Binomial stuff

1

Ok I have 'major' holes in my knowledge i've discovered so yet another question...

The first three terms in the expansion in ascending powers of x of;

(1+px)/(1+qx) and of (1+x)^0.1

...are identical. Find the values of p and q, assuming x is sufficiently small for both expansions to be valid.

And let invisible have a chance to do it first :biggrin:

Reply 1

k@tie

10

jonas123

Ok I have 'major' holes in my knowledge i've discovered so yet another question...

The first three terms in the expansion in ascending powers of x of;

(1+px)/(1+qx) and of (1+x)^0.1

...are identical. Find the values of p and q, assuming x is sufficiently small for both expansions to be valid.

And let invisible have a chance to do it first :biggrin:

Fine! Invisible....the floor is yours!

Reply 2

jonas123

OP

1

Invisible, you have 10 mins to claim your fame...then it goes out to the general public :tongue:

Reply 3

k@tie

10

jonas123

Invisible, you have 10 mins to claim your fame...then it goes out to the general public :tongue:

give him an extension of 4 minutes

Reply 4

jonas123

OP

1

Katie Heskins

give him an extension of 4 minutes

I think we're being too generous :smile:

Reply 5

username9816

14

jonas123

The first three terms in the expansion in ascending powers of x of;
(1+px)/(1+qx) and of (1+x)^0.1
...are identical. Find the values of p and q, assuming x is sufficiently small for both expansions to be valid.

(1+px)/(1+qx) = (1+px)(1+qx)^-1 = (1+px)(1 - qx + q^2x^2...)

= 1 - qx + q^2x^2 + px - pqx^2... = 1+(p-q)x + x^2(q^2 - pq)...

Now: (1+x)^0.1 = 1+0.1x - 0.045x^2...

As the first 3 terms in the expansion of the LHS and RHS are equivalent, therefore:

1+(p-q)x + x^2(q^2 - pq) = 1+0.1x - 0.045x^2

Equating Co-efficients Gives:

--> p-q = 0.1 ---> p = q + 0.1
--> q^2 - pq = -0.045
--> q^2 - q(q + 0.1) = -0.045
--> q^2 - q^2 - 0.1q + 0.045 = 0
--> 0.1q = 0.045
--> q = 0.45

Sub. q = 0.45 into p - q = 0.1:

---> p = q + 0.1 = 0.45 + 0.1 = 0.55
---> p = 0.55

Binomial stuff

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