Mixing
Evaluating the $k$th derivative of composite $gcirc f$ if $f^{prime}(0) = 0$ without using Bruno's theorem
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Suppose $f$ and $g$ are infinitely differentiable. Suppose $f^{prime}(0) = 0$. Consider the composite function $g circ f$. If I wanted to calculate the $k$th derivative of $g circ f$ and evaluate this function at $0$, we know a priori that it will be zero. However, to prove this without appealing to Bruno's Formula, would it be sufficient to observe that
$$frac{d^k}{dx^k} g(f(x)) = frac{d^{k-1}}{dx^{k-1}} big[g^{prime}(f(x)) f^{prime}(x)big]$$
so that
$$frac{d^k}{dx^k} g(f(0)) = frac{d^{k-1}}{dx^{k-1}} ,big[g^{prime}(f(0)) f^{prime}(0)big] = 0$$
Edit: Assuming that $0$ is in the domain of $f$, of course.
derivatives function-and-relation-composition
asked Jan 10 at 5:49
Benedict VoltaireBenedict Voltaire
1,308928
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add a comment |
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Suppose $f$ and $g$ are infinitely differentiable. Suppose $f^{prime}(0) = 0$. Consider the composite function $g circ f$. If I wanted to calculate the $k$th derivative of $g circ f$ and evaluate this function at $0$, we know a priori that it will be zero. However, to prove this without appealing to Bruno's Formula, would it be sufficient to observe that
$$frac{d^k}{dx^k} g(f(x)) = frac{d^{k-1}}{dx^{k-1}} big[g^{prime}(f(x)) f^{prime}(x)big]$$
so that
$$frac{d^k}{dx^k} g(f(0)) = frac{d^{k-1}}{dx^{k-1}} ,big[g^{prime}(f(0)) f^{prime}(0)big] = 0$$
Edit: Assuming that $0$ is in the domain of $f$, of course.
derivatives function-and-relation-composition
asked Jan 10 at 5:49
Benedict VoltaireBenedict Voltaire
1,308928
$endgroup$
add a comment |
$begingroup$
Suppose $f$ and $g$ are infinitely differentiable. Suppose $f^{prime}(0) = 0$. Consider the composite function $g circ f$. If I wanted to calculate the $k$th derivative of $g circ f$ and evaluate this function at $0$, we know a priori that it will be zero. However, to prove this without appealing to Bruno's Formula, would it be sufficient to observe that
$$frac{d^k}{dx^k} g(f(x)) = frac{d^{k-1}}{dx^{k-1}} big[g^{prime}(f(x)) f^{prime}(x)big]$$
so that
$$frac{d^k}{dx^k} g(f(0)) = frac{d^{k-1}}{dx^{k-1}} ,big[g^{prime}(f(0)) f^{prime}(0)big] = 0$$
Edit: Assuming that $0$ is in the domain of $f$, of course.
derivatives function-and-relation-composition
asked Jan 10 at 5:49
Benedict VoltaireBenedict Voltaire
1,308928
$endgroup$
Suppose $f$ and $g$ are infinitely differentiable. Suppose $f^{prime}(0) = 0$. Consider the composite function $g circ f$. If I wanted to calculate the $k$th derivative of $g circ f$ and evaluate this function at $0$, we know a priori that it will be zero. However, to prove this without appealing to Bruno's Formula, would it be sufficient to observe that
$$frac{d^k}{dx^k} g(f(x)) = frac{d^{k-1}}{dx^{k-1}} big[g^{prime}(f(x)) f^{prime}(x)big]$$
so that
$$frac{d^k}{dx^k} g(f(0)) = frac{d^{k-1}}{dx^{k-1}} ,big[g^{prime}(f(0)) f^{prime}(0)big] = 0$$
Edit: Assuming that $0$ is in the domain of $f$, of course.
derivatives function-and-relation-composition
derivatives function-and-relation-composition
asked Jan 10 at 5:49
Benedict VoltaireBenedict Voltaire
1,308928
asked Jan 10 at 5:49
Benedict VoltaireBenedict Voltaire
1,308928
asked Jan 10 at 5:49
Benedict VoltaireBenedict Voltaire
1,308928
asked Jan 10 at 5:49
Benedict VoltaireBenedict Voltaire
1,308928
asked Jan 10 at 5:49
Benedict VoltaireBenedict Voltaire
1,308928
1,308928
add a comment |
add a comment |
1 Answer
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Counterexample: $g(x)=e^x$ and $f(x)=x^2$. If $h:=g circ f$, then $h''(0)=2 ne 0.$
answered Jan 10 at 5:58
FredFred
45.7k1848
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Thank you for your counterexample.
$endgroup$
– Benedict Voltaire
Jan 10 at 6:05
add a comment |
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1 Answer
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$begingroup$
Counterexample: $g(x)=e^x$ and $f(x)=x^2$. If $h:=g circ f$, then $h''(0)=2 ne 0.$
answered Jan 10 at 5:58
FredFred
45.7k1848
$endgroup$
$begingroup$
Thank you for your counterexample.
$endgroup$
– Benedict Voltaire
Jan 10 at 6:05
add a comment |
$begingroup$
Counterexample: $g(x)=e^x$ and $f(x)=x^2$. If $h:=g circ f$, then $h''(0)=2 ne 0.$
answered Jan 10 at 5:58
FredFred
45.7k1848
$endgroup$
$begingroup$
Thank you for your counterexample.
$endgroup$
– Benedict Voltaire
Jan 10 at 6:05
add a comment |
$begingroup$
Counterexample: $g(x)=e^x$ and $f(x)=x^2$. If $h:=g circ f$, then $h''(0)=2 ne 0.$
answered Jan 10 at 5:58
FredFred
45.7k1848
$endgroup$
Counterexample: $g(x)=e^x$ and $f(x)=x^2$. If $h:=g circ f$, then $h''(0)=2 ne 0.$
answered Jan 10 at 5:58
FredFred
45.7k1848
answered Jan 10 at 5:58
FredFred
45.7k1848
answered Jan 10 at 5:58
FredFred
45.7k1848
answered Jan 10 at 5:58
FredFred
45.7k1848
45.7k1848
$begingroup$
Thank you for your counterexample.
$endgroup$
– Benedict Voltaire
Jan 10 at 6:05
add a comment |
$begingroup$
Thank you for your counterexample.
$endgroup$
– Benedict Voltaire
Jan 10 at 6:05
$begingroup$
Thank you for your counterexample.
$endgroup$
– Benedict Voltaire
Jan 10 at 6:05
$begingroup$
Thank you for your counterexample.
$endgroup$
– Benedict Voltaire
Jan 10 at 6:05
add a comment |
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