# Math tricks for everyone

by agentredlum
Tags: math, tricks
 P: 460 Well.....no one is posting any tricks, thats sad. I'll post another trick, hope it motivates some of you. What is i^i and how to show what it is. This trick uses Eulers famous identity.....e^(ix) = cos(x) + isin(x) Notice that when x = pi/2 e^(ipi/2) = cos(pi/2) + isin(pi/2) cos(pi/2) = 0 and sin(pi/2) = 1 e^(ipi/2) = i Now use this for the base in i^i, don't use it for the exponent [e^(ipi/2)]^i When you raise a base with an exponent to another exponent, you multiply the exponents (ipi/2)*i = -pi/2 since i*i = -1 Therefore i^i = e^(-pi/2) This is a real number! What an AMAZING result, an imaginary base to an imaginary power can be real.
 P: 737 $$\int_0^x darctanx=\int_0^x \frac{dx}{x^2+1}=\int_0^x \frac{i}{2(x-i)} - \frac{i}{2(x+i)} dx = \frac{i}{2} ln \left( \frac{x-i}{x+i} \right) \Rightarrow \forall x \in \Re, \: arctanx = \frac{i}{2}ln \left( \frac{x-i}{x+i} \right)$$
P: 460
Math tricks for everyone

 Quote by TylerH $$\int darctanx=\int \frac{dx}{x^2+1}=\int \frac{i}{2(x-i)} - \frac{i}{2(x+i)} dx = \frac{iln \left( \frac{x-i}{x+i} \right)}{2}$$
My browser did not decode so i can't see it...too bad. Can you rewrite it using normal keyboard symbols?

Sorry for the trouble, thanx for posting.
 P: 737 Here's a picture. Attached Thumbnails
P: 460
 Quote by TylerH Here's a picture.
VERY NICE! I like it a lot! POST MORE PLEASE!! THANK YOU!
 P: 2 Dude your formula does not hold good for all quadratic expressions. Instead use x= -b+(b^2-4ac)^1/2 or x= -b-(b^2-4ac)^1/2
P: 460
 Quote by TejasB Dude your formula does not hold good for all quadratic expressions. Instead use x= -b+(b^2-4ac)^1/2 or x= -b-(b^2-4ac)^1/2
give me an example where it fails please, thanx for posting
P: 460
 Quote by agentredlum VERY NICE! I like it a lot! POST MORE PLEASE!! THANK YOU!
Wait a minute TylerH, you got the order wrong, change signs in both denominators, then it works.
PF Gold
P: 1,957
 Quote by TylerH $$\int_0^x darctanx=\int_0^x \frac{dx}{x^2+1}=\int_0^x \frac{i}{2(x-i)} - \frac{i}{2(x+i)} dx = \frac{i}{2} ln \left( \frac{x-i}{x+i} \right) \Rightarrow \forall x \in \Re, \: arctanx = \frac{i}{2}ln \left( \frac{x-i}{x+i} \right)$$
Did you remember to write this as a DEFINITE integral? Because something tells me that 1/2 ln(-1) is not zero.
P: 737
 Quote by Char. Limit Did you remember to write this as a DEFINITE integral? Because something tells me that 1/2 ln(-1) is not zero.
What should it be? Wikipedia says it's correct.
 PF Gold P: 1,957 Oh. All right then.
 Mentor P: 18,346 No, Char is correct. The formula certainly doesn't hold in 0. So there must be some mistake somewhere. (The mistake being that complex integrals do not behave in the way you're describing) The wikipedia is correct though.
 P: 737 I just now noticed they're different. I must have been reading it as what I expected to be there. :)
P: 839
 Quote by agentredlum This is a real number! What an AMAZING result, an imaginary base to an imaginary power can be real.
Theorem: There exists 2 irrational numbers, a and b, such that ab is rational.

Proof: Consider sqrt{2}sqrt{2}. If this number is rational we are done. Suppose not. Define a=sqrt{2}sqrt{2} and b=sqrt{2}. Then ab = (sqrt{2}sqrt{2})sqrt{2} = sqrt{2}2 = 2, completing the proof.
P: 460
 Quote by pwsnafu Theorem: There exists 2 irrational numbers, a and b, such that ab is rational. Proof: Consider sqrt{2}sqrt{2}. If this number is rational we are done. Suppose not. Define a=sqrt{2}sqrt{2} and b=sqrt{2}. Then ab = (sqrt{2}sqrt{2})sqrt{2} = sqrt{2}2 = 2, completing the proof.
Can you send a picture of the proof the way TylerH did above? My browser does not decode TeX sorry. Thanks for posting. POST MORE PLEASE!
P: 460
 Quote by Char. Limit Did you remember to write this as a DEFINITE integral? Because something tells me that 1/2 ln(-1) is not zero.
e^x = -1 is no problem...this is FAMOUS ...x=ipi

e^(ipi) = -1 so ln(-1) = ipi

Put ln(-1) in TI89 calculator in COMPLEX mode

P: 460
 Quote by TylerH I just now noticed they're different. I must have been reading it as what I expected to be there. :)
You can edit your post. Go back and change signs. When it asks you for a reason, write wrong numbers.

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