Taylor's series

A well-behaved function can be expanded into a power series. This means that for all non-negative integers $k$ there are real numbers $a_k$ such that

$f(x)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\cdots$

Let us calculate the first four derivatives using $(x^n{)}^{\prime }=n{x}^{n-1}$:

$f^{\prime }(x)=a_1+2a_{2}x+3a_3{x}^2+4a_4{x}^3+5a_5{x}^4+\cdots$

$f^{″}(x)=2a_2+2\cdot 3a_{3}x+3\cdot 4a_4{x}^2+4\cdot 5a_5{x}^3+\cdots$

$f^{‴}(x)=2\cdot 3a_3+2\cdot 3\cdot 4a_{4}x+3\cdot 4\cdot 5a_5{x}^2+\cdots$

$f^{⁗}(x)=2\cdot 3\cdot 4a_4+2\cdot 3\cdot 4\cdot 5a_{5}x+\cdots$

Setting $x$ equal to zero, we obtain

$f(0)=a_0,f^{\prime }(0)=a_1,f^{″}(0)=2a_2,f^{‴}(0)=2\times 3a_3,f^{⁗}(0)=2\times 3\times 4a_4.$

Let us write $f^{(n)}(x)$ for the $n$-th derivative of $f(x).$ We also write $f^{(0)}(x)=f(x)$ — think of $f(x)$ as the "zeroth derivative" of $f(x).$ We thus arrive at the general result $f^{(k)}(0)=k!a_k,$ where the factorial $k!$ is defined as equal to 1 for $k=0$ and $k=1$ and as the product of all natural numbers $n\le k$ for $k>1.$ Expressing the coefficients $a_k$ in terms of the derivatives of $f(x)$ at $x=0,$ we obtain

$f(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(0)}{k!}{x}^k=f(0)+f^{\prime }(0)x+f^{″}(0)\frac{x^2}{2!}+f^{‴}(0)\frac{x^3}{3!}+\cdots$

This is the Taylor series for $f(x).$

A remarkable result: if you know the value of a well-behaved function $f(x)$ and the values of all of its derivatives at the single point $x=0$ then you know $f(x)$ at all points $x.$ Besides, there is nothing special about $x=0,$ so $f(x)$ is also determined by its value and the values of its derivatives at any other point $x_0$:

$f(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k.$

Examples

cos(x)

$f(x)=\cos (x);f(0)=\cos (0)=1$

$f^{\prime }(x)=-\sin (x);f^{\prime }(0)=-\sin (0)=0$

$f^{″}(x)=-\cos (x);f^{″}(0)=-\cos (0)=-1$

$f^{‴}(x)=\sin (x);f^{‴}(0)=\sin (0)=0$

$f^{⁗}(x)=\cos (x);f^{⁗}(0)=\cos (0)=1$

$a_0=f(0)=1$

$a_1=f^{\prime }(0)=0$

$a_2=\frac{f^{″}(0)}{2}=\frac{-1}{2!}$

$a_3=\frac{f^{‴}(0)}{2*3}=\frac{0}{3!}=0$

$a_4=\frac{f^{⁗}(0)}{2*3*4}=\frac{1}{4!}$

$\cos (x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\cdots$ $=1+(0)x+\frac{-1}{2}{x}^2+0x+\frac{1}{4!}{x}^4+\cdots$ $=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots$

Template:RoundBoxTop Some basic checking:

$\cos (0)=1-\frac{0^2}{2!}+\frac{0^4}{4!}-\frac{0^6}{6!}+\frac{0^8}{8!}-\cdots =1$

$\cos (-x)=1-\frac{(-x{)}^2}{2!}+\frac{(-x{)}^4}{4!}-\frac{(-x{)}^6}{6!}+\frac{(-x{)}^8}{8!}-\cdots$ $=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots =\cos (x)$

$\frac{d(\cos (x))}{dx}=0-\frac{2x}{2!}+\frac{4x^3}{4!}-\frac{6x^5}{6!}+\frac{8x^7}{8!}\cdots$ $=-x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}\cdots$ $=-\sin (x)$

$\sin (x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$

$\frac{d(\sin (x))}{dx}=1-\frac{3x^2}{3!}+\frac{5x^4}{5!}-\frac{7x^6}{7!}+\cdots$ $=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$ $=\cos (x)$ Template:RoundBoxBottom

arctan(x)

$y=\arctan (x)$

$x=\tan (y)$

$\frac{dx}{dy}={\sec }^2(y)$

$y^{\prime }=\frac{dy}{dx}={\cos }^2(y)=\frac{1}{1+x^2}$. See $\cos (\theta )$.

Second derivative y″

Template:RoundBoxTop $(1+x^2)y^{\prime }=1$

$(1+x^2)y^{″}+y^{\prime }(2x)=0$

$y^{″}=\frac{-y^{\prime }(2x)}{1+x^2}$ $=\frac{-2x}{1+x^2}.y^{\prime }$ $=\frac{-2x}{1+x^2}.\frac{1}{1+x^2}$ $=\frac{-2x}{(1+x^2{)}^2}$ Template:RoundBoxBottom

Third derivative y¹¹¹

Template:RoundBoxTop $(1+2x^2+x^4)y^{″}=-2x$

$(1+2x^2+x^4)y^{‴}+y^{″}(4x+4x^3)=-2$

$y^{‴}=\frac{-2-y^{″}(4x+4x^3)}{1+2x^2+x^4}$ $=\frac{-2-y^{″}4x(1+x^2)}{1+2x^2+x^4}$ $=\frac{-2-\frac{-2x}{(1+x^2{)}^2}.4x(1+x^2)}{1+2x^2+x^4}$ $=\frac{-2-\frac{-2x}{1+x^2}.4x}{1+2x^2+x^4}$ $=\frac{-2+\frac{8x^2}{1+x^2}}{1+2x^2+x^4}$ $=\frac{\frac{-2(1+x^2)+8x^2}{1+x^2}}{1+2x^2+x^4}$ $=\frac{\frac{-2-2x^2+8x^2}{1+x^2}}{1+2x^2+x^4}$ $=\frac{\frac{-2+6x^2}{1+x^2}}{1+2x^2+x^4}$ $=\frac{-2+6x^2}{(1+2x^2+x^4)(1+x^2)}$ $=\frac{-2+6x^2}{(1+x^2{)}^3}$ Template:RoundBoxBottom

(continued)

If you continue to calculate derivatives, you will produce the following sequence:

$f(0)=0$

$f^{\prime }(0)=1$

$f^{″}(0)=0$

$f^{‴}(0)=-2=-(2!)$

$f^{⁗}(0)=0$

$f^v(0)=24=4!$

$f^{vi}(0)=0$

$f^{vii}(0)=-720=-(6!)$

$f^{viii}(0)=0$

$f^{viiii}(0)=40320=8!$

$f^x(0)=0$

$f^{xi}(0)=-3628800=-(10!)$

$f^{xii}(0)=0$

$f^{xiii}(0)=479001600=12!$

$\arctan (x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\cdots$

$=0+(1)x+(0)x^2+\frac{-(2!)}{3!}{x}^3+(0)x^4+\frac{4!}{5!}{x}^5+(0)x^6+\frac{-(6!)}{7!}{x}^7+(0)x^8+\frac{8!}{9!}{x}^9+(0)x^{10}+\frac{-(10!)}{11!}{x}^{11}+\cdots$

$=x-\frac{1}{3}{x}^3+\frac{1}{5}{x}^5-\frac{1}{7}{x}^7+\frac{1}{9}{x}^9-\frac{1}{11}{x}^{11}+\cdots$ Template:RoundBoxTop Some basic checking:

$\arctan (0)=0$

$\arctan (-x)=-\arctan (x).$

$\frac{d(\arctan (x))}{dx}$ $=1-\frac{3x^2}{3}+\frac{5x^4}{5}-\frac{7x^6}{7}+\frac{9x^8}{9}-\frac{11x^{10}}{11}+\cdots$ $=1-x^2+x^4-x^6+x^8-x^{10}+\cdots$

Also, $\frac{d(\arctan (x))}{dx}=\frac{1}{1+x^2}$

Show that $\frac{1}{1+x^2}$ $=1-x^2+x^4-x^6+x^8-x^{10}+\cdots$

or that $1=(1+x^2)(1-x^2+x^4-x^6+x^8-x^{10}+\cdots )$

$(1+x^2)(1-x^2+x^4-x^6+x^8-x^{10}+\cdots )$

$=1(1-x^2+x^4-x^6+x^8-x^{10}+\cdots )+x^2(1-x^2+x^4-x^6+x^8-x^{10}+\cdots )$

$=1-x^2+x^4-x^6+x^8-x^{10}+\cdots +x^2-x^4+x^6-x^8+x^{10}-x^{12}+\cdots$

$=1\cdots \pm x^{2n}$

If abs$(x)<1,\underset{n\to \mathrm{\infty }}{\lim }{x}^{2n}=0.$ Template:RoundBoxBottom

File:0504arctan(x).png

Figure 1: Graph of $t(x),$ Taylor series representing $\arctan (x)$ for $x$ close to $0.$

In the diagram to the right, $t(x)$ is the Taylor series representing $\arctan (x)$ for $x$ close to $0.$

In the box above the proof that $t(x)$ is an accurate representation of $\arctan (x)$ is valid for abs$(x)<1.$

When abs$(x)>1,$ the diagram vividly illustrates that the series rapidly diverges.

To be accurate, the line $y=\frac{\pi }{4}$ should be $y=\frac{\pi }{4}$ rad or $y={\frac{\pi }{4}}^c$ meaning $y=\frac{\pi }{4}$ radians. In theoretical work a value such as $\frac{\pi }{4}$ is understood to be $\frac{\pi }{4}$ radians or $45^{\circ }$ meaning $45$ degrees. Template:RoundBoxBottom

In practice

Template:RoundBoxTop The expansion of $\arctan (x)$ above is theoretically valid for $|x|<1.$ However, if $|x|$ is close to $1,$ the calculation of $\arctan (x)$ will take forever.

This section uses $x_0$ so that $|x-x_0|$ is small enough to make time of calculation acceptable.

Let $\theta =36{\frac{3}{4}}^{\circ }.$ To calculate $\tan (\theta ):$

$\tan \frac{11\pi }{60}=\tan 33^{\circ }=\frac{1}{4}[2-(2-\sqrt{3})(3+\sqrt{5})][2+\sqrt{2(5-\sqrt{5})}]$

$\tan (147^{\circ })=-\tan (33^{\circ })$

Using the half-angle formula $\tan (\frac{\theta }{2})=\csc (\theta )-\cot (\theta ),$

calculate $\tan (73\frac{1}{2})=\tan (\frac{147}{2})$ and $\tan (36\frac{3}{4})=\tan (\frac{73\frac{1}{2}}{2})$

$\theta =36{\frac{3}{4}}^{\circ }.x_0=\tan (\theta )=0.7467354177837216717375001402.$

$f(x_0)=\frac{36\frac{3}{4}\pi }{180}=0.6414085001079161195194563572.$

This value $(36{\frac{3}{4}}^{\circ })$ was chosen for $\theta$ because $x_0=\tan (\theta )$ is close to $0.75.$ For $1\ge x\ge 0.5,|x-x_0|\le 0.25$ approx.

$0.25^{50}=7.888609052210118e-31.$ If $|x-x_0|==0.25,$ the code below is accurate to $30$ places of decimals.

This section uses the whole sequence of derivatives:

$y000=\arctan (x)$ where $y000=y^{(0)}=y$

$y001=\frac{1}{U}$ where $U=1+x^2,y001=y^{(1)}=y^{\prime }$

$y002=-\frac{(2)*(x)*(y001)}{U}$ where $y002=y^{(2)}=y^{″}$

$y003=-\frac{(4)*(x)*(y002)+(2)*(y001)}{U}$ where $y003=y^{(3)}=y^{‴}$ and so on.

$y004=-\frac{(6)*(x)*(y003)+(6)*(y002)}{U}$

$y005=-\frac{(8)*(x)*(y004)+(12)*(y003)}{U}$

$y006=-\frac{(10)*(x)*(y005)+(20)*(y004)}{U}$

$y007=-\frac{(12)*(x)*(y006)+(30)*(y005)}{U}$

$y008=-\frac{(14)*(x)*(y007)+(42)*(y006)}{U}$

$y009=-\frac{(16)*(x)*(y008)+(56)*(y007)}{U}$

$y010=-\frac{(18)*(x)*(y009)+(72)*(y008)}{U}$

$\cdots$

$y050=-\frac{(98)*(x)*(y049)+(2352)*(y048)}{U}$

Using $f(x)=f(x_0)(x-x_0{)}^0+f^{\prime }(x_0)(x-x_0{)}^1+\frac{f^{″}(x_0)}{2}(x-x_0{)}^2+\frac{f^{‴}(x_0)}{2\cdot 3}(x-x_0{)}^3+\cdots$

then $f(x)=y000+y001(x-x_0)+\frac{y002}{2}(x-x_0{)}^2+\cdots$ and: Template:RoundBoxTop

File:0508arctan(x).png

Figure 1: Graph of $t(x),$ Taylor series representing $\arctan (x)$ for $x$ close to $0.75.$
$x_0=\tan (36{\frac{3}{4}}^{\circ })=0.7467354177837216717375001402,$ close to $0.75.$
$f(x_0)=\arctan (x_0)=0.6414085001079161195194563572$ rad.${\displaystyle }$

y = 0.6414085001079161195194563572
+(0.6420076723519613087221948458)(x-(0.7467354177837216717375001402))
+(-0.3077848130939266477182675970)(x-0.7467354177837216717375001402)^2
+(0.05934881813852229894809158807)(x-0.7467354177837216717375001402)^3
+(0.05612149216561873709345633871)(x-0.7467354177837216717375001402)^4
+(-0.0659097533448882821311588572)(x-0.7467354177837216717375001402)^5
+(0.02864269115336634046783964776)(x-0.7467354177837216717375001402)^6
+(0.006684824489389227404750195292)(x-0.7467354177837216717375001402)^7
+(-0.01939996954693863883077829889)(x-0.7467354177837216717375001402)^8
+(0.01319629210955273736079467214)(x-0.7467354177837216717375001402)^9
+(-0.001423635337528918097834676738)(x-0.7467354177837216717375001402)^10
+(-0.005690817314508170664127596721)(x-0.7467354177837216717375001402)^11
+(0.005763416294060825609852171147)(x-0.7467354177837216717375001402)^12
+(-0.002009530403041012685757678258)(x-0.7467354177837216717375001402)^13
+(-0.001382413103546475118963526286)(x-0.7467354177837216717375001402)^14
+(0.002355235379425975362106687309)(x-0.7467354177837216717375001402)^15
+(-0.001340525935442931206538139095)(x-0.7467354177837216717375001402)^16
+(-0.0001244720120416846251034920203)(x-0.7467354177837216717375001402)^17
+(0.0008777184853106580549638629701)(x-0.7467354177837216717375001402)^18
+(-0.0007257802485492202930793702930)(x-0.7467354177837216717375001402)^19
+(0.0001539460026510816727324277808)(x-0.7467354177837216717375001402)^20
+(0.0002810020934892180446689969911)(x-0.7467354177837216717375001402)^21
+(-0.0003470330774958963760466009045)(x-0.7467354177837216717375001402)^22
+(0.0001535570475871531716152621841)(x-0.7467354177837216717375001402)^23
+(0.00006313260945054237374661397478)(x-0.7467354177837216717375001402)^24
+(-0.0001488094986598041280906554962)(x-0.7467354177837216717375001402)^25
+(0.00009977993191704606200503722720)(x-0.7467354177837216717375001402)^26
+(-0.000003667561779685224841381874106)(x-0.7467354177837216717375001402)^27
+(-0.00005609286432922550484209657985)(x-0.7467354177837216717375001402)^28
+(0.00005412057738460028511566507574)(x-0.7467354177837216717375001402)^29
+(-0.00001655090242419904039018979491)(x-0.7467354177837216717375001402)^30
+(-0.00001714674178231985986067601799)(x-0.7467354177837216717375001402)^31
+(0.00002588855802866968641107644970)(x-0.7467354177837216717375001402)^32
+(-0.00001372909690493026279553133838)(x-0.7467354177837216717375001402)^33
+(-0.000002866406864208033772447118585)(x-0.7467354177837216717375001402)^34
+(0.00001098036048658105543109288040)(x-0.7467354177837216717375001402)^35
+(-0.000008497717911244361532280438636)(x-0.7467354177837216717375001402)^36
+(0.000001259146436001274560243296183)(x-0.7467354177837216717375001402)^37
+(0.000003992939704019955177003526706)(x-0.7467354177837216717375001402)^38
+(-0.000004497268683100848779169934291)(x-0.7467354177837216717375001402)^39
+(0.000001768945188244137235636524921)(x-0.7467354177837216717375001402)^40
+(0.000001091706749083768850937760502)(x-0.7467354177837216717375001402)^41
+(-0.000002103423912310375893571195410)(x-0.7467354177837216717375001402)^42
+(0.000001301617082996039555612971998)(x-0.7467354177837216717375001402)^43
+(6.937967909808721515382339295E-8)(x-0.7467354177837216717375001402)^44
+(-8.635525611989332402947366709E-7)(x-0.7467354177837216717375001402)^45
+(7.673857631132879175874596987E-7)(x-0.7467354177837216717375001402)^46
+(-1.893140553441536683377149770E-7)(x-0.7467354177837216717375001402)^47
+(-2.944033068289732156296704644E-7)(x-0.7467354177837216717375001402)^48
+(3.930991061512879804635643270E-7)(x-0.7467354177837216717375001402)^49
+(-1.879241426421737899718180888E-7)(x-0.7467354177837216717375001402)^50

A faster version

Template:RoundBoxTop The calculation of $\arctan (x)$ above is suitable as input to application grapher.

The following python code has precision set to $33.$ If it is desired to calculate $\arctan (x)$ for one value of $x,$ the following python code is much faster than the code supplied to grapher above.

python code

data = '''
0.641408500107916119519456357419567
0.642007672351961308722194845349589
-0.307784813093926647718267596858344
0.0593488181385222989480915882567083
0.0561214921656187370934563384765525
-0.0659097533448882821311588570897649
0.0286426911533663404678396477942956
0.00668482448938922740475019519860028
-0.0193999695469386388307782988276083
0.0131962921095527373607946721380589
-0.00142363533752891809783467677853379
-0.00569081731450817066412759667853352
0.00576341629406082560985217113222417
-0.00200953040304101268575767827219472
-0.00138241310354647511896352626325379
0.00235523537942597536210668729668636
-0.00134052593544293120653813909608214
-0.000124472012041684625103492011289663
0.000877718485310658054963862962235904
-0.000725780248549220293079370291315901
0.000153946002651081672732427784261987
0.000281002093489218044668996986822804
-0.000347033077495896376046600902527326
0.000153557047587153171615262184995174
0.0000631326094505423737466139726989411
-0.000148809498659804128090655494778209
0.0000997799319170460620050372271284211
-0.00000366756177968522484138187499300975
-0.0000560928643292255048420965789718678
0.0000541205773846002851156650754617952
-0.0000165509024241990403901897952115479
-0.0000171467417823198598606760175136922
0.0000258885580286696864110764494276036
-0.0000137290969049302627955313384274982
-0.00000286640686420803377244711837064614
0.0000109803604865810554310928802164877
-0.00000849771791124436153228043859514619
0.00000125914643600127456024329626150593
0.00000399293970401995517700352660353714
-0.00000449726868310084877916993424187367
0.00000176894518824413723563652493847337
0.00000109170674908376885093776045369610
-0.00000210342391231037589357119537437086
0.00000130161708299603955561297199526169
6.93796790980872151538233731691180E-8
-8.63552561198933240294736649885292E-7
7.67385763113287917587459691270367E-7
-1.89314055344153668337714983394592E-7
-2.94403306828973215629670453652457E-7
3.93099106151287980463564320621599E-7
-1.87924142642173789971818089630347E-7
'''

from decimal import *
getcontext().prec=33

listOfMultipliers  = [ Decimal(v) for v in data.split() ]

def arctan (x) :
    x = Decimal(str(x))
    if 1.05 >= x >= 0.45 : pass
    else : print ('\narctan(x): input is outside recommended range.',end='')
    y = Decimal(0)
    x0 = Decimal('0.746735417783721671737500140715213') # tan36.75
    x_minus_x0 = x - x0
    X = Decimal(1)
    status = 1
    for p in range(0,51) :
        toBeAdded = listOfMultipliers[p] * X
        if abs(toBeAdded) < Decimal('1e-31') :
            status = 0
            break
        y += toBeAdded
        X *= x_minus_x0
    if status :
        print ('\narctan(x): count expired.', end='')
    str1 = '''
arctan({}) = {}, count = {}
'''.format(x,y,p)
    print (str1.rstrip())
    return y

x close to x₀

x = Decimal('0.75')
arctan(x)

arctan(0.75) = 0.643501108793284386802809228717315, count = 12

Template:RoundBoxBottom When $x$ is close to $x_0,$ result is achieved with only 12 passes through loop. Template:RoundBoxBottom

Testing with known values

Template:RoundBoxTop Check results using known combinations of $x$ and $\arctan (x):$

$\tan \frac{3\pi }{20}=\tan 27^{\circ }=\sqrt{5}-1-\sqrt{5-2\sqrt{5}}.$

For $\tan 27^{\circ }$ and other exact values of $\tan (x),$ see Exact Values for Common Angles. Template:RoundBoxTop

π = "3.14159265358979323846264338327950288419716939937510582097494459230781"
π = Decimal(π)
rt3 = Decimal(3).sqrt()
rt5 = Decimal(5).sqrt()
rt15 = Decimal(15).sqrt()

tan27 = rt5 - 1 - (5 - 2*rt5).sqrt()

tan30 = 1/rt3

v1 = 2 - (2-rt3)*(3+rt5) ; v2 = 2+ (2*(5-rt5)).sqrt()
tan33 = v1*v2/4

tan36 = (5-2*rt5).sqrt()

v1 = (2-rt3)*(3-rt5)-2 ; v2 = 2 - (2*(5+rt5)).sqrt()
tan39 = v1*v2/4

tan42 = ( rt15 + rt3 - (10 + 2*rt5).sqrt() )/2

tan45 = Decimal(1)

values = (
    ( 9*π/60, tan27, 27),
    (10*π/60, tan30, 30),
    (11*π/60, tan33, 33),
    (12*π/60, tan36, 36),
    (13*π/60, tan39, 39),
    (14*π/60, tan42, 42),
    (   π/ 4, tan45, 45),
)

for value in values :
    angleInRadians, tan, angleInDegrees = value
    y = arctan(tan)
    print ('for', angleInDegrees, 'degrees, difference =',  angleInRadians-y)

arctan(0.509525449494428810513706911250666) = 0.471238898038468985769396507491970, count = 41
for 27 degrees, difference = -4.5E-32

arctan(0.577350269189625764509148780501958) = 0.523598775598298873077107230546614, count = 34
for 30 degrees, difference = -3.1E-32

arctan(0.649407593197510576982062911311432) = 0.575958653158128760384817953601229, count = 27
for 33 degrees, difference = 1.3E-32

arctan(0.726542528005360885895466757480614) = 0.628318530717958647692528676655896, count = 17
for 36 degrees, difference = 4E-33

arctan(0.809784033195007148036991374235772) = 0.680678408277788535000239399710521, count = 23
for 39 degrees, difference = 3.7E-32

arctan(0.90040404429783994512047720388537) = 0.733038285837618422307950122765236, count = 33
for 42 degrees, difference = -1.9E-32

arctan(1) = 0.785398163397448309615660845819846, count = 43
for 45 degrees, difference = 3.0E-32

Verifying the recommended limits

tan24 = ( (50+22*rt5).sqrt() - 3*rt3 - rt15 ) / 2
tan46_5 = Decimal('1.05378012528096218058753672331544') # tan(46.5) 

values = (
    (24*π/180,   tan24,   24),
    (93*π/360, tan46_5, 46.5),
)

for value in values :
    angleInRadians, tan, angleInDegrees = value
    y = arctan(tan)
    print ('for x =', float(tan), 'difference =',  angleInRadians-y)

arctan(x): input is outside recommended range.
arctan(0.44522868530853616392236703064567) = 0.418879020478639098461685784437249, count = 47
for x = 0.44522868530853615 difference = 1.8E-32

arctan(x): input is outside recommended range.
arctan(1.05378012528096218058753672331544) = 0.811578102177363253269516207347250, count = 48
for x = 1.0537801252809622 difference = -4.4E-32

Template:RoundBoxBottom For $1.05\ge x\ge 0.45$ the above calculation of $\arctan (x)$ is accurate to more than 30 places of decimals. Template:RoundBoxBottom

Input outside Recommended Limits

Template:RoundBoxTop If input is outside recommended limits, this does not necessarily mean that result is invalid.

If $\tan 51^{\circ }\ge x\ge \tan 9^{\circ },$ result is accurate to precision of python floats, 15 places of decimals. Template:RoundBoxBottom Template:RoundBoxBottom Template:RoundBoxBottom

arcsin(x)

$y=\arcsin (x)$

$x=\sin (y)$

$\frac{dx}{dy}=\cos (y)=\sqrt{1-x^2}$

$\frac{dy}{dx}=\frac{1}{\cos (y)}=\frac{1}{\sqrt{1-x^2}}$

Simple differential equations eliminate the square root and make calculations so much easier.

Let $a=\frac{dy}{dx}$

Then $a^{2}U=1$ where $U=1-x^2$ and $U^{\text{'}}=-2x.$

Differentiating both sides:

$a^2(-2x)+U2a{a}^{\text{'}}=0$

$U2a{a}^{\text{'}}=2x{a}^2$

$a^{\text{'}}=\frac{2xaa}{2aU}=\frac{ax}{U}.$

Let $b=a^{\text{'}}$

Then $bU=ax$

Differentiating both sides:

$b(-2x)+U{b}^{\text{'}}=a+x{a}^{\text{'}}=a+xb$

Let $c=b^{\text{'}}$

Then $c=\frac{a+3bx}{U}$

When $x=0,U=1.$ Calculation of more derivatives yields:

$y=f(x)=\arcsin (x)$

$a=f^{\text{'}}(x)=\frac{1}{\sqrt{1-x^2}}$

$b=f^{{\text{'}}^{\prime }}(x)=ax$

$c=f^{{\text{'}}^{″}}(x)=a+3bx$

$d=f^{{\text{'}}^{‴}}(x)=4b+5cx$

$g=f^v(x)=9c+7dx$

$h=f^{vi}(x)=16d+9gx$

$j=f^{vii}(x)=25g+11hx$

$k=f^{viii}(x)=36h+13jx$

$l=f^{viiii}(x)=49j+15kx$

and so on.

$\cdots =k=h=d=b=0$

$a=1;c=1^{2}a;g=3^{2}c;j=5^{2}g;l=7^{2}j;\cdots .$

$\arcsin (x)=x+\frac{1}{3!}{x}^3+\frac{3^2}{5!}{x}^5+\frac{3^2\cdot 5^2}{7!}{x}^7+\frac{3^2\cdot 5^2\cdot 7^2}{9!}{x}^9+\frac{3^2\cdot 5^2\cdot 7^2\cdot 9^2}{11!}{x}^{11}+\cdots$

$\arcsin (x)=x+\frac{1}{3!}{x}^3+\frac{3}{(2!)(2^2)5}{x}^5+\frac{3\cdot 5}{(3!)(2^3)7}{x}^7+\frac{3\cdot 5\cdot 7}{(4!)(2^4)9}{x}^9+\frac{3\cdot 5\cdot 7\cdot 9}{(5!)(2^5)11}{x}^{11}+\cdots$

As programming algorithm:

$\arcsin (x)=x+\frac{top=1}{(bottom=2)3}(X=x^3)$ $+\frac{top=top(5-2)}{(bottom=bottom(5-1))5}(X=X(x^2))$ $+\frac{top=top(7-2)}{(bottom=bottom(7-1))7}(X=X(x^2))$ $+\cdots$ $+\frac{top=top(n-2)}{(bottom=bottom(n-1))n}(X=X(x^2))$

As implemented in Python:

from decimal import * # Default precision is 28.

π = ("3.14159265358979323846264338327950288419716939937510582097494459230781")
π = Decimal(π)

x = Decimal(2).sqrt()/2 # Expecting result of π/4

xSQ = x*x
X = x*xSQ

top = Decimal(1)
bottom = Decimal(2)

bottom1 = bottom*3
sum = x + X*top / bottom1

status = 1
for n in range(5,200,2) :
    X = X*xSQ
    top = top*(n-2)
    bottom = bottom*(n-1)
    bottom1 = bottom*n
    added = X*top/bottom1
    if (added < 1e-29) :
        status = 0
        break
    sum += added

if status :
    print ('error. count expired.')
else :
    print (x, sum==π/4, n)

0.707106781186547524400844362 True 171

In practice

If $|x|$ is close to $1,$ the calculation of $\arcsin (x)$ will take forever.

If you limit $x$ to $\sin 45^{\circ }>=x>=-\sin 45^{\circ },$ then $x^2<=0.5$ and each term is guaranteed to be less than half the preceding term.

If $x>\sin 45^{\circ },$ let $t=\frac{\pi }{2}-y.$

Then $\sin t=\cos y=\sqrt{1-x^2};t=\arcsin (\cos y);y=\frac{\pi }{2}-t.$

Integral of expression

$\frac{dy}{dx}=a^{-x^2}$

According to the reference "this expression cannot be integrated..." However, if we convert the expression to a Taylor series, the integral of the series is quite easily calculated.

Let $y=f(x)=a^{-x^2}=\frac{1}{a^{x^2}}$

When $x=0,y=1$ and the following sequence can be produced.

$y^{{\text{'}}^{\prime }}=-2Ay$ where $A=\ln (a).$

$y^{(4)}=-6A{y}^{{\text{'}}^{\prime }}$

$y^{(6)}=-10A{y}^{(4)}$

$y^{(8)}=-14A{y}^{(6)}$

$y^{(10)}=-18A{y}^{(8)}$

$y^{(12)}=-22A{y}^{(10)}$ and so on.

Taylor series of $f(x)$ for $x$ close to $0=t(x)=1+c02\cdot x^2+c04\cdot x^4+c06\cdot x^6+\cdots$

where $c02=\frac{y^{{\text{'}}^{\prime }}}{2!};c04=\frac{y^{(4)}}{4!};c06=\frac{y^{(6)}}{6!}\cdots$

For $a=2$ python code produces the following:

c02 = -0.6931471805599453094172321215
c04 = 0.2402265069591007123335512632
c06 = -0.05550410866482157995314226378
c08 = 0.009618129107628477161979071575
c10 = -0.001333355814642844342341222199
c12 = 0.0001540353039338160995443709734
c14 = -0.00001525273380405984028002543902
c16 = 0.000001321548679014430948840375823
c18 = -1.017808600923969972749000760E-7
c20 = 7.054911620801123329875392184E-9
c22 = -4.445538271870811497596408561E-10
c24 = 2.567843599348820514199480240E-11
c26 = -1.369148885390412888089195400E-12
c28 = 6.778726354822545633449104318E-14
c30 = -3.132436707088428621634944443E-15
c32 = 1.357024794875514719311296624E-16
c34 = -5.533046532458242043485546100E-18
c36 = 2.130675335489117996020398479E-19
c38 = -7.773008428857356419088997166E-21
c40 = 2.693919438465583416972861154E-22
c42 = -8.891822206800239171648619811E-24

For $a$ close to $1$ or $x$ close to $0,$ the Taylor series is a quite accurate representation of the original expression. When abs$(x)\le 1.6,$ the abs(maximum difference) between expression and Taylor series is $<1e-15.$

For greater accuracy, greater precision may be specified in python or more terms after $c42$ may be added.

The integral $=I(x)=x+C02\cdot x^3+C04\cdot x^5+C06\cdot x^7+\cdots$

where $C02=\frac{c02}{3};C04=\frac{c04}{5};C06=\frac{c06}{7}\cdots$

File:20200324curve&Taylor.png

Figure 1: Curves of $f(x)$ and $t(x)$ where $t(x)$ is Taylor series representing $f(x)$ for $x$ close to $0$.

In figure to right, $t(x)=1.5+\cdots ,$ separating $t(x)$ from $f(x)$ to illustrate shapes of curves.

The correct value of $t(x)=1+\cdots$.

When $x=1,f(x)=0.5$ and $t(x)=0.4999999999999999999999997281$.

To 24 places of decimals $t(x)=0.5000$_$0000$_$0000$_$0000$_$0000$_$0000$. Template:RoundBoxBottom Template:RoundBoxTop

File:20200324curce&Integral.png

Figure 1: Curves of $f(x)$ and $I(x)$ where $I(x)$ is integral of $t(x)$ and represents integral of $f(x)$ for $x$ close to $0$.
In this example, constant of integration $C=0.$

If it were important to calculate the area under $f(x)$ from $x=0$ to $x=1,I(x)$ returns $0.8100254543909558266292852604,$ accurate to about 26 places of decimals. Template:RoundBoxBottom

sin(x) using (x - x0)

$f(x)=\sin (x)$

$f^{\prime }(x)=\cos (x)$

$f^{″}(x)=-\sin (x)$

$f^{‴}(x)=-\cos (x)$

$f^{(4)}=\sin (x)$

Let $x_0=\frac{\pi }{4}=45^{\circ }$

$f(x_0)=f(\frac{\pi }{4})=\sin (\frac{\pi }{4})=\frac{\sqrt{2}}{2}$

$f^{\prime }(x_0)=f^{\prime }(\frac{\pi }{4})=\cos (\frac{\pi }{4})=\frac{\sqrt{2}}{2}$

$f^{″}(x_0)=f^{″}(\frac{\pi }{4})=-\sin (\frac{\pi }{4})=-\frac{\sqrt{2}}{2}$

$f^{‴}(x_0)=f^{‴}(\frac{\pi }{4})=-\cos (\frac{\pi }{4})=-\frac{\sqrt{2}}{2}$

$f^{(4)}(x_0)=f^{(4)}(\frac{\pi }{4})=\sin (\frac{\pi }{4})=\frac{\sqrt{2}}{2}$

$t(x)=\frac{\sqrt{2}}{2}\cdot (x-x_0{)}^0$ $+\frac{\sqrt{2}}{2}\cdot (x-x_0{)}^1$ $-\frac{\sqrt{2}}{2}\cdot \frac{1}{2!}\cdot (x-x_0{)}^2$ $-\frac{\sqrt{2}}{2}\cdot \frac{1}{3!}\cdot (x-x_0{)}^3$ $+\frac{\sqrt{2}}{2}\cdot \frac{1}{4!}\cdot (x-x_0{)}^4+\cdots$

Let $X=(x-x_0)=(x-\frac{\pi }{4})$

Then $t(x)=\frac{\sqrt{2}}{2}($ $1+X$ $-\frac{X^2}{2!}-\frac{X^3}{3!}$ $+\frac{X^4}{4!}+\frac{X^5}{5!}$ $-\frac{X^6}{6!}-\frac{X^7}{7!}$ $\cdots$ $)$

where $t(x)$ is the Taylor series representing $f(x)=\sin (x)$ for values of $x$ close to $45^{\circ }$ or $\frac{\pi }{4}.$

If $\frac{\pi }{2}(90^{\circ })\ge x\ge 0$, then $t(x)$ containing powers of $X$ through $15$ is sufficient to keep the error to $<1e-15.$

Gallery

• Graph of t(x) representing sin⁡(x) with powers of X limited to 5.
  Graph of $t(x)$ representing $\sin (x)$ with powers of $X$ limited to $5$.
• Graph of t(x) representing sin⁡(x) with powers of X limited to 9.
  Graph of $t(x)$ representing $\sin (x)$ with powers of $X$ limited to $9$.
• 
  Graph of $t(x)$ representing $\sin (x)$ with powers of $X$ limited to $19$.

Almost a sine curve

File:0326t(x)29.png

Figure 1: Graph of $t(x)$ representing $\sin (x)$ for $x$ close to $\frac{\pi }{4}$.

Graph to right was produced by Grapher on a Mac.

A python script produced the following data:

( (2^(0.5))/2 )(
 1  +(x-.785398163397448)

 -((x-.785398163397448)^2)/2
 -((x-.785398163397448)^3)/(2(3))

 +((x-.785398163397448)^4)/(24)
 +((x-.785398163397448)^5)/(120)

 -((x-.785398163397448)^6)/(720)
 -((x-.785398163397448)^7)/(5040)

 +((x-.785398163397448)^8)/(40320)
 +((x-.785398163397448)^9)/(362880)

 -((x-.785398163397448)^10)/(3628800)
 -((x-.785398163397448)^11)/(39916800)

 +((x-.785398163397448)^12)/(479001600)
 +((x-.785398163397448)^13)/(6227020800)

 -((x-.785398163397448)^14)/(87178291200)
 -((x-.785398163397448)^15)/(1307674368000)

 +((x-.785398163397448)^16)/(20922789888000)
 +((x-.785398163397448)^17)/(355687428096000)

 -((x-.785398163397448)^18)/(6402373705728000)
 -((x-.785398163397448)^19)/(121645100408832000)

 +((x-.785398163397448)^20)/(2432902008176640000)
 +((x-.785398163397448)^21)/(51090942171709440000)

 -((x-.785398163397448)^22)/(1124000727777607680000)
 -((x-.785398163397448)^23)/(25852016738884976640000)

 +((x-.785398163397448)^24)/(620448401733239439360000)
 +((x-.785398163397448)^25)/(15511210043330985984000000)

 -((x-.785398163397448)^26)/(403291461126605635584000000)
 -((x-.785398163397448)^27)/(10888869450418352160768000000)

 +((x-.785398163397448)^28)/(304888344611713860501504000000)
 +((x-.785398163397448)^29)/(8841761993739701954543616000000)
)

I highlighted the data, copied it with command C and pasted it into the input area of Grapher. Well done! Grapher. Template:RoundBoxBottom

Integral of 1/x

The Taylor series for $f(x)=\frac{1}{x}$ for $x$ close to $2$ is:

$t(x)=\frac{1}{2}-\frac{1}{2^2}\cdot (x-2)+\frac{1}{2^3}\cdot (x-2{)}^2-\frac{1}{2^4}\cdot (x-2{)}^3\cdots$

The integral of this series is:

$I(x)=\frac{1}{2}\cdot (x-2)-((\frac{1}{2^2})/2)\cdot (x-2{)}^2+((\frac{1}{2^3})/3)\cdot (x-2{)}^3-((\frac{1}{2^4})/4)\cdot (x-2{)}^4\cdots +C$

The integral of $\frac{1}{x}=\ln (x).$

Therefore $\ln (x)=I(x)$ but what is the value of $C?$

Without $C,$ when $x=2,I(x)=0.I(x)$ should be $\ln (2).$

Therefore, for $x$ close to $2:$

$\ln (x)=I(x)$ where $C=\ln (2).$

But what is the value of $\ln (2)?$

Without $C,$ when $x=1,I(x)=-0.693147180559945.I(x)$ should be $0.$

Therefore $C=0.693147180559945$ or $\ln (2).$

For $x$ close to $2:$

$\ln (x)=\ln (2)+\frac{1}{2}\cdot (x-2)-((\frac{1}{2^2})/2)\cdot (x-2{)}^2+((\frac{1}{2^3})/3)\cdot (x-2{)}^3-((\frac{1}{2^4})/4)\cdot (x-2{)}^4\cdots$

where $\ln (2)=0.693147180559945\dots$

File:0328ln(x)closeTo2.png

Figure 1: Graph of $I(x)$ representing $\ln (x)$ for $x$ close to $2$.

y = 0.693147180559945
+ ((1/(2^1))/1)(x - 2)^1
- ((1/(2^2))/2)(x - 2)^2
+ ((1/(2^3))/3)(x - 2)^3
- ((1/(2^4))/4)(x - 2)^4
+ ((1/(2^5))/5)(x - 2)^5
- ((1/(2^6))/6)(x - 2)^6
+ ((1/(2^7))/7)(x - 2)^7
...........................
...........................
- ((1/(2^42))/42)(x - 2)^42
+ ((1/(2^43))/43)(x - 2)^43
- ((1/(2^44))/44)(x - 2)^44
+ ((1/(2^45))/45)(x - 2)^45
- ((1/(2^46))/46)(x - 2)^46
+ ((1/(2^47))/47)(x - 2)^47
- ((1/(2^48))/48)(x - 2)^48
+ ((1/(2^49))/49)(x - 2)^49

Template:RoundBoxBottom Generally, for $x$ close to $x_0:$

$\text{ln}(x)=\text{ln}(x_0)+\frac{1}{1}\cdot \frac{x-x_0}{x_0}$ $-\frac{1}{2}\cdot \frac{(x-x_0{)}^2}{x_0^2}$ $+\frac{1}{3}\cdot \frac{(x-x_0{)}^3}{x_0^3}$ $-\frac{1}{4}\cdot \frac{(x-x_0{)}^4}{x_0^4}$ $+\frac{1}{5}\cdot \frac{(x-x_0{)}^5}{x_0^5}-\cdots$

Calculating ln(x)

This section presents a system for calculating $\ln (x)$ for $10\ge x\ge 1,$ knowing only that $\ln (1)=0.$

# python code
L1 = [1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.2, 2.4, 2.6, 2.8, 
    3.0, 3.3, 3.6, 3.9, 4.2, 4.6, 5.0, 5.5, 6.0, 6.6, 7.2, 7.9, 8.6, 9.3, 10.0]

where L1 is a list containing values of $x_0$ in which each value after the first is $\le 10$% more than the preceding value.

# python code
from decimal import *
getcontext().prec=53 # Preparing for values containing 50 places of decimals.
almostZero = Decimal('1e-' + str( getcontext().prec ))

L1 = [ Decimal(str(v)) for v in L1 ]

def ln_x (x, x0, C=0) :
    '''
    return ln(x) for x close to x0.
    ln_x_ = ln_x (x, x0, C) 
    C is the constant of integration. Usually C = ln(x0).
    '''
    x, x0, C = [ Decimal(str(v)) for v in (x,x0,C) ]
    x_minus_x0 = x-x0;
#    print ('x,x0,x_minus_x0 =',x,x0,x_minus_x0)
    sum = 0
    progressiveValue = 1
    status = 1 ; limit = 4*getcontext().prec
    multiplier = x_minus_x0/x0
    for p in range (1, limit, 2) :
        progressiveValue *= multiplier
        added = progressiveValue / p
        sum += added

        progressiveValue *= multiplier
        added = progressiveValue / (p+1)

        if (abs(added) < almostZero) :
            status = 0
            break
        sum -= added
    if (status) :
        print ('ln_x error: count expired, p =',p)
        exit (95)
    return sum+C

The performance of the above code is better than logarithmic to base $10$. This means, for example, if $x$ contains 60 significant decimal digits, the above code produces a result with fewer than 30 passes through the loop because each iteration of the lop performs two operations.

L1 is designed so that multiplier $=\frac{x\mathrm{\_}minus\mathrm{\_}{x}_0}{x_0}$ is always $\le \frac{1}{10}.$ When $x$ is very close to $x_0,$ time to calculate $\ln (x)$ is greatly reduced.

File:0407ln(x)closeTo(7.9).png

Figure 1: Graph of $I(x)$ representing $\ln (x)$ for $x$ close to $7.9$.
When value $\frac{x-x_0}{x_0}>1,$ the series diverges.
In this case, when $x>15.8.$

y = ln(7.9)
+ ((1/((7.9)^(1)))/(1))((x - 7.9)^(1))
- ((1/((7.9)^(2)))/(2))((x - 7.9)^(2))
+ ((1/((7.9)^(3)))/(3))((x - 7.9)^(3))
- ((1/((7.9)^(4)))/(4))((x - 7.9)^(4))
+ ((1/((7.9)^(5)))/(5))((x - 7.9)^(5))
- ((1/((7.9)^(6)))/(6))((x - 7.9)^(6))
+ ((1/((7.9)^(7)))/(7))((x - 7.9)^(7))
- ((1/((7.9)^(8)))/(8))((x - 7.9)^(8))
+ ((1/((7.9)^(9)))/(9))((x - 7.9)^(9))
.....................
.....................
- ((1/((7.9)^(22)))/(22))((x - 7.9)^(22))
+ ((1/((7.9)^(23)))/(23))((x - 7.9)^(23))
- ((1/((7.9)^(24)))/(24))((x - 7.9)^(24))
+ ((1/((7.9)^(25)))/(25))((x - 7.9)^(25))
- ((1/((7.9)^(26)))/(26))((x - 7.9)^(26))
+ ((1/((7.9)^(27)))/(27))((x - 7.9)^(27))
- ((1/((7.9)^(28)))/(28))((x - 7.9)^(28))
+ ((1/((7.9)^(29)))/(29))((x - 7.9)^(29))

The next piece of code progressively calculates $\ln (1.1),\ln (1.2),\ln (1.3)\dots$ and puts the calculated values in dictionary dict2.

dict2 = dict()
dict2[Decimal('1.0')] = Decimal(0)

for p in range(1, len(L1)) :
    x = L1[p]
    x0 = L1[p-1]
    C = dict2[x0]
#    print ('L1[{}]={}'.format(p,L1[p]))
    ln = ln_x (x, x0, C)
    dict2[x] = ln

print ('dict2 = {')
for x0 in dict2 :
    print ("Decimal('{}'):  +Decimal('{}'),".format( (' '+str(x0))[-4:], dict2[x0]) )
print ('}')

dict2 = {
Decimal(' 1.0'):  +Decimal('0'),
Decimal(' 1.1'):  +Decimal('0.095310179804324860043952123280765092220605365308644199'),
Decimal(' 1.2'):  +Decimal('0.18232155679395462621171802515451463319738933791448698'),
Decimal(' 1.3'):  +Decimal('0.26236426446749105203549598688095439720416645613143414'),
Decimal(' 1.4'):  +Decimal('0.33647223662121293050459341021699209011148337531334347'),
Decimal(' 1.5'):  +Decimal('0.40546510810816438197801311546434913657199042346249420'),
Decimal(' 1.6'):  +Decimal('0.47000362924573555365093703114834206470089904881224805'),
Decimal(' 1.7'):  +Decimal('0.53062825106217039623154316318876232798710152395697182'),
Decimal(' 1.8'):  +Decimal('0.58778666490211900818973114061886376976937976137698120'),
Decimal(' 1.9'):  +Decimal('0.64185388617239477599103597720348932963627777267035586'),
Decimal(' 2.0'):  +Decimal('0.69314718055994530941723212145817656807550013436025527'),
Decimal(' 2.2'):  +Decimal('0.78845736036427016946118424473894166029610549966889947'),
Decimal(' 2.4'):  +Decimal('0.87546873735389993562895014661269120127288947227474225'),
Decimal(' 2.6'):  +Decimal('0.95551144502743636145272810833913096527966659049168941'),
Decimal(' 2.8'):  +Decimal('1.0296194171811582399218255316751686581869835096735987'),
Decimal(' 3.0'):  +Decimal('1.0986122886681096913952452369225257046474905578227494'),
Decimal(' 3.3'):  +Decimal('1.1939224684724345514391973602032907968680959231313936'),
Decimal(' 3.6'):  +Decimal('1.2809338454620643176069632620770403378448798957372364'),
Decimal(' 3.9'):  +Decimal('1.3609765531356007434307412238034801018516570139541836'),
Decimal(' 4.2'):  +Decimal('1.4350845252893226218998386471395177947589739331360929'),
Decimal(' 4.6'):  +Decimal('1.5260563034950493162059934985840084789167789605719180'),
Decimal(' 5.0'):  +Decimal('1.6094379124341003746007593332261876395256013542685177'),
Decimal(' 5.5'):  +Decimal('1.7047480922384252346447114565069527317462067195771619'),
Decimal(' 6.0'):  +Decimal('1.7917594692280550008124773583807022727229906921830047'),
Decimal(' 6.6'):  +Decimal('1.8870696490323798608564294816614673649435960574916489'),
Decimal(' 7.2'):  +Decimal('1.9740810260220096270241953835352169059203800300974917'),
Decimal(' 7.9'):  +Decimal('2.0668627594729758101549540867970467145724397357938367'),
Decimal(' 8.6'):  +Decimal('2.1517622032594620488720831801196593960335348306130377'),
Decimal(' 9.3'):  +Decimal('2.2300144001592102533064181067805187074963279996745685'),
Decimal('10.0'):  +Decimal('2.3025850929940456840179914546843642076011014886287730'),
}

Template:RoundBoxTop A quick check:

ln(2.2) - (ln(1.1) + ln(2.0)) = 0E-50
ln(2.4) - (ln(1.2) + ln(2.0)) = 0E-50
ln(2.6) - (ln(1.3) + ln(2.0)) = 0E-50
ln(2.8) - (ln(1.4) + ln(2.0)) = 0E-50
ln(3.0) - (ln(1.5) + ln(2.0)) = -0E-50
ln(3.3) - (ln(1.1) + ln(3.0)) = 0E-50
ln(3.6) - (ln(1.2) + ln(3.0)) = 0E-50
ln(3.6) - (ln(1.8) + ln(2.0)) = -0E-50
ln(3.9) - (ln(1.3) + ln(3.0)) = 0E-50
ln(4.2) - (ln(1.4) + ln(3.0)) = 0E-50
ln(5.5) - (ln(1.1) + ln(5.0)) = 0E-50
ln(6.0) - (ln(1.2) + ln(5.0)) = 0E-50
ln(6.0) - (ln(2.0) + ln(3.0)) = 0E-50
ln(6.6) - (ln(1.1) + ln(6.0)) = 0E-50
ln(6.6) - (ln(2.2) + ln(3.0)) = 0E-50
ln(6.6) - (ln(3.3) + ln(2.0)) = 0E-50
ln(7.2) - (ln(1.2) + ln(6.0)) = 0E-50
ln(7.2) - (ln(2.4) + ln(3.0)) = 0E-50
ln(10.0) - (ln(5.0) + ln(2.0)) = 0E-50

Put the data from dict2 into 2 tuples Tx0, Tln_x0

Tx0 = tuple(L1)
Tln_x0 = tuple([ dict2[v] for v in Tx0 ])

Calculate the decision points.

L1 = []
for p in range (0, len(Tx0)-1) :
    a,b = Tx0[p], Tx0[p+1]
    dp = 2*a*b/(a+b)
    L1 += [ dp ]
Tdp = tuple(L1)

Display the three tuples.

for T in ('Tx0', 'Tln_x0', 'Tdp') :
    t = eval(T)
    print (T, '= (')
    for v in t :
        print ("""+Decimal('{}'),""".format(v))
    print (')')
    print ()

Previous code was used to produce three tuples. Operational code follows:

Values of $x_0:$

Tx0 = ( Decimal('1'), Decimal('1.1'), Decimal('1.2'), Decimal('1.3'), Decimal('1.4'), Decimal('1.5'), Decimal('1.6'), Decimal('1.7'), Decimal('1.8'), Decimal('1.9'), Decimal('2.0'), Decimal('2.2'), Decimal('2.4'), Decimal('2.6'), Decimal('2.8'), Decimal('3.0'), Decimal('3.3'), Decimal('3.6'), Decimal('3.9'), Decimal('4.2'), Decimal('4.6'), Decimal('5.0'), Decimal('5.5'), Decimal('6.0'), Decimal('6.6'), Decimal('7.2'), Decimal('7.9'), Decimal('8.6'), Decimal('9.3'), Decimal('10.0'), )

Values of $\ln (x_0):$

Tln_x0 = ( +Decimal('0'), +Decimal('0.095310179804324860043952123280765092220605365308644199'), +Decimal('0.18232155679395462621171802515451463319738933791448698'), +Decimal('0.26236426446749105203549598688095439720416645613143414'), +Decimal('0.33647223662121293050459341021699209011148337531334347'), +Decimal('0.40546510810816438197801311546434913657199042346249420'), +Decimal('0.47000362924573555365093703114834206470089904881224805'), +Decimal('0.53062825106217039623154316318876232798710152395697182'), +Decimal('0.58778666490211900818973114061886376976937976137698120'), +Decimal('0.64185388617239477599103597720348932963627777267035586'), +Decimal('0.69314718055994530941723212145817656807550013436025527'), +Decimal('0.78845736036427016946118424473894166029610549966889947'), +Decimal('0.87546873735389993562895014661269120127288947227474225'), +Decimal('0.95551144502743636145272810833913096527966659049168941'), +Decimal('1.0296194171811582399218255316751686581869835096735987'), +Decimal('1.0986122886681096913952452369225257046474905578227494'), +Decimal('1.1939224684724345514391973602032907968680959231313936'), +Decimal('1.2809338454620643176069632620770403378448798957372364'), +Decimal('1.3609765531356007434307412238034801018516570139541836'), +Decimal('1.4350845252893226218998386471395177947589739331360929'), +Decimal('1.5260563034950493162059934985840084789167789605719180'), +Decimal('1.6094379124341003746007593332261876395256013542685177'), +Decimal('1.7047480922384252346447114565069527317462067195771619'), +Decimal('1.7917594692280550008124773583807022727229906921830047'), +Decimal('1.8870696490323798608564294816614673649435960574916489'), +Decimal('1.9740810260220096270241953835352169059203800300974917'), +Decimal('2.0668627594729758101549540867970467145724397357938367'), +Decimal('2.1517622032594620488720831801196593960335348306130377'), +Decimal('2.2300144001592102533064181067805187074963279996745685'), +Decimal('2.3025850929940456840179914546843642076011014886287730'), )

Decision points:

Tdp = ( +Decimal('1.0476190476190476190476190476190476190476190476190476'), +Decimal('1.1478260869565217391304347826086956521739130434782609'), +Decimal('1.248'), +Decimal('1.3481481481481481481481481481481481481481481481481481'), +Decimal('1.4482758620689655172413793103448275862068965517241379'), +Decimal('1.5483870967741935483870967741935483870967741935483871'), +Decimal('1.6484848484848484848484848484848484848484848484848485'), +Decimal('1.7485714285714285714285714285714285714285714285714286'), +Decimal('1.8486486486486486486486486486486486486486486486486486'), +Decimal('1.9487179487179487179487179487179487179487179487179487'), +Decimal('2.0952380952380952380952380952380952380952380952380952'), +Decimal('2.2956521739130434782608695652173913043478260869565217'), +Decimal('2.496'), +Decimal('2.6962962962962962962962962962962962962962962962962963'), +Decimal('2.8965517241379310344827586206896551724137931034482759'), +Decimal('3.1428571428571428571428571428571428571428571428571429'), +Decimal('3.4434782608695652173913043478260869565217391304347826'), +Decimal('3.744'), +Decimal('4.0444444444444444444444444444444444444444444444444444'), +Decimal('4.3909090909090909090909090909090909090909090909090909'), +Decimal('4.7916666666666666666666666666666666666666666666666667'), +Decimal('5.2380952380952380952380952380952380952380952380952381'), +Decimal('5.7391304347826086956521739130434782608695652173913043'), +Decimal('6.2857142857142857142857142857142857142857142857142857'), +Decimal('6.8869565217391304347826086956521739130434782608695652'), +Decimal('7.5337748344370860927152317880794701986754966887417219'), +Decimal('8.2351515151515151515151515151515151515151515151515152'), +Decimal('8.9363128491620111731843575418994413407821229050279330'), +Decimal('9.6373056994818652849740932642487046632124352331606218'), )

At each decision point $x$ is assigned to the next low value or the next high value of $x_0.$ For example, if $x$ is between $1.2,1.3,$ the decision point is $1.248.$ This means that the ratio $\frac{1.248-1.2}{1.2}=\frac{1.3-1.248}{1.3}=0.04$ and the maximum value of abs$(\frac{x-x_0}{x_0})=0.04.$

During creation of Tln_x0 the maximum value of $\frac{x-x_0}{x_0}=\frac{1}{10}.$ During normal operations after creation of Tln_x0, maximum value of abs$(\frac{x-x_0}{x_0})=\frac{1.047619\dots -1}{1}$ between $1.0,1.1.$

Choose a suitable value of x0 with the value of its natural log.

def choose_x0_C (x) :
    '''
    (x0, C) = choose_x0_C (x)
    '''
    if (10 >= x >= 1) : pass
    else: exit (93)

    for p in range (len(Tx0)-2, -1, -1):
        if (x >= Tx0[p]) :
            if (x >= Tdp[p]) : return (Tx0[p+1], Tln_x0[p+1])
            return (Tx0[p], Tln_x0[p])
    exit(92)

Ready to calculate, for example, $\ln (3.456789)$

x = Decimal('3.456789')
(x0, C) = choose_x0_C (x)
ln_x_ = ln_x (x, x0, C)
print ('ln({}) = {}'.format(x, ln_x_.quantize(Decimal('1e-50'))))

ln(3.456789) = 1.24034_01234_96758_02986_53847_82231_30004_00340_53893_89110 # displayed with 50 places of decimals.

Testing ln(x)

Choose random numbers $a,b$ so that $10\ge a\ge b\ge 1.$

Produce values $\ln (a),\ln (b).$

Calculate product $p=a\cdot b.$

Produce value $\ln (p).$

If $p>10,p=(\frac{p}{10})\cdot 10$ and $\ln (p)=\ln (\frac{p}{10})+\ln (10).$

Verify that $\ln (p)=\ln (a)+\ln (b).$

# python code
import random

ln_10 = Tln_x0[-1]
fiftyPlacesOfDecimals = Decimal('1e-50')

def randomNumber() :
    s1 = str(random.getrandbits(getcontext().prec * 4))
    d1 = Decimal(s1[0] + '.' + s1[1:])
    if (d1 == 0) : d1 = randomNumber()
    while (d1 < 1) : d1 *= 10
    return d1

d1 = randomNumber()
d2 = randomNumber()

(x0, C) = choose_x0_C (d1)
ln_d1_ = ln_x (d1, x0, C)

(x0, C) = choose_x0_C (d2)
ln_d2_ = ln_x (d2, x0, C)

product = d1*d2
add_ln10 = 0
if (product > 10) :
    product /= 10
    add_ln10 += 1

(x0, C) = choose_x0_C (product)
ln_product_ = ln_x (product, x0, C)
if (add_ln10) : ln_product_ += ln_10

difference = (ln_product_ - ( ln_d1_ + ln_d2_ )).quantize(fiftyPlacesOfDecimals)

print ('''
d1          = {}
ln_d1_      = {}
d2          = {}
ln_d2_      = {}
ln_product_ = {}
'''.format(
d1,ln_d1_ ,
d2,ln_d2_ ,
ln_product_ ,
))

if difference :  print ('''
difference  = {} ****
'''.format(
difference,
))

For example: During testing, successive invocations of the above code produced:

d1          = 3.300463847393627263496303126765085976697315885228780009201595937
ln_d1_      = 1.1940630184110798505583266934968432937656468440595029
d2          = 4.727915623201914684885711302927600487326893972103794963997766615
ln_d2_      = 1.5534844337520634527664958773360448454701186698422347
ln_product_ = 2.7475474521631433033248225708328881392357655139017377

d1          = 6.56429212435850275252301147228535243835226966080458915176241218
ln_d1_      = 1.8816446762531860392218213681767770852191644273705970
d2          = 8.15468991518212749204100104755219361919087392341006662123706307
ln_d2_      = 2.0985932114606734087366302984138612677420896519457258
ln_product_ = 3.9802378877138594479584516665906383529612540793163228