H2 math

• Occam's Razor

  20 Mar 09, 15:01

  Thanks Wen Shih. I have another Q.
  
  Verify thar r^2(r+1)^2(2r+1) - r^2(r-1)^2(2r-1) = 10r^4 + 2r^2 (can)
  Hence prove that
  n
  ∑    r^4 = 1/30 n(n+1)(2n+1)(3n^2+3n-1).
  r=1
  
  [The result
  n
  ∑   r^2 = 1/6 n(n+1)(2n+1)  can be assumed without proof ]
  r=1

• wee_ws

  20 Mar 09, 22:23

  Hi,

    Take sigma of both sides for n terms.

    Consider method of differences for LHS. Apply the assumed result on RHS.

    Try it, thanks!

  Cheers,
  Wen Shih 

• Occam's Razor

  21 Mar 09, 18:40

  Thanks! =)

• davidche

  22 Mar 09, 13:46

  H2 math is tuff!!! ITs prolly the only subject tat i have problem in

• wee_ws

  22 Mar 09, 14:20

  Hi,

    Keep trying and do clarify whenever necessary to unblock obstacles to your understanding. Jiayou!

  Cheers,
  Wen Shih

• Occam's Razor

  25 Mar 09, 17:24

  Hi, I have another question.
  
  Express 2,1,3,4,7,11,18,... (Lucas sequence) in recurrence relation.

• wee_ws

  25 Mar 09, 18:46

  Hi,

    Observe this pattern:

    T_1 = 2

    T_2 = 1

    T_3 = T_1 + T_2

    T_4 = T_2 + T_3

    T_5 = T_3 + T_4

    T_6 = T_4 + T_5

    T_7 = T_5 + T_6

    So, T_n = T_? + T_? subject to which intial conditions?

    Thanks!

  Cheers,
  Wen Shih

• Occam's Razor

  28 Mar 09, 16:45

  Thanks, I have another Q. Im not sure how to prove (ii) and (iv).
  
  

  
  

• wee_ws

  28 Mar 09, 17:16

  Hi,

    If the sequence converges, both x_n and x_(n+1) approach to a limit L.

    So L = -e^L.

    So L + e^L = 0.

    Is L = alpha and why?

    To prove that x_(n + 1) > x_n , we start by considering x_n - x_(n + 1).

    Now x_n - x_(n + 1) = x_n + e^(x_n).

    When x < alpha, x + e^x < 0 (from the graph), so what can we say about x_n + e^(x_n)?

    Try it, thanks!

  Cheers,
  Wen Shih

• Destined2REIGN

  29 Mar 09, 16:44

  1) The sequence u1,u2,u3,u4... is such that U_n+1 =(5U_n+4)/(U_n+2). Given U_n < 4 for all positive interger n, show U_n < U_n+1 for all positive interger n.
  
  2) U_n+1 = (U_n^2-200)/(2U_n - 30). Show that if U_n > 15. Then U_n+1 > 15 for all positive Integer n.
  
  3) The sequence u1,u2... is such that U_n>4 for every positive interger n. U_n+1 =3U_n/4 + 4/U_n. Prove U_n>U_n+1 for every positve interger n.

   

  Side Qn. How Do I express (n+2)^5-(n-2)^5 as a polynomial?

• wee_ws

  29 Mar 09, 19:48

  Hi,

    We look at the approach to solve Q1.

    U_(n + 1) = (5U_n + 4)/(U_n + 2) = 5 - 6/(U_n + 2).

    Sketch the graphs of y = x and y = 5 - 6/(x + 2). It is easy to verify that they intersect at the point where x = 4.

    Now, from the sketch, we see that the graph of the rectangular hyperbola is higher than the straight-line graph for x < 4. Equivalently, x_(n + 1) > x_n for x_n < 4.

    Do investigate whether the same approach may be applied to Q2 & Q3.

    Use Pascal's triangle to expand (n + 2)^5 - (n - 2)^5 efficiently.

    1

    1 1

    1 2 1

    1 3 3 1

    1 4 6 4 1

    1 5 10 10 5 1

    Thanks!

  Cheers,
  Wen Shih 

• wee_ws

  29 Mar 09, 19:48

  Hi,

    Graphs are helpful in allowing us to visualise behaviours of sequences. Thanks!

  Cheers,
  Wen Shih

• Occam's Razor

  30 Mar 09, 15:33

  Hi, regarding my last Q part (iv), I tried considering x_n - x_(n + 1) and I got X_n+1 < X_n for X_n > alpha.
  
  The question said X_n+1 > X_n for X_n >alpha. When I try to substitute values of X_n >alpha, I got X_n+1 < X_n for X_n > alpha instead.
  
  E.g. when X_n is -0.566, then X_n+1 = -e^-0.566 which is < X_n. So is the question wrong or did I do something wrong? Thanks =)

• wee_ws

  30 Mar 09, 17:21

  Hi,

    Question is indeed problematic.

    From graph, we can conclude that x_n - x_(n + 1) > 0 for x_n > alpha.

    Do highlight it to your teacher. Thanks!

  Cheers,
  Wen Shih

• Destined2REIGN

  10 Apr 09, 21:32

  Inequalities: Need help solving
  
  A)1/{(3-x)(x-1)} > 1 x not equal to 1 or 3 (Done all the way till (3-x)(x-1) - {(3-x)(x-1)}^2 >0. Think I doing smth wrong already).

  *Made a typo in the sign . How do I continue from here? I doubt expanding everything to x^4 is the right way to go.
  B) |x| < 2/( |x| -1)
  
  Both w/o calculator.

  Thx In advance to those teaching me how to do it.

• wee_ws

  10 Apr 09, 21:57

  Hi,

    1 / {(3 -x)(x - 1)} > 1

    Multiply both sides by (3 - x)^2 . (x - 1)^2, we obtain

    (3 - x)(x - 1) > (3 - x)^2 . (x - 1)^2

    => (3 - x)^2 . (x - 1)^2 - (3 - x)(x - 1) < 0, so your sign is incorrect.

    For the other question, let a = |x|. Solve a < 2 / (a - 1) first.

    Thanks!

  Cheers,
  Wen Shih 

• Destined2REIGN

  10 Apr 09, 22:20

  Originally posted by wee_ws:

  Hi,

    1 / {(3 -x)(x - 1)} > 1

    Multiply both sides by (3 - x)^2 . (x - 1)^2, we obtain

    (3 - x)(x - 1) > (3 - x)^2 . (x - 1)^2

    => (3 - x)^2 . (x - 1)^2 - (3 - x)(x - 1) < 0, so your sign is incorrect.

    For the other question, let a = |x|. Solve a < 2 / (a - 1) first.

    Thanks!

  Cheers,
  Wen Shih 

  Thanks. Updated Q A) Made a typo in the sign. Paiseh.

• wee_ws

  11 Apr 09, 07:40

  Hi,

    Do not expand. Factorise instead, i.e.

    (3 - x)^2 . (x - 1)^2 - (3 - x)(x - 1) < 0

    => (3 - x)(x - 1) [(3 - x)(x - 1) - 1] < 0

    :

    Thanks!

  Cheers,
  Wen Shih

• blueftw

  14 Apr 09, 21:12

  need a little asap help on my concepts.

  when a question asks to calculate unbiased estimate, which test statistic to use?

   

• wee_ws

  14 Apr 09, 23:00

  Hi,

    None at all. Formula for unbiased estimate for variance can be found in the formulae list. Thanks!

  Cheers,
  Wen Shih

• wee_ws

  15 Apr 09, 09:18

  Hi,

    No worry, what matters most is that you see the light of concepts :)

    You may like to take a look at the statistics summary I have done up on my website:

    http://www.freewebs.com/weews/statisticssummary.htm

    Thanks!

  Cheers,
  Wen Shih

• Destined2REIGN

  25 Apr 09, 17:22

  Sry ar... hijack back.
  
  Find eqn of the following ellipses?
  Intercepts at (0,+-2), (+-3,  0). Sketched the ellipse already but still dun know how to continue from there....

• wee_ws

  25 Apr 09, 19:01

  Hi,

    Recall that an ellipse with centre at the origin has the general form (x^2)/(a^2) + (y^2)/(b^2) = 1.

    Given the two points, can you find a and b?

    Thanks!

  Cheers,
  Wen Shih

   

• eagle

  25 Apr 09, 19:24

  To add on, if you cannot remember the formula, just look at your friendly graphic calculator :D

• Destined2REIGN

  25 Apr 09, 19:39

  Originally posted by wee_ws:

  Hi,

    Recall that an ellipse with centre at the origin has the general form (x^2)/(a^2) + (y^2)/(b^2) = 1.

    Given the two points, can you find a and b?

    Thanks!

  Cheers,
  Wen Shih

   

  
  Thanks! is the answer: a=3 b = 2?
  
  Got another qn:
  
  A ractangular hyperbola has asymptopes whose intersection point is the same as the intersection point of the curve in the earlier question and passes through (1,1). Find the equation of the hyperbola.

  Earlier Qn: Hyperbola intersects at x axis at x=-1 and x= 1 and y = 0.5
  
  And asymptotes x=-2 and y=-x+2. So which intersection point is the qn referring to?