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 #4
avatar+26400 
+5

(3x^3-5x^2+10x-3)/(3x-4)

 

In mathematics, Horner's method (also known as Horner scheme in the UK or Horner's rule in the U.S.) an algorithm for calculating polynomials

see:  https://en.wikipedia.org/wiki/Horner%27s_method#cite_note-HornerRule-2

$$\small{
\begin{array}{lrrl}
& f_1{(x)} &=& 3x^3-5x^2+10x-3\\
& f_2{(x)} &=& 3x-4\\
&\text{Divide } f_1{(x)} \text{ by } f_2{(x)} \text{ using Honer's method }
\end{array}
}$$

$$\small{\text{$
\begin{array}{r|crcrcrc|r}
3 &\quad& 3 &\quad& -5 &\quad& 10 &\quad& -3 \\
\hline
&&&&&&&\\
4 &\quad& &\quad& 4 &\quad& -\dfrac{4}{3} &\quad& 4\cdot \dfrac{26}{9}\\
\hline
&&&&&&&\\
&\quad& 1 &\quad& -1\cdot \dfrac{1}{3} &\quad&\dfrac{26}{3}\cdot \dfrac{1}{3} &\quad & \dfrac{77}{9}
\end{array}
$}}$$

 

$$1\cdot x^2 - \dfrac{1}{3}\cdot x + \dfrac{26}{9}
+ \dfrac{77}{9(3x-4)}$$

 

 

 

Aug 18, 2015
 #106
avatar+118723 
+5

@@ End of Day Wrap    Tues 18/8/15     Sydney, Australia Time   7:06pm       ♪ ♫   

 

Hello all,

 

Considering how hectic it was on Sunday it has been very quiet since.  I suppose the homework has not started in earnest in the US yet :/

Anyway we still had some really good answers from Geno3141, Cphill, Heureka, MathsGod1 and Rubenhh.  Thank you  

 

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

 

Interest Posts:

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1) Number puzzle                                           Thanks Heureka 

2) Finding smallest common multiple   FTJ+      Thanks Geno3141

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Aug 18, 2015
 #3
avatar+26400 
0

7.6855172413793103 in fractions is?

 

1. Continued fraction of 7.6855172413793103 :

$$\small{\text{$
\begin{array}{rrrrrrrrrr}
[x_0;&x_1,&x_2,&x_3,&x_4,&x_5,&x_6,&x_7,&x_8,&x_9]\\
\[[7;&1,&2,&5,&1,&1,&3,&1,&1,&2]
\end{array}
$
}}$$
 

$$\small{\text{$
\begin{array}{lclcl}
\textcolor[rgb]{1,0,0}{x_0} &=& \textcolor[rgb]{1,0,0}{7}.6855172413793103 \\
&& 1/0.6855172413793103= \\
\textcolor[rgb]{1,0,0}{x_1}&=&\textcolor[rgb]{1,0,0}{1}.4587525150905433549486051115546459473433926066921394\\
&& 1/0.4587525150905433549486051115546459473433926066921394=\\
\textcolor[rgb]{1,0,0}{x_2}&=&\textcolor[rgb]{1,0,0}{2}.1798245614035083186653585718683317691923088345579715\\
&& 1/0.1798245614035083186653585718683317691923088345579715=\\
\textcolor[rgb]{1,0,0}{x_3}&=&\textcolor[rgb]{1,0,0}{5}.5609756097561115779298036883141182477038929421104669\\
&& 1/0.5609756097561115779298036883141182477038929421104669=\\
\textcolor[rgb]{1,0,0}{x_4}&=&\textcolor[rgb]{1,0,0}{1}.7826086956521293714555765605533410043560222722903369\\
&& 1/0.7826086956521293714555765605533410043560222722903369=\\
\textcolor[rgb]{1,0,0}{x_5}&=&\textcolor[rgb]{1,0,0}{1}.2777777777778505015432098790380390517833503327285810\\
&& 1/0.2777777777778505015432098790380390517833503327285810=\\
\textcolor[rgb]{1,0,0}{x_6}&=&\textcolor[rgb]{1,0,0}{3}.5999999999990575000000002144187499999512197343750242\\
&& 1/0.5999999999990575000000002144187499999512197343750242=\\
\textcolor[rgb]{1,0,0}{x_7}&=&\textcolor[rgb]{1,0,0}{1}.6666666666692847222222257391435185232429161265495199\\
&& 1/0.6666666666692847222222257391435185232429161265495199=\\
\textcolor[rgb]{1,0,0}{x_8}&=&\textcolor[rgb]{1,0,0}{1}.4999999999941093750000152199023437106755773194376239\\
&& 1/0.4999999999941093750000152199023437106755773194376239=\\
\textcolor[rgb]{1,0,0}{x_9}&=&\textcolor[rgb]{1,0,0}{2}.0000000000\ldots
\end{array}
$}}$$

 

2. The successive convergents with numerators p and denominators q are given by the

$$\small{ \text{ Formula }
\boxed{
\begin{array}{lcl}
\dfrac{p_{n+1}}{q_{n+1}} = \dfrac{x_{n+1}\cdot p_n + p_{n-1} }{x_{n+1}\cdot q_n + q_{n-1} } \qquad \dfrac{p_0} {q_0} = \dfrac{x_0}{1}=\dfrac{ \textcolor[rgb]{1,0,0}{7} }{1} \qquad \dfrac{p_{-1}} {q_{-1} } = \dfrac{1}{0} \\\\
$ For example $ n=0:~~
\dfrac{p_{1}}{q_{1}}
= \dfrac{x_1\cdot p_0 + p_{-1} }{x_1\cdot q_0 + q_{-1} }
= \dfrac{x_1\cdot x_0+1}{x_1\cdot 1+ 0 }
= \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 7 + 1 }{ \textcolor[rgb]{1,0,0}{1}\cdot 1 + 0} = \dfrac{8}{1} = 8\\\\
$ For example $ n=1:~~
\dfrac{p_{2}}{q_{2}}
= \dfrac{x_2\cdot p_1 + p_0 }{x_2\cdot q_1 + q_0 }
= \dfrac{x_2\cdot 8 + 7}{x_2\cdot 1+ 1 }
= \dfrac{\textcolor[rgb]{1,0,0}{2}\cdot 8 + 7 }{ \textcolor[rgb]{1,0,0}{2}\cdot 1+ 1} = \dfrac{23}{3} = 7.6\overline{6}\\\\
$ For example $ n=2:~~
\dfrac{p_{3}}{q_{3}}
= \dfrac{x_3\cdot p_2 + p_1 }{x_3\cdot q_2 + q_1 }
= \dfrac{x_3\cdot 23 + 8}{x_3\cdot 3+ 1 }
= \dfrac{\textcolor[rgb]{1,0,0}{5}\cdot 23 + 8 }{ \textcolor[rgb]{1,0,0}{5}\cdot 3+ 1} = \dfrac{123}{16} = 7.6875\\\\
$ For example $ n=3:~~
\dfrac{p_{4}}{q_{4}}
= \dfrac{x_4\cdot p_3 + p_2 }{x_4\cdot q_3 + q_2 }
= \dfrac{x_4\cdot 123 + 23}{x_4\cdot 16 + 3 }
= \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 123 + 23 }{ \textcolor[rgb]{1,0,0}{1}\cdot 16 + 3 } = \dfrac{146}{19} = 7.68421052631578947\ldots \\\\
$ For example $ n=4:~~
\dfrac{p_{5}}{q_{5}}
= \dfrac{x_5\cdot p_4 + p_3 }{x_5\cdot q_4 + q_3 }
= \dfrac{x_5\cdot 146 +123}{x_5\cdot 19+16 }
= \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 146 +123 }{ \textcolor[rgb]{1,0,0}{1}\cdot 19+16 } = \dfrac{269}{35} = 7.68571428571428571\ldots \\\\
$ For example $ n=5:~~
\dfrac{p_{6}}{q_{6}}
= \dfrac{x_6\cdot p_5 + p_4 }{x_6\cdot q_5 + q_4 }
= \dfrac{x_6\cdot 269+ 146}{x_6\cdot 35+19 }
= \dfrac{\textcolor[rgb]{1,0,0}{3}\cdot 269+ 146 }{ \textcolor[rgb]{1,0,0}{3}\cdot 35+19 } = \dfrac{953}{124} = 7.68548387096774194\ldots \\\\
$ For example $ n=6:~~
\dfrac{p_{7}}{q_{7}}
= \dfrac{x_7\cdot p_6 + p_5 }{x_7\cdot q_6 + q_5 }
= \dfrac{x_7\cdot 953+ 269 }{x_7\cdot 124 + 35 }
= \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 953+ 269 }{ \textcolor[rgb]{1,0,0}{1}\cdot 124 + 35 } = \dfrac{1222}{159} = 7.68553459119496855\ldots \\\\
$ For example $ n=7:~~
\dfrac{p_{8}}{q_{8}}
= \dfrac{x_8\cdot p_7 + p_6 }{x_8\cdot q_7 + q_6 }
= \dfrac{x_8\cdot 1222 + 953 }{x_8\cdot 159 + 124 }
= \dfrac{\textcolor[rgb]{1,0,0}{1}\cdot 1222 + 953 }{ \textcolor[rgb]{1,0,0}{1}\cdot 159 + 124 } = \dfrac{2175}{283} = 7.68551236749116607\ldots \\\\
$ For example $ n=8:~~
\dfrac{p_{9}}{q_{9}}
= \dfrac{x_9\cdot p_8 + p_7 }{x_9\cdot q_8 + q_7 }
= \dfrac{x_9\cdot 2175 + 1222 }{x_9\cdot 283 + 159 }
= \dfrac{\textcolor[rgb]{1,0,0}{2}\cdot 2175 + 1222 }{ \textcolor[rgb]{1,0,0}{2}\cdot 283 + 159 } = \dfrac{5572}{725} = 7.68551724137931034\ldots \\\\
\end{array}
}
}$$

 

$$7.6855172413793103 \approx \dfrac{5572}{725} = 7.68551724137931034\ldots$$

 

Aug 18, 2015
 #3
avatar+118723 
0
Aug 18, 2015

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