Solution of the transcendental equation a^x+bx+c=0
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Today we will find a general solution to the transcendental equation a^x+bx+c=0. Here's an example 2^x-3x-1=0 • all solutions to 2^x-3...
Check out a detailed the Lambert W function introduction: • Lambert W Function (do...
0:00 solve a^x+bx+c=0
7:02 Check out Brilliant
8:06 bonus part (the conditions on this formula)
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Thank you all!
@blackpenredpen 4 місяці тому
Learn contest math on Brilliant: 👉brilliant.org/blackpenredpen/ (now with a 30-day free trial plus 20% off with this link!)
@matthiaspihusch 4 місяці тому
Question: Why does W(-(e^-1)) give us two real solutions, shouldnt it be just -1?
@ChadTanker 4 місяці тому
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)
@ibn_syahr_09 4 місяці тому
Can you explain about x^4 + ax^2 + bx + c
@ektamge4064 4 місяці тому
@@matthiaspihuschiiiiilllllllllp
@wowyok4507 4 місяці тому
signed up!
@TanmaY_Integrates 4 місяці тому
Lambert W function ❌ bprp fish function ✅
@TramNguyen-pk2ht 4 місяці тому
W(fishe^fish) = fish for a recap
@ulisses_nicolau_barros 4 місяці тому
Underrated comment
@Ralfosaurus 4 місяці тому
"We shall 'save the fish' on both sides"
@elweewutroone 4 місяці тому
W(🐟e^🐟) = 🐟
@BurningShipFractal 4 місяці тому
Where does the letter 「W」 come from ?
@ClickBeetleTV 4 місяці тому
"My car won't start" "Have you tried the Lambert W function?" "Holy shit it worked"
@fsisrael9224 4 дні тому
-"But did you get the fish back?"
@The_NSeven 4 місяці тому
I'm not sure why, but my favorite videos of yours are always the ones with the Lambert W function
@A_literal_cube 4 місяці тому
Did you mean the fish function?
@gswcooper7162 4 місяці тому
I mean, you're not alone, but I don't know why I like the Fish function so much either... :D
@The_NSeven 4 місяці тому
@@A_literal_cube my bad
@tanelkagan 4 місяці тому
This is the balance of the universe at work, because they're my least favourite ones!
@The_NSeven 4 місяці тому
@@tanelkagan That's kinda funny haha
@riccardopesce7264 4 місяці тому
I've just wrapped up a math study session; it's now time to relax by watching some more math.
@ethangibson8645 4 місяці тому
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_ 4 місяці тому
Nice video! You explained it in a way that a lot of people can understand. I appreciate that a lot.
@blackpenredpen 4 місяці тому
Thank you!
@MichaelMaths_ 4 місяці тому
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.
@haydenrobloxgamer3501 4 місяці тому
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!
@eccotom1 4 місяці тому
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.
@omarsayed3874 4 місяці тому
f(x) = x, hope that helps
@eccotom1 4 місяці тому
@@omarsayed3874 x = x+1 + x-1 only for x=0 lol. and the only linear unction satisfying the relation is f(x) = 0
@omarsayed3874 4 місяці тому
@@eccotom1 ah yes i forgot we will get 2x
@alonelyphoenix8942 4 місяці тому
When in doubt, use f(x) = 0
@thethinker6258 4 місяці тому
Teacher, can you integrate or differentiate the Lambert W function?
@-minushyphen1two379 4 місяці тому
You can do it using the formula for the derivative of the inverse function, he made a video about this before
@CarlBach-ol9zb Місяць тому
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.
@lpschaf8943 4 місяці тому
Thank you so much. This was very satisfying.
@lazarusisaacng 4 місяці тому
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 👍.
@shoeman6966 4 місяці тому
This man’s algebraic manipulation ability is superb!
@kenroyadams2762 4 місяці тому
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...😅
@alonelyphoenix8942 4 місяці тому
He himself made the tree, apparently u can buy it lol
@gljdds4164 4 місяці тому
i love how you always use the fish when explaining the lambert w function
@gastonsolaril.237 4 місяці тому
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 4 місяці тому
What if e is raised to a quadratic polynomial. Can that be solved for x?
@gastonsolaril.237 4 місяці тому
@@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.
@mrexl9830 4 місяці тому
Freaking LOVE the lambert W functions
@Zach010ROBLOX 4 місяці тому
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?
@soupisfornoobs4081 4 місяці тому
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_MS 4 місяці тому
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.
@iwilldefeatraymak2536 4 місяці тому
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_MS 4 місяці тому
Since 'a' is the base for the logarithm, this formula would have some restrictions. Mainly 'a' must be greater than 1.
@spoopy1322 4 місяці тому
I love your videos! ❤
@robinsparrow1618 2 місяці тому
i had never heard of the lambert W function before watching your videos! i'm intrigued...
@wafflely9877 4 місяці тому
Make a video on the integral from -1 to 1 of (-e^x^2/3)+e dx!! 🙏
@General12th 4 місяці тому
Hi BPRP! So good!
@phat_khiep 4 місяці тому
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.
@whiteskeleton9453 4 місяці тому
Formula for series in n world for n^y/x^n please make a video for it😊
@user-gm8ir4sd6m 4 місяці тому
can you please make a video talking about the lebesgue integral and also iys connection with the laplas transfromation
@pahandulanga1039 4 місяці тому
Can you make a video of you solving an equation using this formula?
@sebmata135 2 місяці тому
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
@emmanuellaurens2132 2 місяці тому
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 🙂
@lpschaf8943 4 місяці тому
beautiful video
@tambuwalmathsclass 4 місяці тому
Amazing 😊
@mcgamescompany 4 місяці тому
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
@gamerpedia1535 4 місяці тому
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.
@zachansen8293 4 місяці тому
@@gamerpedia1535 en.wikipedia.org/wiki/Lambert_W_function#Numerical_evaluation
@soupisfornoobs4081 4 місяці тому
For small x, W(x) is just x-x² so yes I'd say there is an advantage
@TranquilSeaOfMath 4 місяці тому
Fairly straight forward presentation. Nice example of Lambert W Function with merchandise tie-in.
@elsicup 4 місяці тому
I was trying to solve this thing About 2 weeks ago, thank u😊
@scottleung9587 4 місяці тому
Nice job!
@ton146 4 місяці тому
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?
@trucid2 4 місяці тому
You can rewrite a degree two polynomial in different ways: ax^2+bx+c=(px+q)(rx+s) a(x−h)^2=k
@user-zp9cn2qq8g 4 місяці тому
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?
@MatthisDayer 4 місяці тому
you know what, i was just playing with these kinds of equations yesterday, ab^(cx) + dx = e
@Deejaynerate 4 місяці тому
If you change the equation slightly so that a^x is multiplied by -c, then the formula becomes xlna = 0
@ivantaradin49 4 місяці тому
what if the x, which is multiplied by b, is square rooted??? ( a^x + b*sqrtx +c =0 )
@MichaelRothwell1 4 місяці тому
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
@Halleluyah83 4 місяці тому
Hello) Thank You))
@Max-mx5yc 4 місяці тому
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)
@math_qz_2 4 місяці тому
Excellent 😮
@philip2205 4 місяці тому
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?
@vikrantharukiy7160 3 місяці тому
As for the first one, just divide all terms by a and solve
@shafikbarah9273 4 місяці тому
Is there a general way to get the general formula of any sequence just from the reccursive formula?
@Wouter10123 4 місяці тому
Generating functions
@dfjao97 4 місяці тому
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.
@GomissK 4 місяці тому
formula for a^x^3 + b^x^2 + c^x + d pls
@johnny_eth 4 місяці тому
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?
@Ninja20704 4 місяці тому
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
@guydell7850 4 місяці тому
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
@lawrencejelsma8118 4 місяці тому
@@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.
@table5584 4 місяці тому
Thanks, now I can solve 1^x + 2x - 5 = 0 😊
@deltalima6703 4 місяці тому
Nope, doesnt work if a=1, so you still cant figure out that x=2 is a solution. :-p
@minhdoantuan8807 4 місяці тому
@@deltalima6703in that case, 1^x = 1 for all x, so 2x - 4 = 0, or x = 2
@HimanshuRajOk 4 місяці тому
@@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?
@HimanshuRajOk 4 місяці тому
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 :(
@pierreabbat6157 4 місяці тому
What do you do if you have tuna times exponential of haddock?
@Bhuvan_MS 4 місяці тому
Is the eqn of the form: x^x+px+q=0 also solvable using Lambert-W function?
@vikrantharukiy7160 3 місяці тому
I tried and failed
@Bhuvan_MS 3 місяці тому
@@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...
@Nylspider 3 місяці тому
I always find the fact that you draw fish with eyebrows to be unreasonably funny
@sergeygaevoy6422 4 місяці тому
I think we assume a > 0, a 1 and b 0. Otherwise it is a much simplier (trivial) equation.
@remicou8420 4 місяці тому
he explains at the end why those parameters are disallowed. you can’t compute the result if any of the conditions are broken
@sergeygaevoy6422 4 місяці тому
@@remicou8420 Thank, there is a "post-credit" scene ...
@NullExceptionch 4 місяці тому
Can you please solve this? “Tan(x)=sqrt(x+1)
@nokta9819 4 місяці тому
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 4 місяці тому
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
@nokta9819 4 місяці тому
@@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
@shahar6840 4 місяці тому
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
@DEYGAMEDU 4 місяці тому
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
@klasta2167 4 місяці тому
(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.
@RishabMurthy 4 місяці тому
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
@RishabMurthy 4 місяці тому
Without iteration
@RishabMurthy 4 місяці тому
And is there a way to solve xe^e^x = (number)
@RishabMurthy 4 місяці тому
Or xsin (e^x)
@maxrs07 4 місяці тому
can u calculate W func by hand or its numerical only
@adarshk7484 4 місяці тому
do integral of 1/(1-x^20) dx
@user-cf8os4pd7i 4 місяці тому
What is the invers of f(x)=x4+x3+2 Please solve it
@isjosh8064 4 місяці тому
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
@cheliu9140 4 місяці тому
Can anyone answers these questions? I really need it to explain to my friends. The questions are below: Is this equation: i = √(-1) acceptable? and Is the n-th root of a (n < 0, a is a real number) definable? If it is, is it acceptable?
@soupisfornoobs4081 4 місяці тому
Pretty much yes, yes and yes. Do you want an explanation?
@cheliu9140 4 місяці тому
@@soupisfornoobs4081 yes
@killianobrien2007 4 місяці тому
1. When using the square root symbol it is implied that it is the principal root so yeah 2. So like the -2th root of 1/9 is 3 because 3^-2=1/9? If so that is acceptable.
@aymanadyel3515 4 місяці тому
no, only positive real numbers can be under a radical otherwise you’ll stumble into a lot of problems. i is defined as a “number” that when squared gives -1. here’s a problem with negative numbers under the square root supposing sqrt(-1) even exists, we get: sqrt(-1) = sqrt(-1), no tricks here => sqrt(-1)^2 = sqrt(-1)*sqrt(-1) and, applying the properties of squares and square root, we get: => -1 = sqrt((-1)*(-1)) => -1 = sqrt(1) => -1 = 1 which is absurd, so our hypothesis that sqrt(-1) exists was false. so i wouldn’t consider it acceptable, but we could use it as a notation for something else. however the properties wouldn’t be the same and you’d have to be careful. as such you could define the n-th root of a negative number but again you would have to be careful. for example, cube root of -8 is -2, you can verify this. but to be careful i would put it this way: what’s a number which when elevated to the third power yields -8. you can write it as this equation: x^3 = -8 like this, you’re using polynomials which we know how to deal with very well rather than making your life harder trying to deal with all the nonsense brought by negative numbers under radicals. Hope that helps !
@aymanadyel3515 4 місяці тому
oh my bad i didn’t read the second question correctly @killianobrien2007 pretty much answered it though but again i’d rather translate everything to equations rather than use notations i’m not confortable with: what number yields 1/9 when raised to the -2 power ? x^-2 = 1/9 again i feel like this would be easier to work with before creating the notation to solve weirder problems like x^-6 = pi/3 but looking at it like this i feel like logarithms will do a better job for these than n-th root notations with n
@bivekchaudhari4593 4 місяці тому
Please solve this question integral of 1/1+x⁵ dx
@129140163 4 місяці тому
5:15 ROFL that brief hyper speed-up tickled my funny bone! 😂
@dkdashutsa1575 3 місяці тому
Is there any formula for summation of i = 1 to n of W(i)
@Grassmpl 4 місяці тому
Use newtons method to approximate.
@redroach401 4 місяці тому
can you please solve: (x+1)^x=64.
@Cbgt 4 місяці тому
Please solve (lnx)•(x^x)=1 I just can't do it myself
@IRM321 4 місяці тому
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.
@dethmaiden1991 4 місяці тому
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) 🤷♂️
@vikrantharukiy7160 3 місяці тому
I don’t think it’s possible without numerical methods Could be wrong tho
@jacplanespotting314 4 місяці тому
So, what level of high school or college made is this geared to, in your opinion?
@sayedyousafhashimi6227 4 місяці тому
It would be pretty cool if solve me the following question which I found and I could not solve. limit x approaches 0 of (x^x^^^x -x!)/(x!^x! -1)
@mrpineapple7666 4 місяці тому
What happens if we want complex solutions?
@crowreligion 2 місяці тому
Use other branches of lambert W function There are branches after every integer, and everything except for branch 0 and -1 gives complex solutions
@user-bl3wx6mi4q 4 місяці тому
Sir can do arithmetics for me? Ratio and proportion and linear equations. If you like can please do congruence of triangle
@TranquilSeaOfMath 4 місяці тому
What specifically are you looking for?
@darcash1738 Місяць тому
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
@shyamaldevdarshan 4 місяці тому
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!
@noahblack914 4 місяці тому
6:57 My favorite definition of trancendental lol
@rorydaulton6858 4 місяці тому
You have a minor mistake in your video. Near the end you say that if "-1/e
@MichaelRothwell1 4 місяці тому
Totally agree. I spotted this glitch too.
@necrolord1920 4 місяці тому
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.
@avalon42nds 4 місяці тому
but what about a^x + x root b?
@francescoruffinengo3458 4 місяці тому
What pens are you using?
@NelDora-ih1bd 4 місяці тому
hello what white board is that?
@Catman_321 4 місяці тому
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
@michellauzon4640 3 місяці тому
Nice
@fsisrael9224 4 місяці тому
That moment when you get the fish back 😮 Truly a W moment
@harrymetu2746 4 місяці тому
Cool!
@zhabiboss 4 місяці тому
Fish function
@mcwulf25 4 місяці тому
Looked impossible but now I know 👍
@thatomofolo452 4 місяці тому
Straight line function
@pchevasath 2 місяці тому
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?
@jejnsndn 4 місяці тому
May you integrate sqr of x³+1 ( the square root is all over the expression)
@greatjafar 4 місяці тому
∫√(x³+1)dx
@seroujghazarian6343 4 місяці тому
I=int(sqrt(1+x²)dx) x=tan(θ) dx=sec²(θ)dθ I=int(sqrt(1+tan²(θ))sec²(θ)dθ)=int(sec³(θ)dθ)=sec(θ)tan(θ)-int(sec(θ)tan²(θ)dθ)=sec(θ)tan(θ)-int(sec(θ)(sec²(θ)-1)dθ)=sec(θ)tan(θ)-int(sec³(θ)dθ)+int(sec(θ)dθ)=sec(θ)tan(θ)-I+int(sec(θ)(sec(θ)+tan(θ))/(sec(θ)+tan(θ))dθ) 2I=sec(θ)tan(θ)+int((sec²(θ)+sec(θ)tan(θ))/(tan(θ)+sec(θ))dθ)=sec(θ)tan(θ)+ln|sec(θ)+tan(θ)|+k=xsqrt(x²+1)+ln(x+sqrt(x²+1))+k int(sqrt(x²+1)dx)=(xsqrt(x²+1)+ln(x+sqrt(x²+1)))/2+c
@integraliss 4 місяці тому
@@seroujghazarian6343 hmm I think so too bro
@orisphera 4 місяці тому
What about x**a+bx+c=0 (same with a and the first x swapped)?
@orisphera 4 місяці тому
Perhaps b=ka would be useful
@user-zz3sn8ky7z 4 місяці тому
Then it's just the a-th root of (-bx-c), isn't it?
@orisphera 4 місяці тому
@@user-zz3sn8ky7z But there's x in (-bx-c)
@padmasangale8194 4 місяці тому
Bro pls solve *x²[logx (base 10)]⁵=100* Can we also solve it with Lambert W func?
@gigamasterhd4239 3 місяці тому
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.
@padmasangale8194 3 місяці тому
@@gigamasterhd4239 thanks😊 👍
@gigamasterhd4239 3 місяці тому
@@padmasangale8194 No problem, very happy to help! Have a great rest of your day. 👍
@padmasangale8194 3 місяці тому
@@gigamasterhd4239 ⚡🔥
@RubyPiec 3 місяці тому
my calculator has no lambert w function button. how can i simulate one
@romanbykov5922 4 місяці тому
Hey, great video as usual. Btw, did you see the solution for your integral? Was sent to your email.
@xcoolchoixandanjgaming1076 4 місяці тому
The fact that the shirt youre wearing is also the fish function lol
@bud5 4 місяці тому
bro is obsessed with fish
@steamedeggeggegg 28 днів тому
MVP of this episode: multiply both sides
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