How do you find the area inside the spiral r=θ from θ=0 to 2π?
follow the equation: A=1/2[int (from a to b) r^2 dtheta A= area of that section of the spiral a= pretend to be greek alpha= start angle from 0= 0 radians b= pretend to be greek beta = end angle from 0= 2pi radians r= theta int --> short from integral A.D. --> short for antiderivative A= 1/2[int from 0 to 2pi] theta^2 dtheta A.D. of any variable = x^(n+1)/(n+1) A,D. of r: A.D. [theta^2] dtheta = (theta^(2+1))/(2+1)=(theta^3)/3 + C so integrating we get: 1/2 [int {from 0 to 2pi} theta^2 dtheta] = 1/2 [theta^3/3] evaluated from 0 to 2pi sub theta for 2pi and subtract subbing 0 for theta: 1/2 [({2pi}^3/3) - ({0}^3/3)] 2pi * 2pi * 2pi =(2^3) * (pi^3)= 8pi^3 ---> (8pi^3)/3 0 * 0 * 0 = 0 ---> 0/3 = 0 1/2 [ (8pi^3)/3 - 0] = 1/2 * (8pi^3)/3 = (8pi^3)/6 = (4pi^3)/3 or 41.342 is your area from 0 to 2pi. Sometimes it helps to start by finding what the area would be without any valued bound, or arbitrary bound. in other words solve the integral with the limits of integration as arbitrary letters like a and b. solving this will give you an equation that will find the area of the polar function with any two angle intervals. 1/2 [ int {from a to b} theta^2 dtheta ]= 1/2 [ theta^3/3 ] evaluated from a to b sub out theta for a and b you get: 1/2 [ (b)^3/3 - (a)^3/3 ] = (b^3 - a^3)/3 * 1/2= (b^3 - a^3)/6. Now just sub any two angles in for a and b and as long as b>a, you will find your area. Hope this helps anyone out there B)
Prove the statement: "For every epsilon > 0 and "a" belonging to the real numbers,?
EDIT: brashion's reasoning is correct in spirit, but incorrect nonetheless. if a is irrational, then a + r is irrational if r is rational. guess he/she forgot about the +a part. The second half of ben's answer is essentially what I'm doing, just going into more detail. He uses a cardinality argument in the first half, which I assume you aren't allowed to use, as this is likely an introductory (undergraduate) real analysis course, and you likely haven't studied the cardinal numbers yet, or indeed even defined what a cardinality IS. First, a lemma: For any delta > 0, there is a rational number x of the form x = 1/n with 0 < x < delta Proof: Take x = 1/[least integer greater than(1/delta)]. And another Lemma: For any delta > 0, there is an irrational number x with 0 < x < delta Proof: Let D = delta/sqrt(2). Then, by the previous lemma, there is a rational number y with 0 < y < D = delta/sqrt(2) Let x = sqrt(2)*y. Then x is irrational and 0 < sqrt(2)*y < D*sqrt(2) = delta so that 0 < x < delta, as desired. Now, to find a rational number in the interval (a - epsilon, a + epsilon), we can do the following: step 1: find an integer n with 1/n < epsilon (see lemma 1). Now, we can take m = greatest integer less than (n*a). then m/n <= a < (m+1)/n (you need to prove this) so that |a - m/n| < 1/n (needs proof as well) but then |a - m/n| < 1/n < epsilon, and so m/n E (a - epsilon, a + epsilon). So we now have a rational number in the interval. To get an irrational number in the interval, use the process above to get a rational number y (not equal to zero - need to make sure of this - it may take a little extra work) in the interval (a/sqrt(2) - epsilon/sqrt(2) , a/sqrt(2) + epsilon/sqrt(2)) then a/sqrt(2) - epsilon/sqrt(2) < y < a/sqrt(2) + epsilon/sqrt(2) and so a - epsilon < sqrt(2)*y < a + epsilon But y is rational, not equal to zero, so sqrt(2)*y is irrational, and so we have an irrational number in the interval (a - epsilon , a + epsilon), as desired.
The given equation implies 1/a+1/b+1/c = 1/42, whence (1/a+1/b+1/c)/3 = 1/126, i.e. the harmonic mean of a,b,c is 126. Now it is well-known that the arithmetic mean of three positive numbers is greater than their geometric mean, which is greater than their harmonic mean, with equality being possible only if the numbers arde themselves equal. In other words, (a+b+c)/3 >= 126 and a+b+c>=378. Equality is possible if we let a=b=c=126 — but our problem explicitly requires a,b,c to be different positive integers, so that we actually have strict inequality: a+b+c>378.We can try to find a solution close to this lower bound by arbitrarily assuming b=126 and fiddling around with a and c. Indeed, we then have 1/a+1/c=1/63, whence c=63a/(a-63). Now if a were prime to 63, a-63 would be prime to both 63 and a and couldn’t be a divisor of their product, unless it were 1, i.e. unless a=64 and c=63*64, which would give rise to a huge sum a+b+c. So a must be divisible by 7 and/or by 3. Trying a=7x, we obtain c=63x/(x-9); once again, if x is prime to 9, x-9 will be prime to both x and 9 and must therefore divide 7, i.e. be either 1 or 7. x-9=1 yields a=70 and c=630, while x-9=7 yields a=112 and c=144, whence a+b+c=382. This is one of Amit Alon’s solutions and is remarkably close to the lowest bound of 379.(Of course, this doesn’t prove there are no solutions that sum to 379, 380 or 381… We have to take Amit Alon’s word for that or re-check his programming.)
It's quite easy to see that for [math]n \geq 2[/math] we have [math](n + 1)\pi^{-(n + 1)} < \frac{3}{2 \pi}(n \pi^{-n}) < \frac{1}{2}(n \pi^{-n}) [/math]. The sum of the series is clearly bounded above by the sum of the merely geometric series [math](\frac{2}{\pi 2^n)[/math], which is [math]\frac{2}{\pi}[/math]. Since all terms are positive, the original series does converge (and quickly, too). Say it converges to [math]s[/math].This looks like a plausible candidate for a solution like this:[math](\pi - 1)s = \sum\limits_{n = 1}^{\infty} \frac{n}{\pi^{(n - 1)}} - \frac{n}{\pi^n} [/math][math]= \sum\limits_{n = 1}^{\infty} \frac{n}{\pi^{(n - 1)}} -\sum\limits_{n = 1}^{\infty} \frac{n}{\pi^n}[/math][math] = 1 + \sum\limits_{n = 1}^{\infty} \frac{n + 1}{\pi^n} - \sum\limits_{n = 1}^{\infty} \frac{n}{\pi^n} [/math][math]= 1 + \sum\limits_{n = 1}^{\infty} \frac{1}{\pi^n}[/math][math] = 1 + \frac{\pi}{\pi - 1} = \frac{\pi}{\pi - 1}[/math]Finicky people who mistrust this derivation are welcome to do it more cautiously by considering the sums, not to [math]\infty[/math], but to arbitrary upper bounds which can go to infinity. This gives a straightforward telescoping sum in which everything vanishes but the initial [math]1[/math], a finite geometric series with a familiar limit, and a final term which certainly tends to [math]0[/math]. I merely skipped past that when I could see where it was going.Anyway, we end up with [math]s = \frac{\pi}{(\pi - 1)^2}[/math].
How many sides does a circle have?
There are two sources of ambiguity here. The word CIRCLE sometimes refers to a circular disk (because it is common to talk about "the area of a circle"), but often the word refers only to the boundary of the disk (namely, the points that are equally distant from the point at the center of the figure). The meaning of the word SIDE depends on its context. A polygon in the plane has, by definition, n vertices and each pair of consecutive vertices are joined by a side. It is clear that a polygon with n vertices has n sides. In general, the word side depends on the dimension of the figure -- a side is always part of the boundary. So what is the side of a circle? If you think of the circle as a disk then it it has an up-side and a down-side. If you think of it as a curve, then it has an inside and an outside. If you think of it as the limit of an n-sided regular polygon, then one can justify the answer that the circle has infinitely many infinitesimal sides. Our conclusion: the question, "How many sides does a circle have?", is too ambiguous to have a definite answer. Perhaps the appropriate response is "there is no natural way of saying what the side of a circle is." A circle difinately has 0 corners as the definition of a corner is :the point where two lines meet or intersect. A circle is of course ONE line.
Mathematicians, help! Is Pi irrational?
"We know that Pi carries on forever, and that every possible combination of integers can be found within it." -- We know that Pi's decimal representation is nonterminating. But we do not know that "every possible combination of integers can be found within it." Many strongly suspect this to be true, but the randomness of Pi's decimal representation is an open question. Interestingly, the hexadecimal representation of Pi is *not* random, so there is reason to believe that its decimal representation may also not be random. "So, if you were to go on long enough, wouldn't there come a point where Pi reaches an infinite amount of zeros?" -- No. If the decimal representation of Pi is in fact random, then there is a point where it contains "arbitrarily many" zeroes. That is: pick any finite number, as high as you like, a googleplex or whatever, and there will be that many zeroes in a row somewhere. "We know that 0.9999.... (Irrational) is equal to 1." -- You are mistaken about the irrationality of 0.9999.... A rational number is one that can be accurately represented as a ratio between two whole numbers. That's where the name "rational" comes from. The fact that 0.9999.... = 1 shows that 0.9999.... is rational. Many rational numbers have nonterminating decimal representations: 1/3 = 0.333 .. 1/6 = 0.166... In fact, *most* rational numbers have nonterminating decimal representations. What's characteristic about them, though, is that a rational number's decimal representation at some point repeats. It might repeat a single digit, or it might repeat a combination of digits: 1/7 = 0.142857 142857 142857 ... With rational numbers, that repetition continues to infinity. The converse also holds: if you have a decimal number with a combination of digits that repeats to infinity, that number will be rational. You can write it as a fraction of two whole numbers. We can't do the same with Pi because we know that Pi is irrational. Its decimal representation may contain "arbitrarily long" sequences of 9's. But it does not contain an infinitely long sequence of 9's. If it did, then we could represent Pi accurately as a fraction, and we know that we can't.
There is nothing wrong in this!!!Pi is indeed 4 here. But what happened is you have increased the tolerance for approximation.In some systems 99 might be taken as equal to 100 but in other system even 99.98 may not be accepted as 100. It depends on how accurately you can measure your system and here you are dividing the difference in perimeters of square and circle into so many negligibly small values, there by affecting the accuracy with which you can measure your system to such levels that pi becomes approximately equal to 4.This is what has happened here:The difference between the length of the square and length of the arc of the circle it encloses always reminds the same (4 - 3.14= ) and is .86. Initially there are 4 lines of .215 units each.Then you start doing the above procedure for the first time and it now becomes 8 lines of .1075 units each. Now after each iteration this halves and after quite number of iterations become negligibly small enough for you to ignore them. But what you dont realise that there are actually large number of such lines of negligible length which is not apparent from the picture and when you sum them all it comes to .86.To give you an analogy, you try to measure quantity of water in a room, but you simply took say a bowl of water from the room and splashed it on the floor and then simply ignored the tiny water droplets. Then you measured amount of water in the room without including the bowl of water.So you simply have compromised on accuracy of your system.