Calculus of Taylor and Power Series
Topics You Need To Understand For This Page |
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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.
\(\displaystyle{ \frac{1}{1-x} = \sum_ {n=0}^{\infty}{x^n} }\) |
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converges \( -1 < x < 1 \) |
power series |
\(\displaystyle{ e^x = \sum_{n=0}^{x^n}{\frac{x^n}{n!}} }\) |
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converges \(-\infty < x < \infty\) |
Maclaurin series |
\(\displaystyle{\ln(x) = \sum_{n=0}^{\infty}{ \frac{(-1)^{n+1} (x-1)^n}{n} } }\) |
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Taylor series about \( x = 1 \) |
\(\displaystyle{ \sin(x) = \sum_{n=0}^{\infty}{ \frac{(-1)^n x^{2n+1}}{(2n+1)!} } }\) |
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converges \(-\infty < x < \infty\) |
Maclaurin series |
\(\displaystyle{ \cos(x) = \sum_{n=0}^{\infty}{ \frac{(-1)^n x^{2n}}{(2n)!} } }\) |
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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 |
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Approximate \( \int_{0}^{1}{ x~\cos(x^3) ~dx } \) to within 3 decimal places using the Taylor Series.
Final Answer |
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\(\displaystyle{ \int_{0}^{1}{ x~\cos(x^3) ~dx }\approx 0.440 }\)
Problem Statement
Approximate \( \int_{0}^{1}{ x~\cos(x^3) ~dx } \) to within 3 decimal places using the Taylor Series.
Solution
video by PatrickJMT |
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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 |
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Use the Maclaurin series to evaluate \(\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } }\)
Final Answer |
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\(\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } = 1/3 }\)
Problem Statement
Use the Maclaurin series to evaluate \(\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } }\)
Solution
video by PatrickJMT |
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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
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
video by Patrick Byrnes |
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