Why isn't Un the derivative of Sn in sequence and derivative?

[terryds]

I see that derivative of y with respect to x is just like the ratio of y over x.
But, Why U_n (the formula to find nth term) is not the derivative of S_n (the sum of sequence formula) ??

For example,

1 2 5 10 -> y = x²+1
+1 +3 +5 -> y= 2x-1
+2 +2 -> y=2

I think that the second level is 2x (since it's the derivative of y = x²+1), but it's not...
Derivative of y with respect to x is just like how much y will gain with x gaining, right ?

Is there any way or formula to correlate the derivative of S_n with U_n ??

 

[PeroK]

I see that derivative of y with respect to x is just like the ratio of y over x.
But, Why U_n (the formula to find nth term) is not the derivative of S_n (the sum of sequence formula) ??

For example,

1 2 5 10 -> y = x²+1
+1 +3 +5 -> y= 2x-1
+2 +2 -> y=2

I think that the second level is 2x (since it's the derivative of y = x²+1), but it's not...
Derivative of y with respect to x is just like how much y will gain with x gaining, right ?

Is there any way or formula to correlate the derivative of S_n with U_n ??

The derivative is a continuous measure of change. You are comparing it with discrete changes between $x = 0, 1, 2 \ \dots$

The function $y = x^2 + 1$ is a smooth curve. You are effectively joining up the points $(0, 1), (1, 2), (2, 5)$ with straight lines. That is the difference.

If, instead, you took $x = 2, 4, 6 \dots$ or $x = 10, 20, 30 \dots$ you would find a bigger discrepancy between the derivative at a point and the slope of the straight line to the next point.

Or, if you took $x = 0.1, 0.2, 0.3 \dots$ you would find the slope of the straight lines to be closer to the derivative.

The smaller you take the difference between values of $x$, the closer you will get to the derivative. The derivative is, in fact, the limit of the slope of these straight lines as the difference between your points get arbitrarily small.

 

[Mark44]

I see that derivative of y with respect to x is just like the ratio of y over x.

No, it isn't. The derivative of y with respect to x is the ratio of change in y over change in x. IOW, $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac {\Delta y}{\Delta x}$, one of several formulations for this derivative.

 

[davidmoore63@y]

We have U(n) = S(n+1) - S(n)
which looks similar to the derivative
dy/dx = lim (h->0) of (y(x+h) - y(x)) /h
In fact it is the same expression when h=1