Calculus 2 : Taylor and Maclaurin Series

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #41 : Taylor And Maclaurin Series

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

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

Possible Answers:

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

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

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

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

Correct answer:

\displaystyle (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

\displaystyle \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.

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

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

The derivatives were found using the following rules:

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

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

\displaystyle (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

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

 

Example Question #42 : Taylor And Maclaurin Series

Find the third degree Maclaurin Polynomial of the function

\displaystyle f(x)=e^{3x}

Possible Answers:

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

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

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

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

Correct answer:

\displaystyle 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

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

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

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

First we evaluate 

\displaystyle f(0)=e^{3(0)}=1

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

\displaystyle f'(x)=3e^{3x}

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

\displaystyle f^{(3)}(x)=27e^{3x}

When \displaystyle x=0 we find the derivative values of

\displaystyle f'(0)=3e^{3(0)}=3

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

\displaystyle f^{(3)}(0)=27e^{3(0)}=27

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

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

Example Question #43 : Taylor And Maclaurin Series

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

\displaystyle f(x)=x\ln x

Possible Answers:

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

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

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

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

Correct answer:

\displaystyle 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

\displaystyle \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:

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

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

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

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

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

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

The derivatives were found using the following rules:

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)\cdot f(x)g'(x)\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\displaystyle \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 \displaystyle (n-2) starting at the n=2 derivative. 

The derivative can then be expressed as 

\displaystyle 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

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

Example Question #281 : Series In Calculus

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

\displaystyle 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!}.

Example Question #45 : Taylor And Maclaurin Series

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

Example Question #46 : Taylor And Maclaurin Series

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

 

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We can write \displaystyle y as a Mclaurin series by saying that:

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

Since y is just a polynomial expression:

\displaystyle 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. 

Example Question #3071 : Calculus Ii

Give the Maclaurin series for the function

\displaystyle f(x) = e ^{2x ^{2}}

up to the third term.

Possible Answers:

\displaystyle f(x) = 2x^{2} - \frac{4x^{6}}{3} + \frac{4x^{10}}{15}-...

\displaystyle f(x) = 2x^{2} - 4x^{4}+ 8x^{6}+...

\displaystyle f(x) = 1 + 2x^{2} + 2x^{4}+...

\displaystyle f(x) = 1 - 2x^{4}+ \frac{2x^{8}}{3}...

\displaystyle f(x) = 1 + 2x^{2} + 4x^{4}+...

Correct answer:

\displaystyle f(x) = 1 + 2x^{2} + 2x^{4}+...

Explanation:

The Maclaurin series for \displaystyle e^{x}, taken to the third term, is:

\displaystyle e ^{x} = 1 + x + \frac{x^{2}}{2}+...

Substitute \displaystyle 2x^{2} for \displaystyle x:

\displaystyle f(x) = e ^{2x ^{2}}

\displaystyle e ^{x} = 1 + 2x^{2} + \frac{(2x^{2})^{2}}{2}+...

\displaystyle e ^{x} = 1 + 2x^{2} + \frac{4x^{4}}{2}+...

\displaystyle e ^{x} = 1 + 2x^{2} + 2x^{4} +...

Example Question #291 : Series In Calculus

Give the Maclaurin series for the function

\displaystyle f(x) = \sqrt{3^x}

up to the third term.

Possible Answers:

\displaystyle f(x)= 1 + ( \ln 3 ) x +\left [ \frac{ ( \ln 3 )^{2} }{2} \right ] \; x^{2}+...

\displaystyle f(x)= 1 + ( \ln 3 ) x +( \ln 3 )^{2} x^{2}+...

\displaystyle f(x)= ( \ln 3 ) x +( \ln 3 )^{2} x^{2}+( \ln 3 )^{3} x^{3} +...

\displaystyle f(x)= 1 + \left ( \frac{\ln 3}{2} \right ) x +\left [ \frac{ ( \ln 3 )^{2} }{8} \right ] \; x^{2}+...

Correct answer:

\displaystyle f(x)= 1 + \left ( \frac{\ln 3}{2} \right ) x +\left [ \frac{ ( \ln 3 )^{2} }{8} \right ] \; x^{2}+...

Explanation:

Rewrite this function as 

\displaystyle f(x) = \sqrt{3^x} =\left ( \sqrt{3} \right )^x

\displaystyle = \left ( 3 ^{\frac{1}{2}} \right )^x = \left ( e^{\ln 3 ^{\frac{1}{2}}} \right )^x = \left ( e^{\frac{1}{2} \ln 3} \right )^x= \left ( e^{\frac{1}{2} \ln 3 \; \cdot x } \right )

The Maclaurin series for \displaystyle e^{x}, taken to the third term, is:

\displaystyle e ^{x} = 1 + x + \frac{x^{2}}{2}+...

Substitute \displaystyle \frac{1}{2} \ln 3 \; \cdot x for \displaystyle x:

\displaystyle f(x) = \sqrt{3^x} = 1 + \frac{1}{2} \ln 3 \; \cdot x + \frac{(\frac{1}{2} \ln 3 \; \cdot x)^{2}}{2}+...

\displaystyle = 1 + \left ( \frac{\ln 3}{2} \right ) x +\left [ \frac{ ( \ln 3 )^{2} }{8} \right ] \; x^{2}+...

Example Question #292 : Series In Calculus

Give the Maclaurin series for the function

\displaystyle f(x) = \cos \sqrt {2x}

up to the third term.

Possible Answers:

\displaystyle x^{2} - \frac{1}{6}x^{4} + \frac{1}{90} x^{6}...

\displaystyle 1 - x^{2} + \frac{1}{6}x^{4} -...

\displaystyle x - \frac{1}{6}x^{2} + \frac{1}{90} x^{3}...

\displaystyle 1 +x^{2} + \frac{1}{6}x^{4} -...

\displaystyle 1 - x + \frac{1}{6}x^{2} -...

Correct answer:

\displaystyle 1 - x + \frac{1}{6}x^{2} -...

Explanation:

The Maclaurin series for \displaystyle \cos x is 

\displaystyle \cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...

Substitute \displaystyle \sqrt{2x} for \displaystyle x. The series becomes

\displaystyle \cos \sqrt{2x} = 1 - \frac{\left ( \sqrt{2x} \right )^{2}}{2} + \frac{\left ( \sqrt{2x} \right)^{4}}{24} -...

\displaystyle = 1 - \frac{2x}{2} + \frac{4x^{2}}{24} -...

\displaystyle = 1 - x + \frac{1}{6}x^{2} -...

Example Question #50 : Taylor And Maclaurin Series

Give the Maclaurin series for the function

\displaystyle f(x) = \cos \left (2x^{2} \right )

up to the third term.

Possible Answers:

\displaystyle 1 -2x^{4} + \frac{ 2 }{3}x^{8} -...

\displaystyle 2x^{2} +\frac{4}{3} x^{6} + \frac{4}{15}x^{10} - ...

\displaystyle 2x^{2} -\frac{4}{3} x^{6} + \frac{4}{15}x^{10} - ...

\displaystyle 2x^{4} - \frac{ 2 }{3}x^{8} + \frac{4}{45}x^{12} ...

\displaystyle 1 +2x^{4} + \frac{ 2 }{3}x^{8} +...

Correct answer:

\displaystyle 1 -2x^{4} + \frac{ 2 }{3}x^{8} -...

Explanation:

The Maclaurin series for \displaystyle \cos x is 

\displaystyle \cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...

Substitute \displaystyle 2x^{2} for \displaystyle x. The series becomes

\displaystyle \cos \left (2x^{2} \right ) = 1 - \frac{\left (2x^{2} \right )^{2}}{2} + \frac{\left (2x^{2} \right )^{4}}{24} -...

\displaystyle = 1 - \frac{ 4x^{4} }{2} + \frac{ 16x^{8} }{24} -...

\displaystyle = 1 -2x^{4} + \frac{ 2 }{3}x^{8} -...

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