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1. Find the truncuated error of the Maclaurin series sinx cosx given that is truncated into 3 terms. These are the example that might help..see the picture below

1. Find the truncuated error of the Maclaurin series sinx cosx given that is truncated into 3 terms. These are the example that might help..see the picture below

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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1. Find the truncuated error of the Maclaurin series sinx cosx given that is truncated into 3 terms.

These are the example that might help..see the picture below

 

Truncation Error 1. Step 1: f(x) = e*: • f(x) = e* f'(x) = (e*) = e* f"(x) = (e*) = e* Calculate the derivatives for the function Step 2: Fill in the derivatives you calculated in step 1 with 0 as the input. This gives you the values you need to insert into the Maclaurin series: f(x) = e° = 1 f'(0) = e° = 1 f"(0) = e° = 1 f''0) = e° = 1 f"(0) = e° = 1 f'(0) = e° = 1 d? dx2 d3 f''(x) = (e*) = e* dx3 dx4 (e*) = ex d5 f"(x) = f'(x) = = e* dxs (e*) Step 3: Fill in the Maclaurin formula with the values you calculated in Step 2: Solution: Truncation Error = Actual Value – Truncated Value (1+x+x* + ...) -1+x+ ? Truncation Error =x + ** + e* = 1+(x) +x)? +x)³ + Truncation Error = x5 120 + 6 e* = 1+x + x² + x³ + x* + 20 x5+ %3D 6 120 Since the above equation is truncated into 3, thus, e* = 1+ x + x? 2. Step 1: f(x) = cos x: Step 2: Fill in the derivatives you calculated in step 1 with 0 as the input. This gives you the values you need Calculate the derivatives for the function f(x) = cos x to insert into the Maclaurin series: f'(x) = (cos x) = – sin x f(x) = cos(0) = 1 f'(0) = – sin(0) = 0 f"(0) = – cos(0) = –1 f"(x) = d2 (cos x) = – cos x dx2 *Property of STI Page 3 of 4 03 Handout 2 STI IT1903 f''0) = sin(0) = 0 f"(0) = cos(0) = 1 f'(0) = – sin(0) = 0 f''(x) = (cos x) = sinx dx f"(x) = -(cos x) = cos x dx d5 drs (cos x) =– sin x f'(x) = Step 3: Fill in the Maclaurin formula with the values you calculated in Step 2: Solution: Truncation Error = Actual Value – Truncated Value +유() + (3)2 + 유(2)* + Truncation Error = (1- ;x² + cos x = (0) 24 *" Truncation Error = ... 1 - 5! 24 cos X %3D 1 + (0) + 들x2 + Dx* + x* (0) x5 +... 120 cos x = 1 - x? + *.. Since the above equation is truncated into 3, thus, x? + ** cos x = 1 –
Transcribed Image Text:Truncation Error 1. Step 1: f(x) = e*: • f(x) = e* f'(x) = (e*) = e* f"(x) = (e*) = e* Calculate the derivatives for the function Step 2: Fill in the derivatives you calculated in step 1 with 0 as the input. This gives you the values you need to insert into the Maclaurin series: f(x) = e° = 1 f'(0) = e° = 1 f"(0) = e° = 1 f''0) = e° = 1 f"(0) = e° = 1 f'(0) = e° = 1 d? dx2 d3 f''(x) = (e*) = e* dx3 dx4 (e*) = ex d5 f"(x) = f'(x) = = e* dxs (e*) Step 3: Fill in the Maclaurin formula with the values you calculated in Step 2: Solution: Truncation Error = Actual Value – Truncated Value (1+x+x* + ...) -1+x+ ? Truncation Error =x + ** + e* = 1+(x) +x)? +x)³ + Truncation Error = x5 120 + 6 e* = 1+x + x² + x³ + x* + 20 x5+ %3D 6 120 Since the above equation is truncated into 3, thus, e* = 1+ x + x? 2. Step 1: f(x) = cos x: Step 2: Fill in the derivatives you calculated in step 1 with 0 as the input. This gives you the values you need Calculate the derivatives for the function f(x) = cos x to insert into the Maclaurin series: f'(x) = (cos x) = – sin x f(x) = cos(0) = 1 f'(0) = – sin(0) = 0 f"(0) = – cos(0) = –1 f"(x) = d2 (cos x) = – cos x dx2 *Property of STI Page 3 of 4 03 Handout 2 STI IT1903 f''0) = sin(0) = 0 f"(0) = cos(0) = 1 f'(0) = – sin(0) = 0 f''(x) = (cos x) = sinx dx f"(x) = -(cos x) = cos x dx d5 drs (cos x) =– sin x f'(x) = Step 3: Fill in the Maclaurin formula with the values you calculated in Step 2: Solution: Truncation Error = Actual Value – Truncated Value +유() + (3)2 + 유(2)* + Truncation Error = (1- ;x² + cos x = (0) 24 *" Truncation Error = ... 1 - 5! 24 cos X %3D 1 + (0) + 들x2 + Dx* + x* (0) x5 +... 120 cos x = 1 - x? + *.. Since the above equation is truncated into 3, thus, x? + ** cos x = 1 –
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