Divulgando o Mathematics Stack Exchange

Exemplos do que respondi e perguntei, neste novo site, em dois casos, num como autor de uma resposta e no outro como autor de uma pergunta.

Pergunta de jkottnauer

Adding powers of  i

I’ve been struggling with figuring out how to add powers of i.

For example, the result of i^{3}+i^{4}+i^{5} is 1. But how do I get the result of i^{3}+i^{4}+\ldots +i^{50}? (…)

Minha resposta

Observing that i^{3}+i^{4}+\ldots +i^{50} is a geometric progression with ratio i,  first term i^3 and 50-3+1=48 terms, we have

i^{3}+i^{4}+\ldots +i^{50}=i^{3}\times \dfrac{1-i^{50-3+1}}{1-i}= =i^{3}\times\dfrac{1-i^{48}}{1-i}=i^{2}i\times \dfrac{1-(i^{2})^{24}}{1-i}=

=-i\dfrac{1-(-1)^{24}}{1-i}=-i\dfrac{1-1}{1-i}=0

Minha pergunta

Generalizing \sum_{n=1}^{\infty }n^{2}/x^{n} to \sum_{n=1}^{\infty }n^{p}/x^{n}

For computing the present worth of an infinite sequence of equally spaced payments (n^{2}) I had the need to evaluate

\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}}{x^{n}}=\dfrac{x(x+1)}{(x-1)^{3}}\qquad x>1.

The method I used was based on the geometric series \displaystyle\sum_{n=1}^{\infty}x^{n}=\dfrac{x}{1-x} differentiating each side followed by a multiplication by x, differentiating a second time and multiplying again by x. There is at least a second (more difficult) method that is to compute the series partial sums and letting n go to infinity.

Question: Is there a closed form for

\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{p}}{x^{n}}\qquad x>1,p\in\mathbb{Z}^{+}?

What is the sketch of its proof in case it exists?

Uma das respostas: esta de Qiaochu Yuan

Adenda de 17/9/2010: Uma pergunta de Wade

Aunt and Uncle’s fuel oil tank dip stick problem

(…)

The problem is usually stated in the form of a letter from an Aunt and Uncle:
Dear niece/nephew, How are things going for you and your folks? We hear you are doing quite well it school. Keep it up! Given this success, we were hoping you could help us figure out a little dilemma. As you know, our home is heated by fuel oil, and we have a big tank buried in the side yard. The tank is a cylinder, 20 feet long and 10 feet in diameter, lying on its side five feet deep, with a narrow tube coming to a fill cap at ground level. Your uncle has a 15 foot length of old pipe that we’d like to utilize as a dip stick in order to know when we are getting close to needing a fill-up. We know that 0 feet is empty, 5 feet is half full, and 10 feet is completely full. Trouble is, we don’t know how to mark any other points. We are pretty sure they will not be uniformly spaced. What we really want is to know, within the nearest 0.01 foot, where to mark the dip stick for every multiple of 10% from 0% to 100%. Can you figure this out for us? Of course, we will want to see details of your solution and check it ourselves, and it would especially help if you could draw us a scale model of the dip stick. Love, Auntie Flo and Uncle Jim (…)

Minha resposta

oiltanklevel

oiltanklevelversusoilsticklength

Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.

Lower figure: graph of oil volume/max. volume (in %) versus oil level l (in feet). The horizontal straight lines represent the area/volume ratio A(l)/A(10)=V(l)/V(10) (in %) for every multiple of 10% from 0% to 100%.

Since the tank radius is 5, the oil level with respect to the bottom of the tank is given by l=5-5\cos \dfrac{\theta }{2}, where \theta is the central angle as shown in the figure. The area of the tank cross section filled with oil is A(\theta )=\dfrac{25}{2}\theta -\dfrac{25}{2}\sin \theta

or

A(l)=25\arccos (\dfrac{5-l}{5})-\dfrac{25}{2}\sin (2\arccos (\dfrac{5-l}{5}))

The area ratio A(l)/A(10)=V(l)/V(10) where V(l) is the oil volume.

Let f(l) denote this area ratio in percentage:

f(l)=\dfrac{100}{\pi }\arccos \left( 1-\dfrac{1}{5}l\right) -\dfrac{50}{\pi }\sin \left( 2\arccos \left( 1-\dfrac{1}{5}l\right) \right)

Here is the sequence of f(l) for l=0,1,2,\ldots ,10. The graph of f(l) is shown above.

f(0)=0, f(1)=5.2044, f(2)=14.238, f(3)=25.232, f(4)=37.353, f(5)=50,

f(6)=62.647, f(7)=74.768, f(8)=85.762, f(9)=94.796, f(10)=100

(…)

The oil level marks (in feet) should be placed at

0,1.57,2.54,3.40,4.21,

5,5.79,6.60,7.46,8.44,10

corresponding to the oil volume percentage of

0,10,20,30,40,

50,60,70,80,90,100.

This calculation was based on the following f function values:

f(0)=0.0, f(1.5648)=10.0, f(2.5407)=20.0, f(3.40155)=30.0, f(4.21135)=40.0

f(5)=50.0, f(5.7887)=60.0, f(6.59845)=70.000, f(7.4593)=80.000,

f(8.4352)=90.000, f(10)=100.0

Figure of marks:

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Divulgação, Matemática, Math, Mathematics Stack Exchange com as etiquetas , , , . ligação permanente.

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