Find variables from a multiplication of geometric sequences

Question

Given $$f(x) = \sum_{i=0}^\infty a_ix^i$$ And $$f(x)(1+2x+2x^2+x^3) = \frac{1}{(1-x)^3}$$ Find the values of $a_0$, $a_1$, and $a_2$.

I try to expand them but it doesn't seem to lead me to any solution. How can I isolate a0, a1, and a2 to find them?

Answer 1

Tip:

Notice $1+2x+2x^2+x^3=\left(1+x+x^2\right)(1+x)$

Then $$\begin{split}f(x)&=\frac1{(1-x)^3(1+x)\left(1+x+x^2\right)}\\&=\frac1{1-x}\cdot\frac1{(1-x)(1+x)}\cdot\frac1{(1-x)\left(1+x+x^2\right)}\\&=\frac1{1-x}\cdot\frac1{1-x^2}\cdot\frac1{1-x^3}\\&=\left(1+x+x^2+x^3+...\right)\left(1+x^2+x^4+x^6+...\right)\left(1+x^3+x^6+x^9+...\right)\end{split}$$

Then expansion finishes it all. Try it!

Answer 2

From the second condition we get $$f(x) = \frac{1}{(1-x)^3(1+2x+2x^2+x^3)} \tag 1$$ From the first condition we get $$f(x)=\sum_{i=0}^\infty a_ix^i = a_0 + a_1x+a_2x^2+a_3x^3 + ... \tag 2$$ Hence, $$a_0 = f(0) \overset {(1)}= \dfrac{1}{1\cdot 1} = 1 \tag 3$$ To find $a_1$, differentiate $(2)$ $$f'(x) = a_1 + a_2x + ...$$ Hence, $$a_1 = f'(0)$$ wich you can find from $(1)$ $$f'(x)=-\frac{-3(1-x)^2(1+2x+2x^2+x^3) + (1-x)^3(2+4x+3x^2)}{\text{denominator}^2}$$ which gives $$ a_1 = f'(0) = -\frac{-3+2}{1} = 1$$ Similarly, you can find $$a_2 = f''(0)$$

Answer 3

$f(x)= a_0+a_1x+a_2x^2+O(x^3)$ $f(x)(1+2x+2x^2+x^3)$ $=a_0+(2a_0+a_1)x+(2a_0+2a_1+a_2) x^2 +O(x^3)$ $$\textrm{Again } f(x)(1+2x+2x^2+x^3)=\frac{1}{(1-x)^3}=1+3x+6x^2+O(x^3)$$ Comparing the like powers of x we get:$$a_0=1,a_1=1,a_2=2$$

Answer 4

$$f(x)(1+2x+2x^2+x^3) = \frac{1}{(1-x)^3}\implies $$ $$(1-x-x^2+x^4+x^5-x^6)\,f(x)=1$$

Now, let $f(x)=a+b x+c x^2$ and ignore the terms $x^n$ if $n>3$. This gives $$a+ (b-a)x- (a+b-c)x^2=1$$ Comparing the terms, $a=1$, $b=a=1$, $c=a+b=2$.

Answer 5

Also possible as:

Thanks for all of your help guys! :)