# Details about the function n/ln(n)

1. May 17, 2012

### tuhinrao

From the file attached I would like to know the following.
if Y=n/ln(n), Is there a way of explicitly expressing n in terms of Y.

Relations I found are:

There are 2 values of n for every Y. Except at Y=e , the two values converge to n=e.
If n1 and n2 are the values of n
then
n1^n2=n2^n1.

So is there a way of finding n1, given n2?

What could be the possible type of functions involved?
From the graph it is seen that |n1-e| and |n2-e| are related inversely.
What could be this relation?

#### Attached Files:

• ###### n by ln n.png
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2. May 18, 2012

### haruspex

No, there's no way to express n as a function of y using any standard functions. Of course, it's always possible to invent a new one for the purpose.
Mixtures functions of different types (polynomial, exponential, logarithmic, trigonometric..) are nearly always impossible to invert. E.g. y = x.exp(x), y = sin(x)/x, ...

3. May 20, 2012

### Mute

There is a special function called the Lambert-W function (aka the Product-Log) which you can use to write n in terms of y. The Lambert-W function is the function w=W(x) such that
x = w exp(w).

If you invert your equation so that 1/y = ln(n)/n, and then let n = exp(a), this gives

$$\frac{1}{y} = a e^{-a}$$

We see that if we multiply both sides by -1 this will be in Lambert-W form, giving a = W(-1/y). Inverting n = exp(a), this gives

$$n = \ln W_k\left(-\frac{1}{y}\right).$$

Some very important notes: The Lambert-W function has two real-valued branches, corresponding to k = 0 and k = -1. Usually the k=0 branch corresponds to the desired solution. The other branches gives complex values for W, so inverting ln n = W(-1/y) is more complicated if want solutions for these others branches. However, since you seem mostly interested in real values you don't need to worry about this.

4. May 22, 2012

### tuhinrao

Thanks 'Mute'. Got my way through.

tuhinrao