HUNG FUNG BK 2 P.131 REV3 Q1
a. Let f and g be two continuous real functions. Suppose Xo is a real constant such that f(x) = g(x) for all x is real number except {Xo}
Show that f(Xo)=g(Xo)
b. Let p(x) and q(x) be two real polynomials. Suppose Xo is a real number and h,k are two integers such that p(Xo)≠ 0 , q(Xo) ≠0 and [(x-Xo)^h]p(x)=[(x-Xo)^k]q(x)
Show that h=k and p(x) = q(x) for all x is real number
請問b part點做
solution既方法我唔明
PURE MATHS CONTINUNITY
2007-03-31 4:39 am
回答 (2)
2007-04-01 4:52 am
✔ 最佳答案
[(x-Xo)^h] p(x) = [(x-Xo)^k] q(x)[(x-Xo)^(h-k)] p(x) = q(x)
Note that q is continuous on R, so we have
q(X0) = lim (x -> X0) q(x) = lim (x -> X0) [(x-Xo)^(h-k)] p(x)
since q(X0) is non-zero, and p(X0) = lim (x -> X0) p(x) is also non-zero
so we must have h - k = 0, i.e. h = k
(otherwise, if they are non-equal, the term (x - X0)^(h - k) will be zero or undefined
(depeneds on h > k or h < k) )
So we obtain
[(x-Xo)^h] p(x) = [(x-Xo)^h] q(x)
i.e. p(x) = q(x) except x = X0
Using (a), we have p = q for all real x.
2007-04-15 3:05 am
x is tending to x0, if (x0-x0)^(h-k) then no matter what h and k is (except h=k), it always =0.
when h=k, it is 0^0 which is undefined.
Explain more if you can, please
when h=k, it is 0^0 which is undefined.
Explain more if you can, please
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