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.

Pick out the true statements:

a. If $P$ is a polynomial in one variable with real coefficients which has all its roots real, then its derivative $P'$ has all its roots real as well.

b. The equation $\cos(\sin x) = x$ has exactly one solution in the interval $[0,\pi/2]$.

c. $\cos x > 1 −x^2/2$ for all $x > 0$.

My thoughts:

By Rolles theorem, (a) must be true, but I am not sure about the others.

share|improve this question
add comment

3 Answers 3

(a) Rolles Theorem, like you mentioned.

(b) Consider $f(x) = \cos (\sin x) - x$, then show that $f'(x) = \sin (\sin x) \sin x - 1 < 0$ (Hint: $\sin x < x$ in that range)

(c) The MacLaurin's Expansion of $\cos x$ is ... Use the remainder theorem.

share|improve this answer
add comment

a) Rolle's Theorem is indeed useful here. It works very nicely if all the roots of $P(x)$ have multiplicity $1$. But we need also to deal with polynomials $P(x)$ like $x^3(x-1)^2 (x-2)$. The additional result that we need is that if $(x-a)^k$ divides $P(x)$, where $k\ge 2$, then $(x-a)^{k-1}$ divides $P'(x)$. This is easy to prove. Let $P(x)=(x-a)^kQ(x)$, and use the Product Rule.

Suppose $Q(x)$ has degree $d\ge 1$. Now you will be able to show that all of the $d-1$ complex roots of $P'(x)=0$ are real.

b) One standard approach to a question like this one is to let $f(x)=\cos(\sin x)-x$. Note that $f(0)$ is positive and $f(1)$ is negative, so by the Intermediate Value Theorem there is a root in between. Note that $f'(x)=(\cos x)(-\sin(\cos x))-1$, so $f$ is decreasing steadily on our interval.

c) Use the same basic strategy as in b). Let $f(x)=\cos x-(1-\frac{x^2}{2})$. We have $f(0)=0$. It is enough to show that $f$ is increasing on the interval $(0,\infty)$. We have $f'(x)=-\sin x+x$.

To show that this is $\gt 0$ on $(0,\infty)$, use the same basic strategy. Let $g(x)=x-\sin x$. Then $g(0)=0$ and $g'(x)=1-\cos x$.

share|improve this answer
add comment

HINTS:

b. Try graphing both functions to get an idea of the answer. To prove that it's correct: what does the derivative of $\cos(\sin(x))$ look like on $[0, \pi /2]$?

c. Again, a graph can do wonders. Do you know the power series expansion $\cos(x)=1-x^2/2+x^4/4!-\dots$ ? That can help quite a bit in confirming your answer.

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.