Distribution Theory F_n -> d_o

[MidnightR]

[PLAIN]http://img5.imageshack.us/img5/4661/img8965n.jpg

Problem 1.35. If you need help with the notation let me know but I think it's fairly standard.

For 1. I think this integral is equal to the same integral between 0 and 1 because for x<0 F_n = 0 and for x>1 F_n = 0 but other than that I'm not sure. I'm guessing it has something to do with as n tends to infinity f(nx) = 0

Just starting the problem so any tips please thanks

 

[Dick]

f(x) is nonzero between for x between -1 and 1. Don't ignore the absolute value. Where is F_n(x) nonzero? What's the integral of f(x) and F_n(x)?

 

[MidnightR]

f(nx) = 3/4(1-(n^2)(x^2)) for -1/n <= x <= 1/n and
f(nx) = 0 for |x| > 1/n

Hence we have the integral between -1/n and 1/n of nf(nx)Phi(x).dx

but the integral between -1/n and 1/n of nf(nx) = 1 so we just have

the integral between -1/n and 1/n of Phi(x).dx

This is where I am so far...

 

[Dick]

f(nx) = 3/4(1-(n^2)(x^2)) for -1/n <= x <= 1/n and
f(nx) = 0 for |x| > 1/n

Hence we have the integral between -1/n and 1/n of nf(nx)Phi(x).dx

but the integral between -1/n and 1/n of nf(nx) = 1 so we just have

the integral between -1/n and 1/n of Phi(x).dx

This is where I am so far...

Not quite. You still have the integral of F_n(x)*Phi(x)dx. You can't just set the F_n to 1 because it's integral is 1. You have to use that Phi is continuous. If you want to prove it directly from the definition of continuity you have that for all epsilon>0 there is a delta>0 such that |Phi(x)-Phi(0)|<epsilon if |x-0|<delta. You CAN say that the integral of F_n(x)*Phi(0)*dx=Phi(0).

 

[MidnightR]

How about this [Note: |x| < 0 should read |x|< delta]

 

[Dick]

How about this [Note: |x| < 0 should read |x|< delta]

That looks pretty good to me.

 

[MidnightR]

Uh for the second one I'm getting -1/2 Phi'(0) not -Phi'(0)

Have they made a mistake? I can't see a problem with my method. It's essentially the same as the first question except you use the fact that

int between 1/n and -1/n of \frac{-3n^3}{2}xPhi(x).dx is equal to

int between 1/n and -1/n of [x^2\frac{-3n^3}{4}]'Phi(x).dx is equal to

int between 1/n and -1/n of x\frac{3n^3}{4}Phi'(x).dx

 

[Dick]

Uh for the second one I'm getting -1/2 Phi'(0) not -Phi'(0)

Have they made a mistake? I can't see a problem with my method. It's essentially the same as the first question except you use the fact that

int between 1/n and -1/n of \frac{-3n^3}{2}xPhi(x).dx is equal to

int between 1/n and -1/n of [x^2\frac{-3n^3}{4}]'Phi(x).dx is equal to

int between 1/n and -1/n of x\frac{3n^3}{4}Phi'(x).dx

I guess I would write -3n^3x/2 as F_n'(x). So you can change the integral of F_n'(x)*Phi(x) into the integral of -F_n(x)*Phi'(x) and then use the first part. Or you could write Phi(x)=Phi(0)+x*Phi'(0)+... and do the integral directly. In either case I don't see an extra factor of 2.

 

[MidnightR]

I guess I would write -3n^3x/2 as F_n'(x). So you can change the integral of F_n'(x)*Phi(x) into the integral of -F_n(x)*Phi'(x) and then use the first part.

I don't follow, the question tells you g_n = -3n^3x/2, I can't set g_n' = -3n^3x/2

 

[Dick]

I don't follow, the question tells you g_n = -3n^3x/2, I can't set g_n' = -3n^3x/2

My point is that g_n(x)=-3n^3x/2 is equal to the derivative of the f_n(x) function from the first part. g_n(x)=f_n'(x). Why not use that fact in your integration by parts?

 

[MidnightR]

My point is that g_n(x)=-3n^3x/2 is equal to the derivative of the f_n(x) function from the first part. g_n(x)=f_n'(x). Why not use that fact in your integration by parts?

Ha ha, yes I see what you mean. Hm that was silly.

As an aside if you were to set phi(x) = phi(0) + xphi'(0) + O(x^2)

What would you do with the phi(0), obviously xphi'(0) gives you your result and all O(x^2) terms (and higher) just = 0 as n-> infinity

 

[Dick]

Ha ha, yes I see what you mean. Hm that was silly.

As an aside if you were to set phi(x) = phi(0) + xphi'(0) + O(x^2)

What would you do with the phi(0), obviously xphi'(0) gives you your result and all O(x^2) terms (and higher) just = 0 as n-> infinity

The integral of x*phi(0) doesn't have to approach zero. It IS zero.