A limit problem. (Is something wrong with matlab?)

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SUMMARY

The forum discussion centers on the discrepancies encountered when using MATLAB R2007a to compute limits involving hyperbolic sine functions. The user reports that while the limit of the difference quotient for sin(z) correctly yields cos(z), the analogous calculation for sinh(z) returns an unexpected result of limit((sinh(z+h)-sinh(z))/h,h=0). This issue arises from MATLAB's handling of the limit function, which leads to a NaN output. In contrast, Maxima provides the correct derivative, cosh(z), highlighting a potential limitation in MATLAB's limit algorithm.

PREREQUISITES
  • Familiarity with MATLAB R2007a syntax and functions
  • Understanding of calculus concepts, specifically limits and derivatives
  • Knowledge of hyperbolic functions, particularly sinh and cosh
  • Experience with symbolic computation in MATLAB using the 'syms' command
NEXT STEPS
  • Investigate MATLAB's limit function and its algorithmic limitations
  • Explore the differences in symbolic computation between MATLAB and Maxima
  • Learn how to derive hyperbolic functions using the definition of sinh(x)
  • Examine MATLAB's symbolic toolbox for potential updates or alternatives
USEFUL FOR

This discussion is beneficial for students and educators in mathematics, particularly those studying calculus and symbolic computation. It is also relevant for MATLAB users seeking to understand the nuances of limit calculations involving hyperbolic functions.

oahsen
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With the following code segment ;
limit((sin(z+h)-sin(z))/h,h,0)
matlab gives the result cos(z). (which is the differential of the sin(z)).
However, with the code ;
limit((sinh(z+h)-sinh(z))/h,h,0)
matlab gives an absurd result;
limit((sinh(z+h)-sinh(z))/h,h = 0)
newertheless, the result should be cosh(z). Why does MATLAB makes such a wrong calculation also what is the meaning of this result? Does anybody have an idea about that?
 
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I tried the same code, and i got:

ans=
NaN

I don't know why this answer is given. However, if you are looking for the derivative of a function, try this line:

> syms x; diff(sinh(x))
ans =
cosh(x)
 
Just for the record, Maxima does it right.

(%i1) limit((sinh(x+h)-sinh(x))/h,h,0);

gives exp(-x)*(exp(2x)+1)/2;
 
atqamar said:
I tried the same code, and i got:

ans=
NaN

I don't know why this answer is given. However, if you are looking for the derivative of a function, try this line:

> syms x; diff(sinh(x))
ans =
cosh(x)

actually this is the problem. The teacher asks us to find the derivative with;
syms x; diff(sinh(x)) and after that finding it with;
syms z h; limit((sinh(z+h)-sinh(z))/h,h,0)

and he wants us to comment on the difference.

With my matlab(R2007a) I find the first result as cosh(z) as expected however with the second code it gives me ;
limit((sinh(z+h)-sinh(z))/h,h = 0).

Some of my friend said that Maxima finds the correct result. I wonder if it is something about the limit algorithm of Matlab or I do not know? What kind of comment can I write for the difference?
 
I'm wondering if you didn't accidently enter a "=" instead of "," and MatLab just echoed it back to you.
 
HallsofIvy said:
I'm wondering if you didn't accidently enter a "=" instead of "," and MatLab just echoed it back to you.
I checked it 1000 times. This is the copy of the command window;

>> syms z h;
>> limit((sinh(z+h)-sinh(z))/h,h,0)

ans =

limit((sinh(z+h)-sinh(z))/h,h = 0)
 
If limits have to be used to find the derivative, use the exponents definition of sinh(x) and find the limit:

>> syms z h; limit((((exp(z+h)-exp(-z-h))/2)-((exp(z)-exp(-z))/2))/h,h,0)
ans =
(1/2*exp(2*z)+1/2)*exp(-z)

The answer, when simplified, is cosh(x).
 

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