Recent content by Bob19

Hi i have this following assignment in Analysis Given X \subseteq \mathbb{R}^n which is a nonempty subset of \mathbb{R}^n The set \{ \| | x -y \| | \ | x \in X \} has an infimum such that f(y) = \{ \| | x -y \| | \ | x \in X \} where f: \mathbb{R}^n \rightarrow \mathbb{R}^n...

Okay those two aspects then prove the argument that sup(AuB) = max(sup A, sup B) ? /Bob

My definition of Supremum is a follows: every non-empty, bounded above subset of R has a smallest upper bound. Then sup(A \cup B) has a larger smallest upper bound than sup(A) and sup(B) according to the definition of Supremum ? Does this prove the given argument in my first post? /Bob...

Hello I have two non-empty sets A and B which is bounded above by R. Then I'm tasked with proving that sup(A \cup B) = max(sup A, sup B) which supposedly means that sup(A \cup B) is the largest of the two numbers sup A and sup B. Can this then be written as sup(A) < sup(A \cup...

Hello Hall and Thank You, x = 1 then \frac{e^{-\lamba}\lamba^1}{1!}= \frac{\lamba^1}{(1+ 1)1!} = e^{- \lambda} Which is then P(x \geq 1) = 1- e^{- \lambda} Am I on the right track now ? Sincerely Bob

Hello I'm Presented with the following Poisson distribution question P(X = x) = \frac{e^{-\lambda} \cdot \lambda^{x}}{x!} where x \in (1,2,3,\ldots) and \lambda > 0 Then I'm suppose to show that the above can be re-written if P(X \leq 1) = 1 - e^{- \lambda} Any idears on how I...

Hi Matt and thank You, Sorry that I have been slow to understand your surgestions. Following the first equation, then the second eqaution: z^8 + z^6 + z^4 + z^2 +1 = (z^10 + z^8 + z^6 + z^4 + z^2)/(z^2) Am I on the right path now? Sincerely and Best Regards, Bob

If I don't divide or multiply by z^2 and z^2 is not a root of the polynomial, how in God's name is it transformed into z^2 = 0 ?? The only way I can see that is z^8 + z^6 + z^4 + z^2 +1 = z^8 + z^ 6 + z^4 + 1 and then get z^2 = 0 Sincerely Bob

You meant that z^2 is a root for the polynomial ? thereby (z^2)^4 + (z^2)^3 + (z^2)^2 +1 = 0 /Bob

Okay I misunderstood I have now multiplied it by (z^(2) -1) and got z^16 + z ^12 - 1 = 0 or z^16 + z^12 = 1 Am I on the right track now ? Sincerley Bob

yes but if I divide the equation with z^2 I get z^4 + z ^ 3 + z ^2 + z + 1/z^2 = 0 how do You get z^ 2 from that? Sincerely Bob

Hello Hall, it is that thank Yoy :smile: But there is a second equation which has caused me some trouble too: z^8 + z ^ 6+ z^4 + z^2 +1 = 0 z is the same as in the first one, but how I get i simular result as in the first one? If I multiply left and right by (z-1). I don't get an...

Theorem r^n - s^n = (r-s) \sum_{j=0} ^{n-1} r^{j} s^{n-j-1} where r,s are numbers in the Ring R and n is a natural number. I'm presented with the following polymial z^4 + z^3 + z^2 + z +1 = 0 where z = cos(\frac{2 \pi}{5}) + i sin(\frac{2 \pi}{5}) To show the above I'm told to...

Hello I'm suppose to show Given z^4 + z^3 + z^2 + z + 1 = 0 where z = cos(\frac{2 \pi}{5}) + i sin(\frac{2 \pi}{5}) by using the binomial product formula. r^n - s^n Is that then if r,s = z then z^4 - z^4 = 0 ? Sincerely and Best Regards Bob

Hello I have been looking through my textbook which gives the following procedure on how to show if (cos(\frac{4 \pi}{6}) + i sin(\frac{4 \pi}{6})) Is the six root of unity for z^{6} = 1 Its know that 1 = cos(2 k \pi) + i sin(2 k \pi) where k = 1,2,3,...m According to De Moivre's...