Find the expansion of this term

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SUMMARY

The discussion focuses on the expansion of the term \( e^{m \arctan x} \) and the proof that \( (n+1)a_{n+1} + (n-1)a_{n-1} = ma_n \). Participants emphasize the importance of deriving series coefficients through differentiation rather than arbitrary substitutions. The Maclaurin series expansion is highlighted as a key method for obtaining the terms of the series, specifically by substituting \( y \) with \( \arctan(x) \). This approach leads to a clearer understanding of the expansion process.

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  • Understanding of Maclaurin series expansion
  • Familiarity with differentiation techniques
  • Knowledge of the arctangent function
  • Basic algebraic manipulation skills
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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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