arindameducationusc
  • arindameducationusc
Anbody, who can explain "Big O notation"? (Its a language for controlling growth)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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arindameducationusc
  • arindameducationusc
Check attachment for more details.. Please explain me someone.....
Astrophysics
  • Astrophysics
I think this will help you https://en.wikipedia.org/wiki/Big_O_notation
arindameducationusc
  • arindameducationusc
@Astrophysics Its so hard to understand.... but its interesting... it proves What is Differentiation in a totally new way...

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UsukiDoll
  • UsukiDoll
I've seen it written in some papers...I was like what is this? when I read them
arindameducationusc
  • arindameducationusc
If I understand, I will give in Informative.
arindameducationusc
  • arindameducationusc
@Empty
arindameducationusc
  • arindameducationusc
@phi please explain in terms of some functions like factorial, exponential, polynomial..
arindameducationusc
  • arindameducationusc
Is it Factorial>Exponential >Polynomial>Logarithmic>Constant. Something like this @phi
arindameducationusc
  • arindameducationusc
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
UnkleRhaukus
  • UnkleRhaukus
is that MATH104? With Ghrist?
arindameducationusc
  • arindameducationusc
Yes Yes..... @UnkleRhaukus
nincompoop
  • nincompoop
it does not explain differentiation in a different way. It characterizes your function and that's the basic thing it does.
arindameducationusc
  • arindameducationusc
okay...@nincompoop , characterise in the sense?
UnkleRhaukus
  • UnkleRhaukus
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)
arindameducationusc
  • arindameducationusc
Awesome Explanation @UnkleRhaukus Thumbs up....!
nincompoop
  • nincompoop
characterizes your function as changes occur (if your value of x, y, or z grows bigger or smaller)
arindameducationusc
  • arindameducationusc
Anyway thank you @nincompoop I understood.. Thank you so much... @UnkleRhaukus
nincompoop
  • nincompoop
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.
nincompoop
  • nincompoop
|dw:1438432120419:dw|
nincompoop
  • nincompoop
|dw:1438432161979:dw|
nincompoop
  • nincompoop
|dw:1438432278540:dw|
arindameducationusc
  • arindameducationusc
Okay... interesting
UnkleRhaukus
  • UnkleRhaukus
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
nincompoop
  • nincompoop
our original function did not change, just the sample and O can probably be used to properly characterize those changes.
nincompoop
  • nincompoop
ciao This is a good question
arindameducationusc
  • arindameducationusc
Right! @UnkleRhaukus
arindameducationusc
  • arindameducationusc
Thank you soo much both of you..... @UnkleRhaukus @nincompoop
Empty
  • Empty
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.
arindameducationusc
  • arindameducationusc
Nice @Empty
anonymous
  • anonymous
\(f\in O(g)\) 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 \(C\)
anonymous
  • anonymous
in other words, $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$ is finite
anonymous
  • anonymous
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
arindameducationusc
  • arindameducationusc
@oldrin.bataku I can't understand what you are saying...........
arindameducationusc
  • arindameducationusc
@oldrin.bataku Here what did you say......
arindameducationusc
  • arindameducationusc
@oldrin.bataku Nice... but can you explain in terms of graph? By making a graph in rough

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