# Mean Value Theorem related proof

For what real values does the polynomial

φ (x):= 1+x+x^2+...+x^(2m-1)

take the value 0? What can you say about the sign of φ(x) varies?

Prove that the function

f(x):= 1+x+x^2/2+...+(x^n)/n

has no real roots when n is even. What can you say about the roots of f when n is odd?

• Why do you ask those questions? Commented Feb 27, 2014 at 17:30
• @copper.hat: Obviously? These polynomials are the Taylor approximants to $1-\log(1-x)$. So, for example, their value at $x=-\frac12$ ought to converge to $1-\log\frac32\approx 0.595$. Commented Feb 27, 2014 at 17:36
• @HenningMakholm: I goofed. I misread the question. Need an eye check... Commented Feb 27, 2014 at 17:40
• Related discussion of the same problem: math.stackexchange.com/q/696128/115115 Commented Mar 2, 2014 at 17:32

Note that the derivative of $$φ (x):= 1+x+\frac{x^2}2+...+\frac{x^n}n$$ is $$φ' (x):= 1+x+x^2+...+x^{n-1}=\frac{x^n-1}{x-1}$$ with well known root set. This allows to find the regions of monotonicity of $φ$.
Are you sure it's MVT? Using Intermediate Value Theorem, you can say that if $f$ changes sign in the interval $(a,b)$, that is, if $f(a) < 0$ and $f(b) > 0$ or vice versa, then there exists $c \in (a,b)$ such that $f(c)=0$.