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EscherichiaRinku
 4 years ago
Show that the polynomial
\[L_n(x) = \sum_{k=0}^{n} (1)^k {n \choose k} \frac{x^k}{k!}\]
is an eigenfunction of the symmetric operator
\[D(f(x)) = xf^{\prime \prime}(x)+(1x)f^{\prime}(x)\]
I know that to show this, I need to solve the following equation for lambda:
\[D(L_n) = \lambda L_n\]
So I just plug the polynomial into the equation and... dang, I don't know how differentiate the polynomial :( Can someone suggest a derivative of the polynomial or some workaround?
EscherichiaRinku
 4 years ago
Show that the polynomial \[L_n(x) = \sum_{k=0}^{n} (1)^k {n \choose k} \frac{x^k}{k!}\] is an eigenfunction of the symmetric operator \[D(f(x)) = xf^{\prime \prime}(x)+(1x)f^{\prime}(x)\] I know that to show this, I need to solve the following equation for lambda: \[D(L_n) = \lambda L_n\] So I just plug the polynomial into the equation and... dang, I don't know how differentiate the polynomial :( Can someone suggest a derivative of the polynomial or some workaround?

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2You're dealing with eigenfunctions and you don't know how to differentiate a polynomial? Seriously, what's giving you the problem here?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2If you're really stuck, start with showing this relation for \( L_3 \) and \( L_4 \), and then generalize the algebra.

EscherichiaRinku
 4 years ago
Best ResponseYou've already chosen the best response.0Well, I sort of thought of simply differentiating the general form, as we've done that before. I'm not very good at proving things :(

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Then write out L_3 and L_4 explicitly (i.e., write out each term of the polynomial, calculating explicitly the coefficient, getting rid of choose symbols as well as the summation), then see how it satisfies the equation. Once you see that, you'll have a much easier time proving the general case.
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