x → 0 lim x 2 2 0 1 7 − ∑ k = 1 2 0 1 7 cos k x
The above limit can be expressed in the form b 4 0 3 5 a ( a + 1 ) , where a , b ∈ Z + . Find 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.
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My solution. 2 0 1 7 = 1 + 1 + 1 + . . . 2017 times. We can split the problem into 2017 limits, like this:
x → 0 lim x 2 1 − cos x + x → 0 lim x 2 1 − cos ( 2 x ) + . . . + x → 0 lim x 2 1 − cos ( 2 0 1 7 x ) until the end of that. For each separate limit, just multiply the numerator and denominator by ( 1 + cos k x ) , k = 1 , 2 , 3 , . . . , 2 0 1 7 , to obtain for each limit x 2 ( 1 + cos k x ) 1 − cos 2 k x = x 2 ( 1 + cos k x ) sin 2 k x , k = 1 , . . . , 2 0 1 7 . Split the limit into the familiar x → 0 lim x sin k x and x → 0 lim 1 + cos k x 1 = 2 1 , k = 1 , . . . , 2 0 1 7 . 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 2 1 k = 1 ∑ 2 0 1 6 k 2 = 6 4 0 3 5 ⋅ 2 0 1 7 ⋅ 2 0 1 8 , so a = 2 0 1 7 , b = 1 2 , a + b = 2 0 2 9 .
Awesome approach👍 You also could have used half angle formulas from Trigonometry to shorten it further.
First notice that ( 2 0 1 7 − ∑ k = 1 2 0 1 7 cos k x ) / x 2 = ∑ k = 1 2 0 1 7 ( 1 − cos k x ) / ( x 2 ) . Anticipating that this expression has a defined value, call it α .
Consider now an individual term in this sum ( 1 − cos x k ) / x 2 . If we can calculate lim x → 0 ( 1 − cos k x ) / x 2 , i.e., show it exists and find its value, then we can find the value of α 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 . The function ( cos k x ) / 2 is continuous at zero, so we may solve this limit by substituting x = 0 : we obtain lim x → 0 ( 1 − cos k x ) / x 2 = ( k 2 cos 0 ) / 2 = k 2 / 2 .
Therefore, α = 2 1 ∑ k = 1 2 0 1 7 k 2 . We have a formula for the sum of the first n squares, namely, n ( n + 1 ) ( 2 n + 1 ) / 6 . So α = 1 2 1 ( 2 0 1 7 ⋅ ( 2 0 1 7 + 1 ) ⋅ 4 0 3 5 . From this expression, it is easy to see that a = 2 0 1 7 and b = 1 2 , so that a + b = 2 0 2 9 .
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Relevant wiki: Maclaurin Series
L = x → 0 lim x 2 2 0 1 7 − ∑ k = 1 2 0 1 7 cos k x = x → 0 lim x 2 2 0 1 7 − ∑ k = 1 2 0 1 7 ( 1 − 2 ! ( k x ) 2 + 4 ! ( k x ) 4 − 6 ! ( k x ) 6 + ⋯ ) = x → 0 lim x 2 ∑ k = 1 2 0 1 7 ( 2 ! ( k x ) 2 − 4 ! ( k x ) 4 + 6 ! ( k x ) 6 − ⋯ ) = x → 0 lim k = 1 ∑ 2 0 1 7 ( 2 ! k 2 − 4 ! k 4 x 2 + 6 ! k 6 x 4 − ⋯ ) = 2 1 k = 1 ∑ 2 0 1 7 k 2 = 1 2 2 0 1 7 ( 2 0 1 8 ) ( 4 0 3 5 By Maclaurin series Divide up and down by x 2
⟹ a + b = 2 0 1 7 + 1 2 = 2 0 2 9