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# Binomial expansion watch

1. Hello everyone,

I'm stuck on this question - when I substitute 0.001 into the expanded form, I get 0.997991.... , which is ten times smaller than the actual answer.

I'm just wondering where I went wrong - did I do the binomial expansion wrong?
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2. (Original post by Electrogeek)
Hello everyone,

I'm stuck on this question - when I substitute 0.001 into the expanded form, I get 0.997991.... , which is ten times smaller than the actual answer.

I'm just wondering where I went wrong - did I do the binomial expansion wrong?
You haven't done anything wrong.

The expansion does not directly give you the fifth root you require. Can you see why?
3. (Original post by Electrogeek)
Hello everyone,

I'm stuck on this question - when I substitute 0.001 into the expanded form, I get 0.997991.... , which is ten times smaller than the actual answer.

I'm just wondering where I went wrong - did I do the binomial expansion wrong?
Firstly note this:
Sub , this gives us .

Then,
check

This gives us:

So the answer you are after is indeed 10 times
4. (Original post by rashid.mubasher)
Firstly note this:
Sub , this gives us .

Then,
check

This gives us:

So the answer you are after is indeed 10 times
I understand now! I was getting worried then.

Thanks for the help. 🙂
5. To save me opening another thread, can I have some help with 7b - I don't know how to find which values x is valid for in the new expression.

Also, I don't know if the combined expression is correct either?
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6. In part (b) it looks as though you have copied the first bracketed expression wrongly, with 3 rather than 27 in the denominator of two of the terms. When it comes to multiplying out the two brackets, the approach I would recommend is:

* On one row, multiply the first expression through by the first term of the second expression.

* Start a new row. Multiply the first expression through by the second term of the second expression, lining up column-wise with the first row so the x^1 terms form a column and the x^2 terms form a column. You can stop at the x^2 term.

* Start a third row and do likewise for the third term in the second expression.

* Now you can add up the columns to find the coefficients of x^0, x^1 and x^2.

I find this approach cuts down errors (and it will help the examiner to see what you're doing).
7. (Original post by old_engineer)
In part (b) it looks as though you have copied the first bracketed expression wrongly, with 3 rather than 27 in the denominator of two of the terms. When it comes to multiplying out the two brackets, the approach I would recommend is:

* On one row, multiply the first expression through by the first term of the second expression.

* Start a new row. Multiply the first expression through by the second term of the second expression, lining up column-wise with the first row so the x^1 terms form a column and the x^2 terms form a column. You can stop at the x^2 term.

* Start a third row and do likewise for the third term in the second expression.

* Now you can add up the columns to find the coefficients of x^0, x^1 and x^2.

I find this approach cuts down errors (and it will help the examiner to see what you're doing).
Thanks for the tip and spotting the mistake - I still got the same expression so I guess I just wrote it down wrong.

So how do you find where x is valid for this expression?
8. (Original post by Electrogeek)
So how do you find where x is valid for this expression?
You have two intervals of validity for x. The smallest one wins.
9. (Original post by Mr M)
You have two intervals of validity for x. The smallest one wins.
so would I be right in saying |x| < 1.5?
10. The overall validity is set by whichever is the more restrictive of the individual validities.
11. (Original post by Electrogeek)
so would I be right in saying |x| < 1.5?
Yes.
12. (Original post by old_engineer)
The overall validity is set by whichever is the more restrictive of the individual validities.
Cool - thanks for all the help.

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