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[Another]

Let f2(y)=integral(csc(x) from 1 to y).
f2(y)=-ln(csc(y)+cot(y))+ln(csc(1)+cot(1)).

integral(csc(x) from 1 to y) can only equal y at points y0 that solve equation f2(y0)=y0. Numerically y0=2.95069285315252.

f(x)=y0 independent of x satisfies f(x)=integral of csc(x) from 1 to f(x) for all x.

So f(pi)=2.95069285315252.

[da_nang]

So, let's define a monster function:

Let f(x) equal the integral of csc(x) from 1 to f(x). Find f(pi).

f(x) = ₁∫^f(x)csc(x)dx = ln(|tan(f(x)/2)|) - ln(tan(0.5)), let f(x) = y

y = ln(|tan(y/2)|) - ln(tan(0.5)) = ln(|tan(y/2)|/tan(0.5))
|tan(y/2)|/tan(0.5) = e^(y)
|tan(y/2)| = tan(0.5)*e^(y)

for π*n < y/2 < π/2 + π*n ⇔ 2π*n < y < π + 2π*n :
tan(y/2) = tan(0.5)*e^(y)

for π*n - π/2 < y/2 < π*n ⇔ 2π*n - π < y < 2π*n :
tan(y/2) = -tan(0.5)*e^(y)

Now solve for the roots, which I can only do numerically. However, since the tangent function is periodic, there will be an infinite number of roots. Notice though that the negative root pairs per period will be spaced closer to each other as the exponential function goes to zero, while the positive root pairs will be spaced further from each other as the exponential function goes to infinity. Thus we could approximate the negative roots to be a number close to 2π*n (but never equal) for negative integer n, and the positive roots to be a number close to π + 2π*n (but never equal) for positive integer n.

Because of that, I'll focus on the roots closer to 0.
1. Let z(y) = tan(y/2) - tan(0.5)*e^(y). z'(y) = (tan²(y/2) + 1)/2 - tan(0.5)*e^(y)

Newton's method: y_n+1 = y_n - z(y_n)/z'(y_n), 0 < y_n < π
y₀ = 2
y ≈ 2.95069

2. Let w(y) = tan(y/2) + tan(0.5)*e^(y). w'(y) = (tan²(y/2) + 1)/2 + tan(0.5)*e^(y)
y_n+1 = y_n - w(y_n)/w'(y_n), -π < y_n < 0
y₀ = -2
y ≈ -0.58878

Now you might think I'm finished, but there's one itty bitty problem. Cosecant is undefined for 2π*n for any integer n. y = 2.95069 is currently the only root that doesn't cross that singularity point. At the moment, I can for certain say that root is valid. Other roots need to be examined. However, approximating the original problem for y = f(x) = -0.58878:
₁∫^-0,58878csc(x)dx = -_-0,58878∫¹csc(x)dx ≈ -_-0,58878∫^-0.00001csc(x)dx -_0.00001∫¹csc(x)dx ≈ -0.58878
Thus it should hold approximately for f(x) = -0.58878, possibly for all roots, but that is left for the reader.

So, through this crazy analysis, I would say that f(x) is multivalued (although the value of x doesn't matter), but f(x) = 2.95069 can probably be seen as the principal value. Thus f(π) = 2.95069

[Darvi]

f(x) = ₁∫^xcsc(x)dx

/fixed

Wait no I misread the original statement. Nevermind.

[ed boy]

So, let's define a monster function:

Let f(x) equal the integral of csc(x) from 1 to f(x). Find f(pi).

I've we're interpreting what you said literally, then the answer is infinity (or, more accurately, complex infinity).
However, if you mean then the other answers become relevant.

[Grek]

What I mean is

f(x) = ₁∫^{₁∫₁∫₁∫₁∫₁∫₁∫ ..^.. csc(x) dxcsc(x) dxcsc(x) dxcsc(x) dxcsc(x) dxcsc(x) dx} csc(x) dx

for an infinitely tall tower of integrals on top and for x=π

[ed boy]

Well, that's not a mathematical definition of a function. You have to be very careful when defining a function that involves infinity, or doing something infinitely many times. Infinity is not a number, and you cannot normally treat it as one (there are some cases where you can, but these are few and you have to be very careful in these situations). What you have to do instead is define it for some arbitrary parameter, then see what happens as this parameter gets extremely large.

You can define a sequence of functions such that (you can't use as the variable being integrated over and the limit, that's why I'm using y), but the behaviour as n becomes arbitrarily large depends on what is, which is not defined.

[Another]

Infinitely tall towers can be sometimes greatly simplified by using recursion and an additional variable/function/operator. In your case infinitely tall tower minus its lowest floor is equal to the whole tower, so all the upper floors could be substituted for an expression that should be equal to the whole tower. (While it is understandable what you mean, but variables over which integration is performed sort of "die" in each corresponding integration, so it is better to name them different from each other and from the variables in the result; like x1, x2, ...,xn)

For x=pi  csc(x) is not defined or infinity, but the recursive functional definition of f(x) above is fine for all x. (It is possible to define a function in such a way that it would be undefined for all x).

If you want some kind of limit for your tower, then you have to define how that limit should be approached. (Limits may end at infinities but can not start at infinitely remote step and the value of the limit may change depending on the approach trajectory in some cases.)

[Grek]

so all the upper floors could be substituted for an expression that should be equal to the whole tower.

Precisely. The original statement of the problem was f(x) = ₁∫^f(x) csc(x)dx, which is equivilent to ₁∫^{₁∫f(x) csc(x)dx} csc(x)dx and so so on and so forth, allowing you to expand or contract the tower of integration as far up or down as you like.

The entire point of the exercise is to ask "What happens if we have the integral of a function be a limit in its own integration?"

[ed boy]

so all the upper floors could be substituted for an expression that should be equal to the whole tower.

Precisely. The original statement of the problem was f(x) = ₁∫^f(x) csc(x)dx, which is equivilent to ₁∫^{₁∫f(x) csc(x)dx} csc(x)dx and so so on and so forth, allowing you to expand or contract the tower of integration as far up or down as you like.

Or more accurately, the integral has that property if it exists. Whether or not it exists is something that needs to be determined.

The entire point of the exercise is to ask "What happens if we have the integral of a function be a limit in its own integration?"

In that case, you have to go into integration theory, and it becomes a lot more complicated, and it becomes dependent on what theory of integration one is using, so an answer can't really be given until you specify which theory of integration we are working with.

[Another]

The entire point of the exercise is to ask "What happens if we have the integral of a function be a limit in its own integration?"

It is easy to see that in total majority of cases either f(x)=const will be a solution or there will be no valid functions. In some cases there will be infinitely many constant solutions and in some degenerate cases any function could be a solution.

If you want something that will be non-trivial in the majority of cases - look at things like f(x)=integral(g(x,y)dy,a,f(x)). Of course it will just boil down to an algebraic equation f(x)=G(x, f(x))-G(x,a), where G(x,y) - antiderivative of g(x,y) with respect to y.

If you want cases where integration is essential - put the unknown function under the integral. Take a look here http://en.wikipedia.org/wiki/Integral_equation

[Simmura McCrea]

1. How do you differentiate 1/ln(x)? Wolfram alpha says you get -1/xln^2(x) but I've no idea how.
2. How do you simplify -x^2 - y^2 where x = rcosz and y = rsinz? Wolfram alpha says -r^2 but I get -2r^2.

EDIT: Ahah, got the first one. Chain rule. Derp.

EDIT2: Got the second one too. Need to think more before asking for help.

[palsch]

Ah, nevermind. Didn't show the edit in the preview.

Spoiler: redundant now (click to show/hide)

1. How do you differentiate 1/ln(x)? Wolfram alpha says you get -1/xln^2(x) but I've no idea how.
2. How do you simplify -x^2 - y^2 where x = rcosz and y = rsinz? Wolfram alpha says -r^2 but I get -2r^2

We know this identity;

(1) cos²(z) + sin²(z) = 1;

Take the original equation and substitute for x and y;
- (r cos(z) )² - (r sin(z) )² 

Square both sets of brackets;
-r² cos²(z) - r² sin²(z)

Collect out the factor of -r²;
-r² (cos²(z) + sin²(z))

Substitute in for identity (1);
-r²

[rutsber]

I had a question something like this on a quiz in Precalculus today:
What would be the height and radius of a cylinder that has a volume of 36 in^3, with the least surface area?

We were supposed to enter it in a graphing calculator and find the lowest point, but to do that you need to get rid of the variable height by solving for it. I got the correct answer by plugging in a few values, which gave me a height of 4 and a radius of 1.69 in.. Can anyone help me figure out what the equation to enter into the calculator would be? It doesn't particular matter now, but I would like to know how to do it for the future. My teacher wrote the equations for the volume and surface area of a cylinder on the board, so it has something to do with that.

[Another]

Your volume is fixed, so that is an equation. Every simple equation is a way to get rid of 1 variable. If you want to get rid of height - express it from V=pi*r^2*h and substitute resulting expression for h into formula for surface area (S=2pi*r^2+2pi*r*h). It seems like you know the rest.

[rutsber]

Oh I see. I was trying to get height from the equation for surface area, which was throwing me off. Thanks for the help.