Tài liệu Evaluation of Functions part 14 ppt
Tài liệu Evaluation of Functions part 14 ppt
... evaluated,
i.e., fineness of the mesh.bb=dvector(1,npt);
coff=dvector(0,ncof-1);
ee=dvector(1,npt);
fs=dvector(1,npt);
u=dmatrix(1,npt,1,ncof);
v=dmatrix(1,ncof,1,ncof);
w=dvector(1,ncof);
wt=dvector(1,npt);
5.13 ... **v);
These are double versions of svdcmp, svbksb.
int i,it,j,ncof,npt;
double devmax,e,hth,power,sum,*bb,*coff,*ee,*fs,**u,**v,*w,*wt,*xs;
ncof=mm+kk+1;
npt=NPFAC*ncof; Numbe...
Tài liệu Evaluation of Functions part 3 pptx
... 1970,
Mathematical Methods of Physics
, 2nd ed. (Reading, MA:
W.A. Benjamin/Addison-Wesley),
§
2.3. [2]
5.2 Evaluation of Continued Fractions
Continued fractions are often powerful ways of evaluating functions ... necessarily
including the domain of convergence of the series, however). Sometimes the
continued fraction converges best where the series does worst, although this is...
Tài liệu Evaluation of Functions part 4 pptx
... generalizes to the
evaluation of the polynomial and nd of its derivatives simultaneously:
void ddpoly(float c[], int nc, float x, float pd[], int nd)
Given the
nc+1
coefficients of a polynomial of degree
nc
as ... simultaneously:
p=c[n];
dp=0.0;
for(j=n-1;j>=0;j ) {dp=dp*x+p; p=p*x+c[j];}
174
Chapter 5. Evaluation of Functions
Sample page from NUMERICAL RECIPES IN C: THE ART...
Tài liệu Evaluation of Functions part 5 ppt
... to
the same function, namely (
cof[0]
+
cof[1]
x + ···+
cof[n]
x
N
)/(1 +
cof[n+1]
x + ···+
5.12 Pad´e Approximants
203
Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN ... approximation in the interval of interest, by the methods of §5.8. We would
have obtained just the same lower-order polynomial. The principal lesson is that the rate
of convergence of...
Tài liệu Evaluation of Functions part 15 ppt
... − z)F
= abF − [c − (a + b +1)z]F
(5 .14. 2)
5 .14 Evaluation of Functions by Path Integration
209
Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)
Copyright ... Digital Computers
(New York: Wiley),
Chapter 13. [1]
5 .14 Evaluation of Functions by Path
Integration
In computer programming, the technique of choice is not necessa...
Tài liệu Evaluation of Functions part 6 pptx
... 178
Chapter 5. Evaluation of Functions
Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)
Copyright (C) ... relation is of little or no practical
use in that direction. This is the case, e.g., for (5.5.2) in the direction of increasing
n,whenx<n. You cannot generate Bessel functions of high n by ... Introduction)
[1]
give a list o...
Tài liệu Evaluation of Functions part 7 ppt
... Quadratic and Cubic Equations
The roots of simple algebraic equations can be viewed as being functions of the
equations’ coefficients. We are taught these functions in elementary algebra. Yet,
surprisingly ... b, c, namely
x =
−b ±
√
b
2
− 4ac
2a
(5.6.2)
184
Chapter 5. Evaluation of Functions
Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-4...
Tài liệu Evaluation of Functions part 10 pptx
... quadrature
[1]
. It is often combined with an
adaptive choice of N, the number of Chebyshev coefficients calculated via equation (5.8.7),
which is also the number of function evaluations of f(x). If a ... degree
of approximation (length of
c
to be used), this routine returns the array
cint[0 n-1]
,the
Chebyshev coefficients of the integral of the function whose coefficients are
c
....
Tài liệu Integration of Functions part 6 pptx
... numerical stability is greatly improved if,
instead of using powers of x as a set of basis functions to represent the p
j
’s, one uses some
other known set of orthogonal polynomials π
j
(x), say. Roughly ... Integrals
Integrals of functions of several variables, over regions with dimension greater
than one, are not easy. There are two reasons for this. First, the number of f...
Tài liệu Integration of Functions part 7 pptx
... Integrals
Integrals of functions of several variables, over regions with dimension greater
than one, are not easy. There are two reasons for this. First, the number of function
evaluations needed ... power
of the number needed to do a one-dimensional integral. If you need 30 function
evaluations to do a one-dimensional integral crudely, then you will likely need on
the order of 30...
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