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

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~

Answer 1

[Walter Roberson]

>> 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) )

[Yuji Zhang]

Hi Walter~

Thanks for sharing!

I see, for 1D curve Y-T, this

>>plot( (T(1:end-1)+T(2:end))/2, DY1 );

reflects the definition of derivative strictly.

--------

Actually gradient(Y) is a vector with same length as T - by the math definition of gradient. So

>>plot ( T, gradient(Y) );

should be good. The problem of this approach is, the 1st and last element of gradient(Y) are of error. - which makes sense according to its math definition too.

For my specific purpose, I deal with a long enough numerical curve (>1000 element, not symbolic function). So I think both approach works. The second might be more convenient of the boundary values are not important.

What do you think Walter? Any discussion is appreciated everyone~

[Walter Roberson]

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.

Answer 2

[Azzi Abdelmalek]

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

[Yuji Zhang]

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?

Answer 3

[Iain]

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.

[Yuji Zhang]

Hi Lain~

Thanks for the discussion~ What do y1 and y2 mean?

[Iain]

Two consecutive points on your curve "y" :P

Maybe I should have them put my name in all caps...

[Felipe Padua]

You could also use

middle_of_the_time_between_y1_and_y2 = movemean(t, 2, 'endpoints', 'discard')