[高微]數學問題!!請進

Question

For c >= 0, define the function f(c) = ∫dx/ (1 + e^(cx)). (上界1, 下界0)

We want to prove the existence of a unique value c_* > 0 such that f(c_*) = 1/4, and then compute c_*.

a)Prove that f(c) is continuous and monotone decreasing to 0 as c → ∞.

b)Prove that there is unique value c_* such that f(c_*) = 1/4 and that 0 < c_* < 4.

c)What combination of algorithms would you use to compute c_* ?
Write the codes to do so, and compute c_* to within 10^(-4). Remember that there will be errors in the numbers you calculate for the integral.

請會的大大告訴我詳細的作答
感激不盡

可以的話請告訴我解答~~謝謝

Answer 1

f(c) = ∫01 dx/ (1+ecx)
a) f'(c)=∫01 -ecx/(1+ecx)2 dx<0 => f(c)遞減
c>0時 ∫01 dx/(1+ecx) <∫01 dx/ecx = (1- e-c)/c
lim(c→ ∞) f(c)<=lim(c→ ∞) (1- e-c)/c = 0
故 f(c)可微分(必連續)且遞減至 0
b) f(0)=1, f(c)=∫01 dx/ (1+ecx)=∫01 e-cx/(1+e-cx) dx= [ln2-ln(1+e-c)] / c
f(4)=0.168749... , f(c*)=1/4=0.25
又f(c)為連續遞減函數,
故存在且唯一 c* 使 f(c*)=1/4, 且0< c*<4
c) f(c)=[ln2-ln(1+e-c)] / c = 1/4
use function iteration by Excel:
x0=2.50000, x1=2.45703 , x3=2.44373 , x4=2.43950
x5=2.43815, x6=2.43772 , x7=2.43758 , x8=2.43753
x9=2.43752, x10=2.43751, x11=2.43751
故c*近似值為 2.4375

Answer 2

c以來>0 =，確定那些f 功能(c) = ∫dx /(1 + e ^ (cx))。 ( 上界1 , 下界0)我們想要證明獨特的價值c的存在 _ *>0和f(c _ *)=1/4，然後計算c _ *. 一)證明f ( c ) 連續和減少到作為c 的0的單調 →8.b)證明有獨特的價值c _ *因此f(c _ *)=1/4 和那0
沒翻的不會,我幫你翻好了,你自己慢慢看ㄛ!