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I am self studying the 17th Chapter of "Mathematical Methods for Physics and Engineering", Riley, Hobson, Bence, 3rd Edition. It is about eigenfunction methods for the solution of linear ODEs.
On page 563, it states:
"As noted earlier, the eigenfunctions of a Hermitian operator may be shown to form a complete set over the relevant interval. ... Working in terms of the normalized eigenfunctions \hat{y}_n, we may thus write:"
<br /> f(x)=\sum_{n}\hat{y}_n(x)\int_{a}^{b}\hat{y}_n^*(z)f(z)w(z)dz =<br /> \int_{a}^{b}f(z)w(z)\sum_{n}\hat{y}_n(x)\hat{y}_n^*(z)dz<br />
from the definition of the dirac delta and the above relation, we get:
<br /> w(z)\sum_{n}\hat{y}_n(x)\hat{y}_n^*(z) = \delta(x-z)~~~~~(17.28)<br />
now we are at the stage where I am getting confused:
"We also note that the LHS of (17.28) is a delta function and so it is only non-zero when z=x; thus w(z) on the LHS can be replaced by w(x) if required"
The problem is, I cannot why this replacement can be made, since w(x) can be zero in isolated points. w(x) is the weight function for calculating inner products of functions.
The relevant relations are those in the previous section.
When x is not equal to z, 17.28 is zero on both sides. If we suppose that w(x) is nowhere zero, then it follows that the sum is zero, since the RHS is zero, and we can make the replacement. BUT if we allow w(x) to be zero in a set of points, which is allowed for a weight function as stated earlier in the chapter and in various online sources, I am having problem justifying the validity of this replacement. This is because, if w(z) is zero for a particular z, then the sum can be non zero for this z and a particular x, and 17.28 stlil holds. But if we replace w(z) with w(x), the LHS may become non-zero. What am I missing?
Homework Statement
On page 563, it states:
"As noted earlier, the eigenfunctions of a Hermitian operator may be shown to form a complete set over the relevant interval. ... Working in terms of the normalized eigenfunctions \hat{y}_n, we may thus write:"
<br /> f(x)=\sum_{n}\hat{y}_n(x)\int_{a}^{b}\hat{y}_n^*(z)f(z)w(z)dz =<br /> \int_{a}^{b}f(z)w(z)\sum_{n}\hat{y}_n(x)\hat{y}_n^*(z)dz<br />
from the definition of the dirac delta and the above relation, we get:
<br /> w(z)\sum_{n}\hat{y}_n(x)\hat{y}_n^*(z) = \delta(x-z)~~~~~(17.28)<br />
now we are at the stage where I am getting confused:
"We also note that the LHS of (17.28) is a delta function and so it is only non-zero when z=x; thus w(z) on the LHS can be replaced by w(x) if required"
The problem is, I cannot why this replacement can be made, since w(x) can be zero in isolated points. w(x) is the weight function for calculating inner products of functions.
Homework Equations
The relevant relations are those in the previous section.
The Attempt at a Solution
When x is not equal to z, 17.28 is zero on both sides. If we suppose that w(x) is nowhere zero, then it follows that the sum is zero, since the RHS is zero, and we can make the replacement. BUT if we allow w(x) to be zero in a set of points, which is allowed for a weight function as stated earlier in the chapter and in various online sources, I am having problem justifying the validity of this replacement. This is because, if w(z) is zero for a particular z, then the sum can be non zero for this z and a particular x, and 17.28 stlil holds. But if we replace w(z) with w(x), the LHS may become non-zero. What am I missing?
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