Related
I have the following problem. I have a function f defined in python using numpy functions. The function is smooth and integrable on positive reals. I want to construct the double antiderivative of the function (assuming that both the value and the slope of the antiderivative at 0 are 0) so that I can evaluate it on any positive real smaller than 100.
Definition of antiderivative of f at x:
integrate f(s) with s from 0 to x
Definition of double antiderivative of f at x:
integrate (integrate f(t) with t from 0 to s) with s from 0 to x
The actual form of f is not important, so I will use a simple one for convenience. But please note that even though my example has a known closed form, my actual function does not.
import numpy as np
f = lambda x: np.exp(-x)*x
My solution is to construct the antiderivative as an array using naive numerical integration:
N = 10000
delta = 100/N
xs = np.linspace(0,100,N+1)
vs = f(xs)
avs = np.cumsum(vs)*delta
aavs = np.cumsum(avs)*delta
This of course works but it gives me arrays instead of functions. But this is not a big problem as I can interpolate aavs using a spline to get a function and get rid of the arrays.
from scipy.interpolate import UnivariateSpline
aaf = UnivariateSpline(xs, aavs)
The function aaf is approximately the double antiderivative of f.
The problem is that even though it works, there is quite a bit of overhead before I can get my function and precision is expensive.
My other idea was to interpolate f by a spline and take the antiderivative of that, however this introduces numerical errors that are too big for what I want to use the function.
Is there any better way to do that? By better I mean faster without sacrificing accuracy.
Edit: What I hope is possible is to use some kind of Fourier transform to avoid integrating twice. I hope that there is some convenient transform of vs that allows to multiply the values component-wise with xs and transform back to get the double antiderivative. I played with this a bit, but I got lost.
Edit: I figured out that by using the trapezoidal rule instead of a naive sum, increases the accuracy quite a bit. Using Simpson's rule should increase the accuracy further, but it's somewhat fiddly to do with numpy arrays.
Edit: As #user202729 rightfully complains, this seems off. The reason it seems off is because I have skipped some details. I explain here why what I say makes sense, but it does not affect my question.
My actual goal is not to find the double antiderivative of f, but to find a transformation of this. I have skipped that because I think it only confuses the matter.
The function f decays exponentially as x approaches 0 or infinity. I am minimizing the numerical error in the integration by starting the sum from 0 and going up to approximately the peak of f. This ensure that the relative error is approximately constant. Then I start from the opposite direction from some very big x and go back to the peak. Then I do the same for the antiderivative values.
Then I transform the aavs by another function which is sensitive to numerical errors. Then I find the region where the errors are big (the values oscillate violently) and drop these values. Finally I approximate what I believe are good values by a spline.
Now if I use spline to approximate f, it introduces an absolute error which is the dominant term in a rather large interval. This gets "integrated" twice and it ends up being a rather large relative error in aavs. Then once I transform aavs, I find that the 'good region' has shrunk considerably.
EDIT: The actual form of f is something I'm still looking into. However, it is going to be a generalisation of the lognormal distribution. Right now I am playing with the following family.
I start by defining a generalization of the normal distribution:
def pdf_n(params, center=0.0, slope=8):
scale, min, diff = params
if diff > 0:
r = min
l = min + diff
else:
r = min - diff
l = min
def retfun(m):
x = (m - center)/scale
E = special.expit(slope*x)*(r - l) + l
return np.exp( -np.power(1 + x*x, E)/2 )
return np.vectorize(retfun)
It may not be obvious what is happening here, but the result is quite simple. The function decays as exp(-x^(2l)) on the left and as exp(-x^(2r)) on the right. For min=1 and diff=0, this is the normal distribution. Note that this is not normalized. Then I define
g = pdf(params)
f = np.vectorize(lambda x:g(np.log(x))/x/area)
where area is the normalization constant.
Note that this is not the actual code I use. I stripped it down to the bare minimum.
You can compute the two np.cumsum (and the divisions) at once more efficiently using Numba. This is significantly faster since there is no need for several temporary arrays to be allocated, filled, read again and freed. Here is a naive implementation:
import numba as nb
#nb.njit('float64[::1](float64[::1], float64)') # Assume vs is contiguous
def doubleAntiderivative_naive(vs, delta):
res = np.empty(vs.size, dtype=np.float64)
sum1, sum2 = 0.0, 0.0
for i in range(vs.size):
sum1 += vs[i] * delta
sum2 += sum1 * delta
res[i] = sum2
return res
However, the sum is not very good in term of numerical stability. A Kahan summation is needed to improve the accuracy (or possibly the alternative Kahan–Babuška-Klein algorithm if you are paranoid about the accuracy and performance do not matter so much). Note that Numpy use a pair-wise algorithm which is quite good but far from being prefect in term of accuracy (this is a good compromise for both performance and accuracy).
Moreover, delta can be factorized during in the summation (ie. the result just need to be premultiplied by delta**2).
Here is an implementation using the more accurate Kahan summation:
#nb.njit('float64[::1](float64[::1], float64)')
def doubleAntiderivative_accurate(vs, delta):
res = np.empty(vs.size, dtype=np.float64)
delta2 = delta * delta
sum1, sum2 = 0.0, 0.0
c1, c2 = 0.0, 0.0
for i in range(vs.size):
# Kahan summation of the antiderivative of vs
y1 = vs[i] - c1
t1 = sum1 + y1
c1 = (t1 - sum1) - y1
sum1 = t1
# Kahan summation of the double antiderivative of vs
y2 = sum1 - c2
t2 = sum2 + y2
c2 = (t2 - sum2) - y2
sum2 = t2
res[i] = sum2 * delta2
return res
Here is the performance of the approaches on my machine (with an i5-9600KF processor):
Numpy cumsum: 51.3 us
Naive Numba: 11.6 us
Accutate Numba: 37.2 us
Here is the relative error of the approaches (based on the provided input function):
Numpy cumsum: 1e-13
Naive Numba: 5e-14
Accutate Numba: 2e-16
Perfect precision: 1e-16 (assuming 64-bit numbers are used)
If f can be easily computed using Numba (this is the case here), then vs[i] can be replaced by calls to f (inlined by Numba). This helps to reduce the memory consumption of the computation (N can be huge without saturating your RAM).
As for the interpolation, the splines often gives good numerical result but they are quite expensive to compute and AFAIK they require the whole array to be computed (each item of the array impact all the spline although some items may have a negligible impact alone). Regarding your needs, you could consider using Lagrange polynomials. You should be careful when using Lagrange polynomials on the edges. In your case, you can easily solve the numerical divergence issue on the edges by extending the array size with the border values (since you know the derivative on each edges of vs is 0). You can apply the interpolation on the fly with this method which can be good for both performance (typically if the computation is parallelized) and memory usage.
First, I created a version of the code I found more intuitive. Here I multiply cumulative sum values by bin widths. I believe there is a small error in the original version of the code related to the bin width issue.
import numpy as np
f = lambda x: np.exp(-x)*x
N = 1000
xs = np.linspace(0,100,N+1)
domainwidth = ( np.max(xs) - np.min(xs) )
binwidth = domainwidth / N
vs = f(xs)
avs = np.cumsum(vs)*binwidth
aavs = np.cumsum(avs)*binwidth
Next, for visualization here is some very simple plotting code:
import matplotlib
import matplotlib.pyplot as plt
plt.figure()
plt.scatter( xs, vs )
plt.figure()
plt.scatter( xs, avs )
plt.figure()
plt.scatter( xs, aavs )
plt.show()
The first integral matches the known result of the example expression and can be seen on wolfram
Below is a simple function that extracts an element from the second derivative. Note that int is a bad rounding function. I assume this is what you have implemented already.
def extract_double_antideriv_value(x):
return aavs[int(x/binwidth)]
singleresult = extract_double_antideriv_value(50.24)
print('singleresult', singleresult)
Whatever full computation steps are required, we need to know them before we can start optimizing. Do you have a million different functions to integrate? If you only need to query a single double anti-derivative many times, your original solution should be fairly ideal.
Symbolic Approximation:
Have you considered approximations to the original function f, which can have closed form integration solutions? You have a limited domain on which the function lives. Perhaps approximate f with a Taylor series (which can be constructed with known maximum error) then integrate exactly? (consider Pade, Taylor, Fourier, Cheby, Lagrange(as suggested by another answer), etc...)
Log Tricks:
Another alternative to dealing with spiky errors, would be to take the log of your original function. Is f always positive? Is the integration error caused because the neighborhood around the max is very small? If so, you can study ln(f) or even ln(ln(f)) instead. It would really help to understand what f looks like more.
Approximation Integration Tricks
There exist countless integration tricks in general, which can make approximate closed form solutions to undo-able integrals. A very common one when exponetnial functions are involved (I think yours is expoential?) is to use Laplace's Method. But which trick to pull out of the bag is highly dependent upon the conditions which f satisfies.
I can use the polyfit() method with a 2D array as input, to calculate polynomials on multiple data sets in a fast manner. After getting these multiple polynomials, I want to calculate the roots of all of these polynomials, in a fast manner.
There is numpy.roots() method for finding the roots of a single polynomial but this method does not work with 2D inputs (meaning multiple polynomials). I am working with millions of polynomials, so I would like to avoid looping over all polynomials using a for loop, map or comprehension because it takes minutes in that case. I would prefer a vectoral numpy operation or series of vectoral operations.
An example code for inefficient calculation:
POLYNOMIAL_COUNT = 1000000
# Create a polynomial of second order with coefficients 2, 3 and 4
coefficients = np.array([[2,3,4]])
# Let's say we have the same polynomial multiple times, represented as a 2D array.
# In reality the polynomial coefficients will be different from each other,
# but they will be the same order.
coefficients = coefficients.repeat(POLYNOMIAL_COUNT, axis=0)
# Calculate roots of these same-order polynomials.
# Looping here takes too much time.
roots = []
for i in range(POLYNOMIAL_COUNT):
roots.append(np.roots(coefficients[i]))
Is there a way to find the roots of multiple same-order polynomials using numpy, but without looping?
For the special case of polynomials up to the fourth order, you can solve in a vectorized manner. Anything higher than that does not have an analytical solution, so requires iterative optimization, which is fundamentally unlikely to be vectorizable since different rows may require a different number of iterations. As #John Coleman suggests, you might be able to get away with using the same number of steps for each one, but will likely have to sacrifice accuracy to do so.
That being said, here is an example of how to vectorize the second order case:
d = coefficients[:, 1:-1]**2 - 4.0 * coefficients[:, ::2].prod(axis=1, keepdims=True)
roots = -0.5 * (coefficients[:, 1:-1] + [1, -1] * np.emath.sqrt(d)) / coefficients[:, :1]
If I got the order of the coefficients wrong, replace coefficients[:, :1] with coefficients[:, -1:] in the denominator of the last assignment. Using np.emath.sqrt is nice because it will return a complex128 result automatically when your discriminant d is negative anywhere, and normal float64 result for all real roots.
You can implement a third order solution or a fourth order solution in a similar manner.
I tried to calculate the Pearson's correlation coefficients between every pairs of rows from two 2D arrays. Then, sort the rows/columns of the correlation matrix based on its diagonal elements. First, the correlation coefficient matrix (i.e., 'ccmtx') was calculated from one random matrix (i.e., 'randmtx') in the following code:
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import pearsonr
def correlation_map(x, y):
n_row_x = x.shape[0]
n_row_y = x.shape[0]
ccmtx_xy = np.empty((n_row_x, n_row_y))
for n in range(n_row_x):
for m in range(n_row_y):
ccmtx_xy[n, m] = pearsonr(x[n, :], y[m, :])[0]
return ccmtx_xy
randmtx = np.random.randn(100, 1000) # generating random matrix
#ccmtx = np.corrcoef(randmtx, randmtx) # cc matrix based on numpy.corrcoef
ccmtx = correlation_map(randmtx, randmtx) # cc matrix based on scipy pearsonr
#
ccmtx_diag = np.diagonal(ccmtx)
#
ids, vals = np.argsort(ccmtx_diag, kind = 'mergesort'), np.sort(ccmtx_diag, kind = 'mergesort')
#ids, vals = np.argsort(ccmtx_diag, kind = 'quicksort'), np.sort(ccmtx_diag, kind = 'quicksort')
plt.plot(ids)
plt.show()
plt.plot(ccmtx_diag[ids])
plt.show()
vals[0]
The issue here is when the 'pearsonr' was used, the diagonal elements of 'ccmtx' are exactly 1.0 which makes sense. However, the 'corrcoef' was used, the diagonal elements of 'ccmtrix' are not exactly one (and slightly less than 1 for some diagonals) seemingly due to a precision error of floating point numbers.
I found to be annoying that the auto-correlation matrix of a single matrix have diagnoal elements not being 1.0 since this resulted in the shuffling of rows/columes of the correlation matrix when the matrix is sorted based on the diagonal elements.
My questions are:
[1] is there any good way to accelerate the computation time when I stick to use the 'pearsonr' function? (e.g., vectorized pearsonr?)
[2] Is there any good way/practice to prevent this precision error when using the 'corrcoef' in numpy? (e.g. 'decimals' option in np.around?)
I have searched the correlation coefficient calculations between all pairs of rows or columns from two matrices. However, as the algorithms containe some sort of "cov / variance" operation, this kind of precision issue seems always existing.
Minor point: the 'mergesort' option seems to provide reliable results than the 'quicksort' as the quicksort shuffled 1d array with exactly 1 to random order.
Any thoughts/comments would be greatly appreciated!
For question 1 vectorized pearsonr see the comments to the question.
I will answer only question 2: how to improve the precision of np.corrcoef.
The correlation matrix R is computed from the covariance matrix C according to
.
The implementation is optimized for performance and memory usage. It computes the covariance matrix, and then performs two divisions by sqrt(C_ii) and by sqrt(Cjj). This separate square-rooting is where the imprecision comes from. For example:
np.sqrt(3 * 3) - 3 == 0.0
np.sqrt(3) * np.sqrt(3) - 3 == -4.4408920985006262e-16
We can fix this by implementing our own simple corrcoef routine:
def corrcoef(a, b):
c = np.cov(a, b)
d = np.diag(c)
return c / np.sqrt(d[:, None] * d[None, :])
Note that this implementation requires more memory than the numpy implementation because it needs to store a temporary matrix with size n * n and it is slightly slower because it needs to do n^2 square roots instead of only 2 n.
I'm trying to replicate some Matlab code in python. I could not find an exact equivalent to the Matlab function quantile. What I found most close is python's mquantiles.
Matlab example:
quantile( [ 8.60789925e-05, 1.98989354e-05 , 1.68308882e-04, 1.69379370e-04], 0.8)
...gives: 0.00016958
Same example in python:
scipy.stats.mstats.mquantiles( [8.60789925e-05, 1.98989354e-05, 1.68308882e-04, 1.69379370e-04], 0.8)
...gives 0.00016912
Does anyone know how to exactly replicate Matlab's quantile function?
The documentation for quantile (under the More About => Algorithms section) gives the exact algorithm used. Here's some python code that does it for a single quantile for a flat array, using bottleneck to do partial sorting:
import numpy as np
import botteleneck as bn
def quantile(a, prob):
"""
Estimates the prob'th quantile of the values in a data array.
Uses the algorithm of matlab's quantile(), namely:
- Remove any nan values
- Take the sorted data as the (.5/n), (1.5/n), ..., (1-.5/n) quantiles.
- Use linear interpolation for values between (.5/n) and (1 - .5/n).
- Use the minimum or maximum for quantiles outside that range.
See also: scipy.stats.mstats.mquantiles
"""
a = np.asanyarray(a)
a = a[np.logical_not(np.isnan(a))].ravel()
n = a.size
if prob >= 1 - .5/n:
return a.max()
elif prob <= .5 / n:
return a.min()
# find the two bounds we're interpreting between:
# that is, find i such that (i+.5) / n <= prob <= (i+1.5)/n
t = n * prob - .5
i = np.floor(t)
# partial sort so that the ith element is at position i, with bigger ones
# to the right and smaller to the left
a = bn.partsort(a, i)
if i == t: # did we luck out and get an integer index?
return a[i]
else:
# we'll linearly interpolate between this and the next index
smaller = a[i]
larger = a[i+1:].min()
if np.isinf(smaller):
return smaller # avoid inf - inf
return smaller + (larger - smaller) * (t - i)
I only did the single-quantile, 1d case because that's all I needed. If you want several quantiles, it's probably worth just doing the full sort; to do it per-axis and knew you didn't have any nans, all you should need to do is add an axis argument to the sort and vectorize the linear interpolation bit. Doing it per-axis with nans would be a little trickier.
This code gives:
>>> quantile([ 8.60789925e-05, 1.98989354e-05 , 1.68308882e-04, 1.69379370e-04], 0.8)
0.00016905822360000001
and the matlab code gave 0.00016905822359999999; the difference is 3e-20. (which is less than machine precision)
Your input vector only has 4 values, which is far too few to get a good approximation of the quantiles of the underlying distribution. The discrepancy is probably the result of Matlab and SciPy using different heuristics to compute quantiles on under sampled distributions.
A bit late, but:
mquantiles is very flexible. You just need to provide alphap and betap parameters.
Here, since MATLAB does a linear interpolation, you need to set the parameters to (0.5,0.5).
In [9]: scipy.stats.mstats.mquantiles( [8.60789925e-05, 1.98989354e-05, 1.68308882e-04, 1.69379370e-04], 0.8, alphap=0.5, betap=0.5)
EDIT: MATLAB says that it does linear interpolation, however it seems that it calculates the quantile through piece-wise linear interpolation, which is equivalent to Type 5 quantile in R, and (0.5, 0.5) in scipy.
I'm trying to interpolate some data for the purpose of plotting. For instance, given N data points, I'd like to be able to generate a "smooth" plot, made up of 10*N or so interpolated data points.
My approach is to generate an N-by-10*N matrix and compute the inner product the original vector and the matrix I generated, yielding a 1-by-10*N vector. I've already worked out the math I'd like to use for the interpolation, but my code is pretty slow. I'm pretty new to Python, so I'm hopeful that some of the experts here can give me some ideas of ways I can try to speed up my code.
I think part of the problem is that generating the matrix requires 10*N^2 calls to the following function:
def sinc(x):
import math
try:
return math.sin(math.pi * x) / (math.pi * x)
except ZeroDivisionError:
return 1.0
(This comes from sampling theory. Essentially, I'm attempting to recreate a signal from its samples, and upsample it to a higher frequency.)
The matrix is generated by the following:
def resampleMatrix(Tso, Tsf, o, f):
from numpy import array as npar
retval = []
for i in range(f):
retval.append([sinc((Tsf*i - Tso*j)/Tso) for j in range(o)])
return npar(retval)
I'm considering breaking up the task into smaller pieces because I don't like the idea of an N^2 matrix sitting in memory. I could probably make 'resampleMatrix' into a generator function and do the inner product row-by-row, but I don't think that will speed up my code much until I start paging stuff in and out of memory.
Thanks in advance for your suggestions!
This is upsampling. See Help with resampling/upsampling for some example solutions.
A fast way to do this (for offline data, like your plotting application) is to use FFTs. This is what SciPy's native resample() function does. It assumes a periodic signal, though, so it's not exactly the same. See this reference:
Here’s the second issue regarding time-domain real signal interpolation, and it’s a big deal indeed. This exact interpolation algorithm provides correct results only if the original x(n) sequence is periodic within its full time interval.
Your function assumes the signal's samples are all 0 outside of the defined range, so the two methods will diverge away from the center point. If you pad the signal with lots of zeros first, it will produce a very close result. There are several more zeros past the edge of the plot not shown here:
Cubic interpolation won't be correct for resampling purposes. This example is an extreme case (near the sampling frequency), but as you can see, cubic interpolation isn't even close. For lower frequencies it should be pretty accurate.
If you want to interpolate data in a quite general and fast way, splines or polynomials are very useful. Scipy has the scipy.interpolate module, which is very useful. You can find many examples in the official pages.
Your question isn't entirely clear; you're trying to optimize the code you posted, right?
Re-writing sinc like this should speed it up considerably. This implementation avoids checking that the math module is imported on every call, doesn't do attribute access three times, and replaces exception handling with a conditional expression:
from math import sin, pi
def sinc(x):
return (sin(pi * x) / (pi * x)) if x != 0 else 1.0
You could also try avoiding creating the matrix twice (and holding it twice in parallel in memory) by creating a numpy.array directly (not from a list of lists):
def resampleMatrix(Tso, Tsf, o, f):
retval = numpy.zeros((f, o))
for i in xrange(f):
for j in xrange(o):
retval[i][j] = sinc((Tsf*i - Tso*j)/Tso)
return retval
(replace xrange with range on Python 3.0 and above)
Finally, you can create rows with numpy.arange as well as calling numpy.sinc on each row or even on the entire matrix:
def resampleMatrix(Tso, Tsf, o, f):
retval = numpy.zeros((f, o))
for i in xrange(f):
retval[i] = numpy.arange(Tsf*i / Tso, Tsf*i / Tso - o, -1.0)
return numpy.sinc(retval)
This should be significantly faster than your original implementation. Try different combinations of these ideas and test their performance, see which works out the best!
I'm not quite sure what you're trying to do, but there are some speedups you can do to create the matrix. Braincore's suggestion to use numpy.sinc is a first step, but the second is to realize that numpy functions want to work on numpy arrays, where they can do loops at C speen, and can do it faster than on individual elements.
def resampleMatrix(Tso, Tsf, o, f):
retval = numpy.sinc((Tsi*numpy.arange(i)[:,numpy.newaxis]
-Tso*numpy.arange(j)[numpy.newaxis,:])/Tso)
return retval
The trick is that by indexing the aranges with the numpy.newaxis, numpy converts the array with shape i to one with shape i x 1, and the array with shape j, to shape 1 x j. At the subtraction step, numpy will "broadcast" the each input to act as a i x j shaped array and the do the subtraction. ("Broadcast" is numpy's term, reflecting the fact no additional copy is made to stretch the i x 1 to i x j.)
Now the numpy.sinc can iterate over all the elements in compiled code, much quicker than any for-loop you could write.
(There's an additional speed-up available if you do the division before the subtraction, especially since inthe latter the division cancels the multiplication.)
The only drawback is that you now pay for an extra Nx10*N array to hold the difference. This might be a dealbreaker if N is large and memory is an issue.
Otherwise, you should be able to write this using numpy.convolve. From what little I just learned about sinc-interpolation, I'd say you want something like numpy.convolve(orig,numpy.sinc(numpy.arange(j)),mode="same"). But I'm probably wrong about the specifics.
If your only interest is to 'generate a "smooth" plot' I would just go with a simple polynomial spline curve fit:
For any two adjacent data points the coefficients of a third degree polynomial function can be computed from the coordinates of those data points and the two additional points to their left and right (disregarding boundary points.) This will generate points on a nice smooth curve with a continuous first dirivitive. There's a straight forward formula for converting 4 coordinates to 4 polynomial coefficients but I don't want to deprive you of the fun of looking it up ;o).
Here's a minimal example of 1d interpolation with scipy -- not as much fun as reinventing, but.
The plot looks like sinc, which is no coincidence:
try google spline resample "approximate sinc".
(Presumably less local / more taps ⇒ better approximation,
but I have no idea how local UnivariateSplines are.)
""" interpolate with scipy.interpolate.UnivariateSpline """
from __future__ import division
import numpy as np
from scipy.interpolate import UnivariateSpline
import pylab as pl
N = 10
H = 8
x = np.arange(N+1)
xup = np.arange( 0, N, 1/H )
y = np.zeros(N+1); y[N//2] = 100
interpolator = UnivariateSpline( x, y, k=3, s=0 ) # s=0 interpolates
yup = interpolator( xup )
np.set_printoptions( 1, threshold=100, suppress=True ) # .1f
print "yup:", yup
pl.plot( x, y, "green", xup, yup, "blue" )
pl.show()
Added feb 2010: see also basic-spline-interpolation-in-a-few-lines-of-numpy
Small improvement. Use the built-in numpy.sinc(x) function which runs in compiled C code.
Possible larger improvement: Can you do the interpolation on the fly (as the plotting occurs)? Or are you tied to a plotting library that only accepts a matrix?
I recommend that you check your algorithm, as it is a non-trivial problem. Specifically, I suggest you gain access to the article "Function Plotting Using Conic Splines" (IEEE Computer Graphics and Applications) by Hu and Pavlidis (1991). Their algorithm implementation allows for adaptive sampling of the function, such that the rendering time is smaller than with regularly spaced approaches.
The abstract follows:
A method is presented whereby, given a
mathematical description of a
function, a conic spline approximating
the plot of the function is produced.
Conic arcs were selected as the
primitive curves because there are
simple incremental plotting algorithms
for conics already included in some
device drivers, and there are simple
algorithms for local approximations by
conics. A split-and-merge algorithm
for choosing the knots adaptively,
according to shape analysis of the
original function based on its
first-order derivatives, is
introduced.