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I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

\[\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx\]$$\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx$$

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In an applied setting, the discrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can compute it with the Fourier transform: \[h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))\]$$h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))$$

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform. If I indicate the transform by $\mathcal{T}$, suppose I compute: \[h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))\]$$h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))$$ What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Better still, does it have a simpler interpretation in terms of $X$ and $Y$ (like "addition")? Is it even a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

\[\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx\]

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In an applied setting, the discrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can compute it with the Fourier transform: \[h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))\]

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform. If I indicate the transform by $\mathcal{T}$, suppose I compute: \[h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))\] What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Better still, does it have a simpler interpretation in terms of $X$ and $Y$ (like "addition")? Is it even a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

$$\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx$$

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In an applied setting, the discrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can compute it with the Fourier transform: $$h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))$$

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform. If I indicate the transform by $\mathcal{T}$, suppose I compute: $$h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))$$ What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Better still, does it have a simpler interpretation in terms of $X$ and $Y$ (like "addition")? Is it even a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

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I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

\[\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx\]

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In an applied setting, the discrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can compute it with the Fourier transform: \[h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))\]

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform. If I indicate the transform by $\mathcal{T}$, suppose I compute: \[h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))\] What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Better still, does it have a simpler interpretation in terms of $X$ and $Y$ (like "addition")? Is it even a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

\[\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx\]

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In an applied setting, the discrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can compute it with the Fourier transform: \[h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))\]

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform. If I indicate the transform by $\mathcal{T}$, suppose I compute: \[h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))\] What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Is it a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

\[\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx\]

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In an applied setting, the discrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can compute it with the Fourier transform: \[h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))\]

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform. If I indicate the transform by $\mathcal{T}$, suppose I compute: \[h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))\] What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Better still, does it have a simpler interpretation in terms of $X$ and $Y$ (like "addition")? Is it even a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

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I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

\[\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx\]

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In my applicationan applied setting, the functions $f$ are probability mass functions. If I wanted to adddiscrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables, I would convolve their pmfs. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can be done efficientlycompute it with the Fourier transform: \[h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))\]

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Is there an equivalent operation that Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform enables. If I indicate the transform by $\mathcal{T}$, suppose I compute: \[h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))\] What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Is it a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

\[\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx\]

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) In my application, the functions $f$ are probability mass functions. If I wanted to add two random variables, I would convolve their pmfs, which can be done efficiently with a Fourier transform. Is there an equivalent operation that this transform enables?

I am trying to perform some computations in an environment where it is much easier to compute the hyperbolic tangent function (tanh) than cosines or sines. This prevents me from performing Fourier transforms. With that limitation in mind, I am interested in the behavior of the following transform:

Given an $L_2$-integrable function $f:R\rightarrow R$, define

\[\overline{f}(a,b)=\int_{-\infty}^{\infty} \tanh(ax+b) f(x) dx\]

My questions are:

(1) Does anyone recognize this transform?

(2) Can we recover $f$ if we are given $\overline{f}(a,b)$ for all $a,b$?

(3) Can we recover $f$ if we are given $\overline{f}(1,b)$ for all $b$?

(4) Is there a counterpart to the "variable addition trick" for the discrete Fourier transform? Let me describe what I have in mind.

In an applied setting, the discrete Fourier transform provides an efficient tool for computing the distribution of the sum of two random variables. The technique works as follows: assume we have random variables $X, Y$ and respective pdfs $f,g$. We wish to estimate the pdf of $X+Y$, which we will call $h$. Since $h$ is the convolution of $f$ and $g$, we can compute it with the Fourier transform: \[h=\mathcal{F}^{-1}(\mathcal{F}(f)\mathcal{F}(g))\]

If we wish to do this in practice, we approximate $f$ and $g$ by a finite set of (say) $N$ equally spaced points. Naively, computing the convolution would take $O(N^2)$ steps, but by using the fast Fourier transform algorithm and the argument above, we can approximately compute $h$ in $O(N \log(N))$ steps. Because one frequently wants to analyze $X+Y$ and the computational speed-up is so dramatic, this trick gets used a lot.

My question is: suppose I try the same trick with the tanh() transform. If I indicate the transform by $\mathcal{T}$, suppose I compute: \[h=\mathcal{T}^{-1}(\mathcal{T}(f)\mathcal{T}(g))\] What is $h$? Does it have a simpler interpretation in terms of $f$ and $g$ (like "convolution")? Is it a pdf? (Thanks to Terry Tao's comment on #3 above, we can see that $\mathcal{T}^{-1}$ exists, but is only defined up to a constant; let us assume we set the constant so that $\int h=1$.)

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