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  "body": "\n## Definition\n\nLet $\\varphi$ be the [[scaling function|YH4BERB952]] of an orthonormal [[multiresolution analysis|1GHQMBR7DE]] of $L^2(\\R)$ with real finite filter $(h_n)_{n \\in \\Z}$, so that\n$$\\varphi(x) = \\sqrt{2} \\sum_{n} h_n \\, \\varphi(2x - n).$$\nPut $g_n = (-1)^n h_{1-n}$ and let $\\psi(x) = \\sqrt{2}\\sum_n g_n \\varphi(2x - n)$ be the associated wavelet.\nFor $j, k \\in \\Z$ write $\\varphi_{j,k}(x) = 2^{j/2}\\varphi(2^j x - k)$ and $\\psi_{j,k}(x) = 2^{j/2}\\psi(2^j x - k)$.\nFor $f \\in L^2(\\R)$ put $a_{j,k} = \\ip{f}{\\varphi_{j,k}}$ and $d_{j,k} = \\ip{f}{\\psi_{j,k}}$.\n\nThe *fast wavelet transform* from level $J$ down to level $J_0 \u003c J$ is the algorithm that takes the sequence $(a_{J,k})_k$ and, for $j = J, J-1, \\dots, J_0 + 1$, computes\n$$a_{j-1,k} = \\sum_{m} h_{m-2k} \\, a_{j,m}, \\qquad d_{j-1,k} = \\sum_{m} g_{m-2k} \\, a_{j,m}.$$\nIts output is $(a_{J_0,k})_k$ together with $(d_{j,k})_k$ for $J_0 \\le j \u003c J$.\nThe inverse fast wavelet transform computes, for $j = J_0 + 1, \\dots, J$,\n$$a_{j,m} = \\sum_{k} h_{m-2k} \\, a_{j-1,k} + \\sum_{k} g_{m-2k} \\, d_{j-1,k}.$$\n\n## Remarks\n\nThe formulas follow from $\\varphi_{j-1,k} = \\sum_m h_{m-2k}\\varphi_{j,m}$ and $\\psi_{j-1,k} = \\sum_m g_{m-2k}\\varphi_{j,m}$, which are the refinement equations rescaled.\nEach step is a convolution with a filter followed by downsampling by $2$.\n\nExample: for the Haar wavelet, $h_0 = h_1 = 1/\\sqrt{2}$, so $g_0 = 1/\\sqrt{2}$ and $g_1 = -1/\\sqrt{2}$.\nOne step maps $a_{j,\\cdot}$ to $a_{j-1,k} = (a_{j,2k} + a_{j,2k+1})/\\sqrt{2}$ and $d_{j-1,k} = (a_{j,2k} - a_{j,2k+1})/\\sqrt{2}$, and the inverse step recovers $a_{j,2k}$ and $a_{j,2k+1}$ from these two numbers.\n\nNon-example: if the downsampling is left out and all outputs of the filters are kept, the result is the undecimated (stationary) wavelet transform.\nIt is redundant: it doubles the number of coefficients at each level, while the fast wavelet transform keeps the total number of coefficients equal to the input length.\n\nWith an input of length $N$ and filters of length $L$, the algorithm uses at most about $2LN$ multiplications.\n\n## Explanations\n\n### L1 human\n\nThe fast wavelet transform computes all wavelet coefficients by repeatedly filtering and halving.\nIt never computes an integral: everything comes from the filter coefficients of the scaling function.\n",
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