RE-ARRANGING EQUATION

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DYLANUK11999

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equation.png

Am trying to find how to re-arrange this equation to show that it equals 956N (Although I think it is 955, but hey-ho!)

Basically, I am trying to enter this into excel as this formula will be used often
 
DYLANUK11999 said:
View attachment 33334

Am trying to find how to re-arrange this equation to show that it equals 956N (Although I think it is 955, but hey-ho!)

Basically, I am trying to enter this into excel as this formula will be used often
Click to expand...
You could multiply by T2^2, but then you'd have to solve a cubic equation, which is very hard to do as a formula.

With these particular numbers, there are three solutions, one positive; the easiest way to find them is with software. I graphed your equation in Desmos, and it gave this answer:

1657241756003.png

The actual answer is closer to 955 than to 956, but is neither of them: about 955.278. Perhaps there's a reason someone would want to round up?
 
DYLANUK11999 said:
View attachment 33334

Am trying to find how to re-arrange this equation to show that it equals 956N (Although I think it is 955, but hey-ho!)

Basically, I am trying to enter this into excel as this formula will be used often
Click to expand...
You could use Newton's method of calculating roots of polynomial functions and estimate the value. But prior to doing that, it is a good idea to plot the function (y = x3 + 6863 * x2 - 7134611197) and have rough approximation/s of the answer/s.

1657299810095.png
 
I have done a bit more digging and found the attachment formula. Is anyone able to assist in how the highlighted numbers are derived in the quadratic formula?Quad Formula.png
 
DYLANUK11999 said:
I have done a bit more digging and found the attachment formula. Is anyone able to assist in how the highlighted numbers are derived in the quadratic formula?View attachment 33345
Click to expand...
We might if we knew where this formula came from. You call it a "quadratic" but I suspect it's a formula for a cubic equation, seeing as that's what you are solving. Can you show us the source of this expression?

-Dan
 
DYLANUK11999 said:
I have done a bit more digging and found the attachment formula. Is anyone able to assist in how the highlighted numbers are derived in the quadratic formula?View attachment 33345
Click to expand...
Presumably it's related to the method shown here:

en.wikipedia.org

Cubic equation - Wikipedia

en.wikipedia.org en.wikipedia.org

I tend to forget about the trigonometric approach to cubics! Looking for a simple formula version, I ran across this paper, which includes a relevant comment:

I was once asked a question that came down after a while to solving cubics of the casus irreducibilis type. Naturally, I suggested Newton’s Method. That wasn’t suitable, since the person needed a formula, for use with an early spreadsheet, so I used the trigonometric method. There was even some money in it—not much, but enough for a fair number of bottles of cheap wine. And there was pleasure in seeing an antique formula, probably not used for practical purposes for many years, turn out to be useful.​

So far, I haven't found a good site with a formula that would directly apply to an equation of your form (which is not the "depressed" form that they typically start from) and produce a result like this. But the formula you have clearly involves the [imath]\cos(3\theta)[/imath] that is involved in the method discussed.
 
I found a site with a very usable formula:

Trigonometry/The solution of cubic equations - Wikibooks, open books for an open world

en.wikibooks.org en.wikibooks.org

Here are the values used in that formula for your problem:

1657335699293.png​

This gives all three solutions as found by graphing the cubic equation (the first being the positive solution you want):

1657335942944.png​

The only thing to complain about is that this isn't the same formula used by the site you found, so we can't explain their numbers without digging deeper. I see no reason to do so.
 
Sorry, I haven't got the time to format this more nicely. Nonetheless, it should explain the trig method.
(Although the method post#12 seems much simpler :ROFLMAO: )

Rich (BB code): 
6863 = 7134611197/t^2 - t

6863*t^2 = 7134611197 - t^3

t^3 + 6863*t^2 - 7134611197 = 0

=========

Convert to a depressed cubic (which has no squared term)...

First, make sure the t^3 term has no coefficient (if it does then remove it by division/ multiplication)

Then, substitute t = x + b where "b" is the t^2 coefficient divided by 3, in this case
b = -6863/3

(x - 6863/3)^3 + 6863*(x - 6863/3)^2 - 7134611197 = 0

x^3 - 47100769/3*x +  453870652975/27 = 0

Now we have a depressed cubic in the form of

x^3 + p*x + q = 0  AA

where p = -47100769/3
and q = 453870652975/27

=========

Equation AA will be made similar to this trig identity...

4*cos^3(theta) - 3*cos(theta) - cos(3*theta) = 0  BB

=========

Substitute x = a*cos(theta) into AA, and constrain a to be positive, giving...

a^3*cos^3(theta) + p*a*cos(theta) + q = 0  CC

Multiply by 4/a^3...

4*cos^3(theta) + 4*p/a^2*cos(theta) + 4*q/a^3 = 0  DD

Now the first terms (of BB and DD) match

--

Equate the 2nd term in both DD and BB...

4*p/a^2*cos(theta) = - 3*cos(theta)
4*p/a^2 = -3
a^2 = -4*p/3
therefore a = 2*sqrt(-p/3)

--

Now equate 3rd term in both DD and BB...

cos(3*theta) = -4*q/a^3
      = -4*q/(2*sqrt(-p/3))^3
      = -q/(2*(-p/3)^(3/2))
theta = (2*k*pi ± acos(-q/(2*(-p/3)^(3/2))))/3

Plug in our numbers
      = (2*k*pi ± acos( -(453870652975/27)/(2*(-(-47100769/3)/3)^(3/2)) ))/3
      = (2*k*pi ± acos( -453870652975/646505155294 ))/3
     
--

NEARLY DONE!

Now remember that we set...
x = a*cos(theta)
x = 2*sqrt(-p/3)*cos(theta)    using the value of "a" found above

Also remember that t = x + b (see above)...
t = 2*sqrt(-p/3)*cos(theta) + b

t = 2*sqrt((47100769/3)/3)*cos(theta) + b
t = 13726/3*cos(theta) + b
t = 13726/3*cos(theta) - 6863/3

Finally...

Using theta = (2*k*pi + acos(-453870652975/646505155294))/3

k=0 gives t = 955.27763725558437718745
k=1 gives t = -6704.26681472689808061425
k=2 gives t = -1114.01082252868629661893

It also works for the "minus" values
theta = (2*k*pi - acos(-453870652975/646505155294))/3

k=0 gives t = 955.27763725558437718745
k=1 gives t = -1114.01082252868629657318
k=2 gives t = -6704.26681472689808066001
 
Dr.Peterson said:
I found a site with a very usable formula:

Trigonometry/The solution of cubic equations - Wikibooks, open books for an open world

en.wikibooks.org en.wikibooks.org

Here are the values used in that formula for your problem:


This gives all three solutions as found by graphing the cubic equation (the first being the positive solution you want):


The only thing to complain about is that this isn't the same formula used by the site you found, so we can't explain their numbers without digging deeper. I see no reason to do so.
Click to expand...
Apologies for the late reply, I thought the thread had gone dead as I had not received any notifications. To be cheeky is there any way of having the excel spreadsheet that you used above?

Thankyou for taking the time to reply.

And thankyou to everyone else that commented.
 
DYLANUK11999 said:
is there any way of having the excel spreadsheet that you used above?
Click to expand...
Let's see if I'm allowed to attach it: no.

See if this .csv file works for you (rename it to .csv and try opening it):
 

Attachments

  • FMH134839 cubic by trig.txt
    338 bytes · Views: 3
DYLANUK11999 said:
Apologies for the late reply, I thought the thread had gone dead as I had not received any notifications. To be cheeky is there any way of having the excel spreadsheet that you used above?

Thankyou for taking the time to reply.

And thankyou to everyone else that commented.
Click to expand...
No problem i have done this now. Thanks again!
 
1659211405924.png

1659211493673.png

Sorry to be a pain on this again, I have worked further through the calculations we have been provided and entering the above numbers a slightly different formula is used. I dont suppose anyone is familiar with the above?
 
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