doing derivative using diff(Y)/dT makes the vector shorter

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Hi everybody,
I'm doing derivative of a curve Y-T I think it's:
T = linspace(-t, t, n); Y = somefuction; dT = T(2)-T(1); DY1 = diff(Y)/dT;
But then DY1 is one element shorter than Y. How do people usually deal with this?
I'm currently dealing with this by shorten the T axis:
plot( T(1:end-1), DY1 );
I don't know whether this is the best way... Is there are relative standard way? Let me know. Thanks everyone~

Accepted Answer

Walter Roberson
Walter Roberson on 14 Jun 2013
>> dT = 1;
>> x = 1:dT:3;
>> y = x.^2;
>> diff(y) ./ dT
3 5
>> diff('x^2', 'x')
2 * x
>> subs( diff('x^2', 'x'), 'x', x(1:length(y)) )
2 4
Thus we can see that using diff(y)/dT does not give us the same results as if we worked symbolically.
Question then: at what x values are [3 5] the correct derivative according to symbolic methods ? 2 * xp = [3 5]... by examination, xp must be [3/2 5/2]. Which, by no coincidence at all is the evaluation at the midpoints between x(n) and x(n+1) -- (x(n) + x(n+1))/2, or compactly (x(1:end-1) + x(2:end))/2
If we step back for a few seconds, we can see that using the numeric formula diff(y) ./ dT assigns the entire difference y(n) to y(n+1) as if it were at x(n), but that is not how derivatives work: derivatives are the tangent around x(n) and so y(n-1) must be taken into account, not just y(n) and y(n+1). Easiest resolution is to use x(n) and x(n+1) and y(n) and y(n+1) to construct the slope associated with the midpoint (x(n) + x(n+1))/2
plot( (T(1:end-1)+T(2:end))/2, DY1 )
If, however, you need to a slope at each x(n), then you have problems with the definition of slope at the endpoints. You might, in that case, wish to use the definition predefined:
plot( T, gradient(T) )
  2 Comments
Walter Roberson
Walter Roberson on 15 Jun 2013
I have no recommendation. Both approaches are valid in different situations. When a task requires a derivative at every point, I study why it requires that in the circumstances, and use whatever endpoint calculations are most suitable for the circumstances. More often, perhaps, I would use the interior points only, in order to avoid the problem.

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More Answers (2)

Azzi Abdelmalek
Azzi Abdelmalek on 14 Jun 2013
Edited: Azzi Abdelmalek on 14 Jun 2013
Edit
(x(2)-x(1))/(t(2)-t(1)) correspond to the approximative right derivative at the point(t(1),x(1)). The last point is (t(n-1),x(n-1)), which means that you are doing right
  1 Comment
Yuji Zhang
Yuji Zhang on 15 Jun 2013
Hi Azzi~
Yes, it's just inconvenient cause you need to worry about the length... I think gradient(Y) could be better. What do you think?

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Iain
Iain on 14 Jun 2013
How I normally do it:
average_slope_between_y1_and_y2 = diff(Y)./diff(t);
middle_of_the_time_between_y1_and_y2 = (t(2:end)+t(1:end-1))./2;
Alternatively, fit a curve to your data, and differentiate that.
  3 Comments
Felipe Padua
Felipe Padua on 12 Oct 2021
You could also use
middle_of_the_time_between_y1_and_y2 = movemean(t, 2, 'endpoints', 'discard')

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