Math Is Fun Forum

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,151

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!

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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.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Waterloo!

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

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

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)