# Maths binomial help

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Hi everyone,

I’m stuck with question 4c, I’ve done the rest of it but I’m not sure as to how to attempt that last one.

I’ve put in pictures of both my original workings and the question.

Thanks in advance!

I’m stuck with question 4c, I’ve done the rest of it but I’m not sure as to how to attempt that last one.

I’ve put in pictures of both my original workings and the question.

Thanks in advance!

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#2

(Original post by

Hi everyone,

I’m stuck with question 4c, I’ve done the rest of it but I’m not sure as to how to attempt that last one.

I’ve put in pictures of both my original workings and the question.

Thanks in advance!

**science_geeks**)Hi everyone,

I’m stuck with question 4c, I’ve done the rest of it but I’m not sure as to how to attempt that last one.

I’ve put in pictures of both my original workings and the question.

Thanks in advance!

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|x|<1 I think?

(Original post by

Do you know what the range of validity is for a standard expansion (1+x)^(-1) ?

**RDKGames**)Do you know what the range of validity is for a standard expansion (1+x)^(-1) ?

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#4

(Original post by

|x|<1 I think?

**science_geeks**)|x|<1 I think?

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Sorry completely misread your question!!

I’m not entirely sure what the ranges would be? That’s where I’m getting confused..

I’m not entirely sure what the ranges would be? That’s where I’m getting confused..

(Original post by

Yeah, so what are the ranges for your expansions (1-x)^(-1) and (1+3x)^(-1) ?

**RDKGames**)Yeah, so what are the ranges for your expansions (1-x)^(-1) and (1+3x)^(-1) ?

Last edited by science_geeks; 11 months ago

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#6

(Original post by

Sorry completely misread your question!!

I’m not entirely sure what the ranges would be? That’s where I’m getting confused..

**science_geeks**)Sorry completely misread your question!!

I’m not entirely sure what the ranges would be? That’s where I’m getting confused..

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|x|>1???

(Original post by

Well to go from (1+x)^(-1) with validity |x|<1 we *just replace x by -x* and obtain (1-x)^(-1) with validity (??)

**RDKGames**)Well to go from (1+x)^(-1) with validity |x|<1 we *just replace x by -x* and obtain (1-x)^(-1) with validity (??)

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#8

(Original post by

|x|>1???

**science_geeks**)|x|>1???

But since |-x| = |x| that means the range of validity is exactly the same.

Now try again with (1+3x)^(-1)

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|-3x|<1??

(Original post by

No, you replace x by -x so the validity goes from |x|<1 to |-x|<1.

But since |-x| = |x| that means the range of validity is exactly the same.

Now try again with (1+3x)^(-1)

**RDKGames**)No, you replace x by -x so the validity goes from |x|<1 to |-x|<1.

But since |-x| = |x| that means the range of validity is exactly the same.

Now try again with (1+3x)^(-1)

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#10

(Original post by

|-3x|<1??

**science_geeks**)|-3x|<1??

To go from (1+x)^(-1) to (1+3x)^(-1) we replace x by 3x therefore the validity goes from being |x|<1 to being |3x|<1 which is the same as |x|<1/3.

So to answer part c, you need to see whether x=1/2 satisfies both ranges of validity in your case |x|<1 and |x|<1/3

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(Original post by

Not sure why the minus, though as I have poined out in the above post it has no effect on the range.

To go from (1+x)^(-1) to (1+3x)^(-1) we replace x by 3x therefore the validity goes from being |x|<1 to being |3x|<1 which is the same as |x|<1/3.

So to answer part c, you need to see whether x=1/2 satisfies both ranges of validity in your case |x|<1 and |x|<1/3

**RDKGames**)Not sure why the minus, though as I have poined out in the above post it has no effect on the range.

To go from (1+x)^(-1) to (1+3x)^(-1) we replace x by 3x therefore the validity goes from being |x|<1 to being |3x|<1 which is the same as |x|<1/3.

So to answer part c, you need to see whether x=1/2 satisfies both ranges of validity in your case |x|<1 and |x|<1/3

Sorry, I could just be not getting this or overcomplicating it but would you then have |x|<1/3 and |x|<1 but how would you show if x=1/2 satisfies this?

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#12

(Original post by

Sorry, I could just be not getting this or overcomplicating it but would you then have |x|<1/3 and |x|<1 but how would you show if x=1/2 satisfies this?

**science_geeks**)Sorry, I could just be not getting this or overcomplicating it but would you then have |x|<1/3 and |x|<1 but how would you show if x=1/2 satisfies this?

Is this less than 1 ? If so, then this value is in the range of validity for (1+x)^(-1) and the expansion is valid for this value.

Is it less than 1/3 ?? Proceed to deduce the answer in similar fashion.

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(Original post by

Well... |x| = |1/2| = 1/2.

Is this less than 1 ? If so, then this value is in the range of validity for (1+x)^(-1) and the expansion is valid for this value.

Is it less than 1/3 ?? Proceed to deduce the answer in similar fashion.

**RDKGames**)Well... |x| = |1/2| = 1/2.

Is this less than 1 ? If so, then this value is in the range of validity for (1+x)^(-1) and the expansion is valid for this value.

Is it less than 1/3 ?? Proceed to deduce the answer in similar fashion.

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#14

(Original post by

Okay. I think I've got it.. I've put in a picture of my workings.

**science_geeks**)Okay. I think I've got it.. I've put in a picture of my workings.

Yep

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(Original post by

Yep

**RDKGames**)Yep

I'm also slightly unsure with what I should do with this question (3c) I'll post a picture of my prior workings too.

RDKGames - could you possibly help with this one too please?

Last edited by science_geeks; 11 months ago

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#16

(Original post by

Yay! Finally, sorry that it took so long for me to understand this

I'm also slightly unsure with what I should do with this question (3c) I'll post a picture of my prior workings too.

**science_geeks**)Yay! Finally, sorry that it took so long for me to understand this

I'm also slightly unsure with what I should do with this question (3c) I'll post a picture of my prior workings too.

Then sub that value in the approximation

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(Original post by

You need a valid value of x such that (1+3x) is 1.000003 and (1-3x) is 0.999997.

Then sub that value in the approximation

**RDKGames**)You need a valid value of x such that (1+3x) is 1.000003 and (1-3x) is 0.999997.

Then sub that value in the approximation

(1-3x)= 0.999997 and then solve for x. Then sub back into the binomial expansion??

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#18

(Original post by

So would that be (1+3x) = 1.000003 (then solve)

(1-3x)= 0.999997 and then solve for x. Then sub back into the binomial expansion??

**science_geeks**)So would that be (1+3x) = 1.000003 (then solve)

(1-3x)= 0.999997 and then solve for x. Then sub back into the binomial expansion??

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(Original post by

Yep

**RDKGames**)Yep

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