Welcome to Tough Inductions! Get Ready to Challenge Yourself!

[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 !

 

[Office_Shredder]

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

 

[shirel]

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

Does anyone have a trigonometric induction?

 

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

 

[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?

 

[morphism]

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

Here's another exercise:

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

 

[Werg22]

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

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

 

[shirel]

OK. Thanks.

 

[truewt]

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

 

[Office_Shredder]

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)[/color]

 

[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..

 

[Office_Shredder]

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)

 

[truewt]

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

 

[truewt]

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)[/color]

What do you mean by trivial?

 

[truewt]

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

 

[phoenixthoth]

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 !

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.

 

[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.

 

[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).

 

[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 x_n+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 x_n+1, or:

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

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

 

[prasannapakkiam]

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

 

[phoenixthoth]

I don't know if induction is the best way to do this...
Prove that for n and k integers with 0 <= k <= n, that the binomial coefficient formula n!/(k!(n-k)!) is an integer.

 

[Office_Shredder]

What do you mean by trivial?

It shouldn't be tough at all given the hint I posted...

 

[mathwonk]

that looks good john duck