Maybe he’s using some kind of “always maps area under the curve to 0” degenerate metric. At least they used readable math. =P
Sphlaergh
There might be a reason to why it’s always 0. One of my TAs in a course in applied mathematics concluded one session by saying “And as you can see, the answer for this example is 0. This is because it’s an example from an introductory course in theoretical physics. All problems in that course are deliberately constructed so that the answer will be 0, 1 or pi; simply so that the students will know that they have done it right without the need of a key.”
Lucas James
In fact I checked it and it turns out that this is 0 if and only if s is an integer multiple of 2 (except for s=0)
Lucas James
In fact double-checked it…only true if s=2.
WTF
I’m sorry, but isn’t the function being integrated strictly positive on (0,\infty)?
4.00 (4 votes)
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Maybe he’s using some kind of “always maps area under the curve to 0” degenerate metric. At least they used readable math. =P
There might be a reason to why it’s always 0. One of my TAs in a course in applied mathematics concluded one session by saying “And as you can see, the answer for this example is 0. This is because it’s an example from an introductory course in theoretical physics. All problems in that course are deliberately constructed so that the answer will be 0, 1 or pi; simply so that the students will know that they have done it right without the need of a key.”
In fact I checked it and it turns out that this is 0 if and only if s is an integer multiple of 2 (except for s=0)
In fact double-checked it…only true if s=2.
I’m sorry, but isn’t the function being integrated strictly positive on (0,\infty)?