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Champ's question at Yahoo! Answers (Power series)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

Find a power series for y = e^x^2 about x = 0(Manipulate common Maclaurin series for e^x). Use this power series to show that y = e^x^2 is a solution to the differential equation y' - 2xy = 0.
Please help and show your steps. I've been trying to solve it for hours!
Here is a link to the question:

Find a power series... Calc Help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Champ,

From $e^x=\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ forall $x\in\mathbb{R}$ we immediately get $y=e^{x^2}=\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}$ for all $x\in\mathbb{R}$. Now, using the derivation term by term:

$$\begin{aligned}y'-2xy&=\displaystyle\sum_{n=1}^{\infty}\frac{2nx^{2n-1}}{n!}-2x\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}\\&=\displaystyle\sum_{n=1}^{\infty}\frac{2nx^{2n-1}}{n!}-\displaystyle\sum_{n=0}^{\infty}\frac{2x^{2n+1}}{n!}\\&=\displaystyle\sum_{n=1}^{\infty}\frac{2x^{2n-1}}{(n-1)!}-\displaystyle\sum_{n=1}^{\infty}\frac{2x^{2n-1}}{(n-1)!}\\&=0\end{aligned}$$

If you have further questions you can post them in the Calculus section.
 

calcboi

New member
Mar 31, 2013
16
Hello Champ,

From $e^x=\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ forall $x\in\mathbb{R}$ we immediately get $y=e^{x^2}=\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}$ for all $x\in\mathbb{R}$. Now, using the derivation term by term:

$$\begin{aligned}y'-2xy&=\displaystyle\sum_{n=1}^{\infty}\frac{2nx^{2n-1}}{n!}-2x\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}\\&=\displaystyle\sum_{n=1}^{\infty}\frac{2nx^{2n-1}}{n!}-\displaystyle\sum_{n=0}^{\infty}\frac{2x^{2n+1}}{n!}\\&=\displaystyle\sum_{n=1}^{\infty}\frac{2x^{2n-1}}{(n-1)!}-\displaystyle\sum_{n=1}^{\infty}\frac{2x^{2n-1}}{(n-1)!}\\&=0\end{aligned}$$

If you have further questions you can post them in the Calculus section.
Thanks, I got most of that but the part I was stuck on was how to derive n!. It appears that you when you derived it, n! remained unchanged. It this true?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Thanks, I got most of that but the part I was stuck on was how to derive n!. It appears that you when you derived it, n! remained unchanged. It this true?
Yes, it is true. Take into account that $1/n!$ is a constant.
 

calcboi

New member
Mar 31, 2013
16
Thanks, I see how that works if its a constant. Anyways, I have a question on which test to use for series n=1 to infinity for (-1)^n / (n^3)-ln(n) in order to determine convergence/divergence. I am pretty sure I determined it converges through the Alternating Series Test but I am not sure whether it is conditional or absolute. I tried the Direct Comparison Test but it was inconclusive, and I am stuck now on what to do. Can you please help?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Thanks, I see how that works if its a constant. Anyways, I have a question on which test to use for series n=1 to infinity for (-1)^n / (n^3)-ln(n) in order to determine convergence/divergence. I am pretty sure I determined it converges through the Alternating Series Test but I am not sure whether it is conditional or absolute. I tried the Direct Comparison Test but it was inconclusive, and I am stuck now on what to do. Can you please help?
No problem calcboi we'll help you. But please, ask this question in the Calculus forum, according to rule #8 of MHB. Thank you.
 

calcboi

New member
Mar 31, 2013
16
Oops, sorry. I will ask the question in the Calc thread.