Question

power series representation

Answer 1

for f(x)= ln(5-x)
d/dx [ln(5-x)] = -1/(5-x)

Answer 2

\[-\frac{ \frac{ 1 }{ 5 } }{ 1-\frac{ x }{ 5 }}\]

Answer 3

Cool looks like you're on the right track so far c:
\[\Large\rm \ln(5-x)=-\frac{1}{5}\int\limits \frac{1}{1-\frac{x}{5}}dx\]

Answer 4

okay so i did the geometric series as\[\sum_{n=0}^{\infty} -\frac{ 1 }{ 5}(\frac{ x }{ 5 })^n\]

Answer 5

Mmm yah good!
\[\Large\rm \ln(5-x)=-\frac{1}{5}\int\limits\limits \sum_{n=0}^{\infty}\left(\frac{x}{5}\right)^ndx\]

Answer 6

yeah the ntegration is my problem, i got to this step\[-\frac{ 1 }{ 5 }( \sum_{n=0}^{\infty}\frac{ x ^{(n+1)} }{ 5^n(n+1) +c })\]

Answer 7

oops its shd be the series + C

Answer 8

but from there i dont know how to simlplify it

Answer 9

So this integral is giving you trouble?
\[\Large\rm \int\limits \left(\frac{x}{5}\right)^n dx\]You can do a u-sub if that will help.
Oh oh nevermind I see what you did. you took a 5 out because we get a factor of 5. I see.

Answer 10

See the 1/5 in front of the sum?
I would bring that into the sum so you can also write your denominator as 5^{n+1}

Answer 11

yup, so\[-\sum_{n=0}^{\infty} \frac{ x ^{n+1} }{ 5^{n+1}(n+1)}+c\]

Answer 12

then after?

Answer 13

\[\Large\rm -\sum_{n=0}^{\infty}\frac{1}{n+1}\left(\frac{x}{5}\right)^{n+1}+C\]Hmm I'm not sure if you can do much beyond this..

Answer 14

ohh, so means thats the representation?, my prof simplified it to \[-\frac{ C }{ 5 }-\sum_{n=1}^{\infty} \frac{ x^n }{ n5^n }\]
idk how he got that

Answer 15

Oh Umm yah i guess it's a good idea to have a power of n instead of n+1 on our geometric representation.

So make a substitution:
n+1 = k

or something like that.
So to replace the index in the lower boundary of the sum we have:
n=0 gives,
0+1=k

so our sum would start from 1, yes?

Answer 16

okay so if it starts from one, can you explain why all the +1 in the exponents are gone?

Answer 17

\[\Large\rm -\sum_{\color{orangered}{n=0}}^{\infty}\frac{1}{\color{orangered}{n+1}}\left(\frac{x}{5}\right)^{\color{orangered}{n+1}}+C\]We're making a substitution, \(\Large\rm k=n+1\).
We'll need to replace all of these orange parts using our substitution.\[\Large\rm -\sum_{\color{orangered}{k=1}}^{\infty}\frac{1}{\color{orangered}{k}}\left(\frac{x}{5}\right)^{\color{orangered}{k}}+C\]Yes?
Confused on any of that?

Answer 18

ahhh! haha okay i get it, why is that necessary? anw thanks it very clear to me now! haha

Answer 19

I'm not exactly sure why that's done!
It's common to have the geometric series have a really simple exponent so it's clear where the sum starts from I guess.

Answer 20

Able to make sense of the garbage attached to that C?

Answer 21

yup! it means when x=0, then C=-5ln5 is it that? so ln(5-x) = ln 5 - series

Answer 22

Mmm you seem to have a better understanding of this than I do XD haha
I'm a little rusty.

I was thinking the constant was just arbitrary and we could just absorb the -1/5 into the C.
But that doesn't appear to be the case.

So yes... *cough* what you said.. that's.. probably... correct... :D lol

Answer 23

haha okay thank you so much!! :D