Results 1 to 15 of 15

Math Help - prove these innequalities

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    prove these innequalities



    how am i supposed to prove that the function is bigger then its tailor series??

    the taylor series are written in endless way.
    its should be equal

    how to prove that its bigger??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Let f_n(x) = e^x - \left( 1 + x + ... + \frac{x^n}{n!} \right).
    You need to prove that f_n(x) > 0.
    You can prove this by showing f ' (x) > 0\text{ for }x>0 and f(0) \geq 0.

    Prove this by induction.
    What you need to realize is that f_{n+1}(x) ' = f_n(x).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    regarding the first:
    the base case
    <br />
f_1(x) = e^x - \left( 1 + x\right)\\<br />
    i know that x>0
    but it doesnt prove that
    <br />
f_1(x)>0<br />
    ??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by transgalactic View Post
    regarding the first:
    the base case
    <br />
f_1(x) = e^x - \left( 1 + x\right)\\<br />
    i know that x>0
    but it doesnt prove that
    <br />
f_1(x)>0<br />
    ??
    First f_1(0) = e^0 - (1 + 0) = 0. Second, f_1'(x) = e^x - 1 > 0 for x>0. Thus, the function f_1 is increasing for x>0. But at x=0 we have f_1(0) = 0 which means if x>0 then f_1(x) > f_1(0) = 0. Therefore, e^x - (1+x) > 0 \text{ for }x>0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i wasnt given that x>0
    so my general function are bigger then their taylor series by the remainder of Rn

    this is the logic
    how to formulate it into equation??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by transgalactic View Post
    i wasnt given that x>0
    so my general function are bigger then their taylor series by the remainder of Rn

    this is the logic
    how to formulate it into equation??
    This inequality is not true for x<0.
    Just graph y=e^x and y=1 + x + \tfrac{1}{2}x^2 and see what happens.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by transgalactic View Post


    how am i supposed to prove that the function is bigger then its tailor series??

    the taylor series are written in endless way.
    its should be equal

    how to prove that its bigger??
    Look at the sign of the remainder term for the taylor series?

    CB
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    how to prove that??
    drawing a graph is not considered as prove

    i got the idea that i need to consider two cases x>0 then the function is smaller by the remainder
    and x<0 then the function is bigger by the remainder.


    how to describe it mathematicly
    ??
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by transgalactic View Post
    how to prove that??
    drawing a graph is not considered as prove

    i got the idea that i need to consider two cases x>0 then the function is smaller by the remainder
    and x<0 then the function is bigger by the remainder.


    how to describe it mathematicly
    ??
    Once again, consider the sign of the remainder term in Taylors theorem!

    CB
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    if the sign is positive then the function is bigger then the series
    if the sign is negative then the function is smaller then the series

    ??
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,725
    Thanks
    1478
    How many boards are you posting this on? I feel like I am chasing you around the internet! I explained this to you on "Physics Forums" but since you might confuse others:

    You are NOT asked to prove that "this function is larger than its Taylor series". You are asked to prove that it is larger than its Taylor polynomial of degree n, for any n.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i got your idea that its the function is bigger then the polynomial by the members with powers higher then "n".

    but its only a logic
    how to formulate it into mathematical equations??
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i should prove it by induction??
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by transgalactic View Post
    if the sign is positive then the function is bigger then the series
    if the sign is negative then the function is smaller then the series

    ??
    Yes, now look up one or more of the forms of remainder and determine the sign/s

    CB
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    the remainder formula is
    0<c<x
    <br />
R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}<br />

    so i guess that for x>0 its:
    <br />
R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}<br />

    for x<0
    <br />
R_n(x)=-\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}<br />

    should i prove it by induction??
    if so then if n=1 then
    <br />
e^x=1+x+ \frac{f^{1+1}(c)}{(1+1)!}x^{1+1}<br />

    but i dont know whats the value of c??
    and still it depends on the values of x
    ??
    Last edited by transgalactic; February 15th 2009 at 11:14 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove this ...
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 10th 2009, 09:47 AM
  2. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  3. how to prove these innequalities
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 6th 2009, 05:03 AM
  4. prove that
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 7th 2008, 05:14 PM
  5. prove
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 7th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum