y0b1 xNote y1 nPni1yiand x1 nPni1xiProofIgnoringthesecondnormalequationstartbydividingthe12rstnormalequationbyn1 nnXi1yib0b1 nnXi1xiRearrangingthisequationandnotingthat y1 nPni ID: 827116
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Proof:First,wemultiplytherstnormale
Proof:First,wemultiplytherstnormalequationby�Pni=1xiandthesecondonebyn:�nXi=1xinXi=1yi=�b0nnXi=1xi�b1 nXi=1xi!2nnXi=1xiyi=b0nnXi=1xi+b1nnXi=1x2i:Next,addthetwoaboveequations,notingthatthetermsinvolvingb0cancelout:nnXi=1xiyi�nXi=1xinXi=1yi=b1nnXi=1x2i&#
0;b1 nXi=1xi!2=b124nnXi=1x2i� nXi=1xi
0;b1 nXi=1xi!2=b124nnXi=1x2i� nXi=1xi!235:Solvingforb1isnowastraightforwardmatterofdividingbothsidesbyhnPni=1x2i�(Pni=1xi)2i.4.Onceb1isknown,showthatsolvingthenormalequationsforb0yieldsb0=y�b1x(Note:y=1nPni=1yiandx=1nPni=1xi.)Proof:Ignoringthesecondnormaleq
uation,startbydividingtherstnormale
uation,startbydividingtherstnormalequationbyn:1nnXi=1yi=b0+b1nnXi=1xi:Rearrangingthisequation,andnotingthaty=1nPni=1yiandx=1nPni=1xi,weobtainb0=y�b1xasdesired.5.Provethatthesetwoequationsarevalid:nXi=1xiyi�1nnXi=1xinXi=1yi=nXi=1(xi�x)(yi�y)nXi=
1x2i�1n nXi=1xi!2=nXi=1(xi�x)2T
1x2i�1n nXi=1xi!2=nXi=1(xi�x)2Therefore,ifwedividersxsybys2x,then�1cancelsfromtheprevioustwoexpressionsandweareleftwithrsxsys2x=Pni=1(xi�x)(yi�y)Pni=1(xi�x)2;whichisidenticaltotheformulaforb1inpart6(frompage144).Weconcludethatb1=rsxsys2x=rsysx: