Power Series

[scott73]

Alright, I am given the function (8x)/(14+x) = (Summation) c(n)X^(n), and asked to find the first 4 coefficients and the radius. I am having trouble figuring out how to get the coefficients, I know I cant just plug 1,2,3,etc. in for x and get them, I think you may have to convert another power series that looks like this, and then figure the coefficients. Any Help on this one would be great!,

Also, I have the problem [Summation]((n+1)(x^n)/(9^(n+2))), for -9>x<9, and I am asked to find a formula for the sum of the series. I thought that I figured it out to be abs(9/x) but that was incorrect. What am I doing wrong?

 

[Gene]

I've never understood nor believed in power series but since no one else is jumping in...
a₀ = f(0)/0!
a₁ = f'(0)/1!
a₂ = f''(0)/2!
...
a_n = fⁿ(0)/n!
in a₀ + a₁x + a₂x²...a_nx^n

I don't see a trick on the sumation yet. Still thinking.

 

[galactus]

Alright, I am given the function (8x)/(14+x) = (Summation) c(n)X^(n), and asked to find the first 4 coefficients and the radius. I am having trouble figuring out how to get the coefficients, I know I cant just plug 1,2,3,etc. in for x and get them, I think you may have to convert another power series that looks like this, and then figure the coefficients. Any Help on this one would be great!,

Also, I have the problem [Summation]((n+1)(x^n)/(9^(n+2))), for -9>x<9, and I am asked to find a formula for the sum of the series. I thought that I figured it out to be abs(9/x) but that was incorrect. What am I doing wrong?

 

[Gene]

I suspect that -9>x<9 is realy -9>x>9
and you want a power series.
((n+1)(x^n)/(9^(n+2))) =
((n+1)/9^(n+2))*x^n
f(0)=0
f'(x)=((n+1)/9^(n+2))*n*x^(n-1)=0
In fact since 0^y=0 for all y, I suspect another typo that is harder to correct. I don't recall what 0^0 is which is the only possible end.

 

[pka]

 

[soroban]

Hello, scott73!

I think I've got the second problem . . .
. . I worked out a long and primitive solution.

[Summation]((n+1)(xⁿ)/(9ⁿ⁺²), for |x| < 9

Find a formula for the sum of the series.

Write out the terms of the summation:

. . . . . . .2x . .3x² . .4x³ . .5x⁴
. . S .= . --- + ---- + ---- + ---- + . . . . [1]
. . . . . . .9³ . . 9⁴ . . 9⁵ . . .9⁶

Multiply by x/9:

. . . x . . . . .2x² . .3x³ . .4x⁴
. . . -- S .= .---- + ---- + ---- + . . . . [2]
. . . 9 . . . . .9⁴ . . .9⁵ . . .9⁶

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2x . .x² . . x³ . . x⁴
Subtract [2] from [1]: . (1 - x/9)S .= .--- + --- + --- + --- + . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9³ . .9⁴ . .9⁵ . . 9⁶

. . . . . . . . . . . . . . . . . . . . . . .x . . .x . . x² . . x³ . . x⁴
We have: . [(9 - x)/x]S . = . --- + --- + --- + --- + --- + ...
. . . . . . . . . . . . . . . . . . . . . . 9³ . .9³ . .9⁴ . . 9⁵ . .9⁶
. . . . . . . . . . . . . . . . . . . . . . . . . .\_______________

The last portion is a geometric series with first term, a = x/9³, and common ratio, r = x/9.
. . Its sum is: .x/[81(9 - x)]

. . . . . . . . . . . . . 9 - x . . . . . .x . . . . . .x . . . . . . .x(18 - x)
Now we have: . ------ S . = . --- + ----------- . = . -----------
. . . . . . . . . . . . . . x . . . . . . . 9³ . .81(9 - x) . . . . 9³(9 - x)

. . . . . . . . . . . . . . . .x²(18 - x)
Therefore: . S . = . --------------
. . . . . . . . . . . . . . . .9³(9 - x)²

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Someone check my work . . . please1
There are at least a brazillion* places for errors.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*
"How many are there in a brazillion?"
"Pardon me?"
"I just heard it on the news: 12 brazillion soldiers killed."

 

[galactus]

Hello all. I cheated and ran this through Maple. It gave me a closed form of

x(18-x)
---------=1/(x-9)²-1/81
81(9-x)².

Which is Soroban's answer multiplied by 9/x. This is with n=1 to infinity.

Notice that the answer for the sum from n=1..infinity is the sum n=0..infinity minus 1/81.

 

[Gene]

Neat trick! I don't have time to go over the whole thing right now but (1-x/9) = (9-x)/9 not (9-x)/x
----------------
Gene

 

[galactus]

Good catch Gene!. Here it is:


\(\displaystyle \frac{x(18-x)}{9^{3}(9-x)}(\frac{9}{9-x})=\frac{x(18-x)}{81(9-x)^{2}}=\frac{1}{(x-9)^{2}}-\frac{1}{81}\)