[hO$SSKJr SSKJr \R r\S5rSr \SSj5r Sr \SS j5r S r \S 5r04S jr\S 5r\SSj5r\SSj5r\S5r\S5rg!\a \RrNtf=f))xrange)defunc[U5nURnSUS--n[US-5HnX4X-- nXEU- -US--nM U$)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)intzeror)ctxsndbks W/var/www/auris/envauris/lib/python3.13/site-packages/mpmath/calculus/differentiation.py differencer s] AA A QA AaC[ X  A#YAaC  Hc URS5nURSS5nURSS5nUSU--US--n URn XlURS5n U cMURS 5(a[URU55n OSn UR SU*U- U - 5n OUR U 5n URSS5nU(a$XR U5-n [US-5n U nO[U*US-S5n SU -nU(aUS U -- nU Vs/sHo"X/U --5PM nnUX4Xl$s snf!Xlf=f) Nsingularaddprec directionrrhrelativeg?)getprecrmagldexpconvertsignr)r fxr roptionsrrrworkprecorigr hextramagstepsnormrvaluess rhstepsr*sW{{:&Hkk)R(G K+IQwY1Q3'H 88D KK  9{{:&& O   !dU7]945A AAKK Q/  )$ $A1Q3KEDA2qsA&EaCD  QJA$)*Eq!AcE(E*t%+sCE/EEEEc `^^^^^Sn[T5n[T5mSnU(a1TVs/sHnTRU5PM snm[TTTWU5$UR SS5nTS:Xa3US:wa-UR S5(dT"TRT55$TR n US:Xa3[ TTTTU 40UD6upn U TlTRU T5U T-- n OUS:XaT=R S- slTRUR S S 55mUUUUU4S jnTRUSS TR-/5nUTRT5-S TR-- n O[S U-5eU TlU 7$![a GNzf=fs snf!U Tlf=f)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 FTmethodsteprquadrrradiusg?cR>TTRU5-nTU-nT"U5UT-- $N)expj)treizr r!r r/r"s rgdiff..gs0SXXa[(Gtc1f}$rrzunknown method: %r) list TypeErrorr _partial_diffrrr*rquadtspi factorial ValueError)r r!r"r r#partialorders_r,rr)r(r$vr6r r/s```` @rdiffrCCsRG a G%& 'QS[[^Q 'S!Q88 [[6 *FAv&F"7;;z+B+BQ  88D V %+CAq$%J'%J "F(CHvq)D!G3A v  HHNH[[Xt!<=F % % 1q!CFF(m,ACMM!$$#&&1A1F:; ; 2I7     (.s#FF1CF$ FF$ F-c^^^^^U(dT"5$[U5(dT"U6$Sm[[U55HmUT(dM O UTmUUUUU4SjnSUT'[TXRUT5$)NrcF>^UUU4SjnTR"UTTT40TD6$)Nc.>T"TSTU4-TTS-S-6$Nr)r3r!f_argsis rinner1_partial_diff..fdiff_inner..inners*vbqzQD(6!A#$<79 9rrC)rIrKr r!rJr#orders` r fdiff_inner"_partial_diff..fdiff_inners$ :xxvay%;7;;r)sumrangelenr:)r r!xsr@r#rOrJrNs`` ` @@rr:r:st s v;;"v  A 3v;  !99   1IE<<F1I kvw ??rNc+# Uc URnO [U5nURSS5S:wa0SnXSS-:a%UR"XU40UD6v US- nXSS-:aM%gURS5nU(aURXSSS9v OU"UR U55v US:agX0R:XaS upxOSUS-pUR n [ XX(U 40UD6upn [Xx5H2nXlURX5X-- n XlU 7v XS:dM2 g U[US -S-5p[X5nM}!Xlf=f7f) a 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?) infrrrCrrr*rrmin)r r!r"r r#rrABcallprecyr(r$r s rdiffsr\s_L y GG F{{8V$. a%i((1.g. . FAa%i {{:&HhhqQh.. A1uGG|1!A#1 88"31EWEA $#NN1(472#"Hv#aeAg,1 I $s+AE#BE7EE%#EEEc2^^[T5m/mUU4SjnU$)Nc>[[T5US-5HnTR[T55 M TU$rG)rrSappendnext)rrJdatagens rr!iterable_to_function..f,s5D 1Q3'A KKS "(Awr)iter)rbr!ras` @riterable_to_functionre)s s)C D Hrc## [U5nUS:XaUSHnUv M g[URUSUS-55n[URXS-S55nSnU"U5U"S5-nSn[SUS-5H(n XU - S--U -nXxU"Xi- 5-U"U 5-- nM* Uv US- nMY7f)a 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)rSre diffs_prodr) r factorsNcurBr r ars rrgrg2sH G AAvAG !A!? @ 1!? @ !qt AAAac]1QK1$13Z!A$&&#G FAsB>CcX;aX$U(dSS0US'[US- 5n[S[U555n0n[U5H+upEUSS-4USS-nXc;aX6==U- ss'M'XSU'M- [U5HlupE[U5(dM[ U5HFupxU(dMUSUUS- XGS-S-4-XGS-S-n X;aX9==X-- ss'M@X-X9'MH Mn X1U'X$)z nth differentiation polynomial for exp (Faa di Bruno's formula). TODO: most exponents are zero, so maybe a sparse representation would be better. rrrc34# UHupUS-U4v M g7f)rnNrH).0rjrBs r dpoly..ts 2\EQafQZ\sNr)dpolydict iteritemsrQ enumerate) r _cacheRRapowerscountpowers1rppowers2s rrsrshs {y !Hq  ac A 2Yq\ 22A B"1 !9Q;.6!":- = K5 KwK & #1 6{{ V$CAq !*!FQ3KM'::VaCD\I=K17*K"#'BK %&1I 9rc #&^# [U5mURT"S55nUv SnURS5n[[ U55H.upVXFUR U4Sj[ U555-- nM0 XB-v US- nMd7f)aD 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 rrc3V># UHupU(dMT"US-5U-v M g7f)rNrH)rprr}fns rrqdiffs_exp..s%L5FEQ!ZR!WaZ5Fs )))reexpmpfrursfprodrv)r fdiffsf0rJr rzrjrs @r diffs_exprsX f %B AB H A GGAJ"58,IF 399LYv5FLLL LA-f  Q sBBc ^^^^[[TRTRU555S-S5nXS- S- mUUUU4SjnTR XbU5TR XS- 5- $)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) rc:>^TRUUU4SjTT/5$)Nc&>TU- T-T"U5-$r1rH)r3r!rr"s r-differint....sacAX!_r)r.)r"r r!rx0s`rrdifferint..s#((4r1g>r)maxrceilrerCgamma)r r!r"r rmr6rs`` ` @r differintrs]H C# $Q &*A AA>A 88A! syy~ --rc 0^^^^TS:XaT$UUUU4SjnU$)a Given a function `f`, returns a function `g(x)` that evaluates the nth derivative `f^{(n)}(x)`:: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> cos2 = diffun(sin) >>> sin2 = diffun(sin, 4) >>> cos(1.3), cos2(1.3) (0.267498828624587, 0.267498828624587) >>> sin(1.3), sin2(1.3) (0.963558185417193, 0.963558185417193) The function `f` must support arbitrary precision evaluation. See :func:`~mpmath.diff` for additional details and supported keyword options. rc.>TR"TUT40TD6$r1rM)r"r r!r r#s rr6diffun..gsxx1a+7++rrH)r r!r r#r6s```` rdiffunr s & Av,, Hrc B[UR"XU40UD65nURSS5(a8UVVs/sH(upgURU5UR U5- PM* snn$UVVs/sHupgXpR U5- PM snn$s snnfs snnf)aW 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 chopT)rvr\rrr=)r r!r"r r#rbrJr s rtaylorr"s8 CIIaA11 2C{{64  9<= CMM!,,==/23stq--""s33>3s /B2Bc[U5X#-S-:a [S5eUS:Xa5US:XaUR/UR/4$USUS-UR/4$URU5n[ U5H2n[ [ X2U-S-55HnXU-U- XEU4'M M4 URXS-X#-S-5*nUR XG5nUR/[U5-n S/US--n [ US-5H9nXn [ S[ X65S-5HnXUXU- -- n M XU'M; X4$)ax Computes a Pade approximation of degree `(L, M)` to a function. Given at least `L+M+1` Taylor coefficients `a` approximating a function `A(x)`, :func:`~mpmath.pade` returns coefficients of polynomials `P, Q` satisfying .. math :: P = \sum_{k=0}^L p_k x^k Q = \sum_{k=0}^M q_k x^k Q_0 = 1 A(x) Q(x) = P(x) + O(x^{L+M+1}) `P(x)/Q(x)` can provide a good approximation to an analytic function beyond the radius of convergence of its Taylor series (example from G.A. Baker 'Essentials of Pade Approximants' Academic Press, Ch.1A):: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> one = mpf(1) >>> def f(x): ... return sqrt((one + 2*x)/(one + x)) ... >>> a = taylor(f, 0, 6) >>> p, q = pade(a, 3, 3) >>> x = 10 >>> polyval(p[::-1], x)/polyval(q[::-1], x) 1.38169105566806 >>> f(x) 1.38169855941551 rz%L+M+1 Coefficients should be providedrN)rSr>onematrixrRrWlu_solver8) r rlLMrXjrJrBr"qr}r s rpaderDsZP 1vA~@AAAv 6GG9swwi' 'Tac7SWWI% % 1 A 1Xs1c!e}%A!AhAdG& AsQSU$ %%A QA  DGA QqS A 1Q3Z Dq#a(Q,'A 1a!f A(!  4Kr)rr1)rr) libmp.backendrcalculusrrtruAttributeErroritemsrr*rCr:r\rergrsrrrrrrHrrrs"I  "!HHHT@"GGR 33jB44lF.F.P  044BBB Is A::BB