Question

Help! Will fan and medal!

Assume that
1a_1+2a_2+...+na_n=1,
where the a_j are real numbers.
As a function of n, what is the minimum value of 1a_1^2+2a_2^2+...+na_n^2?

Answer 1

@Bill_Cipher

Answer 2

so this mean that
1a_1 =1
2a_2 =2*2 =4
3a_3=3*3 =9
4a_4 =4*4 = 16
.....
na_n =na_n = n^2

for example for n=2

1a_1 =1*1 =1
2a_2 = 2*2 =2

1a_1 +2a_2 = 1 => 1+2=3=1

1a_1^2 = 1^2 = 1
2a_2^2 = 2*2^2 =8

1a_1^2 +2a_2^2 =9 =3^2 = 1^2 = 1

so from this result that

1a_1^2 +2a_2^2 +... +na_n^2 = 1

Answer 3

1 isn't correct though. (I tried already.)

Answer 4

@ganeshie8 please your opinion about this - thank you in advance

Answer 5

2

Answer 6

How did you get that? BTW it isn't right.

Answer 7

Basically we have...

\[\sum_{i = 1}^{n} i*a_i = 1\]
And you want to find out minimum value of

\[\sum_{i = 1}^{n}i*a_i ^2\]

Answer 8

Hmm... well I may have something in mind..

Answer 9

Have you read about Cauchy Schwarz inequality?

Answer 10

Yes.

Answer 11

Or let me write it like this instead of that...

\[\large{|\sum\sqrt{i}(\sqrt{i}*{a_i})|^2 \le \sum|\sqrt{i}^2| \sum|(\sqrt{i}a_i)^2|}\]

Answer 12

Yeps!! This is much better :D

Answer 13

Now, I know the LHS value... we can find out \(\sum_{i = 1}^{n} i\) and we want to find out the value of the other summation

Answer 14

Can you do after this?

Answer 15

I'm not sure.

Answer 16

See... we have this formula:

\(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\)

Answer 17

Okay.

Answer 18

Also, LHS of the inequality will become 1^2 = 1...

So, we can separate out the last summation as:

\[\large{\sum_{i=1}^{n} i a_i^2 \ge \cfrac{2}{n(n+1)}}\]

Answer 19

From here you can find out the min. value