Definition 1. By the notation [imath]p^k(x)[/imath] we mean the [imath]k[/imath]:th derivative of [imath]p[/imath], the polynomial we get by differentiating [imath]p(x)[/imath] [imath]k[/imath] times.
Now suppose [imath]p(x)=x^n[/imath]. Then the [imath]k:[/imath]th derivative of [imath]x^n[/imath] is
[math]\frac{n!}{(n-k)!}x^{n-k}[/math]
Let [imath]p(x)=a_0+a_1x+...+a_nx^n[/imath] be our usual polynomial. If we differentiate it [imath]k[/imath] times all terms with degree [imath]< k[/imath] vanish and we have
[math]p^k(x)=(a_k)k!+a_{k+1}\left(\frac{(k+1)!}{1!}\right)x+...+a_n\left(\frac{n!}{(n-k)!}\right)x^{n-k}[/math]
If we let [imath]x=0[/imath], then only the constant term "survives",
[math]p^k(0)=k!a_k \quad \text{or} \quad a_k=\frac{p^k(0)}{k!}[/math]Hence we can express the coefficients in [imath]p[/imath] with the help of [imath]p[/imath]:s derivative in 0. We can now use this trick on a special polynomial: [imath]p(x)=(1+x)^n[/imath]. By a previous theorem, the [imath]k:[/imath]th derivative of [imath]p[/imath] is equal to
[math]n(n-1)...(n-k+1)(1+x)^{n-k}=\frac{n!}{(n-k)!}(1+x)^{n-k}[/math]
So
[math]p^k(0)=\frac{n!}{(n-k)!}[/math]
It follows that the coefficient for [imath]x^k[/imath] in [imath](1+x)^n[/imath] is equal to
[math]\frac{p^k(0)}{k!}=\frac{n!}{(n-k)!k!}=\binom{n}{k}[/math] Which is the coefficient for [imath]x^k[/imath] in [imath](1+x)^n[/imath]
What confuses me about this proof is that the author uses the same notation for all polynomials. I.e first that [imath]p(x)=(x+1)^n[/imath] and then as [imath]p(x)=a_0+a_1x+...+a_nx^n[/imath]. Suppose for clarity that we denote [imath]p_*(x)=a_0+a_1x+...+a_nx^n[/imath]. Then the underlying assumption seems to be that
[math]p^k(0)=p_*^k(0)[/math]
This is not entirely obvious to me. Is this true since the polynomial [imath]p_*(x)[/imath] was more general?
Now suppose [imath]p(x)=x^n[/imath]. Then the [imath]k:[/imath]th derivative of [imath]x^n[/imath] is
[math]\frac{n!}{(n-k)!}x^{n-k}[/math]
Let [imath]p(x)=a_0+a_1x+...+a_nx^n[/imath] be our usual polynomial. If we differentiate it [imath]k[/imath] times all terms with degree [imath]< k[/imath] vanish and we have
[math]p^k(x)=(a_k)k!+a_{k+1}\left(\frac{(k+1)!}{1!}\right)x+...+a_n\left(\frac{n!}{(n-k)!}\right)x^{n-k}[/math]
If we let [imath]x=0[/imath], then only the constant term "survives",
[math]p^k(0)=k!a_k \quad \text{or} \quad a_k=\frac{p^k(0)}{k!}[/math]Hence we can express the coefficients in [imath]p[/imath] with the help of [imath]p[/imath]:s derivative in 0. We can now use this trick on a special polynomial: [imath]p(x)=(1+x)^n[/imath]. By a previous theorem, the [imath]k:[/imath]th derivative of [imath]p[/imath] is equal to
[math]n(n-1)...(n-k+1)(1+x)^{n-k}=\frac{n!}{(n-k)!}(1+x)^{n-k}[/math]
So
[math]p^k(0)=\frac{n!}{(n-k)!}[/math]
It follows that the coefficient for [imath]x^k[/imath] in [imath](1+x)^n[/imath] is equal to
[math]\frac{p^k(0)}{k!}=\frac{n!}{(n-k)!k!}=\binom{n}{k}[/math] Which is the coefficient for [imath]x^k[/imath] in [imath](1+x)^n[/imath]
What confuses me about this proof is that the author uses the same notation for all polynomials. I.e first that [imath]p(x)=(x+1)^n[/imath] and then as [imath]p(x)=a_0+a_1x+...+a_nx^n[/imath]. Suppose for clarity that we denote [imath]p_*(x)=a_0+a_1x+...+a_nx^n[/imath]. Then the underlying assumption seems to be that
[math]p^k(0)=p_*^k(0)[/math]
This is not entirely obvious to me. Is this true since the polynomial [imath]p_*(x)[/imath] was more general?
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