Math Quiz Thread

[beero1000]

New York City lies at 40°42′46″N 74°00′21″W and London lies at 51°30′26″N 0°7′39″W. Assuming a perfectly spherical earth, and neglecting altitude, from which angle (measured from due north) should a plane depart New York if it takes the shortest path along the surface? What angle should it depart if it takes the path of constant bearing? What is the difference in distances between the two paths?

 

[Juma]

New York City lies at 40°42′46″N 74°00′21″W and London lies at 51°30′26″N 0°7′39″W. Assuming a perfectly spherical earth, and neglecting altitude, from which angle (measured from due north) should a plane depart New York if it takes the shortest path along the surface? What angle should it depart if it takes the path of constant bearing? What is the difference in distances between the two paths?

Did you miss something? Shortest path is at constant bearing. Is there some map projection involved that you forgot to mention?

 

[beero1000]

New York City lies at 40°42′46″N 74°00′21″W and London lies at 51°30′26″N 0°7′39″W. Assuming a perfectly spherical earth, and neglecting altitude, from which angle (measured from due north) should a plane depart New York if it takes the shortest path along the surface? What angle should it depart if it takes the path of constant bearing? What is the difference in distances between the two paths?

Did you miss something? Shortest path is at constant bearing. Is there some map projection involved that you forgot to mention?

Shortest paths on the sphere do not have constant bearing (the bearing is given by the angle made with lines of longitude).

 

[steve_bnk]

Why are you calculating an area? It is not called for in the way you stated the problem.


Each term in the series you defined is defined as a point on a sine curve, there is no area and no dx. It should be a path integral,the length of the curve.


In terms of area the integral of the sine is -cosine. You have a way of making simple issues convoluted.


The integrands are [0,pi] not[0,5000/pi]



I calculate the series sum as 3183 for5000 points. for 50,000 points the area by Riemann Sums as 2. Thearea by direct integration 2.00.


Area is 2 calulated two ways.




clear;
points = 5000;
integral_sum = 0;
dx = %pi/points;
n = 0;
sum = 0;


area = -cos(%pi) - (-cos(0));


for p = 1oints + 1 ;
s(p) = sin(n*%pi/points);
sum = sum + s(p);
integral_sum = integral_sum +(dx*s(p));
n = n + 1;
end


sum
area_by_riemann_sum
area_by_direct_integral


->sum
ans =

31830.989


-->area_by_riemann_sum
area_by_riemann_sum =

2.


-->area_by_direct_integral
area_by_direct_integral =

2.

 

[steve_bnk]

nm

 

[Juma]

New York City lies at 40°42′46″N 74°00′21″W and London lies at 51°30′26″N 0°7′39″W. Assuming a perfectly spherical earth, and neglecting altitude, from which angle (measured from due north) should a plane depart New York if it takes the shortest path along the surface? What angle should it depart if it takes the path of constant bearing? What is the difference in distances between the two paths?

Did you miss something? Shortest path is at constant bearing. Is there some map projection involved that you forgot to mention?

Shortest paths on the sphere do not have constant bearing (the bearing is given by the angle made with lines of longitude).

Ok.

 

[Bomb#20]

If \(P = \lim_{n \to\infty} \prod_{k=1}^n (1 + \frac{1}{2^k})\), prove (by hand) that \(P\) converges to a number such that \(2.3< P < 2.5\).

By hand? Seriously? Not even a calculator?!? Fine, be that way!

\(P > L, where L = \prod_{k=1}^5 (1 + \frac{1}{2^k})\)

I make L = 2 + 318/1024 + 23/32768 > 2 + 3180/10240 > 2 + 3072 / 10240 = 2.3

\(P = 3/2 Q, where Q = \lim_{n \to\infty} \prod_{k=2}^n (1 + \frac{1}{2^k})\)

Show

log(Q) = log(1+1/4) + log(1+1/8) + log(1+1/16) + ...

log(Q) < 1/4 + 1/8 + 1/16 + ...

log(Q) < 1/2

log(Q²) < 1

Q² < e = 2.71828... < 2.777... = 25/9

Q² < 25/9

Q < 5/3

3/2 Q < 5/2

P < 2.5

 

[ryan]

This is a tricky calculus problem that my students had some problems with.

So does the sum converge or diverge?

\(\sum_{n=1}^{\infty }\frac{n (( l )) n(n)}{(n+1)^{3}}\)

What is the symbol that I put in double brackets? Is it the natural logarithm at n?

 

[beero1000]

If \(P = \lim_{n \to\infty} \prod_{k=1}^n (1 + \frac{1}{2^k})\), prove (by hand) that \(P\) converges to a number such that \(2.3< P < 2.5\).

By hand? Seriously? Not even a calculator?!? Fine, be that way!

\(P > L, where L = \prod_{k=1}^5 (1 + \frac{1}{2^k})\)

I make L = 2 + 318/1024 + 23/32768 > 2 + 3180/10240 > 2 + 3072 / 10240 = 2.3

\(P = 3/2 Q, where Q = \lim_{n \to\infty} \prod_{k=2}^n (1 + \frac{1}{2^k})\)

Show

log(Q) = log(1+1/4) + log(1+1/8) + log(1+1/16) + ...

log(Q) < 1/4 + 1/8 + 1/16 + ...

log(Q) < 1/2

log(Q²) < 1

Q² < e = 2.71828... < 2.777... = 25/9

Q² < 25/9

Q < 5/3

3/2 Q < 5/2

P < 2.5

That's basically it, though the lower bound can be done without that calculation. Once you have an upper bound, we know we have convergence, and the lower bound can be done by rearranging terms in the product as follows \(P = \lim_{n\to\infty}\prod_{k=1}^n (1 + \frac{1}{2^k}) \geq (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots) + (\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots) = 2 + \frac{1}{4}(\frac{4}{3}) = 2.33\)

- - - Updated - - -

This is a tricky calculus problem that my students had some problems with.

So does the sum converge or diverge?

\(\sum_{n=1}^{\infty }\frac{n (( l )) n(n)}{(n+1)^{3}}\)

What is the symbol that I put in double brackets? Is it the natural logarithm at n?

\(\ln(n)\) is the natural log of the index n.

 

[Kharakov]

What does the following infinitely nested radical generate (x and n greater than 1):

\( \sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}} \)

What important number do you get from the following:

x=2, n=2, a= the total number of radicals, do the limit approach, not an infinite amount of nestings
note: corrected the following equation, it was missing ()
\( \sqrt[n]{\left( x-\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}}\right) \ \times (n\ \times x^{n-1})^a \ \)

Note that what you multiply the radicals by is simply the derivative of x^n...

Show

what about this one B>1 x>0:
\(log_B(B^x-x + log_B(B^x-x + log_B(B^x-x+\ldots)))\)

 

[lpetrich]

Compute (by hand) the value of \(\frac{1000000}{\sin(\frac{\pi}{5000}) +\sin(\frac{2\pi}{5000}) + \dots + \sin(\frac{5000\pi}{5000})}\) to the nearest thousandth.

My solution:

First find the denominator's sum in general
\(\sum_{k=1}^n \sin ka = \frac{1}{2\sin ((1/2)a)}\sum_{k=1}^n \cos ((k-1/2)a) - \cos ((k+1/2)a) = \frac{\cos ((1/2)a) - \cos ((n+1/2)a)}{2\sin ((1/2)a) \)

That number becomes
\(1000000 \tan(\frac{\pi}{10000})} \sim 100\pi \left(1 + \frac13 \left( \frac{\pi}{10000} \right)^2 + \cdots \right) \sim 314.159\)

If \(P = \lim_{n \to\infty} \prod_{k=1}^n (1 + \frac{1}{2^k})\), prove (by hand) that \(P\) converges to a number such that \(2.3< P < 2.5\).

\(\ln P = \sum_{k=1}^{\infty} \ln (1 + 2^{-k} ) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \frac{2^{-n}}{1-2^{-n}}\)
Beyond this, I'm stumped.

 

[lpetrich]

New York City lies at 40°42′46″N 74°00′21″W and London lies at 51°30′26″N 0°7′39″W. Assuming a perfectly spherical earth, and neglecting altitude, from which angle (measured from due north) should a plane depart New York if it takes the shortest path along the surface? What angle should it depart if it takes the path of constant bearing? What is the difference in distances between the two paths?

 Spherical trigonometry has the necessary formulas for the shortest path.
One first gets the distance using the law of cosines, then the bearing using either the law of sines or the law of cosines. Spherical triangles have a second law of cosines, from interchanging (side length) <-> pi - (opposite vertex angle).

Assumed Earth radius = 6371 km

Law of cosines:
Distance = 0.874312 radians = 50.0944 degrees = 5570 km

Law of cosines:
Bearing from NYC = 0.893832 radians = 51.2128 degrees
Bearing from London = 1.25088 radians = 71.6701 degrees

The  Haversine formula is a version of the spherical law of cosines that won't lose precision for distances between close points.

 Rhumb line (loxodrome) is for constant bearing. This path looks like a straight line in the Mercator projection.

One makes a loxodrome latitude with arctanh(sin(latitude))
tan(bearing) = (longitude difference) / (loxodrome latitude difference)
distance as angle = (original latitude difference) * sec(bearing)

Bearing = 1.36213 radians = 78.0443 degrees
Distance = 0.909455 radians = 52.1079 degrees = 5794 km

 

[lpetrich]

What does the following infinitely nested radical generate (x and n greater than 1):

\( \sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}} \)

The solution is essentially recursive:
\( y = \sqrt[n]{x^n-x+y} \)
or
\( x^n - y^n = x - y \)
Thus, a solution is y = x.

What important number do you get from the following:

x=2, n=2, a= the total number of radicals, do the limit approach, not an infinite amount of nestings
note: corrected the following equation, it was missing ()
\( \sqrt[n]{\left( x-\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}}\right) \ \times (n\ \times x^{n-1})^a \ \)

Note that what you multiply the radicals by is simply the derivative of x^n...

I find that it converges to a value independent of a, but I cannot proceed any further.

Show

what about this one B>1 x>0:
\(log_B(B^x-x + log_B(B^x-x + log_B(B^x-x+\ldots)))\)

Show

Recursion again.
\(y = \log_B(B^x - x + y)\)
giving
\(B^y - B^x = y - x\)
or y = x again.

 

[Kharakov]

Nice spot. Not all formulas of these types actually converge to x. When using sin or cosine inverse functions, you have to take the output range into account.

Show

As to the second one, I believe it is similar to Viete's formula for Pi, which is what it converges to. What I wonder now is whether all numbers generated by the nested root formula are transcendental.. but that's something I don't know how to prove.

 

[beero1000]

What important number do you get from the following:

x=2, n=2, a= the total number of radicals, do the limit approach, not an infinite amount of nestings
note: corrected the following equation, it was missing ()
\( \sqrt[n]{\left( x-\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\ldots}}}\right) \ \times (n\ \times x^{n-1})^a \ \)

Note that what you multiply the radicals by is simply the derivative of x^n...

Show

\(\pi\). I know the geometric reasoning, where does the derivative come in?

 

[Kharakov]

Hey beero- by your timing, I know you got it before I posted it. Good catch- that was the first way I was taught (indirectly, I thought I figured it out on my own... haha)

Show

Pi

.

I'll answer your question after I hide the answer I posted above...

Ok-

Show

Basically, I was taught that subtracting the inverse functions that converge to x from x if you do not allow them to reach the limit (of infinite nestings), dividing x-<a nestings (a>1)> by x- (a+1 nestings) became closer and closer to the derivative of the interior functions. I asked someone why, and they ended up writing out a Taylor series approximation for the above. Sorry about this poorly worded reply- my friends son has been interrupting my thought stream rather constantly, and I'm pretty tired, so it's hard to balance a kid with this. Hopefully I can get back to you in a bit.

Wow... anyway. Check out Stephen Tashi's second post in this thread over at physicsforums. He's the one who worked out the general Taylor series that showed why the derivatives popped up.

 

[lpetrich]

 Viète's formula, Vieta's Formula for Pi - ProofWiki -- it involves various trigonometric identities: double-angle and half-angle ones.

\(x = \frac{\sin x}{\prod_{k=1}^{\infty} \cos \frac{x}{2^k}}\)

Archimedes's famous calculation was essentially using
\(2^k \sin \frac{x}{2^k} < x < 2^k \tan \frac{x}{2^k}\)
for k -> infinity, also using trig identities.

 

[Kharakov]

Yeah... In the related formulas section on wikipedia it talks about the one I mentioned.

http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formula#Related_formulas

Also you'll notice you can create the golden ratio with x=golden ratio n=2 (x^2-x=1), the plastic constant with x=plastic constant, n=3: (x^3-x=1). So dropping a 1 into the nested roots gives you these numbers.

 

[Playball40]

Is this thread here to make me feel stupid?

 

[steve_bnk]

Is this thread here to make me feel stupid?

When confronted with some area of knowledge you do not have a lot of there are three options.

1. Just appreciate it, much like watching sports or enjoying music without being able to play an instrument.
2. Feel stupid and convince your self you can not do math and science.
3.Embark on a journey of discovery.

You can't go wrong with #3.