Timeline for functions with same area
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 22, 2012 at 6:01 | answer | added | Aaron Meyerowitz | timeline score: 0 | |
| Apr 16, 2012 at 4:31 | comment | added | Igor Rivin | @fedja: next you will tell me 57 is not prime, what's the world coming to... | |
| Apr 15, 2012 at 22:34 | comment | added | fedja | $1+\sin x$ can (and does) take values well outside $[0,1]$. The rest is fine... | |
| Apr 15, 2012 at 18:12 | comment | added | Igor Rivin | @fedja: I am maybe being totally dense (in the non-technical sense...) but if you have your two functions $f_1=\chi_1$ and $f_=\chi_2,$ and $f_3 = 1+ \sin x,$ these three functions will be linearly independent, and then the problem is simple 3-d geometry (when viewed as happening in the span of $f_1, f_2, f_3.$ What I am missing? | |
| Apr 15, 2012 at 18:04 | comment | added | Igor Rivin | @Charles: I had always thought that Hilbert Spaces were complex, but one lives and learns... | |
| Apr 15, 2012 at 11:49 | comment | added | fedja | Be careful, however (especially with Fourier): you can easily go outside the range $[0,1]$ this way! In general there is no simple recipe. Imagine that you have $\alpha_1=\alpha_2=m$. In this case, the only solutions are characteristic functions of sets of measure $m$. They can be moved one to another, no question, but you need more good luck than there is out there doing it with an explicit algebraic formula! | |
| Apr 15, 2012 at 11:16 | comment | added | ConfuseD | @Charles: Done. | |
| Apr 15, 2012 at 11:15 | history | edited | ConfuseD | CC BY-SA 3.0 |
added 66 characters in body
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| Apr 15, 2012 at 11:08 | comment | added | Charles Matthews | @ConfuseD: Could you edit the question to state the new version? | |
| Apr 15, 2012 at 11:07 | comment | added | Charles Matthews | @Igor: Unless you meant a Hilbert space is always "virtual". | |
| Apr 15, 2012 at 10:59 | comment | added | Charles Matthews | @Igor: According to a well-known reference site, "A Hilbert space H is a real or complex inner product space that is also a complete metric space with respect to the distance function induced by the inner product." | |
| Apr 15, 2012 at 1:12 | comment | added | Igor Rivin | @Charles: is not "real Hilbert space" a contradiction in terms? | |
| Apr 15, 2012 at 1:11 | answer | added | Igor Rivin | timeline score: 2 | |
| Apr 14, 2012 at 21:27 | comment | added | ConfuseD | @Pietro: Thanks! This is exactly the kind of solution I am looking for. Now assuming that $f_1$ and $f_2$ have the range $[0,1]$ and we want our family of combination's to lie in the same range, is there any way to ensure that? | |
| Apr 14, 2012 at 18:04 | comment | added | Pietro Majer | You may modify the first path $f_c:=cf_1 + (1-c)f_2$ adding a term $\alpha(c) h$ with $\int_0^T h=0$,and $\int_0^T h^2=1$. Then find $\alpha(c)\in\mathbb{R}$ such that $\int_0^T (f_c +\alpha(c)h)^2 = a_2$ solving the second degree polynomial equation. | |
| Apr 14, 2012 at 16:56 | vote | accept | ConfuseD | ||
| Apr 14, 2012 at 17:00 | |||||
| Apr 14, 2012 at 16:31 | answer | added | Charles Matthews | timeline score: 3 | |
| Apr 14, 2012 at 15:40 | comment | added | ConfuseD | @Charles: Yes I am interested in the parametric information and I just want a simple answer. Looking at functions as points in $L_2$3 does make it easy to visualize and may be also prove some existence results but is not helping in writing a parametrized version. @Mark: Getting an $F$ like you suggested in your first comment would be ideal. And I think there is one plane and one surface of a sphere (after doing a simple normalization), and we are interested in their intersection. After getting this intuition, geometry is no longer helpful to me. I really hope you could get something | |
| Apr 14, 2012 at 14:08 | comment | added | user6976 | At least this is how the question can be understood. As it is formulated now, the answer is obviously "yes" because an empty family of functions satisfies the condition. | |
| Apr 14, 2012 at 14:06 | comment | added | user6976 | There are two hyperplanes, and the question is whether the intersection is (smooth) path-connected. | |
| Apr 14, 2012 at 14:03 | comment | added | Charles Matthews | Abstractly you're intersecting a sphere in a real Hilbert space with a hyperplane. The hyperplane is a closed affine-linear subspace, by Cauchy-Schwarz. So the geometry isn't too bad. You seem to want some parametric information on the intersection. "Combining the functions" to you seems to mean working in the algebra they generate. Which is in the territory of the Stone-Weierstrass theorem, though that's for the complex (uniform) algebra, applying to continuous functions. There must clues in the functional analysis, but your formulation suggests you want a simple answer. | |
| Apr 14, 2012 at 13:41 | comment | added | user6976 | What do you mean by a "family of functions"? A smooth function $F(x,a)$, $a\in [0,1]$ with $F(x,0)=f_1, F(x,1)=f_2$? | |
| Apr 14, 2012 at 12:29 | history | asked | ConfuseD | CC BY-SA 3.0 |