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Dr Frost A level Maths

Someone please help, i've never seen a question like this before...
(edited 4 years ago)
Screenshot 2019-09-04 at 17.01.29.png46.0KB
Original post by MutsaMun
Someone please help, i've never seen a question like this before...
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just managed to figure out how to attach it
Original post by RDKGames
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Original post by CardiBSuperfan3
just managed to figure out how to attach it


Expanding 2x2+5x−10(x−1)(x+2)\dfrac{2x^2 + 5x -10}{(x-1)(x+2)} in ascending powers of xx is the same as expanding 2−1x−1+4x+22-\dfrac{1}{x-1}+ \dfrac{4}{x+2} in ascending powers of xx which is the same as expanding each term individually in ascending powers of xx and adding the common terms together.

So, since we are interested in powers up to and including x2x^2, you should expand 1x−1\dfrac{1}{x-1} and 4x+2\dfrac{4}{x+2} individually first.
Original post by CardiBSuperfan3
Someone please help, i've never seen a question like this before...

use binomial expansion. i dont know what examboard you do but the formula is in the edexcel formula booklet. the 1/(x-1) would become (x-1)^-1 coz its a reciprocal. than you just use the formula and write down all the values up to x^2.
hope that helped if you cant figure out still let me know ill give it a try myself
hi yeah im doing edexcel
i can get up to doing the reciprocals but i dont know where to go from there
(edited 4 years ago)
Original post by CardiBSuperfan3
hi yeah im doing edexcel
i can get up to doing the reciprocals but i dont know where to go from there


Are you familiar with the binomial expansion formula

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+…(1+x)^n = 1+nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \ldots

??
Original post by CardiBSuperfan3
hi yeah im doing edexcel
i can get up to doing the reciprocals but i dont know where to go from there

ive got
2 - (x-1)^-1 4(x 2)^-1

ill give it a try and let you know in a bit...xxx
Original post by RDKGames
Are you familiar with the binomial expansion formula

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+…(1+x)^n = 1+nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \ldots

??


yes
Original post by CardiBSuperfan3
yes


So use that.

Notice that the middle term can be rewritten;
−1x−1=11−x=(1−x)−1-\dfrac{1}{x-1} = \dfrac{1}{1-x} = (1-x)^{-1}
where the difference between this and (1+x)n(1+x)^n is that you firstly replace xx in the formula I gave you by −x-x and set n=−1n=-1. Get the terms up to x2x^2. What do you get?

Repeat for 4(2+x)−14(2+x)^{-1} which requires a slight modification before the formula above can be applied in the same way.
Original post by RDKGames
So use that.

Notice that the middle term can be rewritten;
−1x−1=11−x=(1−x)−1-\dfrac{1}{x-1} = \dfrac{1}{1-x} = (1-x)^{-1}
where the difference between this and (1+x)n(1+x)^n is that you firstly replace xx in the formula I gave you by −x-x and set n=−1n=-1. Get the terms up to x2x^2. What do you get?

Repeat for 4(2+x)−14(2+x)^{-1} which requires a slight modification before the formula above can be applied in the same way.

okay for the first bit i got 1 + x + x^2
and then i got 8 - 4x - x^2
which I'm pretty sure is completely wrong because it said i would somehow have fractions
Original post by CardiBSuperfan3
okay for the first bit i got 1 + x + x^2
and then i got 8 - 4x - x^2
which I'm pretty sure is completely wrong because it said i would somehow have fractions


Yep it's wrong.

The first bit looks good though, it's (1−x)−1=1+x+x2+O(x3)(1-x)^{-1} = 1+x+x^2 + O(x^3).

Then for the second bit, you need to realise that to expand (2+x)−1(2+x)^{-1} using the binomial expansion I posted above, you first need to make sure that rather than being 2+... it's actually 1+... inside the bracket.
How do we do that? Simply factor out a 2 out of the bracket. Then it becomes 4(2+x)−1=4[2(1+x2)]−1=4⋅2−1(1+x2)−14(2+x)^{-1} = 4[2(1+\frac{x}{2})]^{-1} = 4\cdot 2^{-1}(1+\frac{x}{2})^{-1}.
So we reduce the problem to 4(2+x)−1=2(1+x2)−14(2+x)^{-1} = 2(1+\frac{x}{2})^{-1}

Now you need to expand just (1+x2)−1(1+\frac{x}{2})^{-1} using the binomial formula above. Simply replace xx by x2\frac{x}{2} and set n=−1n=-1. Expand up to x2x^2 and then multiply all coefficients by 2 since we seek 2(1+x2)−12(1+\frac{x}{2})^{-1}.
Original post by RDKGames
Yep it's wrong.

The first bit looks good though, it's (1−x)−1=1+x+x2+O(x3)(1-x)^{-1} = 1+x+x^2 + O(x^3).

Then for the second bit, you need to realise that to expand (2+x)−1(2+x)^{-1} using the binomial expansion I posted above, you first need to make sure that rather than being 2+... it's actually 1+... inside the bracket.
How do we do that? Simply factor out a 2 out of the bracket. Then it becomes 4(2+x)−1=4[2(1+x2)]−1=4⋅2−1(1+x2)−14(2+x)^{-1} = 4[2(1+\frac{x}{2})]^{-1} = 4\cdot 2^{-1}(1+\frac{x}{2})^{-1}.
So we reduce the problem to 4(2+x)−1=2(1+x2)−14(2+x)^{-1} = 2(1+\frac{x}{2})^{-1}

Now you need to expand just (1+x2)−1(1+\frac{x}{2})^{-1} using the binomial formula above. Simply replace xx by x2\frac{x}{2} and set n=−1n=-1. Expand up to x2x^2 and then multiply all coefficients by 2 since we seek 2(1+x2)−12(1+\frac{x}{2})^{-1}.

yeah that's what i did but do i not need to then multiply by 4?
Original post by CardiBSuperfan3
yeah that's what i did but do i not need to then multiply by 4?


I've shown you that 4(2+x)−14(2+x)^{-1} is the same as 2(1+x2)−12(1+\frac{x}{2})^{-1}.

So when you expand (1+x2)−1(1+\frac{x}{2})^{-1} you need to multiply by 2 since that's what's in front of it, not 4.

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