A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Kainui

  • 5 years ago

What is the difference between Delta x and dx?

  • This Question is Closed
  1. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    infinity

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  4. Kainui
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  7. Kainui
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  9. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  10. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  13. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  14. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int dy = \int 1dx\] \[y=x+c\]

  15. Kainui
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  17. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  18. Kainui
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  20. Kainui
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    good luck :)

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.