# simple arithmetic problem but..

i am having problem in this proof. i need to find the certain coefficients of this statement on the right side.

given:
$P: \mathbb{C} \Longrightarrow \mathbb{C}, \quad P(x) := 12 − 7x + x^2$
$Q : \mathbb{C} \Longrightarrow \mathbb{C}, \quad Q(x) := 30 + 22x + 10x^2 + 2x^3$
i need to find the coefficients of $c_i$ für $0 \leq i \leq 10$, so that $(P + Q)(x) = \sum_{i=0}^{10}c_i(x-2)^i$.

what i did is: i added $P+Q$ which gave me $2x^3+11x^2+15x+42$ and now i am trying to find a combination of coefficients for $c_i(x-2)^i$ so that if i sum them, it should be equal to $P+Q$. but it seems very tedious, because i need to test for all numbers and sum them. is there any other lighter way for this?

thanks a lot

-
You can cut down on the work considerably by noticing that it would be impossible to have $c_i \neq 0$ for any $i \geq 3$, since then you would have $x^i$ on the right hand side but not on the left. – Antonio Vargas Dec 9 '12 at 8:37
An alternative approach: consider finding the Taylor series of $P(x) + Q(x)$ at $x = 2$. – Antonio Vargas Dec 9 '12 at 8:39
yeah, you are right. i did this. now i am testing only for $c_1(x-2)^1+c_2(x-2)^2+c_3(x-2)^3$ but this is itself also much work – doniyor Dec 9 '12 at 8:40
we didnot have taylor series yet in the lecture, so i cannot use it i think – doniyor Dec 9 '12 at 8:41

$2(x-2)^3+a(x-2)^2+b(x-2)+c=2x^3+11x^2+15x+42$

Put $x-2=y$

$\implies 2y^3+ay^2+by+c=2(y+2)^3+11(y+2)^2+15(y+2)+42$

$=2(y^3+3y^2\cdot2+3y\cdot2^2+2^3)+11(y^2+2y\cdot2+2^2)+15(y+2)+42$

$=2y^3+y^2(2\cdot3\cdot 2+11)+y(2\cdot3\cdot 2^2+11\cdot2\cdot2+15)+2\cdot2^3+11\cdot2^2+15\cdot2+42$

Now compare the coefficients of the different powers of $y$ to determine the values of $a,b,c$.

-
@doniyor, for an identity $\sum_{i\le r\le j}a_rx^r=\sum_{i\le s\le j}b_sx^s\implies a_k=b_k$ for $i\le k\le j$ – lab bhattacharjee Dec 9 '12 at 9:00
@doniyor, we don't need the value of $y$. Comparing the coefficients of $y^2,a=2\cdot3\cdot2+11=23;$ of $y,b=2\cdot3\cdot 2^2+11\cdot2\cdot2+15=83$ and comparing the constant terms, $c=2\cdot2^3+11\cdot2^2+15\cdot2+42=132$ – lab bhattacharjee Dec 9 '12 at 11:17
@doniyor, please verify – lab bhattacharjee Dec 9 '12 at 11:23