how to use fzero function?
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Hello~
I am trying to use fzero function and I am confused about the usage of the function
If I write code like following, it works
syms x
fzero('x.^2-2*x-5',2)
However, if I write code like below, it does not work
syms x
eqn=x.^2-2*x-5;
fzero('eqn', 2) or fzero(eqn,2)
Can anyone tell me the difference between two cases?
thank you
采纳的回答
Walter Roberson
2021-4-7
syms x
fzero('x.^2-2*x-5',2)
The syms x is not having any effect there. What is happening is that you are encountering an undocumented syntax that creates an inline() object for some kinds of character vectors.
syms x
eqn=x.^2-2*x-5;
The syms x is having an effect there, so eqn is created as a symbolic expression.
fzero('eqn', 2)
That undocumented syntax... when you happen to pass in a character vector that looks like an identifier, the internal code turns it into a function handle; for example fzero('cos', 2) would turn into fzero(@cos,2) effectively. But there is no visible function named eqn so that will fail.
fzero(eqn,2)
fzero is not defined for symbolic expressions.
You need to choose between proceeding numerically or symbolically. Numerically you would use
eqn = @(x) x.^2-2*x-5
fzero(eqn,2)
and symbolically you would use
syms x
eqn=x.^2-2*x-5;
vpasolve(eqn,2)
or
syms x
eqn=x.^2-2*x-5;
solve(eqn)
Both of those symbolic processes will return two roots for a polynomial of degree 2.
There is no exact equivalent to fzero() for symbolic work.
更多回答（1 个）
David Hill
2021-4-7
First, fzero finds roots of nonlinear functions and should not be symbolic. If you have a polynomial you should use roots() function. If you have a symbolic use solve() or vpasolve() functions. The documentation for fzero expects a function input that can handle an array of values. The easiest way to satisfy is with an anonymous function as input.
eqn=@(x)x.^2-2*x-5;
fzero(eqn,2);
But for a simple polynomial, better to use roots()
roots([1 -2 -5]);
2 个评论
Duckyoon Go
2021-4-7
Walter Roberson
2021-4-8
fzero() can be useful for polynomials if you want to find a root in a particular range.
... However, it might well be the case that it would be faster to use roots() and filter the results.
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