Consider the curve from the x-values n to n+k. Construct an upper bound rectangle from n to n+k as the width, and f(n) as the height. This will give the inequality:
In my paper, students were given the existence of the E-M constant. Sy's question is proving its existence.Very similar to last year's BOS trial question 15 by carrot.
I expected students to do something similar to what Realise did. Does the existence of the limit need to be specified despite the upper bound that is independent of n?In my paper, students were given the existence of the E-M constant. Sy's question is proving its existence.
Also Sy, how did you expect students to prove the existence of it without the monotone convergence theorem?
Let the equal sides of the triangle be A and the side opposite 108 degrees be B.By considering an isosceles triangle with one of it's angles 108 degrees, prove that:
1. Don't quite get what you mean by the upper bound of gamma being independent of n.I expected students to do something similar to what Realise did. Does the existence of the limit need to be specified despite the upper bound that is independent of n?
The limit should be able to be done simply using squeeze theorem
Ah sorry I thought you meant the e^k question for some reason.1. Don't quite get what you mean by the upper bound of gamma being independent of n.
2. Squeeze Theorem is usually used to find the closed form for some sort of sequence/series/limit. Gamma, as we know so far, has no single closed form.
Yep, that is incorrect. A counter-example would be the curve y=sin(x). It is always less than 1 and greater than -1, but it most certainly does NOT converge to a finite limit.Ah sorry I thought you meant the e^k question for some reason.
For that one, I aimed for students to first consider the upper rectangles of the graph y=1/x from x=1 to x=(n)
Now the lower rectangles from x=2 to x=n
So no matter how n increases, it will always be less than 1, and greater than zero, thus converging to a finite limit.
Is this incorrect?
Ah yep that is trueYep, that is incorrect. A trivial example is to consider the curve y=sin(x). It is always less than 1 and greater than -1, but it most certainly does NOT converge to a finite limit.
That is the monotone convergence theorem, as I mentioned earlier =)Ah yep that is true
Apologies
EDIT: What if I showed
And then made the argument that as n increases H_n - ln(n) keeps decreasing however it is bounded on the lower by 0, then it approaches some finite limit?
Ah ok thanks for the clarificationThat is the monotone convergence theorem, as I mentioned earlier =)
My approach was to split the triangle into two smaller triangles by dividing the 108 angle into 72 and 36 angles and connecting that vertex to the opposite side. Let the equal sides of the larger triangle be 1, and the side opposite the 108 angle be someI expected students to do something similar to what Realise did. Does the existence of the limit need to be specified despite the upper bound that is independent of n?
The limit should be able to be done simply using squeeze theorem
Let the equal sides of the triangle be A and the side opposite 108 degrees be B.
From the sine rule of a triangle:
Now, using the cosine rule on angle theta:
Now on the larger angel:
Now:
Simplifying:
Now, we know that:
Substituting in the value for B/A we get:
I feel like I took a long approach
I got 840 and I think i got a more logical answer now:In the spirit of the AMC coming up next Thursday, here is a nice question from last year's paper:
1.
Well you don't need to know how to prove the factorial thing, think about the definition of a prime number...it is convenient that we have found a function f that "checks divisibility".1.
It is clear through Demoivre's theorem:
Substituting in
We get:
Taking the case of m/d being an integer, we look at the series version of f(m,d) it is clear that:
There is no other case.
2.
I'm going to guess that
g(n) = f(x(n), y(n))
And not an integer for composite n.
Now,from your earlier paper about the formula for primes lie, the property of prime numbers:
I still don't know how to prove this but I'm going to take it as true
Thus, we can see that
Therefore:
3.
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Hopefully that is correct