Finding a Function from Experimental Data

[albert12345]

I've been given a set of data from an experiment:

x= 0, 1, 2, 5, 7, 10
Y=(3), (5), (8,7), (25,4) (40), (66,2)

My mission is to come up with a function that i can use, based on these values I've been given. I've tride to LN the values so I get a straight line (Y=kx +m), but I don't know what the next step is. I am stuck. Really need your help!

Thanks,

 

[genericusrnme]

LN?
Natural Log?

What do you mean when you say y=(8,7)?
Is there two different values for y?
If there is then you're not going to get a function that gives you those values back..

 

[albert12345]

LN?
Natural Log?

What do you mean when you say y=(8,7)?
Is there two different values for y?
If there is then you're not going to get a function that gives you those values back..

When x is 2 then y= 8,7
when x is 5 then y= 25,4

Yes, natural log. You need to come up with a function that works with the values I got. I think it's a power function.

 

[technician]

If you suspect that it is a power function then the starting point is to write a power equation relating y and x. What would you write?

 

[albert12345]

If you suspect that it is a power function then the starting point is to write a power equation relating y and x. What would you write?

I think it shoul look like this: Y = C +ax^n
The problem is to get the values a and n.
C is given since y = 3 when x =0 --> C=3.

I don't know what to do next, I've tried a few things but now I'm just lost :S

 

[technician]

If you have y=3 when x=0 I would rewrite the equation as (y-3) = ax^n
Then you can take Ln(y-3) and Ln(x) to plot a graph.
If you get a straight line then the gradient =n and the intercept = Ln(a)
You cannot take Ln(y) and Ln (C + ax^n)
Hope this helps.
I will have a go with your numbers and see what I get !

 

[technician]

I got a pretty good straight line and my calculated gradient an intercept reproduced the table of values to better than 10%
I will be interested to see what you get.

 

[albert12345]

If you have y=3 when x=0 I would rewrite the equation as (y-3) = ax^n
Then you can take Ln(y-3) and Ln(x) to plot a graph.
If you get a straight line then the gradient =n and the intercept = Ln(a)
You cannot take Ln(y) and Ln (C + ax^n)
Hope this helps.
I will have a go with your numbers and see what I get !

I don't really understand what you will do from here:Then you can take Ln(y-3) and Ln(x) to plot a graph.

What function did you come up with?

 

[technician]

Are you OK with the step
Y = 3 + ax^n ? You cannot take Ln(3 + ax^n) so I rearranged the expression to be
Y-3 = ax^n and this is the power law equatio that can be analyzed
Now take logs...Ln(y-3) = Ln(a) + nLn(x)
This equation is in the form of a straight line graph ( y = mx + C)
So plot Ln( y-3) against Ln(x) and you should get a straight line.
The slope of the line is n and where the line cuts the Ln( y-3) axis is Ln(a)
You need to make another table to get the values ( y-3), Ln(y-3) and Ln(x)
Hope this helps!

 

[albert12345]

I tried it for x =7, y=40.

Ln (40-3) = ln (a) + n ln (7)
3,6109 = ln (a) + n * 1,94

The problem now is that I have two variables? I don't know how to figure them out. You told me to plot the result but It's difficult with two variables. Mabye I am just stupid?

 

[albert12345]

No one who can help me?

 

[technician]

I will write my working out for you and send it...give me about 10mins...dont despair

 

[albert12345]

I will write my working out for you and send it...give me about 10mins...dont despair

Thanks a lot!

 

[technician]

here you go, not brilliant copy but I hope you can see what I did

 

[albert12345]

Thank you. That's similar to what I did the first time I tried but I guess I got a bit confused :) I managed to get a function that is very precise now :D Thank you for your valuable help! It feels great when you finally understand :D