A math problem on Glenn Beck – can you solve it?

As I was flipping through the channels, I was surprised to see a math expression written on one of the chalkboards of the Glenn Beck program.

And it wasn’t something simple: the expression involved an infinite series. Can you solve the problem? (pictured below)


Here is a screen capture that I took from the Fox News website.

I magnified the equation, but you can click on the picture to see a larger image:

The problem reads:

 

Can you figure it out?

I’ll give you a hint: as with many things on Glenn Beck, you should double check the assumptions.



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  • Scott

    Given the use of sin and cos, I suspect some periodicity regarding this series.

    For n=(pi/2), (3pi/4), pi, (5pi/4), (3pi/2), (7pi/4), 2pi, (9pi/4), and (5pi/2);

    2n = pi, (3pi/2), 2pi, (5pi/2), 3pi, (7pi/2), 4pi, (9pi/2), 5pi; and

    sin(2n) = 0, -1, 0, 1, 0, -1, 0, 1, 0

    Likewise:
    cos(n) = 0, (-sqrt(2)/2), -1, (-sqrt(2)/2), 0, (sqrt(2)/2), 1, (sqrt(2)/2), 0;

    cos^4(n) = 0, (1/4), 1, (1/4), 0, (1/4), 1 (1/4), 0; and

    1+cos^4(n) = 1, (5/4), 2, (5/4), 1, (5/4), 2, (1/4), 1;

    This gives sin(2n)/(1+cos^4(n)):

    0, -(4/5), 0, (4/5), 0, (-4/5), 0, (4/5), 0

    The sum of which is, of course, 0.

    However, the any partial sum will only be zero if n is a multiple of (pi/2). For any arbitrary value of n, the sum of that partial series will fluxtuate from (-4/5) to (4/5), which to me says it is not convergent.

  • Faraz

    Since the index spans natural numbers, I would imagine sin and cos are in degrees. Under this assumption:

    Write f(n) = sin(2n)/(1+cos^4(n))

    Put n = k (mod 360) (congruence)

    Clearly, f(n) = f(k).

    Note that for k = 0, 1,… 180, f(k) + f(180-k) =
    f(k) + sin(360-2k)/(1+cos^4(180-k))=

    f(k) + -sin(2k)/(1+cos^4(k)) since the fourth power of cos means the sign of cos k doesn’t matter.

    But this equals zero.

    Likewise, for k = 181, 182… 359,
    f(k) + f(360-k) =
    f(k) + sin(720 – 2k)/(1+cos^4(360-k))=
    f(k) + -sin(2k)/(1+cos^4(k)) = 0.

    Hence SUM f(n) for n = 0,1,… 359 (mod 360) = 0.

    This is kind of a telescoping sum, but it’s not convergent by the true definition (due to the periodicity of f(n)).

    In radians, however, the story would be different. Note that if m and n are natural numbers, it’s not possible that sin (n) = sin (m) if its measured in radians. I’m not sure what the answer would be in that case.

  • sdfdsv

    ahahaha it’s april 1st, check the limit test

  • http://www.youtube.com/user/JuusoAlasuutari Juuso Alasuutari

    Number goes up, another number goes down. I can’t explain that.

  • uri vasas

    It can be solved using the integral test for convergence (I am cheating, because I saw the program and the the kid started it that way, but I believe I could’ve figured it out anyway)

    Take the integral of the stated function from 1 to infinity.substitute u = (cosx)^2 then du = -2cosxsinxdx.
    The numerator of the integrand is equivalent to 2cosxsinx. So the entire integrand is equivalent -1/(1+u^2).

    I will not change the bounds of the integral as this is tricky to do with the upper bound going to infinity and u being a periodic function.

    once you integrate you have arctan((cosx)^2) going from 1 to infinity. This is equal to arctan((cos1)^2) – lim ( t->infinity)arctan((cos(t))^2) =:I

    I just realized that this next step is iffy (and I think this means the sum doesn’t converge) but I will do it anyway. Here it is: because arctan(y) >= -pi/2, (or -arctan(y) <= pi/2) we get that
    I <= arctan((cos1)^2) + pi/2 < infinity. Which means that the sum converges.

    The reason this is iffy is that although IF I exists then it is less than infinity, this does not mean that I exists. In fact as t goes to infinity, cos(t)^2 oscillates from zero to one making arctan(cos(t)^2) oscillate as well making the limit divergent. Which means the sum diverges.

    Way to go, Glenn Beck. You know calculus about as well as you know politics. I wonder if the kid would have spotted the error.

  • guy

    Uri, recall the restrictions must be taken into taken consideration before we can use the integral test for convergence. Our series must be continuous, positive, and monotonically decreasing on [1,inf.) A quick graph of the series shows that it fails two of these criteria, so we cannot use the integral test for convergence in this case.

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Someone found out the context of this question was Glenn Beck was interviewing a 12 year old math genius Jacob Barnett.

    Here is the clip: http://www.youtube.com/watch?v=hBW4S9xcTOk&t=9m39s

    Jacob starts to solve by using the integral test to show it does converge.

    I believe this is a mistake. As the sequence does not converge, the sum does not converge either by the limit test.

    The answer can also be calculated in Wolfram Alpha: http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+n%3Dinf+%28sin%282n%29%2F%281%2Bcos^4%28n%29%29%29

    Special thanks to Reddit comments for the Youtube and Wolfram Alpha links.

  • Pingback: Are Pillsbury Toaster Strudels tastier than Kellogg’s Pop Tarts – ads I overanalyze - Mind Your Decisions

  • http://asad123.wordpress.com Asad

    I showed this to my friend Junaid who explained it very simply. Look at the ranges of the numerator and denominator. The numerator, being a sine, ranges from -1 to 1. The demoninator ranges from 1 to 2. The limit of the series doesn’t approach 0, and it’s not an alternating series where positives and negatives cancel, so it must diverge.

    I noticed on YouTube people are saying this kid is so amazing. I commented, under the handle ibnsina786, that he has his math wrong and directed them to come here for proof. I hope you get some traffic out of it.

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Thank you Asad, that was really nice to post a link on Youtube.

  • Cccil Rousseau

    Previous solvers have had everthing in place except a rigorous proof that f(n) doesn’t approach zero. Let’s review. (i) The integral test doesn’t apply because the integrand is not monotone decreasing.
    (ii) If we prove that f(n) doesn’t approach 0 the game is over and the series diverges, (iii) We might as well change the nth term to sin(n) because the denominator remains beween 1 and 2; sin(n)/(1+ cos^4(n)) is small iff sin(n) is small. (iv) Now for the slightly technical part. The terms of the series are sin(1), sin(2), sin(3), … where 1,2,3,… are radians measure. The sine function is periodic with period 2 pi. Take the domain to be [0, 2 pi]; then the terms are sin(1), … sin(6), sin(7-2 pi), sin(8-2 pi),… Think of a frog jumping from point to point on the interval [0, 2 pi]. If he is at point x, then his next jump will be to x + 1 mod 2 pi. Note that |sin x| is small iff x is in one of the neighborhoods (lilypads)|x| < delta, |x – pi| < delta, or |x- 2 pi| < delta where delta is small. Use contradiction. Suppose sin(n) converges to 0. At some point our frog is on a lilypad and needs to jump to another one a distance of approximately pi or 2 pi away and his only option is a jump of +1. This is impossible. The series diverges.

    Note. Think of the process of going from f(x) to f(n) as sampling; informatin is lost. (In effect that's why you need the monotone condition in the integral test.) The example f(x) = sin(pi x) has f(n) = 0 for all n and yet f(x) does not approach 0. It may seem like overkill but we have to have some kind of argument – we just can't assume that f(n) does not converge to 0 because f(x) does not converge to 0.

  • Cccil Rousseau

    Sorry. I was doing this from memory and I forgot that it was sin(2n). Please withdraw this and I will modify the argument.

  • Cccil Rousseau

    My failure to remember the factor of 2 is easily fixed. With the correct term sin(2n), there are 5 lilypads instead of 3, and these are neighborhoods of 0, pi/2, pi, 3pi/2, and 2pi (the points in [0, 2pi] where sin(2n) = 0). Again the frog finds himself on one of the lilypads and needing to jump to another one with a jump of +1. If the size of the lilypad is small enough then this is impossible. This argument can be extended to sin(kn) and ultimately depends on the irratioanality of pi.

  • Adrian

    the series can be evauluated using the integral test which proves it is divergent simply because at the end you’ll have arctan(cos^2(n)) as n approaches infinity …this already leaves an indeterminate bound between 0 and pi over two since cosine squared oscillates from 0 to 1 hence this is a divergent infinite series…. although its a nice one :)

  • Stephen

    When I saw the problem, I assumed that it was a typo, and they meant sinh and cosh instead of sin and cos.

  • Haoha

    It is not convergent. In fact, the sequence does not even tend to zero. There are infinitely many summands greater than one.
    You cannot use integral criterion because series changes sign

  • Haohan00

    I am very confused. It’s virtually trivial that the sequence sin(2n)/(1+cos^4(n)) doesn’t tend to 0. And therefore its sum cannot be convergent. What’s all this messing around with radians/degrees/integrals??

    (This is clear from the fact that sum(xn) convergent implies xn tends to 0)

  • Haohan00

    Perhaps my comment above about radians/degrees was wrong. 

    But still, to clarify, I say 
    sin(2n)/(1+cos^4(n)) doesn’t tend to 0 because:l sin(2n)/(1+cos^4(n)) l = l sin(2n) l / l 1+cos^4(n) l
                                     >= l sin(2n) l / 2The RHS here clearly doesn’t tend to 0 in degrees as we periodically hit sin(2), which isn’t 0.

    In radians, 2n would have to “tend” to a multiple of pi/2 (I realise this is horrible use of the term “tend”) in order for sin(2n) to converge to 0. But this cannot be true as 2 doesn’t divide pi/2, and vice versa, a whole number of times. (This statement perhaps requires proof, but should be fairly intuitive)

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