 Solution

ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2009/a/2/b/solution

ϕ_{0}, \dots, ϕ_{r + 1}

form a basis of

P_{r + 1}

, ordered by degree. Since

x ϕ_{r}

is monic of degree

r + 1

x ϕ_{r} = ϕ_{r + 1} + \sum_{j = 0}^{r} a_{j} ϕ_{j} .

For

j \leq r - 2

x

gives

(x ϕ_{r}, ϕ_{j}) = (ϕ_{r}, x ϕ_{j}) = 0

, because

de g (x ϕ_{j}) < r

. Only

a_{r}

and

a_{r - 1}

remain. Taking inner products with

ϕ_{r}

yields

a_{r} = (ϕ_{r}, x ϕ_{r}) / (ϕ_{r}, ϕ_{r})

Similarly,

a_{r - 1} = \frac{( x ϕ _{r} , ϕ _{r - 1} )}{( ϕ _{r - 1} , ϕ _{r - 1} )} = \frac{( ϕ _{r} , x ϕ _{r - 1} )}{( ϕ _{r - 1} , ϕ _{r - 1} )} = \frac{( ϕ _{r} , ϕ _{r} )}{( ϕ _{r - 1} , ϕ _{r - 1} )},

since

x ϕ_{r - 1} - ϕ_{r}

has degree at most

r - 1

ϕ_{r + 1} = (x - a_{r}) ϕ_{r} - \frac{∥ ϕ _{r} ∥ _{2}^{2}}{∥ ϕ _{r - 1} ∥ _{2}^{2}} ϕ_{r - 1} .

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