| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I am reading a PDE book, and have a general question. Can all C^oo
functins f:R^2-->R be represented with two C^oo function g,h:R-->R
Such that f(x,y) = g(x)*h(y) ? I see no reason why not, but could
someone more knowledgable elaborate?
Eugene
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1156.1 | COOKIE::PBERGH | Peter Bergh, DTN 523-3007 | Tue Dec 05 1989 10:13 | 6 | |
>> I am reading a PDE book, and have a general question. Can all C^oo >> functins f:R^2-->R be represented with two C^oo function g,h:R-->R >> Such that f(x,y) = g(x)*h(y) ? I see no reason why not, but could >> someone more knowledgable elaborate? A quick counterrexample would appear to be x/(x+y) | |||||
| 1156.2 | HPSTEK::XIA | In my beginning is my end. | Tue Dec 05 1989 11:07 | 6 | |
re .1,
Could you explain a bit more as to why not please? I must have been
missing something.
Eugene
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| 1156.3 | TRACE::GILBERT | Ownership Obligates | Tue Dec 05 1989 11:32 | 15 | |
Suppose f(x,y) = g(x)*h(y). Then f(x1,y1)*f(x2,y2) = f(x1,y2)*f(x2,y1).
Let f(x,y) = a*x + b*y. Now, substituting into the above, we have:
(a*x1 + b*y1)*(a*x2 + b*y2) = (a*x1 + b*y2)*(a*x2 + b*y1)
Expanding and cancelling, we have
a*x1*b*y2 + b*y1*a*x2 = a*x1*b*y1 + b*y2*a*x2
0 = a*b*(x2-x1)*(y2-y1)
The xs and ys are free and we may consider x1 <> x2, y1 <> y2.
So if f(x,y) = a*x + b*y is separable, then either a = 0 or b = 0.
Choosing both a and b non-zero yields a function that's not separable.
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| 1156.4 | HPSTEK::XIA | In my beginning is my end. | Tue Dec 05 1989 12:03 | 3 | |
re.3
Thanks.
Eugene
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