x → 0 lim x 2 1 − k = 1 ∏ 2 3 cos k x = ?
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wow that was quick
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Yes, usually for such limit problems, considering the (Laurent) polynomial expansion makes quick work. That's the generalized approach of L'Hopital.
Let's generalize this for n:
= x → 0 lim x 2 1 − k = 1 ∏ n cos k x
x → 0 lim x 2 1 − cos x ⋅ cos 2 x ⋅ cos 3 x ⋅ . . . . . ⋅ cos n x
= x → 0 lim x 2 1 − cos x + cos x − cos x ⋅ cos 2 x ⋅ cos 3 x ⋅ . . . . . ⋅ cos n x
= x → 0 lim x 2 1 − cos x + x → 0 lim cos x x → 0 lim x 2 ( 1 − cos 2 x ⋅ cos 3 x ⋅ . . . ⋅ cos n x )
= 2 1 2 + x → 0 lim x 2 ( 1 − cos 2 x + cos 2 x − cos 2 x ⋅ cos 3 x ⋅ . . . ⋅ cos n x )
= 2 1 2 + x → 0 lim x 2 1 − cos 2 x + x → 0 lim cos 2 x x → 0 lim x 2 ( 1 − cos 3 x ⋅ cos 4 x ⋅ . . . ⋅ cos n x )
= 2 1 2 + 2 2 2 + x → 0 lim x 2 ( 1 − cos 3 x ⋅ cos 4 x ⋅ . . . ⋅ cos n x )
= 2 1 2 + 2 2 2 + 2 3 2 + . . . . . . . + 2 ( n − 1 ) 2 + x → 0 lim x 2 1 − cos n x
= 2 1 2 + 2 2 2 + 2 3 2 + . . . . . . . + 2 ( n − 1 ) 2 + 2 n 2
= 2 1 2 + 2 2 + 3 2 + . . . + ( n − 1 ) 2 + n 2 = 2 k = 1 ∑ n k 2 = 1 2 n ( n + 1 ) ( 2 n + 1 )
For our question, substituting n = 2 3 gives 1 2 2 3 ⋅ 2 4 ⋅ 4 7 = 2 1 6 2
A simpler way to present it is via induction (to replace the repeated factorization that you have).
Using your idea of taking out a single term, we could modify it to
x → 0 lim x 2 1 − ∏ k = 1 n cos k x = x → 0 lim x 2 1 − cos n x + x 2 cos n x ( 1 − ∏ k = 1 n − 1 cos k x ) = 2 n 2 + 2 1 × ∑ k = 1 n − 1 k 2
Interesting solution and a nice generalization!
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Thank you :D
Actually I saw this question in a competitive exam's paper:
x → 0 lim ( sin x ) 2 1 − cos x cos 2 x cos 3 x
So I thought of generalizing this for n.
We usually do not expect a limit problem to yield a big answer.
L'Hopital is not the savior? Think again.
Substitute x = 0 to the numerator and the denominator. Both result in 0, so we have the indeterminate form 0 0 . By L'Hopital's theorem, we can differentiate both the numerator and the denominator, and the limit will stay equal.
The denominator becomes 2 x . In the numerator, the 1 vanishes. The remaining term, − ∏ cos k x , splits into 23 terms by product rule, once for differentiating cos k x for each k . It becomes ∑ k = 1 2 3 k ⋅ cos x ⋅ cos 2 x ⋅ … ⋅ cos ( k − 1 ) x ⋅ sin k x ⋅ cos ( k + 1 ) x ⋅ … ⋅ cos 2 3 x ; the factor cos k x is differentiated into − k sin k x , and the negative sign cancels.
Now, try substituting again. The numerator is still 0 because each term is zero due to the sin k x term. The denominator is also still 0. So use L'Hopital's theorem again; differentiate again.
The denominator becomes simply 2. In the numerator, each term of the summation splits into 23 terms by product rule. One term comes from differentiating sin k x ; this results in k 2 ⋅ ∏ i = 1 2 3 cos i x (the term that contains sin k x also contains the factor k , since we differentiated cos k x into − k sin k x before). The other comes from differentiating some cos i x , which leaves the sin k x factor intact.
Finally, substitute again. In the numerator, every term that contains sin k x is gone, leaving only the terms in the form k 2 ⋅ ∏ cos i x . In all these terms, the cosines evaluate to 1, so all the terms are in the form k 2 . So the numerator is ∑ k = 1 2 3 k 2 = 6 2 3 ⋅ 2 4 ⋅ 4 7 . The denominator is simply 2. So substituting x = 0 to the expression gives 6 ⋅ 2 2 3 ⋅ 2 4 ⋅ 4 7 = 2 1 6 2 , which must be the answer because the numerator and the denominator are both continuous at x = 0 .
This method can be generalized to evaluate
x → 0 lim x m 1 − ∏ k = 1 n cos k x
for any positive integers n , m , although the solution will become more messy. (But at least we get the denominator generalized too, compared to other solutions that only do m = 2 !) If m is odd, the answer is 0 (there's an odd number of sines remaining, in particular at least one); otherwise, each surviving term in the final summation is in the form coefficient ⋅ ( − 1 ) m / 2 + 1 ⋅ ∏ k = 1 n cos k x , where the coefficient comes from differentiating several of the cosine factors. Each factor must be differentiated an even number of times (to turn them into cosines again). We just need to count the number of ways to get the same coefficient.
For example, for m = 6 , there are three ways to get all cosines:
Thus summing over everything, the final coefficient will be
a ∑ a 6 + 1 5 ⋅ a = b ∑ a 4 b 2 + 1 5 ⋅ $a, b, c$ distinct ∑ a 2 b 2 c 2
And how do we compute it? Well, add 1 4 ∑ a 6 − 1 4 ∑ a 6 , and we can gather those with multiplier 15 as simply ( ∑ a 2 ) 3 . Thus the answer is
1 5 ⋅ ( a = 1 ∑ n a 2 ) 3 − 1 4 ⋅ a = 1 ∑ n a 6
which can be easily computed by Newton's identities.
The title meant that you cannot solve this by directly applying L'Hopital's Rule to the expression. Besides, this question can be solved using a rather shorter method.
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But I did apply L'Hopital's rule directly to the expression. You can actually apply it twice and you immediately get the answer. The numerator will become very long, but it is still purely L'Hopital's rule.
Example where there are three cosines, spelled out fully in L'Hopital's rule:
x → 0 lim x 2 1 − cos x cos 2 x cos 3 x = x → 0 lim 2 x sin x cos 2 x cos 3 x + 2 cos x sin 2 x cos 3 x + 3 cos x cos 2 x sin 3 x = x → 0 lim 2 cos x cos 2 x cos 3 x − 2 sin x sin 2 x cos 3 x − 3 sin x cos 2 x sin 3 x − 2 sin x sin 2 x cos 3 x + 4 cos x cos 2 x cos 3 x − 6 cos x sin 2 x sin 3 x − 3 sin x cos 2 x sin 3 x − 6 cos x sin 2 x sin 3 x + 9 cos x cos 2 x cos 3 x = 2 1 ⋅ 1 ⋅ 1 ⋅ 1 − 2 ⋅ 0 ⋅ 0 ⋅ 1 − 3 ⋅ 0 ⋅ 1 ⋅ 0 − 2 ⋅ 0 ⋅ 0 ⋅ 1 + 4 ⋅ 1 ⋅ 1 ⋅ 1 − 6 ⋅ 1 ⋅ 0 ⋅ 0 − 3 ⋅ 0 ⋅ 1 ⋅ 0 − 6 ⋅ 1 ⋅ 0 ⋅ 0 + 9 ⋅ 1 ⋅ 1 ⋅ 1 = 2 1 − 0 − 0 − 0 + 4 − 0 − 0 − 0 + 9 = 2 1 4 = 7
First and second lines are L'Hopital's rule; third line is substitution due to continuity; the rest is computation. Standard application(s) of L'Hopital's rule. It's easy to generalize that to the actual problem of 23 cosines, if you want to spell out everything.
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@Arkajyoti Banerjee I agree with Ivan. Basically, you're trying to find the first non-zero coefficient in the (Laurent) polynomial expansion of these terms. Both L'hopital's rule (Ivan's solution) and converting to Taylor polynomials (Michael's solution) are the same way to do this.
In particular, because the constant and the linear term are 0 (as in Michael's polynomials), this implies that we have to apply L'hopital twice. Once to remove the constant, once to remove the linear term. Of course, the expansion might be tedious (as Ivan pointed out), which is why I prefer to transition directly to the Taylor polynomial forms for complicated terms.
I just meant that this problem cannot be solved by applying L'Hopital's Rule without algebraic manipulation. You just applied it for the expression till 3 terms and followed the pattern. The original expression contained 23. (Generalisation contained n). So you don't call this a direct application, do you?
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@Arkajyoti Banerjee – I can directly apply it (twice) to the expression; the method is exactly the same. It will simply be tedious (see my example with n = 3 above, and imagine it extending to n = 2 3 ; a single line will take an entire page of paper) and not particularly illuminating (generalization is always good in math since it gives insight; instead of "why 2162" you get "oh, because (\frac{1^2 + 2^2 + \ldots + n^2}{2}").
Yes, we can directly apply, I say..
Think Bhai Think...
Differentiate and then putting x sin x , 2 2 x sin 2 x , . . . = 1 , We get it .
I got it like that
Relevant wiki: Maclaurin Series
L = x → 0 lim x 2 1 − ∏ k = 1 2 3 cos k x By Maclaurin series = x → 0 lim x 2 1 ( 1 − k = 1 ∏ 2 3 ( 1 − 2 ! x 2 + 4 ! x 4 + ⋯ ) ( 1 − 2 ! 2 2 x 2 + 4 ! 2 4 x 4 + ⋯ ) ( 1 − 2 ! 3 2 x 2 + 4 ! 3 4 x 4 + ⋯ ) ⋯ ( 1 − 2 ! 2 3 2 x 2 + 4 ! 2 3 4 x 4 + ⋯ ) ) = x → 0 lim x 2 1 ( 1 − ( 1 − 2 ! ∑ k = 1 2 3 k 2 x 2 + O ( x 4 ) ) ) = x → 0 lim x 2 1 ( 6 ⋅ 2 ! ( 2 3 ) ( 2 4 ) ( 4 6 + 1 ) x 2 − O ( x 4 ) ) = x → 0 lim x 2 1 ( 2 1 6 2 x 2 − O ( x 4 ) ) = x → 0 lim ( 2 1 6 2 − O ( x 2 ) ) = 2 1 6 2
I was expecting a solution from you for this one, and it is breathtakingly awesome as always! :D
A solution using probabilistic method.
The first key observation is the identity k = 1 ∏ n cos ( k θ ) = 2 n 1 e i ∈ { − 1 , + 1 } , 1 ≤ i ≤ n ∑ cos ( ( e 1 + 2 e 2 + ⋯ + n e n ) θ ) this can be proved using the Euler's formula and induction. Next, define the random variables X i , 1 ≤ i ≤ n , such that P ( X i = − 1 ) = P ( X i = + 1 ) = 2 1 , and let X i are all independent. Further define the vectors X = [ X 1 X 2 ⋯ X n ] T , a = [ 1 2 ⋯ n ] T . Then, we get k = 1 ∏ n cos ( k θ ) = E ( cos ( a T X θ ) ) where the expectation is taken with respect to X i , 1 ≤ i ≤ n . Then, the given limit can be evaluated as x → 0 lim x 2 1 − E ( cos ( a T X x ) ) = x → 0 lim E [ x 2 ( 2 sin 2 ( a T X x / 2 ) ) ] = 2 1 E [ ( a T X ) 2 ] where the last step follows from Monotone Convergence Theorem . Now, E [ ( a T X ) 2 ] = a T E ( X X T ) a Note that that the ( i , j ) th entry of the matrix E ( X X T ) is E ( X i X j ) = E ( X i ) E ( X j ) = 0 , whenever i = j , because of independence and the fact that the random variables are zero mean by definition. The diagonal entries of the matrix are E ( X i 2 ) = 1 . Hence, the matrix E ( X X T ) is the identity matrix of order n . Thus, the limit evaluates to a T a / 2 = 2 ∑ i = 1 n a i 2 = 2 ∑ i = 1 n i 2 = 1 2 n ( n + 1 ) ( 2 n + 1 ) , which gives the answer to the given problem, for n = 2 3 , as 2 1 6 2 .
More generally, using the same approach as above, we can evaluate the limit lim x → 0 x 2 1 − ∏ k = 1 n cos ( θ k x ) to be ( ∑ k = 1 n θ k 2 ) / 2 .
I'm an eleventh standard student. But I will look this up for sure lol.
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The expectation that I have used is not anything advanced, and is mostly used to represent things in a more compact way. I am sure you can understand the proof once you know what expectation is, try reading the basic of expectation from here .
Wow, that's really interesting!
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Thanks Calvin! I think this method can be used in more general problems where some kind of product structure is present.
Frankly speaking L'Hopital was the saviour. :P Just diffrentiate the numarator....... you get the limit as i = 1 ∑ n 2 ( i ) 2
Can you be kind enough to elaborate your solution?
Let f ( x ) = c o s x . c o s 2 x . . . . c o s 2 3 x f ′ ( x ) = − s i n x . cos 2 x . c o s 3 x . . . c o s 2 3 x − 2 c o s x . s i n 2 x . c o s 3 x . . . . . c o s 2 3 x − . . . . − 2 3 c o s x . c o s 2 x . . . . s i n 2 3 x f ′ ( x ) = − c o s x . c o s 2 x . . . c o s 2 3 x . t a n x − 2 c o s x . c o s 2 x . . . c o s 2 3 x . t a n 2 x − . . . . . − 2 3 c o s x . c o s 2 x . . . . c o s 2 3 x . t a n 2 3 x f ′ ( x ) = − f ( x ) ( t a n x + 2 t a n 2 x + . . . . . . . + 2 3 t a n 2 3 x ) f ′ ′ ( x ) = − f ′ ( x ) ( t a n x + 2 t a n 2 x + . . . . . . . + 2 3 t a n 2 3 x ) − f ( x ) ( s e c 2 x + 2 2 s e c 2 2 x + 3 2 s e c 2 3 x + . . . . . + 2 3 2 s e c 2 2 3 x )
Now come to th limit
lim x → 0 x 2 1 − f ( x ) = lim x → 0 2 x − f ′ ( x ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . As 0/0 form = lim x t o 0 2 − f ′ ′ ( x ) . . . . . . . . . . . . . . . . . . . . As f’(0)=0 and also 0/0 form = 2 − f ′ ′ ( 0 ) = 2 1 2 + 2 2 + 3 2 + . . . . . . . . . . + 2 3 2 = 2 1 6 2
f ( 0 ) = 1 ; f ′ ′ ( 0 ) = f ′ ( 0 ) × 0 − ( 1 2 + 2 2 + 3 2 + . . . . . + 2 3 2 )
@Kushal Bose - Great Solution and indeed a good method for differentiation for the function. Solved in exactly the same manner
Same way did it
I actually did it with L- Hopital only . Lol
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Since the limit is x → 0 , we'll use the series expansion of the Cosine function, so that we have
x → 0 lim x 2 1 ( 1 − k = 1 ∏ 2 3 ( 1 − 2 ! k 2 x 2 + A x B + 2 ) )
x → 0 lim x 2 1 ( 1 − ( 1 − 2 ! x 2 k = 1 ∑ 2 3 k 2 + A x B + 2 ) )
x → 0 lim x 2 1 ( 1 − ( 1 − 2 ! x 2 4 3 2 4 + A x B + 2 ) )
so that the limit value is 2 1 6 2
A and B > 0 stand for arbitrary nonzero values.