How to find the values for which the expansion is valid

What values of x is the expansion of
(1+x)^{-1}
What about
(1+5x)^{-1}

what do you mean by that

@mqb2766
is it -0.2<x>0.2

No, but thinking along the right lines. Lets do (1+x)^-1 first as above.

You must know the expansion of (1+x)^{-1} if you're doing the question.
The range of values for which its valid is easy to understand if you write it down. Its a standard series.

1-x+x2
just for clarifications we talking about question 8c

When x is 1 this goes
1, 0, 1, 0, 1, 0 ....
It does not converge. Similarly when x=-1 and the function heads off to infinity.
It only converges when |x|<1 or
-1<x<1

How does this now apply to (1+5x)^{-1}?

I don't understand what you mean by 1,0,1,0,10

1 = 1
1 - x = 0
1 - x + x^2 = 1
1 - x + x^2 - x^3 = 0
...
As you increase the order of the series, it does not converge when x=1

how would you find the convergence point?

Loosely speaking, the terms must get smaller as the order increases, i.e
|x|<1
Or
-1<x<1
This is a standard result. Have a read of your notes and make sure you understand?

1-x+x2
just for clarifications we talking about question 8c

ok if the expansion (1+5x)^-1 will the validy be -0.2<x<0.2

Ignore the question for now because you need to understand what the range of validity is for a basic $(1+x)^{-1}$.

Expanding this binomially, this is $(1+x)^{-1} = 1-x+x^2-x^3+x^4-x^5 + \ldots$.

This is the same as writing $(1+x)^{-1} = \displaystyle \sum_{k=0}^{\infty} (-x)^k$.

Hopefully you recognise this as an infinite sum of a geometric sequence, where the common ratio between each term is precisely just $-x$.

Also hopefully you know that an infinite geometric series gives you a finite value ONLY when the common ratio satisfies $|r| < 1$. This implies that in order for our expansion to be valid and not shoot off to infinity, we require that $|-x| < 1$. This is the same as saying that $|x| < 1$, or in other words, $-1 < x < 1$.

Now you apply this to your question.

In $\dfrac{6}{1+5x}$ this is rewritten as $6(1+5x)^{-1}$. What is the range of validity here? Compare $(1+5x)^{-1}$ with $(1+x)^{-1}$. The difference is that we replace $x$ by $5x$ in our well understood case of $(1+x)^{-1}$ by $5x$ to obtain $(1+5x)^{-1}$.

We do the same and replace $x$ in $|x| < 1$ to obtain $|5x| < 1$, hence the range of validity is...??


Then you repeat this and determine what the range of validity is for $4(1-3x)^{-1}$.

Once you get this point, we can finish up.

things I don't understand
infinite geometric sequence
|x|. I don't understand what the lines in front of the x means

I was taught that if for example the question is (1+5x)^-1 I would do -1<x<1 and then do -1<5x<1 and then divide through by 5. is this wrong?

thanks

things I don't understand
infinite geometric sequence
|x|. I don't understand what the lines in front of the x means

I was taught that if for example the question is (1+5x)^-1 I would do -1<x<1 and then do -1<5x<1 and then divide through by 5. is this wrong?

thanks