The best way to solve x³ + bx = c

[logics]

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?

 

[n_kelthuzad]

u mean 606087.936...? or 606087936

 

[logics]

u mean 606087.936...? or 606087936

Thanks, Victor, I corrected the typo, but, 846 or 84.6, I suppose the operations do not change.

 

[n_kelthuzad]

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

 

[n_kelthuzad]

now I look at my answer I got confused... cause the formula does not contain b so it is wrong?

 

[D H]

Your result, x=\exp((\log c)/3) is a convoluted way of writing x=\sqrt[3]c.

What you did with logarithms is wrong.

 

[Mentallic]

ln x + ln (x+ib) + ln (x-ib)=ln c
ln x + ln |x| + iarg(z) + ln |x| - iarg(z)=ln c

\ln(x+ib) should be \ln\sqrt{x^2+b^2}+i\cdot \arg(z)

Because for a complex number x+iy, we can convert it into the mod-arg form r e^{i\theta} and then \ln\left(r\cdot e^{i\theta}\right)=\ln(r)+i\theta where r=\sqrt{x^2+y^2}, \theta=\tan^{-1}\left(\frac{y}{x}\right)

 

[HallsofIvy]

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, (a- b)^3= a^3- 3a^2b+ 3ab^2- b^3 and 3ab(a- b)= 3a^2b- 3ab^2 so that (a- b)^3+ 3ab(a- b)= a^3- b^3. That is, if x=a- b, m= 3ab, n= a^3- b^3, x^3+ mx=n.

Now that will fit x^3+ 7x= 606,087.936 with 3ab= 7, a^3- b^3= 606087.936.<br /> <br /> Of course, that means b=7/a so a^3- 7^3/a^3= 606087.936. Multiplying through by a^3, gives (a^3)^2- 606087.936a^3- 7^3= 0, a quadratic function for a^3. We can solve for a^3 using the quadratic formula: <br /> a^3= \frac{606087.936\pm\sqrt{(606087.936)^2+ 4(7^3)}}{2}<br /> <br /> So computing a^3 involves: <br /> 1) Find (606087.936)^2- one operation<br /> 2) Find 4(7^3)- three operations<br /> 3) Add them- one operation<br /> 4) Take the square root- one operation<br /> 5) Add to or subtract from 606087.936- one operation<br /> 6) Divide by 2- one operation<br /> 7) Take the cube root<br /> <br /> That is, 8 operations are required to find just a. We don&#039;t have to repeat all that to find b^3, we have a^3- b^3= 606087.936 so b^3= a^3- 606087.936, 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.

 

[logics]

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?

 

[logics]

\ln(x+ib) should be \ln\sqrt{x^2+b^2}+i\cdot \arg(z)

Is that method valid, Mentallic, anyway?, could you, please, write the right procedure applying it to x³+9x=149354

 

[Mentallic]

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.

 

[logics]

...We can solve for a^3 using the quadratic formula:
a^3 = \frac{606087.936\pm\sqrt{(606087.936)^2+ 4(7^3 ?)}}{2}

a) Probably that is a typo, shouldn't it be 7³/ 27 ?

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

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? :
\ x^5 + bx^3 = c ,... x^5 + bx = c,... x^5 + bx^3 + cx =d, ...

*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?.

 

[HallsofIvy]

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.

 

[logics]

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
\frac{\sqrt[3]{2b/3}}{\sqrt[3]{\sqrt{3}\sqrt{4b^3+27c^2-9c}}} ...
is that any better?