Page 1 of 2 12 LastLast
Results 1 to 15 of 23

Math Help - Inductin guidance

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    23

    Inductin guidance

    I did not have an issue with the topic when it was relatively straight forward in that it gave f(0) and then wanted f(1),f(2),f(3)....

    The problem arose when they gave f(0) and f(1) and wanted f(2),f(3),f(4),f(5)...

    I'm going to pick a non homework portion of the problem:

    Find f(2),f(3),f(4),F(5) if f is defined recursively by f(0)=f(1)=1 and for n=1,2,..

    With f(n+1)=f(n)f(n-1)

    Thanks for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    So, f(n+1)=f(n)f(n-1), f(0)=f(1)=1

    f(2)=f(1+1)=f(1)f(1-1)=f(1)f(0)=1 \cdot 1=1

    f(3)=f(2+1)=f(2)f(2-1)=f(2)f(1)=1 \cdot 1=1 ( f(2)=1

    Actually, by induction you can prove that for every natural n, f(n)=1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2010
    Posts
    23
    Thanks, I guess it probably really didn't help me understand the process with f(0) and f(1) both = 1. What if like f(0)=2 and f(1)=3? Which is where I am mixed up, attempting to determine the relationship betweeen f(0) and f(1) in determing the remaining functions. Thanks though, your reply gave me some insight, I'm just trying to figure out how to work with these two together. like f(2)=5=5(5-1)=20 f(3)=(21+1)=22(22-21)...? If you were to run through your process through these, or smaller (if others too high) I feel certain I would understand completely. Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2010
    Posts
    23
    So then as I understand it. If the equation becomes f(n+1)=f(n)-f(n-1)

    This is what I get. Is it correct? Y/N is suffice.

    f(0)=f(1)=1
    f(n+1)=f(n)-f(n-1)
    f(2)=f(1)-f(1-1)=1-0=1
    f(3)=f(2)-f(2-1)=2-1=1
    f(4)=f(3)-f(3-1)=3-2=1
    f(5)=f(4)-f(4-1)=4-3=1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Correct! So, what can you say about f(n)?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2010
    Posts
    23
    Quote Originally Posted by Also sprach Zarathustra View Post
    Correct! So, what can you say about f(n)?
    Not really sure of the wording. But f(n) will always be 1 in this case. Not sure if that is what you are asking. Or is it something else? LOL Gimme a hint on it, I'm holding on to the seat of my pants here. On one hand I love the course, haven't had to think this hard in years, on the other I'm not sure I've disliked a course at much at the same time either. (Protective of GPA).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    You wright again! f(n) will always be 1 in this case. Try to prove it via induction!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jul 2010
    Posts
    23
    Basis step: p(1)....f(n+1)=f(n)-f(n-1)=1=f(1)-f(1-1)=1-0=1.....1=1

    Inductinve step: 1+2+..+(k+1)-(k+1-1)
    k(2)=k+1-k=1.
    k(3)=k+1-k=1.....

    Really not so sure I did that right, although I did establish the formula always came back as 1, I'm confused because I thought the 1+2+....had to be added, and that does not balance. Feel free to point me in the right direction.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Jul 2010
    Posts
    23
    One more question if I may (You've been awesome help). I need to give a recursive definition of the sequence {a1},n=1,2,3....if and then four formulas. I will pick one of the ones that are not HW and see if I can grasp this procedure. an=4n-2.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    I'm so sorry! You did some mistake and I am not noticed it!

    I will start from the beginning!

    f(n+1)=f(n)-f(n-1), when f(0)=f(1)=1

    f(0)=1 it's given.

    f(1)=1 it's given.

    f(2)=f(1)-f(0)=0

    f(3)=f(2)-f(1)=0-1=-1

    f(4)=f(3)-f(2)=-1-0=-1

    f(5)=f(4)-f(3)=-1-(-1)=0

    f(6)=f(5)-f(4)=0-(-1)=1

    f(7)=f(6)-f(5)=1-0=1

    f(8)=f(7)-f(6)=1-1=0

    So, what we can see now is a sequence...

    1,1,0,-1,-1,0,1,1,0,...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Jul 2010
    Posts
    23
    No worries, I think I have it down. Which is mostly looking for here. An idea of the process, not someone to do my work for me. I would rather fail that let that be the case.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Nickolase View Post
    No worries, I think I have it down. Which is mostly looking for here. An idea of the process, not someone to do my work for me. I would rather fail that let that be the case.
    Words of wisdom!

    Could you maybe be more specific on your second question?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Jul 2010
    Posts
    23
    Quote Originally Posted by Also sprach Zarathustra View Post
    Words of wisdom!

    Could you maybe be more specific on your second question?
    I can try.

    Give a recursive definition of the sequence {an},n=1,2,3.....if an=4n-2 (The An's are sub n's, just not sure how to do that yet) all the information that is given. If necessary I can post the other similar problem with answer.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Nickolase View Post
    I can try.

    Give a recursive definition of the sequence {an},n=1,2,3.....if an=4n-2 (The An's are sub n's, just not sure how to do that yet) all the information that is given. If necessary I can post the other similar problem with answer.

    a_n=4n-2

    a_1=4\cdot 1 -2=4-2=2
    a_2=4\cdot 2 -2=8-2=6
    a_3=4\cdot 3-2=12-2=10
    a_4=14

    We can obtain that recursive formula of a_n is:

    a_{n+1}=a_{n}+4(*) {an arithmetic sequence}

    Now, we will check if this recursive formula is correct:

    a_n=4n-2

    so, a_{n+1}=4(n+1)-2=4n+4-2=4n+2


    Now we put the two to (*) and see what happens:


    4n+2=(4n-2)+4

    4n+2=4n+2

    0=0

    hence, we guess recursive formula well!
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Jul 2010
    Posts
    23
    Wow, crazy good. Did this stuff come naturally to you? I thought I was very good with math until this course. It is a wake up call to my sensabilities. Thanks so much.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Integration guidance please
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 1st 2011, 03:37 PM
  2. Replies: 0
    Last Post: February 1st 2010, 05:09 AM
  3. Need Guidance...
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 27th 2009, 07:12 PM
  4. Need Some Guidance
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 24th 2009, 05:37 PM
  5. Need some derivatives guidance
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 28th 2007, 02:48 PM

Search Tags


/mathhelpforum @mathhelpforum