We assume a function f is of the form f(t)=k=1nakk(t) where 1,2,,n are given functions on the
interval I=[a,b]. If points {t}=1mI are chosen so that at1<t2<<tmb we can sample f at the points t to
obtain f(t)=k=1nakk(t) Here we assume that m>n. When this is done in a laboratory situation,
errors occur and instead of the exact values of f(t), the sampling procedure produces data y(1m)
approximating f(t). The aim is to choose the constants ak(1kn) so that the function f defined by (1)
best fits the data in the sense that i=1m(yik=1nakk(tc))2 is minimised. Written in terms of matrix
multiplication and Euclidean norms, we see that this is the unconstrained optimisation problem
minimisey2 where y=(y1,y2,,ym)T is the given data, a=(a1,a2,,an)T is the unknown vector of
coefficients, and Mmn(R) has (,k)-th entry ak=k(t). (a) Show that if a is a solution of (2) then Ty=Ta
(b) Suppose I=[1,1],n=3,m=5 and 1(t)=1,2(t)=t,3(t)=t2,t1=1, t2=1/2,t3=0,t4=1/2,t5=1 and y=(2,1,0,1
,1)T. Compute the minimising vector a. (c) Given the solution of part b, plot on the same graph the
data points {(t,y)}=15 and the quadratic function f(t)=a1+a2t+a3t2..
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We assume a function f is of the form ftk1nakkt where .pdf
1. We assume a function f is of the form f(t)=k=1nakk(t) where 1,2,,n are given functions on the
interval I=[a,b]. If points {t}=1mI are chosen so that at1<t2<<tmb we can sample f at the points t to
obtain f(t)=k=1nakk(t) Here we assume that m>n. When this is done in a laboratory situation,
errors occur and instead of the exact values of f(t), the sampling procedure produces data y(1m)
approximating f(t). The aim is to choose the constants ak(1kn) so that the function f defined by (1)
best fits the data in the sense that i=1m(yik=1nakk(tc))2 is minimised. Written in terms of matrix
multiplication and Euclidean norms, we see that this is the unconstrained optimisation problem
minimisey2 where y=(y1,y2,,ym)T is the given data, a=(a1,a2,,an)T is the unknown vector of
coefficients, and Mmn(R) has (,k)-th entry ak=k(t). (a) Show that if a is a solution of (2) then Ty=Ta
(b) Suppose I=[1,1],n=3,m=5 and 1(t)=1,2(t)=t,3(t)=t2,t1=1, t2=1/2,t3=0,t4=1/2,t5=1 and y=(2,1,0,1
,1)T. Compute the minimising vector a. (c) Given the solution of part b, plot on the same graph the
data points {(t,y)}=15 and the quadratic function f(t)=a1+a2t+a3t2.
![We assume a function f is of the form f(t)=k=1nakk(t) where 1,2,,n are given functions on the
interval I=[a,b]. If points {t}=1mI are chosen so that at1<t2<<tmb we can sample f at the points t to
obtain f(t)=k=1nakk(t) Here we assume that m>n. When this is done in a laboratory situation,
errors occur and instead of the exact values of f(t), the sampling procedure produces data y(1m)
approximating f(t). The aim is to choose the constants ak(1kn) so that the function f defined by (1)
best fits the data in the sense that i=1m(yik=1nakk(tc))2 is minimised. Written in terms of matrix
multiplication and Euclidean norms, we see that this is the unconstrained optimisation problem
minimisey2 where y=(y1,y2,,ym)T is the given data, a=(a1,a2,,an)T is the unknown vector of
coefficients, and Mmn(R) has (,k)-th entry ak=k(t). (a) Show that if a is a solution of (2) then Ty=Ta
(b) Suppose I=[1,1],n=3,m=5 and 1(t)=1,2(t)=t,3(t)=t2,t1=1, t2=1/2,t3=0,t4=1/2,t5=1 and y=(2,1,0,1
,1)T. Compute the minimising vector a. (c) Given the solution of part b, plot on the same graph the
data points {(t,y)}=15 and the quadratic function f(t)=a1+a2t+a3t2.](https://image.slidesharecdn.com/weassumeafunctionfisoftheformftk1nakktwhere-230322022513-e9267a93/85/We-assume-a-function-f-is-of-the-form-ftk1nakkt-where-pdf-1-320.jpg)
