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Calculus 2 : Taylor Series

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Example Question #41 : Taylor Series

Write the first three terms of the Taylor series about a= $2\pi$ for the following function:

$f(x)=e^{3x}+\cos(x)$

Possible Answers:

$(e^{2\pi}+1)+3e^{2\pi}(x-2\pi)+\frac{(9e^{2\pi}-1)(x-2\pi)^2}{2}$

$0+(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)$

$(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+(9e^{6\pi}-1)(x-2\pi)^2$

$(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+\frac{(9e^{6\pi}-1)(x-2\pi)^2}{2}$

Correct answer:

$(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+\frac{(9e^{6\pi}-1)(x-2\pi)^2}{2}$

Explanation:

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

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

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

$f'(x)=3e^{3x}-\sin(x)$

$f''(x)=9e^{3x}-\cos(x)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^{ax}=ae^{ax}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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

$(e^{6\pi}+1)\frac{(x-2\pi)^0}{0!}+3e^{6\pi}\frac{(x-2\pi)^1}{1!}+(9e^{6\pi}-1)\frac{(x-2\pi)^2}{2!}$

which simplifies to

$(e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+\frac{(9e^{6\pi}-1)(x-2\pi)^2}{2}$

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Example Question #42 : Taylor Series

Find the third degree Maclaurin Polynomial of the function

$f(x)=e^{3x}$

Possible Answers:

$P(x)=1+3x+\frac{9x^2}{2}+\frac{27x^3}{3!}$

P(x)=1-3x+9x^2-27x^3

$P(x)=3+6x+\frac{9x^2}{2}+\frac{12x^3}{3!}$

$P(x)=1-3x+\frac{9x^2}{2}-\frac{27x^3}{3!}$

Correct answer:

$P(x)=1+3x+\frac{9x^2}{2}+\frac{27x^3}{3!}$

Explanation:

The formula for the Maclaurin Series of a function is defined as follows

$P(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}$ where $f^{(n)}(0)$ is the n-th derivative when x=0

For the third degree polynomial we solve find the sum with the upper bound n=3

$\sum_{n=0}^{3}\frac{f^{(n)}(0)x^n}{n!}$

First we evaluate

$f(0)=e^{3(0)}=1$

We must also solve for the first, second, and third derivative of f(x)

$f'(x)=3e^{3x}$

$f''(x)=9e^{3x}$

$f^{(3)}(x)=27e^{3x}$

When x=0 we find the derivative values of

$f'(0)=3e^{3(0)}=3$

$f''(0)=9e^{3(0)}=9$

$f^{(3)}(0)=27e^{3(0)}=27$

Using these values we find the third degree Maclaurin Polynomial to be

$P(x)=1+3x+\frac{9x^2}{2}+\frac{27x^3}{3!}$

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Example Question #43 : Taylor Series

Find the Taylor series about a=7 for the following function:

$f(x)=x\ln x$

Possible Answers:

$0+0+0+ \sum_{n=0}^{\infty} (-1)^n(n-2)(n-1)(7)^{-n+1}(x-7)^n$

$7\ln 7+\ln7 +1 + \sum_{n=0}^{\infty} (-1)^n(n-2)(n-1)(7)^{-n+1}(x-7)^n$

$7\ln 7+\ln7 +1 + \sum_{n=0}^{\infty} (-1)^n(n-2)(n-1)(7)^{-n+1}(x-7)^{n+1}$

$\sum_{n=0}^{\infty} (-1)^n(n-2)(n-1)(7)^{-n+1}(x-7)^n$

Correct answer:

$7\ln 7+\ln7 +1 + \sum_{n=0}^{\infty} (-1)^n(n-2)(n-1)(7)^{-n+1}(x-7)^n$

Explanation:

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

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

We must take the nth derivative of the given function and determine the trend as n goes to infinity. To determine this, we start at n=0 (the zeroth derivative, or the function itself), and go further:

$f^0(x)=x\ln x$

$f'(x)=\ln(x)+\frac{x}{x}=\ln(x)+1$

$f''(x)=\frac{1}{x}$

$f'''(x)=-\frac{1}{x^2}$

$f^{(4)}(x)=2x^{-3}$

$f^{(5)}(x)=-6x^{-4}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)\cdot f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x}\ln(x)=\frac{1}{x}$

We must now find a pattern for the derivatives we took. The zeroth and first derivatives do not follow a pattern, but the derivatives after do follow a pattern: the sign alternates, the power of x decreases by 1, and the coefficient is the factorial of (n-2) starting at the n=2 derivative.

The derivative can then be expressed as

$f^{(n)}(a)=(-1)^n(n-2)!(x)^{-n+1}$

for n greater than or equal to 2.

For the Taylor series itself, we must evaluate the derivative at x=a=7. When we do this, and write the pattern elements for the derivatives past n=1, we get

$7\ln 7+\ln7 +1 + \sum_{n=0}^{\infty} (-1)^n(n-2)(n-1)(7)^{-n+1}(x-7)^n$

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Example Question #44 : Taylor Series

Find an expression for the Taylor Series of $e^{2x}$ .

Possible Answers:

$\sum_{n=0}^\infty \frac{x^{2n}}{2(n!)}$

$\sum_{n=0}^\infty 2^n\frac{x^{2n}}{n!}$

$\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}$

$\sum_{n=0}^\infty 2\frac{x^{2n}}{n!}$

$\sum_{n=0}^\infty 2^n\frac{x^{2}}{n!}$

Correct answer:

$\sum_{n=0}^\infty 2^n\frac{x^{2n}}{n!}$

Explanation:

To obtain the Taylor Series for $e^{2x}$ , we start with the Taylor Series for $e^{x} = \sum_{n=0}^\infty \frac{x^n}{n!}$ , and substitute on both sides with , and simplify.

$e^{2x} = \sum_{n=0}^\infty \frac{(2x)^n}{n!} = \sum_{n =0}^\infty \frac{2^nx^n}{n!}=\sum_{n=0}^\infty2^n\frac{x^n}{n!}$ .

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Example Question #45 : Taylor Series

Find the expression for the Taylor Series of $x^2\sin(x^2)$ .

Possible Answers:

$\sum_{n=0}^\infty\frac{(-1)^n2^{n+1}x^{n+3}}{n+1}$

$\sum_{n=0}^\infty\frac{(-1)^n2^{n+1}x^{n+1}}{(n+1)!}$

$\sum_{n=0}^\infty\frac{(-1)^nx^{n+3}}{n!}$

$\sum_{n=0}^\infty\frac{(-1)^nx^{2n+4}}{(n+1)!}$

$\sum_{n=0}^\infty\frac{2^{n+1}x^{n+3}}{(n+1)!}$

Correct answer:

$\sum_{n=0}^\infty\frac{(-1)^nx^{2n+4}}{(n+1)!}$

Explanation:

To obtain the Taylor Series for $x^2\sin(x^2)$ , we start with the Taylor Series for $\sin(x) = \sum_{n=0}^\infty \frac{(-1)^nx^{n+1}}{(n+1)!}$ .

$\sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^n(x^2)^{n+1}}{(n+1)!}$ . (Substitute with x^2 on both sides)

$x^2\sin(x^2) = x^2\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{n+1}}{(n+1)!}$ . (Multiply both sides by x^2 ,

$= \sum_{n=0}^\infty x^2\frac{(-1)^nx^{2n+2}}{(n+1)!} =\sum_{n=0}^\infty\frac{(-1)^nx^{2n+4}}{(n+1)!}$ .

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Example Question #46 : Taylor Series

Use Taylor Expansion of around x=0 and determine the value of y''

Possible Answers:

$y''=\sum_{2}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

$y''=\sum_{1}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-1}$

$y''=\sum_{-2}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n+2}$

$y''=\sum_{2}^{\infty} (n)(n-2)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

$y''=\sum_{0}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

Correct answer:

$y''=\sum_{2}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

Explanation:

We can write as a Mclaurin series by saying that:

$y=\sum_{0}^{\infty} \frac{y^{(n)}(0)}{n!}(x)^n$

Since y is just a polynomial expression:

$y''=\sum_{2}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}$

The reason we must bring the lower bound to 2 is because Taylor Series can ONLY be written as a summation of polynomials and no rational components.

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Calculus 2 : Taylor Series

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Taylor Series

Example Question #42 : Taylor Series

Example Question #43 : Taylor Series

Example Question #44 : Taylor Series

Example Question #45 : Taylor Series

Example Question #46 : Taylor Series

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