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Question 29340: The question is: Show that the nth-order finite differences for the function f(x)= x^2 - 4x + 4 of degree n are nonzero and constant. I don't know what procedures to use that's why I couldn't try solving it. I hope that you can reply me back a.s.a.p.. I really appreciate it.. Thank You!!!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! The question is: Show that the nth-order finite differences for the function f(x)= x^2 - 4x + 4 of degree n are nonzero and constant. I don't know what procedures to use that's why I couldn't try solving it. I hope that you can reply me back a.s.a.p.. I really appreciate it.. Thank You!!!
NO..THE QUESTION IS NOT WORDED PROPERLY...YOUR FUNCTION IS 2 ND. DEGREE POLYNOMIAL.SO ITS SECOND ORDER DIFFERENCES ARE CONSTANT..LET US SEE
F(X)=X^2-4X+4
F(X+H)=(X+H)^2-4(X+H)+4...SO
FIRST ORDER DIFFERENCE IS D1(X) =F(X+H)-F(X)=(X+H)^2-4(X+H)+4-(X^2-4X+4)
=X^2+2HX+H^2-4X-4H-X^2+4X-4=2HX+H^2-4H-4....WHICH IS A FIRST DEGREE EXPRESSION..FOR SIMPLICITY,LET US WRITE IT AS
D1(X)=AX+B...WHERE A=2H AND B=H^2-4H-4..SO WE GET
D1(X+H)=A(X+H)+B
D2(X)=D1(X+H)-D1(X)=A(X+H)+B-(AX+B)=AX+AH+B-AX-B=AH...SINCE A=2H...WE GET.
D2(X)=2H^2...WHICH IS INDEPENDENT OF X AND IS A CONSTANT..
IN GENERAL IF F(X)=X^N+.......ETC..THAT IS NTH.DEGREE PLYNOMIAL ,THEN ITS N TH.ORDER DIFFERENCES WILL BE CONSTANT..BEYOND THAT THE DIFFERENCES WILL VANISH..THAT IS EQUAL TO ZERO
PROOF IS ...AS YOU SAW ABOVE WITH EACH ORDER OF DIFFERENCE,THE FUNCTION POLYNOMIAL GETS REDUCED BY 1 DEGREE..THAT IS IF
F(X)=X^N+......
F(X+H)=(X+H)^N+.....
D1(X)=(X+H)^N+....-X^N-......=X^N+N*H*X^(N-1)+....-X^N-....=*H*X^(N-1)+....
WHICH IS CLEARLY POLYNOMIAL OF (N-1) DEGREE...LIKE THIS WHEN WE GO TO N TH. ORDER DIFFERENCE ,WE GET POLYNOMIAL OF N-N =0 DEGREE OR A CONSTANT POLYNOMIAL..
I DONT THINK YOU NEED A MORE RIGOROUS PROOF..IT DEPENDS ON YOUR COURSE OF STUDY..IF YOU INFORM THE DETAILS WE CAN CHECK UP..
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