How to find the values for which the expansion is valid Watch

dasda
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attached below. 8c thanks
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mqb2766
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What values of x is the expansion of
(1+x)^{-1}
What about
(1+5x)^{-1}
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dasda
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(Original post by mqb2766)
What values of x is the expansion of
(1+x)^{-1}
What about
(1+5x)^{-1}
what do you mean by that
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mqb2766
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(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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dasda
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mqb2766
is it -0.2<x>0.2
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mqb2766
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(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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dasda
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(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 ?
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mqb2766
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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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dasda
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(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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mqb2766
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(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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dasda
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(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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mqb2766
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(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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dasda
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(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?
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mqb2766
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(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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dasda
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(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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RDKGames
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(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 (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| &lt; 1. This implies that in order for our expansion to be valid and not shoot off to infinity, we require that |-x| &lt; 1. This is the same as saying that |x| &lt; 1, or in other words, -1 &lt; x &lt; 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} to obtain (1+5x)^{-1}.

We do the same and replace x in |x| &lt; 1 to obtain |5x| &lt; 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.
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mqb2766
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(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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dasda
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(Original post by RDKGames)
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| &lt; 1. This implies that in order for our expansion to be valid and not shoot off to infinity, we require that |-x| &lt; 1. This is the same as saying that |x| &lt; 1, or in other words, -1 &lt; x &lt; 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| &lt; 1 to obtain |5x| &lt; 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
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mqb2766
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(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)
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RDKGames
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(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?
What about the modulus function?

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 -1&lt;x&lt;1 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 (2+5x)^{-1} will not be -1&lt;5x&lt;1.
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