Prove using Leibniz's Product Rule
$begingroup$
Suppose $fin C^{infty}(mathbb R)$. Show that for $x neq 0$
$$frac{1}{x^{n+1}}f^{(n)}(frac{1}{x})=(-1)^{n}frac{d^n}{dx^n}[x^{n-1}f(frac{1}{x})]$$
I am supposed to attempt it from right-hand side to left-hand side using Leibniz product rule for derivative, but I found out that I stuck on simplifying the sigma notation. Any clue for it?
derivatives
asked Dec 7 '18 at 16:09
weilam06weilam06
9511
$endgroup$
add a comment |
$begingroup$
Suppose $fin C^{infty}(mathbb R)$. Show that for $x neq 0$
$$frac{1}{x^{n+1}}f^{(n)}(frac{1}{x})=(-1)^{n}frac{d^n}{dx^n}[x^{n-1}f(frac{1}{x})]$$
I am supposed to attempt it from right-hand side to left-hand side using Leibniz product rule for derivative, but I found out that I stuck on simplifying the sigma notation. Any clue for it?
derivatives
asked Dec 7 '18 at 16:09
weilam06weilam06
9511
$endgroup$
$begingroup$
Did you try induction?
$endgroup$
– Will M.
Dec 7 '18 at 17:23
$begingroup$
Nope. Because I am concerned about the derivative of $f(frac{1}{n})$ of order $n$.
$endgroup$
– weilam06
Dec 7 '18 at 17:25
add a comment |
$begingroup$
Suppose $fin C^{infty}(mathbb R)$. Show that for $x neq 0$
$$frac{1}{x^{n+1}}f^{(n)}(frac{1}{x})=(-1)^{n}frac{d^n}{dx^n}[x^{n-1}f(frac{1}{x})]$$
I am supposed to attempt it from right-hand side to left-hand side using Leibniz product rule for derivative, but I found out that I stuck on simplifying the sigma notation. Any clue for it?
derivatives
asked Dec 7 '18 at 16:09
weilam06weilam06
9511
$endgroup$
Suppose $fin C^{infty}(mathbb R)$. Show that for $x neq 0$
$$frac{1}{x^{n+1}}f^{(n)}(frac{1}{x})=(-1)^{n}frac{d^n}{dx^n}[x^{n-1}f(frac{1}{x})]$$
I am supposed to attempt it from right-hand side to left-hand side using Leibniz product rule for derivative, but I found out that I stuck on simplifying the sigma notation. Any clue for it?
derivatives
derivatives
asked Dec 7 '18 at 16:09
weilam06weilam06
9511
asked Dec 7 '18 at 16:09
weilam06weilam06
9511
edited Dec 7 '18 at 17:27
weilam06
asked Dec 7 '18 at 16:09
weilam06weilam06
9511
asked Dec 7 '18 at 16:09
weilam06weilam06
9511
asked Dec 7 '18 at 16:09
weilam06weilam06
9511
9511
$begingroup$
Did you try induction?
$endgroup$
– Will M.
Dec 7 '18 at 17:23
$begingroup$
Nope. Because I am concerned about the derivative of $f(frac{1}{n})$ of order $n$.
$endgroup$
– weilam06
Dec 7 '18 at 17:25
add a comment |
$begingroup$
Did you try induction?
$endgroup$
– Will M.
Dec 7 '18 at 17:23
$begingroup$
Nope. Because I am concerned about the derivative of $f(frac{1}{n})$ of order $n$.
$endgroup$
– weilam06
Dec 7 '18 at 17:25
$begingroup$
Did you try induction?
$endgroup$
– Will M.
Dec 7 '18 at 17:23
$begingroup$
Did you try induction?
$endgroup$
– Will M.
Dec 7 '18 at 17:23
$begingroup$
Nope. Because I am concerned about the derivative of $f(frac{1}{n})$ of order $n$.
$endgroup$
– weilam06
Dec 7 '18 at 17:25
$begingroup$
Nope. Because I am concerned about the derivative of $f(frac{1}{n})$ of order $n$.
$endgroup$
– weilam06
Dec 7 '18 at 17:25
add a comment |
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$begingroup$
Did you try induction?
$endgroup$
– Will M.
Dec 7 '18 at 17:23
$begingroup$
Nope. Because I am concerned about the derivative of $f(frac{1}{n})$ of order $n$.
$endgroup$
– weilam06
Dec 7 '18 at 17:25