# Problem 40

Show 40 post(s) from this thread on one page
Page 1 of 3 123 Last
• Nov 10th 2007, 08:12 PM
ThePerfectHacker
Problem 40
1)This problem is for the younger kids give them a chance, please. Let $A,B$ be randomly chosen from $\{ 1,2,...,2007 \}$ (they can be the same). What is the probability that it is possible to find two real numbers that have their product equal to $A$ and their sum equal to $B$. (By the way, "real", means non-imaginary).

2)Let $f(x),g(x)$ be continous on $[0,1]$ so that $\int_0^1 x^n f(x) dx = \int_0^1 x^n g(x) dx$ for all $n\geq 0$. Prove that $f(x) = g(x)$. (This is a classic).
• Nov 10th 2007, 09:03 PM
kalagota
f and g are cont on [0,1], then f and g are integrable.

Let $F_0 = \int_0^1 x^n f(x)dx = \int_0^1 x^n g(x)dx$

Let $f_0(x) = x^n f(x)$ and $g_0(x) = x^n g(x)$

then both $f_0, g_0$ are continuous on [0,1], which implies that $F_0$ is differentiable on [0,1], and $F_0'(x) = f_0(x) = g_0(x)$, for all $x \in [0,1]$ (by a corollary to the FTOC)

$\implies x^n f(x) = x^n g(x)$

$\implies f(x) = g(x)$. qed
• Nov 10th 2007, 09:16 PM
ThePerfectHacker
I am sorry, I cannot follow.

Quote:

Originally Posted by kalagota
f and g are cont on [0,1], then f and g are integrable.

Let $F_0 = \int_0^1 x^n f(x)dx = \int_0^1 x^n g(x)dx$

How is this a well-defined function? Maybe for a particular value of $n$?

Quote:

then both $f_0, g_0$ are continuous on [0,1], which implies that $F_0$ is differentiable on [0,1], and $F_0'(x) = f_0(x) = g_0(x)$, for all $x \in [0,1]$ (by a corollary to the FTOC)
That is not the Fundamental Theorem of Calculus.
---

Remember I am saying that:
$\int_0^1 f(x)dx = \int_0^1 g(x) dx$, $\int_0^1 xf(x) dx = \int_0^1 xg(x) dx$, $\int_0^1 x^2f(x) dx = \int_0^1 x^2g(x) dx$, ...
• Nov 11th 2007, 12:29 AM
janvdl
Quote:

Originally Posted by ThePerfectHacker
1)This problem is for the younger kids give them a chance, please. Let $A,B$ be randomly chosen from $\{ 1,2,...,2007 \}$ (they can be the same). What is the probability that it is possible to find two real numbers that have their product equal to $A$ and their sum equal to $B$. (By the way, "real", means non-imaginary).

1.) $(\frac{1}{2007} + \frac{1}{2007}) \times (\frac{1}{2007} + \frac{1}{2007})$

= $\frac{4}{4028049}$

= $\frac{1}{1007012,25}$
• Nov 11th 2007, 12:37 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
2)Let $f(x),g(x)$ be continous on $[0,1]$ so that $\int_0^1 x^n f(x) dx = \int_0^1 x^n g(x) dx$ for all $n\geq 0$. Prove that $f(x) = g(x)$. (This is a classic).

Let $P_n(x);\ n \in \mathbb(Z)_+$ be an orthonormal basis of polynomials for $L^2_{[0,1]}$.

Then by the conditions specified in the problem:

$\langle P_n, f-g \rangle =0;\ \forall n \in \mathbb{Z}_+$

which implies that $f(x)-g(x)=0\ a.e. \in [0,1]$ which as f-g is continuous implies $f(x)-g(x)=0\ x\in [0,1]$

RonL
• Nov 11th 2007, 12:46 AM
angel.white
Quote:

Originally Posted by janvdl
1.) $(\frac{1}{2007} + \frac{1}{2007}) \times (\frac{1}{2007} + \frac{1}{2007})$

= $\frac{4}{4028049}$

= $\frac{1}{1007012,25}$

Hmm, that is much simpler than the method I thought of in my head (I'm a bit too embarrassed to say what my method was, but I'll just say it involves a new pack of pencils, a full ream of paper, a gallon of coffee, a carton of cigarettes, and no obligations for several days.)
• Nov 11th 2007, 12:50 AM
janvdl
Quote:

Originally Posted by angel.white
Hmm, that is much simpler than the method I thought of in my head (I'm a bit too embarrassed to say what my method was, but I'll just say it involves a new pack of pencils, a full ream of paper, a gallon of coffee, a carton of cigarettes, and no obligations for several days.)

Haha, doesn't mean my method is right you know :D
• Nov 11th 2007, 01:10 AM
Jhevon
Quote:

Originally Posted by CaptainBlack
Let $P_n(x);\ n \in \mathbb(Z)_+$ be an orthonormal basis of polynomials for $L^2_{[0,1]}$.

Then by the conditions specified in the problem:

$\langle P_n, f-g \rangle =0;\ \forall n \in \mathbb{Z}_+$

which implies that $f(x)-g(x)=0\ a.e. \in [0,1]$ which as f-g is continuous implies $f(x)-g(x)=0\ x\in [0,1]$

RonL

i always have trouble proving things that seem to me to be obvious.

would it be wrong to do this?

$\int_0^1 x^n f(x)~dx = \int_0^1 x^n g(x)~dx$

$\Rightarrow \int_0^1 x^n f(x)~dx - \int_0^1 x^n g(x)~dx = 0$

$\Rightarrow \int_0^1 x^n [f(x) - g(x)]~dx = 0$

Obviously, the only way this integral can be zero for all x and all n is if $f(x) - g(x) = 0$. the result follows immediately.

Now i can see where the problem here would be, the "obviously" part.
• Nov 11th 2007, 01:49 AM
kalagota
Quote:

Originally Posted by ThePerfectHacker

That is not the Fundamental Theorem of Calculus.
---

i think i clearly state that the reason is from a corollary to FTOC and not to FTOC itself..

anyways, i am not really good at this..

EDIT: now i get what you mean.. thx for that remark..
• Nov 11th 2007, 06:23 AM
CaptainBlack
Quote:

Originally Posted by Jhevon
i always have trouble proving things that seem to me to be obvious.

would it be wrong to do this?

$\int_0^1 x^n f(x)~dx = \int_0^1 x^n g(x)~dx$

$\Rightarrow \int_0^1 x^n f(x)~dx - \int_0^1 x^n g(x)~dx = 0$

$\Rightarrow \int_0^1 x^n [f(x) - g(x)]~dx = 0$

Obviously, the only way this integral can be zero for all x and all n is if $f(x) - g(x) = 0$. the result follows immediately.

Now i can see where the problem here would be, the "obviously" part.

These proofs are thinly disguised use of the Stone-Weierstrass theorem.

RonL
• Nov 11th 2007, 09:07 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
These proofs are thinly disguised use of the Stone-Weierstrass theorem.

You got the main part right! Weierstrass theorem is the secret trick here.

Quote:

Originally Posted by Jhevon
Now i can see where the problem here would be, the "obviously" part.

It is really not that obvious.

Quote:

Originally Posted by janvdl
1.) $(\frac{1}{2007} + \frac{1}{2007}) \times (\frac{1}{2007} + \frac{1}{2007})$

= $\frac{4}{4028049}$

= $\frac{1}{1007012,25}$

How did you get that? My solution is much longer.
• Nov 11th 2007, 10:32 AM
janvdl
Quote:

Originally Posted by ThePerfectHacker
How did you get that? My solution is much longer.

Haha, i guess what matters is if we got the same answer? And if mine is right?
• Nov 11th 2007, 12:34 PM
ThePerfectHacker
Quote:

Originally Posted by janvdl
Haha, i guess what matters is if we got the same answer? And if mine is right?

I have no idea what the answer is, I only created the problem and thus know how to solve it. I did not fully solve it.
• Nov 11th 2007, 12:50 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
You got the main part right! Weierstrass theorem is the secret trick here.

are you saying we should approximate f(x) - g(x) with some other polynomial, say h(x), and show that we must have h(x) = 0 ?
• Nov 11th 2007, 12:56 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
are you saying we should approximate f(x) - g(x) with some other polynomial, say h(x), and show that we must have h(x) = 0 ?

Hint: If $f(x)$ is continous on $[0,1]$ then there exists a sequence of polynomial $p_n(x)$ that converge uniformly to $f(x)$. (That is the Stone-Weierstrass theorem).
Show 40 post(s) from this thread on one page
Page 1 of 3 123 Last