Question

Differentiating and integration power series: find power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once) give the interval of convergence for the resulting series.

g(x)=ln(1-3x). Using g(x)=1/1-3x

Answer 1

a power series is of the form
\[\sum_{n=0}^{\inf}f^{(n)}(a)\frac{(x-a)^n}{n!}\]right?

Answer 2

I don't know is this a formula.

Answer 3

its a general setup yes; where a is the value we are centering around, and f^n(a) defines the value of the derivative at a

Answer 4

Ok

Answer 5

...value of the nth derivative at "a" ; that is

Answer 6

So how can I solve this budd I am very confused

Answer 7

well, since your not familiar with this notation, i have to know what methods youve tried to begin with. otherwise ill just confuse you more

Answer 8

I think is1

Answer 9

find power series representation for g centered at 0
by differentiating or integrating the power series for f

there is nothing defined as an "f" in your posting

give the interval of convergence for the resulting series.
g(x)=ln(1-3x). Using g(x)=1/1-3x

i know how to work an interval of convergence, but the information here is a little garbled.

Answer 10

Well I saw a method in book but it was confusing as hell like it said you should take dx of 1+x^2+x^3+.... Then some how they come up with Series of k=0 as k is app 0 then (k+1)

Answer 11

For f let me see one min

Answer 12

sounds like im on the right track at least.

should we construct the function of "f" from g(x) = ln(1-3x) by chance?

Answer 13

Well according to theorem 9.5 it says let function of f be defined by the power series sum c_k(x-a)^k on its interval of convergence I.

Answer 14

Then it says 1.) f is continues function on i

Answer 15

Then 2.) the power series may be differentiated or integrated term by term, and the resulting power series converges to f prime (x) or integral of f(x)dx+c, respectively, at all points in the interior of I, where c is an arbitrary constant.

Answer 16

to simplify matters, lets use ln(u) such that u=1-3x. when x=0, u=1
\[g=ln(u)\]\[g'=u^{-1}\]\[g''=-u^{-2}\]\[g'''=2u^{-3}\]\[g^{(4)}=-3*2u^{-4}\]\[g^{(5)}=4*3*2u^{-5}\]\[...\]\[g^{(n)}=(-1)^{(n+1)}(n-1)!*u^{-n}\]; and with u=1, that last u^(-n) part becomes trivial
\[f(u)=\sum_{n=0}^{\infty}g^{(n)}(1)\frac{u^n}{n!}\]
\[f(u)=\sum_{n=0}^{\infty}(-1)^{(n+1)}(n-1)!\frac{u^n}{n!}\]
\[f(u)=\sum_{n=0}^{\infty}(-1)^{(n+1)}\frac{u^n}{n}\]
its been ahile, so im not too sure how accurate my thought is, so let me check it out first

Answer 17

Sure thanks for trying I will be waiting

Answer 18

yeah, something went awry there.

the set up for the series should be, and using the k notation to keep it consistent with your material:
\[-\sum_{k=1}^{\infty}\frac{(-3)^k~(-x)^k}{k}\]

Answer 19

So now I should just put the given ln(1-3x) in this formula

Answer 20

no, this formula is the equivalent of ln(1-3x) between x=-1/3 and x=1/3

Answer 21

So okay going back to what the question is asking what do u suggest me to do.

Answer 22

take a look how the graph of the poly conforms to the graph of the logarithm in an interval close to x=0

http://www.wolframalpha.com/input/?i=y%3D%28-1%29%28-3%29%5E%281%29%28-x%29%2B%28-1%29%28-3%29%5E%282%29%28-x%29%5E2%2F2%2B%28-1%29%28-3%29%5E%283%29%28-x%29%5E3%2F3%2B%28-1%29%28-3%29%5E%284%29%28-x%29%5E4%2F4%2C+y%3Dln%281-3x%29

Answer 23

I can't I got to sign up

Answer 24

.... thats odd, wolframalpha is a free website.

Answer 25

heres a screen shot; its good to know why this process is important. working with a poly is much simpler than the log.

Answer 26

Ok saw that

Answer 27

the setup for this would be the same as i did, but using the ln(1-3x) instead of the u-substitution :)

Answer 28

I am sorry not to look for free ride but in general can you please tell me how to solve this step by step thanks

Answer 29

you form a power series of the setup\[f(x)=g(0)+g'(0)x+g''(0)\frac{x^2}{2!}+g'''(0)\frac{x^3}{3!}+...+g^{(n)}(0)\frac{x^n}{n!}+...\]
where you use the values of the successive derivatives
\[g=ln(1-3x)~:~g(0)=0\]\[g'=-3*(1-3x)^{-1}~:~g'(0)=-3\]\[g''=-3*-1*-3*(1-3x)^{-2}~:~g''(0)=-9\]etc....
its a rather long and drawn out process

Answer 30

once you have a general rule for how the function f(x) is being worked out, you can take the summation notation and take the limit of it as a ratio of n/(n-1) terms to find the interval of convergence

Answer 31

... or at least that is how im reading the problem.

Answer 32

Ok one min please

Answer 33

Okay I got the terms now so what's next?

Answer 34

write up a summation notation that defines the polynomial

Answer 35

what terms did you come up with?

Answer 36

The derivatives of the ln(1-3x) then entering the 0 in them then getting ans.

Answer 37

do they match with these?

0, -3, -9, -54, -486, ...

Answer 38

Yes

Answer 39

good, then at least ive got that correctly :)

Answer 40

soo; for the g' s the rule:\[(-1)^{k+1}~(-3)^k~(k-1)!\]and we apply this to the \[\frac{x^k}{k!}\]to get:
\[\sum(-1)^{k+1}~(-3)^k~(k-1)!\frac{x^k}{k!}\]
\[\sum(-1)^{k+1}~(-3)^k~(k-1)!\frac{x^k}{k!}\]
\[\sum(-1)^{k+1}~(-3)^k~\frac{x^k}{k}\]and to clean it up
\[\sum~\frac{(-1)^{k+1}~(-3)^k~x^k}{k}\]is what we get to work with to find the interval of convergence

Answer 41

we find the convergence by taking the limit of the ratio of k/k-1 as k approaches infinity
\[lim ~\frac{(-1)^{k+1}~(-3)^k~x^k}{k}\frac{k-1}{(-1)^{k}~(-3)^{k-1}~x^{k-1}}\]
\[lim ~\frac{(-1)~(-3)~x}{k}\frac{k-1}{1}\]
\[lim ~(-1)~(-3)~x\frac{k-1}{k}\]
since everything without a k in it after the simplification is over has no effect on the limit; take it to the outside and wrap it in absolute values
\[ |(-1)(-3)x|\lim_{k\to\ inf}|\frac{k-1}{k}\]
\[ |3x|\lim_{k\to\ inf}|\frac{k-1}{k}\]
the limit as kapproaches infinity of (k-1)/k is 1, leaving us with |3x|

Answer 42

now for reasons I cant recall, this is defined as the radius of convergence; and the interval takes this |3x| and compares it against 1 for the interval of convergence:

\[|3x|<1\]\[3|x|<1\]\[|x|<\frac13\]or simply giving as interval of convergence as\[-\frac13<x<\frac13\]

Answer 43

allow me to butt in
i think you were supposed to integrate the series for \(\frac{1}{1-3x}\) term by term. of course computing directly will work as well

Answer 44

Satellite 73 can you please explain further.

Answer 45

yes

Answer 46

start with the well known fact
\[\frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{k=1}^{\infty}x^k\] for \(|x|<1\)

Answer 47

this is the standard trick, and if you are confused at that step (it looks like you were above) then you have to take that as a given
but since you have \(\frac{1}{1-3x}\) you have to replace \(x\) by \(3x\) giving
[\frac{1}{1-3x}=1+3x+9x^2+...=\sum_{k=0}^{\infty}(3x)^k\]

Answer 48

hmm
\[\frac{1}{1-3x}=1+3x+9x^2+...=\sum_{k=0}^{\infty}(3x)^k\] thats better

Answer 49

then you integrate this sucker term by term, since it is almost the derivative of
\[\ln(1-3x)\]

Answer 50

How for the last step?

Answer 51

after you integrate term by term (i.e. write a formula for what you would get) multiply all by \(-3\) because
\[\frac{d}{dx}\ln(1-3x)=\frac{-3}{1-3x}\] by the chain rule

Answer 52

I know budd but how can integrate this

Answer 53

it is just another way of doing it, using the gimmick of knowing the power series of the derivative, (you have to know this for the trick to work) and then integrating term by term

Answer 54

\[1\to x\]
\[3x\to \frac{3x^2}{2}\]
\[(3x)^2\to \frac{3^2x^3}{3}\]
\[(3x)^3\to \frac{3^3x^4}{4}\] literally term by term
it really amounts to figuring out the general term that is all

Answer 55

Oh I got you budd you wanted me to put the values back in the integration formula so now what budd

Answer 56

multiply by -3 and you are done

Answer 57

So i got -3x-9x^2/3-9x^3

Answer 58

Oh also what about interval of convergence?