Proof Involving Integration by Parts and a Series of Functions

[Tollschnee]

Homework Statement

Let f be continuous on an interval I containing 0, and define f₁(x) = ∫f(t)dt, f₂(x) = ∫f₁(t)dt, and in general, f_n(x) = ∫f_n-1(t)dt for n≥2. Show that f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt for every n≥0.

ALL INTEGRALS DEFINED FROM 0 to x (I can't format :( )

Homework Equations

Integration by parts: ∫u dv = uv - ∫vdu.

The Attempt at a Solution

I want to do this proof by induction. I am assuming the base step is when n=0 since that is the first possibility. This step is very simple because you can plug 0 into f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt and get f₁(x) = ∫f(t)dt.

Now the induction step is where I get messed up. I assume that f_n = ∫(x-t)^n-1/(n-1)! * f(t)dt. Also, I know that f_n= (x-t)ⁿ/n! * f(t) based on the fact that f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt = ∫f_n(t)dt.

I have tried doing an integral by parts setting u = f_n(t) ; du = f_n-1(t) ; dv = dt ; and v = t... but I keep ending up with either sequences that don't mean anything or some annoying cyclical stuff where I end up right back where I started.

 

[NihilTico]

Homework Statement

Let f be continuous on an interval I containing 0, and define f₁(x) = ∫f(t)dt, f₂(x) = ∫f₁(t)dt, and in general, f_n(x) = ∫f_n-1(t)dt for n≥2. Show that f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt for every n≥0.

ALL INTEGRALS DEFINED FROM 0 to x (I can't format :( )

Homework Equations

Integration by parts: ∫u dv = uv - ∫vdu.

The Attempt at a Solution

I want to do this proof by induction. I am assuming the base step is when n=0 since that is the first possibility. This step is very simple because you can plug 0 into f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt and get f₁(x) = ∫f(t)dt.

Now the induction step is where I get messed up. I assume that f_n = ∫(x-t)^n-1/(n-1)! * f(t)dt. Also, I know that f_n= (x-t)ⁿ/n! * f(t) based on the fact that f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt = ∫f_n(t)dt.

I have tried doing an integral by parts setting u = f_n(t) ; du = f_n-1(t) ; dv = dt ; and v = t... but I keep ending up with either sequences that don't mean anything or some annoying cyclical stuff where I end up right back where I started.

I think you are on the right track.
So $f_1(x)=\int_{0}^{x}{f_0(x)}$ and $f_n(x)=\int_{0}^{x}{f_{n-1}(x)}$ and $f_{n+1}(x)=\int_{0}^{x}{f_{n}(x)}$.

Here's the key insight:
If $x$ is allowed to be any real number, then, first considering the definite integral and letting the indefinite integral of $f_{n}(t)$ be $\int{f_{n}(t)dt}=F_{n}(t)+C$, we see that $f_{n+1}(x)=\int_{0}^{x}{f_{n}(t)}=F_{n}(x)-F_{n}(0)$. This certainly looks as if it were saying that, for any value of $x$, we have that $f_{n+1}(t)=\int{f_{n}(t)dt}+F_{n}(0)$ and, hence, $f_{n+1}'(t)=f_{n}(t)$.

Once you have that, you can integrate by parts, and some things should become clear. I'll try to point you in the right direction.
Letting $f_{n+1}(x)=\int_{0}^{x}$ I integrate by parts and obtain $f_{n+1}(x)=xf_{n}(x)-\int_{0}^{x}{tf_{n}'(t)dt}$. But, don't we know another way of expressing $f_{n}(x)$? We do, in fact, and as we are given that this can be expressed as an integral of another function. What about $f_{n}'(t)$? Well, because of the insight that $f_{n+1}'(t)=f_{n}(t)$ I claim that this can be expressed as $f_{n-1}'(t)$ and that, combining these two things I have stated here, you have something that may look familiar and you should be able to follow it through from there!

 

[vela]

Homework Statement

Let f be continuous on an interval I containing 0, and define f₁(x) = ∫f(t)dt, f₂(x) = ∫f₁(t)dt, and in general, f_n(x) = ∫f_n-1(t)dt for n≥2. Show that f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt for every n≥0.

ALL INTEGRALS DEFINED FROM 0 to x (I can't format :( )

Homework Equations

Integration by parts: ∫u dv = uv - ∫vdu.

The Attempt at a Solution

I want to do this proof by induction. I am assuming the base step is when n=0 since that is the first possibility. This step is very simple because you can plug 0 into f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt and get f₁(x) = ∫f(t)dt.

Now the induction step is where I get messed up. I assume that f_n = ∫(x-t)^n-1/(n-1)! * f(t)dt. Also, I know that f_n= (x-t)ⁿ/n! * f(t) based on the fact that f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt = ∫f_n(t)dt.

It's not a fact that
$$f_{n+1}(x) = \int \frac{(x-t)^n}{n!} f(t)\,dt.$$ That's what you're supposed to prove.

Another problem I see is in your definition $f_{n+1}(x) = \int f_n(t)\,dt$. The lefthand side is a function of x, but x doesn't appear anywhere on the righthand side.

I have tried doing an integral by parts setting u = f_n(t) ; du = f_n-1(t) ; dv = dt ; and v = t... but I keep ending up with either sequences that don't mean anything or some annoying cyclical stuff where I end up right back where I started.

 

[SammyS]

Homework Statement

...

∫[(x-t)ⁿ/n!]f(t)dt​

ALL INTEGRALS DEFINED FROM 0 to x (I can't format :( )

vela must have missed those limits of integration.

You could try sub/super-scripts.

∫₀^x along with larger size to make "∫" into "∫" .

...

I want to do this proof by induction. I am assuming the base step is when n=0 since that is the first possibility. This step is very simple because you can plug 0 into f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt and get f₁(x) = ∫f(t)dt.

That looks OK for the base step.

Now the induction step is where I get messed up. I assume that f_n = ∫(x-t)^n-1/(n-1)! * f(t)dt. Also, I know that f_n= (x-t)ⁿ/n! * f(t) based on the fact that f_n+1(x) = ∫[(x-t)ⁿ/n!]f(t)dt = ∫f_n(t)dt.

In line with what vela said: You are to prove

$\displaystyle f_{n+1}(x) = \int_0^x \frac{(x-t)^{n}}{(n)!} f(t)\,dt\ ,\$ not assume it.​

For the inductive step, I like to state something like:

Assume it's true that: $\displaystyle f_{k}(x) = \int_0^x \frac{(x-t)^{k-1}}{(k-1)!} f(t)\,dt\ ,\$ for some k ≥ 0.​

From that and using algebra & calculus you need to show that you can arrive at:

$\displaystyle f_{k+1}(x) = \int_0^x \frac{(x-t)^{k}}{(k)!} f(t)\,dt\ ,\$​

This is where you are likely to use integration by parts.



Do you see that logic behind the inductive process?

In the base step, you showed the general statement is true for k = 0 ( or n=0, depending on the index you use).


If you prove the inductive step, then the statement being true for k = 0, means that you showed it's true for k = 1.

The statement being true for k = 1, means that you showed it's true for k = 2.

The statement being true for k = 2, means that you showed it's true for k = 3.

etc.


So $\displaystyle f_{n+1}(x) = \int_0^x \frac{(x-t)^{n}}{(n)!} f(t)\,dt\ \$ is true for all n ≥ 0 .

 

[vela]

vela must have missed those limits of integration.

Sure did.