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Students these days are not able to solve this one 

Flammable Maths
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20 сен 2024

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Комментарии : 56   
@steve112285
@steve112285 День назад
When I teach L'Hospital, I teach that for infinity-infinity form you can factor out the dominant behavior. You'll either get infinity*nonzero which will give +-infinity, or infinity*0. In the latter case, move one factor into the denominator as a reciprocal, then apply L'Hospital.
@bobsoup2319
@bobsoup2319 День назад
Squeeze theorem it by completing the square by adding a 1/4 to the sqrt. This can be done for the original limit as the upper bound, then as the lower bound when you multiply the limit it by its conjugate. Upper: sqrt(x^2+x+.25)-x Lower: x/[sqrt(x^2+x+.25)+x] Both evaluate to 1/2
@vincentproplayer
@vincentproplayer День назад
Also an interesting observation is that in the general case, lim(x->inf) (x^n + a x^(n-1) + P((n-2))^(1/n) - x, where P(n-2) is a polinomial of the n-2th order, converges to a/n
@arianlili7252
@arianlili7252 День назад
Just rationalise and divide by x we get ans as 1/2 :)
@JeremyGluckStuff
@JeremyGluckStuff День назад
dang
@impulsiveDecider
@impulsiveDecider День назад
Isn't this just his first approach? Are you being sarcastic?
@keshavchauhan6290
@keshavchauhan6290 День назад
WHY DID YOU SPOIL IT FOR ME? THIS COMMENT IS LITERALLY THE FIRST COMMENT THAT SAW BEFORE TRYING THE QUESTION.
@quentind1924
@quentind1924 День назад
@@keshavchauhan6290 In this channel (and other channels about math problems), comments are often spoilers
@arianlili7252
@arianlili7252 23 часа назад
​​@@impulsiveDecider I just saw the question in the thumbnail and wrote the first approach that came to my mind
@FacultyofKhan
@FacultyofKhan День назад
Flammys drip be bussin frfr How am I doing fellow kids?
@michaelh13
@michaelh13 День назад
Kids would say “This sigma’s aura tends to infinity, very based”
@tomkerruish2982
@tomkerruish2982 День назад
No cap!
@lohkvongoethe
@lohkvongoethe День назад
@FacultyofKhan Get back to work... Jim... Stuart... What's your first name btw?
@ricardoparada5375
@ricardoparada5375 День назад
My old high school math teacher would have my head if I didn’t know how to solve it the first way. I’m glad I was taught to not just memorize formulas
@laurentreouven
@laurentreouven День назад
At 6"28, you should choose another function name and not f, for f(z)=(1+z)^(1/2) because was already used for f(x)=(1+1/x)^(1/2). You cannot use only f, because with z=1/x, it leads to f(1/z)=(1+z)^(1/2), not f(z)=(1+z)^(1/2) as you wrote.
@userchrh
@userchrh 4 часа назад
I’ve found a really elegant non-calculus solution. Notice that √(x² + x) - x is clearly a monotonically increasing function (since √(x² + x) > √(x²) = x). Now suppose √(x² + x) - x ≤ m for all x ≥ 0 where m is minimal. Then x² + x ≤ (m + x)² = m² + 2mx + x² which simplifies to x(1 - 2m) ≤ m². This is clearly always satisfied for m = 1/2 and setting m < 1/2 implies 1 - 2m > 0 but x can be arbitrarily large and so x(1 - 2m) ≤ m² isn’t satisfied for all x. Thus the limit is m = 1/2.
@JohnSmith-eg9lc
@JohnSmith-eg9lc День назад
By doing a transform of variables x = 1/h, you get the limit of [sqrt(1+h) - sqrt(1)]/h, as h -> 0, which is just the derivative of the function sqrt(x) evaluated at x = 1. Use of the Maclaurin expansion of the square root term is therefore circular in this case, since if you know the value of the derivative of sqrt(1+x) for all x where the derivative is defined, you must also know the value of the derivative of sqrt(x) at x = 1.
@henningmorbus5436
@henningmorbus5436 День назад
Erstmal muss ich sagen: Ich hab vor Jahren ein paar Videos von dir zur Laplace-Transformation zur Lösung von Differentialgleichungen gesehen und gleichzeitig noch ein paar von blackpenredpen, diese gelöst. Das hat damals sehr geholfen. Einfach Mal alle möglichen Methoden zu durchlaufen. Man braucht da ja alles mögliche: Polynomdivision, Linearfaktorzerlegung, die Transformation an sich, wie man gewisse Integrale löst etc. etc. Das Problem, was du hier so ein bisschen aufmachst ist ja das Folgende: Hier fehlt einfach ein bisschen das Wissen, was möglich ist - sozusagen die Spielregeln der Mathematik. Also es ist nicht bekannt, was man eigentlich darf, was erlaubt ist. So ähnlich wie der Ansatz ebener Wellen in sämtlichen DGLs in der Physik. Wir wissen den einfach, denken da gar nicht drüber nach und nehmen den, weil er oft gut funktioniert. Natürlich kann man auch Potenzreihen machen (QM-Harmonischer Oszillator oder auch allgemein, Wasserstoffatom etc.), Lösungsmethoden gibt es ja etliche. Aber dafür muss man sie eben kennen. Es ist irgendwie ein bisschen zu Trennen zwischen diesem handwerklichen Wissen, das einfach mit dem regelmäßigen Lösen von Problem kommt und dem tatsächlichen Verständnis von großen Konzepten. Ich finde, dass diese Dinge nicht unbedingt getrennt ablaufen, aber schon etwas unabhängig voneinander sind. Ich hab das Gefühl, dass bildungstechnisch einfach vieles schief läuft. Mach gerade meinen Master in Physik. Denke Mal, dass auch an den Unis unser Lehrplan eigenartig ist. Ich diskutiere da regelmäßig mit einem Kommilitonen drüber und wir sind uns über ein besseres Konzept nicht wirklich einig. Wir dachten zum Einen, dass man vielleicht ein ganzes Jahr die mathematischen Grundlagen komplett wiederholen und erlernen müsste, die für die Physik notwendig wären. Zum anderen denke ich eher, dass sich vieles wiederholt und man in eben das Klein-Klein und spezielles Nischenwissen verfällt. Wenn wir allgemeine Konzepte verstehen wollen, dann sollte man sich wirklich über diese allgemeinen Methoden Gedanken machen. Was ist wirklich relevant, was soll man wirklich verstehen. Ich finde, dass man vieles später in der Anwendung selber lernt. Habe eine Zeit in der experimentellen Teilchenphysik gearbeitet, jetzt bin ich an einem komplett anderen Institut. Die Skills die man braucht, die lernt man dann vor Ort. Und ich finde genau dazu soll einem das Studium und in gewisser Weise auch die mathematische Bildung der Schule vorbereiten. Nicht nur die einfachen Methoden, sondern einfach ein grundlegendes Verständnis, dass es einem so einfach wie möglich macht, verschiedenste Bildungsmöglichkeiten und Ausbildungswege einzuschlagen. Ich denke einfach, dass das eine sehr schwere Gratwanderung ist. Bisschen vergleichbar dazu wäre auch Sport bei Kindern. Welche Sportarten zeigst du deinen Kindern, damit sie selber später frei und individuell entscheiden können, ihre eigenen Interessen zu verfolgen? Man könnte irgendwas total spezifisches wie Tischtennis machen, oder was koordinatives. Oder du machst irgendwas allgemeines, Kraftsport, Cardio, bisschen Koordination - damit man später viele Skills hat und diese in vielen anderen Interessensgebieten anwenden kann. Alles in allem doch sehr komplex. Denke als Bildungswissenschaftler wirst du dir da natürlich auch einige Gedanken dazu haben.
@stefanalecu9532
@stefanalecu9532 День назад
What the hell are you yapping about
@PapaFlammy69
@PapaFlammy69 День назад
Hi
@meanintellectual6593
@meanintellectual6593 День назад
Tbh your students prob didn’t get the correct answer because wolfram alpha gives as a result infinity.
@midorithursday
@midorithursday День назад
@@meanintellectual6593 No it doesn't. I get 1/2 just as it should be.
@meanintellectual6593
@meanintellectual6593 День назад
I may have not been clear with what I said above, so I will detail it a little bit better. Essentially when anybody does a limit on wolfram alpha, you will have a little square pop up to the right of limit, into which you need to write your function. On mobile devices if you put the square root first and the “-x” after (as many do), the limit will not be applied to the “second term”, resulting in a subtraction between an infinity large number and “x” (which gives you infinity). If you put the whole limit into the limit “box” (with the use of brackets for example), you get obviously the right result. The way it’s written could cause this misunderstanding and I above I joked on how reliant and also how unable of use it correctly new generations can be. I guess I am just not great with irony, my bad for any misunderstanding.
@midorithursday
@midorithursday День назад
@@meanintellectual6593 Ah, yeah, that's a lot of subtext. I think putting quotes around "wolfram alpha gives as a result infinity" to indicate that it was the students saying that, not you, would've made it clear that you were making some kind of joke, but I'm not sure I would've gotten it even then without a parenthetical explanation that they had likely typed in the expression incorrectly.
@meanintellectual6593
@meanintellectual6593 День назад
I am not that fluent in English so I need to improve and thank you for the advice, I am going to keep it in mind!
@tomkerruish2982
@tomkerruish2982 День назад
Factor out x to get x(sqrt(1 + 1/x) - 1), then use the binomial theorem to get x(1 + 1/2x + o(1/x) - 1) = ½ + o(1) which goes to ½. Edit: This is the kind of comment left by someone who hasn't finished the video.
@RohitDutta420
@RohitDutta420 День назад
Is there any solution when the outer is positive?
@bbg6295
@bbg6295 День назад
This is just (sqrt(1+1/x) - 1)/(1/x) so the answer is the derivative of sqrt(x) at 1, which is 1/2
@youtubeaccount1718
@youtubeaccount1718 День назад
that was clever im getting rusty man
@m.caeben2578
@m.caeben2578 День назад
1/2. Just factorize the x and apply Taylor.
@error-wi3xb
@error-wi3xb 14 часов назад
Expansion
@m.caeben2578
@m.caeben2578 13 часов назад
@@error-wi3xb I only kept the zeroeth term of the phrase 👍
@LocksVid
@LocksVid День назад
0.5
@BPEREZRobertJamesL
@BPEREZRobertJamesL День назад
I used a different method... It's complicated though... Firstly, I showed that the function is bounded above. Clearly, f(x) = sqrt(x^2 + x) - x < sqrt(x^2+ 2x + 1) - x = x + 1 - x = 0 for any x > 0. Then, I showed that it is increasing on [x,+inf). Let x>0 and ε>0. Then f(x + ε) - f(x) = sqrt((x + ε)^2 + x + ε) - x - ε - (sqrt(x^2 + x) - x) = sqrt((x + ε)^2 + x + ε) - sqrt(x^2 + x) - ε. Now, note that 4x^2 + 4x + 1 = (2x + 1)^2 > 4x^2 + 4x =4(x^2 + x), which implies that 2x + 1 > 2sqrt(x^2 + x) and thus, (x + ε)^2 + x + ε = x^2 + 2xε + ε^2 + x + ε = x^2 + x + ε^2 + (2x + 1)ε > x^2 + x + ε^2 + 2εsqrt(x^2 + x) = (sqrt(x^2 + x) + ε)^2 Thus, sqrt((x + ε)^2 + x + ε) > sqrt(x^2 + x) + ε which implies that sqrt((x + ε)^2 + x + ε) - sqrt(x^2 + x) - ε > 0 and thus, f(x + ε) - f(x) > 0 --> f(x + ε) > f(x) Then, I used the fact that whenever a function is bounded above and increasing, it has a limit at infinity. I leave it as an exercise to the reader (lol) to prove this theorem. So, by modus ponens, we know that L = lim_(x -> inf) f(x) exists. Now, notice that f^2 (x) + 2xf(x) + x^2 = (f(x) + x)^2 = sqrt(x^2 + x)^2 = x^2 + x. Thus, (f(x))^2 + 2xf(x) = x, which implies that ((f(x))^2)/x + 2f(x) = 1 Also, notice that f(x)/x tends to zero when x goes to infinity (because sqrt(x^2 + x)/x goes to 1 when x goes to infinity. I leave the verification of this as an exercise to the reader...) Thus, using the laws of limits, we get: 1 = lim_(x->inf) 1 = lim_(x->inf) (((f(x))^2)/x + 2f(x)) = lim_(x->inf) (((f(x))^2)/x) + 2L = lim_(x->inf) (f(x)/x)L + 2L = 0 + 2L = 2L And thus, 2L = 1 \Rightarrow L = 1/2.
@ginalley
@ginalley День назад
There has been an explosion of physics content especially around quantum mechanics atm. I wonder if you could capitalise on the hype and do some mathematical derivations of physical systems for those easy views.
@dodojef301
@dodojef301 13 часов назад
Kay before I watch the video I got 1/2 from rationalizing
@denizgoksu9868
@denizgoksu9868 День назад
We live in a society fr
@mathmachine4266
@mathmachine4266 День назад
1/2 Taylor's series
@tiziocaio101
@tiziocaio101 День назад
I’d solve it with the Taylor series for (1+1/x)^1/2
@saadaijaz6403
@saadaijaz6403 День назад
Haven't seen the vide but it's one of those bs questions in our quizzes, the answer should be 1/2
@brickie9816
@brickie9816 День назад
my approach was as follows: lim x->inf sqrt(x^2+x)-x = lim x->inf sqrt(x^2(1+1/x))-x = lim x->inf x*sqrt(1+1/x)-x = lim x->inf x*sqrt(1+0)-x = lim x->inf x-x = 0 why do i get an incorrect result? what's wrong with that approach?
@Ayman-sc3fk
@Ayman-sc3fk День назад
infinity minus infinity is undefined
@error-wi3xb
@error-wi3xb День назад
Inf - inf not set....😅
@juliandonker3009
@juliandonker3009 17 часов назад
To elaborate on previous comments: You make a mistake when you plug in 1/x=0 on the second to last equality. That is in fact the limit, but you cannot plug in one limit and leave the others. If you keep the expression you will get : lim x->inf x*sqrt(1+1/x)-x= lim x->inf x*(sqrt(1+1/x)-1) and since sqrt(1+1/x) is never PRECISELY EQUAL TO 1, the limit of the entire expression will not equal 0. In other words, the x’s on your final row aren’t equal to eachother and can therefore not cancel eachother out, even in the limit. Hope this helps:)
@brickie9816
@brickie9816 14 часов назад
@@juliandonker3009 thank you! this helped massively.
@Hadar1991
@Hadar1991 День назад
The more "academical" method is like dropping a nuke to kill a fly. You can, but why? :v Also I will complain about your notation, because you should write lim (sqrt(x^2 + x) - x) otherwise the "- x" part is not affected by the limit.
@kthegreat69420
@kthegreat69420 День назад
Because I like nukes!
@NighthawkRPL
@NighthawkRPL День назад
man your content aged like milk. You used to solve olyampiad level math and now you are doing some elementary school kid's homework/.
@FacultyofKhan
@FacultyofKhan День назад
His meme game has also taken a step back! Long ago I remember him beefing with other youtubers, making fun of people’s pronunciation etc (I too was made fun of lol), dabbing and putting anime references in his thumbnails. He’s becoming too milquetoast for my taste! At least he wears supreme shirts tho
@stefanalecu9532
@stefanalecu9532 День назад
He's mid in general, he was never good, and he stopped giving a shit because of real life (I know he said something about this in a video)
@dylancook3282
@dylancook3282 День назад
Wtf I remembered u being happy. I guess that’s impossible for Germans, but get better soon
@PapaFlammy69
@PapaFlammy69 День назад
What do you mean?
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