Re: Arctangent approximation

• To: mathgroup at smc.vnet.net
• Subject: [mg107670] Re: Arctangent approximation
• From: dh <dh at metrohm.com>
• Date: Mon, 22 Feb 2010 08:32:22 -0500 (EST)
• References: <hlglio\$l5p\$1@smc.vnet.net>

```Hi Sidey,
this is actually a truncated power series where the last term has been
changed a little:
ArcTan[x]= x-x^3/3+x^5/5-x^7/7+x^9/9-x^11/11 + ...
f1[x]= x-x^3/3+x^5/5-x^7/7+x^9/9
f2[x]= x-x^3/3+x^5/5-x^7/7+x^9 9/48   this is your function
f3[x]=x-x^3/3+x^5/5-x^7/7+x^9/9-x^11/11

We create a table with following columns:
x, ArcTan[x],f1[x]-ArcTan[x],f2[x]-ArcTan[x],f2[x]-ArcTan[x]:

x       ArcTan        h1            h2            h3
0.      0.            0.            0.            0.
0.16    0.16          1.3*10^(-10)  -2.8*10^(-10) -2.7*10^(-12)
0.31    0.3           2.5*10^(-7)   4.*10^(-8)    -2.1*10^(-8)
0.47    0.44          0.000019      0.000012      -3.6*10^(-6)
0.63    0.56          0.00041       0.00031       -0.00014
0.79    0.67          0.0042        0.0034        -0.0022
0.94    0.76          0.027         0.023         -0.02
1.1     0.83          0.13          0.11          -0.13
1.3     0.9           0.48          0.43          -0.64
1.4     0.96          1.5           1.4           -2.6
1.6     1.            4.3           3.9           -8.8

Daniel

On 17.02.2010 12:57, sidey wrote:
> A professor (Herbert Medina) came up with this remarkable
> approximation.
>
> Since I don't have Mathematica, could someone run it for
> numberofdigits=9 and 12  please? (and send me the result?)
>
> here is a short result
> h2(x) = x - x^3/3 + x^5/5 - x^7/7 + 5x^9/48..
>
>                   Thank you.
>
> PS reference is: http://myweb.lmu.edu/hmedina/Papers/Arctan.pdf
>
> Clear[h, m, a, numberofdigits]
> numberofdigits=12;
> m=Floor[-(1/5) Log[4,5*0.1^(numberofdigits+1)]]+1;
> Do[a[2j]=(-1)^(j+1) Sum[Binomial[4m,2k] (-1)^k, {k,j+1,2m}];
> a[2j-1]=(-1)^(j+1) Sum[Binomial[4m,2k+1] (-1)^k, {k,j,2m-1}],
> {j,0,2m-1}];
> h[m,x_]:=Sum[(-1)^(j+1) / (2j-1) x^(2j-1), {j,1,2m}] +
> Sum[a[j]/((-1)^(m+1) 4^m (4m+j+1)) x^(4m+j+1), {j,0, 4m-2}];
> Print["h[",m,",x] given below computes ArcTan[x] with ",
> numberofdigits," digits of accuracy for x in [0,1]."]
> Print["h[m,x] = ", h[m,x]]
>

--

Daniel Huber
Metrohm Ltd.
Oberdorfstr. 68
CH-9100 Herisau
Tel. +41 71 353 8585, Fax +41 71 353 8907
E-Mail:<mailto:dh at metrohm.com>
Internet:<http://www.metrohm.com>

```

• Prev by Date: Re: Import data from an email
• Next by Date: ? about MatrixPlot used with ListPlot
• Previous by thread: Re: Arctangent approximation
• Next by thread: Q: Problem importing two sparse arrays from a .mat file (Update)