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Tough induction

  1. May 10, 2007 #1
    tough induction!!

    hey !

    I'm new here... :redface:

    I'm looking for a really tough induction!

    I will appreaciate who will write here some inductions..

    thanks...


    * I JUST LOVVVVVE THOSE SMILIES ! :tongue:
     
  2. jcsd
  3. May 10, 2007 #2

    Office_Shredder

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    Ok

    Prove, by induction, that

    [tex]log(2) = \sum_{n=1}^k \frac{(-1)^{n-1}}{n} +(-1)^k 2\int_0^1 \frac{t^{2k+1}dt}{1+t^2}[/tex]

    A hint: start by showing [tex]log(2) = 2\int_0^1\frac{tdt}{1+t^2}[/tex]

    EDIT: Does anyone here know why my first latex image isn't working?

    EDIT: I appear to have gotten it to work.... never mind
     
    Last edited: May 10, 2007
  4. May 10, 2007 #3
    wow! That's great!
    thank you very much.

    Does anyone have a trigonometric induction?
     
  5. May 11, 2007 #4

    VietDao29

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    Have you solved Office_Shredder's one?

    Ok, here's a pretty good problem in trigonometry. There are many ways to go about solving it, and Induction is a way to go.

    Problem:
    Prove that:
    [tex]1 + 2 \sum_{k = 1} ^ n \left( \cos (k x) \right) = \frac{\sin \left[ \left( n + \frac{1}{2} \right) x \right]}{\sin \left( \frac{x}{2} \right)}[/tex]
     
  6. May 11, 2007 #5
    Thank you very much!

    About Office_Shredder 's induction I think it is wrong.. It must be ln(2) instead of log(2).

    Am I wrong?
     
  7. May 11, 2007 #6

    morphism

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    It's common for mathematicians to use log(x) to mean ln(x).

    Here's another exercise:

    Prove that
    [tex]\sum_{i=1}^n \tan (r\theta) \tan ((r+1)\theta) = \tan ((n+1)\theta) \cot \theta \; - \, n \, - \, 1[/tex]
     
    Last edited: May 11, 2007
  8. May 11, 2007 #7
    Just to point out: that is actually a really nice result and constitutes a clever proof of the integral of sin x.
     
  9. May 12, 2007 #8
    OK. Thanks. :smile:
     
  10. May 22, 2007 #9
    Thanks for the problems :) managed to solve the last 2 problems, but the one involving log2, how do you go about doing it?
     
  11. May 22, 2007 #10

    Office_Shredder

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    Spoiler for the log 2 problem:

    For the log2 problem, it's trivial for k=1. For the inductive step, multiply inside the integral by (1+t2-t2 and see what you can get (if you're at all like me, you spent an hour trying to do integration by parts before feeling stupid :p)
     
  12. May 22, 2007 #11
    Hmmm cos I was thinking induction is prove for all cases of n..

    but i guess this usually comes up as a follow up question to an induction question? something like an inference question after the induction is done..
     
  13. May 22, 2007 #12

    Office_Shredder

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    No truewt... prove that the relation is true for any k. So you show it's true for k=1, then show if it's true for k=r, it's true for k=r+1 via the hint I gave (or on your own of course)
     
  14. May 23, 2007 #13
    Hmmm I see thanks. I'll go try it now.
     
  15. May 23, 2007 #14
    What do you mean by trivial?
     
  16. May 23, 2007 #15
    Thanks. Solved the log2 problem. But it is still really ln 2 :P
     
  17. May 24, 2007 #16
    Here's something related to induction. Find a statement P(n) about the natural number n such that for all n, if P(n) is true then P(n+1) is true but P(n) is false for all n.
     
  18. May 25, 2007 #17

    mathwonk

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    prove by induction, the moduli space of curves of genus g has dimension 3g-3, if g > 1.


    prove by induction the torelli map (taking a curve to its jacobian) from the moduli space of curves of genus g to the moduli space of abelian varieties of dimension g is generically injective, for all g.

    find by induction, the volume of a sphere of dimension d, assuming the area formula for a circle.

    find by induction the number of vertices of an n dimensional cube.
     
    Last edited: May 25, 2007
  19. May 26, 2007 #18

    mathwonk

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    hey, at least the last one is easy. the first is an exercise in intuition about moduli of curves acquiring singularities, and the second an easy version of some research i have done in a similar case. the third one is a good calculus problem.

    the point here is that induction works also in geometry, and is very powerful. these questions give the flavor of the kind of algebraic geometry research i have been involved in.
    realistically, try the 4th one first, and then the third one, (perhaps this one should be called a recursive calculation).
     
    Last edited: May 26, 2007
  20. May 26, 2007 #19
    Ok, for the third one, an n-sphere is given by the equation:

    [tex]\sum_{i=0}^{n} x_{i}^{2} = r^{2}[/tex]

    and the area of a circle (or 1-sphere) is given by pi*r2. Then the basis of induction is to find the volume of a 2-sphere in terms of the volume of a 1-sphere. Consider the equation of the 2-sphere for different values of x2 (note that |x2| cannot be greater than r):

    [tex]x_{0}^{2} + x_{1}^{2} = r^{2} - x_{2}^{2} = R^{2}[/tex]

    It's just the equation of a 1-sphere with radius R given by

    [tex]R = \sqrt{r^{2} - x_{2}^{2}}[/tex]

    The volume of the 2-sphere of radius r is just the sum of the volumes of 1-spheres as we range through possible values of x2, or:

    [tex]\int_{-r}^{r} \pi(r^{2} - x_{2}^{2}) dx_{2}[/tex]

    For the step of induction we have to show that the volume of an (n+1)-sphere can be found in terms of the volume of an n-sphere. Consider the equation of an (n+1)-sphere of radius r as we range through values of xn+1. Again, it's just the equation of an n-sphere with a radius of

    [tex]R = \sqrt{r^{2} - x_{n+1}^{2}}[/tex]

    Then the volume of the (n+1)-sphere is the sum of the volumes of the n-spheres as we range through all possible values of xn+1, or:

    [tex]\int_{-r}^{r} V_{n}( \sqrt{r^{2} - x_{n+1}^{2}}) dx_{n+1}[/tex]

    Where Vn(x) is the volume of an n-sphere of radius x.
     
    Last edited: May 26, 2007
  21. May 27, 2007 #20
    Here is an easy one: Prove by induction, the Binomial Theorem.
     
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