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16 сағат бұрын
Ай бұрын
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@Archimedes_Notes Күн бұрын
This question is for high schoolers. They know only the axioms of R ,algebra 1 ,2 and some precalculus.
@damyankorena 2 күн бұрын
Proof 1: Let f(x)=√x. Suppose x,y are different and f(x)=f(y). Then √x=√y so x=y which is a contradiction. Now f is continuous and injective therefore it is strictly increasing or decreasing. f(1)>f(0) so f is increasing. 2nd proof: df/dx = 1/(2√x) > 0 so f is increasing 3rd proof: Obviously f(x)>=0 for all x. f(x)>f(y) f(x)-f(y)>0 (f(x)-f(y))(f(x)+f(y))>0(f(x)+f(y)) x-y>0 x>y Therefore f(x)>f(y) only when x>y. 4th proof: the inverse of f which is g(x)=x² is well-defined therefore f and g are injections. lim[x->inf] f(x) -> inf, and so by continuity f is increasing.
@JahnviTiwari-uy6sq 8 күн бұрын
Thanx sir for this amazing explanation
@bbprintingshop8759 19 күн бұрын
Help me 1/7=1
@Archimedes_Notes 18 күн бұрын
@@bbprintingshop8759 No it is 1=1/7 Help me
@bbprintingshop8759 19 күн бұрын
Help me 1/7=1
@christopherhelton6728 21 күн бұрын
point on circle (-2,3). slope of radius = (-3/2). radius is perpendicular to tangent line so slope of tangent line = -(1/(-3/2)), or 2/3. Use point (-2,3) to find y-intercept. 3 = (2/3)(-2) + b. b = 13/3. Tangent line y=mx+b is y=(2/3)x + (13/3).
@rezikotrikadze1697 28 күн бұрын
Just graph it out it's simpler
@salim_0 28 күн бұрын
you can just remove the first absolute value because 5 is already positive
@Archimedes_Notes 27 күн бұрын
@@salim_0 No ,that is incorrect. |A|=b means that Either A=b or -A=b , we will lose half the solutions.
@salim_0 27 күн бұрын
@Archimedes_Notes my bad, but you wouldn't lose the solutions, because thhe whole thing cant be -5
@Archimedes_Notes 27 күн бұрын
@@salim_0 It is fine. The issue that i am talking about happens alot. We can lose the solution, To be clear take f(x)=abs(abs(x^(2)-1))-3))) and try to solve f(x)=2. You cant get rid of the first absolute value.graph it to see clearly
@tIhIngan Ай бұрын
You changed the problem. The title says the tangent line is 12x+5y-12=0, you are using 12x+5y-9=0
@Archimedes_Notes Ай бұрын
@@tIhIngan We had 3 problems. All deal with tangents, we will look at that
@Archimedes_Notes Ай бұрын
@@tIhIngan it is fixed now. Thank you
@Saba-em8oz Ай бұрын
did you prove the inequality above on which you said you will prove in next video?
@codeboitutorials Ай бұрын
Very nice solution! I gave this a go before watching and ended up rearranging the original equation, squaring both sides and then through some trig relationships for tan^2, sin2x and cos2x I eventually also reached tan2α=1! Quite interesting that tan2α seems to be the key to this question.
@Archimedes_Notes Ай бұрын
@@codeboitutorials Very intelligent that you give it a shot before watching the video. I still do the same.
@ChimezieFredAnaekwe Ай бұрын
I thought you have take the most extreme for your final answer, as we don't inequalities by just adding the extreme terms.
@adameinstein7324 Ай бұрын
How do you solve this question? Can you add more? ALSO YOUR COMMENT ISNT UNDERSTOOD. WRITE COMPLETE SENTENCES
@c1majesty85 Ай бұрын
I understood how double counting can work in concept but this is the first video I found actually explaining it mathematically were it made sense for me
@salehmoosavi875 Ай бұрын
Great 👍
@Radhakrishnan-uf6vx Ай бұрын
That's a good way to explain it .. loved it
@kakubhai8175 Ай бұрын
Nice video loved the explanation
@medmoded7767 Ай бұрын
Thank you for mentioning my approach great proof btw
@SsG-m2m Ай бұрын
I have nothing to do this topic but i love your dedication and consistent,i have seen that you r video did great try analytics that video like the title was solving mit which i think attracted the audience keep going man❤
@Archimedes_Notes Ай бұрын
Great suggestion!
@fir3pr051 Ай бұрын
Thanks for the explanation❤
@godz5254 Ай бұрын
Why does the last denominator of the telescope sum cancel out, if there is no next term?
@jupski4960 Ай бұрын
damn
@KookyPiranha Ай бұрын
can you explain it more accurately im not sure what you want me to find exactly
@ikbintom Ай бұрын
Nice work. Note that in the title, (-) should be (-1). Thanks!
@Archimedes_Notes Ай бұрын
@@ikbintom thank you
@mathmachine4266 Ай бұрын
(1+∫[0,x] (1-t²/2+t⁴/24-1)/t² dt)^(1/x) (1+∫[0,x] -1/2+t²/24 dt)^(1/x) (1-x/2+O(x³))^(1/x) e^(-1/2)
@jonathandawson3091 Ай бұрын
1/sqrt(e)? Since it evaluates to lim n->\infty (1-1/2n)^n. Solved mentally ~30 seconds so I must be wrong but the answer has to be close.
@Archimedes_Notes Ай бұрын
@@jonathandawson3091 If you have done it mentally, you are a super genius or you are coming from another planet. Please write down the mental process...i will wait for it
@jonathandawson3091 Ай бұрын
@@Archimedes_Notes Oh, I am nothing like that. Mentally, I first thought if the integrand could be evaluated. Then I saw since x->0, all we care about is very small values inside the integral as well. Taking the limit inside (which needs to be justified and I am sure can be), cos(t) = 1 + t^2/2 + O(t^4) from Taylor expansion, so we get Int -1/2 dt which is -x/2. Thus we need lim x->0 (1-x/2)^x. This is the familiar exponential definition; if we say n=1/x then we get lim n->\infty (1-1/2n)^n = lim sqrt((1-1/2n)^2n) = sqrt(1/e).
@Archimedes_Notes Ай бұрын
@jonathandawson3091 YES. THAT IS GREAT.
@TSSPDarkStar 2 ай бұрын
Cool video
@UnrealNarcissist 2 ай бұрын
Simple,. the distance between the two centers minus the sum of the two radii.
@UnrealNarcissist 2 ай бұрын
To understand why, there are several approaches but remember that a line from the center to the edge of the circle is perpendicular to the circle which means that the line between the two points where it crosses the circles is perpendicular to both the circles, making this the shorts distance between the circles
@leochinchillaa 2 ай бұрын
You failed to prove that: 1. lim as x approaches 0 of [(sinx/x)(xlnx)] = (lim as x approaches 0 of sinx/x)(lim as x approaches 0 of xlnx) 2. lim as x approaches 0 from the left hand side of xlnx = 0 (thus failing to prove that a limit exists for the lim as x approaches 0 of xlnx)
@Archimedes_Notes 2 ай бұрын
Do you have any ideas? How to improve the proof? We skipped many steps. The aim was to find the limit of sin x ln x as x approaches 0.
@emilt298 2 ай бұрын
This is a good method, you can also draw a right-angled triangle with side lengths x, 1, and sqrt(1+x^2) by Pythagoras, then arctan(x) is the angle between the sides of lengths 1 and x, it then follows that sin(arctan(x)) = x/sqrt(1+x^2)
@Archimedes_Notes 2 ай бұрын
@@emilt298 What if the angle is negative when x is between [-pi/2 ,0]
@crowbar_the_rogue 2 ай бұрын
You could probably also differantiate it and the global minimum (that's what it's called in English, right?) would turn out to be more than zero.
@Archimedes_Notes 2 ай бұрын
@@crowbar_the_rogue No , this is wrong . See one of the videos in tgis channel.
@Archimedes_Notes 2 ай бұрын
@@crowbar_the_rogue kzbin.infoXRMz06Fvcl4?si=QfEZNUKk61XtyODr
@Archimedes_Notes 2 ай бұрын
Check the link in this short video. It will help
@renesperb 2 ай бұрын
A straightforward way: the limit is the same as √(1+x^2) -x, x-> inf.Write the whole expression as (√(1+1/x^2)-1)/ (1/x) .Set t = 1/x and you have lim t->0 (√(1+t^2) -1 )/t . Now you apply the rule of De L'Hopital ,which gives you immediately the answer 0.
@KookyPiranha 2 ай бұрын
inequality scaling solves this immediately
@amadeus-1011 2 ай бұрын
very cool
@jAKUB-g2y 2 ай бұрын
we see that if x>0 the inequality holds, so assume that x<=0 move x to the other side and we can square both sides and we're done
@Archimedes_Notes 2 ай бұрын
@@jAKUB-g2y Hello Jacob, In the second statement you will have to elaborate more. You said x<=0, move x to other side and square.You will have to add more here....
@jAKUB-g2y 2 ай бұрын
@@Archimedes_Notes x^2 cancels and we have 1>0
@Archimedes_Notes 2 ай бұрын
I see what you mean. My remark was about the writing. This is fine. In writing proofs, you will not tell the reader to finish the proof. I remember when i was taking proof writing, the professor took 5 points because I said the proof is clear now.this is a side note. I got what you mean.
@DevoutSkeptic 2 ай бұрын
This is such an incredibly trivial problem. I have no idea why this would be in a math competition.
@Archimedes_Notes 2 ай бұрын
@@DevoutSkeptic TIME BASED
@GAMEDATA1010 2 ай бұрын
@@Archimedes_Notes time based I could even say it diverges in less than 2 seconds it really is super trivial
@Archimedes_Notes 2 ай бұрын
@GAMEDATA1010 Completey agree. You need to see the other question: Find the points of intersectio of e^x and y=x+2, correct to 12 decimal places. You can try this one in 2 secondes..
@DevoutSkeptic 2 ай бұрын
@@Archimedes_Notes Use the Lambert W function, 4Head
@GAMEDATA1010 2 ай бұрын
@@Archimedes_Notes Answer: -W(-1/e^2) - 2 e^x = x+2 (x+2)e^-x = 1 do a change of variable to cancel away the two x = -u -2 ue^(u+2) = -1 ue^u * e^2 = 1 ue^u = 1/e^-1 W(ue^u) = u = W(-1/e^2) change back to terms of x u = -x - 2 -x - 2 = W(-1/e^2) x = -W(-1/e^2) - 2 now given that the lambert W function is an infinite sum it would likely return a transcendental number so either you would need an approximation or a computer but in most scenarios where you are utilizing the lambert W function you will just be using a calculator or computer program so our closed form result will suffice as an exact solution.
@colinsellars2318 2 ай бұрын
Say dude you should be more fluid.
@dhet1272 2 ай бұрын
You are right
@tomasgarau5338 2 ай бұрын
Another proof: for all x, sqrt(x^2 + 1) > sqrt(x^2) = |x| because sqrt is strictly increasing. So knowing that |x| >= - x, we have 0 < - |x| + sqrt(x^2 + 1) <= x + sqrt(x^2 + 1).
@davidbrisbane7206 2 ай бұрын
Differentiate ... 2[(x-1)+(x-2)+(x-3)+(x-4)+(x-5)]= 10x - 15 A critical point occurs at 10x - 15 = 0. So, x = 3 Differentiate 10x - 15 = 10 > 0, so this is a minimum. The minimum is (-2)² + (-1)² + (0)² + (1)² + (2)² = 10.
@vishalmishra3046 2 ай бұрын
*Simpler solution* Divide by highest power of infinite n from both numerator and denominator polynomials, then eliminate higher than 1 powers in denominator (1/n^2), since exponent is linear in n, then use binomial approximation to simplify to single power of n, and finally use theorem e^x = (1 + x/n)^n limit n to infinity, to get [ (1 + 1/n + 3/n^2) / (1 + 3/n + 5/n^2) ]^n = [ (1 + 1/n) / (1 + 3/n) ]^n = [ (1 + 1/n) (1 - 3/n) ]^n = (1 - 2/n)^n = e^-2 *Simple, right* ?
@davidbrisbane7206 2 ай бұрын
(1+ x/n)^n = e^x. Let x = -2. So, (1 + (-2)/n)^n = e^(-2) = 1/e².
@alexia5595 2 ай бұрын
never stop posting i love watching your videos!
@Archimedes_Notes 2 ай бұрын
Thank you! Will do!
@lionelhuber8639 2 ай бұрын
Well explained thanks
@Archimedes_Notes 2 ай бұрын
So nice of you
@lionelhuber8639 2 ай бұрын
Very interesting Video, thanks man
@Archimedes_Notes 2 ай бұрын
@@lionelhuber8639 My pleasure.