## 17Calculus Infinite Series - Limits, Derivatives and Integrals

##### 17Calculus

Calculus of Taylor and Power Series

Topics You Need To Understand For This Page

Limits, derivatives and integrals of series work just as you would expect. Since we have the sum and difference rules for limits and derivatives and integrals are just limits, we can use them to take limits, derivatives and integrals of each term of a series.

### List of Common Power and Taylor Series

$$\displaystyle{ \frac{1}{1-x} = \sum_ {n=0}^{\infty}{x^n} }$$

converges $$-1 < x < 1$$

power series

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

converges $$-\infty < x < \infty$$

Maclaurin series

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

Taylor series about $$x = 1$$

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

converges $$-\infty < x < \infty$$

Maclaurin series

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

converges $$-\infty < x < \infty$$

Maclaurin series

Series Calculus

As we said above, we can take limits, derivatives and integrals of infinite series just as we have with finite functions. The only difference is that we need to watch our notation.

Okay, time for some practice problems.

Practice

Approximate $$\int_{0}^{1}{ x~\cos(x^3) ~dx }$$ to within 3 decimal places using the Taylor Series.

Problem Statement

Approximate $$\int_{0}^{1}{ x~\cos(x^3) ~dx }$$ to within 3 decimal places using the Taylor Series.

Final Answer

$$\displaystyle{ \int_{0}^{1}{ x~\cos(x^3) ~dx }\approx 0.440 }$$

Problem Statement

Approximate $$\int_{0}^{1}{ x~\cos(x^3) ~dx }$$ to within 3 decimal places using the Taylor Series.

Solution

### PatrickJMT - 369 video solution

video by PatrickJMT

Final Answer

$$\displaystyle{ \int_{0}^{1}{ x~\cos(x^3) ~dx }\approx 0.440 }$$

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Use the Maclaurin series to evaluate $$\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } }$$

Problem Statement

Use the Maclaurin series to evaluate $$\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } }$$

Final Answer

$$\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } = 1/3 }$$

Problem Statement

Use the Maclaurin series to evaluate $$\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } }$$

Solution

### PatrickJMT - 370 video solution

video by PatrickJMT

Final Answer

$$\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } = 1/3 }$$

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Calculate the Maclaurin polynomial of order 3 for $$f(x) = \dfrac{x}{(3-x)^2}$$ using a known series.

Problem Statement

Calculate the Maclaurin polynomial of order 3 for $$f(x) = \dfrac{x}{(3-x)^2}$$ using a known series.

Solution

### Patrick Byrnes - 4311 video solution

video by Patrick Byrnes

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Calculate the Taylor polynomial of order 3 for $$f(x) = \dfrac{x}{(3-x)^2}$$ centered at 2 using a known series.

Problem Statement

Calculate the Taylor polynomial of order 3 for $$f(x) = \dfrac{x}{(3-x)^2}$$ centered at 2 using a known series.

Solution

### Patrick Byrnes - 4313 video solution

video by Patrick Byrnes

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