$$\sum_{n=1}^{\infty}a_nx^{n-1}$$
so now if n=2 would the sum change to
$$\sum_{n=2}^{\infty}a_{n+1}x^n$$?

Question

$$\sum_{n=1}^{\infty}a_nx^{n-1}$$
so now if n=2 would the sum change to 
$$\sum_{n=2}^{\infty}a_{n+1}x^n$$?

anonymous · Answer

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anonymous · Answer

so I did it wrong?

anonymous · Answer

why is it $$a_{n-1}$$?

zepdrix · Answer

Is this for DE, power series? Trying to get the terms to match so you can combine them?

anonymous · Answer

yes

anonymous · Answer

I posted the whole problem earlier, but no one wanted to help me, so I broke it up term by term

zepdrix · Answer

AHHHHH OpenStudy froze again! Lost my message :( gimme a sec to retype... sigh

anonymous · Answer

ok

zepdrix · Answer

I find it helpful to make a substitution for n, instead of trying to do it all in your head.$$\large \text{Let} \quad k=n-1, \quad \text{then} \quad n=k+1$$


$$\large \sum_{n=1}^\infty a_n\cdot x^{n-1} \qquad \rightarrow \qquad \sum_{k=0}^\infty a_{k+1}\cdot x^k$$

zepdrix · Answer

The subscript on A, and the power on X should be pretty straightforward. Do you understand where I plugged those in?

Then for the lower limit of our sum, when n = 1, it looks like based on our substitution,
$$\large k=1-1$$

zepdrix · Answer

We want our X term to be x^n or x^k, no additional +1 or -1 nonsense.
So that is how we determine the substitution. But I'm sure you already knew that, since you were shifting the N value over 1 place.

anonymous · Answer

give me a min to process this (or two ;P)