Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2008-05-27 03:04:38

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 48,413

Waterloo!

For 18 years, I had been believing that the maximum value of x^{1/x} for any value of x is 1.444667861 approximately, that is when x=e.

This afternoon, I realised this is true only for integal values of x.

When x belongs to Real Numbers, there is no upper limit for the value of

This can be seen from the following illustration:-

When







Since 10 is an even number

In general,
when x=-1x10^n, x^{1/x} would be equal to n x 10^n.

Hence, as the negative real number n gets closer to zero, the value of x^{1/x} tends to increase indefinitely!


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

#2 2008-05-27 04:06:13

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Waterloo!

ganesh wrote:

This afternoon, I realised this is true only for integal values of x.

Wrong. It is true for all positive real values of x.

ganesh wrote:

In general,
when x=-1x10^n, x^{1/x} would be equal to n x 10^n.

Wrong.

which is complex (not a real number) for
.

You’re probably thinking of

. It that case it would be equal to [Dickinson]\left(10^n\right)^{10^n}\ (n\in\mathbb{Z}^+)[/Dickinson].

In fact

, so it’t better not to define the function for negative x.

Last edited by JaneFairfax (2008-05-27 04:17:46)

Offline

Board footer

Powered by FluxBB