is it really necessary to say that you have to define that as piecewise(x == 0, undefined, x^2 + 2*x + 1) or can you just define it as x^2 + 2*x + 1 ? If you are writing the derivative of c*x^n for integer n, is it really necessary to define it as piecewise(n == 1 & x == 0, undefined, (c*n)*x^(n-1)) or as piecewise(n == 1 & x == 0, c, (c*n)*x^(n-1)) ? Is the derivative of c*x to be defined differently than the derivative of c*x^1, or do we just have to carry around a whole bunch of special cases to protect against the possibility of evaluating 0^0 when the relevant formula would work fine for any other x^0 ?How to fix this problem 0^0 in Matlab !?
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It is crucial to understand that this expression could be used in problems related to engineering, physics, mathematics, or any other aspect of real life.
Typically, Matlab is used to solve PDE and ODE problems. Perhaps users calculated this term 0^0 incorrectly in the process.

>> % Reviewed by Bewar Yousif Ali
>> % How to fix this problem 0^0 in Matlab !?
>> % Mathematically, x^0=1 if x≠0 is equal 1 else undefined(NaN)
>> 0^0
ans =
1
>> f=@(x,y) x^y;
>> f(0,0)
ans =
1
>> v=[2 0 5 -1];
>> v.^0
ans =
1 1 1 1
1 Kommentar
Walter Roberson
am 11 Dez. 2023
There are a lot of contexts in which
is considered well defined.
For example, if you have
is it really necessary to say that you have to define that as piecewise(x == 0, undefined, x^2 + 2*x + 1) or can you just define it as x^2 + 2*x + 1 ? If you are writing the derivative of c*x^n for integer n, is it really necessary to define it as piecewise(n == 1 & x == 0, undefined, (c*n)*x^(n-1)) or as piecewise(n == 1 & x == 0, c, (c*n)*x^(n-1)) ? Is the derivative of c*x to be defined differently than the derivative of c*x^1, or do we just have to carry around a whole bunch of special cases to protect against the possibility of evaluating 0^0 when the relevant formula would work fine for any other x^0 ?
is it really necessary to say that you have to define that as piecewise(x == 0, undefined, x^2 + 2*x + 1) or can you just define it as x^2 + 2*x + 1 ? If you are writing the derivative of c*x^n for integer n, is it really necessary to define it as piecewise(n == 1 & x == 0, undefined, (c*n)*x^(n-1)) or as piecewise(n == 1 & x == 0, c, (c*n)*x^(n-1)) ? Is the derivative of c*x to be defined differently than the derivative of c*x^1, or do we just have to carry around a whole bunch of special cases to protect against the possibility of evaluating 0^0 when the relevant formula would work fine for any other x^0 ?Antworten (0)
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