Results 1 to 5 of 5

Math Help - sequences and summations...i don't even know where to start...

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    5

    sequences and summations...i don't even know where to start...

    The homeworks are getting overwhelming...I have NO IDEA where to start...I haven't taken a math class in 10 years (calculus II).

    Let a-sub-n be the nth term f the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6,..., constructed by including the integer k exactly k times. Show that a-sub-n = floor(sqrt(2n) + 1/2).

    I honestly don't even know what the first step is or what I should even be thinking about...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jan 2010
    Posts
    5
    This is like...the very first problem of only our 5th homework assignment...they're already too hard for me. Anyone?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello teamtrinity
    Quote Originally Posted by teamtrinity View Post
    The homeworks are getting overwhelming...I have NO IDEA where to start...I haven't taken a math class in 10 years (calculus II).

    Let a-sub-n be the nth term f the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6,..., constructed by including the integer k exactly k times. Show that a-sub-n = floor(sqrt(2n) + 1/2).

    I honestly don't even know what the first step is or what I should even be thinking about...
    OK, here's where to start.

    The last of the terms that has value k-1 is in i^{th} position where:
    i = 1 + 2 + 3 + ... + (k-1) = \tfrac12k(k-1)
    and, similarly, the last of the integers with value k is in the position \tfrac12k(k+1).

    So for \tfrac12k(k-1) + 1\le n \le \tfrac12k(k+1),\; a_n = k.

    Can you continue?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2009
    Posts
    90
    Hi Grandad,
    how did you find out that that was the place to start? Can you explain a bit more, what's going on here? I see what you're saying. You spotted the relationship that:

    1 = 1
    2 = 3
    3 = 6
    4 = 10

    but like teamtrinity, I would have started with the following
    \lfloor \sqrt{2n} + \frac{1}{2} \rfloor \Leftrightarrow x \leq \sqrt{2n} + \frac{1}{2} < x + 1

    and then probably not know how to go further! How did you work it out?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello bmp05
    Quote Originally Posted by bmp05 View Post
    Hi Grandad,
    how did you find out that that was the place to start? Can you explain a bit more, what's going on here? I see what you're saying. You spotted the relationship that:

    1 = 1
    2 = 3
    3 = 6
    4 = 10

    but like teamtrinity, I would have started with the following
    \lfloor \sqrt{2n} + \frac{1}{2} \rfloor \Leftrightarrow x \leq \sqrt{2n} + \frac{1}{2} < x + 1

    and then probably not know how to go further! How did you work it out?
    First, let's just complete the proof. We've shown that the terms with value k are in position n, where:
    \tfrac12k(k-1)+1\le n \le \tfrac12k(k+1)

    \Rightarrow k(k-1) + 2 \le 2n \le k(k+1)
    Completing the square for the LH inequality:
    (k-\tfrac12)^2 +\tfrac74\le 2n

    \Rightarrow k \le \sqrt{2n-\tfrac74}+\tfrac12

    \Rightarrow k < \sqrt{2n} + \tfrac12
    And in the same way for the RH inequality, we get:
    2n\le (k+\tfrac12)^2-\tfrac14
    \Rightarrow \sqrt{2n+\tfrac14} - \tfrac12 < k

    \Rightarrow k +1> \sqrt{2n} +\tfrac12
    So, putting the two together:
    \Rightarrow k < \sqrt{2n} + \tfrac12<k+1

    \Rightarrow k =a_n= \lfloor\sqrt{2n} + \tfrac12\rfloor
    So how did I know that was the place to start? The honest answer is that I didn't. It was simply that this was the only way I could think of to describe what was happening. Clearly the position of the last one of any given set of integers, value k, was the sum of the first k natural numbers which is well known to be \tfrac12k(k+1). So it was just a question of working out where these integers would start, and writing all this down algebraically.

    Until I just worked out the above proof of the actual result involving the floor function, I had no idea what would happen next. But if you begin by writing down an algebraic condition to describe something that you know to be true, then the result should follow automatically. But sometimes, of course, it still isn't that easy!

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Summations help!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 15th 2011, 08:54 AM
  2. Summations
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: December 6th 2010, 08:22 AM
  3. Sequences and Summations (1)
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 4th 2009, 03:42 PM
  4. summations
    Posted in the Algebra Forum
    Replies: 6
    Last Post: December 16th 2008, 02:41 PM
  5. Sequences and Summations Proof (Pre-imaging)
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: October 17th 2008, 06:06 AM

Search Tags


/mathhelpforum @mathhelpforum