### Table 1: Frequency contents of the function f1(t) The representation of a periodic function (or of a function that is de ned only on a nite interval) as the linear combination of sines and cosines, is known as the Fourier series expansion of the function. The Fourier transform is a tool for obtaining such frequency and amplitude information for sequences and functions, which are not necessarily periodic. (Note that sequences are just a special case of functions.) 2

1995

"... In PAGE 2: ... The function f1(t) consists of sines and cosines of 3 frequencies. 1 2 3 4 5 c -6 -4 -2 0 2 4 6 1 2 3 4 5 d -6 -4 -2 0 2 4 6 1 2 3 4 5 a -6 -4 -2 0 2 4 6 1 2 3 4 5 b -6 -4 -2 0 2 4 6 Figure 1: A plot of f1(t), (d), and its components (a; b; c), for t = 0::5 Thus, the frequency analysis of f1(t), can be summarized in a table such as Table1 , which provides for each frequency of f1 the amplitude of the sine wave and of the cosine wave... In PAGE 3: ... Both are easily derived from the Taylor series expansion of cos, sin, and e . Through addition and subtraction they can be rewritten as: cos( ) = ei + e?i 2 sin( ) = ei ? e?i 2i (3) Hence, we can substitute the sin and cos expressions of equation 1 by the respective expressions of equation 3 and get: f(t) = n X k=1[Ak 2 (e2 i!kt + e?2 i!kt) + Bk 2i (e2 i!kt ? e?2 i!kt)] (4) If we denote: Ck = Ak?iBk 2 k gt; 0 Ck = Ak+iBk 2 k lt; 0 C0 = 0 !k = ?!?k k lt; 0 (5) we can again rewrite f(t): f(t) = n X k=?n[Cke2 i!kt] (6) Under this new notation we can rewrite the frequency analysis of Table1 as shown in Table 2. k Frequency (!k) Ck ?3 ?1 2 ?2 ?2 2i ?1 ?1=2 i=4 0 0 0 1 1=2 ?i=4 2 2 ?2i 3 1 ?2 Table 2: Another form of frequency contents of the function f1(t) Further manipulation of equation 6 is based on using the polar notation for complex numbers, that is: x + iy = r(cos( ) + isin( )) = rei where r = jx + iyj = qx2 + y2 and tan( ) = y x... ..."

Cited by 9

### Table 1: Frequency contents of the function f1(t) The representation of a periodic function (or of a function that is de ned only on a nite interval) as the linear combination of sines and cosines, is known as the Fourier series expansion of the function. The Fourier transform is a tool for obtaining such frequency and amplitude information for sequences and functions, which are not necessarily periodic. (Note that sequences are just a special case of functions.) 2

1995

"... In PAGE 3: ... The function f1(t) consists of sines and cosines of 3 frequencies. 1 2 3 4 5 c -6 -4 -2 0 2 4 6 1 2 3 4 5 d -6 -4 -2 0 2 4 6 1 2 3 4 5 a -6 -4 -2 0 2 4 6 1 2 3 4 5 b -6 -4 -2 0 2 4 6 Figure 1: A plot of f1(t), (d), and its components (a; b; c), for t = 0::5 Thus, the frequency analysis of f1(t), can be summarized in a table such as Table1 , which provides for each frequency of f1 the amplitude of the sine wave and of the cosine wave... In PAGE 4: ... Both are easily derived from the Taylor series expansion of cos, sin, and e . Through addition and subtraction they can be rewritten as: cos( ) = ei + e?i 2 sin( ) = ei ? e?i 2i (3) Hence, we can substitute the sin and cos expressions of equation 1 by the respective expressions of equation 3 and get: f(t) = n X k=1[Ak 2 (e2 i!kt + e?2 i!kt) + Bk 2i (e2 i!kt ? e?2 i!kt)] (4) If we denote: Ck = Ak?iBk 2 k gt; 0 Ck = Ak+iBk 2 k lt; 0 C0 = 0 !k = ?!?k k lt; 0 (5) we can again rewrite f(t): f(t) = n X k=?n[Cke2 i!kt] (6) Under this new notation we can rewrite the frequency analysis of Table1 as shown in Table 2. k Frequency (!k) Ck ?3 ?1 2 ?2 ?2 2i ?1 ?1=2 i=4 0 0 0 1 1=2 ?i=4 2 2 ?2i 3 1 ?2 Table 2: Another form of frequency contents of the function f1(t) Further manipulation of equation 6 is based on using the polar notation for complex numbers, that is: x + iy = r(cos( ) + isin( )) = rei where r = jx + iyj = qx2 + y2 and tan( ) = y x... ..."

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### Table 14: The Laplace expansion for the solution function The coe cient The Laplace expansion

1994

"... In PAGE 59: ... If the given boundary function f has the Laplace series expansion f = 1 X n=0 anSn Then the solution of the equation (3.2), @u @n, will have the expansion @u @n = 1 X n=0 bnSn where bn = ?nan The following Table 13 gives the Laplace expansion for the given boundary function f = x, and Table14 is the Laplace expansion for the solution.... ..."

Cited by 1

### Table II illustrates the Doppler frequency spread of an OFDM terniinal that is traveling with a speed (v), v=200km/hrA. For such channels, the time-varying effects in the channel are sufficiently slow, i.e., coherence time is more than 100 times of the interval of OFDM symbol [SI, so that plr(f) in Equation (3) can be represented by the first two terms of its Taylor series expansion at the center of the OFDM symbol.

in Performance Analysis of High QAM OFDM System Over Frequency Selective Time-Varying Fading Channel

2003

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### Table 2: Derivative Approximations. An important aspect to be considered is the validity of the asymptotic expression for the bias of the parameter estimate. In general, the bias may be expressed in series expansion with respect the sampling period h as: B(h) = C1h + C2h2 + C3h3 + O(h4) (9:2) with the coe cients Ci depending on both the process parameters and the deriva- tive approximation. When h tends to zero, the asymptotic expression can be considered:

### Table 5. Behaviour of Taylor apos;s series of the solution with a bad triangulation Order of expansion 0 1 3 5 10 20 30 50 70

"... In PAGE 20: ... Guillaume and M. Masmoudi We give in Table5 the relative errors e1 and e2 for the norm L1: { e1 is the error between Taylor expansion of yF at the point I and the solution yI+V computed on the triangulation 2.5; { e2 is the error between Taylor expansion of yF at the point I and the solution computed on the correct triangulation 2.... ..."

### Table 5: Tests for nonlinearity in conditional volatility - percent rejections and median percent difference in standard deviation in expansions and recessions across all series and per country L SC NL NL-SC

### Table 13: The Laplace expansion for the function f = x The coe cient The Laplace expansion

1994

"... In PAGE 59: ... If the given boundary function f has the Laplace series expansion f = 1 X n=0 anSn Then the solution of the equation (3.2), @u @n, will have the expansion @u @n = 1 X n=0 bnSn where bn = ?nan The following Table13 gives the Laplace expansion for the given boundary function f = x, and Table 14 is the Laplace expansion for the solution.... ..."

Cited by 1

### Table 4. Behaviour of Taylor apos;s series of the cost function with a bad triangulation Order of expansion Triangulation 2.3 Triangulation 2.4 Triangulation 2.5

"... In PAGE 19: ...249 We give in Table4 the results of the computing of j(I + V ) obtained when using Taylor apos;s expansion of j at the point I for these di erent perturbations, which have to be compared on the one hand with the direct computation of j(I + V ) on those perturbations with bad triangulation (triangulations 2.3, 2.... ..."