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2/3*19^(1/2)*cos(1/3*arccos(7/38*19^(1/2))) - 1/3
Could anyone please help me simplifying this?
Thanks in advance,
Stas B.
If any one could solve
X'(y-Xβ)=0
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2.507018644 using a TI-36X Solar calculator
Same answer in radians or degrees, which makes sense if you think about it.
igloo myrtilles fourmis
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Stas B.: Using this bracketing: (((2/3)*(19^(1/2)))*(cos(((1/3)*(acos(((7/38)*(19^(1/2)))))))))-(1/3), Answer = 2.50701864409298
George, Graphically not Manually: Plot of 4*x^3-3*x-7/(2*sqrt(19))
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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Thanks, MIF. From your graph, I get positive solution about 1. And treat√19 = 4.5
Use these approximations, I get an answer 2+2/3.
m is the cosine.
Last edited by George,Y (2006-06-11 18:06:13)
X'(y-Xβ)=0
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As they say: "there are many ways to skin a cat"
(The function plotter is useful, even though it does seem like cheating.)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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I need an exact expression of the answear.
Like mine, only simpler.
Oh and George, if you were planning to find out the answear by solving that cubic aquation, that wouldn't work.
You would have to use the Cardano formula and get the cosine and arc-cosine back.
2/3*19^(1/2)*m-1/3
X'(y-Xβ)=0
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Oh, now I see!
Thanks!
But are you sure your cubic equation is correct?
It's useless.
Finding 'm' using that qubic equation doesn't give us anything because again we get the cosine and the arc-cosine I need to get rid of.
Unless you know another way to solve a cubic equation.
X'(y-Xβ)=0
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For the cubic equation, I don't think it will help, but here are the exact answers ( I bet you didn't thought this ):
IPBLE: Increasing Performance By Lowering Expectations.
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For the cubic equation, I don't think it will help, but here are the exact answers ( I bet you didn't thought this ):
m=-\sqrt{\frac{1}{2}-\frac{1}{8} \sqrt[3]{\frac{1}{38} \left(11+21 i \sqrt{3}\right)}-\frac{1}{8} i \sqrt{3} \sqrt[3]{\frac{1}{38} \left(11+21 i
\sqrt{3}\right)}-\frac{1}{4 2^{2/3} \sqrt[3]{\frac{1}{19} \left(11+21 i \sqrt{3}\right)}}+\frac{i \sqrt{3}}{4 2^{2/3} \sqrt[3]{\frac{1}{19} \left(11+21 i
\sqrt{3}\right)}}}\lor m=-\sqrt{\frac{1}{2}-\frac{1}{8} \sqrt[3]{\frac{1}{38} \left(11+21 i \sqrt{3}\right)}+\frac{1}{8} i \sqrt{3} \sqrt[3]{\frac{1}{38}
\left(11+21 i \sqrt{3}\right)}-\frac{1}{4 2^{2/3} \sqrt[3]{\frac{1}{19} \left(11+21 i \sqrt{3}\right)}}-\frac{i \sqrt{3}}{4 2^{2/3} \sqrt[3]{\frac{1}{19}
\left(11+21 i \sqrt{3}\right)}}}\lor m=\sqrt{\frac{1}{2}+\frac{1}{4} \sqrt[3]{\frac{1}{38} \left(11+21 i \sqrt{3}\right)}+\frac{1}{2 2^{2/3} \sqrt[3]{\frac{1}{19}
\left(11+21 i \sqrt{3}\right)}}}
IPBLE: Increasing Performance By Lowering Expectations.
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