Results 1 to 6 of 6

Math Help - Help with these questions

  1. #1
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648

    Question Help with these questions

    1.Prove that n^4 + 4^n(n \geq2) is never a prime number.


    2. Given f(n) + f(n-1) = nf(n-1) + (n-1)f(n-2)(\mbox{for}\ n\ge 2)
    and f(0)=1 and f(1) =0, prove that
    \frac{f(n)}{n!}=\sum^n_{k=0}\frac{(-1)^k}{k!}


    3. In a given circle, C is the midpoint of arc ACD. CE is perpendicular to BD(B is a point on the circle between A and C).
    If AB= l_2and BE = l_1, find ED in terms of l_1 and  l_2.
    Last edited by CaptainBlack; July 1st 2006 at 09:27 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jun 2005
    Posts
    295
    Awards
    1
    For (1) you could try listing the first few values of the formula in decimal (this is always a good idea with this sort of problem). You might also like to try listing them in base 17.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    #2 is not true
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by malaygoel


    2. Given f(n) + f(n-1) = nf(n-1) + (n-1)f(n-2)(\mbox{for}\ n\ge 2)
    and f(0)=1 and f(1) =0, prove that
    \frac{f(n)}{n!}=\sum^n_{k=0}\frac{(-1)^k}{k!}
    The solution to the difference equation initial value problem:

    f(n) + f(n-1) = nf(n-1) + (n-1)f(n-2)(\mbox{for}\ n\ge 2)

    f(0)=1 and f(1) =0

    is unique, so we need only show that f(n) defined by:

    \frac{f(n)}{n!}=\sum^n_{k=0}\frac{(-1)^k}{k!}\ \ \ (1)

    is a solution.

    ------------------------------------------------------------

    Now suppose:

    f(n)=n!\ \sum^n_{k=0}\frac{(-1)^k}{k!},

    which is just a rewrite of (1), and n \ge 2.

    Then:

    <br />
f(n)=n (n-1)!\ \left[\sum^{n-1}_{k=0}\frac{(-1)^k}{k!}+\frac{(-1)^n}{n!}\right]<br />
    <br />
=n\ f(n-1)+ \frac{n(n-1)!(-1)^n}{n!}=n\ f(n-1)+(-1)^n<br />

    Hence also:

    <br />
f(n-1)=(n-1)\ f(n-2)+(-1)^{n-1}<br />

    So:

    <br />
f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)<br />
.

    Also:

    <br />
f(0)=0! (-1)^0/0!=1,\ \mbox{and}\ f(1)=1!((-1)^0/0!+(-1)^1/1!)=0<br />
,

    That is we have shown that f(n) defined by (1) satisfies the initial value problem
    and hence is the unique solution.

    RonL
    Last edited by CaptainBlack; July 1st 2006 at 10:54 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Did you change the values of f(0) and f(1) around?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Did you change the values of f(0) and f(1) around?
    Not me, they were f(0)=1, f(1)=0 this morning when I first looked at the problem.

    I did edit it this morning to make the TeX for "for x >= 2" clearer, that's
    all honest Guv.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 04:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 04:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 03:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 07:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum