I'm trying to evaluate the limit as h approachs 0 of the sum from k = 0 to n of:
$\frac{1}{h^n}(-1)^{k+n}\binom{n}{k}\frac{1}{(x+kh)^2-2(x+kh)+17}$
If it helps, it's the limit definition of the nth derivative of $\frac{1}{x^2-2x+17}$
Any thoughts?
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1$\begingroup$ Let $c>b^2$, then $f(x)=\frac1{x^2-2bx+c}=\frac{1}{(x-b-i\sqrt{c-b^2})(x-b+i\sqrt{c-b^2})}=\frac1{2i\sqrt{c-b^2}}\Big(\frac1{x-b-i\sqrt{c-b^2}}-\frac1{x-b+i\sqrt{c-b^2}}\Big)$ $$f^{(n)}(x)=\frac{(-1)^nn!}{\sqrt{c-b^2}}\,\Im\frac1{(x-b-i\sqrt{c-b^2})^{n+1}}$$ $$=\frac{(-1)^nn!}{\sqrt{c-b^2}}\left(\frac1{x^2-2b+c}\right)^{\frac{n+1}2}\sin\left((n+1)\arctan\frac{\sqrt{c-b^2}}{|x-b|}\right)$$ $\endgroup$– SvyatoslavCommented Mar 17, 2023 at 3:55
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$\begingroup$ @Svyatoslav Where did the formula for f^(n)(x) come from? Also, quicking plotting this into Desmos shows it's not quite the same as f^(n)(x). Is this an approximation? Is there an error? $\endgroup$– GhullCommented Mar 17, 2023 at 12:24
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1 Answer
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Too long for a comment. Let's check.
$\displaystyle \qquad\sin x=\frac{|\tan x|}{\sqrt{1+\tan^2x}};\,\cos x=\frac1{\sqrt{1+\tan^2x}}\tag{0}$
$\displaystyle n=0$ $$ f^{(0)}(x)=\frac1{\sqrt{c-b^2}}\frac1{\sqrt{x^2-2bx+c}}\sin\left(\arctan\frac{\sqrt{c-b^2}}{|x-b|}\right)$$ $$=\frac1{\sqrt{c-b^2}}\frac1{\sqrt{x^2-2bx+c}}\frac{\frac{\sqrt{c-b^2}}{|x-b|}}{\sqrt{1+\frac{c-b^2}{(x-b)^2}}}=\frac{1}{x^2-2bx+c}\tag{1}$$ $\displaystyle n=1$ $$\qquad f^{(1)}(x)=-\frac1{\sqrt{c-b^2}}\frac1{x^2-2bx+c}2\sin\left(\arctan\frac{\sqrt{c-b^2}}{|x-b|}\right)\cos\left(\arctan\frac{\sqrt{c-b^2}}{|x-b|}\right)$$ $$=-\frac2{\sqrt{c-b^2}}\frac1{x^2-2bx+c}\frac{\frac{\sqrt{c-b^2}}{|x-b|}}{\sqrt{1+\frac{c-b^2}{(x-b)^2}}}\frac1{\sqrt{1+\frac{c-b^2}{(x-b)^2}}}=-2\frac{x-b}{(x^2-2bx+c)^2}$$
In the comment I used $$ \Im\frac {d^{n-1}}{dp^{n-1}}\left(\frac1{p-iq}\right)=\Im\frac{(-1)^{n-1}(n-1)!}{(p-iq)^n}=(-1)^{n-1}(n-1)!\,\Im\left(\frac{1}{\sqrt{p^2+q^2}}e^{i\arctan\frac qp}\right)^n$$ $$=(-1)^{n-1}(n-1)!\frac{\sin(n\arctan\frac qp)}{(p^2+q^2)^\frac n2}$$
answered Mar 17, 2023 at 13:23
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$\begingroup$ Ah I think that makes sense now. I have one more question: why are we only taking the imaginary part? Thanks! $\endgroup$– GhullCommented Mar 17, 2023 at 15:20
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$\begingroup$ The imaginary part appears in a natural way, because$$\frac1{x^2-2bx+c}=\frac{1}{(x-b-i\sqrt{c-b^2})(x-b+i\sqrt{c-b^2})}$$ $$=\frac1{\sqrt{c-b^2}}\,\frac1{2i}\Big(\frac1{x-b-i\sqrt{c-b^2}}-\frac1{x-b+i\sqrt{c-b^2}}\Big)$$ $$=\frac1{\sqrt{c-b^2}}\,\frac1{2i}\Big(\frac1{x-b-i\sqrt{c-b^2}}-\Big(\frac1{x-b-i\sqrt{c-b^2}}\Big)^*\Big)=\frac1{\sqrt{c-b^2}}\,\Im\,\frac1{x-b-i\sqrt{c-b^2}}$$ $\endgroup$ Commented Mar 17, 2023 at 15:29
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