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

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#4

(Original post by

what do you mean by that

**dasda**)what do you mean by that

(1+x)^{-1}

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(Original post by

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

**mqb2766**)No, but thinking along the right lines. Lets do (1+x)^-1 first as above.

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#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.

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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(Original post by

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.

**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.

just for clarifications we talking about question 8c

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#10

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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(Original post by

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}?

**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}?

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#12

(Original post by

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

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

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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(Original post by

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

**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

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#14

(Original post by

how would you find the convergence point?

**dasda**)how would you find the convergence point?

|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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(Original post by

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?

**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?

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#16

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

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#17

(Original post by

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

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

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(Original post by

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.

**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.

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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#19

(Original post by

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

**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

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#20

(Original post by

things I don't understand

infinite geometric sequence

|x|. I don't understand what the lines in front of the x means

**dasda**)things I don't understand

infinite geometric sequence

|x|. I don't understand what the lines in front of the x means

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

thanks

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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