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July 5

zeta function

Why multiplied Riemann equation Gamma by Zeta equation? — Preceding unsigned comment added by 151.236.160.91 (talk) 19:35, 5 July 2015 (UTC)[reply]

Because of the identity

\int _{0}^{\infty }e^{-nx}x^{s-1}\,dx=\Gamma (s)n^{-s}.

So, summing both sides for

n=1

to infinity gives:

\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}-1}}\,dx=\Gamma (s)\zeta (s)

where

\operatorname {Re} (s)>1

and

\zeta (s)=\sum _{n=1}^{\infty }n^{-s}

. Sławomir Biały (talk) 21:23, 5 July 2015 (UTC)[reply]

Measuring an angle in a letter N

Hello. From time to time, I do projects which involve constructing letters of the alphabet. Most of the time, I measure and eyeball the sides and angles. But I got to wondering about calculating an exact angle. The one that causes me difficulty is the letter N. I try to get the diagonal to line up with the verticals, and I can do it by trial-and-error. But even knowing trigonometric functions, I can't figure out what to input. I measured the angle in red by hand and got approximately 30.743°, but can it be calculated knowing only the other measurements shown? Thank you. → Michael J Ⓣ Ⓒ Ⓜ 23:02, 5 July 2015 (UTC)[reply]

The angle is

\alpha =\tan ^{-1}((21+{\sqrt {57}})/48)=30.7436831\ldots ^{\circ }

. To find it, let x be the vertical distance between the two diagonal lines. You have

x={\frac {1}{\sin \alpha }}

, and also

{\frac {3}{7-x}}=\tan \alpha

. Solving this gives the result. -- Meni Rosenfeld (talk) 23:40, 5 July 2015 (UTC)[reply]

I am astonished that your hand measurement was accurate to one-thousandth of a degree! 109.153.225.51 (talk) 23:51, 5 July 2015 (UTC)[reply]

I was about to comment the same thing, but User:109.153.225.51 beat me to it. That is some impressively accurate measuring you are doing there! Good job! :) —SeekingAnswers (reply) 09:22, 6 July 2015 (UTC)[reply]

Thank you. But I still have a question. How does one determine what is x (The vertical distance between the diagonals, as you say) knowing only the measurements marked? (I would like to know in general terms, in case I need to construct letters of different dimensions.) Thank you. → Michael J Ⓣ Ⓒ Ⓜ 11:30, 6 July 2015 (UTC)[reply]

x is also found from these two equations (two equations in two unknowns. Can be reduced to a quadratic equation using trigonometric identities). In this particular case

x=(3{\sqrt {57}}-7)/8=1.95619\ldots

. To generalize to other shapes, try to figure out why the equations are correct - I can't really explain without drawing, but it shouldn't be hard. The numbers 1, 3 and 7 in the equations are taken from the measurements; the 1 is the diagonal thickness, the thickness of the vertical bars is irrelevant. -- Meni Rosenfeld (talk) 13:28, 6 July 2015 (UTC)[reply]

The claim: " I measured the angle in red by hand and got approximately 30.743° " is more than impressive - it is unbelievable. How did you do it? Bo Jacoby (talk) 11:13, 7 July 2015 (UTC).[reply]

Using a graphics program, I created the diagonal bar parallel to the two side bars, then rotated it incrementally, first 1° at a time, then 0.1°, then 0.01° and finally 0.001° (the smallest unit the program allows) until the edge nearly lined up with the corner of the vertical. → Michael J Ⓣ Ⓒ Ⓜ 14:17, 7 July 2015 (UTC)[reply]

That explains it! Thanks. Bo Jacoby (talk) 20:20, 7 July 2015 (UTC).[reply]