oh man, wow! that's just soooo cool. Firstly I would have never thought of splitting a general function in an even and an odd term, that's just so genious and it feels so natural too but secondly splitting up e^x in that way and deriving cosh and sinh from that... man that's just wow (also because it's so simple) One of the few vids where I wish I could like 10 times
How do we know that there's a unique decomposition in even and odd part? Because if that's non the case, we can't refer to cosh as "THE even part of e^x".
If you shift the even part up or down with a +c, you'd still get an even function, but the odd part would need a -c to compensate which would no longer make it odd since it's not antisymmetric around 0, eliminating any other even-odd pair decompositions of a function.
@@badrunna-im it seems to me you only proved that a traslation of the solution is not a solution itself. In our case, cosh+c cant be an even part of exp. But What if there's another function, completely different, which is the even part of exp? You should state and demonstrate that if another even part of the f(x) exists it must be a traslation of the first. Am i wrong?
Suppose there exists another even function g(x) and another odd function h(x) such that f(x) = g(x) + h(x). If the pair is different, for some value of x, E(x) ≠ g(x). We will call this value a. E(a) ≠ g(a) E(a) = g(a) + c, c ≠ 0 Substitute this expression for E(a) in f(a). f(a) = g(a) + c + O(a) But because f(a) = g(a) + h(a), it must be true that c + O(a) = h(a). Given that both O(a) and h(a) are odd: c + O(-a) = h(-a) c - O(a) = -h(a) h(a) = O(a) - c But if h(a) = O(a) + c and h(a) = O(a) - c, then c = 0. This is a contradiction to our condition E(a) = g(a) + c, c ≠ 0, so there must not exist another pair of functions.
You're right, Luca. The right way to do it is to shift in a SINGLE point. Suppose f(x) = E'(x) + O'(x) with E, E' even and O, O' odd. Suppose E'(x) = E(x) + c, f(x) = E(x) + c + O'(x) = E(x) + O(x) and f(-x) = E(x) + c - O'(x) = E(x) - O(x) O'(x) = O(x) - c and O'(x) = O(x) + c which leads to c = 0 which donifies the problem.
What if you try to find the even and odd parts of sqrt(x)? Both E(x) and O(x) don’t work with real numbers, so does that mean complex functions can also be even or odd?
Haven't watched yet, but what I came up with is splitting the taylor series into two parts. Sum(i=0)(inf)(x^i/i!)=Sum(i=0)(inf)(x^(2i+1)/(2i+1)!)+Sum(i=0)(inf)(x^(2i)/(2i)!
Zac Chaney the -x property is the definition. The graphical equivalent is that the even functions have y axis symmetry and the odd functions have origin symmetry (also called “anti symmetry”). There are some calculus properties that can be derived, like the integral from -a to a of an odd function is 0, and the integral from -a to a of an even function is 2 times the integral from 0 to a. And you have combination rules which are interesting, like Odd function times even function is an odd function but odd function times odd function is an even function. Those two are opposite what happens with odd and even numbers: odd number times even number is an even number, and odd number times odd number is odd.
Joshua Hillerup if -x is not in the domain for x that are in the domain then it does not work. So you can’t do this with square roots or logs, not in the reals. Also, if the original function is already even, then the odd part will be zero, and vice versa.
The presumption is that if f is defined at some x, f is also defined at the -x. Even function is symmetrical w.r.t the x=0 axis, and the odd (0,0) which clearly means that a function must not only be defined at the positive x domain in order to be split into even and odds. So no. It doesn’t work since it violates the presumption of the procedure
September starts and everybody's like: "it's already halloween"
Meanwhile blackpenredpen straight up skips to christmas
GlitchedCode lolllll yup!!!!
Lol
Immediately knew it was sinh(x) + cosh(x) because that’s exactly what I’m studying right now.
oh man, wow! that's just soooo cool. Firstly I would have never thought of splitting a general function in an even and an odd term, that's just so genious and it feels so natural too but secondly splitting up e^x in that way and deriving cosh and sinh from that... man that's just wow (also because it's so simple)
One of the few vids where I wish I could like 10 times
It’s one of the cases were the general problem is easier to solve than the particular one. Excellent video.
Jaime Duncan : )
Bonus: write it as the sum of even and odd maclaurin series
Very cool
This can now be used to find the fourier cosine series and sine series for a function separately and then combine them
borg972 : )
this blew my mind in solving a problem, thank you :3
thanks man i just had a silly doubt in it... but now it's damn crystal clear
I love these type of puzzle videos but actually reveal to be a big part of mathematics.
How do we know that there's a unique decomposition in even and odd part? Because if that's non the case, we can't refer to cosh as "THE even part of e^x".
If you shift the even part up or down with a +c, you'd still get an even function, but the odd part would need a -c to compensate which would no longer make it odd since it's not antisymmetric around 0, eliminating any other even-odd pair decompositions of a function.
@@badrunna-im it seems to me you only proved that a traslation of the solution is not a solution itself. In our case, cosh+c cant be an even part of exp. But What if there's another function, completely different, which is the even part of exp? You should state and demonstrate that if another even part of the f(x) exists it must be a traslation of the first. Am i wrong?
Suppose there exists another even function g(x) and another odd function h(x) such that f(x) = g(x) + h(x). If the pair is different, for some value of x, E(x) ≠ g(x). We will call this value a.
E(a) ≠ g(a)
E(a) = g(a) + c, c ≠ 0
Substitute this expression for E(a) in f(a).
f(a) = g(a) + c + O(a)
But because f(a) = g(a) + h(a), it must be true that c + O(a) = h(a).
Given that both O(a) and h(a) are odd:
c + O(-a) = h(-a)
c - O(a) = -h(a)
h(a) = O(a) - c
But if h(a) = O(a) + c and h(a) = O(a) - c, then c = 0. This is a contradiction to our condition E(a) = g(a) + c, c ≠ 0, so there must not exist another pair of functions.
You're right, Luca. The right way to do it is to shift in a SINGLE point. Suppose f(x) = E'(x) + O'(x) with E, E' even and O, O' odd.
Suppose E'(x) = E(x) + c,
f(x) = E(x) + c + O'(x) = E(x) + O(x) and f(-x) = E(x) + c - O'(x) = E(x) - O(x)
O'(x) = O(x) - c and O'(x) = O(x) + c
which leads to c = 0 which donifies the problem.
cmon man..
there is not a unique decomposition! but there always is one for every function
1:40 Aaaaaaaaddddd!
This guy is sweeter than sweet.
Chi to the power. Tutor needed.
Dan Rodriguez ?
+blackpenredpen just another racist comment
greenpenredpen, YAAAAAAAAAAAAY!
BlackPenRedPenGreenPen
Papai Pal you bet!!!
Very good explanation. Thanks
These transitions are getting better 😂
Casey Roberson thank you. I have done tons of them to (somewhat) master it. : )
similar to breaking a rank 2 tensor into skew and symmetric parts
Eij = 1/2(Eij + Eji) + 1/2(Eij - Eji)
Saw it on Papa Flammys channel first :) But very nice anyways :)
just watched this video and the complex relation videl before it. Wow awesome math
yay! nice video
try this integral dx/(tanx+cotx+secx+cscx)
so beautiful proof
Very helpful, thank you!
What if you try to find the even and odd parts of sqrt(x)? Both E(x) and O(x) don’t work with real numbers, so does that mean complex functions can also be even or odd?
Great vide, loved it
but was just wondering why E(-x)=E(x) and O(-x)=O(x) ?
Zakaria Saad those are how we check if functions are even or odd. If you have f(-x)=f(x), then we say f is even. Like cox(-x)=cos(x). So cos is even.
blackpenredpen Thank you :)
The Christmas colors gave the punch line away..
Amazing !
oh yes thank you for this
Haven't watched yet, but what I came up with is splitting the taylor series into two parts. Sum(i=0)(inf)(x^i/i!)=Sum(i=0)(inf)(x^(2i+1)/(2i+1)!)+Sum(i=0)(inf)(x^(2i)/(2i)!
Update! I didn't know at the time, but these are the taylor series of sinh(x) and cosh(x), so we got the same thing, but with different methods
that was so coooool
f(x) = [f(x) + f(-x)]/2 + [f(x) - f(-x)]/2
first is even, second is odd
how could you only divide the f(-x)'s by two but not multiply by 2?
Awsome!
Are there any other properties of even and odd functions, other than how they treat -x? I feel like there's a world of utility there...
Zac Chaney the -x property is the definition. The graphical equivalent is that the even functions have y axis symmetry and the odd functions have origin symmetry (also called “anti symmetry”). There are some calculus properties that can be derived, like the integral from -a to a of an odd function is 0, and the integral from -a to a of an even function is 2 times the integral from 0 to a. And you have combination rules which are interesting, like Odd function times even function is an odd function but odd function times odd function is an even function. Those two are opposite what happens with odd and even numbers: odd number times even number is an even number, and odd number times odd number is odd.
Does this always work? What if you're say trying it on log(x) in the reals?
Joshua Hillerup if -x is not in the domain for x that are in the domain then it does not work. So you can’t do this with square roots or logs, not in the reals. Also, if the original function is already even, then the odd part will be zero, and vice versa.
log|x| is even
yeah that does not answer the question hahs
The presumption is that if f is defined at some x, f is also defined at the -x. Even function is symmetrical w.r.t the x=0 axis, and the odd (0,0) which clearly means that a function must not only be defined at the positive x domain in order to be split into even and odds.
So no. It doesn’t work since it violates the presumption of the procedure
Infact I would say hyper sine and hyper cosine , credit to my undergraduate buddy.
直接想到的是Taylor展開式的奇偶數次項 … 另外我想知道這表示法是唯一的嗎？怎麼證明
how come you always upload so early?
you mean taylor polynomial?
nice!
What the? It feels weird how nothing was on the board in the start like the other vids.
M. Shebl ikr, I didn't know what I could put down first lol
Integrate (1/x (e^x)) dx .
Cosh and sinshhhh
Before Watching: Taylor series
Taylor polynomial lol
Ok, I may respect hyperbolic functions more
Varad Mahashabde : )
whats your name
blackpenredpenbluepengreenpenyellowpenorangepen... #YAY
Guilherme Nery Rocha and the abbreviation is?
@@blackpenredpen it's bprpblpgpypop 😁
hahaha, nice!!!
sinczeks
First!
Amazing!