表达式11: "p" equals StartFraction, left bracket, 0... left parenthesis, "N" minus 1 , right parenthesis , right bracket Over "N" minus 1 , EndFraction pip=0...N−1N−1π
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表达式12: left parenthesis, "p" , "h" Subscript, 2 , Baseline left parenthesis, "p" , right parenthesis , right parenthesisp,h2p
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3. Do a "slow Fourier transform"
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表达式14: "f" Subscript, "D" "C" "T" , Baseline left parenthesis, "k" , "q" , right parenthesis equals "q" left bracket, 1 , right bracket plus "q" left bracket, "N" , right bracket cosine left parenthesis, pi "k" , right parenthesis plus 2 Start sum from "n" equals 1 to "N" minus 2, end sum, "q" left bracket, "n" plus 1 , right bracket cosine left parenthesis, StartFraction, pi Over "N" minus 1 , EndFraction "n" "k" , right parenthesisfDCTk,q=q1+qNcosπk+2N−2∑n=1qn+1cosπN−1nk
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表达式15: "X" equals "f" Subscript, "D" "C" "T" , Baseline left parenthesis, left bracket, 0... "N" minus 1 , right bracket , "h" Subscript, 2 , Baseline left parenthesis, "p" , right parenthesis , right parenthesisX=fDCT0...N−1,h2p
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表达式16: "y" equals StartFraction, "f" Subscript, "D" "C" "T" , Baseline left parenthesis, StartNestedFraction, "N" minus 1 NestedOver pi , EndNestedFraction "x" , "X" , right parenthesis Over 2 left parenthesis, "N" minus 1 , right parenthesis , EndFractiony=fDCTN−1πx,X2N−1
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4. It actually looks alright! But in terms of worst-case error, it's pretty off near zero
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表达式18: absolute value left parenthesis, StartFraction, "f" Subscript, "D" "C" "T" , Baseline left parenthesis, StartNestedFraction, "N" minus 1 NestedOver pi , EndNestedFraction "x" , "X" , right parenthesis Over 2 left parenthesis, "N" minus 1 , right parenthesis , EndFraction minus "h" Subscript, 2 , Baseline left parenthesis, "x" , right parenthesis , right parenthesisabsfDCTN−1πx,X2N−1−h2x
