Please Help!

Stas B. · 2006-05-29 02:44:32

sqrt(4 + sqrt(4 + sqrt(4 - x))) = x
Could anyone please help me solving that equation?
I came up with this:
x^8 - 16x^6 + 88x^4 - 192x^2 + x + 140 = 0
But then again, how do I solve that?

Thanks in advance,
Stast B.

John E. Franklin · 2006-05-29 03:22:04

In the sqrt 4's equation, x could be 2.5070186 or 2.5070187 or thereabouts.
I cheated and used JustBasic to program the equation and loop around.

Stas B. · 2006-05-29 03:34:51

The problem is, I need a mathematical proof, x = 2.5... isn't quite enough...
This is odd, I was sure there is an elegant way to solve that.
How about the second equation, is it actually solvable?

John E. Franklin · 2006-05-29 04:35:45

On your eighth power equation, again I cheated because I don't know how to solve it.
Here some numbers that you probably don't need for x that make the equation zero.
-2.5615528(1 or 2 for the last digit)
-2.2851424(8 or 9 for the eighth digit after the decimal)
-1.6510934(0 or 1)
-1.2218761(6 or 7)
1.2738905(5 or 6)
1.5615528(1 or 2)
2.3772028(5 or 6)
2.5070186(4 or 5)
Notice there are eight answers and it is power eight for highest, x^8
Maybe $2000 Mathematica program can show you how to solve it? I don't have it.

numen · 2006-05-29 05:00:22

There is no general solution to an x^8 equation, in fact, I think a general solution is impossible.

Solving it without the help of a computer would be a big waste of time. Your equation seem right though, I checked it.

Ricky · 2006-05-29 06:16:49

It isn't so much that there is no solution. For an order 8, sometimes, there are solutions. For example:

x^8 = 1

Or a more complex example:

(x - 3)(x + 2)(x - 5)(x + 1)(x - 10)(x +5)(x - 2)(x + 3) = 0

Someone can foil that out if they really want to.

But for any equation of order 5 or more, it is possible that the equation can not be solved by radicals. That is, the answer can not be expressed by using roots and rational numbers alone.

Stas B. · 2006-05-30 00:35:13

My math teacher said it has something to do with trigonometry.
Any ideas?

MathsIsFun · 2006-05-30 01:16:18

2.5070186 is close to sqrt(2π)=2.506628...

Is it related to a series for π?

Jai Ganesh · 2006-05-30 02:44:26

Devilish equation

Stas B. · 2006-05-30 07:05:21

The exact value of 'x' can be expressed using trigonometry, with a sine or a cosine (or some other trigonometric functions) of something and is epproximately equals to 2.5070186.
Any ideas?

John E. Franklin · 2006-05-30 10:13:36

Can't help for long right now but read this.
http://www.rism.com/Trig/values.htm
and then below if need to know what hav() is:
http://www.dynagen.co.za/eugene/where/sphertrg.html

krassi_holmz · 2006-06-02 05:51:25

HA-HA-HA!!!
x ≈ 2.5070186440929762986607999237156780290259
764201303696751265821783529769648210199715760
034086194090715665720271011885426265690772588
1312692288635583724
(150d)

And now, the exact result:

MathsIsFun · 2006-06-02 12:03:53

Monumental effort, krassi! How?

Stas B. · 2006-06-10 23:13:36

You solved it using Maple (or any other math software out there), didn't you?
I already solved the equation myself, but my teacher complains about my answear being long and ugly. (Containing cosines and arc-cosines and stuff.)
Yours looks better.
What does 'i' stand for?

Jai Ganesh · 2006-06-11 01:14:38

Stat B,
i is the the square root of -1.

i is the positive value of the solution of

,

the negative value of the solution being -i.

George,Y · 2006-06-11 01:45:27

Stas B. wrote:

You solved it using Maple (or any other math software out there), didn't you?
I already solved the equation myself, but my teacher complains about my answear being long and ugly. (Containing cosines and arc-cosines and stuff.)
Yours looks better.
What does 'i' stand for?

Maple and Mathematica, which is better? How do you feel about it guys??

In Plotting, I really support Maple.

Stas B. · 2006-06-11 02:47:00

Complex number stuff. I see.
But I need a solution with only real numbers.
Here's mine:
2/3*19^(1/2)*cos(1/3*arccos(7/38*19^(1/2))) - 1/3
Could anyone please help me simplifying it?

Stas B. · 2006-06-11 05:32:12

By the way, here's how I came up with my solution:

sqrt(4 + sqrt(4 + sqrt(4-x))) = x

Therefor:
4 > x > sqrt(4 + sqrt(4))
4 > x > sqrt(6)
(*)4 > x > 2.4

The original equation can be written as:
-140 - x + 192x^2 - 88x^4 + 16x^6 - x^8 = 0

By factorizing it we get:
-(x^2 + x - 4)(x^3 - 2x - 3x + 5)(x^3 + x^2 - 6x - 7) = 0

So we can solve the three equations that form the factors to get all eight roots.
We solve the first, find out that its roots don't satisfy (*).
Same goes for the second (cubic) equation.
Then we solve the third using the Cardano formula.
Only one of the three roots satisfy the condition, it is the one I've described in my previous post.

krassi_holmz · 2006-06-13 02:25:05

Stas B. wrote:

You solved it using Maple (or any other math software out there), didn't you?
I already solved the equation myself, but my teacher complains about my answear being long and ugly. (Containing cosines and arc-cosines and stuff.)
Yours looks better.
What does 'i' stand for?

George,Y wrote:

Stas B. wrote:
You solved it using Maple (or any other math software out there), didn't you?
I already solved the equation myself, but my teacher complains about my answear being long and ugly. (Containing cosines and arc-cosines and stuff.)
Yours looks better.
What does 'i' stand for?
Maple and Mathematica, which is better? How do you feel about it guys??
In Plotting, I really support Maple.

I agree Maple is better in plotting sometimes...
Or maybe just I can't "draw" good...
I'm for Mathematica

Here are some Mathematica results (by me, in this forum):
Search

George,Y · 2006-06-13 13:55:07

If Mathematica wasn't that clever in plotting!

In handling theoritical problems, Mathematica is better.

krassi_holmz · 2006-06-13 21:09:27

Agree.

krassi_holmz · 2006-06-13 21:11:23

But everything which can be done with the one, can be done and with the other (even the plotting in Mathematica, but it requires little more code than in Maple)

krassi_holmz · 2006-06-13 21:17:22

Stas B. wrote:

By the way, here's how I came up with my solution:
sqrt(4 + sqrt(4 + sqrt(4-x))) = x
Therefor:
4 > x > sqrt(4 + sqrt(4))
4 > x > sqrt(6)
(*)4 > x > 2.4
The original equation can be written as:
-140 - x + 192x^2 - 88x^4 + 16x^6 - x^8 = 0
By factorizing it we get:
-(x^2 + x - 4)(x^3 - 2x - 3x + 5)(x^3 + x^2 - 6x - 7) = 0

It's not that easy:

-(x^2 + x - 4)(x^3 - 2x - 3x + 5)(x^3 + x^2 - 6x - 7)=-140 + 55 x + 170 x^2 - 68 x^3 - 72 x^4 + 22 x^5 + 14 x^6 - 2 x^7 - x^8

So, wrong factorisation.

Last edited by krassi_holmz (2006-06-13 21:22:56)

krassi_holmz · 2006-06-13 21:20:08

Want to see Mathematica? Here is it (direct approach):

Last edited by krassi_holmz (2006-06-13 21:21:49)

Math Is Fun Forum

#1 2006-05-29 02:44:32

Please Help!

#2 2006-05-29 03:22:04

Re: Please Help!

#3 2006-05-29 03:34:51

Re: Please Help!

#4 2006-05-29 04:35:45

Re: Please Help!

#5 2006-05-29 05:00:22

Re: Please Help!

#6 2006-05-29 06:16:49

Re: Please Help!

#7 2006-05-30 00:35:13

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#8 2006-05-30 01:16:18

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#9 2006-05-30 02:44:26

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#10 2006-05-30 07:05:21

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#11 2006-05-30 10:13:36

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#12 2006-06-02 05:51:25

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#13 2006-06-02 12:03:53

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#14 2006-06-10 23:13:36

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#15 2006-06-11 01:14:38

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#16 2006-06-11 01:45:27

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#17 2006-06-11 02:47:00

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#18 2006-06-11 05:32:12

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#19 2006-06-13 02:25:05

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#20 2006-06-13 13:55:07

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#21 2006-06-13 21:09:27

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