Math Is Fun Forum

...however I do understand now; for any x ∈ R, x² > 0. You simply have x = 1/a. Your proof is fine.

Consider this...

Since 1/a is the multiplicative inverse of a, it holds that a × 1/a = 1. Since a > 0, we have a positive number times the "unknown" status of 1/a equalling the positive number one. But a positive number must be multiplied by another positive number for the product to also be positive, so it must be that 1/a > 0.

What what?

I had an entertaining mechanics professor who could often refer to objects in a problem as spherical horses. For some reason it cracked me up. There's something special about a spherical horse drowning in glycerol while we do some fluid mechanics problems.

log(2x + 1) ≠ log 2x + log 1.

You can simplify the right hand side using the following rules of logarithms:

b log a = log a[sup]b[/sup].
log a - log b = log(a/b).

Then when you have an expression on the right hand side that is in terms of a single logarithm, you can cancel out the logarithms:

log a = log b  =>  a = b.

So you will end up with 2x + 1 = something. I will leave it up to you to work out the details! Good luck.

I must say

is my favorite as well, it is just so beautiful, displaying 5 of the most important numbers in mathematics, addition, multiplication, exponentiation, and equivalence. Also it is quite simple, which is a main factor in it being well-known.

Other results I am particularly fond of are

and

(of course taking the second expression too literally will give rise to dispute; it is actually an asymptotic relation)

Also the fact that

which shows that i[sup]i[/sup] is in fact a real number, is quite astonishing.

I enjoy infinite sums as well, but usually none stick out significantly from the rest. Here is Zhylliolom's identity for φ (please submit this one, don't forget to mention me ):

(actually, I need to rederive my identity to make sure it is actually correct here, so wait on that!)

Hey guys look

Then

Therefore

It really works!!!!!!11!1!1cos² x + sin² x!1!!1!!1!

*most horrific disemboweling of a theory ever witnessed by mankind removed for the sake of the young ones on the forum*

Just read one of his papers and shake your head, I think it'll be enough for you.

http://www.bookofparagon.com/Mathematic … ineVII.pdf

Since the sample size is 11, you have to have 11 scores in your list. Since the minimum is 61, one of the scores in the list must be 61 and no other score may be lower than 61. Since the maximum is 100, one of the scores in the list must be 100, and no other score may be higher than 100. The mean score is the sum of all scores divided by the sample size. Then if we multiply the mean by the sample size we will know what the sum of the scores is. This value is 81*11 = 891.

We know that 61 and 100 must be in the list, as determined above. Then from the determined relation, we have a + b + c + d + e + f + g + h + i + j + k = 891 (where the letters a through k represent different test scores in the list). So we can say that j = 61 and k = 100 and subtract these values from each side to get a + b + c + d + e + f + g + h + i = 730. So now we have to find out a set of 9 remaining test scores such that the sum of the test scores is 730, when listed from least to greatest the 5th test score is 91 (since the median is 91), and the scores aren't less than 61 or greater than 100. From that last sentence we now know that one of these scores is 91, so we can say that i = 91 and subtract it from each side to get a + b + c + d + e + f + g + h = 639. Now we must be careful. Since the median was 91, four of these remaining scores must be less than 91 and four of them must be greater than 91. Any set of scores satisfying this and a + b + c + d + e + f + g + h = 639 will work then. These values may be found by "guessing": let the last four scores be 95 (which is greater than 91 so we can choose this). Then subtract these four 95's from each side to get a + b + c + d = 259. Now we must find four scores greater than 61 but less than 91 that sum up to 259. If 3 of these scores are 65 and one is 64, then we have 65 + 65 + 65 + 64 = 259, so these values work.

Then one list of test scores that would result in the statistics given is 61, 64, 65, 65, 65, 91, 95, 95, 95, 95, 100. Hopefully you could follow my reasoning somewhat. Also note that this isn't the only possible list, you can easily find others.

There's another "rule" that you don't have up there that helps solve this one: sec² θ - tan² θ = 1. Watch how it solves the problem:

The calculus method should provide some insight into the problem. Define the function f(x) := 3[sup]x[/sup] - 2[sup]x+1[/sup] - 1. The zeros of this function will correspond to solutions to your problem. By your inspection already, we know that x = 2 is a zero. Now our goal is to see if any other zeros exist. First, find the extrema of the function. Taking the derivative, we have f'(x) = 3[sup]x[/sup]ln 3 - 2[sup]x[/sup]2 ln 2. Setting the derivative equal to zero, we find that at the extreme values 3[sup]x[/sup]ln 3 = 2[sup]x[/sup]2 ln 2. Take the natural logarithm of both sides and solve for x to get x = (ln(2 ln 2) - ln(ln 3))/(ln 3 - ln 2) ≈ 0.5736. So there is only one extreme value for this function, and it is a minimum (I leave it to you to test that it is indeed a minimum). Now investigate the end behavior of the function: will it ever cross the x-axis again? There are no other maximums or minimums, so we know that f(x) can't jump above the x-axis and then under it (or vice-versa). So let us look at what happens to the function as x approaches -∞ and also when x approaches +∞.

What does this tell us? We know that f(x) is negative around values such that x < 2. The minimum of f, ~f(0.5736) was also a negative value. Then as x gets increasingly smaller f(x) approaches -1; this tells us that f(x) will never cross the x-axis as x -> -∞. Why is this? If f(x) went above the x-axis for some very negative value of x, then it would need to once again go below it so that as x -> -∞ f(x) -> -1. But this would require f(x) to have a maximum somewhere, which it does not. Likewise, we know that f(x) is positive around values such that x > 2. The limit tells us that f(x) will just get bigger and bigger in that direction. So for a zero to exist past 2, f(x) would need to dive below the x-axis and then rise back up towards ∞ in the limit, which would create a maximum and a minimum. But once again, we know that f(x) only has one minimum and this minimum has already been determined. From this evidence, we may conclude that f(x) has only one zero, at x = 2, and thus the equation 3[sup]n[/sup] = 2[sup]n+1[/sup] + 1 only has the solution n = 2.

Interesting and reasonable, All_Is_Number.

My friend, who is an internet legend, claimed to get 0.999, but the server conveniently was "having trouble," as you guys mentioned sometimes happened, and thus his score wasn't recorded. I've seen him type many times and he is insanely fast, but the alphabet in one second seems pretty extraordinary.