hi, i'm a freshman in hs and i'm wondering about the new SAT's. this topic might've been discussed before but i was wondering what would be an "elite" score on the new SAT? That is, an SAT score that would correspond to a 1400 (not quite elite) - 1600 on the old SAT scale???

if this topic was discussed before, moderators, feel free to delete it.

I think we will know that better after percentile data are released for aggregate scores on the new test. Percentile data on [url=http://www.collegeboard.com/prod_downloads/about/news_info/cbsenior/yr2003/pdf/2003CBSVM.pdf]aggregate scores for the old SAT[/url] are quite interesting: only a few students score high on both the verbal and the math sections, but enough to fill all the entering classes in the few most selective colleges.

THis is just an esitmate, and it may be lower since there are only 3 tests, but a 1400 should correpond to 2100 on the new SAT>  I just multiplied by 3/2, but once the percentiles come out, there's porbably a better score to correspond with 1400.

[quote="mcalderbank"]With the New SAT, the real advantage now lies with taking an SAT class. For the Essay, Collegeboard grades according to a formula, and if you know the formula, then you willl do well. Besides, I think 2 graders spend 2 or 3 minutes looking at each essay.[/quote]

I agree. Without practice and preparation, it is very difficult to come up with a respectable essay in 25 minutes. Especially if you didn't know until that day that you had to write that essay (which will undoubtedly happento some people). On the other hand, if you were already accustomed to writing 25 minute essays, you would be at a huge advantage (I would say I am, but my poor writing compensates for that). A similar argument can be applied to the writing section in general. It's all about practice; I've done a lot with that stuff, like from previous SAT II Writing sections and the Princeton Review book for the new SAT I. I seriously doubt many of my classmates would manage to pull off anything near me if they had not had any practice. It is a matter of seeing things from a different perspective (I guess it's comparable to math in some ways) and you have to learn it (or just be super talented).

Well, I'm getting my new SAT scores in 2 days, so we should know what an "elite" score is by then.

I hate the essay section with a passion. I remember the prompt: Is creativity required in today's world? I sat there for 10 minutes thinking of something, and ended up BS-ing something about medicine. I hope I can pull off a 2200.  :(


If i was trying to express a linear transformation using a script L so that L(U, V) is L:U $ \longrightarrow$ V, how would i do that? I know that the notation doesn't really matter, but I was curious

\mathcal

$ \mathcal{L}: U \rightarrow V$


logarithms
real analysis
real analysis unsolved

Let $ (x_n)_{n \in \mathbb{N}}$ be a sequence, $ x_1\equal{}2, x_2\equal{}4$, and $ x_{n\plus{}1}\equal{}\sqrt{x_nx_{n\minus{}1}}$.

Prove that the sequence $ (x_n)_{n \in \mathbb{N}}$ is convergent, and find its limit.

Taking logarithms (base 2 ;) ) it is equivalent to $ y_1 \equal{} 1, y_2 \equal{} 2$ and $ y_{n \plus{} 1} \equal{} \frac{y_n \plus{} y_{n \minus{} 1}}{2}$ and you can solve explicitly.

In the case $ x_{1}\equal{}2$, $ x_{2}\equal{}4$, you can write $ x_{n}\equal{}2^{y_{n}}$. Hence you get $ x_{n\plus{}1}\equal{}2^{y_{n\plus{}1}}\equal{}\sqrt{x_{n}x_{n\minus{}1}}\equal{}2^{y_{n}\plus{}y_{n\minus{}1}}$. Therefore you just solve $ y_{n \plus{} 1} \equal{} \frac{y_n \plus{} y_{n \minus{} 1}}{2}$.

However JBL`s solution is more general.

Since $ [x_{2i\plus{}1},x_{2i}]\subset[x_{2i\minus{}1},x_{2i\minus{}2}]\subset \cdots \subset [x_{1},x_{2}]$ and $ x_{2i}\minus{}x_{2i\plus{}1}<\frac{1}{2}(x_{2i\minus{}2}\minus{}x_{2i\minus{}1})<\cdots<\left(\frac{1}{2}\right)^{i}(x_{2}\minus{}x_{1})$, the lengths of the intervals converge to 0. So for any $ x_{1},x_{2}$ the sequence $ \left\{x_{n}\right\}$ converges. Now we use JBL`s approach to calculate that limit.


Two cars head toward each other from opposite ends of a highway 444 km long. The speed of the first car is 85 km/hr. The speed of the second car is 100 km/hr. In how many minutes will they meet?

Call the number of minutes it takes them $ x$. Since the first car can can travel $ 85$ km an hour, it can travel $ \frac{17}{12}$ km a minute. The second car can travel $ \frac{5}{3}$ km a minute. Our equation looks like this:

$ \frac{17}{12}x \plus{} \frac{5}{3}x \equal{} 444$

$ \frac{37}{12}x \equal{} 444$

$ x \equal{} \boxed{144}$

...you could just do

$ 444\div(85\plus{}100)*60$.

$ *60$ is to change hour to minutes.

Which is basically solving an equation, just converting to minutes after, not before.

Yes but isabella explained it. I thought it'd be much harder.


Putnam
calculus
Support
college
integration
trigonometry
floor function

Hello everyone, I've posted very sparingly but have been a long-time lurker.  So far I've taken the Putnam twice, 0 on the first one, 12 last year.  My specialty tends to be in enumerative combinatorics and other discrete math topics.

Despite what I consider relative success, I had the displeasure of attending a high school that couldn't support my zeal for mathematics.  As such, I was forced to teach myself all of calculus and then pass out of all the classes when I went to college.  I've noticed though that having no formal training in calculus has hurt me in the Putnam and similar mathematical competitions.

As I said, I know the subject satisfactorily, but not exceptionally as needed for the Putnam.  I have had one semester of Real Analysis and one of Complex Analyis since I've started college.

Are there any books out there that could give me a stronger base in calculus?

Maybe a textbook with challenging problems? (The calculus book I used in high school is "Thomas' Calculus")

How have all of you improved your calculus after doing almost exclusively non-calculus competitions in high school?

Thanks for all of your help

Spivak and Apostol are known for their rigorous approaches and tough proofs.

Some of the standard textbooks may not prepare you for Putnam since their intentions may be to teach you theorems and their applications. I suggest you get college competition old papers such as Putnam volumes and some other ones and try to solve some of the calculus problems. Berkeley Problems in Mathematics (although intended as a preparation book for a first year graduate student) has some calculus problems you can play with.

Here's a small set of problems I assigned as a week's homework for a 1-unit Putnam preparation class. Most of these problems have appeared in some form here on AoPS/Mathlinks; in fact, that's where I got several of them. And two of them are from recent Putnam exams.

[b]1.[/b] Compute (with explanation) $ \int_0^{\frac {\pi}2}\frac {dx}{1 \plus{} (\tan x)^{\sqrt {2}}}.$

[b]2.[/b] Suppose $ g(u)$ is continuous for $ u\in[0,1].$ Prove that
\[ \int_0^{\frac {\pi}2}xg(\sin x)\,dx \equal{} \frac {\pi}4\int_0^{\frac {\pi}2}g(\sin x)\,dx
\]
[b]3.[/b] Compute (with explanation) $ \left\lfloor \sum_{k \equal{} 1}^{2007^2} \frac {1}{\sqrt {k}}\right\rfloor.$

[b]4.[/b] Find the volume of the region of points $ (x,y,z)$ such that
\[ \left(x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 8\right)^{2}\le 36\left(x^{2} \plus{} y^{2}\right).
\]
[b]5.[/b] Prove that for any positive integer $ n,$ $ \int_0^{\infty}\lfloor ne^{ \minus{} x}\rfloor\,dx \equal{} \ln\left(\frac {n^n}{n!}\right).$ Use this to show that $ n! > n^ne^{ \minus{} n}.$

[b]6.[/b] For $ u\ge 0$ let $ F(u) \equal{} \int_0^{\pi}\ln(1 \minus{} 2u\cos x \plus{} u^2)\,dx.$ Compute $ F(u)$ as a function of $ u.$

[b]7.[/b] Evaluate $ \int_0^1\frac {\ln(x \plus{} 1)}{x^2 \plus{} 1}\,dx.$

[quote]Maybe a textbook with challenging problems? (The calculus book I used in high school is "Thomas' Calculus") [/quote]
Stewart's calculus book (the long version, not the shorter "Essential Calculus") is a mainstream textbook much like Thomas is a mainstream textbook, but there's a feature called "Problems Plus" - little sections in between chapters with more challenging problems. That could be a place to start.

Zorich's mathematical analysis is one of the toughest
Especially the execises
Harder than any American analysis textbooks

I can't speak for Zorich but as I said, I think Spivak is also a nice place to start.  Many of the seeminly esoteric problems and proofs are actually theorems or rules that have been referenced on Putnam or similar competitions many times.

Zorich is pretty tough. It's geared at Russian freshmen; I've completed an analysis course at a pretty good university and I still can't solve most of the problems in there. Partly because some of the problems (and at least one entire section on piecewise parametrizable manifolds) are incorrect and mostly because some are the beginnings of research papers :P I would caution though that it's not a book for self study. No one can master it on their own but solving the problems is of far greater value than reading the theorems
There is a book called 'Putnam and Beyond' which is also good.


limit
logarithms
calculus
calculus computations

prove that
$ \lim_{n\to\plus{}\infty}(1\plus{}\frac{1}{n})^{n}\equal{}e$

That is the most common definition of $ e$.  If you want to start with a different definition and prove its equivalence to that one, you'll need to specify which definition you're using.

[hide="Solution"]$ L\equal{}\lim_{n\to\infty}\left(1\plus{}\frac 1n\right)^n$

$ \ln L\equal{}\lim_{n\to\infty}n\ln\left(1\plus{}\frac{1}{n}\right)\equal{}\lim_{n\to\infty}\frac{\ln\left(1\plus{}\frac{1}{n}\right)}{\frac 1n}\equal{}\frac 00$

So now we bash it out with L'Hopitals...

$ \ln L\equal{}\lim_{n\to\infty}\frac{\frac{\minus{}\frac 1{n^2}}{1\plus{}\frac 1n}}{\minus{}\frac 1{n^2}}\equal{}\lim_{n\to\infty}\frac{1}{1\plus{}\frac 1n}\equal{}1$

So $ \ln L\equal{}1\implies L\equal{}e$ $ \Box$[/hide]

Though this assumes that $ e$ is defined as the inverse of $ \ln$, I guess. (As JoeBlow said, we have to assume some definition of $ e$).


algebra
polynomial
function
algebra unsolved

Let $ P(x)$=$ x^4\plus{}x^3\minus{}3x^2\plus{}x\plus{}2$ and $ k$ $ \in$ $ Z^\plus{}$
    prove that $ (P(x))^k$  has one positive root.

Beware that it is not true for $ k\equal{}1$.

In order for $ (P(x))^k\equal{}0$, we must have $ P(x)\equal{}0$, right?

i think by $ P(x)^{k}$ he means $ P(P(P(....P(x))))_{k times}$

Sorry We need prove that:$ (P(x))^k$ always has one negative cofficient.

I think $ P(x)^k$ has a nagative cofficient if $ P(x)^k$ has a positive root.

It is easy followed from surjective for function $ f: R_\plus{}\to R_\plus{}, f(x)\equal{}P(x)$ if $ P(x)\ge 0$, $ f(x)\equal{}0$ if $ P(x)<0$.


What is the greatest four-digit number divisible by 12?

We're looking for the greatest four-digit integer divisible by both $ 3$ and $ 4$. Using divisibility rules, we can see that the answer is $ \boxed{9996}$.


a) Prove that the equation $3^z=x^2+y^2+1$ has infinitely many integer solutions.

a') Prove that the equation $3^z=x^2+y^2-1$ has infinitely many integer solutions.  :lol:

b) Prove that the equation $7^z=x^2+y^2+1$ has only finitely many integer solutions.

For the first two:

The second one needs no comments, of course, as the :lol: above says, but we can use it for the first one. Every $3^{2k}$ can be written as $(3^k)^2+1^2-1$. Now, if we can find $3^{2k}=x^2+y^2+1$, then $3^{4k}-1=(3^{2k}-1)(3^{2k}+1)$, which is a product of two numbers that are sums of two squares, and is thus a sum of two squares, then $3^{8k}-1=(3^{4k}-1)(3^{4k}+1)$, which are both sums of two squares, and so on, so we would be done.

In order to find $3^{2k}-1=x^2+y^2$, we can take $k=1,x=y=2$.

For the third one:

since $7^z-1$ is a sum of squares, we must have that all prime factors dividing it, are $4k+1$, or are raised to an even power.

Now, Modulo $8$, we get $z$ is even. and then we have $3|7^z-1$, Now suppose we have found a sequence of primes $p_1,p_2...,p_k=3$ mod 4, such that $p_m|((7^2)^{p_{m-1}}-1)/(7^2-1)$, and $p_i|7^z-1$, then $p_i|z$ Now, consider $((7^2)^{p_{k}}-1)/(7^2-1)$ it is $3$ mod $4$, so it has a prime factor of the form $4k+3$, different from $p_i$ for $i=1,2,3...k$ and define $p_{k+1}$ to be this prime, so we get an infinite sequence, imposible!


Without giving it away, could someone lead me in the right direction as to how I could find the sum of a series with a changing ratio. The ratio does follow a pattern. Thanks

which kind of pattern??? Because if the pattern varies, there's different method to solve it (sometimes even unsolvable).

But if it is of the form $a$, $a(2r)$, $a(3r^2)$, $a(4r^3)$, ... There's a very easy way to do it if $|r| < 1$ (not sure about other domain).

solution is below:

[hide]
well if we subtract $\frac{a}{1-r}$ from the sum we get $ar$, $2ar^2$, $3ar^3$, ... and then we can subtract $\frac{ar}{1-r}$ from it again! therefore it is $\frac{a + ar + ar^2 + \dots}{1-r}$ or $\frac{a}{(1-r)^2}$
[/hide]

pattern is  $r_k=\frac{2^k}{4^k + 1}$ where kth ratio is the ratio used to get from the $a_k$ term to the $a_{k+1}$ term

Personally, for that pattern i'd turn the thing backwards:

$a_{k-1}=(2^{k-1}+{1\over 2^{k-1}})a_k$

(taking the reciprocal of the ratio, hence summing from the last term backwards instead of the first term forwards)


articles calculus Gauss Support search videos

So, I'm writing a speech, essay like thing.

I am writing about how the US's mathmatical curriculum has...degenerated in....its current state.

I don't think that it's all horrible. Nor that it's all bad. I do, however, think that there are some educational aspects that deserve some mocking.

Uhm, this is going to be rather short as I am at school and my time on the computer is thus limited.

If anyone has some specific ideas that you feel I need to cover or compare or something of that sort, please let me know. I think that there would be some interesting opinions that could be contributed by the people on this forum.

I'll add more later,
~The_Scintillator

I think it would be interesting to compare US and Chinese math education from K-12.  Examine what is taught in each year, and why the Chinese have been successful

In what respect do you say the Chinese have been successful? (My knowledge is obviously inapplicable here since I can just speak of what I see in American institutions, so I'm curious.)

I read an article in the newspaper comparing the international scores of reading and writing, but more interesting, mathematics. The article places the United States at [i]fifteenth[/i] place in the world, in terms of mathematics. I think this would a great addition to your essay (if you could find the source).

@Mystic: I am under the impression that the Chinese schools teach advanced math at a younger age. Also, the class is not held up by [lower-end] students. If a student is behind, he/she is cut. This way, the class moves faster and "No one is left behind."  :D

well yes, I have no doubt that their math classes are more advanced, but by what metric do you measure success when making your statement

It seems that you've made up your mind before even doing any research. What basis do you have for your assertion that the American Math Curriculum has degenerated? Are you basing it on anecdotal evidence? Comparative test scores from different eras? Do you know what the American Math curriculum looked like in the 1800s? 1920s? 1950s? 1970s?

Attached is an article (giving one perspective) on the History of Calculus education in America. There are many other articles about different aspects of American Mathematics Education available on the web.

A few topics in this very forum address the state of mathematics education in the US. Read them.

you mean the ones which are filled with a bunch of middle schoolers whining about how easy their classes are? well, at least until some people who know their stuff give useful evidence such as the article above. of course, as gauss does note, the article is at best one perspective, but the question is by no means already decided.

[quote="gauss202"]It seems that you've made up your mind before even doing any research. What basis do you have for your assertion that the American Math Curriculum has degenerated?[/quote]

The way I see this, you can formulate your essay based on the opinion that the math curriculum has degenerated (and support it), or you can use the various documents available to determine the current state of curriculum and present your findings. Either format is entirely up to you, and you should write it based on what you're comfortable with. (If I were you, I'd do the opinion, because it fits better with a "speech"-like essay.)

[quote="MysticTerminator"]you mean the ones which are filled with a bunch of middle schoolers whining about how easy their classes are? well, at least until some people who know their stuff give useful evidence such as the article above.[/quote]

You have a pointy (as opposed to pointless) point there. But some of the replies in those topics are decent. And this thread isn't doing all that much better, you must admit.

I liked the Lockhart's Lament thingy.

Uhm. Well, this was interesting.
I just going to ask for what you liked about our system and what you didn't.

This is an OPINIONATED essay.
I picked an opinion and will set out to prove it. It's not necessarily right or wrong.

Thank you for the link--I shall read it.

Okay.
I vaguely remember two things that I read about that I can't remember the sources for. I will continue to search, but could someone help me out if they remember? I think they were both from AoPS....

One was about a guy who talked about something. He covered that the atitude towards math was different here, and said that in home country, when he talked to a taxi driver and told him he was a math teacher, the guy said, "That's nice. I enjoyed math." When he came here to teach, he found many didn't, and got the reply "I hated math as a kid". He said it wasn't their fault they hated math--they had bad teachers. And it wasn't really the teachers fault, 'cause....yeah. That's all I can remember. Gah.

Then the next one was about a survey. They went around and taped a bunch of different math classes around the country (US), and thought things would be different, you know--teahcing styles and such. But actually, all the classes were startalingly the same. The teacher would get the students quited down, they'd go over last nights homework, teach new stuff, assign more homewor, and possibly gave ime to do homework in class.

Help?!

Was one a movie with the title containing Million Minutes?

[quote="The_Scintillator"]He covered that the atitude towards math was different here, and said that in home country, when he talked to a taxi driver and told him he was a math teacher, the guy said, "That's nice. I enjoyed math." When he came here to teach, he found many didn't, and got the reply "I hated math as a kid". He said it wasn't their fault they hated math--they had bad teachers. And it wasn't really the teachers fault, 'cause....yeah.[/quote]
That sounds like a quote from Titu Andreescu that appeared in the book [i]Countdown[/i] by Steve Olson.  (See pages 50 and 51.)

[quote="The_Scintillator"]Then the next one was about a survey. They went around and taped a bunch of different math classes around the country (US), and thought things would be different, you know--teahcing styles and such. But actually, all the classes were startalingly the same. The teacher would get the students quited down, they'd go over last nights homework, teach new stuff, assign more homewor, and possibly gave ime to do homework in class.[/quote]
That sounds like the TIMSS video study by James Stigler and colleagues.  It was also discussed in [i]Countdown[/i] (pages 44 through 49).


geometry
geometric transformation
reflection
trigonometry
articles

I did some research for a school project on global warming, but do you think it's really happening? For example, here in Binghamton, NY, we're having a very hot summer. The temperatures broke several previous records. On the other hand, our winter that just passed a while ago was bitterly cold, much colder than usual. It was cold the winter before that. How does global warming explain that?

It seems logical that the carbon dioxide would trap the heat trying to escape, so it might be happening. From what I see now though, it sure doesn't seem like a real gradual rise in temperature.

Of course though, this is made through a long process, but how do you know that the rise in the temperature in the past 100 years wasn't simply a natural trend? There were no factories to warm the earth up after the ice age.

But if this is really happening, the results will be devastating: higher sea levels, unbearable temperatures, more storms and disasters, damaged crop growth, salted rivers (from ocean rise), and a lot of other negative effects.

Also, there has been so many attempts to stop global warming, but do you think the government is really doing anything big to stop this issue?

The movie [i]The Day After Tomorrow[/i] showed a projection that's probably not going to happen, but if anything like that happens, we aren't going to like it.

hm...

haven't you seen the commercial about oil saying:

it took 125 years to use a trillion tons of gas

it will take only 35years to use another trillion

( or something )

the global worming basically happens because of the pollution and CO2 ( carbon dioxide ) that breaks/harms the ozone layer that defends the earth from excessive radiation from sun..

if you go to Beijing or Tokyo or Seul, if you look out the window in the airplane while you're descending to the city, you can see the air all yellow.
( gross! )

In 50-100 years, development will increase dramatically, which will cause new industries to rise. Car pollution will disappear, but the later industries will neutralize the reduction, making the production of enviroment wrecking chemicals the same rate as it is now. If this development is indeed happening, thenif the industries develop even more than the neutralization amount, then trouble begins.

Global Warming is the documented trend that over the last 100 years, the year-round average temperature on the earth has risen by approximately 1°C. That is undisputed. What is disputed is whether this is the complete picture; some allege that the supposed variance in temperature is merely the return of normalcy after a period of unusual coolness. Others claim that it reflects a dramatic increase in the Greenhouse Effect.

The Greenhouse Effect refers to the tendency of heat to become trapped in the Earth's atmosphere. In essence, the Earth behaves like a black body. It gives off heat in the EM distribution prescribed by Wien's Law, primarily as infrared radiation for surface temperatures.  Much of this emission is captured by the so-called Greenhouse Gases, such as carbon dioxide, which are able to absorb incident photons of suitable wavelengths (those corresponding to changes in potential energy states of the gases). Eventually, these particles return to their ground states (lowest potential energy), re-emitting the captured energy. Some of this energy returns to the earth. There is no disputing that, by far, the gas that contributes the most to the Greenhouse Effect is water vapor - not CO2. (In fact, this is why mornings after cloudy nights are generally warmer than those after clear nights.) That said, CO2 contributes a non-trivial part of  the Greenhouse Effect. Hence, what is disputed is the level of human contribution to the Greenhouse Effect; in particular, whether the large increase in CO2 levels is due in significant part to human activity.

But still other factors also play a major role in determining the Earth's energy flux. Albedo, or the Earth's reflectivity, is one such variable. Strictly speaking, it is the fraction of incident radiation that is immediately reflected back to space. But this is closely tied to the Earth's cloudiness, which in turn depends on the Earth's humidity, which of course varies with evaporation rates and the Earth's temperature. Hence, one may be led to think Albedo constitutes a negative feedback loop (opposed to change.) Then again, Global Warming tends to eliminate reflective surfaces such as ice, which should further propel Global Warming. So does Albedo actually constitute a positive feedback loop? Another factor would be the positioning of landmasses on the Earth's surface, since the flux density of incident radiation varies with the cosine of one's latitude. With this concluding paragraph, I hope I have made my point that volumes can be written on the matter and that obtaining a complete answer is no simple task.

I think that in the next 100 years or so, we will see if global warming is really happening. Right now, we can merely guess at the fact. still, are we doing enough to try to slow or stop global warming? There are so many sites out there that give us tips about stopping global warming or slow it. But are we actually following these tips? Is the government doing anything to stop it.

Some say that global warming is more severe than the threat of terrorism. That I agree to, because while terrorism only affects a given group people, global warming can affect the world population. Both of them, however, are very hard to stop, but I think that global warming is harder because you're trying to fight science.

I never knew water vapor was the chief component of the green house effect; I knew it was a greenhouse gas, but I thought carbon dioxide was the major cause of the strong greenhouse effect we are seeing.

The government is probably not helping us try to stop global warming. These days, have you seen any cities or towns restrict industrial development? No. I sure haven't. Industrial plants are still developing all over the world. Laws enforced by the government on deforestation merely goes into effect. Some I've never even heard of. Nature is a big help in our efforts to stop global warming, but the problem is that we are destroying nature fr things we don't need, such as roller skates and DVDs.

Why isn't the government helping, nobody really knows. Maybe it's because all they care about is that economy of theirs, or that huge debt that America has.

[quote="236Factorial"]
I never knew water vapor was the chief component of the green house effect; I knew it was a greenhouse gas, but I thought carbon dioxide was the major cause of the strong greenhouse effect we are seeing.[/quote]

I did a science project on the water cycle, and I found that if we didn't have any water vapor, the temperature on Earth would be well below freezing.

$CO_2$ is indeed a primary reason, since smoke stacks send out that stuff, accompanied by sulfur.

[quote="236Factorial"]
I never knew water vapor was the chief component of the green house effect; I knew it was a greenhouse gas, but I thought carbon dioxide was the major cause of the strong greenhouse effect we are seeing.[/quote]

Atmospheric carbon dioxide is of particular concern because it has been steadily increasing since the late 50's, thereby increasing its role in the Greenhouse Effect. At the famous Mauna Loa site in Hawaii, which is considered to be a reliable indicator for a number of reasons, the atmospheric CO2 levels have risen from 316ppm in 1959 to 377ppm in 2004. The obvious question is: why, where is this CO2 coming from? A strong argument can be made that we humans are the main perpetrators of this rise, and that is cause for concern.

I don't think there is such a consistent trend in water vapor, which is why CO2 comes up more often in discussions of the Greenhouse Effect and Global Warming.

Bush withdrew from the Kyoto Treaty because he didn't want to put the United States economy on the fence. Remember that we are the most Industrialized nation in the world, and we rely heavily on fossil fuels.

What you don't know is that the Bush Administration is acting on other measures. Recently the White House invited top climatologists and scientists to the President's Climate Change Technology Program to discuss possible courses of action.

If you subscribe to Popular Science, there was a recent issue talking about possible solutions that don't require limitation on CO2 emissions. From storing, filtering, and chemically altering CO2 to reflecting solar radiation, there are alternative methods to combat global warming. Global warming is also a fact. I'm not sure why the Bush Administration refused to admit it at first, but facts show that the globe is warming up.

And I'm pretty sure the the basis of the scenario in The Day After Tomorrow was a change in the salinity of the Mid-Atlantic Current.

[quote="hwenterprise"]Nature is a big help in our efforts to stop global warming, but the problem is that we are destroying nature fr things we don't need, such as roller skates and DVDs.
[/quote]

Hmm, though I bet if everyone started using rollar skates and bikes instead of cars, it would help!  Hehehe just a though...

Yes, that is very true. :D

The EPA has a section on Global Warming, and I copied some down for your convenience.

What is being done?

[quote="The Environmental Protection Agency (EPA)"]

 Today, action is occurring at every level to reduce, to avoid, and to better understand the risks associated with climate change. Many cities and states across the country have prepared greenhouse gas inventories; and many are actively pursuing programs and policies that will result in greenhouse gas emission reductions.

At the national level, the U.S. Global Change Research Program  coordinates the world's most extensive research effort on climate change. In addition, EPA and other federal agencies are actively engaging the private sector, states, and localities in partnerships based on a win-win philosophy and aimed at addressing the challenge of global warming while, at the same time, strengthening the economy. For more information, see the US Climate Action Report (U.S. Department of State, May 2002).

At the global level, countries around the world have expressed a firm commitment to strengthening international responses to the risks of climate change. The U.S. is working to strengthen international action and broaden participation under the auspices of the United Nations Framework Convention on Climate Change.
[/quote]

Sea Level Rising:

[quote="EPA"]
Nationwide Impacts of Sea Level Rise
"The Cost of Holding Back the Sea" estimates the area of dry and wet land that would erode or be inundated from a 50, 100, or 200 cm rise in global sea level, for each of seven regions comprising the contiguous 48 states. The study provides state-specific estimates of the cost of nourishing beaches, and region-specific estimates of the cost of protecting low mainland areas with dikes and pumping systems, assuming that densely developed areas will be protected. It also provides nationwide estimates of the value of undeveloped land that would be lost, and the cost of elevating homes and roads in the nation's densely developed areas. Additional details from this study are provided in Chapter 7 (Sea Level) of the 1989 Report to Congress and papers from the underlying "Appendix B," including assessments of the engineering challenges and sand requirements involved in elevating barrier islands, and tables showing projected regional land loss.

Maps of Vulnerable Lands provides state-specific estimates of the amount of land below the 1.5- and 3.5-meter contours along the U.S. Atlantic and Gulf Coasts. As the name suggests, this report provides maps. It also includes a that compares its estimates of the amount of low land with the land-loss estimates reported in "The Cost of Holding Back the Sea."

A Study by the Federal Emergency Management Agency (FEMA) estimated potential increases in flood damages and the area of the coastal floodplain resulting from a 1- or 3-foot rise in sea level. This study uses different assumptions than The Cost of Holding Back the Sea. Therefore, the total cost of sea level rise is probably less than the sum of the cost estimates from each of the two studies. The FEMA study assumes that the only direct effect of sea level rise on flooding is that all flood levels are 1 or 3 feet higher. By contrast, The Cost of Holding Back the Sea assumes that most structures are elevated or protected with dikes, which would decrease flood damages.

Beach Erosion and Barrier Islands
"Sea Level Rise and Barrier Islands" provides a generic explanation of the impact of sea level rise on barrier islands, as well as the four fundamental pathways by which an island can respond: dike, elevate the island, abandon, or engineered retreat. The article includes a cost-benefit study of Long Beach Island, New Jersey, which provided the basis for the assumption by the Report to Congress and Cost of Holding Back the Sea that the most common strategy for developed barrier islands would be to elevate the islands by pumping sand onto the ocean beaches and haul in fill to the low bay sides. Appendix B of the Report to Congress explores the engineering issues in detail.

The National Academy of Sciences Report Responding to Changes in Sea Level (which will be posted on this web site soon) provides a comprehensive analysis of how beaches respond to rising sea level, and the various engineering strategies that can be employed to hold back the sea. Many of the shore protection methods are described with even more detail in Greenhouse Effect and Sea Level Rise--Shore Protection Strategies . The chapters in Greenhouse Effect and Sea Level Rise on Charleston and Galveston also discuss general coastal processes involving barrier islands.. The chapter on "Planning for Sea Level Rise Before and After a Coastal Disaster" examines why it might be a good idea for coastal communities to decide whether they intend to hold back the sea before a serious hurricane strikes. Planning for Sea Level Rise Along the Coast of Maine discusses coastal processes along Maine's open coast.

Potential Impacts of Sea Level Rise on the Beach at Ocean City, Maryland examines the erosion likely to take place there, and makes the case for why the prospect of sea level rise implied that the State's policy that the state had for controlling erosion--rock groins--should be replaced with a policy of beach nourishment. Shortly after the study was released, the policy was revised along those lines.

Sea Level Rise and Saltwater Intrusion
Salinity in the Delaware Estuary provides general explanations of the effects of sea level rise on surface and groundwater. A quantitative assessment focuses on Philadelphia's intake on the Delaware River and aquifers in suburban New Jersey. A final section discusses implications for New York City's water supply and the once-proposed Tocks Island Dam, as well as various anticipatory planning options. Greenhouse Effect and Sea Level Rise--Shore Protection Strategies provides additional detail on engineering solutions. Land Use Policy has two graphics that do not appear elsewhere, which illustrate why sea level rise increases salinity in rivers and aquifers. The California Chapter of the 1989 Report to Congress summarizes a study of the impacts of sea level rise on salinity in San Francisco Bay, as well as the combined impacts of sea level rise and earlier snowmelt from global warming on the infrastructure that supplies freshwater to California's Central Valley. None of these reports examine the potentially important impacts of sea level rise on South Florida's Biscayne aquifer.

The implications of salinity increases for wetlands are discussed in Sea Level Rise and Coastal Wetlands , Saving Louisiana's Coastal Wetland and the California Chapter chapter of the Report to Congress.

Floods and Flood Damages
The FEMA Report provides the most comprehensive assessment of the impact of sea level rise on flood damages from storm surges in the United States. Greenhouse Effect and Sea Level Rise modified storm surge models to estimate changes in flood frequency and severity in Charleston, South Carolina and Galveston, Texas. The Drainage Study analyzes the impact of sea level rise on flooding from rainstorms, resulting from the "backwater effect" by which higher sea level slows the speed at which low areas drain. Appendix B of the 1989 Report to Congress includes a study of how the stormwater management system of Long Beach Island, New Jersey would have to be modified with a substantial rise in sea level.

Impacts on Other Nations
IPCC Response Strategies Report presents the results of an international panel of experts who evaluated the impacts of sea level rise and possible responses. Land Use Policy discusses the general implications of sea level rise with an international context. Many of the other reports on this site examine issues that are applicable to other nations. For example, most of the analysis of legal implications in Maryland Law Review Article are applicable to Commonwealth Nations who inherited the same public trust doctrine from England that the United States inherited. Saving Louisiana's Coastal Wetlands examines how flood-control levees, river bank stabilization, and other human activities have removed the natural processes that would otherwise enable a large river delta to keep pace with sea level rise and it's own tendency to subside. IPCC Response Strategies Report points out that similar situations have--or could--also occur along the Nile, Ganges, Niger, Mekong, Yellow and other major river deltas.

Although coral atolls are very different from U.S. barrier Islands , some of the response strategies discussed in that report may still be applicable. Land Use Policy includes a graphics that illustrates why sea level rise threatens groundwater supplies on these small islands.

Estimates of Future Sea Level Rise
EPA's 1983 Report Projecting Future Sea Level Rise developed the first set of year-by-year scenarios of sea level rise. That analysis is summarized in the chapter on "Estimating Future Sea Level Rise" in Greenhouse Effect and Sea Level Rise Given the exploratory nature of that first analysis, it had very large error bounds, especially for the potential contribution from glaciers. In 1985, the widespread interest in the early EPA studies prompted the National Academy of Sciences' Polar Research Board to convene a panel which developed a narrower uncertainty range for the snow and ice contribution to climate change. Based on those studies, a general convention arose to examine the implications of a one-meter rise in global sea level, as well as higher and lower scenarios for sensitivity analysis. The National Academy of Sciences report Responding to Changes in Sea Level (which will soon be posted on this site) recommended that engineers analyze the implications of 50, 100, and 150 cm, and Chapter 7 of the Report to Congress assessed the impacts of rise of 50, 100, and 200 cm rise in global sea level.

Starting in 1990, the Intergovernmental Panel on Climate Change began publishing scenarios of global sea level rise. EPA revised its own projections in the 1995 Publication The Probability of Sea Level Rise. Unlike the IPCC results, the EPA study estimates the probability of the various scenarios, and relative sea level rise at specific locations. Chapter 7 provides probability distributions of global sea level rise consistent with the IPCC sea level rise scenarios released at about the same time; Chapter 9 provides a procedure for converting the global estimates into a probability distribution of future sea level rise at specific locations. Chapter 8 explains why the projections of future sea level rise have gradually been lowered over the last two decades. "The Risk of Sea Level Rise" is a shorter peer-review journal article about the same study; but because it was published a year later it includes a discussion of comments from experts who objected to some of the aspects about how the study was conducted. A report by the Aspen Global Change Institute  provides a detailed critique of the EPA study.

Coastal Wetlands
Greenhouse Effect, Sea Level Rise, and Coastal Wetlands provides the first comprehensive overview of how sea level rise might threaten coastal wetlands. Other chapters assessed vulnerability of wetlands around Charleston, South Carolina and Southern Ocean County, New Jersey based on field surveys. That report also estimated the possible nationwide loss of wetlands in a crude elevation mapping exercise using topographic maps. Those estimates were refined in an improved elevation mapping study which used LANDSAT imagery, and was reported in Appendix B of the EPA Report to Congress as well as in The Cost of Holding Back the Sea. Saving Louisiana's Coastal Wetland discusses an overall strategy, but provides no quantitative assessment.

Outside Louisiana, one possible key to wetland survival might be policies to allow wetlands to migrate inland. "Greenhouse Effect and Coastal Wetland Policy: How Americans Could Abandon an Area the Size of Massachusetts at Minimum Cost" evaluates the alternative policies under various criteria, providing the first comprehensive evaluation of rolling easement policies. The Maryland Law Review article "Rising Seas, Coastal Erosion, and the Takings Clause: How to Save Wetlands and Beaches Without Hurting Property Owners" refines the same policy analysis, expands it to include beaches and public access, and evaluates the legal ramifications in detail. Virtually all of "Coastal Wetland Policy" is either science that is described almost verbatim in Coastal Wetlands overview or the policy analysis that is described almost verbatim in Part II-C of the Maryland Law Review article. While the Maryland Law Review article focuses on what states can do to allow wetlands to migrate inland, an article in Golden Gate Law Review entitled Does the U.S. Government Realize that the Sea is Rising? provides a parallel analysis of how the federal wetlands programs could protect wetlands as sea level rises. That article also provides a table that summarizes in detail regarding shoreline structures that threaten wetlands as sea level rises. It also includes a table that overlays the elevation data from "Maps of Vulnerable Lands" with some land use data, to show that the percentage of land 5-10 feet above sea level that has been set aside in conservation areas, is far smaller than the land below 5 feet.

EPA research on wetlands vulnerability continues and is briefly described in Forthcoming Sea Level Rise Mapping Products

Do We Need to Do Anything About Sea Level Rise Now?
The primary mission of the EPA Sea Level Rise reports has been identifying instances where it is rational to prepare now for the coastal consequences of climate change, even though most of those consequences are decades away--or more. The Overview Chapter in the 1984 publication Greenhouse Effect and Sea Level Rise: A Challenge for this Generation outlines the case for taking adaptive actions: At least some global warming and accelerated sea level rise are inevitable because of historic emissions, the inertia of the oceans, and the economy's current dependence on fossil fuels. (In the mid-1980s there was not yet a consensus that the global surface temperature is rising.) Therefore, communities and individuals will have to eventually adapt to consequences of global warming such as sea level rise. The primary question is whether to wait or take action now. The chapter on "...Before and After a Coastal Disaster" shows that the decision whether to rebuild after a coastal disaster might depend on whether property owners believe that they can rely on the government to fund efforts to hold back the sea. The Economics Chapter estimates the possible costs of sea level rise for Charleston and Galveston, and concluded that anticipating sea level rise could cut in half the eventual costs. In the Forward, EPA's original Administrator, William Ruckelshaus, points out that democracies often find it difficult to prepare for looming problems until a catastrophe occurs. "The ultimate danger is that by remaining reliant on "the catastrophe theory of planning" in an era producing catastrophes of a magnitude greater than in the past, we can place our institutions in situations where precipitate action is the sole option – and it is then that our institutions themselves can be imperiled and individual rights overrun."

If nothing else, Greenhouse Effect and Sea Level Rise demonstrated that research to improve forecasts of sea level rise has value. Society can save money by preparing for sea level rise, but such preparation is impossible without reasonably reliable projections of how much the sea might rise. Although the Overview and Independent Reviewer chapters provided a extensive explanations of the benefits of planning for sea level rise, this book did not systematically analyze whether these benefits exceed the costs of acting sooner rather than later.

Subsequent reports investigated the specifics of adaptation. The Ocean City study showed that the prospect of sea level rise justified a shift in the town's erosion control strategy from groins to beach nourishment; a change that the State of Maryland adopted shortly thereafter. The Delaware Estuary Salinity Study showed that the risk of future salinity increases could, in theory, justify setting aside land for possible dam construction, while acknowledging that other problems caused by dams might be important enough to keep the dam from actually being built. The drainage study showed that storm drains being rebuilt should probably be designed for higher sea level, because the cost of subsequently retrofitting the system later would be large compared with using somewhat larger pipers today. Saving Louisiana's Coastal Wetlands showed that a strategy had to be developed soon to deal with the combined impact of human activities and relative sea level rise in Louisiana.

A series of assessments showed that wetlands are generally at risk, including case studies of Charleston, South Carolina and Ocean County, New Jersey, as well as a nationwide assessment of wetland vulnerability. In theory, wetland ecosystems can migrate inland, but with increasing coastal development, the wetlands may be squeezed between development and the rising sea. If society wants the wetlands to migrate inland, "Coastal Wetland Policy" showed that such a landward migration needs to be planned decades before an area is flooded to have any reasonable chance of being successful.

Many people have suggested that the long-term planning necessary to allow wetlands to survive sea level rise would be difficult because of the uncertainty of future sea level rise and the long-time horizon. In response to this concern, "Coastal Wetland Policy" and other reports developed the concept of "rolling easement" and demonstrated that it would be cost-effective for private and public conservancies, as well as regulatory agencies, to implement rolling easement policies today. Planning for Sea Level Rise Along the Coast of Maine shows that rolling easements are a more cost-effective way to preserve natural shores than setbacks and other mesures that prevent development. An article in Maryland Law Review examined many of the related legal issues and showed that rolling easements do not impair property rights, unlike setbacks, which can run afoul of the "takings" clause of the U.S. Constitution's. The Maryland Law Review also showed that policies to protect intertidal ecosystems are most urgent along estuarine beaches, which are being destroyed more rapidly than marshes and swamps. That article also showed that the most important near-term decision is simply deciding which areas will be protected and which areas will retain the naturally migrating shorelines. Current efforts are focussed on accelerating the process by which state and local governments address this fundamental question.

Studies of decisions affected by sea level rise have also identified situations where no action is necessary. "Responding to Changes in Sea Level" (which will be posted soon) points out that port facilities and many other coastal structures are rebuilt frequently enough, and the impact of sea level rise is small enough, so that new facilities need merely consider current sea level. Moreover, many structures can be erected rapidly enough so that anticipating sea level rise in unnecessary. For example, there is no need to build a dike today to protect an area that will not be flooded for 50 years. Similarly, the FEMA Report concluded that sea level rise does not necessitate modifications of the federal flood insurance program, because rates can be increased as properties become more risky. "The Cost of Holding Back the Sea" makes the point that most engineering activities can focus on current sea level, while land-use decisions may require a longer-term consideration of future sea level rise.

[/quote]

The seriousness is expressed here:

[quote="EPA"] Rising global temperatures are expected to raise sea level, and change precipitation and other local climate conditions. Changing regional climate could alter forests, crop yields, and water supplies. It could also affect human health, animals, and many types of ecosystems. Deserts may expand into existing rangelands, and features of some of our National Parks may be permanently altered.

Most of the United States is expected to warm, although sulfates may limit warming in some areas. Scientists currently are unable to determine which parts of the United States will become wetter or drier, but there is likely to be an overall trend toward increased precipitation and evaporation, more intense rainstorms, and drier soils.

Unfortunately, many of the potentially most important impacts depend upon whether rainfall increases or decreases, which can not be reliably projected for specific areas. [/quote]

UNFORTUNATELY, I wasn't able to get the links, so here are the links to the sources:

ARTICLE ONE: http://yosemite.epa.gov/oar/globalwarming.nsf/content/actions.html

ARTICLE TWO: http://yosemite.epa.gov/oar/globalwarming.nsf/content/ResourceCenterPublicationsSeaLevelRiseIndex.html

ARTICLE THREE: http://yosemite.epa.gov/oar/globalwarming.nsf/content/impacts.html

Sorry for the lengthy articles.

[quote="236factorial"]The sun won't destroy us for at least another 5 billion years or so, although i've heard we're crashing into another galaxy in about 2 million years, there's a huge asteriod coming at us, projected to hit in about 800 years, so many ways to end the earth and its life  :P[/quote]1) Isn't the sun supposedly going to turn into a red giant star and engulf Earth in sooner than 1 billion year?  :huh:

2) Yep, Andromeda

3) More like $59287491837694\pi$ asteroids. :D

4) I've also heard that the earth's core might become inactive like Mars', which causes the magnetosphere to disappear and all life to end.

5) etc, etc, etc, etc

Yeah, lots of ways to end earth and its life.  :(

I did a report on this last year! It was more of a speech. My thesis was that it was happening...but I really didn't have many rebuttals to the argument that it was natural.

[quote="236factorial"]Yeah, i went to China in April, and all the buildings looked very dirty because of all the smog, and it was so thick that it felt like it was very foggy all day long. Especially in Beijing, I didn't think it would be that bad there.

Yes, the point is, the global warming rate, if industries are the reason, would rise so that the termperture may rise up to 10 degrees, and sea level as much as 2.5 feet. That's when things get scary....[/quote]

China's growth rate is so high for that very reason: unlike the U.S., they have almost no environmental regulation. Yum!

I couldn't help but post this link: [url=http://qntm.org/destroy]How to destroy the Earth[/url]

I took portions of this article about how jet contrails are affecting the climate of the land below them.  If this doesn't make you think, I'm really sorry...

All below quotes are from:  http://www.airliners.net/articles/read.main?id=85

"A NASA study conducted in the USA between 1975 and 1994 found the average temperature to have increased by 1°F. Though a single degree may seem trivial, the incredibly large scale that it applies to makes it significant since just 9°F separates our current average temperature from the last Ice Age."

"The terrorist attacks of September 11, 2001 was the aforementioned event, and it was likely to have excited meteorological researchers involved in contrail impact studies. The national airspace was shut down for three days, something that had not yet occurred since the jet age began in the 1960s and is not likely to occur ever again. Scientists took advantage of this unique three day period in history that lacked contrails. What they learned was shocking and is enough evidence to effectively silence any counterargument to their case. One measure of climate is the average daily temperature range (DTR). For thirty years this had been recorded and extra cirrus clouds in the atmosphere would reduce this range by trapping heat. “September 11 – 14, 2001 had the biggest diurnal temperature range of any three-day period in the past 30 years,” said Andrew M. Carleton. Not in three decades had there been such a large temperature spread between the daytime highs and the nighttime lows. Furthermore, the increase in DTR during those three days was more than double the national average for regions of the United States where contrail coverage was previously known to be most abundant, such as the Midwest, northeast, and northwest regions. The specific increase in the range was 2°F, which in three days was twice the amount the average temperature had increased by over thirty years time. This is evidence that contrails do alter the climate of the land they drift above. "

[quote="h78reg"]in some places, there has been a bitterly cold winter, like in the northeastern US. but, in europe, there are severe droughts in spain, portugal, and italy. many as a matter of fact believe we're entering another ice age. i personally would go for the ice age theory. it seems to me that in 50-100 years, most vehicles won't run on gasoline. so, the environment won't always be a problem. but, the sun will get larger and destroy the earth eventually. so......either theory seems acceptable in some parts of the world in different periods of time.[/quote]

Don't forget that cars that run on [i]water[/i] are being tested this year. I also agree about the bitterly cold winter. I was thinking "Did the sun cool up already?". I couldn't believe the droughts. They were only in the summer here. (I live in NY.) I dissagree about the iceage thing. I have felt 85 degrees this year. Yes, the sun is getting bigger.

But I think that we could cut down dramaticly on the $co_2$ made by planting more plants and making cars run on water.

The cars mentioned don't run on water but on hydrogenium. And guess how this is primary produced (well, there are efforts to change this, but right now): still indirectly from coal; we need simply more regeneratable sources of energy (sun seems best for big amounts).

[quote="ZetaX"]hydrogenium[/quote]

In English, just "hydrogen."  The way we acquire hydrogen is by using electrolosys to split water into its constituent parts.  The energy needed to do this must be generated somewhere, so as ZetaX said, we've just moved the problem somewhere else, we haven't solved it.  In fact, there are probably three seperate energy problems: one is running out of oil, basically the only thing which powers transportation right now.  That problem is more-or-less immediate.  Another is running out of all non-renewable forms of energy, including coal and uranium, but this less-immediate.  A third is global warming, and it is probably the most significant of the three in the short term, because in twenty years it may already be entirely out of our control.

The way that global warming, seemingly paradoxically, could cause an ice-age is as follows: there's a lot of water stored in the arctic in the form of massive glaciers and ice-sheets.  Rising air temperatures will cause more and more of this ice to melt.  Melted ice is cold water -- the flow of cold water into the North Atlantic could interfere substantially with climate.  Europe will be particularly at risk, since much of northern Europe is really only inhabitable because of the gulf stream which brings warm air over across the Atlantic.  If the melted Arctic ice cools that air down substantially, it could make for cooler summers world-wide and thus spark another ice age.  In fact, we are currently in what is known as an inter-glacial period -- a relatively short period of warmth that is part of a natural cycle of freezing and unfreezing.  So we should be due for an ice age within a relatively short time (I think a couple of thousand years, but I'm not positive), anyhow.  What global warming might do is make it happen sooner and faster.

And, in case anyone is wondering, global warming and an ice age do not somehow "cancel each other out."

There's also a theory that if the ice reaches a latitude of about 40 degrees North or so, the Earth reflects too much of the Sun's rays back and the ice age cannot be stopped eventually resulting in a snowball Earth. Scientists belived this happened 600 million years ago.

And how do they think the ice melted? Well the lack of vegetation and water solvent meant over a few million years, CO2 levels rose to about 7%-8% causing massive greenhouse effects melting the ice, which caused much of the CO2 to dissolve in the water stabilising our Earth's temperature.

Vicious cycle isn't it?


This is particularly for all people who consider going into finance. There is an interesting article about the future of Financial services, technology, customer relations, retail banking, corporate, wholesale and private banking trends. This article can be found [url=http://www.globalchange.com/technobank.htm]here.[/url] Other articles on [url=http://www.globalchange.com/cyberb_index.htm]Cyberbanking[/url] can be found here.

How do you see the future ? Do you agree with the article ?


Which is larger? $ 100^{101}$ or $ 101^{100}$?  please show workings too. thanks!:D

Hint
Let, $ f(x)\equal{}\frac{\log x}{x}$.

i believe that there is another approach to that?

Raising both numbers to $ \frac{1}{100 \cdot 101}$, we are then to compare $ 100^{1/100}$ with $ 101^{1/101}$. Now, is the function $ f(x) \equal{} x^{1/x}$ increasing or decreasing?

[quote="Carcul"]Raising both numbers to $ \frac {1}{100 \cdot 101}$, we are then to compare $ 100^{1/100}$ with $ 101^{1/101}$. Now, is the function $ f(x) \equal{} x^{1/x}$ increasing or decreasing?[/quote]

$ f(x) \equal{} x^{1/x}$ is neither increasing nor decreasing. it oscillates.

edit: now i get it! thanks...its increasing then decreasing:D

[quote="Jakey"][quote="Bugi"]Another solution:
For $ a\ge b \ge e$ holds $ b^a \ge a^b$
So $ 100^{101}$ is greater.[/quote]

could you show a proof for that? :)[/quote]

Let, $ f(x)\equal{}\frac{\log x}{x}$.
So, $ e<a<b \Rightarrow f(a)>f(b)$.
Thus, $ \frac{\log a}{a} > \frac{\log b}{b} \iff b \log a > a \log b \iff \log a^b > \log b^a$.
$ \therefore \ a^b > b^a$

[quote="Kouichi Nakagawa"]Let, $ f(x) \equal{} \frac {\log x}{x}$.
So, $ e < a < b \Rightarrow f(a) > f(b)$.
Thus, $ \frac {\log a}{a} > \frac {\log b}{b} \iff b \log a > a \log b \iff \log a^b > \log b^a$.
$ \therefore \ a^b > b^a$[/quote]

Umm, you still need to prove that $ f(x) \equal{} \frac {\ln(x)}{x}$ implies that for $ e < a < b$, $ f(a) > f(b)$.

The only way that I know to do this is to use calculus, which is beyond high school basics.

[hide="calculus solution"]Note that $ f'(x) \equal{} \frac1{x}\cdot\left(\frac {1}{x}\right) \plus{} \ln(x)\cdot\left( \minus{} \frac1{x^2}\right)$

$ f'(x) \equal{} \frac {1}{x^2}\left(1 \minus{} \ln(x)\right)$

Thus, $ f'(x) < 0$ whenever $ \ln(x) \minus{} 1 > 0$, or when $ x > e$.[/hide]


geometry
circumcircle
trigonometry
geometry proposed

Let $ ABC$ be an acute triangle with circumcircle $ \Gamma$. The tangents at $ A$ and $ B$ to $ \Gamma$ meet the tangent at $ C$ to $ \Gamma$ at points $ D$ and $ E$ respectively. Let $ AE$ meet $ BC$ at $ P$ and $ BD$ meet $ AC$ at $ R$. Points $ Q$ and $ S$ are defined as the midpoints of $ AP$ and $ BR$ respectively. Prove that $ \angle ABQ \equal{} \angle BAS$.

let $ x'\equal{}\angle ABQ$ and $ y'\equal{}\angle BAP$. It is easy to show that $ \frac{\frac{AB}{BP}\plus{}\cos{\beta}}{\sin{\beta}}\equal{}\frac{1}{\tan{x'}}$ and similar for $ y'$. Hence it is enough to prove  $ \frac{\frac{AB}{BP}\plus{}\cos{\beta}}{\sin{\beta}}\equal{} \frac{\frac{AB}{AR}\plus{}\cos{\alpha}}{\sin{\alpha}}$. It is well known that $ AE$ and median from $ A$ are isogonal conjugate hence we obtain that last one is equivalent to $ \frac{\frac{\sin{\beta\plus{}x}}{\sin{x}}\plus{}\cos{\beta}}{\sin{\beta}}\equal{}\frac{\frac{\sin{\alpha\plus{}y}}{\sin{y}}\plus{}\cos{\alpha}}{\sin{\alpha}}$ where $ x\equal{}\angle CAE'$ and $ y\equal{}\angle CBR'$, where $ E'$ and $ R'$ are midpoints of $ BC$ and $ AC$. However last one simplify to $ \sin^2{b}\plus{}\sin{\gamma}\sin{\alpha}\cos{\beta}\equal{}\sin^2{\alpha}\plus{}\sin{\beta}\sin{\gamma}\cos{\alpha}$, which is trivial.


$Question 1$ $w, x, y$ and $z$ are positive integers. It is also known that $w/x$ < $y/z$. What is your opinion regarding the value of $(w+y)/(x+z)$? The options are:
A. Always < $w/x$
B. Always between $w/x$ and $y/z$
C. Always < $y/z$
D. None of the above

$Question 2$ There is a square of any given side. You place any point arbitrarily inside the area of the square. Find out the probability of this arbitrarily chosen point to be nearer to the centre of the square (where the 2 diameter intersect) than to any point on the circumference of the square.

[hide="2"]
since the point must be closer to the middle than the sides, then the distance the point can be from the edge is $\frac{1}{4}x$ where x is the side length of the square

thus there is a 1 in 4 probability that the point will be nearer to the center of the square
[/hide]

[hide="1"]
$w/x<y/z\Rightarrow wz<xy$

Adding $zy$ to both sides, we have

$wz+zy<xy+zy \Rightarrow(w+y)z<(x+z)y \Rightarrow (w+y)/(x+z)<y/z$

Similarly, adding $wx$, we have $w/x<(w+y)/(x+z)$.

So, the answer is [b]B[/b].
[/hide]


Which email should I tell my proctor to send a hand grading request? Are there any costs for this? This is for AIME ii

There doesn't seem to be any information about this (that I could find upon a quick glance).  I would just call the AMC office and ask them.  The phone number is on page 2 of the Teacher's Manual.

I'm not sure what you mean by "hand scored".  Do you believe that your paper was misscanned?  While I guess I wouldn't totally rule this out, it seems very unlikely.  Also note that if there is any discrepancy between what numbers you wrote for an answer and what numbers you bubbled in, the question is automatically marked wrong and there is no appeal.

The answers I copied down from teh booklet after taking the test has 7 correct answers (1,2,4,5,6,7,10), while the official score says 5.

I'll ask my teacher to call AMC tomorrow...

I'll repeat the contact information posted recently:

Contacting the AMC Office
1. Please send all emails about the AMC to
amcinfo@maa.org
Ms. Jensen in our office answers that email daily and re-directs the email to the proper person for an answer if she does not have the immediate answer.

For almost all questions, it speeds up the answer immensely if you include your school CEEB number, your school name (or the name of the location where you took a contest if not at your school), school address, city and state.

It also helps to include your full name (as used on the AMC answer form) and your address, including city and state. Many names are similar, and additional identifying information helps us find the proper record in order to correctly answer your question.

You can also call the phone number listed on all of publications during regular business hours (Central time zone).

We are a small office, and I am making it a deliberate and conscious policy to direct all inquiries through these two access points so that they are answered promptly and consistently by the responsible person.

[quote] send a hand grading request? Are there any costs for this? [/quote]

Two comments:
1.  I have personally examined 10 papers with a regrading request in the last couple of weeks.  In 7 cases, the person clearly mis-marked the answers, either filling in the wrong circle, or bubbling in the wrong column.  In the remaining 3 cases, incomplete erasures caused the scanner to read the wrong answer.  In no cases have I found a scanner error.
2.  Until this year, the number of re-scoring requests on AIME has been minimal, I only recall one or two each year.  This year the number of requests has greatly increased.  Before, this, we never felt a need to include a re-scoring request form in the AIME Teacher's Manual,  we got many form the more numerous AMC takers. Given the number of AIME requests this year, perhaps next year we need to include one.

[[quote]My teacher got records of everyone's actual answers (in an excel spreadsheet), so you should be able to compare and figure out which ones there are inconsistencies with...

To my knowledge this is the first year this has ever been done, so its possible that a lot of people don't even know about it.[/quote]

Yes, this is a new feature we have added to the emailed reports, a feature added at the request of a teacher who wanted to generate her own statsitics and reports from her school's data.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="#H34N1"]By any chance, if we misbubbled but have teh correct answer on teh top, will we get credit? (I don't think this is in my case though...)[/quote]

You don't. It says so on the booklet (or manual not sure which).

Meh I told my teacher and he said he'll do it sometime this week

I thought your teacher was a woman? o.O?


let n be positive integer and p be a prime. Prove that n! divides
p^n*(p^n-1)*(p^(n-1)-1)*...(p^2-1)*(p-1).

[quote="pleurestique"]someone else posted the same problem here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83000[/quote]

many thnks!!!


LaTeX
trigonometry
logarithms
real analysis
real analysis unsolved

Can anyone help me figure this out:

Show that f_n(x) = e^-n|1-sin x| converges in measure to f(x) = 0 on [a,b].

The convergence seems uniform to me, thus in measure too.

MissMel, you really need to be clearer with your syntax. Ideally, use $\text{\LaTeX},$ but at the very least, use parentheses where neccessary to clarify your intent.

julien_santini clearly read your expression as $e^{-n}|1-\sin x|.$

I suspect that you might actually have meant $e^{-n|1-\sin x|},$ which would be a different problem.

This sequence converges pointwise almost everywhere to zero, but not everywhere.

Given $\epsilon>0,$ what is the set on which $e^{-n|1-\sin x|}\ge\epsilon?$

For that, we need $n(1-\sin x)\le|\ln\epsilon|$, or $(1-\sin x)\le\frac{|\ln\epsilon|}n.$

For the sake of the argument, let's say $[a,b]=[0,2\pi].$ (Other intervals are just finite unions of similar intervals.) $(1-\sin x)\le\delta$ on an interval around $\frac{\pi}2$ of length that we can estimate as approximately $C\sqrt{\delta}.$

Thus, $m\{x: |f_n(x)|\ge\epsilon\}\le C\sqrt{\frac{|\ln\epsilon|}n}.$

We can choose $n$ (after we already know $\epsilon$) to make the right hand side of that last expression as small as we please - and that's convergence in measure.

Thanks. I that really helped me figure it out.


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that:
\[ \frac {1}{3a \plus{} b \plus{} c} \plus{} \frac {1}{3b \plus{} a \plus{} c} \plus{} \frac {1}{3c \plus{} a \plus{} b}\leq\frac {3}{5}
\]

[hide="remark"]With the same conditions $ \frac {1}{4a \plus{} b \plus{} c} \plus{} \frac {1}{4b \plus{} a \plus{} c} \plus{} \frac {1}{4c \plus{} a \plus{} b}\leq\frac {1}{2}$ is wrong already.  [/hide]

with pqr it's equivalent to :
$ \frac{27(3\plus{}4q\plus{}9r)^2}{25q} \geq (4q\plus{}7)^2$ and now we are free to use schur  :)

Here is my solution:
Firstly, we claim that the following inequality holds for all $ a,b,c \ge 0$ (the proof will be left to the readers):
\[ \sum \frac {a}{3a \plus{} b \plus{} c} \ge \frac {4}{5} \minus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{5(a \plus{} b \plus{} c)^2}
\]
Now, we will apply this inequality to our problem:
Re write the inequality as
\[ \sum \frac {a \plus{} b \plus{} c}{3a \plus{} b \plus{} c} \le \frac {3(a \plus{} b \plus{} c)}{5}
\]

\[ \sum \left( 1 \minus{} \frac {a \plus{} b \plus{} c}{3a \plus{} b \plus{} c}\right) \ge 3 \minus{} \frac {3(a \plus{} b \plus{} c)}{5}
\]

\[ 2\sum \frac {a}{3a \plus{} b \plus{} c} \ge 3 \minus{} \frac {3(a \plus{} b \plus{} c)}{5}
\]
Applying the above inequality, we can deduce the inequality to
\[ 2\left( \frac {4}{5} \minus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{5(a \plus{} b \plus{} c)^2}\right) \ge 3 \minus{} \frac {3(a \plus{} b \plus{} c)}{5}
\]
Setting $ p \equal{} a \plus{} b \plus{} c \ge 3$, then the above inequality becomes
\[ 2\left( \frac {4}{5} \minus{} \frac {3(p^2 \minus{} 6)}{5p^2}\right) \ge 3 \minus{} \frac {3p}{5}
\]

\[ \frac {(p \minus{} 3)(3p^2 \minus{} 4p \minus{} 12)}{5p^2} \ge 0
\]
which is obviously true since
\[ 3p^2 \minus{} 4p \minus{} 12 \ge 5p \minus{} 12 > 0
\]
:)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and let $ \mathbf{I}$ denote the set that consists of the indices of all normal subgroups of $ G$. Suppose that $ G$ possesses a proper subgroup $ H$ such that $ H \cap gHg^{ \minus{} 1} \equal{} 1$ for every $ g \in G \backslash H$. Prove that $ \vert H \vert \in \mathbf{I}$.

-- Hope you guys like this new problem proposal of mine. In any case, I look forward to reading your replies quite soon.

Best of luck, ;)

P.S. Last month's proposal can be found [url=http://www.artofproblemsolving.com/viewtopic.php?t=291125][color=orange]here[/color][/url].

This is just Frobenius's theorem on Frobenius groups.  Proofs are in most textbooks on character theory.  In Isaacs, it is theorem 7.2, page 100.

Note that H is necessarily a Hall subgroup, so the existence of a normal subgroup is equivalent to the existence of a normal complement to H, which in turn must clearly be the identity along with set of elements not contained in any conjugate.

Isaacs's Finite Group Theory has some reduced cases that do not require character theory in chapter 6, especially pages 185–186.

I got an opposite conclusion, i.e. $ |H|\notin I$. which means, if I haven't misunderstood, there is no normal subgroup of size |H|.

it suffices to prove that, $ \forall g \cdot$  $ |C(g)|\geq |H|$, if $ g\neq 1$.
denote $ K\equal{}(G\minus{}\bigcup_{g\in G}gHg^{\minus{}1})\bigcup${1}. we have $ |K|\equal{}|C(H)|\equal{}\frac{|G|}{|N_{G}(H)}\equal{}\frac{|G|}{|H|}$,
$ \forall k\in K$, $ k\neq 1$, then $ \forall h\neq g\in H$, we have $ hkh^{\minus{}1}\neq gkg^{\minus{}1}$, else $ h^{\minus{}1}gkg^{\minus{}1}h\equal{}k\Rightarrow k(h^{\minus{}1}g)k^{\minus{}1}\equal{}h^{\minus{}1}g\in H$ contradict to $ k\notin H$. So $ C(k)\geq |H|$.

$ \forall h\in H$, C(h)=C(H)=|K|. prove that |K|>|H|. But we already have $ \forall k\cdot |H|\leq |C(k)|\cdot$ and $ C(k)\subseteq K$-{1}.

[quote="mathiq"]I got an opposite conclusion, i.e. $ |H|\notin I$. which means, if I haven't misunderstood, there is no normal subgroup of size |H|.[/quote]

I is defined as the set of normal subgroups N of index [G:N]=|H|, not size |N|=|H|.

Your proof is fine, but you might be careful that many people use C(g) to denote the centralizer rather than the conjugacy class of g, so that |G|/|C(g)| in one convention is equal to |C(g)| in the other.

Needed to make the following remark on the previous entry of [b]Jack[/b].

Where it reads

[quote="JackSchmidt"][b]I[/b] is defined as the set of normal subgroups N of ...[/quote]

it should read

[b]I[/b] is defined as the set of indices of all normal subgroups of G.

Please, excuse my nitpickiness. It's all about preciseness, you know...


$ f: N^\plus{}\rightarrow N^\plus{}$ satisfies $ f(1) \equal{} 1, f(2n) < 6 f(n)$, and $ 3 f(n) f(2n\plus{}1) \equal{} f(2n) \plus{} 3 f(2n) f(n)$ for all $ n$. Find all $ m, n$ such that $ f(m) \plus{} f(n) \equal{} 293$.

Obviosly $ 3f(n)|f(2n),$ because $ f(2n)<6f(n)$ we get $ f(2n)\equal{}3f(n)$ and $ \ f(2n\plus{}1)\equal{}f(2n)(1\plus{}\frac{1}{3f(n)})\equal{}f(2n)\plus{}1.$
It give for $ x\equal{}a_k2^k\plus{}...\plus{}a_0$ $ f(x)\equal{}a_k3^k\plus{}...\plus{}a_0$. Because all digits 0 or 1, then we get, that $ f(m)\plus{}f(n)$ had digits $ a_i\plus{}b_i$, were $ a_i,b_i$ are digits $ f(m)$ and $ f(n)$.
$ 293\equal{}2\plus{}3*97\equal{}2\plus{}3*(1\plus{}3*32)\equal{}2\plus{}3\plus{}2*3^2\plus{}3^3\plus{}3^5$.
Therefore $ m\equal{}1\plus{}a_1*2\plus{}4\plus{}a_3*8\plus{}a_5*32,n\equal{}1\plus{}(1\minus{}a_1)*2\plus{}4\plus{}(1\minus{}a_3)*8\plus{}(1\minus{}a_5)*32$ - 8 variants of solution.


modular arithmetic
Divisibility
Digits
decimal representation
IMO
IMO 1975
digit sum

When $4444^{4444}$ is written in decimal notation, the sum of its digits is $ A.$ Let $B$ be the sum of the digits of $A.$ Find the sum of the digits of $ B.$ ($A$ and $B$ are written in decimal notation.)

Let $n$ be a positive integer. We will denote by $\sigma(n)$ its sum of digits in base 10. It is very easy to show that $\sigma(n) \leq 9(\log n + 1).$
So, $A = \sigma(4444^{4444}) \leq 9(\log(4444^{4444})+1) \leq 9(4444 \log(10^5)+1) = 199989.$
Then, $B = \sigma(A) \leq 46,$ then $\sigma(B) \leq 12.$ So, $1 \leq \sigma(B) \leq 12.$
Besides, we know that modulo 9 : $\sigma(b) \equiv B \equiv A \equiv 4444^{4444}.$ Finally, $4444^{4444} = (9 \cdot 443 + 7)^{4444} \equiv 7^{4444} \pmod 9.$ Moreover, $7^3 \equiv 1 \pmod 9$ and so $7^{4444} \equiv 7 \pmod 9.$ Eventually, $1 \leq \sigma(B) \leq 12$ and $\sigma(B) \equiv 7 \pmod 9$ which implies $\sigma(B) = 7.$

We have $ 4444^{4444}\leq10000^{5000}$ so $ 4444^{4444}$ has fewer than $ 20,000$ digits. So $ A$ is less than $ 180,000$. Among the numbers less than $ 180,000$ the number with the largest sum of digits is $ 179,999$ with the sum $ 44$. So $ B\leq44$, and the sum of the digits of $ B$ is at most the sum of the digits of $ 39$ which is $ 12$.
But every number is congruent to the sum of its digits modulo $ 9$. Hence $ 4444$ is congruent to $ 7$, modulo $ 9$ and we have $ 7^{3}\equiv7(mod 9)$.
It follows that $ 4444^{4444}$ is congruent to $ 7$ modulo $ 9$.
Hence the answer is $ 7$.

[quote="manjil"]and we have $ 7^{3}\equiv7(mod 9)$.[/quote]No, we don't.

Approach by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=5787]ZetaX[/url]:

when $ Q(n)$ is the digit sum: $ n \equal{} \sum n_i 10^i \equiv \sum n_i \equal{} Q(n) \mod 9$
So we only have to find $ 4444^{4444} \mod 9$, but $ 4444 \equal{} 9 \cdot 494 \minus{} 2 \equiv \minus{} 2\mod 9$, so:
$ 4444^{4444} \equiv ( \minus{} 2)^{4444} \mod 9$. But now as said in the script, $ ( \minus{} 2)^3 \equiv \minus{} 8 \equiv 1 \mod 9$ and so $ ( \minus{} 2)^{4444} \equal{} ( \minus{} 2)^{3 \cdot 1481 \plus{} 1} \equiv ( \minus{} 2)^1 \equiv 7 \mod 9$, and its done.

We know that $ B \leq 54$ and that $ B \equal{} Q(A) \equiv A \equal{} Q( 4444^{4444} ) \equiv 4444^{4444} \equiv 7 \mod 9$. But there are only finitly many numbers with that property, and writing down all of them (namely $ B \equal{} 7,16,25,34,43,52$) we see that all of them fulfill $ Q(B) \equal{} 7$, so there is no other possibility left.

Sorry I wanted to mean $ 7^{3}$ is congruent to $ 1$ modulo $ 9$. That was a typo!

[hide="A103"]Note that the sum of the digits of $B$ is congruent to $4444^{4444}$ modulo 9, and that $4444^{4444} \equiv 2^{4444} \equiv 2^{4} \equiv 7 \pmod 9$. Note that $4444^{4444}<10000^{4444}$, so it has less than $17777$ digits. Therefore, its digit sum must be less than $159993$, so $A<159993$. We can weaken the bound to $A<159999+1$. Therefore, $B \le 1+5+9+9+9+9=42$. Let $C$ be the desired answer. The largest digit-sum of a number less than or equal to 42 is $3+9=12$, so $C \le 12$. Now, it is easy to see that the only possible answer is $\boxed{7}$. [/hide]

[hide=My solution]The number of digits of $4444^{4444}$ is less than the number of digits of $10^{4\cdot 4444}$, which is $17776$. Hence $\max A$ is $9\cdot 17776=159984$. So, $B$ maximum value is $45$. And the sum of digits of $B$ is less than $10$ (and greater than $0$). Now \[ 4444^{4444}\equiv 7^{4444}\equiv 7\pmod 9 \]Hence it is $7$.[/hide]

[u][b]Solution:[/b][/u] $d(n) \equiv$ sum of digits of $n$. We want $d(d(d(4444^{4444})))$
Note:
1. $d(\#\#) \leq d(99) = 18$.(\# is denoting digits)
2. Warning: $m>n$ doesn't imply $d(m) > d(n)$ (Not always.) Example: $1111>7$ but $d(1111)=4<d(7)=7$.
3. We know that $100<125<1000$, so, if we apply $\log_{10}$, (note that the inequality will maintain, as $\log$ is an increasing function) we will get: $2<\text{log}_{10}125<3$. It basically tells us (about) the number of digits of a number.
[u]Check:[/u] $\log_{10}(4444^{4444}) = 4444 \times \log_{10}4444 < 4444 \times 4 =17776$.
So, $4444^{4444}$ will have no more than $17776$ digits. We are looking for an upper bound on the digits' sum.
Now we can say
\[ d(4444^{4444}) < \overbrace{999...9}^\text{17776 9's} = 9 \times 17776 = 159984\]
We have $d(4444^{4444})<159984$.
So,$$d(d(4444^{4444}))<d(199999)=46$$
$$\implies d(d(d(4444^{4444}))) < d(49)=13$$
[i]Question.[/i] Why are we keeping the leftmost digit of the numbers in the RHS of the inequalities intact?
[i]Ans:[/i] Let us take an example, $46>39$ but $d(46)=10 \not >d(39)=12$. So, by keeping the leftmost digit intact we are ensuring that the (upper) bound on the digit sum is correctly maintained and properly established. And secondly, we are also establishing a sharper bound on the digit sum, by doing so; as $9$ is the largest digit in the set of digits $(0,1,2...9)$, so, without replacing the leftmost digit by $9$ (which may be correct, but then it would make the digit sum larger) we are keeping it intact and replacing the following digits with $9$.
So, $$\boxed{d(d(d(4444^{4444}))) < 13} -(*)$$
[u]FACT:[/u] $n \equiv d(n) \ (mod\ 9)$
Note that $4444 \equiv 7 \ (mod\ 9)$.
$\implies 4444^2 \equiv 7^2 \equiv 4 \ (\text{mod}\ 9) \implies 4444^4 \equiv 4^2 \equiv 7 \ (\text{mod}\ 9)$
$\implies 4444^8 \equiv 7^2 \equiv 4 \ (\text{mod}\ 9) \implies 4444^{16} \equiv 7\ (\text{mod}\ 9)$
$\implies 4444^{32} \equiv 4\ (\text{mod}\ 9) \implies 4444^{64} \equiv 7\ (\text{mod}\ 9)$
$\implies 4444^{128} \equiv 4\ (\text{mod}\ 9) \implies 4444^{256} \equiv 7\ (\text{mod}\ 9)$
$\implies 4444^{512} \equiv 4\ (\text{mod}\ 9) \implies 4444^{1024} \equiv 7\ (\text{mod}\ 9)$
$\implies 4444^{2048} \equiv 4\ (\text{mod}\ 9) \implies 4444^{4096} \equiv 7\ (\text{mod}\ 9)$
So, $$4444^{4444} \equiv 4444^{4096} \times 4444^{256} \times 4444^{64} \times 4444^{16} \times 4444^{8} \times 4444^{4} \equiv 7 \cdot 7 \cdot 7 \cdot 7 \cdot 4 \cdot 7\ (\text{mod}\ 9) $$
$$\boxed{4444^{4444} \equiv 7 (\text{mod}\ 9)}$$
$$\implies d(4444^{4444}) \equiv 7 (\text{mod}\ 9) \implies d(d(4444^{4444})) \equiv 7 (\text{mod}\ 9)$$
$$ \implies \boxed{d(d(d(4444^{4444}))) \equiv 7 (\text{mod}\ 9)}-(**)$$
By $(*)$ and $(**)$ we have $$\boxed {d(d(d(4444^{4444}))) = 7}$$
[u][b]Verification:[/b][/u] According to Wolfram Alpha, the digit sum of $4444^{4444}$ is $72601$. $\implies d(4444^{4444})=72601 \implies d(72601) = 16 \implies d(16)=7$.

[hide= :)] A tribute to [b][u][i][color=#000]blackpen[/color][color=#f00]redpen[/color][/i][/u][/b]. :D  :)  [/hide]

here we need to compute:
s(s(s(4444^{4444}))), using the inequality :
s(n) \le 9([logn]+1) several times, we have :
s(4444^{4444})\le 9([log 4444^{4444}] +1)<9\times20000=180000
now , again we have :
s(s(4444^{4444}))\le 9([log s(4444^{4444})] +1)\le9(log 180000 +1)\le 63.
so, s(s(s(4444^{4444})))\le 14 (as among all the numbers 1 to 63 , maximum value of sum of digits is 14)

on the other hand , s(s(s(n)))\equivs(n)\pmod{9}
 again , s(n)\equiv n\pmod{9} and since  4444^{4444}\eqviv7^{4444}\eqviv 7\times 7^{3\times1481\equiv 7\pmod{9}

Let $S(n)$ denote the sum of the digits of $n$, so we want $S(S(S(4444^{4444}))) = S(S(A))= S(B)$.

[color=#6FA8DC][b]Claim 1:[/b][/color] $S(n) \equiv n$ mod $9$ for all $n$

Let
$$n = \overline{a_k} \; \overline{a_{k-1}} \cdots \overline{a_1} = 10^{k-1} \cdot a_k + 10^{k-2} \cdot a_{k-1} + \cdots + 10^0 \cdot a_1.$$

Then, when taken mod $9$, we get that
$$n \equiv 1^{k-1} \cdot a_k + 1^{k-2} \cdot a_{k-1} + \cdots + 1^0 \cdot a_1 \equiv a_k + a_{k-1} + \cdots + a_1 = S(n),$$
as desired.

[color=#6FA8DC][b]Claim 2.1:[/b][/color] $0 < A \le 148005$

Trivially $0 < A$.

For some arbitrary $n$ with $\mathcal P$ digits, $S(n)$ is maximized when $\underbrace{n = 999\cdots999}_{\mathcal P ~ 9\text{s}}$, so if $n$ has $\mathcal P$ digits, then $S(n) < 9 \cdot \mathcal P$. This also means that $S(n)$ is maximized when $n$ has as many digits as possible.

Note that $4444^{4444} < \dfrac{10000^{4444}}{2^{4444}}$. If we find a lower bound for $2^{4444}$, we can thus find an upper bound for $4444^{4444}$.

Evidently $2^{4444} > 2^{4440}$. Since $2^{10} > 10^3$, we get that $2^{4440} > 10^{444 \cdot 3}$, so $2^{4444} > 10^{1332}$.

Thus, $4444^{4444} < \dfrac{10000^{4444}}{2^{4444}} < \dfrac{10000^{4444}}{10^{1332}}$, so thus it has at most $16445$ digits and $S(4444^{4444}) \le 16445 \cdot 9$, so thus $A \le 148005$, as desired.

[color=#6FA8DC][b]Claim 2.2:[/b][/color] $0 < B \le 45$

Trivially $0 < B$.

We have that $A \le 148005$. If $A$ has $5$ digits, then $B$ is at most $45$.

If $A$ has $6$ digits, then the first digit of $A$ will increase by $1$ (from $99999$), but then its second digit will decrease by more than $1$ (from $99999$), so this will not give a higher upper bound for $B$.

Thus, $B \le 45$, as desired.

[color=#6FA8DC][b]Claim 2.3:[/b][/color] $0 < S(B) \le 12$

Trivially $0 < S(B)$.

Since $B \le 45$, we get that $S(B)$ is at most $12$ (for $B = 39$), as desired.

[color=#6FA8DC][b]Claim 3:[/b][/color] $4444^{4444} \equiv 7$ mod $9$

Note that $4444^{4444} \equiv 7^{4444}$ mod $9$, and since $7^3 \equiv 1$ mod $9$, we get that $4444^{4444} \equiv (7^3)^{1481} \cdot 7 \equiv 1^{1481} \cdot 7 \equiv 1 \cdot 7 \equiv 7$ mod $9$, as desired.

[color=#6FA8DC][b]Ending:[/b][/color]

By [color=#6FA8DC][b]Claim 1[/b][/color] and [color=#6FA8DC][b]Claim 3[/b][/color], we get that $S(B) \equiv 7$ mod $9$, and by [color=#6FA8DC][b]Claim 2.3[/b][/color], we get that $0 < S(B) \le 12$. Thus, $S(B) = \boxed{7}$.

[color=#6FA8DC][b]Remark:[/b][/color] The bounding in Claim 2 didn't have to be [i]that[/i] strict, but whatever.

Let $S(n)$ be the sum of the digits of $n$ and let $C=S(B)$. We know that $S(n)\equiv n\pmod9$. Then:
$$A\equiv4444^{4444}\equiv(-2)^{4444}\equiv2^{4444}\equiv2\cdot8^{1481}\equiv2\cdot(-1)^{1481}\equiv-2\equiv7\pmod9$$
and
$$C\equiv B\equiv A\equiv7\pmod9.$$
Now we bound $C$. Since $4444^{4444}\le10000^{4444}=10^{22220}$, $A\le22220\cdot9=199980$. Since $A\le10^7$, $B\le7\cdot9=63$ and $C\le6+9=15$. But in order for $C$ to be $7\pmod9$, it must be exactly $\boxed7$.

Use a program

[pywindow]
n = 4444
Preon = n ** n
Preon = str(Preon)
s = 0
for i in range(len(Preon)):
   s += int(Preon[i])
s = str(s)
s1 = 0
for i in range(len(s)):
   s1 += int(s[i])
s2 = 0
s1 = str(s1)
for i in range(len(s1)):
   s2 += int(s1[i])
print(s2)
[/pywindow]

Simply use the bound $$S(n) \leq 9(\log_{10}(n) + 1).$$ Observe that $$S(4444^{4444}) \leq 9(\log_{10}(4444^{4444}) + 1) < 9(4444 \cdot 4 + 1) = 159993.$$ Thus $$S(S(4444^{4444})) \leq S(99999) = 35.$$ All we will need is $$S(S(S(4444^{4444}))) < 16,$$ which is evident.

Now notice that $S(n) \equiv n \pmod 9$, and one can check that $4444^{4444} \equiv 7 \pmod 9$. Thus, $$S(S(S(n))) \equiv 7 \pmod 9,$$ and by our earlier inequality it must be precisely $\boxed 7$.


trigonometry
trig identities
Law of Cosines
geometry
power of a point

im not sure if this has been posted yet but...

[quote="tarquin"]$\angle QTS=\angle SRQ$ and $\angle RST = \angle RQT$ because they intercept the same arcs and so $\triangle PQT$ is similar to $\triangle PSR$.

$\frac {PT}{PR} = \frac {PQ}{PS}$ so $PT \cdot PS = PR \cdot PQ$[/quote]

what do you by "they intercept the same arcs"? :?

It means that $\angle QTS$ and $\angle SRQ$ both intercept arc $QS$. And it means that $\angle RST$ and $\angle RQT$ both intercept arc $TR$.

I think for the second one we can construct RS and QS. We know that angle P is congruent to itself. QRS is an inscribed angle, so it's measure is QS/2. Since PS is a tangent, we know that angle QSP is arc QRS/2. The arc measures that angles QRS and QSP subtend add up to 360, so the angles are supplementary, so QSP is congruent to SRP. By AA similarity, triangle QSP is similar to triangle SRP. We can then set the proportion PS/PR=PQ/PS, so PS*PS=PR*PQ

EDIT: I think you're missing the secant-secant case as well as the tangent-tangent case.

The first "power of a point" formula is cool. It can be used to prove the Law of Cosines, as in the animation of the navigation bar.


Besides the first term having $ ( \minus{} 1)!$ in the denominator, I'm not sure what this means. Nor am I sure what the point of the question is? Could you clarify, binomial 4eva, so I can try to help?

See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1709657#1709657

Unrelatedly, $ (\minus{}1)!$ will never appear in that summation.

Yes, I substituted for $ n$ instead of $ i$ and then read $ i$ as $ 1$: I wasn't thinking straight. :blush:

[quote="binomial_4eva"]so what is the answer?[/quote]

Are you reading any of these posts?! They're telling you that there's a $ ( \minus{} 1)!$ in the denominator. Now, what does that tell you?

EDIT: Lol, JBL. Sorry.

Poincare, I don't think [i]you[/i] were reading all the posts.  :roll:

[quote="binomial_4eva"]thats how they judge everything!without knowing well he talks for himself.
does it have an answer anyone ? or have i written it wrong ?[/quote]

Well for one thing, unless I'm missing something, doesn't this depend on the definition of $ a_i$?


algebra
polynomial
inequalities
inequalities proposed

Probably too easy for this forum: For every $x\in \mathbb{R}$, prove that
\[x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\frac{2}{5}>0. \]

:blush:  mmm, what a pity... i'll continue trying to find a solution in that way...  :oops:

Another solution (not originally mine): The inequality
$x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+2/5 > 0$

is true if $x\leq 0$, so it is sufficient to prove the case $x > 0$. The inequality is equivalent to the equality $(x^{7}+1)/(x+1)-3/5 > 0$ or equivalently $5x^{7}-3x+2 > 0$. Because of what campos said, it is sufficient to prove the case $0<x<1$. This can be proven by AM-GM-inequality by applying the inequality to the polynomial
\[5 x^{7}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\, \] as SplashD told in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124753]here.[/url]
Namely $5x^{7}+2>7(5x^{7}\cdot (1/3)^{6})^{1/7}=7(5\cdot (1/3)^{6})^{1/7}x.$ The inequality follows since $7(5\cdot (1/3)^{6})^{1/7}>3$.


function
logarithms
integration
geometry
geometric transformation
reflection
complex analysis

Find all entire functions $f$ such that if $|z|=1$ then $|f(z)|=1$.

It's been discussed before.

Anyway, let $\tilde f(z)=f(z)\prod_{i=1}^k\frac{1-\bar\alpha_iz}{z-\alpha_i}$, where $\alpha_i,\ i=\overline{1,k}$ are the roots of $f$ inside the unit disk (counting multiplicities). Then $\tilde f$ is a function with the same property as $f$ (it fixes the unit circle), except that it has no zeros inside the unit disk. Since $\ln\left|\tilde f(z)\right|$ is a harmonic function in the unit disk, we have $\ln\left|\tilde f(0)\right|=\frac 1{2\pi}\int_0^{2\pi}\ln\left|\tilde f(e^{i\theta})\right|\ \mbox d\theta=0$, meaning that $\left|\tilde f(0)\right|=1$. This means that $\left|\tilde f\right|$ has a local maximum in the unit disk, and is thus constant; denote this constant by $\alpha$ ($|\alpha|$=1, of course). Now we have $f(z)=\alpha\prod_{i=1}^k\frac{z-\alpha_i}{1-\bar\alpha_iz}$, and from the fact that $f$ is entire we get $\alpha_i=0,\ \forall i\Rightarrow f(z)=\alpha z^k$.

Nice. I did almost the same, but to show that a non constant function $f$ that satisfies $|f(z)|=1$ on the unit circle has at least a zero inside the unit disk, one can suppose the contrary and apply the maximum modulus principle to $\frac{1}{f}$. But of course, it is almost the same as your solution..

I have another solution and I'm in doubt if it's correct. Suppose that $f(z_{0})=0$. Then you could use reflection principle to prove that $f(\frac{1}{\bar{z_{0}}})=\infty$ (Consider the ring shaped set $R=\{z\in\mathbb C|1\leq|z|\leq |z_{0}|\}$ you see on the boundry $|z|=1$ then $|f(z)|=1$. (You see there exist a mobius transformation $g$ which $g\circ f\circ g^{-1}$) sends real line to real line. With the reflection principle now you can extend $g\circ f\circ g^{-1}a$.) So you can extend $f: R\longrightarrow \hat{\mathbb C}$ to $g: R\cup R'\rightarrow \hat{\mathbb C}$ which $R'$ is inversion of $R$ in the unit circle, and also $g(\frac{1}{\bar{z}})=\frac{1}{\overline{g(z)}}$ for each $z\in R$. Now you could see $f=g$ on $R$. So they are equal in $R\cup R'$. So $f(\frac{1}{\bar{z_{0}}})=g(\frac{1}{\bar{z_{0}}})=\infty$. This is contradiction with that $f$ is an entire function.

Nobody checks if my solution is correct or not?


Given triangle $ABC,P$ is inside it $PA\cap BC=A'$ and let $PA'=w_{a}$ similarly we have $w_{b},w_{c}$ prove that
$PBPCw_{a}+PCPAw_{b}+PAPBw_{c}\ge 12w_{a}w_{b}w_{c}$ :)

Nobdy want to do it  :( , here is my solution
We need to prove $\sum\frac{PBPC}{w_{b}w_{c}}\ge 12$, assume $P(x,y,z)$ because $P$ inside triangle therefore $x,y,z>0$ we have $\frac{PA}{w_{a}}=\frac{y+z}{x}$ now we will show $\sum\frac{(x+y)(x+z)}{yz}\ge 12$ forall $x,y,z>0$ indeed
$\sum\frac{(x+y)(x+z)}{yz}-3=\sum\frac{x(x+y+z)}{yz}=\frac{(x^{2}+y^{2}+z^{2})(x+y+z)}{xyz}\ge 9$ trivial by AM-GM
Now from here we can show
$\sum\frac{w_{b}w_{c}}{PBPC}\ge \frac{3}{4}\Leftrightarrow PAw_{b}w_{c}+PBw_{c}w_{a}+PCw_{a}w_{b}\ge \frac{3}{4}PA\cdot PB\cdot PC$ by the inequality $\sum\frac{yz}{(x+y)(x+z)}\ge \frac{3}{4}$ :)


function
limit
analytic geometry
graphing lines
slope
calculus
derivative

:?: I have been doing some research on my exponential functions project. while doing that, one function that came into my mind again and again was none other than x^x.
first thing  wondered about it was that how will its graph look like. when i tried on my graphing calc. what I got was a really wierd figure that starts at (0,1).
but this point kinda freaked me out cuz what my told us about exponential functions was that "a" (f(x)= a^x) can never be zero but "x" can. and in this case "a" is actually "x" and can it be zero? (my teacher told me that it cant be but I wanna know that why the graph has the point (0,1)? and what is 0^0? if it is undefined then  why the graphing software shows that 0^0=1? I am really confused by this thing] :huh:  :maybe:  :!:

It's not exactly true that $ 0^0 \equal{} 1$.  What is true is that $ \lim_{x \to 0^{\plus{}}} x^x \equal{} 1$; in other words, the graph of $ x^x$ can get arbitrarily close to the point $ (0, 1)$.  This is what is known as a removable singularity, and for some purposes you can just pretend it's an actual point on the graph.

Depending on how you define the exponential, by the way, the graph has an additional set of points left of the $ y$-axis.  $ x^x$ can be defined for negative rational $ x$ if $ x$ has odd denominator when written in lowest terms.  Of course, you probably don't want to care about these points.

As long as you've got that graphing calculator ....

Enter the function $ f(x) \equal{} |x|^x.$

(That way, it's defined on both sides of the axis.) At some ordinary magnification, it looks like a pretty smooth curve, doesn't it?

Now center your cursor on $ (0,1)$ and issue a "zoom in" command. It's really best for this exercise if that command preserves angles and slopes by zooming in the same amount vertically as horizontally - if that's under your control, make it do that.

And keep zooming, and zooming, and zooming, watching the picture the whole time. If we do this for a differentiable function, we should at very high magnifications see what appears to be a straight line whose slope is the derivative at the point we're zooming towards.

But does that happen here? At this point, I'm not telling you what happens - I'm just asking you to carry out the process and be observant. What do you observe, and what comments do you have?


Prove that $ 2(a^5\plus{}b^5)\plus{}c^5\ge \frac 54 abc(a\plus{}b)^2,\ \forall\ a,b,c\ge 0$.

[quote="Mateescu Constantin"]Prove that $ V \equal{} 2(a^5 \plus{} b^5) \plus{} c^5\ge \frac 54 abc(a \plus{} b)^2,\ \forall\ a,b,c\ge 0$.[/quote]
It's very easy by AM-GM!
We have $ 2(a^5 \plus{} b^5) \geq 4( \frac {a \plus{} b}{2})^5$
$ \equal{} > V \geq 4( \frac {a \plus{} b}{2})^5 \plus{} c^5 \geq 5( \frac {a \plus{} b}{2})^4c \geq \frac {5}{4}abc(a \plus{} b)^2$ :P


LaTeX
binomial coefficients
number theory unsolved
number theory

let p be a prime, and n, k be two nonnegative integers, with

$ n \equal{} n[0] \plus{} n[1]p \plus{} n[2]p^2 \plus{} ... \plus{} n[t]p^t$

and

$ k \equal{} k[0] \plus{} k[1]p \plus{} k[2]p^2 \plus{} ... \plus{} k[t]p^t$

be their base p representations. Prove that

$ (nCk) \equal{} (n[0]Ck[0])(n[1]Ck[1])...(n[t]Ck[t])(modp)$

NOTE: here n[0] is actually n with a subscript 0, sorry couldn't figure out latex command for that one, so ...

hence, n[0], n[1], ..., n[t], k[0], k[1], ..., k[t] are all natural numbers, meant (n subscript zero), (n subscript 1) and so on.

also, prove that the number of the binomial coefficients among (nC0),
(nC1), ... , (nCn) that leave a remainder 1 when divided by 3 is more than
the number of those that leave a remainder 2 when divided by 3, for any given positive integer n.

the deadline for this was over today, so please help!!!

I've no proof that the deadline is over...
Locked untill  know better...


Would you say "This, That and Something Else" or "This, That, and Something Else"?

Yes, this is an annoying question. I think the correct grammar is "spoon, plate, and fork," though articles in newspapers generally leave off the last comma in a list, making "spoon, plate and fork."

The last comma is necessary unless you are only listing two objects.  One big thing people mess up though which annoys me is distinguishing between less and fewer.

i had some amish grammar book and only if it was a compound sentence was it necessary i think....but it's optional in this case(i know)

EDIT dues to funcia's no spell thing...

I think of it as putting parentheses. Watch:

This, that, and something else

This, (that and something else)

[quote="mustafa"][quote="funcia"]i had some amish grammar book and only if it was a compound sentence was it necessary i think....but it's optional in this case(i know)

[/quote]

The Amish are always right.[/quote]
Oh yeah!!

By the Axiom Of Amish Education

(in case anyone was wondering why I would make an assumption)

My local [url=http://www.heb.com/welcome/index.jsp]HEB[/url] has the lanes labeled "10 items or fewer"

[quote="Ignite168"]This is stylistic; either is accepted. I personally prefer "This, That, and Something Else" but it's a matter of personal choice. The more important thing is to be consistent of which one you pick.[/quote]

Actually, the serial comma is needed when describing a series of objects. However, if you are not trying to list a series, then you don't use it. For example, one of Frost's poems contains the line "the woods are lovely, dark and deep." Dark and deep are meant to be taken as a separate phrase.