Timeline for Understanding quadrature rule of a function multiplied by another $C^{\infty}$ function
Current License: CC BY-SA 4.0
2 events
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| Nov 5 at 18:40 | comment | added | Sam | $n$ is tending to infinity. If what you are saying is true for every $n$, then due to smoothness of $g$, it would be a zero function @gerw. Moreover, $f$ can't be a polynomial due to constraint, $f$ is not $C^{m+1}[-1,1]$, it's regular upto $C^m$ only. | |
| Nov 5 at 13:47 | history | answered | gerw | CC BY-SA 4.0 |