x 的 x 次方 的 x 次方 = 3. 求 x

飛天魏國大將軍張遼，你實在太大意了！你竟然將x^(x^x)當咗做(x^x)^x都有ge！！！http://hk.knowledge.yahoo.com/question/?qid=7007022702260呢題有三層次方你都冇計錯，但係點解去到呢題只有兩層次方你竟然「失魂魚」？


其實我嘅答案先至係正確：

We can solve it by means of Newton' s method of approximation of roots as follows:

Consider f(x) = x^(x^x) – 3 , then the root for f(x) = 0 can be found as follows:

x_(n + 1) = x_n – f(x_n)/f ' (x_n)

Observe that f(1) < 0 and f(2) > 0, we can be sure that the roots is between 1 and 2.

Then for f ' (x), we use the natural log method differentiation as follows:

f ' (x) = d/dx (x^(x^x))

Let y = x^(x^x)
ln y = (x^x)ln x

Let z = x^x
ln z = x ln x
(1/z) dz/dx = d/dx (x ln x) = ln x + x × 1/x = ln x + 1
dz/dx = z(ln x + 1) = (x^x)(ln x + 1)

∴(1/y) dy/dx = d/dx ((x^x)ln x) = (x^x)(ln x + 1)ln x + (x^x)(1/x) = (x^x)ln x(ln x + 1) + x^(x – 1)
dy/dx
= y[(x^x)ln x(ln x + 1) + x^(x – 1)]
= (x^(x^x))[(x^x)(ln x)² + (x^x)ln x + x^(x – 1)]
= [x^(x^x + x – 1)][x(ln x)² + x ln x + 1]

So, we start with x_0 = 1.5:

x_1 = x_0 – f(x_0)/f ' (x_0) = 1.68681
x_2 = x_1 – f(x_1)/f ' (x_1) = 1.64164
x_3 = x_2 – f(x_2)/f ' (x_2) = 1.63519
x_4 = x_3 – f(x_3)/f ' (x_3) = 1.63508
x_5 = x_4 – f(x_4)/f ' (x_4) = 1.63508

So, correcting to 4 decimal places, the approximate root is x = 1.6351

飛天魏國大將軍張遼嘅回答錯咗啲乜？

其實佢由ln y = x ln (x^x)呢行起開始計錯。
因為好似佢咁計法，
y = x^x^x
我估佢當咗x^x^x係(x^x)^x
先至令佢下一步寫成咁：
ln y = x ln (x^x)
跟住下一步佢自然會咁寫：
= x² ln x

飛天魏國大將軍張遼，乜你計到呢步嘅時候唔覺得有可疑個咩？
得出ln y = x² ln x這個結果即係表示用y = x^x^x做出嚟嘅結果竟然同用y = x^x^2做出嚟嘅結果完全吻合。這是絕對不合理的！
因為只有(x^x)^x先至會= x^(x × x) = x^x^2
更何況，如果x^x^x真係= (x^x)^x嘅話，
點解唔索性寫做x^x^2就算呢？
由此可見，x^x^x ≠ (x^x)^x，否則x^x^x嘅存在將會失去意義。

再一次提醒所有人：
x的x次方的x次方即係x^x^x，亦即係x^(x^x)。
因為

計算次序
←───
　　ｘ
　ｘ
ｘ

例如：
3^3^3 = 3^(3^3) = 3^27 ≈ 7.6256 × 10¹²
but 3^3^3 ≠ (3^3)^3 = 27^3 = 19683

希望從此以後，唔會再出現有人好似飛天魏國大將軍張遼今次嘅錯法啦！

As follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Aug07/Crazynew1.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Aug07/Crazynew2.jpg

X = 1.566
X^X^X
1.566^1.566^1.566
=3.003991751
=3.00(取三位有效數字)