这是一份leeds利兹大学MATH260101作业代写的成功案例
Numerical Methods and Data Analysis
$$
\mathrm{W}{\mathrm{i}}=\mathrm{t}{0} / \mathrm{N}=\mathrm{t}(\mathrm{N}) / \mathrm{N} \equiv \delta \text {. }
$$
This means that our discrete Fourier transform can be written as
$$
\mathrm{F}\left(\mathrm{z}{\mathrm{k}}\right)=\delta \sum{i=0}^{N-1} \mathrm{f}\left(\mathrm{t}_{\mathrm{j}}\right) \mathrm{e}^{2 \pi i z(\mathrm{j} \delta)} .
$$
Let us proceed with the detailed implementation of the FFT. First we must calculate the weights $\mathrm{W}{\mathrm{j}}$ that appear in by means of so that $$ \mathrm{W}{\mathrm{j}}=\delta=4 / 2^{3}=1 / 2 .
$$
The first sub-division into sub-transforms involving the even and odd terms in the series specified by is
$$
\mathrm{F}{\mathrm{k}}=\delta\left(F{\mathrm{k}}^{0}+\mathrm{Q}{\mathrm{k}}^{1} F{\mathrm{k}}^{1}\right)
$$
MATH260101 COURSE NOTES ：
$$
\mathrm{S}\left(\mathrm{x}{\mathrm{i}}\right)=\frac{1}{\mathrm{~N}} \sum{\mathrm{j}=1}^{\mathrm{i}} \mathrm{n}\left(\mathrm{x}{\mathrm{j}}<\mathrm{x}\right) . $$ This is to be compared with the cumulative probability distribution of the parent population, which is $$ \mathrm{p}(\mathrm{x})=\int{0}^{\mathrm{x}} \mathrm{f}(\mathrm{z}) \mathrm{dz} .
$$
The statistic which is used to compare the two cumulative probability distributions is the largest departure $\mathrm{D}{0}$ between the two cumulative probability distributions, or $$ \mathrm{D}{0} \equiv \operatorname{Max}\left|\mathrm{S}\left(\mathrm{x}{\mathrm{i}}\right)-\mathrm{p}\left(\mathrm{x}{\mathrm{i}}\right)\right|, \forall \mathrm{x}_{\mathrm{i}}
$$