Science:Math Exam Resources/Courses/MATH103/April 2006/Question 02 (e)

MATH103 April 2006

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Question 02 (e)
The Taylor series for the function $\displaystyle {\frac {1}{1+x^{2}}}$ around $x=0$ is $\displaystyle {\frac {1}{1+x^{2}}}=\sum _{n=0}^{\infty }(-1)^{n}x^{2n}=1-x^{2}+x^{4}-x^{6}+x^{8}-...$ Use this fact to find the Taylor series for $\displaystyle \arctan(x)$ around $x=0$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Notice that ${\frac {1}{1+x^{2}}}$ is the derivative of $\displaystyle \arctan(x)$.

Hint 2
Take the integral of the Taylor Series of ${\frac {1}{1+x^{2}}}$ to find the Taylor Series for $\displaystyle \arctan(x)$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Notice that ${\frac {1}{1+x^{2}}}$ is the derivative of $\displaystyle \arctan(x)$. Take the integral of its Taylor Series to find the Taylor Series for $\displaystyle \arctan(x)$. ${\begin{aligned}\arctan(x)&=\int {\frac {1}{1+x^{2}}}\,dx\\&=\int \left(1-x^{2}+x^{4}-x^{6}+x^{8}\mp \dots \right)\,dx\\&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+{\frac {x^{9}}{9}}\mp \dots \end{aligned}}$ Looking at the patterns in these terms, you can see that $\arctan(x)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{2n+1}}$

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