Putnam Problem of the Day

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10578703]

[math]
\text{Let }A\text{ be a positive real number. What are the possible values of }\sum_{j=0}^\infty x_j^2,
\\
\text{given that }x_0,x_1,\ldots\text{ are positive numbers for which }\sum_{j=0}^\infty x_j=A\text{?}
[/math]

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10578704]

Previous Thread >>10576533

[Anonymous No.10578730]

>>10578703
this ones easy isnt it
any positive number less than A^2

[Anonymous No.10578732]

>>10578730
No work=no credit

[Anonymous No.10578735]

>>10578730
Can't be zero either lad.

[Anonymous No.10578738]

>>10578735
>any POSITIVE number

[Anonymous No.10578753]

>>10578738
Sorry lad.
>>10578732
I could constructively prove that we can get arbitrarily close to zero in two lines, and then prove convexity.
But constructive proofs suck to type on mongolian image boards.

[Anonymous No.10578765]

>>10578730
>any positive number
including rational numbers? and algebraic numbers?

[Anonymous No.10578811]

Just use the Cauchy-Schwarz inequality ?

[Anonymous No.10578828]

>>10578703
Thread theme
https://youtube.com/watch?v=Ssh71hePR8Q

[Anonymous No.10578887]

I have a marvelous proof that the only possible sum is 12, which the post size limit of this board is too small to contain.

[Anonymous No.10579056]

>>10578887
fucking autist

[Anonymous No.10579357]

x_j for j is an element of [1,inf) must be a convergent sum.

summation from j=1 to inf of x_j = chi

design a convergent sum such that: A = x_0 + chi = 2*chi; x_0 = chi

For some value n, A = n*x_0 + chi / n, n > 0

summation of x_j^2 = chi (n^2 + 1/n^2) = A/2 * (n^2 + 1/n^2)

This shows that the summation is at least A to infinity.
let 0<m<1
Let x_j be the convergent sum (1-m)^j*m
lim m-> 1 of summation x_j^2 = 1^2 + 0*1^2 = 1
lim m->0 of summation of x_j^2 = 0^2+1*x^2 = 0

[Anonymous No.10579360]

>>10579357
conclusion: summation of x_j^2 >0

[Anonymous No.10579408]

>>10579357
woops, the second part of the proof only justifies (0,A^2). I'm not sure how to approach (A^2,A]

[Anonymous No.10579447]

>>10578703
Clearly by choosing x1=p, x2=p^2, x3=p^3, etc we get sum(xi)=p/(1-p), let B = A*(1-p)/p; then sum(B*xi) = B*sum(xi) = A for any A. Then sum((B*xi)^2)=B^2*sum(xi^2)=A^2*(1-p)^2/p^2 * p^2/(1-p^2)= A^2 * (1-p)^2/(1-p^2). Choosing p close to 1 we get limit (1-p)^2/(1-p^2) tends to zero and is p tends to zero limit is 1,A^2 is fixed so possible values are (0, A^2). This was an easy one, probably the first in respective year.

[Anonymous No.10579458]

>>10579447
Yeah I also forgot to say that the sum is clearly positive and not higher that A^2 (we can get it from square of the sum formula).

[Anonymous No.10579464]

>>10579447
you would need the limiting sum to be valid to do that

[Anonymous No.10579476]

>>10579464
Well it’s clearly valid, for any finite n: sum((B*xi)^2, i= 1 to n)=B*sum((xi)^2). We can see that partial sums converge. This applies to any series in my proof.

[Anonymous No.10579757]

>>10578703
Is it bad I can't solve any of these problems?

[Anonymous No.10579872]

>>10579757
yes

[Anonymous No.10580156]

>>10579757
The median score on the Putnam, a competition for undergrad math autists, is typically zero. So no.

[Anonymous No.10580300]

>>10579476
you cant limit p to 1 or 0
if you could 0 and A^2 would also be valid

[Anonymous No.10580372]

>>10580156
everyone being a retard doesn't mean it's ok to be one

[Anonymous No.10580380]

>>10580300
Thx for response mate. I can limit p to 1 or 0 for S(p)=(1-p)^2/(1-p^2) because these limits exist, I never said I can limit p for the sum because it's meaningless: by construction 0< p < 1 or series and B diverge. But we can choose such p that the sum is close to 0 or A^2 as we want, i.e. 0 < sum < A^2.

[Anonymous No.10580473]

Like this folk said:
>>10578730
Proof:
(A,0,0,...) shows A^2 is reached.
(A/N,A/N,...,A/N,0,0,...,) shows you get arbitrarily close to 0. Fiddling with continuity you get the interval (0,A].
There are no others:
Normalize and assume A=1, but then [math]x_i^2\leq x_i[/math]

[Anonymous No.10580481]

>>10580473
Did not see that 0 can't happen. Put some epsilons and get (0,A^2).

[Anonymous No.10580487]

>>10580473
Zero is not positive number

[Anonymous No.10580516]

>>10580487
Yes, see correction
>>10580481

[Anonymous No.10580948]

>>10579476
>>10579476
>>10580380
limiting p to 0 makes terms in your series 0, which is not valid
you've only shown that you can get your series arbitrarily close to a number, not equal to that number

[Anonymous No.10581029]

>>10580948
wrong

[Anonymous No.10581035]

>>10580948
Wrong. Series valid for any 0 < p < 1. Sum is fixed since B = (1-p)/p increases.
Seems like you just don't undertand the proof. Read it correctly.