Question

Find a power series that represents 1/(3x) and give its interval of convergence.

Answer 1

\[\frac{ 1 }{ 3 } (\frac{ 1 }{ 1+(x-1) })\]

Answer 2

Is this the full description of your problem? a power series that represents 1/(3x)? that sounds very vague to me. I am not sure if they want the power series to converge to 1/(3x) or if it's supposed to do something else.

Answer 3

Sorry, yes, for the power series to converge to 1/(3x). I agree, the problem is written vaguely.

Answer 4

\[\Large \frac{1}{1-x}= \sum_{n=0}^{\infty} x^n , \forall |x| < 1\]

Answer 5

so I would say we substitute \(-x +1\) on both sides and multiply by \(1/3\)

Answer 6

does that already give you some input that you can work with?

Answer 7

Yes! Thank you so much!

Answer 8

Great, you can discuss the interval of convergence with a proper substitution. Also make sure that you discuss the boundary points on which convergences becomes critical.

Answer 9

In this case, is the interval of convergence 1<x<2?

Answer 10

I get \[\large 0<x<2 \]

Answer 11

Can you recheck? otherwise I will try again.

Answer 12

No, you are right. I used a wrong value on one side. Thank you! My calc teacher never even taught me the substitution trick. It's so helpful.

Answer 13

It's essential for a lot of Taylor Approximation you will encounter later on. Here is a nice example: If we substitute \(-x^2\) in the above we get:
\[\Large \frac{1}{1+x^2}= \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n} \]
We can integrate both sides, and we can do that within the boundaries in which the power series does absolutely converge and we get \[ \int \frac{1}{1+x^2}dx=\arctan(x)= \int \sum_{n=0}^{\infty} (-1)^n x^{2n}= \sum_{n=0}^\infty (-1)\frac{x^{2n+1}}{2n+1}+C \]
Which is yet another application of similar manner and a nice clean result for a taylor approximation of arctan :)