violetendncy

0.99999999... = x Therefore 9.99999999... = 10x Subtract them and preserve equality! (10x) - (x) = (9.99999999...) - (0.999999999...) Thus, 9x = 9 x = 1 and 1 = 0.99999999...

In physics, until we figure out whether or not the expansion of the universe is decelerating (this would be the non-euclidean elliptic geometry), Hyperbolic geometry could be the most accurate way to define the geometries of fields (like gravity) etc.

coolness, thanks

Hey there, I recently did found myself wanting to work out the area of a circle by hand, and here was what I came up with... Tools: A compass, a pen/pencil/stylus, a straightedge, paper or otherwise neat drawing surface, and oodles of free time which may or may not drive you into a crushing torpor. Method: Take the compass and make for yourself a nice circle of some arbitrary radius (make it big enough to really see everything -- about hand-sized). Be sure to fix this radius in the compass (perhaps you've a nice one with a screw-lock or something else handy). Do not swallow your compass. Now taking your compass, carefully stick the pointy end on an arbitrary point somewhere on the circle (perhaps your favorite spot?). Trace-out your second circle here and be sure to note how it should cross directly over the center of the first circle you made. It also should intersect the first circle at exactly two points. Choose your favorite of these two and stick the center of a third circle (same radius you've been using again) on that point. It should intersect at two points on the original circle, and you can just pick up your compass and move around that way again. After doing this carefully for a while, you'll notice that you've made a total of five lovely circles--one mother circle in the middle and four clones radiating around her. Take a step back and examine what you've done. You should definitely notice how these four circles intersect each other outside and inside the mother circle in two places for every pair of circles. There should be four intersections outside the momma circle, and four inside (all at the exact center of the momma). Connect these intersection points by making them straight through the center (normal to the mother circle). At this time, you should have exactly five circles and two straight lines drawn (via-straightedge). These lines should be perpendicular to each other and make a cross. If you will notice, these lines make new intersection points on the circle....you may continue to divide the circle this way (by powers of two) if you repeat the above method by placing your compass on these new points and trace around the momma circle again until all new points on the circle have become home to a new circle center (same radii of course). Now, while that is interesting certainly, it is only a mere step (necessary of course) in the right direction to answering your original question..."how do i work out the area of a circle without a calculator?" So here we go.... Please notice your original two lines making the cross intersect the mother circle in four places that are perfectly (at least in your head they are perfect...because this is all an exercise to emphasize the validity of mathematical ideas here) and evenly spaced around the circular perimeter/circumference. Please draw a square inside this circle (a circumscription, you might say) with vertices at those four points. Now, I hope you've made your circle big enough to really see all of what comes next. Focus your attention on your favorite quadrant of the circle....I'll use the second just because I'm weird like that. You should see a very nice 45-45-90 triangle there...one fourth of the area of the square (which happens to be the biggest square that can fit). The legs of this triangle are each a magnitude R of the radius you randomly picked, making the hypotenuse a line of magnitude [(2RR)^(1/2)]. This simplifies to R times the square root of 2. Now notice the line you drew before (unless you stopped after making the cross...if so, keep dividing the circle's angles in half as described in the beginning) cuts this triangle in half, bisecting the hypotenuse and running through the center of the momma circle. That line should have intersected the mother circle 45 degrees above the quadrant you're looking at....in other words, it should have split the cross's 90 degree angle in half there. Draw a new line from the point where that splitting line intersects the mother circle to one of the two vertices of the 45-45-90 triangle that is NOT at the center of the mother circle. This should make for a new right triangle...one that has a long leg and a very short leg. You will notice that you could repeat this for all the quadrants and you'd have made for yourself a nice little octagon. By continuing to divide the circle's angles in half and making these new little triangles on top of the hypotenuses of the old right triangles, you can successfully construct polygons with powers of two as their side number (square-->2^2 : octogon-->2^3 : 16-sides-->2^4 etc.). You know the area of a square is base times height.....in this case R squared divided by two gets our triangle, and 4 times that for the entire polygon. Area =(( (2^2)(R^2) )/ 2) + ...next polygons... The base of the triangle sitting on top of the triangle that is part of the big square is just half of the old hypotenuse we just found a second ago....R times the square root of 2. So side B for this triangle is R times the square root of 2 all divided by 2. We'll refer to that as B. The other leg of this new triangle, leg A, is at present unstated. Notice, however, that there is another right triangle that has half the area in the original 45-45-90 triangle. It is kinda flipped with it's hypotenuse as the radius of the circle. Well, we already know that it's hypotenuse is R, and one of it's sides is just half of the 45-45-90's hypotenuse...this is side B of the new triangle, too. Now we can find the other leg (actually it is obvious in this special case....but a couple polygons down the line, it would not have been so obvious because everything becomes a radical mess!). The other leg is the square root of all of R squared minus B squared. Now, this is the missing leg of that triangle, but as for the smallest leg of the new triangle--the one whose hypotenuse and side A are unknown--we can say that R minus the square root of all of R squared minus B squared is that side length....side A for the new triangle (the one piggy-backing on the 45-45-90). With this knowledge, we can find the hypotenuse of that triangle, and simply multiply A and B and divide by 2 to get the area of the triangle. Multiply by the next power of two (with respect to whatever polygon that hypotenuse is a part of) and you've an area that is even closer to that of a circle... Area ==( (2)(R^2) ) + ( (2)(R^2)(-1+sqrt(2)) + ...next polygons... You can repeat this method to get closer and closer to the area of a circle, and if you assume R is one, closer and closer to a hand estimation of the elusive Pi constant. Hope this helped!

Hey, I don't know if anyone will read this....but by SMERSH, are you referring to the James Bond novels....because those are awesome!

I think that's what I did, the program finds triangles to double their area and then multiply by some power of two. Another thing that bothers me about this method is that I'm relying on the Pythagorean Theorem so heavily, but I really don't know how or why it was originally developed....I mean, did that school of thought or whatever just arbitrarily stumble upon the fact?

Stupid smiley faces.....that should be a colon and then a capital letter "D"

Hello friends, it has occured to me that I know depressingly little about mathematics in general. I mean, I can do all the stuff in class they want you to do, but I really am not comfortable with any of the stuff. In saying this, I mean that I feel as though there are certain things I haven't learned....probably because they were attempted to be taught to me by southern women entertaining too many children.... Anywise, I was fiddling around with a circle the other day, and I wrote a TI-BASIC program that computes the Pi constant from scratch. If any of you are interested, here is the code (Ti-83 plus): :PROGRAM:CIRCLE :ClrHome :Input "Radius:",R :0-->F :R-->A :A-->B :3-->K :sqrt( (AA) + (BB))-->C :2AB-->E :F+E-->F :While K<200 :0.5©-->A :(R-sqrt( (RR) - (AA) ))-->B :sqrt( (AA) + (BB) )-->C :(2^(K-1))(AB)-->E :(1+K)-->K :F+E-->F :End :Disp "Area=",F :Disp "Pi=",(F/(RR)) Now, you can follow the program flow for a bit and realise how close it gets. Actually, I think I've worked it so that it calculates EXACTLY pi. But I am just unsure. As I said, I'm probably making it all too simple....but basically, It divides a circle into polygons and then solves for the right triangles that emerge from continuously halving your angles, beginning with 360 degrees. I wish I could post a picture to illustrate this more properly, but alas that's not the point. What I want to know is if any of you have better ways of explaining pi (and/or calculating it). Also, I'd like to know if any of you can show some manner of dissection for the volume of a sphere, a priori. Thanks!

Oh, I don't mean to seem contrapuntal....I was a little confused on the original question myself, and so I was merely confirming my findings initially. As for the gamma stuff, I really don't know enough about it to make any sort of intelligent comment. Lol. I'm sure you're probably right, though.

I'm right....just look o'er my answer. It has to be the sqrt(bsquared minus csquared), and not [b minus csquared] because otherwise it's just addition and root-extraction...and i assume Nadine wasn't confused about that. it's all set equal to a......

Hi, from what I've read, It's basically the theory of decision making in games...

Hey Dave, That's what I was thinking too....but I've only ever used matrix determinants to solve systems of linear equations with variables to the first power.....something like: [ [Ax + By + Cz = K1], [Dx + Ey + Fz = K2], [Gx + Hy + Iz = K3] ]. it happens oodles of times in physics crud where you're pluggin and chuggin for systems where you've assumed conservation of linear momentum and kinetic energy (like on good pool tables) to get velocities and isolate unknown vectors....but of course the energy stuff has a Vsquared term so i usually have to jetison my matrix strategy due to lack of understanding. for simple inelastic collision problems, it works great though so you get to take the reduced row echelon form and stuff.....what is this Gaussian elimination (I assume it's not when Gauss took a poop)? I'd love to know anything you feel like tellin' me! Thanks

I've also seen "and" used to express the location of the decimal point, bringing us to "forty-five thousand nine-hundred fifteen."

It helps in situations like this to make a Free-Body Diagram (drawing of all the forces coming from a point mass....the dot in your diagram representing the block). It REALLY helps and simplifies everything to then rotate your drawing and look at everything as if the surface of the ramp were perfectly horizontal. This way, it is VERY clear which force is responsible for the acceleration down the ramp--the weight--, and the normal force is pointing straight up. We know the magnitude of the weight force is Mg.....the mass of the block(usu. in kilograms) times the assumed and famous "g" or about 9.80665 meters/second/second. If we know the height of the ramp (at its highest point, perhaps) from the PERFECTLY level ground, and if we know the length of the ramp (that is the length of the side of the ramp that is resting on the ground...the ramp's bottom), we can easily calculate the angle of inclination as the inverse tangent of the height (Y) divided by the length (X). Call it slope....the inverse tangent of "m", then (interestingly enough, "m" is used to denote slope because the French word for to rise is "moter"....if you happen to know why it came from France, I'd love to know). Now then, we have at this point a little picture of a ramp, a square box on top of it, and two vectors....two forces, that is. One is the normal force extending from the surface of the ramp directly under the block perpendicularly outward. The other is the weight force of the block extending from the block's centre of mass directly to the centre of the earth (interestingly enough, there is more mass in the southern hemisphere of the earth, but you know of course the earth we're talking about is highly idealized, etc. etc.). This weight force is attempting to pull the block through the ramp and into China. Because this is obviously NOT happening, and because the block isn't floating or flying into space, we can say that the SUM OF ALL THE FORCES IN THE VERTICAL DIRECTION IS ZERO NEWTONS (vertical meaning the up-down axis along which the normal force is acting---remember we rotated our picture to make the normal force go straight up and down....to simplify things). Now to calculations.... We know Newton's second law as the Impulse-momentum thingy, but we can do some hand waving to come up with the more familiar F=Ma. Therefore, we'll use this equation in conjuction with equations made by observation to isolate for variables that aren't so immediately obvious and solve for them. Bust it up into components.... Fy=Forces perpendicular to the ramp surface=Ma=N-Mgcos(angle of incl.) We said this equals zero. Notice also that I did NOT just call the normal force Mg....only weight can be defined that way for situations like this...but as i said the Y-component of the Normal force IS Mg, indeed! ergo... Fy=MAy=N-Mgcos(angle of incl.)=0 thus, N=Mgcos(angle of incl.) We now have an expression for the magnitude of N...the normal force. This is because the ENTIRE normal force is directed in opposition to the Y-component of the Weight force. We see that there is NO acceleration in this direction (Ay=0). Now make for X stuff (parallel to the surface of the ramp)... Fx=MAx=Mgsin(angle of incl.) Here we can simply divide-out by the mass of the block/trolley we have the acceleration down the ramp... Fx/M=Ax=gsin(angle of incl.). I've found that solving stuff like this by experiment is more difficult than it sounds due to the soup of CHAOS we are swimming around in. Don't let it discourage you though, because usually you can figure out what's messing you up the most and erradicate it from everything else you do.

Nadine: I assume you meant : "the square root of a quantitiy equal to B squared minus C squared as a positive real number" I may be mistaken if you wanted : "the difference of the square root of a quantity equal to B squared with a quantity equal to C squared." With that in mind, I press on...hopefully not in vein! The key to understanding this is Pythagoras. That wiley Grecian stated (per your modus operandi) " 'A' squared plus 'B' squared is equal to 'C' squared".....if 'A' and 'B' are the legs of a right triangle (arbitrary in order) and 'C' is the hypotenuse(the one characteristic longest side) of that right triangle. Thus, we've: "XX" meaning X times X, or X squared... "sqrt()" meaning the square root of whatever is inside the parentheses... (CC) = (AA) + (BB) C = sqrt( (AA) + (BB) ) Which finds the hypotenuse side length. In order for us to get something like what you got, we need to be looking for the magnitude (value or length size) of a LEG, rather than the HYPOTENUSE. Thus thinking, we make the following: (CC) = (AA) + (BB) (CC) - (AA) = (BB) taking a root for the solution of B (one of the legs); sqrt( (CC) - (AA) ) = B. Now, you mentioned that you wanted the positive solution for B. It should then be noted that the only way for the square-root of a number to be negative is if you are finding some "complex solution".....meaning you are dealing with imaginary numbers. Assuming you want REAL SOLUTIONS for B, you must understand that C squared must be greater than A squared. Ostensibly, you might think that is a BIG limitation on the whole formula, however realize please (and fret not) that you can choose which value is C and which is A. In short, make absolutely certain that you make C the hypotenuse ALWAYS, because in order for you to call it the hypotenuse in the first place, you must observe it to be the longest side of the right triangle in question. To answer your question numerically, we make: B = sqrt( 31.36 + 19.36 ) = sqrt( 50.72 ) =>> 7.1217975259059422378186235125897.....ad nausaem