Taylor expansion

lalo · 19 Apr 2005

beautiful!

what's a Taylor series ? 😀 ..and who's Taylor ? 😀

not the duran duran's bass player i guess 😀

lalo

Shifrin · 19 Apr 2005

You could check out :-

http://en.wikipedia.org/wiki/Taylor%27s_theorem

and :-

http://en.wikipedia.org/wiki/Taylor_expansion

The maths isn't as hard as it might look. I'd have a go at trying to explain myself but it's a very long time since I did such stuff. I wish I'd kept my hand in with maths (did it at uni) but I've not looked at it for several years now - still got all my lecture notes etc though!

Shif.

NothanUmber · 2 May 2005

Just rediscovered MuPad for those cases (good for people who don't own one of the "bigger" math packages). Download and install

ftp://ftp.mupad.de/MuPAD/distrib/window ... ab_253.exe

and enter e.g.

taylor(5^x,x=0,10)

in order to develop the taylor series for 5^x around zero of order 10. (Disable "Pretty Print" for easier copy'n'paste and ignore the O(somevalue) at the end - that's just the order of the supressed rest that doesn't bother you). Then replace all evental ln-constants with their approximated values and wrap the result inside a "simplify"

e.g. 1 + x*ln(5) + 1/2*x^2*ln(5)^2 + 1/6*x^3*ln(5)^3 + 1/24*x^4*ln(5)^4 + 1/120\ *x^5*ln(5)^5 + 1/720*x^6*ln(5)^6 + 1/5040*x^7*ln(5)^7 + 1/40320*x^8*ln(5)^\ 8 + 1/362880*x^9*ln(5)^9 with ln(5) = 1,6094379124341003746007593332262

becomes:

simplify(1 + x*1.6094379124341003746007593332262 + 1/2*x^2*1.6094379124341003746007593332262^2 + 1/6*x^3*1.6094379124341003746007593332262^3 + 1/24*x^4*1.6094379124341003746007593332262^4 + 1/120\ *x^5*1.6094379124341003746007593332262^5 + 1/720*x^6*1.6094379124341003746007593332262^6 + 1/5040*x^7*1.6094379124341003746007593332262^7 + 1/40320*x^8*1.6094379124341003746007593332262^\ 8 + 1/362880*x^9*1.6094379124341003746007593332262^9)

which gives the result:

1.609437912*x + 1.295145197*x^2 + 0.694818594*x^3 + 0.2795668469*x^4 + 0.0\ 8998909649*x^5 + 0.02413864393*x^6 + 0.005549949814*x^7 + 0.001116537455*x\ ^8 + 0.0001996664124*x^9 + 1

which you can directly use in SynthMaker 🙂 (write x^3 as x*x*x etc.)

(Perhaps there is an even simpler way, e.g. in Maple there is an approximate-function that automatically approximates values like the constant lns with a given precision. So MuPad-gurus step forward 😉 )

NothanUmber

Sk · 2 May 2005

If you want to have fun with Taylor series, try the "taylortool" in Matlab 🙂

stefan · 4 May 2005

This was a useful page to me right now http://www3.sympatico.ca/mcomeau/webpub ... eg/reg.htm

To make a polynomial estimation of a given series of numbers. Very useful imo.

ake · 7 May 2005

If you want to have fun with math 🙂 maxima is a great free tool

http://maxima.sourceforge.net/

Åke

hansdump · 6 Aug 2009

In the meantime, it was so boring here, that I made a little asm optimized taylor exp version of yours here!

Feel free to use it, tell me if there´s still some possibility to optimize...

bye H.

//asm code here: streamin x;streamout y; float x2=0; float x3=0; float x4=0; float x5=0; float x6=0; float x7=0; float F1=1; float F0P5=0.5; float F0P166667=0.166667; float F0P0416667=0.0416667; float F0P00833333=0.00833333; float F0P00138889=0.00138889; float F0P000198413=0.000198413;

movaps xmm0,x; mulps xmm0,x; movaps x2,xmm0; mulps xmm0,x; movaps x3,xmm0; mulps xmm0,x; movaps x4,xmm0; mulps xmm0,x; movaps x5,xmm0; mulps xmm0,x; movaps x6,xmm0; mulps xmm0,x; movaps x7,xmm0;

movaps xmm0,F1; addps xmm0,x;

movaps xmm1,F0P5; mulps xmm1,x2; addps xmm0,xmm1;

movaps xmm1,F0P166667; mulps xmm1,x3; addps xmm0,xmm1;

movaps xmm1,F0P0416667; mulps xmm1,x4; addps xmm0,xmm1;

movaps xmm1,F0P00833333; mulps xmm1,x5; addps xmm0,xmm1;

movaps xmm1,F0P00138889; mulps xmm1,x6; addps xmm0,xmm1;

movaps xmm1,F0P000198413; mulps xmm1,x7; addps xmm0,xmm1;

movaps y,xmm0;

hansdump · 11 Nov 2009

I only wanna say that you can further improve this by the horner scheme but sry I´m too lazy now to implement this 😇

exp(x) approximatly = 1+(1+(1/2+(1/6+(1/24+(1/120+(1/720+1/5040*x)*x)*x)*x)*x)*x)*x

H.

MichaelBenjamin · 12 Nov 2009

heres the asm:

streamin x; streamout exp; float F1=1; float F0P5=0.5; float F0P16666667=0.16666667; float F0P041666668=0.041666668; float F0P0083333338=0.0083333338; float F0P0013888889=0.0013888889; float F0P0001984127=0.0001984127;

movaps xmm0,F0P0001984127; movaps xmm7,x;

mulps xmm0,xmm7; addps xmm0,F0P0013888889; mulps xmm0,xmm7; addps xmm0,F0P0083333338; mulps xmm0,xmm7; addps xmm0,F0P041666668; mulps xmm0,xmm7; addps xmm0,F0P16666667; mulps xmm0,xmm7; addps xmm0,F0P5; mulps xmm0,xmm7; addps xmm0,F1;

movaps exp,xmm0;

hansdump · 13 Nov 2009

hehe! I like this forum! 😈

H.

MegaHurtz · 14 Nov 2009

is this a more efficient way, as opposed to (e)x?

infuzion · 16 Nov 2009

MegaHurtz wrote: is this a more efficient way, as opposed to (e)x?

Is there such a function in SM? I don't think so offhand.

MB wins the award! Should go in the WiKi. 😉 😇

MegaHurtz · 17 Nov 2009

I don`t get wich power of what this is? Somebody?

hansdump · 18 Jan 2010

octave says:

exp(1)=2.7183

wikipedia says:

tanh(x)=1-2/(exp(2x)+1)

http://mathworld.wolfram.com/MaclaurinSeries.html

H.