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I was thinking about the piety too, Ricky. Good work.
Waypoint 1: Sine. Don't know what you'll do with that though.
Final cache: ∫dx/x = ln x + c, I guess they're talking about a rotten log here. Who knows, maybe the c comes into play too.
Should I be solving more than these two things? That Google hint makes me feel like I'm wasting my Ph.D here.
Or you could make them definite integrals so you'd just have to type in a number, and possibly "pi" or some other common constant.
I used to live in California, so I've had some beach experiences... not much stands out though. There was this one time I went and caught a bunch of crabs and I was told that I had to let them go before we left for home, but I kept one with me and it got loose in my uncle's house. Its reign of terror ended, but not before my uncle sustained injuries trying to get it out.
How dare you.
You know, despite that stereotype, I don't think I've seen too many of those emo MySpaces of legend. Of course, I mainly see my friends' MySpaces, and I don't have any emo friends. Maybe I just don't look hard enough for this stuff.
Hmm... does the Integration Bee count as a game? It's like a spelling bee, but you integrate. You get a minute and 30 seconds to solve a given integral. It is one of the hottest things I have ever enountered. I've never actually had the chance to be in one, but how could it not be the most enjoyable event? Maybe I'll find one in college hosted by the math department. Until then, I'll practice my integration of obscure and challenging functions. You know, if that advanced board ever gets created, I might make something like an integration bee thread, I even have a list of integrals on this desk here which are perfect for the task. Maybe I could even post this on the Puzzles board. Ok, I'm too excited about all this, I'll stop typing.
I have one. You'll have to go look for it though. Look for a guy with "math" and "physics" as part of his interests and Euler's formula in his pictures, there probably aren't that many on MySpace, so it'll most likely be me.
Wow, good work. I really wish I could be learning Chinese. I guess I'll take it at MIT in a few years.
I'd like to write some calculus-type exercises, but I feel like if I did that now, it wouldn't really fit yet, it's a big jump from the current level of material. Maybe I should try some lighter stuff first?
Haha, Euler's World is a cute name. Anything Euler (Euler's Lair? Euler's Dungeon? Those are pretty scary) would be great, but that's coming from his biggest fan...
This seems like a good idea though. It would be nice to have a place to go here where I can look forward to discussing higher-level mathematics. Maybe having such a section would help to attract more members who are pretty focused on math, like math majors in college or kids like me who enjoy nothing more than sitting at home reading a math textbook, dreaming of the day they can take a "real" math course. Anyway, I hope this section becomes reality someday.
The discovery of another method of finding ζ(2) has led me to post on this topic again. I find this one enjoyable, as it doesn't seem like a standard approach.
We start with the Maclaurin series for sin x:
Now divide by x to obtain
This function will have zeros at nπ, so factor it according to that fact:
This simplifies to
Now if some crazy old man were to multiply this all out, the coefficient of x² would be
But in our original expression for sin(x)/x, the coefficient for x² was -1/3! = -1/6, so the sum must be equal to -1/6:
Now, simplify:
As this board increases in both the amount of exercise topics and branches of mathematics covered, perhaps subboards should be created in order to help sort out the exercises. For example, opening the "Exercises" board leads the user to another directory, with different boards for different subjects, such as "Arithmetic", "Algebra", "Geometry", "Calculus", etc. I suggest this because as the exercise board grows and some topics start getting kicked a page or two back, people are going to be less likely to notice them and they'll be "wasted". Also, it'll help users find what they need easier.
...if there doesn't exist a real number inbetween a and b, then a = b.
Yes.
I've managed to convince myself that 3.999... does in fact equal 4 by the following method:
Now from basic mathematics we know that this infinite sum is a geometric series and converges to
so
This method satisfies me completely, I wouldn't dare argue against this now. What I quoted you on is also very logical and convincing, as well is the rest of your post. Good work.
Edit:
Another way to look at it is this:
4.000000...
-3.999999...
0.000000...
Somebody could argue that there's a 1 at the end of it, but there are an infinite amount of zeros before it, so the 1 will never be reached. I don't know, this way of thinking isn't as formal as the rest of this post, but it's just another view of the problem.
The only explaination for the phenomenon that comes to my mind immediately is this:
For one of the steps, you multiply n by 10, and assume you have a new digit from nowhere in saying 10n = 39.999... In reality, when we multiply a number by 10, we lose a decimal place, like so:
10 × 3.99 = 39.9.
So with this thinking, in our problem here we would have 10n = 39.999...9, where this 9 at the end is one decimal place closer to the ones place than the original "final" 9. (of course this is a foolish thing to say, how can there be a "final" 9 in an infinite sequence of 9's? But this is all for the sake of argument, and in the end it seems to be a valid explaination, so follow me still) Now this is a different number than the 39.999... erroneously declared earlier. If we subtracted 3.999... from this number, we'd get 35.999...91. (once again, I realize that it's foolish to assume this can be done with an infinite decimal, but I feel that it shows the real workings of this problem) Divide this by 9 and you get n = 3.999..., and everyone is happy and the world of mathematics is safe and sound once again. Now, the procedure I described is something we aren't sure works for an infinite amount of digits (or maybe it does and I don't know that is does for sure), so let's see how this works for a finite amount of digits. Let m = 3.99. Then 10m = 39.9, and subtracting m, 9m = 35.91. Divide both sides by 9 and you get m = 3.99, as expected. No matter how many decimal places of 9's we have after the 3, we can still work this out and m doesn't magically round up, unless you're using a calculator that rounds up after so many digits. Assuming that this trend continues into the realm of infinite 9's after the decimal is a lot safer than assuming 3.999... = 4. I think this is the correct way to look at the problem.
It works, and it's great. Fun to just spin the hands around.
Excellent work Zach. I never thought I'd see the term "troll" on this forum.
It looks like "Amplitude of", with two non-breaking spaces? I don't know, I'm just throwing things out there.
Consider this:
If we were to graph the values of
we would get 0 for every value of x except for x such that cos x = ±1, since ±1[sup]∞[/sup] is indeterminate. Thus, we would have a line tracing the x-axis with point discontinuities at all x = nπ for n ∈ Z.
Something that is myelinated has a myelin sheath. The myelin on the axon of our brain's neurons helps to speed up electrochemical impulses in the brain. I'm not actually sure if more myelin produces more of a speed-up though, I was just using the term to look cool.
You could say that "ideas" have a speed... we could assign the speed at which electrochemical impulses travel through the brain as the speed of an idea. Of course, these impulses cannot exceed the speed of light, no matter how myelinated our neurons are.
1986? I wasn't even born yet!
Anyway, I think the problem here is that people are treating infinity as an integer by trying to assign parity to it. You should be considering infinity a concept, not a number. Infinity isn't some definite quantity. Consider this: the sum of the reciprocals of the integers (the harmonic series) is divergent, and the sum of the reciprocals of the primes is as well. You can be comfortable in saying that the integers are a much denser set of numbers than the primes, so the sum of the reciprocals of the integers add up to something greater than the sum of the reciprocals of the primes, even though both sums are infinite. I suppose what I am trying to say is that you can't say you're alternating something an infinite amount of times and ask what it will end on because there's no way to have a definite value for infinity that we could assign parity to, and an infinite process would of course never be completed.
Ricky: I feel a little silly now that I didn't approach it that way. But I arrived at the same answer, so I suppose it isn't too bad. I like your method better though.
George, Y: I know what the hyperbolic functions are for, but in application to infinite series, I'm not too educated on their applications. All I've got in that field are the Taylor series, and I have seen that there are some identities(actually I'm sure I've only seen one) for infinite series that involve hyperbolic functions. What's going on with the hyperbolic functions for you, George?
I wonder if I could find a textbook or website for techniques of evaluating infinite sums. I was trying to find any methods in my textbooks, and they said that most infinite series can be given a numerical answer that they converge to, but then they went on to say it was outside the scope of the text. That always happens! I suppose I'll continue my search.
If you feel like taking the extra time, LaTeX makes it look prettier. Things like "√" from plain text aren't the most exciting displays of mathematical notation. The only thing on those pages that would have a big improvement thanks to LaTeX are the "√"'s though. Some of the fractions could be "\frac"ed too, for optimal looks.
Ricky, your sum is divergent. Is i supposed to start at 1 instead? I'll assume this is the case and try out the problem. Also, I don't know what you mean by an alternating series, don't alternating series have (-1)[sup]i[/sup] in them?
Let's start by evaluating the partial sums:
After these first few partial sums, a pattern is noted:
We will now prove this by induction. We start with the base case:
So the base case holds. Now assume that
Then we now want to show that the equality holds for n + 1:
So by mathematical induction,
Now the question is, "what do the partial sums converge to?" To find this out we take the limit as n approaches infinity:
So we conclude that the series
converges to the value of 1.