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How eΠi =(is equal to) -1

bhai is ko zara bata tu do k likha kia hay... symbols say hat k...

jahan tak mera andaza hey

e= exponential
pi = 3.14
i = iota (under root of -1, its a complex number)

i have no idea,,, how to prove it

however i can show it

look

eΠi = -1



agar aap sab ko nazar aa raha hey,,,, tu iska matlab hey ke main ne yeah show kerwa diya

tahir bhai actully i means to say

dat show

how

eΠi =(is equal to) -1

alrite....

well its a power series equation and i want its complete mathematical calculation u can may b use Cos/sign teta to get the rite calculation of this riddle

can we solve it by physical calculations?

match equation shud b solve by mathematichal calculation n e how if u think u can solve this using phycial equ. den go ahead

bhai yah bohat ziada mathematical hay,,, hmaain tu maths parhay hoaay bhi 13 years hu gaay haain , ab kahan say complete karain isay...

chlain dekhtey hain kisi aur ne detail se solve na kia to main khud bata don ga iss ki calculation

yaar waisey yeah koi riddle tu na hua,,,, kal ku tum ne kahna hey schrodinger equation prove kerain,,,,, yeah koi uar iss tarah ki cheez

anyway,,,, according to Euler'formula

exi = cos x + isin x (where trignometric function are described in radians)

in this case

x = Π,, therefore

eΠi = cosΠ + isinΠ

as cosΠ= -1 and sinΠ = 0 for trignometric functions in radians,,,, hence

eΠi = -1 + i*0 = -1






















All credits goes to googlee mooglee of Π,,, which lead me tu eulers identity

iss riddle ki mathematical calculation yeh rahi

This is a power series equation.
e^(i*Π) = 1 + (i*Π)/1! + (i*Π)^2/2! + (1*Π)^3/3! + (i*Π)^4/4! + ...
= 1 + i(Π/1!) - Π^2/2! - i(Π^3/3!) + Π^4/4! + Π^5/5! - ...
= 1 - Π^2/2! + pi^4/4! - ... + i[Π/1! - Π^3/3! + Π^5/5! - ...]
= cos[Π] + i*sin[Π]
cos(Π) = 1
sin(Π) = 0 e^(Πi) = -1

by d way Euler'formula ka method b durast hai