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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?

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  • $\begingroup$ Why do you ask those questions? $\endgroup$ Feb 27, 2014 at 17:30
  • $\begingroup$ @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 $1-\log\frac32\approx 0.595$. $\endgroup$ Feb 27, 2014 at 17:36
  • $\begingroup$ @HenningMakholm: I goofed. I misread the question. Need an eye check... $\endgroup$
    – copper.hat
    Feb 27, 2014 at 17:40
  • $\begingroup$ Related discussion of the same problem: math.stackexchange.com/q/696128/115115 $\endgroup$ Mar 2, 2014 at 17:32

2 Answers 2

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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 $φ$.

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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 $f(c)=0$.

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  • $\begingroup$ I should say I've only taken Real and I may not even be able to tell how far over my head this is. We didn't cover Taylor series in my Real class. Generally speaking, if it's homework, use the most recently covered topics to as a guide to the likely best methods of proving it. $\endgroup$
    – Tyler
    Feb 27, 2014 at 17:59

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