Results 1 to 4 of 4

Math Help - Modulus Functions

  1. #1
    Member classicstrings's Avatar
    Joined
    Mar 2006
    Posts
    175
    Awards
    1

    Modulus Functions

    Hello again! I was hoping someone can help me with dervatives of mod functions, thanks!

    Considering suitable domains find the rule for the derivative function:
    y = mod(sinx), y = mod(sin2x), y = mod((e^(-0.1x))sinx) and y = mod((e^(-0.2x)sin(3x)).

    Consider y = 5(mod(e^(-ax)sin(bx)). What is the effect of changing the value of a and b?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by classicstrings View Post
    Hello again! I was hoping someone can help me with dervatives of mod functions, thanks!

    Considering suitable domains find the rule for the derivative function:
    y = mod(sinx), y = mod(sin2x), y = mod((e^(-0.1x))sinx) and y = mod((e^(-0.2x)sin(3x)).

    Consider y = 5(mod(e^(-ax)sin(bx)). What is the effect of changing the value of a and b?
    y= |sin(x)| = sgn(sin(x)) sin(x)

    Where sgn(u)=1 if u>=0 and sgn(u)=-1 if u<0

    so between the zeros of sin(x) sgn(sin(x)) is a constant so:

    dy/dx = sgn(sin(x)) d/dx sin(x) = sgn(sin(x)) cos(x).

    Now we need to check if the derivative exists when sin(x)=0, but we
    know that |sin(x)| has corners at these points so the derivative does
    not exist;

    so:

    d/dx[|sin(x)| = sgn(sin(x)) cos(x) for x in R-{k*pi, k in Z}

    This may now be simplified by observing that when u!=0 we may
    write:

    sgn(u) = u/|u|,

    so we may write:

    d/dx[|sin(x)| = [sin(x)/|sin(x)|] cos(x) for x in R-{k*pi, k in Z}

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member classicstrings's Avatar
    Joined
    Mar 2006
    Posts
    175
    Awards
    1
    Hi Ron!

    I have a few questions...

    Also - you wront mod(sinx) as being equal to (mod(sinx))*sinx, how did you get that?

    I have plotted the derivative of y = sgn(x), and I see that it could be said to be a composite function with cos(x) and -cos(x) being defined for particular values of x. However I am having trouble finding those values where they are defined.

    I actually have never seen your method, but I think I need to write my derivatives as a composite of two functions where there is no mod at all. Is your method just a simplified version of a possible composite function?
    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 classicstrings View Post
    Hi Ron!

    I have a few questions...

    Also - you wront mod(sinx) as being equal to (mod(sinx))*sinx, how did you get that?
    That is not what I did as far as I recall, if I did it is a mistake. What I did
    do was define a function sgn(u) which is equal to 1 if u>=0, and -1 if u<0.
    Essentialy this is the sign if u.

    Then |u| = sgn(u) u, since if u<0 sgn(u)=-1.
    (also sgn(u)|u| = u as 1/sgn(u)= sgn(u)).

    (here sgn is an abreviation for signum or sign)

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Holomorphic functions with constant modulus
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: June 21st 2010, 09:22 PM
  2. Modulus functions
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 18th 2010, 12:29 AM
  3. Modulus Functions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 15th 2010, 12:02 AM
  4. Sketching of graphs with modulus functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 7th 2008, 10:23 PM
  5. Inverse and Modulus Functions help please! D:
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: March 24th 2008, 04:53 AM

Search Tags


/mathhelpforum @mathhelpforum