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    1.
    show that y=1+int e^t^2 dt lim x and 0 is the solution of the differential equation dy/dx=e^x^2 for which y=1 at x=0

    2.
    evaluate (1+i)^n+(1-n)^n when n=20
    using
    (cosx+isinx)^n=cos nx+isin nx

    3

    A is a matrix of the form
    abc
    0de
    00f
    A maps the points
    1 2 2 1
    0 1 3 -1
    1 and 1 on to the points 4 and 4 respectively
    (a) given that det A=8 find A (I can do this)

    B
    0-11
    2-12
    1 0 1

    I
    100
    010
    001
    (b)Show that B^3=I ,and hence find B^-1(I can also do this)

    The matrix C satisfies the equation
    B(C+I)=(B+I)^2
    (c)
    Find
    p,q,r such that the image of
    p
    q
    r under the transformation represented by
    C^-1 B^1 is
    1
    1
    -3

    I cant do the last part
    I need explanation especially the equation B(C+I)=(B+I)^2

    4.
    Show by induction that if n is an integer and n>1, 7^n-6n-1 is divisible by 36

    sorry lots of questions
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    Show by induction that if n is an integer and n>1, 7^n-6n-1 is divisible by 36

    When n = 2:

    7^2 - 6*2 - 1 = 49 - 12 - 1 = 36

    So this is true.

    Now assume this works for every integer n. Now it should work for n+1:

    7^(n+1) - 6(n+1) - 1
    = 7^n.7^1 - 6n + 6 - 1
    = 7.7^n - 6n + 6 - 1

    This should equal 36a, where a is a positive integer, showing that the expressions is a factor of 36:

    7.7^n - 6n + 6 - 1 = 36a
    a = (7/36)*7^n - n/6 + 1/6 - 1/36

    Prove that whenever n is an integer, so is a:

    erm.. dead end! Someone help a P1 mathematician out!
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    (Original post by totaljj)
    4.
    Show by induction that if n is an integer and n>1, 7^n-6n-1 is divisible by 36
    Assume true for k

    7^k+1 - 6(k+1) -1 = 7^k+1 - 6k - 7 = 6.7^k - 6 + (7^k-6k-1).

    Now (7^k-6k-1) is divisible by 36 as assumed, and 6.7^k-6 = 6(7^k-1). 7^k-1 is divisible by 6 so 6(7^k-1) is divisible by 36, so the whole thing is divisible by 36. True for n = 2, true for n = k => true for n = k+1, so true for all n > 1 by induction.
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    Does ne one have the solomon p6 papers???.....
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    (Original post by totaljj)
    1.
    show that y=1+int e^t^2 dt lim x and 0 is the solution of the differential equation dy/dx=e^x^2 for which y=1 at x=0
    Let f(x) = (int from 0 to x) e^(t^2) dt. Then f'(x) = e^(x^2). But y'(x) = e^(x^2). So y(x) = f(x) + constant. [Two functions with the same derivative can differ by only a constant.] Since y(0) = 1 and f(0) = 0, it follows that y(x) = 1 + f(x).
    (Original post by totaljj)
    2.
    evaluate (1+i)^n+(1-n)^n when n=20
    using
    (cosx+isinx)^n=cos nx+isin nx
    I think the question should say (1 + i)^n + (1 - i)^n.

    (1 + i)^20 + (1 - i)^n
    = [sqrt(2)(cos(pi/4) + i sin(pi/4))]^20
    + [sqrt(2)(cos(pi/4) - i sin(pi/4))]^20
    = 2^10 [(cos(pi/4) + i sin(pi/4))^20 + (cos(pi/4) - i sin(pi/4))^20]
    = 2^10 [(cos(5pi) + i sin(5pi)) + (cos(5pi) - i sin(5pi))]
    = 2^11 cos(5pi)
    = 2^11 cos(pi)
    = -2^11.
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    (Original post by beauford)
    Assume true for k

    7^k+1 - 6(k+1) -1 = 7^k+1 - 6k - 7 = 6.7^k - 6 + (7^k-6k-1).

    Now (7^k-6k-1) is divisible by 36 as assumed, and 6.7^k-6 = 6(7^k-1). 7^k-1 is divisible by 6 so 6(7^k-1) is divisible by 36, so the whole thing is divisible by 36. True for n = 2, true for n = k => true for n = k+1, so true for all n > 1 by induction.
    Nice. Your proof suggests a generalization of the result.

    Let A >= 2 be an integer. Define f(k) = A^k - (A - 1)k - 1. Then, for all k >= 1,

    f(k + 1) - f(k)
    = (A - 1)A^k - (A - 1)
    = (A - 1)(A^k - 1)
    = (A - 1)^2 * (A^(k - 1) + ... + A^2 + A + 1)

    is divisible by (A - 1)^2. Since f(1) = A - (A - 1) - 1 = 0 it follows that f(k) is divisible by (A - 1)^2 for all k >= 1.

    So we've proved

    2^k - k - 1 is divisible by 1 (big deal),
    3^k - 2k - 1 is divisible by 4,
    4^k - 3k - 1 is divisible by 9,
    5^k - 4k - 1 is divisible by 16,
    6^k - 5k - 1 is divisible by 25,
    7^k - 6k - 1 is divisible by 36,

    and so on.
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    For 3 do you mean C^-1.B^-1 ? If so, call the vector (p, q , r) X and call the result, (1, 1, -3) Y. Then C^-1.B^-1(X) = (Y) ie X = BC(Y).

    From the equation B(C+I) = (B+I)^2 we have that BC + B = B^2 + 2B + I so BC = B^2 + B + I. You know that B^-1 = B^2 so you can compute BC. Use this to find p, q and r
    • Thread Starter
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    am i suppose to know B^-1=B^2?
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    (Original post by totaljj)
    am i suppose to know B^-1=B^2?
    B^3 = I implies B^(-1) = B^2.
 
 
 
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