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Reversing substitution |
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| Nov24-12, 04:31 PM | #1 |
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Reversing substitution
Hi guys... I'm probably missing something pretty basic here but I can't seem to figure this out. I was working on a problem recently: for the complex functions f(z)=ez and g(z)=z, find their intersections. This post is not about the problem, it is about something I noticed while tackling it (incorrectly).
Anyways, here's what I noticed: If you set these functions equal to each other, you get ez=z So, naturally: z=ln(z) From here I saw that a basic substitution was applicable, so the equation can be rewritten: ez=ln(z) Basically, what I have shown is that the function h(z)=ez-z has the same zeroes as the function i(z)=ez-ln(z). Now here's what's troubling me: what if the original problem that I gave you was to find the zeroes of i(z)? Originally, we obtained i(z) from h(z) by using a substitution, but is there some way that we can go in reverse from i(z) to h(z) using a "reverse substitution"? I'm sorry if this is rather unclear. Is there anything fundamental that I am missing? Thanks a lot |
| Nov24-12, 05:05 PM | #2 |
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You can do, pretty much any whacky thing you want because everything you do starts from the assumption that z is a real number such that [itex]e^z= z[/itex] and there is NO such number.
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| Nov24-12, 06:57 PM | #3 |
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| Nov24-12, 11:57 PM | #4 |
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Reversing substitution
This is related to the Lambert W function.
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| Nov25-12, 12:28 AM | #5 |
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| Nov25-12, 01:56 PM | #6 |
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I guess this is kind of unclear. Here is the problem stated more clearly:
Prove that the system of f(z)=ez and g(z)=ln(z) has the same solutions as the system of f(z)=ez and h(z)=z |
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