Question

calculus help

Answer 1

What?

Answer 2

@AZ

Answer 3

\[\sum_{n=0}^{\infty}a_n x^n\] has a radius of convergence of 25

That means |x| < 25 is when the power series converges
and |x| > 25 is when the power series diverges

You can use the ratio test to determine the convergence of \( a_n x^n\) and then equate that to 25
And then determine the convergence of \( a_n x^{2n}\) and use that to find out what the radius of convergence would be

The ratio test:
\[L = \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert \]
If L < 1, then it's absolutely convergent

Answer 4

I haven't worked out the ratio test but there should just be a more elegant solution in that for the first power series:
|x| < 25

then for the other power series
|x^2| < 25
and to determine the radius of convergence, you have to have it in the form of |x| < R
where R is your radius of convergence

Answer 5

@imqwerty @surjithayer can you confirm?

Answer 6

\[First ~series: ~\sum_{n=0}^{\infty}a_n\color{lime}{\left(x\right)}^n~~converges~for ~R=25.\]That means that the series will converge if the thing written in lime has an absolute value less than \(R=25\). So it converges if \(\color{\lime}{\left|x\right|} < 25\)
\[Second~series: \sum_{n=0}^{\infty}a_n\color{lime}{\left(x\right)}^{2n}~can~be~written ~as~\sum_{n=0}^{\infty}a_n\color{lime}{\left(x^2\right)}^n.\]\[\sum_{n=0}^{\infty}a_n\color{lime}{\left(x^2\right)}^n ~looks ~very~similar~to ~the~first~series.\]And we know that this type of series converges if the thing written in lime has an absolute value less than 25.
\(\color{\lime}{\left|x^2\right|} < 25\)
\(\color{\lime}{\left|x\right|}^2 < 25\)
Therefore, \(\color{\lime}{\left|x\right|} < 5\) for the second series to converge.