I spent way too long thinking about why I was wrong last night. Would you care to enlighten us?
I'm assuming that your numerator for the five-seater table was essentially attempting to count the no. of ways to arrange people onto those seats (ignoring rotational symmetry, since that's in the denominator) such that at least one of the people is a woman. Am I correct in this assumption? If so, that numerator isn't the right answer for counting this number of arrangements. You may want to find the right number as another perms and combs exercise.
You can't combine the partial sum terms prior to taking limits. The two terms, when merged, result in something that is still strictly between -pi/2 and pi/2. But the two terms, when their numerical values computed, add to make something greater than pi/2.Well I got this, but wolfram says 3pi/4. I'm not sure if my working in the previous part is wrong or that the limit should have been pi, cos that would give the answer
\quad \sum _{ k=1 }^{ n } \tan ^{ -1 }{ \frac { 2 }{ k^{ 2 } } } \\ =\tan ^{ -1 }{ 2 } -\tan ^{ -1 }{ 0 } +\tan ^{ -1 }{ 3 } -\tan ^{ -1 }{ 1 } +\tan ^{ -1 }{ 4 } -\tan ^{ -1 }{ 2 } ...+\tan ^{ -1 }{ n } -\tan ^{ -1 }{ (n-2) } +\tan ^{ -1 }{ (n+1) } -\tan ^{ -1 }{ (n-1) } \\ =-\frac { \pi }{ 4 } +\tan ^{ -1 }{ n } +\tan ^{ -1 }{ (n+1) } \\ =-\frac { \pi }{ 4 } +\tan ^{ -1 }{ (\frac { 2n+1 }{ 1-n-n^{ 2 } } } )\quad by\quad same\quad methods\quad as\quad above\\ c)\quad \sum _{ k=1 }^{ \infty } \tan ^{ -1 }{ \frac { 2 }{ k^{ 2 } } } \\ =-\frac { \pi }{ 4 } +\lim _{ n->\infty }{ \tan ^{ -1 }{ (\frac { 2n+1 }{ 1-n-n^{ 2 } } } ) } \\ =-\frac { \pi }{ 4 } +\lim _{ n->\infty }{ \tan ^{ -1 }{ (\frac { \frac { 2 }{ n } +\frac { 1 }{ n^{ 2 } } }{ \frac { 1 }{ n^{ 2 } } -\frac { 1 }{ n } -1 } } ) } \\ =-\frac { \pi }{ 4 } +\tan ^{ -1 }{ 0 } \\ =-\frac { \pi }{ 4 } \\ \\ )
They're referring to the zeroes of the original function. These are -1 and 2.This was from the Year 12 Cambridge textbook opening chapter.
Is there a mistake in this example? It says 'So there are zeroes at x=2 and x=-1, and (after testing) turning points at (0,4) ( a maximum) and (2,0) ( a minimum), and a point of inflexion at (1,2). '
My thinking is it should have said 'So there are zeroes at x=2,x=0 '? (I also think x=1 is the point of inflexion, not sure if we call this a zero or not)
Specifically for SHM, the particle is rest at the two endpoints.
What's x.
Only if velocity is constant.
That won't work but nvm I misread the question.
Unles I'm misreading something, the answer should be equal to 4a = 12 because a equals 3 in this case.
