Quod Erat Demonstrandum

2012/01/12

Core Math 某題

Filed under: mathematics,NSS — johnmayhk @ 10:25 下午

Set 1: {a,b,d,e}
Set 2: {a,b,c,d,e}

(其中 a < b < c < d < e

\sigma_1, \sigma_2 分別是 Set 1,Set 2 的標準差(standard deviation),即

\sigma_1^2=\frac{a^2+b^2+d^2+e^2}{4}-(\frac{a+b+d+e}{4})^2

\sigma_2^2=\frac{a^2+b^2+c^2+d^2+e^2}{5}-(\frac{a+b+c+d+e}{5})^2

\sigma_1^2 > \sigma_2^2

結論頗直觀,但若直觀不能說服人,唯用低級方法(即應該有更好的方法):「拆開」(expand)來證明。

不過「拆開」前,先做一些簡化。考慮

Set 3: {a-c,b-c,d-c,e-c}
Set 4: {a-c,b-c,0,d-c,e-c}

易知 Set 3 及 Set 4 的標準差,根本也是 \sigma_1\sigma_2

那麼我們把問題轉為考慮

Set 1: {u,v,x,y}
Set 2: {u,v,0,x,y}

(其中 u < v < 0 < x < y

結論也同樣有效。我們有

\sigma_1^2=\frac{u^2+v^2+x^2+y^2}{4}-(\frac{u+v+x+y}{4})^2

\sigma_2^2=\frac{u^2+v^2+x^2+y^2}{5}-(\frac{u+v+x+y}{5})^2

考慮

\sigma_1^2-\sigma_2^2

=\frac{1}{400}(11(u^2+v^2+x^2+y^2)-18(uv+ux+uy+vx+vy+xy))

=\frac{1}{400}(11(u^2+v^2+x^2+y^2)-18(u+v)(x+y)-18(uv+xy))

=\frac{1}{400}(11((u-v)^2+(x-y)^2)-18(u+v)(x+y)+4(uv+xy))

> 0

(因 u < v < 0 < x < y,故 -18(u+v)(x+y) > 04(uv+xy) > 0

Q.E.D.

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