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)
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.
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.
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.
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.
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.
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 👍.
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)]
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.
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
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...😅
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.
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
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 🙂
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
@@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 :(
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)
@@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...
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?
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
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.
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.
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.
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
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!
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?
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?
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
Learn contest math on Brilliant: 👉brilliant.org/blackpenredpen/ (now with a 30-day free trial plus 20% off with this link!)
Question: Why does W(-(e^-1)) give us two real solutions, shouldnt it be just -1?
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)
Can you explain about x^4 + ax^2 + bx + c
@@matthiaspihuschiiiiilllllllllp
signed up!
Lambert W function ❌ bprp fish function ✅
W(fishe^fish) = fish for a recap
Underrated comment
"We shall 'save the fish' on both sides"
W(🐟e^🐟) = 🐟
Where does the letter 「W」 come from ?
I'm not sure why, but my favorite videos of yours are always the ones with the Lambert W function
Did you mean the fish function?
I mean, you're not alone, but I don't know why I like the Fish function so much either... :D
@@A_literal_cube my bad
This is the balance of the universe at work, because they're my least favourite ones!
@@tanelkagan That's kinda funny haha
I've just wrapped up a math study session; it's now time to relax by watching some more math.
Sums up my whole life lmao
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.
Nice video! You explained it in a way that a lot of people can understand. I appreciate that a lot.
Thank you!
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.
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.
f(x) = x, hope that helps
@@omarsayed3874 x = x+1 + x-1 only for x=0 lol. and the only linear unction satisfying the relation is f(x) = 0
@@eccotom1 ah yes i forgot we will get 2x
When in doubt, use f(x) = 0
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.
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!!!
What if e is raised to a quadratic polynomial. Can that be solved for x?
@@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.
Teacher, can you integrate or differentiate the Lambert W function?
You can do it using the formula for the derivative of the inverse function, he made a video about this before
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.
This man’s algebraic manipulation ability is superb!
6:02 i love how you almost wrote down the plus
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 👍.
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)]
Since 'a' is the base for the logarithm, this formula would have some restrictions. Mainly 'a' must be greater than 1.
I always find the fact that you draw fish with eyebrows to be unreasonably funny
Now try a tetrated to x + b^x + cx + d could be a fun video ❤
i love how you always use the fish when explaining the lambert w function
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.
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
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...😅
He himself made the tree, apparently u can buy it lol
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.
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
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 🙂
Can you make a video of you solving an equation using this formula?
i had never heard of the lambert W function before watching your videos! i'm intrigued...
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
Thanks, now I can solve 1^x + 2x - 5 = 0 😊
Nope, doesnt work if a=1, so you still cant figure out that x=2 is a solution. :-p
@@deltalima6703in that case, 1^x = 1 for all x, so 2x - 4 = 0, or x = 2
@@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 :(
Am i jealous of his t-shirt
Of course i am i need it😮
Thank you so much. This was very satisfying.
e^x>0 --> x+1 -x=1+(1/e^k)
k=1+(1/e^k)
k-(1/e^k)=1 --> ke^k-1=e^k
(k-1)e^k=1
(k-1)e^(k-1)=1/e
(k-1)=W(1/e)
k=1+W(1/e)
As k is positive value of x (x
Freaking LOVE the lambert W functions
I love your videos! ❤
Amazing 😊
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?
As for the first one, just divide all terms by a and solve
Is there any formula for summation of i = 1 to n of W(i)
Please please make this question a isoceles Triangle having equal sides 12cm height is 7.5cm find the area of Triangle
Make a video on the integral from -1 to 1 of (-e^x^2/3)+e dx!! 🙏
Can we call this completing fishes?
You have a minor mistake in your video. Near the end you say that if "-1/e
Totally agree. I spotted this glitch too.
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)
What is the invers of f(x)=x4+x3+2
Please solve it
Formula for series in n world for n^y/x^n please make a video for it😊
you know what, i was just playing with these kinds of equations yesterday, ab^(cx) + dx = e
Is the eqn of the form: x^x+px+q=0 also solvable using Lambert-W function?
I tried and failed
@@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...
can you please make a video talking about the lebesgue integral and also iys connection with the laplas transfromation
0:28 WHY DOES THE FISH HAVE HORNSSSS?
Edit : Edited timestamp
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?
You can rewrite a degree two polynomial in different ways:
ax^2+bx+c=(px+q)(rx+s)
a(x−h)^2=k
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
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
Use newtons method to approximate.
So, what level of high school or college made is this geared to, in your opinion?
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.
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.
(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.
Can you please solve this?
“Tan(x)=sqrt(x+1)
beautiful video
Is there a general way to get the general formula of any sequence just from the reccursive formula?
Generating functions
How about this function :
X^a + bX + C = 0, instead of a^X, how about this X^a?
I don't think there is a general solution for that.
I was trying to solve this thing About 2 weeks ago, thank u😊
can you please solve: (x+1)^x=64.
do integral of 1/(1-x^20) dx
Excellent 😮
What do you do if you have tuna times exponential of haddock?
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.
Hi BPRP!
So good!
Please solve this question integral of 1/1+x⁵ dx
what if the x, which is multiplied by b, is square rooted??? ( a^x + b*sqrtx +c =0 )
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
Without iteration
And is there a way to solve xe^e^x = (number)
Or xsin (e^x)
Nice job!
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.
@@gamerpedia1535 en.wikipedia.org/wiki/Lambert_W_function#Numerical_evaluation
For small x, W(x) is just x-x² so yes I'd say there is an advantage
Bro pls solve
*x²[logx (base 10)]⁵=100*
Can we also solve it with Lambert W func?
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.
@@gigamasterhd4239 thanks😊 👍
@@padmasangale8194 No problem, very happy to help! Have a great rest of your day. 👍
@@gigamasterhd4239 ⚡🔥
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?
Excelent!
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 ✓
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
@@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
formula for a^x^3 + b^x^2 + c^x + d pls
Hello) Thank You))
Do you think you could you solve a^(x^2)+b^x+c=0 for x?
What a monster
@@kaviramyead7987 Heh heh. thanks! >:D
What happens if we want complex solutions?
Use other branches of lambert W function
There are branches after every integer, and everything except for branch 0 and -1 gives complex solutions
@@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(?)
I think we assume a > 0, a 1 and b 0.
Otherwise it is a much simplier (trivial) equation.
he explains at the end why those parameters are disallowed. you can’t compute the result if any of the conditions are broken
@@remicou8420 Thank, there is a "post-credit" scene ...
6:57 My favorite definition of trancendental lol
but what about a^x + x root b?
hello what white board is that?
If you change the equation slightly so that a^x is multiplied by -c, then the formula becomes xlna = 0
5:15 ROFL that brief hyper speed-up tickled my funny bone! 😂
The fact that the shirt youre wearing is also the fish function lol
Fairly straight forward presentation. Nice example of Lambert W Function with merchandise tie-in.
How to solve: (e^x)-3=ln(x)
Does x= ln(-b-c-a) not work?
can u calculate W func by hand or its numerical only
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!
Please solve (lnx)•(x^x)=1
I just can't do it myself
What about x**a+bx+c=0 (same with a and the first x swapped)?
Perhaps b=ka would be useful
Then it's just the a-th root of (-bx-c), isn't it?
@@user-zz3sn8ky7z But there's x in (-bx-c)
Sir can do arithmetics for me? Ratio and proportion and linear equations. If you like can please do congruence of triangle
What specifically are you looking for?
May you integrate sqr of x³+1 ( the square root is all over the expression)
∫√(x³+1)dx
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
@@seroujghazarian6343 hmm I think so too bro
my calculator has no lambert w function button. how can i simulate one
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?
Sir I would like you to check this and give ur thoughts please 🙏🏼
Wow bro ur right , it can be your own constant 👍
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)
Fish function
Sir Why Pythagoras theorem is not a law?
Laws describe physical observations, theorems are mathematical statements
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?
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