## Kainui 4 years ago What is the difference between Delta x and dx?

1. amistre64

infinity

2. anonymous

$\Delta x$ means the change in x where dx is the derivative of x

3. amistre64

when Delta x reaches infinity it becomes dx; the ghost of a departed ratio

4. Kainui

So when you multiply something by dx, what does that mean essentially?

5. amistre64

it means that you are relating it back to when it had a real value

6. amistre64

you might need to example that multiply by dx bit tho for clarity

7. Kainui

So dx is a "ghost" ratio, basically a ratio with 0 in the denominator... but not quite... and multiplying that times something gives you a value based on... I mean, I know that's the integral but I'm still not sure I understand it.

8. amistre64

2x dx for example is meaningless on its own we need more to the notation

9. amistre64

{S} () dx is an operator on a derivative telling us that "x" was the important stuff

10. amistre64

{S} (2x) dx means that we can undo this derivative to find out where it came from

11. amistre64

or at least a family of functions from which it came; {S} (2x) dx = x^2 + C where C is some arbitrary constant relating this to a family of functions

12. Kainui

$dy=\int\limits_{}^{}dx$$dy/dx = 1$ Those are essentially the same thing, but why does it just work like dividing or multiplying?

13. amistre64

$\frac{dy}{dx}=1$ becasue its a ration at heart can be split into the form $dy=1 dx$ which is meaningless without the $$\int$$ symbol

14. amistre64

$\int dy = \int 1dx$ $y=x+c$

15. Kainui

Hmm so why does there have to be an S symbol if it's multiplied by dx? I mean, I know you need some way of showing that there is a place to go from for definite integrals, but why have the symbol for indefinite integrals? What does the S symbol mean by itself with no dx around?

16. amistre64

it represents an elongated "s" reminding us that an integration is the sum of the derivative parts - with respect to (*): d*

17. amistre64

$\sum_{0}^{\inf}f(x)\Delta x\implies\int_{0}^{inf}f(x)dx$ the summation is for discrete functions; the integral is for continuous functions

18. Kainui

Ok, that makes sense, so it is literally just the exact same thing as saying the limit of an infinite riemann sum of infinitely small values...?

19. amistre64

yes, whereas the Reimann is a discrete count; the integration is a contiuous count

20. Kainui

Ok, I think that makes sense, leibniz notation is kinda weirding me out still, I'll sit and think about it a little longer. Thanks.

21. amistre64

good luck :)