1. Standard memberclandarkfire
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    14 Oct '10 20:222 edits
    Find the next term in this series:

    3/17, 22.5, -45/3, 1.25pi, 4e

    Hint: write a formula for term n
  2. Joined
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    27 Nov '10 21:27
    Originally posted by clandarkfire
    Find the next term in this series:

    3/17, 22.5, -45/3, 1.25pi, 4e

    Hint: write a formula for term n
    Looks like original thinking from someone with hard brains is required! 😳

    Anyway, if there is a formula for a general term of this sequence, F(n), say, then we must have for some natural number k,

    F(k) = 3/17, F(k+1) = 45/2, F(k+2) = -45/3, F(k+3) = 5/4pi, F(k+4) = 4e.

    One thing that can be said about numbers pi and e is that they are transcendental, ie they cannot be roots of (non-constant) polynomial equations with rational coefficients. So F(n) must be either
    a) a polynomial in n with at least some of the coefficients irrational numbers (but most probably not, because of the three rationals in the sequence), or
    b) a non-polynomial expression in n. Two 45s and pi would kind of suggest that trigonometric functions might be involved. Pi and e - may be infinite sums? Integrals? Logarithms? Maybe Euler's identity is used in some way?

    No clue here. Waiting impatiently for the answer to be revealed!
  3. Joined
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    28 Nov '10 20:21
    Originally posted by clandarkfire
    Find the next term in this series:

    3/17, 22.5, -45/3, 1.25pi, 4e

    Hint: write a formula for term n
    1.25pi is an interesting number
    sin(1.25pi) = cos(1.25pi)
    tan(1.25pi) = 1
  4. Standard memberclandarkfire
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    28 Nov '10 20:51
    I'm surprised this problem went so long unanswered.
    As it turns out, it's a bit of a trick: the numbers really have nothing to do with each other. Even so, it is still possible to write a formula for term n, using a sneaky trick of Joseph LaGrange's.

    Here is a formula for term n:

    Tn= (3/17)(n-2)(n-3)(n-4)(n-5)/24 + (22.5)(n-1)(n-3)(n-4)(n-5)/-6 + (-15)(n-1)(n-2)(n-4)(n-5)/4 + (1.25pi)(n-1)(n-2)(n-3)(n-5)/-6 + (4e)(n-1)(n-2)(n-3(n-4)/24

    Well, the numbers that seem to be picked at random, (they were), now are all connected through the formula. So, to find term 6, just plug 6 in for n.

    Term 6 = approximately -247.23

    😞
  5. Joined
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    28 Nov '10 23:20
    That was cheap... By the way, sequence and series are two different things. But thanks for your puzzle!

    Here's a new one.

    Find another answer to this problem. Ie, find a formula which gives the first 5 terms of the OP sequence exactly the same, but the 6th one (and all consecutive terms) is different.

    Hint: there are infinitely many of them 😕
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    28 Nov '10 23:24
    Are there? How so?
  7. Joined
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    29 Nov '10 09:212 edits
    Originally posted by kalvinator
    Are there? How so?
    I think the given solution is an order N polynomial that goes through N+1 points.

    Polynomials usually have a few solutions, so there are probably a few more order 4 polynomials that would go through those 5 points, not an infinite number though.

    But, because as you raise the order the curve can have more bends, there will be infinitely many order N+1 polynomials that go through those points, and infinitely many order N+2 polynomials etc.
  8. Joined
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    02 Dec '10 22:54
    Yes, basically any F(n), polynomial or not, such that F(1) = 3/17, ..., F(5) = 4e, as long as it is valid for all natural numbers n, is a potential source of such answers. The first 5 terms will be the same as in our sequence, but the 6th one will differ in general. Consider this slightly altered clandarkfire's formula:

    F(n,p,r,s,t) = (3/17)sin[(n-2)(n-3)(n-4)(n-5)]/sin(24) + (22.5)[(n-1)(n-3)(n-4)(n-5)/-6]^p + (-15)[(n-1)(n-2)(n-4)(n-5)/4]^r + (1.25pi)[(n-1)(n-2)(n-3)(n-5)/-6]^s + (4e)[(n-1)(n-2)(n-3)(n-4)/24]^t

    Setting n=6 and p=r=s=t=1 gives clandarkfire's answer, although from different (and non-polynomial) formula. For other (allowed) p, r, s, t values it will give different formula's, and, in general, different answers, but the first 5 terms will remain the same in each case.

    Perhaps an easier way to think of it is as follows. If the formula for 5 terms can be constructed, then it can be done for 6 (and 7, 8, ...). So we can pick some arbitrary real number R, and build the new formula using the same sneaky Lagrange polynomials:

    G(n) = (3/17)(n-2)(n-3)(n-4)(n-5)(n-6)/-120 + (22.5)(n-1)(n-3)(n-4)(n-5)(n-6)/24 + (-15)(n-1)(n-2)(n-4)(n-5)(n-6)/-12 + (1.25pi)(n-1)(n-2)(n-3)(n-5)(n-6)/12 + (4e)(n-1)(n-2)(n-3)(n-4)(n-6)/-24 + R(n-1)(n-2)(n-3)(n-4)(n-5)/120

    Then G(1) = 3/17, ..., G(5) = 4e, G(6) = R, ...

    But R is arbitrary - so, in fact, any real number can be the answer to the OP problem.
    This can be applied to any "What is the next term in this or that sequence" type problem. For instance, what is the next term in the sequence F(1) = 1, F(2) = 2, F(3) = 3 ? One might think that F(n) = n, and the answer is 4. But what about

    F(n) = (n-2)(n-3)(n-4)/-6 + 2*(n-1)(n-3)(n-4)/2 + 3*(n-1)(n-2)(n-4)/-2 + 7.34*(n-1)(n-2)(n-3)/6 ?

    Now F(1) = 1, F(2) = 2, F(3) = 3, but F(4) = 7.34. F(4) can be any number, using the same argument as above.

    This is a bit confusing, it has to be said. Perhaps a way I would try to explain it is this. When we deal with non-random number sequences in some mathematical context, we expect them to be unique. To assure us of that we are usually given the index, with which the sequence begins, and a rule, from which the sequence can be determined uniquely - a formula for a general term, a recurrence relation, a generating function for that sequence, whatever. And if there is no such rule specified, then the sequence must be random? In which case, the next term is just a random number.

    😕😕
  9. Joined
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    03 Dec '10 13:052 edits
    Originally posted by clandarkfire
    Find the next term in this series:
    3/17, 22.5, -45/3, 1.25pi, 4e

    ...

    I'm surprised this problem went so long unanswered.

    Term 6 = approximately -247.23
    Is this the only answer?

    If so, it's not a well defined problem, if it's not stated explicitly.

    The question was "Find the next term" as if the answer was an unique one.
    The question should be "Find any next term" or something.
  10. Joined
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    04 Dec '10 00:44
    Is there an integer solution?
  11. Joined
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    04 Dec '10 00:45
    oops, of course there is, sorry
  12. Joined
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    06 Dec '10 16:53
    Originally posted by FabianFnas
    Is this the only answer?

    If so, it's not a well defined problem, if it's not stated explicitly.

    The question was "Find the next term" as if the answer was an unique one.
    The question should be "Find any next term" or something.
    It's sort of a parody, though, isn't it? "What comes next" questions are asked all the time as if there was a "natural" next answer. This shows that there's always other possibilities.
  13. Standard memberPalynka
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    06 Dec '10 17:36
    Originally posted by mtthw
    It's sort of a parody, though, isn't it? "What comes next" questions are asked all the time as if there was a "natural" next answer. This shows that there's always other possibilities.
    Yep. I tend to avoid the "what comes next" questions because of this. You're basically trying to second guess what the other thought but there's never a unique solution.
  14. Joined
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    06 Dec '10 17:592 edits
    Originally posted by Palynka
    Yep. I tend to avoid the "what comes next" questions because of this. You're basically trying to second guess what the other thought but there's never a unique solution.
    I remember being faced with one at school (I was about 11 or 12) where for some reason I missed the blatantly obvious answer, but managed to come up with a convoluted algorithm that fit the series given.

    1, 8, 27, 64, ?

    It's obviously 117, isn't it? 🙂

    (so, can anyone work out what reason I came up with?)
  15. Standard memberRevRSleeker
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    06 Dec '10 22:06
    Originally posted by mtthw
    I remember being faced with one at school (I was about 11 or 12) where for some reason I missed the blatantly obvious answer, but managed to come up with a convoluted algorithm that fit the series given.

    1, 8, 27, 64, ?

    It's obviously 117, isn't it? 🙂

    (so, can anyone work out what reason I came up with?)
    cubes of 1 through 4..the 5th, 5x5x5, 125 is the answer.
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