H2 math

last SG · 2009-03-19T19:16:48+00:00

Hi, I have a Q Express the general term of the series 2/(3)(5) - 3/(5)(7) + 4/(7)(9) - 5/(9)(11) + ... in partial fractions (can) and deduce its sum...

wee_ws SG

Hi Rudin,

It is an Olympiad question from 29th IMO in 1988.

You may be able to read up details of the solution by searching the web.

I'll just outline the steps:

1. Let k = (a^2 + b^2)/(ab + 1).

2. We show that among all pairs of nonnegative integers p, q with the property that (p^2 + q^2)/(pq + 1) = k and p >= q, the one with p + q minimal has q = 0. In this case k = p^2, a perfect square.

3. Let p, q (where p >= q >= 0) be the pair that minimizes p + q and suppose that q > 0. Consider (x^2 + q^2)/(xq + 1) = k which simplifies to the quadratic equation in x, i.e. x^2 - qkx + q^2 - k = 0. One root is p and the other root is qk - p, since their sum is qk in the quadratic equation.

4. Show that 0 <= qk - p < p contradicts the minimality of p + q.

Thanks!

Cheers,
Wen Shih

Rudin SG

Hi Rudin,

It is an Olympiad question from 29th IMO in 1988.

You may be able to read up details of the solution by searching the web.

I'll just outline the steps:

1. Let k = (a^2 + b^2)/(ab + 1).

2. We show that among all pairs of nonnegative integers p, q with the property that (p^2 + q^2)/(pq + 1) = k and p >= q, the one with p + q minimal has q = 0. In this case k = p^2, a perfect square.

4. Show that 0 <= qk - p < p contradicts the minimality of p + q.

Thanks!

Cheers,
Wen Shih

Thanks Mr Wee...

Gosh this is hard...

Destined2REIGN SG

A)if Y=sin x / x, prove that d2y/dx2 + 2/x.dy/dx + y = 0

I manipulated to get yx = sinx and differentiated to get dy/dx and d2y/dx2. So how do I go on from there?

B) If y = ae^(Sin x), a = constant, prove d2y/dx2 = dy/dx (Cos x - Tan x)

Qn A and B are unrelated.

Forbiddensinner SG

A) You should get dy/dx = (1/x^2)(xcosx-sinx)

and d^2y/dx^2 = (1/x^3)(-x^2(sinx)+2sinx-2xcosx)

With y = sinx/x, you just need to put them in the [d2y/dx2 + 2/x.dy/dx + y] order, and you will see how it becomes to zero.

B) dy/dx = acosxe^(sinx) and d^2y/dx^2 = a(cosx)^2.e^(sinx) - asinx.e^(sinx)

d2y/dx2 = dy/dx (Cos x - Tan x) is the same as proving

[d2y/dx2] / [dy/dx] = (Cos x - Tan x)

You should be able to see what comes next.

SBS1984E SG

A) You should get dy/dx = (1/x^2)(xcosx-sinx)

and d^2y/dx^2 = (1/x^3)(-x^2(sinx)+2sinx-2xcosx)

With y = sinx/x, you just need to put them in the [d2y/dx2 + 2/x.dy/dx + y] order, and you will see how it becomes to zero.

erm. I must be doing smth wrong. I got dy/dx as 1/2x^2.tan y

as for B) managed to solve it with your help. Thanks =)

Forbiddensinner SG

erm. I must be doing smth wrong. I got dy/dx as 1/2x^2.tan y

Did you use the correct formula?

dy/dx = [v.du/dx - u.dv/dx] / v^2,

where u is the numerator and v is the denominator.

Destined2REIGN SG

Did you use the correct formula?

dy/dx = [v.du/dx - u.dv/dx] / v^2,

where u is the numerator and v is the denominator.

I brought the power up then I used the muliplication rule. ie the keep one diff other + keep other diff one that one.

Forbiddensinner SG

I brought the power up then I used the muliplication rule. ie the keep one diff other + keep other diff one that one.

So what did you get for dy/dx and d^2y/dx^2, and where are you having problems with?

SBS1984E SG

I got dy/dx =1/2x^2.tan y

Forbiddensinner SG

I got dy/dx =1/2x^2.tan y

y = sinx / x

In this case, u = sinx and v = x

dy/dx = [ v.du/dx - u.dv/dx ] / v^2

= [ x( cosx ) - sinx( 1 ) ] / x^2

= 1/x^2 [ xcosx - sinx ]

SBS1984E SG

Checking of ans: Intergration By Parts.

1) x^2 sin x dx

ANS: cos x (-x^2 +2) +2x sin x +C

2)e^x cos x dx

Ans: (Cos x e^x) /2 + (sinx e^x)/2 +C

3)ln x dx

Ans: ln x - x + C

4) tan ^-1 x dx

Ans: x tan^-1 x - 0.5ln (1 + x^2) + C

Thanks =) Just want to check if my answers are correct. I dun have e answer key.

wee_ws SG

Hi,

1) is correct.

2) is correct.

3) is incorrect. Answer is x ln x - x + c.

4) is correct.

Good work!

In future, you may use Wolfram Mathematica Online Integrator (which I did so) to help you check:

http://integrals.wolfram.com/index.jsp

Thanks!

Cheers,
Wen Shih

eagle SG

Give you something to try
Past prelim question

integrate x^3 cos (x^2)

SBS1984E SG

integrate x^3 cos (x^2)

I stumped at intergrating the cos x^2 part.

wee_ws SG

Hi SBS1984E,

Cambridge examiners would usually begin the question with:

Differentiate sin(x^2) with respect to x.

Try to integrate x^3 cos (x^2) again by applying the above result.

Thanks!

Cheers,
Wen Shih

OHSheet SG

I stumped at intergrating the cos x^2 part.

if ima not wrong, use u=x^2 and subt and diff

wee_ws SG

Hi,

In H2 maths, the substitution will be given :) Thanks!

Cheers,
Wen Shih

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