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Can you reverse engineer a number?

  1. Jan 12, 2015 #1
    If you create a sequence such as 1/4 +1/5+1/6...1/n it will tend to a limit and result in a specific number or something very close to a specific number.

    Is there a way using some method, algorithm, or formula to do the reverse? Can I start with a number and find the sequence that creates it? For example if I give you...

    1.363829

    is it possible to find a sequence of some kind that tends to that number?

    tex
     
  2. jcsd
  3. Jan 12, 2015 #2
    Depends on what you know about the number. Any number with a finite number of decimals is trivial. If it's a repeating decimal then it can be represented by a geometric series.

    In general, a number [itex]0. a_1a_2 ... = \sum a_i 10^{-i}[/itex] but that doesn't really help much.
     
  4. Jan 12, 2015 #3

    HallsofIvy

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    You can "reverse engineer" a number if the function that determines the number is 'invertible'.
     
  5. Jan 12, 2015 #4

    Mark44

    Staff: Mentor

    What you wrote above is not a sequence - it is a finite sum. You get the sum by adding the elements of this sequence: {1/4, 1/5, 1/6, ..., 1/n}.
     
  6. Jan 12, 2015 #5

    Stephen Tashi

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    That example doesn't tend to a limit. It gets arbitrarily large.

    Distinguish between a "sequence" and a "series". A sequence is an ordered set of numbers, with no implication the numbers are added together.

    If we interpret your example as a sequence, it is the ordered set of numbers
    {1/4, (1/4+1/5), (1/4 + 1/5+1/6), ....}

    Another sequence is
    {1/4, 1/5, 1/6, 1/7,....}

    An example of a finite series is

    S = 1/4 + 1/5 + 1/6.

    An example of an infinite series is the "harmonic" series:

    S = 1/1 + 1/2 + 1/3 + 1/4 + ....

    The harmonic series does not approach a limit. As you consider more terms of it, the sum becomes arbitrarily large.

    The series

    S = 1/4 + 1/5 + 1/6 + 1/7 + .....

    Also becomes arbitrarily large as you consider more terms of it.


    A simple way to take a sequence that approaches 1, such as:

    {1/2, 2/3, 3/4, 4/5, ....} and multiply each term by the number to want a sequence to approach.
     
  7. Jan 13, 2015 #6
    If you want a sequence that converges to a given number, you can just take the rounded off versions of it to higher and higher accuracy.

    So,

    3, 3.1, 3.14, 3.141, 3.1415, 3.14159..

    will converge to pi, if you keep going that way. Same for any number. But there are infinitely many sequences that will converge to it. Just add any sequence that converges to 0 to get another one.

    It's not really what I'd call "reverse-engineering" the number, though.

    The term "reverse-engineering a number" reminds me of RSA encryption where you have a function that's relatively easy to compute, but really hard to compute the inverse of. Basically, the idea is that multiplication is easy, but the inverse operation of factorization is hard. So, that's a case where at least no one knows how to reverse-engineer the numbers quickly. It's not that hard to figure out a brute force way. You just list all the prime numbers and try dividing by each until you get no remainder, then repeat until you are done, but for big numbers, this would take way too long to be practical. So, this is the secret behind a lot of internet security. Seems a lot more fitting for the idea of "reverse-engineering" than finding a sequence because you can really think of any natural number as being built out of primes, and to figure out how it was built, you need to know the building blocks from which it is made.
     
  8. Jan 13, 2015 #7

    chiro

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    If you have a function that maps some value of n to f(n) (i.e. the actual value when you compute the series) and f(n) is invertible (i.e. basically increasing or decreasing in the interval which it is in this case) then you can invert the function and use that to get the value of n.

    If your function is smooth (always differentiable) then you can use results to obtain the inverse function using derivatives if you can't get it using symbolic tricks.
     
  9. Jan 13, 2015 #8
    as everyone said this is not the case here of course ... ( it is not a converging sequence )


    for this case here , however , you can use many different ways to make a series converge into the number you want to get ...
    i am gonna show you here a very simple method here :
    as we know
    (1-X)^-1 = 1+x+x^2+......( this series goes on forever )
    here's the trick :
    let (1-X)^-1 = 1.363829
    from this we get the value of X= 0.2667702476 = 0.2668 (i simplified it for typing's sake)
    now we get the series we want
    1.363829 = 1+ (0.2668) + (0.2668)^2 + (0.2668)^3 +......... here you get your infinite series that tends to 1.363829

    now, the same number may be obtained from lots of different series ( especially if it's a rational number like the one you have given here for instance )
     
  10. Jan 13, 2015 #9

    jbriggs444

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    If you are going to "cheat", taking a convergent series and manipulating it with a chosen constant to get it to sum to the value you want, why not do it the easy way? Just use the geometric series {x, 0, 0, 0, 0...} which converges (quite quickly!) to x.
     
  11. Jan 13, 2015 #10

    Mark44

    Staff: Mentor

    What thetexan wrote is not a sequence, as has already been pointed out, but since it is a sum with a finite number of terms, it converges. I'm interpreting literally what the OP wrote. It's possible he meant to write this infinite series -- 1/4 + 1/5 + 1/6 + ... + 1/n + ... -- which is one that does not converge.
     
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