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I have a series of terms as follows:

$$e^{6x\pi.0} + e^{6x\pi.2} + e^{6x\pi.4} + e^{6x\pi.6}$$

Obviously the first term is just 1 but is there a way to specify the terms in one single term or shorten it somehow other than just 1 + ...?

i is an unknown in the expression

Noel Evans
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  • I would suggest that you use something other than $i$, such as $x$, in general. If you do anything with complex numbers this is going to be very confusing. Also, is there any particular form you want? – Joe Tait Feb 27 '14 at 15:18
  • @JoeTait I've updated the is to xs. I'm not after a particular format. Just a way where I can follow the steps of the reduction so I can apply it to other problems – Noel Evans Feb 27 '14 at 15:23

2 Answers2

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If the ".k" is multiply by $k$ then:

$$S=1+e^{6x\pi 2}+e^{6x\pi 4}+ e^{6x\pi 6}$$

$$Se^{6x\pi 2}=e^{6x\pi 2}+e^{6x\pi 4}+e^{6x\pi 6}+ e^{6x\pi 8}$$

Substract the two expressions:

$$Se^{6x\pi 2}-S=(e^{6x\pi 2}+e^{6x\pi 4}+e^{6x\pi 6}+ e^{6x\pi 8})-(1+e^{6x\pi 2}+e^{6x\pi 4}+ e^{6x\pi 6})$$

The only terms which remains are $1$ and $e^{6x\pi 8}$: $$S(e^{6i\pi 2}-1)=e^{6i\pi 8}-1$$

$$S = \frac{e^{6i\pi 8}-1}{e^{6i\pi 2}-1}$$

rlartiga
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  • I can see this gives correct results. First you are multiplying both sides by e^{6i.pi.2}. But how do make the next step? – Noel Evans Feb 27 '14 at 15:39
  • Ahhh! So when you say "Subtract the two expressions" you mean subtract the LHS of expr 1 from the LHS of expr 2. And do the same for the RHS expressions. Wow, I get it! Thanks for your help – Noel Evans Feb 27 '14 at 16:23
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Hint $\ $ For $\ z = e^{\,\large 12\,\pi\,x}\,$ it is $\ 1 + z + z^2 + z^3 = \dfrac{z^4-1}{z-1},\ $ a Geometric Series.

Bill Dubuque
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