Calculus of even and odd functions[Re-upload]

In this video, I showed the interaction between even and odd functions and how they are changed by integration or differentiation

[punditgi]

Nothing odd about Prime Newtons. He's even better than most teachers! 🎉😊

[davidgagen]

Brilliant. A sketch graph of an odd function may have helped show this clearly.

[holyshit]

What happened that you re-uploaded this video ?

[kappasphere]

You can still use even and odd numbers as an analogy for multiplying even and odd functions - but only if you say that your function has an even/odd *power* . Multiplying two values is the same as adding their powers, so even function *times* odd function is the same as even number *plus* odd number - the result will be odd, and if you instead add two numbers of the same parity, the result is even.

Also, assuming that neither function is 0, even function plus odd function will not just be unknown, but you can know that the result is neither even nor odd:
Let f be even, and g be odd. For f+g to be even, you would need f(x)+g(x)=f(-x)+g(-x) <=> g(x)=g(-x). Because g is odd, this implies g(x)=-g(x), and so g must be 0. Analogous reasoning shows that if f+g is odd, then f=0.

[glorrin]

This version is a lot better good job :)

[sudiptoatutube]

Waiting eagerly for the next video. Watching your video while having my morning breakfast, has added more taste to my food!

[AurynBeorn]

Wait, did you say... Fourier? I can't wait! 😊

[ingiford]

Posted this on the pre-release title card:
My favorate even/odd functions are:
even (e ^x) = cosh(x)
odd (e^x) = sinh(x)

and since cosh(x) + sinh(x) = e^x and that d/dx of sinh is cosh and the d/dx of cosh is sinh the derivative of the initial equation on both sides is sinh(x) + cosh(x) = e^x. and each derivative only swaps the even and odd side of the initial equation.

Also:
even (e^(ix)) = cos(x)
odd (e^(ix)) = i * sin(x)

I think those are correct.

[benjaminhogan]

I like those looks into the camara brother. 😀Keep up the good work. Great video!!!!

[Amoeby]

11:04 Actually if we integrated f(x) = 0 which is an even function, then we'd get C which is also an even function. So technically an integral of an even function isn't always an odd function.

[edwynlamb]

about the example of sinx*cosx, you could make it be more clearly by turning it into 1/2*sin(2x), which is an odd function too!

[danobro]

Dang it, i was first on this video but then he reuploaded it :(

[nanamacapagal]

There's actually some really neat proofs about everything discussed here, and some more.

For the whole comment, suppose E and F are even functions, and O and P are odd functions.

Pf1) Product rules.
E * F should be even.
(E * F)(-x)
= E(-x) * F(-x)
= E(x) * F(x)
= (E * F)(x).
O * P should be even.
(O * P)(-x)
= O(-x) * P(-x)
= -O(x) * -P(x)
= O(x) * P(x)
= (O * P)(x).
E * O should be odd.
(E * O)(-x)
= E(-x) * O(-x)
= E(x) * -O(x)
= -(E * O)(x).
You are not mistaken when you made that negatives and positives comparison, it is literally built into the proofs.

Pf2) Calculus rules.
I will start with the derivatives.
E' is an odd function.
E'(-x)
= lim_h->0((E(-x+h) - E(-x))/h)
= lim_h->0((E(x-h) - E(x))/h)
[let j = -h]
= lim_j->0((E(x+j) - E(x))/-j)
= -lim_j->0((E(x+j) - E(x))/j)
= -E'(x).
O' is an even function.
O'(-x)
= lim_h->0((O(-x+h) - O(-x))/h)
= lim_h->0((-O(x-h) + O(x))/h)
[let j = -h]
= lim_j->0((-O(x+j) + O(x))/-j)
= lim_j->0((O(x+j) - O(x))/j)
= O'(x).
EDIT: Forgot to include that the integral versions and the special integral properties...
The integral versions follow immediately from FTC1. If E' is some odd O, then Int[O] is some even E, and vice versa.
Now the special function property. Using this general property:
Int[f(x)dx, x in (-a, a)] = Int[f(x) + f(-x)dx, x in (0, a)]
Assume f is O.
Int[O(x)dx, x in (-a, a)]
= Int[O(x) + O(-x)dx, x in (0, a)]
= Int[O(x) - O(x)dx, x in (0, a)]
= Int[0, x in (0, a)]
= 0
Assume f is E.
Int[E(x)dx, x in (-a, a)]
= Int[E(x) + E(-x)dx, x in (0, a)]
= Int[E(x) + E(x)dx, x in (0, a)]
= 2 * Int[E(x)dx, x in (0, a)]

Bonus: Function decomposition (from Dr. Trefor Bazett, on hyperbolic trig)
An even function + an odd function usually can't be even or odd. But what that means is that if I carefully pick my even and odd functions, I could potentially turn their sum into any function I want. Here's how:
Suppose I had a function H(x) that I want to decompose into E(x) and O(x).
Let's take advantage of their properties.
H(x) = E(x) + O(x)
H(-x) = E(-x) + O(-x) = E(x) - O(x)
Now it's possible to solve for E(x) and O(x):
E(x) = (1/2)(H(x) + H(-x))
O(x) = (1/2)(H(x) - H(-x))
Notice that letting H(x) = e^x leads to E(x) and O(x) being the analytic definitions of hyperbolic cosine and sine, respectively.
You can check that E(x) and O(x) really are even and odd, and that their sum really is H(x).

[holyshit]

13:42
Int(f(x), x=-a..a) = Int(f(x)+f(-x), x=0..a)
Is there example for f(x) that this equation is false ?