[xXxiKillxXx]

(1+x)ⁿ

I know that if n is a positive integer then the expansion is valid for all values for x. If n is not a positive integer (for example if its -1) then how do you know for what range it is valid?

[Zii]

|x| < 1

[xXxiKillxXx]

(Original post by Zii)
|x| < 1

Care to elaborate please?

[xXxiKillxXx]

(Original post by WarriorInAWig)
Well, for negative integers, do you agree that I can rewrite the above as 1/((1+x)^n)? Then what does that tell you about the range of validity?

Yeah I agree but not sure about the 2nd part of your post..

[WarriorInAWig]

Sorry, I was being dumb for a moment there. Someone above me has given you the right answer.

[Zii]

(Original post by xXxiKillxXx)
Care to elaborate please?

Undergrad mathematics. Use the ratio test.

[xXxiKillxXx]

(Original post by Zii)
Undergrad mathematics. Use the ratio test.

I haven't completed A2 Maths yet.. I am doing C4 atm so I can't use the ratio test lol

[Zii]

(Original post by xXxiKillxXx)
I haven't completed A2 Maths yet.. I am doing C4 atm so I can't use the ratio test lol

Then you aren't expected to prove that for (1+x)^n to converge, for any number n which isn't a positive integer, then |x| < 1.

[xXxiKillxXx]

(Original post by Zii)
Then you aren't expected to prove that for (1+x)^n to converge, for any number n which isn't a positive integer, then |x| < 1.

Thanks.. For (1+2x)^-3, for what range of values would it be valid?

[Zii]

(Original post by xXxiKillxXx)
Thanks.. For (1+2x)^-3, for what range of values would it be valid?

|2x| < 1

which is equivalent to

|x| < 1/2

[xXxiKillxXx]

(Original post by Zii)
|2x| < 1

which is equivalent to

|x| < 1/2

So its always the 'x' value that is inside the modulus? (and smaller than 1)?

[Zii]

(Original post by xXxiKillxXx)
So its always the 'x' value that is inside the modulus? (and smaller than 1)?

In general you may have



you must rewrite this by doing the following



it all hinges on it being "1 + something" inside the brackets.

then the modulus of the "something" (in our general case, ) must be less than one, so the original general case converges if

.

Edit: and remember, this is only for values of n which are anything but non-negative integers! If n is a non-negative integer, there is no issue of convergence.

[xXxiKillxXx]

(Original post by Zii)
In general you may have



you must rewrite this by doing the following



it all hinges on it being "1 + something" inside the brackets.

then the modulus of the "something" (in our general case, ) must be less than one, so the original general case converges if

.

Edit: and remember, this is only for values of n which are anything but non-negative integers! If n is a non-negative integer, there is no issue of convergence.

Convergence means when a series gets closer to a certain number.. And if n is a negative number, convergence occurs..

But why is the expansion of

(1+x)^-1 (where n is negative)

an infinite expansion? I mean n is negative so convergence should occur (so how can it be infinite)?

[Zii]

(Original post by xXxiKillxXx)
Convergence means when a series gets closer to a certain number.. And if n is a negative number, convergence occurs..

But why is the expansion of

(1+x)^-1 (where n is negative)

an infinite expansion? I mean n is negative so convergence should occur?

Yeah I'm not sure you even know what convergence means. It's nothing to do with the expansion being finite or not. You are completely misinformed and you need to speak to your maths teacher ASAP.

[xXxiKillxXx]

(Original post by Zii)
Yeah I'm not sure you even know what convergence means. It's nothing to do with the expansion being finite or not. You are completely misinformed and you need to speak to your maths teacher ASAP.

My maths teacher is useless, if he was good I would be asking him

[EEngWillow]

(Original post by xXxiKillxXx)
But why is the expansion of

(1+x)^-1 (where n is negative)

an infinite expansion? I mean n is negative so convergence should occur (so how can it be infinite)?

The expansion of is:



We'll compare the expansion of and :

When your n is a positive integer, you get the following




As you can see, every term after the x^2 term is going to include that multiplication by 0, and hence the series terminates. You can keep expanding, but every term is going to have a multiplication by 0 so there's no point

When you have the negative index though:




i.e. you aren't going to get to a point in this series where the rest of the terms evaluate to 0.

An infinite series can either converge or diverge. You can hopefully see the need for |x| < 1: you need a value between 0 and 1 for the terms to get smaller in value each time (take the example x = 0.5, so x^2 = 0.25, x^3 = 0.125 etc) -> under these circumstances the series converges. If you had instead x = 2, then each term would get larger (x = 2, x^2 = 4, x^3 = 8) and it diverges.

[xXxiKillxXx]

(Original post by EEngWillow)

An infinite series can either converge or diverge. You can hopefully see the need for |x| < 1: you need a value between 0 and 1 for the terms to get smaller in value each time (take the example x = 0.5, so x^2 = 0.25, x^3 = 0.125 etc) -> under these circumstances the series converges. If you had instead x = 2, then each term would get larger (x = 2, x^2 = 4, x^3 = 8) and it diverges.

Thanks a LOT! Regarding the above, just to clarify if |x|<1, this means a series is convergent right? Also convergent means a series tends towards a certain value right?

Thanks