Mathematica Real Part of Solve output

[trow4]

I am trying to solve 0.125 + 0.5 (1-x)^3 - (12.5/y)==0 for x, when y is real and y>0. I thus want to find x= 1- 0.63 ((100-y)/y)^(1/3), so that if y=100, x=1. Mathematica's Solve yields 3 roots:

sol=Solve[0.125 + 0.5 (1 - x)^3 - (12.5/y) == 0, x]

Root 1:
1.+((0.+0. I) y^(1/3))/(-100.+1. y)^(1/3)-((0.31498+0.545562 I) (-100.+1. y)^(1/3))/y^(1/3),

Root 2:
1.+((0.+0. I) y^(1/3))/(-100.+1. y)^(1/3)-((0.31498-0.545562 I) (-100.+1. y)^(1/3))/y^(1/3),

Root 3:
1.+(0. y^(1/3))/(-100.+1. y)^(1/3)+(0.629961 (-100.+1. y)^(1/3))/y^(1/3)



If I now evaluate root 1 at y=100, I get an Infy error because of division by zero. How can I drop the imaginary part of this function? I have tried (without succes) a bunch of things such as:

xroot1=x/.sol[[1]]

*ComplexExpand[xroot1]

*Assuming[p > 0, {Simplify[xroot1]}]

 

[phyzguy]

Try:

Limit[xroot1,y->100]

 

[trow4]

Marvelous, that indeed does the trick. However, if the polynomial is of a higher order (see below) I again run into trouble.

For instance, if I want to find the real root of 0.1 + 0.5 (1-x)^4 - (1/y)=0, (which you can easily solve for x=1- [ 2 (1-0.1 y ) / y ]^(1/4) ), I again use

sol = Solve[0.1 + 0.5 (1 - x)^4 - (1/y) == 0, x];

If I now plot the real part of these roots, none of them look like the one found above.

e.g. Plot[Re[(x /. sol[[1]])], {y, 0, 10}]

Is the solve algorithm not suited for solving equations of this order or am I doing something very very wrong?

EDIT1: It seems to be a Mathematica 7 problem. In Mathematica 8 the plots are just fine. Is this a known bug?

 

[phyzguy]

Interesting. I think what you are looking at is numerical round-off error. When you use a number like 0.1 in Mathematica, it only keeps a certain number of digits, I think 6. However, if you use the number 1/10, then it treats it as an exact mathematical quantity and there is no round-off error. If I use:

sol = Solve[1/10 + 1/2 (1 - x)^4 - (1/y) == 0, x]
Plot[Re[(x /. sol[[1]])], {y, 0, 10}]

Then everything looks fine. i am using Mathematica 7.0.1.

 

[alphy]

I would first analyze the problem and the provided Mathematica output to understand the issue at hand. In this case, the problem involves solving an equation for x when y is a real number and positive. The desired solution is x= 1- 0.63 ((100-y)/y)^(1/3), which is a real number for all values of y.

The Mathematica output provides three roots, which are complex numbers. This is because the equation has three solutions, and the complex numbers represent the real and imaginary parts of each solution. The error occurs when trying to evaluate the first root at y=100, as it involves division by zero.

To drop the imaginary part of the first root, we can use the Re function in Mathematica, which extracts the real part of a complex number. So, we can modify the code to:

xroot1 = Re[x /. sol[[1]]]

This will give us the real part of the first root, which can then be evaluated at y=100 without any errors. Similarly, we can use the Re function for the other two roots as well.

Another approach would be to use the Reduce function in Mathematica, which can handle complex numbers and provide a simplified solution. So, we can modify the code to:

sol = Reduce[0.125 + 0.5 (1 - x)^3 - (12.5/y) == 0, x]

This will give us a simplified solution without any complex numbers. We can then use the x= 1- 0.63 ((100-y)/y)^(1/3) equation to find the corresponding values of x for different values of y.

In conclusion, as a scientist, it is important to analyze and understand the output of any software or tool used in research. In this case, we were able to identify the issue and find a solution by utilizing the functions available in Mathematica.