Simple maths 2

[Drej]

I am trying to simplify the following

[(1/x)]-x

into a single variable of x i.e.:

[(1/x)]-x = ?x?

Any ideas? (This is for my work and is NOT a school homework question.)

Cheers,

-- drej --

 

[yates]

(Hope the symbols work out OK !)
[(1/x)]-x = (xe-1)e-x = xex or x to the power x

QED

 

[Drej]

Thanks, Yates.

Cheers,

-- drej --

 

[JStephen]

You don't need "e" for that- just use the definition of a negative exponent. Assuming, of course, that the "-x" is an exponent, which is not real clear in the post.

 

[NickE]

Normally ^ is used to indicate an exponent.

Thus:

[(1/x)]-x=(1/x)^-x ????

 

[zekeman]

x^x

 

[johnwm]

Or x^x

That's [ignore]x^x
[/ignore]

Good Luck
johnwm
________________________________________________________
To get the best from these forums read faq731-376 before posting

UK steam enthusiasts:

http://www.essexsteam.co.uk

 

[Drej]

The equation reads:

[(the inverse of x)] - x

or

1
--- - x
x

This is clearly shown by the square brackets.

 

[IRstuff]

The square brackets are redundant, which is why everyone got confused.

For your updated definition, what exactly are you looking for? The only other plausible output is:

(1-x²)/x

Not sure if that's any simpler

TTFN

 

[NickE]

Drej- Then try this....

(1/x)-x = ?
to
(1/x) = x
so
1/x = x

Thus:

x=1.

There is only a point solution to this equation. And I'm wondering what sort of engineer cant do algebra.

 

[Cockroach]

Yeah, symbolic logic needs to be addressed as correctly pointed out above. The exponent is typically listed like x^2 for the square of x, so buddy got it right 1/x minus x on the left hand side. I assume equal x on the right hand side? Whatever.

This seems simplistic, multiply through both sides by x to clear the inverse and get:

1 - x^2 = x^2

to which adding x^2 to both sides would ultimately result:

1 = 2 x^2

I would get x = sqrt(1/2) = +/- 0.70711.

This assumes that you need (1/x) - x = x, perhaps a very big assumption.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[ivymike]

Nick - be careful with "wise-crack" answers - since when is
(1/x)-x = ? equivalent to (1/x)-x = 0? Only when ?=0, right?

 

[NickE]

ivymike- yep you're right. I should not have made that last remark.

I hereby apologize to anyone I've offended. Esp. Drej, because I didn't do any better with my substitution of 0 for ?.

As da Cockroach states there is another interpretation.. It's still just algebra though...

 

[ivymike]

some other equivalent expressions:

-(x+1)(x-1)/x
(1-x^2)/x (as mentioned above)

 

[zekeman]

Drei,
{1/x}-x is not an equation; it is simply an expression.
What is wrong with 1/x-x?
Do you want another?
(1-x^2)/x
If you think that simplifies the expression, then go for it.