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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\Delta x \] means the change in x where dx is the derivative of x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when Delta x reaches infinity it becomes dx; the ghost of a departed ratio

Kainui
 5 years ago
Best ResponseYou've already chosen the best response.0So when you multiply something by dx, what does that mean essentially?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it means that you are relating it back to when it had a real value

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1you might need to example that multiply by dx bit tho for clarity

Kainui
 5 years ago
Best ResponseYou've already chosen the best response.0So 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.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.12x dx for example is meaningless on its own we need more to the notation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1{S} () dx is an operator on a derivative telling us that "x" was the important stuff

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1{S} (2x) dx means that we can undo this derivative to find out where it came from

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1or 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

Kainui
 5 years ago
Best ResponseYou've already chosen the best response.0\[dy=\int\limits_{}^{}dx\]\[dy/dx = 1\] Those are essentially the same thing, but why does it just work like dividing or multiplying?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[\frac{dy}{dx}=1\] becasue its a ration at heart can be split into the form \[dy=1 dx\] which is meaningless without the \(\int\) symbol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[\int dy = \int 1dx\] \[y=x+c\]

Kainui
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it represents an elongated "s" reminding us that an integration is the sum of the derivative parts  with respect to (*): d*

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[\sum_{0}^{\inf}f(x)\Delta x\implies\int_{0}^{inf}f(x)dx\] the summation is for discrete functions; the integral is for continuous functions

Kainui
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, 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...?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yes, whereas the Reimann is a discrete count; the integration is a contiuous count

Kainui
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, I think that makes sense, leibniz notation is kinda weirding me out still, I'll sit and think about it a little longer. Thanks.
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