Maths binomial help

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#1
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.

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11 months ago
#2
(Original post by 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.

Do you know what the range of validity is for a standard expansion (1+x)^(-1) ?
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#3
|x|<1 I think?
(Original post by RDKGames)
Do you know what the range of validity is for a standard expansion (1+x)^(-1) ?
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11 months ago
#4
(Original post by science_geeks)
|x|<1 I think?
Yeah, so what are the ranges for your expansions (1-x)^(-1) and (1+3x)^(-1) ?
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#5

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

(Original post by 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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11 months ago
#6
(Original post by science_geeks)

I’m not entirely sure what the ranges would be? That’s where I’m getting confused..
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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#7
|x|>1???

(Original post by 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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11 months ago
#8
(Original post by science_geeks)
|x|>1???
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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#9
|-3x|<1??

(Original post by 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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11 months ago
#10
(Original post by science_geeks)
|-3x|<1??
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
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#11
(Original post by 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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11 months ago
#12
(Original post by 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?
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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#13
(Original post by 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.
Okay. I think I've got it.. I've put in a picture of my workings.
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11 months ago
#14
(Original post by science_geeks)
Okay. I think I've got it.. I've put in a picture of my workings.

Yep
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#15
(Original post by RDKGames)
Yep
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.

RDKGames - could you possibly help with this one too please?
Last edited by science_geeks; 11 months ago
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11 months ago
#16
(Original post by 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.
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
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#17
(Original post by 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
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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11 months ago
#18
(Original post by 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??
Yep
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#19
(Original post by RDKGames)
Yep
Awesome - that makes a lot of sense! Thanks so much
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