What is the solution to r^(n+1)*(1-r) in calculus?

[Roscoe1989]

1. can someone tell me what r^(n+1)*(1-r)=? I need it to solve a proof, but my math is so rusty! thanks!



2. r^(n+1)*(1-r)

 

[flyingpig]

What is the question...? Like the real question.

 

[Roscoe1989]

prove by induction on n that
1+r+r^2...+r^=1-r^n+1/1-r
if r doesn't equal 1

 

[flyingpig]

Just to be clear, you have

r^n + \frac{1}{1 -r} on your right hand side?

 

[Roscoe1989]

oopps, sorry. it should be

1+r+...+r^n=1-r^(n+1) / 1-r

 

[flyingpig]

What is the base case? Let's start with that

 

[flyingpig]

Also, shouldn't |r|<1?

 

[Roscoe1989]

the what case?

 

[flyingpig]

Base Case. You know that step you do in Induction

 

[Roscoe1989]

i don't know?

 

[flyingpig]

What can n be and cannot be?

 

[flyingpig]

edit; mistake

 

[Roscoe1989]

is it n can't be 0? but can be anything else?

 

[Dick]

oopps, sorry. it should be

1+r+...+r^n=1-r^(n+1) / 1-r

If you don't know induction, just multiply out (1+r+...+r^n)*(1-r). Many terms cancel.

 

[Roscoe1989]

my textbook is saying 1-r^n+1

 

[flyingpig]

my textbook is saying 1-r^n+1

Oh sorry I didn't see the 1 in front earlier

 

[Dick]

my textbook is saying 1-r^n+1

Your textbook is right.

 

[flyingpig]

is it n can't be 0? but can be anything else?

try pluggin in 0

 

[Roscoe1989]

n has to be greater than 0

 

[flyingpig]

n has to be greater than 0

Why?

 

[flyingpig]

1 + r + r^2 + ... + r^n = \frac{1 - r^{n+1}}{1-r}

Is what you have. What happens if n = 0?

 

[Roscoe1989]

then the right hand side will equal 1. right?

 

[flyingpig]

then the right hand side will equal 1. right?

And what about the left hand side? What happens there?

 

[Dick]

Why?

Why not? I would interpret 1+r+...+r^n for n=0 to be 1. (1-r^(0+1))/(1-r)=(1-r)/(1-r)=1. I'm not sure you are leading this in a helpful direction, flyingpig.

 

[Roscoe1989]

honestly, i don't know

 

[flyingpig]

Why not? I would interpret 1+r+...+r^n for n=0 to be 1. (1-r^(0+1))/(1-r)=(1-r)/(1-r)=1. I'm not sure you are leading this in a helpful direction, flyingpig.

Nononon, Ros asked if it can be 0 before

 

[Roscoe1989]

does it mean that it's "proven"?

 

[flyingpig]

honestly, i don't know

1 + r + r^2 + ... + r^n = \frac{1 - r^{n+1}}{1-r}

You were originally given this. You got it right when you said the rhs is 1 when n = 0

 

[flyingpig]

does it mean that it's "proven"?

By Induction? No

If only though, if only...

 

[Roscoe1989]

but what does it mean on the LHS?

 

[flyingpig]

It's probably that I tell you what Induction is (in brief form) first.

(1) Base Case. Usually we take n = 1 or 0. We show that it (S(n)) is true first

(2) Inductive Step. Assume that (1) is true and then you will write your Inductive hypothesis and show that it is true for S(n+1).

 

[Dick]

By Induction? No

If only though, if only...

You don't really need induction for an informal proof. See my long buried comment in post 14.

 

[flyingpig]

but what does it mean on the LHS?

It's a sum and on the right hand side, it is a formula of the sum. The powers on the LHS keep going up and you just set the right hand side with n = 0.

Where n is the power.

 

[flyingpig]

You don't really need induction for an informal proof. See my long buried comment in post 14.

Yeah I probably wouldn't do that [induction] either. But Ros asked for

prove by induction on n that
1+r+r^2...+r^=1-r^n+1/1-r
if r doesn't equal 1

in post 3

 

[fluidistic]

Dick and flyingpig, you are missing, I think, that roscoe1989 doesn't know what happens in the LHS when n=0.
When n=0, you don't get 1+r²+r³+...+r⁰.
1+r²+r³+...+r^n is just a notation. When n=0 you simply get r⁰ and this is worth 1. So when n=0, both the LHS and RHS is worth 1, so the equality holds, meaning that it does makes sense to consider n=0 (it's a possible/allowed value for n).

 

[Roscoe1989]

ok...so when the right side equals the left side... what does that mean?

 

[Dick]

Yeah I probably wouldn't do that [induction] either. But Ros asked for



in post 3

Ok, missed the "induction" bit in there.

 

[flyingpig]

Dick and flyingpig, you are missing, I think, that roscoe1989 doesn't know what happens in the LHS when n=0.
When n=0, you don't get 1+r²+r³+...+r⁰.
1+r²+r³+...+r^n is just a notation. When n=0 you simply get r⁰ and this is worth 1. So when n=0, both the LHS and RHS is worth 1, so the equality holds, meaning that it does makes sense to consider n=0 (it's a possible/allowed value for n).

Yeah and I am trying to lead Ros to that.

 

[Roscoe1989]

ooooooh, ok

 

[flyingpig]

ok...so when the right side equals the left side... what does that mean?

That means the equal sign worked!

It's probably that I tell you what Induction is (in brief form) first.

(1) Base Case. Usually we take n = 1 or 0. We show that it (S(n)) is true first

(2) Inductive Step. Assume that (1) is true and then you will write your Inductive hypothesis and show that it is true for S(n+1).

 

[flyingpig]

ooooooh, ok

I just led to the base case, so we just did (1).

 

[Roscoe1989]

so then to prove (2), we would have to do n+1, which, in our case, is 0+1=1 and that would give us 1-r^2 / 1-r right?

 

[flyingpig]

Where did you get the 0 from...? I have to go to sleep soon...

 

[Roscoe1989]

sorry, i meant 1+1=2. you've been really helpful, and don't let me keep you awake! :)

 

[flyingpig]

I have to leave now because by the time I get back, I would be very late already. I will leave you with this.

When I said "n + 1" , I mean for a general n. The base case is when we take the most fundamental n. It just happened to be 0.

So you have to show this

1 + r + r^2 +...+r^{n+1} = \frac{1 - r^{n + 1 + 1}}{1 - r}

 

[Roscoe1989]

thanks flyingpig, you've been a great help!