B <ZO@sddlmZddlmZy ejZWnek r<ejZYnXeddZddZ ed dd Z d d Z ed!d dZ ddZ eddZifddZeddZed"ddZed#ddZeddZeddZd S)$)xrange)defuncCsXt|}|j}d|d@}x8t|dD](}||||7}||||d}q(W|S)z Given a sequence `(s_k)` containing at least `n+1` items, returns the `n`-th forward difference, .. math :: \Delta^n = \sum_{k=0}^{\infty} (-1)^{k+n} {n \choose k} s_k. r)intZzeror)ctxsndbkr >lib/python3.7/site-packages/mpmath/calculus/differentiation.py difference s  rc s|d}|dd}|dd}|d||d} |j} z| |_|ddkr|d rpt|} nd} |d| || n ||dd}|rƈ||9t|d} } nt| |dd} d} |rd 7fd d | D}|| | fS| |_XdS) Nsingularaddprec directionrrhZrelativeg?csg|]}|qSr r ).0r )frxr r =szhsteps..)getprecrZmagZldexpconvertZsignr)rrrr roptionsrrrworkprecZorigZ hextramagZstepsnormvaluesr )rrrrhstepss4          r!c s^d}yt}td}Wntk r0YnX|rXfddDt||S|dd}dkr|dkr|d sSj}z|dkrt|f|\} } } | _| | } nz|dkr@jd 7_|d d fd d} | ddj g}| dj } n t d|Wd|_X| S)a Numerically computes the derivative of `f`, `f'(x)`, or generally for an integer `n \ge 0`, the `n`-th derivative `f^{(n)}(x)`. A few basic examples are:: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> diff(lambda x: x**2 + x, 1.0) 3.0 >>> diff(lambda x: x**2 + x, 1.0, 2) 2.0 >>> diff(lambda x: x**2 + x, 1.0, 3) 0.0 >>> nprint([diff(exp, 3, n) for n in range(5)]) # exp'(x) = exp(x) [20.0855, 20.0855, 20.0855, 20.0855, 20.0855] Even more generally, given a tuple of arguments `(x_1, \ldots, x_k)` and order `(n_1, \ldots, n_k)`, the partial derivative `f^{(n_1,\ldots,n_k)}(x_1,\ldots,x_k)` is evaluated. For example:: >>> diff(lambda x,y: 3*x*y + 2*y - x, (0.25, 0.5), (0,1)) 2.75 >>> diff(lambda x,y: 3*x*y + 2*y - x, (0.25, 0.5), (1,1)) 3.0 **Options** The following optional keyword arguments are recognized: ``method`` Supported methods are ``'step'`` or ``'quad'``: derivatives may be computed using either a finite difference with a small step size `h` (default), or numerical quadrature. ``direction`` Direction of finite difference: can be -1 for a left difference, 0 for a central difference (default), or +1 for a right difference; more generally can be any complex number. ``addprec`` Extra precision for `h` used to account for the function's sensitivity to perturbations (default = 10). ``relative`` Choose `h` relative to the magnitude of `x`, rather than an absolute value; useful for large or tiny `x` (default = False). ``h`` As an alternative to ``addprec`` and ``relative``, manually select the step size `h`. ``singular`` If True, evaluation exactly at the point `x` is avoided; this is useful for differentiating functions with removable singularities. Default = False. ``radius`` Radius of integration contour (with ``method = 'quad'``). Default = 0.25. A larger radius typically is faster and more accurate, but it must be chosen so that `f` has no singularities within the radius from the evaluation point. A finite difference requires `n+1` function evaluations and must be performed at `(n+1)` times the target precision. Accordingly, `f` must support fast evaluation at high precision. With integration, a larger number of function evaluations is required, but not much extra precision is required. For high order derivatives, this method may thus be faster if f is very expensive to evaluate at high precision. **Further examples** The direction option is useful for computing left- or right-sided derivatives of nonsmooth functions:: >>> diff(abs, 0, direction=0) 0.0 >>> diff(abs, 0, direction=1) 1.0 >>> diff(abs, 0, direction=-1) -1.0 More generally, if the direction is nonzero, a right difference is computed where the step size is multiplied by sign(direction). For example, with direction=+j, the derivative from the positive imaginary direction will be computed:: >>> diff(abs, 0, direction=j) (0.0 - 1.0j) With integration, the result may have a small imaginary part even even if the result is purely real:: >>> diff(sqrt, 1, method='quad') # doctest:+ELLIPSIS (0.5 - 4.59...e-26j) >>> chop(_) 0.5 Adding precision to obtain an accurate value:: >>> diff(cos, 1e-30) 0.0 >>> diff(cos, 1e-30, h=0.0001) -9.99999998328279e-31 >>> diff(cos, 1e-30, addprec=100) -1.0e-30 FTcsg|]}|qSr )r)r_)rr rrszdiff..methodsteprquadrrradiusg?cs&|}|}||S)N)Zexpj)tZreiz)rrr r&rr rgszdiff..grzunknown method: %rN) list TypeError _partial_diffrrrr!rZquadtsZpi factorial ValueError)rrrr rpartialordersr#rr rrvr)r r )rrr r&rrdiffCs8i  r2csr|s St|s|Sdxtt|D]|r,Pq,W|fdd}d|<t|||S)Nrcs&fdd}j|fS)Ncs&d|fddS)Nrr )r')rf_argsir rinnersz1_partial_diff..fdiff_inner..inner)r2)r3r5)rrr4rorder)r3r fdiff_innersz"_partial_diff..fdiff_inner)sumrangelenr,)rrZxsr0rr7r )rrr4rr6rr,sr,Nc ksZ|dkr|j}nt|}|dddkr^d}x,||dkrX|j|||f|V|d7}q.WdS|d}|r|j||dddVn|||V|dkrdS||jkrd \}}nd|d}}x|j} t||||| f|\} } } xNt||D]@}z| |_|| || |} Wd| |_X| V||krdSqW|t|d d}}t ||}qWdS) ad Returns a generator that yields the sequence of derivatives .. math :: f(x), f'(x), f''(x), \ldots, f^{(k)}(x), \ldots With ``method='step'``, :func:`~mpmath.diffs` uses only `O(k)` function evaluations to generate the first `k` derivatives, rather than the roughly `O(k^2)` evaluations required if one calls :func:`~mpmath.diff` `k` separate times. With `n < \infty`, the generator stops as soon as the `n`-th derivative has been generated. If the exact number of needed derivatives is known in advance, this is further slightly more efficient. Options are the same as for :func:`~mpmath.diff`. **Examples** >>> from mpmath import * >>> mp.dps = 15 >>> nprint(list(diffs(cos, 1, 5))) [0.540302, -0.841471, -0.540302, 0.841471, 0.540302, -0.841471] >>> for i, d in zip(range(6), diffs(cos, 1)): ... print("%s %s" % (i, d)) ... 0 0.54030230586814 1 -0.841470984807897 2 -0.54030230586814 3 0.841470984807897 4 0.54030230586814 5 -0.841470984807897 Nr#r$rrrT)r)rrgffffff?) infrrr2rrr!rrmin)rrrr rr rABZcallprecyrrr r r rdiffss>&    r@cstgfdd}|S)Ncs2x(tt|dD]}tqW|S)Nr)rr:appendnext)r r4)datagenr rr,sziterable_to_function..f)iter)rDrr )rCrDriterable_to_function)srFc cst|}|dkr*x|dD] }|VqWnt||d|d}t|||dd}d}xn|||d}d}xFtd|dD]4} ||| d| }||||| || 7}qW|V|d7}qdWdS)aV Given a list of `N` iterables or generators yielding `f_k(x), f'_k(x), f''_k(x), \ldots` for `k = 1, \ldots, N`, generate `g(x), g'(x), g''(x), \ldots` where `g(x) = f_1(x) f_2(x) \cdots f_N(x)`. At high precision and for large orders, this is typically more efficient than numerical differentiation if the derivatives of each `f_k(x)` admit direct computation. Note: This function does not increase the working precision internally, so guard digits may have to be added externally for full accuracy. **Examples** >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> f = lambda x: exp(x)*cos(x)*sin(x) >>> u = diffs(f, 1) >>> v = mp.diffs_prod([diffs(exp,1), diffs(cos,1), diffs(sin,1)]) >>> next(u); next(v) 1.23586333600241 1.23586333600241 >>> next(u); next(v) 0.104658952245596 0.104658952245596 >>> next(u); next(v) -5.96999877552086 -5.96999877552086 >>> next(u); next(v) -12.4632923122697 -12.4632923122697 rrNr)r:rF diffs_prodr) rZfactorsNcur1r rar r r rrG2s$  rGc CsB||kr||S|s ddi|d<t|d}tddt|D}i}xPt|D]D\}}|ddf|dd}||kr|||7<qP|||<qPWxt|D]\}}t|sqxxt|D]l\}}|r|d||d||ddf||dd} | |kr|| ||7<q|||| <qWqW|||<||S)z nth differentiation polynomial for exp (Faa di Bruno's formula). TODO: most exponents are zero, so maybe a sparse representation would be better. )rrrcss|]\}}|d|fVqdS))rNr )rrIr1r r r tszdpoly..Nr)dpolydict iteritemsr8 enumerate) r _cacheRZRapowerscountZpowers1r pZpowers2r r rrMhs.   4 rMc#st||d}|Vd}x^|d}x>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> >>> def diffs_loggamma(x): ... yield loggamma(x) ... i = 0 ... while 1: ... yield psi(i,x) ... i += 1 ... >>> u = diffs_exp(diffs_loggamma(3)) >>> v = diffs(gamma, 3) >>> next(u); next(v) 2.0 2.0 >>> next(u); next(v) 1.84556867019693 1.84556867019693 >>> next(u); next(v) 2.49292999190269 2.49292999190269 >>> next(u); next(v) 3.44996501352367 3.44996501352367 rrc3s&|]\}}|r|d|VqdS)rNr )rr rU)fnr rrLszdiffs_exp..N)rFZexpZmpfrOrMZfprodrP)rZfdiffsZf0r4rrSrIr )rVr diffs_exps, ( rWrcsXtt|dd}||dfdd}|||||S)a Calculates the Riemann-Liouville differintegral, or fractional derivative, defined by .. math :: \,_{x_0}{\mathbb{D}}^n_xf(x) = \frac{1}{\Gamma(m-n)} \frac{d^m}{dx^m} \int_{x_0}^{x}(x-t)^{m-n-1}f(t)dt where `f` is a given (presumably well-behaved) function, `x` is the evaluation point, `n` is the order, and `x_0` is the reference point of integration (`m` is an arbitrary parameter selected automatically). With `n = 1`, this is just the standard derivative `f'(x)`; with `n = 2`, the second derivative `f''(x)`, etc. With `n = -1`, it gives `\int_{x_0}^x f(t) dt`, with `n = -2` it gives `\int_{x_0}^x \left( \int_{x_0}^t f(u) du \right) dt`, etc. As `n` is permitted to be any number, this operator generalizes iterated differentiation and iterated integration to a single operator with a continuous order parameter. **Examples** There is an exact formula for the fractional derivative of a monomial `x^p`, which may be used as a reference. For example, the following gives a half-derivative (order 0.5):: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> x = mpf(3); p = 2; n = 0.5 >>> differint(lambda t: t**p, x, n) 7.81764019044672 >>> gamma(p+1)/gamma(p-n+1) * x**(p-n) 7.81764019044672 Another useful test function is the exponential function, whose integration / differentiation formula easy generalizes to arbitrary order. Here we first compute a third derivative, and then a triply nested integral. (The reference point `x_0` is set to `-\infty` to avoid nonzero endpoint terms.):: >>> differint(lambda x: exp(pi*x), -1.5, 3) 0.278538406900792 >>> exp(pi*-1.5) * pi**3 0.278538406900792 >>> differint(lambda x: exp(pi*x), 3.5, -3, -inf) 1922.50563031149 >>> exp(pi*3.5) / pi**3 1922.50563031149 However, for noninteger `n`, the differentiation formula for the exponential function must be modified to give the same result as the Riemann-Liouville differintegral:: >>> x = mpf(3.5) >>> c = pi >>> n = 1+2*j >>> differint(lambda x: exp(c*x), x, n) (-123295.005390743 + 140955.117867654j) >>> x**(-n) * exp(c)**x * (x*c)**n * gammainc(-n, 0, x*c) / gamma(-n) (-123295.005390743 + 140955.117867654j) rcsfddgS)Ncs||S)Nr )r')rrrr rsz-differint....)r%)r)rrrXx0)rrrYszdifferint..)maxrZceilrer2Zgamma)rrrr rZmr)r )rrrXrZr differintsD r^c s"dkr Sfdd}|S)a3 Given a function `f`, returns a function `g(x)` that evaluates the nth derivative `f^{(n)}(x)`:: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> cos2 = diffun(sin) >>> sin2 = diffun(sin, 4) >>> cos(1.3), cos2(1.3) (0.267498828624587, 0.267498828624587) >>> sin(1.3), sin2(1.3) (0.963558185417193, 0.963558185417193) The function `f` must support arbitrary precision evaluation. See :func:`~mpmath.diff` for additional details and supported keyword options. rcsj|fS)N)r2)r)rrr rr rr)szdiffun..gr )rrr rr)r )rrr rrdiffun sr_c sJtj|||f|}|ddr4fdd|DSfdd|DSdS)a Produces a degree-`n` Taylor polynomial around the point `x` of the given function `f`. The coefficients are returned as a list. >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> nprint(chop(taylor(sin, 0, 5))) [0.0, 1.0, 0.0, -0.166667, 0.0, 0.00833333] The coefficients are computed using high-order numerical differentiation. The function must be possible to evaluate to arbitrary precision. See :func:`~mpmath.diff` for additional details and supported keyword options. Note that to evaluate the Taylor polynomial as an approximation of `f`, e.g. with :func:`~mpmath.polyval`, the coefficients must be reversed, and the point of the Taylor expansion must be subtracted from the argument: >>> p = taylor(exp, 2.0, 10) >>> polyval(p[::-1], 2.5 - 2.0) 12.1824939606092 >>> exp(2.5) 12.1824939607035 chopTcs$g|]\}}||qSr )r`r-)rr4r )rr rr@sztaylor..csg|]\}}||qSr )r-)rr4r )rr rrBsN)rPr@r)rrrr rrDr )rrtaylor"s rac CsZt|||dkrtd|dkrT|dkr<|jg|jgfS|d|d|jgfS||}xHt|D]<}x6tt|||dD]}|||||||f<qWqhW|||d||d }|||}|jgt|} dg|d} x^t|dD]N}||} x6tdt||dD]}| | ||||7} q"W| | |<qW| | fS)a Computes a Pade approximation of degree `(L, M)` to a function. Given at least `L+M+1` Taylor coefficients `a` approximating a function `A(x)`, :func:`~mpmath.pade` returns coefficients of polynomials `P, Q` satisfying .. math :: P = \sum_{k=0}^L p_k x^k Q = \sum_{k=0}^M q_k x^k Q_0 = 1 A(x) Q(x) = P(x) + O(x^{L+M+1}) `P(x)/Q(x)` can provide a good approximation to an analytic function beyond the radius of convergence of its Taylor series (example from G.A. Baker 'Essentials of Pade Approximants' Academic Press, Ch.1A):: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> one = mpf(1) >>> def f(x): ... return sqrt((one + 2*x)/(one + x)) ... >>> a = taylor(f, 0, 6) >>> p, q = pade(a, 3, 3) >>> x = 10 >>> polyval(p[::-1], x)/polyval(q[::-1], x) 1.38169105566806 >>> f(x) 1.38169855941551 rz%L+M+1 Coefficients should be providedrN)r:r.ZoneZmatrixr9r<Zlu_solver*) rrKLMr=jr4r1rqrUrr r rpadeDs((    rf)r)N)rr)r)Z libmp.backendrZcalculusrrNrOAttributeErroritemsrr!r2r,r@rFrGrMrWr^r_rarfr r r rs.     $   I 6 ! 7 H  "