Question

how do I write a power series for function(s) ?

Answer 1

My function is f(x) = x/(1-3x)

Answer 2

division might help

Answer 3

otherwise you want to determne the rule for the coefficients of:

\[f(x)=a_o+a_1x+a_2x^2+a_3x^3+...\]
and take successive derivatives to determine the as

Answer 4

can we go slower please.

Answer 5

how do I get a_0, is it f'(0) ?

Answer 6

a_0 = f(0)

\[f(0)=a_o+a_1(0)+a_2(0)+a_3(0)+...\]

Answer 7

oh. so let me get the a(0) first.

Answer 8

\[a(0)=\frac{0}{1-3(0)}=0\]

Answer 9

then a(1) is ?

Answer 10

a_1=f(1)?

Answer 11

we take the derivative of both sides, since they have to move in the same fashion their derivatives will equal

Answer 12

\[f'(x)=0+a_1+2a_2x+3a_3x^2+4a_4x^3+...\]
\[f'(0)=0+a_1+2a_2(0)+3a_3(0)^2+4a_4(0)^3+...\]
\[f'(0)=a_1\]

Answer 13

a_1=1

Answer 14

are you familiar with power series notation?

Answer 15

what do you mean by power series notation?

Answer 16

probably am, but not with the wording

Answer 17

the first term of the power series is a constant, so we are looking for it after we take derivatives

\[f(x)=\sum_0a_nx^n~:~f(0)=a_0\]
\[f'(x)=\sum_1a_nnx^{n-1}~:~f'(0)=a_1(1)\]
\[f''(x)=\sum_2a_nn(n-1)x^{n-2}~:~f''(0)=a_2(2)\]
\[f'''(x)=\sum_3a_nn(n-1)(n-2)x^{n-3}~:~f''(0)=a_3(6)\]

Answer 18

f'''(0) = 6 a_3 that is, missed a tic

Answer 19

but heres a trick, we can sometimes form the power series by simply dividing in a special order

x +3x^2 +3^2x^3 + ....
----------------------
1-3x | x
-(x -3x^2)
---------
3x^2
-(3x^2 -3^2x^3)
--------------
3^2x^3
-(3^2x^3 -3^3x^3)
------------------
3^3x^3

its simpler than trying to takes derivatives to me

Answer 20

our a_n = 3^(n-1), for n>=1

Answer 21

I am struggling through this part, but I will review it... tnx for feedback.

Answer 22

which part is the struggle? knowing why it works? or taking the derivatives? or something else?