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Re: continued fraction->simple fraction
Posted:
Feb 21, 2013 11:05 AM
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On 20.01.13 16:58, bar@robox.up.krakow.pl wrote:
> Hello,
> Is a way i Matlab to convert formula looks like continued fraction
> to the sum of simple fraction ?
>
>
> example:
>
> 1/((-0.89 + 2.31 I) +
>
> 0.9/((1.41 + 1.41 I) +
> 0.99/((1 - 1.73 I) + 0.99/((-1 - I) + z))))/((1.41 +
> 1.41 I) + 0.99/((1 - 1.73 I) + 0.99/((-1 - I) + z)))^2
If your input is a symbolic object, try simplifyFraction:
>> syms z
>> f = 1/((-0.89 + 2.31i) + 0.9/((1.41 + 1.41i) + ...
0.99/((1 - 1.73i) + 0.99/((-1 - i) + z))))/((1.41 +1.41i) + ...
0.99/((1 - 1.73i) + 0.99/((-1 - i) + z)))^2;
>> simplifyFraction(f)
ans =
-(1000000*(z*(100 - 173*i) - 174 + 73*i)^2)/(9*(z*(16131 - 3431*i) -
14909 - 8047*i)*(z*(343098 - 3512620*i) - 2663758 + 2508796*i))
ans =
-(1.11e5*(z*(100.0 - 173.0*i) - 174.0 + 73.0*i)^2)/((z*(3.43e5 -
3.51e6*i) - 2.66e6 + 2.51e6*i)*(z*(1.61e4 - 3433.0*i) - 1.49e4 - 8055.0*i))
I'm not sure what you mean by ?the sum of simple fractions,? though. You
can expand the combined fraction into a sum, but I'm not sure that is an
improvement:
>> vpa(expand(simplifyFraction(f)), 3)
ans =
(z^2*(- 2.21e9 - 3.84e9*i))/(z*(6.77e10 - 9.92e10*i) + z^2*(6.52e9 +
5.78e10*i) - 5.99e10 + 1.6e10*i) + (z*(- 1.06e9 + 8.31e9*i))/(z*(6.77e10
- 9.92e10*i) + z^2*(6.52e9 + 5.78e10*i) - 5.99e10 + 1.6e10*i) + (2.77e9
- 2.82e9*i)/(z*(6.77e10 - 9.92e10*i) + z^2*(6.52e9 + 5.78e10*i) -
5.99e10 + 1.6e10*i)
> answer: (accuracy about 10^-10)
>
> (-0.0698942 + 0.0304086 I) + (
> 0.0406374 + 0.0181059 I)/((-0.782731 - 0.665337 I) + 1. z) - (
> 0.0510353 + 0.0306458 I)/((-0.780845 - 0.68207 I) + 1. z)
I fail to see how that is the same as the partial sum of a continued
fraction above. It might be an earlier partial sum, of course.
HTH,
Christopher
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