The best way to solve x³ + bx = c

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What is the quickest way to solve a Khayyàm equation when b is not a square number?

Consider: x³ + 7x = 606,087.936,
( x³ + 5x = 132906...)

using the best algorithm, how many operations are required to find the solution ( x = 84.6, 51... ) ?
If we regard it as a reduced/depressed form, is it possible to find the original equation (or are there more than one) having the same solution?
 
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  • #2
u mean 606087.936...? or 606087936
 
  • #3
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u mean 606087.936...? or 606087936
Thanks, Victor, I corrected the typo, but, 846 or 84.6, I suppose the operations do not change.
 
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  • #4
x^3+bx=c
x(x^2+b)=c
x(x+ib)(x-ib)=c
ln (x(x+ib)(x-ib))=ln c
ln x + ln (x+ib) + ln (x-ib)=ln c
ln x + ln |x| + iarg(z) + ln |x| - iarg(z)=ln c
ln x + ln x + ln x = ln c
3ln x = ln c
ln x = (ln c)/3
x=e^(ln c)/3
 
  • #5
now I look at my answer I got confused... cause the formula does not contain b so it is wrong???
 
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  • #6
D H
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Your result, [itex]x=\exp((\log c)/3)[/itex] is a convoluted way of writing [itex]x=\sqrt[3]c[/itex].

What you did with logarithms is wrong.
 
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  • #7
Mentallic
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ln x + ln (x+ib) + ln (x-ib)=ln c
ln x + ln |x| + iarg(z) + ln |x| - iarg(z)=ln c
[itex]\ln(x+ib)[/itex] should be [tex]\ln\sqrt{x^2+b^2}+i\cdot \arg(z)[/tex]

Because for a complex number x+iy, we can convert it into the mod-arg form [tex]r e^{i\theta}[/tex] and then [tex]\ln\left(r\cdot e^{i\theta}\right)=\ln(r)+i\theta[/tex] where [tex]r=\sqrt{x^2+y^2}, \theta=\tan^{-1}\left(\frac{y}{x}\right)[/tex]
 
  • #8
HallsofIvy
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What is the quickest way to solve a Khayyàm equation when b is not a square number?

Consider: x³ + 7x = 606,087.936,
( x³ + 5x = 132906...)

using the best algorithm, how many operations are required to find the solution ( x = 84.6, 51... ) ?
If we regard it as a reduced/depressed form, is it possible to find the original equation (or are there more than one) having the same solution?
For any numbers, a and b, [itex](a- b)^3= a^3- 3a^2b+ 3ab^2- b^3[/itex] and [itex]3ab(a- b)= 3a^2b- 3ab^2[/itex] so that [itex](a- b)^3+ 3ab(a- b)= a^3- b^3[/itex]. That is, if x=a- b, m= 3ab, [itex]n= a^3- b^3[/itex], [itex]x^3+ mx=n[/itex].

Now that will fit [itex]x^3+ 7x= 606,087.936[/itex] with 3ab= 7, [itex]a^3- b^3= 606087.936.

Of course, that means b=7/a so [itex]a^3- 7^3/a^3= 606087.936[/itex]. Multiplying through by [itex]a^3[/itex], gives [itex](a^3)^2- 606087.936a^3- 7^3= 0[/itex], a quadratic function for [itex]a^3[/itex]. We can solve for [itex]a^3[/itex] using the quadratic formula:
[tex]a^3= \frac{606087.936\pm\sqrt{(606087.936)^2+ 4(7^3)}}{2}[/tex]

So computing [itex]a^3[/itex] involves:
1) Find [itex](606087.936)^2[/itex]- one operation
2) Find [itex]4(7^3)[/itex]- three operations
3) Add them- one operation
4) Take the square root- one operation
5) Add to or subtract from 606087.936- one operation
6) Divide by 2- one operation
7) Take the cube root

That is, 8 operations are required to find just a. We don't have to repeat all that to find [itex]b^3[/itex], we have [itex]a^3- b^3= 606087.936[/itex] so [itex]b^3= a^3- 606087.936[/itex], a ninth operation, and then we must take the cube root, a tenth operation, to find b itself. Finally, x=a- b after a total of eleven operations.
 
  • #9
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That is, 8 operations are required . Finally, x=a- b after a total of eleven operations.
Thanks a lot, Sir.
What is the name of this method?, only 11, actually 9 [if we consider 7³ just one] operations is definitely an eccellent achievement, it should apply also to equations where b is negative, in the region where there is only one solution i.e.:
x³ -9x = 148400 (x = 53, c> 10,3...) or maybe even when x > 3.
Can we say this algorithm gives always the or one of the solutions of a depressed cubic?
 
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  • #10
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[itex]\ln(x+ib)[/itex] should be [tex]\ln\sqrt{x^2+b^2}+i\cdot \arg(z)[/tex]
Is that method valid, Mentallic, anyway?, could you, please, write the right procedure applying it to x³+9x=149354
 
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  • #11
Mentallic
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Is that method valid, Mentallic, anyway?, could you, please, write the right procedure applying it to x³+9x=149354
No, it doesn't get us any closer to solving the problem at hand. I was just showing n_kelthuzad why his solution was invalid.
 
  • #12
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...We can solve for [itex]a^3[/itex] using the quadratic formula:
[tex]a^3 = \frac{606087.936\pm\sqrt{(606087.936)^2+ 4(7^3 ?)}}{2}[/tex]
a) Probably that is a typo, shouldn't it be 7³/ 27 ?

[itex]a^3 = \left(\frac{x + \sqrt{x^2 + 4*b}}{2}\right)^3 = \frac{c+\sqrt{c^2 + 4 * b^3/3^3}}{2}[/itex]

b) What happens if b is negative: x³ - bx = c, how do we find the other two solutions?

c) * Is there a similar brilliant solution for quintic equations with only one solution? :
[itex]\ x^5 + bx^3 = c ,... x^5 + bx = c,... x^5 + bx^3 + cx =d[/itex], .....

*should I make a new thread for this question?
If we regard it as a reduced/depressed form, is it possible to find the original equation (or are there more than one) having the same solution?
Can we find an equivalent original equation x³ + bx² +cx.... having only one solution : 84.6?, does it depend on the relation between b and c?.
 
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  • #13
HallsofIvy
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Thanks a lot, Sir.
What is the name of this method?
That is "Cardano's cubic formula"
http://en.wikipedia.org/wiki/Cubic_function

only 11, actually 9 [if we consider 7³ just one] operations is definitely an eccellent achievement, it should apply also to equations where b is negative, in the region where there is only one solution i.e.:
x³ -9x = 148400 (x = 53, c> 10,3...) or maybe even when x > 3.
Can we say this algorithm gives always the or one of the solutions of a depressed cubic?
I don't know what you could mean by "the" solution of a cubic.
 
  • #14
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I don't know what you could mean by "the" solution of a cubic.
I consider the question from the point of view of logics:
If, [on a "function explorer"], you look at the curve x³+ bx = c you see a curve similar, almost identical, to x³= c.. Could one ever say that x³ = c has one real and two imaginary solutions? Am I missing something?
I suppose that if, historically, "casus irriducibilis" had not occurred, we would not be here to talk about imaginary solutions (or numbers).

I know the fundamental theorem says "nth power-n zeroes", but I suppose it is a general frame and it is understood "at most" with b> 0, and only odd powers : x³+x;.. x^5 + x³+ x, etc..., there is only one solution, am I wrong?.
That is why I ventured to say that x³ + bx= 0 is a cubic equation with only one solution. How should one identify these equations?
On the other hand, also a cubic with b<0 or with even power has 2/more roots in an infinitesimal region of the y-axis, elsewhere we find the solution.
I hope I made myself clear.
Is there an algorithm to solve a depressed quintic x^5+.... when there is only one solution?

P.S: at wolfram, sometimes they show the algorithm, a formula starting with
[itex]\frac{\sqrt[3]{2b/3}}{\sqrt[3]{\sqrt{3}\sqrt{4b^3+27c^2-9c}}}[/itex] ....
is that any better?
 
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