More fun with quadratic forms

Algebra Level 5

Find the smallest positive integer n n such that there exist real numbers x 0 , x 1 , , x n x_0,x_1,\ldots,x_n with k = 0 n x k 2 = 1 \displaystyle \sum_{k=0}^{n}x_k^2=1 and k = 1 n x 0 x k > 1 \displaystyle \sum_{k=1}^{n}x_0x_k>1 .

If you come to the conclusion that no such n n exists, enter 666.


The answer is 5.

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1 solution

Rishabh Jain
Jan 26, 2016

Applying Cauchy Shwarz inequality \color{forestgreen}{\text{Applying Cauchy Shwarz inequality}} S = k = 1 n x 0 x k ( n x 0 2 ) k = 1 n x k 2 = ( n x 0 2 ) ( 1 x 0 2 ) S=\displaystyle \sum_{k=1}^{n}x_0x_k\leq \sqrt{(n x_0^2)\displaystyle \sum_{k=1}^{n}x_k^2}=\sqrt{(nx_0^2)(1-x_0^2)} Applying G M A M \color{forestgreen}{\text{Applying}~GM\leq AM} n × ( x 0 2 + 1 x 0 2 ) 4 \leq\sqrt{n\times \frac{(x_0^2+1-x_0^2)}{4}} = n 4 > 1 o r n > 4 =\sqrt{\dfrac{n}{4}}>1~ or ~ \Large n>4 Hence smallest integral n= 5 \boxed 5 For equality x 1 = x 2 = . . . . = x n = 1 2 n a n d x 0 = 1 2 \color{#D61F06}{\text{For equality}~x_1=x_2=....=x_n=\dfrac{1}{\sqrt{2n}} ~and~x_0=\dfrac{1}{\sqrt{2}}} For n= 5 5 , we get x 1 = x 2 = . . . . = x n = 1 10 a n d x 0 = 1 2 x_1=x_2=....=x_n=\dfrac{1}{\sqrt{10}} ~and~x_0=\dfrac{1}{\sqrt{2}}

Yes, nicely done! (+1) For completeness, can you show us that there is a solution for n = 5 n=5 ?

Try this more challenging problem !

Otto Bretscher - 5 years, 4 months ago

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Thanks... I have edited the solution. Now is it complete??

Rishabh Jain - 5 years, 4 months ago

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yes, perfect!

Otto Bretscher - 5 years, 4 months ago

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