Inspired by Calvin Lin - Part I

Algebra Level 5

f n ( x ) = x 2 + ( x + 1 ) 2 + ( x + 2 ) 2 + … + ( x + n ) 2 \large f_n(x)=x^2 + ( x + 1) ^2 +(x+2)^2+\ldots+(x+n)^2

The function f n ( x ) f_n(x) is defined above for x x ranges over all real values with natural number n n .

If the minimum value of f n ( x ) f_n(x) is ( 2 10 − 12 ) (2^{10}-12) then find the value of n n .


The answer is 22.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

f n ( x ) = x 2 + ( x + 1 ) 2 + ( x + 2 ) 2 + . . . + ( x + n ) 2 f_n(x) = x^2 + (x+1)^2 + (x+2)^2 + ... + (x+n)^2

Differentiating and equating to zero, we have:

f ′ ( x ) = 2 x + 2 ( x + 1 ) + 2 ( x + 2 ) + . . . + 2 ( x + n ) = 2 ( n + 1 ) x + n ( n + 1 ) 2 f ′ ( x ) = 0 ⇒ 2 ( n + 1 ) x = − n ( n + 1 ) 2 x = − n 2 \begin{aligned} f'(x) & = 2x+2(x+1)+2(x+2)+...+2(x+n) \\ & = 2(n+1)x + \frac{n(n+1)}{2} \\ f'(x) & = 0 \\ \Rightarrow 2(n+1)x & = - \frac{n(n+1)}{2} \\ x & = -\frac{n}{2} \end{aligned}

We note that the lower limit of f n ( x ) ≥ 0 f_n(x) \ge 0 is when x = − n 2 x=-\dfrac{n}{2} . Therefore, minimum f n ( x ) f_n(x) is:

f n ( − n 2 ) = ( − n 2 ) 2 + ( 1 − n 2 ) 2 + ( 2 − n 2 ) 2 + . . . + ( n 2 ) 2 = { 2 [ 1 2 + 2 2 + 3 3 + . . . + ( n 2 ) 2 ] for even n 2 [ ( 1 2 ) 2 + ( 3 2 ) 2 + ( 5 2 ) 2 + . . . + ( n 2 ) 2 ] for odd n \begin{aligned} f_n\left(-\frac{n}{2}\right) & = \left(-\frac{n}{2}\right)^2 + \left(1-\frac{n}{2}\right)^2 + \left(2-\frac{n}{2}\right)^2 + ... + \left(\frac{n}{2}\right)^2 \\ & = \begin{cases} 2 \left[ 1^2+2^2+ 3^3 + ... + \left( \frac{n}{2}\right)^2 \right] & \text{for even } n \\ 2 \left[ \left( \frac{1}{2}\right)^2+\left( \frac{3}{2}\right)^2+ \left( \frac{5}{2}\right)^2 + ... + \left( \frac{n}{2}\right)^2 \right] & \text{for odd } n \end{cases} \end{aligned}

Now 2 10 − 12 = 1012 2^{10}-12 = 1012 , assuming the required n n is even, then:

2 [ 1 2 + 2 2 + 3 3 + . . . + ( n 2 ) 2 ] = 1012 2 n 2 ( n 2 + 1 ) ( 2 n 2 + 1 ) 6 = 1012 n ( n + 2 ) ( n + 1 ) 12 = 1012 n ( n + 1 ) ( n + 2 ) = 22 × × 23 × 24 ⇒ n = 22 \begin{aligned} 2 \left[ 1^2+2^2+ 3^3 + ... + \left( \frac{n}{2}\right)^2 \right] & = 1012 \\ \frac{\frac{2n}{2}(\frac{n}{2}+1)(\frac{2n}{2}+1)}{6} & = 1012 \\ \frac{n(n+2)(n+1)}{12} & = 1012 \\ n(n+1)(n+2) & = 22\times \times 23 \times 24 \\ \Rightarrow n & = \boxed{22} \end{aligned}

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Moderator note:

This approach is not correct. It has not proven that "the minimum occurs exactly at − n 2 - \frac{n}{2} . (Though yes, the statement is true)

What he did was:
1) Show that f ( x ) ≥ g ( x ) f(x) \geq g(x)
2) The minimum of g ( x ) g(x) occurs at x ∗ x^* .

However, because equality doesn't occur in the Cauchy Schwarz that he set up, all that we have is f ( x ) > g ( x ) f(x) > g(x) , and hence we are unable to conclude that the minimum of f f occurs at x ∗ x^* .

The approach would have worked if we had the additional condition 3) f ( x ∗ ) = g ( x ∗ ) f(x^*) = g( x^* ) .
In this case, we can conclude that f ( x ) ≥ g ( x ) ≥ g ( x ∗ ) = f ( x ∗ ) f( x) \geq g(x) \geq g(x^* ) = f ( x^*) , and hence the minimum of f f is at x ∗ x^* .

Umm sir, there are some minor typos in your solution .

I think it should be 2 10 2^{10} in the 6th line from the bottom , and you have put one too many × \times in the 2nd last line .

A Former Brilliant Member - 6 years ago

Sorry, my mistake. I wasn't sure. I used calculus actually.

Chew-Seong Cheong - 6 years ago

I solved it using the fact :

The minimum value of a x 2 + b x + c ax^2 + bx + c is c − b 2 / 4 a c - b^2/4a

Dev Sharma - 5 years, 5 months ago

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