Can't figure out the steps to this problem

[eastbay_cutie]

Hello everyone on the Math Help forums!

As part of a larger problem, I can't figure out the steps to go from:

P = 1 - (L/(1+i))

to​

P=(1+i)/L

Is there some kind of trick involved? Thanks for your help!


​

 

[tkhunny]

I hope the "larger problem" is not very tricky, since only basic algebra is requred for your actual question.

P = 1 - (L/(1+i))

P = (1+i)/(1+i) - (L/(1+i))

P = (1+i-L)/(1+i)

That's about it. Since this has nothing to do with the result you indicated, let's check it out. Pick L = 1 and i = 0

P = 1 - (L/(1+i))
P = 1 - 1/(1+0) = 1 - 1/1 = 1-1 = 0

P = (1+i-L)/(1+i)
P = (1+0-1)/(1+0) = 0/1 = 0 -- Mine checks out.

P = (1+i)/L
P = (1+0)/1 = 1/1 = 1 -- Whoops. That version does not check out.

So, how long were you planning to struggle before figuring out that the transformation was simply incorrect?

 

[eastbay_cutie]

Hi, thanks for your help. I believed that the math was incorrect, and I needed outside confirmation since I didn't trust myself. Its part of the set-up to a larger problem, and I believed that the author's exposition contained an error.

I get:

P = [(1 + i - L)/(1 + i)]


P = [(1 + i - L)/(1 + i)]


P = [((1 + i)/(1 + i) - L/(1 + i)


P = [1 - (L/(1 + i)]

which the author states simplifies to:

P=(1+i)/L

​

 

[eastbay_cutie]

but maybe my mistake is in these steps. Is this valid?

P = {Sum from j=0 to infinity of} ((1 + i - L)^j)/((1 + i)^j)


P = [(1 + i - L)/(1 + i)]

 

[Deleted member 4993]

but maybe my mistake is in these steps. Is this valid?

P = {Sum from j=0 to infinity of} ((1 + i - L)^j)/((1 + i)^j)


P = [(1 + i - L)/(1 + i)]

No - certainly not....

\(\displaystyle P \ = \ \sum_{j=0}^{\infty}\dfrac{(1+i-L)^j}{(1+i)^j}\)

this would look like:

\(\displaystyle P \ = \ 1 + \dfrac{(1+i-L)^1}{(1+i)^1} \ + \ \dfrac{(1+i-L)^2}{(1+i)^2} \ + \ ....\)

This is a geometric series of the form...

\(\displaystyle P \ = \ 1 \ + \ r \ + \ r^2 \ + \ r^3 \ .....\)

Now continue.....