# Quod Erat Demonstrandum

## 2010/02/28

### 最大可能和期望值

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 12:04 上午

$X$ ~ $B(n , p)$
$P_k = P(X = k)$ （其中 $k = 0, 1, \dots n$

$P_{k + 1} > P_k$
$\Leftrightarrow C_{k + 1}^np^{k + 1}(1 - p)^{n - k - 1} > C_{k}^np^{k}(1 - p)^{n - k}$
$\Leftrightarrow \frac{p}{k + 1} > \frac{1 - p}{n - k}$
$\Leftrightarrow k < (n + 1)p - 1$

$P_0 < P_1 < P_2 < \dots < P_{[(n + 1)p] - 1} < P_{[(n + 1)p]} \ge P_{[(n + 1)p] + 1} \ge \dots \ge P_n$

$[np]$$[(n + 1)p]$ 兩值可以是很接近，比如 $n$ 值很大 $p$ 值很小的時候。

[OT] 之前同學問，如何解以下聯立方程：

$2^{x^2 + y} + 2^{x + y^2} = 8$ ………… (1)
$\sqrt{x} + \sqrt{y} = 2$ ……….. (2)

（其中 $x, y$ 為實數）

## 6 則迴響 »

1. Hi,

The last question you proposed is a very good one! Thanks for sharing :)

Cheers,
Wen Shih

迴響 由 wenshih — 2010/02/28 @ 8:12 上午 | 回應

2. Let a = sqrt{x}, b = sqrt{y}, then we have a, b > 0. (otherwise, it is easy to see that there is no solution)

Now, we have (8/2)^2 = [(2^{a^4 + b^2} + 2^{a^2 + b^4})/2]^2 >= 2^{a^4 + b^4 + a^2 + b^2} By AM-GM.
This means, 4 >= a^4 + b^4 + a^2 + b^2.

Now, by power mean, we have (a^4 + b^4)/2 >= [(a + b)/2]^4 = 1, and likewise, (a^2 + b^2)/2 >= [(a + b)/2]^2 = 1.
This means, 4 >= a^4 + b^4 + a^2 + b^2 >= 2 + 2 = 4.
Therefore, by the equality condition of the power means and AM-GM, we have a = b = 1. Done.

迴響 由 koopakoo — 2010/03/02 @ 2:44 上午 | 回應

3. By the way, what does [OT] mean?

迴響 由 koopakoo — 2010/03/02 @ 2:45 上午 | 回應

• Thank you Koopa!

OT = off topic

迴響 由 johnmayhk — 2010/03/02 @ 7:48 上午 | 回應

4. a symmetric equation, easily we have x=y=1.

迴響 由 code_vest — 2010/03/18 @ 9:53 下午 | 回應

• It seems that the uniqueness of solution cannot be guaranteed by symmetry.

迴響 由 johnmayhk — 2010/03/18 @ 10:40 下午 | 回應