Re: [連續]
Because f(c)=1/2 and f(x)= sinx/x
so that we can rewrite f(c)=1/2 as following equation:
2sinC = C
Now we can define a new function g:(0,pi] --> R:
g(x)=2sinx - x
We can say g(x) is continue on (0,pi]
g(pi/2)= 2 sin(pi/2) - pi/2 = 2 - pi/2 > 0
g(pi)=2sin(pi) - pi = 2*0 -pi = -pi<0
so g(pi/2)g(pi) <0---[*]
from [*] and I.V.T.
There exists a value cε(0,pi] such that g(c)= 0
and we know g(c)= 2sinC-C = 0
that is, sinC/C = 1/2
※ 引述《Qmmm (．．Ｑ3Ｍ．．)》之銘言：
: 題目:
: sinx 1
: 設f(x)= ---- , x ε (0,π] 試證存在一數c ε(0,π)使得f(c)= ---
: x 2
: =============================================================================
: 我想問這題中間值定理不是要在閉區間連續嗎?
: 可是這題在(0,π] 連續
: 所以有沒有違反定理??怎麼說呢?
: 如果沒有可不可以用我的解法
: 令F(x) = sinx/x - 1/2
: 則 F(0) = lim x->0 sinx/x - 1/2 = 1/2 >0
: {
: F(π)= sinπ/π - 1/2 = -1/2 <0
: 由中間值定理知
: F(0)F(π)<0
: 存在一數c ε(0,π)
: 使得 F(c)=0 => sinc/c - 1/2 即 f(c) = 1/2
: 因為王氏微積分說不能這樣解...
: 謝謝!
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