# Thread: [SOLVED] Manipulation of power series

1. ## [SOLVED] Manipulation of power series

Find a power series representation of the function:

$f(x) = ln(5-x)$

Here is my work:

$\int ln(5-x) = \frac{1}{5-x} = 5(\frac{1}{1-\frac{x}{5}})$

$\sum^{\infty}_{n=0} (\frac{x}{5})^n$

Integrate.

$= \frac{(\frac{x}{5})^{n+1}}{n+1}$

Multiply by -1 then the series is:

$\sum^{\infty}_{n=0} \frac{(-1)(\frac{x}{5})^{n+1}}{n+1}$

$= \sum^{\infty}_{n=1} \frac{(-1)^{n-1} (\frac{x}{5})^n}{n}$

Here is where I fall apart...Do I need to integrate again? No idea....
Thanks!!

2. Originally Posted by mollymcf2009
Find a power series representation of the function:

$f(x) = ln(5-x)$

Here is my work:

$\int ln(5-x) = \frac{1}{5-x} = 5(\frac{1}{1-\frac{x}{5}})$

$\sum^{\infty}_{n=0} (\frac{x}{5})^n$

Integrate.

$= \frac{(\frac{x}{5})^{n+1}}{n+1}$

Multiply by -1 then the series is:

$\sum^{\infty}_{n=0} \frac{(-1)(\frac{x}{5})^{n+1}}{n+1}$

$= \sum^{\infty}_{n=1} \frac{(-1)^{n-1} (\frac{x}{5})^n}{n}$

Here is where I fall apart...Do I need to integrate again? No idea....
Thanks!!
first mistake. $\int \ln (5 - x)~dx \ne \frac 1{5 - x}$. that is closer to finding the derivative, and even that is wrong.

also, $\frac 1{5 - x} \ne 5 \cdot \frac 1{1 - \frac x5}$

note that $\ln (5 - x) = - \int \frac 1{5 - x}~dx$

now, find the power series for $- \frac 1{5 - x}$ and integrate term by term

you also took the integral of the power series incorrectly (if you can't do it in your head, use substitution)

3. The DERIVATIVE of $\ln x$ is 1/x not the INTEGRAL.
And do not forget the chain rule.

4. Originally Posted by Jhevon
first mistake. $\int \ln (5 - x)~dx \ne \frac 1{5 - x}$. that is closer to finding the derivative, and even that is wrong.

also, $\frac 1{5 - x} \ne 5 \cdot \frac 1{1 - \frac x5}$

note that $\ln (5 - x) = - \int \frac 1{5 - x}~dx$

now, find the power series for $- \frac 1{5 - x}$ and integrate term by term

you also took the integral of the power series incorrectly (if you can't do it in your head, use substitution)

Ok, I got the right answer, but it is entirely possible that it was pure luck LOL Can y'all check my work?

$f(x) = ln(5-x)$

$-\int\frac{1}{5-x}$

My power series is:

$-\frac{1}{5} \sum^{\infty}_{n=0} (\frac{x}{5})^n$

$-\int \frac{1}{5} + \sum^{\infty}_{n=0} \int (\frac{x}{5})^n$

$
= - ln(5) + \sum^{\infty}_{n=0} \frac{(x)^{n+1}}{n+1} \cdot \frac{(\frac{1}{5})^{n+1}}{n+1}$
***IS THIS LEGIT FOR THIS INTEGRATION?

$= - ln(5) + \sum^{\infty}_{n=1} \frac{x^n (\frac{1}{5})^n}{n}$

$= ln(5) - \sum^{\infty}_{n=1} x^n \cdot \frac{1}{5^nn}$

$
= ln(5) - \sum^{\infty}_{n=1}\frac{x^n}{5^nn}$

5. Originally Posted by mollymcf2009
Ok, I got the right answer, but it is entirely possible that it was pure luck LOL Can y'all check my work?

$f(x) = ln(5-x)$

$-\int\frac{1}{5-x}$

My power series is:

$-\frac{1}{5} \sum^{\infty}_{n=0} (\frac{x}{5})^n$

$-\int \frac{1}{5} {\color{red}+} \sum^{\infty}_{n=0} \int (\frac{x}{5})^n$
this is wrong. where did that + sign come from??

$
= {\color{red}- ln(5)} + \sum^{\infty}_{n=0} \frac{(x)^{n+1}}{n+1} \cdot {\color{blue}\frac{(\frac{1}{5})^{n+1}}{n+1}}$
***IS THIS LEGIT FOR THIS INTEGRATION?
huh?... $\int k~dx = kx + C$, for any constant $k$

and the thing in blue is just wrong. where you got that from, i don't know. i think you again thought that you could integate a constant the same way you do a variable. this is wrong!

6. Originally Posted by Jhevon
this is wrong. where did that + sign come from??

huh?... $\int k~dx = kx + C$, for any constant $k$

and the thing in blue is just wrong. where you got that from, i don't know

Ok, I officially don't understand this. The correct answer is:

$= ln(5) - \sum^{\infty}_{n=1}\frac{x^n}{5^nn}$

So, how did the +/- sign after the log get there? I am really tired, I know that $\frac{1}{5}$ is just a constant, and I'm actually pretty good at my integrations, I'm just foggy brained and basically going through the motions at this point. Do you mind just helping me get through this problem so I can at least understand what I am doing? I really am understanding this better than yesterday, thanks to you Jhevon, but like I said yesterday, Molly & series are not seeing eye to eye.

7. Originally Posted by mollymcf2009
Ok, I got the right answer, but it is entirely possible that it was pure luck LOL Can y'all check my work?
so it was luck...and a little bit of cheating on your part

$= {\color{red}-} ln(5) + \sum^{\infty}_{n=1} \frac{x^n (\frac{1}{5})^n}{n}$

$= ln(5) {\color{red}-} \sum^{\infty}_{n=1} x^n \cdot \frac{1}{5^nn}$
how did you change the signs here? you seem to just be doing things to make your work look like the answer. naughty!

anyway, here's the solution:

$\ln (5 - x) = \ln 5 \left( 1 - \frac x5 \right)$

$= \ln 5 + \ln \left( 1 - \frac x5 \right)$

$= \ln 5 - \frac 15 \int \frac 1{1 - \frac x5}~dx$

$= \ln 5 - \frac 15 \int \sum_{n = 0}^\infty \left( \frac x5 \right)^n~dx$

$= \ln 5 - \frac 15 \sum_{n = 0}^\infty \int \left( \frac x5 \right)^n~dx$

$= \ln 5 - \frac 15 \sum_{n = 0}^\infty \frac {5\left( \frac x5 \right)^{n+1}}{n + 1}$

$= \ln 5 - \sum_{n = 1}^\infty \frac {\left( \frac x5 \right)^n}n$

$= \ln 5 - \sum_{n = 1}^\infty \frac {x^n}{5^nn}$

Now you can go to bed

8. Originally Posted by Jhevon
so it was luck...and a little bit of cheating on your part

Hey, you gotta love Webassign and multiple choice

how did you change the signs here? you seem to just be doing things to make your work look like the answer. naughty!
BUSTED!!! LOL At this point, I guess I am doing that just because I can't figure out any other way to get this to sink in. I figure if I work enough of these, and have someone (ha ha...um...YOU) blast my every move, that eventually I'll get it?!?!? or NOT?!?!? LOL

anyway, here's the solution:

$\ln (5 - x) = \ln 5 \left( 1 - \frac x5 \right)$

$= \ln 5 + \ln \left( 1 - \frac x5 \right)$

$= \ln 5 - \frac 15 \int \frac 1{1 - \frac x5}~dx$

$= \ln 5 - \frac 15 \int \sum_{n = 0}^\infty \left( \frac x5 \right)^n~dx$

$= \ln 5 - \frac 15 \sum_{n = 0}^\infty \int \left( \frac x5 \right)^n~dx$

$= \ln 5 - \frac 15 \sum_{n = 0}^\infty \frac {5\left( \frac x5 \right)^{n+1}}{n + 1}$

$= \ln 5 - \sum_{n = 1}^\infty \frac {\left( \frac x5 \right)^n}n$

$= \ln 5 - \sum_{n = 1}^\infty \frac {x^n}{5^nn}$

Now you can go to bed
You are THE BEST!!!!!!! Again, I can not tell you how much I appreciate your help on all of this! Now I AM going to bed! You get some rest too, as tomorrow and next week I'll be harassing you with Taylor & MacLaurin