MHB Remainder converges uniformly to 0

[mathmari]

Hey!

We have the function $f (x) = e^{\lambda x}$ on an interval $[a, b] , \ \lambda \in \mathbb{R}$.

I want to show that the remainder $R_n (x) = f (x)- p_n (x)$ at the lagrange interpolation of $f (x)$ with $n+1$ points from $[a, b]$ for $n \rightarrow \infty$ converges uniformly to $0$.

For each $n$ the points can be chosen arbitrarily in $[a, b]$.
The remainder is defined as $\displaystyle{R_n(x)=f(x)-p_n(x)=\frac{f^{(n+1)}(\xi_x)}{(n+1)!}\prod_{i=0}^n(x-x_i)=\frac{\lambda^{n+1}e^{\lambda \xi_x}}{(n+1)!}\prod_{i=0}^n(x-x_i)}$, right?

Then we have the following:

$\displaystyle{|R_n(x)|=|f(x)-p_n(x)|=\left |\frac{\lambda^{n+1}e^{\lambda \xi_x}}{(n+1)!}\prod_{i=0}^n(x-x_i)\right |=\left |\frac{\lambda^{n+1}e^{\lambda \xi_x}}{(n+1)!}\right |\prod_{i=0}^n|x-x_i|}\\ \displaystyle{\leq \frac{|\lambda|^{n+1}\sup_{\xi_x\in [a,b]}e^{\lambda \xi_x}}{(n+1)!} \cdot \sup_{x\in [a,b]}\prod_{i=0}^n|x-x_i|}$

right?

How could we continue? (Wondering)

 

[I like Serena]

Hey mathmari!

Isn't $|x-x_i| \le b-a$? (Wondering)

 

[mathmari]

Isn't $|x-x_i| \le b-a$? (Wondering)

Ah yes! So, we have the following:

\begin{align*}|R_n(x)|&=|f(x)-p_n(x)|=\left |\frac{\lambda^{n+1}e^{\lambda \xi_x}}{(n+1)!}\prod_{i=0}^n(x-x_i)\right |=\left |\frac{\lambda^{n+1}e^{\lambda \xi_x}}{(n+1)!}\right |\prod_{i=0}^n|x-x_i| \\ & \leq \frac{|\lambda|^{n+1}\sup_{\xi_x\in [a,b]}e^{\lambda \xi_x}}{(n+1)!} \cdot \sup_{x\in [a,b]}\prod_{i=0}^n|x-x_i|\leq \frac{|\lambda|^{n+1}\sup_{\xi_x\in [a,b]}e^{\lambda \xi_x}}{(n+1)!} \cdot \prod_{i=0}^n|b-a|=\frac{|\lambda|^{n+1}\sup_{\xi_x\in [a,b]}e^{\lambda \xi_x}}{(n+1)!} \cdot |b-a|^n\end{align*}

Can we say also something for $\sup_{\xi_x\in [a,b]}e^{\lambda \xi_x}$ ? (Wondering)

 

[I like Serena]

Can we say also something for $\sup_{\xi_x\in [a,b]}e^{\lambda \xi_x}$ ?

Isn't it a constant that is independent from $x$? (Wondering)

 

[mathmari]

Isn't it a constant that is independent from $x$? (Wondering)

Ah, so is $\xi_x$ independent from $x$ ?

 

[I like Serena]

Ah, so is $\xi_x$ independent from $x$ ?

Nope. But that supremum is either in $a$ or in $b$, isn't it? (Wondering)

 

[mathmari]

Nope. But that supremum is either in $a$ or in $b$, isn't it?

Ah ok! So, say $c=\sup_{\xi_x\in [a,b]}e^{\lambda \xi_x}$ we get $$\frac{|\lambda|^{n+1}\cdot c}{(n+1)!} \cdot |b-a|^n$$ But how do we know that this converges to $0$ ? Doesn;t this hold only when $\lambda$ and $|b-a|$ are less than $1$ ? (Wondering)

 

[I like Serena]

Ah ok! So, say $c=\sup_{\xi_x\in [a,b]}e^{\lambda \xi_x}$ we get $$\frac{|\lambda|^{n+1}\cdot c}{(n+1)!} \cdot |b-a|^n$$

Shouldn't that be $\frac{|\lambda|^{n+1}\cdot c}{(n+1)!} \cdot |b-a|^{n+1}$? (Wondering)

But how do we know that this converges to $0$ ? Doesn;t this hold only when $\lambda$ and $|b-a|$ are less than $1$ ? (Wondering)

Which increases asymptotically faster, $c^n$ or $n!$? (Wondering)

 

[mathmari]

Shouldn't that be $\frac{|\lambda|^{n+1}\cdot c}{(n+1)!} \cdot |b-a|^{n+1}$? (Wondering)

Oh yes, I forgot to counter $i=0$. (Blush)

Which increases asymptotically faster, $c^n$ or $n!$? (Wondering)

The factorial? How can we see that? (Wondering)

 

[I like Serena]

The factorial? How can we see that?

A couple of years ago someone explained it here. (Clapping)

 

[mathmari]

A couple of years ago someone explained it here. (Clapping)

Ooh yes! (Tmi)(Blush) Thank you very much! (Yes)