# How to find the values for which the expansion is validWatch

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#1
attached below. 8c thanks
0
5 months ago
#2
What values of x is the expansion of
(1+x)^{-1}
(1+5x)^{-1}
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#3
(Original post by mqb2766)
What values of x is the expansion of
(1+x)^{-1}
(1+5x)^{-1}
what do you mean by that
0
5 months ago
#4
(Original post by dasda)
what do you mean by that
Ok, take a step back, what is the series expansion for
(1+x)^{-1}
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#5
mqb2766
is it -0.2<x>0.2
0
5 months ago
#6
(Original post by dasda)
mqb2766
is it -0.2<x>0.2
No, but thinking along the right lines. Lets do (1+x)^-1 first as above.
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#7
(Original post by mqb2766)
No, but thinking along the right lines. Lets do (1+x)^-1 first as above.
I don't know but would it be -1<X>1 ?
0
5 months ago
#8
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.
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#9
(Original post by mqb2766)
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
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5 months ago
#10
(Original post by dasda)
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}?
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#11
(Original post by mqb2766)
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
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5 months ago
#12
(Original post by dasda)
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
Last edited by mqb2766; 5 months ago
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#13
(Original post by mqb2766)
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?
0
5 months ago
#14
(Original post by dasda)
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?
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#15
(Original post by mqb2766)
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?
ok if the expansion (1+5x)^-1 will the validy be -0.2<x<0.2
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5 months ago
#16
(Original post by dasda)
1-x+x2
just for clarifications we talking about question 8c
Ignore the question for now because you need to understand what the range of validity is for a basic .

Expanding this binomially, this is .

This is the same as writing .

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

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

Now you apply this to your question.

In this is rewritten as . What is the range of validity here? Compare with . The difference is that we replace by in our well understood case of to obtain .

We do the same and replace in to obtain , hence the range of validity is...??

Then you repeat this and determine what the range of validity is for .

Once you get this point, we can finish up.
Last edited by RDKGames; 5 months ago
0
5 months ago
#17
(Original post by dasda)
ok if the expansion (1+5x)^-1 will the validy be -0.2<x<0.2
Yes, when x=-0.2, this mskes the function and series head off to infinity and the series oscillates when x=0.2
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#18
(Original post by RDKGames)
Ignore the question for now because you need to understand what the range of validity is for a basic .

Expanding this binomially, this is .

This is the same as writing .

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

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

Now you apply this to your question.

In this is rewritten as . What is the range of validity here? Compare with . The difference is that we replace by in our well understood case of by to obtain .

We do the same and replace in to obtain , hence the range of validity is...??

Then you repeat this and determine what the range of validity is for .

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
0
5 months ago
#19
(Original post by dasda)
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
Yes, its right, for the last question (convergence)
0
5 months ago
#20
(Original post by dasda)
things I don't understand
infinite geometric sequence
|x|. I don't understand what the lines in front of the x means
Have you ever come across a geometric sequence?

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
It's not wrong but in order to understand why we have you need to be aware of my explanation above.

Also you need to be careful if you're simply following this rule because the range of validity for will not be .
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