Help with Fourier Transform problem

[bobhays]

I have a fourier transform problem and I don't understand a step. The picture is the solution

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The difference between the highlighted and boxed part is that they changed the limits of the summation from n=0 to n=1. Effectively they pulled out the n=0 term. The problem is that the coefficients also changed. At n=0 the term in the sum is equal to 1*e^0 = 1. So despite changing the limits to n=1 the coefficient should stay the same. Is there a reason why it was changed to 5?

[kingkickolas]

Pretty weird, took me more than a little bit . I'll remember this little trick though!

 

[bobhays]

2 minutes ago, kingkickolas said:

Pretty weird, took me more than a little bit . I'll remember this little trick though!

 

Perfect! thank you so much

 

EDIT: How did you write the formulas?

[kingkickolas]

30 minutes ago, bobhays said:

Perfect! thank you so much

 

EDIT: How did you write the formulas?

Any LaTeX editor will do. If it's just a couple lines, this site is fine, otherwise ShareLaTeX is great or you can download a LaTeX editor. For those specific formulas it was just the following, if that's what you were asking:

\begin{align*}
e^{-j\omega} \sum_{n=0}^\infty \left(\frac{1}{5} \right)^n e^{-j\omega n}
		&= \left( 5 \cdot \frac{1}{5} \right ) \cdot e^{-j\omega} \sum_{n=0}^\infty \left(\frac{1}{5} \right)^n e^{-j\omega n} \\
	&= 5 \sum_{n=0}^\infty \left(\frac{1}{5} \right)^{n+1} e^{-j\omega (n+1)} \\
	&= 5 \sum_{n=1}^\infty \left(\frac{1}{5} \right)^{n} e^{-j\omega n}
\end{align*}