Coefficent in a infinite power series

[pac1337]

How do I find a coefficent of x⁹ in a power series like this: (1+x³+x⁶+x⁹+...)³
or this: (x²+x³+x⁴+x⁵+...)³

 

[lex]

If they are literally 'like these', then you can use [MATH](1+X+X^2+X^3....)=\frac{1}{1-X} \text{ when }|X|<1[/MATH]So [MATH](1+x^3+x^6+x^9+...)^3 = \left(\frac{1}{1-x^3}\right)^3=(1-x^3)^{-3}[/MATH]which you can then expand up to [MATH]x^9[/MATH]
[MATH](x^2+x^3+x^4+...)^3=(x^2(1+x+x^2+..))^3=x^6(1-x)^{-3}[/MATH]which you can expand up to [MATH]x^3[/MATH]
or you can expand them quite easily manually up to [MATH]x^9[/MATH]or you can ask wolframalpha to tell you: here

 

[Deleted member 4993]

In the response 2, remember that |x| < 1

 

[pka]

How do I find a coefficent of x⁹ in a power series like this: (1+x³+x⁶+x⁹+...)³or this: (x²+x³+x⁴+x⁵+...)³

Look at this expansion. see the \(10x^9\) it is true for any upper limit \(k\ge 3\).
Now look at this sum.

 

[Steven G]

If they are literally 'like these', then you can use [MATH](1+X+X^2+X^3....)=\frac{1}{1-X} \text{ when }|X|<1[/MATH]So [MATH](1+x^3+x^6+x^9+...)^3 = \left(\frac{1}{1-x^3}\right)^3=(1-x^3)^{-3}[/MATH]which you can then expand up to [MATH]x^9[/MATH]

I understood all this until you said to expand up to x⁹. Isn't (1-x^3)^^-3 a rational expression?

 

[pka]

If they are literally 'like these', then you can use [MATH](1+X+X^2+X^3....)=\frac{1}{1-X} \text{ when }|X|<1[/MATH] So [MATH](1+x^3+x^6+x^9+...)^3 = \left(\frac{1}{1-x^3}\right)^3=(1-x^3)^{-3}[/MATH] which you can then expand up to [MATH]x^9[/MATH]

I understood all this until you said to expand up to x⁹. Isn't (1-x^3)^^-3 a rational expression?

Where was it posted that \(|X|<1~?\)

 

[lex]

I understood all this until you said to expand up to x⁹. Isn't (1-x^3)^^-3 a rational expression?

Negative Binomial Series -- from Wolfram MathWorld

The series which arises in the binomial theorem for negative integer -n, (x+a)^(-n) = sum_(k=0)^(infty)(-n; k)x^ka^(-n-k) (1) = sum_(k=0)^(infty)(-1)^k(n+k-1; k)x^ka^(-n-k) (2) for |x|<a. For a=1, the negative binomial series simplifies to (3)

mathworld.wolfram.com

 

[lex]

Where was it posted that \(|X|<1~?\)

[MATH](1+X+X^2+X^3....) \text{ only converges to }\frac{1}{1-X} \text{ when }|X|<1[/MATH]This is where |X|<1 first appears.
The power series expansion of [MATH]\frac{1}{1-X}[/MATH] is [MATH]\hspace1ex 1+X+X^2+X^3[/MATH].... (and this converges when |X|<1).
[MATH]\frac{1}{(1-x^3)^3}[/MATH] produces the same power series as in the original post, (and it converges when |x|<1) and so can be used to generate the coefficient of [MATH]x^9[/MATH].

 

[lex]

[MATH](1-x^3)^{-3}=1 + (-3)(-x^3)+\frac{(-3)(-4)}{2!}(-x^3)^2+\frac{(-3)(-4)(-5)}{3!}(-x^3)^3+...+ \tfrac{\prod\limits_{i=0}^{k-1} (3+i)}{k!} \;x^{3k}+...[/MATH]\begin{align*}
\text{So the coefficient of } x^9 \text{ is: }&\frac{(-3)(-4)(-5)}{3!}\\
&=10\\
\end{align*}