1. goddamn equasions
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how do i solve this? with matrix?
you have 4 unknown variables and 4 equasions.
i need to know values of a,b,c,d.
#e is 2.71 or something like that.
^ means power (3^2=3D9)
135=3Da*1000^b*#e^(c*(0.2))*(0.1)^d
360=3Da*950^b*#e^(c*(0.4))*1^d
765=3Da*900^b*#e^(c*(0.6))*5^d
130=3Da*850^b*#e^(c*(0.2))*(0.1)^d
i tried it with Derive but it solves a shit
is there any other program i can get that will solve it for me?
here is verions of four equasions in C:
135=3Da*pow(1000,b)*exp(c*(0.2))*pow(0.1,d)
360=3Da*pow(950,b)*exp(c*(0.4))*pow(1,d)
765=3Da*pow(900,b)*exp(c*(0.6))*pow(5,d)
130=3Da*pow(850,b)*exp(c*(0.2))*pow(0.1,d)
and in BASIC:
135=3Da*1000^b*EXP(c*(0.2))*(0.1)^d
360=3Da*950^b*EXP(c*(0.4))*1^d
765=3Da*900^b*EXP(c*(0.6))*5^d
130=3Da*850^b*EXP(c*(0.2))*(0.1)^d
and in FORTRAN:
135=3Da*1000**b*EXP(c*(0.2))*(0.1)**d
360=3Da*950**b*EXP(c*(0.4))*1**d
765=3Da*900**b*EXP(c*(0.6))*5**d
130=3Da*850**b*EXP(c*(0.2))*(0.1)**d
and in PASCAL:
135=3Da*POW(1000,b)*EXP(c*(0.2))*POW(0.1,d)
360=3Da*POW(950,b)*EXP(c*(0.4))*POW(1,d)
765=3Da*POW(900,b)*EXP(c*(0.6))*POW(5,d)
130=3Da*POW(850,b)*EXP(c*(0.2))*POW(0.1,d)
ok, enough.
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<DIV><FONT face=3DVerdana size=3D2>how do i solve this? with =
matrix?</FONT></DIV>
<DIV><FONT face=3DVerdana size=3D2>you have 4 unknown variables =
and 4=20
equasions.</FONT></DIV>
<DIV><FONT face=3DVerdana size=3D2>i need to know values of =
a,b,c,d.</FONT></DIV>
<DIV><FONT face=3DVerdana size=3D2>#e is 2.71 or something like =
that.</FONT></DIV>
<DIV><FONT face=3DVerdana size=3D2>^ means power (3^2=3D9)</FONT></DIV>
<DIV><FONT face=3DVerdana size=3D2></FONT> </DIV>
<DIV><FONT face=3DVerdana =
<DIV> </DIV>
<DIV><FONT face=3DVerdana =
<DIV> </DIV>
<DIV><FONT face=3DVerdana =
<DIV> </DIV>
<DIV><FONT face=3DVerdana =
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2>i tried it with Derive but it solves =
a=20
shit</FONT></DIV>
<DIV><FONT face=3DVerdana size=3D2>is there any other program i can get =
that will=20
solve it for me?</FONT></DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2>here is verions of four equasions in=20
C:</FONT></DIV>
<DIV><FONT face=3DVerdana=20
<DIV> </DIV>
<DIV><FONT face=3DVerdana=20
<DIV> </DIV>
<DIV><FONT face=3DVerdana=20
<DIV> </DIV>
<DIV><FONT face=3DVerdana=20
<DIV> </DIV>
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2>and in BASIC:</FONT></DIV>
<DIV><FONT face=3DVerdana =
<DIV> </DIV>
<DIV><FONT face=3DVerdana =
<DIV> </DIV>
<DIV><FONT face=3DVerdana =
<DIV> </DIV>
<DIV><FONT face=3DVerdana =
<DIV> </DIV>
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2>and in FORTRAN:</FONT></DIV>
<DIV><FONT face=3DVerdana size=3D2> =20
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2> =20
360=3Da*950**b*EXP(c*(0.4))*1**d</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2> =20
765=3Da*900**b*EXP(c*(0.6))*5**d</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2> =20
130=3Da*850**b*EXP(c*(0.2))*(0.1)**d</FONT></DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2>and in PASCAL:</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DVerdana=20
<DIV> </DIV>
<DIV><FONT face=3DVerdana=20
<DIV> </DIV>
<DIV><FONT face=3DVerdana=20
<DIV> </DIV>
<DIV><FONT face=3DVerdana=20
<DIV> </DIV>
<DIV><FONT face=3DVerdana size=3D2></FONT> </DIV>
<DIV><FONT face=3DVerdana size=3D2>ok, =
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2. Re: goddamn equasions
koda wrote:
>>
solve:
135=a*1000^b*#e^(c*(0.2))*(0.1)^d
360=a*950^b*#e^(c*(0.4))*1^d
765=a*900^b*#e^(c*(0.6))*5^d
130=a*850^b*#e^(c*(0.2))*(0.1)^d
<<
I was working on a Euphoria program to solve non lin's but haven't gotten
back to it in a while. i hope to take a break from windows programming
soon and get back to it.
Offhand, this looks unsolvable in general because of all the multiplys.
Are you sure there are not any additions such as:
135 = a*1000^b + e^(c*(0.2)) + (0.1)^d ???
These can often be solved to some degree using non linear equation
solving methods. Can't solve them using linear algebra in general.
Also, still need n equations to solve for n variables in general.
--Al
3. Re: goddamn equasions
=A9koda wrote:
>
> Part 1.1 Type: Plain Text (text/plain)
> Encoding: quoted-printable
135=3Da*1000^b*#e^(c*(0.2))*(0.1)^d
360=3Da*950^b*#e^(c*(0.4))*1^d
765=3Da*900^b*#e^(c*(0.6))*5^d
130=3Da*850^b*#e^(c*(0.2))*(0.1)^d
First, use the nat. Logarithm on all eqs. for they are all multipicativ.
That wil make ti possible to solve a linear System with Derive.
I'm right now in a hurry. Rolf
4. Re: goddamn equasions
©koda wrote:
>>
135=a*1000^b*#e^(c*(0.2))*(0.1)^d
360=a*950^b*#e^(c*(0.4))*1^d
765=a*900^b*#e^(c*(0.6))*5^d
130=a*850^b*#e^(c*(0.2))*(0.1)^d
<<
I'm assuming he meant:
let n be subscripts: {1,2,3,...,N=number of equations} and
let An,Bn,Cn,Dn be subscripted constants: {A1,A2,..B1,B2..etc} and
let a,b,c,d be variables, then
the system:
An = a*Bn^b * e^(c*Cn) * Dn^d
isn't solvable.
Changing the form to a commonly encountered form
by insert additions however and its a different story:
An = a*Bn^b + e^(c*Cn) + Dn^d
This system can often be solved using non linear equation
solving methods.
--Al
5. Re: goddamn equasions
Al Getz wrote:
> Are you sure there are not any additions such as:
> 135 = a*1000^b + e^(c*(0.2)) + (0.1)^d ???
No, my original equasion is right.
> Also, still need n equations to solve for n variables in general.
Yes, that's what i know also and so it must be solvable.
Thanks to everybody who helped.
Matthew Lewis solved it with Excel functions, don't ask me how, cause it's
nothing clear to me.
Anybody interseted, he found this:
a=45.27
b=.232
c=1.203
d=.327
6. Re: goddamn equasions
Skoda,
Just to clear one thing up:
it's not that you need n equations to be able to solve a set of equations
with n variables , it's that you need n INDEPENDENT equations. Graphical=
ly
that means the equations give n DIFFERENT lines (or curves), all
intersecting at one (or more) common point(s).
x=3D y + 1
2x =3D 2y + 2
are NOT 2 independent equations.
Dan
----- Original Message -----
From: "=A9koda" <tone.skoda at SIOL.NET>
To: <EUPHORIA at LISTSERV.MUOHIO.EDU>
Sent: Thursday, September 14, 2000 12:36 PM
Subject: Re: goddamn equasions
> Al Getz wrote:
>
> > Are you sure there are not any additions such as:
> > 135 =3D a*1000^b + e^(c*(0.2)) + (0.1)^d ???
>
> No, my original equasion is right.
>
> > Also, still need n equations to solve for n variables in general.
>
> Yes, that's what i know also and so it must be solvable.
>
> Thanks to everybody who helped.
> Matthew Lewis solved it with Excel functions, don't ask me how, cause i=
t's
> nothing clear to me.
> Anybody interseted, he found this:
>
> a=3D45.27
> b=3D.232
> c=3D1.203
> d=3D.327
7. Re: goddamn equasions
Koda,
Sorry, i didnt realize you were asking such a trivial question.
I had thought you stuck in any old constants and such just to
illustrate the nature of the question you had, especially since
you never see the number 1 raised to a power in any real life
situation.
You happen to be lucky here, the system you are trying to solve
looks like a textbook example, as it can be shown to have only one
independent variable, a very rare occurance in real life. It can
be solved using simple algebra and natural logarithms.
The specific system:
[1] 135= a * 1000^b * e^(.2c) * (0.1)^d
[2] 360= a * 950^b * e^(.4c) * 1^d
[3] 765= a * 900^b * e^(.6c) * 5^d
[4] 130= a * 850^b * e^(.2c) * (0.1)^d
together with the relationship:
A^b=e^(B*b)
can be converted into the system:
135= a * e^(B1*b + C1*c + D1*d)
360= a * e^(B2*b + 2*C1*c + 0 )
765= a * e^(B3*b + 3*C1*c + D3*d)
130= a * e^(B4*b + C1*c + D1*d)
where:
B1=ln(1000)
B2=ln(950)
B3=ln(900)
B4=ln(850)
C1=.2
D1=ln(.1)
D3=ln(5)
Now divide all by a and taking the log of both sides leads to:
[5] ln(135/a)= B1*b + C1*c + D1*d
[6] ln(360/a)= B2*b + 2*C1*c
[7] ln(765/a)= B3*b + 3*C1*c + D3*d
[8] ln(130/a)= B4*b + C1*c + D1*d
Now subtracting eq[5]-eq[8] and solving for b in terms of a yields:
[9] b = (ln(135/a)-ln(130/a))/(B1-B4)
Now substitute b into eq[6] and solve for c in terms of a:
[10] c = (ln(360/a)-B2*b)/2/C1
Now substitute b and c into eq[7] and solve for d in terms of a:
[11] d = (ln(765/a)- B3*b - 3*C1*c)/D3
This shows "a" is the only independent variable in that system.
Now let variable "a" range in value numerically,
solve for b,c,d in terms of this a,
then substitute into eq[1] and calculate the right side numerically.
When the left side equals the right side you have the solution
to a,b,c and d.
Since you have only one independent variable, you can also
use Newtons method for roots for that last step:
rearrange eq[1] and let:
f(x)= 135 - a * 1000^b * e^(.2c) * (0.1)^d
where f(x) is an error function of x created by subtracting
the left side of eq[1] from the right side of eq[1], and
letting a = x and calculating b,c,d from eq[9], eq[10], eq[11].
Now choosing x such that f(x)=0 and we have a solution.
To help arrive at this x, pick a guess, say, 20, then:
first pass:
a = x - f(x)/f'(x) (Newtons)
= 20 - (-25.96575)/(1.733109)
= 34.98218
second pass:
a = x - f(x)/f'(x)
= 34.98218 - ( -7.707)/(0.878447)
= 43.75578
third pass:
a = x - f(x)/f'(x)
= 43.75578 - ( -.99323)/(.66925)
= 45.23984
fourth pass:
a = x - f(x)/f'(x)
= 45.23984 - ( -0.0199605)/(.6426765)
= 45.27090
fifth pass:
a = x - f(x)/f'(x)
= 45.27090 - ( -8E-6)/(.642)
= 45.270915
after which var a doesnt change much because the second term
approaches zero.
On my calculator i came up with the following values:
a=45.270915
b=0.232221121
c=1.20305991
d=0.326645325
f'(x) is the numerical derivative taken at x, and can be approx:
f'(x)= [ f(x)- f(x + delta) ]/(delta + delta)
where delta is typically set equal to some small value, like 0.0001.
Its sometimes necessary to analyze the equations using calculus
to see if more solutions exist.
Oh and yes, the n equations have to be independent as dan was saying too.
(thanks dan)
Have fun with it,
--Al
8. Re: goddamn equasions
Koda,
Sorry in my last post i quickly typed in the wrong approximation to the
first derivative.
The correct approximation to the first derivative is:
f'(x)= [ f(x + delta)- f(x) ]/(delta + delta)
where delta is typically set equal to some small value, such as
0.0001.
9. Re: goddamn equasions
Thanks Al Getz .
> Oh and yes, the n equations have to be independent as dan was saying too.
> (thanks dan)
I knew that also...
> Have fun with it,
> --Al
I'll try... 
10. Re: goddamn equasions
koda,
One additional point with that special system:
[5] ln(135/a)= B1*b + C1*c + D1*d
[6] ln(360/a)= B2*b + 2*C1*c
[7] ln(765/a)= B3*b + 3*C1*c + D3*d
[8] ln(130/a)= B4*b + C1*c + D1*d
with the help of the relationship:
ln(A/B)=ln(A)-ln(B)
you can convert to an "almost" total linear system:
ln(135)= B1*b + C1*c + D1*d + ln(a)
ln(360)= B2*b + 2*C1*c + ln(a)
ln(765)= B3*b + 3*C1*c + D3*d + ln(a)
ln(130)= B4*b + C1*c + D1*d + ln(a)
and solve by the usual method of simultaneous solutions:
x = M^(-1) * v
{ x,v vectors, M square matrix }
Then:
a = e^(ln (a) )
constants repeated here:
B1=ln(1000)
B2=ln(950)
B3=ln(900)
B4=ln(850)
C1=.2
D1=ln(.1)
D3=ln(5)
Usually you need to use numerical analysis with these types of equations.
This was a very special case.
--Al
