Derivation of Lagrange Family of Interpolation functions

[bugatti79]

Folks,

I am puzzled how the linear interpolation functions (see attached) were determined based on the following equation below

$\displaystyle \psi_i=\frac{(\xi-\xi_1)(\xi-\xi_2)...(\xi-\xi_{i-1})(\xi-\xi_{i+1})...(\xi-\xi_n)}{(\xi_i-\xi_1)(\xi_i-\xi_2)...(\xi_i-\xi_{i-1})(\xi_i-\xi_{i+1})...(\xi_i-\xi_n)}$

What do the dots represent above?

and

\psi_i(\xi_j)= 1 if $i=j$ and $0$ if $i\ne j$

For example how is $\psi_1(\xi)$ determined?

thanks

 

[mathman]

Folks,

I am puzzled how the linear interpolation functions (see attached) were determined based on the following equation below

$\displaystyle \psi_i=\frac{(\xi-\xi_1)(\xi-\xi_2)...(\xi-\xi_{i-1})(\xi-\xi_{i+1})...(\xi-\xi_n)}{(\xi_i-\xi_1)(\xi_i-\xi_2)...(\xi_i-\xi_{i-1})(\xi_i-\xi_{i+1})...(\xi_i-\xi_n)}$

What do the dots represent above?

and

\psi_i(\xi_j)= 1 if $i=j$ and $0$ if $i\ne j$

For example how is $\psi_1(\xi)$ determined?


thanks

The dots mean fill in all the like terms for indices between the given. In this case, 3 to i-2 and i+2 to n-1.

For the second question, when j = i, the numerator and denominator are the same. When j ≠ i, one term of the numerator product = 0.

 

[bugatti79]

The dots mean fill in all the like terms for indices between the given. In this case, 3 to i-2 and i+2 to n-1.

For the second question, when j = i, the numerator and denominator are the same. When j ≠ i, one term of the numerator product = 0.

Sorry, I still don't follow...

I don't see how $\psi_1(\xi)=\frac{1}{2}(1-\xi)$ is obtained...?

 

[Mark44]

Folks,

I am puzzled how the linear interpolation functions (see attached) were determined based on the following equation below

$\displaystyle \psi_i=\frac{(\xi-\xi_1)(\xi-\xi_2)...(\xi-\xi_{i-1})(\xi-\xi_{i+1})...(\xi-\xi_n)}{(\xi_i-\xi_1)(\xi_i-\xi_2)...(\xi_i-\xi_{i-1})(\xi_i-\xi_{i+1})...(\xi_i-\xi_n)}$

What do the dots represent above?

The dots (...) are called an ellipsis, and appear as three periods. The ellipsis means "continuing in the same fashion." The first missing factor in both the numerator and denominator would be (x - x₃) and the next would be (x - x₄), and so on. (I don't see any purpose in writing $\xi$ when plain old x will do just fine.)

 

[rbj]

I am puzzled how the linear interpolation functions (see attached) were determined based on the following equation below

$\displaystyle \psi_i=\frac{(\xi-\xi_1)(\xi-\xi_2)...(\xi-\xi_{i-1})(\xi-\xi_{i+1})...(\xi-\xi_n)}{(\xi_i-\xi_1)(\xi_i-\xi_2)...(\xi_i-\xi_{i-1})(\xi_i-\xi_{i+1})...(\xi_i-\xi_n)}$

What do the dots represent above?

like any other dots like that which you see in a mathematical expression. it means "fill in the blank" with the continuing pattern.

it's not as good as my Applied Engineering Mathematics book from Kreyszig, but the wikipedia article on Lagrange polynomials should have the answers to your question.

 

[bugatti79]

Folks,



$\displaystyle \psi_i=\frac{(\xi-\xi_1)(\xi-\xi_2)...(\xi-\xi_{i-1})(\xi-\xi_{i+1})...(\xi-\xi_n)}{(\xi_i-\xi_1)(\xi_i-\xi_2)...(\xi_i-\xi_{i-1})(\xi_i-\xi_{i+1})...(\xi_i-\xi_n)}$



$\psi_1(\xi)=\frac{1}{2}(1-\xi)$ is obtained...?

$\displaystyle \psi_1=\frac{(\xi-\xi_1)(\xi-\xi_2)(\xi-\xi_{0})(\xi-\xi_{2})(\xi-\xi_1)}{(\xi_1-\xi_1)(\xi_1-\xi_2)(\xi_1-\xi_{0})(\xi_1-\xi_{2})(\xi_1-\xi_1)}$...?

 

[mathman]

$\displaystyle \psi_1=\frac{(\xi-\xi_1)(\xi-\xi_2)(\xi-\xi_{0})(\xi-\xi_{2})(\xi-\xi_1)}{(\xi_1-\xi_1)(\xi_1-\xi_2)(\xi_1-\xi_{0})(\xi_1-\xi_{2})(\xi_1-\xi_1)}$...?

ξ₁ - ξ₁ is NOT a term in the denominator. Look at the definition of ψ_i, the ith term is left out of the numerator and denominator.

Where did you get the expression for ψ₁(ξ)?

 

[bugatti79]

ξ₁ - ξ₁ is NOT a term in the denominator. Look at the definition of ψ_i, the ith term is left out of the numerator and denominator.

Where did you get the expression for ψ₁(ξ)?

Its the first equation shown in the picture in first post...

 

[mathman]

Its the first equation shown in the picture in first post...

The pictures are slightly criptic. The important thing is that the definition of ψ_i depends on n. The particular item you ask about is for n = 2.
In that case:
ψ₁(ξ) = (ξ-ξ₂)/(ξ₁-ξ₂), where ξ₁= -1 and ξ₂=+1.

 

[bugatti79]

The pictures are slightly criptic. The important thing is that the definition of ψ_i depends on n. The particular item you ask about is for n = 2.
In that case:
ψ₁(ξ) = (ξ-ξ₂)/(ξ₁-ξ₂), where ξ₁= -1 and ξ₂=+1.

OK, thanks