Proving Induction and Integral

[Benny]

Hi, I'm having trouble with the following question.

Q. Prove that

<br /> \int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{n - 1} } {f\left( {x_n } \right)} } } dx_n ...dx_2 dx_1 = \frac{1}{{\left( {n - 1} \right)!}}\int\limits_0^x {\left( {x - t} \right)^{n - 1} } f\left( t \right)dt<br />

The hint says to use induction. So I tried showing that it's true for n = 1. For that particular case the statement is a single integral with terminals 0 (lower) and x (upper) I think. But that doesn't matter too much for my purposes. After assuming that the result is true for some integer k, I'm having trouble verifying the result for n = k + 1. I wrote down

<br /> I = \int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{k - 1} } {\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)dx_{k + 1} dx_k } } } } ...dx_2 dx_1 <br />...(1)

Since the result that I've assumed has f(x_k) as the integrand rather than
f(x_(k+1)), I tried to use parts to get rid of the f(x_(k+1)). I did this by differenting f(x_(k+1)) and integrating "1". But doing this leads to integrals which cancel, leading me back to equation (1) above. Can someone help me out? Thanks.

 

[HallsofIvy]

Hi, I'm having trouble with the following question.

Q. Prove that

<br /> \int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{n - 1} } {f\left( {x_n } \right)} } } dx_n ...dx_2 dx_1 = \frac{1}{{\left( {n - 1} \right)!}}\int\limits_0^x {\left( {x - t} \right)^{n - 1} } f\left( t \right)dt<br />

The hint says to use induction. So I tried showing that it's true for n = 1. For that particular case the statement is a single integral with terminals 0 (lower) and x (upper) I think. But that doesn't matter too much for my purposes.

Yes, in case n= 1, that becomes
\int_0^x f(x)dx= \int_0^x f(t)dt[/itex]<br /> which is true since it is the same integral with different &quot;dummy&quot; variables.<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> After assuming that the result is true for some integer k, I&#039;m having trouble verifying the result for n = k + 1. I wrote down<br /> <br /> &lt;br /&gt; I = \int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{k - 1} } {\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)dx_{k + 1} dx_k } } } } ...dx_2 dx_1 &lt;br /&gt;...(1)<br /> <br /> Since the result that I&#039;ve assumed has f(x_k) as the integrand rather than<br /> f(x_(k+1)), I tried to use parts to get rid of the f(x_(k+1)). I did this by differenting f(x_(k+1)) and integrating &quot;1&quot;. But doing this leads to integrals which cancel, leading me back to equation (1) above. Can someone help me out? Thanks. </div> </div> </blockquote> So your k+1 integral is just the k integral with <br /> fx_(x_k)= \int_{o}^{x_k}f(x_{k+1})dx_{k+1}.<br /> How about integration by parts with <br /> u=f_(x_k)= \int_{o}^{x_k}f(x_{k+1})dx_{k+1}.<br /> and dv= dx_k?

 

[Benny]

I'm not sure what you mean by

<br /> fx\left( {x_k } \right) = \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} <br />

Is f applied to x_k or x(x_k)?

Your suggestion looks like what I tried before.

<br /> \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} = \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } - \int\limits_0^{x_k } {x_{k + 1} f&#039;\left( {x_{k + 1} } \right)dx_{k + 1} } <br />

<br /> = \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } - \left[ {x_{k + 1} f\left( {x_{k + 1} } \right)} \right]_0^{x_k } + \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} <br />

Which is the "n = k+1" integral I started with. Unless I missed something.

 

[HallsofIvy]

Sorry, I had an extra "x" in that formula!

No, my suggestion is not exactly what you did before. You tryed to integrate
\int_0^{x_k}f(x_{k+1})dx_{k+1}
by parts. My suggestion was to integrate
\int_0^{x_{k-1}}\int_0^{x_k}f(x_{k+1})dx_{k+1}dx_k
taking
u= \int_0^{x_k}f(x_{k+1})
and du= dx_k+1.

(If you use integration by parts twice swapping what you use for u and dv in the middle, of course you just get back to what you started with- I'm saying use integration by parts once to get rid of that innermost integral.)

Look at a simple example:
\int_0^x\int_0^y f(t)dt dy
Let u= \int_0^y f(t)dt and dv= dy. Then du= f(y) and v= y. We have
uv\|_0^x- \int_0^xvdu= y\int_0^xf(t)dt- \int_0^x yf(y)dy
If we change the dummy variable y in the second integral to t, we have
y\int_0^xf(t)dt- \int_0^x tf(t)dt= \int_0^x(y-t)f(t)dt
Do you see how that fits into your induction?

 

[Benny]

Oh ok, I can see that what I did was different to what you suggested. I misread it before.

I'm not aware of any other requirements. The question doesn't refer to other conditions which are required.

I'll see what I can come up with, thanks.

 

[Benny]

I've thought about this and I still can't get it out.

Firstly, if u = \int\limits_0^y {f\left( t \right)dt} ,dv = dy then du = f\left( y \right)dy,v = y by FTC. Ok I get that part. But what about uv evaluated at 0 and x?

In the example you gave, the outer most integral was done wrty so I would've thought that would mean in evaluating uv at 0 and x, where the variable 'y' appear in uv, it would be replaced by x. That is,

<br /> \left[ {uv} \right]_0^x = \left[ {y\int\limits_0^y {f\left( t \right)dt} } \right]_0^x = x\int\limits_0^x {f\left( t \right)dt} <br />

In any case, considering the following integral "I" and integrating by parts as you suggested, with

u = \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} ,dv = dx_k I obtain:

<br /> I = \int\limits_0^{x_{k - 1} } {\int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} } dx_{k + 1} dx_k <br />

<br /> = \left[ {x_k \int\limits_0^{x_k } {f\left( {x_{k + 1} } \right)} dx_{k + 1} } \right]_0^{x_{k - 1} } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)} dx_k <br />

<br /> = x_{k - 1} \int\limits_0^{x_{k - 1} } {f\left( {x_{k + 1} } \right)dx_{k + 1} } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)dx_k } <br />

<br /> = x_{k - 1} \int\limits_0^{x_{k - 1} } {f\left( {x_k } \right)dx_k } - \int\limits_0^{x_{k - 1} } {x_k f\left( {x_k } \right)dx_k } <br />

<br /> = \int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - x_k } \right)f\left( {x_k } \right)dx_k } <br />

I'm not sure how I can use the above and the assumption (that the statement is true for n = k) to show that the n = k + 1 case is true.

 

[HallsofIvy]

Remember that you can change "dummy" variables at will. You need to change that x_k+1 to "t".

 

[Benny]

I can rewrite the last line as:

<br /> \int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - x_k } \right)} f\left( {x_k } \right)dx_k = \int\limits_0^{x_{k - 1} } {\left( {x_{k - 1} - t} \right)} f\left( t \right)dt<br />

But I don't see how I can use the the assumption to prove the result from the above.

 

[HallsofIvy]

All those x_k are confusing the devil out of me! That's why earlier I gave an example with x and y. I showed earlier that
\int_0^x\int_0^y f(t)dt dy= \int_0^x(y-t)f(t)dt
That can be put in terms of x_k and x_k+1 by simply make x= x_k-1 and y= x_k:
\int_0^{x_{k-1}}\int_0^{x_k}f(t)dt dy= \int_0^{x_{k-1}(x_k-t)f(t)dt

Now apply your induction hypothesis, that
\int\limits_0^x {\int\limits_0^{x_1 } {...\int\limits_0^{x_{k - 1} } {f\left( {x_n } \right)} } } dx_k ...dx_2 dx_1 = \frac{1}{{\left( {k - 1} \right)!}}\int\limits_0^x {\left( {x - t} \right)^{n - 1} } f\left( t \right)dt
with f(x) replaced by (x_k-x)f(x).

 

[Benny]

Ok I'll try that thanks.