フィルターのクリア

How can I get a better simplification of my results

3 ビュー (過去 30 日間)
El Vapor
El Vapor 2020 年 6 月 24 日
コメント済み: Walter Roberson 2023 年 12 月 23 日
I have been working on a program that gives me some results as a function of "k" and "m" (symbolic variables). However, this result is shown as this:
When, it is much simpler, the solution is just this:
How can I manage to make Matlab give me the result in a more digestible way? (I've already tried "simplify")
Thank you very much guys

採用された回答

Takumi
Takumi 2020 年 6 月 24 日
syms k m
s = [-(3*10^(1/2)*(k*m)^(1/2))/(10*m);
(3*10^(1/2)*(k*m)^(1/2))/(10*m)];
assume([m,k],'positive');
s_sign = sign(s); % signum
assume([m,k],'clear') % reset assume
S = s.^2;
S = simplify(S);
s = s_sign.*sqrt(S)
s = 
  2 件のコメント
Ahmet Anil
Ahmet Anil 2023 年 12 月 23 日
Whats your purpose at that lists?
assume([m,k],'positive');
s_sign = sign(s); % signum
assume([m,k],'clear')
Walter Roberson
Walter Roberson 2023 年 12 月 23 日
syms k m
s = [-(3*10^(1/2)*(k*m)^(1/2))/(10*m);
(3*10^(1/2)*(k*m)^(1/2))/(10*m)];
s_sign = sign(s); % signum
S = s.^2;
S = simplify(S);
s = s_sign.*sqrt(S)
s = 
So the assume() is there to have the sign() vanish

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeNumbers and Precision についてさらに検索

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