Quick Inverse Function

nVid

What is the inverse 1/x+1

( one, divided by x plus one)

Cheers

Fremen

1/x+1 = y

Now solve for x, and that's the inverse as a function of y.

red_fox

Perhaps I'm overthinking it, but are you trying to show

[LATEX] \displaystyle \sum_{i=0}^{\infty} x^i = \frac{1}{1-x}[/latex] for [latex] -1 < x < 1 [/LATEX]

i.e. a convergent geometric series. If not, then nevermind

nVid

Fremen wrote: »

1/x+1 = y

Now solve for x, and that's the inverse as a function of y.

I have forgot if the +1 stays under the divide line.

x = 1/y+1 ?

too late at night :rolleyes:

Fremen

Try it with actual numbers and see which way works

nVid

Fremen wrote: »

Try it with actual numbers and see which way works

I have a feeling your familiar with problem based learning.

I realy need to brush up on the basic manipulation of formulae.

Anywhere online that has a page about it, or you seem proficient can you give a few examples.

AB = C -> A = C/B

Somthing like that?

Dirac

nVid wrote: »

What is the inverse 1/x+1

( one, divided by x plus one)

Cheers

I probably shouldn't be doing this - it's b etter for you if you can figure it out yourself, but here goes:

1/(x+1) = y

Now cross-multiply to get

1 = y(x+1)

Multiply out the right-hand-side to get:

1 = xy + y

Now, you want the term involving x to be on it's own on one side, therefore you get:

1 - y = xy

Finally divide both sides by y (to get rid of the y on the right, leaving just x) to get

(1 - y) / y = x

That's your answer.