series convergence/divergence tests (calc II)

This is a discussion on series convergence/divergence tests (calc II) within the A Brief History of Cprogramming.com forums, part of the Community Boards category; Alright, so tell me if this is bunk. I have gone through every problem in my book. I can look ...

  1. #1
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    series convergence/divergence tests (calc II)

    Alright, so tell me if this is bunk. I have gone through every problem in my book. I can look at the series for around 30 seconds or less, and decide whether it converges or diverges.

    Yet we have to use all these stupid tests...

    What gives?

  2. #2
    Software Developer jverkoey's Avatar
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    What do you mean? Knowing integral convergence test, nth term test, alternating series test, and etc...? All these theorems are there mainly to take in to account most of the general equations you'll run in to, so they're good to know.

    Also, they're the formal way of proving that the series converges/diverges. Just "knowing" that a series conv/div isn't good enough in academia.
    Last edited by jverkoey; 04-12-2006 at 03:10 AM.

  3. #3
    S Sang-drax's Avatar
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    Quote Originally Posted by lschmidt
    Alright, so tell me if this is bunk. I have gone through every problem in my book. I can look at the series for around 30 seconds or less, and decide whether it converges or diverges.
    You've misunderstood the purpose of the problems in your book. Their purpose is to help you understand the theory, not to learn by heart to avoid the theory.
    In real life/other courses you won't have any benefits from being able to look at a series and tell if it converges. If you learn all the 'stupid tests' though, you'll gain useful knowledge of calculus.
    Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling

  4. #4
    aoeuhtns
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    Does this series converge? You have 30 seconds.

    |cos(1)|/1 + |cos(2)|/2 + |cos(3)|/3 + ...

    Okay, that was easy. How about this one?

    1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + ...
    There are 10 types of people in this world, those who cringed when reading the beginning of this sentence and those who salivated to how superior they are for understanding something as simple as binary.

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    I hope you are kidding.

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    Software Developer jverkoey's Avatar
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    No, we never kid here.

  7. #7
    aoeuhtns
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    Okay, so both of those fairly obviously diverge. You probably couldn't prove it easily, though, unless you had the right tools available. And if somebody sat there convinced that the sum of the reciprocals of primes converged, how would you convince them?

    In math, you can't get away with going by your gut feeling, because at one point in time or another, you'll be wrong. For example, the sum of

    1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + - ...

    is log(2). Well all I'm going to do is move a few terms around. What's the value of the sum,

    1 - 1/2 + 1/3 + 1/5 - 1/4 + 1/7 + 1/9 - 1/6 + 1/11 + 1/13 - 1/8 + + - ...

    And is it possible to take a solid ball, split it into ten subsets of points, and then only by moving those subsets around rigidly, form a solid ball with half the radius? The gut feeling says no.
    There are 10 types of people in this world, those who cringed when reading the beginning of this sentence and those who salivated to how superior they are for understanding something as simple as binary.

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    How about

    SUM (1/x)

    Where x are the numbers without the digit 9?


    Or
    1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + 21/256...
    ?


    Besides, math isn't about gut feelings. It's about formal, solid, axiomatic proof.
    Code:
    #include <stdio.h>
    
    void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){
    puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9
    /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i]
    ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][
    t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}

  9. #9
    aoeuhtns
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    Interesting ones!
    Quote Originally Posted by jafet
    SUM (1/x)

    Where x are the numbers without the digit 9?
    Is less than 10*(1+1/2+1/3+1/4+1/5+1/6+1/7+1/8).

    Quote Originally Posted by jafet
    Or
    1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + 21/256...
    ?
    Is less than phi^2.
    There are 10 types of people in this world, those who cringed when reading the beginning of this sentence and those who salivated to how superior they are for understanding something as simple as binary.

  10. #10
    S Sang-drax's Avatar
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    Quote Originally Posted by Rashakil Fol
    Is less than 10*(1+1/2+1/3+1/4+1/5+1/6+1/7+1/8).
    Why?
    Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling

  11. #11
    aoeuhtns
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    You can put parentheses around pieces of the series, grouping together the fractions whose denominators have the same number of digits. E.g.

    (1/1 + ... + 1/8) + (1/10 + ... + 1/88) + (1/100 + ... + 1/888) + ...

    Since 1/(10n + r) < 1/(10n), for 1 <= r <= 8, we know that

    ((1/10 + 1/11 + 1/12 + ... + 1/18) + (1/20 + 1/21 + 1/22 + ... + 1/28) + ... + (1/80 + 1/81 + 1/82 + ... + 1/88)) <
    ((1/10 + 1/10 + 1/10 + ... + 1/10) + (1/20 + 1/20 + 1/20 + ... + 1/20) + ... + (1/80 + 1/80 + 1/80 + ... + 1/80)) =
    9*(1/10 + 1/20 + ... + 1/80) =
    (9/10) * (1/1 + 1/2 + ... + 1/8).

    This argument applies for the rest of the parenthesized groups. So the sum for each parenthesized group is less than 9/10 the sum of the previous parenthesized group. From there we know that the overall sum is less than that of a geometric series with ratio 9/10, whose first term is 1+1/2+...+1/8.
    There are 10 types of people in this world, those who cringed when reading the beginning of this sentence and those who salivated to how superior they are for understanding something as simple as binary.

  12. #12
    Rad gcn_zelda's Avatar
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    How does knowing whether a series converges or diverges come into play in "the real world"?

    I'm not doubting that it does. I just can't think of any occasion in which it would be of any assistance.

  13. #13
    aoeuhtns
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    If you're trying to invent an algorithm that approximates something, you'll want to know whether it approximates that thing well, and how quickly it can come up with that approximation. Your approximation method might only work for certain inputs, so you'd need to know which inputs these work on. Or it might only work quickly for certain inputs. With certain restrictions of input, you can have a guarantee of how long it takes to get a certain degree of approximation.

    The 'something' could be a real number, a picture, a music file, ....
    There are 10 types of people in this world, those who cringed when reading the beginning of this sentence and those who salivated to how superior they are for understanding something as simple as binary.

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