A good problem for a test.

Calculus Level 4

lim x 0 2017 k = 1 2017 cos k x x 2 \large \lim_{x \to 0} \frac{2017 - \sum_{k=1}^{2017} \cos kx}{x^2}

The above limit can be expressed in the form 4035 a ( a + 1 ) b \dfrac{4035a(a+1)}{b} , where a , b Z + a, b \in \mathbb{Z}^+ . Find a + b a + b .

Note. The intention is for you to solve the problem without L'Hopital's rule or Taylor Series. Think Calculus I, limits unit, before you learned differentiation.


The answer is 2029.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Chew-Seong Cheong
Nov 13, 2017

Relevant wiki: Maclaurin Series

L = lim x 0 2017 k = 1 2017 cos k x x 2 By Maclaurin series = lim x 0 2017 k = 1 2017 ( 1 ( k x ) 2 2 ! + ( k x ) 4 4 ! ( k x ) 6 6 ! + ) x 2 = lim x 0 k = 1 2017 ( ( k x ) 2 2 ! ( k x ) 4 4 ! + ( k x ) 6 6 ! ) x 2 Divide up and down by x 2 = lim x 0 k = 1 2017 ( k 2 2 ! k 4 x 2 4 ! + k 6 x 4 6 ! ) = 1 2 k = 1 2017 k 2 = 2017 ( 2018 ) ( 4035 12 \begin{aligned} L & = \lim_{x \to 0} \frac {2017-\sum_{k=1}^{2017}\color{#3D99F6}\cos kx}{x^2} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \frac {2017-\sum_{k=1}^{2017}\color{#3D99F6} \left(1 - \frac {(kx)^2}{2!} + \frac {(kx)^4}{4!} - \frac {(kx)^6}{6!} + \cdots \right) }{x^2} \\ & = \lim_{x \to 0} \frac {\sum_{k=1}^{2017} \left(\frac {(kx)^2}{2!} - \frac {(kx)^4}{4!} + \frac {(kx)^6}{6!} - \cdots \right) }{x^2} & \small \color{#3D99F6} \text{Divide up and down by }x^2 \\ & = \lim_{x \to 0} \sum_{k=1}^{2017} \left(\frac {k^2}{2!} - \frac {k^4x^2}{4!} + \frac {k^6x^4}{6!} - \cdots \right) \\ & = \frac 12 \sum_{k=1}^{2017} k^2 \\ & = \frac {2017(2018)(4035}{12} \end{aligned}

a + b = 2017 + 12 = 2029 \implies a +b = 2017+12 = \boxed{2029}

Hobart Pao
Nov 25, 2017

My solution. 2017 = 1 + 1 + 1 + . . . 2017 = 1 + 1 + 1 + ... 2017 times. We can split the problem into 2017 limits, like this:

lim x 0 1 cos x x 2 + lim x 0 1 cos ( 2 x ) x 2 + . . . + lim x 0 1 cos ( 2017 x ) x 2 \displaystyle \lim_{x \to 0} \dfrac{1 - \cos x}{x^2} + \lim_{x \to 0} \dfrac{1 - \cos(2x)}{x^2} + ... + \lim_{x \to 0} \dfrac{1 - \cos (2017 x)}{x^2} until the end of that. For each separate limit, just multiply the numerator and denominator by ( 1 + cos k x ) , k = 1 , 2 , 3 , . . . , 2017 (1 + \cos kx ), k = 1, 2, 3, ..., 2017 , to obtain for each limit 1 cos 2 k x x 2 ( 1 + cos k x ) = sin 2 k x x 2 ( 1 + cos k x ) , k = 1 , . . . , 2017 \dfrac{1 - \cos^2 kx}{x^2 (1 + \cos kx)} = \dfrac{\sin^2 kx}{x^2 (1 + \cos kx )}, k = 1, ..., 2017 . Split the limit into the familiar lim x 0 sin k x x \displaystyle \lim_{x\to 0} \dfrac{\sin kx}{x} and lim x 0 1 1 + cos k x = 1 2 \displaystyle \lim_{x\to 0} \dfrac{1}{1+ \cos kx} = \dfrac{1}{2} , k = 1 , . . . , 2017 k = 1, ..., 2017 . I omit how to solve each of these limits, because it is obvious. After evaluating each limit, multiply everything back together and we end up with 1 2 k = 1 2016 k 2 = 4035 2017 2018 6 \displaystyle \dfrac{1}{2} \sum_{k=1}^{2016} k^2 = \dfrac{4035 \cdot 2017 \cdot 2018}{6} , so a = 2017 , b = 12 , a + b = 2029 a = 2017, b = 12, a + b = \boxed{2029} .

Awesome approach👍 You also could have used half angle formulas from Trigonometry to shorten it further.

Amritaansh Narain - 1 year, 11 months ago
Nicolas Bryenton
Nov 13, 2017

First notice that ( 2017 k = 1 2017 cos k x ) / x 2 = k = 1 2017 ( 1 cos k x ) / ( x 2 ) (2017 - \sum_{k=1}^{2017}\cos kx)/x^2 = \sum_{k=1}^{2017} (1-\cos kx)/(x^2) . Anticipating that this expression has a defined value, call it α \alpha .

Consider now an individual term in this sum ( 1 cos x k ) / x 2 (1 - \cos xk)/x^2 . If we can calculate lim x 0 ( 1 cos k x ) / x 2 \lim_{x \to 0} (1-\cos kx)/x^2 , i.e., show it exists and find its value, then we can find the value of α \alpha by summing the limits, as the limit of a sum is the sum of the limits, when these limits exist. Applying L'Hopital's rule twice (careful: why are we allowed to do this?) gives that this limit is the same as lim x 0 ( k 2 cos k x ) / 2 \lim_{x \to 0} (k^2\cos kx )/ 2 . The function ( cos k x ) / 2 (\cos kx)/2 is continuous at zero, so we may solve this limit by substituting x = 0 x = 0 : we obtain lim x 0 ( 1 cos k x ) / x 2 = ( k 2 cos 0 ) / 2 = k 2 / 2 \lim_{x \to 0} (1- \cos kx)/x^2 = (k^2\cos0)/2 = k^2/2 .

Therefore, α = 1 2 k = 1 2017 k 2 \alpha = \frac{1}{2}\sum_{k=1}^{2017} k^2 . We have a formula for the sum of the first n n squares, namely, n ( n + 1 ) ( 2 n + 1 ) / 6 n(n+1)(2n+1)/6 . So α = 1 12 ( 2017 ( 2017 + 1 ) 4035 \alpha = \frac{1}{12} (2017 \cdot (2017+1)\cdot 4035 . From this expression, it is easy to see that a = 2017 a = 2017 and b = 12 b = 12 , so that a + b = 2029 a+b = 2029 .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...