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AP Calculus BC : Maclaurin Series for exponential and trigonometric functions

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Example Questions

Example Question #1 : Maclaurin Series For Exponential And Trigonometric Functions

Write out the first three terms of the Taylor series about a=1 for the following function:

f(x)=x^2+e^x

Possible Answers:

(1+e^1)+[(2+e^1)(x-1)]+(x-1)^2

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

0+(1+e^1)+[(2+e^1)(x-1)]

0+0+(x-1)^2

Correct answer:

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

Explanation:

The Taylor series about x=a for a function is given by

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

For the first three terms (n=0, 1, 2) we must find the zeroth, first, and second derivative of the function. The zeroth derivative is just the function itself.

f'(x)=2x+e^x

f''(x)=2+e^x

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$

Now, using the above formula, write out the first three terms:

$\frac{(1+e^1)(x-1)^0}{0!}+\frac{[(2+e^1)(x-1)^1]}{1!}+\frac{(2+e^1)(x-1)^2}{2!}$

which simplified becomes

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

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Example Question #61 : Taylor And Maclaurin Series

Find the Maclaurin Series of the function

$f(x)=sin\,x$

up to the fifth degree.

Possible Answers:

$x+\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!}+\frac{x^5}{5!}$

DNE

$1-\frac{x^2}{2!}+\frac{x^4}{4!}$

$x-\frac{x^3}{3!}+\frac{x^5}{5!}$

Correct answer:

$x-\frac{x^3}{3!}+\frac{x^5}{5!}$

Explanation:

The formula for an i-th degree Maclaurin Polynomial is

$f(x)=\sum_{n=0}^{i}\frac{f^{(n)}(0)}{n!}x^n$

For the fifth degree polynomial, we must evaluate the function up to its fifth derivative.

$\begin{matrix} f(0)=sin(0)=0\\ f'(0)=cos(0)=1\\ f''(0)=-sin(0)=0\\ f'''(0)=-cos(0)=-1\\ f^{(4)}=sin(0)=0 \\ \end{matrix}$

$f^{(5)}(0)=cos(0)=1$

The summation becomes

$\sum_{n=0}^{5}\frac{f^{(n)}(0)}{n!}x^n=\frac{f(0)}{0!}x^0+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\frac{f^{(5)}(0)}{5!}x^5$

And substituting for the values of the function and the first five derivative values, we get the Maclaurin Polynomial

$f(x)=sin(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}$

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AP Calculus BC : Maclaurin Series for exponential and trigonometric functions

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Maclaurin Series For Exponential And Trigonometric Functions

Example Question #61 : Taylor And Maclaurin Series

All AP Calculus BC Resources

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