taylor series of tan^-1(x)/x
- Thread starter taipenx
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You want to find the Taylor series of [imath]\displaystyle g(x) = \frac{\tan^{-1}x}{x}[/imath].
First question comes to mind, can we reduce this problem to a smaller one? Yes. We want to find the Taylor series of [imath]\displaystyle f(x) = \tan^{-1}x[/imath], then when we are done we will divide the series by [imath]\displaystyle x[/imath].
If you want to find the Taylor series of a function, you have to tell about what point this series will be. Otherwise, these mathematicians will assume a point that you don't like. And because I am not a mathematician, I will assume a peaceful point, [imath]\displaystyle x = 0[/imath], which happens to be that we are finding the Maclaurin series of [imath]\displaystyle f(x)[/imath].
Don't worry about the name. The Maclaurin series is just a Taylor series about [imath]\displaystyle x = 0[/imath].
Now comes the difficult (fun) part. We will introduce the machine that produces the Maclaurin series:
[imath]\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n[/imath]
This notation is a little difficult to understand, so I will simplify it.
[imath]\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdot\cdot\cdot \ \cdot [/imath]
Basically, you just need to take the derivatives of the function [imath]\displaystyle f(x) = \tan^{-1}x[/imath] and plug them above after substituting [imath]x = 0[/imath]. But this machine produce infinite derivatives! Now is the important question, how many of them do you need? The answer is, until you recognize a pattern. Don't panic, usually [imath]2[/imath] to [imath]3[/imath] derivatives will be enough to see the pattern. Try it!
After you recognize the pattern, you will create an infinite series.
[imath]\displaystyle \tan^{-1}x = \sum_{n=0}^{\infty}a_nx^{k(n)}[/imath]
The final step is to divide by [imath]x[/imath]:
[imath]\displaystyle \frac{\tan^{-1}x}{x} = \sum_{n=0}^{\infty}\frac{a_nx^{k(n)}}{x} = \sum_{n=0}^{\infty}a_nx^{k(n)-1}[/imath]
At this point, you have successfully found the Taylor series of [imath]g(x)[/imath] about [imath]x = 0[/imath].
Bonus question. Does this series converge for all values of [imath]x[/imath]?
First question comes to mind, can we reduce this problem to a smaller one? Yes. We want to find the Taylor series of [imath]\displaystyle f(x) = \tan^{-1}x[/imath], then when we are done we will divide the series by [imath]\displaystyle x[/imath].
If you want to find the Taylor series of a function, you have to tell about what point this series will be. Otherwise, these mathematicians will assume a point that you don't like. And because I am not a mathematician, I will assume a peaceful point, [imath]\displaystyle x = 0[/imath], which happens to be that we are finding the Maclaurin series of [imath]\displaystyle f(x)[/imath].
Don't worry about the name. The Maclaurin series is just a Taylor series about [imath]\displaystyle x = 0[/imath].
Now comes the difficult (fun) part. We will introduce the machine that produces the Maclaurin series:
[imath]\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n[/imath]
This notation is a little difficult to understand, so I will simplify it.
[imath]\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdot\cdot\cdot \ \cdot [/imath]
Basically, you just need to take the derivatives of the function [imath]\displaystyle f(x) = \tan^{-1}x[/imath] and plug them above after substituting [imath]x = 0[/imath]. But this machine produce infinite derivatives! Now is the important question, how many of them do you need? The answer is, until you recognize a pattern. Don't panic, usually [imath]2[/imath] to [imath]3[/imath] derivatives will be enough to see the pattern. Try it!
After you recognize the pattern, you will create an infinite series.
[imath]\displaystyle \tan^{-1}x = \sum_{n=0}^{\infty}a_nx^{k(n)}[/imath]
The final step is to divide by [imath]x[/imath]:
[imath]\displaystyle \frac{\tan^{-1}x}{x} = \sum_{n=0}^{\infty}\frac{a_nx^{k(n)}}{x} = \sum_{n=0}^{\infty}a_nx^{k(n)-1}[/imath]
At this point, you have successfully found the Taylor series of [imath]g(x)[/imath] about [imath]x = 0[/imath].
Bonus question. Does this series converge for all values of [imath]x[/imath]?
Are you following a class/textbook?