使用倍频程求解方程 [英] solving equation using octave

问题描述

我想解决一个简单的方程式

I have a simple equation I'm trying to solve

num1=-2
num2=-3

x+num2=num1
x+-3=-2
x=1

如何以八度为单位进行此操作.在matlab中，我可以执行 y = solve('x-3 = -2')，但这在我正在使用的版本的 octave 3.8.1中不起作用 >.如何获得八度来求解这些类型的方程式?

How can I do this in octave. In matlab I can do y = solve('x-3 = -2') but this doesn't work in octave 3.8.1 which is the version I'm using. How can I get octave to solve these types of equations?

我对解决方案的数值很感兴趣.

I'm interested in the numeric value for a solution.

推荐答案

我假设您问题中的方程式是一个示例.如果您对数值解法感兴趣，通常不需要使用符号数学.在八度(或Matlab)中，您可以使用 fzero 查找用单变量自由变量表示的非线性方程的实根/零.对于简单的线性示例，使用匿名函数来代表你的方程式:

I'm assuming that the equation in your question is an example. If you're interested in a numeric solution, there is often no need to use symbolic math. In Octave (or Matlab), you can can use fzero to find a real root/zero of a nonlinear equation in terms of a single-variable free variable. For your simple linear example, using an anonymous function to represent your equation:

num1 = -2;
num2 = -3;
f = @(x)x+num2-num1;
x0 = 0; % Initial guess for x
x = fzero(f,x0)

如果方程式具有多个根/零，则需要在每个零附近尝试不同的初始猜测，以找到确切的值.

If an equation has multiple roots/zeros you'll need to try different initial guesses in the vicinity of each zero to find the exact value.

Octave还具有Matlab的 fsolve 求解多变量非线性方程组.如果您的方程是线性的(例如A*x = b)，则应查看 linsolve .

Octave also has a version of Matlab's fsolve to solve systems of nonlinear equations in multiple variables. If your equations are linear (e.g., A*x = b), you should look at linsolve.

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