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

- arindameducationusc

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

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- arindameducationusc

Check attachment for more details..
Please explain me someone.....

##### 1 Attachment

- Astrophysics

I think this will help you https://en.wikipedia.org/wiki/Big_O_notation

- arindameducationusc

@Astrophysics
Its so hard to understand....
but its interesting... it proves What is Differentiation in a totally new way...

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## More answers

- UsukiDoll

I've seen it written in some papers...I was like what is this? when I read them

- arindameducationusc

If I understand, I will give in Informative.

- arindameducationusc

@Empty

- arindameducationusc

@phi please explain in terms of some functions like factorial, exponential, polynomial..

- arindameducationusc

Is it Factorial>Exponential >Polynomial>Logarithmic>Constant.
Something like this @phi

- 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

is that MATH104? With Ghrist?

- arindameducationusc

Yes Yes..... @UnkleRhaukus

- nincompoop

it does not explain differentiation in a different way. It characterizes your function and that's the basic thing it does.

- arindameducationusc

okay...@nincompoop , characterise in the sense?

- 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

Awesome Explanation @UnkleRhaukus
Thumbs up....!

- nincompoop

characterizes your function as changes occur (if your value of x, y, or z grows bigger or smaller)

- arindameducationusc

Anyway thank you @nincompoop
I understood..
Thank you so much... @UnkleRhaukus

- 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

|dw:1438432120419:dw|

- nincompoop

|dw:1438432161979:dw|

- nincompoop

|dw:1438432278540:dw|

- arindameducationusc

Okay... interesting

- 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

our original function did not change, just the sample and O can probably be used to properly characterize those changes.

- nincompoop

ciao
This is a good question

- arindameducationusc

Right! @UnkleRhaukus

- arindameducationusc

Thank you soo much both of you..... @UnkleRhaukus
@nincompoop

- 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

Nice @Empty

- 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

in other words, $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$ is finite

- 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

@oldrin.bataku
I can't understand what you are saying...........

- arindameducationusc

@oldrin.bataku Here what did you say......

- arindameducationusc

@oldrin.bataku Nice... but can you explain in terms of graph? By making a graph in rough

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