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Calculus 2 : Series in Calculus

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : 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:

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

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

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

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

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

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 And Maclaurin Series

Consider the following series:

$f(t) = \sum_{ n=0}^{\infty} \left(\frac{1+t)^n}{2^n n!}\right) = 1 + \frac{1 + t}{2} + \frac{(1+t)^2}{8} + ....$

What is the evaluation of this series at t=1 ?

Possible Answers:

$The\ series\ diverges$

cos(1)

sin(1)

$\pi$

Correct answer:

Explanation:

Recall the important McLaurin series for e^x :

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

The series in this question is the same, only replacing x with $\frac{1+t}{2}$ .

So if we have $f(t) = e^{\frac{1+t}{2}}$ , we need only evaluate at t=1 , and thus we obtain .

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Example Question #1 : Taylor And Maclaurin Series

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

f(x)=x^2+3x+2

Possible Answers:

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

2+5(x-1)+(x-1)^2

6+5(x-1)+(x-1)^2

2-5(x-1)+(x-1)^2

Correct answer:

6+5(x-1)+(x-1)^2

Explanation:

The general form for the Taylor series (of a function f(x)) about x=a is the following:

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

Because we only want the first three terms, we simply plug in a=1, and then n=0, 1, and 2 for the first three terms (starting at n=0).

The hardest part, then, is finding the zeroth, first, and second derivative of the function given:

$f^{0}(1)=f(1)=6$

f'(1)=2(1)+3=5

f''(1)=2

The derivative was found using the following rules:

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

Then, simply plug in the remaining information and write out the terms:

$\frac{6(x-1)^0}{0!}+\frac{5(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}=6+5(x-1)+(x-1)^2$

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Example Question #2 : Taylor And Maclaurin 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 #1 : Taylor 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

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

0+0+0

(2e^1+2)+2e^1(x-1)+e^1(x-1)^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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Example Question #1 : Maclaurin Series

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

f(x)=10x^2+x

Possible Answers:

0+0+0

11+21x+10x^2

0+21(x-1)+10(x-1)^2

11+21(x-1)+10(x-1)^2

Correct answer:

11+21(x-1)+10(x-1)^2

Explanation:

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

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

We must find the zeroth, first, and second derivative of the function (for n=0, 1, and 2). The zeroth derivative is just the function itself.

f'(x)=20x+1

f''(x)=20

The derivatives were found using the following rule:
$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, follow the above formula to write out the first three terms:

$\frac{11(x-1)^0}{0!}+\frac{21(x-1)^1}{1!}+\frac{20(x-1)^2}{2!}$

which simplified becomes

11+21(x-1)+10(x-1)^2

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Example Question #3041 : Calculus Ii

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:

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

0+0+(x-1)^2

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

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

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 #12 : Taylor Series

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

$f(x)=3\cos(x)$

Possible Answers:

$3+0-3\frac{(x-\pi)^2}{2}$

$0+0+3\frac{(x-\pi)^2}{2}$

0+0+0

$-3+0-3\frac{(x-\pi)^2}{2}$

Correct answer:

$3+0-3\frac{(x-\pi)^2}{2}$

Explanation:

The Taylor series about x=a for a function is

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

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

f^0=3cos(x)

$f'(x)=-3\sin(x)$

$f''(x)=-3\cos(x)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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

$\frac{3(x-\pi)^0}{0!}-3\cdot \frac{0(x-\pi)^1}{1!}-\frac{3(x-\pi)^2}{2!}$

which simplified becomes

$3+0-3\frac{(x-\pi)^2}{2}$ .

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Example Question #12 : Taylor Series

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

$f(x)=\cos(x)\sin(x)$

Possible Answers:

$0-(x-\pi)$

0+0

$0+(x-\pi)$

0+1

Correct answer:

$0+(x-\pi)$

Explanation:

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

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

Now, for the first two terms (n=0, 1) we must find the zeroth and first derivative of the function, the zeroth derivative being the function itself:

$f'(x)=-\sin^2(x)+\cos^2(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

Now, use the above formula to write out the first two terms:

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

which simplified becomes

$0+(x-\pi)$

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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #3 : Taylor And Maclaurin Series

Example Question #1 : Taylor And Maclaurin Series

Example Question #1 : Taylor And Maclaurin Series

Example Question #2 : Taylor And Maclaurin Series

Example Question #5 : Taylor And Maclaurin Series

Example Question #1 : Taylor Series

Example Question #1 : Maclaurin Series

Example Question #3041 : Calculus Ii

Example Question #12 : Taylor Series

Example Question #12 : Taylor Series

All Calculus 2 Resources

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