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is there any method to constrain the fitting according to this using csapi? no. the spline is uniquely determined by your (x,y)...
9 hours 前 | 0
by a "linear approximating function" you mean a function l with l(x1,x2,x3) = a b*x1 c*x2 d*x3 where a, b, c and d are ...
11 hours 前 | 0
adjust the sizes of the arrays involved in the calculation of "basestationthroughputs" (see above). further, use "fcn2optimexpr...
15 hours 前 | 0
this function sets adiabatic boundary conditions at both ends. function [pl,ql,pr,qr] = heatbc(xl,ul,xr,ur,t) pl = 0;%ul-293...
16 hours 前 | 0
my guess is that the x below is one of many other solutions. a = [1 1 1 1 1 1;-0.52 -0.52 0.3 0.3 0.64 0.64;0.94 -0.74 0.94 -0....
16 hours 前 | 0
i already showed you how to do that. if xvec and tvec are your given vectors of grid points and associated temperatures, set ...
1 day 前 | 1
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as you can see above, you get the same plot with a "professional" matlab solver (ode45). so i guess there is something wrong wit...
2 days 前 | 0
d = 1.5e-7; % diffusivity r = 0.05; % radius ca0 = 0.2; % initial conc. cas = 0.0; ca_target = 0.1*ca0; % target conc. n = ...
3 days 前 | 0
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your function is not vectorized. thus you have to use integral_value = integral(f, 0, inf, 'arrayvalued',true) instead of int...
3 days 前 | 1
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%for plot clear clc f='diesel';%input('gasoline or diesel: ','s'); c=6;%input('number of engine cylinders '); t='yes';%inpu...
4 days 前 | 0
then my guess is that some of the x-values (maybe starting around 60) are nan values. remove them before applying "cumtrapz".
4 days 前 | 2
because i got the task to do it like that and not to use: data(:,i 1) = data(:,i) * p. then use for i = 2:l
5 days 前 | 0
you will have to reorder the "output" array, but i think the code should be faster. my_fun = @(t,param) t.^2; t = 2; omega = ...
5 days 前 | 0
as you correctly write in your code, the line x_best_so_far = x_best_so_far(1); % initialize x_best_so_far as a scalar sets x_...
6 days 前 | 0
n = 2000; dy = 1/n; y = linspace(0,1,n 1).'; ml2v = @(y)1000*(y-0.5).^2; e = -1/(2*dy^2)*ones(n-1,1); d = 1/dy^2 ml2v(y(2...
6 days 前 | 1
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use "gradient" three times to get approximations for y', y'' and y''' in the points of your t-vector. let diff_y, diff_y2, diff...
6 days 前 | 0
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t1 = [1 6]; y1 = [1 1]; t2 = [6 7]; y2 = [0.45 0.45]; t3 = [7 8]; y3 = [1 1]; t4 = [8 9]; y4 = [0.4 0.4]; t5 = [9 11]; y5 = ...
7 days 前 | 1
t = [0:0.2:3].'; y = [3.43101641262231;-0.527550212558249;3.43697613505530;0.186721377613475;5.11301585254667; 6.786264197...
7 days 前 | 0
you can't use t(k), t(k 1), t(k 2), y(k), y(k 1) and y(k 2) to define y(k). the values for these y's and t's are not yet known. ...
7 days 前 | 0
i substituted 1/sqrt(x) by y in your equation. as you can see, there are only negative solutions such that resubstituting gives ...
7 days 前 | 0
why do you think the red line is not a curve ? for this point cloud of experimental data, i think the red line is the best you ...
7 days 前 | 1
a = [11 0.001 3 11 0.001 4 12 0.003 5 9 0.002 6 8 0.000 7 10 0.004 8 8 0.000 9 9 0...
7 days 前 | 0
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your equations are not suited to be solved with pdepe. both contain no second derivatives in space which is necessary for an eq...
8 days 前 | 0
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use plot(1:365,[pr;ones(1,365)*pr_ave]); instead of plot(pr,pr_ave); if pr is a row vector, or plot((1:365).',[pr,ones(365...
8 days 前 | 0
did you read somewhere that "vpasolve" can handle inequalities ? anyway, it seems the problem is too hard for a symbolic soluti...
8 days 前 | 0
t=[0 5]; initz=[0; 0]; [t,z]=ode15s(@f3, t, initz, odeset('reltol',1e-12,'abstol',1e-12)); for i=1:numel(t) dz = f3(t(i),z...
8 days 前 | 0
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piecewise1111copy() function piecewise1111copy x1=1; u=3; teta=zeros(3,1); teta(1)=0; for i=1:3 teta(i 1)=2*i; end ...
8 days 前 | 0
works for me (see above).
9 days 前 | 1
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explicitly listing the sizes of the matrices involved (see above), i come to the conclusion that yd(t) must be a 4x1 column vect...
9 days 前 | 1
try a solution with c, k, m, x0 and v0 being symbolic variables. then you only need to "subs" the numerical values for the para...
9 days 前 | 0
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