
#1
Jun2711, 02:39 PM

P: 460

I would like to have this thread dedicated to showing math tricks from all areas of mathematics. Hopefully the title has aroused your interest and you have an interesting trick you would like to share with everyone. Let me start by showing one of my favorite tricks, perhaps something that has not occured to many of you?
Start with a general quadratic, do not set it equal to zero, set it equal to bx+c ax^2 = bx + c multiply everything by 4a 4(ax)^2 = 4abx + 4ac subtract 4abx from both sides 4(ax)^2  4abx = 4ac add b^2 to both sides 4(ax)^2  4abx + b^2 = b^2 + 4ac factor the left hand side (2ax  b)^2 = b^2 + 4ac take square roots of both sides 2ax  b = +sqrt(b^2 + 4ac) add b to both sides 2ax = b +sqrt(b^2 + 4ac) divide by 2a, a NOT zero x = [b + sqrt(b^2 + 4ac)]/(2a) This quadratic formula works perfectly fine for quadratic equations, just make sure you isolate the ax^2 term BEFORE you identify a, b, and c 1) Notice that this version has 2 less minus signs than the more popular version 2) The division in the derivation is done AT THE LAST STEP instead of at the first step in the more popular derivation, avoiding 'messy' fractions. 3) In this derivation there was no need to split numerator and denominator into separate radicals 4) Writing a program using this version, instead of the more popular version, requires less memory since there are less 'objects' the program needs to keep track of. (Zero is absent, 2 less minus symbols) I hope you find this interesting and i look forward to seeing your tricks. The method of completing the square... multiplying by 4a and adding b^2 i learned from NIVEN AND ZUCKERMAN in their book ELEMENTARY NUMBER THEORY however it was an example they used on a congruence, they did not apply it to the quadratic formula. 



#2
Jun2911, 05:36 AM

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. 



#3
Jun2911, 06:45 AM

P: 737

[tex]\int_0^x darctanx=\int_0^x \frac{dx}{x^2+1}=\int_0^x \frac{i}{2(xi)}  \frac{i}{2(x+i)} dx = \frac{i}{2} ln \left( \frac{xi}{x+i} \right) \Rightarrow \forall x \in \Re, \: arctanx = \frac{i}{2}ln \left( \frac{xi}{x+i} \right)[/tex]




#4
Jun2911, 06:53 AM

P: 460

Math tricks for everyoneSorry for the trouble, thanx for posting. 



#5
Jun2911, 06:59 AM

P: 737

Here's a picture.




#6
Jun2911, 07:11 AM

P: 460





#7
Jun2911, 07:11 AM

P: 2

Dude your formula does not hold good for all quadratic expressions. Instead use
x= b+(b^24ac)^1/2 or x= b(b^24ac)^1/2 



#8
Jun2911, 07:22 AM

P: 460





#9
Jun2911, 07:46 AM

P: 460





#10
Jun2911, 05:41 PM

PF Gold
P: 1,930





#11
Jun2911, 06:29 PM

P: 737





#13
Jun2911, 06:57 PM

Mentor
P: 16,690

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. 



#14
Jun2911, 07:01 PM

P: 737

I just now noticed they're different. I must have been reading it as what I expected to be there. :)




#15
Jun2911, 08:32 PM

Sci Advisor
P: 779

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 a^{b} = (sqrt{2}^{sqrt{2}})^{sqrt{2}} = sqrt{2}^{2} = 2, completing the proof. 



#16
Jun2911, 11:33 PM

P: 460





#17
Jun2911, 11:53 PM

P: 460

e^(ipi) = 1 so ln(1) = ipi Put ln(1) in TI89 calculator in COMPLEX mode 



#18
Jun3011, 12:02 AM

P: 460




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