Recursive function, diff under the integral sign

[heinerL]

Hey

i stumbled over a problem which i can't solve:

f is a continuous function on [a,b] and F_n is recursive given through:

$$F_1(x)=\int_a^x f(t) \ dt$$
and
$$F_{n+1}=\int_a^x F_n(t) \ dt$$

I have to proof that F_n looks like this:

$$F_n(x)=\frac{1}{(n-1)!}\int_a^x (x-t)^{n-1} f(t) \ dt$$

For n=1 it's no big deal but then i have no clue. I think it must have something to do with differentiation under the integral sign because that's where i found it.

Hope anybody can help me! thx

 

[D H]

Hint: Use induction. You have the base case since "For n=1 it's no big deal."

 

[heinerL]

;)

actually i used induction but for n+1 I am stuck:

n=1:
$$\frac{1}{0!} \int_a^x (x-t)^{1-1} f(t) \ dt = \int_a^x f(t) \ dt$$

n=n+1
$$\frac{1}{(n+1-1)!} \int_a^x (x-t)^{n+1-1} f(t) \ dt = \frac{1}{n!}\int_a^x (x-t)^{n} f(t) \ dt$$

 

[D H]

Since you found this under differentiation under the sign, don't you think you should be doing that?

 

[heinerL]

sure which i did:

$$\frac{dF_{n+1}}{dx}=\cdots =\frac{1}{(n-1)!}\int_a^x (x-t)^(n-1) f(t) \ dt$$

but if i look at how F_n+1 is defined i should get:

$$F_n+1=\int_a^x F_n(t) \ dt= \int_a^x \frac{1}{(n-1)!}*\int_a^t (t-t)^{n-1} f(t) \ dt = \int_a^x \frac{1}{(n-1)!}$$

 

[D H]

First, what is $F_{n+1}(x)$ given

$$F_n(x)=\frac{1}{(n-1)!}\int_a^x (x-t)^{n-1} f(t) \ dt$$

(Just rewrite with $n\to n+1$).

Now try both differentiations again. In particular, what is $$\frac{dF_{n+1}(x)}{dx}$$ given

1. The formulation of $F_{n+1}(x)$ from the above, and
2. $$F_{n+1}=\int_a^x F_n(t)\, dt$$

 

[heinerL]

i'm sorry i have absolutly no idea, :-(

i thought I did the n->n+1 and the differentiation correct.

 

[D H]

Try doing what I asked you to do in post #6, and be explicit in your steps. You skipped over a lot of steps, so it is hard to say where you went wrong.

 

[heinerL]

So here's what i did:

$$F_n(x)=\\frac{1}{(n-1)!}\\int_a^x (x-t)^{n-1} f(t) \\ dt$$

so n+1:
$$\frac{1}{(n+1-1)!} \\int_a^x (x-t)^{n+1-1} f(t) \\ dt = \\frac{1}{n!}\\int_a^x (x-t)^{n} f(t) \\ dt$$

then $$\frac{dF_{n+1}(x)}{dx}$$ :

$$\frac{1}{n!} \int_a^x \frac{d}{dx} f(t) (x-t)^n dt - f(a)(x-a)^n \frac{d}{dx}(a)+0^n f(x) \frac{d}{dx}(x) = \frac{1}{n!} \int_a^x n*(x-t)^{n-1} f(t) dt = \frac{1}{(n-1)!} \int_a^x (x-t)^(n-1)f(t) dt$$

 

[D H]

So here's what i did:

That needs some help.

Ahh! You are using a lot of double backslashes.

$$F_n(x)=\frac{1}{(n-1)!}\int_a^x (x-t)^{n-1} f(t) \ dt$$

so n+1:
$$\frac{1}{(n+1-1)!} \int_a^x (x-t)^{n+1-1} f(t) \ dt = \frac{1}{n!}\int_a^x (x-t)^{n} f(t) \ dt$$

What are you trying to do here, and how does the second line follow from the first?

Please try following my suggestions in post #6.