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Odd & Even Functions (was thread Fourier Series )

  1. Sep 13, 2013 #1
    Odd & Even Functions (was thread "Fourier Series")

    1. The problem statement, all variables and given/known data
    determine if the functions below are odd even or neither:
    a) f(x)=x^2+2
    b) f(x)=(x^2+2)tan(x^2)
    c) f(x) = (x^2+2)sin(x)tan(x^2)


    2. Relevant equations

    even - f(x) = f(-x)
    odd - f(-x)=-f(x)



    3. The attempt at a solution

    I've managed the first algebraically:

    x=1
    f(1)=1+2=3
    x=-1
    f(-1)=1+2=3

    so f is even

    I reckon I need to do the same for the next function however do not know how to use tan in that equation.

    Any help is appreciated.
     
  2. jcsd
  3. Sep 13, 2013 #2

    DrClaude

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    Don't substitute numbers:
    $$
    \begin{align}
    f(-x) &= (-x)^2 +2 \\
    &= x^2 + 2 \\
    &= f(x)
    \end{align}
    $$

    Try that with the other functions.
     
  4. Sep 13, 2013 #3
    so next would be:

    f(x)=(-x^2)tan(-x^2)
    =x^2+2tanx^2
    =f(x)

    so f would be odd?
     
  5. Sep 13, 2013 #4

    Zondrina

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    Not quite. I'm not sure what you're doing there ( You're using the wrong function and you're omitting a negative sign ), but this is what it should look like:

    b) ##f(x)=(x^2+2)tan(x^2)##

    ##f(-x) = (-x^2 + 2)tan(-x^2)##
    ##f(-x) = (x^2 + 2) + tan(x^2)##
    ##f(-x) =f(x)##

    Try the third one now.
     
  6. Sep 13, 2013 #5

    DrClaude

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    What is that?

    $$
    \begin{align}
    f(-x) &= ((-x)^2 + 2)\tan((-x)^2) \\
    &= (x^2 + 2)\tan(x^2) \\
    &= f(x)
    \end{align}
    $$
    That one is even again.
     
  7. Sep 13, 2013 #6
    so last one would be:

    f(x)=(x^2+2)sin(x)tan(x^2)
    =((-x)^2+2)sin(x)tan((-x)^2))
    = f(x)
     
  8. Sep 13, 2013 #7
    Think about why you have such conditions for odd or even functions, must mean they are symmetrical in relation to something.
    When you multiply integers together, your products are also even or odd, no different with products of different functions. If you are in doubt of your calculations, consider the factors and see if they are odd or even functions.
    If you multiply 2 odd numbers your result is an odd one.
    For example b) you already showed a is an even function. Doesn't matter if the other factor is even or odd, the result will be even regardless.
    You can easily double check yourself like this, hope that helps :=)
     
  9. Sep 13, 2013 #8

    vela

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    The second line should say "f(-x) = ..." Why didn't you change sin x to sin (-x)? Try again.
     
  10. Sep 13, 2013 #9

    epenguin

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    :surprised won't an even function X an odd one be odd?

    The rules are just the same as for number addition (not multiplication) , even X even or odd X odd are even and odd X even is odd - and it should be very simple to see this.
     
  11. Sep 16, 2013 #10


    f(x)=(x^2+2)sin(x)tan(x^2)
    =((-x)^2+2)sin(-x)tan((-x)^2))
    = f(-x)

    so function is odd?
     
  12. Sep 16, 2013 #11

    epenguin

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    :surprised. f(x) = f(-x)

    f is odd, even or neither?
     
  13. Sep 16, 2013 #12

    DrClaude

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    This makes no mathematical sense. Why is there a minus sign appearing in the second line? You have the definition of ##f(x)##, and therefore you start from ##f(-x)## and use that definition:
    $$
    \begin{align}
    f(-x) &= ((-x)^2 +2) \sin(-x) \tan((-x)^2) \\
    &= (x^2 + 2) (-\sin(x)) \tan(x^2) \\
    &= - (x^2 + 2) \sin(x) \tan(x^2) \\
    &= -f(-x)
    \end{align}
    $$
    So the is function is even, odd, or neither?
     
  14. Sep 16, 2013 #13
    Pardon me, I don't know what I was thinking when I wrote that, of course it simple to see why.

    Post #7 is a wheelbarrow full of manure and should be considered as such.
     
  15. Sep 16, 2013 #14

    epenguin

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    I will not cast the first stone. :biggrin:
     
    Last edited: Sep 16, 2013
  16. Sep 16, 2013 #15

    vela

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    This is incorrect because you changed x to -x only on the righthand side of the equation. Remember that you have to treat both sides the same way.

    So to expand on what DrClaude did, you have
    $$f(x) = (x^2+2)(\sin x)(\tan x^2).$$ That was given to you. Now you're interested in what f(-x) equals to see how it compares to f(x). If we change x to -x on the lefthand side of the equation (so that we have f(-x) which is what we're interested in), we have to change x to -x on the righthand side of the equation as well.
    $$f(-x) = ((-x)^2+2) (\sin (-x)) (\tan (-x)^2).$$ At this point, you don't know how f(x) compares to f(-x). They could be equal, differ by a sign, or neither. After a little simplification, however, you can show that
    $$f(-x) = -(x^2+2)(\sin x)(\tan x^2).$$ Everything after the minus sign on the righthand side happens to be exactly what f(x) equals, so we can say that
    $$f(-x) = -[(x^2+2)(\sin x)(\tan x^2)] = -f(x).$$ So we have that f(-x)=-f(x). So what does that make f(x) — even, odd, or neither?

    Try looking over posts 2 and 4 again. You should see that DrClaude and Zondrina followed the same basic pattern.
     
  17. Sep 17, 2013 #16
    Just read back over this thread and now it makes sense sometimes you cannot see the wood for the trees thanks all
     
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