Yes.

>>4207964
Proof plz.

This boils down to finding (xn)nN such that the coefficients of ni=0(X−xi)  are all nonzero.

I believe (n)nN works because no power of can be written as a sum of other powers of ; the terms of the sequence won't cancel out and their exponents either.

>>4208029
No, that would make the sequence of coefficients for the polynomial depend on n.

A possible strategy: Construct sequences a0a1a2 and x0x1x2 such that

pn(xk)0 for kn and k even
pn(xk)0 for kn and k odd

If this can be done, the existence of the roots is guaranteed by the intermediate value theorem.

If we choose all the x's positive, give the a's alternating signs, and make sure that

an+1xkn+1anxnk for nk,

that will force all the pn(xk) for n > k to have the same sign as pk(xk).

By making the x's increasing, we only need to check the last and largest one:

an+1xnn+1anxnn

That and we need each pn(xn) to have the right sign.

So what remains is showing we can construct sequences satisfying these two conditions at every step.

>>4208090
So we start with a0=1 and x0=1.

And given the series up to n, we proceed as follows to get an+1 and xn+1:

First we choose an an+1 small enough to satisfy the

an+1xnn+1anxnn

condition.

an+1=−anxnn2xnn+1 works.

Then we have to choose an xn+1 large enough that pn+1(xn+1) has the sign of its last and highest order term. Asymptotically, xn+1 increases faster than xn and all the other terms, so we will can find an xn+1 that does this no matter how small we make an+1, as long as we didn't make it zero (which we didn't).

That completes the proof.

Just look at the Maclaurin expansion for Unknown control sequence '\sinx'.

>>4208183
*sinx+cosx

for every n degree polynomial with n roots, is there some n+1 degree term you can add to it that makes the function cross the x axis one more time? it seems like it, but idk

>>4208183
yeah, that was my first instinct. i don't see why that doesn't satisfy the problem.

>>4208229
http://www.wolframalpha.com/input/?i=sin%28x%29+%2B+cos%28x%29+maclaurin+expansion
It does. Not sure why this question is so easy.

>>4208463
From the graphs, that seems to be returning n+1 roots, not n.

>>4208470
You must be looking at it wrong then since that's completely impossible. Any (non-zero) polynomial will have at most n real roots.

>>4208523
Posting to remove that stupid comment from first page.

>>4208463
>>4208229
I guess the hard part would be to formally prove that it has n distinct roots, but even that's not too difficult: you just have to show that there are n distinct inflection points, each alternating in sign, which can be done easily by taking some derivatives.

>>4209064
*n-1

>>4209064
How does that prove that the function actually crossed the axis before turning back around?

>>4209455
odd multiplicity in the factors
i.e.
x^22 only touches the x-axis while
x^23 crosses it

in a slightly more complex case,
a polynomial (x-3)^2(x-1) has two roots (3 and 1), 3 has a power of two, which is even, so it does not cross the x-axis, while 1 has a power of 1, which is odd, so it does cross the x-axis

sorry, english is not my native language

>>4209455
Any polynomial is a continuous function. If I have a continuous function that's greater than zero at x = a and less than zero at x = b, then between a and b the function has to cross the x axis at least once, by the intermediate value theorem. So if there are n-1 inflection points, alternating in sign, then between the inflection points p_n has to cross the x-axis (that they're inflection points is irrelevant; they're just easier to find). This shows that there are n-2 roots. To get the final two roots, just show that limxpn(x) is a different sign than the rightmost inflection point. Doing this on the left side gives the final two roots.

this is a good problem

Well. I see I don't know my math today either

>>4209064
>I guess the hard part would be to actually solve the problem

I personally find that remembering to put the stickers on all my problem sheets is the hardest part.

>>4209543
But can't you have all the inflection points e.g. above the x axis yet still have n distinct roots? (e.g. all the inflection points of sin(x)+1/2 are positive)

While it seems desirable to take a specific function like sin(x)+cos(x) and look at its maclaurin series, it's probably more straightforward to do this by induction. Assume it's true for n, and construct a value a_{n+1} that will cause exactly one more root by choosing a small enough value that the influence over the subset of the domain where the existing roots are is negligible, and looking at the limits as x goes to +/- infinity and the parity of n+1 to determine the sign of a_{n+1}, such that the new polynomial has an additional root outside of that subset of the domain.