wolfram alpha operation in matlab
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fima v
on 12 Mar 2017
Commented: Walter Roberson
on 12 Mar 2017
how can i enter an expression and get the value of the variable that creates it?(like in wolfram alpha attached photo)
Thanks
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/161612/image.jpeg)
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More Answers (2)
Stephen23
on 12 Mar 2017
>> fun = @(x)(1+x)/(1-x)-1.08393;
>> fzero(fun,0)
ans =
0.0402748652785842
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John D'Errico
on 12 Mar 2017
Edited: John D'Errico
on 12 Mar 2017
It seems silly to throw a computer at something that is done using pencil and paper. I think sometimes we forget how to think, allowing the computer to do that for us. That can get to be a habit, and once we lose a skill, it won't be grown again easily.
k = (1+x)/(1-x)
First, look at the plot that is most easily drawn using ezplot:
ezplot(@(x) (1+x)./(1-x))
grid on
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/173225/image.jpeg)
Here we see that if we were to swap the axes, trying to produce x as function of k, we would still have a hyperbolic form.
Quick pencil and paper shows (as long as x is not 1, in which case there is no solution anyway):
x = (k-1)/(k+1)
Here we see that k may never be -1 either, in which case no solution could have existed since we would have needed to divide by zero. As I suggested, the result should again have a clearly hyperbolic form, which it does. Always check your results against your knowledge of the process. Any computation, whether done using paper or computer, should always be checked for sanity.
In MATLAB, for vector or array k, the solution is now trivial.
x_k = @(k) (k-1)./(k+1);
x_k(0:0.25:2)
ans =
-1 -0.6 -0.33333 -0.14286 0 0.11111 0.2 0.27273 0.33333
x_k(1.08393)
ans =
0.040275
format long g
x_k(1.08393)
ans =
0.0402748652785842
Note the use of the ./ operator.
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