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arindameducationusc

  • one year ago

Anbody, who can explain "Big O notation"? (Its a language for controlling growth)

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  1. arindameducationusc
    • one year ago
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    Check attachment for more details.. Please explain me someone.....

  2. Astrophysics
    • one year ago
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    I think this will help you https://en.wikipedia.org/wiki/Big_O_notation

  3. arindameducationusc
    • one year ago
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    @Astrophysics Its so hard to understand.... but its interesting... it proves What is Differentiation in a totally new way...

  4. UsukiDoll
    • one year ago
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    I've seen it written in some papers...I was like what is this? when I read them

  5. arindameducationusc
    • one year ago
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    If I understand, I will give in Informative.

  6. arindameducationusc
    • one year ago
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    @Empty

  7. arindameducationusc
    • one year ago
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    @phi please explain in terms of some functions like factorial, exponential, polynomial..

  8. arindameducationusc
    • one year ago
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    Is it Factorial>Exponential >Polynomial>Logarithmic>Constant. Something like this @phi

  9. arindameducationusc
    • one year ago
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    Is that the reason we use logarithmic functions in s=k logw ( Chemistry)... To calculate in terms of small numbers.... What about factorial? where is it used then? and why? @phi

  10. UnkleRhaukus
    • one year ago
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    is that MATH104? With Ghrist?

  11. arindameducationusc
    • one year ago
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    Yes Yes..... @UnkleRhaukus

  12. nincompoop
    • one year ago
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    it does not explain differentiation in a different way. It characterizes your function and that's the basic thing it does.

  13. arindameducationusc
    • one year ago
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    okay...@nincompoop , characterise in the sense?

  14. UnkleRhaukus
    • one year ago
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    Big Omicron notation is used to describe how a function behaves as the input becomes very large, or very small. The ‘O’, stands for Order of growth. f(x) = 5 + 2x is said to be of ‘linear order’ as x increases, which we can write as f(x) is O(x), [if x was hundreds , f(x) would be hundreds ] just as g(x) = 5 - 2x + 20x^2 is said to be of ‘quadratic order’ (with increasing x) g(x) is O(x^2) [if x was thousands , g(x) would be millions] _ Basically only the dominant term appears in the expression for the order of growth. _ If instead we consider x to be decreasing towards zero: f(x) = 5 + 2x is now said to be of constant order; O(1), the 5 is now the dominant term [if x was a few percents, f(x) would be of constant order (still about 5) ] and g(x) = 5 - 2x + 20x^2 would, likewise be of constant order as x decreases (to zero)

  15. arindameducationusc
    • one year ago
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    Awesome Explanation @UnkleRhaukus Thumbs up....!

  16. nincompoop
    • one year ago
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    characterizes your function as changes occur (if your value of x, y, or z grows bigger or smaller)

  17. arindameducationusc
    • one year ago
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    Anyway thank you @nincompoop I understood.. Thank you so much... @UnkleRhaukus

  18. nincompoop
    • one year ago
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    This is probably a good way to see some statistical analysis such as when the curve is not normalized (difficult to analyze with high degree of certainty) and by increasing the number of sample, it goes towards normalization over time, but with ever decreasing deviation and ever increasing steepness of curve.

  19. nincompoop
    • one year ago
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    |dw:1438432120419:dw|

  20. nincompoop
    • one year ago
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    |dw:1438432161979:dw|

  21. nincompoop
    • one year ago
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    |dw:1438432278540:dw|

  22. arindameducationusc
    • one year ago
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    Okay... interesting

  23. UnkleRhaukus
    • one year ago
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    a good use for this notation is when truncating an infinite series such as taylor series for example the taylor expansion of the sine function \[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots \] Using the Omicron notation \[\sin(x)=x-\frac{x^3}{3!}+\text O(x^5)\] if x is much smaller than 1, clearly the terms of order 5 and greater will be insignificant

  24. nincompoop
    • one year ago
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    our original function did not change, just the sample and O can probably be used to properly characterize those changes.

  25. nincompoop
    • one year ago
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    ciao This is a good question

  26. arindameducationusc
    • one year ago
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    Right! @UnkleRhaukus

  27. arindameducationusc
    • one year ago
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    Thank you soo much both of you..... @UnkleRhaukus @nincompoop

  28. Empty
    • one year ago
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    If you're familiar with L'Hopital's rule you can see that employing this method is much faster for evaluating limits: \[\lim_{x \to \infty} \frac{3x^5-7x^2+7}{9x^5+12x^4+14x^2-199}\] Instead of differentiating like 5 times or whatever, you can just realize the other terms are trash as you approach infinity and all that really matter are those \(x^5\) terms so the limit for this stupid example is 1/3.

  29. arindameducationusc
    • one year ago
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    Nice @Empty

  30. anonymous
    • one year ago
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    \(f\in O(g)\) as \(x\to\infty\) just means that \(|f|\) is *eventually* (i.e. there exists some real \(x_0\) such that the following holds for \(x\ge x_0\)) bounded above by \(C|g|\) for some positive real \(C\)

  31. anonymous
    • one year ago
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    in other words, $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$ is finite

  32. anonymous
    • one year ago
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    so for example when we deal with convergence of power series \(\sum a_n x^n\) we're interested in what values of \(R\) satisfy \(a_n\in O(R^{-n})\), here taken as limits in \(n\): $$\lim_{n\to\infty}\frac{a_n}{R^{-n}}=\lim_{n\to\infty}\left(R\sqrt[n]{a_n}\right)^n$$ the least upper bound of \(R\)s that work here is called the radius of convergence

  33. arindameducationusc
    • one year ago
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    @oldrin.bataku I can't understand what you are saying...........

  34. arindameducationusc
    • one year ago
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    @oldrin.bataku Here what did you say......

  35. arindameducationusc
    • one year ago
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    @oldrin.bataku Nice... but can you explain in terms of graph? By making a graph in rough

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