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GRE Subject Test: Math : Taylor's Theorem

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Example Questions

Example Question #1 :Taylor Series

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

Possible Answers:

$f(x)= x^{2}+\sqrt{x}+1$

$f(x)=\ln|x+3|$

$f(x)=x+3\sin(x)$

$f(x)= \frac{x^{5}}{3}+2$

Correct answer:

$f(x)= \frac{x^{5}}{3}+2$

Explanation:

Recall the Maclaurin series formula:

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

$=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

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Example Question #2 :Taylor And Maclaurin Series

Let $f(x)=(1+\frac{x}{2})^{1/3}$

Find the the first three terms of the Taylor Series forcentered at.

Possible Answers:

$T_{2}=0+\frac{x}{6} -\frac{x^2}{18}$

$T_{2}(x)=1+\frac{x}{6}-\frac{x^2}{36}$

$T_3=1+\frac{x}{3}+4x^2$

$T_{3}=0+0+0$

$T_{2}(x)=1+\frac{x}{6}+\frac{x^2}{36}$

Correct answer:

$T_{2}(x)=1+\frac{x}{6}-\frac{x^2}{36}$

Explanation:

Using the formula of a binomial series centered at 0:

$(1+x)^k =\sum_{n=0}^{\infty } \binom{k}{n}x^n$ ,

where we replacewith $\frac{x}{2}$ and $k=\frac{1}{3}$ , we get:

$\sum_{n=0}^{2}\binom{\frac{1}{3}}{n}\left(\frac{x}{2}\right)^n$ for the first 3 terms.

Then, we find the terms where,

$T_2(x)=\binom{\frac{1}{3}}{0}(\frac{x}{2})^0 + \binom{\frac{1}{3}}{1}(\frac{x}{2})^1+\binom{\frac{1}{3}}{2}(\frac{x}{2})^2=1+\frac{x}{6}- \frac{x^2}{36}$

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Example Question #3 :Taylor And Maclaurin Series

Determine the convergence of the Taylor Series forat x=5 where $f(x)=\sum_{n=0}^{\infty }\frac{x^n}{n!}$ .

Possible Answers:

Does not exist.

Conditionally Convergent.

Absolutely Convergent.

Divergent.

Inconclusive.

Correct answer:

Absolutely Convergent.

Explanation:

By the ratio test, the series $\sum_{n=0}^{\infty }\frac{5^n}{n!}$ converges absolutely:

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{5^{n+1}}{(n+1)!} \cdot \frac{n!}{5^n} \right |=\lim_{n\rightarrow \infty }\left | \frac{5}{n+1} \right |=0<1$

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Example Question #4 :Taylor And Maclaurin Series

Find the interval of convergence forof the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

Possible Answers:

$\left ( \frac{-1}{5}, \frac{1}{5} \right )$

x=5

$(-\infty, \infty)$

x=0

$\left ( -5, 5\right )$

Correct answer:

x=5

Explanation:

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

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Example Question #5 :Taylor And Maclaurin Series

Suppose that the derivative of a function, denoted f'(x) , can be approximated by the third degree Taylor polynomial, centered at:

$f'(x) \approx 1 + (x-2) + \frac{4}{3} (x-2)^2 + \frac{8}{9}(x-2)^3$

If f(2) = 7 ,find the third degree Taylor polynomial forcentered at.

Possible Answers:

$(x-2) + 2(x-2)^2 + {4}(x-2)^3$

$7 + x + (x-2)^2 + \frac{4}{9}(x-2)^3$

7 + (x-2) + 2(x-2)^2 + 4(x-2)^4

$(x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3 + \frac{8}{27}(x-2)^4$

$7 + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

Correct answer:

$7 + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

Explanation:

To get f(x) , we need to find the antiderivative of f'(x) by integrating the third degree polynomial term by term.

We only want up to a third degree polynomial, so we can disregard the fourth order term:

$f(x) = c + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

Since f(2) = 7 , substitute c=7 for the final f(x) .

$f(x)=7 + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

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Example Question #11 :Applications Of Derivatives

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

f(x)=x^2+2x+1

Possible Answers:

4+4(x-1)+(x-1)^2+0

0+4+4(x-1)+(x-1)^2

4+4(x-1)^2+(x-1)^3+0

0+0+0+0

Correct answer:

4+4(x-1)+(x-1)^2+0

Explanation:

The Taylor series about x=a of any function is given by the following:

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

所以,我们必须找到第0,首先,其次,third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms):

$f^{0}(x)=f(x)$

f'(x)=2x+2

f''(x)=2

f'''(x)=0

The derivatives were found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following:

$\frac{4(x-1)^0}{0!}+\frac{4(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}+\frac{0(x-1)^3}{3!}$

which when simplified is equal to

4+4(x-1)+(x-1)^2+0 .

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Example Question #9 :Taylor And Maclaurin Series

Find the first two terms of the Taylor series about a=3 for the following function:

f(x)=2x+5

Possible Answers:

$\infty +2(x-3)$

11+2(x-3)

11+7(x-3)

11-2(x-3)

Correct answer:

11+2(x-3)

Explanation:

The general formula for the Taylor series about x=a for a function is

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

First, we must find the zeroth and first derivative of the function.

The zeroth derivative of a function is just the function itself, so we only have to find the first derivative:

f'(x)=2

The derivative was found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, write the first two terms of the sequence (n=0 and n=1):

$\frac{[2(3)+5](x-3)^0}{0!}+\frac{2(x-3)^1}{1!}=11+2(x-3)$

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Example Question #10 :Taylor And Maclaurin Series

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

f(x)=2e^x+2

Possible Answers:

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

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

0+0+0

$0+(x-1)+\frac{(x-1)^2}{2}$

Correct answer:

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

Explanation:

The general formula for the Taylor series of a given function about x=a is

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

We were asked to find the first three terms, which correspond to n=0, 1, and 2. So first, we need to find the zeroth, first, and second derivative of the given function. The zeroth derivative is just the function itself.

f'(x)=2e^x

f''(x)=2e^x

The derivatives were found using the following rules:

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

Now use the above formula to write out the first three terms:

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

Simplified, this becomes

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

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GRE Subject Test: Math : Taylor's Theorem

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 :Taylor Series

Example Question #2 :Taylor And Maclaurin Series

Example Question #3 :Taylor And Maclaurin Series

Example Question #4 :Taylor And Maclaurin Series

Example Question #5 :Taylor And Maclaurin Series

Example Question #11 :Applications Of Derivatives

Example Question #9 :Taylor And Maclaurin Series

Example Question #10 :Taylor And Maclaurin Series

All GRE Subject Test: Math Resources

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