# P3 Algebra Questions

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#1
Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!
0
15 years ago
#2
(Original post by TheQueen1986)
Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

Thank you!!!
Chain rule means when you differentiate...you leave the differential of whats in the brackets on top of the fraction...and divide it by f(x)

I think..it's been awhile since i did maths G
0
15 years ago
#3
(Original post by TheQueen1986)
Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!
ln x is the natural logarithm of x. If you are having problem with a basic function like the logarithm function then you need to go and revise quite a bit.

Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3
then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

If you need more help just ask or pm me.
0
15 years ago
#4
(Original post by AntiMagicMan)
ln x is the natural logarithm of x. If you are having problem with a basic function like the logarithm function then you need to go and revise quite a bit.

Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3
then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

If you need more help just ask or pm me.

taylor series is not taught in P3 only binomial expansion can be used for this
0
15 years ago
#5
It isn't... meh they must have changed the syllabus since I did it.
0
#6
OK thanks, I get about the logarithm stuff.

But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.
0
15 years ago
#7
(Original post by TheQueen1986)
OK thanks, I get about the logarithm stuff.

But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.
For number 2- split the part in the brackets into partial fractions by long division of the term...this should give you something like "1 + 1/x+1" ....something like that...i haven't done it so that's probably wrong..but i'd imagine it was in that form.

Then you can use binomial expansion to figure it out..i think

G
0
15 years ago
#8
(Original post by TheQueen1986)
Hi, I'm stuck with several things.

1. v = ln (2x-3) dv/dx=2/2x-3

I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

If you could give me some clues on how to tackle 2 & 3, would help so much.

2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

Thank you!!!
For 2. Binomial expansion can bedone by converting into
[(1+x)^(1/3)][(1-x)^(-1/3)]

Then expand the two separately and multiply them.
If you need help on the actual expansion then PM me.
0
15 years ago
#9
Q3:
(1+ px+qx^2)^-2.
Put px + qx^2 = t. Using binomial expansion:
(1 + t)^-2 = 1 + (-2)t + (-2)(-3)t^2/2! + ...
= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
= 1 + (-2p)x + (-2q + 3p^2)x^2 +...

So you got the power of x and x^2. Replace then solve.
Try this.
0
#10
Sorry, still have some probs.

Q2:

I did what you did, but I still have troubles.

Ok I expand the numerator and the denominator. But to what x^??

Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

Q3:

How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...
0
15 years ago
#11
(Original post by TheQueen1986)
2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.
2.

[(1+x)/(1-x)]^(1/3) = [(1+x)^(1/3)][(1-x)^(-1/3)]

Now using binomial expansiona dn slittping the two expressions seperately:

(1+x)^(1/3) = 1 + x/3 + [(1/3)(-2/3)x^2]/2 + ...

(1-x)^(-1/3) = 1 + x/3 + [(-1/3)(-4/3)x^2]/2 + ...

Now multiplying the two expanded expressions and collecting the terms only upto x^2

[(1+x)/(1-x)]^(1/3) = [1 + x/3 + (x^2)/9 + ...][1 + x/3 + (2/9)(x^2) + ...]

= 1 +x/3 + (2/9)(x^2) + x/3 + (1/9)(x^2) + (1/9)(x^2)

= 1 + (2/3)x + (4/9)(x^2)

[(1+x)/(1-x)]^(1/3) = 1 + (2/3)x + (4/9)(x^2) + ...

================================ ==========================
0
15 years ago
#12
Does the above post makes sense to u TheQueen1986?
0
15 years ago
#13
(Original post by TheQueen1986)
Sorry, still have some probs.

Q2:

I did what you did, but I still have troubles.

Ok I expand the numerator and the denominator. But to what x^??

Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

Q3:

How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...
Hic, it's the formula for binomial expansion in P2 but u must have condition:
|px + qx^2| < 1. But in ur question, i didnt find that, maybe there's another way for this (not Taylor series).
But if it uses binomial expansion, as u know
(1 + x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3! + .....
But i think it's right, cuz it's P2 question.  0
#14
Yeah I got Q2 and Q3 sorted completely now. Thanks guys.
0
15 years ago
#15
A fellow A level student trying to finish their A2 maths before the year starts? Heh.

Q1: Use the chain rule, don't make false assumptions:
V = ln(2x-3)
V = ln(t), t=2x-3
dv/dt = 1/t, dt/dx =2
dv/dx = 2/t,
dv/dx = 2/(2x-3)
0
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