Recent content by rbpl

Thank you guys for your help. Mistermath that totally makes sense, when I was trying to see the pattern in the results for x^2 + y^2 I made a stupid miscalculation and for every 1^2 * 10^2... and 10^2 * 1^2... I wrote 100... instead of 101... which made other numbres just as attractive as...

There are 100 combinations for X + Y and 100 combinations for X^2 + Y^2. Since U and V could be any of the results we can assume that the correct solution is one of the 100 combination for 100 combinations. Whatever the result is for X + Y the guy has to check numbers for at least 1 result...

Game Theory a problem which is a bit similar to the "Impossible Puzzle" From numbers 1 to 10, two integers X, and Y (not necessarily distinct) are chosen by a referee . The referee informs secretly to Joe the integer U where U = X + Y . The referee informs secretly to Bob the integer V where V...

Let f_n : [0,1] → [0,1] be a sequence of Riemann integrable functions, and f : [0, 1] → [0, 1] be a function so that for each k there is N_k so that supremum_(1/k<x≤1) of |f_n(x) − f(x)| < 1/k , for n ≥ N_k . Prove that f is Riemann integrable and ∫ f(x) dx = lim_n→∞ ∫ f_n(x) dx I am really...

Is this kind of manipulations permissible: f'(x)=(n=1 to infinity)∑ (x^n-1)/(3^n) from this it follows that f'(x)=(n=1 to infinity)∑ (x^n-1)/(3^n-1+1) f'(x)=(n=1 to infinity)∑ (1/3)*((x/3)^(n-1)) f'(x)=(n=0 to infinity)∑ (1/3)*(x/3)^(n) If it is then I'm not sure if I see anything special here...

Thank you for the quick response. So we have that: f'(x)=(n=1 to infinity)∑ (x^n-1)/(3^n) from this it follows that f'(x)=(n=0 to infinity)∑ (x^n-1)/(3^n-1) f'(x)=(n=0 to infinity)∑ (x/3)^(n-1) f'(x)=(n=1 to infinity)∑ (x/3)^(n) I got confused with limits of the summation am I correct? When...

Consider the power series (n=1 to infinity) \Sigma (x^n)/(n*3^n). (a) Find the radius of convergence for this series. (b) For which values of x does the series converge? (include the discussion of the end points). (c) If f(x) denotes the sum of the series, find f'(x) as explicitly as...

Thank you for your help.

Yes, I understand it makes sense now. So, I guess that in my w2 in the post where I used Gram-Schmidt, I schould have done this: (1 0) - (-1 0) (-3 1)(-1 0) (0 0) (1 0) . (-3 1) = ------------- (0 0) (-3 1)(-3 1) (0 0) (0 0) (-1 0) - (3/11) . (-3 1) = (1 0)...

Since S= (a b) such that a + 3b - c = 0 (c d) We have that S= (-3b-c b) (c d) Thus the basis for the Gram-Schmidt Orthogonalization is: (-3 1),(-1 0),(0 0) (0 0) (1 0) (1 0) This gives us: w1= (-3 1) (0 0) w2= (-1/10 -3/10) (1 0) w3= (1/2 0)...

let u=(a1 b1), v=(a2 b2), w=(a3 b3) (c1 d1) (c2 d2) (c3 d3) (a1 b1) . (a2 b2) = a1a2 + 2b1b2 + c1c2 + 2d1d2 (c1 d1) (c2 d2) There are four properties that: 1. <u,v>=<v,u> (a1 b1) . (a2 b2) = a1a2 + 2b1b2 + c1c2 + 2d1d2 (c1 d1) (c2 d2) (a2 b2) . (a1 b1) =...

Let M_2x2 denote the space of 2x2 matrices with real coeffcients. Show that (a1 b1) . (a2 b2) (c1 d1) (c2 d2) = a1a2 + 2b1b2 + c1c2 + 2d1d2 defines an inner product on M_2x2. Find an orthogonal basis of the subspace S = (a b) such that a + 3b - c = 0 (c d) of M_2x2...

I was trying to follow an example from my old calculus book the example was: lim x^x goes to lim(x^x)=lim(xlnx)=lim(lnx/x^-1)=lim(x^-1/-x^-2)=lim(-x)=0 So I tried to apply it there

Yes, you are right I can't believe I forgot about that, in that case what I get is 3. Since: ln(((n+1)^(n-1))/(n^n)) =ln(n+1)^(n-1)-ln(n^n) =((n-1)ln(n+1))-((n)ln(n)) =[ln(n+1)]/(n-1)^-1-[ln(n)/(n^-1)] =(n+1)^-1/(-(n-1)^-2)-(n^-1)/(-(n)^-2) =(-(n+1)^-1)((n-1)^2)+n =(-(n-1^2)/(n+1))+n...

Ok, so if \lim_{n\rightarrow\infty} \frac{(n+1)^{n-1}}{n^n} then \lim_{n\rightarrow\infty} {(n+1)^{n-1}}{(1/(n^n))}={(n+1)^{n-1}}{((n^{-n}))} then...