Proof Involving Integration by Parts and a Series of Functions

In summary: I am using the product rule for derivatives and Leibniz's rule for differentiating under the integral sign.Then, since you will have assumed in the inductive hypothesis that ##f_k(x) = \int_0^x \frac{(x-t)^{k-1}}{(k-1
  • #1
Tollschnee
3
0

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, fn(x) = ∫fn-1(t)dt for n≥2. Show that fn+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 fn+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 fn = ∫(x-t)n-1/(n-1)! * f(t)dt. Also, I know that fn= (x-t)n/n! * f(t) based on the fact that fn+1(x) = ∫[(x-t)n/n!]f(t)dt = ∫fn(t)dt.

I have tried doing an integral by parts setting u = fn(t) ; du = fn-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.
 
Physics news on Phys.org
  • #2
Tollschnee said:

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, fn(x) = ∫fn-1(t)dt for n≥2. Show that fn+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 fn+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 fn = ∫(x-t)n-1/(n-1)! * f(t)dt. Also, I know that fn= (x-t)n/n! * f(t) based on the fact that fn+1(x) = ∫[(x-t)n/n!]f(t)dt = ∫fn(t)dt.

I have tried doing an integral by parts setting u = fn(t) ; du = fn-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 [itex]f_1(x)=\int_{0}^{x}{f_0(x)}[/itex] and [itex]f_n(x)=\int_{0}^{x}{f_{n-1}(x)}[/itex] and [itex]f_{n+1}(x)=\int_{0}^{x}{f_{n}(x)}[/itex].

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

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 [itex]f_{n+1}(x)=\int_{0}^{x}[/itex] I integrate by parts and obtain [itex]f_{n+1}(x)=xf_{n}(x)-\int_{0}^{x}{tf_{n}'(t)dt}[/itex]. But, don't we know another way of expressing [itex]f_{n}(x)[/itex]? We do, in fact, and as we are given that this can be expressed as an integral of another function. What about [itex]f_{n}'(t)[/itex]? Well, because of the insight that [itex]f_{n+1}'(t)=f_{n}(t)[/itex] I claim that this can be expressed as [itex]f_{n-1}'(t)[/itex] 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!
 
  • #3
Tollschnee said:

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, fn(x) = ∫fn-1(t)dt for n≥2. Show that fn+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 fn+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 fn = ∫(x-t)n-1/(n-1)! * f(t)dt. Also, I know that fn= (x-t)n/n! * f(t) based on the fact that fn+1(x) = ∫[(x-t)n/n!]f(t)dt = ∫fn(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 = fn(t) ; du = fn-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.
 
  • #4
Tollschnee said:

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.

0x 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 fn+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 fn = ∫(x-t)n-1/(n-1)! * f(t)dt. Also, I know that fn= (x-t)n/n! * f(t) based on the fact that fn+1(x) = ∫[(x-t)n/n!]f(t)dt = ∫fn(t)dt.
In line with what vela said: You are to prove
[itex]\displaystyle f_{n+1}(x) = \int_0^x \frac{(x-t)^{n}}{(n)!} f(t)\,dt\ ,\ [/itex] not assume it.​

For the inductive step, I like to state something like:
Assume it's true that: [itex]\displaystyle f_{k}(x) = \int_0^x \frac{(x-t)^{k-1}}{(k-1)!} f(t)\,dt\ ,\ [/itex] for some k ≥ 0.​

From that and using algebra & calculus you need to show that you can arrive at:
[itex]\displaystyle f_{k+1}(x) = \int_0^x \frac{(x-t)^{k}}{(k)!} f(t)\,dt\ ,\ [/itex]​

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 [itex]\displaystyle f_{n+1}(x) = \int_0^x \frac{(x-t)^{n}}{(n)!} f(t)\,dt\ \ [/itex] is true for all n ≥ 0 .
 
Last edited:
  • #5
SammyS said:
vela must have missed those limits of integration.
Sure did. :redface:
 

What is integration by parts?

Integration by parts is a technique used in calculus to evaluate integrals of products of functions. It is based on the product rule of differentiation and allows us to break down a complex integral into simpler integrals.

What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are two functions and du and dv are their differentials.

How do you choose which function to use as u and which to use as dv?

The acronym "LIATE" can be used to determine which function should be used as u and which as dv. LIATE stands for logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential. The function that comes first in this list should be used as u, and the other function should be used as dv.

What is a series of functions?

A series of functions is a sum of an infinite number of functions. It can be represented using sigma notation, where the terms of the series are written in a general form and the index n indicates the position of the term.

How do you use integration by parts with a series of functions?

To use integration by parts with a series of functions, we can apply the formula for integration by parts to each term in the series. This will result in a new series, which can then be evaluated using other techniques, such as the comparison test or the ratio test.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
495
  • Calculus and Beyond Homework Help
Replies
23
Views
834
  • Calculus and Beyond Homework Help
Replies
9
Views
756
  • Calculus and Beyond Homework Help
Replies
12
Views
917
  • Calculus and Beyond Homework Help
Replies
5
Views
751
  • Calculus and Beyond Homework Help
Replies
1
Views
454
  • Calculus and Beyond Homework Help
Replies
1
Views
531
  • Calculus and Beyond Homework Help
Replies
15
Views
735
  • Calculus and Beyond Homework Help
Replies
1
Views
646
  • Calculus and Beyond Homework Help
Replies
24
Views
617
Back
Top