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hey, induction: 2k+3>2* sqrt{(k+1)(k+2)} can you just square this and move it to the other side to make it >0 If they don't require you to use induction, for example: "By mathematical induction or otherwise" x,y>0, {(x+y)/2}^n or = 1 <=== don't quite get this prove by induction: for n>or =...

case bash case 1: atmost 1 I in the word ---> 8*6*7*5 case 2: 2Is in the word -> 7*6 * 4*3 add them together

They are classic questions.. very classic... Usually what they do for q8 is to take something classic like this and specialise it to some case...

you don't need a tutor.. you just have to approach it the right way

shit course teaches you jackshit especially when you get the lecturer who teaches out form his little how to do excel book. If I want to learn some maths I would do a maths course not a finance course...

(z1)i = z2

you aren't reading carefully enough.. You were supposed to make integral of B_1 0, what you did was setting the value of B_1 at 1 to 0

i. easy ii.slightly sneaky argument. L_{k+1} has to go in one of the first k envelopes. suppose it goes into E_{n}. Then you are left with k letters and k envelopes.. Then you break it up into 2 cases.. a) L_{n} goes into E_{k+1} b) L_{n} doesn't go into E_{k+1} the first case (a) gives u_{k-1}...

I will use z and w for z1 and z2 and z' w' to denote their conjugates (z+w)/(z-w) = (z+w)(z'-w') / |z-w|^2 = (|z|^2 - |w|^2 - zw' + wz') / |z-w|^2 ----since |z| = |w|----- = (wz' - zw')/|z-w|^2 = [wz' - (wz')']/|z-w|^2 which must be pure imaginary because it's the difference between a number...

it's false.. the sum is either 0 or n. [1-w][1+w^1+w^2+.......w^(n-1)] = 1-w^n = 0 so if w is not 1, then 1-w is not 0 so [1+w^1+w^2+.......w^(n-1)] must be 0, else the sum is n.

It's not an elemetary integral. You can reduce it to the partial perimeter of an ellipse and look up a table.

Q has coordinates (-acos@,-bsin@) the tangent line at P is (x - acos@)(-b cos@)/(a sin@) = (y - bsin@) simplifies to (a sin@)y + (b cos@)x = ab now R = (-acos@, b(1 + cos^2 @)/sin@) So QR = [b(1 + cos^2 @)/sin@ ]+ bsin@ = 2b/sin @ using that as base, the correspnding height would be...

There was a comment after that "after multiplying through by a^(n-2). you have something similar with b. now add the 2 equations up and rearrange." for example if you nultiply that equation by a, you get a^3 - a^2 + a = 0

there are 11 alphas and betas all together. the probability of a particular one coming first is 1/11 and there are 6 alphas. so 6/11

If you need something a little more comprehensive "Friendly Introduction to Mathematical Logic" Christopher C. Leary:

i) method one: There are 15! arrangements (imagine the letters are all different first)... if you assign equal probability to these.. you will see that the probably that an alpha comes before a beta is 6/11 so you get 15! * 6 / 11 then divide by (4!5!6!) to mod out the permutations to get...

i. You should be able to do this. ii. since a^2 - a + 1 = 0, you have a^n - a^(n-1) + a^(n-2) = 0 after multiplying through by a^(n-2). you have something similar with b. now add the 2 equations up and rearrange. iii. check the first 2 cases, then work on part ii's recurrence formula.. you will...

yes but that is not the point.. the argument goes: Since ln is a monotonic, strictly increasing function, the x minimizing ln(f) is the same as the one which minimizes f.

Best way to do the question is to take logs first.. minimizing f is same as minimizing ln(f) = ln(a+b+x) - ln(x)/3 + ln(constant) differentiating gives 1/[a+b+x] - 1/[3x] not hard to get x = (a+b)/ 2 from there. the 2nd derivative is -1/[a+b+x]^2 + 1/(3x^2) and if you substitute x = (a+b)/...

Go study it yourself.