OpenEuphoria

2. Re: Algebra problem

• Posted by ags <eu at 531pi.co.nz> Jul 07, 2005
• 629 views

Derek Parnell wrote:
> 
> My high school algebra has deserted me 
> 
>  Given the formula 
> 
>    A^b + Cb = d
> 
> how do I solve for 'b'?

Just pieced together from the web... could be quite wrong, sad eh? 

A^b + Cb = d

-> log(A)b + Cb = d

-> b(log(A) + C) = d

... b = d / (log(A) + C))

Gary

4. Re: Algebra problem

• Posted by "Juergen Luethje" <j.lue at gmx.de> Jul 07, 2005
• 624 views

Derek Parnell wrote:

> My high school algebra has deserted me 
>
>  Given the formula
>
>    A^b + Cb = d
>
> how do I solve for 'b'?

Just for the sake of clarity, I assume you mean the same as:
    A^b + C*b = d ,
right?

This is not simple, because b is not only exponent of A, but also
factor of C.

If we know the values of A, C, and d, we can go the following way:
Change the equation to
    A^b = -C*b + d

Plot the graphs of the two functions
    f(b) = A^b
    g(b) = -C*b + d

The b values of all points where both graphs intersect are solutions
of the original equation.

( Note:
  It is not actually necessary to plot 2 functions.
  We can also e.g. change the equation to
    A^b + C*b - d = 0 ,

  then plot the function
    f(b) = A^b + C*b - d

  and look for the points where the function graph intersects with
  the x-axis. )

                   -----------------

Example for
    A = 2
    C = 3
    d = 5

The equation to solve is then (using x instead of b, because the program
that I used for plotting only "knows" x):
    2^x + 3*x = 5

The two functions are then
    f(x) = 2^x
    g(x) = -3*x + 5

On the plot at <http://home.arcor.de/luethje/temp/equ.jpg>, we see that
there is one intersection at x = 1.

I hope someone else knows a simpler and more general way ...

Regards,
   Juergen


PS: Rob, I wanted to upload the "attachment" to www.rapideuphoria.com
   (as you once have offered). But I forgot the name of the directory,
   where the file then is. Maybe you can add some short instructions to
   the page <http://www.rapideuphoria.com/cgi-bin/usercont.exu?dbId=new>?

6. Re: Algebra problem

• Posted by Alexander Toresson <alexander.toresson at gmail.com> Jul 07, 2005
• 635 views

Derek Parnell wrote:
> 
> My high school algebra has deserted me 
> 
>  Given the formula 
> 
>    A^b + Cb = d
> 
> how do I solve for 'b'?
> 
> -- 
> Derek Parnell
> Melbourne, Australia
> 

I dunno really if this is solvable. Anyway, here are my attempts:

-----------------------

A^b + Cb = d

log(A^b) = log(d - Cb)

b * log(A) = log(d - Cb)

log(b * log(A)) = log(log(d - Cb))

log(b) + log(log(A)) = log(log(d - Cb))

e^(log(b) + log(log(A))) = log(d - Cb)

e^(e^(log(b) + log(log(A)))) = d - Cb

e^(e^(log(b)) + e^(log(log(A)))) = d - Cb

e^(b + log(A)) = d - Cb

--------------------

A^b + Cb = d

A^b = d - Cb

-A^b + d = Cb

(-A^b + d)/C = b

log((-A^b + d)/C) = log(b)

log(-A^b + d) - log(C) = log(b)

--------------------

A^b + Cb = d

A = (d - Cb)^(1/b)

log(A) = log((d - Cb)^(1/b))

log(A) = (1/b) * log(d - Cb)

log(A) * b = log(d - Cb)


Regards, Alexander Toresson

8. Re: Algebra problem

• Posted by don cole <doncole at pacbell.net> Jul 07, 2005
• 696 views

Derek Parnell wrote:
> 
> My high school algebra has deserted me 
> 
>  Given the formula 
> 
>    A^b + Cb = d
> 
> how do I solve for 'b'?
> 
> -- 
> Derek Parnell
> Melbourne, Australia
> 

Insufficient data to solve.

Say for example b=1 then you have

  A+C=d

A,C or d cannot be solved without knowing the value of A,C or d.

You could say A=d-C

or 

C=d-A

or d=A+C

But this really doesn't solve anything.

don cole
SF

10. Re: Algebra problem

• Posted by Al Getz <Xaxo at aol.com> Jul 07, 2005
• 609 views

Hi there,


This looked interesting so i entered two forms of the solution
into my Scientific Calculator programmed in Euphoria of course :) 
My first thought was to simply use a numerical solver to try this,
but then i decided to use the other solution also and compare results.

For reference the calculator's built in solver has this simple syntax:
  x=Solve(LeftSide,RightSide,x,guess,max_iterations,errorlimit)
  where
    x is the var being solved for
    LeftSide is the left side of the equation (A^b+C*b)
    RightSide is the right side of the equation (D) (d made cap D for clarity)
    guess is the initial guess
    max_iterations is the max number of iter's to attempt to find the solution
    errorlimit is the desired max error of the value of x returned

I entered these lines exactly like this:

A=2
C=3
D=5
b1=Solve(A^b+C*b,D,b,1,100,1e-6)
lambertw(X)=Solve(w*e^w,X,w,1,100,1e-6)
b2=-(lambertw(ln(A)/C*e^(D*ln(A)/C))*C-D*ln(A))/ln(A)/C

(i copy and pasted the right side of this last line assuming it was correct)
then selected all six lines and hit "In Place Eval" and i got two 
answers...b1 is using the direct built in solver, and b2 is using
the built in solver to calculate lambertw(X) to whatever precision
the errorlimit is set to and then find the solution given in the previous
post.

Using the direct built in solver, b1=1 exactly (to 16 digits).
Using the alternate solution, b2=0.999999673312148.

 From examining the alternate solution for other error limits, i could
quickly see that the precision of the answer depended mostly on the
error in calculating lambertw(X).  If i decreased the error limit,
i got a better answer.

Actually i entered a second form for lambertw(X) also and got the
same results as for the previous lambertw(X)...the higher the accuracy
in calculating lambertw the better the final answer.

I havent investigated the numerical stability of either method
any further however, and one should keep in mind that functions like
this can get quickly out of hand when dealing with the limit of 16
digits on most computers without additional numerical support.


Take care,
Al

And, good luck with your Euphoria programming!

My bumper sticker: "I brake for LED's"

12. Re: Algebra problem

• Posted by ags <eu at 531pi.co.nz> Jul 07, 2005
• 616 views
• Last edited Jul 08, 2005

Juergen Luethje wrote:
> -> log(A^b + Cb) = log(d)
> 
> And now ...?

I try to remember for next time :)

Gary

14. Re: Algebra problem

• Posted by "Juergen Luethje" <j.lue at gmx.de> Jul 08, 2005
• 620 views

Derek Parnell wrote:

> Erik-Jan van Kampen wrote:
>
>> Derek Parnell wrote:
>>>
>>> My high school algebra has deserted me 
>>>
>>>  Given the formula
>>>
>>>    A^b + Cb = d
>>>
>>> how do I solve for 'b'?
>>>
>>> -- 
>>> Derek Parnell
>>> Melbourne, Australia
>>>
>>
>> Hi Derek,
>>
>>  From the Matlab 6.5 Symbolic Toolbox:
>>
>>>> solve(A^b+C*b-d,b)
>>
>> ans =
>>
>> -(lambertw(log(A)/C*exp(d*log(A)/C))*C-d*log(A))/log(A)/C
>>
>
> Thanks. I thought it would be something simple like that ;-0
>
> This solution also implies that 'A' must > 1 and 'C' must be > 0
> otherwise there is no valid value for 'b'.

No generally.
A must not be = 1, but it can be < 1, e.g. for
   A = 0.6
   C = 3
   d = 5

there are the two solutions:
   b1 =  1.512753283
   b2 = -6.184691234

sequence s
atom A, C, d, b

A = 0.6
C = 3
d = 5
s = {1.512753283, -6.184691234}

-- check results
for i = 1 to length(s) do
   b = s[i]
   printf(1, "%f^%f + %f*%f = %f (expected: %f)\n", {A,b,C,b,power(A,b)+C*b,d})
end for

Also e.g.
   A =  2
   C = -3
   d =  5

works for me.

Not only positive arguments are allowed for the lambertw() function --
although I made this restriction in my "quick and dirty" Eu
implementation yesterday. I'm very sorry if that might have given you
a wrong idea.

The function lambertw(x) doesn't have a real value for x < -1/e,
and it apparently has two values when x is in the interval [-1/e, 0[
(see http://mathworld.wolfram.com/LambertW-Function.html).

> For those curious about this formula, it is being used in project task
> estimations.

[snipped interesting explanation]

Regards,
   Juergen