Find the expansion of this term

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Homework Help Overview

The discussion revolves around finding the expansion of the term \( e^{m \arctan x} \) and proving a relationship involving its coefficients. The problem is situated within the context of series expansions and involves concepts from calculus and mathematical analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the series expansion of \( e^{m \arctan x} \) and question how to derive the terms of the series. There are attempts to relate the series to known expansions, such as the Maclaurin series, and concerns are raised about substituting arbitrary functions into the series.

Discussion Status

The discussion is ongoing, with participants exploring different methods to derive the series coefficients. Some guidance has been offered regarding the need for careful differentiation to obtain the coefficients, and there is recognition of the challenges involved in the problem.

Contextual Notes

Participants note the importance of correctly deriving series coefficients rather than substituting arbitrary functions, indicating a potential misunderstanding of series expansion techniques.

utkarshakash
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Homework Statement


IF e^{m \arctan x}=a_0 + a_1x + a_2x^2 + a_3x^3...+a_nx^n+...
prove that (n+1)a_{n+1} + (n-1)a_{n-1}=ma_n
and hence obtain the expansion of e^{m \arctan x}.

Homework Equations



The Attempt at a Solution


$$e^{m \arctan x} = 1+m \arctan x + (m \arctan x)^2/2! + (m \arctan x)^3/3! + ...$$
But I need to simplify this expansion in terms of x and I'm clueless how to do that.
 
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How did you get the terms of that series?
 
Simon Bridge said:
How did you get the terms of that series?
The Maclaurin series expansion of emy is
$$1 + my + \frac{(my)^2}{2!} + \dots + \frac{(my)^n}{n!} + \dots$$

Replace y with arctan(x) to get the expansion that utkarshakash shows.

I've been looking at this problem, but no joy as yet.
 
You cannot just substitute any arbitrary function for y ... consider: what if y=ln|x|/m ?
You have to work out the series coefficients by actually doing the differentiation. It's not as hard as it's, at first, seeming.
 
Simon Bridge said:
You cannot just substitute any arbitrary function for y ... consider: what if y=ln|x|/m ?
You have to work out the series coefficients by actually doing the differentiation. It's not as hard as it's, at first, seeming.

Got it! Thanks a lot!
 
No worries - if a shortcut does not get you where you need to be, try going the long way round ;)
 

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