maths

The approximation of this equation can be done by a stepwise process called Newton's method which is
x(n+1) = xn - [F(xn)/F'(xn)] where xn is the value of root after nth iteration.

Thus, we have to convert the equation into:
x^x-9 = 0 with x > 1 (since we are sure that the root is greater than 1)
that the roots to be found is the answer of this question.

Newton's method is applicable to this equation since f(x) = x^x - 9 is strictly increasing.

Now, concerning f'(x), we have to use logarithmic differentiation:
ln[f(x)] = xlnx
f'(x)/f(x) = x(1/x) + lnx
f'(x) = f(x) + f(x)lnx = x^x + x^xlnx
which is positive for all x>1.

Starting with x=2, i.e. x0=2, we have:

f(2) = -5, f'(2) = 6.7726
Thus, x1 = 2.7383

Continuing the iteration:
x2 = 2.5243
x3 = 2.4564
x4 = 2.4510
x5 = 2.4510

So to speak, the answer is 2.4510 (corr. to 4 d.p.)

In fact, you may choose the precision scale on your own and check for each iteration value against its previous iteration value. If the difference is less than half of the precision scale you prefer, this is the answer you need.

當x=1, x^x=1,
當x=2, x^x=4,
當x=3, x^x=27,

所以x唔係一個整數.

(約至小數後4個位)
當x=2.45, x^x=8.9837
當x=2.451, x^x=9.0008

當x=2.4509, x^x=8.9991
當x=2.45095, x^x=8.9999
當x=2.45096, x^x=9.0001

當x=2.450955, x^x=9.0000

故x^x=9, x約等於2.450955

2006-12-19 22:46:15 補充：
可能真係有呢個x, 但係機會好微.呢度只可以俾個約數你.

no answer

2006-12-19 22:44:52 補充：
To:a123more3的3次方是3x3x3answer is:27

X既X次方等於9, 那X= 3