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# p6 questions watch

1. 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
2. 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!
3. (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.
4. Does ne one have the solomon p6 papers???.....
5. (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.
6. (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.
7. 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
8. am i suppose to know B^-1=B^2?
9. (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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