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rectangle
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A computer screen shows a $98 \times 98$ chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result,  the colors in the selected rectangle switch (black becomes white, white becomes black). Find, with proof, the minimum number of mouse clicks needed to make the chessboard all one color.

Neat problem. Indeed, the answer is $98$, but it's nicer to have a proof posted, isn't it? :) I just hope this one's correct.

Let's prove a moregeneral assertion: given an $m\times n$ board, in the same conditions, the answer to the same question is $\left\lfloor\frac m2\right\rfloor+\left\lfloor\frac n2\right\rfloor\ (*)$.

I won't bother to describe an example which shows that $(*)$ does indeed work. It's a bit hard to explain in writing, but it should be obvious after playing around with a board. You take your rectangles to be $\left\lfloor \frac m2\right\rfloor$ horizontal rectangles of height $1$ and width $n$, and $\left\lfloor \frac n2\right\rfloor$ vertical rectangles of height $m$ and width $1$.

Conversely, we have to prove that $(*)$ is minimal. Notice that any collection of rectangles with the desired property must also have the property that every segment on the board, except for those on the outer edges, must belong to the boundary of at least one rectangle in the collection. It thus suffices to show by induction on $m+n$ that a collection $\mathcal C$ with this latter property must contain at least $(*)$ rectangles. The small cases are easy to check, so assume now that we've proven the assertion for values of $m+n$ smaller than the current one.

Assume that $m\le n$, and let the rectangle be named $ABCD$, with $BC$ and $AD$ being the lower and upper horizontal sides of length $n$ respectively. If $\mathcal C$ contains no rectangles with one side on $AD$ and one side on $BC$, it means that we need at least $\left\lfloor \frac n2\right\rfloor$ rectangles to cover the vertical segments of length one with one endpoint on $BC$, and $\left\lfloor \frac n2\right\rfloor$ more to cover those which have an endpoint on $AD$, i.e. we have $\ge 2\left\lfloor \frac n2\right\rfloor\ge (*)$ rectangles in total. On the other hand, if we do have a rectangle $R$ with the upper horizontal side on $AD$ and the lower one on $BC$, we can assume that its vertical sides are different from $AB,CD$. For each one of its vertical sides, merge the two columns to the right and to the left of the respective side, and eliminate $R$. We get a new problem, with an $m\times (n-2)$ table, which requires at least $(*)-1$ rectangles by the induction hypothesis. We add $R$ back into the picture, and we reach the desired conclusion: $\mathcal C$ must have $\ge (*)$ elements.

i feel really weird posting this

We will prove that on an n*n board, there are n clicks required when n is even.

Induct

Base
On a 2*2, there are clearly 2 click required.

Inductive step
WLOG, let the top left square be black. On an n*n board, do the n-2 clicks for the previous case. Then we are left with an (n-1)*(n-1) black square. and then 2 white strips with a black square in the bottom right corner. it kinda likes like
[code]
b b b ... b w
b b b ... b w
b b b ... b w
. . .    . .
. . .    . .
. . .    . .
b b b ... b w
w w w ... w b
[/code]

then just click the bottom 1*98 rectangle and the 98*1 rectangle on the side and you're done.

To prove this is the shortest way. use the fact that an (n-2)*(n-2) requires atleast n-2 clicks to show that n-1 clicks isnt possible.

[quote="abacadaea"]
To prove this is the shortest way. use the fact that an (n-2)*(n-2) requires atleast n-2 clicks to show that n-1 clicks isnt possible.[/quote]

Unfortunately, I think this is the non-trivial part (and probably isn't true), but I think you might have misunderstood the problem anyway (or have I?).

It may be simpler to consider border tiles (tile is a pair of adjacent squares). For nxn board we have 4(n-1) border tiles. At most we can make four tiles into matching ones (choose a column of length n that doesn't contain a corner). So we need at least n-1. However corners screw this up so the minimum is n.

Hmm, I think this works.
[hide="Solution"][b]Lemma:[/b] It takes at least 49 moves to make all the squares of a 1x98 rectangle equal. (given similar alternated coloring and operations)
[b]Proof:[/b] 49 is easily achieved (example: pick all white squares). Now consider how many boundaries have opposite colored squares on either side. We initially start with 97 such boundaries. Any click reduces this count by at most 2, so we must make at least 49 moves.

Now we consider the 98x98 rectangle. 98 clicks is achievable by choosing every other row then every other column. Now consider each 1x98 edge of the 98x98 rectangle (yes, they overlap on the corners). For the whole board to end up one single color, we need each edge to end up as one single color. For this to happen, each edge must have 49 operations that alter at least one element of the edge without altering all elements of the edge at the same time (since this does not contribute to reducing the opposite colored boundaries, as in the proof of our lemma). Let us say that each edge has a "need" of 49 clicks. Thus we have a total need of $ 4\cdot49\equal{}196$.
We now show that any move can satisfy at most 2 needs. Suppose it satisfied a need of at least 3 edges. Then it must satisfy the needs of two opposite edges, and intersect at least one other edge. But this implies that it envelops that whole edge, meaning that it does not satisfy its need, which is a contradiction.
Since we can satisfy at most 2 needs with one move, we need at least $ 196/2\equal{}98$ moves, and we are done.[/hide]

My solution. I think it works..

[hide="Solution"]Call a "happy column" a column with all squares colored the same color and a "happy row" a row with all squares colored the same color.

Lemma 1: It takes a minimum of $49$ clicks to convert a column from its initial coloring to a happy coloring.
Proof: Choose any column $C$. We can forget about the rest of the board since whenever we create a rectangle with other squares that involves $C$, it is the same as choosing a sequence of adjacent square(s) in $C$. At the start, the column has 49 black squares and 49 white squares that alternate. Choose a sequence of adjacent squares and perform a click. For every initially black square in the sequence, the click decreases the number of black squares by $1$ and increases the number of white squares by $1$. The reverse holds true for an initially white square. WLOG we want to get the sequence to become all black (so we want $98$ black squares). Each sequence can at most decrease the number of white squares (and increase the number of black squares) by $1$, since we can at most choose one more initially white than initially black square due to the alternating nature of the chessboard coloring. Thus, it takes a minimum of $98-49=49$ clicks.
This completes the lemma.

Lemma 1 obviously extends itself to any row $R$: it takes a minimum of $49$ clicks to convert a row from its initial coloring to a happy coloring.

Now consider the $98 \times 98$ board.
WLOG we want the entire board to be black.
We need to convert the bottom-most row to black, so create $49$ rectangles such that in $49$ clicks, the bottom-most row is black. Since we want the entire board black, we can maximize the utility of these rectangles by converting other rows black. This can be done by making the heights of the $49$ rectangles be $98$.
We now have $98$ rows, each of which has a uniform color, and whose colors alternate.
Note that the columns are virtually intact (at most, the colors in their rectangles have been switched).
We now need to convert the left-most column to black. Since by Lemma 1 it will take at least $49$ clicks anyway, if we can find a way to finish coloring the entire board black in $49$ clicks, then we have found a coloring scheme that produces the minimum number of clicks required. This is easy, just convert all $49$ white rows to black.
Thus, the minimum number of clicks for a $98 \times 98$ board is $49+49=98$.
$Q.E.D.$[/hide]

Consider any even $n$ in place of 98. There are $4n-2$ grid vertices adjacent to oddly many black squares (all external ones except two white corners), call these bad, and good otherwise. Consider a move on rectangle $R$. If grid vertex $v$ is not a vertex of $R$, it is adjacent to evenly many squares in $R$, so after the move $v$ retains its state as good/bad. Thus only the 4 vertices of $R$ can change their state.

Now every move decreases $S$, the number of bad vertices, by at most 4. But in the end $S=0$, thus we need at least $n$ moves. Take $\frac{n}{2}$ rows to get $\frac{n}{2}$ black columns, then take these columns. Thus $n$ is indeed the minimum.

We consider a $2n\times 2n$ grid instead of $98$. Now, you can do it in $2n$ click by first selecting all odd rows in first $n$ moves then odd column in next $n$ moves. Now, we show that this bound is optimal.

WLOG, assume that the colour we end up with is the colour of the bottom left square(say white). Now, we ignore the complete grid except the bottom row and left most column. We only consider their action on these squares. Now, the number of rectangles acting on the corner must be even. So, we can randomly pair them. Now, for any pair of these rectangles, take the resultant of their action on the row and the column, as the resultant of these two is equivalent to two different rectangles acting on just the row and just the column(we can ignore the common part), we can replace these with the new rectangles.

Now, every rectangle just acts on the row or the column. Again observe that any two intersecting rectangles can be replaced with two non-intersecting rectangles.

Now, as any rectangle now does not intersect with any other rectangle, it cannot have any white squares. So, every rectangle is a single black square. Now, there are $n$ black squares in the rows and $n$ black squares in the columns. Thus, we get a total of $2n$ different rectangles.

[hide=Note]
This idea naturally extends to $m\times n$ grid to get the answer as $\left\lfloor\frac{m}{2}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor$
[/hide]

Solution from [i]Twitch Solves ISL[/i]:

The answer is $98$. One of several possible constructions is to toggle all columns and rows with even indices.
In the other direction, let $n = 98$ and suppose that $k$ rectangles are used, none of which are $n \times n$ (else we may delete it). Then, for any two orthogonally adjacent cells, the edge between them must be contained in the edge of one of the $k$ rectangles.
We define a [i]gridline[/i] to be a line segment that runs in the interior of the board from one side of the board to the other. Hence there are $2n-2$ gridlines exactly. Moreover, we can classify these rectangles into two types: [list] 	[*][i]Full length rectangles[/i]: 	these span from one edge of the board to the other. 	The two long sides completely cover two gridlines, 	but the other two sides of the rectangle do not. 	[*][i]Partial length rectangles[/i]: 	each of four sides can partially cover ``half a'' gridlines. [/list] See illustration below for $n = 6$.
[asy] 	unitsize(0.5cm); 	picture base; 	for (int i=1; i<=5; ++i) { 		draw(base, (0,i)--(6,i), black ); 		draw(base, (i,0)--(i,6), black ); 	}
picture full; 	picture partial;
add(full, base); 	add(partial, base); 	label(full, "Full length", (3,0), dir(-90), red); 	label(partial, "Partial length", (3,0), dir(-90), blue);
filldraw(full, box( (0,3), (6,1) ), lightred+opacity(0.3), red+1.4 ); 	filldraw(partial, box( (2,2), (4,5) ), lightcyan+opacity(0.3), blue+1.4 ); 	add(shift(0,0)*base); 	add(shift(7,0)*full); 	add(shift(14,0)*partial); [/asy]
Since there are $2n-2$ gridlines; and each rectangle can cover at most two gridlines in total (where partial-length rectangles are ``worth $\frac{1}{2}$'' on each of the four sides), we immediately get the bound $2k \ge 2n-2$, or $k \ge n-1$.

To finish, we prove that:
[b][color=red]Claim:[/color][/b]  	If equality holds and $k = n-1$, then $n$ is odd.

[i]Proof.[/i]  	If equality holds, then look at the horizontal gridlines 	and say two gridlines are [i]related[/i] if some rectangle 	has horizontal edges along both gridlines. 	(Hence, the graph has degree either $1$ or $2$ at each vertex, 	for equality to hold.) 	The reader may verify the resulting graph consists only of 	even length cycles and single edges, 	which would mean $n-1$ is even. $\blacksquare$
Hence for $n = 98$ the answer is indeed $98$ as claimed.

The desired minimum is 98.

To show that it is achievable, first we make a move on all even columns and then on all odd rows.

Now we have to show that this is indeed the desired minimum. We observe that each edge between squares has to be on the border of a rectangle we make a move on at least once, since otherwise the squares on the two sides of it will be different colors. Call a special edge an edge of the grid(between two squares) which is connected to the border of the board but not on the border. There are a total of $4(98-1)$ such special edges. Also each rectangle we make a move on has at most 4 special edges on it's border. But we have to change the color of two of the corner squares, and the rectangles we make a move on in order to do this contain at most 2 special edges, implying that we need at least $(4(98-1)-4)/4+2 = 98$ rectangles in order to have made a move on all of the special edges.

The answer is 98. In general, for an $n\times n$ board with even $n$, the answer is $n$. For the construction, hit all of the odd numbered rows and columns.

To prove optimality, note that the edge between any pair of adjacent squares must be in the perimeter of one of the flipping rectangles. Consider the $2(n-1)$ edges attached to the top/bottom of the board and the $2(n-1)$ edges attached to the left/right side. Each rectangle can hit at most:

(1) 4 vertical edges\\
(2) 1 vertical edge and 1 horizontal edge\\
(3) 4 horizontal edges.

Clearly, at least $\frac{2(n-1)+2(n-1)}{4}=n-1$ are needed. However, $n-1$ is also not achievable because in order for each rectangle to hit 4 edges, all of them must be (1) or (3), and since there are $2\cdot 97\equiv 2\pmod{4}$ top edges, this is impossible. Thus, there are at least $n$ flipping rectangles and we're done. $\blacksquare$.

[hide = Solution] For an arbitrary $n \times n$ board I claim the minimum required moves is $\left \lceil \frac{n}{2} \right \rceil + \left \lfloor \frac{n}{2} \right \rfloor = n$. We prove by induction.

The base case $n = 2$ is obvious. Now assume for $n = k - 1$ this holds. Now if $k$ is odd, then $k - 1$ is even so it would take $\frac{k - 1}{2} + \frac{k - 1}{2} = k - 1$ moves to obtain a $k \times k$ board covered with all but one row or column remaining. Thus it would take $k - 1 + 1 = k$ moves.

If $k$ is even, then in the same way, we obtain at the end one extra move where we eliminate the final entire row or column once more. Whatever the case may be, it takes $k$ moves regardless. Hence for $n = 98$, it would take $\boxed{98}$ moves. $\blacksquare$ [/hide]
[hide = Remark(s)] If we replaced $n \times n$ by an $m \times n$ board we would have the minimum required moves is $\frac{m}{2} + \frac{n}{2}$. [/hide]

a bit jank

The answer is $98$, which works by toggling every other row and every other column. To prove that this is minimal, consider the following problem: the $98 \times 98$ board starts off completely white, and by performing the same color-toggle operation we wish to color it into a checkerboard pattern such that the top left square is white. This is clearly equivalent to the original.

[color=#00f][b]Claim:[/b][/color] Consider a $1 \times 98$ section of the board. If $k$ selected rectangles intersect it, then the black tiles form at most $k$ black connected components. Furthermore, if the end tiles of the section are opposite color, we actually have at most $k-a$ connected components, where $a$ of the intersecting rectangles intersect the entirety of the section.
[color=#00f]Proof:[/color] We first prove the first part of the claim by induction on $k$, with the base case of $k=0$ being obvious. Our inductive step consists of proving that flipping some rectangle part of the section increases the number of connected components by $1$. We consider these three cases:
[list]
[*] If both endpoints of the rectangle are black, and the rectangle contains $m$ black components, then upon being flipped there are $m-1$ black components. Since the endpoints could have been connected to larger black components that don't get deleted, there is a guaranteed loss of $m-2$ black components, so the number of components increases by at most $1$.
[*] If both endpoints of the rectangle are white, and the rectangle contains $m$ black components, then upon being flipped there are $m+1$ black components, so the increase is at most $1$.
[*] If one endpoint is black and one is white, then the flipped rectangle contains the same number of black components as before (both $m$). Since the black endpoint could've been part of a larger component, there is a guaranteed loss of $m-1$ black components, so the number of components increases by at most $1$.
[/list]
Now to prove the "extended claim", we work backwards. The actual order of the rectangles doesn't matter, so apply the "completely covering" rectangles last. Note that the inverse application of a rectangle on a checkerboard pattern in a $1 \times 98$ triangle doesn't change the number of connected components (since $98$ is even), so after applying all of those last we are left with $k-a$ rectangle intersections which can produce at most $k-a$ black connected components.

To finish the problem, we consider the left, right, top, and bottom $1 \times 98$ sections of the chessboard. After all the rectangles are drawn, these rectangles should each have $49$ connected black components (considered independently, not together). We thus need a total of $49 \cdot 4=196$ rectangle intersections which aren't "complete covers". On the other hand, any given rectangle can only intersect two of the sections without covering them, since if a rectangle covers two opposite sections and also intersects a third, then the third section must be completely covered. Thus we require at least $196/2=98$ rectangles, which is the desired bound. $\blacksquare$


calculus derivative limit trigonometry logarithms function algebra

[color=darkblue][b]Examine if these limits exist[/b] and in affirmative case ascertain :

$ 1\blacktriangleright\lim_{x\to\infty}\arcsin\frac{x+1}{x}$.

$ 2\blacktriangleright\lim_{x\to\infty}\frac{1-\cot x}{e^{x}}$.

$ 3\blacktriangleright\lim_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt x)$.

$ 4\blacktriangleright\lim_{x\tox_{0}}x\cdot[\frac{1}{x}]$, where $ x_{0}\in\{-\infty , 0,\infty\}$, where $ [x] = z\Longleftrightarrow\{\begin{array}{c}z\in Z\\ \\ z\le x < z+1\end{array}$.

==============================================================

$ 5\blacktriangleright\lim_{x\to\frac{\pi}{2}}\frac{\ln\ [\sin(4m+1)x]}{\ln\ [\sin(4n+1)x]}$, where $ \{m,n\}\subset Z$.

$ 6\blacktriangleright\lim_{x\to 1}\frac{x^{n}-nx+n-1}{(x^{a}-1)\cdot\ln x}$, where $ n\in N^{*}$ and $ a\in\mathcal R^{*}$.

$ 7\blacktriangleright\lim_{x\to 1}\frac{nx^{n+1}-(n+1)x^{n}+1}{(e^{x}-e)\cdot\sin\pi x}$. where $ n\in N^{*}$.

$ 8\blacktriangleright\lim_{x\to\infty}(\tan\frac{\pi x}{2x+1})^{\frac{1}{x}}$ (carefully, it is difficulty !).

==================================================

$ 9\blacktriangleright\lim_{x\to 0}\frac{x^{2}+x+1-\sin x-\cos x}{(e^{x}-1)^{2}}$ (carefully, it is more difficultly !).

$ 10\blacktriangleright\lim_{x\to 0}\frac{\sqrt [m]{1+x}+\sqrt [m]{1-x}-2}{x^{2}}$, where $ m\in Z$, $ m\ge 2$ (carefully, it is very difficulty !).

Please, [b]without the l'Hospital's rules[/b] ! Successfully !

[b]Remark.[/b] These limits appear in the Romanian book "[b]Analiza matematica, Editura TEORA, Bucuresti 1999"[/b], page 135.[/color]

every your ("cerefully, it is difficultly") limits is a standart limit which easly solved by Taylor

[color=darkred]Sorry, [b]Extremal ![/b] I mentioned "[b]without the l'Hospital's rules[/b]", i.e. [b][u]without derivatives[/u][/b], now is it more clearly !? For example the exercise $ 10$ for $ m \equal{} 3$ and the limit $ \lim_{x\to 0}\frac{x\minus{}\sin x}{x^{2}}$ being mentioned deliberately "[b]without the l'Hospital's rules[/b]", were given in a [b]Romanian districtual contest[/b] some years ago.[/color]

[color=darkblue]For you, [b]Extremal[/b], here is a generalization of the exercise $ 10$ :[/color]

[color=darkred]$ 11\blacktriangleright\sum_{k \equal{} 1}^{n}a_{k}\equal{} 0$ and $ p\in Z$, $ p\ge 2$ $ \implies$ $ \lim_{x\to 0}\frac{1}{x^{2}}\cdot\left(\sum_{k \equal{} 1}^{n}\sqrt [p]{1\plus{}a_{k}x\plus{}b_{k}x^{2}\plus{}\ldots}\minus{}n\right) \equal{}\frac{1}{p}\cdot\sum_{k \equal{} 1}^{n}b_{k}\minus{}\frac{p\minus{}1}{2p^{2}}\cdot\sum_{k \equal{} 1}^{n}a^{2}_{k}$.

I don't like to use the derivatives, the l'Hospital's rules or the Taylor's series  ([b][u]is inaesthetically ![/u][/b]) when the exercise is solved easily with these.[/color]

[color=darkblue]Here is an interesting and very nice exercise for to use [b]only[/b] the derivatives. Try to solve with derivatives or Taylor's series the following exercise :[/color]

[color=darkred]$ 12\blacktriangleright$ Ascertain $ \mathcal Im(f)$ for the function $ f:\left(0,\frac{\pi}{2}\right]\rightarrow\mathcal R$, where $ f(x) \equal{}\frac{\sin x\minus{}x\cos x}{x\minus{}\sin x}$. In fact, you must study the its monotony.[/color]
[hide="Answer."]
http://www.mathlinks.ro/Forum/viewtopic.php?t=70753 , http://www.mathlinks.ro/Forum/viewtopic.php?t=44733 ,  http://www.mathlinks.ro/Forum/viewtopic.php?t=45414 .[/hide]

mm, but " the l'Hospital's rules" and Taylor is difference
:
for example your (generalization problem #10)
By Taylor I find asymptotical expression in point 0 of your expression.
indeed I can do it:
$ \sum_{i \equal{} 1}^{n}(1\plus{}a_{k}x\plus{}b_{k}x^{2}\plus{}...)^{\frac{1}{p}}\minus{}n$
every my step I will be explain
$ (1\plus{}a_{k}x\plus{}b_{k}x^{2}\plus{}...)^{\frac{1}{p}}\equal{} 1\plus{}\frac{a_{k}x\plus{}b_{k}x^{2}}{p}\plus{}\frac{1}{2p}(\frac{1}{p}\minus{}1)a_{k}^{2}x^{2}\plus{}O(x^{3})$(By Taylor)
therefore after summation we have
$ \sum (1\plus{}\frac{a_{k}x\plus{}b_{k}x^{2}}{p}\plus{}\frac{1}{2p}(\frac{1}{p}\minus{}1)a_{k}^{2}x^{2}\plus{}O(x^{3})) \equal{} n\plus{}\frac{x}{p}\sum a_{k}\plus{}x^{2}\sum(\frac{b_{k}}{p}\plus{}\frac{1}{2p}(\frac{1}{p}\minus{}1)a_{k}^{2}\plus{}O(x))$
therefore we have (because $ \sum a_{k}\equal{} 0$)
$ \lim_{x\to 0}\frac{1}{x^{2}}x^{2}\sum(\frac{b_{k}}{p}\plus{}\frac{1}{2p}(\frac{1}{p}\minus{}1)a_{k}^{2}\plus{}O(x)) \equal{}\sum(\frac{b_{k}}{p}\plus{}\frac{1}{2p}(\frac{1}{p}\minus{}1)a_{k}^{2})$

I don't understand you.
I said that every this problem can be solved by Taylor (without l'Hospital's rules).
in first you don't say nothing about Taylor. after my post (solution of your generalization problem) and you edit your post ,  and then you say about Taylor.

[color=darkblue][b]O.K. ! You are right.[/b] But I wanted to say without derivatives in generally, i.e. l'H.'s, Taylor's, a.s.o. !

The your proof with the Taylor's series is very nicely and quickly !

And yet I wish to seek, if possible, a proof without .... It is the my entertainment. Otherwise, I am bored with Taylor's & Co. [/color]

[quote="Extremal"]mm, but " the l'Hospital's rules" and Taylor is difference[/quote]

[color=darkred][b]Extremal[/b], allow me a rhetorical issue : is truly that the Taylor's series are proved with the l'Hospital's rules ?![/color]

[quote="Virgil Nicula"][color=darkblue]The your proof with the Taylor's series is very nicely and quickly !

And yet I wish to seek, if possible, a proof without .... It is the my entertainment. Otherwise, I am bored with Taylor's & Co. [/color]

[quote="Extremal"]mm, but " the l'Hospital's rules" and Taylor is difference[/quote]

[color=darkred][b]Extremal[/b], allow me a rhetorical issue : is truly that the Taylor's series are proved with the l'Hospital's rules ?![/color][/quote]
In my class we don't use l'Hospital's rules to prove that every "good" function can be written in point $ x_{0}$ as $ f(x) \equal{}\sum\frac{f^{(i)}(x_{0})(x\minus{}x_{0})^{i}}{i!}$
if you want I can post proof from Fihtengol'c ,math analysis. (in Russian language).
Idea is next:
at first we prove it for polynom $ p(x)$ and after we consider $ f(x)\minus{}p(x)$ and prove that it is small.

[color=darkblue][b]Not even at least the estimate of the rest ?[/b]

[b]P.S.[/b] What is the your opinion w.r.t. the exercises $ 1$ and $ 2$ ?[/color]

[hide="6"]
$ \lim_{x\to 1}\frac{x^{n}\minus{}nx\plus{}n\minus{}1}{\frac{(e^{a})^{ln(x)}}{ln(x)}\cdot ln^{2}(x)}\equal{}\lim_{x\to 1}\frac{(x^{n}\minus{}nx\plus{}n\minus{}1)}{a(x\minus{}1)^{2}}\cdot\left(\frac{x\minus{}1}{ln(x)}\right)^{2}\equal{}\lim_{x\to 1}\frac{(x\minus{}1)(x^{n\minus{}1}\plus{}..\plus{}x\minus{}(n\minus{}1))}{a(x\minus{}1)^{2}}$

$ \equal{}\lim_{x\to 1}\frac{(x^{n\minus{}1}\minus{}1)\plus{}(x^{n\minus{}2}\minus{}1)\plus{}..\plus{}x\minus{}1}{a(x\minus{}1)}\equal{}$

$ \lim_{x\to 1}\frac{(x\minus{}1)(x^{n\minus{}2}\plus{}2x^{n\minus{}3}\plus{}3x^{n\minus{}4}\plus{}..\plus{}(n\minus{}1)\cdot 1)}{a(x\minus{}1)}\equal{}\frac{n(n\minus{}1)}{2a}$

For the first and the second step i used $ \lim_{f(x)\to 0}\frac{c^{f(x)}\minus{}1}{f(x)}\equal{} ln(c)$ and $ \lim_{x\to 0}\frac{ln(x\plus{}1)}{x}\equal{} 1$
[/hide]

[hide="9"]

$ \lim_{x\to 0}\frac{x^{2}\plus{}x\plus{}1\minus{}\sin(x)\minus{}\cos(x)}{x^{2}\cdot\left(\frac{e^{x}\minus{}1}{x}\right)^{2}}\equal{}\lim_{x\to 0}\frac{x^{2}\plus{}x\plus{}1\minus{}\sin(x)\minus{}\cos(x)}{x^{2}}$

$ \equal{} 1\plus{}\lim_{x\to 0}\frac{1}{x}\left(1\minus{}\frac{sinx}{x}\right)\plus{}\lim_{x\to 0}\left(\frac{1}{x^{2}}\minus{}\frac{1\minus{}2\sin^{2}\left(\frac{x}{2}\right)}{4\cdot\left(\frac{x}{2}\right)^{2}}\right)$

$ \equal{} 1\plus{}\frac{1}{2}\plus{}\lim_{x\to 0}\frac{1}{x}\left( 1\minus{}\frac{\sin(x)}{x}\right) \equal{}\frac{3}{2}$

In order to prove the last limit we may use $ x\minus{}\frac{x^{3}}{6}\leq sinx\leq x$ when $ x\to 0$
[/hide]

[hide="1"]
This isn't good; the function $ \arcsin\frac{x\plus{}1}{x}$ isn't even defined for positive $ x$.
[/hide]
[hide="2"]
The limit cannot exist. For any positive integer $ N$, the function takes on 0 at $ \pi\left( N\plus{}\frac{1}{4}\right)$, but can be made sufficiently large at $ N\pi\minus{}\epsilon_{N}$, where $ \epsilon_{N}$ is a small enough positive real number.
[/hide]
[hide="3"]
We have $ \sqrt{x\plus{}1}\equal{}\frac{\sqrt{x\plus{}1}\plus{}\sqrt{x}}{2}\plus{}\frac{\sqrt{x\plus{}1}\minus{}\sqrt{x}}{2}$ and $ \sqrt{x}\equal{}\frac{\sqrt{x\plus{}1}\plus{}\sqrt{x}}{2}\minus{}\frac{\sqrt{x\plus{}1}\minus{}\sqrt{x}}{2}$. As a result,
\[ \sin\sqrt{x\plus{}1}\minus{}\sin\sqrt{x}\equal{}2\cos\left(\frac{\sqrt{x\plus{}1}\plus{}\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x\plus{}1}\minus{}\sqrt{x}}{2}\right)\]
Now $ \frac{\sqrt{x\plus{}1}\minus{}\sqrt{x}}{2}\equal{}\frac{1}{2\left(\sqrt{x}\plus{}\sqrt{x\plus{}1}\right)}$ so the above function is bounded above in absolute value by $ 2\left|\sin\left(\frac{1}{2\left(\sqrt{x}\plus{}\sqrt{x\plus{}1}\right)}\right)\right|$ so it tends to $ \boxed{0}$ as $ x\rightarrow\infty$.
[/hide]

Someone can post some things about l'Hospital's rules ?. I think it is famous method , but i don't know it  :blush:

[quote="scorpius119"][b]This isn't good[/b] ; the function $ f(x) =\arcsin\frac{x+1}{x}$ isn't even defined for positive $ x$.[/quote]
[color=darkblue]O.K., you didn't fall in a trap ! However, what means "[b]this isn't good[/b]" ?!
More correctly, the [b]mathematical object[/b] $ 1\blacktriangleright\lim_{x\to\infty}\arcsin\frac{x+1}{x}$ [b]hasn't sense[/b]
because the [b]definition domain[/b] of the function $ f$ is ${ \mathcal D =(-\infty},-\frac{1}{2}]$ and ${ \infty\not\in\mathcal D'}=[-\infty,-\frac{1}{2}]$- the [b]derivate[/b] of $ \mathcal D$--
the set of the accumulation points $ \alpha$ w.r.t. the set $ \mathcal D$, i.e. exists a sequence $ x_{n}\ ,\ n\in N^{*}$ so that $ \{\begin{array}{c}(\forall ) n\in N^{*}\,\ x_{n}\in\mathcal D\\ \\ (\forall )n\in N^{*}\ ,\ x_{n}\ne\alpha\\ \\ x_{n}\rightarrow\alpha\end{array}$.
Unlike it, the mathematical object $ 2\blacktriangleright\lim_{x\to\infty}\frac{1-\cot x}{e^{x}}$ [b]has sense, but it doesn't exists [/b]!

In conclusion, [b]the chain of a work[/b] with a mathematical object is :

$ 1\blacktriangleright$  [b]Has it a sense ?[/b] If NOT, then STOP.

$ 2\blacktriangleright$ [b]Exists it ?[/b] If NOT, then STOP.

$ 3\blacktriangleright$ [b]Is it uniquelly ?[/b] If NOT, how many ?

$ 4\blacktriangleright$ [b]Can construct at least one from these objects ?[/b]

Thanks for the your interest ! By the way, what means  "Location [b]..., PA[/b]" ? (exists m.p.).[/color]

If we want to "hairsplit", we could consider the principal value of the complex arcsine, in which case that limit (1) would be equal to $ \frac{\pi}{2}$.
[quote="Virgil Nicula"][color=darkred][b]Extremal[/b], allow me a rhetorical issue : is truly that the Taylor's series are proved with the l'Hospital's rules ?![/color][/quote]
[url=http://planetmath.org/?op=getobj&from=objects&id=2814]Here[/url] is a proof via Rolle's Theorem.

[color=darkblue]Thank you, [b]Perfect_Radio[/b] ! O.K., [b]Extremal[/b], I should say "[b]with derivatives[/b]" instead of "[b]with the l'H.'s rules[/b]".

From it aroused the all misunderstanding. One think and another write, it is the my "tiny" sin ![/color]

[quote="scorpius119"][b]This isn't good[/b] ; the function $ f(x) \equal{}\arcsin\frac{x\plus{}1}{x}$ isn't even defined for positive $ x$.[/quote]
[/quote]
I think he mean this
Name 	Usual notation 	Definition 	Domain of x for real result 	Range of usual principal value
arcsine 	y = arcsin(x) 	x = sin(y) 	−1 to +1 	−π/2 ≤ y ≤ π/2
maybe problem is next:
find limit
$ f(x) \equal{}\arcsin\frac{x\minus{}1}{x}$
then answer is $ \frac{\pi}{2}$


algorithm percent calculus derivative function puzzles

You are trapped in a strange, strange, world. Once you figure out how the world works, you escape.

You will need to ask several questions in order to solve this. Please only ask yes/no questions.

Is this world entirely real? Am I alone in this world?

Is this world entirely real? [b]Can't answer[/b]
Am I alone in this world? [b]No[/b]

Are there other humans here?
Are there other animals here?
Are there other supernatural beings here?
Are there any other creatures that somehow don't fit in the above categories?

Are there other humans here? [b]No[/b]
Are there other animals here? [b]Yes[/b]
Are there other supernatural beings here? [b]Yes[/b]
Are there any other creatures that somehow don't fit in the above categories? [b]Yes[/b]

teehee

Is there a class structure in this world? like hierachy?

is there oxygen?

Can I ask any questions? or only questions that "I" can possibly get the answer if I were in that world?

If I can ask any question, here are the questions:
Is the world related to computer?
Or crazy experiment?
Is the whole world controlled by one person(or being)?
or a group of person?
He/She/It/They would make profit from this world?
He/She/It/They do this only because of curiosity?
Am I human in the world?

Is there a class structure in this world? like hierachy? [b]No[/b]
is there oxygen? [b]Yes[/b]
Can I ask any questions? [b]No[/b]
or only questions that "I" can possibly get the answer if I were in that world? [b]No[/b]

Is the world related to computer? [b]No[/b]
Or crazy experiment? [b]No[/b]
Is the whole world controlled by one person(or being)? [b]No[/b]
or a group of person? [b]No[/b]
He/She/It/They would make profit from this world? [b]No[/b]
He/She/It/They do this only because of curiosity? [b]No[/b]
Am I human in the world? [b]Yes[/b]

teehee

Assume nothing!

Is it somewhere on the Earth?
Can the animals there be found in the normal world?

Do I exist? [b]No[/b]
Is it somewhere on the Earth? [b]No[/b]
Can the animals there be found in the normal world? [b]No[/b]

Contradictions are your most valuable clue

Will you always tell the truth?
Will you always tell the truth?

[quote="randomdragoon"]Do I exist? [b]No[/b][/quote]

...  :o

Okay, so

Does anything exist?

Is this world virtual?

Could this world co-exist with the real universe?

Is this world a fragment of my imagination?

Has any other human ever been here?

Is it possible to get out without physically doing anything?

Well, Temperal, consider this:
[quote="lingomaniac88"]Do your responses have anything to do with the structure of the question?[/quote]
[quote="randomdragoon"]Do your responses have anything to do with the structure of the question? :D[b]No[/b][/quote]

I'm not saying you're wrong.  I'm thinking that the  :D  is a clue that it does have something to do with the structure of the question.  But that's just a hunch.

Fine, we can actually find out.

Do your responses have anything to do with the structure of the question?

Do your answers have anything to do with the structure of the question?

Do your answers have anything to do with how the question is worded?

Your answers do have something to do with the wording of the question, right?

Your answers depend on the structure of the questions, correct?

 :D

How about we all protest and quit posting a question, so randomdragoon will eventually have to tell us the answer or give hints..or stop lying! :mad:  Do you guys want to do that?

[quote="DonkeyKong"]How about we all protest and quit posting a question, so randomdragoon will eventually have to tell us the answer or give hints..or stop lying! Do you guys want to do that?[/quote]

hints or answers spoil the fun

Okay here are some more questions

 :D ?
Is this a ninja :ninja: ?
ddr :ddr: ?

I want to know more about how he reacts to odd symbols

oh.. fine..

Some questions:
Do you wish to be in this strange world?If you do, do you want to stay forever?
Do the residents in this world have green card or any kind of identification?
Do they(residents) have a goverment?
Do they(residents) have armies?
Do they(residents) work?
Do they(residents) have soul?
Do they(residents) have transportation?
Do they(residents) have house to live in?
Do they(residents) have physical features?
Do they(residents) have emotions?
Are they(residents) weird? or depends on how you look at them..?

If all the answer to my questions are no, What the heck is wrong with that world??

Cool, I guess I was kind of on the right track with having the answers to the questions relating to the structure of the questions.  Although the answers have to do with the structure of the previous question...

[quote="randomdragoon"][quote]Looking at the previous list, we can see that the answer is "yes" if the previous letter is a vowel and "no" if it is a consonant. [/quote]
By God, you've escaped! Congratulations!

if it is the first question: answer [b]can't answer[/b]
if the previous question ended with a consonant: answer [b]no[/b]
if the previous question ended with a vowel: answer [b]yes[/b]
if the previous question ended with a digit: answer [b]I don't know[/b][/quote]
Whose quote is that?
Is the game over??
If it is, wow... it really took us long time to figure this out.
I think we should to something special for the person who got right...!

[quote="Temperal"]I guess I was mistaken.  :D

But if the game's over, what's this?

if it is the first question: answer can't answer
if the previous question ended with a consonant: answer no
if the previous question ended with a vowel: answer yes
if the previous question ended with a digit: answer I don't know[/quote]
I think randomdragoon is telling us the "world" that he is been trying to tell us.

that was pco's quote. donkey kong- your idea, what should we do? PM please

@ProtestanT and Co.: The game has already ended; randomdragoon explained the world to everyone in his previous post.

mathnerd 314, what does your sig mean?
if anyone wants some good riddles, go to funny pages in puzzles and brainteasers.


calculus integration function limit LaTeX inequalities

here is a nice property of integrals i came across.
1)Let $ f(x)$ be a strictly increasing convex function on $ (a,b)$ .Prove that
$ (b - a)f(\frac {a + b}{2})\leq \int_{a}^{b}f(x) dx\leq (b - a)(\frac {f(a) + f(b)}{2})$

i got a proof of the lhs but ,for rhs ive only  got the geometric interpretation.can someone provide a rigorous proof of the r.h.s using only calculus.

2)Prove that:

$ \lim_{n\to \infty} (\int_{a}^{b}(f(x))^{n}dx)^{\frac {1}{n}}$ is the maximum value of the function on $ [a,b]$.

[color=blue]MOD EDIT:I latexified it[/color]
u can learn $ \text{\LaTeX}$ [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]here[/url]

Define $ \Psi(x) = \int_{0}^{x} f(t) dt$
$ \frac {\Psi(b) - \Psi(a)}{b - a} \leq \frac {f(a) + f(b)}{2}$
by LMVT there exists an $ \alpha \in (a,b)$ such that $ \Psi'(\alpha) = \frac {\Psi(b) - \Psi(a)}{b - a} = f(\alpha)$
so it suffices to prove that for $ \alpha \in (a,b)$ we have
$ f(\alpha) \leq \frac {f(a) + f(b)}{2}$
now by the definition of convexity of $ f \in [a,b]$ we have for every $ t \in [0,1]$ we have the inequality
$ f(at + (1 - t)b) \leq tf(a) + (1 - t)f(b)$
let $ t = \frac {b - \alpha}{b - a}$ clearly $ t \in [0,1]$
so we have
 ${ f(\alpha) \leq \frac {b - \alpha}{b - a}f(a) + \frac {\alpha - a}{b - a}f(b)\leq \frac {f(a) + f(b)}{2} \leftrightarrow (a + b - 2\alpha}f(a) \leq(b + a - 2\alpha)f(b)$

which follows from increasing nature...

also Alter_ego how do u know my place of study??

for 2 i think some extra conditions are needed like the function has to be positive evrywhere otherwise the limit will be meaningless :)

I think I got an answer, but its too painful to sit n type it out :(

don't worry just type in the heart of the proof it i will latexify it.... :)

Too lazy to... Just woke up.. Will post it here in sometime :) Sorry!

Edit : Just realized that its wrong... Working on it again.....

no problem....actually i feel i have lost some part of the question while latexifying it probably :)

Oh.... What exactly is this latex? The link doesnt work on my computer....

Edit: No consecutive posts allowed for less than 10 minutes..... So this is what I did

 I used Riemann's summation method, and the right hand side limit could be proved to be greater than f(b)+f(a) / 2.... Saw it in one of my books earlier....

why not ....
actually $ \text{\LaTeX}$ helps u type mathematically

Oh.. I tried to post again, and it said no consecutive posts in 10 mins... No idea

can u pot ur problem [url=http://www.mathlinks.ro/index.php?f=145]here[/url]

Now its ok.. But I couldnt spot the problem there actually.....

:lol:  :lol:  :lol: that was actually post not spot :rotfl:  :rotfl:

Its just something I use often... No offense! Somewhat like saying its your fault!

Whats with you guys? Lol.... Anyhow, I am just bored beyond words, and we really shouldnt be using this thread to chat....

[quote="pardesi]
so it suffices to prove that for $ \alpha \in (a,b)$ we have
$ f(\alpha) \leq \frac {f(a) + f(b)}{2}$

i think this is not true for all alpha in (a,b).we may choose alpha to be arbitarily close to b and then this statement wont hold....infact the last step in your proof is  not correct.
${ f(\alpha) \leq \frac {b - \alpha}{b - a}f(a) + \frac {\alpha - a}{b - a}f(b)\leq \frac {f(a) + f(b)}{2} \leftrightarrow (a + b - 2\alpha}f(a) \leq(b + a - 2\alpha)f(b)$.

the last step(the very last one ) will only work if $ b+a>2\alpha$. however this condition need not be satisfied as alpha can be anywhere in $ (a,b)$.
[color=blue]
MOD EDIT:I latexified and quoted myself :D  [/color]

right how could i do such asilly mistake...
alter ego just add  dollar sign before and the ned of a $ \text{\LaTeX}$ statemnet...
for ex if u wan to write $ x^{2}$ the code should be [code]$x^{2}$[/code]

actually my last line should be $ (2\alpha \minus{}a\minus{}b) f(a) \leq (2\alpha \minus{}a\minus{}b)f(b)$ do simplify...my siplmification was wrong but from lhs it follows that $ f(\alpha) \geq f(\frac{a\plus{}b}{2})$ so $ 2\alpha \minus{}a\minus{}b \geq 0$ so now we are done


Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she left after she buys all the CDs?

A. $\displaystyle \frac{1}{5}$

B. $\displaystyle \frac{1}{3}$

C. $\displaystyle \frac{2}{5}$

D. $\displaystyle \frac{2}{3}$

E. $\displaystyle \frac{4}{5}$

[hide]
$\frac{1}{5}\cdot3=\frac{3}{5}$
So she has $\frac{2}{5}$ left. So the answer is $C$[/hide]


What integer $ n$ has the property that $ 5^{96}$ greater than $ n^{72}$ and $ 5^{96}$ is less than $ (n\plus{}1)^{72}$?

I don't know of a way to do it by hand, but with a calculator, it is quickly obvious that the answer is $ \boxed{8}$

Here's another way: [hide]You know that $n^{72} < 5^{96} < (n+1)^{72}$. We can see that $72 = \frac{3}{4} \times 96$, so $5^{96} = (5^{\frac{4}{3}})^{72}$. This means that $n < 5^{\frac{4}{3}} = 625^{\frac{1}{3}} < n + 1$. So we're looking for an integer $n$ such that $n^{3} < 625 < (n+1)^{3}$. Looking at the integers, we see that $8^{3} = 512 < 625 < (8+1)^{3} = 9^{3} = 729$, so we get that $n = \boxed{8}$, which is the answer to our problem.[/hide]


integration
function
calculus
algebra
binomial theorem
combinatorics

Express  \[ \sum_{k=0}^n (-1)^k (n-k)!(n+k)!  \] in closed form.

I don't have the time to slice through it right now, but this is solved by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14982]this[/url].

That's strange because it works perfectly fine for me. Anyway, it points to the topic named "Sum", with source "I forgot...", posted by Bruno Holanda in the Proposed and Own Problems subsection of the Algebra section.

The link doesn't work for me either, but here's a solution (sorry if it's the same as the one on Grobber's link, for those of you who can view it):

$\frac {1}{2n+2}*( (n+k)! (n-k+1)! + (n+k+1)! (n-k)! ) = \frac {1}{2n+2} (n+k)! (n-k)! ((n - k +1) + (n+k+1)) = (n-k)! (n+k)!$.  Therefore,

$\sum_{k=0}^n (-1)^k (n-k)!(n+k)! = \frac {1}{2n+2}\sum_{k=0}^n (-1)^k ((n+k)! (n-k+1)! + (n+k+1)! (n-k)!) = \frac {1}{2n+2}(n! (n+1)! + (-1)^n (2n+1)!)$, a valid closed form expression.

Yes it is the same I gave for the other problem (and the link works for me  :)  )

Pierre.


calculus
integration
limit
logarithms
trigonometry
function
calculus computations

Find the following limit value.

\[ \lim_{n\to\infty} \frac{1}{\ln n}\int_{\pi}^{n\pi} \frac{|\sin t|}{t}dt\ (n=2,3,\cdots).  \]

The answer is 0. I don't have any elemntary solution right now.
Ussually Kunny's problem has a trick  :D
We can use the fact that the improper integral $\frac{\sin x}{x}$ is conditionally convergen.

$\frac{\int_0^{\pi} |sin(t)|dt}{\pi}=\frac{2}{\pi}=C$. So it is natural to think that $f(n)=\int_{\pi}^{n \pi} \frac{|sin(t)|dt}{t}$ will be equivalent to $C \int_{\pi}^{n \pi}\frac{dt}{t}$: one will prove that $f(n)$ is equivalent to $Cln(n)$.

$\int_{\pi}^{n \pi} \frac{|sin(t)|dt}{t}-C \int_{\pi}^{n \pi}\frac{dt}{t}=\int_{\pi}^{n \pi}\frac{g(t)dt}{t}$

where $g(t)=|sin(t)|-C$. Because of the choice of $C$, $G(x)=\int_{0}^{x}g(t)dt$ is bounded by a constant $K$.
This is why: $\int_{\pi}^{n \pi}\frac{g(t)dt}{t}=(\frac{G(n \pi)}{n \pi}-\frac{G(\pi)}{\pi}) + \int_{\pi}^{n \pi} \frac{G(t)dt}{t^2}$. Hence $|\int_{\pi}^{n \pi}\frac{g(t)dt}{t}|$ is bounded. Conclusion follows.

I've also have the reasoning like you before, But i don't know if we can take the form
$C \int_{0}^{n \pi} \frac{1}{t} \, dt$. Since the function $|\sin x|$ is not monotone in interval $[0, n \pi]$ while $n \rightarrow \infty$ ( :P  second kind of MVT for integral )

I used $C = |\sin \xi |$ for $\xi$ is a number on $[0, n \pi]$.
Anyway that just first kind of MVT for integral.

I miss in this
$\int_{0}^{\pi} |\sin t| \, dt = 0$  :blush:

[quote="xirti"]You can find from $MVT$a number $C$ such that it respects the desired ecuation. But actually you find $C_n$ because $C$ depends on $n$.

$\int_0^1|\sin t|dt=|\sin t||_0^1=\sin 1\neq 0.$ What did you mean?[/quote]

Ok I' meant $\int_{0}^{\pi} |\sin x| \, dx$ which is equal to 2.
(BTW, you've forgot to integrate $\sin x$.... )

I said that's why i gave wrong answer, my solution is quite bit like Alek's. But since I misscalculate that (fatal mistake), then i gave the answer zero (C=0) which is wrong answer.

Another things is that i mix the condition of MVT for integral second kind with the first one.
$\int_{a}^b f(x) \varphi(x) \, dx = \varphi(a) \int_{a}^{\xi} f(x) \, dx + \varphi(b) \int_{\xi}^{b} f(x) \, dx$. This need $\varphi(x)$ to be monotone decreasing.

Since the first $\int_{a}^b f(x) \varphi(x) \, dx = \varphi(\xi) \int_{a}^b f(x) \, dx$ don't need that.

uuhhh Holiday dull my mind.....


basically what the questions means is that:

is it a reqwuirement of a statement to be true that it must be possible to prove to be true?

does there exist a statement which is true but cannot be proven?

If so, is it really true?

maybe bordering philosophy but I find this really interesting.

look for "Kurt Gödel" and you´ll find answers.

This isn't really linear algebra, it is philosophy. :P

It really depends on your own views. The great mathematician Brouwer (mr. fixed point) was pretty much the founder of intuitionism, which is one of the main "branches" of the "philosophy of math." (The other's are Hilbert's formalism and what's called Platonism -- I think Gödel was this.) Intuitionism, very simplistically put, says that the claim that a statement is true is equivalent to the claim that you can prove it. They do not consider non-constructive proofs to be legitimate. They don't use law of excluded middle. (That is, if you can prove "A or B" and you can prove C from A and also you can prove C from B, intuitionists say you haven't proven C.)

Interestingly, Gödel proved that intuitionism (formally stated) is not a simplification of classical mathematics; it is in fact an extension. This is not his most well-known proof. The idea is to state all theorems that we're used to in the negation (to a non-intuitionist, "$\exists x$ such that the statement $P(x)$ is true" is equivalent to "not $\forall x$ $P(x)$ is false"), and reserve the positive statement for when the proof is constructive. (At least, I that's what I gather; I haven't read the proof.)

What [b]sqrt4[/b] probably was referring to was Gödel's Incompleteness Theorem(s), which (paraphrased) show that any axiomatic system sophisticated enough to state its own consistency is either inconsistent (you can prove the statement $A$ and its negation) or incomplete (there is a true statement that you cannot prove). Think of the statement "There is no proof for this statement." If you can prove it, that's a contradiction. If you can't, it's true, but you can't prove it!

In fact, Peano Arithmetic (the widely accepted axiomatic foundation upon which this forum rests) falls in the scope of Gödel's theorem. Almost all mathematicians believe that PA is incomplete, rather than inconsistent. There is overwhelming evidence that this is true (nobody has found an inconsistency yet!) but it is not proven.

If you're interested in further reading, send me a PM and I'll give you some references.


Find all continuous function $ R\to R$ such that $ f(x^2).f(x) \equal{} 1$, $ \forall x \in R$

[hide]From $ f(x^2)f(x) \equal{} 1$ follows that $ f(x) \neq 0$ for all $ x$; since $ f$ is continuous, it has constant sign. For $ x \equal{} 0$ we have $ f(0)^2 \equal{} 1$, hence $ f(0) \equal{} \pm 1$. Now, from $ f(x^2)f(x) \equal{} 1 \equal{} f(x^4)f(x^2)$ follows $ f(x) \equal{} f(x^4)$. Take $ |x| < 1$, and by iterating $ f(x) \equal{} f(x^{4^n})$.
But then $ \lim_{n \to \infty} x^{4^n} \equal{} 0$, so $ f(x) \equal{} f(0)$.
Similarly, for $ x \equal{} 1$ we have $ f(1) \equal{} \pm 1$, and from $ f(x) \equal{} f(|x|^{1/4^n})$ and $ \lim_{n \to \infty} |x|^{1/4^n} \equal{} 1$ when $ |x| > 1$, follows $ f(x) \equal{} f(1)$.
In conclusion, $ f$ is the constant function $ 1$ or $ \minus{} 1$.[/hide]

$ f(x) \equal{} f( \minus{} x)$ for all $ x$.
We have that $ f(x^2)f(x) \equal{} 1$ and $ f(x^2)f(x^4) \equal{} 1$, so $ f(x) \equal{} f(x^4)$.
Let $ z > 0$. So, $ f(z) \equal{} f(z^{1/4}) \equal{} f(z^{1/16}) \equal{} ... \equal{} f(z^{1/4^n})$.
$ f$ is continuous, so $ f(1) \equal{} \lim _{n\rightarrow \infty} {f(z^{1/4^n})} \equal{} f(z)$.
So, $ f(x)$ is constant for $ x > 0$, and from $ f(x) \equal{} f( \minus{} x)$ we have that $ f(x) \equal{} C$ on $ R\minus{}\{0\}$. From continuously we have that $ f(0) \equal{} C$, too.
So, $ C \equal{} 1$, or $ C \equal{} \minus{} 1$.
Answer: $ f(x) \equal{} 1$ for all $ x$, or $ f(x) \equal{} \minus{} 1$ for all $ x$.

How you got  $ f(x^2)f(x)\equal{}f(x^4)f(x^2)$  ?

[quote="enndb0x"]How you got  $ f(x^2)f(x) \equal{} f(x^4)f(x^2)$  ?[/quote]
Take $ x\rightarrow x^2$ and you wiil have $ f(x^4)f(x^2)\equal{}1\equal{}f(x^2)f(x)$.


If $a$, $b$ and $c$ are greater than $1$, then prove that $1< \log_{ab} a+\log_{bc} b+\log_{ca} c <2$.

[hide="my answer"]
Put $A=\log_{10}a , B=\log_{10}b ,C=log_{10}c .$
 $a>1,b>1,c>1 \Rightarrow A>0,B>0,C>0$

then we have $\log_{ab}a+\log_{bc}b+\log_{ca}c=\frac{A}{A+B}+\frac{B}{B+C}+\frac{C}{C+A}(=P)$
so
         $P>\frac{A}{A+B+C}+\frac{B}{A+B+C}+\frac{C}{A+B+C}=1\cdots (\ast)$

and
         $P=(1-\frac{B}{A+B})+(1-\frac{C}{B+C})+(1-\frac{A}{C+A})$
　　　　   $=3-(\frac{B}{A+B}+\frac{C}{B+C}+\frac{A}{C+A})$
　　　　　$<3-(\frac{B}{A+B+C}+\frac{C}{A+B+C}+\frac{A}{A+B+C})$
　　　　　$=3-\frac{A+B+C}{A+B+C}=2\cdots(\ast\ast)$

therefore,$1< \log_{ab} a+\log_{bc} b+\log_{ca} c <2$

Is this right? :P
[/hide]


Let $ a,b,c$ be  distinct real numbers such that  $ a^{3}\equal{}3(b^{2}\plus{}c^{2})\minus{}25$ , $ b^{3}\equal{}3(a^{2}\plus{}c^{2})\minus{}25$ , $ c^{3}\equal{}3(b^{2}\plus{}a^{2})\minus{}25$ . Find the value of $ abc$.

Subtracting of both sides of two equations gives $ a\plus{}b\plus{}c\equal{}\minus{}3$.

Let :
1) $ a^3 \equal{} 3(b^2 \plus{} c^2) \minus{} 25$
2) $ b^3 \equal{} 3(a^2 \plus{} c^2) \minus{} 25$
3) $ c^3 \equal{} 3(b^2 \plus{} a^2) \minus{} 25$

Now sub 1) - 2) : $ a^3 \minus{} b^3 \equal{} 3(b^2 \minus{} a^2)$ or $ (a \minus{} b)(a^2 \plus{} ab \plus{} b^2) \equal{} \minus{} 3(a \minus{} b)(a \plus{} b)$. This means that $ a \equal{} b$ or $ a^2 \plus{} ab \plus{} b^2 \plus{} 3(a \plus{} b) \equal{} 0$. The same for 2) - 3) : $ b \equal{} c$ or $ b^2 \plus{} bc \plus{} c^2 \plus{} 3(b \plus{} c) \equal{} 0$. And for 1) - 3) $ a \equal{} c$ or $ a^2 \plus{} ac \plus{} c^2 \plus{} 3(a \plus{} c) \equal{} 0$.

So, if $ a \equal{} b \equal{} c$ then we should find $ a$ from here : $ a^3 \minus{} 6a^2 \plus{} 25 \equal{} 0,$ which gives $ a \equal{} 5$ and $ abc \equal{} a^3 \equal{} 125.$

If $ a\equal{}b$ and $ b^2 \plus{} bc \plus{} c^2 \plus{} 3(b \plus{} c) \equal{} 0$ get : $ 27(6a^2\minus{}25)\equal{}(a^3\minus{}3a^2\plus{}25)^3$ which has solution $ a\equal{}5.$ This gives $ c\equal{}5$. So $ abc\equal{}125$ again.

After finding $ a\plus{}b\plus{}c\equal{}\minus{}3$, then adding of both sides of 3 equations gives $ abc\equal{}2\minus{}7(ab\plus{}bc\plus{}ca)$.

Maybe $ ab\plus{}bc\plus{}ca$ can be found by multiplying both sides of 3 equations.

sorry, i don't know what is distinct :blush: now ok

... $ (a\plus{}b\plus{}c\equal{}\minus{}3 , ab\plus{}ac\plus{}bc\equal{}0 )$ -->  $ abc\equal{}2$ :)


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

$p$ a prime number.
A subgroup $G$ of $S_p$ is 'good' if:
* $G$ acts transitively on $\{1,2,...,p\}$
* $G$ is solvable.

$H=\{x\to ax+b |a \in (\frac{Z}{pZ})^*, b\in \frac{Z}{pZ}\}$ is a group of automorphisms of $\frac{Z}{pZ}$. It can be seen as a subgroups of $S_p$.
Prove that any "good" subgroup of $S_p$ is conjuated to a subgroup of $H$..

How about subgroups of $S_p$ generated by a cycle of length $p$? They're abelian and thus soluble and certainly act transitively on the set, but they are not conjugates of $H$. :?

aaa.. Missed that :) It even took me a while to see what you meant. Thank you.

Edit:

[b]Lemma:[/b] If $p$ is prime and $G$ is a transitive subgroup of $S_p$, then any non-trivial normal subgroup $N$ of $G$ is also transitive.

[b]Proof:[/b] There is a proof in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=orbit&t=47737]this topic[/url] that if $G$ acts transitively on a set, then the action of any normal subgroup of $G$ on that set has orbits of equal size. Since this size must be a divisor of $p$ which is also $>1$ in our case, the lemma is proven.

We now prove the main result by induction on $|G|$. If $G$ is abelian, then we can conclude that it's cyclic of order $p$: in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=abelian&t=46589]this topic[/url] it was shown that whenever $n$ is square-free, the only transitive abelian subgroups of $S_n$ are the groups generated by a cycle of length $n$, so this must hold in our case ($n=p$) as well, and we're done. Otherwise, there must be a normal non-trivial subgroup $N$ of $G$, different from $G$. We can apply the induction hypothesis on $N$, since $N$ is soluble, being a subgroup of a soluble group, and it's transitive because this is what the lemma above states. By the induction hypothesis, $N$ is a transitive conjugate of a subgroup of $H$, meaning that it has a characteristic subgroup $P$ of order $p$. Since $P$ is characteristic in $N$ and $N$ is normal in $G$, we conclude that $P$ is normal in $G$, so $G$ is a subgroup of the normalizer of $P$ in $S_p$. It's easy to see, however, that the normalizer of a subgroup of order $p$ of $S_p$ is a conjugate of $H$, and the induction step is done.


algebra
polynomial
number theory unsolved
number theory

Let $ k \in \mathbb{N}$. A polynomial is called [i]$ k$-valid[/i] if all its coefficients are integers between 0 and $ k$ inclusively. (Here we don't consider 0 to be a natural number.)

[b]a.)[/b] For $ n \in \mathbb{N}$ let $ a_n$ be the number of 5-valid polynomials $ p$ which satisfy $ p(3) = n.$ Prove that each natural number occurs in the sequence $ (a_n)_n$ at least once but only finitely often.

[b]b.)[/b] For $ n \in \mathbb{N}$ let $ a_n$ be the number of 4-valid polynomials $ p$ which satisfy $ p(3) = n.$ Prove that each natural number occurs infinitely often in the sequence $ (a_n)_n$ .

For a:
Notice that $a_{n}=a_{\left[\dfrac{n}{3}\right]}+a_{\left[\dfrac{n}{3}\right]+1}$
Then it can be proven $a_{n}-a_{n-1}=0$ or$1$


Each of the digits 2, 4, 5, 6, 8 and 9 is placed in exactly one of the boxes of the subtraction problem shown. What is the least possible positive difference?

\begin{tabular}{c}  \Box \Box \Box - \Box \Box \Box \end{tabular}

You want the LEAST possible positive difference.
The least difference you will be have when you put two numbers in the hundreds place is 1.
These are your choices:
5_ _ - 4_ _
6_ _ - 5_ _
9_ _ - 8_ _
Now look at the other four numbers.
Take the first one for example. You have 2, 6, 8, and 9 left.
Create the smallest two-digit number possible. 26
Now create the greatest two-digit number possible of the remaining ones. 98.
Subtract, and you get -72. You want this difference as small as possible. At the end, you will add this number to 100, which you already have calculated.
So you have:
526-498 = 28
624-598 = 26
924-865 = 59.
Therefore, the LEAST positive difference is 26.


parameterization
inequalities
function
absolute value
algebra proposed
algebra

Hi, all mathlinkers...
Happy New Year
I wish you all the best!

I'm sorry for my abscence but I was too busy recently...
I found some interesting problems on the Net and I want to share them with you ... feel them like Christmas gifts with some delay...

Solve in reals the following systems:

1. $ \{\begin{aligned} (x^{2} + y^{2} - xy)(x - y) = 1 + y^{3} \\
(x^{2} + y^{2} + xy)(x + y) = 1 - y^{3}\end{aligned}$

2. $ \{\begin{aligned} \frac {2x}{1 - x^{2}} = \sqrt {\frac {1 - y}{1 + y}} \\
x^{2} + 1 = \frac {1}{y^{2}}\end{aligned}$

3. $ \{\begin{aligned} x + y^{2} = z^{3} \\
x^{2} + y^{3} = z^{4} \\
x^{3} + y^{4} = z^{5} \end{aligned}$

4. For what values of real parameters a, b the following system of reals have exactly three (pairs of) solutions?
$ \{\begin{aligned} x^{2} + y^{2} = a^{2} \\
|y| = x + b\end{aligned}$

5. Real non-zero numbers $ a,b,c$ satisfies the following conditions:
$ a^{2} - b^{2} = bc$
$ b^{2} - c^{2} = ca$
Prove that: $ a^{2} - c^{2} = ab$

6. Prove that if the following inequalities for real number are true:
$ a_{1} - 3a_{2} + 2a_{3} \geq 0 \\
a_{2} - 3a_{3} + 2a_{4} \geq 0 \\
... \\
a_{n - 2} - 3a_{n - 1} + 2a_{n} \geq 0 \\
a_{n - 1} - 3a_{n + 2}a_{1} \geq 0 \\
a_{n - 3}a_{1} + 2a_{2} \geq 0$
Then $ a_{1} = ... = a_{n}$

7. Solve the following system of reals:
$ (1 + 4x_{1})^{4} = 1 + 4x_{2} \\
(1 + 4x_{2})^{4} = 1 + 4x_{3} \\
... \\
(1 + 4x_{n - 1})^{4} = 1 + 4x_{n} \\
(1 + 4x_{n})^{4} = 1 + 4x_{1}$

I've solutions to some of the problems ... I also have answers to all of the problems. I believe you will like these problems ... enjoy them.

[hide="1"]Suppose that $ x \equal{} y$. Then the first equation gives that $ x^3 \equal{} \minus{}1$ and the second equation gives that $ x^3 \equal{} 1/7$. Contradiction. We get the same information for $ x \equal{} \minus{}y$, so contradiction there too. Therefore we can divide by $ x\plus{}y$, $ x\minus{}y$ without losing solutions.

Multiply the first equation by $ x\plus{}y$ and the second by $ x\minus{}y$.
\[ (x^3 \plus{} y^3)(x\minus{}y) \equal{} (1\plus{}y^3)(x\plus{}y)\]
\[ (x^3 \minus{} y^3)(x\plus{}y) \equal{} (1\minus{}y^3)(x\minus{}y)\]

Multiplying both equations together and cancelling $ (x\plus{}y)(x\minus{}y)$ gives

\[ x^6 \minus{} y^6 \equal{} 1 \minus{} y^6\]

So $ x^6 \equal{} 1$ and $ x \equal{} \pm 1$. If $ x \equal{} 1$, the first equation gives $ 2y^3 \minus{} 2y^2 \plus{} 2y \equal{} 0$ and the second equation gives $ 2y^3 \plus{} 2y^2 \plus{} 2y \equal{} 0$. Subtracting the two gives $ y^2 \equal{} 0$, so $ y \equal{} 0$. We get the same information (from different equations) for $ x \equal{} \minus{}1$. So there are two solutions: $ (x,y) \in \{(\minus{}1,0), (1,0)\}$.[/hide]

[hide="4"]If $ (x,y)$ is a solution, $ (x,\minus{}y)$ is a solution. So for an odd number of solutions we need $ y \equal{} 0$ to give a solution. This implies $ x^2 \equal{} a^2$ and $ x \equal{} \minus{}b$. So we require $ b^2 \equal{} a^2$. Suppose $ b < 0$. For there two be two other solutions besides $ (x,y) \equal{} (\minus{}b,0)$, we need a solution with $ |y| > 0$. But then the second equation gives that $ x > \minus{}b$, implying that $ x^2 > b^2 \equal{} a^2$ contradicting the first equation. So $ b \geq 0$. We can see that if $ a$ is any nonzero real and $ b$ is the absolute value of $ a$, then $ (x,y)$ can only be $ (\minus{}b,0), (0,b), (0,\minus{}b)$, all of which are distinct solutions.[/hide]

[hide="5"]Not true: $ (a,b,c) \equal{} (0,1,\minus{}1)$. The needed restriction is $ a \neq 0$. Proof using that:

Add the two equations to get $ a^2 \minus{} c^2 \equal{} c(b\plus{}a)$. Multiply both sides by $ a\minus{}b$ to get $ (a\minus{}b)(a^2 \minus{} c^2) \equal{} c(a^2 \minus{} b^2) \equal{} bc^2$. Expand the LHS: $ a^3 \minus{} ba^2 \minus{} ac^2 \plus{} bc^2 \equal{} bc^2$, so $ a^3 \minus{} ba^2 \minus{} ac^2 \equal{} 0$. This implies that $ a(a^2 \minus{} ab \minus{} c^2) \equal{} 0$. Since $ a \neq 0$, $ a^2 \minus{} ab \minus{} c^2 \equal{} 0$, as desired.[/hide]

[hide="6"]Assuming the last two equations have typos (overactive bracing :P ) and that the whole thing is cyclic. Assume $ a_k \equal{} a_{n\plus{}k}$. Suppose $ a_i < a_{i\plus{}1}$. Then the $ i$th equation shows that $ a_{i\plus{}1} < a_{i\plus{}2}$. Thus, $ a_i < a_{i\plus{}1} < a_{i\plus{}2} < \ldots < a_{n\plus{}i} \equal{} a_i$, contradiction. So the sequence must have all equal terms.[/hide]

[hide="7"]Let $ f(x) \equal{} 1 \plus{} 4x$ (note it is bijective). We see that $ f(x_1)^{4^n} \equal{} f(x_2)^{4^{n\minus{}1}} \equal{} \ldots \equal{} f(x_n)^4 \equal{} f(x_1)$. So $ f(x_1)$ can only be $ 0,1$, and this gets $ f(x_1) \equal{} f(x_2) \equal{} \ldots \equal{} f(x_n)$. $ x_1$ is $ \minus{}1/4,0$ when $ f(x_1)$ is $ 0,1$ respectively. This gives the following two solutions:
$ x_1 \equal{} x_2 \equal{} \ldots \equal{} x_n, x_1 \in \{\minus{}1/4,0\}$.[/hide]

Here is solution of this problems.
1. $ x^3 \minus{} y^3 \minus{} 2xy (x \minus{} y) \equal{} 1 \plus{} y^3$ so $ x^3 \minus{} 2y^3 \minus{} 2xy(x \minus{} y) \equal{} 1$ (1)
    $ x^3 \plus{} 2y^3 \plus{} 2xy (x \plus{} y ) \equal{} 1$ (2)
Consider $ (1) \minus{} (2)$ we have :
$ 4 (y^3 \plus{} x^2y) \equal{} 0,\Leftrightarrow y (x^2 \plus{} y^2) \equal{} 0$
[u]Case 1 :[/u] $ y \equal{} 0$ then $ x \equal{} 1$
[u]Case2 [/u] $ x^2 \plus{} y^2 \equal{} 0$ so $ x \equal{} y \equal{} 0$ this is not solution of this system .
So $ (x,y) \equal{} (0,1)$
2. Easy to check $ 1 \minus{} y^2 \equal{} (xy)^2$ and we enough when $ x\geq 0$ the case $ x < 0$ the same.
$ \frac {2x}{1 \minus{} x62} \equal{} |\frac {xy}{y \plus{} 1}|$
Then the equation quite easy to solve.
3) We have
$ (x \plus{} y^2)(x^3 \plus{} y^4) \equal{} (x^2 \plus{} y^3)^2$
$ \Leftrightarrow xy^2(x \minus{} y)^2 \equal{} 0$
[u]Case 1[/u] $ x \equal{} 0$
$ (y,z) \equal{} (0,0),(1,1)$
[u]Case 2 [/u] $ y \equal{} 0$ then
$ (x,z) \equal{} (0,0 ) ,(1,1),( \minus{} 1, \minus{} 1)$
[u]Case 3[/u] $ x \equal{} y$
$ z^4 (x^2 \plus{} x) \equal{} z^3(x^3 \plus{} x^2)$
If $ x \equal{} \minus{} 1$ then $ z \equal{} \minus{} 1$
$ z \equal{} 0$ then $ x \equal{} 0$ or $ x \equal{} \minus{} 1$
If $ z(x \plus{} 1)$ is not equal 0 then $ x \equal{} z$
So $ x(x^2 \minus{} x \minus{} 1) \equal{} 0$
easy to find out all solution of this equation.
4.From the equation :
$ x^2 \plus{} (x \plus{} b)^2 \equal{} a^2$
Because have exactly three solution then  this equation have root $ x \plus{} b \equal{} 0$ or $ x \equal{} \minus{} b$
So $ b^2 \equal{} a^2$
The equation become $ x(x \plus{} b) \equal{} 0$
so $ x \equal{} 0,x \equal{} \minus{} b$
So $ |y| \equal{} b$ it mean that $ b > 0$
Result $ b > 0,a \equal{} b$ or $ b > 0,a \equal{} \minus{} b$
5. If $ c \equal{} 0$ it is true.
Consider $ c$ is different from 0 .
Let $ x \equal{} \frac {a}{c},y \equal{} \frac {b}{c}$
So
(1) $ x^2\equal{}y^2\plus{}y$
(2)$ y^2 \minus{} 1 \equal{} x$
So $ (y^2 \minus{} 1)x^2 \equal{} (y^2 \plus{} y)x$
The case $ x \equal{} 0$ or $ y \equal{} \minus{} 1$ it is easy to check it is true.
so $ (y \minus{} 1)x \equal{} y$
So $ a^2 \minus{} c^2 \equal{} ab$
Problem claim.
5. Posted before.
6. $ x_k\geq \frac { \minus{} 1}{4}$
So the funtion
$ f(t) \equal{} (4t \plus{} 1)^4$ is monotone on $ [\frac { \minus{} 1}{4}, \plus{} \infty)$
So $ x_1 \equal{} .. \equal{} x_n$
Consider root of the equation
$ (4x \plus{} 1)^4 \equal{} 4x \plus{} 1$ then $ x \equal{} 0$ or $ x \equal{} \frac { \minus{} 1}{4}$
This is solution of this system.
 :) Sorry for my poor english

Thank you guys! Excellent solutions! Did you know these problems in advance?


algebra
polynomial
function
limit
induction
calculus
calculus computations

If F is a polynomial, show that

[img]http://img292.imageshack.us/img292/4382/sumeq03ax0.gif[/img]

consider the function $ g_{i}(x) \equal{} c_{i}x^{i}$
clearly we will prove that $ g_{i}(x)$ is continious at all $ a \in \mathbb{R}$
so we need to prove $ \lim_{x \to a}g_{i}(x) \equal{} g_{i}(a) \equal{} a^{i}$
to prove this we need to show that for each choice of $ \epsilon \in \mathbb{R}^{ \plus{} }$ such that $ |c_{i}x \minus{} a| < \epsilon$ there exists a $ \delta \in \mathbb{R} ^{ \plus{} }$ such that $ |(c_{i}x)^{i} \minus{} a^{i}| < \delta$
note that $ |(c_{i}x)^{i} \plus{} ( \minus{} a^{i})| < |c_{i}x|^{i} \plus{} | \minus{} a^{i}| \equal{} |(c_{i}x)^{i}| \plus{} |a^{i}|$
also $ |c_{i}x| \minus{} |a| < |c_{i}x \minus{} a| < \epsilon \Rightarrow |c_{i}x| < \epsilon \plus{} |a|$
taking powers since both sides are positive we get
$ |(c_{i}x)^{i} \plus{} ( \minus{} a^{i})| < (\epsilon \plus{} |a|)^{i} \plus{} |a|^{i} \equal{} \delta$
hence we get
$ \boxed{\lim_{x \to a}g_{i}(x) \equal{} g_{i}(a) \equal{} a^{i}}$
now any polynomial is of the form $ p(x) \equal{} \sum_{i \equal{} 0}^{n}c_{i}x^{i} \equal{} \sum_{i \equal{} 0}^{n}g_{i}(x)$
hence it is a sum of continious functions...so it itself is continious

that was very helpful. Thank you so much. with all appreciation.

We need some building blocks, some theorems we can use. Here's a collection of theorems. Check to see if you've already proved them, or if you could prove them readily.

1. $ \lim_{x\to a}c\equal{}c.$ (The limit of a constant is itself.)

2. $ \lim_{x\to a}x\equal{}a.$

3. If $ \lim_{x\to a}f(x)\equal{}L$ and $ \lim_{x\to a}g(x)\equal{}M,$ then $ \lim_{x\to a}(f(x)\plus{}g(x))\equal{}L\plus{}M.$ (The limit of a sum is the sum of the limits, provided they both exist.)

4. If $ \lim_{x\to a}f(x)\equal{}L$ and $ \lim_{x\to a}g(x)\equal{}M,$ then $ \lim_{x\to a}f(x)g(x)\equal{}LM.$ (The limit of a product is the product of the limits, provided they both exist.)

The trickiest of these proofs is #4, but it's still straightforward enough to give as an exam question.

With those ingredients, and with free use of mathematical induction, we should be able prove what we want, that polynomials are continuous.

I'll provide one key piece of that proof here:

Lemma: for any $ n\in\mathbb{N},\ \lim_{x\to a}x^n\equal{}a^n.$

Proof by induction: This is true for $ n\equal{}1,$ by #2 above.

Assume it it true for some $ n.$ Then

$ \lim_{x\to a}x^{n\plus{}1}\equal{}\lim_{x\to a}x^n\cdot x\equal{}
\left(\lim_{x\to a}x^n\right)\left(\lim_{x\to a}x\right),$ provided both of those limits exists. (Quoting #4 above.)

$ \equal{}a^n\cdot a$ by the inductive hypothesis and by #2

$ \equal{}a^{n\plus{}1}.$ That is was we wanted, so the lemma is proved by induction.


superior algebra
superior algebra solved

Prove that the only isomarphism of $(\mathbb{R},+,\cdot)$ is the identity.

$f(1)=1$ so $f(q)=q$ for any $q \in \mathbb{Q}$. Moreover, $f(x^2)=f(x)^2 \geq 0$. This is why $f$ is increasing. So one can conclude that $f$ is the identity.

In a book was written that identity is the only [b]continuous[/b] isomorphism. And, as a remark was written that we should not think that this is the only isomorphism. I did the same proof but I was not sure if it works..

It's right. The key is that the order on $\mathbb{R}$ is encoded as a strictly algebraic property, and the order is what determines the topology.

It's possible to construct subfields of $\mathbb{R}$ for which this doesn't work and there are many automorphisms. If $K$ is a field of all algebraic numbers $x$ with the property that all algebraic conjugates of $x$ are real, then $K$ has many automorphisms- the field of all such numbers is Galois over $\mathbb{Q}$.

There are more fields with that property, e.g. the $p$-adic fields $\mathbb{Q}_p$ have just the trivial automorphisms.


[size=150]12345678910111213.....19941995   are written on the board and we name this number as  N_1...[/size]

[b]we will fulfil the following operation to N_1 :[/b] Delete the digits begining from 2 and skipping 3, deleting 4 skipping 5, deleting 6 skipping 7,
                                                               deleting 8 skipping 9, deleting 1 skipping 0 and etc...  and name this new number as N_2
                                                                N_2 = 1357901234567890123456789.......

and [b]we will fulfil the following operation to N_2 :[/b] Delete the digits begining from 1 and skipping 3, deleting 5 skipping 7, deleting 9 skipping 0,
                                                               deleting 1 skipping 2, deleting 3 skipping 4 and etc...  and name this new number as N_3
                                                                N_3 = 370246802468.....

When N is of the form N_2k+1 we start deleting digits from the second digit, and when N is of the form N_2k we start deleting digits from the first digit

....   If we continue with this procedure, what is the last digit written on the board?

$ N_2$ starts with the 1-st digit of $ N_1$
$ N_3$ starts with the 3-rd digit of $ N_1$
$ N_4$ starts with the 3-rd digit of $ N_1$
$ N_5$ starts with the 11-th digit of $ N_1$
and so on.
For $ m\geq 2$,
$ N_{2m \minus{} 1}$ and $ N_{2m}$ both start with the $ \left(1 \plus{} 2\frac {4^{m \minus{} 1} \minus{} 1}3\right)$-th digit of $ N_1$.

We need to find the largest number of this form, not exceeding the length of $ N_1$ equal
\[ 1\cdot 9 \plus{} 2\cdot (99 \minus{} 9) \plus{} 3\cdot (999 \minus{} 99) \plus{} 4\cdot (1995 \minus{} 999) \equal{} 6873.\]
This integer is $ 1 \plus{} 2\frac {4^{m \minus{} 1} \minus{} 1}3$ for $ m \equal{} 7$, that equals 2731.
Therefore, the last digit is given by $ N_{14}$ and equals the 2731-th digit of $ N_1$, that is 9.

THANKS A LOT... THIS IS THE CORRECT ANSWER...

MAY I ASK YOU HOW DID YOU FIND THE FORMULA $ \left(1\plus{}2\frac{4^{m\minus{}1}\minus{}1}{3}\right)$

[quote="epsilon07"]MAY I ASK YOU HOW DID YOU FIND THE FORMULA $ \left(1 \plus{} 2\frac {4^{m \minus{} 1} \minus{} 1}{3}\right)$[/quote]

I GOT IT , NO NEED MORE EXPLANATION, THANKS AGAIN FOR THE SOLUTION


geometry
trapezoid
ratio
geometry unsolved

A trapezoid $ ABCD$ with parallel bases $\overline{ AB}$ and $\overline{ DC}$ is inscribed in a circle $(O).$ $O$ is outside $ ABCD.$ Diagonals of $ ABCD$ intersect at $ E.$ Consider the circle $ \mathcal{C}(r)$ tangent to the diagonals and internally tangent  to the arc $ BC$ of $(O).$ $ M,N$ are midpoints of $\overline{ AB}$ and the arc $ CD$ of $(O),$ respectively. Show that

1) $\frac {1}{r} \equal{} \frac {1}{EM} \plus{} \frac {1}{EN}$

2) If $\overline{ AB}$ is a diameter, find the ratio $ \frac {_{EM}}{^{EN}}$ such that $ \mathcal{C}(r)$ and the circle tangent to the diagonals of the trapezoid and  internally tangent to the  small arc $ CD$ are congruent.

For (1), see [url]http://www.mathlinks.ro/viewtopic.php?t=106321[/url]. For (2), let $ C(r')$ be (the smaller) circle through $ N$ tangent to $ AC, BD.$ $ \frac {EN}{1 \plus{} \frac {EA}{MA}} \equal{} r' \equal{} r \equal{} \frac {EN}{1 \plus{} \frac {EN}{EM}}$ $ \Longleftrightarrow$ $ \frac {EA}{MA} \equal{} \frac {EN}{EM}.$ When $ MA \equal{} R \equal{} EM \plus{} EN,$ then $ x^2 \equal{} \frac {EN^2}{EM^2} \equal{} \frac {EA^2}{R^2} \equal{} \frac {EM^2}{R^2} \plus{} 1 \equal{} \frac {1}{\left(1 \plus{} \frac {EN}{EM}\right)^2} \plus{} 1 \equal{} \frac {1}{(1 \plus{} x)^2} \plus{} 1$ $ \Longleftrightarrow$ $ x^4 \plus{} 2x^3 \minus{} 2x \minus{} 2 \equal{} 0.$ Resolvent cubic $ y^3 \plus{} 4y \plus{} 4 \equal{} 0$ of this quartic does not have constructible roots and neither has the quartic.


topology
advanced fields
advanced fields unsolved

This question came up during an independent project I've been working on.  I don't know the question's status.  If it is treated in the literature, I'd be interested in references.  Otherwise, any input would be appreciated.

We have the following:

A continuously deformable, nonselfintersecting, boundariless, two-dimensional surface cannot be embedded in any three-dimensional (Euclidean) space if it contains a Moebius band.
...examples: Klein bottle, projective plane.

Is there an analogous criterion in the next higher case?

i.e., Under what condition(s) will a continuously deformable, nonselfintersecting, boundariless, three-dimensional surface be impossible to embed in any four-dimensional (Euclidean) space?

[NB.  The (somewhat redundant) nonselfintersecting property is pre-included for convenience.]

Thanx,
Lou Piciullo

O_O

Put it in english and I might be able to read it :P .

Wait, is a mobius band the same thing as a mobius strip?

[quote="Treething"]Wait, is a mobius band the same thing as a mobius strip?[/quote]

Yes -- this is a topology question, one which I am quite ill-equipped to offer any suggestions on.

From my very limited knowledge of topology, isn't a Moebius band three dimensional? How can a two dimensional surface have a three dimensional... thing? And doesn't a Moebius band have boundaries?

Forgive my ignorance.

[quote="Lou Piciullo"]A continuously deformable, nonselfintersecting, boundariless, two-dimensional surface cannot be embedded in any three-dimensional (Euclidean) space if it contains a Moebius band.
...examples: Klein bottle, projective plane.[/quote]

Note the boundariless. Basically a three-dimensional object with only one surface cannot be made. In theory, you could have a infinite plane that would seem two sided but would only have one side.


function
algebra
polynomial
algebra proposed

Hello Jutaro,
[quote="Jutaro"]Given the positive real numbers $a_{1}<a_{2}<\cdots<a_{n}$, consider the function
\[f(x)=\frac{a_{1}}{x+a_{1}}+\frac{a_{2}}{x+a_{2}}+\cdots+\frac{a_{n}}{x+a_{n}}\]
Determine the sum of the lengths of the disjoint intervals formed by all the values of $x$ such that $f(x)>1$.[/quote]

Very nice question.

Let $f(x)=\frac{P(x)}{Q(x)}$ with $P(x)=\sum_{i}a_{i}\prod_{j\neq i}(x+a_{j})$ and $Q(x)=\prod_{i}(x+a_{i})$

$f(x)=1$ $\Leftrightarrow$ $Q(x)-P(x)=0$ and this polynomial have exactly n roots $-a_{n}<r_{n}<-a_{n-1}<r_{n-1}<\cdots<-a_{1}<r_{1}$

The number requested is $\sum_{i}(r_{i}+a_{i})= \sum_{i}r_{i}+\sum_{i}a_{i}$
$\sum_{i}r_{i}$ is the opposite of the coefficient of $x^{n-1}$ in $Q(x)-P(x)$
Coefficient of $x^{n-1}$ in $Q(x)$ is $\sum_{i}a_{i}$
Coefficient of $x^{n-1}$ in $P(x)$ is $\sum_{i}a_{i}$
Hence $\sum_{i}r_{i}=0$

The the number requested is $\sum_{i}a_{i}$

--
Patrick

[quote="pco"]

...
The number requested is $\sum_{i}(r_{i}+a_{i})= \sum_{i}r_{i}+\sum_{i}a_{i}$
...[/quote]

Sorry, but I'm having trouble understanding why it is this sum. I know I'm wrong, but I thought it would be something like $(r_1 - r_2) + (r_3 - r_4) + ...$ (that or $(r_2-r_3) + (r_4-r_5) + ... $).

EDIT: Thanks pco!

[quote="modularmarc101"][quote="pco"]

...
The number requested is $\sum_{i}(r_{i}+a_{i})= \sum_{i}r_{i}+\sum_{i}a_{i}$
...[/quote]

Sorry, but I'm having trouble understanding why it is this sum. I know I'm wrong, but I thought it would be something like $(r_1 - r_2) + (r_3 - r_4) + ...$ (that or $(r_2-r_3) + (r_4-r_5) + ... $).[/quote]

No, you need to go back to the original expression of $f(x)$ :

$f((-\infty,-a_n))=(-\infty,0)$

$f((-a_n,r_n))=(1,+\infty)$ and so we get a "good" interval $(-a_n,r_n)$ whose length is $r_n+a_n$
$f((r_n,-a_{n-1}))=(-\infty,1)$

$f((-a_{n-1},r_{n-1}))=(1,+\infty)$ and so we get another "good" interval $(-a_{n-1},r_{n-1})$ whose length is $r_{n-1}+a_{n-1}$
$f((r_{n-1},-a_{n-2}))=(-\infty,1)$
...
$f((-a_{n-k},r_{n-k}))=(1,+\infty)$ and so we get a "good" interval $(-a_{n-k},r_{n-k})$ whose length is $r_{n-k}+a_{n-k}$
$f((r_{n-k},-a_{n-k-1}))=(-\infty,1)$

Hence the result $\sum_{i}(r_{i}+a_{i})$


let $ S\equal{}\{E_1, E_2, \cdots\}$ be a dosjoint collection of subset of $ \Re.$

Find the $ \sigma\minus{}$algebra generated by $ S.$

[quote="ssheep"]
\[ \sigma(S) \equal{} \{\cup_{i \in I} E_i \ | \ I \subset \{1, 2, \ldots\}\}.
\]
[/quote]

thats not true, is it? what about $ E_1^c$??

note I is a subset of the naturals.. not all of them.

The objection was that the complements should also belong to the $ \sigma$-algebra and $ E_j$ is just a family of disjoint sets, not a partition of $ \mathbb R$, so it may be impossible to represent $ \mathbb R\setminus E_1$ as a union of $ E_j$. I noticed that ssheep changed his answer from the original version but it is still incorrect: how about $ \mathbb R\setminus (E_1\cup E_2)$ ;)

What about this... find the $ \sigma$-algebra generated by $ S$ if $ S\equal{}\{E_1, E_2,\ldots E_n\}$?


How many licence plates can be made using 2 English letters in uppercase followed by a 3-digit number? zThe first digit of a licence plate should not be aero. How many of those licence plates have 2 võels followed by 3 identical digits?

[quote]How many licence plates can be made using 2 English letters in uppercase followed by a 3-digit number? zThe first digit of a licence plate should not be aero. [/quote]

There are 26x26 possible ways to get the 2 letters. Then, there are 9x10x10 possibilities for the 3-digit number. You multiply everything, and you get 608400.

How many of those licence plates have 2 vowels followed by 3 identical digits?

There are 5 vowels (assuming you don't include y), so there are 5x5 possiblities for the first 2 spots. If the 3-digit number can't start with a zero, then there are 9 different 3-digit numbers. So, there are 5x5x9= 225 ways to make that arrangement.

[/quote]


If $ n$ is a prime, then $ n$ is not divisible by the divisors of [$ \sqrt{n}$]

can you prove this for me? I know this is a good tool but I still don't get why

I'm not sure you've got the statement right.  The correct statement is this:  if $ n$ is not divisible by the positive integers less than or equal to $ \lfloor \sqrt{n} \rfloor$, then $ n$ is prime.

The proof begins like this:  if $ d$ divides $ n$, then so does $ \frac{n}{d}$.  And...

and $ \frac{n}{d}$ is going to be less than $ \sqrt{n}$

If and only if $ d$ is greater than $ \sqrt{n}$.  What you meant is $ \text{min} \left( d, \frac{n}{d} \right) \le \lfloor \sqrt{n} \rfloor$.


1.  A combination lock has numbers from 1 to 40.  This brand of lock must use three different numbers in its combination.  How many possible combinations exist for one of these locks?
2.  In a group of ten girls and eight boys on the prom committee two of each must be chosen for the decoration committee.  One of them will then be chosen as leader of the committee.  In how many ways can this committee be chosen?
3.  How many four digit numbers are there in which the four digits are in decreasing order?
4.  How many 4 digit numbers have at least one 7 in them?
5.  The C & P basketball team has 10 players, 3 of which are all-stars.  How many 5 player starting line ups can be chosen that include exactly 2 of the all- stars?

wow, i suck at reading questions.. i misread number 5.. its 3C2*7C3=105, wow same answer...


I have trouble with these followings:

According to Shelly’s analysis, the level of the water on the beach for the day of her trip will fit this equation:
w(t)=20sin29t
In this equation w(t) represents how far the waterline is above or below its average position. The distance is measured in feet and t represents the number of hours elapsed since midnight.

4)Suppose Shelly wants to build her castle 10 feet below the average waterline. What is the maximum amount of time she can arrange to have to maker her castle?

5)Suppose Shelly decides she needs only two hours to build and admire her castle. What is the lowest point on the beach she can build it?

6)Suppose that she wants to walk during the time when the waterline is 14 feet or more below the average waterline... What is the time period during which she can take her walk?

THANKS in advance.

[hide="4"]w(t)=20sin29t

So we want -20 <= w(t) <= -10

Which is double the time when w(t) = -10 to when w(t) hits -20 because itll take the same amount of time to go from -20 to -10.

Find t when w(t) = -10
$-10 = 20\sin29t$
$-1/2 = \sin29t$
$7\pi/6 = 29t$
$7\pi/174 = t$

Find t when w(t) = -20

$-20 = 20 \sin 29t$
$-1 = \sin 29t$
$3\pi/2 = 29t$
$3\pi/58 = t$

Now subtract the result from when w(t) = -10 from when w(t) = -20 and double it.[/hide]

The other problems have a similar concept.


inequalities
induction
algebra proposed
algebra

Let $ a_1,a_2,\dots$ be sequence of real numbers such that $ a_1\equal{}1$, $ a_2\equal{}\dfrac{4}{3}$, and \[ a_{n\plus{}1}\equal{}\sqrt{1\plus{}a_na_{n\minus{}1}}, \quad \forall n \ge 2.\] Prove that for all $ n \ge 2$, \[ a_n^2>a_{n\minus{}1}^2\plus{}\dfrac{1}{2}\] and \[ 1\plus{}\dfrac{1}{a_1}\plus{}\dfrac{1}{a_2}\plus{}\dots\plus{}\dfrac{1}{a_n}>2a_n.\]
[i]Fajar Yuliawan, Bandung[/i]

The first inequality.
[hide]We will prove by induction.
The initial cases are trivial and assume that we may use them now. Note that $ a_n$ are positive reals.
If this inequality is true for $ k$ and $ k\minus{}1$, then $ a_k^2>a_{k\minus{}1}^2\plus{}\dfrac{1}{2}$ and $ a_{k\minus{}1}^2>a_{k\minus{}2}^2\plus{}\dfrac{1}{2}$, then by Cauchy Schwarz Inequality \[ a_{k\plus{}1}^2\equal{}1\plus{}a_ka_{k\minus{}1}>1\plus{}\sqrt{\left(a_{k\minus{}1}^2\plus{}\dfrac{1}{2}\right)\left(a_{k\minus{}2}^2\plus{}\dfrac{1}{2}\right)} \ge 1\plus{}a_{k\minus{}1}a_{k\minus{}2}\plus{}\dfrac{1}{2}\equal{}a_k^2\plus{}\dfrac{1}{2}.\]
Q.E.D.[/hide]

The second inequality.
[hide]Note that from the first inequality, we obtain \[ \dfrac{1}{a_n}>2(a_n\minus{}a_{n\minus{}1})\] but this is telescoping series and the next is straightforward. Q.E.D.[/hide]

Any solution?

Could you post the detail for the second inequality

thank's

Well, second inequality immediately follows from the first one (just observe that $ \frac{1}{a_n}> 2(a_n\minus{}a_{n\minus{}1})$ is eqvivalet to $ 1\plus{}2a_n a_{n\minus{}1}> 2a_n ^2$ or,using reccurence formula, $ 2a_{n\plus{}1} ^2 \minus{}1>2 a_n ^2$ which is exactly the first inequality from the problem).

It is obvious that the order of our sequence is $ \sqrt{n}$ (the lower bound follows from simple induction and the first inequality,the upper bound  comes from simple induction).


Let $O,A',B',C'$ be the circumcentres of triangles $ABC,BOC,COA,AOB$, respectively. Prove that $AA',BB'$ and $CC'$ are concurrent.

I can not say for sure if the problem is correct. If it is you could prove it (in no more than half a page) using complex numbers. You easily observe that $A'(\frac{bc}{b+c}$ (as it is the middle of the segment between O and the point of intersection of the tangents in B and C to the circumcenter). Now you put the condition that the three lines are concurrent and see if your problem holds. I am a little lazy for computations right now, but they aren't ugly :)

They concur at the [url=http://mathworld.wolfram.com/KosnitaPoint.html]Kosnita point[/url].  You can find some literature on this on the internet.

it's easy :). You must to use 4 times trigonometric version of Ceva theory. Write condtions for concur of AA' BA' CA' etc. And from this 3 equation make equation for concur of AA', BB' , CC'


algebra
polynomial
function
integration
calculus
induction
algebra solved

proove that every polynomial P \in Q can be written as a sum of two strict /  polynomials from Q

I doubt that any polynomial can be written as a sum of 2 strictly increasing polinomials. Because the sum of increasing functions is an increasing function and not all polynomials are increasing.
However it can be expressed as the difference of 2 strictly increasing polynomials.
The solution I know is not elementary:
P= 1/2 (\int (P' 2+P'+1)dx -  \int (P' 2 -P'+1)dx)
The functions 1/2 \int (P' 2  \pm P'+1)dx are strictly increasing since P' 2  \pm P'+1>0.

sorry... my mistake... i was in a hurry
you are right mihai, this was the problem
and i like very very much your proof...
this was proposed at  Laurentiu Duican last year at 10th grade
my sol was something like yours ... but not so nice and easy and not complet  :D
thx for the solution

Kuba, you gave this solution with the integrals, this is not elementary for 10th graders, if you have an elementary solution, pls post it!

I like the solution I posted too, but unfortunately it's not mine. A teacher shown it to me 3 years ago and it holds for any [b]differentiable[/b] function not only polynomial ones. Sorry for using this word in this section.

Hi all,

Posted by Amfulger solution is very nice and elementary (why integral is not elementary?). If you want to prove without analysis, then use mathematical induction (For polynomial, we have degree!). I think it is easy to realize this proof.

Namdung

i don't have an elementary solution Lagrangia... i mean one without the integrals... and mine(with integrals to) is much more complicated and it's not completed... i took 5 pt from 7 at the contest :(
but i think amfulger's solution  is VERY SIMPLE and efficient :D
thx amfulger :D


function
induction
algebra unsolved
algebra

Find all continuous functions $f: \Re \longrightarrow \Re$ such that for all $x,y \in \Re$ :
$f(x+y)f(x-y)=f(x)^2-f(y)^2$

[quote="mffmartinelli"]and we can show by induction that $f(n)=an$, for all natural $n$.
[/quote]

We can not show something like that because $f(n)=an$ in not true :? Can you explain more?

[quote="jastrzab"]@mffmartinelli

True...in your equation you can compute $f(n+1)$ using terms $f(n)$ and $f(n-1)$ but to do this by induction you need also $f(2)$!!![/quote]

agreed. :lol:

[quote="mffmartinelli"]We can view that $f(x)=ax$ and $f(x)=sinx$ are solutions. And really my previous tentative was wrong.[/quote]

And $f(x)=sinh(x)$ is a solution too.


I am practicing for our quiz, but there is no way to check if my answers are correct since the book that I got, has no solutions on it. Maybe you can help me with my answers. I can't also get numbers 5 and 6. Thanks.

1.Show that $(a+1)(a-1) = a^{2}-1$
[b]MY SOLUTION[/b]

[hide]   $=a(a-1) = a^{2}-a$distributivity
   $=1(a-1) = a - 1$
   $=a^{2}-{a+a}- 1$
   $=a^{2}-1$[/hide]

2.Show that the operation of division is not commutative.
[b]MY SOLUTION[/b]
[hide]Let $a$ and $b$ be real numbers. $a$ cannot equal $b$, otherwise, $a/b = 1$.
Suppose that the operation of division is commutative,
Then $a/b = b/a$ which implies that $ab^{-1}= ba^{-1}$. Multiplying both sides by ab, then we have, $a^{2}= b^{2}$.  It follows that $a = b$ which is a contradiction.[/hide]

3. If $(x-a)(x-b)(x-c) = 0$, then what can you conclude about $x$

My Solution

[hide][b]Case 1.[/b] If $a = x$, then, $x -{x}=0$, which implies that the equation is 0.
[b]Case 2.[/b] If a is not equal to $x$, then b or c, must equal x, to satisfy the equation.
Hence, $x$ should be equal to either $a, b$ or $c$. [/hide]

4.Prove that if $a^{2}> b^{2}$, then $a>b$.
[b]MY SOLUTION[/b]
[hide]Case 1: a is not equal to b, since that would contradict the condition.
Case 2: If $a^{2}> b^{2}$ and $a < b$, then multiplying a on both sides, will yield $a^{2}<ab$. Multiplying b on both sides, will also yield, $ab <b^{2}$. By transitivity, then $a^{2}< b^{2}$, which is a contradiction. Again, by trichotomy, if $a$ is not equal to $b$, and $a$ is not less than $b$, then $a > b$.[/hide]

5. Prove that the set of rational numbers $Q$ is closed under multiplication and division, that is if $x = n/m$ and $y = p/q$ for integers, $n, m, p, n and q$, then $x(y)$ can be expressed as the quotient of two integers.

6.Show that if a is a real number which is its own multiplicative inverse, then  can only be $1$ or $-1$.

[quote="guile"]1.Show that $(a+1)(a-1) = a^{2}-1$
[b]MY SOLUTION[/b]

$=a(a-1) = a^{2}-a$distributivity
   $=1(a-1) = a - 1$
   $=a^{2}-a+a - 1$
   $=a^{2}-1$[/quote]

So, I know exactly what you're thinking, but what you've written isn't great.  The most obvious flaw is those equal signs at the left of your equations -- you put one of those there when the thing you've written is equal to the thing above it.  But that isn't the case!  That's very sloppy.  You want to do this in two separate parts, but (1) there's no good reason to and (2) you haven't justified that you [i]can[/i] do that.  There's also another application of the distributive property that you did and never mentioned!

$(a+1)(a-1) = (a+1)a+(a+1)(-1)$ by distributivity.
$= (a\cdot a+1 \cdot a)+(a \cdot (-1)+1\cdot(-1))$ by two more applications of distributivity
$= (a^{2}+a)+(-a-1)$ where the simplifications are "knowing what $a^{2}$ means," multiplication by identity, "knowing that $(-1)\cdot x =-x$" (which your class has either proven or ignored) and another multiplication by identity.  (I wouldn't have put explanatory line here if I were writing this up on a quiz, I just would have simplified the 4 expressions without comment.)
$= a^{2}+a-a-1$ by associativity of addition
$= a^{2}-1$ by properties of additive inverses (actually, I really did six or more individual axiomatic steps in the last two lines, but they're all very minor -- a lot of associativity, one additive inverses, one additive identity).

Your solution to 3 seems to me totally unconvincing -- if x isn't a, it has to be b or c?  Why?  You probably want to start with an associative move, $(x-a)((x-b)(x-c)) = 0$.  Then you have to say something like, "We know that if $m\cdot n = 0$ either $m = 0$ or $n = 0$."  (I don't know if you've proven this; if not, it's not too bad at all, using the fact that if neither $m$ nor $n$ were zero, they'd have multiplicative inverses, etc.)

Your solutions to 2 and 4 are your best, but they aren't perfect.  In question 2, you don't want to say that "a cannot equal b," you want to say "pick positive a and b such that they aren't equal to each other."  The other potential problem with your solution to 2 is that it uses the fact "$a^{2}= b^{2}$ and $a, b > 0 \Longrightarrow a = b$."  Has your class proven this?  (From questions 4 and 6, it seems maybe not.)  If not, you'll need to justify that last step.

I have only very minor complaints about your solution 4: in case 1, you say "since that would contradict the condition."  But you don't [i]prove[/i] that it would contradict the condition.  So instead perhaps say, "If $a = b$ then $a^{2}= b^{2}$ which contradicts the condition $a^{2}> b^{2}$."
Then, your final use of trichotomy is not actually part of case 2 -- it should be on a new line.

6)  I'm not sure this is a valid proof, but here goes.  The multiplicative inverse of a real number $a$ is defined as $a^{-1}$ such that $a\left(a^{-1}\right) = 1$. So, if $a$ is its own multiplicative inverse, then we have $a=a^{-1}\Rightarrow (a)(a) = 1$.  We now have three cases:
Case 1:  If $(a>1)$, we have $(a)(a) > a$, which is a contradiction.

Case 2: (i) If $(0\le a<1)$, we have $0<(a)(a)<a$, a contradiction.
            (ii) If $-1<a<0$, we have $-a>a\cdot a>0$, another contradiction

Case 3: If $(a<-1)$, we have $(a)(a)>-a$, again a contradiction.

So, we must have $a=1$ or $a=-1$, checking both of these in the original equation, we see that they are in fact solutions.

[quote="Jimmy"]Case 3: If $(a<-1)$, we have $(a)(a)<-a$, again a contradiction.[/quote]

You're not being nearly careful enough with your signs!  If $a <-1$, $a\cdot a >-a$.  Your case 2 is also problematic -- you should divide it into the positive and negative halves of the interval.  Alternatively, note that since $(-1)^{2}= 1$, $(-a)^{2}= a^{2}$ and $a$ is its own inverse iff $-a$ is its own inverse.  Then you only have to consider the positive cases.

Thanks, I was really careless there :).  Anyway, I think I fixed the problems and it should make sense now.


[quote="thanhnam2902"]
5) $ \frac {1}{a^2.m_a} \plus{} \frac {1}{b^2.m_b} \plus{} \frac {1}{c^2.m_c} > \frac {3}{abc}$
[/quote]
Because
$ m_a\equal{}\frac{\sqrt{2b^2\plus{}2c^2\minus{}a^2}}{2}<\frac{\sqrt{2b^2\plus{}2c^2\minus{}(b\minus{}c)^2}}{2}\equal{}\frac{b\plus{}c}{2}$
$ m_b<\frac{c\plus{}a}{2},m_c<\frac{a\plus{}b}{2}$
then
$ LHS>\frac {2}{a^2.(b\plus{}c)} \plus{} \frac {2}{b^2.(c\plus{}a)} \plus{} \frac {2}{c^2.(a\plus{}b)}$
$ >\frac{2(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})^2}{(b\plus{}c)\plus{}(c\plus{}a)\plus{}(a\plus{}b)}\equal{}\frac{(ab\plus{}bc\plus{}ca)^2}{a^2b^2c^2(a\plus{}b\plus{}c)}$
We need to prove that
$ \frac{(ab\plus{}bc\plus{}ca)^2}{a^2b^2c^2(a\plus{}b\plus{}c)}\geq \frac{3}{abc}\Leftrightarrow (ab\plus{}bc\plus{}ca)^2\geq 3abc(a\plus{}b\plus{}c)$. OK!

OK! Hi [b]tdl[/b]!  I want become your friend.


can you find the shaded region?
http://www.angelfire.com/ab7/whodoshoe/gifgallery/Msinshadow.bmp
http://www.angelfire.com/ab7/whodoshoe/gifgallery/gifgallerys.html

[hide]Graphing the lines with the shorter leg on the x-axis and the longer on the y-axis we have the equations

$y=-\frac 23x + 4$

$y = -\frac 83x + 8$

Their intersection is at $(2, \ \frac 83)$. The other points of the shaded triangle are $(0, \ 4)$ and $(3, \ 0)$. By the shoelace method the area of the shaded region is then equal to $\fbox {2}$.[/hide]


[quote="mysmartmouth"]When I go to add Resources, the button is no longer anywhere in site...I haved talked to a couple of other RManagers, who are experiencing the same problem.  Is this a temporary glitch?  Thanks![/quote]Yeah, I have no idea why the button is not there anymore, but will check.


[quote="linzk3"]how do you get the equation of a line on a graph if you know x/5+y/3=1

please and thanks[/quote]

just simplify to get:
$y=-\frac{3}{5}x+3.$

[quote="nutz_for2.718281828"][hide="solution"]First, we multiply each side by 15 and get

$3x+5y=15$
$y=-\frac{3x}{5}+5$[/hide][/quote]
Incorrect. Y-intercept is 3.

[quote="linzk3"]how do you get the equation of a line on a graph if you know x/5+y/3=1

please and thanks[/quote]

double intercept form
y-intercept=3
x-intercept=5

its also y=-3/5x+3

[quote="i_like_pie"][quote="linzk3"]how do you get the equation of a line on a graph if you know x/5+y/3=1

please and thanks[/quote]
$\frac{x}{5}+\frac{y}{3}=1$ is the standard form of the line.

The y-intercept form is $y=-\frac{3}{5}x+3.$

The slope is $-\frac{3}{5},$ and the y-intercept is $3.$[/quote]
Um, generally mathbooks demand that A, B, and C are integers in standard form, and that A is positive. Standard form is Ax+By=C where a, b, and c are constants. So standard form would be
$3x+5y=15$


LaTeX
Support
inequalities
Functional Analysis
real analysis
real analysis unsolved

Let $ \{a_n\}_{n \geq 1}$ be a bounded sqeuence of real numbers. Prove that there exists a unique bounded sequence $ \{b_n\}$ of real numbers such that $ b_n \equal{} 0, a_n \equal{} b_{n\plus{}1} \plus{} 4b_n \plus{} b_{n\minus{}1}, n \geq 1$.

There's a hint that I should use apply Contraction Mapping Theorem on the space of bounded sequence, but I didn't know how to apply.. Any other methods are welcome  :)

The space $ \ell^{\infty}$ of bounded sequences with the norm $ \Vert x\Vert_{\infty} \equal{} \sup_{j\ge 1}|x_j|$ is a complete normed space (a Banach space), hence in particular a complete metric space.

Let $ (a_n)$ be already given. Define $ T: \ell^{\infty}\to\ell^{\infty}$ by

$ (Tx)_n \equal{} \frac14(a_n \minus{} x_{n \plus{} 1} \minus{} x_{n \minus{} 1}).$

To maintain consistency, define $ (Tx)_1 \equal{} \frac14(a_1 \minus{} x_2).$

Suppose $ x,y\in\ell^{\infty}.$ Then

$ \Vert Tx \minus{} Ty\Vert_{\infty} \equal{} \frac14\,\max_{n}| \minus{} x_{n \plus{} 1} \minus{} x_{n \minus{} 1} \plus{} y_{n \minus{} 1} \plus{} y_{n \plus{} 1}|\le\frac12\Vert x \minus{} y\Vert_{\infty}$

Hence $ T$ is a contraction mapping, to which the Banach fixed point theorem applies.

Do you see the finish from there?

[color=green][Note: there's a LaTeX rendering problem here. I used the code \| in several places and it's showing up as | instead of ||. I'll flag this in the support forum. (It works there?)][/color]

[color=green][Re-edit: following a stevem suggestion, I changed \| to \Vert.][/color]

Thx for the enlightement Kent  :)

One thing that I'm still confused is about the norm u defined, because the sum can be infinite, right ?
Is it valid to compare infinite sum to get the final inequality ?

EDIT : Should your norm be $ \max_{n} |x_n|$ in the first place ?

sup, not max, but yes. And I've edited. That was actually a one-letter typo: I thought "sup" and my fingers said "sum."

This is an interesting approach to the problem.  The value of $ b_1$ determines the entire sequence, so in principle it should be possible to solve this problem directly by figuring out how to define $ b_1$ in terms of the $ a_n$, but the approach by the contraction principle is certainly nice.


Let f be a polynomial such that f(x^2+1) = x^4+4x^2. Determine f(x^2-1).

it says the answer is x^4-4

Help? explanation?

[hide="Solution"]$ f(x^2 \plus{} 1) \equal{}  x^4 \plus{} 4x^2 \Rightarrow f(u\plus{}1)\equal{}u^2 \plus{} 4u$.

$ f(x^2 \minus{} 1) \equal{} f(u\minus{}1) \equal{} f((u\minus{}2)\plus{}1) \equal{} (u\minus{}2)^2 \plus{} 4(u\minus{}2) \equal{} u^2\minus{}4u\plus{}4\plus{}4u\minus{}8\equal{}u^2\minus{}4\equal{}\boxed{x^4\minus{}4}$.[/hide]

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First find f(x). The input has an $ x^2$ term and the output has an $ x ^4$ term, so you can figure that there's some squaring done. If f(x) = $ x^2$, then f($ x^2 \plus{} 1$) would be $ x^4 \plus{} 2x^2 \plus{} 1$. The x^2 term's coefficient is off by two, but luckily, the input has an x^2 term you can add. So you try f(x) = $ x^2 \plus{} 2x$. Then, f($ x^2 \plus{} 1$) = $ x^4 \plus{} 4x^2 \plus{} 3$. Now, all you have to do is subtract 3 to make it match up. So you have f(x) = $ x^2 \plus{} 2x \minus{} 3$.
Plugging in f($ x^2 \minus{} 1$) = $ (x^2 \minus{} 1)^2 \plus{} 2(x^2 \minus{} 1) \minus{} 3 \equal{} x^4 \minus{} 4$.[/hide]

[hide="Yet another way to do it"]You can also do it with a more "direct" substitution.

Let $ u\equal{}x^2\plus{}1\implies x\equal{}\sqrt{u\minus{}1}$.
Then $ f(u)\equal{}(u\minus{}1)^2\plus{}4(u\minus{}1)$ by simply plugging in.
And if you let $ u\equal{}x^2\minus{}1$, you have $ f(u)\equal{}f(x^2\minus{}1)\equal{}(x^2\minus{}2)^2\plus{}4(x^2\minus{}2)\equal{}x^4\minus{}4x^2\plus{}4\plus{}4x^2\minus{}8\equal{}\boxed{x^4\minus{}4}$.[/hide]