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When we have two polynomials written in this form $\sum_{i=0}^{n} a_{i}x^{i} + \sum_{j=0}^{m} b_{j}x^{j}$, how can their sum be equal to $\sum_{k=0}^{MAX(n,m)} (a_{k}+b_{k})x^{k}$?

If we take for example $n=3$ and $m=2$ we get:

$\sum_{i=0}^{3} a_{i}x^{i}= a_{0}+ a_{1}x + a_{2}x^{2} + a_{3}x^{3}$

$\sum_{j=0}^{2} b_{j}x^{j}= b_{0}+ b_{1}x + b_{2}x^{2}$

When we add first and second sum:

$(a_{0}+ a_{1}x + a_{2}x^{2} + a_{3}x^{3}) + (b_{0}+ b_{1}x + b_{2}x^{2})$

we get:

$a_{0}+b_{0} + x(a_{1}+b_{1}) + x^{2}(a_{2}+b_{2})+a_{3}x^{3}$

How can that be same as $\sum_{k=0}^{3} (a_{k}+b_{k})x^{k}$ ( if our max is 3)? when this sum is equal to :

$\sum_{k=0}^{3} (a_{k}+b_{k})x^{k}=a_{0}+b_{0} + x(a_{1}+b_{1}) + x^{2}(a_{2}+b_{2})+x^{3}(a_{3}+b_{3})$?

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  • $\begingroup$ We are simply assuming $b_3=0$, since the degree of $\sum_{k\geq 0}b_k x^k$ is two. $\endgroup$ – Jack D'Aurizio Oct 19 '16 at 19:37
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    $\begingroup$ It is understood/convention that $a_k=0$ for $k>n$ and $b_k=0$ for $k>m$. $\endgroup$ – Hagen von Eitzen Oct 19 '16 at 19:38
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    $\begingroup$ a polynomial is defined as a sequence that becomes, at one point, stationary equal to 0. the degree is defined as the largest index whose term is non-zero $\endgroup$ – Aseed Oct 19 '16 at 19:39

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