curve fitting tool plot is different from my plot

Dear all,

I want to correlate y to x through the curve fitting tool by using a Power2 equation: f(x) = a*x^b+c

x = [1.34673	1.35171	1.3531	1.35402	1.35636	1.35875	1.3608	1.36278	1.36491	1.36728	1.36855	1.37048	1.37468	1.37505	1.37507	1.37917	1.38045	1.38387	1.38449	1.38473	1.38898	1.39084	1.3925	1.39367	1.39505	1.39716	1.39903	1.40043	1.4022	1.40435	1.40562	1.40767	1.40918	1.41063	1.41219	1.41371	1.41533	1.41688	1.41856	1.4197	1.42117	1.42366	1.42503	1.4265	1.42797	1.4293	1.4306	1.432	1.43331	1.43479	1.437	1.43919	1.4416	1.44366	1.44573];
y = [10.41769233	11.9053698	13.1164268	14.25963171	15.00649903	16.35618123	17.45629085	18.49632079	19.61280955	20.63621936	21.69999059	22.64564288	23.56982874	24.47325866	25.41513541	26.47571957	27.50695474	28.56386269	29.46143426	30.3373697	31.04306928	31.78480751	32.53522304	33.29272229	33.98758977	34.84724419	35.51048907	36.16058495	36.79791688	37.42285508	38.07618905	38.65659969	39.24568484	39.82374683	40.35343406	40.91105232	41.47658408	42.04911521	42.61072281	43.12758811	43.63750215	44.4509815	44.93754224	45.39985008	45.91731273	46.36414975	46.80366364	47.26624048	47.66262453	48.12476471	48.83962785	49.53505656	50.21183273	50.87069688	51.4912325];

I tried to plot plot f(x) with the a,b,c values from the curve fitting tool results, but my curve is different from the one I see in the automatic plot.

Has anyone else had the same problem?

Best regards,

Paris

[Paris Pasqualin]

I understand my mistake now, thanks for the answers

Answer 1

[Star Strider]

I haven’t.

Try this, and compare it with your (unposted) code —

x = [1.34673	1.35171	1.3531	1.35402	1.35636	1.35875	1.3608	1.36278	1.36491	1.36728	1.36855	1.37048	1.37468	1.37505	1.37507	1.37917	1.38045	1.38387	1.38449	1.38473	1.38898	1.39084	1.3925	1.39367	1.39505	1.39716	1.39903	1.40043	1.4022	1.40435	1.40562	1.40767	1.40918	1.41063	1.41219	1.41371	1.41533	1.41688	1.41856	1.4197	1.42117	1.42366	1.42503	1.4265	1.42797	1.4293	1.4306	1.432	1.43331	1.43479	1.437	1.43919	1.4416	1.44366	1.44573];
y = [10.41769233	11.9053698	13.1164268	14.25963171	15.00649903	16.35618123	17.45629085	18.49632079	19.61280955	20.63621936	21.69999059	22.64564288	23.56982874	24.47325866	25.41513541	26.47571957	27.50695474	28.56386269	29.46143426	30.3373697	31.04306928	31.78480751	32.53522304	33.29272229	33.98758977	34.84724419	35.51048907	36.16058495	36.79791688	37.42285508	38.07618905	38.65659969	39.24568484	39.82374683	40.35343406	40.91105232	41.47658408	42.04911521	42.61072281	43.12758811	43.63750215	44.4509815	44.93754224	45.39985008	45.91731273	46.36414975	46.80366364	47.26624048	47.66262453	48.12476471	48.83962785	49.53505656	50.21183273	50.87069688	51.4912325];

a = -1300;
b = -9.032;
c = 98.01;

y_fit = a.*x.^b + c;

figure
plot(x, y, '.')
hold on
plot(x, y_fit, '-r')
hold off
grid

EDIT — (15 Jun 2021 at 14:30)

‘This is strange, I ploted as:

x_vet = [1.3467 1.4457];

a = -1300;

b = -9.032;

c = 98.01;

y = @(x) a.*x.^b+c;

plot(x_vet,y(x_vet),'k')’

Not at all strange.

That code calculates only 2 points, so it only plots 2 points and the values fitted to them.

.

Answer 2

[John D'Errico]

I predict there will be no problem here. So let me see what you MAY have done wrong. I say may, since we do not see how you created that plot. First, let me see what the CFTB does on its own.

x = [1.34673	1.35171	1.3531	1.35402	1.35636	1.35875	1.3608	1.36278	1.36491	1.36728	1.36855	1.37048	1.37468	1.37505	1.37507	1.37917	1.38045	1.38387	1.38449	1.38473	1.38898	1.39084	1.3925	1.39367	1.39505	1.39716	1.39903	1.40043	1.4022	1.40435	1.40562	1.40767	1.40918	1.41063	1.41219	1.41371	1.41533	1.41688	1.41856	1.4197	1.42117	1.42366	1.42503	1.4265	1.42797	1.4293	1.4306	1.432	1.43331	1.43479	1.437	1.43919	1.4416	1.44366	1.44573];
y = [10.41769233	11.9053698	13.1164268	14.25963171	15.00649903	16.35618123	17.45629085	18.49632079	19.61280955	20.63621936	21.69999059	22.64564288	23.56982874	24.47325866	25.41513541	26.47571957	27.50695474	28.56386269	29.46143426	30.3373697	31.04306928	31.78480751	32.53522304	33.29272229	33.98758977	34.84724419	35.51048907	36.16058495	36.79791688	37.42285508	38.07618905	38.65659969	39.24568484	39.82374683	40.35343406	40.91105232	41.47658408	42.04911521	42.61072281	43.12758811	43.63750215	44.4509815	44.93754224	45.39985008	45.91731273	46.36414975	46.80366364	47.26624048	47.66262453	48.12476471	48.83962785	49.53505656	50.21183273	50.87069688	51.4912325];

mdl = fittype('a*x.^b + c','indep','x');

estmdl = fit(x',y',mdl,'start',[-1000 -9 100])

estmdl = 
     General model:
     estmdl(x) = a*x.^b + c
     Coefficients (with 95% confidence bounds):
       a =       -1300  (-1522, -1079)
       b =      -9.032  (-9.805, -8.26)
       c =       98.01  (92.48, 103.5)

So it looks as it the estimated fit is the same as yours.

plot(estmdl)
hold on
plot(x,y,'bo')

The fit seems good enough, passing through the data nicely. Now, let me plot the fit myself.

figure
plot(x,y,'bo',x,estmdl.a*x.^estmdl.b + estmdl.c)

I don't see the problem. What you may have done to produce the straight line there is just wild random conjecture. As much as I try to read your mind, or to see inside your computer, I get nothing.

Answer 3

[Steven Lord]

Show us the code you used to create your plot.

Did you use the full double precision values of the coefficients stored in the fit object (obtained using the coeffvalues function) or did you use the approximations printed out in the results windows?

In some cases the decimal places that aren't printed are irrelevant. If you wanted to know if a coefficient was greater than 1 and the displayed value was 0.001 then you don't need to know the places after the 1 to get the answer. But in some cases they are very relevant like if the displayed coefficient is 1.0000 -- is that actually 0.99999999 or 1.000000001?

format short
x = [0.99999999; 1.000000001]

x = 2×1

1.0000 1.0000

In your particular example, a slight change in the exponent can result in a significant change in the answer. Using roughly the values you have for a and b and the x coordinate around where you see a difference in your plots:

y1 = 1000*(1.37^(-9))

y1 = 58.8186

y2 = 1000*(1.37^(-9.01))

y2 = 58.6338