Absolutely interesting limit

Calculus Level 4

We can define a family of real functions g n g_n so that, for instance, g 3 ( x ) = ∣ x − 3 ∣ + ∣ x − 2 ∣ + ∣ x − 1 ∣ + ∣ x ∣ + ∣ x + 1 ∣ + ∣ x + 2 ∣ + ∣ x + 3 ∣ , g_3(x) = |x - 3| + |x - 2| + |x -1| + |x| + |x+1| + |x+2| + |x+3|, and more generally for integers n ≥ 0 n \geq 0 , g n ( x ) = ∑ k = − n n ∣ x + k ∣ . g_n(x) = \sum_{k = -n}^{n} |x + k|. Now let f ( x ) = lim ⁡ n → ∞ ( g n ( x ) − n ( n + 1 ) ) . f(x) = \lim_{n\to\infty} \left(g_n(x) - n(n+1)\right). Evaluate f ( 7.2 ) f(7.2) .

Hint : For integer values of x x , f ( x ) f(x) turns out to be a familiar function.


The answer is 52.

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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2 solutions

Arjen Vreugdenhil
Aug 29, 2020

It turns out that for integer values of x x , f ( x ) = x 2 f(x) = x^2 . You can easily check this with a few examples; I will also prove it below.

Also, each of the functions in the problem is piecewise linear; to be precise, linear on each interval [ k , k + 1 ] [k, k+1] between successive integers.

Thus the solution point lies on the line segment between ( 7 , 49 ) (7, 49) and ( 8 , 64 ) (8, 64) , which has a slope of 15. Thus f ( 7.2 ) = 49 + 0.2 × 15 = 52 f(7.2) = 49 + 0.2\times 15 = \boxed{52} .

Proof that f ( x ) = x 2 f(x) = x^2 for integer values of x x :

Let x x be an integer such that − n ≤ x ≤ n -n \leq x \leq n , g n ( x ) = ( n − x ) + ( n − x − 1 ) + ⋯ + 2 + 1 + 0 + 1 + 2 + ⋯ + ( n + x ) = ( 1 + 2 + ⋯ + ( n − x ) ) + ( 1 + 2 + ⋯ + ( n + x ) ) = 1 2 ( n − x ) ( n − x + 1 ) + 1 2 ( n + x ) ( n + x + 1 ) = 1 2 ( n 2 + x 2 − 2 n x + n − x ) + 1 2 ( n 2 + x 2 + 2 n x + n + x ) = n 2 + x 2 + n = x 2 + n ( n + 1 ) . \begin{aligned} g_n(x) & = (n-x) + (n-x-1) + \cdots + 2 + 1 + 0 + 1 + 2 + \cdots + (n+x) \\ & = \left(1 + 2 + \cdots + (n-x)\right) + \left(1 + 2 + \cdots + (n+x)\right) \\ & = \tfrac12(n-x)(n - x +1 ) + \tfrac12(n+x)(n+x+1) \\ & = \tfrac12(n^2 + x^2 - 2nx + n - x) + \tfrac12(n^2 + x^2 + 2nx + n + x) \\ & = n^2 + x^2 + n = x^2 + n(n+1). \end{aligned} Therefore, for all − n ≤ x ≤ n -n \leq x \leq n , g n ( x ) − n ( n + 1 ) = x 2 , g_n(x) - n(n+1) = x^2, and since by taking the limit we always accomplish n ≥ x n \geq x in the end, this means that f ( x ) = x 2 . f(x) = x^2.

2 Helpful 1 Interesting 5 Brilliant 0 Confused

There is no need for linear interpolation. f ( x ) = ( 2 ⌊ x ⌋ + 1 ) x − ⌊ x ⌋ ( ⌊ x ⌋ + 1 ) f(x) = (2 \lfloor x \rfloor + 1)x - \lfloor x \rfloor (\lfloor x \rfloor + 1)

Atomsky Jahid - 9 months, 2 weeks ago

Hello, I arrived at a different solution when solving this specific case. What I did was remove the absolute value as follows:

[ n − 7.2 ] + [ ( n − 1 ) − 7.2 ] + . . . + [ 9 − 7.2 ] + [ 8 − 7.2 ] + [ 7.2 − 7 ] + . . . + [ 7.2 − 1 ] + 7.2 + ∑ k = 1 n ( 7.2 + k ) [n-7.2]+[(n-1)-7.2]+...+[9-7.2]+[8-7.2]+[7.2-7]+...+[7.2-1]+7.2+\sum_{k=1}^{n}(7.2+k)

Here, we can clearly see that we can collect all the integers like this: n ( n + 1 ) − 28 n(n+1)-28 ; and, after some algebra, we will get 7.2 ( 15 ) 7.2(15) . With this in mind and without any rigorous proof, I intuitively arrived at the general formula f ( x ) = ( 2 ⌊ x ⌋ + 1 ) x − ⌊ x ⌋ ( ⌊ x ⌋ + 1 ) 2 f(x)=(2\lfloor{x}\rfloor+1)x-\dfrac{\lfloor{x}\rfloor(\lfloor{x}\rfloor+1)}{2} .

However, when doing a rigorous proof and checking the answers posted for this problem the formula is: f ( x ) = ( 2 ⌊ x ⌋ + 1 ) x − ⌊ x ⌋ ( ⌊ x ⌋ + 1 ) f(x)=(2\lfloor{x}\rfloor+1)x-\lfloor{x}\rfloor(\lfloor{x}\rfloor+1) . I have spent some time thinking about it but I can't wrap my head around it since I believe I didn't make any mistakes in my first solution, where the answer is 80 instead of 52.

Saúl Huerta - 9 months, 1 week ago

I think something goes wrong when you collect the integers. The first few terms (up to [ 8 − 7.2 ] [8 - 7.2] ) give n + ( n − 1 ) + ( n − 2 ) + ⋯ + 8 = 1 2 n ( n + 1 ) − 28 ; n + (n-1) + (n-2) + \cdots + 8 = \tfrac12n(n+1) - 28; the next few terms give ( − 7 ) + ( − 6 ) + ⋯ + ( − 1 ) = − 28 ; (-7) + (-6) + \cdots + (-1) = -28; the last terms are ∑ k = 1 n k = 1 2 n ( n + 1 ) . \sum_{k=1}^n k = \tfrac12n(n+1). The total, then, is ( 1 2 n ( n + 1 ) − 28 ) + ( − 28 ) + 1 2 n ( n + 1 ) = n ( n + 1 ) − 56. (\tfrac12n(n+1) - 28) + (-28) + \tfrac12n(n+1) = n(n+1) - 56. I suspect that you forgot to add in one of the − 28 -28 terms when adding?

Arjen Vreugdenhil - 9 months, 1 week ago

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Ah yes, silly mistake. I see why now. Thanks!

Saúl Huerta - 9 months, 1 week ago
Atomsky Jahid
Aug 31, 2020

For x ≥ n x \geq n ,

g n ( x ) = ∑ k = − n n ∣ x + k ∣ = ∑ k = − n n ( x + k ) = ∑ k = − n n x + ∑ k = − n n k = ( 2 n + 1 ) x \begin{aligned} g_n(x) &= \sum_{k=-n}^{n} |x+k| \\ &= \sum_{k=-n}^{n} (x+k) \\ &= \sum_{k=-n}^{n} x + \sum_{k=-n}^{n} k \\ &= (2n+1) x \end{aligned}

Similarly, for x ≤ − n x \leq -n , g n ( x ) = − ( 2 n + 1 ) x g_n(x) = -(2n+1)x

But, the case we're most interested in is − n < x < n -n < x < n . From the definition of the floor function, ⌊ x ⌋ ≤ x < ⌊ x ⌋ + 1 \lfloor x \rfloor \leq x < \lfloor x \rfloor + 1 . For legibility and ease, we'll use j = ⌊ x ⌋ j = \lfloor x \rfloor throughout the solution.

g n ( x ) = ∑ k = − n n ∣ x + k ∣ = ∑ k = − n − j − 1 ∣ x + k ∣ + ∑ k = − j n ∣ x + k ∣ = − ∑ k = − n − j − 1 ( x + k ) + ∑ k = − j n ( x + k ) = − ∑ k = − n − j − 1 x − ∑ k = − n − j − 1 k + ∑ k = − j n x + ∑ k = − j n k = − ( n − j ) x − j ( j + 1 ) 2 + n ( n + 1 ) 2 + ( n + j + 1 ) x − j ( j + 1 ) 2 + n ( n + 1 ) 2 = ( 2 j + 1 ) x − j ( j + 1 ) + n ( n + 1 ) \begin{aligned} g_n(x) &= \sum_{k=-n}^{n} |x+k| \\ &= \sum_{k=-n}^{-j-1} |x+k| + \sum_{k=-j}^{n} |x+k| \\ &= - \sum_{k=-n}^{-j-1} (x+k) + \sum_{k=-j}^{n} (x+k) \\ &= - \sum_{k=-n}^{-j-1} x - \sum_{k=-n}^{-j-1} k + \sum_{k=-j}^{n} x + \sum_{k=-j}^{n} k \\ &= - (n-j)x - \frac{j(j+1)}{2} + \frac{n(n+1)}{2} + (n+j+1)x - \frac{j(j+1)}{2} + \frac{n(n+1)}{2} \\ &= (2j+1)x - j(j+1) + n(n+1) \end{aligned}

So, f ( x ) = lim ⁡ n → ∞ ( g n ( x ) − n ( n + 1 ) ) = lim ⁡ n → ∞ ( ( 2 j + 1 ) x − j ( j + 1 ) + n ( n + 1 ) − n ( n + 1 ) ) = lim ⁡ n → ∞ ( ( 2 j + 1 ) x − j ( j + 1 ) ) = ( 2 j + 1 ) x − j ( j + 1 ) ⟹ f ( x ) = ( 2 ⌊ x ⌋ + 1 ) x − ⌊ x ⌋ ( ⌊ x ⌋ + 1 ) \begin{aligned} f(x) &= \lim_{n \to \infty} ( g_n(x) - n(n+1) ) \\ &= \lim_{n \to \infty} ( (2j+1)x - j(j+1) + n(n+1) - n(n+1) ) \\ &= \lim_{n \to \infty} ( (2j+1)x - j(j+1) ) \\ &= (2j+1)x - j(j+1) \\ \implies f(x) &= (2\lfloor x \rfloor + 1)x - \lfloor x \rfloor (\lfloor x \rfloor + 1) \end{aligned}

It's interesting to note that f ( x ) f(x) and x 2 x^2 touch when x x is an integer.

Fig: f(x) in red and x^2 in green Fig: f(x) in red and x^2 in green

Lastly, f ( 7.2 ) = ( 2 × 7 + 1 ) 7.2 − 7 ( 7 + 1 ) = 52 f(7.2) = (2 \times 7 + 1)7.2 - 7(7+1) = \boxed{52}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

By writing x = ⌊ x ⌋ + { x } x = \lfloor x \rfloor + \{x\} (so that 0 ≤ { x } < 1 0 \leq \{x\} < 1 is the fractional part of x x ), you can further simplify to f ( x ) = ( 2 ⌊ x ⌋ + 1 ) ( ⌊ x ⌋ + { x } ) − ⌊ x ⌋ ( ⌊ x ⌋ + 1 ) = ⌊ x ⌋ 2 + ( 2 ⌊ x ⌋ + 1 ) { x } , f(x) = (2\lfloor x \rfloor + 1)(\lfloor x\rfloor + \{x\}) - \lfloor x \rfloor (\lfloor x \rfloor + 1) = \lfloor x \rfloor^2 + (2\lfloor x \rfloor + 1)\{x\}, which describes the linear extrapolation, especially if you consider that 2 ⌊ x ⌋ + 1 = ⌊ x + 1 ⌋ 2 − ⌊ x ⌋ 2 2\lfloor x \rfloor + 1 = \lfloor x + 1 \rfloor^2 - \lfloor x \rfloor^2 .

Arjen Vreugdenhil - 9 months, 2 weeks ago

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Yes. If we use floor, ceiling, and fractional functions, we can also write it as f ( x ) = ⌊ x ⌋ 2 + ( ⌈ x ⌉ 2 − ⌊ x ⌋ 2 ) { x } f(x) = \lfloor x \rfloor ^2 + ( \lceil x \rceil ^2 - \lfloor x \rfloor ^2) \{x\}

Atomsky Jahid - 9 months, 2 weeks ago

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That is pretty.

Arjen Vreugdenhil - 9 months, 2 weeks ago

Or, more symmetrically f ( x ) = ⌊ x ⌋ 2 ( ⌈ x ⌉ − x ) + ⌈ x ⌉ 2 ( x − ⌊ x ⌋ ) . f(x) = \lfloor x\rfloor^2(\lceil x \rceil - x) + \lceil x\rceil^2(x - \lfloor x \rfloor).

Arjen Vreugdenhil - 9 months, 2 weeks ago

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