0
$\begingroup$

I have $$A(x) = \sum_{n=0}^\infty a_n x^n$$ and $$A(x) = (1+x)/(1+7x+6x^2)$$

I need to find $a_0,a_1,a_2,a_3$.

I multiply on each side and I get

$$ (1+7x+6x^2) (a_0+a_1x+a_2x^2+a_3x^3+\ldots)=1+x $$

Now I get $$ a_0+a_1x+7a_0x+\ldots = a_0 + x(a_1+7a_0) = 1+x $$

and I can compare the two sides which gives me $$ a_0 = 1, \qquad a_1=1-7a_0=-6, \qquad a_2=0, \qquad a_3=0. $$

This makes sense to me but according to my homework the correct result should be $a_0 = 1$, $a_1 = -6$, $a_2 = 36$ and $a_3 = -216$.

What am I doing wrong?

$\endgroup$
3
$\begingroup$

Expand what you wrote $$(1+7x+6x^2) (a_0+a_1x+a_2x^2+a_3x^3+\ldots)$$ to get $$a_0+(7 a_0+a_1) x+(6 a_0+7 a_1+a_2) x^2+(6 a_1+7 a_2+a_3) x^3+(6 a_2+7 a_3) x^4+6 a_3 x^5$$ and now compare with the rhs which is $1+x$; so $$a_0=1$$ $$7a_0+a_1=1$$ $$6 a_0+7 a_1+a_2=0$$ $$6 a_1+7 a_2+a_3=0$$

I am sure that you can take from here.

$\endgroup$
0
$\begingroup$

$$1+7x+6x^2=(1+x)(1+6x)$$

So assuming $1+x\ne0,$ we need $$(1+6x)^{-1}=1+(-1)(6x)+\frac{(-1)(-1-1)}{2!}(6x)^2+\cdots=\sum_{r=0}^\infty(-6x)^r$$

assuming $|-6x|<1$ using Newton's generalized binomial theorem

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.