AoPS Academy
to 
have the digit 
and are divisible by 
?
\end{table}
be a three times differentiable function (defined on 
and real-valued) such that has at least five distinct real zeros. Prove that 
has at least two distinct real zeros.
. The problem asks to show having five distinct real roots implies 
has at least two. But by Rolle's Theorem on ![\[ g(x) = e^{\frac12 x} f(x) \implies 2g'(x) = e^{\frac12 x}(f(x) + 2f'(x)) \]](http://latex.artofproblemsolving.com/1/6/9/1693c667210ed467a5a36ed18f1847021dcdacc7.png)
we see that having 
roots gives 
having 
roots, so done.
Then by the product rule, 
are exactly the zeros of 
So 
has at least 5 distinct zeros on the real line. By Rolle's Theorem, between any two zeros of a function must lie at least one zero of the derivative. Hence, 
has at least 4 distinct zeros, 
has at least 3 distinct zeros, and 
has at least 2 distinct zeros. The zeros of are exactly the zeros of 
and we are done.
your experienced input would be highly appreciated.
- comes from differential equations. The interplay among exponentials, polynomial-style manipulations of differential operators, and characteristic polynomials is embedded into thinking about linear constant coefficient differential equations.
because I'm not that clever. That was a self-correcting mistake.
is the standard differential operator, then the expression can be written as 
. I couldn't ever get a solution utilizing this (darn I forgot all my DiffyQ), but I think this is one of the reasons why the solution comes out so nicely.
thing by multiplying by . Personally I proved the same thing but I wasn't clever enough to think to multiply by . Instead, I "brute forced" it: naming all of the roots so 
and 
, defining the function 
, and observing that changes sign between 
and 
for any 
.
has 
distinct real roots, then 
must have at least 
. . . . Now consider the real numbers ![\[s_1<s_2<\ldots <s_{n-1}<r_n.\]](http://latex.artofproblemsolving.com/5/4/1/5418074e233b2b15e94fcf9761b82b9e451fe16b.png)
If 
, then by IVT there must exist a root in 
. thing by multiplying by . ... defining the function , and observing that changes sign between and for any . has real roots then has real roots. This is true. But you go on to prove this by applying the Intermediate Value Theorem to something related to 
. The trouble is that even is is differentiable and hence continuous, the function need not be continuous (for example the function 
). So one cannot apply IVT (or at least one needs to modify the statement somehow).
trick, because I absolutely did not trust myself to deal with these real-analysis type issues.
have to be continuous since, if is three times differentiable it follows that is differentiable, and therefore continuous?
have to be continuous since, if is three times differentiable it follows that is differentiable, and therefore continuous?
has three real roots. But at the very last step, you need 
to be continuous to get 
to have three real roots.
intervals had a root.we see that having roots gives having roots, so done.
has an anti-derivative (see below), so ignore my (now hidden) earlier comments.
is Darboux (beyond that 
are Darboux). (which has an anti-derivative), but I failed to realize this earlier.
has the Darboux property, and is a positive function, so the root-finding method above applied to 
will give roots carrying over to . But logically speaking, this fails to avoid using in the first place.)
(hence 
). Anyways...
be a three times differentiable function (defined on and real-valued) such that has at least five distinct real zeros. Prove that has at least two distinct real zeros.
Then has atleast 
distinct zeroes, and by Rolle's Theorem, 
have atleast 
distinct real zeroes, respectively. Since, 
and is never zero, so we're done. 
removes at most one zero.
. We show that 
has a zero in this interval WLOG let it be second and let the other sign be positive. is strictly positive in the interval. It follows that 
, which means that the value of halves by at most 
every one, and is never zero, contradiction. 
type of stuff and especially their derivative, something the 
in the exponent can be changed with other functions according to needs. For lower degree derivates, you can also solve linear differential equation for your desired function. One very long method (that i did painfully) was just take 
and differentiating it 3 times (which is hectic to do of course) and then in our final expression, we would choose 
according to our needs (in this case 
) and then adjust by some constant. Then we take Rolle (as we need to show 2 roots of a 3rd deg derivative, and noticing we were already given to have 5 distinct roots, this was kind of motivated too.
and differentiating it 3 times (which is hectic to do of course)
and 
intersect at the point 
.
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
.
.
,
has at least 5 distinct real roots, and so 
has at least 2 distinct real roots. Since 
,
has at least 2 distinct real roots as desired.
and 
and 
were getting absolutely nowhere; quite unsurprisingly in fact, since this wouldn't actually say much about 
,
to give what I wanted. But then 
,
to get the right coefficients. Then we want 
to all be the same; this motivates 
,
,
instead.
.
in 
for which 
.
has even multiplicity, so it's tangent to the 
then 
and vice versa, so we have a root there too. Hence we have at least 
has at least four real roots, so 
has at least three, so 
has at least two. 
.
.