this morning i found that for f(x)=ln(x^2+1), that the (4n+2)th derivatives all vanish at x=1. it took me an hour to figure out why, using geometric series, but surely there must be an easier way to see this. does anyone have an elegant proof of this? erich
log(x^2 + 1) = log(x + i) + log(x - i), so it suffices to show that the (4n + 2)th derivatives of log(x + i) are all pure-imaginary when x = 1. Let x = 1 + y, so we are interested in the Taylor series of: log(y + (1 + i)) about the point y = 0. We want to show that the (4n + 2)th coefficients are all pure-imaginary. But log(y + (1 + i)) is just the derivative of 1/(y + (1 + i)), so it suffices to show that the (4n + 1)th coefficients of the Taylor series of 1/(y + (1 + i)) about 0 are all pure-imaginary. This follows from the claim that the coefficient in y^n of the series expansion of 1/(y + a) is [a real multiple of] a^-n, which is true by dimensional analysis. Sincerely, Adam P. Goucher
Sent: Thursday, November 12, 2015 at 3:24 PM From: "Erich Friedman" <erichfriedman68@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] ln(x^2+1)
this morning i found that for f(x)=ln(x^2+1), that the (4n+2)th derivatives all vanish at x=1. it took me an hour to figure out why, using geometric series, but surely there must be an easier way to see this. does anyone have an elegant proof of this?
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Nice proof. —Dan
On Nov 12, 2015, at 8:05 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
log(x^2 + 1) = log(x + i) + log(x - i), so it suffices to show that the (4n + 2)th derivatives of log(x + i) are all pure-imaginary when x = 1.
Let x = 1 + y, so we are interested in the Taylor series of:
log(y + (1 + i))
about the point y = 0. We want to show that the (4n + 2)th coefficients are all pure-imaginary.
But log(y + (1 + i)) is just the derivative of 1/(y + (1 + i)),
I think calculus is taught differently in the U.K. from in the U.S. (:-)>
so it suffices to show that the (4n + 1)th coefficients of the Taylor series of 1/(y + (1 + i)) about 0 are all pure-imaginary.
This follows from the claim that the coefficient in y^n of the series expansion of 1/(y + a) is [a real multiple of] a^-n, which is true by dimensional analysis.
participants (3)
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Adam P. Goucher -
Dan Asimov -
Erich Friedman