I was hoping that at least my math teacher would be honest with me, devoid of malicious intent. But once again I find out. Another person hiding the truth from me. Why do I always finding myself in this type of relationships?
Interestingly enough, exponential quotients and logarithmic quotients also present this behavior. (ln^2 (x) + 1)/ln(x) has a curved asymptote of ln(x). If we distribute, we'll see that this function is equivalent to ln(x) + 1/ln(x), which at infinity is asymptotically equivalent to ln(x)
Oh, I remember finding these non-linear asymptotes in high school, when I was _not listening_ to the math class lol. I was playing with polynomial division, plotting what I got and comparing that with corresponding rational functions. After trying that with like a third degree numerator divided by a first degree denominator, I got some really nice shapes and there was just no coming back! We had vertical, horizontal and diagonal asymptotes as part of the curriculum. But getting a parabolic one was just so much cooler and more interesting, so that is what I went to explore instead. edit: Long division for polynomials is also pretty cool. And so is the Horner's scheme!
In italy we tend to stress the importance of asymptotes when they are linear. In that case we show students that you can find their equation just by calculating lim f(x)/x (which gives you the angular coefficient of the line, lets call it "m") and then lim f(x)-mx which will give the intercept.
It is also part of curriculum in Romania I had the idea about just dividing the polynomials too during summer break. But the Division by X limit will work for relations with square roots and everything @Flammy your division hurts me. I would just go with A(X)=B(X)*Q(X)+R(X) So R(X) is a normal polynomial not a fraction. We are interested in A(X)/B(X)=Q(X)+R(X)/B(X) Not sure why you using R(X) as a fraction straight up was so disturbing for me
When he talks about dying in Mexico, I picture the 🎥 movie _The Boys From_ _Brazil,_ and all the expatriated Germans that skedaddled to Argentina. Gregory Peck was in it.
I don’t think that’s the standard approach to remainders. The remainder shouldn’t be multiplied by the quotient (I.e., you should have p(x) = q(x)g(x) + r(x), which would then imply that p(x)q(x) = g(x) + r(x)/q(x)). You directly found r(x)/q(x) and called it the remainder. Not a big deal but it sort of confuses the standard notion of a remainder. Also, in one of you early examples with an asymptote of 0, it could have been cool to point out that it asymptotically approaches 3/x (I might be misremembering what the constant was). The point being the inverse case isn’t two different than the case you focused on. These sorts of asymptotic equivalences are especially pretty important in engineering and physics.
Also, I guess you didn’t want to stress the polynomial division, but if the denominator is a monomial, it’s easy to just split the numerator by each term and get an immediate result. (Maybe you me approach was based on the intended audience of this video?)
You goofed the unary sign on the remainder. It should be… - (2/3x) . Just the same, no matter what unary sign you use, + or - , it doesn’t change the value of the overall limit. I did find it interesting that asymptotes do not have to be straight lines. I also find it interesting that the result of the polynomial division ends-up being the *line equation* of the slanted asymptote.
23:07 - LOL! Greetings from Mexico hahahaha 👋 BTW, about the result in 18:38, for single term divisors like this, we can also use the shortcut of just splitting the divisor into both terms of the dividend, just like a fraction denominator but of course we would have missed the full explanation. Thanks for the great content and keep it up! [Edit: you meant -2/3x in 18:59 but it ends up not affecting]
it's actually pretty fun, you can take a sum f(x)+a(x) with literally any function f(x), like cosh(x), and add to it a(x)=1/x, 1/x^2 or something like this and get an interesting asymptote. if you want the asymptote to go really close to the function do 1/(ax^n) with some large value of a.
Or, cosh(x) itself has an even more interesting curvilinear asymptote, approaching the function cosh(x)+1/x ...? If curvilinear asymptotes are a thing, then you cannot distinguish between which curve approaches what other.
I click to learn parabolic math, ambiguities and dualities, electromagnetic applications, stable and unstable particles joining, and connections to SSS solving triangles/Big Bounce physics. You are a spooky dude, who immediately tried to muddle the mathematics and physics with he vs. she thinking. Get your head right!
18:58 I'm pretty sure you meant to write "-" instead of a "+" there. 🤔
Yup, my bad! Thanks a bunch =)
@@PapaFlammy69 Snarky comments of denial. Mental!
I was hoping that at least my math teacher would be honest with me, devoid of malicious intent.
But once again I find out. Another person hiding the truth from me. Why do I always finding myself in this type of relationships?
Interestingly enough, exponential quotients and logarithmic quotients also present this behavior. (ln^2 (x) + 1)/ln(x) has a curved asymptote of ln(x). If we distribute, we'll see that this function is equivalent to ln(x) + 1/ln(x), which at infinity is asymptotically equivalent to ln(x)
Oh, I remember finding these non-linear asymptotes in high school, when I was _not listening_ to the math class lol. I was playing with polynomial division, plotting what I got and comparing that with corresponding rational functions. After trying that with like a third degree numerator divided by a first degree denominator, I got some really nice shapes and there was just no coming back!
We had vertical, horizontal and diagonal asymptotes as part of the curriculum. But getting a parabolic one was just so much cooler and more interesting, so that is what I went to explore instead.
edit: Long division for polynomials is also pretty cool. And so is the Horner's scheme!
In italy we tend to stress the importance of asymptotes when they are linear. In that case we show students that you can find their equation just by calculating lim f(x)/x (which gives you the angular coefficient of the line, lets call it "m") and then lim f(x)-mx which will give the intercept.
Here in Poland it's pretty much like that as well.
Same in Greece
In Baden Württemberg (a German state) this is part of the curriculum as well.
It is also part of curriculum in Romania
I had the idea about just dividing the polynomials too during summer break. But the Division by X limit will work for relations with square roots and everything
@Flammy your division hurts me.
I would just go with
A(X)=B(X)*Q(X)+R(X)
So R(X) is a normal polynomial not a fraction. We are interested in
A(X)/B(X)=Q(X)+R(X)/B(X)
Not sure why you using R(X) as a fraction straight up was so disturbing for me
Math departments HATE this one simple trick!
:D
I always watch from beginning to End.
My teacher taught us how to find those oblique asymptots!!! Kudos to him 😂
nice!!!
*@[**06:17**]:* Omitting the non-leading terms is the convenient strategy for this, by the way.
When he talks about dying in Mexico, I picture the 🎥 movie _The Boys From_ _Brazil,_ and all the expatriated Germans that skedaddled to Argentina. Gregory Peck was in it.
I remember doing this in the 80's when learning how to hand draw various equations
I don’t think that’s the standard approach to remainders. The remainder shouldn’t be multiplied by the quotient (I.e., you should have p(x) = q(x)g(x) + r(x), which would then imply that p(x)q(x) = g(x) + r(x)/q(x)). You directly found r(x)/q(x) and called it the remainder. Not a big deal but it sort of confuses the standard notion of a remainder.
Also, in one of you early examples with an asymptote of 0, it could have been cool to point out that it asymptotically approaches 3/x (I might be misremembering what the constant was). The point being the inverse case isn’t two different than the case you focused on. These sorts of asymptotic equivalences are especially pretty important in engineering and physics.
Also, I guess you didn’t want to stress the polynomial division, but if the denominator is a monomial, it’s easy to just split the numerator by each term and get an immediate result.
(Maybe you me approach was based on the intended audience of this video?)
You goofed the unary sign on the remainder.
It should be…
- (2/3x) .
Just the same, no matter what unary sign you use, + or - , it doesn’t change the value of the overall limit. I did find it interesting that asymptotes do not have to be straight lines. I also find it interesting that the result of the polynomial division ends-up being the *line equation* of the slanted asymptote.
23:07 - LOL! Greetings from Mexico hahahaha 👋 BTW, about the result in 18:38, for single term divisors like this, we can also use the shortcut of just splitting the divisor into both terms of the dividend, just like a fraction denominator but of course we would have missed the full explanation. Thanks for the great content and keep it up! [Edit: you meant -2/3x in 18:59 but it ends up not affecting]
yup, my bad!
Finally a vid I understood literally anything in since this happens to the the exact topic we are currently covering in maths
Thanks papa
very nice! :)
Hmm I tend to partial decomposition, the results are similar, but it helps find O's
Please, a lecture about Puiseux expansion and others expansions at x → ∞
Handsome teacher
Asymptotes were always pretty fun to compute in school :D
Awesome papa flammy
I'll probably die somewhere in Mexico (I'm mexican)
r i p
This video is asymptotically cool
5:01 you gotta get your asshole checked 🥶
:D
that was really cool, I only think the introduction was a bit too long before it got to the actually interesting stuff
Thx for the feedback! That's why I added the timestamps =)
it's actually pretty fun, you can take a sum f(x)+a(x) with literally any function f(x), like cosh(x), and add to it a(x)=1/x, 1/x^2 or something like this and get an interesting asymptote. if you want the asymptote to go really close to the function do 1/(ax^n) with some large value of a.
Or, cosh(x) itself has an even more interesting curvilinear asymptote, approaching the function cosh(x)+1/x ...?
If curvilinear asymptotes are a thing, then you cannot distinguish between which curve approaches what other.
Teachers: there is no war in Ba Sing Se
what about your first video
? wdym
what about your first video
what about your first video
what about your first video
@@PapaFlammy69search "oh no Daddy anata wa wuck desu" and that's your video that uploaded on 1970
I click to learn parabolic math, ambiguities and dualities, electromagnetic applications, stable and unstable particles joining, and connections to SSS solving triangles/Big Bounce physics. You are a spooky dude, who immediately tried to muddle the mathematics and physics with he vs. she thinking. Get your head right!
What?
@@PapaFlammy69 The intro to your video.
asymp-toe 🤤💀
Get out 🔥 🗣️❗️
You look and act exactly like Justin Hammer. Why?
arent you a math teacher