/sci/ - Science & Math

>>14963157
If you expand the square of a general polynomial you get the following:
[eqn] \left(c_{0}+c_{1}x+c_{2}x^{2}+\dotsb+c_{n}x^{n}\right)^{2}\\=\left(c_{0}c_{0}\right)+\left(c_{0}c_{1}+c_{1}c_{0}\right)x+\left(c_{0}c_{2}+c_{1}c_{1}+c_{2}c_{0}\right)x^{2}+\dotsb+\left(\sum_{i+j=k}c_{i}c_{j}\right)x^{k}+\dotsb+\left(c_{n}c_{n}\right)x^{2n}\\=\sum_{k=0}^{2n}\left(\sum_{i+j=k}c_{i}c_{j}\right)x^{k} [/eqn]

The second sum might be confusing a bit, it mean that you should take the sum of all coefficient whose degrees sum to k. The example below illustrates this:
For your polynomial you should determine each coefficient separately

[math] c_{0}=(1\cdot1)=1\\c_{1}=(1\cdot1)+(1\cdot1)=2\\c_{2}=(1\cdot2)+(1\cdot1)+(2\cdot1)=5\\c_{3}=(1\cdot3)+(1\cdot2)+(2\cdot1)+(3\cdot1)=10\\c_{4}=(1\cdot3)+(2\cdot2)+(3\cdot1)=10\\c_{5}=(2\cdot3)+(3\cdot2)=12\\c_{6}=(3\cdot3)=9 [/math]

So

[math] \left(1+x+2x^{2}+3x^{3}\right)^{2}=1+2x+5x^{2}+10x^{3}+10x^{4}+12x^{5}+9x^{6} [/math]