我正在尝试计算Matlab和Mathematica中该软件无法象征性完成的功能的积分。
到目前为止,这是我的MatLab代码,但我了解它可能并没有太大帮助。
f = @(t) asin(0.5*sin(t));
a = @(t) sin(t);
F = int(f,t) % Matlab can't do this
F =
int(asin(sin(t)/2), t)
A = int(a,t) % This works
A =
-cos(t)
dt = 1/(N-1); % some small number
for i=1:N
F(i) = integral(f,(i-1)*dt,i*dt);
A(i) = integral(a,(i-1)*dt,i*dt);
end
for循环中的两个计算都给出了f
或a
的近似值,而不是它们与dt
相乘后的积分。
在数学堆栈交换上,我发现了一个question,它得出了一个有限差分,如积分方法。但是,当我在Matlab中进行计算时,它会输出f
的缩小版本,这在绘制后很明显(请参阅上文,我的意思是缩小)。我认为这是因为对于较小的时间间隔,积分基本上将函数近似于不同程度的精度(再次参见上文)。
我正在尝试获取积分的符号方程或每个位置上函数的积分的近似值。
因此,我的问题是如果我具有MatLab和Mathematica无法轻易采用的积分的功能
int
,integral
,trapz
)或
答案 0 :(得分:1)
您的代码几乎可以了
for i=1:N
F(i) = integral(f,0,i*dt);
end
您也可以这样做
F(1)=integral(f,0,dt)
for i=2:N
F(i) = F(i-1)+integral(f,(i-1)*dt,i*dt);
end
第二种选择肯定更有效
因为原语实际上是F(x)= int(f(x),0,x)(0定义了一个常数)并且对于足够小的dx,您已经证明f(x)= int(f(x ),x,x + dx)/ dx i。您已经证明MATLABintégral函数可以发挥作用。
答案 1 :(得分:0)
The accepted answer in general is by far the best method I would say but if certain restrictions on your functions are allowable then there is a second method.
For two functions f
and g
see below
T = 1; % Period
NT = 1; % Number of periods
dt = 0.01; % time interval
time = 0:dt:NT*T; % time
syms t
x = K*sin(2*pi*t+B); % edit as appropriate
% f = A/tanh(K)*tanh(K*sin(2*pi*t+p))
% g = A/asin(K)*asin(K*sin(2*pi*t+p))
formulas found here
f = A1/tanh(K1)*(2^(2*1)-1)*2^(2*1)*bernoulli(2*1)/factorial(2*1)*x^(2*1-1);
% |K1|<pi/2
g = A2/asin(K2)*factorial(2*0)/(2^(2*0)*factorial(0)^2*(2*0+1))*x^(2*0+1);
% |K2|<1
there are no such limitations in the accepted answer
N = 60;
for k=2:N
a1 = (2^(2*k)-1)*2^(2*k)*bernoulli(2*k)/factorial(2*k);
f = f + A1/tanh(K1)*a1*x^(2*k-1);
a2 = factorial(2*k)/(2^(2*k)*factorial(k)^2*(2*k+1));
g = g + A2/asin(K2)*a*x^(2*k+1);
end
MATLAB can calculate sin^n(t)
for n being an integer.
F = int(f,t);
phi = double(subs(F,t,time));
G = int(g,t);
psi = double(subs(G,t,time));