o "`O@sddlmZddlmZzejZWn eyejZYnweddZddZ ed dd Z d d Z ed!d dZ ddZ eddZifddZeddZed"ddZed ddZeddZeddZd S)#)xrange)defuncCsTt|}|j}d|d@}t|dD]}||||7}||||d}q|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 A/usr/lib/python3/dist-packages/mpmath/calculus/differentiation.py difference s  rc s |d}|dd}|dd}|d||d} |j} zm| |_|ddurF|d r8t|} nd} |d| || n||dd}|rc||9t|d} } n t| |dd} d} |rxd 7fd d | D}|| | fW| |_S| |_w) Nsingularaddprec directionrrhZrelativeg?csg|] }|qSr r ).0r frxr r =zhsteps..)getprecrZmagZldexpconvertsignr)rrrr roptionsrrrworkprecZorigZ hextramagZstepsnormvaluesr rrhstepss4          r$c spd}z t}td}Wn tyYnw|r+fddDt||S|dd}dkrE|dkrE|d sESj}zk|dkrjt|fi|\} } } | _| | } n<|dkrjd 7_|d d fd d} | ddj g}| dj } n t d|W|_| SW|_| S|_w)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&|}|}||SN)Zexpj)tZreizrrr r*rr rgszdiff..grzunknown method: %r) list TypeError _partial_diffrrrr$rZquadtsZpi factorial ValueError)rrrr r partialordersr'rr#r"r!vr/r r r.rdiffCsDi    r8csp|sSt|s |Sdtt|D]|rnq|fdd}d|<t|||S)Nrcs*fdd}j|fiS)Ncs&d|fddSNrr r,)rf_argsir rinners&z1_partial_diff..fdiff_inner..innerr8)r;r=rrr<r Zorder)r;r fdiff_innersz"_partial_diff..fdiff_inner)sumrangelenr2)rrZxsr6r r@r r?rr2sr2Nc ksh|dur |j}nt|}|dddkr5d}||dkr3|j|||fi|V|d7}||dksdS|d}|rG|j||dddVn|||V|dkrUdS||jkr_d \}}nd|d}} |j} t||||| fi|\} } } t||D]$}z| |_|| || |} W| |_n| |_w| V||krdSq~|t|d d}}t ||}qg) ad Returns a generator that yields the sequence of derivatives .. math :: f(x), f'(x), f''(x), \ldots, f^{(k)}(x), \ldots With ``method='step'``, :func:`~mpmath.diffs` uses only `O(k)` function evaluations to generate the first `k` derivatives, rather than the roughly `O(k^2)` evaluations required if one calls :func:`~mpmath.diff` `k` separate times. With `n < \infty`, the generator stops as soon as the `n`-th derivative has been generated. If the exact number of needed derivatives is known in advance, this is further slightly more efficient. Options are the same as for :func:`~mpmath.diff`. **Examples** >>> from mpmath import * >>> mp.dps = 15 >>> nprint(list(diffs(cos, 1, 5))) [0.540302, -0.841471, -0.540302, 0.841471, 0.540302, -0.841471] >>> for i, d in zip(range(6), diffs(cos, 1)): ... print("%s %s" % (i, d)) ... 0 0.54030230586814 1 -0.841470984807897 2 -0.54030230586814 3 0.841470984807897 4 0.54030230586814 5 -0.841470984807897 Nr'r(rrrT)r)rrgffffff?) infrrr8rrr$rrmin)rrrr r r rABZcallprecyr"r!r r r rdiffssF&      rIcstgfdd}|S)Ncs.tt|dD] }tq |Sr9)rrCappendnext)r r<datagenr rr,sziterable_to_function..f)iter)rNrr rLriterable_to_function)srPc cst|}|dkr|dD]}|Vq dSt||d|d}t|||dd}d} |||d}d}td|dD]} ||| d| }||||| || 7}qC|V|d7}q2)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)rCrP diffs_prodr) rZfactorsNcur7r rar r r rrQ2s$$ rQc Cs4||vr||S|sddi|d<t|d}tddt|D}i}t|D]"\}}|ddf|dd}||vrE|||7<q'|||<q't|D]C\}}t|sWqNt|D]5\}}|r|d||d||ddf||dd} | |vr|| ||7<q[|||| <q[qN|||<||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)rVNr )rrSr7r r r tszdpoly..Nr)dpolydict iteritemsrA enumerate) r _cacheRZRapowerscountZpowers1r pZpowers2r r rrXhs2   4 rXc#s|t||d}|Vd} |d}tt|D]\}}|||fddt|D7}q||V|d7}q)a Given an iterable or generator yielding `f(x), f'(x), f''(x), \ldots` generate `g(x), g'(x), g''(x), \ldots` where `g(x) = \exp(f(x))`. At high precision and for large orders, this is typically more efficient than numerical differentiation if the derivatives of `f(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** The derivatives of the gamma function can be computed using logarithmic differentiation:: >>> 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 r`fnr rrWs&zdiffs_exp..)rPZexpZmpfrZrXZfprodr[)rZfdiffsZf0r<rr^rSr rar diffs_exps, & rcrcsXtt|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||Sr+r r:)rrrr rsz-differint....)r)rrrrdx0rfrrerzdifferint..)maxrZceilrer8Zgamma)rrrr rhmr/r rgr differintsD rlc s"dkrSfdd}|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|fiSr+r>rfrrr r r rr/szdiffun..gr )rrr r r/r rmrdiffun srnc sJtj|||fi|}|ddrfdd|DSfdd|DS)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$g|]\}}||qSr )ror3rr<r r&r rr@s$ztaylor..csg|] \}}||qSr )r3rpr&r rrBs)r[rIr)rrrr r rNr r&rtaylor"s rqc CsFt|||dkrtd|dkr*|dkr|jg|jgfS|d|d|jgfS||}t|D]}tt|||dD]}|||||||f<q@q3|||d||d }|||}|jgt|} dg|d} t|dD]#}||} tdt||dD]}| | ||||7} q| | |<q{| | 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)rCr4ZoneZmatrixrBrEZlu_solver0) rrULMrFjr<r7rqr`rr r rpadeDs*(     rv)rr+)rr)Z libmp.backendrZcalculusrrYrZAttributeErroritemsrr$r8r2rIrPrQrXrcrlrnrqrvr r r rs<     $   I  5! 6 H  !