Solve the equation [tex]2(x-1)^9(x+1)^8=1[/tex]

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Solve the equation [tex]2(x-1)^9(x+1)^8=1[/tex]

Postby Guest » Thu Jan 07, 2016 3:20 am

Find all real roots:
[tex]2(x-1)^9(x+1)^8=1[/tex]
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Re: Solve the equation [tex]2(x-1)^9(x+1)^8=1[/tex]

Postby Guest » Wed Jan 27, 2016 4:41 am

x=1.4217911
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Re: Solve the equation [tex]2(x-1)^9(x+1)^8=1[/tex]

Postby Guest » Sat Jan 30, 2016 6:02 am

How do you calculate it?
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Re: Solve the equation [tex]2(x-1)^9(x+1)^8=1[/tex]

Postby Guest » Sun Jan 31, 2016 7:08 am

A plot of the graph [tex]y=2(x-1)^9(x+1)^8-1[/tex] should convince you there is only one root (a root for this graph corresponds to a solution of the original equation). Formally proving this fact involves finding the turning points which are at [tex]x=-1,-1/17[/tex] and [tex]1[/tex] and showing that they all have [tex]y[/tex] values below [tex]0[/tex], and since [tex]y[/tex] is a polynomial of odd degree and the coefficient of the monomial of highest degree is positive, it can only have crossed the [tex]x[/tex]-axis once.

The root is between [tex]1[/tex] and [tex]2[/tex] (as the [tex]y[/tex] value changes from negative to positive). To get a more accurate answer you can use Newton-Raphson with an initial guess of 1.5:
[tex]x_0=1.5[/tex]
[tex]x_{n+1} = x_{n} -\frac{2(x_n-1)^9(x_n+1)^8-1}{2(17x_n+1)(x_n-1)^8(x_n+1)^7}[/tex]

Which gives (using a computer):
[tex]x_{1} = 1.4607439698113207[/tex]
[tex]x_{2} = 1.434299330659228[/tex]
[tex]x_{3} = 1.4233741563856475[/tex]
[tex]x_{4} = 1.421818929333896[/tex]
[tex]x_{5} = 1.4217910964455613[/tex]
[tex]x_{6} = 1.421791087714604[/tex]
[tex]x_{7} = 1.4217910877146032[/tex]
[tex]x_{8} = 1.4217910877146032[/tex]
[tex]x_{9} = 1.4217910877146032[/tex]
[tex]x_{10} = 1.4217910877146032[/tex]

As you can see the value converges and appears to be accurate to 16 decimal places after the 7th iteration.

I'm afraid there isn't an exact answer (involving simple operations like square roots and arithmetic operators on integers), the best we can do is give a numerical approximation.

Hope this helped,

R. Baber.
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Re: Solve the equation [tex]2(x-1)^9(x+1)^8=1[/tex]

Postby Guest » Sun Jan 31, 2016 7:45 am

In case you are interested, I did the computation using higher precision fixed point numbers (a computational representation of decimal numbers), and after the 10th iteration the value is in fact accurate to 224 decimal places (a testament to the sheer power of Newton-Raphson)

[tex]x_{10}[/tex] = 1.
42179 10877 14603 20455 54001 73768 98707 32097 77707 76988 78250 48530 16226 94221 24320 15284 09841 99596 69618 71790 54170 74141 15090 63168 91861 29773 05289 33980 41326 64231 16603 40210 50468 10364 94767 52329 65067 74774 34816 51475 30846 65350 70438 46528 3878

(Spaces were added to stop the number being automatically edited by the forum post reply editor.)

R. Baber.
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Re: Solve the equation [tex]2(x-1)^9(x+1)^8=1[/tex]

Postby Guest » Thu Feb 04, 2016 3:38 pm

Just apply intermediate value theorem on function [tex]f(x)=2(x-1)^9(x+1)^8[/tex] over interval [1.4, 1,5]...
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