Symbolic limit does not perform the calculation

6 ビュー (過去 30 日間)
John
John 2020 年 4 月 14 日
コメント済み: Walter Roberson 2020 年 4 月 14 日
I'm trying to calculate a symbolic limit but, somehow, the calculation is not being performed...it returns the same function. I haven't been able to figure out why that is happening. Can anyone spot the reason why?
clear all;
clc;
syms n lambda
assumeAlso(n,'integer')
assumeAlso(n>=3)
assumeAlso(lambda,'real')
assumeAlso(lambda<=1)
assumeAlso(lambda>=0)
r = sqrt(4-(1+lambda)^2)
a = (2-r)/(1+lambda)
b = 2/(r*(1-a^n))
l11 = 2*b*(a+a^(n-1))
liml11 = limit(l11,lambda,1,'left')

採用された回答

Walter Roberson
Walter Roberson 2020 年 4 月 14 日
編集済み: Walter Roberson 2020 年 4 月 14 日
[N,D] = numden(l11);
limit(D,lambda,1,'left')
simplify(expand(limit(N,lambda,1,'left')))
The denominator goes to 0 (but possibly at a different speed than the numerator). We are therefore risking an infinite or nan value.
The numerator goes to 32*2^n .
With n being an integer, 2^n is positive, so -32*2^n is negative, so the numerator is negative.
The limit is therefore positive/zero which would be +infinity
However, really you need to examine the rate at which both parts go to zero. There is a chance that the infinity should be negative. Maple thinks the limit should be signum(-4/0)*infinity which would get you -infinity .
  1 件のコメント
Walter Roberson
Walter Roberson 2020 年 4 月 14 日
I have some typos about negative in the above that I will correct when I get back to my desk.

サインインしてコメントする。

その他の回答 (0 件)

製品


リリース

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by