Exercise 2 (Gauss-Legendre schemes; 4 points) Consider the...

Exercise 2 (Gauss-Legendre schemes; 4 points) Consider the interval [0, 1]. a) Find the roots $a_1$ and $a_2$ of the Legendre polynomial $P_2(2x - 1)$. b) Compute $c_j = \int_0^1 L_j(x) dx$, $j = 1, 2$, where the $L_j$ are Lagrange polynomials. c) Finally, compute $b_{ij} = \int_0^{a_i} L_j(x) dx$, $i = 1, 2$. d) Write down the corresponding Butcher tableau. What method does this Butcher tableau belong to?

Transcript

00:01 This is the column number 25.

00:03 This column said that use the grams emit procedure and difference is given that is problem number a .4 in the appendix to ortho -normalize the function 1 x, x, square and x -q on the interval minus 1 to plus 1.

00:22 You may recognize the results.

00:24 They are apart from the normalization that is legend the polymer 4 .1 so let's do the solution of this this is the base on the gram cement procedure this is problem number 25 and the problem based on the normalize the this one x x square and x q using the gram cement method or procedure and he here the basis that are not normalized are given are that is e1 is equal to 1, e2 is equal to x, e3 is equal to x square and e4 is equal to x.

02:13 And this is a procedure explained by the gramssemit.

02:17 In this procedure we have to follow some steps that the step number one that is e1 is equal to 1 now we have to normalize this one as the normalization state can be presented by this prime is equal to this e1 and the magnitude of this even and where e1 this e1 magnitude square is equal to e1 and e1 and we can write this that is minus in 1 2 plus 1 this is the interval and e1 is equal to 1 here 1 d x so the value of this will be 2 hence the value of this is that is that is mod of e1 is equal to root 2 hence the normalization state basis vector that is prime is equal to 1 upon now we have to find out the e2 normalization that is normalized 2 second is equal to x now there's some procedure in this procedure we have to find out first em e1 normalization e1 and e2 the value of this is 1 by root 2 minus 1 2 plus 1 and e 2 is equal to x and d x and this value will be 0 now we have find out e 2 e2 that is magnitude of this e 2 e 2 is equal to 2 is equal to minus 1 to plus 1 e 2 is equal to x that is x square b x and this value will be x2 by 3 and this is minus 1 this plus 1 then it will be 2 by 3 now normalize e 2 a 2 prime is equal to e 2 and divide by square root of this that is we can like this like this e2 that means it is a 2 is x and this is square root of 2 by 3 then it will be e2 normalize gase square root of 3 by 2 and x we have normalized 2 eigenve vector that is a 2 now we have to normalize the e3 3 3 is e3 is equal to x square now we have to find normalize this with e1 the first e1 normalize and e3 is equal to e1 1 1 is 1 by root 2 1 by 2 minus 1 plus 1 and e 3 is x square and d x this will be 1 upon root 2 and this will be x2 by 3 and minus 1 to plus 1 it will be 1 by root 2 into 2 by 3 now point a2 and e3 e 2 is root 3 by 2 e 2 is root 3 by 2 minus infinity 2 plus infinitive and this is x and e3 is x square d x it is x q and the value of this under root 3 by 2 and this will be x 4 minus 1 plus 1 and it is 0 now e 3 double prime is equal to e3 minus e1 normalized with e3 and e1 so here e3 is equal to x squared minus this value is e1 this value is 1 by root 2 and 2 by 3 1 by root 2 and 2 by 3 and e 1 and this value reason by root 2...

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