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They are certainly not the same thing.

You might sometimes see them appear in the same context because transforms of Laplace-Fourier type are immensely useful for analyzing linear differential operators like the Laplacian. But the Fourier transform has better analytic properties, so that's the one you are more likely to see used.

Here's some intuition you might find helpful.

The discrete Laplacian computes the difference between a node's averaged neighbors and the node itself. It's often used in image processing and that gives an easy way to visualize it. The 1D case where the kernel is [1 -2 1] is especially simple:

In an area of constant color the Laplacian is zero. Indeed, even if you have linear variation it remains zero, e.g. if in the neighborhood is [1 2 3] then the Laplacian's value at the center point is

$$1 \cdot 1 + (-2) \cdot 2 + 3 \cdot 1 = 0.$$

But quadratic and higher-order variation excites the Laplacian and generally results in non-zero values. Thus it's especially useful for detecting 'discontinuities'. jumps'. That's why it's the weapon of choice in edge detection. It's often combined with a Gaussian to pre-filter out any small-scale features or noise that might cause spurious edges to be detected.

I should mention that the Laplacian in two dimensions and higher is significantly more rich richer than the one-dimensional case indicatesmight suggest. For one, not all two-dimensional images with a uniformly zero Laplacian are linear. But qualitatively a lot of the same intuition holds true as to how the Laplacian reacts to variation.

They are certainly not the same thing.

You might sometimes see them appear in the same context because transforms of Laplace-Fourier type are immensely useful for analyzing linear differential operators like the Laplacian. But the Fourier transform has better analytic properties, so that's the one you are more likely to see used.

Here's some intuition you might find usefulhelpful.

The discrete Laplacian computes the difference between a node's averaged neighbors and the node itself. It's often used in image processing and that gives an easy way to visualize it. The 1D case where the kernel is [1 -2 1] is especially simpleyet representative:

In an area of constant color the Laplacian is zero. Indeed, even if you have linear variation it remains zero, e.g. if the neighborhood is [1 2 3] then the Laplacian's value at the center point is

$$1 \cdot 1 + (-2) \cdot 2 + 3 \cdot 1 = 0.$$

But quadratic and higher-order variation excites the Laplacian and generally results in non-zero values. Thus it's especially useful for detecting 'discontinuities'. That's why it's the weapon of choice in edge detection. It's often combined with a Gaussian to pre-filter out any small-scale features or noise that might cause spurious edges to be detected.

I should mention that the Laplacian in two dimensions and higher is significantly more rich than the one-dimensional case indicates. For one, not all two-dimensional images with a uniformly zero Laplacian are linear. But qualitatively a lot of the same intuition holds true as to how the Laplacian reacts to variation.

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They are certainly not the same thing.

You might sometimes see them appear in the same context because transforms of Laplace-Fourier type are immensely useful for analyzing linear differential operators like the Laplacian. But the Fourier transform has better analytic properties, so that's the one you are more likely to see used.

Here's some intuition you might find useful.

The discrete Laplacian computes the difference between a node's averaged neighbors and the node itself. It's often used in image processing and that gives an easy way to visualize it. The 1D case where the kernel is [1 -2 1] is especially simple yet representative:

In an area of constant color the Laplacian is zero. If Indeed, even if you have an area of linearly varying color then the Laplacian will also be linear variation it remains zero, e.g. if the neighborhood is [1 2 3] then the Laplacian's value at the center point is

$1$1 \cdot 1 + (-2) \cdot 2 + 3 \cdot 1 = 0$.0.$\$

But quadratic and higher-order variation excites the Laplacian and generally results in non-zero values. Thus it's especially useful for detecting 'discontinuities'. That's why it's the weapon of choice in edge detection. It's often combined with a Gaussian to filter pre-filter out any small-scale features or noise that might cause spurious edges to be detected.

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