Question

Find a power seriesssss

Answer 1

For x^x and radius of convergence XD

@ganeshie8

Answer 2

what is the definition of \(x^x\) ?

Answer 3

Hint : \[e^{x\ln x}\]

Answer 4

\[x^x = e^{xlnx} = \sum \frac{ (xlnx)^n }{ n! }\]

Answer 5

This?

Answer 6

that looks good !
for radius of convergence try ur fav convergence test : ratio test ?

Answer 7

Really it's that easy..wow.
Yeah I'll try that right now

Answer 8

power of power series !

Answer 9

do u really need to use a test ?

Answer 10

I got infinity lol i messed up I think

Answer 11

you know that it converges to x^x no matter what the value of x is, so..

Answer 12

There are issues with \(e^{x\ln(x)}\) that don't exist with \(x^x\) when \(x\geq 0\).

Answer 13

Well, there is an issue with \(0\) in either case, but not for negative \(x\).

Answer 14

we let 0^0 = 1

Answer 15

atleast while working with series i think..

Answer 16

It is infinity yeah :D

Answer 17

wait a sec, what we got over there is not really a power series as the coefficients are function of x right ?

Answer 18

I guess

Answer 19

you can fix that by plugging in power seroes of lnx over there
but then u need to shift the center too hmm

Answer 20

it may not be correct to call it a power series, but its okay to call it infinnite series expansion @wio what do u think

Answer 21

Power series must be in the form of: \[
\sum c_n(x-a)^n
\]

Answer 22

Exactly ^

Answer 23

Yeah the power series converges for all x

Answer 24

In this case \(a=0\) and \(c_n = [\ln(x)]^n / n!\)

Answer 25

Is it allowed for that \(x\) to be there? I am not so sure.

Answer 26

we dont want x terms in coefficient right ?

Answer 27

Using ratio test I got 0 which means for it converges everywhere so it's infinite xD

Answer 28

if it is allowed, we are okays yeah

Answer 29

I think it is allowed

Answer 30

I'm pretty sure that \(x\) can't be in the coefficient. \(c_n\) is supposed to be a constant; allowing \(c_n\) to contain \(x\) turns it into a function.

Answer 31

Also, if we're to find a Taylor expansion for \(x^x\), it would make more sense to find it about the point \(x=1\) (not \(x=0\) since \(x^x\) and it's derivatives aren't defined when \(x=0\)).

Answer 32

yeah it wont be a polynomial anymore if we allow x's in coeffiecient function

Answer 33

agree we need to center the taylor expansion at x=1
not sure why wolfram is fooling us
http://www.wolframalpha.com/input/?i=power+series+x%5Ex