i ! \mid i! \mid

Calculus Level 3

i ! = a sinh ( a ) \mid i! \mid = \sqrt {\frac {a}{\sinh(a)}}

If the above holds true for real a a , find a e \dfrac {a}{e} .

Notations :


The answer is 1.15572734979.

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1 solution

Dwaipayan Shikari
Dec 28, 2020

Γ ( 1 + i ) = i ! i Γ ( i ) = i ! \Gamma{(1+i)} =i! \implies ∣i\Gamma{(i)}∣=∣i!∣ Γ ( 1 i ) Γ ( i ) = π sin ( i π ) i Γ ( i ) 2 = 2 i π e π e π \Gamma{(1-i)}\Gamma{(i)} = \frac{π}{\sin(iπ)} \implies-i ∣\Gamma{(i)}∣^2 =\frac{2iπ}{e^{-π} -e^{π}} Γ ( i ) = π e π e π 2 Γ ( i ) = π sinh ( π ) ∣\Gamma{(i)}∣=\sqrt{\frac{π}{\frac{e^π-e^{-π}}{2}}}\implies ∣\Gamma{(i)}∣=\sqrt{\frac{π}{\sinh(π)} } i Γ ( i ) = i ! = π sinh ( π ) ∣i\Gamma{(i)} ∣=∣i!∣= \sqrt{\frac{π}{\sinh(π)}} Answer is π e = 1.155 \color{#20A900}\boxed{\frac{π}{e}=1.155} EDIT ; i = 1 \color{#E81990}\textrm{EDIT} ; ∣i∣= 1

@Dwaipayan Shikari I didn't consider that approach, so nice! I got the answer using the Euler form of the gamma function and then relating complex conjugates, which was much harder. Any idea how to calculate the real and imaginary parts?

N. Aadhaar Murty - 5 months, 2 weeks ago

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Did you mean Γ ( s ) = e γ s s n = 1 e s n ( 1 s n ) 1 \Gamma{(s)} = \frac{e^{-\gamma s}}{s} \prod_{n=1}^∞ e^{\frac{s}{n}} (1-\frac{s}{n})^{-1}

Dwaipayan Shikari - 5 months, 2 weeks ago

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No, I meant -

Γ ( z ) = 1 z n = 1 ( 1 + 1 n ) z ( 1 + z n ) 1 \Gamma(z) = \frac {1}{z} \prod_{n=1}^{\infty} (1+\frac {1}{n})^z (1+\frac {z}{n})^{-1}

You can prove that the conjugates are conserved under the 4 arithmetic operations.

N. Aadhaar Murty - 5 months, 2 weeks ago

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