Question

write the following function in the form of a series

ln(5-x)

so

d/dx ln(5-x) = -1/5(1-(x/5))

I get,

\[\ln(5-x) + c=\frac{-1}{5}\sum_{n=0}^{\infty} \int\limits_{}^{}(\frac{x}{5})^{n}\]

Set x = 0, I get c = -ln(5)

\[\ln(5-x) =\ln(5) - \frac{1}{5}\sum_{n=0}^{\infty} \int\limits_{}^{}(\frac{x}{5})^{n} \]

Can some show me how to solve this and clarify if possible if I'm doing this correctly?

Answer 1

I'd use maclaurin series. to figure this out.

Answer 2

ok then, so I'm way off?

Answer 3

I think you can keep doing what you are doing, but I think it's just easier doing the maclarurin series.

Answer 4

I'm suppose to learn that but, I haven't gotten around to it yet still trying to crack this :\

Answer 5

Just keep going with the internal.

Answer 6

\[\sum_{\infty}^{n=0} \int\limits_{}^{} x^n/5^{n+1}\]

Answer 7

thanks :), I think my method is correct it is just different from what my textbook is doing, I keep getting the same radius for all the answers but the series appear different

Answer 8

I will probably just break down and go down to the math help centre at my university though to make sure.

Answer 9

here is what i would do, but it might be silly:
start with something well known, \(\log(1+x)\) and adjust

Answer 10

I got the answer

\[\ln(5)-\sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)5^{n+1}}\]

with R = 5

Answer 11

my reasoning behind what I'm doing is that we know the function

1/(1-x) = \[\frac{1}{1-x}=\sum_{n=2}^{\infty}(x)^{n}\]

Answer 12

Yeah, that's right.

Answer 13

so I can just take the derivative of the ln(x) function to get something I want and then just integrate the series and the derivative of the ln(x) function to get back to where I was

Answer 14

Then again I'm not 100% sure another member of this website said I was incorrect