Question

Hey mates:
Suppose I want to approximate f(x) = (1-3x)^(-5) using a MacLaurin Series. The way I did is to integrate until F(x) = (1/C)*(1-3x)^(-1) and then approximate it using geometric series and differentiate it back again. The answer matches, but I was wondering, is it the only way? It looks very "kludgy" to solve it like that.

Answer 1

thats actually a pretty nifty way to explore it.

otherwise you have to go thru the ordeal of trying to find a pattern in the derivatives to define the same rule by

Answer 2

0: 1-3x.-5
1: -5.-3. 1-3x. -6
2: -5.-6.-3.-3. 1-3x. -7
3: -5.-6.-7.-3.-3.-3. 1-3x . -8
4: -(5.6.7.8) .-3^4 . 1-3x . -9
5: -(9.8.7.6.5) (-3)^5 (1-3x)^(-10)

\[f^{(n)}(x) = -\frac{n!}{4!}(-3)^n(1-3x)^{-5-n}\]

Answer 3

seem to work out for n>0

Answer 4

(n+4)!

Answer 5

Hello @amistre64 Thanks for helping me out again :-)
One way I thought also was to rewrite it as:\[(1/(1-3x))^{5}\] and approximate it using the geometric series raised by 5, but it didn't work. I will fiddle around with your answer also.

Answer 6

yeah, i tried that a few days ago when I was fiddling with it; dint pan out then either :)

Answer 7

Wolfram gives a interesting alternate form for your solution: http://www.wolframalpha.com/input/?i=-1%2F8+3%5E%28-1%2Bn%29+e%5E%28i+n+%CF%80%29+%281-3+x%29%5E%28-5-n%29+%284%2Bn%29%21&lk=1&a=ClashPrefs_*Math-
Also, it seems that it works fine. Will check its radius of convergence.

Answer 8

http://www.wolframalpha.com/input/?i=%28n%2B4%29%21%2F4%21+%28-3%29%5En+%281-3x%29%5E%28-5-n%29

seems to be a fair match except for the sign changes

Answer 9

that of course is the function that defines the coeffs of the series; and not the series representation itself

Answer 10

well, i spose the coeffs are the main part of the series lol; the rest just being x^n/n!

Answer 11

3^n x^n is a series for (1-3x)^(-1)

if you tried (3^n x^n)^5 its gonna fail simply becasue that is not a correct description of (1-3x)^(-5)

\[\left(\sum_{inf}^{0}3^nx^n\right)^5\]would be it

which is actually an infinitly long polynomial raised to the 5th.

Answer 12

\[3^nx^n = \frac{1}{1-3x}\]
\[[3^nx^n]'=3^n\ nx^{n-1}\]
\[[3^nx^n]''=3^n\ n(n-1)x^{n-2}\]
\[[3^nx^n]'''=3^n\ n(n-1)(n-2)x^{n-3}\]
\[[3^nx^n]^{(4)}=3^n\ n(n-1)(n-2)(n-3)x^{n-4}=\frac{(-3)^4}{(1-3x)^5}\]

Answer 13

or rather, it should; but the wolf says that would be:\[\frac{1}{x^3(3x-1)^5}\]

Answer 14

if I change x^(n-4) to x^n ; its goes close

\[\frac{1}{(3x-1)^5}\]

which means im off by a negative