In this video I present the Lagrange Inversion Theorem. It's an interesting new take on Taylor series. For more videos including an example of this theorem, visit: • Lambert W Function
I figured out how to do this with linear algebra. Starting with series you want to invert, make a lower triangular Toeplitz matrix T from the series, constant term first. So the matrix for the series of the identity function f(x) = x would be an identify matrix. Then find the Krylov subspace initializing with a vector like the column of an identity matrix, and the column the series. Each subsequent column of the Krylov matrix will be T^n v, which will be the series raised to the nth power. Now invert the matrix. Your second column is your Lagrange inverted series. There's also other interesting stuff, like if find the matrix square root, the resulting series is the functional square root such that g(g(x)) = f(x) It's also good for finding the series resulting from composing multiple functions like h(g(f(x)) without ever calculating a single derivative
Sadly, I haven't been able to find a clear demonstration of this. I know that the argument roughly goes that you can write the funny coefficient as the residue of an nth order pole and then show that the residue is equal to to normal Taylor series coefficient.
thank you for your explanation but i don't know how to solve this equation 2*cos(x) - beta*sin(x) + beta*sin(alpha*x) = 0 can you help me please there is another alternative to solve it with another method .... alpha and beta are two parameters a need to solve x =f(alpha , beta)
I know this came out 2 years ago, but might i ask where the (X-X0)^n in the original Taylor series comes from, not the Lagrange inversion one, but just the original equation you wrote down at the beginning.
A Maclaurin series is just a Taylor Series centered at x=0 so x^n suffices; whereas a general Taylor Series is just for (x - x_0)^n. For example the Taylor Series expansion for ln(x) centered at x = 1 (with a radius of convergence 0
@@physicsandmathlectures3289 i got into this so much that i started imagining how the math police will look like....lol.......anyway, Thanks for the lecture sir...
Well yeah, he said that explicitly multiple times. This is supposed to build intuition for people that are unfamiliar with the Lagrange Inversion Theorem as he also stated.
I figured out how to do this with linear algebra. Starting with series you want to invert, make a lower triangular Toeplitz matrix T from the series, constant term first. So the matrix for the series of the identity function f(x) = x would be an identify matrix.
Then find the Krylov subspace initializing with a vector like the column of an identity matrix, and the column the series. Each subsequent column of the Krylov matrix will be T^n v, which will be the series raised to the nth power. Now invert the matrix. Your second column is your Lagrange inverted series.
There's also other interesting stuff, like if find the matrix square root, the resulting series is the functional square root such that g(g(x)) = f(x)
It's also good for finding the series resulting from composing multiple functions like h(g(f(x)) without ever calculating a single derivative
Very nice, thank you!
Glad you enjoyed it!
Pretty well explained for a difficult theorem ;), do you know where i can find a clear demonstration ?
Sadly, I haven't been able to find a clear demonstration of this. I know that the argument roughly goes that you can write the funny coefficient as the residue of an nth order pole and then show that the residue is equal to to normal Taylor series coefficient.
Hy thanks for this opportunity
Can you check these équation end how We can resolve
X^5 +ax^3-2ba=0
Where did you get the equation from?
why would you need a series representation of a polynomial?
@Asyam Abyan is that a sugondese sawcon or a subondese sawcon?
This is a classic case of "In the limit, it all works out to be the same!"
Does anyone know where I could find a proof of this?
hi! i have this function 2cos(x)-B*(sin(x)-sin(A*x))=0 and i want to use this theorem can you help me
thank you for your explanation but i don't know how to solve this equation 2*cos(x) - beta*sin(x) + beta*sin(alpha*x) = 0 can you help me please there is another alternative to solve it with another method .... alpha and beta are two parameters a need to solve x =f(alpha , beta)
I know this came out 2 years ago, but might i ask where the (X-X0)^n in the original Taylor series comes from, not the Lagrange inversion one, but just the original equation you wrote down at the beginning.
A Maclaurin series is just a Taylor Series centered at x=0 so x^n suffices; whereas a general Taylor Series is just for (x - x_0)^n. For example the Taylor Series expansion for ln(x) centered at x = 1 (with a radius of convergence 0
0:38 I thought it was y(x)=sum_n=0^Infinity y^(n)(x_0)(x-x_0)^n/n!
Like... Not indexed at n=1
But that 0th term just collapses to some y_0 = y(x_0), so it's all good.
Math police hahaha!!!!
I'm glad someone found it funny, haha!
@@physicsandmathlectures3289 i got into this so much that i started imagining how the math police will look like....lol.......anyway,
Thanks for the lecture sir...
this was my first video watching of yours and will be my last.
everything you did is illegal. I will not be watching anymore of your videos.
Please don't call the math police 😰
Well yeah, he said that explicitly multiple times. This is supposed to build intuition for people that are unfamiliar with the Lagrange Inversion Theorem as he also stated.