Question:
I am working on a problem involving the product of two polynomials and their derivatives, and I need some help understanding a specific part of the derivation.
Suppose we have two polynomials:
[math]p(x)= a_0 + a_1x + a_2x^2 + \cdots + a_nx^n[/math]
[math]q(x)= b_0 + b_1x + b_2x^2 + \cdots + b_nx^n[/math]
We want to prove that:
[math](pq)'(x)= p'(x)q(x) + p(x)q'(x)[/math]
To be more specific, we have shown that the coefficient of [imath]x^{k-1}[/imath] in the product [imath]pq[/imath] is equal to
[math]k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)[/math]
However, the part I am struggling with involves finding the coefficient of [imath]x^{k-1}[/imath] in [imath]p'(x)q(x)+p(x)q'(x)[/imath]. I want to understand why the same result holds for the expression [imath]p'(x)q(x)+p(x)q'(x)[/imath]. I have tried expanding both products [imath]p'(x)q(x)[/imath] and [imath]p(x)q'(x)[/imath] and then combining the terms, but I'm not sure how to prove that the sum of these terms also yields the coefficient [imath]k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)[/imath].
My solution attempt is the following:
Suppose
Let [imath]p'(x)=c_0+c_1x+\cdots +c_{n-1}x^{n-1}[/imath] where [imath]c_0=a_1,c_1=2a_2, \cdots, c_{n-1}=na_n [/imath]
and [imath]q'(x)=d_0+d_1x+\cdots d_{n-1}x^{n-1} [/imath] for [imath]d_0=b_1, d_1=2b_2,\cdots, d_{n-1}=nb_n[/imath]
Now consider the product [imath]p'(x)q(x)[/imath]. Its terms will be of the form [imath]c_ib_jx^{i+j}[/imath], and for the exponent of [imath]x[/imath] to equal [imath]x^{k-1}[/imath] we must have [imath]i+j=k-1[/imath]. Hence the coefficient is equal to
[math]c_0b_{k-1}+c_1b_{k-2}+\cdots+c_{k-1}b_0[/math]
By substitution, we find that the coefficient for [imath]x^{k-1}[/imath] in the product [imath]p'(x)q(x)[/imath] is equal to
[math]a_1b_{k-1}+2a_2b_{k-2}+\cdots+ka_kb_0[/math]
By a similar method, we find the coefficient for [imath]x^{k-1}[/imath] in the product [imath]p(x)q'(x)[/imath] is equal to
[math]ka_0b_k+\cdots+2a_{k-2}b_2+a_{k-1}b_1[/math]
Here I'm stuck as I fail to see how the addition of these two expressions equals [math]k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)[/math]
I am working on a problem involving the product of two polynomials and their derivatives, and I need some help understanding a specific part of the derivation.
Suppose we have two polynomials:
[math]p(x)= a_0 + a_1x + a_2x^2 + \cdots + a_nx^n[/math]
[math]q(x)= b_0 + b_1x + b_2x^2 + \cdots + b_nx^n[/math]
We want to prove that:
[math](pq)'(x)= p'(x)q(x) + p(x)q'(x)[/math]
To be more specific, we have shown that the coefficient of [imath]x^{k-1}[/imath] in the product [imath]pq[/imath] is equal to
[math]k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)[/math]
However, the part I am struggling with involves finding the coefficient of [imath]x^{k-1}[/imath] in [imath]p'(x)q(x)+p(x)q'(x)[/imath]. I want to understand why the same result holds for the expression [imath]p'(x)q(x)+p(x)q'(x)[/imath]. I have tried expanding both products [imath]p'(x)q(x)[/imath] and [imath]p(x)q'(x)[/imath] and then combining the terms, but I'm not sure how to prove that the sum of these terms also yields the coefficient [imath]k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)[/imath].
My solution attempt is the following:
Suppose
Let [imath]p'(x)=c_0+c_1x+\cdots +c_{n-1}x^{n-1}[/imath] where [imath]c_0=a_1,c_1=2a_2, \cdots, c_{n-1}=na_n [/imath]
and [imath]q'(x)=d_0+d_1x+\cdots d_{n-1}x^{n-1} [/imath] for [imath]d_0=b_1, d_1=2b_2,\cdots, d_{n-1}=nb_n[/imath]
Now consider the product [imath]p'(x)q(x)[/imath]. Its terms will be of the form [imath]c_ib_jx^{i+j}[/imath], and for the exponent of [imath]x[/imath] to equal [imath]x^{k-1}[/imath] we must have [imath]i+j=k-1[/imath]. Hence the coefficient is equal to
[math]c_0b_{k-1}+c_1b_{k-2}+\cdots+c_{k-1}b_0[/math]
By substitution, we find that the coefficient for [imath]x^{k-1}[/imath] in the product [imath]p'(x)q(x)[/imath] is equal to
[math]a_1b_{k-1}+2a_2b_{k-2}+\cdots+ka_kb_0[/math]
By a similar method, we find the coefficient for [imath]x^{k-1}[/imath] in the product [imath]p(x)q'(x)[/imath] is equal to
[math]ka_0b_k+\cdots+2a_{k-2}b_2+a_{k-1}b_1[/math]
Here I'm stuck as I fail to see how the addition of these two expressions equals [math]k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)[/math]
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