Timeline for Exponential of a Differential Operator
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 23, 2017 at 16:08 | comment | added | Jens | @wondering You're simply not doing it right, | |
| Aug 23, 2017 at 14:35 | comment | added | wondering | But if I calculate the first three terms of the exponential expansion, i.e. $\Exp[x+\partial] f=(1+(x+\partial_x)+(x+\partial_x)^2+(x+\partial_x))^3+\dots)f(x)$, then I don't get the result above. How comes that you get terms like $x$ or $1/2$ in the series, if $f(x)$ has to appear in every term as a derivative of some order ($0,1,2,...$)? | |
| Aug 19, 2017 at 15:26 | comment | added | Jens |
@wondering Yes, it's correct. It doesn't matter whether or not the terms in dop commute, as long as dop is a linear operator.
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| Aug 19, 2017 at 13:16 | comment | added | wondering | Is the result for $operatorExp[dOp2, 3][f[x]]$ correct? $x$ and $D[f, x]$ are non-commuting operators. | |
| Jun 27, 2014 at 17:20 | comment | added | user85503 | Perfect! Your second approach is exactly what I needed. | |
| Jun 27, 2014 at 17:18 | vote | accept | user85503 | ||
| Jun 25, 2014 at 19:24 | history | edited | Jens | CC BY-SA 3.0 |
Another example
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| Jun 25, 2014 at 19:05 | history | edited | Jens | CC BY-SA 3.0 |
Added justification of the approach.
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| Jun 25, 2014 at 18:15 | history | answered | Jens | CC BY-SA 3.0 |