The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Maths binomial help

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! :smile: E18686DF-93CB-4E00-98A3-D90FF95DEEBB.jpg.jpeg90C10660-7F96-46CD-95DE-C607C92C5892.jpg.jpeg2C569B30-4D28-4C2D-8648-88BE8954E50D.jpg.jpeg
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.

Thanks in advance! :smile:


Do you know what the range of validity is for a standard expansion (1+x)^(-1) ?
|x|<1 I think?
Original post by RDKGames
Do you know what the range of validity is for a standard expansion (1+x)^(-1) ?
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) ?
Sorry completely misread your question!!

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) ?
(edited 4 years ago)
Original post by science_geeks
Sorry completely misread your question!!

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 (??)
|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 (??)
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)
|-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)
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
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?
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.
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. IMG_3669.JPG
Original post by science_geeks
Okay. I think I've got it.. I've put in a picture of my workings. IMG_3669.JPG



Yep
Original post by RDKGames
Yep

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

I'm also slightly unsure with what I should do with this question (3c) I'll post a picture of my prior workings too. Screenshot 2019-12-16 at 10.14.34.pngBinomial.JPG

@RDKGames - could you possibly help with this one too please?
(edited 4 years ago)
Original post by science_geeks
Yay! Finally, sorry that it took so long for me to understand this :colondollar:

I'm also slightly unsure with what I should do with this question (3c) I'll post a picture of my prior workings too. Screenshot 2019-12-16 at 10.14.34.pngBinomial.JPG


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
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??
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
Original post by RDKGames
Yep

Awesome - that makes a lot of sense! Thanks so much :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you