 + ... = inf,n=0 ∑1/n! (x^2)^n = 1+x^2 + 1/2x^4 + 1/6x^6 + ... = inf,n=0 ∑1/n!(x^2n) series for 1/1-x converges when -1<x<1 x-> (-2x) series for 1/(1+2x) converges when -1<-2x<1 1/2 > x > -1/2 d/dx e^x = d/dx(1+x+1/2x^2 + 1/6x^3 + 1/24x^4 + ...) = d/dx inf,n=0∑1/n!(x^n) e^x= 0+1+x+1/2x^2+1/6x^3 + ... = inf, n=0 ∑n/n!(x^n-1) = inf, n=0 ∑1/(n-1)! (x^n-1) ∫e^x^2 dx <-- impossible to do this ∫e^x^2 dx = ∫(1+x^2+1/2x^4 + 1/6x^6 + ...) dx = ∫ inf, n=0 ∑1/(n-1)! (x^2n) dx = C+1/3x^3 + 1/10x^5 + 1/42 x^7 + .... = C + inf,n=0 1/n!(2n+1) (x^2n+1)

TS for e^x is inf,n=0 ∑1/n!(x^n). Is e^x = inf,n=0 ∑1/n!(x^n) ?

e^x= 1+x+1/2x^2 + 1/6x^3 + 1/24x^4 + ... 1 is not very close to e^x 1+x is kind of close 1+x+1/2x^2 is closer 1+x+1/2x^2 + 1/6x^3 is clear

Taylor (and Maclaurin) Series

f(x) = f(a) + f'(a)/1 (x-a) + f''(a)/2! (x-a)^2 + f'''(a)/3! (x-a)^3 +... this series is called the Taylor Series for f centered at x=a. The Taylor series centered at x=0 is common enough that it's also called the Maclaurin series for f

Find the Taylor series at x=0 for e^x, sin(x), and cos(x)

f(x)=e^x Cn=f^(n)(n)/ n! = f^(n)(0)/n! n: 0, 1, 2, 3, 4 f^(n)(x): (0!=1) e^x, e^x, e^x, e^x f^(n)(0): 1, 1, 1, 1 Cn: 1, 1, 1/2, 1/6, 1/24 Taylor series for e^x 1+1x+1/2x^2 + 1/6x^3 + 1/24x^4 + .... notice: f^(n) (0) = 1 Cn=f^(n) (0)/ n! = 1/n! inf,n=0 ∑1/n! (x^n)

theorem for power series representation

if f has a power series representation centered at a, then the coefficients are given by the formula Cn=(f^(n) (a))/n! that is, if f has a power series representation, then it is f(x)=f(a) + f'(a)/1 (x-a) + f''(a)/2! (x-a)^2 + f'''(a)/3! (x-a)^3 +... you can prove this by differentiating the power series n times, and plugging in x=a.

inf, n=0 ∑(x+2)^n

inf,n=0∑(x-(-2))^n centered at x=-2 geo, r= x+2 converges when -1<x+2<1 -3<x<-1 sum= 1/(1-(x+2)) = 1/(-x-1)

A power series ∑Cn(x^n) is a function of x. Can we give a better formula for this function? Usually not: That would require finding the sum, not just convergence testing.

instead, reverse this question. Given a function f (x), can we find a power series representation for it? That is, can we find a power series so that ∑Cn(x^n) = f(x) ?

terminology review

interval of convergence: the set of all x that make your power series converge radius of convergence: half the width of the IoC. How far you can get from the center point and have the series still converge.

Find the Taylor series at x=0 for sin(x), and cos(x)

n: 0, 1, 2, 3, 4, 5 f^(n)(x): sin(x), cos(x), -sin(x), -cos(x), sin(x), cos(x) f^(n)(0): 0, 1, 0, -1, 0, 1 Cn: 0, 1, 0, -1/6, 0, 1/120 Taylor series for sin(x): x-1/6x^3 + 1/120 x^5 - 1/7! x^7 +... - even terms not present - signs are alternating - 1/(exponent)! more formal inf,n=0 ∑[(-1)^n/(2n+1)!](x^2n+1)

Find the Taylor series at x=0 for cos(x)

n: 0, 1, 2, 3, 4, 5, 6 f^(n)(x): cos(x), -sin(x), -cos(x), sin(x), cos(x), -sin(x), -cos(x) f^(n)(0): 1, 0, -1, 0, 1, 0, -1 Cn: 1, 0, -1/2, 0, 1/24, 0, -1/720 Taylor series for cos(x): 1-1/2x^2 + 1/24 x^4 - 1/720 x^6 + ... -only even exponents show - signs still alternate - 1/(exponent)! inf, n=0 ∑[(-1)^n/(2n)!)]x^2n note that these require finding a pattern in the derivatives of the function to work well!

sec(x)

n: 0, 1, 2 f^(n)(x): sec(x), sec(x)tan(x), (secxtanx)tanx + secx(sec^2x) = secxtan^2x + sec^e3x n: 0, 1, 2 f^(n)(0): 1, 0, 1 Cn= f^(n)(0)/n! 1, 0, 1/2 sec(x) = 1+ 0x + 1/2x^2 + - write out what you have and don't necessarily have a pattern and that's okay - not going to have pattern - stop when you want to stop - derivs get very complicated and will probably not get far enough where you see a pattern where there is one

The Taylor series for 1/x centered at x=2

n: 0, 1, 2, 3 f^(n)(x): 1/x -1/x^2=-x^-2 2x^-3 -6x^-4 = -6/x^4 f^(n) (2) 1/2 -1/4 2/8=1/4 -6/16= -3/8 Cn=f^(n)(2)/n! 1/2 -1/4 1/8 -1/16 1/2-1/4(x-2)+1/8(x-2)^2 - 1/16(x-2)^3 +...

Application 2: Compound Interest A loan nominally charges 6% annual interest, compounded monthly. Approximate the actual annual interest rate you pay on this loan. Translation: Approximate (1 + 0.005)^12.

n: 0, 1, 2, 3 f^(n+1)(x): (1+x)^k, k(1+x)^k-1, k(k-1)(1+x)^k-2, k(k-1)(k-2)(1+x)^k-3 P: principal After 1 month: P(1.005) 2 months: P(1.005)^2 12 months: P(1.005)^k k= actual rate you pay (1+0.005)^12 ~ 1+12(0.005) + 66(0.005)^2 = 1.06165 ~ 6.165% actual rate (1+x)^k x=0.005 k=12 1+12x + 12(11)/2! x^2 1+ 12x + (12(11)/2!) x^2 ~ 1+ 12x + 66x^2

sqrt(x+4)

n: 0, 1, 2, 3, 4 f^n)(x): (x+4)^1/2 1/2(x+4)^-1/2 -1/4(x+4)^-3/2 3/8(x+4)^-5/2 -15/16(x+4)^-7/2 f^(n)(0): 2 1/2(4)^-1/2 = 1/2(2)^-1 = 1/4 -1/4(4)^-3/2 = -1/4(2)^-3 = 1/32 3/8(2)^-5 = 3/256 - 15/16(2)^-7 = -15/2048 Cn: 2, 1/4, -1/64, -1/512, -5/16384 sqrt(x+4) = 2+1/4x - 1/64x^2+ 1/512 x^3 - 5/16384 x^4 + ...

aside: defining e

one way to define the number e works like this: - define a function exp(x)= 1+x+1/2x^2 + 1/6x^3 + ... - by the ratio test, this converges for all x, so the domain of exp is all real numbers - define e=exp(1) (then you should also prove things like exp(a+b) = exp(a) (exp(b)) this is hard

the ratio test

power series have terms with n in an exponent, so the ratio test is usually the best tool to analyze them. (rarely, the root test) example: for what values of x does inf,n=0 ∑ n(x^n) converge? - the ratio of consecutive terms approaches x - the series converges if |x|<1, and diverges if |x|>1. - if |x| = 1, then we have to use a different test.

example 2 for which values of x does inf,n=0 ∑(1/n!)x^n converge?

since aₙ=1/n!(x^n), compute the limit of ratios: aₙ+1/aₙ = [(1/(n+1)!(x^(n+1)))/ (1/n!(x^n))] = [(n!/(n+1)!)(x^n+1 / x^n)] = 1/n+1(x) lim n->inf aₙ+1/aₙ = lim n-> inf x/n+1 = x(lim n->inf 1/ n+1) = x(0) = 0 converges regardless of x!

disclaimer: when it goes wrong

sometimes Taylor Series converge, but not to f(x). This is hard to detect: even if f is infinitely differentiable for all real numbers, it may have a badly behaved Taylor series. This is thankfully very rare, and you shouldn't worry about it.

disclaimer: when it goes wrong Taylor series converge

sometimes Taylor series converge, but not to f(x). This is hard to detect: even if f is infinitely differentiable for all real numbers, it may have a badly behaved Taylor series very rare

convergence

the Taylor series for e^x, sin(x), and cos(x) converge for all x. But how do we know their sums are equal to the original functions? what if e^x, sin(x), and cos(x) don't have power series representations? they do, but it's hard to prove. You can use: theorem: the functions e^x, sin(x), and cos(x) are all equal to the sum of their Taylor series.

caution for radius and interval of convergence

the concepts of radius and interval of convergence apply only to power series. A series like ∑1/n is not a power series, so it doesn't make sense to talk about its radius or interval of convergence.

story 2: e^-1/x^2

the following function behaves very strangely: f(x) = {e^-1/x^2, x≠0 ; 0, x=0} one can show that its Taylor Series is 0+0x+0x^2+0x^3+0x^4+0x^5 +... But clearly this is not equal to f(x) what if x is not real? f(ai) = e^-1/(ai)^2 = e^-1/a^2i^2 = e^-1/a^2(-1) = e^1/a^2 lim x->0 f(x) =0 lim x->0 f(ai) = lim a-> 0 e^1/a^2 = inf I/a^2 -> inf

interval of convergence

the interval of convergence of a power series is the interval consisting of all values of x for which the series converges. ex: the interval of convergence of inf, n=0∑(x-1)^n is (0,2). unless the radius of convergence is 0, the interval of convergence is one of: - (-inf, inf) - (a-R, a+R) - (a-R, a+R] - [a-R, a+R) - [a-R, a+R]

radius of convergence

the number R in the theorem is called the series' radius of convergence. Also, by convention: - if the series converges only for x=a, then R=0. - if the series converges for all x, then R=inf

What is e?

the number e is special in many ways: - the horizontal asymptote of the graph of f(x) = (1+ 1/x)^x is y=e - inf, n=0 ∑1/n! = e - the function f(x)=e^x is its own derivative - the ideal number of options at each level of a phone menu is e some of these can even be used as definitions of e. inf,n=0 ∑1/n! (x^n) = exp(x) (=e^x) exp(1) = e e^a+b = e^a(e^b) exp(a+b) = exp(a) (exp(b) ) exp([1. 3]) = ? 2 4

uses of the ratio test

the ratio test is great for series where the terms contain factorials it is also good for series involving exponential expressions. CAUTION: - the ratio test does not tell you the value of the sum! It only tells you if it converges or diverges! - watch out for the "inconclusive" case! Then you need to use a different test. - the ratio test is NOT good for rational functions; it'll usually be inconclusive

process for analyzing power series

there are two steps to analyzing a power series: - apply the Ratio Test. This will tell you the radius of convergence - if R is finite and nonzero, then test the endpoints of the intervals of convergence. Plug in these values of x, and use another test to determine if the resulting series converges In the 2nd step, don't use the Ratio Test or Root Test; they will both be inconclusive. Commonly useful tests include the Alternating Series Test, Limit Test, and Limit Comparison Test

substituting

we can substitute into the geometric series. For example: 1/(1-t) = 1+t+t^2+t^3+t^4+...= inf,n=0 ∑(t^n) substitue t=2x to obtain: 1/(1-2x) = 1+2x+4x^2 +8x^3 + 16x^4 + ... = inf,n=0 ∑2^n(x^n) = inf,n=0 ∑(2x^n) or substitute t=-x^2 to obtain: 1/(1+x^2) = 1-x^2 + x^4 - x^6 +x^8 - ... = inf,n=0∑(-x^2)^n = inf,n=0∑(-1)^n(x^2n) now we have power series representations for 1/1-2x and 1/1+x^2 ! good t=cx^k don't do this t=e^x 1/(1-e^x) = 1+e^x+e^2x+e^3x+... - not a PS, not fixable probably don't do this t=x-1 1/(1-(x+1)) = 1+(x+1) + (x+1)^2 + (x+1)^3 + ... still a PS, but centered at x=-1

three possible outcomes

when testing a power series for convergence, there are only three possible outcomes theorem a power series ∑Cn(x-a)^n has one of the following behaviors: - it converges fo x=a, and diverges for all other x. - it converges (absolutely) for all x - there is a positive number R such that the series converges (absolutely) when |x-a| < R, and diverges when |x-a| > R. in the third case, anything can happen when |x-a| = R; the Ratio Test will be inconclusive.

convergence and sums

whether or not a power series converges may depend on x. If it does converge, its sum may also depend on x. Example: 1+x+x^2+x^3+x^4+x^5+.. this converges if -1<x<1, and diverges otherwise. When it converges, its sum is 1/1-x

Use a combination of substitution and integration/differentiationto represent the following functions as power series: ln(1+x)

∫(1-x+x^2-x^3+...)= ∫inf,n=0∑(-1)^n(x^n) ln(1+x) = C+x-1/2x^2 + 1/3x^3 - 1/4x^4 + .... = C + inf,n=0 ∑((-1)^n)/n+1 (x^n+1) what should C be? plug x=0 ln(1+0) = C+0+1/2(0) + 1/3(0) + ... 0=C ln(1+x) = inf,n=0 ∑((-1)^n / (n+1))(x^n+1) for -1 < x < 1