How to solve this system of equations?

[betito123]

I have 4 points and 4 equations of the form a(x-c)^b+d. I want to search for the parameters a,b,c and d. But i have no idea how to solve that. I tried using NSolve in Mathematica but to results popped up.

 

[lev888]

I have 4 points and 4 equations of the form a(x-c)^b+d. I want to search for the parameters a,b,c and d. But i have no idea how to solve that. I tried using NSolve in Mathematica but to results popped up.

Please post everything.

 

[betito123]

Please post everything.

1) 1 = a (1 - c)^b + d
2) 8 =a (16 - c)^b + d
3) 20 = a (31 - c)^b + d
4) 82 =a (45 - c)^b + d

 

[Deleted member 4993]

1) 1 = a (1 - c)^b + d
2) 8 =a (16 - c)^b + d
3) 20 = a (31 - c)^b + d
4) 82 =a (45 - c)^b + d

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment

Start with taking "logs"

 

[Romsek]

Searching 4D space for solutions to non-linear equations is a daunting task even for Mathematica.

Do you have some expected ranges of values for a,b,c,d? Something that could be useful for an initial guess?

 

[betito123]

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment

Start with taking "logs" - so that the equations become linear in log.

Thanks for your answer, since i have a d term, i cant see how using logs would help me solve this system.

 

[betito123]

Searching 4D space for solutions to non-linear equations is a daunting task even for Mathematica.

Do you have some expected ranges of values for a,b,c,d? Something that could be useful for an initial guess?

probably a=1, b=2,c=0,d=0. i tried using Findroot using the same method but it gives me some weird numbers

 

[betito123]

Searching 4D space for solutions to non-linear equations is a daunting task even for Mathematica.

Do you have some expected ranges of values for a,b,c,d? Something that could be useful for an initial guess?

using a graph of data, im trying to fit 4 points in a function that looks like a(x-c)^b+d

 

[firemath]

Using a graph was a good choice, but taking logs will work better and be much faster.
What math are you taking?

 

[Romsek]

using a graph of data, im trying to fit 4 points in a function that looks like a(x-c)^b+d

that 5D graph paper is mighty nifty!

 

[topsquark]

1) 1 = a (1 - c)^b + d
2) 8 =a (16 - c)^b + d
3) 20 = a (31 - c)^b + d
4) 82 =a (45 - c)^b + d

Creative use of logs or not this one is a nightmare. Is this the given problem or did you derive the equations from a larger problem?

-Dan

Adenedum: Sorry. I missed a post. Since you are graphing, do you know that an exact solution exists or are you trying to find a best fit of the four parameters, or something of that nature?

 

[Romsek]

using a graph of data, im trying to fit 4 points in a function that looks like a(x-c)^b+d

Have you considered the Mathematica function NonlinearModelFit?

I suspect it will solve this right quick for you now that I know what you're doing.

 

[Romsek]

Have you considered the Mathematica function NonlinearModelFit?

I suspect it will solve this right quick for you now that I know what you're doing.

your data doesn't fit this model very well.

 

[betito123]

Creative use of logs or not this one is a nightmare. Is this the given problem or did you derive the equations from a larger problem?

-Dan

Adenedum: Sorry. I missed a post. Since you are graphing, do you know that an exact solution exists or are you trying to find a best fit of the four parameters, or something of that nature?

Im trying to simply find the parameters of the function that i want to use. i have 4 points and i want to use a(x-c)^b+d.

 

[betito123]

Another question arises which is: Why is it so hard to find the parameters? i know i can do a polynomial based on those 4 points but like why is it so hard to solve that system?

 

[Romsek]

I suspect it has to do with extreme sensitivity of the solution to the parameter b

 

[Steven G]

1) 1 = a (1 - c)^b + d
2) 8 =a (16 - c)^b + d
3) 20 = a (31 - c)^b + d
4) 82 =a (45 - c)^b + d

Ah, so you do NOT have 4 equations of the form a(x-c)^b+d. You have only have one and it is equations of the form a(x-c)^b+d