JavaScript has two number types: `Number`

and `BigInt`

.

The most frequently-used number type, `Number`

, is a 64-bit floating point IEEE 754 number.

The largest exact integral value of this type is `Number.MAX_SAFE_INTEGER`

, which is:

- 2
^{53}-1, or
- +/- 9,007,199,254,740,991, or
- nine quadrillion seven trillion one hundred ninety-nine billion two hundred fifty-four million seven hundred forty thousand nine hundred ninety-one

To put this in perspective: one quadrillion bytes is a petabyte (or one thousand terabytes).

"Safe" in this context refers to the ability to represent integers exactly and to correctly compare them.

From the spec:

Note that all the positive and negative integers whose magnitude is no
greater than 2^{53} are representable in the `Number`

type (indeed, the
integer 0 has two representations, +0 and -0).

To safely use integers larger than this, you need to use `BigInt`

, which has no upper bound.

Note that the bitwise operators and shift operators operate on 32-bit integers, so in that case, the max safe integer is 2^{31}-1, or 2,147,483,647.

```
const log = console.log
var x = 9007199254740992
var y = -x
log(x == x + 1) // true !
log(y == y - 1) // also true !
// Arithmetic operators work, but bitwise/shifts only operate on int32:
log(x / 2) // 4503599627370496
log(x >> 1) // 0
log(x | 1) // 1
```

Technical note on the subject of the number 9,007,199,254,740,992: There is an exact IEEE-754 representation of this value, and you can assign and read this value from a variable, so for *very carefully* chosen applications in the domain of integers less than or equal to this value, you could treat this as a maximum value.

In the general case, you must treat this IEEE-754 value as inexact, because it is ambiguous whether it is encoding the logical value 9,007,199,254,740,992 or 9,007,199,254,740,993.

There's a simple trick for this problem:

```
bool IsPowerOfTwo(ulong x)
{
return (x & (x - 1)) == 0;
}
```

Note, this function will report `true`

for `0`

, which is not a power of `2`

. If you want to exclude that, here's how:

```
bool IsPowerOfTwo(ulong x)
{
return (x != 0) && ((x & (x - 1)) == 0);
}
```

### Explanation

First and foremost the bitwise binary & operator from MSDN definition:

Binary & operators are predefined for the integral types and bool. For
integral types, & computes the logical bitwise AND of its operands.
For bool operands, & computes the logical AND of its operands; that
is, the result is true if and only if both its operands are true.

Now let's take a look at how this all plays out:

The function returns boolean (true / false) and accepts one incoming parameter of type unsigned long (x, in this case). Let us for the sake of simplicity assume that someone has passed the value 4 and called the function like so:

```
bool b = IsPowerOfTwo(4)
```

Now we replace each occurrence of x with 4:

```
return (4 != 0) && ((4 & (4-1)) == 0);
```

Well we already know that 4 != 0 evals to true, so far so good. But what about:

```
((4 & (4-1)) == 0)
```

This translates to this of course:

```
((4 & 3) == 0)
```

But what exactly is `4&3`

?

The binary representation of 4 is 100 and the binary representation of 3 is 011 (remember the & takes the binary representation of these numbers). So we have:

```
100 = 4
011 = 3
```

Imagine these values being stacked up much like elementary addition. The `&`

operator says that if both values are equal to 1 then the result is 1, otherwise it is 0. So `1 & 1 = 1`

, `1 & 0 = 0`

, `0 & 0 = 0`

, and `0 & 1 = 0`

. So we do the math:

```
100
011
----
000
```

The result is simply 0. So we go back and look at what our return statement now translates to:

```
return (4 != 0) && ((4 & 3) == 0);
```

Which translates now to:

```
return true && (0 == 0);
```

```
return true && true;
```

We all know that `true && true`

is simply `true`

, and this shows that for our example, 4 is a power of 2.

## Best Solution

try

`1 / (1 + e^(1-x))`

it's the logistic function shifted by 1 unit

If you want it to approach faster, you can change e to something higher

Edit:to have f(0) = 0 you could use

`1 - 2^(-x)`