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Odd & Even Functions (was thread Fourier Series )

  • Thread starter eddievic
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  • #1
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Odd & Even Functions (was thread "Fourier Series")

Homework Statement


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)


Homework Equations



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



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.
 

Answers and Replies

  • #2
DrClaude
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I've managed the first algebraically:

x=1
f(1)=1+2=3
x=-1
f(-1)=1+2=3
Don't substitute numbers:
$$
\begin{align}
f(-x) &= (-x)^2 +2 \\
&= x^2 + 2 \\
&= f(x)
\end{align}
$$

Try that with the other functions.
 
  • #3
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so next would be:

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

so f would be odd?
 
  • #4
Zondrina
Homework Helper
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so next would be:

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

so f would be odd?
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.
 
  • #5
DrClaude
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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.
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.
 
  • #6
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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.
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)
 
  • #7
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10
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 :=)
 
  • #8
vela
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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)
The second line should say "f(-x) = ..." Why didn't you change sin x to sin (-x)? Try again.
 
  • #9
epenguin
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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.
: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.
 
  • #10
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The second line should say "f(-x) = ..." Why didn't you change sin x to sin (-x)? Try again.


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

so function is odd?
 
  • #11
epenguin
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f(x)=(x^2+2)sin(x)tan(x^2)
=((-x)^2+2)sin(-x)tan((-x)^2))
= f(-x)

so function is odd?
:surprised. f(x) = f(-x)

f is odd, even or neither?
 
  • #12
DrClaude
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f(x)=(x^2+2)sin(x)tan(x^2)
=((-x)^2+2)sin(-x)tan((-x)^2))
= f(-x)

so function is odd?
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?
 
  • #13
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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.
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.
 
  • #14
epenguin
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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.
I will not cast the first stone. :biggrin:
 
Last edited:
  • #15
vela
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f(x)=(x^2+2)sin(x)tan(x^2)
=((-x)^2+2)sin(-x)tan((-x)^2))
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.
 
  • #16
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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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