Sentence examples for of the fitted function from inspiring English sources

We used such fitting in order to avoid any assumption about the form of the fitted function.

The location of the maximum of the fitted function provides the required subpixel motion estimate.

An example of the fitted function can be seen in Figure3.

The value of the standard deviation σ of the fitted function was less than 0.22 for all samples with the distribution confined to a limited range.

The observed asymmetric distribution of the nanoparticles can be fitted by assuming a log-normal distribution function, ( f(d)=frac{1}{{left(2pi right)}^{1/2} dsigma} exp left[-frac{{left( ln d- ln leftlangle drightrangle right)}^2}{2{sigma}^2}right] ), where is the mean diameter and σ is the standard deviation of the fitted function.

The modeling assumes that there are no missing events at time zero (the present day) under the present global volcano monitoring systems and the present day recurrence rate (the eruption frequency; events/ka) is calculated as the intercept of the fitted function at time zero.

For 500 μM Ca2+, in the absence of complexin the fitted function is f(t) = 0.0014 + 2.66e−t/0.11 + 0.22e−t/1.94 and it is 10 times more likely compared to a single exponential fit, and in the presence of 5 μM complexin the fitted function is f(t) = 0.0006 + 0.42e−t/0.81 + 0.05e−t/4.76 and it is 10 times more likely compared to a single exponential fit.

For 500 μM Ca2+, in the absence of complexin the fitted function is f(t) = 0.0015 + 0.096e−t/1.93 and it is 10 times more likely compared to a two-exponential fit, and in the presence of 5 μM complexin the fitted function is f(t) = 0.0021 + 0.084e−t/2.38 + 0.18e−t/0.36 and it is 10 times more likely compared to a single exponential fit.

For 250 μM Ca2+, in the absence of complexin the fitted function is f(t) = 0.0026 + 0.027e−t/2.98 and it is three times more likely compared to a two-exponential fit, and in the presence of 5 μM complexin the fitted function is f(t) = 0.0005 + 0.10e−t/0.89 + 0.011e−t/14.5 and it is 10 times more likely compared to a single exponential fit.

The top panel of Figure2 shows the deviation of the theoretical predictions from the fitted function.

The functional form of the fitted functions for all the three cases, i.e., considering results with inside and outside nodes and also with the result taking their average is compared in Figs. 8 and 9 using 3-dimensional plots considering two different sets of combination of the tube geometry.