What is wrong with this proof showing that  e^(pi*i)=-1 is not true?
358699
Please visit http://mathforum.org/library/drmath/view/52240.html
652728
a=b=i^(1/3)\ni=-1\nab+a^2b+ab^2 = [i^(1/3)*i^(1/3)] + [i^(2/3)*i^(1/3)] + [i^(1/3)*i^(2/3)]\n= [i^(2/3)] + i + i\n= -2+(-1)^(2/3)
454578
mrniceguy's answer is correct, but he has one little eror in the work.  i^7 = -i
758984
i^(-1) = 1/i = -i\n\nsummation from i = 1 to n i^(-1) = -ni\n\nput if you mistyped and meant which is likely\n\ni^(-n) it is geometic series\n\n= -(i+i^2+i^3+..... )\n= -(1-i^(n+1))/(1-i)
589852
Sigma sign with k=0 on bottom and 4 on top; (-1)^k over (2k+1)^2
83972
R*exp(i*theta) = R [ cos(theta) +i*sin(theta) ]\nthis is the polar representation (raduis R ,angle theta) transformed to the cartisienne form (X=R cos(theta), Y=R sin(theta))  where i is imaginary to matk that y is perpendic to x\n\nR=1 theta=pi/2\nexp(i*pi/2) = cos(pi/2) + i sin(pi/2)\nexp(i*pi/2) = 0 +i *1 =i\n\nso i=exp(i*pi/2)=e^(i*pi/2)    where e=2.7818281828\ni^i=e^(i*pi/2)^i\ni^i=e^(i^2*pi/2)\ni^i=exp(-pi/2)\n=0.207879576
1042552
Here is my proof. Suppose P(x) is a polynomial in odd degree, for example 4x^5 + 3x^4 - 5x^3 + 2x^2 - 6x + 9. For large enough values of x, the term in x^5 will be so much larger than any of the other terms that it will swamp the smaller powers of x, and x^5 alone will determine the value of P. \n\nNow, if P is of odd degree and x is very large and positive, P will be positive; if x is very large and negative, then P will be negative. Therefore, somewhere between these two values of P there must be at least one point where P = 0. QED.
736311
It equals (2/3)*pi.   It does not equal (-1/3)*pi  -- that would be (-inverse cosine (-1/2)).
817918
This is an oldie:  What's i^i, where i = sqrt(-1)?   Ans:  By Euler's formula, we know that e^pi i = -1.  Therefore e^(1/2) pi i = i, or e^-(1/2) pi = i^i = 0.20788.
487411
e^3x+1=e^-2x-6\nor 3x+1=-2x-6\nor 3x+2x=-6+1\nor e=-5/5=-1
956637
Sorry tried a lot, searched a lot & surfed a lot but found no solution, I think Pascal is right.\n\nP.s- Saket it is ∫e^sinx dx & not ∫(e^x)*sin x dx, if it was that, it would have been trivial.
510030
OK, found the error:\n\nThe sum you have defined is sum(i=1 to n)i(n-i+1)  <== Note "+1"\n\n1. Let n = 1.\nThen 1(1) = 1\nand (1/6)(1)(1+1)(1+2) = 1.\n\n2. Let this be true for n.\nThen sum(i=1 to n)i(n-i) = (1/6)(n)(n+1)(n+2)\n\nSo for sum(i=1 to n+1)(i(n+1-i+1))  <== Note "+1"\n= sum(i=1 to n+1)(i(n-i+1)) + sum(i=1 to n+1)(i)\n= sum(i=1 to n)(i(n-i+1) + 0 + sum(i=1 to n+1)(i)\n\n2a) = n(n+1)(n+2)/6 + (n+1)(n+2)/2\n= (n+1)(n+2)/6 [ n + 3 ]\n= (n+1)(n+2)(n+3) / 6  ==> proof since this is what you expect from the theorem.\n\nEdit: Missing the +1 in the original summation is the issue.
1092202
looks like you've defined e using a series.  I think you need some periods of ellipsis to show that that series is infinite, since e is an irrational number
338816
As in many other pseudo-demonstrations of this kind, the error is to divide by zero (2-2) both sides of the identity. You can't do this because the division by zero is an undefined operation.\n\nFrom a well-known logic theorem, starting from a false statement, you can consistently derive any other false statement. This is the common source of many apparent paradoxes.
350908
no this is how to work it out. I am assuming you know how to take the logarithm of a coplx number and find the modulus and argand of it.\n\ni^i = e^(logi^i) = e(ilogi)\n\nnow log i = log|i| + i(argi)\n= 0 + iπ/2 + i2kπ\n\nwhere k is an integer\n\nso substituting this is\ni^i = e^(i(iπ/2 + 2ikπ)\n= e^(-π/2-2kπ)]\n= e^(π[-1/2-2k])
821031
I=I1+I2; I1=int{for x=pi/6 until pi/2} of x*dx = x^2/2{for x=pi/6 until pi/2}= pi^2/9;\nI2=int{for x=pi/6 until pi/2}of  25*dx/(sinx)^2; \nLet t=tg(x), then 1/(sinx)^2=1/t^2+1 and dx=(1+t^2)*dt;\nI2=int{for t=1/sqrt(3) until +inf}of 25*t^(-2)*dt = -25/t = 25*sqrt(3)
815664
Imaginary part of\n(e^(i*x)+ e^(2*i*x) + ...)\n=Imag(e^(i*x)/(1- e^(i*x))\n=Imag(e^(i*x)/\n  (1-cos(x)-i*sin(x)))\n=Imag(e^(i*x)\n   (1-cos(x)+i*sin(x))/\n   ((1-cos(x))^2+(sin(x)^2))\n=((cos(x))\n(sin(x))+(sin(x))\n(1-cos(x)))/(2-2*cos(x)\n=sin(x)/(2(1-cos(x)))