Prove, without using a computer, that

$ F_{1000}> 10^{200}$

where $ F_{1000}$ denotes the 1,000th Fibonacci number.


calculus
integration
trigonometry
function
derivative
search
calculus computations

[quote="kunny"]Find the value of $m\ (0<m<1)$ for which $\int_{0}^{\pi}|\sin x-mx|\ dx$ is minimized.[/quote]

Let $f(m)=\int_{0}^{\pi}|\sin x-mx|\ dx$. Fix $\alpha\in (0,\pi )$ such that $\sin\alpha=m\alpha$. Now we have \[f(m)=\int_{0}^{\pi}|\sin x-mx|\ dx = \int_{0}^{\alpha}(\sin x-mx) dx+\int_{\alpha}^{\pi}(mx-\sin x) dx\]  \[= m\left(\frac{\pi^{2}}{2}-\alpha^{2}\right)-2\cos\alpha\] to find the extrema, set $f^\prime (m) = \frac{\pi^{2}}{2}-\alpha^{2}=0$. So choose $\alpha = \frac{\pi}{\sqrt{2}}$ but we also require that $\sin\alpha=m\alpha$ and thus it follows that the minimum occurs for $m=\frac{1}{\alpha}\sin\alpha = \frac{\sqrt{2}}{\pi}\sin\frac{\pi}{\sqrt{2}}$, and the minimum value is then \[f\left( \frac{\sqrt{2}}{\pi}\sin\frac{\pi}{\sqrt{2}}\right) = \frac{\pi}{\sqrt{2}}\sin\frac{\pi}{\sqrt{2}}\]

[color=darkblue][b]Kunny[/b], your problem is nicely and is easily remembered.
A similar (but ugly) example : $\min_{m>0}I(m)\ ,$ where ${I(m)=\int_{0}^{1}}\left|\sqrt{x-x^{2}}-mx\right|\ dx\ .$
If have leisure, then "attack" and this problem !

[hide="A little helper."]
$\int \sqrt{x-x^{2}}\ dx=\frac{1}{4}\cdot (2x-1)\sqrt{x-x^{2}}+\frac{1}{8}\cdot\arcsin\ (2x-1)+C\ .$[/hide]
[hide="Answer."]
The minimum point for the function $I(m)$ is $m_{0}=\sqrt{-1+\sqrt 2}\ .$[/hide][/color]

I had asked a similar question in a class and this was a natural question in return; when differentiating  that function of $m$, why don't you consider $\alpha$ as a function of $m$. I think answering to these  questions,  in this tutorial section, is unavoidable. :)

[quote="Virgil Nicula"]Kunny, your problem is nicely and is easily remembered.
A similar (but ugly) example : $\min_{m>0}I(m)\ ,$ where ${I(m)=\int_{0}^{1}}|\sqrt{x-x^{2}}-mx|\ dx\ .$[/quote]

Thank you very much, Virgil Nicula! :)

Here is the solution to your problem.

The graph of $y=\sqrt{x-x^{2}}$ is the semi circle with centerd at $(\frac{1}{2},\ 0)$ and  radius $\frac{1}{2},$ as $m>0,$ two graphs $y=\sqrt{x-x^{2}}$ and  the line $y=mx$ have an intersection point except the origin. Thus let $t$ such that $\sqrt{t-t^{2}}-mt=0\ (0<t<1),$

$I(m)=\int_{0}^{t}|\sqrt{x-x^{2}}-mx|\ dx$

$=\int_{0}^{t}|\sqrt{x-x^{2}}-mx|\ dx-\int_{t}^{1}|\sqrt{x-x^{2}}-mx|\ dx$

$\geq\frac{1}{2}(t-1)\sqrt{t-t^{2}}+\frac{1}{4}\sin^{-1}(2t-1)-\frac{\pi}{4}-\frac{1}{2}(2t^{2}-1)m.$

Thus determinimg the value of $t$ such that this equation takes the constant number independent of the value of $m,$ we obtain $2t^{2}-1=0\ (0<t<1) \Longleftrightarrow t=\frac{1}{\sqrt{2}},$ yielding $m=\sqrt{\sqrt{2}-1}.$

I made your problem several years ago, but this problem wasn't suited for Japanese university entrance exam. Because Japanese high school students don't study $\sin^{-1}x.$

[quote="benorin"][quote="kunny"]Find the value of $m\ (0<m<1)$ for which $\int_{0}^{\pi}|\sin x-mx|\ dx$ is minimized.[/quote]

 and thus it follows that the minimum occurs for $m=\frac{1}{\alpha}\sin\alpha = \frac{\sqrt{2}}{\pi}\sin\frac{\pi}{\sqrt{2}}$, and the minimum value is then \[f\left( \frac{\sqrt{2}}{\pi}\sin\frac{\pi}{\sqrt{2}}\right) = \frac{\pi}{\sqrt{2}}\sin\frac{\pi}{\sqrt{2}}\] [/quote]

 Could you explain this more cleary, for whose who still don't get a point? Especially I'm intresting in you final conclusion.

[quote="The Captain"][quote="benorin"] Fix $\alpha\in (0,\pi )$ such that $\sin\alpha=m\alpha$.

to find the extrema, set $f^\prime (m) = \frac{\pi^{2}}{2}-\alpha^{2}=0$. [/quote]
I don't think that your solution is true, although the final answer is true. To me, that kind of differentiation is meaningless. You have fixed $\alpha$ without taking this into account that $\alpha$ is a function of $m$. If I were you I would write the result of integration as a function of $\alpha$, then I would differentiate it with respect to $\alpha$ and finally search for its zeros; $\alpha = \tan \alpha$ or $\frac{\pi^{2}}{2}=\alpha^{2}$.[/quote]

Your right! So rather let $\alpha : m\rightarrow (0,\pi )$ according to the rule $\sin\alpha (m) = m\alpha (m)$, so that putting $\alpha=\alpha (m)$ in the prior work gives the value of the integral as being \[f(m) = m\left( \frac{\pi^{2}}{2}-\alpha^{2}(m)\right)-2\cos\alpha (m)\] upon differentiating w.r.t. m we have \[f^{\prime}(m) = \left( \frac{\pi^{2}}{2}-\alpha^{2}(m)\right)-2m\alpha (m)\alpha^{\prime}(m)+2\alpha^{\prime}(m)\sin\alpha (m)\] but since $\sin\alpha (m) = m\alpha (m)$ we have \[f^{\prime}(m) = \left( \frac{\pi^{2}}{2}-\alpha^{2}(m)\right)-2m\alpha (m)\alpha^{\prime}(m)+2\alpha^{\prime}(m)m\alpha (m) = \frac{\pi^{2}}{2}-\alpha^{2}=0\] where it was set equal to zero to find the extrema, which occur for those values of $m$ such that $\alpha (m) = \frac{\pi}{\sqrt{2}}$ (as in the previous work). The rest is all the same...

Thanks Capt'n,

     --Ben


calculus
integration
derivative
real analysis

I've been having trouble grasping the idea of substitution when evaluating indefinite integrals. I know how to do them, I just don't understand what is going on behind the scenes.. and I would like to clear up some misconceptions.

Let's say we have the indefinite integral:

$\int(3x+4)^4dx$

from what I have read, understood, and been told: dx is part of the notation. the dx tells us to intergate with respect to x.

Now if we let $u=3x+4$. Then the substitution method continues by differentiating, $du/dx = 3$, solving for dx, and substituing, resulting in:
$\int\frac{u^4du}{3}$.

I don't understand how one can differentiate and solve for dx. What exactly is dx when alone? What exactly is dy when alone? If dx is related to differentials at all, and represents a small value for the change in x, then I don't understand how this is used as a reversal of the chain rule.

Also, if the dx in the integration notation is simply just a part of the notation, then how can we 'substitute?' What exactly are we substituing?

I've been trying to understand this concept for a while now, and I feel like I have missed something crucial.

If someone could please explain why u-substitution works, how it is in fact the reverse of the chain rule, or why dx can be solved for, i would appreciate it. Thanks.

I regret to say it, but Myth's response is actually pretty accurate -- you need to take a real analysis course in order to understand what's going on there.  It might help to try thinking of the dx, in all contexts, as limits of changes.  In the case of the derivative, $\frac{du}{dx}$ is the limit of ( :tri: u)/( :tri: x) while in the integral, you need to multiply by it in order to later divide by it when taking derivatives or in order to cancel out when integrating a derivative.

I was afraid that would be the answer... I guess I'll have to wait until college, .or hopefully I'll have a mathematical epiphany and realize some things all of a sudden.

Thanks anyway.

Their responses are much more pessimistic than they need to be. Sure, you may need real analysis to [i]prove[/i] these results, but you can sometimes do better than to prove a result rigorously when it comes to understanding. Write out the chain rule for derivatives. Now integrate everything. What do you see?


ΝΔΟ
$ \sqrt {n^n} < n! < \left(\frac {n \plus{} 1}{2}\right)^n$
οπου $ n\equal{}3,4,...$

Απο Cauchy εχουμε
$ \frac {1 \plus{} 2 \plus{} ... \plus{} n}{n} \equal{} \frac {\frac {n(n \plus{} 1)}{2}}{n} \equal{} \frac {n \plus{} 1}{2} > \sqrt [n]{n!}\Leftrightarrow$

$ \left(\frac {n \plus{} 1}{2}\right)^n > n!$

Επισης

[color=red]λαθος[/color]

$ n! > \sqrt {n^n}$

Μια λίγο πιο λυκειακή λύση:

για το δεξί σκέλος αρκεί να δείξουμε ότι:

$ \forall k \le \frac {n}{2}, k \in \mathbb{N}, (n \minus{} k)k \le (\frac {n \plus{} 1}{2})^2 \Leftrightarrow$

$ nk \minus{} k^2 \minus{} (\frac {n \plus{} 1}{2})^2 < 0$

αρκεί τώρα να δείξουμε ότι $ \forall k \le \frac {n}{2}, k \in \mathbb{R}, nk \minus{} k^2 \minus{} (\frac {n \plus{} 1}{2})^2 < 0$

έστω $ g(k) \equal{} nk \minus{} k^2 \minus{} (\frac {n \plus{} 1}{2})^2$ τότε $ g'(k) \equal{} n \minus{} 2k \ge 0$ άρα η g είνα αύξουσα (γνησίως εκτός από το $ k \equal{} \frac {n}{2}$)

άρα $ g(k) \le g(\frac {n}{2}) \equal{} \frac {n^2}{4} \minus{} \frac {(n \plus{} 1)^2}{4} \equal{} \frac {(2n \plus{} 1)( \minus{} 1)}{4} < 0 \Leftrightarrow nk \minus{} k^2 < (\frac {n \plus{} 1}{2})^2$

q.e.d. :P

(όμοια για το αριστερό μέλος η συνάρτηση έχει τύπο $ n(n \minus{} k) \minus{} k^2$ και είναι πάλι αύξουσα άρα είναι μεγαlύτερη από την τιμή στο 0 που είναι $ n^2 > 0$)

Mailo  οσον αφορά το πρώτο συμφωνώ για το δευτερο όμως δεν συμφωνώ αφού γίνεται $ (n!)^2\equal{}1^2\cdot2^2\cdot.........n^2$

[quote="kritikos"]Mailo  οσον αφορά το πρώτο συμφωνώ για το δευτερο όμως δεν συμφωνώ αφού γίνεται $ (n!)^2 \equal{} 1^2\cdot2^2\cdot.........n^2$[/quote]

ninjaed  :ninja:

Sorry Μανιέ αλλά την ώρα που το εβαλες το έβαλα και εγώ οπότε δεν το είδα. :blush:

Δημήτρη, εγώ δεν είδα τη δικιά σου.

Ξέχασα στη λύση μου για το αριστερό σκέλος να βγάλω το ίσον :blush:  αλλά αυτό είναι προφανές. :lol:

Τελικα πως μπορουμε να αποδειξουμε οτι $ n!>\sqrt{n^n}$ ?

Δες στη λύση μου την υποσημείωση στο τέλος.

Για να δείξουμε ότι $ n!\geq{\sqrt {n^n}}$ αρκεί ισοδύναμα να δείξουμε ότι $ (n!)^2\geq{n^n}$.Θα αποδείξουμε το ζητούμενο με επαγωγή.
Για $ n \equal{} 1$ έχουμε την ισότητα.
Ας υποθέσουμε ότι έχουμε το ζητούμενο για κάποιον $ n\in{N}$ δηλαδή $ (n!)^2\geq{n^n}\Leftrightarrow{{(n!)^2}{(n \plus{} 1)^2}\geq{n^n}{(n \plus{} 1)^2}}\Leftrightarrow{{{[(n \plus{} 1)!]}^2}\geq{{n^n}{(n \plus{} 1)^2}}} (1)$ και θα δείξουμε ότι ισχύει και για τον $ n \plus{} 1$ δηλαδή: $ {{[(n \plus{} 1)!]}^2}\geq{(n \plus{} 1)^{n \plus{} 1}}$.Αρκεί σύμφωνα με την $ (1)$ να δείξουμε ότι: $ {{n^n}{(n \plus{} 1)^2}}\geq{(n \plus{} 1)^{n \plus{} 1}}\Leftrightarrow{{n^n}\ge{(n \plus{} 1)^{n \minus{} 1}}}\Leftrightarrow{{(\frac {n}{n \plus{} 1})^{n}}\geq{\frac {1}{n \plus{} 1}}}\Leftrightarrow{{(1 \minus{} {\frac {1}{n \plus{} 1}})^n}\geq{\frac {1}{n \plus{} 1}}}$
το οποίο είναι προφανές από τη ανισότητα του Bernoulli.

Χμμ, οντως με επαγωγη βγαινει πολυ ευκολα... ωραιος!

Μπορούμε να αποδείξουμε και το δεξί μέλος της ανισότητας πολύ εύκολα επίσης με επαγωγή.
Για $ n \equal{} 3$ ισχύει αφού $ 3! \equal{} 6$ και $ (\frac{3 \plus{} 1}{2})^3 \equal{} 8 > 3$.
Υποθέτουμε ότι ισχύει για κάποιον $ n\in{N}$ δηλαδή $ n! < {(\frac {n \plus{} 1}{2})^n} \Leftrightarrow {(n \plus{} 1)! < \frac {(n \plus{} 1)^{n \plus{} 1}}{2^n}}$ και θα δείξουμε ότι ισχύει και για $ n \plus{} 1$ δηλαδή $ (n \plus{} 1)! < (\frac {n \plus{} 2}{2})^{n \plus{} 1}$ τότε από την (1) αρκεί $ \frac {(n \plus{} 1)^{n \plus{} 1}}{2^n} < {(\frac {n \plus{} 2}{2})^{n \plus{} 1}} \Leftrightarrow {2 < (1 \plus{} \frac {1}{n \plus{} 1})^{n \plus{} 1}}$ που είναι προφανές από την ανισότητα Bernoulli.

[quote="mailo"][quote="maniopas"]Μια λίγο πιο λυκειακή λύση:

για το δεξί σκέλος αρκεί να δείξουμε ότι:

$ \forall k \le \frac {n}{2}, k \in \mathbb{N}, (n \minus{} k)k \le (\frac {n \plus{} 1}{2})^2 \Leftrightarrow$

$ nk \minus{} k^2 \minus{} (\frac {n \plus{} 1}{2})^2 < 0$
[/quote]

Βασικα δεν καταλαβα την λυση σου...
Γιατι περνεις $ k \le \frac {n}{2}$ και γιατι αρκει να δειξουμε οτι $ (n \minus{} k)k \le (\frac {n \plus{} 1}{2})^2$ ? :maybe:
Τι σχεση εχει αυτο με το $ n! < \left(\frac {n \plus{} 1}{2}\right)^n$ ?[/quote]

γιατί είναι $ (1 \cdot 2...n)^2 \equal{} 1(n \minus{} 1) \cdot 2(n \minus{} 2) ... (n \minus{} 2)2 \cdot (n \minus{} 1)1 \cdot n$

$ \equal{} (1(n \minus{} 1) \cdot 2(n \minus{} 2) ...)^2 \Leftrightarrow$

$ n! \le \sqrt {1(n \minus{} 1) \cdot 2(n \minus{} 2) ... (n \minus{} 2)2 \cdot (n \minus{} 1)1 \cdot n} \le$

$ \sqrt {(\frac {n \plus{} 1}{2})^2(\frac {n \plus{} 1}{2})^2....(\frac {n \plus{} 1}{2})^2(\frac {n \plus{} 1}{2})^2(\frac {n \plus{} 1}{2})^2} \equal{}$

$ (\frac {n \plus{} 1}{2})^n$

Για το αριστερό σκέλος βλέπω ότι η λύση μου έχει χοντρό typo. Έτσι είναι το σωστό:

αρκεί να δείξουμε ότι:

$ \forall 1 \le k \le \frac {n}{2}, k \in \mathbb{N}, (n \minus{} k)k \le n \Leftrightarrow$

$ n(k \minus{} 1) \minus{} k^2 < 0$

αρκεί τώρα να δείξουμε ότι $ \forall 1 \le k \le \frac {n}{2}, k \in \mathbb{R}, n(k \minus{} 1) \minus{} k^2 < 0$

έστω $ g(k) \equal{} n(k \minus{} 1) \minus{} k^2$ τότε $ g'(k) \equal{} n \minus{} 2k \ge 0$ άρα η g είνα αύξουσα

άρα $ g(k) \le g(1) \equal{} 0 \Leftrightarrow nk \minus{} k^2 < n$

q.e.d.


You know the big science companies that make science-ee gear like test tubes and centrifuges and fume hoods?

The people who work the company are at least somewhat science-ee because they presumably know what science materials and tools to make. But they also run the business.

So did they study business and science in college necessarily?

Even the sales representatives for those companies have some scientific background - often a bachelors or even sometimes a masters degree in biology, chemistry, or physics. If you're making a bid to sell a GC-mass spec (and the service contract on it), it would help if you had some idea what a GC-mass spec does and why someone would want one.


geometry articles calculus ratio percent

I was wondering what career opportunities exist after getting a degree in Accounting and Economics, including those related to mathematical fields?

While I'm not an accountant, I work with university-level Accounting students and faculty and have followed the careers of many former Accounting students.  From this perspective, Accounting careers can be categorized into roughly 5 areas including:

1.  Financial Accounting - producing financial statements, including income (profit/loss) statements, balance sheets (assets, liabilities, owner's equity) for use by stockholders, managers and others to determine the financial status of the firm.  These financial statements are used by investors and potential investors to help predict future earnings (stock price), credit-worthiness, etc.

2.  Public Accounting / auditing - working for a firm (possibly your own) that audits the financial statements of other firms for accuracy and compliance with generally accepted accounting principles (GAAP).  Since it's cost prohibitive to audit every transaction that impacts these financial statements, accountants auditing the statements often use statistical and mathematical techniques to determine which accounts to examine during an audit.  Poor audit practices used by some public accounting firms in the 1990's led to infamous corporate meltdowns (i.e. Enron & Arthur Andersen) and huge investment losses.  The resulting Sarbanes-Oaxley legislation has created increased needs for auditors, especially those within firms (non-public "internal" auditors).

3.  Cost Accounting - Cost/managerial accountants determine cost structures associated with products, services, or processes used by a firm so that managers can deploy resources effectively and develop profitable pricing strategies.

4.  Tax Accounting - Preparing tax returns and associated documentation for individuals and corportations.

5.  Governmental / Not-for-Profit Accounting - developing information that will help government and not-for-profit agencies manage and deploy resources effectively.

Many Accounting careers involve setting up systems to collect data used to produce and audit financial statements, cost statements and other numerical measures of the health of a company.   Detail-oriented rule-followers who are comfortable with using numbers to make decisions can typically be trained to be successful accountants.  Many good mathematics students have these characteristics.

Communicating numerical information to those less comfortable with numbers (the vast majority of folks) is vitally important in accounting as well as most applied mathematics disciplines.  Many good mathematics students have difficulties communicating the value of their applied analyses with those who use the work and their work may not have much impact as a result.  A good business school (home to most B.S. Accounting programs in the U.S.) will develop these skills and force you to work and communicate with many who are not particularly mathematically talented.

Much of the "theory" associated with financial accounting tends to revolve around developing and modifying the Generally Accepted Accounting Principles (in the US) and similar sets of rules in the EU and elsewhere.  Unlike mathematics, these rules are not derivable from a set of axioms, and must be studied and to some extent "memorized" in order to be successfully applied.  Further, the rules change over time thus requiring accountants to continually study and update their knowledge for success.

There are often practical situations where accountants must use professional and ethical judgements in applying GAAP.  While it's a far cry from deriving theorems, many find accounting to be stimulating, satisfying and a bit more creative than is often sterotyped.

Thank you very much for your response, gt59. I am thankful for your description. I have a pretty good general idea as I have visited KPMG and I wrote an article on it in my college paper, I am giving the link here:

http://www.thetriangle.org/media/paper689/news/2005/05/20/EdOp/Accounting.Offers.Much.More.Than.The.Number.Game-953300.shtml

Also, I am supposed to have an internship for 6 months from September to March and I have already received an offer from InterDigital Communications, which is a electronics company that has a lot of patents. I will be in the Accounting Department, and am really excited about getting the chance to work. I am also planning to major in Economics, I didnt mention that I recently completed my freshmen year.

How important do you think excel is to the accounting profession, all the interviews I went to asked my knowledge of excel, I am now trying to study it, some say that it can be learned on the spot.

Excel is a spreadsheet used extensively for analyzing and presenting accounting data.  You are expected be proficient with electronic spreadsheets for nearly any career in financial services or accounting (or business for that matter.)  There are several tutorial-type books that could help you develop "user specialist-" and "expert-" level skills with Excel.  Course Technology ( [url]http://www.course.com[/url] ) publishes several such books. Some business schools and community colleges offer spreadsheet courses as well.  Take a look at some of the sample spreadsheets provided with Excel - some of the more useful and complex analyses probably won't be learned "on the spot" - you'll need to practice these skills on your internship and future coursework.

MS Visual Basic is the programming language used to customize Excel spreadsheets and is worth learning while you master spreadsheets.

I'd also recommend learning about database management systems (such as Microsoft Access) - useful tools that are often underutilized in managing data.  Often spreadsheets are used in situations that might be more appropriately be handled with databases because the analyst/user doesn't know anything about databases.

As for the double major - not a bad idea.  You might consider building skills in Finance (1st cousin to Economics but more applied) and / or information systems through double majors or minors.  Either would nicely complement an Accounting degree.  At my school, a double major in either Finance or IS takes just one extra semester when planned well.

If you're thinking about graduate school, keep taking math courses - at least through the calculus sequence and linear algebra.

Most states require 150 hours of college coursework before the Certified Public Accountant (CPA) designation can be earned.  The CPA is required for auditors at public accounting firms like KPMG.  The extra hours will give you some flexibility on the skills you choose to develop over the next 4 years of school.  Many schools move students into master's programs to complete the 150 hour requirement.  The double major in Econ will probably give you all that you need.  I'd look closely at the Economics curriculum to see if you'll get the skills that you need - especially if you like working in public accounting.

Thank you very much for your suggesstions. I am interested in being a CPA, I know about the 150 hour requirement and it looks like the bachelor's degree might not be enough to get it. This was discussed when I visited KPMG. Pennsylvania is not yet a 150 hour state but it soon will be. Also, I believe there is like a 1.5 year working in a public accounting firm required before being able to take the exam. What college or organization do you work for, sorry for asking? I dont think I said that I am in Drexel University in Philly, I recently completed my first year.

the Big Five just became the Big Four after Arthur Andersen tanked, and may become the Big Three if the government exercises its harsher options (criminal litigation) against KPMG's tax-shelter business.  This means that there is already a glut of unemployed accountants on the market, and it may get worse.

Fleeting guest, how wrong u r. First of all, ARthur Anderson was recently acquited even thuogh it wont recouperate, this does not create less jobs but more. After these scandels with Enron, the SArbanes-Oxely act was passed which is just great for accountants providing them with much more work and now more CPA's are needed then ever before. There r more then 1 million accountants in the US and even more are needed.

There were articles in the past year in Wall Street Journal about the employment situation after the Arthur Andersen collapse (the legal situation doesn't matter, as their business is kaput).  My assertions or my recollections of the articles may well turn out to be wrong, but unless you absolutely love the profession (and how would you know at this stage whether other professions might not suit you better) it would be a really good idea to try and understand the potential effects of the the Andersen and KPMG scandals on the field by the time you're ready to enter it.

Generally, if a zillion unemployed people in profession X are dumped on the market in a short time, it depresses the wages.  If the zillion is the entire labor force of 2 of the 5 largest companies, the wage depression can be expected to last for some time to come, even if some people eventually leave the business or are absorbed by other firms.

Wait fleeting guest. The Big Four do not compromise most of Accountants in the US, not even close to a million accountants work there. Even though there will always be a few accounting companies and tons of work.

"zillion" is a figure of speech meaning "a large number", I didn't mean 1000000 or any specific number of unemployed accountants.  (Or maybe you mean the total number of US accountants is about a million?)  The WSJ articles indicated the situation was bad for the Andersen ex-employees.  The effect on the job market of adding a large number of unemployed has more to do with the ratio of unemployed to positions for hire, not to the total number employed in the profession, which is to say that it is likely to be a Big Deal that Andersen and maybe KPMG are going under.

Anyway, I understand that there is some need for accounting in the US economy independent of the state of the Big $N \leq 4$ firms; but the Andersen and possibly the KPMG situations are likely to have fallout.

I think the employees that worked in Anderson have just moved to another big four or other such company after the scandal and are not worse off.

I don't know why you think so, as my recollection is that Wall Street Journal reported the exact opposite within the last few months.    I read it in the paper version, so unfortunately cannot produce a link at the moment.    Either way, the situation may vary over time, and I repeat again in case you are not aware, KPMG is currently in deep trouble with the US government and negotiating for its survival.

As a footnote, things look quite bad for KPMG.   Whether or not the Big 4 becomes the Big 3, expect plenty of people to jump as the ship takes on water.

1. USD 456 million settlement with US government, admitting fraudulent sale of tax shelters.   This is the largest tax evasion case in US history.
2. KPMG former clients exposed to collection action by IRS re: 2 billion dollars in evaded taxes
3. Criminal indictment of 8 former KPMG partners and one lawyer (paid USD 50K per letter to advise clients that shelters were "probably" legal)
4. Today's edition of New York Times reports that US government is planning criminal indictment of "at least a dozen" more KPMG executives (current or former is not specified), in addition to the 8 former partners already named.

Yes, I heard about the tax shelters even though I am not too familiar with what they are. Is that like suggesting to your clients not to pay taxes? But the thing is that there will always be some Big Four or BIg Three because all major corporations need an auditor, so there will be companies that audit them. Are you suggesting that it will just be minor companies? Also, there are a lot of accounting jobs in major corporations, not just the Big 4. Btw, KPMG will not be liquidated despite their horrible condition.

Tax shelters are schemes for placing money in (combinations of) transactions whose end result is to reclassify taxable income as non-taxable or taxable at a lower rate.   Typically this involves exploiting differences in laws between countries (or committing outright fraud) and international barriers to the documentation and investigation of the underlying transactions, and obscuring the ownership of the entities to whom the "losses" are due (i.e. that the "loss" is actually a payment from the tax-evader to himself as he controls both the paying and the receiving entities).

re: major and minor companies, it was just reported that the absorption of Arthur Andersen business by one of the Big 4 would have been illegal had it happened in the economically equivalent form of a merger before AA collapsed.   There is an economic index used by regulators to evaluate the legality (antitrust) of corporate mergers where a score of 50 or so blocks the transaction, and the hypothetical subsumption of AA rated about 450 points.   So if AA and KPMG business goes to the big firms the regulatory climate may well move in the direction of ever-greater control over the remaining Big 3 and encouragement of a decentralization.

If the dozen or more additional KPMG partners up for indictment include current partners, they may collapse anyway despite the settlement as people start to jump ship.  It is unclear at the moment.

Sorry, I didnt quite follow your second paragraph. Can you explain it in simpler language, maybe its just me, but nevertheless.

There is a statistic ("Herfindahl-Hirschman Index") that US antitrust regulators use as a measure of oligopoly in a market, which varies between 0 and 10000.  Proposed mergers that would raise this index by 50 or more draw the concern of regulators.

http://www.usdoj.gov/atr/public/testimony/hhi.htm

According to the following paper (not published, but cited in the Sept 6 Washington Post article), the H-H index increase due to elimination of Arthur Andersen is 455.   That is, had one or more accounting firms tried to acquire Andersen through a merger rather than absorbing the business of dead Anderson, the transaction would probably have violated antitrust regulations.

http://papers.ssrn.com/sol3/papers.cfm?abstract_id=691445

The abstract is below.  Note that the "catastrophic civil litigation" adumbrated in the last sentense is already in progress; KPMG is the defendant in a class-action suit by clients to whom it sold the tax shelters.  Its settlement with the government, in which it admitted criminal wrongdoing and fraudulence of the tax products it sold, may strengthen the civil cases against it.   If the Big 4 become an even more oligopolistic Big 3, stricter regulation of the accounting industry could develop on antitrust grounds, above and beyond the Enron and KPMG scandals.

It is premature to predict for certain the effect on the accounting job market, but anybody thinking of entering the accounting business would be well advised to be aware of the earthquake now in progress.

"Abstract:
Andersen's demise in an already concentrated market for the provision of auditing services to large public firms is expected to increase market power and the prices of auditing services. This paper first documents the increase in concentration. Since Andersen's exit was driven by a choice of public policy, an appropriate reference point for appraising the concentration changes is the Department of Justice's 1992 Merger Guidelines' delineation of acceptable concentration levels and changes. These suggest that Andersen's disappearance, viewed as if it were a horizontal merger, would have violated those Guidelines. Consistent with this prediction, we find evidence of anticompetitive price changes for auditing services. Percent changes in audit fees are found to be significantly correlated with percent changes in market concentration, and former Andersen clients are shown to have faced significantly larger increases in their audit fees than other firms that changed auditors within the same time period. The paper concludes that the policy of putting Andersen out of business has foreclosed future policy options for the government in its dealings with the accounting industry, the structure of which is now more vulnerable to both criminal and catastrophic civil litigation."

Are you refering to the KAtrina earthquake, or the KPMG problem as the earthquake, I think KPMG but not 100% sure.

I think that accountants will be needed for the basic reason that I said. Any major or even average corporation needs to get audited, and they need an accounting company to do it. Therefore, there have to be accounting comapnies that always keep working.

Also, people need the help of an accountant to do tax etc etc.

I was reading in the businessweek that some consider these tax shelters as a transaction just to save taxes, if its just that I dont think that breaks any logical laws, at least those that should be in place in a free market economy where voluntary action of both parties is all that matters in a transaction.

[quote="hello"]I was reading in the businessweek that some consider these tax shelters as a transaction just to save taxes, if its just that I dont think that breaks any logical laws, at least those that should be in place in a free market economy where voluntary action of both parties is all that matters in a transaction.[/quote]

The IRS has a different opinion, which is that if a transaction has "no economic substance" it is likely to be an illegal tax shelter.  Other things may come into consideration, but an otherwise vacuous tax-saving transaction that fails the economic substance test comes under suspicion.

As for people needing accountants, etc ---- people always need programmers, doctors and farmers, too, but that is not to say that events of the day do not move the market for those jobs up and down.   The collapse of 1-2 of the top 5 firms in a concentrated market should certainly have some noticeable effects, and it would be naive to assume that only positive changes can result.

I believe that it isn not free market to make tax shelters illegal just because they "have no economic substance" people are using their brains to cut their taxes so why not?

Doctors are btw a high level profession. I dont believe that the number of accountants will really go down.


[quote="Rijul saini"] How many permutations of $ (1,2,\cdots,n)$, $ n \ge 3$, have exactly [b]two[/b] inversions?[/quote]
$ \frac{(n\minus{}2)(n\plus{}1)}{2}$
See http://oeis.org/A008302 and around.

[quote="Rijul saini"]Let $ \sigma \equal{} (a_1 , a_2 , \cdots , a_n)$ be permutation of $ (1 ,2 ,\cdots, n)$. A pair $ (a_i,a_j)$ is said to correspond to an inversion of $ \sigma$, if $ i < j$ but $ a_i > a_j$. How many permutations of $ (1,2,\cdots,n)$, $ n \ge 3$, have exactly [b]two[/b] inversions?

For example, In the permutation $ (2,,4,5,3,1)$, there are $ 6$ inversions corresponding to the pairs $ (2,1),(4,3),(4,1),(5,3),(5,1),(3,1)$.[/quote]

If call $ T(n,k)$ is the number of permutations of $ (1,2,\cdots,n)$, $ n > k$, have exactly $ k$ inversions?
We can prove that $ T(n,k) \equal{} T(n \minus{} 1,k) \plus{} T(n,k \minus{} 1)$

But we have $ T(n,0) \equal{} 1$ and $ T(n,1) \equal{} n \minus{} 1$

Exactly what i tried. I just wasn't sure of the answer.

[quote=Deviprasad]I did it by  a simple divide and conquer approach

We can achieve exactly two inversion by the following two processes
(1) We can flip any two consequtive numbers(the consequtive number sets must be disjoint) or
(2) we can take any three consequtive number subsets which are disjoint and perform any one of the two operations

   bring the first element to the second the second to the third and the third to the first
or
  take the first element to the last , the second to the frst and the last to the second position respectively.

These are the only possible ways to achieve two inversions(exactly)

Now we count the number of ways to do it

for the first case it is
(n-3)*(n-2)/2
and for the second it is
2(n-2)

so summing up the two we should get the result.[/quote]

How do we know these are the only ways in which our requirement is achieved?

[quote=tarzanjunior][quote=phymaths]I have proof a by Recurrence ..[/quote]
I too have 1. Here's it!!
Let $\sigma_n=(1,2,...,n)$
Let $a_n$ be the no. of permutations of $\sigma_n$ which have exactly $2$ inversions.
Let $b_n$ be the no. of permutations of $\sigma_n$ which have exactly $1$ inversion.
Here, we will look at the location of $1$, the minimal element.
If $1$ is in the first position in the permutation, it is equivalent to finding $a_{n-1}$
If $1$ is n the second position, one permutaion is $(a,1)$ where $a$ is in the first place. Note that if $a>3$, there would exist more than $2$ permutations.
If $a=2$, it is equivalent to finding $b_{n-2}$
If $a=3$, it forces the 3rd term to be $2$ and the other terms $(4,5,...,n)$
This gives $b_{n-2} +1 $ ways.
If $1$ is in the third position, it forces the permutation to be $(2,3,1,4,5,6,...,n)$
This gves only $1$ way.
$1$ cannot be in the third, 4th,.... position as this would mean more than $2$ inversions.
The above relations give $\boxed{ a_n=a_{n-1}+b_{n-2}+2}$
Now we calculate $b_n$.
If $1$ is in the first position, we get $b_{n-1}$ ways
If $1$ is in the second position, we only get one permutation, namely $(2,1,3,4,...,n)$
Hence $\boxed{b_n=b_{n-1}+1}$
Note that $b_3$ is $2$.
By induction, we can easily prove that $b_n=n-1$
Hence we get $ a_n=a_{n-1}+n-1$
We can easily check that $a_3=2$
Using the relation above, we get $\boxed{a_n= \frac{(n-2)(n+1)}{2}}$[/quote]

@tarzanjunior ( or anyone), why would we get $b_{n-2}+1$ ways in the 10th  and 11th line? Won’t it be just 1 way?

[quote=Rijul saini]Let $ \sigma = (a_1, a_2, \cdots , a_n)$ be permutation of $ (1, 2 ,\cdots, n)$. A pair $ (a_i, a_j)$ is said to correspond to an [b]inversion[/b] of $\sigma$ if $ i<j$ but $ a_i>a_j$. How many permutations of $ (1,2,\cdots,n)$, $ n \ge 3$, have exactly [b]two[/b] inversions?

For example, In the permutation $(2,4,5,3,1)$, there are 6 inversions corresponding to the pairs $ (2,1),(4,3),(4,1),(5,3),(5,1),(3,1)$.[/quote]

Lol we did this at Gujarat INMOTC this year.
The answer is $\binom{n-1}{2}+\binom{n-2}{1}=\frac{(n-2)(n+1)}{2}$
Proof-
There are 2 possibilities- either we do $a_k\leftrightarrow a_{k+1}$ and $a_m\leftrightarrow a_{m+1}$, or we do $a_{k}\leftrightarrow a_{k+2}$
The first one can be done in $\binom{n-1}{2}$ ways and the latter in $\binom{n-2}{1}$ ways. Hence the result

And btw,
[url=https://artofproblemsolving.com/community/c6h365838p2012330]Of Course![/url]


Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs. These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have differing radii r1 and r2. The discs are given angular velocities of magnitudes ω1 and ω2, respectively, and brought into contact at their edges. After the discs interact via friction, it is found that both discs come exactly to a halt. Which of the following must hold? Ignore effects associated with the vertical rods.
[img]http://img121.imageshack.us/img121/6263/81245.jpg[/img]
$ A. \omega_1^2 r_1 \equal{} \omega_2^2 r_2$
$ B. \omega_1 r_1 \equal{} \omega_2 r_2$
$ C. \omega_1 r_1^2 \equal{} \omega_2 r_2^2$
$ D. \omega_1 r_1^3 \equal{} \omega_2 r_2^3$
$ E. \omega_1 r_1^4 \equal{} \omega_2 r_2^4$

Let $ \rho$ be the mass density. Then $ m_1\equal{}\rho\pi h r_1^2$ and $ m_2\equal{}\rho\pi h  r_2^2$. Since $ I\equal{}\nu mr^2$, $ I_1\equal{}\nu \rho\pi h r_1^4$ and $ I_2\equal{}\nu \rho\pi h r_2^4$. Since $ \tau \equal{} I\alpha$, $ \alpha \equal{} \tau / I$. The frictional force $ f$ between them would be the same by Newton's third. This means that $ \alpha_1 \equal{} \dfrac{fr_1}{\nu \rho\pi h r_1^4}$. Let $ \minus{}\dfrac{f}{\nu \rho\pi h}\equal{}C$. $ C$ is positive in this situation because friction is negative in my sign convention for this problem. This constant $ C$ is the same for both of the spinning thingies. So now $ \alpha_1\equal{}C/r_1^3$ and $ \alpha_2\equal{}C/r_2^3$. This means that since $ t\equal{}\dfrac{\omega_f\minus{}\omega_i}{\alpha}$, $ t$ is the same for both to stop as they stop simulataneously, and since $ \omega_f$ for both is 0, lots of sign stuff happens and we get that $ t\equal{}\omega_1/\alpha_1 \equal{} \omega_2/\alpha_2$. Quickly re-arranging and plugging in, we see that $ t\equal{}\dfrac{\omega_1}{C/r_1^3}\equal{}\dfrac{\omega_2}{C/r_2^3}$ so by algebra $ r_1^3\omega \equal{} r_2^3 \omega$. Thus the answer is $ \boxed{D}$.

I know your solution is correct, but how come I can't say that momentum is conserved and that
\[ m_1 v_1 r_1 \equal{} m_2 v_2 r_2\]
?
Then we know $ m_1 \equal{} \left(\frac {r_1}{r_2}\right)^2 \cdot m_2$, so we have
\[ v_1 r_1^3 \equal{} v_2 r_2^3 \implies \omega_1 r_1^4 \equal{} \omega_2 r_2^4.\]
Can somebody point out the error in my reasoning? Thanks.

The problem with your solution is straightforward. As you know
the conservation of angular momentum can be applied when
there's no torque on the system. Since there's a torque caused
by friction  force on each disc, the conservation rule can no longer
be applied.

I thought that the torque was within the system, because it was caused by the discs. How do you differentiate between internal/external?

[quote="#H34N1"]Let $ \rho$ be the mass density. Then $ m_1 \equal{} \rho\pi h r_1^2$ and $ m_2 \equal{} \rho\pi h r_2^2$. Since $ I \equal{} \nu mr^2$, $ I_1 \equal{} \nu \rho\pi h r_1^4$ and $ I_2 \equal{} \nu \rho\pi h r_2^4$. Since $ \tau \equal{} I\alpha$, $ \alpha \equal{} \tau / I$. The frictional force $ f$ between them would be the same by Newton's third. This means that $ \alpha_1 \equal{} \dfrac{fr_1}{\nu \rho\pi h r_1^4}$. Let $ \minus{} \dfrac{f}{\nu \rho\pi h} \equal{} C$. $ C$ is positive in this situation because friction is negative in my sign convention for this problem. This constant $ C$ is the same for both of the spinning thingies. So now $ \alpha_1 \equal{} C/r_1^3$ and $ \alpha_2 \equal{} C/r_2^3$. This means that since $ t \equal{} \dfrac{\omega_f \minus{} \omega_i}{\alpha}$, $ t$ is the same for both to stop as they stop simulataneously, and since $ \omega_f$ for both is 0, lots of sign stuff happens and we get that $ t \equal{} \omega_1/\alpha_1 \equal{} \omega_2/\alpha_2$. Quickly re-arranging and plugging in, we see that $ t \equal{} \dfrac{\omega_1}{C/r_1^3} \equal{} \dfrac{\omega_2}{C/r_2^3}$ so by algebra $ r_1^3\omega \equal{} r_2^3 \omega$. Thus the answer is $ \boxed{D}$.[/quote]

Is this the only solution? Because I have never heard of mass density, I don't know where $ m_{1}\equal{}\rho\pi h r_{1}^{2}$ nor $ I\equal{}\nu mr^{2}$ came from. Or any of the other equations you have given.
I'm just wondering, because I mean, I may not be national physics olympiad material, but I've retained a lot from AP Physics and am able to solve many of the other problems, but have never seen any of this.

I'm pretty sure both mass density and moment of inertia is taught in AP Physics...
In fact, you would use mass density in the derivation of moment of inertia.

Ahh, $ I\equal{}\nu mr^{2}$ is moment of inertia? Ah, sorry, I didn't recognize it becuase of the $ \nu$. What is that anyways?
We've never derived moment of inertia...

The moments of inertia are usually multiples of $ mR^2$ or $ mL^2$. In case we do not know the moment of inertia for a specific object, just label it as $ \nu mR^2$ and hope that it cancels out in the end.

Ahh, I've figured it out. It's similar to the posted solution, it's just that I could not figure out any of the notation that #H34N1 used.

I have $ \tau \equal{} I\alpha \equal{}Fr$ so $ f\equal{}\frac{I\alpha}{r}$
Since frictional forces are equal, one can say that
$ \frac{I_1\alpha_1}{r_1}\equal{}\frac{I_2\alpha_2}{r_2}$
Moment of inertia is proportional to some constant, so $ I\equal{}kmr^2$. And mass is proportional to the volume (in this case, it's a cylinder, so it's proportional to the area), so it is $ m\equal{}jr^2$. Since $ \omega\equal{}\omega_0\plus{}\alpha t$, one can rearrange to say that $ \alpha\equal{}\frac{\omega}{t}$.

Combining all this, one gets
$ \frac{kjr_1^4\frac{\omega_1}{t}}{r_1}\equal{}\frac{kjr_2^4\frac{\omega_2}{t}}{r_2}$
The constants, $ k,j$ are equal, and $ t$ is equal as previously explained, so those cancel out leaving $ \boxed{D r_{1}^{3}\omega \equal{} r_{2}^{3}\omega}$, as needed.

It's been a while since I've done (and thought about) this problem, but I believe that energy would not work because you their rotational energies are not the same. Work = Fx would only be applicable to linear (translational) motion energy. For rotational energy, you want it to relate to the angular distance (the radians), which is not the same.

Sorry if this explanation is not satisfactory (or even correct), as I haven't done these types of problems in a long time!

[quote="buhbuhbuhbowser"]Hi, I understand the previous solution posted, but I'm wondering why using energy to solve the problem wouldn't work.

Since both disks are in contact, the linear distance X their surfaces slide across each other is the same for both, as is the force F between them. Thus the work done on each would be -FX, and their initial energies are the same. With some algebra, the answer would be C[/quote]

Actually, I'm going to clear up my answer a little.
You assumed that the linear distance x (at the circumference) is the same, but this is not true. The force is the same on both. However, since torque ($\tau$) = moment of inertia ($I$) * rotational acceleration ($\alpha$), and torque ($\tau$) = force ($F$) * radius ($r$), you can say that force is proportional to $rm\alpha$. Since mass is clearly not equal, then the linear acceleration is not going to be equal. Thus, they do not travel equal distances at the circumference.

[quote="ahaanomegas"]If I am not mistaken, the only thing that is conserved here is angular momentum. Is that correct?[/quote]
Angular momentum is NOT conserved. This should answer your question:
[quote="#H34N1"]The post two posts above doesn't answer your question. There is a friction force which acts in the opposite direction of translational velocity of both of these discs. The equal and opposte force as well as this friction force would cause the discs to fly apart. In order to hold the discs together there needs to be an external force on the system. This creates an external torque regardless of where you choose the point about which you take the angular momentum. This means that ang. momentum isn't conserved.[/quote]

[quote="Bassir"]What is the linear analogue to this type of question?[/quote]
I'm not sure if there's a linear analogue... it would have to be two flat objects in space hitting each other and sliding, eventually coming to a stop.

If you consider the points on the cylinder that are at the point of contact, each point is moving away from each other. The relative velocity of the two points should be the same.
So I think there is no problem with that reasoning.

I think the problem can be thought of this way: Take a wheel that's spinning very quickly, and touch the rim to the rim of another stationary wheel. From the perspective of the stationary wheel, the spinning wheel exerts some torque T for some time t, the angular acceleration is a product of T and t times some constant. No matter how fast the moving wheel is spinning, it never exerts more than a certain torque on the stationary wheel. If, however, one considers it from an energy perspective, the energy dissipated on the moving wheel is greater per unit time if the wheel moves faster. This must be because the moving wheel slips and generates heat, making invalid the conservation of energy

i hope i'm right about why i was wrong =D

Thanks pascal12. So, if there was no friction, angular momentum (L) would be conserved?

Correct, since if there was no friction, there would be no external forces on the spinning discs.


There are 10 types of postcards in a post office.  How many ways are there to buy 6 postcards?

This is just

$ \dbinom{10}{6}=\dbinom{10}{4}=\frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1}=\frac{10 \cdot 7}{1} \cdot \frac{8}{4 \cdot 2} \cdot \frac{9}{3}=70 \cdot 1 \cdot 3=210$.

[quote="BOGTRO"]This is just

$ \dbinom{10}{6} = \dbinom{10}{4} = \frac {10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} = \frac {10 \cdot 7}{1} \cdot \frac {8}{4 \cdot 2} \cdot \frac {9}{3} = 70 \cdot 1 \cdot 3 = 210$.[/quote]

But you can have 6 postcards of <insert postcard title>, so there is more.  I was wondering whether it is $ \binom{15}{5}$ or $ \binom{15}{6/9}$..("/" implies either)

Sorry, but I'm not going to post all of my cases out again- i did all of them, and then my computer failed :mad: . Anyway, I worked out everything with casework, and counted the ways to get each one using a mix of combinations and permutations. The answer is 5005, 15C6. How did you come up with that quick and easy answer, math007?

Oops?

So we use balls and urns. We have 10 "urns"

and 5 "balls". So $ \binom{15}{5}$ should be right. I'm not sure about this though, I'm not good wiht balls and urns. I'll talk more about this tomorrow, I have to go now.

Where do you get the 15 in your solution?

10 balls and 5 dividers. 10+5=15 objects, and we are choosing 5.

Hmm.... When I do out all the cases (I had to because I'm terrible at balls and urns application), I keep getting 5005, which happens to be 15C6. I'm pretty certain of this. You probably have the right idea, but there might be a slight complication with the problem somewhere.


what is the Pinacol-pinacolone Rearrangemnet product of the compound 1-phenyl 2 Methyl propane 1,2 diol?

Here's a useful link:

http://en.wikipedia.org/wiki/Pinacol_rearrangement

P.S. I think only one of the hydroxyl group can be protonated (due to the need for an alkyl shift).

i think hydride shift takes place, not methyl shift as the order of migration is H>aryl>alkly. also the tertiary carbocation formed is more stable than the benzylic carbocation.

but the book(Girijesh dubey) takes the Benzyllic one  to be more stable

Also the rearrangement order has aryl at first

thats bcoz hes indian. ive referred boyd and other books(paula bruice for eg) which give the order tertiary>benzylic. but if the benzylic carbocation is also tertiary. then i guess the order given in your book is rite. also migration ability of h>aryl if aryl group does not have any elecrton releasing substituents

yes i know t-Butyl>Ar-C but what will u write in an examination hall set by an indian

[quote="Carcul"]Begin by noting that the products of this rearrangement also sometimes depend on reaction conditions, and several products are often obtained. However, in this case I would vote on 3-phenylbutanone as the main product.[/quote]

Can you elaborate on that?

Thanks.

P.S. What influence the order of migration?

Elaborate on what? Can you be more specific?

I am not sure now :P . Maybe just the order of migration (what influence it and etc.).


We call a number $ x\in [0,1]$ satanic if the decimal expansion of $ x$ contains somewhere the sequence $ 666$; in other words, if $ x\equal{}\sum_1^{\infty}a_k10^{\minus{}k}$, where $ a_k\in\{0,1,...,9\}$ and there exists $ n$ such that $ a_n\equal{}a_{n\plus{}1}\equal{}a_{n\plus{}2}\equal{}6.$ Show that almost all numbers in $ [0,1]$ are satanic, i.e. that $ m([0,1]\minus{}S)\equal{}0$, where $ S$ is the set of satanic numbers and $ m$ is the measure of the set.

Divide $ [0,1]$ into intervals of form $ [k/1000^{n},(k\plus{}1)/1000^n)$. Count those that do not consist entirely of satanic numbers. There are at most $ 999^n$ such intervals because we have $ 999$ choices for the first triple of digits, $ 999$ for the next triple of digits, and so on. Thus, the measure of the set of non-satanic numbers is at most $ 0.999^n$. But $ n$ is arbitrary.


inequalities
trigonometry
calculus
derivative
geometry
geometric transformation
reflection

How many inequality proposed by Jack Garfunkel ?

I think is ''a lot of'' :D But I don't know ''how many''.

I hope we can list some of one's inequality. I heared that almost of them is very nice and difficult. Specialy it has a beatiful form!

The most beautiful of his (I think)
\[\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\le \frac{5}{4}\sqrt{a+b+c}.\]
He has "a lot of" trig inequalities (with $sin$ and $cos$) but I don't like this kind because it is almost easy. He has very fewer algebraic inequalities.

Let $a+b+c=S$ if we denote $x=\frac{a}{S}$ ,$y=\frac{b}{S}$ and $z=\frac{c}{S}$ it's enough to prove that \[\sum \frac{x}{\sqrt{x+y}}\leq \frac{5}{4}\]
with $x+y+z=1$

Hmm, not easy! Please post your solution! I know a solution in Crux, it is very long!

I didn't reach a solution.It was just a start.

[b]A rezult[/b]: If triangle $ABC$ has at least one angle $\ge \frac{\pi}{3}$ then: $h_{a}+h_{b}+h_{c}\ge 7r+R$

Continue: in triangle $ABC$, we have: $\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\ge\frac{2}{\sqrt{3}}(\sin A+\sin B+\sin C)$

The last post in day:
$\left(\frac{\cos A}{\cos\frac{A}{2}}+\frac{\cos B}{\cos\frac{B}{2}}+\frac{\cos C}{\cos\frac{C}{2}}\right)^{2}\ge 1$

[quote="N.T.TUAN"]Can you post prove of this assertion?[/quote]

I won't because my solution using derivative is almost the same as Harazi's in the link (famous)

http://www.mathlinks.ro/Forum/viewtopic.php?t=1459

There is also another solution of Hungkhtn in Mathematic Reflection (issue in 2006) using entirely mixing variable but I'm to lazy to search it. I think you know or can find this issue.

[b]Problem 6[/b]. Let $x,y,z,t\in\mathbb{R}$ such that $x+y+z+t=2\pi$ then $\sin (x+y)+\sin (y+z)+\sin (z+t)+\sin (z+x)\ge sin 2x+\sin 2y+\sin 2z+\sin 2t$

[quote="GacKiem"]

There is also another solution of Hungkhtn in Mathematic Reflection (issue in 2006) using entirely mixing variable but I'm to lazy to search it. I think you know or can find this issue.[/quote]

It's here:

[b]Problem 7[/b]:
$\sin A\sin B+\sin B\sin C+\sin C\sin A\le \frac{7}{4}+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

You can find many inequalities of Jack Garfunkel in "The biggest collection of inequalities..." e-book. :lol:

Just asking - Jack Garfunkel was from Forest Hills and so was Art Garfunkel and Art Garfunkel studied Mathematics in college. Also Art's dad's name was Jack. Were they the same person?


Isn't it the definition of the Kosnita point? Or how do you define it? :maybe:

In 2003, it was proven by Grinberg (I believe Darij) that the Kosnita point is the isogonal conjugate of the nine-point centre.

[quote="Sofeto_89"]Please... Can somebody give the proof of the statement "Kosnita point and the nine-point center are issogonal conjugate". Please i've been looking everywhere but i haven't seen it untill now. 10q:)[/quote]
And perhaps you want to know what happends if fermat point
[url]http://mathworld.wolfram.com/FermatPoints.html[/url]
 is the same as Kosnita point  :wink:

[quote="JeriC"]See
http://forumgeom.fau.edu/FG2003volume3/FG200311index.html
Yes, it was proved by Darij Grinberg :)[/quote]
Sorry, but I can't find Darij's proof in the Forum Geometricorum 3. Can you show me his proof, please ? Here's my proof:
Let $ f$ is the inversion has $ A$ to pole and $ AB.AC$ to constant. So the inversive image of $ (ABC)$ is the line $ d$ passes through $ f(B)$ and $ f( C)$, and the inversive image of $ O$ ( the circumcenter of $ \triangle ABC$ ) is the symmetric point of $ A$ wrt $ d$. We have $ A,X$ ( the circumcenter of $ \triangle BOC$ ),and $ S$-the center of the circle passes through $ f(B),f( C),f(O)$ are collinear.
Now, note that $ f(B), f(C)$ are the symmetric points of $ C,B$ wrt to $ At$ - the bisectors of angle $ \angle BAC$, so $ f(O)$ is the symmetric point of $ A'$ wrt to $ At$ ($ A'$ is the symmetric point of $ A$ wrt to $ BC$ ).  Let $ O_{1}$ is the symmetric point of $ O$ wrt to $ BC$, we have $ O_{1}$ is the symmetric point of $ S$ wrt to $ At$. It means that $ AO_{1}$ is the isogonal line of $ AX$ wrt the angle $ \angle BAC$. Easy to see that the nine-point center is the midpoint of $ AO_{1}$. So we have Kosnita point and the nine-point center are isogonal conjugate.

Dear Mathlinkers,

see also
http://jl.ayme.pagesperso-orange.fr/Docs/Le%20point%20de%20Kosnitza.pdf

Sincerely
Jean-Louis

A useful property of isogonal lines is this: If $PB=PC$ and $QB=QC$ then $AP$ and $AQ$ are isogonal iff $P$ and $Q$ invert to each other wrt the circumcircle.

For the 9-pt-circle, the intersection of the cevian with the perpendicular bisector is the reflection of $O$ over $BC$ (well-known). Hence, the intersection of its isogonal cevian with the perpendicular bisector is the circumcenter of $OBC$ which is what I believe the Kosnita point is defined as.


How many real numbers $x$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$?

[hide="Solution"]
The range of $\sin (x)$ is $[-1,1]$. Hence, we only need to consider $\left|\frac{1}{5}\log_2 x\right| \le 1$. This is satisfied for $\frac{1}{32} \le x \le 32$. First let's consider $\frac{1}{32} \le x < 1$. In this interval, the left hand side is negative while the right hand side is negative only in $[1/5,2/5]$ and $[3/5,4/5]$, so there are 4 solutions.

When $1< x \le 32$, the left hand side is positive, and the right hand side is positive only in $[6/5,7/5]$, $[8/5,9/5]$, ..., and $[158/5,169/5]$. There are then 2 points of intersection in each of these intervals, and there are 77 intervals.

When $x=1$ both sides of the equation are 0, so in all we have $4 + 77 \cdot 2 + 1 = \boxed {159}$ solutions.
[/hide]

[hide="Solution"]1. On interval $(0,1)$ function $y=5\sin(5\pi x)$ has two negative half-periods, on $(1/5,2/5)$ and $(3/5, 4/5)$, reaching value $-5$ in $x_1=3/10$ and $x_2=7/10$. Since $\log_2(3/10)>-5$ and $\log_2(7/10)>-5$, the sinusoid will intersect the logarithm curve in four points (two for each half-period).

2. In the point $x=1$ the equation is satisfied, since $\log_21=5\sin(5\pi\cdot 1)$

3. On interval $(1,32)$ function $y=5\sin(5\pi x)$ has $77$ positive half-periods — $5$ on each interval $(2n-1, 2n+1)$, plus $2$ on $(31,32)$ — reaching value $5$ in the midpoints of these half-periods. Since $\log_2 x\leq 5$ for $1\leq x\leq 32$, it follows that the logarithmic curve will intersect the sinusiod in $2$ points for each positive half-period, giving $154$ points.

In total, we have $4+1+154=159$ points.[/hide]

[hide]
$y=\frac{1}{5}\log_2 x (a)$ - $(A)$ = $\sin (5\pi x)$ - $(B)$ can only be true when $0<y=<1$.
$\frac{1}{5}\log_2 x=1\Rightarrow x=32$.  $B$ has a period of $\frac{2\pi}{5\pi}=\frac{2}{5}$. Therefore the graph $A$ passes through $B$ $\frac{32}{2/5}=80$ times.
For each period, $A$ passes through $B$ 2 times except first one.
So therefore there are $80*2-1=\boxed{159}$ such points where $A$ and $B$ intersect.. [/hide]

[quote=HumanMathematician]you just had to bump a post THAT ancient?!
can u don't[/quote]
they're just posting a solution, meanwhile you're being of no help at all

also you bump old posts too


Find
$ \sin^220^{\circ}\plus{}\cos^280^ {\circ}\plus{}\sqrt 3\sin20^{\circ}\cos80^ {\circ}$

[hide="ans."]

$ \frac 1 4$

[/hide]

Another arithmetic manipulation.
But, it is a trifling difference.

\begin{eqnarray*} \sin^2 20^\circ + \cos^2 80^\circ + \sqrt {3} \sin 20^\circ \cos 80^\circ & = & - \frac {1}{2}(\cos 40^\circ - 1) + \frac {1}{2}(\cos 160^\circ + 1) + \frac {\sqrt {3}}{2} (\sin 100^\circ - \sin 60^\circ) \\
& = & - \frac {1}{2}(\cos 40^\circ + \cos 20^\circ) + 1 + \frac {\sqrt {3}}{2} \sin 80^\circ - \frac {3}{4} \\
& = & - \cos 30^\circ \cos 10^\circ + \frac {\sqrt {3}}{2} \cos 10^\circ + \frac {1}{4} \\
& = & - \frac {\sqrt {3}}{2} \cos 10^\circ + \frac {\sqrt {3}}{2} \cos 10^\circ + \frac {1}{4} \\
& = & \boxed{\frac {1}{4}} \end{eqnarray*}


trigonometry
integration
logarithms
calculus
calculus computations

Compute the following sum $ \sum_{n\equal{}\minus{}\infty}^{\plus{}\infty} \frac{\sin(\frac{n\pi}{4})}{2n\pi} \cdot \frac{\sin(\frac{n\pi}{6})}{5n\pi}$

It suffices to calculate the sum where $ n$ runs over the set of positive integers, which can be computed as follows:

\begin{eqnarray*} \sum_{n = 1}^{\infty} \frac {\sin(\frac {n\pi}{4})}{2n\pi} \cdot \frac {\sin(\frac {n\pi}{6})}{5n\pi} & = & \frac {1}{20\pi^2} \sum_{n = 1}^{\infty} \frac {1}{n^2} \left[ \cos \left( \tfrac{n\pi}{12} \right) - \cos \left( \tfrac{5n\pi}{12} \right) \right] \\
& = & \frac {1}{20\pi^2} \sum_{n = 1}^{\infty} \frac {1}{n} \int_{\frac {n\pi}{12}}^{\frac {5n\pi}{12}} \sin n\theta \, d\theta \\
& = & \frac {1}{20\pi^2} \int_{\frac {n\pi}{12}}^{\frac {5n\pi}{12}} \text{Im}\left[ - \log ( 1 - e^{i\theta} ) \right] \, d\theta \\
& = & \frac {1}{20\pi^2} \int_{\frac {n\pi}{12}}^{\frac {5n\pi}{12}} \frac {\pi - \theta}{2} \, d\theta \\
& = & \frac {1}{160} \end{eqnarray*}

Here, interchanging the order of infinite sum and integral is justified by uniform convergence, which follows from Dirichlet's test in this case.


it's not true if $ a,b$ and $ c$ are negatives , for $ a,b$ and $ c$ positives it's easy from $ x^2 \plus{} 1 \ge 2x$ for $ x \equal{} a,b,c$.and the rest is nesbitt's inequality.

[quote="tchebytchev"][b]it's not true if $ a,b$ and $ c$ are negatives [/b], for $ a,b$ and $ c$ positives it's easy from $ x^2 \plus{} 1 \ge 2x$ for $ x \equal{} a,b,c$.and the rest is nesbitt's inequality.[/quote]

Thats the part that was kind of hard to cover...rest id got it down to the nesbitt part.
Still no way to cover a,b,c <0 ?

it's obviously false in this case , if $ a,b,c <0$ the quantity $ (a^2\plus{}1)/(b\plus{}c) \plus{} (b^2\plus{}1)/(c\plus{}a) \plus{} (c^2\plus{}1)/(a\plus{}b)$ is negative.

It is obviously true for $ a,b,c$ positive. Can u state what are $ a,b,c$?

a^2+1>=2a

so,

sum of (a^2+1)/b+c>= sum of   2a/(b+c)

2a/(b+c)+2b/(a+c)+2c/(a+b)>3

which is obviusly true from harmonic,arithmetic inequality

[quote="Zed_darkx"]Prove

(a^2+1)/(b+c) + (b^2+1)/(c+a) + (c^2+1)/(a+b) >= 3[/quote]

well it's not false for all negative a,b,c. let a get arbitrarily large with b positive and c negative. on the other hand, if b and c are both negative, that makes the inequality false

i'm guessing that the original stipulations were just $ a,b,c>0$


limit
induction
linear algebra
linear algebra unsolved

easy one:
$E$ a vectorial space (infinite dimension). Prove there is no continuous endomorphisms $u$ and $v$ such that $uv-vu=I$.
(try in the finite dimensional case first  :rotfl: )

I was leafing through a really cool Romanian analysis book, and I saw three problems in a row leading to a proof of this (although the authors didn't mention they were trying to give hints for this problem :)).

One can prove a more general result: $(*)$ given a normed algebra $\mathcal A$ with unit, no two elements $a,b$ satisfy $ab-ba=1$. We can apply $(*)$ to our problem by taking the normed algebra to be that of the bounded linear operators on $V$.

[i]Lemma 1[/i] Let $D$ be a derivation on $\mathcal A$, i.e. a continuous linear map on $A$ with the property that $D(xy)=xDy+yDx$, and assume $b\in\mathcal A$ satisfies $D^2b=0$. Then $D^n(b^n)=n!(Db)^n$.

This is a simple application of the multinomial formula $D^n(a_1\cdot\ldots\cdot a_k)=\sum_{n_1+\ldots+n_k=n}C_{n_i}D^{n_1}a_1\cdot\ldots\cdot D^{n_k}a_k$, where $C_{n_i}$ is the multinomial coefficient $\frac{n!}{n_1!\cdot\ldots\cdot n_k!}$, to $k=n$ and $a_i=b,\ i=\overline{1,k}$.

[i]Lemma 2[/i] If $a,b\in\mathcal A$ satisfy $a(ab-ba)=(ab-ba)a$, then $r(ab-ba)=0$, where $r(\cdot)$ denotes the spectral radius.

Consider the derivation $D$ defined by $D(x)=ax-xa$. The hypothesis tells us that $D^2b=0$, so by Lemma 1, $(Db)^n=\frac 1{n!}D^n(b^n)$. This means that $r(ab-ba)=r(Db)=\lim_{n\to\infty}\left(\|(Db)^n\|^{\frac 1n}=\frac 1{\sqrt[n]{n!}}\|D^n(b^n)\|^{\frac 1n}\right)$. Now, $\frac 1{\sqrt[n]{n!}}\to 0$, while $\|D^n(b^n)\|^{\frac 1n}$ is bounded from above by $\|D\|\cdot\|b\|$, and this gives the desired result.

Now we can finally prove that $ab-ba\ne 1$. Assuming the contrary, we would have $a(ab-ba)=(ab-ba)a$, so, by Lemma 2, we would have to have $r(1)=r(ab-ba)=0$, which is a contradiction.

I hope it's Ok.

I'll try to write a solution non-using notions of Banach algebra:
Firstly, as far as i understand $E$ was supposed to be a normed linear space.
Then it is not diffiuclt ot get(for example by induction) that $u^{n}v-vu^{n}=nu^{n-1}(*)$. Now we just need to pass to norm in $(*)$, where the norm is the usual operator norm in $\mathbb{L}(E)$  $\parallel u\parallel =sup_{x\neq 0}\frac{\parallel ux\parallel }{\parallel x\parallel }$.It is easy to get that $\parallel uv\parallel \leq \parallel u\parallel \parallel v\parallel $ for any $u,v \in \mathbb{L}(E)$. Then if we take norm in $(*)$ we get $n\parallel u^{n-1}\parallel \leq 2\parallel u^{n}\parallel \parallel v\parallel \leq 2\parallel u^{n-1}\parallel \parallel u\parallel \parallel v\parallel $ which implies that $n \leq 2\parallel u\parallel \parallel v\parallel $ which is surely impossible to take place for any $n \in \mathbb{N}$.
Is it right?

just to complete eugene solution , you must assume that u isn't nilpotent , but if u was so, you can show by induction that u is the nul map wish gave a contradiction too.


algebra
polynomial
combinatorial inequality
binomial coefficients
Inequality
IMO
IMO 1985

For any polynomial $P(x)=a_0+a_1x+\ldots+a_kx^k$ with integer coefficients, the number of odd coefficients is denoted by $o(P)$. For $i-0,1,2,\ldots$ let $Q_i(x)=(1+x)^i$. Prove that if $i_1,i_2,\ldots,i_n$ are integers satisfying $0\le i_1<i_2<\ldots<i_n$, then: \[ o(Q_{i_1}+Q_{i_2}+\ldots+Q_{i_n})\ge o(Q_{i_1}).  \]

The coefficient of $x^m$ in $(1+x)^n$ is odd iff al the $1$'s in the binary representation of $m$ appear in the binary representation of $n$ as well. If to each positive integer $p$ we associate a finite subset $S(p)$ of $\mathbb N$ whose characteristic vector is the binary representation of $p$ (and we make $S(0)=\emptyset$), then that condition is written as $S(n)\supseteq S(m)$.

The only thing for which we need $i_1<i_j,\ \forall j>1$ is that we can conclude $\forall j>1,\ A(i_1)\not\supseteq A(i_j)$. Let $A_j=A(i_j),\ \forall j\in\overline{1,n}$. For a system of finite sets $M_1,\ldots,M_k$ we denote by $o(M_1,\ldots,M_k)$ the number of subsets of $\bigcup_{i=1}^kM_i$ which are contained in an odd number of sets $M_i$. The problem is now this:

$(*)$ Given finite sets $A_1,\ldots,A_n$ s.t. $\forall i>1,\ A_1\not\supseteq A_i$, we have $o(A_1,\ldots,A_n)\ge o(A_1)$.

Suppose we prove $(*)$ in the particular case $A_1=\emptyset$ (we call this restricted statement $(**)$). Then we can extend our proof to one for the general case as follows:

For any subset $M$ suppose $A_1,\ldots,A_p$ are the $A_i$’s which contain $M$. Then we apply $(**)$ to the system $\emptyset=\bar A_1,\bar A_2,\ldots,\bar A_p$, where $\bar A_j=A_j\setminus A_1$. The conclusion is that we can find $S_M\subset\bigcup\bar A_j$ contained in an odd number of $\bar A_j$’s. The collection consisting of all the sets of the form $S_M\cup M,\ M\subseteq A_1$ has at least $2^{|A_1|}=o(A_1)$ elements, all contained in an odd number of $A_i$’s, and we’re done.

All that’s left now is to prove $(**)$, i.e. that for $A_1=\emptyset$ we have $o(A_1,\ldots,A_n)\ge 1$. We start with the maximal elements of the multiset $\{A_i\}$. If every maximal element (wrt inclusion) appears an even number of times, we take all of them out, and continue with the maximal elements in the multiset we’re left with. We repeat this procedure, shrinking the multiset we’re working with each time, until we first meet a maximal element (in the respective truncated multiset) which appears an odd number of times, and choose that one. We must reach a stage when there is a maximal element appearing an odd number of times, because in the worst case scenario we take all the $A_i$’s out except for $A_1=\emptyset$, and because of the condition $\forall i>1,\ A_1\not\supseteq A_i,\ \emptyset=A_1$ appears just once (and thus an odd number of times) in each of our multisets.

Let $D(n)$ denote te positions in which $n$ has digit $1$'s in base $2$. So $n = \sum_{x \in D(n)} 2^x$, for example $D(4)=\{2\}$, $D(5)=\{2,0\}$. Notice that the coefficient of $x^k$ in $Q_n$ is odd iff $D(k) \subseteq D(n)$. I will create an injection $f$ from the $k$'s that satisfy $D(k) \subseteq D(i_1)$ to the $k$'s that satisfy $D(k) \subseteq D(i_j)$ for an odd number of $j$'s. This will end the problem.

First of all, $f(k)$ will satisfy that $D(f(k)) \cap D(i_1) = D(k)$ (that is, $f(k)$ will be $k$ plus some other digits $1$ that are not in $i_1$). This will guarantee injectivity, since if $f(k_1)=f(k_2)$ then $D(k_1) = D(f(k_1)) \cap D(i_1) = D(f(k_2)) \cap D(i_1) = D(k_2)$ so $k_1=k_2$.

Now we must prove we can construct such an $f$ such that $D(f(k)) \subseteq D(i_j)$ for an odd number of $j$'s. To do this, take $k$. Let $1 \textless j_1, ..., j_m$ be such that $D(k) \subseteq D(i_{j_t})$ for all $ 1 \le t \le m$.

If $m$ is even, then $D(f(k)) \subseteq D(i_r)$ for an odd number of $r$'s, so let $f(k)=k$, which satisfies $D(f(k)) \cap D(i_1) = D(k)$ since $D(k) \subseteq D(i_1)$.

If $m$ is odd, this is where the difficulty of the problem lies :P Let $D$ be the set of positions in binary that are $0$ in $i_1$, but that are $1$ in $i_r$ for some $r$. (This second restriction is only included in order to make $D$ finite). Since $i_1$ is the least of all $i$'s, then $i_{j_t}>i_1$, and so $i_{j_t}$ must contain a digit $1$ in a position where $i_1$ doesn't (otherwise we would have $i_{j_t} \le i_1$). For each $1 \le t \le m$ let $X_t$ be the set of nonempty sets $Y$ that satisfy $Y \subseteq D \cap D(i_{j_t})$. Notice that $D \cap D(i_{j_t}) \ge 1$ since $i_{j_t}$ has digits $1$ that are $0$ in $i_1$. Since $Y$ cannot be empty, we have that $|X_t|$ must equal a power of $2$ minus 1, and therefore $X_t$ is odd for all $1 \le t \le m$.

But recall that $m$ is odd. Therefore the sum of all $|X_t|$ is odd. But the sum of all $|X_t|$ equals $\sum_{Y \subseteq D} g(Y)$, where $g(Y)$ is the number of $t$'s such that $Y \in X_t$. Therefore, there must exist a nonempty subset $Y$ of $D$ such that $g(Y)$ is odd. We will have $f(k)$ be such that $D(f(k))=D(k) \cup Y$. Clearly $D(f(k)) \cap D(i_1) = D(k)$.

Now we must prove $D(f(k)) \subseteq D(i_r)$ for an odd number of $r$'s. Notice that $D(f(k)) \subseteq D(i_r)$ is only possible if $r=j_t$ for some $t$, or $r=1$. Since $Y$ is a nonempty set of positions in binary which are $0$ in $i_1$, we have that $r=1$ does not work. Therefore, the number of $r$'s for which $D(f(k)) \subseteq D(i_r)$ is equal to $g(Y)$ (by definition), which is odd!

Therefore we have constructed $f$ successfully! Huzzah! We are done.


geometry
geometric transformation
combinatorics unsolved
combinatorics

All squares of a $ 20\times 20$ table are empty. Misha* and Sasha** in turn put chips in free squares (Misha* begins). The player after whose move there are four chips on the intersection of two rows and two columns wins. Which of the players has a winning strategy?

[i]Proposed by A. Golovanov[/i]

[b]US Name Conversions: [/b]

[i]Misha*: Naoki
Sasha**: Richard[/i]

Naoki and Richard, huh? Is this a thing of translation or did the author make a tribute to artofproblemsolving.com?

Richard wins. Partition the columns into pairs. Whenever Naoki plays, play in the same row and the paired column. This forbids any more moves in either of the paired columns, even if the row played on is not empty. So after 20 moves by both players, Naoki has no more moves left without giving Richard a win.

Before I realized I misread it, I was trying to solve the problem where making the configuration of four chips was a loss and not a win. This is much harder. :|

[quote="MellowMelon"]Naoki and Richard, huh? Is this a thing of translation or did the author make a tribute to artofproblemsolving.com?
[/quote]
   They were Misha and Sasha. Or, if these names are too difficult, Michael and Alexander (as these two people are called in official documents). I know both, they will soon appear and restore their identity by any means however violent.

[quote="MellowMelon"]Before I realized I misread it, I was trying to solve the problem where making the configuration of four chips was a loss and not a win. This is much harder. :|[/quote]I misread it too. I was really confused when I read your solution. :rotfl:


The ninth and tenth terms of an arithmetic sequence are $ \frac14$ and $ \frac13$, respectively. What is the second term?

Since it is arithmetic sequence, that means that each of the terms differ by some x. We find x by subtracting the two fractions since they are consecutive.

$ \frac {1}3 \minus{} \frac {1}4 \equal{} \frac {4}{12} \minus{} \frac {3}{12} \equal{} \frac {1}{12}$

So, to get the ninth term, we must start with some base term a and add x eight times. So,

$ a \plus{} 8x \equal{} \frac {1}4$

$ a \plus{} 8*\frac {1}{12} \equal{} \frac {3}{12}$

$ a \equal{} \minus{}\frac {5}{12}$

So, the second term is $ a \plus{} x$ which is $ \minus{}\frac { 4}{12} \equal{} \minus{}\frac {1}{3}$


Please calculate sum: $ \sum_{k\equal{}1}^{n}2^{k^2}$.

it can't be calculated, only approximation is available.

No, there is no neat proof (and, frankly speaking, I'm not sure is there is any proof: there is a huge theory about absence of elementary antiderivatives but the difference equations are harder and I haven't seen much about them). There are plenty of indirect indications that the explicit formula does not exist (for instance, it is possible to show that $ \sum_{n\equal{}0}^\infty z^{n^2}$ is not an elementary function of $ z$) but I do not see any simple reduction here. :(


topology
function
real analysis
real analysis unsolved

Why is the metric space $ \frac {d(x,A)}{d(x,A) \plus{} d(x,B)}$ important? Here $ A$ and $ B$ are $ \in\mathbb{R}$. I always see it in books in topology and analysis...I think this is somehow useful in dynamical systems of some sort? I would be glad just to know a few applications of it!

Thanks in advance,

Los

We can take a metric space $ M$ and $ A,B\subset M$, not just $ A,B\in \mathbb{R}$.

This function is continuous and gets value $ 0$ if $ x\in A$, $ 1$ if $ x\in B$. So a continuous function can tell if we are not in a particular set.

is it particularly important in functional analysis, for example? It seems I'll have to do more reading  :roll:


vector
pigeonhole principle
superior algebra
superior algebra unsolved

If $ V$ is a vector space over an infinite ground field . Then $ V$ is not the union of a finite number
of subspaces $ V_i \subset  V$, $ V_i \neq V$.
How to prove this?
I am thinking of proving by contradiction...
thanks!

Proof by contradiction seems like a good start.  Suppose that we have such a union.

We can assume that for any $ i$, there is some $ v_i\in V_i$ such that $ v_i\notin V_j$ for $ i\neq j$. (if any $ V_i$ does not satisfy this condition, we can throw it out)

Pick any $ i\neq j$, and consider the infinite set $ \{v_i \plus{} \alpha v_j | \alpha \in F\}$.  By assumption, every member of this set lies in some $ V_k$, so by pigeonhole, two members lie in the same $ V_k$, say $ v_i \plus{} \beta v_j, v_i \plus{} \gamma v_j \in V_k$.  But then $ (v_i \plus{} \beta v_j) \minus{} (v_i \plus{} \gamma v_j) \equal{} (\beta \minus{} \gamma)v_j \in V_k$, so $ v_j \in V_k$, which implies $ k\equal{}j$.  But then $ v_i \plus{} \beta v_j\in V_j$, so $ v_i \in V_j$, contradiction.


function
integration
trigonometry
calculus
calculus computations

For real valued continuous function $ f(t)$ on $ \mathbb{R}$, define the functional sequence $ u_n(t) ,\ n\equal{}1,\ 2,\ \cdots,$ as

$ \left\{
\begin{array}{ll}
u_1(t)\equal{}f(t) &\quad  \\
u_{n\plus{}1}(t)\equal{}\int_0^t \sin (t\minus{}s)u_n(s)\ ds &\quad
\end{array}
\right.$ for $ t\in\mathbb{R}$.

(1) Prove that $ u(t)\equal{}\sum_{n\equal{}1}^{\infty} u_n(t)$ uniformly converges for $ t\in{[\minus{}T,\ T]}$ with $ \forall{T}>0$.

(2) Express $ u(t)$ by $ f(t)$.

(3) Find $ f(t)$ such that $ u(t)$ become the constant function $ 1$.

May I do problem number $ 3$ ?
[hide]
$ u(t)\equal{}u_1(t)\plus{}u_2(t)\plus{}u_3(t)\plus{}\cdots$
$ \int_0^t \sin (t \minus{} s)u(t)\ ds\equal{}\int_0^t \sin (t \minus{} s)u_1(t)\ ds \plus{} \int_0^t \sin (t \minus{} s)u_2(t)\ ds \plus{} \int_0^t \sin (t \minus{} s)u_3(t)\ ds\plus{}\cdots \equal{} u_2(t)\plus{}u_3(t)\plus{}u_4(t)\plus{}\cdots$
Then :
$ u(t)\minus{}\int_0^t \sin (t \minus{} s)u(t)\ ds\equal{}u_1(t)\equal{}f(t)$
If $ u(t)\equal{}1$ is a constant function, then :
$ 1\minus{}\int_0^t \sin (t \minus{} s)\ ds\equal{}f(t)$
$ 1\minus{}(cos(0)\minus{}cos(\minus{}t))\equal{}f(t)$
$ cos(\minus{}t)\equal{}cos(t)\equal{}f(t)$.
[/hide]
Sorry if I am wrong.

Here's an approach for 1):
I was looking at the interval $ [0,T]$ for some $ T > 0$(for simplicity). If the fourier series of $ f(t) \minus{} f(0)$ converges uniformly, then, expanding it as $ \sum a_k \sin\frac {\pi k t}{T}$ and subsituting into the expression for $ u_1$, we roughly get that the coefficients for $ u_1$ behave as $ a_k\frac {1}{k^2}$. Using this recursive relation, we can directly bound the coefficients of $ u(t)$(basically, i want $ f$ to be smooth enough to allow the ablsolute convergence of coefficients to interchange the order of summation when needed).
For arbitrary $ f$, just start with, say, $ u_2$, which is already smooth enough for our purposes.


MATHCOUNTS
geometry
trapezoid
area of a triangle

for trapezoid ABCD, AB is parallel to CD, AB=8 cm and CD=20 cm.  the area of a triangle BCD is 100 square centimeters. what is the number of square centimeters in the area of trapezoid ABCD?


inequalities
logarithms
limit
inequalities unsolved

hello !!!

prove that

$ \LARGE \displaystyle \forall x \in ]0,\plus{} \infty[ \backslash \{1\} \quad \frac{\ln x }{x\minus{}1} \leq \frac{1\plus{} \sqrt[3]{x}}{x\plus{} \sqrt[3]{x}}$

[quote="*210*"]hello !!!

prove that

$ \LARGE \displaystyle \forall x \in ]0, \plus{} \infty[ \backslash \{1\} \quad \frac {\ln x }{x \minus{} 1} \leq \frac {1 \plus{} \sqrt [3]{x}}{x \plus{} \sqrt [3]{x}}$[/quote]
Replace $ x\rightarrow x^3.$
Hence, we need to prove that $ f(x) > 0$ where $ f(x) \equal{} \frac {(x^3 \minus{} 1)(x \plus{} 1)}{3(x^3 \plus{} x)} \minus{} \ln x$ for $ x > 1$
and $ f(x) < 0$ for $ 0 < x < 1,$
which is true because $ f'(x) \equal{} \frac {(x \minus{} 1)^4(x^2 \plus{} x \plus{} 1)}{3x^2(x^2 \plus{} 1)^2} > 0$ and $ \lim_{x\rightarrow1}\frac{\ln x}{x\minus{}1}\equal{}1.$


algebra
polynomial
calculus
integration
algebra proposed

Find all polynomials $ f(x)$ with rational coefficients, whose degree are as least as possible such that: $ f(3^{\frac{1}{3}}\plus{}9^{\frac{1}{3}}) \equal{} 3\plus{}3^{\frac{1}{3}}$.
Does there exist a polynomial f with integral coefficients such that: $ f(3^{\frac{1}{3}}\plus{}9^{\frac{1}{3}}) \equal{} 3\plus{}3^{\frac{1}{3}}$?

P.S. I think this problem is not difficult. Goal here is to be solved by many people and they to spend good time.

Please continuou here http://www.mathlinks.ro/Forum/viewtopic.php?p=786138#786138 .  :wink:


Prove that $2x^{4}+21x^{3}-6x^{2}+9x-3$ cannot be factorized into two polynomials with integer coefficients.

Thanks

this is a direct application of:

[url]http://mathworld.wolfram.com/EisensteinsIrreducibilityCriterion.html[/url]

[quote="mathgeniuse^ln(x)"]Wait, so what would I do to prove this?

What prime should I chose?[/quote]

what prime divides all the coefficients except for the leading one...?

OMG.

3.

Thanks Man, but how do I prove the Eisenstein thingy.


Let $ ABC$ be an acute triangle and $ P$ a point inside.

Let $ L,M,N$ be the orthogonal projections of $ P$ at $ BC , AC , AB$
Find $ P$ such that : $ BL^2 \plus{} CM^2 \plus{} AN ^2$ is minimal.

We have
\begin{align*}BL^2 + CM^2 + AN^2 & = CL^2 + AM^2 + BN^2 \\
& = \frac {BL^2 + CL^2}{2} + \frac {CM^2 + AM^2}{2} + \frac {AN^2 + BN^2}{2} \\
& \ge \left(\frac {BL + CL}{2}\right)^2 + \left(\frac {CM + AM}{2}\right)^2 + \left(\frac {AN + BN}{2}\right)^2 \\
& = \frac {BC^2 + CA^2 + AB^2}{4}\end{align*}
Therefore equality occurs iff $ L,M,N$ are the midpoints of the sides, which happens when $ P$ is the circumcenter of triangle $ ABC$.


In the land of Hexagonia, the six cities are connected by a rail network
such that there is a direct rail line connecting each pair of cities. On
Sundays, some lines may be closed for repair. The passengers’ rail
charter stipulates that any city must be accessible by rail from any
other (not necessarily directly) at all times. In how many different
ways can some of the lines be closed subject to this condition?


Okay, so here's the puzzle (which is actually a great magic trick once you know it, but I'm giving it to you as a puzzle)

I think of a number between 0 and 15. You get 7 yes-or-no questions to ask me, and I have to answer them truthfully except that I can lie once, if I so choose. I don't have to lie, but I get to lie once. Yet no matter when I choose to lie (or if I don't), you can determine my number. How?

not hard, just say is it less than 7? if he says no, ask is it less than 7 again?, then just keep going on

That doesn't work.  You ask, "is it less than or equal to 7?" and I say "no."  You ask, "is it less than or equal to 7?" and I say "no."  You ask, "is it less than or equal to 11?" and I say "no."  You ask, "is it less than or equal to 11?" and I say "no."  You ask, "is it less than or equal to 13?" and I say "no."  You ask, "is it less than or equal to 13?" and I say "yes."  You have one question left.

This is called [i]Ulam's game[/i] (or the Liar game).  I actually did a little research on a variant of it, called [i]half liar[/i], in which the answerer may lie if the answer to a question is "no," but not if the answer is "yes."  I know that Professor Joel Spencer, at NYU, has done a lot of research on a number of variations of this game (e.g., a larger number of allowed lies than just 1) and gotten good asymptotic results for how many questions the guesser needs to win, starting with a set of large size.  In general, adding a lie typically doubles the number of questions necessary for the guesser to win.

Here's my analysis of it:

[hide]Let the position $ (a, b)$ indicate that there are $ a$ numbers such that if the thinker is thinking of one of those numbers they haven't yet lied and that there are $ b$ numbers such that if the thinker is thinking of one of those numbers they have lied.  Similarly, let the question $ (c, d)$ be a question of the form, "Is it any of these $ c$ numbers from the set of numbers for which, if you're thinking of one of them you haven't lied, or any of these $ d$ numbers from the set of numbers for which if you're thinking of one of them you have already lied?"  Thus, the initial position is $ (16, 0)$, and if the first question is, "Is your number less than or equal to 6?" it is of the form $ (7, 0)$.  A winning position is of the form $ (a, b)$ with $ a\plus{}b \equal{} 1$.  (A position of the form $ (0, 0)$ would indicate that the answerer had cheated.)

Given position $ (a, b)$ and question $ (c, d)$, the answer "yes" moves us to position $ (c, d\plus{}a\minus{}c)$ and the answer "no" moves us to the position $ (a\minus{}c, b\minus{}d\plus{}c)$.  Note that asking a question of the form $ (c, d)$ is equivalent to asking a question of the form $ (a\minus{}c, b\minus{}d)$.  (This is not true in half-liar, which is one of the reasons analyzing it is rather complicated.)

Consider the following sequence of questions and answers:
Begin at $ (16, 0)$.
Ask $ (8, 0)$.  Six questions left.  Both answers are of the form $ (8, 8)$.
Ask $ (4, 4)$.  Five questions left.  Both answers are of the form $ (4, 8)$.
Ask $ (2, 4)$.  Four questions left.  Both answers are of the form $ (2, 6)$.
Ask $ (1, 3)$.  Three questions left.  Both answers are of the form $ (1, 4)$.
Ask $ (1, 1)$.  If the answer is "yes," the position is $ (1, 1)$.  If the answer is "no," the new position is $ (0, 4)$.  Two questions left.  In the second case, we have a binary search and we win.  In the first case, ask $ (1, 0)$.  If the answer is "yes," the position is $ (1, 0)$ and we win.  If the answer is "no," the position is $ (0, 2)$ and we have a binary search and we win.

In all cases, the question-asker can win regardless of the answers given to him.[/hide]

Two interesting-seeming expositions concerning this problem are at http://www.cs.mcgill.ca/~psavad/cs251/ulam.html and [url]http://cadigweb.ew.usna.edu/~wdj/montague/montague_mathhonors1998-1999.html[/url].  The second link contains an actual strategy for solving the problem posed by Hamster1800 without feedback.

So here's my version of the solution:

[hide]
Let's start by formulating four questions which would allow us to find the number without lies.

1) Is it greater than 7?
2) Is it 4, 5, 6, 7, 12, 13, 14,or 15?
3) Is it 2, 3, 6, 7, 10, 11, 14, or 15?
4) Is it odd?

Then we can simply interpret Y as 1, N as 0 and read it as a binary number. You can check for youself:
NNNN=0, NNNY=1, etc.

Now we want a nice way to correct errors. Well, we have 7 questions. Let's arrange them like this:
ab1c234
where a,b,c are our added questions. Now, we ideally want to be able to read off the location of the error in binary similarly to before, hence why we put them at powers of two.

Now, we just have a be the xor of the odd locations (1,2,4)
b be the xor of those with a 1 in the 2's place (1,3,4)
c be the xor of those with a 1 in the 4's place (2,3,4)

This gives:
a) Is it 4, 6, 8, 9, 10, 11, 13, or 15?
b) Is it 1, 2, 5, 6, 8, 11, 12, or 15?
c) Is it 1, 2, 4, 7, 9, 10, 12, or 15?

So now, we just ask the questions:
a) Is it one of 4, 6, 8, 9, 10, 11, 13, or 15?
b) Is it one of 1, 2, 5, 6, 8, 11, 12, or 15?
1) Is it greater than 7?
c) Is it one of 1, 2, 3, 7, 9, 10, 12, or 15?
2) Is it one of 4, 5, 6, 7, 12, 13, 14, or 15?
3) Is it one of 2, 3, 6, 7, 10, 11, 14, or 15?
4) Is it odd?

We then get a string of Ns and Ys, say
NNYNNYN

Our original message looks like this, assuming there were no lies:
YNYN
The answers to a, b, and c from this would be
Y, N, Y
But they are actually N, N, N. So, both a and c were wrong, giving an error location of 5 ($ 101_{2}$).

Thus, the correct message was
NNYNYYN => YYYN => The number was 14.

This is a well known single-error fixing code, but I forget the name. I think it's Hamming Code or something.
[/hide]

It's related to the Hamming code (I'm not sure if it's the Hamming code exactly) -- an essentially (maybe exactly?) equivalent solution is in the second link I posted above, along with other interesting discussion.

It is exactly the Hamming code. A version of the solution that makes this explicit:

Create a sheet with the Hamming code identifications and give a copy of it to the answerer:
[hide="Code"]0000000 - 0
1100001 - 1
1010010 - 2
0110011 - 3
0110100 - 4
1010101 - 5
1100110 - 6
0000111 - 7
1111000 - 8
0011001 - 9
0101010 - 10
1001011 - 11
1001100 - 12
0101101 - 13
0011110 - 14
1111111 - 15[/hide]

Ask the following seven questions:
Is the [nth] bit of the code for your number 1?
Take the code corresponding to their answers. It's either one of the codes on your list, or it's one bit away from exactly one of the codes on your list. Either way, you know which number it was.

This form of the solution is also completely unsuitable for use as a magic trick.

This problem was a homework problem in Conjecture & Proof at the Budapest Semesters in Mathematics program. At CTY and MathZoom this summer my classes really enjoyed tackling it.

[hide="my solution; somewhat implementable"]
Questions 1 through 4: ask for the binary digits of $ n$.

Question 5: Did you lie on questions $ \{1,2,3\}$?

If yes on 5: the lie has been used somewhere in $ \{1,2,3,5\}$. Two questions, four possibilities, and in any case once the lie's location is determined, either flip that bit (or do nothing, if it was #5).

If no on 5: then #5 was answered truthfully, so all that remains is to determine whether or not #4 was a lie. Use questions 6 and 7 to ask if #4 was a lie. You need to ask twice, because a "yes" on 6 might be a lie.
[/hide]

An interesting counter-problem is to look at things from the other angle. Since most people won't be able to come up with a foolproof set of questions at first, how should the "guessee" choose where to lie (if at all) to maximize the "possible" set of numbers at the end of the 7 questions? Obviously this depends on the questions themselves, but I found that a well-chosen lie will often result in there being maybe 3 plausible values for $ n$ after 7 questions.


Prove that the set $ \{1,2,...,432\}$ can be partitioned in $ 144$ subsets, each with $ 3$ elements, such that in each subset one of the numbers equals the sum of the other two.

A bit more general was posted before in http://www.mathlinks.ro/Forum/viewtopic.php?p=1054571#1054571

Daniel

I don't think that post includes a complete solution to this instance of the problem. Call $ n$ "good" if $ \{1, \ldots, n\}$ can be split into triplets where one element is the sum of the other two. Xell observed that if $ n$ is good then so are $ 4n$ and $ 4n\plus{}3$. Together with the obvious $ n\equal{}3$ starting point this allows $ n\equal{}3324$ to be proven good, but I don't see how this solves $ n\equal{}432\equal{}27 \times 16$. Note that $ n\equal{}6$ is definitely not good, so the fact that 27 is good (which is true) cannot be derived from the $ 4n\plus{}3$ construction.

To show that 27 is good we can use the triplets

$ (1, 26, 27), (2, 23, 25), (3, 14, 17), (4, 20, 24), (5, 11, 16), (6, 12, 18), (7, 15, 22), (8, 13, 21), (9, 10, 19)$

and now we can proceed to show that 432 is good using the $ n \mapsto 4n$ construction. However this is not a very elegant solution, and I'm still looking for one.


What is the sum of the first four positive perfect cubes?

Well, the first 4 positive cubes are $ (1^3) \plus{} (2^3) \plus{} (3^3) \plus{} (4^3) \equal{} 1 \plus{} 8 \plus{} 27 \plus{} 64 \equal{} \boxed {100}$

Alternately, the sum of the first $ n$ cubes is the sum of their bases squared.  For example, if we had $ 1^3\plus{}2^3\plus{}\dots\plus{}n^3$, that would be equal to $ (1\plus{}2\dots\plus{}n)^2$.  So in this case, we would have $ (1\plus{}2\plus{}3\plus{}4)^2\equal{}(10)^2\equal{}\boxed{100}$, just like what mrsaturns got.


geometry
rectangle
circumcircle
geometry proposed

Circles (O),(I) intersects at point M,N.Let Ab be the line tangent to these circles at A,B respectively (A lies on (O),B lies on (I) :!: )so that M lies closer to AB than N.E lies on OI such that BE is perpendicular to AM.BF is a diameter of (I).prove that M,E.F are collinear

AB, OI meet at the external similarity center H of the circles (O), (I). Let HM meet the circle (I) again at K, let KL be a diameter of (I), and let AD be a diameter of (O). The quadrilateral BKFL is a rectangle with diagonals BF, KL intersecting at the center I of its circumcircle (I), hence $KF \parallel BL.$ The triangles $\triangle ADM \sim \triangle BFK$ are centrally similar with the same similarity center H, hence their sides are parallel, in particular $MD \parallel KF \parallel BL.$ Since AD is a diameter of (O), $MD \perp AM$ and consequently, $BL \perp AM$ as well. Considering the cyclic pentagon BKFLM as a degenerate cyclic hexagon BBKFLM with the double vertex B, $BB \equiv AB$ is a tangent of (I) at B passing through H, and the intersections $H \equiv BB \cap KM,$ $I \equiv BF \cap KL,$ $E \equiv BL \cap MF$ are collinear. The points F, E, M are now collinear by definition, $E \in HI \equiv OI,$ and $BE \equiv BL \perp AM.$

[quote="comet_vn"][quote="weiquan"]This forum is not for homework![/quote]
I see this problem is good
so i want to post it
certainly "forum is not for homework"
I solved It two weeks ago  :P[/quote]

I am sorry then, but then you shouldn't post it in the "Unsolved" section, lest people think you haven't solved it :P


In a collection of red, blue, and green marbles, there are $ 25\%$ more red marbles than blue marbles, and there are $ 60\%$ more green marbles than red marbles. Suppose that there are $ r$ red marbles. What is the total number of marbles in that collection?

$ \textbf{(A)}\ 2.85r \qquad \textbf{(B)}\ 3r \qquad \textbf{(C)}\ 3.4r \qquad \textbf{(D)}\ 3.85r \qquad \textbf{(E)}\ 4.25r$

[hide]Let $ r$ be the number of red marbles and $ g$ be the number of green marbles so we have $ r\equal{}5/4 b$ and $ g\equal{}8/5r$ then $ b\equal{}4/5r$ and the total amount is $ r\plus{}b\plus{}g\equal{}r\plus{}4/5r\plus{}8/5r\equal{}17/5r\equal{}3.4r$

hence the answer is $ \boxed{C}$.[/hide]


Ring Theory
calculus
integration
algebra
function
domain
factorial

An integral domain R is a unique factorization domain if and only if every nonzero prime ideal in R contains a nonzero principal ideal that is prime.

$ \implies$ Assume that $ R$ is an UFD and let $ \mathfrak{p}$ be any non-zero prime ideal in $ R$. Take $ a \in \mathfrak{p}$ and write $ a \equal{} \pi_1 \cdots \pi_n$. At least one of the $ \pi_i$ must lie in $ P$ such that $ (\pi_i) \subset P$.

Doesn't that theorem only hold for Noetherian rings?

same solution than mine....no yannick......it not only holds for noetherian rings

[quote="myan"]For the converse, use that an integral domain $ R$ is factorial iff every height-$ 1$-prime is principal.[/quote]

That's true for Noetherian domains, but the converse fails generally since there are non-field domains
that have no prime ideals of height 1, e.g. see example 34 in Hutchins, Examples of Commutative Rings.


Let $ABC$ be an acute triangle with $\angle{B}> \angle{C}$, and let $\Gamma$ be its circumencircle. The line passing through $A$ and perpendicular to $BC$ meets $\Gamma$ at $D$. The line passing through $D$ and perpendicular to $AB$ meets $AB$ at $Q$ and $\Gamma$ at $P$.
Show that $\angle{PBC}=2 \angle{BPQ}$

we have:$ <PBC=<QBC-<PBQ=180-<ADQ-<PBQ=180-2<ABC=2(90-<ABC)=2<BAD=2<BPQ.$
And we are done :lol:

Yes... That's what I did on the exam too  :lol:


rotation
linear algebra
matrix
trigonometry
analytic geometry

If you rotate the axes by an angle $\alpha$, then

$[\begin{matrix}x'\\ y'\end{matrix}]=[\begin{matrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{matrix}]\cdot[\begin{matrix}x\\ y\end{matrix}]$

or, in complex notation,

$x'+iy'=(\cos\alpha+i\sin\alpha)(x+iy)$

or, in the more direct form

$x+iy=\frac{x'+iy'}{\cos\alpha+i\sin\alpha}$

From there you get the old coordinates $x,y$ expressed through the new coordinates $x',y'$, so just plug those expressions into $y=f(x)$, simplify and you're done.

[quote="Allan Z"]can anyone teach me how to deal with the problems with rotating the axis..

for example:
what is the equation of the graph when you rotate $y=x^{2}$ counter-clockwise by $15^\circ$[/quote]

You could convert the graphs to polar coordinates then change every instance of $\theta$ to, say, $\theta+15$...then convert back


There are positive integers $k$, $n$, and $m$ such that

\[ \frac{19}{20} < \frac{1}{k} + \frac{1}{n} + \frac{1}{m} < 1. \]
What is the smallest possible value of $k+n+m$?

I think [hide]that the answer is 12, with k=2, n=3, and m=7.  1/2+1/3+1/7=41/42, which is bigger than 19/20.[/hide]

Is there a better way to do a problem like this besides guess and check? For example:

The sum of the reciprocals of three consecutive integers is 47/60. What is the sum of the three integers?

The problem is not complete guess and check because you can set up an upper bound with any single solution.


limit
abstract algebra
calculus
calculus computations

For $ a \in \mathbb{R}$ let $ f: (a,\infty) \in C^1$. If $ \alpha, \beta \in \mathbb{R}$, $ f(x) \rightarrow A$ and $ f'(x) \rightarrow B$ as $ x\rightarrow \infty$. Prove $ B \equal{} 0$.

I guess $ f\in C^{1}(a, \infty)$: $ f(x)\to \alpha, \ f'(x)\to \beta$ as $ x \to \infty$ ?

[hide="Hint"]Use Lagrange's Mean Value Theorem.[/hide]

[hide="Solution"]
Suppose that $ \beta \neq 0$.
Since $ \lim_{x\to \infty}f(x) \equal{} \alpha \in \mathbb{R}$ we have $ \lim_{n\to\infty}(f(n \plus{} 1) \minus{} f(n)) \equal{} 0$.
From Lagrange's Theorem:
$ \forall n\in \mathbb{N}, n > a \ \exists c_{n} \in (n, n \plus{} 1) \quad f'(c_{n}) \equal{} f(n \plus{} 1) \minus{} f(n)$
Of course $ \lim_{n\to \infty}c_{n} \equal{} \infty$, and we obtain a contradiction: $ \lim_{n\to \infty}f'(c_{n}) \equal{} 0 \neq \beta$ [/hide]


Prove that there exist two strictly increasing sequences $a_{n}$ and $b_{n}$ such that $a_{n}(a_{n} +1)$ divides $b_{n}^2 +1$ for every natural $n$.

Consider the equation:
$ b^2\plus{}1\equal{}5a(a\plus{}1)$
$ \Leftrightarrow 4(b^2\plus{}1)\equal{}5(2a\plus{}1)^2\minus{}5$
$ \Leftrightarrow 4b^2\minus{}5(2a\plus{}1)^2\equal{}\minus{}9$
Let $ 2b\equal{}3x,2a\plus{}1\equal{}3y$
Then:
$ x^2\minus{}5b^2\equal{}\minus{}1$
It is the Pell's equation:A root sequence:
$ x_1\equal{}2,y_1\equal{}1$
$ x_2\equal{}38,y_2\equal{}17$
$ x_{n\plus{}1}\equal{}9x_n\minus{}x_{n\minus{}1}$
$ y_{n\plus{}1}\equal{}9y_n\minus{}y_{n\minus{}1}$
$ x_n\equiv 0(\mod 2),\forall n\in N$
$ \{y_n (\mod 2)\}$ is period
$ y_{n}\equal{}y_{n\plus{}6}$ so it contain infinite odd number.
Now con sider :
$ b_n\equal{}\frac{3x_n}{2},a_n\equal{}\frac{3y_n\minus{}1}{2}$
Easy to check they satisfty the relative.

For reference: another solution can be found [url=http://www.kalva.demon.co.uk/short/soln/sh99n3.html]here[/url].

another solution:
Lemma:if $ n$ is not a multiple of $ 4$,and every odd prime divisor of $ n$ have form $ 4k \plus{} 1$,then there exist infinaitly many $ x$ such that $ n|x^2 \plus{} 1$
Now suppose that $ c_n$ be strictly increasing sequence  such that for all natural $ n$:every prime divisor of $ c_n$ have the form $ 4k \plus{} 1$.
let $ a_n \equal{} c_n^2$...therefore $ a_n(a_n \plus{} 1)$ is not a multiple of $ 4$ and every odd prime divisor of $ a_n(a_n \plus{} 1)$ have form $ 4k \plus{} 1$.(why?)
therefore by Lemma,exist infinitly many $ x$ such that $ a_n(a_n \plus{} 1)|x^2 \plus{} 1$
now just define $ b_n \equal{}$
arbitary natural number $ x$,such that $ a_n(a_n \plus{} 1)|x^2 \plus{} 1$ and $ x > b_k$ for $ k < n$...