Write-an-expression-involving-e-and-pi-which-when-evaluated-gives-result-1-

Question Number 8644 by prakash jain last updated on 19/Oct/16

Write an expression involving

e and π which when evaluated

gives result 1 .

Commented by Yozzias last updated on 19/Oct/16

{ee}^{- 1} π π^{- 1} = 1

Commented by prakash jain last updated on 19/Oct/16

Valid answer but i was expecting

like - e^{i π} = 1

Commented by Yozzias last updated on 20/Oct/16

2 \sqrt{(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \dots)} [e Γ (1 / 2) \underset{n = 1}{\sum^{3000}} n] {\int_{0}^{e^{- 1}} π^{Γ (1 / 2)} dx} {[\int_{0}^{3001} π^{\frac{3}{2} + Γ (1 / 2)} ⌊ x ⌋ dx]}^{- 1} (\frac{d}{du} (u \sqrt{\frac{2}{π}})) {\frac{π e^{- 1}}{4} \lim_{n \to \infty} (\frac{n}{n^{2}} + \frac{n + 1}{n^{2}} + \dots + \frac{3 n}{n^{2}})} e^{{\sin \frac{π}{\sqrt{4}} + \cos \frac{π}{\sqrt{4}}} {\cosh \frac{1}{π} + \sinh \frac{1}{π}} e^{- π^{- 1}}} \log_{π} {\frac{2 π \sinh^{- 1} 2}{\ln (9 + 4 \sqrt{5})}}

= 1

Commented by prakash jain last updated on 20/Oct/16

o k . I w i l l t r y t o s o l v e .

\sinh^{- 1} 2 = \ln (2 + \sqrt{5})

\frac{2 π \ln (2 + \sqrt{5})}{\ln (9 + 4 \sqrt{5})} = \frac{π \ln {(2 + \sqrt{5})}^{2}}{\ln (9 + 4 \sqrt{5})} = π

\log_{π} {\frac{2 π \sinh^{- 1} 2}{\ln (9 + 4 \sqrt{5})}} = 1

c o n t i n u e