Timeline for How to Simplify exponents?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2016 at 7:38 | vote | accept | Jerry Guern | ||
| Feb 21, 2015 at 12:59 | history | edited | m_goldberg | CC BY-SA 3.0 |
added 16 characters in body
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| Feb 21, 2015 at 2:23 | answer | added | 2012rcampion | timeline score: 12 | |
| Feb 21, 2015 at 1:16 | comment | added | Jerry Guern | @belisarius Okay, I did that. | |
| Feb 21, 2015 at 1:13 | history | edited | Jerry Guern | CC BY-SA 3.0 |
added more detail
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| Feb 21, 2015 at 0:54 | answer | added | David G. Stork | timeline score: -1 | |
| Feb 21, 2015 at 0:46 | comment | added | Dr. belisarius | perhaps you should write down the expected result | |
| Feb 21, 2015 at 0:46 | history | edited | David G. Stork | CC BY-SA 3.0 |
added 3 characters in body
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| Feb 21, 2015 at 0:41 | comment | added | Jerry Guern | @belisarius No, that's just dividing back out the wide gaussian that I just multiplied in. What I want is for each of those exponents to get individually simplified. For example, in the 2nd and 3rd terms, the x^2 term of the exponent polynomials will cancel out. | |
| Feb 21, 2015 at 0:36 | comment | added | Dr. belisarius | $\frac{1}{3} e^{\frac{(7-2 x)^2}{8 (v+2)}} \left(e^{-\frac{(x-5)^2}{2 (v+2)}}+e^{-\frac{(x-2)^2}{2 (v+2)}}+\frac{e^{-\frac{(x+2)^2}{2 (v+1)}}}{\sqrt{\frac{v+1}{v+2}}}\right)$ | |
| Feb 21, 2015 at 0:34 | comment | added | Dr. belisarius |
h[x, v] // Simplify ?
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| Feb 21, 2015 at 0:25 | history | asked | Jerry Guern | CC BY-SA 3.0 |