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@matthiaspihusch8 ай бұрын
Question: Why does W(-(e^-1)) give us two real solutions, shouldnt it be just -1?
@ChadTanker8 ай бұрын
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)
@santri_kelana_918 ай бұрын
Can you explain about x^4 + ax^2 + bx + c
@ektamge40648 ай бұрын
@@matthiaspihuschiiiiilllllllllp
@wowyok45078 ай бұрын
signed up!
@ClickBeetleTV8 ай бұрын
"My car won't start" "Have you tried the Lambert W function?" "Holy shit it worked"
@fsisrael92244 ай бұрын
-"But did you get the fish back?"
@pinkibiswas76914 күн бұрын
This exchange humorously suggests that the Lambert W function, a somewhat obscure mathematical function, could somehow magically fix a car. It's a playful way of blending the abstract world of mathematics with everyday problems, where the idea is that something as complex as the Lambert W function might be the unexpected solution to a practical issue like a car not starting. The humor lies in the absurdity of the situation, as the Lambert W function has nothing to do with cars but is instead related to solving certain types of equations.
@TanmaY_TalK8 ай бұрын
Lambert W function ❌ bprp fish function ✅
@TramNguyen-pk2ht8 ай бұрын
W(fishe^fish) = fish for a recap
@ulisses_nicolau_barros8 ай бұрын
Underrated comment
@Ralfosaurus8 ай бұрын
"We shall 'save the fish' on both sides"
@elweewutroone8 ай бұрын
W(🐟e^🐟) = 🐟
@BurningShipFractal8 ай бұрын
Where does the letter 「W」 come from ?
@The_NSeven8 ай бұрын
I'm not sure why, but my favorite videos of yours are always the ones with the Lambert W function
@A_literal_cube8 ай бұрын
Did you mean the fish function?
@gswcooper71628 ай бұрын
I mean, you're not alone, but I don't know why I like the Fish function so much either... :D
@The_NSeven8 ай бұрын
@@A_literal_cube my bad
@tanelkagan8 ай бұрын
This is the balance of the universe at work, because they're my least favourite ones!
@The_NSeven8 ай бұрын
@@tanelkagan That's kinda funny haha
@riccardopesce72648 ай бұрын
I've just wrapped up a math study session; it's now time to relax by watching some more math.
@ethangibson86458 ай бұрын
I like watching your channel as a computer science college student because they have made me realize that somewhere in all of the calculus, vectors, etc I've gotten a little rusty at the basics.
@grave.digga_8 ай бұрын
Nice video! You explained it in a way that a lot of people can understand. I appreciate that a lot.
@blackpenredpen8 ай бұрын
Thank you!
@haydenrobloxgamer35018 ай бұрын
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!
@eccotom18 ай бұрын
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.
@omarsayed38748 ай бұрын
f(x) = x, hope that helps
@eccotom18 ай бұрын
@@omarsayed3874 x = x+1 + x-1 only for x=0 lol. and the only linear unction satisfying the relation is f(x) = 0
@omarsayed38748 ай бұрын
@@eccotom1 ah yes i forgot we will get 2x
@alonelyphoenix89428 ай бұрын
When in doubt, use f(x) = 0
@MichaelMaths_8 ай бұрын
I was looking into generalizing a formula for this a few years ago and it is very cool how it parallels solving quadratics. Instead of completing the square, we want to get the xe^x form, and there are even discriminant cases for the different branches of the Lambert W function.
@thethinker62588 ай бұрын
Teacher, can you integrate or differentiate the Lambert W function?
@-minushyphen1two3798 ай бұрын
You can do it using the formula for the derivative of the inverse function, he made a video about this before
@CarlBach-ol9zb5 ай бұрын
It can be differentiated. I saw a video doing that. And, of course, all continuous functions can be integrated AFAIK, so this one can be too.
@gastonsolaril.2378 ай бұрын
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!!!
@trucid28 ай бұрын
What if e is raised to a quadratic polynomial. Can that be solved for x?
@gastonsolaril.2378 ай бұрын
@@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.
@phat_khiep8 ай бұрын
There are n multiple choice questions, each question has i options to choose from. Step 1: Randomly choose the mth option (with m less than or equal to i and m greater than or equal to 1) in the first multiple choice question Step 2: Repeat the option in the 1st multiple-choice question in the next (k-1) multiple-choice questions. Step 3: To choose the option in the (k+1) multiple choice question, we will choose in the following way for each case: Case 1: If the option chosen in the kth multiple choice question is the mth option (with m smaller than i), then choose the (m+1)th option. Case 2: If the option chosen in the kth multiple choice question is the ith option, then choose the 1st option. Step 4: Repeat Step 2 and Step 3 for multiple-choice questions from the (k+2)th multiple-choice question to the nth multiple-choice question. Each multiple choice question has only 1 correct answer. Let t be the number of multiple-choice questions answered correctly in n multiple-choice questions, t follows the Bernouli distribution. Find k to t max.
@iwilldefeatraymak25368 ай бұрын
Another way a^x + bx+c=0 Subtract both sides by c and divide both sides by b (1/b)×a^x +x =-c/b Do (a) power both sides a^((1/b)×a^x)×a^x=a^(-c/b) Change the first a to e^ln(a) (a^x)×e^[ln(a)(1/b)×a^x]=a^(-c/b) Multiply both sides by ln(a)×(1/b) (ln(a)/b)×(a^x)×e^[(ln(a)/b)×(a^x)]=ln(a)/b ×a^(-c/b) Now you can use the w function (Ln(a)/b)× a^x= w[ln(a)/b ×a^(-c/b)] Divide both sides by ln(a)/b then take log base (a) from both sides x=log (base a)[ b(W[ln(a)/b× a^(-c/b)])/ln(a)]
@Bhuvan_MS8 ай бұрын
Since 'a' is the base for the logarithm, this formula would have some restrictions. Mainly 'a' must be greater than 1.
@kenroyadams27628 ай бұрын
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...😅
@alonelyphoenix89428 ай бұрын
He himself made the tree, apparently u can buy it lol
@Zach010ROBLOX8 ай бұрын
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?
@soupisfornoobs40818 ай бұрын
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)
@Bhuvan_MS8 ай бұрын
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.
@rorydaulton68588 ай бұрын
You have a minor mistake in your video. Near the end you say that if "-1/e
@MichaelRothwell18 ай бұрын
Totally agree. I spotted this glitch too.
@shoeman69668 ай бұрын
This man’s algebraic manipulation ability is superb!
@philip22058 ай бұрын
What about (1) ax^a + bx^b + c = 0, (2) ax^a + bx^b + cx^c = 0 or (3) the general case ax^a + bx^b + ... + nx^n?
@vikrantharukiy71607 ай бұрын
As for the first one, just divide all terms by a and solve
@lazarusisaacng8 ай бұрын
I met your video that it is the first Lambert W function. And now this video can tell us about more information like quadratic equation, I must give you 👍.
@dkdashutsa15758 ай бұрын
Is there any formula for summation of i = 1 to n of W(i)
@Bhuvan_MS8 ай бұрын
Is the eqn of the form: x^x+px+q=0 also solvable using Lambert-W function?
@vikrantharukiy71607 ай бұрын
I tried and failed
@Bhuvan_MS7 ай бұрын
@@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...
@gswcooper71628 ай бұрын
Do you think you could you solve a^(x^2)+b^x+c=0 for x?
@kaviramyead79878 ай бұрын
What a monster
@gswcooper71628 ай бұрын
@@kaviramyead7987 Heh heh. thanks! >:D
@mcgamescompany8 ай бұрын
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
@gamerpedia15358 ай бұрын
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.
For small x, W(x) is just x-x² so yes I'd say there is an advantage
@Nylspider7 ай бұрын
I always find the fact that you draw fish with eyebrows to be unreasonably funny
@isjosh80648 ай бұрын
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
@pahandulanga10398 ай бұрын
Can you make a video of you solving an equation using this formula?
@table55848 ай бұрын
Thanks, now I can solve 1^x + 2x - 5 = 0 😊
@deltalima67038 ай бұрын
Nope, doesnt work if a=1, so you still cant figure out that x=2 is a solution. :-p
@minhdoantuan88078 ай бұрын
@@deltalima6703in that case, 1^x = 1 for all x, so 2x - 4 = 0, or x = 2
@HimanshuRajOk8 ай бұрын
@@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?
@HimanshuRajOk8 ай бұрын
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 :(
@shahar68408 ай бұрын
ax(x^2 + bx/a + c/a + d/x/a) = 0 d/x/a = da/x if d =! 0 then the root can't be 0. If d = 0 then one of the roots is 0. If x = 0 then d = 0 0(x^2 + bx/a + c/a + da/x) 0(0+0++0 + 0/0 * 0) = 0(0/0) = 0 0/0 * 0 = 0
@ton1468 ай бұрын
When I was at UCT 55 years ago the lecturer showed us two other quadratic formulas involving an a,b and c which also gave the roots as well. I have never seen them again or been able to derive them. Does anyone else have a clue?
@trucid28 ай бұрын
You can rewrite a degree two polynomial in different ways: ax^2+bx+c=(px+q)(rx+s) a(x−h)^2=k
@AyushTomar-wp3is8 ай бұрын
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?
@AyushTomar-wp3is8 ай бұрын
Sir I would like you to check this and give ur thoughts please 🙏🏼
@Voiduser-ds3pg8 ай бұрын
Wow bro ur right , it can be your own constant 👍
@qubyy17142 ай бұрын
Now try a tetrated to x + b^x + cx + d could be a fun video ❤
@tambuwalmathsclass8 ай бұрын
Amazing 😊
@Max-mx5yc8 ай бұрын
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)
@IRM3218 ай бұрын
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.
@MichaelRothwell18 ай бұрын
This is the solution I wrote before seeing the video, and so before seeing the conditions on a and b. It agrees with the solution in the video, except that I point out that if a^(-c/b)(ln a)/b=-1/e then there is only one solution (as the values given by W₋₁ and W₀ coincide in this case). It is clear that we want to use the Lambert W function here. It is also clear that we are going to have to consider several cases besides the "nice" case in which a>0, a≠1, b≠0, i.e.a=1 or a=0 or a
@johnny_eth8 ай бұрын
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?
@Ninja207048 ай бұрын
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
@guydell78508 ай бұрын
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
@lawrencejelsma81188 ай бұрын
@@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.
@user-rk4nm9yf7d7 ай бұрын
Can we call this completing fishes?
@CuberSourav8 ай бұрын
Integrate the Cubic formula Math for Fun 😂
@rupeshrupesh28678 ай бұрын
Got liked it's coming
@zhabiboss8 ай бұрын
Fish function
@jacplanespotting3148 ай бұрын
So, what level of high school or college made is this geared to, in your opinion?
@gljdds41648 ай бұрын
i love how you always use the fish when explaining the lambert w function
@spoopy13228 ай бұрын
I love your videos! ❤
@nokta98198 ай бұрын
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 ✓
@soupisfornoobs40818 ай бұрын
This looks like another product log situation. You could probably get from that to a more general case of this video with a substitution like a^x = u
@nokta98198 ай бұрын
@@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
@sebmata1356 ай бұрын
Pretty cool that there's a general solution for the intersection of an exponential and a line! Very interesting manipulations to get to Lambert W on lines 2, 3 and 4
@emmanuellaurens21326 ай бұрын
There's a general solution because mathematicians decided they wanted one badly enough, and so just named it the Lambert W function. 🙃 Well, okay, it's a bit more complicated than that, but now they can pretend they can solve this kind of equations exactly rather than just to an arbitrary degree of precision 🙂
@robinsparrow16186 ай бұрын
i had never heard of the lambert W function before watching your videos! i'm intrigued...
@lpschaf89438 ай бұрын
Thank you so much. This was very satisfying.
@darcash17385 ай бұрын
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
@wafflely98778 ай бұрын
Make a video on the integral from -1 to 1 of (-e^x^2/3)+e dx!! 🙏
@necrolord19208 ай бұрын
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.
@user-cf8os4pd7i8 ай бұрын
What is the invers of f(x)=x4+x3+2 Please solve it
@mrexl98308 ай бұрын
Freaking LOVE the lambert W functions
@dfjao978 ай бұрын
Can you help me solve this? A right triangle have a base length of 3x, a height of 4x and a hypotenuse of 5x. Find x.
@sergeygaevoy64228 ай бұрын
I think we assume a > 0, a 1 and b 0. Otherwise it is a much simplier (trivial) equation.
@remicou84208 ай бұрын
he explains at the end why those parameters are disallowed. you can’t compute the result if any of the conditions are broken
@sergeygaevoy64228 ай бұрын
@@remicou8420 Thank, there is a "post-credit" scene ...
@klasta21678 ай бұрын
(sin^(8-x)(cos(2x)))/(x^(8-e^(8-x))) Can you solve this? My professor gave this in internals for 5 marks, its kinda easy but do try.
@whiteskeleton94538 ай бұрын
Formula for series in n world for n^y/x^n please make a video for it😊
@user-gm8ir4sd6m8 ай бұрын
can you please make a video talking about the lebesgue integral and also iys connection with the laplas transfromation
@NullExceptionch8 ай бұрын
Can you please solve this? “Tan(x)=sqrt(x+1)
@BryndanMeyerholtTheRealDeal8 ай бұрын
I thought that there was a typo and should have been ax^2 instead of a^x…
@TranquilSeaOfMath8 ай бұрын
Fairly straight forward presentation. Nice example of Lambert W Function with merchandise tie-in.
@NelDora-ih1bd8 ай бұрын
hello what white board is that?
@khalidisab17126 ай бұрын
Does x= ln(-b-c-a) not work?
@shafikbarah92738 ай бұрын
Is there a general way to get the general formula of any sequence just from the reccursive formula?
@Wouter101238 ай бұрын
Generating functions
@ivantaradin498 ай бұрын
what if the x, which is multiplied by b, is square rooted??? ( a^x + b*sqrtx +c =0 )
@redroach4018 ай бұрын
can you please solve: (x+1)^x=64.
@Grassmpl8 ай бұрын
Use newtons method to approximate.
@tanuj6557 ай бұрын
Please please make this question a isoceles Triangle having equal sides 12cm height is 7.5cm find the area of Triangle
@user-zp9cn2qq8g8 ай бұрын
I love you video very much, and I also have a very very very hard question for you, if 2^x + 3^x = 4^x, can you find the x?
@RishabMurthy8 ай бұрын
Is there a way to solve x^e^x = (numb) or ln (x) / e^x = sin (x) or solving complex equatkons with sin (x) like x^(sin (x)) = numb
@RishabMurthy8 ай бұрын
Without iteration
@RishabMurthy8 ай бұрын
And is there a way to solve xe^e^x = (number)
@RishabMurthy8 ай бұрын
Or xsin (e^x)
@mrpineapple76668 ай бұрын
What happens if we want complex solutions?
@crowreligion6 ай бұрын
Use other branches of lambert W function There are branches after every integer, and everything except for branch 0 and -1 gives complex solutions
@Xnoob5452 ай бұрын
@@crowreligion and also you do not need to follow all of the conditions he mentioned I think a can be anything except 1 and inside can be anything(?)
@xcoolchoixandanjgaming10768 ай бұрын
The fact that the shirt youre wearing is also the fish function lol
@MatthisDayer8 ай бұрын
you know what, i was just playing with these kinds of equations yesterday, ab^(cx) + dx = e
@DEYGAMEDU8 ай бұрын
Sir I have a question how to solve the lambart W function. I mean if there is not xe^x so how it will be solved by the calculator or us
@General12th8 ай бұрын
Hi BPRP! So good!
@orenawaerenyeager7 ай бұрын
Am i jealous of his t-shirt Of course i am i need it😮
@1291401638 ай бұрын
5:15 ROFL that brief hyper speed-up tickled my funny bone! 😂
@adarshk74848 ай бұрын
do integral of 1/(1-x^20) dx
@lpschaf89438 ай бұрын
beautiful video
@padmasangale81948 ай бұрын
Bro pls solve *x²[logx (base 10)]⁵=100* Can we also solve it with Lambert W func?
@gigamasterhd42398 ай бұрын
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.
@padmasangale81948 ай бұрын
@@gigamasterhd4239 thanks😊 👍
@gigamasterhd42398 ай бұрын
@@padmasangale8194 No problem, very happy to help! Have a great rest of your day. 👍
@padmasangale81948 ай бұрын
@@gigamasterhd4239 ⚡🔥
@pchevasath6 ай бұрын
I want to find the critical points of the graph f(x) = 2^x - x^2. So I find f’(x) = (2^x)(ln2) - 2x and set this to 0. Using the above formula, I get x = (-1/ln2)(W(1/2)), which only offers one solution. But I know that the graph f(x) = 2^x - x^2 has 2 critical points. What did I do wrong?
@gigamasterhd42393 ай бұрын
You‘re missing the other real solution on the k=-1 branch. Your first solution is -W(-1/2*log^2(2))/log(2) ≈ 0.4851 and the second is -W_(-1)(-1/2*log^2(2))/log(2) ≈ 3.2124. Remember that you always have two real solutions (one on the k=0 and the k=-1 branch) if -1/e
@javierferrandizlarramona65882 ай бұрын
Excelent!
@bivekchaudhari45938 ай бұрын
Please solve this question integral of 1/1+x⁵ dx
@karhi42718 ай бұрын
How to solve: (e^x)-3=ln(x)
@Cbgt8 ай бұрын
Please solve (lnx)•(x^x)=1 I just can't do it myself
@pierreabbat61578 ай бұрын
What do you do if you have tuna times exponential of haddock?
@shyamaldevdarshan8 ай бұрын
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!
@elsicup8 ай бұрын
I was trying to solve this thing About 2 weeks ago, thank u😊
@nihilisticgacha5 ай бұрын
when the thumbnail of this video enters my peripheral vision, i thought you were holding a vape lol
@Deejaynerate8 ай бұрын
If you change the equation slightly so that a^x is multiplied by -c, then the formula becomes xlna = 0
@maxrs078 ай бұрын
can u calculate W func by hand or its numerical only
@RubyPiec7 ай бұрын
my calculator has no lambert w function button. how can i simulate one
@scottleung95878 ай бұрын
Nice job!
@jejnsndn8 ай бұрын
May you integrate sqr of x³+1 ( the square root is all over the expression)
What about x**a+bx+c=0 (same with a and the first x swapped)?
@orisphera8 ай бұрын
Perhaps b=ka would be useful
@user-zz3sn8ky7z8 ай бұрын
Then it's just the a-th root of (-bx-c), isn't it?
@orisphera8 ай бұрын
@@user-zz3sn8ky7z But there's x in (-bx-c)
@steamedeggeggegg5 ай бұрын
MVP of this episode: multiply both sides
@aubrieque8 ай бұрын
but what about a^x + x root b?
@noahblack9148 ай бұрын
6:57 My favorite definition of trancendental lol
@bioroomproductions8 ай бұрын
Now try a^(1/x)+bx+c=0
@math_qz_28 ай бұрын
Excellent 😮
@dethmaiden19918 ай бұрын
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) 🤷♂️
@vikrantharukiy71607 ай бұрын
I don’t think it’s possible without numerical methods Could be wrong tho