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Wrong value for the derivative of the solution of an ODE in MATLAB


By : Abinadi Cordova
Date : October 14 2020, 08:10 PM
Does that help GCC is an array of pairs of values, starting for a more exact integration with something like
code :
[  0.00000000e+00,   1.00000000e+00],
[  9.99957927e-05,   9.99915855e-01],
[  1.99983171e-04,   9.99831715e-01],
[  2.99962136e-04,   9.99747579e-01],
[  3.99932687e-04,   9.99663448e-01],
 0                         1
 1.66652642851402e-05      1.00001666526429
-1.40231141152723e-05      0.999985976885885
 1.66638620438405e-05      1.00001666386204
-1.40217119017434e-05      0.999985978288098
eqGreen = @(t, g)[g(2), - N(g(1), g(2))];
Cc = [0, a0];
GCC(1:5,:) =

                     0                     1
  9.99957927208439e-05     0.999915855174488
  0.000199983171186392     0.999831714893813
  0.000299962135851046     0.999747579156327
  0.000399932687169042     0.999663447960381

di(1:5,:)=

  9.99957927208439e-05  -8.41448255122224e-05
  9.99873784655483e-05  -8.41402806746050e-05
  9.99789646646540e-05  -8.41357374861129e-05
  9.99705513179961e-05  -8.41311959463020e-05
  9.99621384254097e-05  -8.41266560547282e-05


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First Derivative filter Matlab


By : ljbeas
Date : March 29 2020, 07:55 AM
With these it helps First of all, your code is syntatically incorrect:
There is no end statement to any of your loops, and besides, you don't even need loops here. You seem to read your image into the variable myimage, but you're using an undefined variable X when attempting to calculate the derivative. The order of your assignment statements is reversed. The variable you wish to assign to should be written in the left hand part of the assignement.
code :
a(2:end) - a(1:end-1)
X(2:end, :) - X(1:end-1, :)
X(:, 2:end) - X(:, 1:end-1)

Doing a derivative in MatLab


By : Javaw0cky
Date : March 29 2020, 07:55 AM
To fix the issue you can do There is little you can do. This topic has already been discussed ad nauseum at the MathWorks web forums.
@noah has provided an example of numerical differentiation (ie: differencing) on a finite data set, whereas you are looking for a means of doing symbolic differentiation of continuous-domain-continuous-range function applied to a discrete-domain data set.

Plotting a solution and its derivative, of a first order ODE


By : user3396108
Date : March 29 2020, 07:55 AM
will be helpful for those in need From odeint documentation:
code :
import matplotlib.pyplot as plt
from numpy import gradient, squeeze, sqrt
from scipy.integrate import solve_ivp


def fun(t, y):
    l = 2
    m = 1.15
    k = 3 * sqrt(1 - 1 / m)
    return (-y * (y - 1) / t - y * (2 / t + t / m / (1 - t ** 2 / m)
                                    * (2 * sqrt(1 - t ** 2 / m) - k)
                                    / (k - sqrt(1 - t ** 2 / m)))
            - 1 / (1 - t ** 2 / m) * (-l * (l + 1) / t + 3 * t / m
                                      * (k + 2 * sqrt(1 - t ** 2 / m))
                                      / (k - sqrt(1 - t ** 2 / m)))
            + 4 * t ** 3 / m ** 2 / (1 - t ** 2 / m)
            / (k - sqrt(1 - t ** 2 / m)) ** 2)


sol = solve_ivp(fun, t_span=[1e-10, 1], y0=[2], method='BDF',
                dense_output=True)
if sol.success is True:
    print(sol.t.shape, sol.y.shape)
    plt.plot(sol.t, squeeze(sol.y), color='xkcd:avocado',
             label='Scipy Solution')
    plt.plot(sol.t, fun(sol.t, squeeze(sol.y)), linestyle='dashed',
             color='xkcd:purple', label='Derivative Using the Function')
    plt.plot(sol.t, gradient(squeeze(sol.y), sol.t), linestyle='dotted',
             color='xkcd:bright orange', label='Derivative Using Numpy')
    plt.legend()
    plt.tight_layout()
    plt.savefig('so.png', bbox_inches='tight', dpi=300)
    plt.show()

First derivative in matlab


By : Dona Caryl
Date : March 29 2020, 07:55 AM
hope this fix your issue Assuming you can evaluate the function easily, here is a vary simple way to estimate the derivative. (Assuming the function behaves nicely)

Derivative of a function in Matlab


By : sabari salome
Date : March 29 2020, 07:55 AM
I hope this helps you . There is no mistake here. The derivative of a second degree polynomial is a first degree polynomial... hence the variable x is still present in the result and you cannot evaluate it numerically unless you give a value to x:
code :
vpa(subs(diff(A),x,4)) % evaluates the derivative for X=4, yields 5.9
vpa(diff(A,2)) % this returns: -3.2
vpa(diff(A)) % this returns: 18.7 - 3.2*x 
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