如何确定常数C p (x) =美元美元残雪^ {-}$ ?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换

如何确定常数C p (x) =美元美元残雪^ {-}$ ?

2016 - 04 - 16 - t04:45:38z

Given a distribution obeying the power-law (fractal) relation, such as the cumulative distribution function $L_{cf}(> X) = CR^{-D}$, if $X$ is given, how does one find the constant $C$ from a given data set?

For example, given this table:

[Source: C. Li et al. / Geomorphology 130 (2011)]

how does one determine how the values for $L_{cf}$ were obtained? It appears that they used the method of least squares fitting to obtain the values for $D$, but I don't seem to understand where the values for $C$ are coming from. Any help would be appreciated.

由戈登斯坦格回答如何确定常数C p (x) =美元美元残雪^ {-}$ ?

2016 - 04 - 16 - t06:07:13z

< p > C是纯粹的经验对于任何给定的情况。当心这样的方程!利用系数6个有效数字给出了< em > < / em >错觉精度高时,事实上,整个方法是极其“橡胶”,高度依赖于当地水文地质/土壤类型。表达式是唯一有效的位置校准。排水系统的一个小改变可以使一个巨大的变化方程,就我个人而言,我认为这种随机简化复杂的流程是不值得写的那张纸!智能导数值的潜在滑坡条件下,以地质和排水,可能只是一样好导游(通常不校准)方程。< / p >

由丹尼尔回答。heydebreck对于如何确定常数C p (x) =美元美元残雪^ {-}$ ?

2016 - 04 - 16 - t19:34:48z

< p >有两个解决方案使用最小二乘法计算C和D $美元美元。这两种方法会产生不同的结果为常数。没有正确<强> < / >强的方法。< / p > < h2 >最小二乘法< / h2 > < p >我们定义最小平方误差如下:$ $ \文本{伦敦}= \ sum_{我}{\离开(y_i - f (x_i) \右)^ 2}$ $ $ y_i x_i美元美元是我们的数据通过,我们要适应一个函数f (x)美元。其目的是减少我们的错误文本\{伦敦}美元。< / p > < p >解决方案(最小\文本{伦敦})美元对于一个线性函数f (x) =美元\ cdot x + b美元描述< a href = " https://en.wikipedia.org/wiki/Linear_least_squares_%28mathematics%29 " rel = " nofollow noreferrer " > < / >。计算最低的基本概念\文本{伦敦}是设置美元美元\部分\文本{伦敦}/ \部分美元和美元\部分\文本{伦敦}/ \ b部分$ = $ 0 $和解决由此产生的方程系统和b美元美元。< / p > < p >已经说过我们两个解计算C和D美元美元$ f (x) = C \ cdot x ^ {-} $ < / p > < h2 >解决方案1:查对数函数< / h2 > < p >我们重写$ $ y_i = C \ cdot x_i ^ {-} $ $ $ $ \ ln {y_i} = \ ln{\离开(C \ cdot x_i ^{-} \右)}= \ ln {C} - D \ cdot \ ln {x_i} $ $ < / p > < p >现在我们有一个函数的形式\ \波浪符号{y_i} =美元cdot \波浪号{x_i} + b与美元\波浪号{y_i} = \ ln {y_i} $, $ \波浪号{x_i} = \ ln {x_i} $ $ = - D $和$ b = \ ln {C} $。因此,我们查对数测量x_i y_i值美元和美元放进线性最小平方误差法的公式。从产生的美元美元b我们比计算D美元和加元美元。

Solution 2: Insert the Function as it is

We set $f(x) = C\cdot x^{-D}$, insert it into our $\text{lse}$ formula: $$ \text{lse} = \sum_i{\left(y_i - C\cdot x^{-D}\right)^2}$$ and minimize the resulting formula with respect to $C$ and $D$ for the given set of $x_i$ and $y_i$. This is a tricky and not as straight-forward as it is in the linear case. You could make $\partial \text{lse}/\partial C$ and $\partial \text{lse}/\partial D$ and look how far you come.

I personally, prefer the answer of Gordon Stanger :-) .