Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on a optimization problem and need to show that \begin{equation} \frac{1}{2}x^4 - x^3 -x + 100 = 0 \end{equation} has no real solution in order to prove certain properties about the function I am working on. I don't need to know the zeros just show that they don't exist. I'm doing this because I would like to show that \begin{equation} \frac{1}{2}x^4 - x^3 > x - 100 \end{equation} for all $x$. What would be the best way to go about this?

Thanks in advance.

share|improve this question
    
see math.stackexchange.com/questions/494243/… there I use Sturm's theorem to show this. This method is straightforward and is implemented in some CAS (e.g. Maxima) to find the numbers of real roots in an interval –  miracle173 Feb 3 at 0:02
add comment

5 Answers 5

up vote 5 down vote accepted

We can do it by looking at the expression for different ranges of $x$. It is clearly positive if $x\le 0$.

If $0 < x \le 4$, then $x^3+x\lt 100$, so our expression is positive even without the aid of $\frac{1}{2}x^4$. If $x\gt 4$, then $\frac{1}{2}x^4\gt 2x^3$, so $\frac{1}{2}x^4-x^3-x\gt 0$.

share|improve this answer
    
Thank you. Much appreciated. –  user245465 Feb 2 at 18:27
    
You are welcome. –  André Nicolas Feb 2 at 18:31
add comment

This is a quartic, its discriminant tells all you need.

share|improve this answer
add comment

For negative $x$, all summands are $\ge 0$ so $f(x)\ge 100$. For $0\le x \le 4$, $f(x)\ge 100-x^3-x\ge 100-64-4>0$. Note that $$ f(x+1)=\frac12x^4+x^3-2x+\frac{197}2\ge (x^2-2)x+\frac{197}2>0$$ for $x>\sqrt 2$, so $f(x)>0$ for $x>\sqrt 2-1$. This covers all of $\mathbb R$.

share|improve this answer
add comment

Find the derivative ov the left side of your first equation and use it's zeros to find the minima of that function. They will, hopefully, all be positive :)

share|improve this answer
add comment

Sketch $\frac{1}{2}x^4 - x^3 -x + 100$. There are two changes in concavity between $x=-2$ and $x=3$, so the graph can't "change direction" outside those values, and the minimum must be between them. Direct calculation shows that within that interval, the function exceeds 50 at each integer-valued $x$.

For there to be a zero within $0.5$ units of any of these integer points, the function would have to have an average rate of change of more than $50$ on an interval of length less than $0.5$, and by the mean value theorem, the absolute value of the derivative would have to exceed $\frac{50}{0.5}=100$ somewhere. It doesn't. The absolute value of the derivative in that interval is easily bounded above by $2|3|^3+3|3|^2+1=82$ (the sum of the absolute values of the terms of the derivative at the largest $|x|$-value.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.