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Math tricks for everyone

  1. Jun 27, 2011 #1
    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. jcsd
  3. Jun 29, 2011 #2
    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


    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.
  4. Jun 29, 2011 #3
    [tex]\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)[/tex]
    Last edited: Jun 29, 2011
  5. Jun 29, 2011 #4
    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.
  6. Jun 29, 2011 #5
    Here's a picture.

    Attached Files:

  7. Jun 29, 2011 #6
  8. Jun 29, 2011 #7
    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
  9. Jun 29, 2011 #8
    give me an example where it fails please, thanx for posting
  10. Jun 29, 2011 #9
    Wait a minute TylerH, you got the order wrong, change signs in both denominators, then it works.
  11. Jun 29, 2011 #10

    Char. Limit

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    Gold Member

    Did you remember to write this as a DEFINITE integral? Because something tells me that 1/2 ln(-1) is not zero.
  12. Jun 29, 2011 #11
    What should it be? Wikipedia says it's correct.
  13. Jun 29, 2011 #12

    Char. Limit

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    Gold Member

    Oh. All right then.
  14. Jun 29, 2011 #13
    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.
    Last edited: Jun 29, 2011
  15. Jun 29, 2011 #14
    I just now noticed they're different. I must have been reading it as what I expected to be there. :)
  16. Jun 29, 2011 #15


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    Science Advisor

    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.
  17. Jun 29, 2011 #16
    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!
  18. Jun 29, 2011 #17
    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

  19. Jun 30, 2011 #18
    You can edit your post. Go back and change signs. When it asks you for a reason, write wrong numbers.
  20. Jun 30, 2011 #19

    The integral is the area under the curve 1/(x^2 + 1)

    This is defined for all real numbers. z is a 'dummy' variable of integration and is replaced by x when the definite integral is computed.

    So the trick works for all real numbers x.
  21. Jul 3, 2011 #20
    All right, heres a simple trick, i hope i'm not the only one who finds this interesting...

    Consider the matrix

    [0 1]
    [1 1]

    Powers of this matrix generate 3 Fibbonacci numbers. The nth power generates F(n-1), F(n), F(n+1)

    Example, when n=4 it generates 2,3,5 which are the 3rd, 4th and 5th Fibbonacci numbers, cool huh?
    Last edited: Jul 3, 2011
  22. Jul 6, 2011 #21
    Not really a trick, but something to think about if you've never heard of fractional calculus:

    for a general polynomial of the form [itex]x^{k}[/itex]

    the nth derivative is given by the formula

    [itex]\frac{d^{n}(x^{k})}{dx^{n}}[/itex]= [itex]\frac{k!x^{k-n}}{(k-n)!}[/itex]

    Selecting n = 1/2, gives

    [itex]\frac{d^{1/2}(x^{k})}{dx^{1/2}}[/itex]= [itex]\frac{k!x^{k-1/2}}{(k-1/2)!}[/itex]

    Applying another derivative with n=1/2 gives

    [itex]\frac{d^{1/2}\frac{(k!x^{k-1/2})}{(k-1/2)!}}{dx^{1/2}}[/itex]= [itex]\frac{k!(k-1/2)!x^{k-1}}{(k-1/2)!(k-1)!}[/itex] = [itex]\frac{k!x^{k-1}}{(k-1)!}[/itex]

    which is the same as letting n=1 and doing a standard derivative.
  23. Jul 7, 2011 #22
    WOW! I never heard of fractional calculus but i googled it and it is a legitimate subject. Thank you for the post. PLEASE POST MORE!

    Can you please post the calculation as a picture? My browser does not decode TeX and I would love to follow the calculation.

    Thanx again
  24. Jul 7, 2011 #23
  25. Jul 7, 2011 #24
    came in here expecting 1+1 = 11
    saw posts
    everything went better than expect :D
  26. Jul 7, 2011 #25
    Thank you. I invite you to post a trick. Any trick you find interesting, or something not well known like HALF derivatives. :smile:
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