MCHE 301-Test 1
In general, which would have better truncation error, the Backward Difference formula using a step size of 0.0000001 or the Central Difference Formula with a step size of 0.001? A. Backward B. Central
Solution: *Example formula: F(x) = x^2 at x=2 *Backward Difference Formula: F'(3)= [(3^2 - 2.9999999^2)/.0000001] = 5.999999 *Central Difference Formula: F'(3)= [(3.001^2 - 2.999^2)/0.002] =5.99999 **Answer: A. Backward
What is the base-10 equivalent of the unsigned integer stored in binary as shown below? 01001101
Solution: 128 64 32 16 8 4 2 1 0 1 0 0 1 1 0 1 = 1+4+8+64= 77
Use the rectangular rule to integrate the function f=6x² from x=2 to x=5 using 4 strips. Report your answer to two decimal places.
Solution: h = (5-2)÷4 = 0.75 I = 0.75 [6(2)² + 6(2.75)² + 6(3.5)² + 6(4.25)² ] = 188.44
Assume that we are storing real numbers in 8 bits, using one sign bit, a 3 bit exponent, and a 4 bit fraction. What is the biggest number we can store? 01101111
Bias is 3. Exponent is 6. Fraction is 1/2 + 1/4 +1/8 +1/16 = .9375 (+1)[2^(6-3)](1+.9375) = 15.5
What color is represented by the hexadecimal number: FF00FF? A. Red B. Yellow C. Purple D. Turquoise
C. Purple When a color is stored in a database through a hexadecimal code, it is arranged in the order of Blue, Green, and then Red. So in reference to the spectrum provided at the bottom of the page, it can be observed that in the code FF00FF, Blue and Red are set at their max values while Green is set at zero, ultimately creating an equal mixture of the two colors producing a shade of purple.
Use the backward difference formula to compute the derivative of the function f(x) = exp^(x) - 10, at the point x = 4. Use a step size of 0.001. Report the answer to two decimal places.
F(x) = exp^(x) - 10 ; x=4 ; h=0.001 F(x) = [(e^4 - 10)-(e^3.999 - 10)]/0.001 F(x) = 54.57086 F(x) = 54.57
In general, which would have better truncation error, the Forward Difference formula using a step size of 0.0000002 or the Central Difference Formula with a step size of 0.002?
Forward Difference formula! *The forward difference formula is a first order scheme since the error goes as the first power of h. The truncation error is bounded by (Mh)/2 where M is a bound on f '(t) for t near x. Thus the formula is more and more accurate with decreasing h since the truncation error is then smaller.
What ASCII character is represented by the unsigned integer stored in binary as shown below? 01100001 a) = b) N c) a d) ]
-ANSWER: c) a -SOLUTION: *Binary interpretation of 01100001 shown below: 0 1 1 0 0 0 0 1 128 64 32 16 8 4 2 1 Which gives 64 + 32 + 1 = 97 Which is the decimal number for the character "a"
What is the base 10 equivalent of the real number stored in single precision according to the IEEE standard, as shown below? (Assume there is one sign bit, a bias of 127, and 8 bit exponent, and a 23 bit fraction)? 01100000011100000000000000000000
(+1)(1+ ½+ ¼+ 1/8) (2^((128+64)-127)) =(+1)(1.875)(2^65) =6.92x10^19
Assume that we are storing real numbers in 32 bits, using one sign bit, a 8 bit exponent, and a 23 bit fraction. What is the biggest number we can store? 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
(+1)×(2^(254-127) )×(1 + 1/2 + 1/4 + 1/8 + ⋯+ 1/8,388,608) = 3.402823669x10^38
What is the next biggest number that can be stored in base 10, in single precision of the real number stored below? 0 10000000 11000000010000000000000
(1) (2^(128-127))[1+(2^-1)+(2^-2)+(2^-10)+(2^-23)] = 3.501953363
Assume we are using the 'Forward Difference' formula to compute the derivative of the formula f(x) = x^3 + 6x at the point x=3. (using step size h = 0.01). Assuming we are using single precision to store our numbers, what type of error is dominating, and how could we minimize the total error of the derivative?
(Single precision) → E_T≈ 10^(-7) & (Forward Difference) → E_T≈ o(h) *Since h = 0.01 is >> than E_T≈ 10^(-7) than we can state that the truncation error will dominate the derivative. (10^-2 > 10^-7). A good way to minimize the error is to decrease the step size. Decreasing the step size will make the approximation more accurate, and will also adjust the relationship between the truncation error (E_T) and the step size (h).
Assume we are going to approximate the value of the function f=x4 at the point x=2.5, using the first four terms of the Taylor Series expanded about the point x=2.0. Calculate the truncation error.
0.0625
What ASCII character is represented by the unsigned integer stored in binary as shown below? 00100101
128 64 32 16 8 4 2 1 0 0 1 0 0 1 0 1 32+4+1=37 #37 on the ASCII Chart is %
Using the rectangular rule to integrate the function f= x^5 from x=1 to x=5 using 5 strips. Report your answer in two decimal places.
1519.98
Convert the number 2546 from base 6 to a base 10 number.
2*(6^2) + 5*(6^1)+4*(6^0) = 72 + 30 + 4 = 106
Use the trapezoidal rule to ingrate f=x^4 from x=1 to x=4 using 4 strips. Report your answer to two decimal places.
216.38
Assume that we are storing real numbers in 32 bits, using one sign bit, a 12 bit exponent and a 19 bit fraction. How big should the bias be?
2^(12-1) - 1 = 2047
What is the base-10 equivalent of the unsigned integer stored in binary as shown below? 100110110
2^8 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 Binary: 1 0 0 1 1 0 1 1 0 Base10:256 0 0 32 16 0 4 2 0 *Add up all of the result for base-10 we get: 2 + 4 + 16 + 32 + 256 = 310
Assume we are using the forward difference formula to compute the derivative of the function f(x)=x^5-8x+7, at the point x = 3, using a step size of 0.01. Assume we are using single precision to store our numbers and that we can calculate the second derivative analytically. Calculate the optimum step size. Report your answer to five decimal places.
Solution: h(opt) = √[(±4e)/(f''(x)] ; e = 1 × 10^-7 (single precision) f'(x) = (5x^4)-8 ; f''(x)=20x^3 h(opt) = √[(4( 1×10^-7))/(20(3^3))] h(opt) = 6.824×10^-6 *Answer: h(opt) = 2.722 × 10^-5
Approximate the value of the function f(x) = x^5 + 17x at point x = 4.7 using the first five terms of the Taylor Series expanded about the point x = 4. Round to 3 decimal places.
Solution: *Use an equation of the form R_n= (f^5 (x))/n!*h^5. *After deriving our original function five times we find that f^5 (x)=120. *We know that h= x_i-x so we solve to get h = 4.7 - 4 = .7 *Now we simply plug our values into our equation: R (base n) = (120/5!)*(.7^5) = .168
Convert the number 2675 from base 5 to a base 10 integer.
Solution: 2(5^3 ) + 6(5^2 ) + 7( 5^1) + 5(5^0 ) = 440 440 ... base 10 4(10^2 ) + 4(10^1 ) + 0(10^0 ) = 440 ... base 10
Assume we are storing real numbers in 32 bits, using one sign bit, an 8 bit exponent, and a 23 bit fraction. What is the "machine epsilon," Ꜫ?
Solution: *The machine epsilon is equal to 2^(-x), with x being the number of fraction bits. *So the machine epsilon for 32 bits is: 2^(-23) = 1.19209e-7
Use the central difference formula to compute the derivative of the function f(x)=x^7 - 10, at the point x = 4. Use a step size of 0.001. Report the answer to three decimal places.
Solution: f(x) = x^7 - 10 f'(x) = 7x^6 f'(4) = 7(4)^6 =28672 f'(x) = [f(x(i+1))-f(xi)/(2h)] + O(h^2 ) = [f((4.001^7)-10) - f((3.999^7)-10)]/(2(.001)) = 28672.00748
Use the central difference formula to compute the derivative of the function f(x)=x^7 - 10, at the point x = 4. Use a step size of 0.001. Report the answer to three decimal places.
Solution: f(x) = x^7 - 10 f'(x) = 7x^6 f'(4) = 7(4)^6 = 28672 f'(x) = [(f(xi+1))-f(xi)]/(2h) + O(h^2 ) = [(4.001^7 - 10)-(3.999^7 - 10)]/[2(.001)] = 28672.009
Assume that we are storing real numbers in 32 bits, using one sign bit, a 10 bit exponent, and a 21 bit fraction. What is the "machine epsilon," ε?
Solution: n = bits in fraction ε = 1/(2^n) ε = 1/(2^21) ε = 4.76837x10^-7
What ASCII character is represented by the unsigned integer stored in binary as shown below? 00011011
Solve: 00011011 = 1(2^0) + 1(2^1) + 0(2^2) + 1(2^3) + 1(2^4) + 0(2^5) + 0(2^6) + 0(2^7) = 27 So answer is Escape
What type of major error occurs whenever you use an approximation in the place of an exact mathematical procedure?
Truncation Error
If we are using the forward difference formula to approximate a function of x (in single precision), and using a step size of .1, what kind of error is dominating would we expect and how could we lessen that error.
Truncation error and we could reduce the step size to a value such as .001 or .0001
If you have a problem where there are lots of approximations in the function values, what type of error would be likely to dominate when calculating numerical derivatives? a) Round-off error b) Truncation error
b) Truncation error
Which sort of error stems from using approximation methods instead of accurate calculations? a) Roundoff error b) Truncation error c) Total engineering error
b) Truncation error
Find the ASCII character represented by the binary number below: 100110 a) % b) { c) ' d) &
d) &
Use the trapezoid rule to integrate the function f=x^3 from x = 2 to x = 6 using 3 strips. Report your answer to two decimal places.
f = x^3 xo = 2 xf = 6 n = 3 h = (b-a)/n I = integral of (fn(x)) dx from a to b h = (6-2)/3 = 4/3 I = 334.22*
Use backward difference formula to compute the derivative of f(x) = 2x^2 + 3x at point x=4. Use a step size of 0.01. Report answer to 2 decimal places. f'(x) = [f(x_i )-f(x_i-1)]/h
f'(x) = [f(x_i )-f(x_i-1)]/h f'(4) = [f(4)-f(3.99)]/0.01 = (44-43.81)/0.01 = 18.98
Use the backward difference formula to compute the derivative of the function f(x) = x^4-8, at the point x = 4. Use a step size of 0.01. Report the answer to two decimal places. f'(x) = [f(xi )-f(xi-0.01)]/[(xi)-(xi-0.01)]
f'(x) = [f(xi )-f(xi-0.01)]/[(xi)-(xi-0.01)] = [f(4)-f(3.99)]/(4-3.99) = 2.5504/0.01 = 255.04
Use the forward difference formula to compute the derivative of the function f(x) = x^2+5, at the point x = 10 Use a step size of 0.01. Report the answer to two decimal places.
f'(x)= [(10.01^2+5)-(10^2+5)]/0.01 = 20.01
Use the central difference formula to compute the derivative of the function f(x) = 9x^4 - 9 at the point x=2. Use a step size of 0.01. Report answers to four decimal places.
f'(xi) = [f(xi+1) - f(xi-1)]/(2h) f(1.99) = 132.1415 f(2) = 135 f(2.01) = 137.9017 f'(xi) = [f'(2.01) - f'(1.99)]/(2(0.01)) = 288.0072
Assume we are using the forward difference formula to compute the derivative of the function f(x) =x^4 - 10, at the point x = 4, using a step size of 0.01. Assume we are using single precision to store our numbers and that we can calculate the second derivative analytically. Calculate the optimum step size. Report your answer to five decimal places. h(opt) = √[(±4e)/(f''(x)]
f(x) = x^4 - 10 f'(x) = 4x^3 f''(x) = 12x^2 f''(4) = 12(4^2) = 192 h(opt) = √{[(4(e-7)]/(192)} =0.00005
Approximate the value of the function f=x6 at the point x=2.5, using the first four terms of the Taylor Series expanded about the point x=2.0.
f(xi+1) = xi^6 + 6xi^5(x-xi) + 30xi^4[(x-xi)^2/2!] + 120xi^3[(x-xi)^3/3!] f(xi+1) = 2^6 + 6(2^5)[2.5-2] + 30(2^4)[(2.5-2)^2/2] + 120(2^3)[(2.5-2)^3/6] f(xi+1) = 240
Use the rectangular rule to integrate the function f=x^6 from x = 1 to x = 4 using 5 strips. Report your answer to two decimal places.
h = (4-1)/5 h = 0.6 I (approx) = 0.6(1^6 + 1.6^6 + 2.2^6 + 2.8^6 + 3.4^6) I (approx) = 1294.71
Convert E9B3A_16 from a base 16 number to a base 10 number.
〖E9B3A〗_16=E(〖16〗^4 )+9(〖16〗^3 )+B(〖16〗^2 )+3(〖16〗^1 )+A(〖16〗^0 ) =14(〖16〗^4 )+9(〖16〗^3 )+11(〖16〗^2 )+3(〖16〗^1 )+10(〖16〗^0 ) =917504+36864+2816+48+10 =957242
Using the backward difference formula, compute the derivative of the function f(x)=x^2 + 8 at the point x = 2. Use a step size of 0.001. Report the answer to two decimal places. *Backwards difference formula: x = [f(x)-f(x-h)]/h
*Backwards difference formula: x = [f(x)-f(x-h)]/h x = [(2^2 + 8)-(1.999^2 + 8)]/.001 x = 3.999 *Exact Answer x=4
Assume real numbers, in 8 bits, are being stored, using 1 sign bit, a 5 bit exponent, and a 2 bit fraction. What is the biggest number that can be stored?
*Begin with the sign bit. In order to reach the biggest number, this must be positive (0) so we set it equal to: (-1)^0 *Next, move on to the exponent. Since the exponent has 5 bits, the last digit to the right must be a zero (saved for infinity). Therefore the exponent looks like 11110 and the process to solve this is as shown. Exp=16(1)+8(1)+4(1)+2(1)+1(0)=30 *The bias would be 15, so exponent is written as: 2^(30-15)=2^15 *Next would be the fraction. For the fraction we assume all numbers are one, or 111 and solve it like this: Fraction=1 + 1/2(1) + 1/4(1) = 1.75 *Summing up the results our number looks like 01111011 and to solve it this is how: (-1^0)(2^15)( 1.75)≌ 3.5x10^15
Assume that we are storing real numbers in 16 bits, using one sign bit, 4-bit exponent, and an 11-bit fraction. What is the biggest the bias should be?
*Bias=(n-1) *n= the largest exponent 0 0 0 0 0 1 1 1 1 128 64 32 16 8 4 2 1 Bias = (8-1) = 7
Use the forward difference formula to compute the derivative of the function f(x)= x4-12, at the point x = 5. Use a step size of 0.02. Report the decimal to two decimal places. *Forward Difference Formula: f'(xi) = [f(xi+1)-f(xi)]/h
*Forward Difference Formula: f'(xi) = [f(xi+1)-f(xi)]/h f(x)=x4-12 x=5 h=0.02 f'(x) = [(5.02^4-12) - (5^4-12)] /0.02 f'(x) = 503.01
What is the base-10 equivalent of the real number stored in single precision according to the IEEE standard, as shown below? (Assume there is one sign bit, a bias of 127, an 8 bit exponent, and a 23 bit fraction). 00011101001000100010110110100101
*Sign bit is zero, so = (+1) *8 exponent bits are 00111010. Using the chart given in class we find this to equal: 2^5+2^4+2^3+2^1=58 *23 fraction bits are 01000100010110110100101. This can be solved for similarly to find a decimal equaling 0.25850892066955. *Next we plug these values into the formula: Number = (-1signbit) (1+fractionbits) (2exp-bias) Giving: Number=(+1)(1+0.26701796)(2^(58-127) ) *After calculations, this formula gives us a final answer of 2.15E-21
What ASCII character is represented by the unsigned integer stored in binary as shown below? 00110101
1 + 0 + 4 + 0 + 16 + 32 + 0 + 0 = 53 53 on the chart is equal to the character number 5
What ASCII character is represented by the unsigned binary integer shown below? 01001110
1) Convert above number from binary to base-10. 0 + 1(64) + 0 + 0 + 1(8) + 1(4) + 1(2)+ 0 = 78 2) Refer to ASCII chart to see which character is represented by 78 base-10. = N
Convert 3302 from a base 4 to a base 10 integer.
:3(4^3) + 3(4^2) + 0(4^1) + 2(4^0) = 242
Assume we are using the forward difference formula to compute the derivative of the function f = (1/x^14) -5 at the point x = 2, using a step size of 0.01. Assume we are using single precision to store our numbers. What type of error is dominating and what should we do to minimize the total error in the derivative? a) Truncation error dominates and we should reduce the step size to decrease the error b) Truncation error dominates and we should increase the step size to decrease the error c) Round-off error dominates and we should reduce the step size to decrease the error d) Round-off error dominates and we should increase the step size to decrease the error
???
Assume we are using the forward difference formula to compute the derivative of the function f(x) = x^2 + 7, at the point x = 1, using a step size of 0.1. Assume we are using single precision to store our numbers. What type of error is dominating and what should we do to minimize the total error in the derivative? [(1.1^2 + 7)-(1^2 + 7)]/(1.1-1) = 2.1 so.....[(1.01^2 + 7)-(1^2 + 7)]/(1.01-1) =2.01 Select one: a. Round-off error dominates and we should reduce the step size to decrease the error. b. Round-off error dominates and we should increase the step size to decrease the error. c. Truncation error dominates and we should increase the step size to decrease the error. d. Truncation error dominates and we should reduce the step size to decrease the error.
???
Given the real number stored in single precision according to the IEEE standard as shown below (Assume there is one sign bit, a bias of 127, an 8 bit exponent, and a 23 bit fraction): 11010010011000000000000000000000 What is the next biggest number the computer can store (in base 10)?
ANSWER: 27 26 25 24 23 22 21 20 128 64 32 16 8 4 2 1 1 0 1 0 0 1 0 0 = 164 The sign bit is a 1 therefore the sign is negative. The exponent, 164 is subtracted by the 127 bias. The current fraction is 1.75 to get the next biggest number; we add the smallest possible fraction (2-23) to the existing one of 1.75. This leads to the final answer of: (-1)(2164-127)(1.75+2-23)= -2.405e11
Convert the number 5678 (base 4) from base 4 to base 10 integer.
Answer: 5678 (base 4) = 5(4^3) + 6(4^2) + 7(4^1) + 8(4^0) = 320 + 96 + 28 +8 5678 (base 4) = 452 **Can't be right because 5 and 6 and 7 and 8 are not in base 4 number system.**
What is the base-10 equivalent of the unsigned integer stored in binary as shown below? 10110010
Answer: 2^7+ 0 + 2^5 + 2^4 + 0 + 0 + 2^1 + 0=178
Assuming we're storing real numbers in 32 bits, (one sign bit, a 9 bit exponent, and a 22 bit fraction). What is the machine epsilon ( Ɛm)?
Answer: Machine epsilon = 1/ (2^n) (where n= # of bits in fraction section) Ɛm =(1/(2^22)) = 1/4194304 Ɛm= 2.38421*10^(-7)
Use the central difference formula to compute the derivative of the function f(x)= (1/3 x^4 - 6), at point x = 5. Use a step size of 0.01. Report your answer to 4 decimal places. Central Difference formula: f'(x)= [f(xi+1) - f(xi-1)]/(2h)
Central Difference formula:f'(x)=[f(xi+1)-f(xi-1)]/(2h) = {[(1/3)(5.01^4)−6] − [(1/3)(4.99^4)−6)]} / 2(0.01) =𝟏𝟔𝟔.𝟔𝟔𝟕
Assume that we are storing real numbers in 32 bits, using one sign bit, a 8 bit exponent, and a 23 bit fraction. What is the biggest number we can store? bias=2^7-1=127
Due to all 1's in the 8 bit exponent being identically infinity we put 1's in the first seven bits and a 0 in the last bit 11111110=2^7+2^6+2^5+2^4+2^3+2^2+2^1+0=254 (sign bit)(2^exp-bias)(1+fraction bit) fraction bit=1+∑_(23 to 0)▒〖1/(2^n)〗≈2 (+1)(2^(254-127) )(2)=3.40282366921X10^38
What color is represented by the following hexadecimal number? 0064C2
Hexadecimal numbers: 0 ← Least intense 8 1 9 2 A 3 B 4 C 5 D 6 E 7 more intense→F *The RBG color system is represented this way RRGGBB RR: Hex red value GG: Hex green value BB: Hex blue Value *Since the first two digits are 00 there is no red color, the next two digits 6 and 4 but those are lower in value than the last two digits C and 2 therefore the color is a Blue 0064C2 is Blue
Use the trapezoidal rule to integrate the function f=x^2 from x = 1 to x = 4 using 6 strips. Report your answer to two decimal places.
I = (.5/2)[1^2 + 2(1.5^2) + 2(2^2) + 2(2.5^2) + 2(3^2) + 2(3.5^2) + 4^2] = 21.125
Use Simpson's 1/3 Rule to integrate the function f=9x^2 from x=1 to x=4 using 4 stops. Report your answer to 2 decimal places.
I = (h/3)[f(x1) + 4f(x2) + 2f(x3) + 4f(x4) + f(x5)] I = (.75/3)[9(1^2) + 4(9(1.75^2)) + 2(9(2.5^2)) + 4(9(3.25^2)) + 9(4^2)] I = 189
Assume that we are storing real numbers in 32 bits, using one sign bit, a 7 bit exponent, and a 24 bit fraction. What is the biggest number we can store?
Largest Number = (+1)(1.99999994039535)[2^(126-63)] = 1.84467E+19
Assume that we are storing real numbers in 16 bits, using one sign bit, an 8 bit exponent and a 7 bit fraction. How big should the bias be?
NUMBER: 128 64 32 16 8 4 2 1 BIT #: 8 7 6 5 4 3 2 1 *Since an 8 bit exponent would be 128, the bias is #-1 so, *Bias = 128-1 = 127 Solution
If you have a problem where the function values are inherently inaccurate, what type of error would be likely to dominate when calculating numerical derivatives?
Round-off error
Which would give you better accuracy, the rectangular rule with h=10^-7, or Simpson's 1/3 rule with h=10^-2 ?
Simpson's 1/3 rule
Assume we are using the central difference formula to compute the derivative of the function f(x) = x^4 - 7 at the point x = 4, using a step size of 0.01. Assume we are using single precision to store our numbers. What type of error is dominating and what should we do to minimize the total error in the derivative? a. Truncation error dominates, and we should reduce the step size to decrease the error. b. Round-off error dominates and we should reduce the step size to decrease the error. c. Truncation error dominates and we should increase the step size to decrease the error. d. Round-off error dominates and we should increase the step size to decrease the error.
a. Truncation error dominates, and we should reduce the step size to decrease the error. *The function is using an approximation in the central difference formula, which thus causes the truncation error.*
If we approximate a function using the first time three terms of the Taylor Series, the resulting approximation will be: a. Cubic b. Quadratic c. Constant d. Linear
b.) Quadratic
Which type of error occurs from an approximation in place of an exact mathematical procedure? a) Round-off Error b) Truncation Error
b: Truncation Error. *Truncation error occurs from using approximations instead of exact procedures.
If we approximate a function using the first two terms of the Taylor Series, the resulting approximation will be: a. Constant b. Cubic c. Linear d. Quadratic
c. Linear
Use the trapezoid rule to integrate f=x^2 using the points x= (2, 3, 4, 5, 6, 7). Report to two decimal places.
Δx= (7-2)/5 Δx= 1 Integral of x^2 from 2 to 7 = .5[f(2) + 2f(3) + 2f(4) + 2f(5) + 2f(6) + f(7)] Integral of x^2 from 2 to 7 = .5( 2 + 18 + 32 + 50 + 72 + 49) Integral of x^2 from 2 to 7 = .5(225) Integral of x^2 from 2 to 7 = 112.50 ←Answer
A binary number contains 32 numbers that consists of 1 sign bit, an 8 bit sign exponent and a 23 bit fraction. What is the machine epsilon of this number?
ε = 1 / 2^n This is where n is equal to the number of bits in the fraction. ε = 1/ 2^23 ε = 1.192092896 x 10^-7 ←Answer
Assume that we are storing real numbers in 32 bits, using one sign bit, a 8 bit exponent, and a 23 bit fraction. What is the "machine epsilon," ε?
εm = 1/2 (bits in fraction) εm = 1/2^23 Answer: εm = .000000119