60 |
Mathematics |
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2.14Numerical methods
Straight-line fittinga
Data |
{xi},{yi} |
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(2.570) |
y |
y = mx+ c |
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Weightsb |
{wi} |
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(2.571) |
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y = mx+ c |
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(2.572) |
(0,c) |
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Model |
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Residuals |
di = yi − mxi − c |
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(2.573) |
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(x,y) |
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wixi , |
wiyi! |
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Weighted |
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(x,y) = wi |
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centre |
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(2.574) |
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Weighted |
D = wi(xi − x)2 |
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(2.575) |
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moment |
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Gradient |
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2 ) |
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(2.576) |
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D |
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w |
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var[m] |
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widi |
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(2.577) |
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D − 2 |
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n |
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c = y − mx |
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− 2 |
(2.578) |
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Intercept |
var[c] |
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1 |
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x2 |
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widi2 |
(2.579) |
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bLeast-squares fit of data to |
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mx+ c. Errors on y-values only. |
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If the errors on yi are uncorrelated, then wi = 1/var[yi]. |
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Time series analysisa
Discrete |
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M/2 |
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(r s)j = |
k=−( 2)+1 |
sj−krk |
(2.580) |
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convolution |
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M/ |
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Bartlett |
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(triangular) |
wj = 1 |
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j − N/2 |
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(2.581) |
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window |
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N/2 |
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Welch |
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2 |
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(quadratic) |
wj = 1 |
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(2.582) |
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window |
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N/2 |
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Hanning |
wj = |
1 |
1 − cos |
2πj |
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(2.583) |
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Hamming |
wj = 0.54 − 0.46cos |
2πj |
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window |
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N |
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ri |
response function |
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time series |
Mresponse function duration
wj windowing function
Nlength of time series
1 |
Welch |
Hamming |
0.8 |
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w 0.6 |
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Bartlett |
0.4 |
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0.2 |
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Hanning |
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0.2 0.4 0.6 0.8 1 |
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j/N |
aThe time series runs from j = 0...(N − 1), and the windowing functions peak at j = N/2.
2.14 Numerical methods |
61 |
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Numerical integration
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h |
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2 |
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f(x) |
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x |
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x0 |
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xN |
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xN |
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= (xN − x0)/N |
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f(x) dx |
(f0 + 2f1 + 2f2 + · · · |
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Trapezoidal rule |
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fi |
width) |
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+ 2fN−1 + fN ) |
(2.585) |
fi = f(xi) |
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N |
number of |
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subintervals |
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xN |
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x0 |
f(x) dx |
(f0 + 4f1 + 2f2 + 4f3 |
+ · · · |
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Simpson’s rulea |
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+ 4fN−1 + fN ) |
(2.586) |
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aN must be even. Simpson’s rule is exact for quadratics and cubics. |
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Numerical di erentiationa
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df |
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[−f(x+ 2h) + 8f(x+ h) − 8f(x− h) + f(x− 2h)] |
(2.587) |
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dx |
12h |
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[f(x+ h) − f(x− h)] |
(2.588) |
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2h |
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d2f |
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[−f(x+ 2h) + 16f(x+ h) − 30f(x) + 16f(x− h) − f(x− 2h)] |
(2.589) |
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dx2 |
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12h2 |
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[f(x+ h) − 2f(x) + f(x− h)] |
(2.590) |
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h2 |
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d3f |
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[f(x+ 2h) − 2f(x+ h) + 2f(x− h) − f(x− 2h)] |
(2.591) |
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dx3 |
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2h3 |
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aDerivatives of f(x) at x. h is a small interval in x.
Relations containing “ ” are O(h4); those containing “ ” are O(h2).
Numerical solutions to f(x) = 0
Secant method |
xn+1 = xn |
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f(xn) (2.592) |
f |
function of x |
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f(x∞ ) = 0 |
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− f(xn) − f(xn−1) |
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Newton–Raphson |
xn+1 = xn |
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f(xn) |
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(2.593) |
f |
= df/dx |
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method |
− f (xn) |
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62 Mathematics
Numerical solutions to ordinary di erential equationsa
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if |
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dy |
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(2.594) |
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dx |
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Euler’s method |
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h = xn+1 − xn |
(2.595) |
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then |
yn+1 |
= yn + hf(xn,yn) + O(h2) |
(2.596) |
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Runge–Kutta method (fourth-order)
if |
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dy |
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(2.597) |
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dx |
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and |
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h = xn+1 − xn |
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(2.598) |
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k1 |
= hf(xn,yn) |
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(2.599) |
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k2 |
= hf(xn + h/2,yn + k1/2) |
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k3 |
= hf(xn + h/2,yn + k2/2) |
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k4 |
= hf(xn + h,yn + k3) |
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(2.602) |
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k1 |
k2 |
k3 |
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then |
yn+1 |
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(2.603) |
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aOrdinary di erential equations (ODEs) of the form ddxy = f(x,y). Higher order equations should be reduced to a set of coupled first-order equations and solved in parallel.