# Tough induction

1. May 10, 2007

### shirel

tough induction!!

hey !

I'm new here...

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. May 10, 2007

### Office_Shredder

Staff Emeritus
Ok

Prove, by induction, that

$$log(2) = \sum_{n=1}^k \frac{(-1)^{n-1}}{n} +(-1)^k 2\int_0^1 \frac{t^{2k+1}dt}{1+t^2}$$

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

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
3. May 10, 2007

### shirel

wow! That's great!
thank you very much.

Does anyone have a trigonometric induction?

4. May 11, 2007

### VietDao29

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:
$$1 + 2 \sum_{k = 1} ^ n \left( \cos (k x) \right) = \frac{\sin \left[ \left( n + \frac{1}{2} \right) x \right]}{\sin \left( \frac{x}{2} \right)}$$

5. May 11, 2007

### shirel

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?

6. May 11, 2007

### morphism

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

Here's another exercise:

Prove that
$$\sum_{i=1}^n \tan (r\theta) \tan ((r+1)\theta) = \tan ((n+1)\theta) \cot \theta \; - \, n \, - \, 1$$

Last edited: May 11, 2007
7. May 11, 2007

### Werg22

Just to point out: that is actually a really nice result and constitutes a clever proof of the integral of sin x.

8. May 12, 2007

### shirel

OK. Thanks.

9. May 22, 2007

### truewt

Thanks for the problems :) managed to solve the last 2 problems, but the one involving log2, how do you go about doing it?

10. May 22, 2007

### Office_Shredder

Staff Emeritus
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)

11. May 22, 2007

### truewt

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..

12. May 22, 2007

### Office_Shredder

Staff Emeritus
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)

13. May 23, 2007

### truewt

Hmmm I see thanks. I'll go try it now.

14. May 23, 2007

### truewt

What do you mean by trivial?

15. May 23, 2007

### truewt

Thanks. Solved the log2 problem. But it is still really ln 2 :P

16. May 24, 2007

### phoenixthoth

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.

17. May 25, 2007

### mathwonk

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
18. May 26, 2007

### mathwonk

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
19. May 26, 2007

### JohnDuck

Ok, for the third one, an n-sphere is given by the equation:

$$\sum_{i=0}^{n} x_{i}^{2} = r^{2}$$

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):

$$x_{0}^{2} + x_{1}^{2} = r^{2} - x_{2}^{2} = R^{2}$$

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

$$R = \sqrt{r^{2} - x_{2}^{2}}$$

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:

$$\int_{-r}^{r} \pi(r^{2} - x_{2}^{2}) dx_{2}$$

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

$$R = \sqrt{r^{2} - x_{n+1}^{2}}$$

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:

$$\int_{-r}^{r} V_{n}( \sqrt{r^{2} - x_{n+1}^{2}}) dx_{n+1}$$

Where Vn(x) is the volume of an n-sphere of radius x.

Last edited: May 26, 2007
20. May 27, 2007

### prasannapakkiam

Here is an easy one: Prove by induction, the Binomial Theorem.