2^x = 4x

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2^x = 4x

Рет қаралды 40,146

26 күн бұрын

This problem has two solutions. The second solution, if done by algebra will require a special function called the Lambert W function. In my solution, I used a power series approximation to estimate the W(t)
This is the video I mentioned
• Lambert W Function

Пікірлер: 113

@johnka5407 25 күн бұрын

'if the value inside, the argument here, is close to 0' when did you become a physicist? 😆

@PrimeNewtons 24 күн бұрын

Haha. There's a lot of engineering in me.

@sethdurais2477 25 күн бұрын

Something that is seriously overlooked in your videos are your straight lines when dividing the board! I know its just a side note but you have to admit that Mr Prime draws some of the best straight lines! It is extremely satisfying to see 💯

@davidsousaRJ 25 күн бұрын

2^x = 4x = 2²x, therefore x = 2^(x-2). Raising the both sides to the (1/(x-2))-th power, we have x^(1/(x-2)) = 2. Note that if we square both sides, we have x^(2/(x-2)) = 4, or x^(2/(x-2)) = 4^(2/(4-2)). By comparison, x = 4 is a solution. The other one is only possible to get using the Lambert function.

@apexgoblin 25 күн бұрын

blackpenredpen derived a formula for a^x + bx + c = 0 you can use the formula here too!

@lubiemuze6368 25 күн бұрын

yep, I ve done that like that

@jimmywatson7950 24 күн бұрын

😮😮😮 can you please tell the formula

@horev8822 23 күн бұрын

@@jimmywatson7950 search bprp solution for transcendal equation

@reminderIknows 23 күн бұрын

@@jimmywatson7950It's the quadratic formula. (-b +/- sqrt(b^2 - 4ac))/2a BUT. BPRP did not invent this formula.

@reminderIknows 23 күн бұрын

The quadratic formula was not derived by bprp.

@samzied 2 күн бұрын

One of the reason I enjoy math is that it transcends our petty egotic drives. Respect for the matter should involve a minimalistic attitude regarding self-promotion when presenting a topic. In any case, one should always make very sure his discourse is blunder-free before thinking he can afford wasting some focus on posing.

@user-du8rw6tb6r 5 күн бұрын

The most ASMR voice ever!!!

@domsunny3715 18 күн бұрын

That’s is actually so cool, great video man

@DaniDy01 25 күн бұрын

I didnt know there was a formula for the w function wow

@YAWTon 25 күн бұрын

It is a formula for an approximation of W, not a formula for W.

@zandergall9895 25 күн бұрын

I think its the Taylor series of the w function, hence why it only works for small x. You need infinite terms for it to be exact

@BangkokBubonaglia 25 күн бұрын

There's a Taylor expansion for any function. You just have to be able to calculate all the derivatives. It looks like in this case you can continue the series and get a better approximation by adding more and more terms of the form (-1)^(n-1) * n^(n-2) * t^n / (n-1)!. It should be pretty easy to prove since W(x) has a nice expression for its derivative: W'(x) = W(x) / x*(1+W(x)). You can calculate the Taylor expansion around any value too. Not just zero.

@frimi8593 9 күн бұрын

⁠@@BangkokBubonagliathere is not a Taylor expansion for any function, though you’re right that there is one for the Lambert W function

@ritwikgupta3655 25 күн бұрын

You celebrate Math. Great to watch.

@CalculusReviser 25 күн бұрын

Excellent, clearly explained video :)

@alpmuslu3954 25 күн бұрын

Love your work man:)

@jakehobrath7721 24 күн бұрын

Damn KZfaq policy!! Now I’ll never know what the flower is called!

@PrimeNewtons 24 күн бұрын

Phi

@jakehobrath7721 24 күн бұрын

@@PrimeNewtonsI figured it couldn’t have been phi for KZfaq to flag it, lol. I can’t imagine what it thought you were saying. Anyways Great video, thank you much!

@the_real_nayak 25 күн бұрын

better way - use iterations ; just start with x = 2^x/4 and put x = 0 , then keep on putting the result values again in the expession till the value of x is almost equals to the expresison of 2^x/4 ; that'd be your answer

@TheFrewah 24 күн бұрын

Well, that would be numerical rather than analytical.

@the_real_nayak 24 күн бұрын

@@TheFrewah since u already know there are 2 solutions , one is 4 and other is somewhere near 0 , better to solve like this instead of going to wolframalpha for W values

@TheFrewah 24 күн бұрын

@@the_real_nayak In practice it may be if you havethis problem as an engineer

@betterbee980 25 күн бұрын

I literally love his videos ❤

@yiutungwong315 53 минут бұрын

π in the Riemann Paradox and Sphere Geometry System Incorporated So Tau = 2π = π^2 = 4 So 2^Tau = 4Tau = 2^4 = 4 × 4 = 16 X can be Solved for 4 and Tau

@TheBedLump_Sans 25 күн бұрын

love from Dubai!

@adamnyback 23 күн бұрын

9:19 "Come on!"

@murdock5537 25 күн бұрын

Really awesome, many thanks, Sir!

@didier3821 25 күн бұрын

Congrats from France

@albajasadur2694 25 күн бұрын

Thank you sir. I have two questions. (1) How can we determine the number of real roots ? (2) Can we get the solution x=4 from Lambert W function method ?

@marianl8718 25 күн бұрын

The Lambert function calculator gives two solutions for (-ln 2) / 4 : - 2.772589 and - 0.214811 These two solutions divided by - ln 2 will give us 4 and 0.3099... .

@YAWTon 25 күн бұрын

(1): 2^x-4x is positive for x< 0, negative for x=3 and positive for x=5. ==> there must be at least two real roots. The second derivative of 2^x-4x is positive for all values of x ==> there are at most 2 real roots. (2) Yes: x=4 is the solution on the second branch of W. (c.f. Wikipedia article on Lambert W function, and Prime Newtons excellent clip, link in the description of this video).

@SG49478 24 күн бұрын

You can use calculus to figure that out. Set f(x)=2^x-4x. Then the first derivative is f'(x)=ln2*2^x-4. To assess for maximum or minimum points we set the first derivative to 0. ln2*2^x-4=0. This equation is easily solvable, 2^x=4/ln2, x=ln(4/ln2)/ln2. The second derivative is f''(x)=(ln2)^2*2^x. This is positive for all real x, therefor x=ln(4/ln2)/ln2 is a local minimum. The value for f for this minimum is negative. However for x=0 f(x) is positive and for x=5 f(x) is positive as weOur minimum is in between these two values and this is the only extreme point we have as f'(x) can become zero only at this one point. Therefor our equation must have exactly 2 solutions.

@marianl8718 24 күн бұрын

@@SG49478 The reasoning is mostly correct, but it is not sufficient proof that we have only two solutions. By trial, two values were found for which the function is positive, 0 and 5, but this is not part of the demonstrative mathematical rigor that is required. My view is that one cannot show that there are only two solutions except by actually solving the ecuation f(x) = 0.

@SG49478 24 күн бұрын

@@marianl8718 Well then explain to me how a steady function with exactly one local minimum where f(x) is negative at that minimum and no local maximum and two values where one is smaller and one is greater than the x value of the local minimum with each of them with f(x) being positive could have by any means more than 2 zero points. That is simply not possible. If the graph turns around and cuts the x-axis a third time, the function would have to have at least one local maximum. However with the first and the second derivative we have proven, that this function can not have a local maximum. Therefor in my opinion the proof is sufficient.

@light_fizz 11 күн бұрын

Great video man

@Ron_DeForest 24 күн бұрын

Just curious. Instead of using the appropriate approach you did, can’t you just use the actual lambert W function? You’ve shown it a few times. That would get you the number you’re looking for regardless of how close to zero the answer is or not, wouldn’t it?

@laydenhalcomb4559 25 күн бұрын

Why did it blur the phi

@XanderAnimations 18 күн бұрын

Yeah really weird

@Mam_Otazku 15 күн бұрын

Yeah really weird

@softwet4341 3 күн бұрын

Yeah really weird

@davannaleah 25 күн бұрын

Of course, you could just use the solver function on your calculator, but where's the fun in that 🎉

@BRUBRUETNONO 24 күн бұрын

Hi, Thanks for your insterestin problem, that I solved that way here below. Tell me if you like it. Of course, I didn't look at your solution. Greetings and keep up the good job. BEGIN Let's name (i) the equation to solve 2^x=4x Let the function f(x)=2^x-4x from R to R So the question is to find the roots of f(x) We can say that f(x) (being the sum of two continuous functions) is as weel continuous on R. Let's evaluate the behavior of f(x). The derivate of f is f'(x)=ln(2).2^x-4 f'(x)=ln(2).2^x-2^2 Then f'(x)=2^2.[ln(2).2^(x-2)-1] Let's see for what values of x, f is increasing so that f'>0. So that ln(2).2^(x-2)-1>0 So ln(2).2^(x-2) > 1 So if x verifies (ii) 2^(x-2) > 1/ln(2) then f(x) is strictly increasing Moreover, as ln(2)>0 (ln(2)#0,693) and the function 2^x is strictly positive on R and the logarythm function is strictly increasing on R+, we can then take the ln on both sides of inequation (ii) and it gives ln[ln(2).2^(x-2)] > ln(1) ln(ln(2))+ln[2^(x-2)] > 0 (x-2)ln(2) > -ln(ln(2)) x > 2 - ln(ln(2))/ln(2) Let following equation and value m (iii) m = 2 - ln(ln(2))/ln(2) we know as well from inequation (ii) that 2^(m-2) = 1/ln(2) that we name equation (iv) We can say that for x € [m ; +inf[ we have f'(x) > 0 and f(x) is strictly increasing for x € [-inf ; m[ we have f'(x) < 0 and f(x) is strictly decreasing Then f(x) has got a minimum value for x=m Let's evaluate f(x) at -infinite and + infinite. We can say that for x --> -inf, 2^x --> 0+ and 4x --> -inf Then for x --> -inf, f(x)=2^x-4x --> +inf We can say that for x --> +inf, f(x)=2^x-4x is equivalent to 2^x Then as for x --> +inf, 2^x --> +inf Then for x --> +inf, f(x) --> +inf Let's evaluate the minimum value of f, being f(m). If f(m) is negative we can say that we will have two solutions. So we have f(m)=2^m-4m we can write as well f(m)=2^m-2^2.m=2^2.[2^(m-2)-m] from (iii) and (iv) we have f(m)=2^2.[1/ln(2)-2+ln(ln(2))/ln(2)]=2^2.[1-2ln(2)+ln(ln(2))]/ln(2) So f(m)=2^2.[ln(e)-ln(2^2)+ln(ln(2))]/ln(2) So f(m)=2^2.ln[e.ln(2)/4]/ln(2) As we know ln(2)#0,693 > 0, then f(m) and ln[e.ln(2)/4] have got the same sign Then Let's see if ln[e.ln(2)/4] < 0 Let's see if e.ln(2)/4 < e^0 Let's see if e.ln(2)/4 < 1 Let's see if e < 4/ln(2) With a calculator we have 4/ln(2)#5,771 and e#2,718 Then e < 4/ln(2) is confirmed and so f(m) < 0 Let's evaluate the value of m = 2 - ln(ln(2))/ln(2) Let n=ln(ln(2))/ln(2). Then m = 2 - n We have 1/2 < ln(2)#0,693 < 1 Then ln(1/2)

@NhaNguyen-cx1ri 9 күн бұрын

2^×=4^× >>2^×-4^×=0 2^×(1-2^×)=0 1=2^× X=0 X⁰=1

@thegamer7537 21 күн бұрын

just divide both sides by zero

@marianondrejkovic2084 25 күн бұрын

If not applying inspection for solution x=4, is it possible to find 4 by algebra via product log function?

@frimi8593 9 күн бұрын

You’ll notice at one point he refers to his formula as “the principle branch of the Lambert function.” Just as sqrt(x) gives us only one of the up to two possible solutions for x=φ^2 (thus we sometimes call it “the principle root”) W(x) only gives us one of the possible solutions for x=φe^φ. It is possible to evaluate one of the other solutions (which in this case would be 4), but it would not use this formula which gives us the “principle branch”

@Th3OneWhoWaits 25 күн бұрын

Pretty sure your voice got muted or something when you were talking about phi sir. Maybe an issue with youtube?

@user-bo1ve3zx3h 24 күн бұрын

Is there any way to find those other lambert w function branches without using product log calculators?

@PrimeNewtons 24 күн бұрын

I doubt it

@shivx3295 24 күн бұрын

Did it by contoured method and solutions coming are 4 and approximately 0.309905

@user-dp1uj6db5z 23 күн бұрын

Gostei muito e obrigado

@RyanLewis-Johnson-wq6xs 20 күн бұрын

X=W(Ln(4th root of 2))/-Ln(2)

@sriharivithalapur7435 25 күн бұрын

If there are multiple solutions, then which solution is achieved by using the Lambert W function? More specifically... In this case can the solution x=4 be achieved using the Lambert W function?

@CarlBach-ol9zb 24 күн бұрын

There are multiple branches of Lambert W function. Each branch of Lambert W is represented using W subscript number. And W_0 and W_-1 provide the real solutions

@frimi8593 9 күн бұрын

⁠@@CarlBach-ol9zbpiggybacking off of this, you’ll notice that he describes the approximation as giving “the principle branch” which will be the one that any calculator will give you if unspecified. You may or may not have heard sqrt called “the principle root” before. This is because the equation x^2=φ may have more than one solution, but the principle root just gives us the positive solution. In this case you may think of “principle” as meaning the “default” answer, even if it’s not the only one

@hasanjakir360 25 күн бұрын

Don't have access to the internet, but can watch on youtube 🎉🎉

@RubyPiec 24 күн бұрын

why did you round to 0.309? The actual answer according to wolfram alpha is 0.3099 which rounds to 0.310

@vecenwilliams8172 24 күн бұрын

I didn't hear round (could have missed it) but he could have truncated it to estimate. Also when he wrote it on the board it was from an estimated method and he said the exact answer from the calculator was 0.31

@RubyPiec 24 күн бұрын

@@vecenwilliams8172 ahh ok

@alexandermorozov2248 23 күн бұрын

x≈0,30990693238069

@ryansullivan3085 23 күн бұрын

5:26 "let's not call it x, let's call it... x" I had to go back and make sure I heard him right lol

@hd.1cool803 24 күн бұрын

Is there any way to get a value for x in the equation 3^x^x = 10? Just like to know because the only way I gotten a value was from a graphing calculator.

@kemosabe761 24 күн бұрын

3^x^x=10 Let x^x=y x.ln x=ln y ln x.e^ln x=ln y W(ln x.e^ln x)=W(ln y) ln x=W(ln y) x=e^W(ln y) Now 3^x^x=3^y=10 y.ln3=ln10 y=ln10/ln3 x=e^W(ln(ln10/ln3)) x~1.5918

@hd.1cool803 23 күн бұрын

@@kemosabe761 thanks!

@sciphyskyguy4337 25 күн бұрын

How quickly would we have gotten to a reasonable answer just using Newton’s method from the start?

@TheFrewah 24 күн бұрын

That wouldn’t be a mathematical way, it would be a numerical method. This channel os about math

@sciphyskyguy4337 24 күн бұрын

@@TheFrewah True, but he just used a truncated power series to estimate a numerical solution to the product-log function.

@TheFrewah 23 күн бұрын

@@sciphyskyguy4337 Yes but still analytical, power series is what you end up with if you want to calculate e to a high degree of decimals.

@sciphyskyguy4337 23 күн бұрын

Newton-Raphson is based on a Taylor series expansion and has a region of convergence. Sounds pretty analytic to me. :-)

@user-ff5ve5ek6f 20 күн бұрын

Omg… Is it a BLACKMATH???

@KlubPenguin 24 күн бұрын

Prove the MacLaurin expansion of the lambert function next

@jeeconquer 21 күн бұрын

X=4 is the answer Take log base 2 in both side and solve further

@yiutungwong315 53 минут бұрын

4 and Tau This is Because π = 2 in the Riemann Paradox and Sphere Geometry System Incorporated

@RyanLewis-Johnson-wq6xs 20 күн бұрын

2^4=4*4 x=4 I didn’t graph use a calculator or anything I did it in my head.

@movavi5096 25 күн бұрын

"1? 2? 3? 4? Ye 4."

@kinshuksinghania4289 25 күн бұрын

Why does the W function not give x=4 as the solution?

@YAWTon 25 күн бұрын

Actually, it does give x=4. W is a multivalued function. For x between -1/e and 0 there are two real branches W_0 and W_-1. In the clip, he shows the solution for the first branch. x=4 is the solution for the second branch. For details read the article on "Lambert W function" in Wikipedia. Also I recommend Prime Newton's clip about the W function (link is in the description of this clip).