This discussion is now closed.

Check out other Related discussions

Hateful C4 Binomial Question; HELP! D; D;

I don't have a clue of what to do, everything I've tried has failed. It's a 4 part question and the first 3 I've got right and verified with the answers at the back, so I won't waste your time writing the stuff I don't think is relevant.

c) Hence or otherwise, find the binomial expansion of

f(x)= 1/[(1-x)(1-2x)]

in ascending power of x, up to and including the term in x^3.

The answer I have and the book states is:

1 + 3x + 7x^2 + 15x^3

[Q i'm stuck on]

d)determine the coefficient of x^n in the series expansion of f(x)

¬.¬

Reply 1

Necro Defain

11

allieRAWR!

I don't have a clue of what to do, everything I've tried has failed. It's a 4 part question and the first 3 I've got right and verified with the answers at the back, so I won't waste your time writing the stuff I don't think is relevant.

c) Hence or otherwise, find the binomial expansion of

f(x)= 1/[(1-x)(1-2x)]

in ascending power of x, up to and including the term in x^3.

The answer I have and the book states is:

1 + 3x + 7x^2 + 15x^3

[Q i'm stuck on]

d)determine the coefficient of x^n in the series expansion of f(x)

¬.¬

Rewrite f(x) = 1/[(1-x)(1-2x)] using partial fractions, then use the series expansion of 1/(1-x) and 1/(1-2x) together to calculate the required co-efficients

Reply 2

allieRAWR!

OP

14

(and there's another question)

Find the binomial expansion of (1-x)^(1/3) up to and including x^4.

Both me and the book got: 1 - x/3 - (x^2)/9 - (5x^3)/81 - (10x^4)/243

b) by substituting in x=0.1 in your expansion from a), find a 5 significant figure approximation for:

i) (0.9)^1/3 (which I rightly got as 0.96529)

ii) (900)^1/3. This one I'm stuck on. The answer is 9.6529, but I can't see the relation betwwen it and i), i.e., why you would move the decimal place to the right by one.

Reply 3

allieRAWR!

OP

14

Necro Defain

Rewrite f(x) = 1/[(1-x)(1-2x)] using partial fractions, then use the series expansion of 1/(1-x) and 1/(1-2x) together to calculate the required co-efficients

Oh, I got that....that was parts a) and b). Perhaps I should write the whole question:

3a) find the binomial expansion of

i) (1-x)^-1 and ii) (1-2x)^-1

in ascending powers of x, up to and including x^3.

I've correctly done this.

B) Show that 2/(1-2x) - 1/(1-x) is that same as 1/[(1-x)(1-2x)]

I've correctly done this too.

c) Hence or otherwise, find the binomial expansion of

f(x)= 1/[(1-x)(1-2x)]

in ascending power of x, up to and including the term in x^3.

^^All the above I've done right. Just d I'm stuck on. The book says the answer is:

[2^(n+1) - 1]x^n, which is actually right, as if you plug in values of n, you get the expansion. Just, I don't know how they've got that coefficient

Reply 4

iainfs

900^(1/3) = (1000 x 0.9)^(1/3) = 1000^(1/3) x 0.9^(1/3) = 10 x 0.9^(1/3)

Reply 5

allieRAWR!

OP

14

iainfs

900^(1/3) = (1000 x 0.9)^(1/3) = 1000^(1/3) x 0.9^(1/3) = 10 x 0.9^(1/3)

Thanks

. Some of my sanity now remains in tact :smile:

D:

Reply 6

iainfs

Part d), the expansion using the partial fraction form is

2 \sum_{n=0}^{\infty} 2^n x^n - \sum_{n=0}^{\infty} x^n

by inspection. You could prove it, maybe by induction or something, but I doubt it's necessary.

\sum_{n=0}^{\infty} 2^{n+1} x^n - \sum_{n=0}^{\infty} x^n

\sum_{n=0}^{\infty} (2^{n+1}-1)x^n

So the coefficient on x^n is

2^{n+1}-1

Reply 7

allieRAWR!

OP

14

iainfs

Part d), the expansion using the partial fraction form is

2 \sum_{n=0}^{\infty} 2^n x^n - \sum_{n=0}^{\infty} x^n

by inspection. You could prove it, maybe by induction or something, but I doubt it's necessary.

\sum_{n=0}^{\infty} 2^{n+1} x^n - \sum_{n=0}^{\infty} x^n

\sum_{n=0}^{\infty} (2^{n+1}-1)x^n

So the coefficient on x^n is

2^{n+1}-1

D: I vaguely remember that notation, not enough to understand it though....it is C2? Ugh...is that the simplest way to do it? There's another similar question in the book so I'm assuming I oughta know how to tackle this for the exam.

Reply 8

iainfs

It looks more complicated than it is. Go back to your working for the part where you first found the expansion.

You probably wrote something like this:

\left[2+4x+8x^2+16x^3 + \cdots\right] - \left[1+x+x^2+x^3 + \cdots \right]

Look at the sequence of coefficents. The first part is a geometric series, first term (a) 2, common ratio (r) 2. So the kth term of that series is

2 \times 2^{k-1} = 2^k

.

On the right, you have coefficients all equal to 1.

Then, by subtracting, the kth term of the series of overall coefficients is

2^k - 1

.

Finally, we have to make a small correction, because what we called the first term in our series is actually the coefficient on x^0 -- in other words, when k=1, n=0; when k=2, n=1 and so on (where n is the power of x). So we can write n+1=k.

Which gives the result that the coefficient on x^n is

2^{n+1} - 1

Hateful C4 Binomial Question; HELP! D; D;

Related discussions

Latest

Trending

Trending

Articles for you