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Proof that math is wrong

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Originally posted by: DanFungus
Originally posted by: Grminalac
Hmm, wouldn't
cos(x)=([1] x/2pi + [1] -x/2pi);
cos(x)=(x/2pi+ -x/2pi);
cos(x)=(0/2pi);
cos(x)=0
Click to expand...

the x/2pi is an exponent...1 to the any exponent = 1
Click to expand...

what about my post??? 1^0.5 is the square root of 1 which is plus or minus 1... so 1 to any exponent is not necessarily 1

<edit>

and moreso... the only ONLY ONLY way to guarantee 1^(x/2pi) = 1 is for x to equal 0 or any multiple of 2pi, because then you would elimate any root problems (because the exponent is an integer). So... the proof follows, cos(x) = 1 for x in any of those cases.
 
hmmmmm....I am in calc 2 and well, I am scared now. I hate expotential functions! did i even spell expotential right?
 
Refer to something called Euler's formula. For those who don't know it, here's a "take it at face value" explaination.

Picture the regular x and y axis but the x is called the 'real axis' and the y is the 'imaginary axis'.

So the number 1 will lie on positive 1 on the x and the number i will like on the positive 1 on the y. Moreso, 1+i will like at the coordinates (1,1)... get it?

e^(i*theta) Theta is in radians (0 to 2pi where 2pi is 360 degrees).

If you draw a line from the center to your complex number (i or 1 or 1+i), theta is equal the angle between the line and the positive x axis. So 1 = e^(i*0), i = e^(i*pi/2), -1 = e^(i*pi), -j = e^(i*3pi/2)



 
Originally posted by: hdeck
what's wrong with saying cos(x) = 1???? plug in 0 for x and that is a true statement...
Click to expand...

It's not true for all values of x.

In any case, there's no domain stated for the function, so your point is not so bad after all: assume the domain is a single point, {0}. 😀
 
take the inverse cosine of both sides, leaving x = .....

following the same logic, you can prove x = 0..... 🙂

There is actually a mathematical mistake here... it has to do with Euler's Equation and its domain. Most good number theory textbooks point out this solution.
 
Originally posted by: Mday
this one is my fave

e^(i*pi) = -1
e^(2*i*pi) = 1
ln(e^(2*i*pi)) = ln(1)
2*i*pi = 0

=)
Click to expand...

Yes, it's more or less the same flawed technique 🙂

As I said above:
This has something to do with the funny way the logarithm function (as an inverse of the power function) is defined in a complex domain, but the argument would be too technical for ATOT

 
This has something to do with the funny way the logarithm function (as an inverse of the power function) is defined in a complex domain, but the argument would be too technical for ATOT
Click to expand...

bla bla bla, why don't you knock yourself out.
 
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