**I just announced the new ***Spring 5* modules in REST With Spring:

*Spring 5*modules in REST With Spring:

**1. Introduction**

In this article, we’ll give an **introduction to the mathematics of big-O notation, as well as show an example of a big-O proof.**

**2. Formal Definition**

**Definition:** *f(x) = O(g(x))* means that there exist two positive constants, *x _{1}* and

*c*, such that

*0 ≤ f(x) ≤ cg(x)*for all

*x ≥ x*.

_{1}**3. The First Positive Constant: ***x*_{1}

*x*

_{1}When saying *f(x) = O(g(x))*, we say** “ f of x is big-O g of x“.**

Don’t let the equals sign fool you: *f(x)* is a function and *O(g(x))* is a set. You can’t have a function equalling a set. That’s like saying that the Earth equals the Solar System. It’s more like, the Earth *belongs* in (the set of planets which comprise) the Solar System. Similarly, ** f(x) belongs in a set of functions called O(g(x))** (

*big-O*

*g*of

*x*).

We have two functions now – *f(x)* and *g(x)*. Functions grow at different speeds. For example, a quadratic function will grow quicker than a linear function. But **sometimes, it takes some time for the quadratic to catch up! **

For example, if we have *a(x) = 2x ^{2} + 2x + 1* and

*b(x) = 10x*, we’ll have

*a(1) = 5*and

*b(1) = 10*. But say now we choose

*x = 15*. Now we have

*a(15) = 450 + 30 + 1 = 481*and

*b(x) = 150*. What’s more is that

**for any value greater than**(Now is a good time to refer back to the formal definition; this is the

*x = 15*, say*x = 20*,*a(x)*will be bigger than*b(x).**x ≥ x*part).

_{1}Importantly for the case of understanding big-O formally, **we don’t particularly care at which point a(x) began outgrowing b(x).** Just that it did, and that from that point onwards, it continues to be larger than

*b(x)*.

**4. The Second Positive Constant: ***c*

*c*

We saw above how it took a while for *a(x)* to catch up to *b(x). *Eventually, it did, but it took a while.

The reason it eventually grows faster is because of the *2x ^{2} *term. Or, more precisely, the

*x*part of it.

^{2}**This tells us that**Using the symbols of the original definition:

*a(x) = O(x*.^{2})*f(x) = a(x)*and

*g(x) = x*.

^{2}For big-O, we don’t care about the other terms in *a(x)*, or the constant *2* in the *2x ^{2}* term.

Looking again at our definition from section 2, this is where the constant *c* comes in. **We can scale g(x) by any positive constant** – as long as

*f(x)*stays smaller than the scaled version (past a certain point,

*x ≥ x*).

_{1}If we take *d(x) = 3x ^{3} + 2x + 10,* we’ll need to find values of

*c*and

*x*, such that

_{1}*d(x) ≤ cx*for all values greater than

^{3}*x*This is exactly what we do in section 4.

_{1.}**5. Putting the Pieces Together**

In order to prove *f(x) = O(gx)*, we need to find two positive constants, *c* and *x _{1}*, such that

*0 ≤ f(x) ≤ cg(x)*for all

*x ≥ x*.

_{1}**We need to find values for**

*c*and*x*such that the inequality holds._{1}What it means, is that past a certain point, a scaled version of *g(x)* will always be bigger than *f(x)*.

**6. Example**

Let d*(x) = 3x ^{3} + 2x + 10*.

Suppose we wish to prove that d*(x) = O(x ^{3})*. This means we need to find two positive integers,

*c*and

*x*such that

_{1}*0 ≤ d(x) ≤ cx*for all

^{3}*x ≥ x*.

_{1}Well, we know that

d*(x) = 3x ^{3} + 2x + 10 ≤ 3x^{3} + 2x^{3} + 10x^{3} = 15x^{3}*

so,

*d(x) ≤ 15x ^{3}*.

So if we set *x _{1} = 1 *and

*c = 15,*then we have that for any

*x ≥ x*

_{1}*, 0 ≤ d(x) ≤ cx*. So

^{3}*d(x) = O(x*.

^{3})We could have found other values for *x _{1}* and

*c*which satisfy the above condition. All that matters is that

**there exist values of**

*x*and_{1}*c*which satisfy the condition.**7. Conclusion**

In this article, we focused on an **introduction to the theory of big-O notation**.

A more practical look at this topic can be found here.

## Leave a Reply