#Sequences problem I can’t solve
48 messages · Page 1 of 1 (latest)
What are the variations of Pn(X) in [0, 1]?
lemme think about it...
give me some time
yeah... i think that i got it
lemme show you
i don't know how to use TexIt, but i will try my best
$let p(x) = x + x^2 + x^3 + ... + x^n, (i will talk about the -1 later)$
Waleed Al-Thqfi
so let's do some algebra
$p(x) = x(1 + x + x^2 + ... + x^(n-1))$
Waleed Al-Thqfi
$P_n(x)=-2+\sum_{k=0}^n x^k$
SWR
$p(x) = x(1 + p(x) - x^n)$
$p(x) = x + x*p(x) - x^(n+1)$
Waleed Al-Thqfi
$p(x) - xp(x) = x - x^{n+1}$
Waleed Al-Thqfi
$p(x)(1-x) = x-x^{n+1}$,
$p(x) = {(x-x^{n+1})}/{(1-x)}$
Waleed Al-Thqfi
now.. let $p(x) = 0$
Waleed Al-Thqfi
$0 = x-x^{n+1}$, x not equal to 1
Waleed Al-Thqfi
let's take n = 6 as an example
$0 = x-x^{7}$, n + 1 = 7
Waleed Al-Thqfi
$0=x(1-x^{6})$
Waleed Al-Thqfi
$0=x(1+x)(x^2-x+1)(1-x)(x^2+x+1)$
Waleed Al-Thqfi
this feels overcomplicated, it says to deduce not prove
i fcked it up, right ?
think so
deduce? i don't know what does that means
pretty sure you're supposed to use the inequality to show that the sequence Un+1 < Un therefore its decreasing
yeah we find 7 values of x for n = 6, but one of them is 1, so we assumed from the beggining that x doesn't equal to 1
i think i read your question wrongly, my bad
no worries
but i think i take it much crazier to find all the solutions for x+x^2+...+x^n in one signle formula lmao
@zinc vessel do you still need help?
yes
@zinc vessel can you use calculus?
yes