[Math] Find the 2nd-degree polynomial that approximates with the method of the least squares the:$f(x)=\frac{1}{10}x^2-2x+10$

approximationorthogonal-polynomialspolynomials

It is known that a rectangular set of polynomials $\phi_k(x), k=0,1,\cdots,n$ for each $x\in[a,b]$ as to a weight function $w(x)$ can be constructed with the use of the following recursive type (Gram-Schmidt process):

$\phi_0(x)=1$

$\phi_1(x)=x- a_0$, where $a_0=\frac{(x\phi_0, \phi_0)}{(\phi_0,\phi_0)}$

and

$\phi_{k+1}(x)=(x-a_k)\phi_k(x)-\beta_k\phi_{k-1}(x),
k=1,2,\cdots,n-1$,

where $a_k=\frac{(x\phi_k, \phi_k)}{(\phi_k,\phi_k)}$, and $b_k=\frac{(x\phi_k, \phi_{k-1})}{(\phi_{k-1},\phi_{k-1})}$.

Using the Gram-Schmidt process above to find: the second-degree polynomial $p_2(x)$ that approximates with the method of the least squares the
$f(x)=\frac{1}{10}x^2-2x+10$

in points $(x_i,f_i), i=1,2,\cdots,6$

with $x_i=10+\frac{i-1}{5}$, $f_i=f(x_i)$ and $w(x)=1.$

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First make a table of your data. $$\begin {array}{r|r|r} i&x_i&f(x_i)\\ \hline 1&10&0\\2&10.2&0.004\\3&10.4&0.016\\4&10.6&0.036\\5&10.8&0.064\\6&11&0.1 \end {array}$$ Then, as the instructions say, $\phi_0=1$ and $\phi_1=x-\frac {(x\phi_0,\phi_0)}{\phi_0,\phi_0}=x-\frac {(x,1)}{(1,1)}$ The terms in parentheses are sums over the products at the points. So $(x,1)=10*1+10.2*1+\ldots 11*1=63, (1,1)=6$, which gives $\phi_1(x)=x-10.5$. Now you need to find $\phi_2$, which will be a quadratic.
Then you can express $f(x)=a_0\phi_0+a_1\phi_1+a_2\phi_2$ You can either find the $a$'s by expanding out the right side and equating like coefficients of $x$, or use the orthogonality to write $a_i=\frac {(f(x_i),\phi_i)}{(\phi_i,\phi_i)}$

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[Math] Find taylor polynomial that approximates e^x with accuracy at least 1.

If this is a fairly standard calculus course, you are working with the Lagrange form of the remainder. Notations tend to differ, but if $R_m(x)$ is the remainder when we truncate the Maclaurin series just after the $x^m$ term, then $$|R_m(x)|=\frac{|f^{(m+1)}(\xi)|}{(m+1)!} |x|^{m+1},$$ where $\xi$ is a number between $0$ and $x$.

In our case, all derivatives are $e^x$. If our $x$ is in the interval $[-2,2]$, then $|f^{(m+1)}(\xi)|\lt e^2$. And since $|x|\le 2$, we have $|x|^{m+1}\le 2^{m+1}$.

So in our situation, we have $$|R_m(x)|\lt \frac{e^2 2^{m+1}}{(m+1)!}.$$ Now you need to hunt around for an $m$ such that the above quantity is less than $1$. The expression is not a very nice one, so calculator experimentation is probably the best approach.

Remark: The problem is rather artificial, since in real work one would use the Maclaurin series for $x$ reasonably close to $0$, certainly not as far out as $x=2$.

Algorithm for orthogonalizing polynomials with specific inner product

I can't tell from what you said what you implemented, this is an example called [Chebfun]. 1. It'd be better if you posted your code. Note this is some code for the Legendgre polynomials with pseudo-code.

x = chebfun('x',[-1 1]);
w = exp(pi*x);
N = 5;
P = OrthPoly(w,N);

    function P = OrthPoly(w,N)
        if isnumeric(w), w = chebfun(w,[-1 1]); end
        d = w.ends;                     % the domain
        x = chebfun('x',d);             % linear chebfun
        P = chebfun(1./sqrt(sum(w)),d); % the constant (normalised)
        for k = 1:N;
            xk = x.*P(:,k);
            P(:,k+1) = xk;
            for j = 1:k       % Subtract out the components
                C = sum(w.*xk.*P(:,j));
                P(:,k+1) = P(:,k+1) - C*P(:,j);
            end
            P(:,k+1) = P(:,k+1)./sqrt(sum(w.*P(:,k+1).^2)); % normalise
        end
    end

It's actually the lanczos iteration..

Note this is from Trefethen..

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[Math] Find taylor polynomial that approximates e^x with accuracy at least 1.

Algorithm for orthogonalizing polynomials with specific inner product

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