- #26

- 218

- 27

My thinking behind writing "for smallest solution... equality holds" was that, suppose we are givenLooking at the solution to 13 first, I think I'm confused by the first couple steps.

This is actually not immediately obvious to me.

$$x \leq y$$

And we have to find the minimum possible value of ##y## for which the above relation is true for all possible values of ##x## (both ##x,y\in \mathbb{R}## are variables)

Now clearly in this case for every value of ##x## there are infinite values of ##y## but to find the minimum value of ##y## we have to first find the maximum value of ##x## and then we can say that if ##y## is

**equal**to that value of ##x##, then the inequality ##x \leq y## will always be satisfied no matter what value ##x## takes.

Yes, you right, I was horribly wrong there, instead the second line should be,The sum in the first line has a as its lowest degree term, but the sum in the second line on the left has a 1 I think? I think the right hand side is unchanged but the left hand side you just added 1 to it? (this isn't super crippling, if you add 1 to the left hand side and still get that it's equal to the right hand side, then that's pretty good, but not obviously minimizing c(n).

$$

\begin{align}

a+a^2+a^3+\ldots +a^{2n-1}+a^{2n}&=c(n)\cdot \left(1+a^{2n+1}\right)\nonumber\\

\sum_{r=0}^{2n-1} a^{2n-r}&=c(n)\cdot \left(1+a^{2n+1}\right)\nonumber\end{align}

$$

Yes, I differentiated both sides w.r.t ##n##.I don't understand what happened in this step. Are you trying to differentiate with respect to n?

I didn't think about all these, I just differentiated both sides hoping that everything should be fine but now I feel stupid .I don't understand what c′(n) is - c(n) only takes values on the natural numbers, so isn't differentiable? Not to mention that I don't know how you differentiate the sum being from 0 to 2n, even if this is kind of handwavy it feels like you need to do something about the fact the bounds are expanding (to take a more rigorous example, ddx∫0xxtdt≠∫0xtdt, you can't just move the differentiation inside the integral when the bounds include the variable)

I'll give that problem a second try, but is my first thinking correct? i.e., for the smallest solution of ##c(n)## we first maximize (somehow) the L.H.S, and then equate both sides?