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GRE Subject Test: Math : Calculus

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Example Questions

Example Question #1 : Taylor And Maclaurin Series

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

Find the the first three terms of the Taylor Series for centered at .

Possible Answers:

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

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

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

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

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

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 replace with $\frac{x}{2}$ 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 #1 : Taylor's Theorem

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

Possible Answers:

Conditionally Convergent.

Inconclusive.

Divergent.

Does not exist.

Absolutely Convergent.

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 #1 : Taylor And Maclaurin Series

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

Possible Answers:

x=5

$(-\infty, \infty)$

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

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

x=0

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 #1 : 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 for centered at .

Possible Answers:

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

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

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

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

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

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 #1 : Taylor Series

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:

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

So, we must find the zeroth, first, second, and 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 #5 : 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:

11+7(x-3)

11+2(x-3)

$\infty +2(x-3)$

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 #6 : 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+0+0

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

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

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

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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Example Question #1 : Integrals

Integrate:

$\int \frac{18x-6}{3x^2-2x}dx$

Possible Answers:

$6\: ln\left | 3x^2-2x\right |+C$

$3\: ln\left | 3x^2-2x\right |+C$

$\frac{1}{3}\: ln\left | 3x^2-2x\right |+C$

$\frac{1}{6}\: ln\left | 3x^2-2x\right |+C$

$\frac{1}{2}\: ln\left | 3x^2-2x\right |+C$

Correct answer:

$3\: ln\left | 3x^2-2x\right |+C$

Explanation:

This problem requires U-Substitution. Let u=3x^2-2x and find .

$du=6x-2\:dx$

Notice that the numerator in $\int \frac{18x-6}{3x^2-2x}dx$ has common factor of 2, 3, or 6. The numerator can be factored so that the term can be a substitute. Factor the numerator using 3 as the common factor.

$\int \frac{18x-6}{3x^2-2x}dx= \int\frac{3(6x-2)}{3x^2-2x}dx=3\int \frac{(6x-2)}{3x^2-2x}dx$

Substitute and terms, integrate, and resubstitute the term.

$3\int \frac{du}{u} = 3\: ln\left | u\right |+C=3\: ln\left | 3x^2-2x\right |+C$

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Example Question #1 : Integrals

Evaluate the following integral:

$\int5x^4(x^5-9)^6dx$

Possible Answers:

$\frac{(x^5+9)^7}{7}+c$

$\frac{(x^5-9)^7}{7}+c$

$\frac{(x^5+9)^6}{6}+c$

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

$\frac{(x^5-9)^7}{7}$

Correct answer:

$\frac{(x^5-9)^7}{7}+c$

Explanation:

To calculate this integral, we could expand that whole binomial, but it would be very time consuming and a bit of a pain. Instead, let's use u substitution:

Given this:

$\int5x^4(x^5-9)^6dx$

We can say that

u=x^5-9

du=5x^4dx

Then, plug it back into our original expression

$\int u^6du$

Evaluate this integral to get

$\frac{u^7}{7}+c$

Then, replace u with what we substituted it for to get our final answer. Note because this is an indefinite integral, we need a plus c in it.

$\frac{(x^5-9)^7}{7}+c$

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Example Question #3 : Integrals

Integrate the following using substitution.

$\int \frac{1}{4-5x}dx$

Possible Answers:

$-\frac{1}{5}\ln\left | 4-5x\right |+c$

$\ln\left | 4-5x\right |+c$

$\frac{-5}{(4-5x)^2}+c$

$\frac{-5}{4-5x}+c$

Correct answer:

$-\frac{1}{5}\ln\left | 4-5x\right |+c$

Explanation:

$\int \frac{1}{4-5x}dx$

Using substitution, we set u=4-5x which means its derivative is du=-5dx .

Substituting for 4-5x , and $- \frac{1}{5}du$ for we have:

$\int \frac{1}{4-5x}dx=-\frac{1}{5} \int \frac{1}{u} du$

Now we just integrate:

$-\frac{1}{5} \int \frac{1}{u} du=-\frac{1}{5}\ln\left | u\right |+c$

Finally, we remove our substitution u=4-5x to arrive at an expression with our original variable:

$-\frac{1}{5}\ln\left | 4-5x\right |+c$

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GRE Subject Test: Math : Calculus

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Taylor And Maclaurin Series

Example Question #1 : Taylor's Theorem

Example Question #1 : Taylor And Maclaurin Series

Example Question #1 : Taylor And Maclaurin Series

Example Question #1 : Taylor Series

Example Question #5 : Taylor And Maclaurin Series

Example Question #6 : Taylor And Maclaurin Series

Example Question #1 : Integrals

Example Question #1 : Integrals

Example Question #3 : Integrals

All GRE Subject Test: Math Resources

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