# Thread: Three problems: Integration, inequality and matrice

1. ## Three problems: Integration, inequality and matrice

Hi,

Here I have three problems I can't get solved no matter what I try. Truthfully I don't have a clue on how to begin. Hopefully someone here is able to help me! All other homework in the course was handed in a long time ago..

Any help is appreciated!

1. Examine the integral of e^(-x)^2 from 0 to infinity.

2. Prove the inequality 1 + ln(x+sqrt(1+x^2)) > sqrt(1+x^2), 0<x<1

3.

P = [1-a a
b 1-b]

S= [-a a
b -b ]

a) show that S^n=k^(n-1)*S, n>2, k=-(a+b)
b) show that P^n=I+((1+k)^n-1)/k*S, where I is the identity matrix.

I'm sorry, don't know how to use that fancy latex-code for the symbols!

2. Hello,

What is the real formula ?

1. Examine the integral of e^(-x)^2 from 0 to infinity.
Is it $\displaystyle e^{(-x)^2}$ (in which case it's quite strange) or is it $\displaystyle (e^{-x})^2$ ?

If it's the second one, you just have to transform it in $\displaystyle e^{-2x}$ and find an antiderivate (it's easy because the derivate of -2x is -2)

3. Originally Posted by Moo
Hello,

What is the real formula ?

Is it $\displaystyle e^{(-x)^2}$ (in which case it's quite strange) or is it $\displaystyle (e^{-x})^2$ ?

If it's the second one, you just have to transform it in $\displaystyle e^{-2x}$ and find an antiderivate (it's easy because the derivate of -2x is -2)
Actually it is e^(-x^2), I got it a bit wrong the first time, sorry.

4. In my memory, this integral can't be defined with usual functions...

Maybe what they mean by examine is to check if it converges, that is to say continuous on $\displaystyle ]0; + \infty[$ and that it has a determined limit at its extremities -> study the limit of the function in the integral in 0 and $\displaystyle + \infty$

Yep, take a look at this : Erf -- from Wolfram MathWorld

5. Originally Posted by ManKKe
Actually it is e^(-x^2), I got it a bit wrong the first time, sorry.
The definite integral can be found under certain circumstances. Lucky for you one of those circumstances is when integrating from 0 to infinity. The usual approach is to square the integral and convert to polar coordinates. Read this (note that by symmetry your integral will be half the integral from -infty to +infty).

6. Originally Posted by ManKKe
[snip]

2. Prove the inequality 1 + ln(x+sqrt(1+x^2)) > sqrt(1+x^2), 0<x<1

[snip]
Do you know the Mean Value Theorem? If so, then I think the following is OK:

Let $\displaystyle f(x) = 1 + \ln \left( x + \sqrt{1 + x^2} \right), ~ 0 < x < 1$.

Then:

$\displaystyle f'(x) = \frac{1}{(x + \sqrt{1 + x^2})} \, \left( 1 + \frac{x}{\sqrt{1 + x^2}}\right) = \frac{1}{(x + \sqrt{1 + x^2})} \, \left( \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}}\right)$

$\displaystyle = \frac{1}{\sqrt{1 + x^2}}$.

Now let x lying in 0 < x < 1 be given. Then from the Mean Value Theorem:

$\displaystyle \frac{f(x) - f(0)}{x - 0} = f'(a) \,$ for some 0 < a < x.

But $\displaystyle f'(a) = \frac{1}{\sqrt{1 + a^2}} > \frac{1}{\sqrt{1 + x^2}}$ for 0 < a < x.

Therefore:

$\displaystyle \frac{f(x) - f(0)}{x - 0} = f'(a)$

$\displaystyle \Rightarrow \frac{1 + \ln (x + \sqrt{1 + x^2} - 1}{x - 0} > \frac{1}{\sqrt{1 + x^2}}$

$\displaystyle \Rightarrow \ln (x + \sqrt{1 + x^2}) > \frac{x}{\sqrt{1 + x^2}}$

$\displaystyle \Rightarrow 1 + \ln (x + \sqrt{1 + x^2}) > 1 + \frac{x}{\sqrt{1 + x^2}}$.

It's not hard to show that $\displaystyle 1 + \frac{x}{\sqrt{1 + x^2}} > \sqrt{1 + x^2} ~$ for 0 < x < 1 and so the result follows.

7. Thank you very much, you have been very helpful! It was very hard to figure out the Gaussian integral since I didn't know about it. To use the MVT on the inequality was also something I wouldn't have come up with myself.

8. The matrix still remains, does anyone have the time to solve it?

Thank you.