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

@PrimeNewtons

26 күн бұрын

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

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 💯

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.

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

@lubiemuze6368

26 күн бұрын

yep, I ve done that like that

@jimmywatson7950

26 күн бұрын

😮😮😮 can you please tell the formula

@horev8822

25 күн бұрын

​@@jimmywatson7950 search bprp solution for transcendal equation

@reminderIknows

24 күн бұрын

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

@reminderIknows

24 күн бұрын

The quadratic formula was not derived by bprp.

That’s is actually so cool, great video man

Excellent, clearly explained video :)

Love your work man:)

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.

You celebrate Math. Great to watch.

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?

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

@YAWTon

26 күн бұрын

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

@zandergall9895

26 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

26 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

11 күн бұрын

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

love from Dubai!

The most ASMR voice ever!!!

I literally love his videos ❤

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

26 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

26 күн бұрын

(1): 2^x-4x is positive for x 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

26 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

26 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

26 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.

Why did it blur the phi

@XanderAnimations

20 күн бұрын

Yeah really weird

@Mam_Otazku

17 күн бұрын

Yeah really weird

@softwet4341

5 күн бұрын

Yeah really weird

Once you can estimate that the 2nd solution for x is near zero, do a first order Taylor series for 2^x ( = 1 +xln2) and solve that = 4x to directly get x = 1/ (4-ln2) ~ 0.302. Using the 2nd term would get you closer at the expense of solving a quadratic equation.

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

@PrimeNewtons

26 күн бұрын

I doubt it

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

@frimi8593

11 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”

Really awesome, many thanks, Sir!

Congrats from France

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

26 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

11 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

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

@PrimeNewtons

26 күн бұрын

Phi

@jakehobrath7721

26 күн бұрын

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

Great video man

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

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

26 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

25 күн бұрын

@@kemosabe761 thanks!

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

26 күн бұрын

Well, that would be numerical rather than analytical.

@the_real_nayak

25 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

25 күн бұрын

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

@justliberty4072

Күн бұрын

@@TheFrewah Well, how about this: if you are going to use an expansion for W, just use a Taylor Series expansion around 0 for 2^x and get the answer to 2 decimal places almost instantly.

@TheFrewah

23 сағат бұрын

@@justliberty4072 That would work.

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

9:19 "Come on!"

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

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) 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] Let's see if e.ln(2)/4 Let's see if e.ln(2)/4 Let's see if e With a calculator we have 4/ln(2)#5,771 and e#2,718 Then e 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 Then ln(1/2)

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

@TheFrewah

26 күн бұрын

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

@sciphyskyguy4337

26 күн бұрын

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

@TheFrewah

25 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

25 күн бұрын

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

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

@vecenwilliams8172

26 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

26 күн бұрын

@@vecenwilliams8172 ahh ok

@alexandermorozov2248

24 күн бұрын

x≈0,30990693238069

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

@YAWTon

26 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).

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

just divide both sides by zero

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

Gostei muito e obrigado

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

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

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

@yiutungwong315

Күн бұрын

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

Prove the MacLaurin expansion of the lambert function next

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

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

Omg… Is it a BLACKMATH???

❤

W function

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

four

4

How’s the approximation of the function found looks like some Taylor series stuff

x=4

@yiutungwong315

Күн бұрын

X can be Solve For 4 and Tau This is Because π = 2 in the Riemann Paradox and Sphere Geometry System Incorporated Tau = 2^π = 4

By inspection, x=4 as 2^x=2⁴=16 and 4x=4(4)=16

This use of a case specific function to get a numerical approximation seems to support my suspicion that maths is a branch of engineering ;)

There is a math error at 9:10. You forgot to divide the LHS by 4. So the solution is not -W(-ln(2))/ln(2) but -4 * W(-ln(2)) / ln(2)

X =4. I did it in my mind.😅

4