Simple question about symbolic limits

Question

Tristan Potter 2020-6-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/545183-simple-question-about-symbolic-limits

编辑： Tristan Potter 2020-6-13

Hello,

I am checking results derived by hand in a MATLAB (2019a) live script and have encountered the following problem (MWE below): When I try to take the limit of the expression

as (symbolic) variable l approaches 0, restricting

, MATLAB cannot find the limit I obtain by hand. However, when I choose arbitrary values for

instead of using an assumption, I get the same limit I derived by hand (which is independent of c). I suspect there is an obvious explanation for this that I am overlooking.

MWE:

%Declare symbolic variables:
syms l real;    
syms t real;    assumeAlso(t>1);
syms a real;    assumeAlso(0<a<1);
syms p real;    assumeAlso(0<p<1);
syms e real;    assumeAlso(e>0);        
syms c real;    assumeAlso(c>0);        
syms w real;    assumeAlso(0<w<1);
%Functions:
phi = e/(1+c)*l^(1+c);                                                         
z   = t/l*(1-exp(-l/t));                                                     
n   = p*(1+c)/(p*(1+c)+z*l/phi*w);                                                                 
pn  = t*p*(1-n)*exp(-l/t)/n*(1-exp(-1*(t*p*(1-n)*exp(-l/t)/(a*n))^(-1))); 
%Limits:
limit(pn,l,0,'right')

MATLAB cannot reduce this limit. However, when I instead impose, e.g.,

, via

syms l real;    
syms t real;    assumeAlso(t>1);
syms a real;    assumeAlso(0<a<1);
syms p real;    assumeAlso(0<p<1);
syms e real;    assumeAlso(e>0);        
syms c real;    assumeAlso(c>0);        
syms w real;    assumeAlso(0<w<1);
%Functions:
phi = e/(1+0.1)*l^(1+0.1);                                                         
z   = t/l*(1-exp(-l/t));                                                     
n   = p*(1+0.1)/(p*(1+0.1)+z*l/phi*w);                                                                 
pn  = t*p*(1-n)*exp(-l/t)/n*(1-exp(-1*(t*p*(1-n)*exp(-l/t)/(a*n))^(-1))); 
%Limits:
limit(pn,l,0,'right')

I get the result I derived by hand. The same holds for seemingly all other positive values of c. Any idea what I'm overlooking?

Thanks in advance.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Ayush Gupta 2020-6-12

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/545183-simple-question-about-symbolic-limits#answer_450069

The limit specified is also dependent on c if c does not have small values. For small values of c, a simplified expression is yielded. Refer to the following link to know more about limit:

https://www.mathworks.com/help/symbolic/limit.html

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Tristan Potter 2020-6-13

编辑：Tristan Potter 2020-6-13

在 MATLAB Online 中打开

Thanks for your response. However, I don't see that this resolves the issue for several reasons:

1) I don't see the dependence on c in a manual derivation of the limit via L'Hopital's rule (the limit I obtain manually corresponds to what MATLAB gives when I use a particular value of c, so I don't think I'm making a mistake in the manual derivation).

2) If I substitute the expression for n into the expression for

above, I obtain:

pn = (t^2*(w-b)/e*exp(-l/t)*(1-exp(-l/t))/l^(c+1)*exp(-a/t*exp(-l*z)/exp(-l/t))*(1-exp(-a*(t^2*(w-b)/e*exp(-l/t)*(1-exp(-l/t))/l^(c+1))^(-1)));

When I evaluate this limit, seemingly any value of c gives the same limit that I get when I do the derivation manually. For example, I can choose

and I get the same answer.

3) Finally, I tried writing

where

and applying L'Hopital's rule manually by taking the derivative of the numerator and denominator. I then used MATLAB to take the limit of the resulting ratio of the derivatives. When I simplify the ratio of derivatives and take the limit in MATLAB, I get the same result that I get manually. When I don't simplify, the problem remains:

n0 = 1-exp(-a*(t^2*(w-b)/e*exp(-l/t)*(1-exp(-l/t))/l^(c+1))^(-1));
d0 = 1/(t^2*(w-b)/e*exp(-l/t)*(1-exp(-l/t))/l^(c+1));
n1 = diff(n0,l);
d1 = diff(d0,l);
limit(n1/d1,l,0,'right')
limit(simplify(n1/d1),l,0,'right')

Am I missing something?

请先登录，再进行评论。