How does one use integral2 (double integral) symbolically?
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I would like to get some help on how to use the integral2 symbolically?
E.g. I have a Joint PDF function -> "@(x,y) exp(-(x+y))"
To verify that it adds up to one, I did the following -> "integral2(pdf,0,inf,0,inf)" with the result being 1 as expected.
Now to calculate the Joint CDF symbolically I defined x and y as symbols and tried to do -> "cdf=@(u,v) exp(-(u+v))" and "integral2(cdf,0,x,0,y)". The error I got was -> "Error using integral2. XMAX must be a floating point scalar."
What I want to get is -> "(1-exp(-y))*(1-exp(-x))" !!
Your help will be much appreciated.
1 件のコメント
Dipak kumar Panigrahy
2020 年 5 月 21 日
clear
clc
syms rtstar
syms rt1 v delta R do %or assign numeric values to the variables
p=10;
cdf1=p*tan(delta)*R^2;
cdf2=p*tan(delta)*R^2;
eqn = ss == (int(int(cdf1,rt,rt1,rtstar),e,0,v)-int(int(cdf2,rt,rtstar,rt2),e,0,v));
sol = solve(eqn, rtstar)
can ayone tell me how to solve this eqn?
採用された回答
Shashank Prasanna
2013 年 8 月 22 日
From the documentation of integral2:
integral2
Numerically evaluate double integral
It is not meant for symbolic computation. If you are interested in symbolic results you need to have Symbolic Toolbox installed
Then you can use the int function:
3 件のコメント
Shashank Prasanna
2013 年 8 月 23 日
編集済み: Shashank Prasanna
2013 年 8 月 23 日
Sir, you have to use functions from the Symbolic Toolbox.
>> syms u v
>> cdf = exp(-(u+v))
>> int(int(cdf,v,0,v),u,0,u)
その他の回答 (1 件)
Okae
2024 年 5 月 14 日
編集済み: Walter Roberson
2024 年 5 月 14 日
how do i type j=\iint\of V_s\ [f\left(\varepsilon_{\left(\left(p+q\right)\right)}-f\left(\varepsilon_p\right)\delta\left(\varepsilon_{\left(\left(p+q\right)\right)}-\varepsilon_p-\hbar\omega q\right)\right]dp_{\left(x\right)d}p_y

in symbolic math toolbox?
0 件のコメント
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