2^x = 4x
HTML-код
- Опубликовано: 21 авг 2024
- 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
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 💯
'if the value inside, the argument here, is close to 0' when did you become a physicist? 😆
Haha. There's a lot of engineering in me.
blackpenredpen derived a formula for a^x + bx + c = 0 you can use the formula here too!
yep, I ve done that like that
😮😮😮 can you please tell the formula
@@jimmywatson7950 search bprp solution for transcendal equation
@@jimmywatson7950It's the quadratic formula. (-b +/- sqrt(b^2 - 4ac))/2a
BUT. BPRP did not invent this formula.
The quadratic formula was not derived by bprp.
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.
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 ?
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... .
(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).
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.
@@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.
@@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.
I didnt know there was a formula for the w function wow
It is a formula for an approximation of W, not a formula for W.
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
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.
@@BangkokBubonagliathere is not a Taylor expansion for any function, though you’re right that there is one for the Lambert W function
You celebrate Math. Great to watch.
That’s is actually so cool, great video man
Why did it blur the phi
Yeah really weird
Yeah really weird
Yeah really weird
X=W(Ln(4th root of 2))/-Ln(2)
The most ASMR voice ever!!!
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
Well, that would be numerical rather than analytical.
@@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
@@the_real_nayak In practice it may be if you havethis problem as an engineer
@@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.
@@justliberty4072 That would work.
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.
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.
Damn RUclips policy!! Now I’ll never know what the flower is called!
Phi
@@PrimeNewtonsI figured it couldn’t have been phi for RUclips to flag it, lol. I can’t imagine what it thought you were saying. Anyways Great video, thank you much!
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?
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
@@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
9:19 "Come on!"
Love your work man:)
love from Dubai!
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
Excellent, clearly explained video :)
I literally love his videos ❤
why did you round to 0.309? The actual answer according to wolfram alpha is 0.3099 which rounds to 0.310
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
@@vecenwilliams8172 ahh ok
x≈0,30990693238069
Congrats from France
just divide both sides by zero
If not applying inspection for solution x=4, is it possible to find 4 by algebra via product log function?
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”
Of course, you could just use the solver function on your calculator, but where's the fun in that 🎉
2^×=4^×
>>2^×-4^×=0
2^×(1-2^×)=0
1=2^×
X=0
X⁰=1
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?
Really awesome, many thanks, Sir!
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)
How quickly would we have gotten to a reasonable answer just using Newton’s method from the start?
That wouldn’t be a mathematical way, it would be a numerical method. This channel os about math
@@TheFrewah True, but he just used a truncated power series to estimate a numerical solution to the product-log function.
@@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.
Newton-Raphson is based on a Taylor series expansion and has a region of convergence. Sounds pretty analytic to me. :-)
Pretty sure your voice got muted or something when you were talking about phi sir. Maybe an issue with youtube?
Did it by contoured method and solutions coming are 4 and approximately 0.309905
2^4=4*4 x=4 I didn’t graph use a calculator or anything I did it in my head.
4 is just one of the solutions.
It says to find all of them.
Omg… Is it a BLACKMATH???
Please take out the text because it covers your writing on board
Great video man
X=4 is the answer
Take log base 2 in both side and solve further
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
Why does the W function not give x=4 as the solution?
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).
Is there any way to find those other lambert w function branches without using product log calculators?
I doubt it
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
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.
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
@@kemosabe761 thanks!
Don't have access to the internet, but can watch on youtube 🎉🎉
"1? 2? 3? 4? Ye 4."
4
W function
x=4
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
four
Gostei muito e obrigado
❤
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)
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 ;)
X =4. I did it in my mind.😅
How’s the approximation of the function found looks like some Taylor series stuff
4