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Any smooth function with a single rounded maximum, if raised to higher and higher powers, goes into a Gaussian
The Next CEO of Stack OverflowTaylor series of ln(1/(1-z)) around 0Why does each successive term in a Taylor series need to be much less than the previous term?Finding certain coefficients in Taylor expansion of $ log(1 +qx^2 + rx^3)$Use taylor expansion to study graph geometrically
$begingroup$
Jaynes (2003) writes on p. 220:
Any smooth function with a single rounded maximum, if raised to higher and higher powers, goes into a Gaussian function.
Here I presume that he means that the function is also non-negative, as this is stated in the context of probability theory. (Otherwise the power operation might cause sign-flipping and the location of the maximum could depend on the evenness of the integer power.)
Assume $f(x)$ has a single rounded maximum at $x=0$ and $f(x) geq 0$. We expand $g(x) = log f(x)^alpha$ around the maximum:
$$ log f(x)^alpha = alpha log f(0) + sum_n=2^infty g^(n)(0) fracx^nn! labela$$
Now if the terms for which $n geq 3$ in this expansion would become smaller for increasing $alpha$, we have local Gaussian behavior around the maximum and the quote is proved. (At least for my purposes.)
However, I do not find this behavior. The expansion coefficients up to fourth order are:
beginalign
g^(1)(0) &= 0 \
g^(2)(0) &= alpha fracf^(2)(0)f(0) < 0\
g^(3)(0) &= alpha fracf^(3)(0)f(0) \
g^(4)(0) &= alpha fracf^(4)(0) + [f^(2)(0)]^2f(0)^2
endalign
I am unsure whether the qualifier "rounded" (from "rounded maximum") might imply something here.
My question: How can one (dis)prove the quote? Can one say something about $f(x)^alpha$ far away from the maximum?
Edit: Intuitively, I feel this statements should hold in many useful cases. When raising $f(x)$ to higher powers, the peak not only gets higher but also thinner. Therefore, in the Taylor expansion around the maximum the deviation from zero, i.e. $x$, lives on a smaller and smaller scale for each increment in the power $alpha$. This will suppress the higher order terms in the Taylor expansion, not by virtue of the coefficients $g^(n)(0)$ but instead by multiplication with $x^n$. This is more or less what happens in the "Concepts" subsection of Wiki's General theory of Laplace's method.
probability-theory taylor-expansion
asked Mar 19 at 11:31
marnixmarnix
13
$endgroup$
add a comment |
$begingroup$
Jaynes (2003) writes on p. 220:
Any smooth function with a single rounded maximum, if raised to higher and higher powers, goes into a Gaussian function.
Here I presume that he means that the function is also non-negative, as this is stated in the context of probability theory. (Otherwise the power operation might cause sign-flipping and the location of the maximum could depend on the evenness of the integer power.)
Assume $f(x)$ has a single rounded maximum at $x=0$ and $f(x) geq 0$. We expand $g(x) = log f(x)^alpha$ around the maximum:
$$ log f(x)^alpha = alpha log f(0) + sum_n=2^infty g^(n)(0) fracx^nn! labela$$
Now if the terms for which $n geq 3$ in this expansion would become smaller for increasing $alpha$, we have local Gaussian behavior around the maximum and the quote is proved. (At least for my purposes.)
However, I do not find this behavior. The expansion coefficients up to fourth order are:
beginalign
g^(1)(0) &= 0 \
g^(2)(0) &= alpha fracf^(2)(0)f(0) < 0\
g^(3)(0) &= alpha fracf^(3)(0)f(0) \
g^(4)(0) &= alpha fracf^(4)(0) + [f^(2)(0)]^2f(0)^2
endalign
I am unsure whether the qualifier "rounded" (from "rounded maximum") might imply something here.
My question: How can one (dis)prove the quote? Can one say something about $f(x)^alpha$ far away from the maximum?
Edit: Intuitively, I feel this statements should hold in many useful cases. When raising $f(x)$ to higher powers, the peak not only gets higher but also thinner. Therefore, in the Taylor expansion around the maximum the deviation from zero, i.e. $x$, lives on a smaller and smaller scale for each increment in the power $alpha$. This will suppress the higher order terms in the Taylor expansion, not by virtue of the coefficients $g^(n)(0)$ but instead by multiplication with $x^n$. This is more or less what happens in the "Concepts" subsection of Wiki's General theory of Laplace's method.
probability-theory taylor-expansion
asked Mar 19 at 11:31
marnixmarnix
13
$endgroup$
$begingroup$
I would be stunned if this were true as written. There must be some other restrictions on the function.
$endgroup$
– Randall
Mar 19 at 11:51
$begingroup$
Yes, I think so too. There are no other restrictions stated, though. Does a very smooth maximum imply something about the derivatives at the maximum $f^(n)(0)$?
$endgroup$
– marnix
Mar 19 at 12:20
$begingroup$
The integrals of powers of $f$ behave like integrals of an appropriate Gaussian. See Laplace's method.
$endgroup$
– Mike Earnest
Mar 19 at 16:18
$begingroup$
Thanks, you've pointed me to a more intuitive understanding. (See the edit.)
$endgroup$
– marnix
Mar 19 at 22:32
add a comment |
$begingroup$
Jaynes (2003) writes on p. 220:
Any smooth function with a single rounded maximum, if raised to higher and higher powers, goes into a Gaussian function.
Here I presume that he means that the function is also non-negative, as this is stated in the context of probability theory. (Otherwise the power operation might cause sign-flipping and the location of the maximum could depend on the evenness of the integer power.)
Assume $f(x)$ has a single rounded maximum at $x=0$ and $f(x) geq 0$. We expand $g(x) = log f(x)^alpha$ around the maximum:
$$ log f(x)^alpha = alpha log f(0) + sum_n=2^infty g^(n)(0) fracx^nn! labela$$
Now if the terms for which $n geq 3$ in this expansion would become smaller for increasing $alpha$, we have local Gaussian behavior around the maximum and the quote is proved. (At least for my purposes.)
However, I do not find this behavior. The expansion coefficients up to fourth order are:
beginalign
g^(1)(0) &= 0 \
g^(2)(0) &= alpha fracf^(2)(0)f(0) < 0\
g^(3)(0) &= alpha fracf^(3)(0)f(0) \
g^(4)(0) &= alpha fracf^(4)(0) + [f^(2)(0)]^2f(0)^2
endalign
I am unsure whether the qualifier "rounded" (from "rounded maximum") might imply something here.
My question: How can one (dis)prove the quote? Can one say something about $f(x)^alpha$ far away from the maximum?
Edit: Intuitively, I feel this statements should hold in many useful cases. When raising $f(x)$ to higher powers, the peak not only gets higher but also thinner. Therefore, in the Taylor expansion around the maximum the deviation from zero, i.e. $x$, lives on a smaller and smaller scale for each increment in the power $alpha$. This will suppress the higher order terms in the Taylor expansion, not by virtue of the coefficients $g^(n)(0)$ but instead by multiplication with $x^n$. This is more or less what happens in the "Concepts" subsection of Wiki's General theory of Laplace's method.
probability-theory taylor-expansion
asked Mar 19 at 11:31
marnixmarnix
13
$endgroup$
Jaynes (2003) writes on p. 220:
Any smooth function with a single rounded maximum, if raised to higher and higher powers, goes into a Gaussian function.
Here I presume that he means that the function is also non-negative, as this is stated in the context of probability theory. (Otherwise the power operation might cause sign-flipping and the location of the maximum could depend on the evenness of the integer power.)
Assume $f(x)$ has a single rounded maximum at $x=0$ and $f(x) geq 0$. We expand $g(x) = log f(x)^alpha$ around the maximum:
$$ log f(x)^alpha = alpha log f(0) + sum_n=2^infty g^(n)(0) fracx^nn! labela$$
Now if the terms for which $n geq 3$ in this expansion would become smaller for increasing $alpha$, we have local Gaussian behavior around the maximum and the quote is proved. (At least for my purposes.)
However, I do not find this behavior. The expansion coefficients up to fourth order are:
beginalign
g^(1)(0) &= 0 \
g^(2)(0) &= alpha fracf^(2)(0)f(0) < 0\
g^(3)(0) &= alpha fracf^(3)(0)f(0) \
g^(4)(0) &= alpha fracf^(4)(0) + [f^(2)(0)]^2f(0)^2
endalign
I am unsure whether the qualifier "rounded" (from "rounded maximum") might imply something here.
My question: How can one (dis)prove the quote? Can one say something about $f(x)^alpha$ far away from the maximum?
Edit: Intuitively, I feel this statements should hold in many useful cases. When raising $f(x)$ to higher powers, the peak not only gets higher but also thinner. Therefore, in the Taylor expansion around the maximum the deviation from zero, i.e. $x$, lives on a smaller and smaller scale for each increment in the power $alpha$. This will suppress the higher order terms in the Taylor expansion, not by virtue of the coefficients $g^(n)(0)$ but instead by multiplication with $x^n$. This is more or less what happens in the "Concepts" subsection of Wiki's General theory of Laplace's method.
probability-theory taylor-expansion
probability-theory taylor-expansion
asked Mar 19 at 11:31
marnixmarnix
13
asked Mar 19 at 11:31
marnixmarnix
13
edited Mar 19 at 22:31
marnix
asked Mar 19 at 11:31
marnixmarnix
13
asked Mar 19 at 11:31
marnixmarnix
13
asked Mar 19 at 11:31
marnixmarnix
13
13
$begingroup$
I would be stunned if this were true as written. There must be some other restrictions on the function.
$endgroup$
– Randall
Mar 19 at 11:51
$begingroup$
Yes, I think so too. There are no other restrictions stated, though. Does a very smooth maximum imply something about the derivatives at the maximum $f^(n)(0)$?
$endgroup$
– marnix
Mar 19 at 12:20
$begingroup$
The integrals of powers of $f$ behave like integrals of an appropriate Gaussian. See Laplace's method.
$endgroup$
– Mike Earnest
Mar 19 at 16:18
$begingroup$
Thanks, you've pointed me to a more intuitive understanding. (See the edit.)
$endgroup$
– marnix
Mar 19 at 22:32
add a comment |
$begingroup$
I would be stunned if this were true as written. There must be some other restrictions on the function.
$endgroup$
– Randall
Mar 19 at 11:51
$begingroup$
Yes, I think so too. There are no other restrictions stated, though. Does a very smooth maximum imply something about the derivatives at the maximum $f^(n)(0)$?
$endgroup$
– marnix
Mar 19 at 12:20
$begingroup$
The integrals of powers of $f$ behave like integrals of an appropriate Gaussian. See Laplace's method.
$endgroup$
– Mike Earnest
Mar 19 at 16:18
$begingroup$
Thanks, you've pointed me to a more intuitive understanding. (See the edit.)
$endgroup$
– marnix
Mar 19 at 22:32
$begingroup$
I would be stunned if this were true as written. There must be some other restrictions on the function.
$endgroup$
– Randall
Mar 19 at 11:51
$begingroup$
I would be stunned if this were true as written. There must be some other restrictions on the function.
$endgroup$
– Randall
Mar 19 at 11:51
$begingroup$
Yes, I think so too. There are no other restrictions stated, though. Does a very smooth maximum imply something about the derivatives at the maximum $f^(n)(0)$?
$endgroup$
– marnix
Mar 19 at 12:20
$begingroup$
Yes, I think so too. There are no other restrictions stated, though. Does a very smooth maximum imply something about the derivatives at the maximum $f^(n)(0)$?
$endgroup$
– marnix
Mar 19 at 12:20
$begingroup$
The integrals of powers of $f$ behave like integrals of an appropriate Gaussian. See Laplace's method.
$endgroup$
– Mike Earnest
Mar 19 at 16:18
$begingroup$
The integrals of powers of $f$ behave like integrals of an appropriate Gaussian. See Laplace's method.
$endgroup$
– Mike Earnest
Mar 19 at 16:18
$begingroup$
Thanks, you've pointed me to a more intuitive understanding. (See the edit.)
$endgroup$
– marnix
Mar 19 at 22:32
$begingroup$
Thanks, you've pointed me to a more intuitive understanding. (See the edit.)
$endgroup$
– marnix
Mar 19 at 22:32
add a comment |
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$begingroup$
I would be stunned if this were true as written. There must be some other restrictions on the function.
$endgroup$
– Randall
Mar 19 at 11:51
$begingroup$
Yes, I think so too. There are no other restrictions stated, though. Does a very smooth maximum imply something about the derivatives at the maximum $f^(n)(0)$?
$endgroup$
– marnix
Mar 19 at 12:20
$begingroup$
The integrals of powers of $f$ behave like integrals of an appropriate Gaussian. See Laplace's method.
$endgroup$
– Mike Earnest
Mar 19 at 16:18
$begingroup$
Thanks, you've pointed me to a more intuitive understanding. (See the edit.)
$endgroup$
– marnix
Mar 19 at 22:32