Compute the integral from zero to infinity of the function "f" with respect to x with function "f" equal to one over e to the x times the cube root of x. (e is Euler's number)
Hello bprp, I was hoping you could solve the equation f(x)= f(x-1) + f(x+1) for f(x). Even though it looks so bare-bones, WolframAlpha says the solution is f(x) = e^(-1/3 i π x) (c_2 + c_1 e^((2 i π x)/3)) (where c_1 and c_2 are arbitrary parameters) which is pretty crazy. It seems very weird how the solution has the whole math trio (pi, e, and i). Thanks for everything you do on the channel and happy holidays!
it's because the resultant family of functions are sinusoids, and are especially known for preserving this sort of convoluting condition (notice how sin(x + pi/2) + sin(x- pi/2) = 0.) an easy example f(x) = sin(x * pi/3) can be obtained by solving sin(a) = sin(2a) for a.
You know... a couple of weeks ago you published a problem of that format on Instagram. And I deduced the EXACT same formula, with the difference that I extended the "linear exponent" to add extra features. Like this: "A exp(Bx + C) + Dx + E = 0" The formula is deduced with the same exact way. There are one or two more thingies inside the Lambert as a result, but... it's the same. It's a beautiful exercise, by the way. Keep up with the good work, bprp!!!
@@trucid2: wow, good challenge. Don't know! I guess we should try it! lol At a first glance (not entirely proven), I feel feasible to say that the polynomial at the exponent of "e" needs to be the same degree as the one that's outside the "e" so that one could align some transformation of such polynomial to the exponential's coefficient and apply Lambert's W: "A exp(p(x)) + q(x) = 0" where "degree(p) = degree(q)" But then one could also seize the fact that any polynomial of degree "n" has a "n+1" powered term, but it's just that its coefficient is zero. Perhaps that could be used for the general case.
This video is amazing! Excellent explanation as per usual. I am absolutely loving the Lambert W function. It is VERY cool. Functions such as these are the reason I love Mathematics. On another note, I need to know where you got that pic of the 'Christmas tree' pleeease...😅
Ooo i love your videos with the Lambert W function! One thing I was curious about was the remaining W(..) term because before you simplified it, it was soooo close to being fish*e^fish, but that c threw things off. Could you explain why/how the C term throws off the formula, and why simplifying it becomes so much harder?
You can see in the derivation that the c is what forces us to multiply by e^whatever, as it doesn't depend on x. As for the W being so close to sinplifying, it's that way also without the c where you get W(lna*e^-lnb)
It's just like saying to solve equations of the form: ax³+bx²+cx=0 ax³+bx²+cx+d In the first eqn, you can factor out the x and reduce the cubic into a monomial and quadratic, which is easily solvable In the second eqn, when an additional 'd'(constant similar to c in quadratic) is present, it becomes so complicated that it took mathematicians several centuries, or even a millennium to arrive at a general solution of a cubic because of a constant. It just shows us how one extra term could change our method so drastically.
@@minhdoantuan8807Can you please check if I'm correct 1^x=5-2x e^(2inπx)=5-2x where n is an integer (e^(-2inπx))(5-2x)=1 Multiply some equal stuff on each side (5inπ-2inπx)(e^(5inπ-2inπx))=(inπ)(e^(5inπ)) Take Lambert W function and solve for x x=2.5 - (W(inπe^(5inπ)))/2inπ Is it correct?
I checked it and it x is indeed 2 when n=1/2 (not integer but still satisfies as exp(2iπ*nx) is exp(2iπ)) but I do not know how to calculate other values of x here in the complex domain since wolfram does not calculate this much :(
@@vikrantharukiy7160 Yes. Apparently we have to multiply both sides by x^something (I don't remember that value) which does not help us to solve the problem. The px term is such a pain...
If the inside is equal to -1/e, we actually only get 1 solution because are exactly at the minimum of xe^x. So we have, with y being the argument: y < -1/e 0 real sol. (under the graph of xe^x) y = -1/e 1 real sol. (at bottom of bump) -1/e < y < 0 2 real sol. (on either side of the bump) y ≥ 0 1 real sol. (in the strictly inc. positive part of the graph)
If a transcendental number is a number that can’t be the value of an equation that it should be impossible to find an equation for e because it’s a transcendental number. Put it answer this value: x^(1/pi*i) + 1 = 0 x = e
What about x*a^x + b*x + c = 0? I ran into this while trying to solve (x+1)^x = 64. Where you eventually get u*e^u - u - ln(64) = 0, where u = ln(64)/x.
Oh nice. I made one for when the exponent is the same as the term before. It doesn’t really work out nicely if the x exponential is different and that’s not the case 😂 A^Bx+Bx = C We get: [-W(A^C lnA)/lnA + C]/B
I've been thinking lately about fractional polinomiais. If a quadratic has two roots (zeros), how many roots does a 2.5 polinomial have? How would we go around solving it?
A polynomial by definition can only have non-negative integer powers of the variable so there is no such thing as a 2.5 degree polynomial. But if you really want, you could substitute t=sqrt(x) which would give you a degree 5 polynomial in terms of t, and then solve for t numerically(there is no general method/formula for solving a degree 5+ polynomial so you have better chances using a numerical method than trying to solve it exactly). Then lastly solve for x
Functions with fractional powers are not considered polynomials, only functions with whole number powers which aren't negative are considered polynomials. Hence for a function with a 2.5 power for example, the fundamental theorem of algebra does not apply (which states that the degree of a polynomial is equal to the number of solutions) as a fractional power isnt a polynomial. As such, as far as my knowledge goes you cant really make conclusive statements about how many solutions a fractional power would have. Hope that makes sense
@@guydell7850... I think the previous commenter stated it accurately. It has to be converted to an integer by the least prime multiple, a factor of 2 in this case, to solve: ax^(2 + 0.5) + bx^(1 + 0.5) + cx^(0.5) type polynomial into a new understandable ax^5 + bx^3 + cx polynomial still but expanding out to have redundant roots as people use of the √ symbol producing only a primary root and the secondary root produces false results for math majors. In electrical engineering physics √x = +/- results not + results because of "right hand rule" electricity flow provisions to enforce positive √x or primary root results that mathematicians defined for calculations. If electrical engineering only relied on a primary root in "flux directionality" and/or power to a "load" received from a source providing that power then electronic circuit designs wouldn't exist as we see today. The electrical engineering "right hand rule" of positive and negative current and voltage direction to the load assumptions led to wave diodes, wave rectifiers, etc. because of A.C. to D.C. fixed voltages needs where it would be ideal if the source fluctuating source voltages and currents would be only positive.
Regarding the computation of the solutions (numerically), do you know if there would be any advantage of using this formula over just solving for a^x+bx+c=0 using something like the newton-raphson method? Like, maybe the lambert w function can be compiten faster and/or with more precision thus this formula would make sense. Regardless, this is a cool mental excercise to familiarize with "weird" functions and inverse functions too
The Lambert W function is generally better explored vs similar computation via other methods. Eg. For certain values, we can tell ahead of time how many iterations we need of the Quadratic-Rate formula to achieve certain precisions. Check out Wikipedia's page on numerical evaluation for the Lambert W Function.
Thanks for the video bprp, btw if you want I have an equation too (ik the answer but it's quite fun to solve): can you solve the equation ~ a x^b + c log_d(f x^g) + h = 0 ~ well I know it's a bit complicated but not hard to solve so I hope you give it a try ✓
@@soupisfornoobs4081 yeah it's another W equation but I think you shouldn't do any substitution it would cause some troubles, I made it and I solve it so I know the answer I just asked for it cuz it's actually fun to solve for me
The equation i.e ((1/√(x!-1)+1/x^2)! It surprisingly approaches to 0.999. For x>2 lim x→∞ I would really appreciate you if you check it and I would like to ask can this be constant which is mine?
10:16 technically, there is only 1 real solution if inside = -1/e. Therefore, to be precise you would write that there is 1 real solution if inside = -1/e or inside >= 0. There are 2 real solutions if -1/e < inside < 0.
Yes, you can solve that using the Lambert W function. Just take the substitution y=log_10(x) which yields the equation 100^y*y=100 which can be solved using the Lambert W function. The equation you brought up can be solved a lot easier than this though (over the reels): Just write log_10(x)^5 as ln(x)^5/ln(10)^5 and multiply both sides by ln(10)^5 giving: x^2*ln(x)^5=100*ln(10)^5=10^2*ln(10)^5 which obviously yields x=10.
I appreciate your effort brother🔥😎🙏❣️👍..As i can see you reply every appreciable question from your comments!😊..so , I would also like to have you look to my question.... Integration of (X^2 + 1){(X^4 + 1)^(3/2)} dx .. Please i want you to give solution!🙏🙂 Thankyou to read!
Inspired by this video, I found the value of a for which y = ax is tangent to y = x^x (nice exact formula using Lambert W). Struggling to find a way to solve x^x = ax for a greater than the above value - stuck at e^ln(x-1)*ln(x) = ln(a) 🤷♂️
Random person:"Math is so hard, everything is so complicated and almost nothing makes logic." Blackpenredpen:"If you have a fish times e to the fish and if you take Lambert W function, you get the fish back. You can save the fish."
Mathematicians always try to generalize the result. Mathematician(M),physicist(P) P:Asks a formula from M M:Consider a space of n dimensions P:But i only want n=3 M:Ya substitute n=3 -Richard Feynman
Could you (try) to solve this transcendental equation? x^x - x - 1 = 0 This number x fascinates me but I can't seem to find an explicit formula for it myself
There‘s no closed form for the solution of this equation. However, there‘s something called Hyper Lambert Functions, might wanna look that up if you‘re interested in solving exponential tower equations.
It is not defined. ArcTangent at x=i is a logarithm type branch point (similar to log(0)=-infinity), that links to a branch cut for the imaginary axis when Im(x)>1. A similar branch point happens for x=-i, and a branch cut for Im(x)
