Calculating the Delta of a Function with Zeros

[LocationX]

I need to show that: $$\delta(g(x)) = \sum_k \frac{\delta(x-x_k)}{|g'(x_k)|}$$

where the set $${x_k}$$ are the zeros of g(x) and $$g'(x_k) \neq 0$$

I'm not really sure where to start for this problem, any clues would be much appreciated!

 

[gabbagabbahey]

Hint: what is the defining property of the Dirac Delta function?

 

[LocationX]

Ok, so i have to show that $$\sum_k \frac{\delta(x-x_k)}{|g'(x_k)|}=0$$ for $$x \neq x_k$$ which I think a trivial step i think...

The final step would be to show that $$\int_{-\inf}^{\inf} \delta(g(x)) dx = \int_{-\inf}^{\inf} \sum_k \frac{\delta(x-x_k)}{|g'(x_k)|} dx =1$$ when $$x= x_k$$ which is what i think would the the hard step...

 

[gabbagabbahey]

Ok, so i have to show that $$\sum_k \frac{\delta(x-x_k)}{|g'(x_k)|}=0$$ for $$x \neq x_k$$ which I think a trivial step i think...

Right.

The final step would be to show that $$\int_{-\inf}^{\inf} \delta(g(x)) dx = \int_{-\inf}^{\inf} \sum_k \frac{\delta(x-x_k)}{|g'(x_k)|} dx =1$$ when $$x= x_k$$ which is what i think would the the hard step...

Not quite. If there are more than one $x_k$, when you integrate from $-\infty$ to $\infty$, you will enclose all of them, and each will contribute 1 to the integral:

$$\int_{-\infty}^{\infty} \delta(g(x)) dx = \int_{-\infty}^{\infty} \sum_k \frac{\delta(x-x_k)}{|g'(x_k)|} dx=\sum_{k}(1)$$

 

[LocationX]

Oh, that makes sense. What I'm confused about now is that the denominator will be different (the g'(x) term) and the sum will not come out to one?

 

[gabbagabbahey]

Sorry, I was mistaken in my previous comment,

$$\int_{-\infty}^{\infty} \delta(g(x)) dx \neq \sum_{k}(1)$$

for much the same reason that

$$\delta(kx)=\frac{1}{|k|}\delta(x)\neq\delta(x)$$

If you have proven this property, you can use the same method to prove your problem statement...

 

[facenian]

a) suppose there is only a finite number of x_k, k=1,2,...n
b) Since g'(x_k)<>0 then there is en interval I_k where g' is monotonous and so invertible
c)choose I_k so that $I_k \bigcap I_j=\phi, \quad for \quad k \neq j$
d)let C be the complement of unions of I_k (k=1,2,...n)
e)Now $\int_{-\infty}^\infty=\int_C + \sum_k \int_{I_k}$, so that the first integral of second member vanishes since its argument<>0
f)Integrate each term of the sum changing the integration variable $x=g^{-1}(u)$

I think that's it

 

[LocationX]

Sorry, I was mistaken in my previous comment,

$$\int_{-\infty}^{\infty} \delta(g(x)) dx \neq \sum_{k}(1)$$

for much the same reason that

$$\delta(kx)=\frac{1}{|k|}\delta(x)\neq\delta(x)$$

If you have proven this property, you can use the same method to prove your problem statement...

I'm not sure i follow, I think what you had was correct, that is:

$$\int_{-\infty}^{\infty} \delta(g(x)) dx = \sum_{k}(1)$$

I'm not sure how to apply your statement below $\left( \delta(kx)=\frac{1}{|k|}\delta(x)\neq\delta(x) \right)$

I see that the constant k factors below the delta, but in the problem, we have a function g(x) instead of a constant. Am I thinking about this correctly?

 

[gabbagabbahey]

Let's look at $\int_{-\infty}^{\infty} \delta(g(x)) dx$... First break the integral into a bunch of intervals; some that just barely enclose exactly one of the zeroes of g(x) and some that enclose no zeroes. Obviously, the intervals that enclose no zeroes will integrate to zero (because the delta function will be zero over the entire interval), leaving you with something like:

$$\int_{-\infty}^{\infty} \delta(g(x)) dx=\sum_{k}\int_{x_k-\epsilon}^{x_k+\epsilon} \delta(g(x)) dx$$

For some adequately small (by adequate, i mean small enough that each interval encloses only one zero of g(x)) positive number $\epsilon$

Now make the substitution $u=g(x)$...what do you get?

 

[LocationX]

Ok this is starting to make sense... I get:

$$\int \frac{\delta(u) du}{g'(x)}$$

Am I assuming that g'(x) -> |g'(x)|?

 

[gabbagabbahey]

Well, what about the limits of integration?

 

[LocationX]

$$\int ^{g(x_k + \epsilon)} _{g(x_k - \epsilon)} \frac{\delta(u) du}{g'(x)}$$

something like that? So basically the evaluated integral is $$\frac{1}{g'(x_k)}$$

 

[gabbagabbahey]

$$\int ^{g(x_k + \epsilon)} _{g(x_k - \epsilon)} \frac{\delta(u) du}{g'(x)}$$

something like that? So basically the evaluated integral is $$\frac{1}{g'(x_k)}$$

Well, you know the expected result is

$$\int ^{g(x_k + \epsilon)} _{g(x_k - \epsilon)} \frac{\delta(u) du}{g'(x)}=\frac{1}{|g'(x_k)|}=\left\{\begin{array}{lr}\frac{1}{g'(x_k)} & ,g'(x_k)>0\\\frac{-1}{g'(x_k)} & ,g'(x_k)<0\end{array}\right.$$

Right?

Is there anything in the integrand that suggests this result? If not, then you need to take a closer look at the limits of integration...

Hint: You are free to choose $\epsilon$ to be infinitesimally small, when you do so, $g(x_k\pm\epsilon)\approx$____?

 

[LocationX]

$$g(x_k\pm\epsilon)\approx 0$$

Honestly, I'm not really sure where to go with this. I was thinking that the delta would give a -1 for g'<0 but that's not the case. I'm now thinking this:

$u=g(x) \rightarrow x=g^{-1}(u)$

$$\int ^{g(x_k + \epsilon)} _{g(x_k - \epsilon)} \frac{\delta(u) du}{g'(x)} = \int ^{g(x_k + \epsilon)} _{g(x_k - \epsilon)} \frac{\delta(u) du}{g'(g^{-1}(u))}$$

make epsilon infinitesimally small...

$$\int ^{0} _{0} \frac{\delta(u) du}{g'(g^{-1}(u))} = \frac{\delta(0)}{g'(g^{-1}(0))} = \frac{1}{g'(x_k)}$$

I'm lost?

 

[gabbagabbahey]

$$g(x_k\pm\epsilon)\approx 0$$

Honestly, I'm not really sure where to go with this. I was thinking that the delta would give a -1 for g'<0 but that's not the case. I'm now thinking this:

I think you can make a little better approximation than that...perhaps a Taylor series?

 

[LocationX]

I get:

$$g(x) \approx g(a)+g'(a)(x-a)$$
expand around x_k
$$g(x_k - \epsilon ) \approx g(x_k)+g'(x_k)((x_k - \epsilon ) - x_k)^2$$

$$\int ^{g'(x_k)\epison ^2} _{g'(x_k)\epison ^2} \frac{\delta(u) du}{g'(x)}$$

?? Still a bit lost (sorry..)

 

[gabbagabbahey]

I get:

$$g(x) \approx g(a)+g'(a)(x-a)$$
expand around x_k
$$g(x_k - \epsilon ) \approx g(x_k)+g'(x_k)((x_k - \epsilon ) - x_k)^2$$

$$\int ^{g'(x_k)\epison ^2} _{g'(x_k)\epison ^2} \frac{\delta(u) du}{g'(x)}$$

?? Still a bit lost (sorry..)

Surely you must mean:

$$g(x_k \pm \epsilon ) \approx g(x_k)+g'(x_k)((x_k \pm \epsilon ) - x_k)=\pm g'(x_k)\epsilon$$

right?

This leaves you with:

$$\int ^{g'(x_k)\epsilon} _{-g'(x_k)\epsilon } \frac{\delta(u) du}{g'(x)}$$

Now, what happens if g'(x_k) is negative...are you still integrating from a negative number to a positive number? If not, what happens to the sign of the integral?

 

[LocationX]

I get it now...

if g'(x_k)>0 then this integral still holds true:
$$\int ^{g'(x_k)\epsilon} _{-g'(x_k)\epsilon } \frac{\delta(u) du}{g'(x)}$$

if g'(x_k)<0, then this integral becomes:
$$\int ^{-g'(x_k)\epsilon} _{g'(x_k)\epsilon } \frac{\delta(u) du}{g'(x)}=-\int ^{g'(x_k)\epsilon} _{-g'(x_k)\epsilon } \frac{\delta(u) du}{g'(x)}$$

Thus the integral will still come out to be positive -g'(x_k) where g'(x_k)<0

 

[gabbagabbahey]

Right!