Math Problem Statement

find the taylor polynomial T2 and compute the error |f(x)-T2| for the given values of a and x. f(x)= e^(sinx), a=pi/2, x=1.3. round each answer to 6 decimal places

Solution

To find the Taylor polynomial T2(x)T_2(x) of the function f(x)=esinxf(x) = e^{\sin x} centered at a=π2a = \frac{\pi}{2} and then compute the error f(x)T2(x)|f(x) - T_2(x)| at x=1.3x = 1.3, we can follow these steps:

1. Calculate the derivatives of f(x)f(x):

Given f(x)=esinxf(x) = e^{\sin x}, we need the first and second derivatives:

  • First derivative: f(x)=esinxcosxf'(x) = e^{\sin x} \cdot \cos x

  • Second derivative: f(x)=esinx(cos2xsinx)f''(x) = e^{\sin x} \cdot (\cos^2 x - \sin x)

2. Evaluate the function and its derivatives at a=π2a = \frac{\pi}{2}:

  • f(π2)=esin(π2)=e1=ef\left(\frac{\pi}{2}\right) = e^{\sin\left(\frac{\pi}{2}\right)} = e^1 = e
  • f(π2)=esin(π2)cos(π2)=e0=0f'\left(\frac{\pi}{2}\right) = e^{\sin\left(\frac{\pi}{2}\right)} \cdot \cos\left(\frac{\pi}{2}\right) = e \cdot 0 = 0
  • f(π2)=esin(π2)(cos2(π2)sin(π2))=e(01)=ef''\left(\frac{\pi}{2}\right) = e^{\sin\left(\frac{\pi}{2}\right)} \cdot \left(\cos^2\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)\right) = e \cdot (0 - 1) = -e

3. Write the Taylor polynomial T2(x)T_2(x) at a=π2a = \frac{\pi}{2}:

The second-degree Taylor polynomial is given by: T2(x)=f(a)+f(a)(xa)+f(a)2(xa)2T_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2 Substituting the values, we get: T2(x)=e+0(xπ2)+e2(xπ2)2=ee2(xπ2)2T_2(x) = e + 0 \cdot (x - \frac{\pi}{2}) + \frac{-e}{2}(x - \frac{\pi}{2})^2 = e - \frac{e}{2}(x - \frac{\pi}{2})^2

4. Compute T2(1.3)T_2(1.3):

First, compute xπ2x - \frac{\pi}{2} at x=1.3x = 1.3: 1.3π21.31.570796=0.2707961.3 - \frac{\pi}{2} \approx 1.3 - 1.570796 = -0.270796 Now, compute T2(1.3)T_2(1.3): T2(1.3)=ee2(0.270796)2T_2(1.3) = e - \frac{e}{2}(-0.270796)^2 T2(1.3)ee2×0.073331e(10.036665)e×0.9633352.620557T_2(1.3) \approx e - \frac{e}{2} \times 0.073331 \approx e(1 - 0.036665) \approx e \times 0.963335 \approx 2.620557

5. Compute the actual function value f(1.3)f(1.3):

f(1.3)=esin(1.3)e0.9635582.621273f(1.3) = e^{\sin(1.3)} \approx e^{0.963558} \approx 2.621273

6. Calculate the error f(1.3)T2(1.3)|f(1.3) - T_2(1.3)|:

f(1.3)T2(1.3)2.6212732.6205570.000716|f(1.3) - T_2(1.3)| \approx |2.621273 - 2.620557| \approx 0.000716

Final Answers:

  • The Taylor polynomial T2(x)T_2(x) is approximately T2(x)=ee2(xπ2)2T_2(x) = e - \frac{e}{2}(x - \frac{\pi}{2})^2.
  • T2(1.3)2.620557T_2(1.3) \approx 2.620557
  • The error f(1.3)T2(1.3)|f(1.3) - T_2(1.3)| is approximately 0.0007160.000716.

All values are rounded to 6 decimal places.

Would you like further details or have any questions? Here are some related questions you might be interested in:

  1. How does the error bound for Taylor polynomials relate to the remainder term?
  2. What happens to the Taylor series if we include more terms?
  3. How does the choice of aa (expansion point) affect the accuracy of the Taylor polynomial?
  4. Can you explain the geometric interpretation of Taylor polynomials?
  5. How does the error behave as xx moves farther from aa?

Tip: When calculating Taylor series, always check the convergence interval to ensure that your approximation is valid for the given xx.

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Math Problem Analysis

Mathematical Concepts

Taylor series
Taylor polynomial
Exponential function
Trigonometric functions

Formulas

Taylor polynomial formula

Theorems

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Suitable Grade Level

Advanced High School