Question

Calculus. Finding a power series representation for an integral with natural log

Answer 1

\[f(x)= \int\limits \frac{ \ln(1-t) }{ t }dt\]

Answer 2

\[f ' (x) = \frac{ 1 }{ 1-t }\]=\[\sum_{n=0}^{\infty}t^n\]=\[1+t+t^2...\]

Answer 3

i know thats not the antiderivative... im just feeling like the derivative is where i should start?

Answer 4

Eqautions huh :)

Answer 5

Ok, So lets think this through

Answer 6

Do i take the antiderivative of the sum i found with the derivative?

Answer 7

or does the quotient rule come into play here too somehow?

Answer 8

how did you get that f'?

Answer 9

also is it f(x)=? or f(t)=?

Answer 10

i was thinking ln (1-t) derivative is 1/1-t and then the deriv of t is just 1

Answer 11

they gave me both. f(x)= integral ln(1-t) / t

Answer 12

power series

Answer 13

but you have the integral there
you can't differentiate the integral thing like that
hmmm... that seems weird because f(x) is a function of x and you have int ln(1-t)/t is a function of t

Answer 14

here is the exact question "Evaluate the indefinite integral as a power series"

Answer 15

\[f(t)=\int\limits t^2 dt \text{ \implies } f'(t)=t^2 \\ \text{ you can take the long way around on this one } \\ f(t)=\frac{t^3}{3}+C \text{ \implies still } f'(t)=t^2 \]

Answer 16

so \[f'(t)=\frac{\ln(1-t)}{t}\] in your case

Answer 17

im not sure why they have f(x) and then use t... that is how they have it...

Answer 18

i can show you the final answer.... i just dont know how they get it

Answer 19

\[C- \sum_{n=1}^{\infty}\frac{ t^n }{ n^2 }\] this is the answer

Answer 20

let's look at ln(1-t)

Answer 21

\[g(t)=\ln(1-t) \\ g'(t)=\frac{-1}{1-t}\]

Answer 22

do you know how to find a power series representation for g'(t)

Answer 23

Does the neg one come from -du=dx?

Answer 24

I think they want you to represent the result of the integral
as a power series.
one way is to write down the taylor series for ln(1-t)
see https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions

divide by t to get a new series, and integrate term by term to get the answer

Answer 25

chain rule
derivative of inside/inside when we have ln( )

Answer 26

yes phi thats what it is

Answer 27

they havent taught us taylor yet..

Answer 28

but you are allow to know:
\[g'(t)=- \sum_{n=0}^{\infty} t^n\]
right?
you can integrate term by term to find g(t)=ln(1-t)

Answer 29

freckles is it \[\sum_{n=0}^{\infty}(x)^n\]

Answer 30

idk if i should use x or t... they got me messed up using both

Answer 31

let's just use t

Answer 32

it doesn't matter
you choose your favorite letter and we will use that one

Answer 33

t is fine

Answer 34

ok
well
\[\sum_{n=0}^{\infty}t^n=\frac{1}{1-t} \\ \text{ so } - \sum_{n=0}^{\infty} =\frac{-1}{1-t}\]

Answer 35

so so far we have ...\[f(t)=\frac{ \ln(1-t) }{ t }, f' (t)= \frac{ -1 }{ 1-t }, \sum_{n=0}^{\infty}t^n\]

Answer 36

\[g(t)=\ln(1-t) \rightarrow g'(t)=\frac{(1-t)'}{1-t}=\frac{0-1}{1-t}=\frac{-1}{1-t} \\ g'(t)=\frac{-1}{1-t}=- \sum_{n=0}^\infty t^n\]

Answer 37

ok no that isn't f' above

Answer 38

to find g(t) integrate g'(t) term by term

Answer 39

and i integrating the derivative of the function f(t)?

Answer 40

do you know how to integrate t^n ?

Answer 41

w.r.t t?

Answer 42

am i

Answer 43

that would be t^n / n

Answer 44

right? integrating t^n= t^n/ n ?

Answer 45

t^(n+1)/(n+1) by power rule

Answer 46

oh yea, add one to the exponent and then drop the new exponent to denom...

Answer 47

\[\text{ so far we have this: } f(t)=\int\limits \frac{\ln(1-t)}{t} dt \\ f'(t)=\frac{\ln(1-t)}{t} \\ \text{ where we wanted \to play with } \ln(1-t) \\ \text{ so I made a new function name } \text{ Let } g(t)=\ln(1-t) \\ \text{ so } g'(t)=\frac{-1}{1-t}=- \sum_{n=0}^\infty t^n \\ \text{ now integrating term by term we can find g from g' } \\ g(t)=- \sum_{n=0}^\infty \frac{t^{n+1}}{n+1} =\ln(1-t) \\ \text{ now } f'(t)=\frac{g(t)}{t}=- \frac{1}{t} \sum _{n=0}^\infty \frac{t^{n+1}}{n+1}\]
this is what we have so far

Answer 48

do you have questions with this so far

Answer 49

we still have to do one more thing

Answer 50

i am reading everything, trying to find some sense... the last post you made is making a lot more sense, where you summed everything up..

Answer 51

\[f'(t)=-\sum_{n=0}^\infty \frac{t^{n+1}}{n+1} \cdot t^{-1} =-\sum_{n=0}^\infty \frac{t^n}{n+1}\]
now the last step I was talking about is find f from f'

Answer 52

i am trying to understand how n+1 becomes n^2

Answer 53

well you still need to find f given f'

Answer 54

can you do that last step
and I will show you the result we get is equivalent to what you have way above

Answer 55

hint: to find f given f' integrate
and put your +C
I guess we should have put C on that one thing when need from g' to g :p

Answer 56

\[\int\limits \sum_{n=0}^{\infty}\frac{ t^n }{ n+1 }= \sum_{n=0}^{\infty}\frac{ t^{n+1} }{ n+1(n+1) }\]?

Answer 57

yeah sorta you need paranthesis around the other (n+1)

Answer 58

-C

Answer 59

\[-\int\limits \sum_{n=0}^{\infty}\frac{ t^{n+1} }{ (n+1)^2 }= C-\sum_{n=1}^{\infty}\frac{ t^n }{ n^2 }\] because we drop the +1 on the n's when we change the index?

Answer 60

let me give you another summary so far:
\[\text{ so far we have this: } f(t)=\int\limits \frac{\ln(1-t)}{t} dt \\ f'(t)=\frac{\ln(1-t)}{t} \\ \text{ where we wanted to play with } \ln(1-t) \\ \text{ so I made a new function name } \text{ Let } g(t)=\ln(1-t) \\ \text{ so } g'(t)=\frac{-1}{1-t}=- \sum_{n=0}^\infty t^n \\ \text{ now integrating term by term we can find g from g' } \\ g(t)=- \sum_{n=0}^\infty \frac{t^{n+1}}{n+1}+C =\ln(1-t) \\ g(0)=0+C=\ln(1-0) \\ \text{ so this integration constant is C=0}\\ \\ \text{ now } f'(t)=\frac{g(t)}{t}=- \frac{1}{t} \sum _{n=0}^\infty \frac{t^{n+1}}{n+1}

\]

\[f'(t)=\frac{g(t)}{t}=- \sum_{n=0}^\infty \frac{t^n}{n+1} \\ f(t)=- \sum_{n=0}^\infty \frac{t^{n+1}}{(n+1)^2}+C \\ \text{ or } f(t)=C- \sum_{n=0}^\infty \frac{t^{n+1}}{(n+1)^2}\]
now our series starts at n=0 and theirs starts at n=1
we can play with what we have to get their exact series thing
now we have n=0 on bottom
add on one both sides n+1=1 <--we have this ...
so this means if we replace old (n+1)'s with (n)'s
and start the series at n=1 instead we will get their exact series looking thing
though you could leave it like this and it is still correct well you might still need to find the constant of integration
so replace (n+1)'s with n's and start series at n=1 we have:
\[f(t)=C- \sum_{n=1}^\infty \frac{t^n}{n^2}\]

Answer 61

wow... this one is really hard!

Answer 62

it doesn't matter if you say
\[f(t)=C-\sum_{n=0}^\infty \frac{t^{n+1}}{(n+1)^2} \text{ or if you say } f(t)=C-\sum_{n=1}^\infty \frac{t^n}{n^2}\]
it is the exact same thing

Answer 63

thank you for the summary! i will need to look over this awhile before i feel i got the hang of this one! i am so grateful tho, thank you!

Answer 64

you can even compare the first terms if you want if you don't see it

Answer 65

so, i take the derivative 1st, and THEN i integrate the power series representation we found by taking the derivative?

Answer 66

here is another example if rewriting the same series with a different starting number:
\[\sum_{i=2}^{5} 5^{i+3} \\ \text{ now we have } \\ i=2 \\ \text{ add say } 3 \text{ on both sides } i+3=2+3 \\ \text{ so } i+3 =5 \\ \text{ so we can write } \\ \sum_{i=2}^{5}5^{i+3} \text{ as } \sum_{i=5}^{5+3} 5^i\]

Answer 67

like when dealing with infinities we don't have to worry about changing infinity to infinity+1 because it is still infinity
but if you have a ending number that is finite then you would have to change it
like start at n=2 and ending at n=5
means if you change it to n+3=2+3=5 and ending at n+3=5+3=8
then you start the series at n=5 and end it at n=8 and change all the (n+3)'s to n's

Answer 68

i don't think we found the derivative to integrate?
what do you mean

Answer 69

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx by the way paul is one of my favorite things to look at for notes
if you want to look too here is a link

---
I took derivative first of the integral function
and we got ln(1-t)/t
I knew the derivative of ln(1-t) was -1/(1-t) which had a famous power series representation
so I played with ln(1-t) first then I came back and divided by t so we could have power series representation for ln(1-t)/t
then seen we wanted int (ln(1-t)/t) I integrate the power series of ln(1-t)/t

Answer 70

int means integral

Answer 71

\[-\frac{ 1 }{ t }\int\limits \sum_{n=0}^{\infty}\frac{ t^{n+1} }{ n+1}= -\int\limits \sum_{n=0}^{\infty}\frac{ t^n }{ n+1 }=C-\sum_{n=0}^{\infty}\frac{ t^{n+1} }{ (n+1)^2 }\]
This is the part that messed me up the most that you really cleared up for me

Answer 72

i think i meant to ask the question like this--> " If i am to sum up how we performed this in words, I would say we took the derivative of f(t) first to find the power series representation of the function. Then we integrated the power series we found

Answer 73

yea

Answer 74

ok... i think im finally understanding this. and thanks for the pauls note link!

Answer 75

im def gonna re study this a bit... but you helped a whole bunch!

Answer 76

well yeah that -1/t is suppose to be inside the integral sign
but coolness :p
I'm glad it makes more sense to you