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Question :

The equation $x^7+x^5+x^3+1=0$ is given. Prove that this equation has only one solution on $\mathbb R$.

Note 1 : You should use Rolle's theorem in your proof. That's what the question wants ! its not optional !!!

Note 2 : The question is taken from "Real Analysis : A first course " : Russel Gordon

Note 2 : I don't have any idea... I've never proved something like that ! Any help would be great !

Arman Malekzadeh
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  • Imagine a graph of the function - that would take some time to plot I guess, but can you work out anything about the gradient that would help you imagine what the graph might look like. Rolle's theorem is a clue perhaps - what is it that it says again? – Cato Dec 08 '16 at 13:07
  • Rolle's theorem would be quite pointless here. Intermediate value theorem would be useful, though. –  Dec 08 '16 at 13:08
  • The number of change-sign of $f(x)=0$ and $f(-x)=0$ is $0$ and $1$ respectively. Then $7-(0+1)=6$. Hence $f(x)=0$ has at least $6$ non real roots. The other root is necessarily real because the degree is odd. – Piquito Dec 08 '16 at 13:43

3 Answers3

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If Rolle's theorem must be used, then this proof should do it.

First, use the fact that $x^7+x^5+x^3+1$ is and odd polynomial. What do we we know about complex roots of a polynomial? They come in pairs, right? So how many complex roots can $x^7+x^5+x^3+1$ have? Clearly, either zero, or two or four or six. Hence, the other roots are real, so $f$ has any one of one,three, five or seven roots. But we are assuming that there is more than one root, so there must be at least three roots.

Suppose that $f(x)=f(y)=f(z)=0$ for $x< y<z$. This implies by Rolle's theorem that there is some $c \in (x,y)$ and some $d \in (y,z)$ such that $f'(c)=f'(d)=0$, and $c<d$. We can compute $f'(x)$, it is $f'(x)= 7x^6 + 6x^4+3x^2$. We can factorize this as $$f'(x) = x^2(7x^4+6x^2+3) = (x^2)\left(7\left(x^2+\frac 37\right)^2 +\frac {12}7\right) = 0$$.

As you can see in the factorization, the term $\left(7\left(x^2+\frac 37\right)^2 +\frac {12}7\right)$ can never be zero, hence it is only possible that $x^2=0$ i.e. $x=0$. But then this means that $c=d=0$, contradicting the fact that they are different.

There are easier proofs, given by factorization IVT etc. but it is always good to see different proofs of the same thing.

Sarvesh Ravichandran Iyer
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Hint. The derivative of the polynomial $P(x)=x^7+x^5+x^3+1$ is always non-negative: $$P'(x)=x^2(7x^4+5x^2+3)\geq 0.$$ Moreover $P$ is a continuous function and $\lim_{\pm\infty}P(x)=\pm\infty$.

What may we conclude?

Robert Z
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Hint: The derivative of $f(x) = x^7 + x^5 + x^3 + 1$ is nonnegative and has only one root in $x = 0$. Use this and the fact that $f(x) > 0$ for all $x \ge 0$.

Edit: To clarify:

We have $f(0) = 1$ and $f(-1) = -2$, so by the intermediate value theorem $f$ has a root between $-1$ and $0$.

Assume $f$ has two different zeroes, $x_1$, $x_2$; Wlog $x_1 < x_2$. Because $f(x) \ge 1 > 0$ for all $x > 0$, both roots must be negative. Now Rolle's theorem implies that there is a $\xi \in (x_1, x_2)$ with $f'(\xi) = 0$.

But $f'(x) = 7x^6 + 5x^4 + 3x^2 = (7x^4 + 5x^2 + 1)x^2$ only has one root in $x = 0$. This is a contradiction to $f'(\xi) = 0$, since $\xi < x_2 < 0$.

Dominik
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