Inverse differentiation? - Solution
y(t)=t^3-2*t+1 for -2<=t<2
The integral, with t0=0, of the derivative of y(t)=t^3-2*t+1
Plot after saving derivative values to a file, reading them back in again and displaying only the integral with initial value t0=0.
The two graphs differ by a shift downwards in the y direction (i.e. by a constant value).
Why do the graphs differ?
We are solving the differential equation:
df/dt=y'
and this gives:
f=y+C where C is the constant of integration.
plotXpose performs numerical integration by assuming that the integral is zero at some initial value, and in the graph above we have used t0=0. This is equivalent to the initial condition that f=0 at t=0. This initial condition, we have shown, is not consistent with our original function - as the two graphs above differ.
To get back the original function of t^3-2*t+1 we need to use a correct initial condition. As plotXpose calculates the integral with reference to a value of t where the integral is 0 then we need to use a value of t^3-2*t+1 where the function is zero for our initial condition.
We can possibly solve t^3-2*t+1=0 by using a guess. From the graph of t^3-2*t+1 it appears that t=1 is a zero. Substituting t=1 in y= t^3-2*t+1 we get y =1-2+1=0. Hence t=1 is a solution to t^3-2*t+1=0.
We now know that we need to solve df/dt=y' where f=0 when t=1, so we change the integration to use t0=1.
y(t)=t^3-2*t+1 for -2<=t<2
The integral, with t0=1, of the derivative of y(t)=t^3-2*t+1
Plot after saving derivative values to a file, reading them back in again and displaying only the integral with initial value t0=1.
The two plots now appear to be the same
Conclusion
We have shown that integration can be inverse differentiation, if the correct initial value is chosen.Note: As y(t)=t^3-2*t+1 is a cubic function it will have 3 roots and, in this case, from the graph, we can see that they are all the roots are real as the function crosses the t axis in three places. We can find the other two roots by dividing t^3-2*t+1 by the know factor of (t-1) and then solving the resulting quadratic, or we can use Newton-Raphson to solve numerically for the other two roots.
The other two roots, to 8 significant figures, are -1.618034 and 0.61803399. Using either of these values for t0 when integrating the derivative, will also result in the initial function (to within rounding errors).
The other two roots, to 8 significant figures, are -1.618034 and 0.61803399. Using either of these values for t0 when integrating the derivative, will also result in the initial function (to within rounding errors).
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plotXpose app is a companion to the book Mathematics for Electrical Engineering and Computing by Mary Attenborough, published by Newnes, 2003.