Recent content by 3.14lwy

thank you first . actually , the question is asking me to show : $ \prod_{1}^{n-1} \sin{(\frac{k\pi}{n})} = \frac{n^{0.5}}{2^{n-1}} by finding the roots of \ x^{2n} - 1 = 0 is the question wrong or I have made misstake?

Given that n = Π[2 – 2cos(kπ/ n)] ... (where Π is the product sign , from k = 1 to n-1 ) as cos2@ = 1 – 2(sin@)^2 then 2 – 2cos(kπ/ n) = 4[sin(kπ/ 2n)]^2 , for k = 1 , 2 , 3 , … n-1 then n = Π[4[sin(kπ/ 2n)]^2] = [4^(n-1)] Π[sin(kπ/ 2n)]^2 but the book then said...

thank you! then , is i^i = e^[(i)*(pi/2)*(i)] = e^(-pi/2) ??

(1) what is the definition of a^i ? (where a is a real number , i = (-1)^(1/2) ) is this still a complex number ? (2) where can I find the prove of "a n degree polynomial has n roots" ??

if Xn = x^n + x^(n-2) + x^(n-4) + ... + 1/[x^(n-4)] + 1/[x^(n-2)] + 1/(x^n) (where n is a non-negative integer) then , X1 = x + 1/x X3 = x^3 + x + 1/x + 1/(x^3) What s the value of X2?? X2 = x^2 + x^0 + 1/(x^2) = x^2 + 1 + 1/(x^2) or X2 = x^2 + x^0 + 1/(x^0) +...

what is the snell's law? and how to show this n = sin(i)/sin(r) (where n = refractive index i = angle of incident r = angle of refraction ) ?? thank you!

S(n) = f(n)[1 + S(n-1)] S(n) - S(n-1) = f(n)[1 + S(n-1)] - S(n-1) = f(n) + [f(n) - 1]S(n-1) ____________________________________________________ S(1) = 2/1 = 2 S(2) = 4/3 + 8/3 = 12/3 = 4 S(3) = 6/5 + (6*4)/(5*3) + (6*4*2)/(5*3*1) = 2*3/5 + 2*4/5 + 2*4*2/5 =...

I don't know will this method work , but i will try to find (sin 1 + sin 2 + sin 3 + ... + sin 90)(cos 1) = (sin 1)(cos 1) + (sin 2)(cos 1) + (sin 3)(cos 1) + ... + (sin 90)(cos 1) or (sin 1 + sin 2 + sin 3 + ... + sin 90)(sin 1) this is just a suggection , may not work...

Oh! :-p it should be S(n) = f(n)[1 + S(n-1)] but still don't know :confused: if it is in this from: X(n) = aX(n-1) + b i can find X(n) in term of a , b and X(1) , but now the a and the b is changing........ :confused:

I find S(n) = f(n)*S(n-1) and f(n) = (2^2n)(n!)/(2n+1)! then , S(n) = product of (2^2r)(r!)/(2r+1)! from r = 1 to n am I right ? however i still don't know how to do?

how to show this: if f(x) = 2x/(2x-1) then show f(n) + f(n)f(n-1) + f(n)f(n-1)f(n-2) + ... + f(n)f(n-1)f(n-2)...f(1) = 2n . I have tried to use induction , and I can show it. but , I can't show it without using induction. can anyone tell how to show it without using induction...

thank you :smile: :smile: :smile: :smile: I will try the method. :smile:

TenaliRaman you are right , but the problem ask me to do it by M.I. I can't finish it by M.I. :frown:

thanks you for all your replies , I will try uart's methods , thank you again :smile: ...... I have try the 2nd method for some time , I stop at there : let P(a) be the prosition ' (a+1)(a+2)...(a+n) is divisible by n! ' when a = 0 , it is true , P(0) is true , assume P(a) is...

how to prove : the product of n consecutive positive integers is divisible by n! by using Mathematical induction , you can assume nCk is an integer ?? it is in urgent , please help , thank you! :smile: