Re: [math-fun] Are these two numbers equal?
Thanks! (That HAD to be true since no two real numbers can be as close as these were.) —Dan
According to maple they are symbolically equal. I did not dig into it to try to see why.
On Mon, Aug 20, 2018 at 5:19 PM Dan Asimov <dasimov@earthlink.net> wrote:
Let f(x) = ln(2-2*cos(x))
Then are these two numbers equal?
A = -Integral from 0 to π/3 of f(t) dt
B = Integral from π/3 to π of f(t) dt
I can't tell. But they're extremely close.
—Dan
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Use 2-2*cos(x) = 4*(sin(x/2))^2, then B-A = 2*pi*log(2) + 2*\int_0^{\pi} log(sin(x/2)) dx. For the last integral, see https://socratic.org/questions/integrate-log-sinx-from-0-to-pi-2 On Tue, Aug 21, 2018 at 8:28 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Thanks! (That HAD to be true since no two real numbers can be as close as these were.)
—Dan
According to maple they are symbolically equal. I did not dig into it to try to see why.
On Mon, Aug 20, 2018 at 5:19 PM Dan Asimov <dasimov@earthlink.net> wrote:
Let f(x) = ln(2-2*cos(x))
Then are these two numbers equal?
A = -Integral from 0 to π/3 of f(t) dt
B = Integral from π/3 to π of f(t) dt
I can't tell. But they're extremely close.
—Dan
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_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I tried this (for free) on Wolfram Alpha, and it says the same thing. It gives the following expressions for the two integrals: -1/18 i (π^2 - 36 Li_2((-1)^(1/3))) 1/18 i (π^2 - 36 Li_2((-1)^(1/3))) which clearly sum to zero. (It says Li_n is the polylogarithm function.) Tom Dan Asimov writes:
Thanks! (That HAD to be true since no two real numbers can be as close as these were.)
—Dan
According to maple they are symbolically equal. I did not dig into it to try to see why.
On Mon, Aug 20, 2018 at 5:19 PM Dan Asimov <dasimov@earthlink.net> wrote:
Let f(x) = ln(2-2*cos(x))
Then are these two numbers equal?
A = -Integral from 0 to π/3 of f(t) dt
B = Integral from π/3 to π of f(t) dt
I can't tell. But they're extremely close.
—Dan
The following link shows the indefinite integral: http://www.wolframalpha.com/input/?i=int+ln(2-2*cos(x))+dx The following two links generate the definite integrals (at first it just shows the numeric value, but it you give it a minute to update, it shows the symbolic form): http://www.wolframalpha.com/input/?i=int+ln(2-2*cos(x))+dx,+x%3D0..pi%2F3 http://www.wolframalpha.com/input/?i=int+ln(2-2*cos(x))+dx,+x%3Dpi%2F3..pi Tom Tom Karzes writes:
I tried this (for free) on Wolfram Alpha, and it says the same thing. It gives the following expressions for the two integrals:
-1/18 i (π^2 - 36 Li_2((-1)^(1/3))) 1/18 i (π^2 - 36 Li_2((-1)^(1/3)))
which clearly sum to zero. (It says Li_n is the polylogarithm function.)
Tom
Dan Asimov writes:
Thanks! (That HAD to be true since no two real numbers can be as close as these were.)
—Dan
According to maple they are symbolically equal. I did not dig into it to try to see why.
On Mon, Aug 20, 2018 at 5:19 PM Dan Asimov <dasimov@earthlink.net> wrote:
Let f(x) = ln(2-2*cos(x))
Then are these two numbers equal?
A = -Integral from 0 to π/3 of f(t) dt
B = Integral from π/3 to π of f(t) dt
I can't tell. But they're extremely close.
—Dan
participants (3)
-
Dan Asimov -
Tom Karzes -
Warut Roonguthai