Please help me to prove the convergence of $\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+…+\frac{1}{n}-\ln n$

Please help me to prove that $\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}-\ln n$ converge and it's limit $\gamma \in [0,1]$. Then, using $\gamma_{n}$ find the sum: $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$.

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Welcome to MSE! I'd suggest though you'd ask question in a less imperative way and in a more friendly way. Plus, have you managed to find something? Any approaches? What I mean is, you're not a book, you're human, so act like one, it makes us more willing to answer. =) –  Patrick Da Silva Aug 10 '11 at 11:49
The following question addresses the convergence –  Sasha Aug 10 '11 at 11:56
possible duplicate of What is $\lim_{n \to \infty}\sum_{k=1}^n 1/k$ –  user17762 Aug 10 '11 at 12:14
Sorry because my question was imperative. My english is not very good and I translated word by word from my language. I will be more carefully next time. –  NumLock Aug 10 '11 at 12:28
In the sense that if you read the answer given, all the good stuff's in there. Although they don't compute the alternated sum. I don't agree it's a duplicate. –  Patrick Da Silva Aug 10 '11 at 12:56

The following question addresses the convergence, so assume $\gamma_n$ converges and denote the limit as $\gamma$.

Now consider $\sum_{n=1}^{2m} (-1)^{n+1} \frac{1}{n}$ and split it into even and odd terms $\sum_{n=1}^m \frac{1}{2n-1} - \sum_{n=1}^m \frac{1}{2n}$. Then complete the sum over odd integers to sum over consequtive integers: $\sum_{n=1}^{2m} \frac{1}{n} - 2 \sum_{n=1}^m \frac{1}{2n}$.

Then subtract logarithms to form $\gamma_{2m} - 2 \gamma_m + \log(2)$, like so $$( \sum_{n=1}^{2m} \frac{1}{n} - \log(2m)) - ( \sum_{n=1}^m \frac{1}{n} - \log(m)) + \log(2).$$

In the limit $m \to \infty$ it becomes $\gamma - \gamma + \log(2) = \log(2)$.

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Thank you for your help too! –  NumLock Aug 10 '11 at 12:34

You can use the mean value theorem to prove first that $$\frac{1}{n+1}<\ln(n+1)-\ln n<\frac{1}{n}$$ then, $\ln(n+1)<1+\ldots+\frac{1}{n}<1+\ln n$, and finally $$\ln(n+1)-\ln n<\gamma_n<1$$ which implies the convergence of $\gamma_n$ and its limit $\gamma\in[0,1]$.

For the second part, I am not sure if you can use $\gamma_n$ to find the sum $\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}$!

I suggest to use the Taylor-Young formula for $x\mapsto\ln(x+1)$ on $[0,1]$ to prove that $$\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}=\ln2.$$

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Thank you !I learnt mean value theorem in highschool but I completely forgot about it. –  NumLock Aug 10 '11 at 12:34
@amine it's mercator series time... –  user38268 Aug 10 '11 at 12:58

The Mean Value Theorem says $$\frac{1}{k+1}<\log(k+1)-\log(k)<\frac{1}{k}$$ Summing from $m$ to $n-1$ ($m < n$), we get \begin{align} \sum_{k=m}^{n-1}\frac{1}{k+1}<\log(n)-\log(m)<\sum_{k=m}^{n-1}\frac{1}{k} \end{align} Subtracting $\displaystyle{\sum_{k=m}^{n-1}\frac{1}{k+1}=\sum_{k=m+1}^n\frac{1}{k}=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^m\frac{1}{k}}$ from all parts, we get $$0<\left(\log(n)-\sum_{k=1}^n\frac{1}{k}\right)-\left(\log(m)-\sum_{k=1}^m\frac{1}{k}\right)<\frac{1}{m}-\frac{1}{n}$$ This proves the existence of $\displaystyle{\lim_{n\to\infty}\left(\log(n)-\sum_{k=1}^n\frac{1}{k}\right)}$. If we set $m=1$ and subtract all sides from $1$, we get $$1>\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)>\frac{1}{n}$$Thus, we have shown that $\displaystyle{\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)}$ exists, and that $0\le\gamma\le 1$.

Note that \begin{align} \sum_{k=1}^{2n}(-1)^{k+1}\frac{1}{k}&=\sum_{k=1}^{2n}\frac{1}{k}-2\sum_{k=1}^n\frac{1}{2k}\\ &=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\\ &=\left(\sum_{k=1}^{2n}\frac{1}{k}-\log(2n)\right)-\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)+\log(2) \end{align} Taking the limit as $n\to\infty$, we get $$\sum_{k=1}^\infty(-1)^{k+1}\frac{1}{k}=\gamma-\gamma+\log(2)=\log(2)$$

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I apologize if this is close to Sasha's answer, but I have been working on this, during breaks, for quite a while, and posted it before I noticed. Since I show more of the details, I will leave it for now. I can remove it if requested. –  robjohn Aug 10 '11 at 17:26
Sorry, for the convergence, I forgot to say that $\gamma_n$ is decreasing because $$\gamma_{n+1}-\gamma_n=\frac{1}{n+1}-(\ln(n+1)-\ln n)<0$$ since it is bounded, it is convergent! –  amine Aug 10 '11 at 21:49

Do we have a "proof without words" question here? I don't find it. There is a lovely diagram found in certain calculus texts that does this for us. Maybe I will animate it, for fun.

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I am not getting this. Do you mean pictorial proof, which suggests the proof ? –  Sasha Aug 10 '11 at 14:15
It's something like the diagram at the top right here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant but then you slide all the blue regions to the left to see that they all fit inside a $1 \times 1$ square. –  GEdgar Aug 10 '11 at 14:54


# Math Help - word problem.

1. ## word problem.

hi ! im having a difficult time answering this word problem.

"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."

I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???

Thank you very much !!

You can check:

In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms

In 5 hours, Ben travels: 4+5+6+7+8=30kms

that means they were together after 5 hours.

3. Nope. Ben started 2 hours after

4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.

5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:

$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.

We also know that the distance traveled by the first is just 6t.

I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:

$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$

Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$

For that, I get one positive root of t = 10 (also t = -1 which is obviously not useful). Voila! Thanks for the problem, I had fun with that.

6. Hello, dugongster!

Alvin set out from a certain point and travelled at the rate of 6 kph.
After Alvin had gone for 2 hours, Ben set on to overtake him and went 4km the first hour,
5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour.

After many hours were they together?

. . In the next $\,h$ hours, he travels another $6h$ km.
Alvin has traveled $6h + 12$ km in the first $\,h$ hours.

During the same $\,h$ hours, Ben has travelled:
. . $4 + 5 + 6 + \hdots + (h+3) \:=\:\dfrac{h(h+7)}{2}$ km. .**

The two distances are equal: . $\dfrac{h(h+7)}{2} \:=\:6h + 12$

. . $h^2 + 7h \:=\:12h + 24 \quad\Rightarrow\quad h^2 - 5h - 24 \:=\:0$

. . $(h - 8)(h + 3) \:=\:0 \quad\Rightarrow\quad h \:=\:8,\:-3$

Answer: . $\text{8 hours}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

We want the sum of the numbers from $\,4$ to $h\!+\!3.$

The sum of the numbers from $\,1$ to $h\!+\!3$ is: . $\dfrac{(h+3)(h+4)}{2}$

The sum of the numbers from $\,1$ to $\,3$ is: . $1+2+3 \:=\:6$

Hence,the sum of the numbers from $\,4$ to $h\!+\!3$ is:

. . $\dfrac{(h+3)(h+4)}{2} - 6 \;\;=\;\; \dfrac{h^2 + 7x + 12 - 12}{2} \;\;=\;\;\dfrac{h(h+7)}{2}$

7. great. thanks to both of you. grep, you rock !!!

8. Originally Posted by Soroban
Hello, dugongster!

Hi Soroban! With due respect (which is considerable, I might add, you being one of the most helpful and, let's face it, awesome people here), I took into account the 2 hour headstart by using (t - 2) as the time in the arithmetic progression (n = (t - 2)), and starting at 4 ( $a_1 = 4$).


# Do the 12 lines of a bingo card have equal chance of winning?

You have a $$5 \times 5$$ bingo card filled with $$25$$ distinct numbers, one per square. There is also a pot containing each number once and you draw them out one by one without replacement.

A line is any of the $$5$$ rows, $$5$$ columns, or $$2$$ main diagonals. A line is completed when its $$5$$ numbers have been drawn. A line wins if it is the first line to be completed.

Question: Do the $$12$$ lines have equal chance of winning? If not, which lines have higher chance of winning?

Why I ask: Define $$T_l$$ to be the time to completion for line $$l$$. It is obvious that all $$T_l$$ are equi-distributed, and thus all $$E[T_l]$$ are equal. However, because the lines overlap, the different $$T_l$$'s are dependent, and it is not clear to me that they have equal chances of being first.

In particular, imagine that the $$5$$-subsets are not arranged in rows, columns and diagonals, but are clustered in some non-uniform way. Then a subset that shares a lot of elements with other subsets might have a lower chance to win than a subset that does not share a lot of elements with other subsets. (I can provide a simple example if there is interest.)

On the bingo card, the $$5$$-subsets are pretty uniform but not exactly uniform, due to the diagonals. So my suspicion is that the line win chances are almost equal but not exactly equal. And I am curious as to which lines have higher chances.

I imagine (but would be happy to be proven wrong) that calculating the exact win prob for a line might be difficult/tedious, so qualitative arguments based on e.g. symmetry are also welcome.

Clarifications: A drawn number can complete multiple lines, so that needs some special handling. However what I am interested in is the question of equality, so I will accept any reasonable way to handle such "shared" wins, i.e. if $$N>1$$ lines are completed at the same draw (and no line has been completed before this draw), then you can treat this as if...

• they all win (in which case the sum of the $$12$$ win probabilities exceed $$1$$, but that doesn't matter since I am interested in which are higher/lower), or,

• the whole experiment is repeated from the beginning (i.e. we condition on such shared wins not happening), or,

• you flip an $$N$$-sided die to determine the winner (i.e. this effectively counts as $$1/N$$ win for each involved line), etc.

• After this, you can do BINGO cards with a "free" space in the middle. – GEdgar Mar 8 at 20:37
• I don't see why you would think any particular line would have a higher chance of winning if there is an equal probability of each square being picked every time. A line requires 5 numbers being picked and the probability of each 5 number combinations are exactly the same. – Matthew Liu Mar 8 at 20:53
• @MatthewLiu Because when one line gets a number, it also helps other lines. The diagonal are at a disadvantage. Look at my solution for $3\times3$ bingo. – saulspatz Mar 8 at 21:24
• @MatthewLiu what you said is why all the $T_l$ are identically-distributed, but they are not independent. A line $L$ wins if $T_L < T_l \forall l \neq L$. – antkam Mar 8 at 22:15
• Ok you guys are correct. The diagonal lines are at a strict disadvantage. – Matthew Liu Mar 11 at 17:14

I must admit, when I read the title of the question, I thought "Of course they all have an equal chance of winning. What a silly question." But then, I read the body, and was delighted to see that you had a valid point, and I was the silly one. I don't know how to solve for $$5\times5$$ bingo in any reasonable time, but for $$3\times3$$ bingo we have only $$9!$$ permutations of the numbers to consider. I wrote a python script to exhaustively test this.

from itertools import permutations
from fractions import Fraction

score = 9*[0]
card = 8*[0]

for p in permutations(range(9)):
card[0] = {0,1,2}
card[1] = {3,4,5}
card[2] = {6,7,8}
card[3] = {0,3,6}
card[4] = {1,4,7}
card[5] = {2,5,8}
card[6] = {0,4,8}
card[7] = {2,4,6}
for number in p:
for line in card:
line -= {number}
winners = len([line for line in card if not line])
if winners:
for idx, line in enumerate(card):
if not line: score[idx] += Fraction(1, winners)
break

for idx in range(8):
print(idx, score[idx])

This produced the output

0 47520
1 45600
2 47520
3 47520
4 45600
5 47520
6 40800
7 40800

This means that the diagonals are worst, and the lines along the edges of the card are best, with the horizontal and vertical lines through the center in the middle.

I should mention that in the case of ties, I awarded an equal fraction of the game to each winner, so that the sum of the scores does work out to $$9!.$$

• What does it produce if you give a full win to each line if 2/3/4 lines win at once? – Banbadle Mar 8 at 21:58
• awesome! i had thought the diagonals are worst but all the rows and columns are equal, but you showed that the row/col passing through the center are penalized. interesting... and now i wonder if the 1st & 2nd rows are different in the 5x5. ooh, yeah, i would like to see the idea of @Banbadle tested too. would the multiple full wins cancel the effect of sharing? – antkam Mar 8 at 22:00
• @Banbadle I just modified saulspatz's code to try other scoring modes. If a full win is given to each simultaneously winning line, the win counts are 54000 > 53290 > 46656 for the edges, center row/col, diagonals respectively. If we condition on no shared wins (i.e. those permutations award no wins to anybody), then the win counts are 41040 > 38880 > 35424, again for edge, center row/col, diagonals. So the order is the same in all 3 win-counting models. – antkam Mar 8 at 22:44

I simulated a standard bingo game, one million games with ten random cards for each game. I recorded the number of times each line triggered bingo, including bingo on multiple cards and multiple lines winning simultaneously on the same card.

The results:

• Horizontal through center wins $$16.2\%$$ of games
• Vertical through center wins $$16.1\%$$ of games
• Diagonals each win $$15.9\%$$ of games
• Other horizontal lines each win $$5.8\%$$ of games
• Other vertical lines each win $$5.6\%$$ of games

Note that this totals over $$100\%$$, accounting for multiple wins in a single game (there were nearly $$1.1$$ million wins in one million games)

After ensuring that these results are indeed accurate and giving some thought to why the diagonals win less than the central row and column, it is clear that this is due to intersecting lines. Each of the squares on the diagonals, when drawn, has the potential to trigger a win on either of two intersecting lines, while each square in the central row or column can only trigger a win on a single intersecting line.

I have also run the simulation without the free center square:

• Rows win $$9.42\%$$
• Diagonals win $$9.17\%$$
• Columns win $$9.05\%$$
• when you say "standard bingo game" do you mean the center square is free (i.e. already marked off)? if so that must be why lines through center win much more coz they only require 4 cells. – antkam Mar 8 at 22:04
• @antkam Yes, center square is free in my simulation. – Daniel Mathias Mar 8 at 22:05
• Even in the free-center case it is open (at least, in my mind) whether the diagonals are at a disadvantage. If you found that diagonals win 10% vs the center row/col winning 20% each, that would be quite conclusive. But at 15.9.% vs 16.1% I am not sure this is a real effect or just within the error bar. – antkam Mar 8 at 22:18
• @antkam A run of ten million games: $16.18\%$, $16.09\%$, $15.91\%$, $15.90\%$ for center row, center column, downward diagonal, upward diagonal. Of course, this only proves that my simulation produces consistent results. It is entirely possible that the inequality is a result of some unintended bias, though I have found no potential cause for bias. – Daniel Mathias Mar 8 at 23:31
• @DanielMathias Thanks. I knew very little about bingo before reading this. Now I have upgraded my knowledge from "very little" to just "little". – badjohn Mar 9 at 14:54


# If $N= (323232 \cdots 50 \text{ digits})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?

If $N= (\underbrace{323232 \cdots}_{50 \text{ digits}})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?

I am looking for a "fast" approach that could be used to solve this in less than $2$ minutes.

• This is computing $N\mod 8$. Since the $9\mod 8 \equiv 1$, this is just the sum of digits of the number $N$, i.e. $(3+2)\times 25\mod 8 \equiv (3+2)\times 1\mod 8 = 5$. Did I miss the question ? – Sasha Aug 26 '11 at 16:45
• @Sasha:No,you didn't miss the question.I just can't see this solution before :/ – Quixotic Aug 26 '11 at 17:26

The same way we find the remainder of dividing a number by $9$ when the number is written in base 10: add the digits.

Why?

Because remember that writing, say, $(38571)_9$ "really" means $$1 + 7\times 9 + 5\times 9^2 + 8\times 9^3 + 3\times 9^4.$$ When you divide $9$ by $8$, the remainder is $1$; when you divide $9^2$ by $8$, the remainder is $1^2 = 1$; then you divide $9^3$ by $8$, the remainder is $1^3=1$, etc. So the remainder of this number when divided by $8$ is the same as the remainder of $$1 + 7\times 1 + 5\times 9^2 + 8\times 1^3 + 3\times 1^4$$ which is the same as the remainder of $1+7+5+8+3 = 24$, which is $0$.

In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.

(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).

(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).

• I feel like stupid now,the first general result is not new to me and could be easily be shown as follows:$N = (a_1a_2\cdots a_k)_r = a_k + \cdots +a_2 \times r^{k-2} + a_2 \times r^{k-1}$ and $S = a_1 + a_2 + \cdots a_k$ then it could be easily shown that $\frac{N-S}{r-1} = I \text{ (an integer) }+\frac{S}{r-1}$,which proves that result. But I don't know why I can't see this direct application before! :/ – Quixotic Aug 26 '11 at 17:23
• Anyways,the more general result is new to me,and perhaps a very useful one,could you hint me a easy proof for the same? – Quixotic Aug 26 '11 at 17:25
• @FoolForMath The key idea is to learn modular arithmetic and, more generally, modular reductions of problems in quotient algebras. It's one way to algebraically divide and conquer. You can find further discussion of such in some of my posts here if you follow the link I gave, and its links... – Bill Dubuque Aug 26 '11 at 17:28
• I understood it now :D and it was before Bill Dubuque's post :-) but yes Bill you are right,my understanding is due to modular arithmetic. – Quixotic Aug 26 '11 at 17:31
• @Foo One key thing to keep in mind is that radix notation has polynomial form, so this can be viewed as a special case of well-known results about polynomials. Always be on the lookout for analogies between numbers and functions (here polynomials). – Bill Dubuque Aug 26 '11 at 17:55

HINT $\$ Employ the radix $\:9\:$ analog of casting out nines in decimal, namely

$$\rm\quad mod\ 8:\ \ \ 9 \equiv 1\ \ \Rightarrow\ \ d_n\: 9^n +\:\cdots\:+d_1\:9 + d_0\ \equiv\ d_n +\:\cdots\:+d_1+d_0$$

The same idea works generally to cast out $\rm\:b\pm1\:$'s in radix $\rm\:b\:$, e.g. see here for many links.

It may be viewed as a number-theoretic specialization of the Polynomial Remainder Theorem

$\rm\quad f(x)\ mod\ (x-r)\:=\:f(r)\:,\$ thus $\rm\ \ f(x)\ mod\ (x-1)\:=\:f(1)\: =\: f_n +\:\cdots\:+f_1 + f_0$


location:

61.
Let $S=1!2!3!\dots 99!100!$ (the product of the first 100 factorials). Prove that there exists an integer $k$ for which $1\le k\le 100$ and $S/k!$ is a perfect square. Is $k$ unique? (Optional: Is it possible to find such a number $k$ that exceeds 100?)
Solution 1. Note that, for each positive integer $j$, $\left(2j-1\right)!\left(2j\right)!=\left[\left(2j-1\right)!\right]{}^{2}·2j$. Hence

$S=\underset{j=1}{\overset{50}{\Pi }}\left[\left(2j-1\right)!\right]{}^{2}\left[2j\right]={2}^{50}50!\left[\underset{j=1}{\overset{50}{\Pi }}\left(2j-1\right)!{\right]}^{2} ,$

from which we see that $k=50$ is the required number.
We show that $k=50$ is the only possibility. First, $k$ cannot exceed 100, for otherwise $101!$ would be a factor of $k!$ but not $S$, and so $S/k!$ would not even be an integer. Let $k\le 100$. The prime $47$ does not divide $k!$ for $k\le 46$ and divides $50!$ to the first power. Since $S/50!$ is a square, it evidently divides $S$ to an odd power. So $k\ge 47$ in order to get a quotient divisible by 47 to an even power. The prime $53$ divides each $k!$ for $k\ge 53$ to the first power and divides $S/50!$, and so $S$ to an even power. Hence, $k\le 52$.
The prime $17$ divides $50!$ and $S/50!$, and hence $S$ to an even power, but it divides each of $51!$ and $52!$ to the third power. So we cannot have $k=51$ or 52. Finally, look at the prime 2. Suppose that ${2}^{2u}$ is the highest power of 2 that divides $S/50!$ and that ${2}^{v}$ is the highest power of 2 that divides $50!$; then ${2}^{2u+v}$ is the highest power of 2 that divides $S$. The highest power of $2$ that divides $48!$ and $49!$ is ${2}^{v-1}$ and the highest power of 2 that divides $46!$ and $47!$ is ${2}^{v-5}$. >From this, we deduce that 2 divides $S/k!$ to an odd power when $47\le k\le 49$. The desired uniqueness of $k$ follows.
Solution 2. Let $p$ be a prime exceeding 50. Then $p$ divides each of $m!$ to the first power for $p\le m\le 100$, so that $p$ divides $S$ to the even power $100-\left(p-1\right)=101-p$. From this, it follows that if $53\ge k$, $p$ must divide $S/k!$ to an odd power.
On the other hand, the prime 47 divides each $m!$ with $47\le m\le 93$ to the first power, and each $m!$ with $94\le m\le 100$ to the second power, so that it divides $S$ to the power with exponent $54+7=61$. Hence, in order that it divide $S/k!$ to an even power, we must make $k$ one of the numbers $47,\dots ,52$.
By an argument, similar to that used in Solution 1, it can be seen that $2$ divides any product of the form $1!2!\dots \left(2m-1\right)!$ to an even power and $100!$ to the power with exponent

$⌊100/2⌋+⌊100/4⌋+⌊100/8⌋+⌊100/16⌋+⌊100/32⌋+⌊100/64⌋=50+25+12+6+3+1=97 .$

Hence, 2 divides $S$ to an odd power. So we need to divide $S$ by $k!$ which 2 divides to an odd power to get a perfect square quotient. This reduces the possibilities for $k$ to 50 or 51. Since

$S={2}^{99}·{3}^{98}·{4}^{97}\dots {99}^{2}·100=\left(2·4\dots 50\right)\left({2}^{49}·{3}^{49}·{4}^{48}\dots 99\right){}^{2}=50!·{2}^{50}\left(\dots \right){}^{2} ,$

$S/50!$ is a square, and so $S/51!=\left(S/50!\right)÷\left(51\right)$ is not a square. The result follows.
Solution 3. As above, $S/\left(50!\right)$ is a square. Suppose that $53\le k\le 100$. Then 53 divides $k!/50!$ to the first power, and so $k!/50!$ cannot be square. Hence $S/k!=\left(S/50!\right)÷\left(k!/50!\right)$ cannot be square. If $k=51$ or 52, then $k!/50!$ is not square, so $S/k!$ cannot be square. Suppose that $k\le 46$. Then 47 divides $50!/k!$ to the first power, so that $50!/k!$ is not square and $S/k!=\left(S/50!\right)×\left(50!/k!\right)$ cannot be square. If $k=47,48$ or 49, then $50!/k!$ is not square and so $S/k!$ is not square. Hence $S/k!$ is square if and only if $k=50$ when $k\le 100$.

62.
Let $n$ be a positive integer. Show that, with three exceptions, $n!+1$ has at least one prime divisor that exceeds $n+1$.
Solution. Any prime divisor of $n!+1$ must be larger than $n$, since all primes not exceeding $n$ divide $n!$. Suppose, if possible, the result fails. Then, the only prime that can divide $n!+1$ is $n+1$, so that, for some positive integer $r$ and nonnegative integer $K$,

$n!+1=\left(n+1\right){}^{r}=1+\mathrm{rn}+{\mathrm{Kn}}^{2} .$

This happens, for example, when $n=1,2,4$: $1!+1=2$, $2!+1=3$, $4!+1={5}^{2}$. Note, however, that the desired result does hold for $n=3$: $3!+1=7$.
Henceforth, assume that $n$ exceeds 4. If $n$ is prime, then $n+1$ is composite, so by our initial comment, all of its prime divisors exceed $n+1$. If $n$ is composite and square, then $n!$ is divisible by the four distinct integers $1,n,\sqrt{n},2\sqrt{n}$, while is $n$ is composite and nonsquare with a nontrivial divisor $d$. then $n!$ is divisible by the four distinct integers $1,d,n/d,n$. Thus, $n!$ is divisible by ${n}^{2}$. Suppose, if possible, the result fails, so that $n!+1=1+\mathrm{rn}+{\mathrm{Kn}}^{2}$, and $1\equiv 1+\mathrm{rn}$ (mod ${n}^{2}$). Thus, $r$ must be divisible by $n$, and, since it is positive, must exceed $n$. Hence

$\left(n+1\right){}^{r}\ge \left(n+1\right){}^{n}>\left(n+1\right)n\left(n-1\right)\dots 1>n!+1 ,$

a contradiction. The desired result follows.

63.
Let $n$ be a positive integer and $k$ a nonnegative integer. Prove that

$n!=\left(n+k\right){}^{n}-\left(\genfrac{}{}{0}{}{n}{1}\right)\left(n+k-1\right){}^{n}+\left(\genfrac{}{}{0}{}{n}{2}\right)\left(n+k-2\right){}^{n}-\dots ±\left(\genfrac{}{}{0}{}{n}{n}\right){k}^{n} .$

Solution 1. Recall the Principle of Inclusion-Exclusion: Let $S$ be a set of $n$ objects, and let ${P}_{1}$, ${P}_{2}$, $\dots$, ${P}_{m}$ be $m$ properties such that, for each object $x\in S$ and each property ${P}_{i}$, either $x$ has the property ${P}_{i}$ or $x$ does not have the property ${P}_{i}$. Let $f\left(i,j,\dots ,k\right)$ denote the number of elements of $S$ each of which has properties ${P}_{i}$, ${P}_{j}$, $\dots$, ${P}_{k}$ (and possibly others as well). Then the number of elements of $S$ each having none of the properties ${P}_{1}$, ${P}_{2}$, $\dots$, ${P}_{m}$ is

$n-\sum _{1\le i\le m}f\left(i\right)+\sum _{1\le i

We apply this to the problem at hand. Note that an ordered selection of $n$ numbers selected from among $1,2,\dots ,n+k$ is a permutation of $\left\{1,2,\dots ,n\right\}$ if and only if it is constrained to contain each of the numbers 1, 2, $\dots$, $n$. Let $S$ be the set of all ordered selections, and we say that a selection has property ${P}_{i}$ iff its fails to include at least $i$ of the numbers $1,2,\dots ,n$ $\left(1\le i\le n\right)$. The number of selections with property ${P}_{i}$ is the product of $\left(\genfrac{}{}{0}{}{n}{i}\right)$, the number of ways of choosing the $i$ numbers not included and $\left(n+k-i\right){}^{n}$, the number of ways of choosing entries for the $n$ positions from the remaining $n+k-1$ numbers. The result follows.
Solution 2. We begin with a lemma:

$\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{m}=\left\{\begin{array}{cc}\hfill 0\hfill & \hfill \left(0\le m\le n-1\right)\hfill \\ \multicolumn{0}{c}{\left(-1\right){}^{n}n!}& \hfill \left(m=n\right) .\hfill \end{array}$

We use the convention that ${0}^{0}=1$. To prove this, note first that $i\left(i-1\right)\dots \left(i-m\right)={i}^{m+1}+{b}_{m}{i}^{m}+\dots +{b}_{1}i+{b}_{0}$ for some integers ${b}_{i}$. We use an induction argument on $m$. The result holds for each positive $n$ and for $m=0$, as the sum is the expansion of $\left(1-1\right){}^{n}$. It also holds for $n=1,2$ and all relevant $m$. Fix $n\ge 3$. Suppose that it holds when $m$ is replaced by $k$ for $0\le k\le m\le n-2$. Then

$\begin{array}{cc}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{m+1}\hfill & =\sum _{i=1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)i\left(i-1\right)\dots \left(i-m\right)-\sum _{k=0}^{m}{b}_{k}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{k}\hfill \\ \multicolumn{0}{c}{}& =\sum _{i=m+1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)i\left(i-1\right)\dots \left(i-m\right)-0\hfill \\ \multicolumn{0}{c}{}& =\sum _{i=m+1}^{n}\left(-1\right){}^{i}\frac{n!i!}{i!\left(n-i\right)!\left(i-m-1\right)!}=\sum _{j=0}^{n-m-1}\left(-1\right){}^{m+1+j}\frac{n!}{\left(n-m-1-j\right)!j!}\hfill \\ \multicolumn{0}{c}{}& =\sum _{j=0}^{n-m-1}\left(-1\right){}^{m+1}\left(-1\right){}^{j}\frac{n\left(n-1\right)\dots \left(n-m\right)\left[\left(n-m-1\right)!\right]}{\left(n-m-1-j\right)!j!}\hfill \\ \multicolumn{0}{c}{}& =\left(-1\right){}^{m+1}n\left(n-1\right)\dots \left(n-m\right)\sum _{j=0}^{n-m-1}\left(-1\right){}^{j}\left(\genfrac{}{}{0}{}{n-m-1}{j}\right)=0 .\hfill \end{array}$

(Note that the $j=0$ term is 1, which is consistent with the ${0}^{0}=1$ convention mentioned earlier.) So $\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{m}=0$ for $0\le m\le n-1$. Now consider the case $m=n$:

$\sum _{i=1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{n}=\sum _{i=1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)i\left(i-1\right)\dots \left(i-n+1\right)-\sum _{k=0}^{n-1}{b}_{k}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{k} .$

Every term in the first sum vanishes except the $n$th and each term of the second sum vanishes. Hence $\sum _{i=1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{n}=\left(-1\right){}^{n}n!$.
Returning to the problem at hand, we see that the right side of the desired equation is equal to

$\begin{array}{cc}\left(n+k\right){}^{n}\hfill & -\left(\genfrac{}{}{0}{}{n}{1}\right)\left(n+k-1\right){}^{n}+\left(\genfrac{}{}{0}{}{n}{2}\right)\left(n+k-2\right){}^{n}-\dots +\left(-1\right){}^{n}\left(\genfrac{}{}{0}{}{n}{n}\right)\left(n+k-n\right){}^{n}\hfill \\ \multicolumn{0}{c}{}& =\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(n-i+k\right){}^{n}=\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(n-i\right){}^{j}{k}^{n-j}\hfill \\ \multicolumn{0}{c}{}& =\sum _{i=0}^{n}\sum _{j=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(\genfrac{}{}{0}{}{n}{j}\right)\left(n-i\right){}^{j}{k}^{n-j}=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){k}^{n-j}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(n-i\right){}^{j}\hfill \\ \multicolumn{0}{c}{}& =\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){k}^{n-j}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{n-i}\right)\left(n-i\right){}^{j}\hfill \\ \multicolumn{0}{c}{}& =\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){k}^{n-j}\sum _{i=0}^{n}\left(-1\right){}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{j} .\hfill \end{array}$

When $0\le j\le n-1$, the sum $\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{n-i}\right)\left(n-i\right){}^{j}=\sum _{i=0}^{n}\left(-1\right){}^{n-i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{j}$ vanishes, while, when $j=n$, it assunes the value $n!$. Thus, the right side of the given equation is equal to $\left(\genfrac{}{}{0}{}{n}{n}\right){k}^{0}n!=n!$ as desired.
Solution 3. Let $m=n+k$, so that $m\ge n$, and let the right side of the equation be denoted by $R$. Then

$\begin{array}{cc}R\hfill & ={m}^{n}-\left(\genfrac{}{}{0}{}{n}{1}\right)\left(m-1\right){}^{n}+\left(\genfrac{}{}{0}{}{n}{2}\right)\left(m-2\right){}^{n}-\dots +\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(m-i\right){}^{n}+\dots +\left(-1\right){}^{n}\left(\genfrac{}{}{0}{}{n}{n}\right)\left(m-n\right){}^{n}\hfill \\ \multicolumn{0}{c}{}& ={m}^{m}\left[\sum _{j=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\right]-\left(\genfrac{}{}{0}{}{n}{1}\right){m}^{n-1}\left[\sum _{i=1}^{n}\left(-1\right){}^{i}i\left(\genfrac{}{}{0}{}{n}{i}\right)\right]+\left(\genfrac{}{}{0}{}{n}{2}\right){m}^{n-2}\left[\sum _{i=1}^{n}\left(-1\right){}^{i}{i}^{2}\left(\genfrac{}{}{0}{}{n}{i}\right)\right]+\dots \hfill \\ \multicolumn{0}{c}{}& +\left(-1\right){}^{n}\left(\genfrac{}{}{0}{}{n}{n}\right)\left[\sum _{i=1}^{n}\left(-1\right){}^{i}{i}^{n}\left(\genfrac{}{}{0}{}{n}{i}\right)\right] .\hfill \end{array}$

Let

${f}_{0}\left(x\right)=\left(1-x\right){}^{n}=\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){x}^{i}$

and let

${f}_{k}\left(x\right)=x{\mathrm{Df}}_{k-1}\left(x\right)$

for $k\ge 1$, where $\mathrm{Df}$ denotes the derivative of a function $f$. Observe that, from the closed expression for ${f}_{0}\left(x\right)$, we can establish by induction that

${f}_{k}\left(x\right)=\sum _{i=0}^{n}\left(-1\right){}^{i}{i}^{k}\left(\genfrac{}{}{0}{}{n}{i}\right){x}^{i}$

so that $R=\sum _{k=0}^{n}\left(-1\right){}^{k}\left(\genfrac{}{}{0}{}{n}{k}\right){m}^{n-k}{f}_{k}\left(1\right)$.
By induction, we establish that

${f}_{k}\left(x\right)=\left(-1\right){}^{k}n\left(n-1\right)\dots \left(n-k+1\right){x}^{k}\left(1-x\right){}^{n-k}+\left(1-x\right){}^{n-k+1}{g}_{k}\left(x\right)$

for some polynomial ${g}_{k}\left(x\right)$. This is true for $k=1$ with ${g}_{1}\left(x\right)=0$. Suppose if holds for $k=j$. Then

$\begin{array}{cc}{f}_{j}\text{'}\left(x\right)\hfill & =\left(-1\right){}^{j}n\left(n-1\right)\dots \left(n-j+1\right){x}^{j-1}\left(1-x\right){}^{n-j}-\left(-1\right){}^{j}n\left(n-1\right)\dots \left(n-j+1\right)\left(n-j\right){x}^{j}\left(1-x\right){}^{n-j-1}\hfill \\ \multicolumn{0}{c}{}& -\left(n-j+1\right)\left(1-x\right){}^{n-j}{g}_{j}\left(x\right)+\left(1-x\right){}^{n-j+1}{g}_{j}\text{'}\left(x\right) ,\hfill \end{array}$

whence

$\begin{array}{cc}{f}_{j+1}\left(x\right)\hfill & =\left(-1\right){}^{j+1}n\left({n}_{1}\right)\dots \left(n-j\right){x}^{j}\left(1-x\right){}^{n-\left(j+1\right)}+\left(1-x\right){}^{n-\left(j+1\right)+1}\left[\left(-1\right){}^{j}n\left(n-1\right)\dots \left(n-j+1\right){x}^{j}\hfill \\ \multicolumn{0}{c}{}& -\left(n-j+1\right){\mathrm{xg}}_{k}\left(x\right)+x\left(1-x\right){g}_{j}\text{'}\left(x\right)\right]\hfill \end{array}$

and we obtain the desired representation by induction. Then for $1\le k\le n-1$, ${f}_{k}\left(1\right)=0$ while ${f}_{n}\left(1\right)=\left(-1\right){}^{n}n!$. Hence $R=\left(-1\right){}^{n}{f}_{n}\left(1\right)=n!$.

64.
Let $M$ be a point in the interior of triangle $\mathrm{ABC}$, and suppose that $D$, $E$, $F$ are respective points on the side $\mathrm{BC}$, $\mathrm{CA}$, $\mathrm{AB}$, which all pass through $M$. (In technical terms, they are cevians.) Suppose that the areas and the perimeters of the triangles $\mathrm{BMD}$, $\mathrm{CME}$, $\mathrm{AMF}$ are equal. Prove that triangle $\mathrm{ABC}$ must be equilateral.
Solution. [L. Lessard] Let the common area of the triangles $\mathrm{BMD}$, $\mathrm{CME}$ and $\mathrm{AMF}$ be $a$ and let their common perimeter be $p$. Let the area and perimeter of $\Delta \mathrm{AME}$ be $u$ and $x$ respectively, of $\Delta \mathrm{MFB}$ be $v$ and $y$ respectively, and of $\Delta \mathrm{CMD}$ be $w$ and $z$ respectively.
By considering pairs of triangles with equal heights, we find that

$\frac{\mathrm{AF}}{\mathrm{FB}}=\frac{a}{v}=\frac{2a+u}{v+a+w}=\frac{a+u}{a+w} ,$

$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{a}{w}=\frac{2a+v}{u+a+w}=\frac{a+v}{a+u} ,$

$\frac{\mathrm{CE}}{\mathrm{EA}}=\frac{a}{u}=\frac{2a+w}{u+a+v}=\frac{a+w}{a+v} .$

>From these three sets of equations, we deduce that

$\frac{{a}^{3}}{\mathrm{uvw}}=1 ;$

${a}^{2}+\left(w-u\right)a-\mathrm{uv}=0 ,$

${a}^{2}+\left(u-w\right)a-\mathrm{vw}=0 ,$

${a}^{2}+\left(v-u\right)a-\mathrm{uw}=0 ;$

whence

${a}^{3}=\mathrm{uvw} \mathrm{and} 3{a}^{2}=\mathrm{uv}+\mathrm{vw}+\mathrm{uw} .$

This means that $\mathrm{uv},\mathrm{vw},\mathrm{uw}$ are three positive numbers whose geometric and arithmetic means are both equal to ${a}^{2}$. Hence ${a}^{2}=\mathrm{uv}=\mathrm{vw}=\mathrm{uw}$, so that $u=v=w=a$. It follows that $\mathrm{AF}=\mathrm{FB}$, $\mathrm{BD}=\mathrm{DC}$, $\mathrm{CE}=\mathrm{EA}$, so that $\mathrm{AD}$, $\mathrm{BE}$ and $\mathrm{CF}$ are medians and $M$ is the centroid.
Wolog, suppose that $\mathrm{AB}\ge \mathrm{BC}\ge \mathrm{CA}$. Since $\mathrm{AB}\ge \mathrm{BC}$, $\angle \mathrm{AEB}\ge {90}^{ˆ}$, and so $\mathrm{AM}\ge \mathrm{MC}$. Thus $x\ge p$. Similarly, $y\ge p$ and $p\ge z$.
Consider triangles $\mathrm{BMD}$ and $\mathrm{AME}$. We have $\mathrm{BD}\ge \mathrm{AE}$, $\mathrm{BM}\ge \mathrm{AM}$, $\mathrm{ME}=\frac{1}{2}\mathrm{BM}$ and $\mathrm{MD}=\frac{1}{2}\mathrm{AM}$. Therefore

$p-x=\left(\mathrm{BD}+\mathrm{MD}+\mathrm{BM}\right)-\left(\mathrm{AE}+\mathrm{ME}+\mathrm{AM}\right)=\left(\mathrm{BD}-\mathrm{AE}\right)+\frac{1}{2}\left(\mathrm{BM}-\mathrm{AM}\right)\ge 0$

and so $p\ge x$. Since also $x\ge p$, we have that $p=x$. But this implies that $\mathrm{AM}=\mathrm{MC}$, so that $\mathrm{ME}\perp \mathrm{AC}$ and $\mathrm{AB}=\mathrm{BC}$. Since $\mathrm{BE}$ is now an axis of a reflection which interchanges $A$ and $C$, as well as $F$ and $D$, it follows that $p=z$ and $p=y$ as well. Thus, $\mathrm{AB}=\mathrm{AC}$ and $\mathrm{AC}=\mathrm{BC}$. Thus, the triangle is equilateral.

65.
Suppose that $\mathrm{XTY}$ is a straight line and that $\mathrm{TU}$ and $\mathrm{TV}$ are two rays emanating from $T$ for which $\angle \mathrm{XTU}=\angle \mathrm{UTV}=\angle \mathrm{VTY}={60}^{ˆ}$. Suppose that $P$, $Q$ and $R$ are respective points on the rays $\mathrm{TY}$, $\mathrm{TU}$ and $\mathrm{TV}$ for which $\mathrm{PQ}=\mathrm{PR}$. Prove that $\angle \mathrm{QPR}={60}^{ˆ}$.
Solution 1. Let $\frakR$ be a rotation of ${60}^{ˆ}$ about $T$ that takes the ray $\mathrm{TU}$ to $\mathrm{TV}$. Then, if $\frakR$ transforms $Q\to Q\text{'}$ and $P\to P\text{'}$, then $Q\text{'}$ lies on $\mathrm{TV}$ and the line $Q\text{'}P\text{'}$ makes an angle of ${60}^{ˆ}$ with $\mathrm{QP}$. Because of the rotation, $\angle P\text{'}\mathrm{TP}={60}^{ˆ}$ and $\mathrm{TP}\text{'}=\mathrm{TP}$, whence $\mathrm{TP}\text{'}P$ is an equilateral triangle.
Since $\angle Q\text{'}\mathrm{TP}=\angle \mathrm{TPP}\text{'}={60}^{ˆ}$, $\mathrm{TV}P\text{'}P$. Let $\frakT$ be the translation that takes $P\text{'}$ to $P$. It takes $Q\text{'}$ to a point $Q\text{'}\text{'}$ on the ray $\mathrm{TV}$, and $\mathrm{PQ}\text{'}\text{'}=P\text{'}Q\text{'}=\mathrm{PQ}$. Hence $Q\text{'}\text{'}$ can be none other than the point $R$ [why?], and the result follows.
Solution 2. The reflection in the line $\mathrm{XY}$ takes $P\to P$, $Q\to Q\text{'}$ and $R\to R\text{'}$. Triangles $\mathrm{PQR}\text{'}$ and $\mathrm{PQ}\text{'}R$ are congruent and isosceles, so that $\angle \mathrm{TQP}=\angle \mathrm{TQ}\text{'}P=\angle \mathrm{TRP}$ (since $\mathrm{PQ}\text{'}=\mathrm{PR}$). Hence $\mathrm{TQRP}$ is a concyclic quadrilateral, whence $\angle \mathrm{QPR}=\angle \mathrm{QTR}={60}^{ˆ}$.
Solution 3. [S. Niu] Let $S$ be a point on $\mathrm{TU}$ for which $\mathrm{SR}\mathrm{XY}$; observe that $\Delta \mathrm{RST}$ is equilateral. We first show that $Q$ lies between $S$ and $T$. For, if $S$ were between $Q$ and $T$, then $\angle \mathrm{PSQ}$ would be obtuse and $\mathrm{PQ}>\mathrm{PS}>\mathrm{PR}$ (since $\angle \mathrm{PRS}>{60}^{ˆ}>\angle \mathrm{PSR}$ in $\Delta \mathrm{PRS}$), a contradiction.
The rotation of ${60}^{ˆ}$ with centre $R$ that takes $S$ onto $T$ takes ray $\mathrm{RQ}$ onto a ray through $R$ that intersects $\mathrm{TY}$ in $M$. Consider triangles $\mathrm{RSQ}$ and $\mathrm{RTM}$. Since $\angle \mathrm{RST}=\angle \mathrm{RTM}={60}^{ˆ}$, $\angle \mathrm{SRQ}={60}^{ˆ}-\angle \mathrm{QRT}=\angle \mathrm{TRM}$ and $\mathrm{SR}=\mathrm{TR}$, we have that $\Delta \mathrm{RSQ}\equiv \Delta \mathrm{RTM}$ and $\mathrm{RQ}=\mathrm{RM}$. (ASA) Since $\angle \mathrm{QRM}={60}^{ˆ}$, $\Delta \mathrm{RQM}$ is equilateral and $\mathrm{RM}=\mathrm{RQ}$. Hence $M$ and $P$ are both equidistant from $Q$ and $R$, and so at the intersection of $\mathrm{TY}$ and the right bisector of $\mathrm{QR}$. Thus, $M=P$ and the result follows.
Solution 4. [H. Pan] Let $Q\text{'}$ and $R\text{'}$ be the respective reflections of $Q$ and $R$ with respect to the axis $\mathrm{XY}$. Since $\angle \mathrm{RTR}\text{'}={120}^{ˆ}$ and $\mathrm{TR}=\mathrm{TR}\text{'}$, $\angle \mathrm{QR}\text{'}R=\angle \mathrm{TR}\text{'}R={30}^{ˆ}$. Since $Q,R,Q\text{'},R\text{'},$ lie on a circle with centre $P$, $\angle \mathrm{QPR}=2\angle \mathrm{QR}\text{'}R={60}^{ˆ}$, as desired.
Solution 5. [R. Barrington Leigh] Let $W$ be a point on $\mathrm{TV}$ such that $\angle \mathrm{WPQ}={60}^{ˆ}=\angle \mathrm{WTU}$. [Why does such a point $W$ exist?] Then $\mathrm{WQTP}$ is a concyclic quadrilateral so that $\angle \mathrm{QWP}={180}^{ˆ}-\angle \mathrm{QTP}={60}^{ˆ}$ and $\Delta \mathrm{PWQ}$ is equilateral. Hence $\mathrm{PW}=\mathrm{PQ}=\mathrm{PR}$.
Suppose $W\ne R$. If $R$ is farther away from $T$ than $W$, then $\angle \mathrm{RPT}>\angle \mathrm{WPT}>\angle \mathrm{WPQ}={60}^{ˆ}⇒{60}^{ˆ}>\angle \mathrm{TRP}=\angle \mathrm{RWP}>{60}^{ˆ}$, a contradiction. If $W$ is farther away from $T$ than $R$, then $\angle \mathrm{WPT}>\angle \mathrm{WPQ}={60}^{ˆ}⇒{60}^{ˆ}>\angle \mathrm{RWP}=\angle \mathrm{WRP}>{60}^{ˆ}$, again a contradiction. So $R=W$ and the result follows.
Solution 6. [M. Holmes] Let the circle through $T,P,Q$ intersect $\mathrm{TV}$ in $N$. Then $\angle \mathrm{QNP}={180}^{ˆ}-\angle \mathrm{QTP}={60}^{ˆ}$. Since $\angle \mathrm{PQN}=\angle \mathrm{PTN}={60}^{ˆ}$, $\Delta \mathrm{PQN}$ is equilateral so that $\mathrm{PN}=\mathrm{PQ}$. Suppose, if possible, that $R\ne N$. Then $N$ and $R$ are two points on $\mathrm{TV}$ equidistant from $P$. Since $\angle \mathrm{PNT}<\angle \mathrm{PNQ}={60}^{ˆ}$ and $\Delta \mathrm{PNR}$ is isosceles, we have that $\angle \mathrm{PNR}<{90}^{ˆ}$, so $N$ cannot lie between $T$ and $R$, and $\angle \mathrm{PRN}=\angle \mathrm{PNR}=\angle \mathrm{PNT}<{60}^{ˆ}$. Since $\angle \mathrm{PTN}={60}^{ˆ}$, we conclude that $T$ must lie between $R$ and $N$, which transgresses the condition of the problem. Hence $R$ and $N$ must coincide and the result follows.
Solution 7. [P. Cheng] Determine $S$ on $\mathrm{TU}$ and $Z$ on $\mathrm{TY}$ for which $\mathrm{SR}\mathrm{XY}$ and $\angle \mathrm{QRZ}={60}^{ˆ}$. Observe that $\angle \mathrm{TSR}=\angle \mathrm{SRT}={60}^{ˆ}$ and $\mathrm{SR}=\mathrm{RT}$.
Consider triangles $\mathrm{SRQ}$ and $\mathrm{TRZ}$. $\angle \mathrm{SRQ}=\angle \mathrm{SRT}-\angle \mathrm{QRT}=\angle \mathrm{QRZ}-\angle \mathrm{QRT}=\angle \mathrm{TRZ}$; $\angle \mathrm{QSR}={60}^{ˆ}=\angle \mathrm{ZTR}$, so that $\Delta \mathrm{SRQ}=\Delta \mathrm{TRZ}$ (ASA).
Hence $\mathrm{RZ}=\mathrm{RQ}⇒\Delta \mathrm{RQZ}$ is equilateral $⇒\mathrm{RZ}=\mathrm{ZQ}$ and $\angle \mathrm{RZQ}={60}^{ˆ}$. Now, both $P$ and $Z$ lie on the intersection of $\mathrm{TY}$ and the right bisector of $\mathrm{QR}$, so they must coincide: $P=Z$. The result follows.
Solution 8. Let the perpendicular, produced, from $Q$ to $\mathrm{XY}$ meet $\mathrm{VT}$, produced, in $S$. Then $\angle \mathrm{XTS}=\angle \mathrm{VTY}={60}^{ˆ}=\angle \mathrm{XTU}$, from which is can be deduced that $\mathrm{TX}$ right bisects $\mathrm{QS}$. Hence $\mathrm{PS}=\mathrm{PQ}=\mathrm{PR}$, so that $Q,R,S$ are all on the same circle with centre $P$.
Since $\angle \mathrm{QTS}={120}^{ˆ}$, we have that $\angle \mathrm{SQT}=\angle \mathrm{QSR}={30}^{ˆ}$, so that $\mathrm{QR}$ must subtend an angle of ${60}^{ˆ}$ at the centre $P$ of the circle. The desired result follows.
Solution 9. [A.Siu] Let the right bisector of $\mathrm{QR}$ meet the circumcircle of $\mathrm{TQR}$ on the same side of $\mathrm{QR}$ at $T$ in $S$. Since $\angle \mathrm{QSR}=\angle \mathrm{QTR}={60}^{ˆ}$ and $\mathrm{QS}=\mathrm{QR}$, $\angle \mathrm{SQR}=\angle \mathrm{SRQ}={60}^{ˆ}$. Hence $\angle \mathrm{STQ}={180}^{ˆ}-\angle \mathrm{SRQ}={120}^{ˆ}$. But $\angle \mathrm{YTQ}={120}^{ˆ}$, so $S$ must lie on $\mathrm{TY}$. It follows that $S=P$.
Solution 10. Assign coordinates with the origin at $T$ and the $x-$axis along $\mathrm{XY}$. The the respective coordinates of $Q$ and $R$ have the form $\left(u,-\sqrt{3}u\right)$ and $\left(v,\sqrt{3}v\right)$ for some real $u$ and $v$. Let the coordinates of $P$ be $\left(w,0\right)$. Then $\mathrm{PQ}=\mathrm{PR}$ yields that $w=2\left(u+v\right)$. [Exercise: work it out.]

$\begin{array}{cc}‖\mathrm{PQ}‖{}^{2}-‖\mathrm{QR}‖{}^{2}\hfill & =\left(u-w\right){}^{2}+3{u}^{2}-\left(u-v\right){}^{2}-3\left(u+v\right){}^{2}\hfill \\ \multicolumn{0}{c}{}& ={w}^{2}-2\mathrm{uw}-4v\left(u+v\right)={w}^{2}-2\mathrm{uw}-2\mathrm{vw}\hfill \\ \multicolumn{0}{c}{}& ={w}^{2}-2\left(u+v\right)w=0 .\hfill \end{array}$

Hence $\mathrm{PQ}=\mathrm{QR}=\mathrm{PR}$ and $\Delta \mathrm{PQR}$ is equilateral. Therefore $\angle \mathrm{QPR}={60}^{ˆ}$.
Solution 11. [J.Y. Jin] Let $\frakC$ be the circumcircle of $\Delta \mathrm{PQR}$. If $T$ lies strictly inside $\frakC$, then ${60}^{ˆ}=\angle \mathrm{QTR}>\angle \mathrm{QPR}$ and ${60}^{ˆ}=\angle \mathrm{PTR}>\angle \mathrm{PQR}=\angle \mathrm{PRQ}$. Thus, all three angle of $\Delta \mathrm{PQR}$ would be less than ${60}^{ˆ}$, which is not possible. Similarly, if $T$ lies strictly outside $\frakC$, then ${60}^{ˆ}=\angle \mathrm{QTR}<\angle \mathrm{QPR}$ and ${60}^{ˆ}=\angle \mathrm{PTR}<\angle \mathrm{PQR}=\angle \mathrm{PRQ}$, so that all three angles of $\Delta \mathrm{PQR}$ would exceed ${60}^{ˆ}$, again not possible. Thus $T$ must be on $\frakC$, whence $\angle \mathrm{QPR}=\angle \mathrm{QTR}={60}^{ˆ}$.
Solution 12. [C. Lau] By the Sine Law,

$\frac{\mathrm{sin}\angle \mathrm{TQP}}{‖\mathrm{TP}‖}=\frac{\mathrm{sin}{120}^{ˆ}}{‖\mathrm{PQ}‖}=\frac{\mathrm{sin}{60}^{ˆ}}{‖\mathrm{PR}‖}=\frac{\mathrm{sin}\angle \mathrm{TRP}}{‖\mathrm{TP}‖} ,$

whence $\mathrm{sin}\angle \mathrm{TQP}=\mathrm{sin}\angle \mathrm{TRP}$. Since $\angle \mathrm{QTP}$, in triangle $\mathrm{QTP}$ is obtuse, $\angle \mathrm{TQP}$ is acute.
Suppose, if possible, that $\angle \mathrm{TRP}$ is obtuse. Then, in triangle $\mathrm{TPR}$, $\mathrm{TP}$ would be the longest side, so $\mathrm{PR}<\mathrm{TP}$. But in triangle $\mathrm{TQP}$, $\mathrm{PQ}$ is the longest side, so $\mathrm{PQ}>\mathrm{TP}$, and so $\mathrm{PQ}\ne \mathrm{PR}$, contrary to hypothesis. Hence $\angle \mathrm{TRP}$ is acute. Therefore, $\angle \mathrm{TQP}=\angle \mathrm{TRP}$. Let $\mathrm{PQ}$ and $\mathrm{RT}$ intersect in $Z$. Then, ${60}^{ˆ}=\angle \mathrm{QTZ}={180}^{ˆ}-\angle \mathrm{TQP}-\angle \mathrm{QZT}={180}^{ˆ}-\angle \mathrm{TRP}-\angle \mathrm{RZP}=\angle \mathrm{QPR}$, as desired.

66.
(a) Let $\mathrm{ABCD}$ be a square and let $E$ be an arbitrary point on the side $\mathrm{CD}$. Suppose that $P$ is a point on the diagonal $\mathrm{AC}$ for which $\mathrm{EP}\perp \mathrm{AC}$ and that $Q$ is a point on $\mathrm{AE}$ produced for which $\mathrm{CQ}\perp \mathrm{AE}$. Prove that $B,P,Q$ are collinear.
(b) Does the result hold if the hypothesis is weakened to require only that $\mathrm{ABCD}$ is a rectangle?
Solution 1. Let $\mathrm{ABCD}$ be a rectangle, and let $E$, $P$, $Q$ be determined as in the problem. Suppose that $\angle \mathrm{ACD}=\angle \mathrm{BDC}=\alpha$. Then $\angle \mathrm{PEC}={90}^{ˆ}-\alpha$. Because $\mathrm{EPQC}$ is concyclic, $\angle \mathrm{PQC}=\angle \mathrm{PEC}={90}^{ˆ}-\alpha$. Because $\mathrm{ABCQD}$ is concyclic, $\angle \mathrm{BQC}=\angle \mathrm{BDC}=\alpha$. The points $B$, $P$, $Q$ are collinear $&lrArr;\angle \mathrm{BQC}=\angle \mathrm{PQC}&lrArr;\alpha ={90}^{ˆ}-\alpha &lrArr;\alpha ={45}^{ˆ}&lrArr;\mathrm{ABCD}$ is a square.
Solution 2. (a) $\mathrm{EPQC}$, with a pair of supplementary opposite angles, is concyclic, so that $\angle \mathrm{CQP}=\angle \mathrm{CEP}={180}^{ˆ}-\angle \mathrm{EPC}-\angle \mathrm{ECP}={45}^{ˆ}$. Since $\mathrm{CBAQ}$ is concyclic, $\angle \mathrm{CQB}=\angle \mathrm{CAB}={45}^{ˆ}$. Thus, $\angle \mathrm{CQP}=\angle \mathrm{CQB}$ so that $Q$, $P$, $B$ are collinear.
(b) Suppose that $\mathrm{ABCD}$ is a nonquare rectangle. Then taking $E=D$ yields a counterexample.
Solution 3. (a) The circle with diameter $\mathrm{AC}$ that passes through the vertices of the square also passes through $Q$. Hence $\angle \mathrm{QBC}=\angle \mathrm{QAC}$. Consider triangles $\mathrm{PBC}$ and $\mathrm{EAC}$. Since triangles $\mathrm{ABC}$ and $\mathrm{EPC}$ are both isosceles right triangles, $\mathrm{BC}:\mathrm{AC}=\mathrm{PC}:\mathrm{EC}$. Also $\angle \mathrm{BCA}=\angle \mathrm{PCE}={45}^{ˆ}$. Hence $\Delta \mathrm{PBC}~\Delta \mathrm{EAC}$ (SAS) so that $\angle \mathrm{PBC}=\angle \mathrm{EAC}=\angle \mathrm{QAC}=\angle \mathrm{QBC}$. It follows that $Q$, $P$, $B$ are collinear.
Solution 4. [S. Niu] Let $\mathrm{ABCD}$ be a rectangle and let $E,P,Q$ be determined as in the problem. Let $\mathrm{EP}$ be produced to meet $\mathrm{BC}$ in $F$. Since $\angle \mathrm{ABF}=\angle \mathrm{APF}$, the quadrilateral $\mathrm{ABPF}$ is concyclic, so that $\angle \mathrm{PBC}=\angle \mathrm{PBF}=\angle \mathrm{PAF}$. Since $\mathrm{ABCQ}$ is concyclic, $\angle \mathrm{QBC}=\angle \mathrm{QAC}=\angle \mathrm{PAE}$. Now $B,P,Q$ are collinear

$&lrArr;\angle \mathrm{PBC}=\angle \mathrm{QBC}&lrArr;\angle \mathrm{PAF}=\angle \mathrm{PAE}&lrArr;\mathrm{AC} \mathrm{right} \mathrm{bisects} \mathrm{EF}$

$&lrArr;\angle \mathrm{ECA}=\angle \mathrm{ACB}={45}^{ˆ}&lrArr;\mathrm{ABCD} \mathrm{is} a \mathrm{square} .$

Solution 5. [M. Holmes] (a) Suppose that $\mathrm{BQ}$ intersects $\mathrm{AC}$ in $R$. Since $\mathrm{ABCQD}$ is concyclic, $\angle \mathrm{AQR}=\angle \mathrm{AQB}=\angle \mathrm{ACB}={45}^{ˆ}$, so that $\angle \mathrm{BQC}={45}^{ˆ}$. Since $\angle \mathrm{EQR}=\angle \mathrm{AQB}=\angle \mathrm{ECR}={45}^{ˆ}$, $\mathrm{ERCQ}$ is concyclic, so that $\angle \mathrm{ERC}={180}^{ˆ}-\angle \mathrm{EQC}={90}^{ˆ}$. Hence $\mathrm{ER}\perp \mathrm{AC}$, so that $R=P$ and the result follows.
Solution 6. [L. Hong] (a) Let $\mathrm{QC}$ intersect $\mathrm{AB}$ in $F$. We apply Menelaus' Theorem to triangle $\mathrm{AFC}$: $B$, $P$, $Q$ are collinear if and only if

$\frac{\mathrm{AB}}{\mathrm{BF}}·\frac{\mathrm{FQ}}{\mathrm{QC}}·\frac{\mathrm{CP}}{\mathrm{PA}}=-1 .$

Let the side length of the square be 1 and the length of $\mathrm{DE}$ be $a$. Then $‖\mathrm{AB}‖=1$. Since $\Delta \mathrm{ADE}~\Delta \mathrm{FBC}$, $\mathrm{AD}:\mathrm{DE}=\mathrm{BF}:\mathrm{BC}$, so that $‖\mathrm{BF}‖=1/a$ and $‖\mathrm{FC}‖=\sqrt{1+{a}^{2}}/a$. Since $\Delta \mathrm{ADE}~\Delta \mathrm{CQE}$, $\mathrm{CQ}:\mathrm{EC}=\mathrm{AD}:\mathrm{EA}$, so that $‖\mathrm{CQ}‖=\left(1-a\right)/\sqrt{1+{a}^{2}}$. Hence

$\frac{‖\mathrm{FQ}‖}{‖\mathrm{CQ}‖}=1+\frac{‖\mathrm{FC}‖}{‖\mathrm{CQ}‖}=1+\frac{1+{a}^{2}}{a\left(1-a\right)}=\frac{1+a}{a\left(1-a\right)} .$

Since $\Delta \mathrm{ECP}$ is right isosceles, $‖\mathrm{CP}‖=\left(1-a\right)/\sqrt{2}$ and $‖\mathrm{PA}‖=\sqrt{2}-‖\mathrm{CP}‖=\left(1+a\right)/\sqrt{2}$. Hence $‖\mathrm{CP}‖/‖\mathrm{PA}‖=\left(1-a\right)/\left(1+a\right)$. Multiplying the three ratios together and taking account of the directed segments gives the product $-1$ and yields the result.
Solution 7. (a) Select coordinates so that $A~\left(0,1\right)$, $B~\left(0,0\right)$, $C~\left(1,0\right)$, $D~\left(1,1\right)$ and $E~\left(1,t\right)$ for some $t$ with $0\le t\le 1$. It is straightforward to verify that $P~\left(1-\frac{t}{2},\frac{t}{2}\right)$.
Since the slope of $\mathrm{AE}$ is $t-1$, the slope of $\mathrm{AQ}$ should be $\left(1-t\right){}^{-1}$. Since the coordinates of $Q$ have the form $\left(1+s,s\left(1-t\right){}^{-1}\right)$ for some $s$, it is straightforward to verify that

$Q~\left(\frac{2-t}{1+\left(1-t\right){}^{2}},\frac{t}{1+\left(1-t\right){}^{2}\right)}\right) .$

It can now be checked that the slope of each of $\mathrm{BQ}$ and $\mathrm{BP}$ is $t\left(2-t\right){}^{-1}$, which yields the result.
(b) The result fails if $A~\left(0,2\right)$, $B~\left(0,0\right)$, $C~\left(1,0\right)$, $D~\left(1,2\right)$. If $E~\left(1,1\right)$, then $P~\left(\frac{3}{5},\frac{4}{5}\right)$ and $Q~\left(\frac{3}{2},\frac{1}{2}\right)$.


# Math Help - Proving that this is a group

1. ## Proving that this is a group

Hello, I would really appreciate any help with the following problem:

Prove that $(G, *)$ is a group, where $G=<-1,1>$ and $a*b=\dfrac{a+b}{1+ab}$,.... $\forall a, b \in G$.

_______________________________

To prove that this is a group, $(G,*)$ must satisfy closure, associativity, existence of an identity element and existence of inverse elements.

Could the property of associativity be demonstrated like this?

$(a*b)*c=(\dfrac{a+b}{1+ab})*c=...=\dfrac{a+b+c+abc }{1+ab+ac+bc}$

$a*(b*c)=a*(\dfrac{b+c}{1+bc})=...=\dfrac{a+b+c+abc }{1+ab+ac+bc}$

Therefore, $(a*b)*c=a*(b*c)$, so the property of associativity is satisfied.

_______________________________

Is the identity element $0$?

From $a*0= \dfrac{a+0}{1+a \cdot 0}=0*a=a$

it follows $a*0=0*a=a$

_______________________________

Is the inverse element of $a$, $-a$?

Because $a*(-a)=\dfrac{a-a}{1-a^2}=\dfrac{0}{1-a^2}=(a \in <-1,1>)=(-a)*a = 0$

_______________________________

If the above is correct (and I'm not certain that it is), the only property left to be demonstrated is closure, i.e. if $a, b \in <-1,1>$ then $a*b =\dfrac{a+b}{1+ab} \in <-1,1>$.

And here I need your help! How to demonstrate the property of closure?

There is a hint in the text: "Observe that $|1+ab|=1+ab$, and prove that $|a+b| \leq 1+ab$ if and only if $0 \leq (1-a)(1-b)$ and $0 \leq (1+a)(1+b)$".

So, how to prove the statements from the hint, and, more importantly, how to apply them to demonstrate the closure property of $(G,*)$?

Many thanks!

2. Originally Posted by gusztav
Hello, I would really appreciate any help with the following problem:

Prove that $(G, *)$ is a group, where $G=<-1,1>$ and $a*b=\dfrac{a+b}{1+ab}$,.... $\forall a, b \in G$.

_______________________________

To prove that this is a group, $(G,*)$ must satisfy closure, associativity, existence of an identity element and existence of inverse elements.

Could the property of associativity be demonstrated like this?

$(a*b)*c=(\dfrac{a+b}{1+ab})*c=...=\dfrac{a+b+c+abc }{1+ab+ac+bc}$

$a*(b*c)=a*(\dfrac{b+c}{1+bc})=...=\dfrac{a+b+c+abc }{1+ab+ac+bc}$

Therefore, $(a*b)*c=a*(b*c)$, so the property of associativity is satisfied.

_______________________________

Is the identity element $0$?

From $a*0= \dfrac{a+0}{1+a \cdot 0}=0*a=a$

it follows $a*0=0*a=a$

_______________________________

Is the inverse element of $a$, $-a$?

Because $a*(-a)=\dfrac{a-a}{1-a^2}=\dfrac{0}{1-a^2}=(a \in <-1,1>)=(-a)*a = 0$

_______________________________

If the above is correct (and I'm not certain that it is), the only property left to be demonstrated is closure, i.e. if $a, b \in <-1,1>$ then $a*b =\dfrac{a+b}{1+ab} \in <-1,1>$.

And here I need your help! How to demonstrate the property of closure?

There is a hint in the text: "Observe that $|1+ab|=1+ab$, and prove that $|a+b| \leq 1+ab$ if and only if $0 \leq (1-a)(1-b)$ and $0 \leq (1+a)(1+b)$".

So, how to prove the statements from the hint, and, more importantly, how to apply them to demonstrate the closure property of $(G,*)$?

Many thanks!

First ,what do you mean by $G=<-1,1>$ ?? The subset of the reals (or complex) with those two elements? If so zero cannot be anything there since it doesn't belong to the set!
Also, the product of $1\,,\,-1$ isn't defined: $1*-1:= \frac{1+(-1)}{1+1(-1)}=\frac{0}{0}$ ...!!

If this is so the above isn't a group.

But perhaps you meant the open interval $(-1,1)$ which, by some misterious reason, you denote by $<-1,1>$...(!)

Then you must show $a,b\in(-1,1)\Longrightarrow \frac{a+b}{1+ab}\in (-1,1)\Longleftrightarrow \left|\frac{a+b}{1+ab}\right|<1$ $\Longleftrightarrow (a+b)^2<(1+ab)^2\Longleftrightarrow a^2+2ab+b^2

$\Longleftrightarrow a^2(b^2-1)-(b^2-1)>0\Longleftrightarrow (a^2-1)(b^2-1)>0$ , and this last inequality is obviously true.

Tonio

3. Thank you very much, Tonio!

Originally Posted by tonio
But perhaps you meant the open interval $(-1,1)$
Yes, I meant the open interval $\{x \in \mathbb{R} : -1

Originally Posted by tonio
Then you must show $a,b\in(-1,1)\Longrightarrow \frac{a+b}{1+ab}\in (-1,1)\Longleftrightarrow \left|\frac{a+b}{1+ab}\right|<1$ $\Longleftrightarrow (a+b)^2<(1+ab)^2\Longleftrightarrow a^2+2ab+b^2

$\Longleftrightarrow a^2(b^2-1)-(b^2-1)>0\Longleftrightarrow (a^2-1)(b^2-1)>0$ , and this last inequality is obviously true.
Aha! Now everything is clear.
(Because $a \in (-1,1) \Longrightarrow |a|<1 \Longrightarrow a^2<1 \Longrightarrow a^2-1<0$ and, similarly, $b^2-1<0$, so $(a^2-1)( b^2-1)>0$ )


# Area under quarter circle by integration

How would one go about finding out the area under a quarter circle by integrating. The quarter circle's radius is r and the whole circle's center is positioned at the origin of the coordinates. (The quarter circle is in the first quarter of the coordinate system)

From the equation $x^2+y^2=r^2$, you may express your area as the following integral $$A=\int_0^r\sqrt{r^2-x^2}\:dx.$$ Then substitute $x=r\sin \theta$, $\theta=\arcsin (x/r)$, to get \begin{align} A&=\int_0^{\pi/2}\sqrt{r^2-r^2\sin^2 \theta}\:r\cos \theta \:d\theta\\ &=r^2\int_0^{\pi/2}\sqrt{1-\sin^2 \theta}\:\cos\theta \:d\theta\\ &=r^2\int_0^{\pi/2}\sqrt{\cos^2 \theta}\:\cos\theta \:d\theta\\ &=r^2\int_0^{\pi/2}\cos^2 \theta \:d\theta\\ &=r^2\int_0^{\pi/2}\frac{1+\cos(2\theta)}2 \:d\theta\\ &=r^2\int_0^{\pi/2}\frac12 \:d\theta+\frac{r^2}2\underbrace{\left[ \frac12\sin(2\theta)\right]_0^{\pi/2}}_{\color{#C00000}{=\:0}}\\ &=\frac{\pi}4r^2. \end{align}
• Yes, we have, for $0<x<r$, $\frac{d\theta}{dx}=\frac{1}{\sqrt{r^2-x^2}}>0$, $0=\arcsin (0/r) \leq \theta (r)\leq \arcsin (r/r)=\pi/2$. Thanks! Oct 8 '15 at 18:45
Here is a quicker solution. The area can be seen as a collection of very thin triangles, one of which is shown below. As $d\theta\to0$, the base of the triangle becomes $rd\theta$ and the height becomes $r$, so the area is $\frac12r^2d\theta$. The limits of $\theta$ are $0$ and $\frac\pi2$. $$\int_0^\frac\pi2\frac12r^2d\theta=\frac12r^2\theta|_0^\frac\pi2=\frac14\pi r^2$$
let circle: $x^2+y^2=r^2$ then consider a slab of area $dA=ydx$ then the area of quarter circle $$A_{1/4}=\int_0^r ydx=\int_0^r \sqrt{r^2-x^2}dx$$ $$=\frac12\left[x\sqrt{r^2-x^2}+r^2\sin^{-1}\left(x/r\right)\right]_0^r$$
$$=\frac12\left[0+r^2(\pi/2)\right]=\frac{\pi}{4}r^2$$ or use double integration: $$=\iint rdr d\theta= \int_0^{\pi/2}\ d\theta\int_0^R rdr=\int_0^{\pi/2}\ d\theta(R^2/2)=(R^2/2)(\pi/2)=\frac{\pi}{4}R^2$$


# Find the 2 digits in front of the number 14, so that the 4-digit number we get is divisible by 26.

I know that 26=13.2. The last digit of the number is 4, it is divisible by 2. I have to remove the last digit and check out if the remaining digits are divisible by 13 or?

• what do you mean by '2 digits in front of the number 14'? – oliverjones Apr 20 '17 at 18:55
• Do you mean $14xx$ or $xx14$ or something different? – Mark Bennet Apr 20 '17 at 18:56
• gotta be xx14 -- two digits in front of the number 14 – gt6989b Apr 20 '17 at 18:56
• I wanted to say xx14. – Elena Apr 20 '17 at 18:58

You are being asked to find an integer $x$ such that $10 \leq x \leq 99$ and $$100 x + 14 \equiv 0 \pmod{26}.$$ Solve this equation, and $x$ will be the two digits you want. There are multiple possible answers.

A good first step is to rewrite the equation in a more convenient form. Since it is modulo $26,$ you can replace any of the numbers by something equivalent modulo $26.$ For example, $100 \equiv 22 \pmod{26},$ so you can write $22$ instead of $100.$ I think it's more convenient to use the fact that $100 \equiv -4 \pmod{26},$ however; it lets you work with smaller numbers.

• i think the OP is asking how to solve this equation – gt6989b Apr 20 '17 at 19:00
• We know that $100x + 14 \equiv 0\pmod{26} \iff 100x + 14 = 26m$, for some integer $m$. Thus, $$x = {26m-14\over 100} = {13m - 7 \over 50}.$$ Since we require that $10\le x \le 99$, choosing $m$ appropriately should give us the desired result. For example, we can choose $m=39$. Then we have $500/50=10$, which implies $x=10$ would be a good starting point, and so $1014$ will be a satisfactory answer. – Decaf-Math Apr 20 '17 at 19:05
• @gt6989b I think that looking at the question this way is fundamentally different from what the OP was trying to do: "remove the last digit and check if the remaining digits are divisible by ..." etc. But I added some detail in hope that it may help. – David K Apr 20 '17 at 19:06
• How many answers do I have to get? 2? The first one is 1014 – Elena Apr 20 '17 at 22:03
• I don't know if you need more than one answer, but you have a choice. Since 1300 is the smallest multiple of 100 divisible by 26, you can add any positive multiple of 1300 to 1014 and get another solution, unless the multiple is so large that the result is greater than 9999. It turns out there is room for seven separate solutions between 1000 and 9999, starting with 1014 and ending with 8814. – David K Apr 20 '17 at 22:21

Starting with $100 x + 14 \equiv 0 \pmod{26}$ you can take $100 \pmod{26}$ to get $74, 48, 22, -4$, etc. $-4$ is the smallest, so let's try that.

Now the equivalence to solve is $-4x + 14 \equiv 0 \pmod{26}$. It'll be easier to see the solution if the LHS is positive, so add $26$ to get $-4x + 40 \equiv 0 \pmod{26}$. Is it possible to subtract 4 from 40 a few times to get $26$? If you subtract $4*3=12$ you still have $2$ to go to get to $26$.

So add another $26$ to the equation to get $$-4x + 40 \equiv 0 \pmod{26}$$. Note that $66-26=40=4*10$. So $x$ is $10$.

You would then verify $x=10$ in the original problem. Is $1014 \equiv 0 \pmod{26}$?

• Yes, it is. But I think a have more than 1 answer because (100, 26)=2 – Elena Apr 20 '17 at 22:09
• I don't understand your comment. There will be $7$ answers. The answer can be any value of the form $10+13x$, with $0 \leq x \leq 6$. I interpreted the problem as requiring just one answer. – Χpẘ Apr 20 '17 at 22:24
• I guess not. David K also specifies $7$ answers in the comments to his answer. – Χpẘ Apr 20 '17 at 22:29

Let $1\le a\le 9$ and $0\le b\le9$ be those two front numbers ($ab14$). So, our desired number is $10^3a+10^2b+14$ where $$10^3a+10^2b+14\equiv0\pmod{26}\implies 12a-4b\equiv12\pmod{26}\implies 3a-3=3(a-1)\equiv b\pmod{13}$$ All choices that satisfy this (just plug in $a=1,2,\ldots9$ and see what happens):

$1014$ $2314$ $3614$ $4914$ $6214$ $7514$ $8814$

Notice how in the congruence, I used $$a\equiv b\pmod{m}\implies \frac{a}{c}\equiv \frac{b}{c}\pmod{\frac{m}{\gcd(m,c)}}$$ as long as $c\vert a$ and $c\vert b$. Also notice that $a\equiv9$ is not possible because that gives an invalid number for $b$

• I thought we only had 1 answer - 1014? Or? – Elena Apr 20 '17 at 19:49
• You can check for yourself that all $7$ of those integers work just fine. – user12345 Apr 20 '17 at 22:51

One way to find a solution is to start from the useful and memorable fact that $7 \times 11 \times 13 = 1001$. Hence $26$ divides any even multiple of $1001$, in particular $4004$.

If we can find $x10$ divisible by $26$, then $4004 + x10$ will yield a solution. $x10$ is divisible (for any $x$) by $2$, so it suffices to find $x1$ divisible by $13$. Since $91 = 7 \times 13$, a solution is $4004 + 910 = 4914$.


# Does the infinite series $\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}$ converge?

I have been wondering if this infinite series converges $$\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}$$

I tried to put it in wolfram alpha but it says that the ratio test is inconclusive, but when I do the ratio test I get: Let first $a_k :=\frac{1}{(4+(-1)^k)^k}$, then we have that $\frac{1}{5^k}\le a_k \le \frac{1}{3^k}$. We also see from here that a_k is always positive and now we have $\frac{1}{5}\le\sqrt[k]{a_k}\le\frac{1}{3}<1, \forall k\in\mathbb{N}$. From that we should actually have that the series converges or am I missing something?

One more thing that I also noticed is that, if I use my inequality we can also have that $$\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}\le\sum_{k=1}^{\infty} \frac{1}{3^k}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$$

Does that mean it converges to $\frac{3}{2}$?

• I'm sure it converges. Your inequality says it all – Yuriy S May 1 '16 at 20:03
• Comparison test. – Robert Israel May 1 '16 at 20:03
• It's approximately $\frac{5}{12}$ by numerical estimation – Yuriy S May 1 '16 at 20:04

Notice that the terms of this series are positive and we have

$$\frac{1}{(4+(-1)^k)^k}\le 3^{-k}$$ and the geometric series $\sum 3^{-k}$ is convergent. Use comparison to conclude.

• Does it also mean it converges to $\frac{3}{2}$ like the series of $3^{-k}$ does? – HeatTheIce May 1 '16 at 20:11
• No, but certainly the sum of this series is $<\frac32$. – user296113 May 1 '16 at 20:12
• Do not forget that in addition to being bounded above by the $3^{-k}$ series (guaranteeing convergence), you are also bounded by $5^{-k}$ from below (which helps you narrow down your value). Therefore, you know the series converges to some value $x$ such that $5/4\leq x \leq 3/2$. – AmateurDotCounter May 1 '16 at 20:36
• @LetEpsilonBeLessThanZero Note that $\sum_1^\infty 5^{-k} = 1/4$. – stochasticboy321 May 1 '16 at 20:38
• @HeatTheIce To compute the sum, try splitting the series into two - one with the odd terms, and another with the even terms. You'll notice that both are convergent geometric series. – stochasticboy321 May 1 '16 at 20:38

Does that mean it converges to $\frac{3}{2}$?

Here is a closed form of the series:

$$\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}=\color{blue}{\frac5{12}}.$$

Proof. By the absolute convergence, one is allowed to write \begin{align} \sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}&=\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^{2k})^{2k}}+\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^{2k-1})^{2k-1}} \\\\&=\sum_{k=1}^{\infty} \frac{1}{25^k}+3\sum_{k=1}^{\infty} \frac{1}{9^k} \\\\&=\frac1{24}+\frac38 \\\\&=\frac5{12}. \end{align}

Use the $\;k\,-$ th root test:

$$\lim\sup_{k\to\infty}\sqrt[k]{\frac1{4+(-1)^k)^k}}=\lim\sup_{k\to\infty}\frac1{4+(-1)^k}=\frac13<1$$

and thus the series converges (observe it is a positive series)


Solving systems of linear equations. This calculator solves Systems of Linear Equations using Gaussian Elimination Method, Inverse Matrix Method, or Cramer's rule.Also you can compute a number of solutions in a system of linear equations (analyse the compatibility) using Rouché–Capelli theorem.. First, we need to find the inverse of the A matrix (assuming it exists!) Section 2.3 Matrix Equations ¶ permalink Objectives. Think of “dividing” both sides of the equation Ax = b or xA = b by A.The coefficient matrix A is always in the “denominator.”. Enter coefficients of your system into the input fields. The dimension compatibility conditions for x = A\b require the two matrices A and b to have the same number of rows. Using the Matrix Calculator we get this: (I left the 1/determinant outside the matrix to make the numbers simpler) Then multiply A-1 by B (we can use the Matrix Calculator again): And we are done! System Of Linear Equations Involving Two Variables Using Determinants. In such a case, the pair of linear equations is said to be consistent. Solution: Given equation can be written in matrix form as : , , Given system … Characterize the vectors b such that Ax = b is consistent, in terms of the span of the columns of A. The matrix valued function $$X (t)$$ is called the fundamental matrix, or the fundamental matrix solution. Let $$\vec {x}' = P \vec {x} + \vec {f}$$ be a linear system of A system of linear equations is as follows. Let the equations be a 1 x+b 1 y+c 1 = 0 and a 2 x+b 2 y+c 2 = 0. row space: The set of all possible linear combinations of its row vectors. Example 1: Solve the equation: 4x+7y-9 = 0 , 5x-8y+15 = 0. 1. The solution is: x = 5, y = 3, z = −2. Typically we consider B= 2Rm 1 ’Rm, a column vector. a 11 x 1 + a 12 x 2 + … + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + … + a 2 n x n = b 2 ⋯ a m 1 x 1 + a m 2 x 2 + … + a m n x n = b m This system can be represented as the matrix equation A ⋅ x → = b → , where A is the coefficient matrix. To solve nonhomogeneous first order linear systems, we use the same technique as we applied to solve single linear nonhomogeneous equations. Key Terms. If the rows of the matrix represent a system of linear equations, then the row space consists of all linear equations that can be deduced algebraically from those in the system. Understand the equivalence between a system of linear equations, an augmented matrix, a vector equation, and a matrix equation. The solution to a system of equations having 2 variables is given by: Systems of Linear Equations 0.1 De nitions Recall that if A2Rm n and B2Rm p, then the augmented matrix [AjB] 2Rm n+p is the matrix [AB], that is the matrix whose rst ncolumns are the columns of A, and whose last p columns are the columns of B. Consistent System. To sketch the graph of pair of linear equations in two variables, we draw two lines representing the equations. Solve the equation by the matrix method of linear equation with the formula and find the values of x,y,z. Developing an effective predator-prey system of differential equations is not the subject of this chapter. Solve several types of systems of linear equations. How To Solve a Linear Equation System Using Determinants? Theorem 3.3.2. Theorem. The whole point of this is to notice that systems of differential equations can arise quite easily from naturally occurring situations. A necessary condition for the system AX = B of n + 1 linear equations in n unknowns to have a solution is that |A B| = 0 i.e. Find where is the inverse of the matrix. the determinant of the augmented matrix equals zero. Whole point of this chapter example 1: solve the equation by the matrix method of linear is. 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What's wrong in my calculation of $\int \frac{\sin x}{1 + \sin x} dx$?

I have the following function:

$$f: \bigg ( - \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg ) \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{\sin x}{1 + \sin x}$$

And I have to find $$\displaystyle\int f(x) dx$$. This is what I did:

$$\int \dfrac{\sin x}{1 + \sin x}dx= \int \dfrac{1+ \sin x - 1}{1 + \sin x}dx = \int dx - \int \dfrac{1}{1 + \sin x}dx =$$ $$= x - \int \dfrac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx$$

$$= x - \int \dfrac{1 - \sin x}{1 - \sin ^2 x} dx$$

$$= x - \int \dfrac{1 - \sin x}{\cos^2 x} dx$$

$$= x - \int \dfrac{1}{\cos^2x}dx + \int \dfrac{\sin x}{\cos^2 x}dx$$

$$= x - \tan x + \int \dfrac{\sin x}{\cos^2 x}dx$$

Let $$u = \cos x$$

$$du = - \sin x dx$$

$$=x - \tan x - \int \dfrac{1}{u^2}du$$

$$= x - \tan x + \dfrac{1}{u} + C$$

$$= x - \tan x + \dfrac{1}{\cos x} + C$$

The problem is that the options given in my textbook are the following:

A. $$x + \tan {\dfrac{x}{2}} + C$$

B. $$\dfrac{1}{1 + \tan{\frac{x}{2}}} + C$$

C. $$x + 2\tan{\dfrac{x}{2}} + C$$

D. $$\dfrac{2}{1 + \tan{\frac{x}{2}}} + C$$

E. $$x + \dfrac{2}{1 + \tan{\frac{x}{2}}} + C$$

None of them are the answer I got solving this integral. What is the mistake that I made and how can I find the right answer? By what I've been reading online, you can get different answers by solving an integral in different ways and all of them are considered correct. They differ by the constant $$C$$. I understand that, but I don't see how to solve this integral in such a way to get an answer among the given $$5$$. And, even more importantly, how can I recognize the right answer in exam conditions if the answer provided by my solution is not present among the given options? Is solving in a different manner my only hope?

• This is one of the faults of multiple choice questions. In this case they expected you to use the substitution $t=\tan\left(\frac{x}{2}\right)$. Your answer will be equivalent to one of the choices. – John Wayland Bales Dec 21 '19 at 0:57
• If you encounter this on exam, then you should be able to convert your answer into one of the ones they give you, by possibly adding a constant and applying trig identities. – Robo300 Dec 21 '19 at 1:02
• A deeper fault of multiple-choice for a question like that is that one could just differentiate the proposed answers and check the result at 0. Correct answer (E), but not proving that the student mastered the underlying concept or technique. – Catalin Zara Dec 21 '19 at 1:39
• @CatalinZara true ....... – Aryadeva Dec 21 '19 at 2:46

Required answer is $$E$$. Observe that $$\frac{1}{\cos x}-\tan x=\frac{1-\sin x}{\cos x}$$ $$=\frac{(\cos\frac{x}{2}-\sin\frac{x}{2})^2}{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}$$ $$=\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}$$ $$=\frac{1-\tan\frac{x}{2}}{1+\tan\frac{x}{2}}$$ $$=\frac{2}{1+\tan\frac{x}{2}}-1$$ Also, note that you could have directly got this answer if you have integrated $$\frac{1}{1+\sin x}$$ in a different manner, as follows. $$\int\frac{1}{1+\sin x}dx=\int\frac{1}{(\cos\frac{x}{2}+\sin\frac{x}{2})^2}dx$$ $$=\int \frac{1}{\cos^2\frac{x}{2}(1+\tan\frac{x}{2})^2}dx$$ Now substitute $$\tan\frac{x}{2}$$ and you are done.

• How did you get to $\dfrac{2}{1+\tan{\frac{x}{2}}}-1$ from $\dfrac{1-\tan {\frac{x}{2}}}{1+ \tan {\frac{x}{2}}}$? I didn't understand that bit. – user592938 Dec 22 '19 at 1:06
• @user2502, $\frac{1-\tan\frac{x}{2}}{1+\tan\frac{x}{2}}=\frac{2-(1+\tan\frac{x}{2})}{1+\tan\frac{x}{2}} = \frac{2}{1+\tan\frac{x}{2}}-1$ – Martund Dec 22 '19 at 6:28

Your answer is correct and it matches with choice (E). You have to use half angle formulas: $$\sin A = \frac{2 \tan \frac{A}{2}}{1+\tan^2 \frac{A}{2}} \quad \cos A = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}}$$ Observe that \begin{align*} \frac{1}{\cos x}-\tan x&=\frac{1-\sin x}{\cos x}\\ & = \frac{\left(1-\tan \frac{x}{2}\right)^2}{1-\tan^2 \frac{x}{2}}\\ & = \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\\ & = 1+\frac{2}{1+\tan \frac{x}{2}}\\ \end{align*}

• How did you to the last line from your second to last line? I didn't understand that final bit. – user592938 Dec 21 '19 at 22:49

Beside verifying your answer is equivalent to one of the listed results, you may also integrate in $$\tan\frac x2$$ since all the choices are in terms of it.

So, use $$\sin x =\frac{2\tan\frac x2}{1+\tan^2\frac x2}$$ to integrate,

$$\int \dfrac{1}{1 + \sin x}dx = \int \dfrac{1}{1 + \frac{2\tan\frac x2}{1+\tan^2\frac x2}}dx =\int \dfrac{1+\tan^2\frac x2}{1+\tan^2\frac x2 + 2\tan\frac x2}dx$$ $$=\int \dfrac{\sec^2\frac x2}{(1+\tan\frac x2)^2} =\int \dfrac{2d(\tan\frac x2)}{(1+\tan\frac x2)^2} =-\dfrac{2}{1+\tan\frac x2}$$


# easy, easy question

• Jul 14th 2009, 07:06 AM
billym
easy, easy question

How many ways are there to seat 10 people, consisting of 5 couples, in a row of seats (10 seats wide) if all couples are to get adjacent seats?
• Jul 14th 2009, 07:20 AM
Soroban
Hello, billym!

Quote:

How many ways are there to seat 10 people, consisting of 5 couples,
in a row of seats (10 seats wide) if all couples are to get adjacent seats?

For reference, let the couples be: . $(A,a),\:(B,b),\:(C,c),\:(D,d),\:(E,e)$

Duct-tape the couples together.

We have 5 "people" to arrange: . $\boxed{Aa}\;\boxed{Bb}\;\boxed{Cc}\;\boxed{Dd}\;\b oxed{Ee}$

There are: . $5! \,=\,120$ permutations.

But for each permutation, the couples can be "swtiched".

. . $\boxed{Aa}$ could be $\boxed{aA}$, $\boxed{Bb}$ could be $\boxed{bB}$, and so on.

There are: . $2^5 \,=\,32$ possible switchings.

Therefore, there are: . $120\cdot32 \:=\:{\color{blue}3840}$ seating arrangements.

• Jul 14th 2009, 07:55 AM
HallsofIvy
Quote:

Originally Posted by Soroban
Hello, billym!

For reference, let the couples be: . $(A,a),\:(B,b),\:(C,c),\:(D,d),\:(E,e)$

Duct-tape the couples together.

I've always suspected you had an evil streak, Soroban!(Giggle)

I have a friend who swears you can do anything with duct-tape. Now I know you can even solve math problems with it!

Quote:

We have 5 "people" to arrange: . $\boxed{Aa}\;\boxed{Bb}\;\boxed{Cc}\;\boxed{Dd}\;\b oxed{Ee}$

There are: . $5! \,=\,120$ permutations.

But for each permutation, the couples can be "swtiched".

. . $\boxed{Aa}$ could be $\boxed{aA}$, $\boxed{Bb}$ could be $\boxed{bB}$, and so on.

There are: . $2^5 \,=\,32$ possible switchings.

Therefore, there are: . $120\cdot32 \:=\:{\color{blue}3840}$ seating arrangements.

• Jul 14th 2009, 08:38 AM
Soroban
Hello, HallsofIvy!

Quote:

I have a friend who swears you can do anything with duct tape.
Duct tape is like The Force.

It has a light side and dark side
and it holds the universe together.


# Two methods of implicit differentiation don't correspond

I recently attempted a question on implicit differentiation twice. I differentiated using one method in the first attempt and then another method in the second attempt but they do not correspond when I plug in values for the variables x and y. Please look at the two methods and tell me what I am doing wrong.

In the first attempt I just took the first derivative of each side of the equation with respect to x. In the second attempt, I took the ln of both sides of the equation first and then found the derivatives of either side of the equation.

Errata: The answer for the first attempt is y' = y(y - e^(x/y)) / (y ^ 2 - xe^(x/y))

• I think you accidentally introduced an extra $y$ into the second term near the end of the first attempt. This aside, the results may look different but really be the same, if you consider that you could replace $e^{x/y}$ with the very different-looking expression $x-y$ because they are the same according to the original equation. – MPW May 19 '16 at 22:06
• Oh yes the extra y – rert588 May 19 '16 at 22:08
• $$e^{x/y}=x-y$$ – Simply Beautiful Art May 19 '16 at 22:08
• Yes. That is the question. – rert588 May 19 '16 at 22:11
• I put the y there by mistake when i was writing out the answers neatly but even without that y, the two attempts do not correspond. – rert588 May 19 '16 at 22:14

On the second to last line of your first attempt, you had $y'=\frac{y(y-e^\frac{x}{y}y)}{y^2-e^{\frac{x}{y}}x}$ where actually it should be $y'=\frac{y(y-e^\frac{x}{y})}{y^2-e^{\frac{x}{y}}x}$. And then if you substitute $e^{\frac{x}{y}}$ with $x-y$, you will get $y'=\frac{y(y-(x-y))}{y^2-(x-y)x}=\frac{2y^2-xy}{y^2-x^2+xy}$, which is the same as the result from your second approach.
• I am pretty sure their first attempt is correct. That $e^{\frac x y}y$ term comes from distributing the $e^{\frac x y}$ over the numerator of the derivative of $\frac x y$. EDIT: Oh, wait, I see. They factored out a $y$, but then forgot to remove the $y$ from the $e^{\frac x y}y$ term. You're right. – Noble Mushtak May 19 '16 at 22:24
• The key thing to notice here is that your derivative with the first method and second method look different, but are actually the same because $e^{x/y}=x-y$, so we can substitute that into the derivative using the first method to show that it's equivalent to the derivative using the second method. – Noble Mushtak May 19 '16 at 22:31