Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Deck of 52 cards, 3 card rank/order repeat pattern probability. A standard deck of 52 playing cards is well shuffled and drawn 1 card at a time without replacement. What is the probability that the ranks of the 3 most recently drawn cards (starting at drawn card 6) have already been seen in the exact same order in the already drawn cards? For example, when the 6th card is drawn, we need to check if cards 4, 5, and 6 are the same exact ranks as cards 1, 2, and 3 in the same exact order. If not, then we draw card 7 and check if cards 5, 6, and 7 have been seen in that same order in cards 1, 2, 3 or 2, 3,and 4... So basically we are checking the last 3 cards drawn against all previously drawn triples, including overlapping triples (such as cards 2,3,4 that overlap cards 1,2,3)
So for example, using A, 2, 3, 4, 5, 6, 7, 8, 9, T, J, Q, K as the 13 possible card ranks, the following partial deck would be a pattern found scenario:
5, Q, 7, 3, K, K, 8, Q, 7, 3 because the Q, 7, 3 triple was previously seen in that exact order.
$UPDATE$: For clarification, you need not scan into the last 3 cards drawn for a match. When I said overlapping triples I did not mean into the last 3 cards drawn, but rather overlapping triples such as scanning cards 1,2,3 then 2,3,4, then 3,4,5... from the last 3 cards drawn (start scanning after card 6 is drawn and after every other drawn card after that such as 7, 8,... 51, 52).
So my question is what is the probability this will happen at least once in a well shuffled deck? That is, at least one triple pattern of ranks will repeat exactly in order in the same deck as the cards are drawn one at a time, as described above.
| Any 3 card sequence can be:
No pair, one pair, 3 of a kind.
With probabilities $\frac{52\cdot48\cdot 44}{52\cdot 51\cdot 50}, \frac{52\cdot3\cdot 48\cdot 3}{52\cdot 51\cdot 50}, \frac{52\cdot3\cdot 2}{52\cdot 51\cdot 50}$ respectively.
Verify that these sum up to 1.
If the sequence is no pair. There is a probability that a given 3 card sequence in the deck matches rank for rank is:
$\frac {3\cdot 3\cdot 3}{49\cdot 48\cdot 47}$
If a sequence has one pair, it is less likely.
$\frac {2\cdot 1\cdot 3}{49\cdot 48\cdot 47}$
If a sequence is 3 of a kind, it is impossible to have a second sequence that matches the first.
That two sequences of 3 cards match
$\frac {52\cdot 48\cdot 44\cdot 27+52\cdot 3\cdot 48\cdot 3\cdot 6}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47}$
How many non-overlapping pairs sequences of 3 cards are there in a 52 card deck? ${48\choose 2}$
How to explain this? If we think of the first sequence of sequence $A$ and the second sequence as sequence $B.$
We can burn some amount of cards before the beginning sequence $A$, and some amount of cards after the end of sequence $B$ lets call these $a,b$ respectively. The sum of the cards burnt must be less than or equal to 46.
The number of pairs of non-negative integers such that $a + b \le 46$ is ${48\choose 2}$
(remember $a,b$ could equal $0.$)
$\frac {52\cdot 48\cdot 44\cdot 27+52\cdot 3\cdot 48\cdot 3\cdot 6}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47}\cdot {48\choose 2} \approx 23.8\%$
Update....
As has been pointed out by the original poster, I have the expected number of duplicated sequences...
Inclusion-Exclusion
Taking a shortcut....
$p = \frac {52\cdot 48\cdot 44\cdot 27+52\cdot 3\cdot 48\cdot 3\cdot 6}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47}$
$p{48\choose 2} - p^2{48\choose 2}{42\choose 2} + p^3{48\choose 2}{42\choose 2}{43\choose 2} \approx 20.34\%$ which is $0.2034$.
| {
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"url": "https://math.stackexchange.com/questions/2936417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that $\dfrac{n^3 + 2n}{3}$ is an integer Show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer
My try:
Let P(n) be the statement that $n^3 + 2n$ is divisible by $3$.
Base step:
When $n = 0$ we have $0^3 + 0 = 0 = 3 \times 0$
So, the base case true.
Inductive hypothesis:
Assume that $P(k)$ is true.
which means $\dfrac{k^3 + 2k}{3}$ is divisible by $3$ and $\dfrac{k^3 + 2k}{3}=p$ for some integer $p$.
Now we need to show that $P(k+1)$ is true.
$(k+1)^3+2(k+1)$ and we will show that this divisible by $3$.
Proof:
$(k+1)^3+2(k1)=k^3+3k^2+3k+1+2k+2$
$.$
$.$
$.$
$.$
$=3(p+k^2+k+1)$
As $p+k^2+k+1$ is an integer we have that $(k+1)^3+2(k+!)$ is divisible by $3$.
Is my above attempt correct?
Did I show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer?
| Write $$n^3+2n$$ as $$n(n^2+2)$$
If $$n\equiv 0 \mod 3$$ then all is clear.
If $$n\equiv 1 \mod 3$$ then $$n^2+2\equiv 0\mod 3$$
If $$n\equiv 2\mod 3$$ then $$n^2+2\equiv 0\mod 3$$
and another idea
:
$$n^3-n+3n=n(n^2-1)+3n=(n-1)n(n+1)+3n$$
| {
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"url": "https://math.stackexchange.com/questions/2936874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve linear congruence with two variable that is not system congruence? This problem from Number theory-James K. Strayer
The congruence is
$$2x+3y\equiv 4 \pmod{7}$$
since it is not system congruence the eliminate method does not work
Can anyone help me?
| $2x\equiv (4 -3y) mod 7$
$x\equiv (2-(3/2)y) mod 7$ (For $k=1$ since $\frac{3+7k}{2}=5$ then $(3/2)\equiv 5 mod7 $
$x\equiv (2-5y) mod 7$.
Since $-5\equiv 2 mod 7$, we get $x\equiv (2+2y) mod 7$. Then $x\equiv (2(1+y)) mod 7$.
Then $y=\bar{-1}=\bar6$ for $x=\bar0$, and $y=\bar{3}$ for $x=\bar1$ (Put $x=1$ in the $x=2(1+y)$ then we have $y=\frac{-1}{2}$, for $k=1$, $y=\frac{-1+7k}{2}=3 $), and $y=\bar{0}$ for $x=\bar2$ (we can calculate by using same argument), and $y=\bar{4}$ for $x=\bar3$, and $y=\bar{1}$ for $x=\bar4$, and $y=\bar{5}$ for $x=\bar5$, and $y=\bar{2}$ for $x=\bar6$.
So the answer is the set of binaries $\{(\bar0,\bar6), (\bar1,\bar3),(\bar2,\bar0),(\bar3,\bar4),(\bar5,\bar5),(\bar6,\bar2) \}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Double summation with improper integral So my friend sent me this really interesting problem. It goes:
Evaluate the following expression:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx .$$
Here is my approach:
First evaluate the integral:
$$ \frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx.$$
This can be done using integration by parts and we get:
$$ \frac{1}{b!} \frac{b}{a} \int_0^\infty \frac{x^{b-1}}{e^{ax}}\ dx.$$
We can do this $ b $ times until we get:
$$ \frac{1}{b!} \frac{(b)(b-a).....(b-b+1)}{a^b} \int_0^\infty \frac{x^{b-b}}{e^{ax}}\ dx.$$
and hence we end up with:
$$ \frac{1}{b!} \frac{b!}{a^b}\qquad\left(\frac{-1 \ e^{-ax}}{a}\Big|_0^\infty\right) = \frac{1}{a^{b+1}}.$$
Now we can apply the sum of GP to infinity formula and we get:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \frac{1}{a^{b+1}} = \sum_{a=2}^\infty \frac{\frac{1}{a^{2}}}{1-\frac{1}{a}}.$$
This is a telescoping series and we end up with $$ \frac{1}{a-1} = \frac{1}{2-1} = 1.$$
Do you guys have any other ways of solving this problem? Please do share it here.
| The integral is of the Gamma type,
$$\int_{0}^\infty \frac{x^{b}}{e^{ax}} \ dx=\frac1{a^{b+1}}\int_{0}^\infty t^be^{-t}\ dx =\frac{b!}{a^{b+1}}.$$
Then
$$\sum_{a=2}^\infty \sum_{b=1}^\infty\frac1{a^{b+1}}=\sum_{a=2}^\infty \frac1{a^2\left(1-\dfrac1a\right)}=\sum_{a=2}^\infty \frac1{a(a-1)}$$ is indeed a telescoping sum, giving
$$\frac1{2-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Solving $y^\prime=(y-1)^2$ with $y(0)=1.01$
$$y^\prime=(y-1)^2, \qquad y(0)=1.01$$
I have tried:
$$\begin{align}
\int \frac{dy}{(y-1)^2} &= \int dx \\[4pt]
-(y-1)^{-1} &= x \\
y &= 1-\frac{1}{x}+C
\end{align}$$
Then I cannot find the $C$, because if $x=0$, then $y= -\infty$.
What's wrong?
| A modified approach is to let $f(t) = y(t) - 1$ to obtain a solution to the modified problem
$$f' = f^2 \, \hspace{5mm} \, f(0) = \frac{1}{100}.$$
With this then
\begin{align}
\frac{d f}{dt} &= f^2 \\
\int \frac{df}{f^2} &= \int dt \\
- \frac{1}{f} &= t + c_{0}.
\end{align}
When $t = 0$ this leads to $c_{0} = - 100$ and
$$f(t) = - \frac{1}{t - 100}.$$
In terms of $y(t)$ this result provides
$$y(t) = 1 + \frac{1}{100 - t} = \frac{101 - t}{100 - t}.$$
Check:
$$y'(t) = \frac{1}{(100 - t)^2} = (y-1)^2$$
and $y(0) = 101/100$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit of $\begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}$
Find the following limit: \begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}
I have tried to divide the numerator and denominator by $\sqrt{x}$, but it did not work.
I have tried to multiply by the conjugates of the numerator and denominator simultaneously but it did not work.
I have tried to multiply by the conjugates of the numerator only but it did not work.
So what shall I do?
| If you want to go even beyond the limit itself, let $x=y+4$ to make
$$A=\frac{\sqrt{2 x+1}-3}{\sqrt{x}-2}=\frac{\sqrt{2 y+9}-3}{\sqrt{y+4}-2}$$ and use the binomial expansion or (better) Taylor series around $y=0$. This would give
$$A=\frac{\frac{y}{3}-\frac{y^2}{54}+O\left(y^3\right) }{\frac{y}{4}-\frac{y^2}{64}+O\left(y^3\right)}=\frac{4}{3}+\frac{y}{108}+O\left(y^2\right)$$ Back to $x$,
$$A=\frac{4}{3}+\frac{x-4}{108}+O\left((x-4)^2\right)$$
Try using $x=5$; the exact value is $A=\left(2+\sqrt{5}\right) \left(\sqrt{11}-3\right)\approx 1.34124$ while the expansion gives $\frac{145}{108}\approx 1.34259$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $|x -a| < \min\{|a|/2,\, \varepsilon a^2 /2\}$ then...
Suppose $a \neq 0$ and $\varepsilon >0$. If $$|x-a| < \min \left \{ \frac{|a|}{2}, \frac{\varepsilon a^2 }{2}\right \},$$
then show that $x \neq 0$ and $\left |\frac{1}{x}-\frac{1}{a}\right| < \varepsilon$.
Proof:
Let $\delta = \min \left \{ \frac{|a|}{2}, \frac{\varepsilon a^2 }{2}\right \}$, where $\delta \leq \frac{|a|}{2}, \frac{\varepsilon a^2}{2}$ , $0< |x - a| < \delta$.
Then $|\frac{1}{x} - \frac{1}{a}| = |\frac{a-x}{ax}|$....
I was able to solve this when x and a are fixed positive. But the question only specifies that $x,a \neq 0$.
| Case 1: $a>0, |a|\le\varepsilon a^2$
In this particular case:
$$\varepsilon a\ge1$$
$$|x-a|<\frac a2\implies$$
$$-\frac{a}{2}<x-a<\frac a2$$
$$\frac{a}{2}<x<\frac {3a}2 \quad \\ \text{(notice: all values are positive)}$$
$$\frac{2}{a}>\frac 1x>\frac {2}{3a}$$
$$\frac{2}{a}-\frac 1a>\frac 1x-\frac 1a>\frac {2}{3a}-\frac 1a$$
$$\frac{1}{a}>\frac 1x-\frac 1a>-\frac {1}{3a}$$
$$\left|\frac 1x-\frac 1a\right|<\frac 1a=\frac
\varepsilon{a\varepsilon}\le\varepsilon$$
Case 2: $a>0, |a|>\varepsilon a^2$
In this particular case:
$$\varepsilon a<1$$
$$|x-a|<\frac {\varepsilon a^2}2\implies$$
$$-\frac {\varepsilon a^2}2<x-a<\frac {\varepsilon a^2}2$$
$$a-\frac {\varepsilon a^2}2<x<a+\frac {\varepsilon a^2}2\\ \text{(notice: all values are positive)}$$
$$\frac{2}{a(2-\varepsilon a)}-\frac 1a>\frac 1x-\frac 1a >\frac{2}{a(2+\varepsilon a)}-\frac 1a$$
$$\frac{\varepsilon}{2-\varepsilon a} > \frac 1x-\frac 1a > -\frac{\varepsilon}{2+\varepsilon a}$$
$$\frac{\varepsilon}{2-\varepsilon a} > \left|\frac 1x-\frac 1a \right|\tag{1}$$
But $\varepsilon a<1 \implies 2-\varepsilon a>1$ or:
$$\frac1{2-\varepsilon a}<1$$
$$\frac\varepsilon{2-\varepsilon a}<\varepsilon\tag{2}$$
Combine (1) and (2) and you get:
$$\varepsilon > \left|\frac 1x-\frac 1a \right|$$
Case 3: $a<0, |a|\le\varepsilon a^2$
Intorduce $b=-a$ so that $b$ is positive. In this particular case:
$$\varepsilon b>=1$$
We have to prove the following implications:
$$|x+b|<\frac b2\implies$$
$$-\frac b2<x+b<\frac b2$$
$$-b-\frac b2<x<\frac b2 - b$$
$$-\frac {3b}2<x<-\frac b2\\ \text{(notice: all numbers are negative)}$$
$$-\frac {2}{3b}+\frac 1b>\frac 1x + \frac 1b>-\frac 2b+\frac 1b$$
$$-\frac {2}{3b}+\frac 1b>\frac 1x + \frac 1b>-\frac 2b+\frac 1b$$
$$\frac {1}{3b}>\frac 1x + \frac 1b>-\frac 1b$$
$$\left|\frac 1x + \frac 1b\right|<\frac 1b=\frac {\varepsilon}{\varepsilon b}\le\varepsilon$$
$$\left|\frac 1x - \frac 1a\right|<\varepsilon$$
Case 4: $a<0, |a|>\varepsilon a^2$
Introduce $b=-a$ so that $b$ is positive. In this particular case:
$$\varepsilon b<1$$
We have to prove the following implications:
$$|x+b|<\frac {\varepsilon b^2}2\implies$$
$$-\frac {\varepsilon b^2}2<x+b<\frac {\varepsilon b^2}2$$
$$-b-\frac {\varepsilon b^2}2<x<-b+\frac {\varepsilon b^2}2\\ \text{(notice: all values are negative)}$$
$$-\frac{2}{b(2+\varepsilon b)}+\frac 1b>\frac 1x+\frac 1b >\frac{2}{b(-2+\varepsilon b)}+\frac 1b$$
$$\frac{\varepsilon}{2+\varepsilon b} > \frac 1x+\frac 1b > -\frac{\varepsilon}{2-\varepsilon b}$$
$$\left|\frac 1x+\frac 1b \right|<\frac{\varepsilon}{2-\varepsilon b}\tag{3}$$
But $\varepsilon b<1 \implies 2-\varepsilon b>1$ or:
$$\frac1{2-\varepsilon b}<1$$
$$\frac\varepsilon{2-\varepsilon b}<\varepsilon\tag{4}$$
Combine (3) and (4) and you get:
$$\left|\frac 1x+\frac 1b \right|=\left|\frac 1x-\frac 1a \right|<\varepsilon$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find $\lim\limits_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}$ Find $\displaystyle \lim_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}$
I tried using L'hopital's rule but it got very messy very fast
UPDATE- So reading about the Taylor series this is what I have so far
$$\lim_{x\to0}\frac{(x-x^3/3!+x^5/5!-\cdots)-(x+x^3/3+2x^5/5+17x^7/315+\cdots)}{(x+x^2/6+3x^5/40+\cdots)-(x-x^3/3+x^5/5-\cdots)}$$
But I'm still stuck
| This is a nice exercise which shows that a lot more can be achieved using standard limits than what most beginners would think.
We are going to use the following standard limits $$\lim_{x\to 0} \frac{\sin x} {x} =1,\lim_{x\to 0} \frac{1-\cos x} {x^2}=\frac{1}{2},\lim_{x\to 0}\frac{\arctan x} {x} =1$$ All of these are immediate consequences of the first limit.
The numerator of the given expression can be written as $$\frac{\sin x} {x} \cdot x^3\cdot\frac{\cos x-1}{x^2}\cdot\frac{1}{\cos x} $$ and using the standard limit we can replace the above with $-x^3/2$.
The denominator needs a little more effort. Using the identities $$\arcsin x=\arctan\frac{x} {\sqrt{1-x^2}},\arctan x-\arctan y=\arctan\frac{x-y} {1+xy}$$ we can write the denominator as $$\arctan\dfrac{\dfrac{x} {\sqrt{1-x^2}}-x } {1+\dfrac{x^2}{\sqrt{1-x^2}}} $$ and using the limit $\lim_{x\to 0}(\arctan x) /x=1$ the above expression can be replaced by $$\dfrac{\dfrac{x} {\sqrt{1-x^2}}-x } {1+\dfrac{x^2}{\sqrt{1-x^2}}}=\frac{x(1-\sqrt{1-x^2})}{x^2+\sqrt{1-x^2}}$$ Thus the desired limit is equal to the limit of $$\dfrac{-x^3/2}{\dfrac{x(1-\sqrt{1-x^2})}{x^2+\sqrt{1-x^2}}}=-\frac{x^2(x^2+\sqrt{1-x^2})}{2(1-\sqrt{1-x^2})}$$ Multiplying numerator and denominator by $1+\sqrt{1-x^2}$ we can simply the expression above as $$-\frac{(x^2+\sqrt{1-x^2})(1+\sqrt {1-x^2})} {2} $$ and this clearly has limit $-1$ as $x\to 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all combinations that sum up to a specific number How to find all the combinations that adds up to a certain number, given the number of elements?
For example if target $T$ is $ 100 $ and number of elements is $2$.
Then the result is $101$. ($100+ 0$, $99 + 1$, $98 + 2$,… $1 + 99$, $0+ 100$.)
What would the formula be, if the given number of elements is $ 3, 4$, etc?
| The more general way to do this is to use generating functions. The number of ways to write $100$ as the sum of two non-negative integers is the coefficient of $x^{100}$ in the product
$$
(1 + x + x^2 + \cdots + x^{100})(1 + x + x^2 + \cdots + x^{100})
$$
and the number of ways to write $100$ as the sum of $5$ non-negative integers is the coefficient of $x^{100}$ in
$$
(1 + x + x^2 + \cdots + x^{100})^{5}.
$$
This method is more flexible because you can control which terms appear in the sum. For example, if you want the first term to be even, the second term to be a multiple of $4$, and the third term to be a multiple of $6$, you want the coefficient of $x^{100}$ in
$$
(1 + x^2 + x^4 + \cdots + x^{100})(1 + x^4 + x^8 + \cdots + x^{100})(1 + x^6 + x^{12} + \cdots + x^{96}).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that: $\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$ Prove that for nonegative $x,y,z$ we have: $$\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$$
I prove that using the tangent line method. We may assume that $x+y+z=1$, so you we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$
A tangent on $f(x)=\sqrt{1-x}$ at $x={1\over 3}$ is $$y=-{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$
So we have, for all $x\in[0,1]$:
$$\sqrt{1-x} \leq -{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$ and we are done...
I wonder if there is elegant method avoiding calculus?
| By your idea:
Let $x+y+z=3$.
Thus, we need to prove that
$$\sum_{cyc}\sqrt{3-x}\leq3\sqrt2$$ or
$$\sum_{cyc}\left(\sqrt2-\sqrt{3-x}\right)\geq0$$ or
$$\sum_{cyc}\frac{x-1}{\sqrt2+\sqrt{3-x}}\geq0$$ or
$$\sum_{cyc}\left(\frac{x-1}{\sqrt2+\sqrt{3-x}}-\frac{x-1}{2\sqrt2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2}{\left(\sqrt2+\sqrt{3-x}\right)^2}\geq0.$$
Also, by Jensen we obtain:
$$\sum_{cyc}\sqrt{x+y}\leq3\sqrt{\frac{\sum\limits_{cyc}(x+y)}{3}}=\sqrt{6(x+y+z)}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How check if the sequence $x_n=(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing or not? I want to check if the sequence $(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing. I tried computing $\frac{x_{n+1}}{x_n}$ to check if the ratio is less than 1 or greater than 1, but I am unable to simplify:
$$\frac{x_{n+1}}{x_n} = \frac{n^{\frac{1}{n}}\left(n+2\right)^{\frac{1}{n+1}}}{\left(n+1\right)^{\frac{2n+1}{n\left(n+1\right)}}}$$
I also tried $x_{n+1} - x_n = \left(\frac{n+2}{n+1}\right)^{\frac{1}{n+1}}-\left(\frac{n+1}{n}\right)^{\frac{1}{n}}$
Is there any other way to check the monotonic behaviour of the sequence $x_n$?
Also I would like to know if I can check $y_n=(1-\frac{1}{n})^{\frac{1}{n}}$ using similar arguments?
| Considering$$x_n=\left(1+\frac{1}{n}\right)^{\frac{1}{n}}\implies \log(x_n)=\frac{1}{n}\log\left(1+\frac{1}{n}\right)$$
$$\log(x_{n+1})-\log(x_n)=\frac{1}{n+1}\log\left(1+\frac{1}{n+1}\right)-\frac{1}{n}\log\left(1+\frac{1}{n}\right)$$ Now, use Taylor expansion for large values of $n$ to get
$$\log(x_{n+1})-\log(x_n)=-\frac{2}{n^3}+\frac{9}{2 n^4}+O\left(\frac{1}{n^5}\right)$$ Continue with Taylor
$$\frac{x_{n+1}}{x_n}=e^{\log(x_{n+1})-\log(x_n)}=1-\frac{2}{n^3}+\frac{9}{2 n^4}+O\left(\frac{1}{n^5}\right)\implies \frac{x_{n+1}}{x_n} <1 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How is the sequence $x_{n+1} = \frac{(x_n)^{2} + 5}{ 6}$ going to converge to 1 Having some trouble understanding that if $x_{1} = 4$ and the sequence where n is defined as $x_{n+1} = \frac{(x_n)^{2} + 5}{ 6}$ how is it going to converge to 1.
I have solved using the L as limit and using the quadratic i get two possibilites that are 5, or 1.
| Since
$$
x_{n+1}=\frac{x_n^2+5}6\tag1
$$
we have
$$
x_{n+2}-x_{n+1}=\frac{x_{n+1}^2-x_n^2}6=(x_{n+1}-x_n)\,\frac{x_{n+1}+x_n}6\tag2
$$
This is a contraction when $\left|\frac{x_{n+1}+x_n}6\right|\lt1$. That is.
$$
-1\lt\frac{(x_n+3)^2-4}{36}\lt1\tag3
$$
Inequality $(3)$ is satisfied when
$$
-3-2\sqrt{10}\lt x_n\lt-3+2\sqrt{10}\tag4
$$
$x_1=4$ is outside the range specified in $(4)$.
| {
"language": "en",
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Factoring $x^5-1$ over $\mathbb{Z_5}$ How do I completely factor $x^5-1$ over $\mathbb{Z_5}$?
I saw one was a root so I divided by $x-1$ and got $(x-1)(x^4+x^3+x^2+x+1)$ which duh is the 5th cyclotomic polynomial and this is the same factorization over the integers. Do I just look for roots from here or is there a better way to go about this? I mean I know I could try to factor it as arbitrary quadratics also, but i'm wondering if there was a more fundamental insight that i'm missing that would make this simpler. Thanks!
| Since $1$ is a root, we can divide out $x-1$ by the factor theorem: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
$1$ is a root of $x^4+x^3+x^2+x+1$, so we can divide it out again: $x^4+x^3+x^2+x+1=(x-1)(x^3+2x^2+3x+4)$
$x^3+2x^2+3x+4=(x-1)(x^2+3x+1)$.
Lastly,
$x^2+3x+1=(x-1)(x-1)$. So we can write $x^5-1$ as the product $(x-1)^5$ in $\mathbb{Z}[x]$.
| {
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Why the $\lim_{n\to\infty} (\frac{n}{n^2+1}+\frac{n}{n^2+4}+.....+\frac{n}{n^2+n^2})= \frac{\pi}{4}$? Why the $\lim_{n\to\infty} (\frac{n}{n^2+1}+\frac{n}{n^2+4}+.....+\frac{n}{n^2+n^2})= \frac{\pi}{4}$?
I read somewhere that it is related to $f(x)=\frac{1}{1+x^2}$ but dont know why...
| $$\sum_{i=1}^n \dfrac{n}{n^2+i^2}=\sum_{i=1}^n \dfrac{n^2}{n^2+i^2}\dfrac{1}{n}=\sum_{i=1}^n \dfrac{1}{1+(0+i/n)^2}\dfrac{1}{n}$$
Now reimagine: $\Delta x = \dfrac{1-0}{n}, x_i = 0+i\Delta x = \dfrac{i}{n}$
Then your sum is equal to
$$\sum_{i=1}^{n} f(x_i) \Delta x$$
Now take the limit as $n\to\infty$
| {
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Closed form formula for an expression involving sums of products. I am stuck trying to simplify or find a closed form formula for the following expression:
$$\sum_{i=0}^{n-1} (\prod_{y=i+1}^{n-1}\frac{y+1}{y+2} \div \prod_{x=i+1}^{k-2}\frac{x+1}{x+2})$$
The function is dependent on 2 variables $n$, and $k$. $k$ can range from 1 to n. What I am essentially trying to do is find the product of $\frac{i+2}{i+3} \cdot \frac{i+3}{i+4} \cdot... \cdot\frac{n}{n+1}$. However when the variable $k$ is larger than $i+2$, remove the product of the terms $\frac{i+2}{i+3} \cdot \frac{i+3}{i+4} \cdot... \cdot\frac{k-1}{k}$ by dividing them out of the original $\frac{i+2}{i+3} \cdot \frac{i+3}{i+4} \cdot... \cdot\frac{n}{n+1}$.
If anyone has any advice or a more efficient way to approach my desired goal of removing the i+2 to k-2 product values when k is larger than i+2 it would be much appreciated. Thank you.
| When looking at
\begin{align*}
\sum_{i=0}^{n-1}\left (\prod_{y=i+1}^{n-1}\frac{y+1}{y+2} \div \prod_{x=i+1}^{k-2}\frac{x+1}{x+2}\right)\qquad 1\leq k\leq n\tag{1}
\end{align*}
we observe the right-hand product $\prod_{x=i+1}^{\color{blue}{k-2}}\frac{x+1}{x+2}$ is empty resulting in $1$ if $k\in\{1,2\}$. We consider therefore two cases.
Case I: $k\in\{1,2\}$
We obtain from (1)
\begin{align*}
\color{blue}{\sum_{i=0}^{n-1}\prod_{y=i+1}^{n-1}\frac{y+1}{y+2}}
&=\sum_{i=0}^{n-1}\frac{\prod_{y=i+1}^{n-1}(y+1)}{\prod_{y=i+1}^{n-1}(y+2)}\tag{2}\\
&=\sum_{i=0}^{n-1}\frac{\prod_{y=i+1}^{n-1}(y+1)}{\prod_{y=i+2}^{n}(y+1)}\tag{3}\\
&=\sum_{i=0}^{n-1}\frac{i+2}{n+1}\tag{4}\\
&=\frac{1}{n+1}\left(\frac{1}{2}(n-1)n+2n\right)\tag{5}\\
&\,\,\color{blue}{=\frac{n(n+3)}{2(n+1)}}
\end{align*}
Comment:
*
*In (2) we rearrange the expression by writing products for numerator and denominator.
*In (3) we shift the index of the product of the denominator to prepare for cancellation.
*In (4) we do the cancellation.
*in (5) we factor out $\frac{k}{n+1}$ and apply finite summation formulas.
Case II: $3\leq k\leq n$
We obtain:
\begin{align*}
\color{blue}{\sum_{i=0}^{n-1}}&\color{blue}{\left(\prod_{y=i+1}^{n-1}\frac{y+1}{y+2}\div\prod_{x=i+1}^{k-2}\frac{x+1}{x+2}\right)}\\
&=\sum_{i=0}^{n-1}\prod_{y=i+1}^{n-1}\frac{y+1}{y+2}\cdot \prod_{x=i+1}^{k-2}\frac{x+2}{x+1}\tag{6}\\
&=\sum_{i=0}^{k-3}\prod_{y=i+1}^{n-1}\frac{y+1}{y+2}\cdot \prod_{x=i+1}^{k-2}\frac{x+2}{x+1}
+\sum_{i=k-2}^{n-1}\prod_{y=i+1}^{n-1}\frac{y+1}{y+2}\tag{7}\\
&=\sum_{i=0}^{k-3}\frac{\prod_{y=i+1}^{n-1}(y+1)}{\prod_{y=i+1}^{n-1}(y+2)}
\cdot \frac{\prod_{x=i+1}^{k-2}(x+2)}{\prod_{x=i+1}^{k-2}(x+1)}
+\sum_{i=0}^{k-3}\frac{\prod_{y=i+1}^{n-1}(y+1)}{\prod_{y=i+1}^{n-1}(y+2)}\tag{8}\\
&=\sum_{i=0}^{k-3}\frac{\prod_{y=i+1}^{n-1}(y+1)}{\prod_{y=i+2}^{n}(y+1)}
\cdot \frac{\prod_{x=i+2}^{k-1}(x+1)}{\prod_{x=i+1}^{k-2}(x+1)}
+\sum_{i=0}^{k-3}\frac{\prod_{y=i+1}^{n-1}(y+1)}{\prod_{y=i+2}^{n}(y+1)}\tag{9}\\
&=\sum_{i=0}^{k-3}\frac{i+2}{n+1} \cdot \frac{k}{i+2}+\sum_{i=k-2}^{n-1}\frac{i+2}{n+1} \tag{10}\\
&=\frac{1}{n+1}\left(k\sum_{i=0}^{n-1} 1+\sum_{i=k-2}^{n-1}(i+2)\right)\\
&=\frac{1}{n+1}\left(kn+\frac{1}{2}(n-1)n+2n-\frac{1}{2}(k-3)(k-2)-2(k-2)\right)\tag{11}\\
&\,\,\color{blue}{=\frac{1}{2}n+1+\frac{k(k-3)}{2(n+1)}}
\end{align*}
Comment:
*
*In (6) we take the reciprocal of the right-hand product and use multiplication ($\cdot$) instead of division ($\div$).
*In (7) we split the sum respecting the empty product similarly to the first case.
*In (8) we rearrange the expression by writing products for numerator and denominator.
*In (9) we shift indices similarly to (3).
*In (10) we do the cancellation.
*In (11) we apply finite summation formulas similarly as in (5).
| {
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How to prove $f(x)=x^4$ is concave up by definition? I know $f(x)=x^4$ is concave up, by calculating its second derivative. However, how to prove that $f(x)=x^4$ is concave up by definition, say $f(\frac{x_1+x_2}{2})<(1/2)f(x_1)+(1/2)f(x_2)$ for all $x_1,~x_2$? I'd tried binomial theorem, but can't get anything.
| I think you mean that $f$ is a convex function.
Id est, we need to prove that:
$$\frac{a^4+b^4}{2}\geq\left(\frac{a+b}{2}\right)^4$$ or
$$8(a^4+b^4)\geq a^4+4a^3b+6a^2b^2+4ab^3+b^4$$ or
$$7a^4-4a^3b-6a^2b^2-4ab^3+7b^4\geq0$$ or
$$7a^4-14a^3b+7a^2b^2+10a^3b-20a^2b^2+10ab^3+7a^2b^2-14ab^3+7b^4\geq0$$ or
$$(a-b)^2(7a^2+10ab+7b^2)\geq0$$ or
$$(a-b)^2(2a^2+2b^2+5(a+b)^2)\geq0,$$ which is obvious.
| {
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$\frac37x + \frac{3\times6}{7\times10}x^2 + \frac{3\times6\times9}{7\times10\times13}x^3+ \cdots$ Convergence of the series for $x>0$:
$$\frac37x + \frac{3\times6}{7\times10}x^2 + \frac{3\times6\times9}{7\times10\times13}x^3+ \cdots$$
The general tem is $a_n = \frac{3\times6\times \cdots \times(3+(n-1)3)}{7\times10 \times\cdots \times(7+(n-1)3)} x^n$.
I am thinking of using Raabe's Test. but cannot proceed.
We have to find $$\lim _{n \to \infty} n\left(\frac{a_n}{a_{n+1} } -1\right) = \lim _{n \to \infty} n \left(\frac{7+3n}{3+3n} \frac1x -1\right)$$
so the limit depends upon $x$.
| $$a_n = \frac{3\times6\times \cdots \times(3+(n-1)3)}{7\times10 \times\cdots \times(7+(n-1)3)} x^n$$
Then $$\frac{a_{n+1}}{a_n} =\frac{3n+3}{3n+7} x = \frac{3+\frac3n}{3 + \frac7n}x$$
From Ratio test we can conclude that when $|x| <1$ the series is convergent and when $|x|>1$ the series is divergent. Ratio Test can't conclude anything for $x = 1$. Here we use Raabe's Test.
$$\lim _{n \to \infty} n\left(\frac{a_n}{a_{n+1} } -1\right) = \lim _{n \to \infty} n \left(\frac{7+3n}{3+3n} -1\right) = \lim _{n \to \infty} \frac{4n}{3+3n} = \frac43 >1$$ and hence the series is convergent.
| {
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Coordinate vector in a polynomial basis Let $\left\{x+1,\:x^2+x,\:2x-1\right\}$ be a basis of $P2$. Find the coordinate vector of $p(x)$ $=$ $x^2$ $+$ $1$ relative to this basis.
So I set up the matrix as follows:
$\begin{pmatrix}1&0&-1&1\\ 1&1&2&0\\ 0&1&0&1\end{pmatrix}$
Which row reduced to:
$\begin{pmatrix}1&0&0&\frac{1}{3}\\ 0&1&0&1\\ 0&0&1&-\frac{2}{3}\end{pmatrix}$
This led me to think that the coordinate vector was:
$\begin{pmatrix}\frac{1}{3}\\ 1\\ -\frac{2}{3}\end{pmatrix}$
But I am not entirely sure if this vector should be written in terms of numbers or polynomials, for example:
$\begin{pmatrix}\frac{1}{3}\\ x\\ -\frac{2}{3}x^2\end{pmatrix}$
I think my first vector with numbers is correct but I just want to verify.
Thank you!
| The coordinate vector contains the coefficients of the basis vectors. So in this case, the vector
$$ \begin{pmatrix} \frac{1}{3} \\ 1 \\ \frac{1}{3} \end{pmatrix} $$
stands for the vector
$$ \frac{1}{3}\big( x + 1 \big) + 1 \big(x^2 + x\big) + \frac{1}{3}\big(2x - 1\big)$$
(As an aside, you can use this fact to check your answer :)
| {
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Computing the value of $\frac{1}{3^2+1} + \frac{1}{4^2+2} + \frac{1}{5^2+3}\ldots=$? I have tried converting this series into a telescopic sum whose terms could cancel out but haven't succeeded in that effort. How should I proceed further?
| Hint: $$\frac{1}{(n+1)(n+4)}=\frac{1}{3} \Bigg[\frac{1}{n+1}-\frac{1}{n+4}\Bigg]$$
By Cancelling terms we have remaining $$ \frac{1}{3} \Bigg[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\Bigg]= \frac{1}{3} \cdot\frac{13}{12}=\frac{13}{36}$$
| {
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What does calculating the inverse of a matrix mean? Assume I have 3 equations, $x+2y+z=2, 3x+8y+z=12, 4y+z=2$ which could be represented in matrix form ($Ax = b$) like this:
$\begin{pmatrix}
1 & 2 & 1\\
3 & 8 & 1\\
0 & 4 & 1
\end{pmatrix}\bigl .\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} = \begin{pmatrix}
2\\
12\\
2
\end{pmatrix}$
Then, the inverse of $A$, $A^{-1}$, would be: $\begin{pmatrix}
2/5 & 1/5 & -3/5\\
-3/10 & 1/10 & 1/5\\
6/5 & -2/5 & 1/5
\end{pmatrix} $.
So, my question is, what does this even mean? We know that $A$ is a coefficients matrix that represents the 3 equations above, so what does $A^{-1}$ mean with respect to these 3 equations? What have I done to the 3 equations is exactly my question.
Please note that I understand very well how to find the inverse of a matrix, I just don't understand the intuition of what's happening and sort of the meaning of the manipulations I am applying to the equations when they are in matrix form.
| Let's look at your original three equations.
$$x+2y+z=2$$
$$3x+8y+z=12$$
$$4y+z=2$$
Now let's multiply by $\frac{2}{5}$, $\frac{1}{5}$, and $\frac{-3}{5}$ respectively. We get
$$\frac{2x}{5}+\frac{4y}{5}+\frac{2z}{5}=\frac{4}{5}$$
$$\frac{3x}{5}+\frac{8y}{5}+\frac{z}{5}=\frac{12}{5}$$
$$\frac{-12y}{5}+\frac{-3z}{5}=\frac{-6}{5}$$
Now add the three equations together. We get
$$x + 0y + 0z = 2$$
or
$$x=2$$
Now multiply the same three equations by $\frac{-3}{10}$, $\frac{1}{10}$, and $\frac{1}{5}$ respectively. We get
$$\frac{-3x}{10}+\frac{-6y}{10}+\frac{-3z}{10}=\frac{-6}{10}$$
$$\frac{3x}{10}+\frac{8y}{10}+\frac{z}{10}=\frac{12}{10}$$
$$\frac{4y}{5}+\frac{z}{5}=\frac{2}{5}$$
Summing
$$0x + y + 0z = \frac{10}{10}$$
or
$$y = 1$$
Now multiply the same three equations by $\frac{6}{5}$, $\frac{-2}{5}$, and $\frac{1}{5}$ respectively. We get
$$\frac{6x}{5}+\frac{12y}{5}+\frac{6z}{5}=\frac{12}{5}$$
$$\frac{-6x}{5}+\frac{-16y}{5}+\frac{-2z}{5}=\frac{-24}{5}$$
$$\frac{4y}{5}+\frac{z}{5}=\frac{2}{5}$$
Summing
$$0x + 0y + z = \frac{-10}{5}$$
or
$$z = -2$$
And if you look at the numbers by which we multiplied, they are from
$$\begin{pmatrix}
2/5 & 1/5 & -3/5\\
-3/10 & 1/10 & 1/5\\
6/5 & -2/5 & 1/5
\end{pmatrix} = A^{-1}$$
We essentially did the matrix multiplication of $A \cdot A^{-1}$ to get $I$ manually when we could have just done
$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} = \begin{pmatrix}
2/5 & 1/5 & -3/5\\
-3/10 & 1/10 & 1/5\\
6/5 & -2/5 & 1/5
\end{pmatrix}\begin{pmatrix}
2\\
12\\
2
\end{pmatrix}=\begin{pmatrix}
2\\
1\\
-2
\end{pmatrix}$$
and gotten the same answer. $A^{-1}$ is essentially the numbers by which we multiply the equations so we can add them together and get the solutions. Solving for the inverse is determining those numbers. Of course, if you're working with the equations, it would be easier to substitute in than to come up with all nine numbers. The convenient thing here is that we don't have to multiply $A\cdot A^{-1}$, as we already know the result. We can just do the right side multiplication.
| {
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Trigonometry problem $\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$
What is the value of:
$$\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$$
I've done trigonometry in my earlier years of high school but I forgot a lot of rules. This is where I'm stuck on this problem:
$\large{{\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ}=\\{\sin(90^\circ+10^\circ)+\cos(60^\circ+10^\circ)\over\cos(90^\circ-10^\circ)-\cos(30^\circ+10^\circ)}=\\{\sin90^\circ\cos10^\circ+\cos90^\circ\sin10^\circ+\cos60^\circ\cos10^\circ-\sin60^\circ\sin10^\circ\over\cos90^\circ\cos10^\circ+\sin90^\circ\sin10^\circ-\cos30^\circ\cos10^\circ+\sin30^\circ\sin10^\circ}=\\{\cos10^\circ+{1\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over\sin10^\circ-{\sqrt3\over2}\cos10^\circ+{1\over2}\sin10^\circ}=\\{{3\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over{3\over2}\sin10^\circ-{\sqrt3\over2}\cos10^\circ}}$
Not sure what I should do further with this.
| Since $\cos x=\sin(90-x)$, $$\frac{\sin100+\cos70}{\cos80-\cos20}=\frac{2\sin\frac{100+20}2\cos\frac{100-20}2}{2\sin\frac{80+20}2\sin\frac{20-80}2}=-\frac{\sin60\cos40}{\sin50\sin30}=-\frac{\frac{\sqrt3}2\sin50}{\frac12\sin50}=-\sqrt3$$ using $$\sin x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2\\\cos x-\cos y=-2\sin\frac{x+y}2\sin\frac{x-y}2$$
| {
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Prove that $3\cdot 5^{2n+1} +2^{3n+1}$ is divisible by $17$ for all $n ∈ \mathbb{N}$
Use mathematical induction to prove that $3\cdot 5^{2n+1} +2^{3n+1}$
is divisible by $17$ for all $n ∈ \mathbb{N}$.
I've tried to do it as follow.
If $n = 1$ then $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3\cdot 5^{2p+1} +2^{3p+1} = 17k $ where $k ∈ \mathbb{N} $. Consider now $n=p+1$. Then
\begin{align}
&3\cdot 5^{2(p+1)+1} +2^{3(p+1)+1}=\\
&3\cdot 5^{2p+1+2} + 2^{3p+1+3}=\\
&3\cdot5^{2p+1}\cdot 5^{2} + 2^{3p+1}\cdot 2^{3}.
\end{align}
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
| Step $n+1:$
$3(5^2)5^{2p+1}+(2^3)2^{3p+1}=$
$3(17+8)5^{2p+1} +8 \cdot 2^{3p+1}=$
$8[3 \cdot 5^{2p+1}+2^{3p+1}]+ 17\cdot 3 \cdot 5^{2p+1}.$
The first term in the above sum is divisible by $17$ (hypothesis), the second term is a multiple of $17$.
| {
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For any primitive pythagorean triple $(a,b,c)$ either $a$ or $b$ must be a multiple of $3$ I'm reading "Friendly Introduction to Number Theory". Now I'm working on Primitive Pythagorean Triples Exercises 2.1 (a) on P18.
We showed that in any primitive Pythagorean triple $(a, b, c)$, either $a$ or $b$ is even. Use the same sort of argument to show that either $a$ or $b$ must be a multiple of 3.
(1) $a^2 + b^2 = c^2$ with a odd, b even, a,b,c having no common factors
(2) $a^2 = c^2 - b^2 = (c-b)(c+b)$
(3) $c + b = s^2$ and $c - b = t^2$
(4) $c = \frac{(s^2 + t^2)}{2}$ and $b = \frac{(s^2 - t^2)}{2}$
(5) $a = \sqrt{(c-b)(c+b)} = st$
(6) $a = st$, $b = \frac{(s^2 - t^2)}{2}$, $c = \frac{(s^2 + t^2)}{2}$
https://www.math.brown.edu/~jhs/frintch1ch6.pdf
I have no idea how I start doing this. Can you give me a hint? I think I need to show that both the following (1) and (2) are satisfied.
$X \neq 0$
(1) $a\equiv 0\pmod 3$ and $b\equiv X\pmod 3$
(2) $b\equiv 0\pmod 3$ and $a\equiv X\pmod 3$
| Using https://en.m.wikipedia.org/wiki/Pythagorean_triple or https://mathcs.clarku.edu/~djoyce/java/elements/bookX/propX29.html
$a=2mn,b=m^2-n^2$ where $m,n$ are coprime integers and not both are odd
$ab=2mn(m^2-n^2)=2(m^3-m)n-2m(n^3-n)$
Now use
The product of $n$ consecutive integers is divisible by $n$ factorial as $m^3-m=(m-1)m(m+1)$
| {
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Finding eigenpairs of matrix A How do you find the eigenpairs of this?
Given two matrices $A=\begin{bmatrix} a& b\\b&-a\end{bmatrix}$ and $B=\begin{bmatrix} a& -b\\b&a\end{bmatrix}$ where $b \neq 0$, find all eigenpairs of $A$ and $B$, and comment if the matrices are diagonalizable or not.
I plugged in the characteristic equation $\det(A-\lambda I)=0$ and is unable to solve for $\lambda $. I got $\lambda ^2-a^2-b^2=0$ and it won't let me factorize. Any idea?
| The characteristic polynomial of
\begin{bmatrix} a & b \\ b & -a \end{bmatrix}
is indeed $\lambda^2-a^2-b^2$. The eigenvalues are $\sqrt{a^2+b^2}$ and $-\sqrt{a^2+b^2}$.
Since
$$A-\sqrt{a^2+b^2}\,I=
\begin{bmatrix}
a-\sqrt{a^2+b^2} & b \\
b & -a-\sqrt{a^2+b^2}
\end{bmatrix}
$$
has rank one, an eigenvector $\begin{bmatrix} x \\ y\end{bmatrix}$ satisfies
$$
\begin{bmatrix} a-\sqrt{a^2+b^2} & b \end{bmatrix}
\begin{bmatrix} x \\ y\end{bmatrix} = 0
$$
so we can take
$$
\begin{bmatrix} x \\ y\end{bmatrix}=
\begin{bmatrix} b \\ \sqrt{a^2+b^2}-a\end{bmatrix}
$$
and this is nonzero, so it is a basis for the eigenspace. The other eigenvectors are the non zero scalar multiples of this one.
Similarly for the other eigenvalue.
For the matrix
\begin{bmatrix} a & -b \\ b & a \end{bmatrix}
the characteristic polynomial is $\lambda^2-2a\lambda+(a^2+b^2)$, whose roots are
$a+ib$ and $a-ib$.
Assuming $a$ and $b$ real, the first matrix is diagonalizable, because it has distinct real eigenvalues. The second matrix is diagonalizable over the complex numbers, but not over the reals.
| {
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Evaluate lim $a_n$, $a_n=\frac{3^n}{n!}$. I've come here as I need some help.
I have to find $\lim_{x\to \infty}a_{n}$,
where $a_{n}=\frac{3^n}{n!}$
I've tried to use Squeeze theorem ,but I couldn't find another two strings that are lower and bigger than $a_{n}$ and have the same limit.
I can consider $a_{n}\leq3^n,$ which has $\lim_{n\to \infty}3^n=\infty$,but maybe there is another proof for this exercise.
I've thought that $$\frac{3^n}{n!}=\frac{3\cdot3\cdot3\cdot3\cdots3}{1\cdot2\cdot3\cdots n}$$ but again,I do not know how to handle it.
Please give me some advice.
| Hint. Note that for $n\geq 3$,
$$\frac{3^n}{n!}=\frac{\overbrace{3\cdot 3\cdot 3\cdot 3\cdots 3}^{n}}{1\cdot 2\cdot 3 \cdot 4\cdots n}=
\frac{3\cdot 3}{1\cdot 2}\cdot \overbrace{\frac{3}{3}\cdot\frac{3}{4} \cdots \frac{3}{n-1}}^{n-3}\cdot \frac {3}{n}\leq \frac{3\cdot 3}{1\cdot 2 }\cdot \frac {3}{n}.$$
In a similar way, for any $a>0$, for $n\geq a$,
$$\frac{a^n}{n!}\leq \frac{a^{\lfloor a\rfloor+1}}{\lfloor a\rfloor!}\cdot \frac{1}{n}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim\limits_{x \to +\infty}\left[\sqrt[4]{x^4+x^3+x^2+x+1}-\sqrt[3]{x^3+x^2+x+1}\cdot \frac{\ln(x+e^x)}{x}\right].$ Problem
Evaluate
$$\lim_{x \to +\infty}\left[\sqrt[4]{x^4+x^3+x^2+x+1}-\sqrt[3]{x^3+x^2+x+1}\cdot \frac{\ln(x+e^x)}{x}\right].$$
Solution
Substitute $x$ for $\dfrac{1}{t}$，namely，$x=\dfrac{1}{t}$. Then
\begin{align*}
&\lim_{x \to +\infty}\left[\sqrt[4]{x^4+x^3+x^2+x+1}-\sqrt[3]{x^3+x^2+x+1}\cdot \frac{\ln(x+e^x)}{x}\right]\\
=&\lim_{t \to 0+}\left[\frac{\sqrt[4]{1+t+t^2+t^3+t^4}}{t}-\frac{\sqrt[3]{1+t+t^2+t^3}}{t}\cdot t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)\right].
\end{align*}
On one hand,according to Taylor's formula, we may have
$$\sqrt[4]{1+t+t^2+t^3+t^4}=1+\frac{1}{4}t+o(t);$$
$$\sqrt[3]{1+t+t^2+t^3}=1+\frac{1}{3}t+o(t).$$
On the other hand,notice $$\lim_{t \to 0+} t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)=1,$$
then we may denote
$$t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)=1+\alpha(t),$$
where $\alpha(t) \to 0(t \to 0+)$.
Futher, we have
$$\lim_{t \to 0+}\frac{\alpha(t)}{t}=\lim_{t \to 0+}\left[\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)-\frac{1}{t}\right]=0,$$
thus, we may denote
$$t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)=1+\alpha(t)=1+o(t).$$
Therefore
\begin{align*}
&\lim_{t \to 0+}\left[\frac{\sqrt[4]{1+t+t^2+t^3+t^4}}{t}-\frac{\sqrt[3]{1+t+t^2+t^3}}{t}\cdot t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)\right]\\
=&\lim_{t \to 0+}\left[\frac{1+\frac{1}{4}t+o(t)}{t}-\frac{1+\frac{1}{3}t+o(t)}{t}\cdot (1+o(t))\right]\\
=&\lim_{t \to 0+}\left[-\frac{1}{12}+\frac{o(t)}{t}+\frac{o(t)}{3}+\frac{o(t^2)}{t}\right]\\
=&-\frac{1}{12}.
\end{align*}
Please correct me if I am wrong. Hope to see other solutions.
| Your approach is fine. But you don't need to deal with higher order terms, just handling till $o(t) $ is sufficient.
Here is another solution. If the expression under limit is $a-bc$ then we can write it as $a-b+b(1-c)$. As you show $a-b\to-1/12$ and further $$1-c=-\frac{\log(1+xe^{-x})}{x}$$ and $b/x\to 1$ so that limit of $b(1-c) $ is same as that of $x(1-c)$ and clearly it is $0$. Thus the answer is $-1/12$.
The limit of $a-b$ can be handled by using your substitution $x=1/t$ and then adding subtracting $1/t$ to get $(a-1/t)-(b-1/t)$. Using the standard limit $$\lim_{z\to k} \frac{z^n-k^n} {z-k} =nk^{n-1}$$ we can see that $a-1/t\to 1/4$ and $b-1/t\to 1/3$.
| {
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Where is the mistake in finding $c$ and $n$ of $\sqrt{x^3+4x} - \sqrt{x^3+x} \sim cx^n$ for $x \to 0$ and $x \to +\infty$ First we simplify in the following way:
$$\frac{3}{\sqrt{x}\left( (1+\frac{4}{x^2})^{1/2}+(1+\frac{1}{x^2})^{1/2} \right)}$$
For $x \to 0$:
We then postulate that since the above must be equivalent to some constant times $x$ to some power, then the inverted fraction is equivalent to some $\frac{1}{cx^n}$, I actually think that this step is flawed, because for $x \to 0$, $\frac{1}{cx^n}$ is indeterminate for $x \neq 0$. We then have
$$\lim_{x \to 0} \frac{\sqrt{x}}{3}\left(\left(1+\frac{4}{x^2}\right)^{1/2}-1 \right) + \lim_{x \to 0} \left( \left(1+\frac{1}{x^2} \right)^{1/2}-1 \right) = \\ \frac{\sqrt{x}}{6} \left( \frac{4}{x^2} + \frac{1}{x^2} \right)$$
And so $c=6/5$ and $n=1.5$, because we have to invert back.
For $x \to +\infty$:
We invert again, but this time also divide by $cx^n$. So that we end up with:
$$\lim_{x \to +\infty} \frac{c}{3} x^n \sqrt{x} \left(\left( 1+\frac{4}{x^2} \right)^{1/2} + \left( 1+\frac{1}{x^2}\right)^{1/2} \right) = \\
\frac{2c}{3}(x^n \sqrt{x})$$
that means that $c=2/3$ and $n=0.5$.
But one or both of these solutions are, in fact, incorrect.
| For $x\to0$, $$\sqrt{x^3+4x} - \sqrt{x^3+x}=\frac{3x}{ \sqrt{x^3+4x} +\sqrt{x^3+x}}=x^{1/2}\frac3{ \sqrt{x^2+4} +\sqrt{x^2+1}}\\\to x^{1/2}$$
because the denominator of the fraction tends to $3$ (the $x^2$ become negligible).
And for $x\to\infty$, $$\sqrt{x^3+4x} - \sqrt{x^3+x}=\frac{3x}{ \sqrt{x^3+4x} +\sqrt{x^3+x}}=\frac32x^{-1/2}\frac{2x^{3/2}}{ \sqrt{x^3+4} +\sqrt{x^3+1}}\\\to \frac32x^{-1/2}$$
because the denominator of the fraction tends to $2x^{3/2}$ (the additive constant become negligible; you can as well consider $y:=\dfrac1x\to0^+$).
| {
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How to prove $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$ So, I was watching this video by blackpenredpen where he mentions that $$\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$$ so I wanted to try and prove it myself.
Let $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=x$
But$\sqrt {a-\sqrt {a+\underbrace{\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}_x}}=x$
$\therefore \sqrt {a-\sqrt {a+x}}=x$
$a-\sqrt {a+x}=x^2$
$x^2-a=-\sqrt {a+x}$
$x^4-2ax^2+a^2=a+x$
$x^4-2ax^2-x+a^2-a=0$
Note that this is of the form $y^4+py^2+qy+r=0$ so we can use Ferrari-Cardano.
We need to find a $z$ such that $(2z-p)y^2-qy+(z^2-r)$ has a discriminant of $0$. The discriminant of $(2z-p)y^2-qy+(z^2-r)$ is equal to $q^2 - 4(2z - p)(z^2 - r),$ which simplifies to $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$
Substituting values from $x^4-2ax^2-x+a^2-a=0$ into $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$ gives us
$8z^3-4\cdot(-2a)\cdot z^2-8\cdot(-a)\cdot z+\left(4\cdot (-2a) \cdot (-a) - (-1)^2 \right)=0 \implies 8z^3+8az^2+8az+(8a^2-1)=0$
Using Cardano's formula, or in my case Wolfram Alpha, we get that $$z_1 = \frac {\sqrt [3]{-16 a^3 - 144 a^2 + 3 \sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}}{6\sqrt[3]2} - \frac {192 a - 64 a^2}{48\cdot 2^{\frac 23} \sqrt [3]{-16 a^3 - 144 a^2 + 3\sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}} - \frac a3$$
I simply can not solve that quintic and continue as it is already too cluttered. Was there a mistake in my problem or is there any other way to do it? Also, I am sincerely sorry but I am not sure how to tag this question.
Edit $1:$
After Ross Millikan's answer, I snooped around in the comment section of the video and found someone who found that it is true using alternating root series. Was his proof correct as $\frac {\sqrt {4a-3}-1}2$ does not seem to have real values for $a \lt \frac 34$? Thank you!
| Here's an easier solution that doesn't require you going out to a fourth degree polynomial. Let $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=x$, $-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\ldots}}}}}}=y$.
Then $x^2 - y = a$, and $y^2-x = a$. Subtracting these two equations, we have that $x^2 - y^2 + x-y =0 \implies (x-y)(x+y+1) = 0$. Since $x \neq y$ unless both are equal to zero, $x+y = -1$. Adding these equations, we have $x^2 +y^2 - (x+y) = 2a$, and substituting everything in, we have $x^2 + x = a-1$, which we can evaluate via the quadratic formula to $x=\frac{-1+\sqrt{4a-3}}{2}$, where we have ignored the other solution since it is always negative.
| {
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How to prove $\frac{1}{2n+2}<\int_0^{\frac{\pi}{4}}\tan^nx\,dx<\frac{1}{2n}$ How to prove $$\frac{1}{2n+2}<\int_0^{\frac{\pi}{4}}\tan^nx\,dx< \frac{1}{2n}$$
Set $A_n=\int_0^{\frac{\pi}{4}}\tan^nx\,dx$, then we have $A_n+A_{n+2}=\frac{1}{n+1}$ and we have $A_{n+2} < A_n$ ,so we can get $$\frac{1}{2n+2}< \int_0^{\frac{\pi}{4}}\tan^nx\,dx < \frac{1}{2n-2}$$
But how to show that$$\int_0^{\frac{\pi}{4}}\tan^nx\,dx < \frac{1}{2n}$$
| Change variable to $t = \tan x$, we have
$$I_n \stackrel{def}{=}\int_0^{\pi/4} \tan^n x dx = \int_0^1 \frac{t^n}{1+t^2} dt$$
Notice for $t \in (0,1)$, we have $\frac{1 + t^2}{2} < 1$. This implies
$$I_n > \int_0^1 \frac{t^n}{1+t^2}\cdot\frac{1+t^2}{2} dt = \frac12\int_0^1 t^n dt = \frac{1}{2(n+1)}$$
On the other direction, AM $\ge GM$ tell us $t = \sqrt{1 \cdot t^2} \le \frac{1+t^2}{2}$ and the inequality is strict when $t \ne 1$. This leads to
$$I_n = \int_0^1 \frac{t^{n-1}}{1+t^2} t dt < \int_0^1 \frac{t^{n-1}}{1+t^2}\cdot \frac{1+t^2}{2} dt = \frac12 \int_0^1 t^{n-1} dt = \frac{1}{2n}$$
| {
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Find a complex number with a real part of $24$
Let $a_1=5-3i$ and let $a_2=2-8i$ be a geometric series.
a. Show $\arg(a_{n+8})=\arg(a_n)$
b. let there be an element with a real part of $24$ find its imaginary part and index
c. how much elements do we need to get to a sum of $-515-133i$
a. $$q=\frac{a_2}{a_1}=\frac{a_1\cdot q}{a_1}=1-i$$
now $a_n=a_1\cdot q^n$ and $a_{n+8}=a_1\cdot q^{n+8}=a_1\cdot q^n\cdot q^8$
but $q^8=16$ so $\arctan(\frac{y_n}{x_n})=\arctan(\frac{16y_n}{16x_n})$
b. how can I find both $n$ and $y$ in $$(5-3i)(1-i)^n=24+yi$$
| $$1-i=\sqrt2\left(\cos\dfrac\pi4-i\sin\dfrac\pi4\right)$$
$\implies$ $$(1-i)^n=2^{n/2}\left(\cos\dfrac{n\pi}4-i\sin\dfrac{n\pi}4\right)$$
Real part of $(5-3i)(1-i)^n=2^{n/2}(5-3i)\left(\cos\dfrac{n\pi}4-i\sin\dfrac{n\pi}4\right)$
will be $$2^{n/2}\left(5\cos\dfrac{n\pi}4-3\sin\dfrac{n\pi}4\right)$$
We need to eliminate $5$ as $\cos\dfrac{n\pi}4=\pm\dfrac1{\sqrt2},0,\pm1\left(\implies\cos\dfrac{n\pi}4=0\implies n=2(2m+1)\right)$ and $n$ must be even to make the product $24$
| {
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Solve for the coefficient of an even generating function Using a generating function, find the number of ways to select 10 candies from a huge pile of red, blue, and green lollipops if the selection has an even number of blue lollipops.
I started, but I don't understand how to continue
$(1 + x + x^2 + x^3 ...)^2 \cdot (1 + x^2 + x^4 + x^6 ...)$
$=(\frac{1}{1-x})^2 \cdot \frac{1}{1-x^2}$
$=\big(1 + \binom{1 + 2 - 1}{1}x + \binom{2 + 2 - 1}{2}x^2 ...\big) \cdot ???$
How do you find coefficient to $x^{10}$?
Also, our textbook tells us $(1 + x^2 + x^4 + x^6 ...)$ = $\frac{1}{1-x^2}$ how?
| The original problem is $(1 + x + x^2 + x^3 ...)^2 \cdot (1 + x^2 + x^4 + x^6 ...)$
$\frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$
Which can be shown by,
$1 = (1 + x + x^2 + x^3 ...)(1-x)$
$1 = (1 + x + x^2 + x^3 ...) - x \cdot (1 + x + x^2 + x^3 ...)$
$1 =1$
To show, $\frac{1}{1-x^2} = (1 + x^2 + x^4 + x^6 ...)$
we substitute $y = x^2$
$\frac{1}{1-y} = (1 + y + y^2 + y^3 ...)$ which is equivalent due to $\frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$ shown above.
To solve the equation,
$=(\frac{1}{1-x})^2 \cdot \frac{1}{1-x^2}$
$=\frac{1}{(1-x)^2} \cdot \frac{1}{1-x^2}$
$=\big(1 + \binom{1 + 2 - 1}{1}x + \binom{2 + 2 - 1}{2}x^2 ...\big) \cdot (1 + x^2 + x^4 + x^6 ...)$
Thus the coefficient of $x^{10}$ is the coefficient of
$=1 \cdot x^{10} + \binom{3}{1}x^2 \cdot x^{8} + \binom{5}{1}x^4 \cdot x^{6} + \binom{7}{1}x^6 \cdot x^{4} + \binom{9}{1}x^8 \cdot x^{2} + \binom{11}{1}x^{10} \cdot 1$
| {
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Can we proof that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?; given $0 < a < b < 1$ I have found in some test problem:
Given $0 < a < b < 1$, can we conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?
I divide the combination of a and b into a few cases. Then, see what happens.
case 1 : $a \rightarrow 0$ and $b \rightarrow 0$
When $a$ approaches to zero and $b$ approaches to $a$ (which is zero), the expression will be evaluated as $\sqrt{0 + 0}$ and $\sqrt{0} + \sqrt{0}$. Since $0 = 0$, this case cannot conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 2 : $a \rightarrow 1$ and $b \rightarrow 1$
When $b$ approaches to one and $a$ approches to $b$ (which is one), the expression will be evaluated as $\sqrt{1 + 1}$ and $\sqrt{1} + \sqrt{1}$. Since
$\sqrt{2} < 2$, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 3 : $a \rightarrow 0$ and $b \rightarrow 1$
When $a$ approaches to zero and $b$ approches to one, the expression will be evaluated as $\sqrt{0 + 1}$ and $\sqrt{0} + \sqrt{1}$. Since
$1 = 1$, then this case cannot concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 4 : $a \rightarrow b$
When $a$ approaches to $b$ the expression will be evaluated as $\sqrt{b + b}$ and $\sqrt{b} + \sqrt{b}$. Since $\sqrt{2b} < 2\sqrt{b} $, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
From above cases, am I still missing some point? If not, How should I write the conclusion from that pieces of thinking mathematically? Because I instinctively believe that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$ should be true (and not $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$).
| Since $f(x)=x^2$ is strictly increasing for $x\ge 0$ we have
$$A<B \iff A^2<B^2 \quad A,B\ge 0$$
and therefore
$$\sqrt{a + b} < \sqrt{a} + \sqrt{b}\iff (\sqrt{a + b})^2 < (\sqrt{a} + \sqrt{b})^2 \iff a+b <a+b+2\sqrt{ab}$$
that is
$$2\sqrt{ab}>0$$
which is true.
| {
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Prove inequality $(1+\frac{1}{2n})^n<2$ How can I prove inequality
$$
(1+\frac{1}{2n})^n<2
$$
for $n\in\mathbb{N}$ by using induction proof method?
I have tried to write that:
for $n=1$ it's true. So let's suppose that $(1+\frac{1}{2n})^n<2$. and now I don't know what do nect because I know that $(1+\frac{1}{2(n+1)})^{n+1} > (1+\frac{1}{2n})^n$ but I don't know how to show that $(1+\frac{1}{2(n+1)})^{n+1} < 2$.
In this task I cannot use the definition of $e$ number.
| It is enough to show that $\log(2)\geq \frac{1}{2}$, since
$$ 2^{1/n} = e^{\frac{\log 2}{n}} > 1+\frac{\log 2}{n} \stackrel{\color{red}{?}}{\geq} 1+\frac{1}{2n}.$$
is granted by the convexity of $e^x$. On the other hand $\log(2)>\frac{1}{2}$ is equivalent to $e<4$, and
$$ e = \sum_{k\geq 0}\frac{1}{k!}=\frac{8}{3}+\sum_{k\geq 4}\frac{1}{k!}<\frac{8}{3}+\frac{1}{6}\sum_{k\geq 4}\frac{1}{4^{k-3}}=\frac{49}{18}. $$
As an alternative,
$$ 0 < \int_{0}^{1}x^3(1-x^3)e^{-x}\,dx = \frac{1158}{e}-426 $$
directly gives $e<\frac{193}{71}$. Or, more elementary:
$$ \log(2)-\frac{1}{2}=\int_{0}^{1}\frac{dx}{1+x}-\int_{0}^{1}\frac{dx}{1+1}=\frac{1}{2}\int_{0}^{1}\frac{1-x}{1+x}\,dx > 0.$$
Yet another approach: for any $n\geq 1$, $2^{1/n}$ is the geometric mean of $\frac{n+1}{n},\frac{n+2}{n+1},\ldots,\frac{2n}{2n-1}$. By the AM-GM inequality it follows that
$$\begin{eqnarray*} 2^{1/n} < \frac{1}{n}\sum_{k=0}^{n-1}\left(1+\frac{1}{n+k}\right)&=&1+\frac{1}{n^2}+\frac{1}{n}\sum_{k=1}^{n-1}\frac{1}{k+n}\\&<&1+\frac{1}{n^2}+\frac{1}{n}\sum_{k=1}^{n-1}\frac{1}{\sqrt{k+n}\sqrt{k+n-1}}\end{eqnarray*} $$
and by the Cauchy-Schwarz inequality (see also page 8 of my notes) we have
$$ \sum_{k=1}^{n-1}\frac{1}{\sqrt{k+n}\sqrt{k+n-1}}\leq \sqrt{\sum_{k=1}^{n-1}1\sum_{k=1}^{n-1}\left(\frac{1}{k+n-1}-\frac{1}{k+n}\right)}=\frac{n-1}{\sqrt{n(2n-1)}},$$
so $2^{1/n}< 1+ \frac{1}{n\sqrt{2}}+\frac{1}{n^2}$ and actually $2^x< 1+\frac{x}{\sqrt{2}}+x^2$ for any $x$ in a pointed neighbourhood of the origin. By replacing $x$ with $-x$ and reciprocating we get
$$ 2^x > \frac{1}{1-\frac{x}{\sqrt{2}}+x^2} $$
and
$$ 2^{1/n} > \frac{1}{1-\frac{1}{n\sqrt{2}}+\frac{1}{n^2}} $$
still is stronger than needed.
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"answer_id": 3
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Solve recurring sequence using a generating function I have the sequence $a_n=3a_{n-1}-3a_{n-2}+a_{n-3}$, $\forall\ n \ge 3$, with $a_0=2$, $a_1=2$, $a_2=4$ being the known terms, and I want to find a non-recursive equation for $a_n$ using a generating function.
What I have done:
$$
\begin{align}
A(x)
& = \sum_{n\ge0}{a_nx^n}
= a_0+a_1x+a_2x^2+\sum_{n\ge 3}\left({3a_{n-1}-3a_{n-2}+a_{n-3}}\right)x^n\\
& = 2+2x+4x^2+3x\sum_{n\ge 3}{a_{n-1}x^{n-1}}
-3x^2\sum_{n\ge 3}{a_{n-2}x^{n-2}}
+x^3\sum_{n\ge 3}{a_{n-3}x^{n-3}}\\
& = 2+2x+4x^2+3xA(x)-3x^2A(x)+x^3A(x)\\
& = \frac{2+2x+4x^2}{1-3x+3x^2-x^3}\\
& = \frac{4x^2+2x+2}{(1-x)^3}
\end{align}
$$
As shown above, I have reached a solution for $A(x)$, but I'm not sure how to use it to find a solution for $a_n$. Any tips pointing me in the right direction would be greatly appreciated.
| Your calculation is correct up to
$$
\begin{align}
A(x)
& = \sum_{n\ge0}{a_nx^n}
= a_0+a_1x+a_2x^2+\sum_{n\ge 3}\left({3a_{n-1}-3a_{n-2}+a_{n-3}}\right)x^n\\
& = 2+2x+4x^2+3x\sum_{n\ge 3}{a_{n-1}x^{n-1}}
-3x^2\sum_{n\ge 3}{a_{n-2}x^{n-2}}
+x^3\sum_{n\ge 3}{a_{n-3}x^{n-3}}
\end{align}
$$
But then
$$
\begin{align}
\sum_{n\ge 3}{a_{n-1}x^{n-1}} &= \sum_{n\ge 2}{a_{n}x^{n}} = A(x) - 2 - 2x \\
\sum_{n\ge 3}{a_{n-2}x^{n-2}} &= \sum_{n\ge 1}{a_{n}x^{n}} = A(x) - 2
\end{align}
$$
so that
$$
A(x) = 2+2x+4x^2 + 3x(A(a) - 2-2x) -3x^2(A(x) - 2) + x^3A(x)
$$
which gives
$$
A(x) = \frac{2-4x + 4x^2}{(1-x)^3} \, .
$$
Differentiating the geometric series $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$
twice gives
$$
\frac{2}{(1-x)^3} = \sum_{n=2}^\infty n(n-1) x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2)x^n \, ,
$$
therefore
$$
\begin{align}
A(x) &= (1-2x + 2 x^2) \sum_{n=0}^\infty (n+1)(n+2)x^n \\
&= \sum_{n=0}^\infty (n+1)(n+2)x^n - 2\sum_{n=0}^\infty (n+1)(n+2)x^{n+1} +2\sum_{n=0}^\infty (n+1)(n+2)x^{n+2} \\
&= \sum_{n=0}^\infty (n+1)(n+2)x^n - 2\sum_{n=1}^\infty n(n+1)x^n +2\sum_{n=2}^\infty (n-1)nx^n \\
&= \sum_{n=0}^\infty (n+1)(n+2)x^n - 2\sum_{n=0}^\infty n(n+1)x^n +2\sum_{n=0}^\infty (n-1)nx^n \\
&= \sum_{n=0}^\infty \bigl( (n+1)(n+2) - 2n(n+1) +2(n-1)n\bigr) x^n \\
&= \sum_{n=0}^\infty (n^2 -n+2) x^n
\end{align}
$$
and $a_n = n^2-n+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solve $x^3-x-6=0$
Solve for x, $$x^3-x-6=0\tag1$$
let $x=y-2$
$$(y-2)^3-(y-2)-6=0\tag2$$
$$y^3-6y^2+11y-12=0\tag3$$
let $x=y+3$
$$(y+3)^3-(y+3)-6=0\tag4$$
$$y^3+9y^2+26y+18=0\tag5$$
$(5)-(4)$:
$$15y^2+15y+30=0\tag6$$
$$y^2+y+2=0\tag7$$
This approach, I have tried, it is not working.
How can I solve $(1)$?
| Guess the solution $x = 2$ by using the Rational Root Theorem. By synthetic division, we have
$$x^{3} - x - 6 = (x - 2)(x^{2} + 2x + 3)$$
Using the quadratic equation on the second equation, we obtain the solution set
$$\{2, -1 + i\sqrt{2}, -1 - i\sqrt{2}\} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluating Derivative of $\sqrt{1+\sqrt{1+\sqrt{1-2x}}}$ Did I evaluate the following derivative correctly? I know I have not simplified to the utmost extent, but I want to know if this method is correct.
Consider, $\sqrt{1+\sqrt{1+\sqrt{1-2x}}}$
Let A = $1+\sqrt{1+\sqrt{1-2x}}$
Let B = $1+\sqrt{1-2x}$
Let C = $1-2x$
$$\left(\sqrt{1+\sqrt{1+\sqrt{1-2x}}}\right)' = \frac{1}{2}A^{-\frac{1}{2}} \cdot \frac{1}{2}B^{-\frac{1}{2}} \cdot \frac{1}{2}C^{-\frac{1}{2}} \cdot -2 = \frac{-1}{4\sqrt{ABC}}$$
| Another way to see it is that with $f(x)=\sqrt{1+x}$, and $g(x)=-2x$, you are calculating the derivative of
$$
f\circ f\circ f\circ g(x)
$$
then you apply the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that the sequence $a_1=3$ and $a_{n+1}=\frac{3(1+a_n)}{3+a_n}$ converges to $\sqrt3$ First of all let´s see how it goes: $3,\frac{3(1+3)}{3+3}=\frac{12}{6}=2,\frac{3(1+2)}{3+2}=\frac{9}{5}=1.8,...$ We see that it's decreasing.
What i need to show first is that the sequence decreases. To do that, i´d say:
$$a_n-a_{n+1}>0\Rightarrow a_n-{\frac{3(1+a_n)}{3+a_n}}>0\Rightarrow a_n^2>3\Rightarrow a_n>\sqrt{3}$$ But i think this is like say that $a_n$ converges to $\sqrt3$, which is asked to be proved. So, How can i show this sequence decreases?. And to show it's bounded?
| We have $a_1>0$ and from the definition of $a_{n+1}$ we have $a_n>0\implies a_{n+1}>0 $ for any $n$. So by induction we have $a_n>0$ for all $n\in \Bbb N.$
We have $a_1>\sqrt 3$. If $a_n>\sqrt 3$ then, since $a_{n+1}>0,$ we have $$a_{n+1}>\sqrt 3\iff a_{n+1}^2>3\iff$$ $$\iff \frac {3^2(1+a_n)^2}{(3+a_n)^2}>3\iff$$ $$\iff 3^2(1+a_n)^2 -3(3+a_n)^2>0\iff$$ $$\iff 6(a_n^2-3)>0.$$ So by induction we have $a_n>\sqrt 3$ for all $n\in \Bbb N.$
Since $3+a_n>a_n>0$ we have $$a_n>a_{n+1} \iff (3+a_n)\left(a_n-\frac {3(1+a_n)}{3+a_n}\right)>0 \iff$$ $$\iff 3a_n+a_n^2> 3+3a_n\iff a_n^2>3,$$ and we have already shown that $a_n^2>3.$ So $a_n>a_{n+1}$ for all $n.$
Since $(a_n)_n$ is a decreasing sequence of positive terms it has a limit $L\geq 0$ . And we have $$L=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\frac {3(1+a_n)}{3+a_n}= \frac {3(1+L)}{3+L}.$$ So $L=\frac {3(1+L)}{3+L}, $
which implies $L^2=3.$ Now $(L\geq 0\land L^2=3)\implies L=\sqrt 3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proofing an inequality Assume we have three points $ a , b , c $ where $a < b < c$ and $b-a=c-b=h $
How can I proof for every $x \in [a,b]$ we have $|(x-a)(x-b)(x-c)| \leq \frac{2\sqrt{3}}{9}h^{3}$
| $$f(x)=|(x-a)(x-b)(x-c)|=(x-a)(x-b)(x-c)=(x-a)(b-x)(c-x)\tag{1}$$
Introduce substitution:
$$b-x=u, \quad u\in[0,h]\tag{2}$$
$$x-a=(b-a)-(b-x)=h-u\tag{3}$$
$$c-x=(c-b)+(b-x)=h+u\tag{4}$$
Now replace (2,3,4) into (1):
$$f(u)=(h-u)u(h+u)=u(h^2-u^2)=uh^2-u^3\tag{5}$$
$$f'(u)=h^2-3u^2=0\implies u=\frac{h\sqrt{3}}{3}\tag{6}$$
$$f(u)|_{u=\frac{h\sqrt{3}}{3}}=\frac{2\sqrt{3}}{9}h^{3}\tag{7}$$
Obviously $f''(u)=-3u^2<0$ for all values of $u\ne0$. This proves that (7) is indeed the maximum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
area between two curves 10 Find the area of the following curves:
$$y=7\cos(2x) ,\quad y=7-7\cos(2x) ,\quad [0,\pi/2].$$
I found the boundaries (calculations not shown but they are correct)
$7\cos(2x) > 7-7\cos(2x), \quad[0,\frac{\pi}{6}]$
$7-7\cos(2x) > 7\cos(2x),\quad [\frac{\pi}{6},\frac{5\pi}{6}]$
\begin{align*}
&\int 7 \cos(2x) - 7-7\cos(2x)~dx\\
=& \frac{7}{2}\sin(2x)-7x-\frac{7}{2} \sin(2x).
\end{align*}
\begin{align*}
&\left[{\frac{7}{2}\sin(2x)-7x-\frac{7}{2} \sin(2x)}\right]_{0}^{\frac{\pi}{6}}\\
=& -\frac{7\pi}{2} - 0 =-\frac{7\pi}{2}.
\end{align*}
\begin{align*}
&\left[7x-\frac{7}{2} \sin(2x)- \frac{7}{2}\sin(2x)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\\
=& \frac{35\pi}{6}+\frac{7\sqrt{3}}{2}-\frac{7\pi}{6}-\frac{7\sqrt{3}}{2}.
\end{align*}
\begin{align*}
&\left[{\frac{7}{2}\sin(2x)-7x-\frac{7}{2} \sin(2x)}\right]_{0}^{\frac{\pi}{6}}+\left[7x-\frac{7}{2} \sin(2x)- \frac{7}{2}\sin(2x)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\\
=&-\frac{7\pi}{2} + \frac{35\pi}{6}+\frac{7\sqrt{3}}{2}-\frac{7\pi}{6}-\frac{7\sqrt{3}}{2}\\
=& \frac{7\pi}{2}.
\end{align*}
This answer was wrong. I double checked my work and have no clue why its not right. To me it seems correct. Can anyone verify?
| Check your signs:
$$\int 7 \cos(2x) - \Big(7 - 7\cos(2x) \Big)$$
$$= \int 7 \cos(2x) - 7 \color{red}{+} 7\cos(2x)$$
$$= \frac{7}{2} \cos(2x) - 7x \color{red}{+} \frac{7}{2} \sin(2x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a limit of $\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3}$ I have no idea how can i solve this. When i'm trying to transform a multiplication I always get a $0*\infty$ ambiguity.
I found only that
$\lim_{n \to \infty}{(\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3})} = \lim_{n \to \infty}(\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3*...}}})$
EDIT: Solution
$\lim_{n \to \infty}{(\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3})} = \lim_{n \to \infty}(\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3*...}}}) = \lim_{n \to \infty}{3^{(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^n})}}$
$\sum_{r=1}^n{\frac{1}{3^n}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}$
So $\lim_{n \to \infty}{3^{(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^n})}} = \lim_{n \to \infty}{3^\frac{1}{2}} = \sqrt{3}$
| Hint:
$$\prod_{r=1}^n3^{(1/3)^r}=3^{\sum_{r=1}^n(1/3)^r}$$
Now $\displaystyle\sum_{r=1}^n\left(\dfrac13\right)r=\dfrac13\left(\dfrac{1-\left(\dfrac13\right)^n}{1-\dfrac13}\right)$^
Finally, $\displaystyle\lim_{n\to\infty}\left(\dfrac13\right)^n=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}=4\sqrt{3}$ I'm trying to calculate the expression: $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ and show that it is equal $4\sqrt{3}$.
I was trying to group the summands and calculate sums of $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}} \hspace{0.5cm}\text{and} \hspace{0.5cm} -\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ where we get $$\frac{2\cos\frac{2\pi}{15}+1}{\sin\frac{2\pi}{15}}-\frac{2\cos\frac{4\pi}{15}-1}{\sin\frac{8\pi}{15}}$$ but unfortunately this sum is not simplified.
How to prove this equality?
| I use degrees:
$$\frac{1}{\sin 12^\circ}+\frac{1}{\sin 24^\circ}-\frac{1}{\sin48^\circ}+\frac{1}{\sin96^\circ}=\\
\frac{\sin 96^\circ+\sin 12^\circ}{\sin 12^\circ\sin 96^\circ}+\frac{\sin 48^\circ-\sin 24^\circ}{\sin 24^\circ\sin 48^\circ} =\\
\frac{2\sin 54^\circ\cos 42^\circ}{\sin 12^\circ\sin 96^\circ}+\frac{2\cos 36^\circ\sin 12^\circ}{\sin 24^\circ\sin 48^\circ} =\\
\frac{2\cos 36^\circ\require{cancel} \cancel{\sin 48^\circ}}{2\sin 12^\circ\cancel{\sin 48^\circ}\cos 48^\circ}+\frac{2\cos 36^\circ\cancel{\sin 12^\circ}}{2\cancel{\sin 12^\circ}\cos 12^\circ\sin 48^\circ} =\\
\cos 36^\circ\cdot \frac{\cos 12^\circ\sin 48^\circ+\sin 12^\circ\cos 48^\circ}{\sin 12^\circ\cos 12^\circ\sin 48^\circ\cos 48^\circ} =\\
\frac{4\cos 36^\circ\sin 60^\circ}{\sin 24^\circ\sin 96^\circ}=\\
\frac{4\cos 36^\circ\sin 60^\circ}{\frac12(\cos 72^\circ-\cos 120^\circ)}=\\
\frac{4\cos 36^\circ\sin 60^\circ}{\frac12((\cos 36^\circ-\frac12)+\frac12)}=4\sqrt{3}.
$$
You can see see here: $\cos72^\circ = \cos36^\circ-\frac12$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of $ \frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+ \cdots$
Finding series sum of $$ \frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+\frac{1}{1\cdot 3\cdot 5\cdot 7\cdot 9}+ \cdots$$
Try: Let $\displaystyle a_{k}=\frac{1}{1\cdot 3\cdot 5\cdot 7\cdots (2k+1)}=\frac{2\cdot 4\cdot 6\cdots 2k}{(2k+1)!}$
So we have $\displaystyle a_{k}=\frac{2^k\cdot k!}{(2k+1)!}$
So our desired sum is $$\sum^{\infty}_{k=1}\frac{2^k\cdot k!}{(2k+1)!}$$
Now i am struck at that point
I did not understand how can i solve further,
Could some help me plaese , thanks
| Let $$f(x)=\frac {x^3}{1.3}+\frac{x^5}{1.3.5}+...$$
We want the value of $f(1)$. Note that
$$\frac{df}{dx}=xf(x)+x^2,\\f(0)=0$$
| {
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"url": "https://math.stackexchange.com/questions/3003371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability when arranging $N$ different numbers in a line we would like to arrange the numbers: $ 1, 2, \ldots , N$ in a line.
*
*what is the probability that digits $1, 2$ are not arranged next to each other?
*what is the probability that digits $1, 2$ are not arranged next to each other, and also the digits $3, 4$ are arranged next to each other?
my attempt:
This is mostly a combinatorics question, as the "Probability", meaning: $$\Omega = N!$$
*
*I treated $1, 2$ as a unit, looking for the number of different ways to arrange them next to each other - which would be the Complementary event.
$$P(A)^C = \frac{(N-1)! \cdot 2 }{N!}$$
where there's in fact $N-1$ needed to be arranged, ans 2 is the number of permutations.
so that $$P(A) = 1- \frac{(N-1)! \cdot 2 }{N!} = 1 - \frac{2}{N}$$
for the second sections - I'm not sure.
suppose I'd be looking the number of different ways to arrange $1, 2, \ldots N$ in a line so that the digits $1, 2$ and $ 3, 4$ are always arranged next to each other, then that will be:
$$\frac{(N-2)! \cdot 2 \cdot 2}{N!}$$ where 4 is the number of permutations (2 for each unit of digits)
but does that mean that there's $$\left(N! - {(N-2)! \cdot 2 \cdot 2}\right) \cdot \left((N-2)! \cdot 2 \right)$$ ways to arrange the numbers?
that doesn't seem right.
| Your answer to the first question is correct.
If the integers $1, 2, 3, \ldots, N$ are arranged in a line, what is the probability that the numbers $3$ and $4$ are arranged next to each other but the numbers $1$ and $2$ are not arranged next to each other.
As you observed, there are $N!$ ways to arrange the numbers in a row.
For the favorable cases, we begin by placing the numbers $3$ and $4$ in a box and arranging the $N - 3$ objects consisting of the box and the $N - 4$ numbers larger than $4$ in a row. The objects can be arranged in $(N - 3)!$ ways. Doing so creates $N - 2$ spaces in which we can place the $1$ and $2$, $N - 4$ between successive objects and two at the ends of the row.
$$\square \boxed{3, 4} \square 5 \square 6 \square \ldots \square N - 1 \square N \square$$
To separate $1$ and $2$, we must place them in two of these $N - 2$ spaces. There are $N - 2$ ways to place $1$, which leaves $N - 3$ ways to place $2$. The numbers $3$ and $4$ can be arranged within the box in $2!$ ways. Hence, the number of arrangements of the $N$ positive integers $1, 2, 3, \ldots, N$ in which $3$ and $4$ are next to each other but $1$ and $2$ are not next to each other is
$$(N - 3)!2!(N - 2)(N - 3) = (N - 2)!2!(N - 3)$$
Therefore, the probability that $3$ and $4$ are next to each other but $1$ and $2$ are not next to each other is
$$\frac{(N - 2)!2!(N - 3)}{N!}$$
| {
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"url": "https://math.stackexchange.com/questions/3004962",
"timestamp": "2023-03-29T00:00:00",
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When is $\prod_{n=2}^{k} n-\frac{1}{n}$ not an integer? For which values of $k$ does the following product not evaluate to an integer: $$\prod_{n=2}^{k} n-\frac{1}{n}$$
I find it somewhat surprising that it not only can evaluate to an integer, but also that it does so most of the time.
The only values of $k$ that I could find so far, such that it does not evaluate to an integer are $2$ and $18$. Do these numbers have any special characteristics which would explain this?
| It is surprising that this is often an integer. But because it is surprising there is probably a simple reason that will be become clear when we try a few examples.
$(2 - \frac 12)(3 - \frac 13) = 2*3 - \frac 32 - \frac 23 + \frac 1{2*3}$.
Okay, one thing that becomes clear is that if we have $(n - \frac 1n)(m - \frac 1n)$ our terms are going to be $nm$ (which is clearly an integer) $-\frac nm, -\frac nm$ and $\frac 1{mmn}$. When we add $-\frac mn -\frac nm$ we get $-(\frac {m^2}{mn} + \frac {n^2}{mn})=-\frac {m^2 + n^2}{nm}$ and we get $mn - \frac {m^2 + n^2-1}{mn}$.
Well, no reason that $mn|m^2 + n^2 -1$ but if $m$ and $n$ are consecutive, i..e. $(n = m+1)$ we get $$m(m+1) - \frac {m^2 + (m+1)^2 - 1}{m(m+1)} = m(m+1) - \frac {2m^2 + 2m}{m(m+1)} = m(m+1) - \frac {2m(m+1)}{m(m+1)} = m(m+1) -2$$ and that.... is an integer.
So $(m - \frac 1m)((m+1) - \frac 1{m+1}$ must be an integer and as integers multiply together to integers we have for a product of even number of terms:
$\prod_{n=2}^{2m+ 1} (n-\frac 1n) = \prod_{k=1}^m [(2k-\frac 1{2k})((2k+1)-\frac 1{2k+1})] = \prod_{k=1}^m (2k(2k+1) -2)=K\in \mathbb Z$.
But if we have a product of odd terms:
$\prod_{n=2}^{2m+ 2}(n-\frac 1n) =(\prod_{n=2}^{2m+ 1}(n-\frac 1n))((2m+2)-\frac 1{2m+2})=K( 2m+2 - \frac 1{2m+2})$ and as $( 2m+2 - \frac 1{2m+2})$ is not an integer $K( 2m+2 - \frac 1{2m+2})$ won't be an integer.....
.... unless $2m+2|K$.
Okay.... we are halfway done.
When will $2m+2|K$?
I'm embarrassed to admit how long the following took me to figure out but:
$K=\prod_{k=1}^m(2k(2k+1) -2)$ and for $2m(2m+1) -2=2(m+1)(2m -1)$ so $2m+2$ will always divide $K$ and this will always be an integer. (Unless we only multiply one term $(2 - \frac 12)$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Largest Digit Sum in bases besides base 10 So I know that in base 10, the largest digit sum a number N can have will occur when all the digits are equal to 9.
For example, 999 adds to 27, which adds to 9. 9999 adds to 36, which adds to 9 as well.
My question is, does this same logic apply to other bases, such as base 2 or base 3? My initial thought is that it does not because if you look at 1111 in base 2, this sums to 4?
| Well in base $b$ the highest digit is $b-1$ so the highest digit sum will occur when all the digits are $b-1$. So the highest sum you can have for an $n$ digit number is $(b-1)(b-1)...(b-1)$ and the sum is $(b-1) + .... = n(b-1)$.
You have a second statement that the sum of those digits is $9$. That's mostly true. But consider $99,999,999,999$. The sum of those digits is $99$. And the sum of those is $18$ and not $9$. But the sum of the digits of $18$ is $1+8=9$ so we do get to $9$ eventually after many steps.
And this is true of base $b$.
A number $N = \sum_{i=0} a_i b^i$ and the sum of the digits of $N$ is $\sum_{i=0} a_i$. So if you subtract the two you get:
$\sum_{i=1} a_i (b^i - 1)$. And $b^i - 1 = (b-1)\times(b^{i-1} + b^{i-2} + ... + b + 1)$. So the difference between the number $N$ and sum of the digits is:
$\sum_{i=1} a_i(b-1)(\sum_{k=0}^{i-1}b^k) = (b-1)[\sum_{i=1} a_i(\sum_{k=0}^{i-1}b^k)]$ which is a multiple of $b-1$.
So $N$ is a multiple of $b-1$ if and only if the sum of the digits are.
Now it might not happen on your first step that you get $b-1$ but it will happen eventually.
Consider $1111_2 = 15_{10}$. The sum of those digits is $4$. And $4$ is a multiple of $2-1$. In base $2$, we have $4_{10} = 100$ and then sum of those is $1 + 0 + 0 =1$. So we are done. But it took us $2$ steps.
Consider base $4$ as a better example: $33333$ has the sum of the digitis is $3+3+3+3+3 = 15_{10} = 3*4 + 3 = 33_{4}$. And the sum of the digits of $33$ is $3+3 = 6_{10} = 4 + 2 = 12_4$. ANd the sum of the digits of $12_4$ is $1+2 = 3$.
Or base $3$. $222222_3$ will have sum $2+2+2+2+2+2 = 12_{10} = 9 + 3 + 1 = 110_3$ and then sum of the digits of $110$ are $1+1+0 = 2$
This WILL always work.
| {
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"timestamp": "2023-03-29T00:00:00",
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solve $\lim_{x\to\infty}{\sqrt{x^4+x^2}+\sqrt{x^2+5x}-x^2-x}$ I tried solve this limit.
$\lim_{x\to\infty}{\sqrt{x^4+x^2}+\sqrt{x^2+5x}-x^2-x}$
I tried multiply by $\sqrt{x^4+x^2}-\sqrt{x^2+5x}$, and apply L'Hospital. but this led to alot of work.. and this question seems to had a very easy and fast way to do. I know the answer is 3.
thanks any help.
| \begin{align}
\lim_{x\to\infty}{\sqrt{x^4+x^2}+\sqrt{x^2+5x}-x^2-x}
&= \lim_{x\to\infty}\sqrt{\left(x^2+\dfrac12\right)^2-\dfrac14} \\
&+\sqrt{\left(x+\frac52\right)^2-\frac{25}{4}} -x^2-x \\
&= \lim_{x\to\infty}{\left(x^2+\dfrac12\right)+\left|x+\frac52\right|-x^2-x} \\
&= \lim_{x\to+\infty}{\dfrac12+x+\frac52-x} \\
&= \color{blue}{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $\sum _{n=1}^{\infty \:\:}\left(\frac{-3}{4}\right)^{n+1}$ converge? This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?
| If $|x| < 1$ then
$\displaystyle\sum_{n=0}^{\infty} x^n
=\frac1{1-x}$.
Therefore
$\begin{align}\\
\sum_{n=1}^{\infty} \left(\frac{-3}{4}\right)^{n+1}
&=\left(\frac{-3}{4}\right)^2\sum_{n=1}^{\infty} \left(\frac{-3}{4}\right)^{n-1}\\
&=\frac{9}{16}\sum_{n=0}^{\infty} \left(\frac{-3}{4}\right)^{n}\\
&=\frac{9}{16}\frac1{1-\left(\frac{-3}{4}\right)}\\
&=\frac{9}{16}\frac{4}{4+3}\\
&=\frac{9}{4\cdot 7}\\
&=\frac{9}{28}\\
\end{align}
$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Alternative proof that if $a,b,c \in \mathbb{R}$ and $(a+b+c)c<0$ then $b^2-4ac>0$? A problem from a Moscow Olympiad states : let $a,b,c$ be real numbers such that $(a+b+c)c<0$. Show that $b^2-4ac>0$. There is a well known proof of this applying the IVT on the polynomial $f(x) = ax^2+bx+c$. Is there any proof that does not involve any calculus, including IVT?
The question emerged from a reddit post
https://www.reddit.com/r/math/comments/9znumt/theorems_of_single_variable_calculus/
| If all you know is completing the square:
Suppose $b^2-4ac\leq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.
By completing the square we can rewrite $ax^2+bx+c=a\left(\left(x+\frac{b}{2a}\right)^2+d^2\right)$ for some real $d$.
We have $c=\frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:
$$\begin{align}(a+b+c)c&=\left(a+b+\frac{b^2}{4a}+ad^2\right)\left(\frac{b^2}{4a}+ad^2\right)\\
&=\left(a+b+\frac{b^2}{4a}\right)\frac{b^2}{4a}+\left(ad^2\frac{b^2}{4a}+\left(a+b+\frac{b^2}{4a}\right)ad^2\right)+\left(ad^2\right)^2\\
&=\frac{b^2}{4}\left(1+\frac{b}{a}+\frac{b^2}{4a^2}\right)+d^2\left(a^2+ab+\frac{b^2}{2}\right)+\left(ad^2\right)^2\\
&=\frac{b^2}{4}\left(1+\frac{b}{2a}\right)^2+d^2\left(a+\frac{b}{2}\right)^2+\left(ad^2\right)^2\\
&\geq0\text{.}
\end{align}$$
Thus by contrapositive, $(a+b+c)c<0\,\Rightarrow\, b^2-4ac>0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Recurrence $k$-th pattern by substitution The recurrence is $T(n) = 9 T(\frac{n}{3}) + {n^2}$.
Simplified to ${3^2} T(\frac{n}{3}) + {n^2}$. Then
$$\begin{align}
T(n)&= 3^4T(\frac{n}{3^2}) + 4n\\
&= 3^6T(\frac{n}{3^3}) + 10n\\
&= 3^8T(\frac{n}{3^4}) + 28 n.
\end{align}$$
I am trying to find the $k$-th pattern of this recurrence.
| In my opinion something is wrong in your approach.
If $T(n)={3^2} T(\frac{n}{3}) + {n^2}$, then $T(\frac{n}{3^k})={3^2} T(\frac{n}{3^{k+1}}) + (\frac{n}{3^k})^2$ and
$$\begin{align}T(n)&={3^2} T(\frac{n}{3}) + {n^2}={3^2} \left({3^2} T(\frac{n}{3^2}) + \frac{n^2}{3^2}\right) + n^2\\
&={3^4} T(\frac{n}{3^2}) + 2n^2
={3^4} \left({3^2} T(\frac{n}{3^3}) + \frac{n^2}{3^4}\right) + 2n^2\\
&={3^6} T(\frac{n}{3^3}) + 3n^2=\dots={9^k} T(\frac{n}{3^k}) + kn^2.\\
\end{align}$$
P.S. Do you know the Master Theorem?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum or value of following expression In the problem I am not able to derive difference term. I know for solving summation we have to make difference term.
How to proceed with this problem?
$$\frac{{\displaystyle \sum_{n=1}^{99}} \sqrt{10+\sqrt{n}}}{{\displaystyle \sum_{n=1}^{99}} \sqrt{10-\sqrt{n}}}$$
| This can be done using an identity I learned from math.SE.
For any $a > 0$ and $0 \le b \le a^2$, we have
$$\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}} = \sqrt{2}\sqrt{a + \sqrt{a^2-b}}$$
To prove this identity, just take squares on both sides and use the fact
$$\begin{align}
\left(\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}\right)^2 &= \left(a + \sqrt{b}\right) + \left(a - \sqrt{b}\right) + 2\sqrt{a^2-b}\\
&= 2\left(a + \sqrt{a^2-b}\right)\end{align}$$
Let $a$ be any integer $> 1$ and $c = a^2-b$. When we sum $b$ from $1$ to $a^2-1$, above identity tell us
$$\sum_{b=1}^{a^2-1} \sqrt{a + \sqrt{b}} + \sum_{b=1}^{a^2-1}\sqrt{a - \sqrt{b}}
= \sqrt{2}\sum_{b=1}^{a^2-1}\sqrt{a + \sqrt{a^2-b}}
= \sqrt{2}\sum_{c=1}^{a^2-1}\sqrt{a + \sqrt{c}}$$
This leads to
$$\frac{\sum_{b=1}^{a^2-1}\sqrt{a+\sqrt{b}}}{\sum_{b=1}^{a^2-1}\sqrt{a-\sqrt{b}}}
= \frac{1}{\sqrt{2}-1} = \sqrt{2}+1
$$
Substitute $a$ by $10$, this reduces to the ratio at hand. i.e. The ratio we seek equals to $\sqrt{2}+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Link between polynomial and derivative of polynomial I can't seem to solve this problem, can anyone help me please? The problem is:
Let real numbers $a$,$b$ and $c$, with $a ≤ b ≤ c$ be the 3 roots of the polynomial $p(x)=x^3 + qx^2 + rx + s$. Show that if we divide the interval $[b, c]$ into six equal parts, then one of the root of $p'(x)$ (the derivative of the polynomial $p(x)$) will be in the 4th part.
What I did was:
Because we know the roots of $p(x)$ are $a$,$b$ and $c$, we can write the polynomial $p(x)$ like this: $p(x) = (x-a)(x-b)(x-c)$
So we have $p(x) = x^3 + qx^2 + rx + s = (x-a)(x-b)(x-c)$
We find the value of $q$,$r$ and $s$:
$q = -(a+b+c)$
$r = (ab + ac + bc)$
$s = -abc$
We have that $p'(x) = 3x^2 + 2qx + r$
We want to find the roots of $p'(x)$, so if we apply the quadratic formula, we get:
$(-2q ± 2*\sqrt{q^2 - 3r})/6$
Because we know the value of q,r and s, we can rewrite this expression like this:
$(2(a+b+c) ± 2\sqrt{a^2 + b^2 + c^2 - ab - bc -ca})/6$
But I am stuck here, any help would be great. Thank you in advance.
| I have an inelegant brute force solution. Since the problem is translation invariant, we may assume $a=0$ to simplify calculations so that $0\leq b\leq c$. It should be clear that the root of $p'(x)$ we are interested in is the largest of the two i.e. the one with the plus sign. We need to prove that
$$\dfrac{3b+3c}{6}\leq\dfrac{2(b+c)+2\sqrt{b^2+c^2-bc}}{6}\leq\dfrac{2b+4c}{6}$$
On the left we obtain
$$b+c\leq 2\sqrt{b^2+c^2-bc}$$
where both sides are non-negative, squaring gives
$$b^2+c^2+2bc\leq 4(b^2+c^2-bc)\Rightarrow 3b^2+3c^2-6bc\geq 0$$
which holds. On the right we have
$$2\sqrt{b^2+c^2-bc}\leq 2c$$
so that
$$b^2+c^2-bc\leq c^2\Rightarrow b(b-c)\leq 0$$
which holds since $0\leq b\leq c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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General solution of $\sin 2x+\cos x=0$
Solve the trigonometric equation $\sin 2x+\cos x=0$
My Attempt
$$
2\sin x\cos x+\cos x=0\implies\cos x=0 \text{ or }\sin x=\frac{-1}{2}=\sin\frac{-\pi}{6}\\
x=(2n+1)\frac{\pi}{2} \text{ or }x=n\pi+(-1)^n\frac{-\pi}{6}\\
x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}
$$
Reference
$$
\cos x=-\sin 2x=\cos\Big(\frac{\pi}{2}+2x\Big)\implies x=2n\pi\pm\Big(\frac{\pi}{2}+2x\Big)\\
-x=2n\pi+\frac{\pi}{2}\text{ or }3x=2n\pi-\frac{\pi}{2}\implies x=2m\pi-\frac{\pi}{2}\text{ or }x=\frac{2m\pi}{3}-\frac{\pi}{6}
$$
But my reference gives the solution $x=2n\pi-\dfrac{\pi}{2}$ or $x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}$. I understand how it is reached and both represent the same solutions. But, how do I derive the solution in my reference from what I found in my attempt ? i.e.,
How to derive
$$
\bigg[x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}\bigg]\\
\implies \bigg[x=2n\pi-\dfrac{\pi}{2}\text{ or }x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}\bigg]
$$
| There are two ways to solve the equation.
Your method: $2\sin x\cos x+\cos x=0$, so $\cos x(2\sin x+1)=0$. Thus we have either $\cos x=0$ or $\sin x=-1/2$. Thus
\begin{align}
x&=\frac{\pi}{2}+2n\pi &\text{or}&& x&=-\frac{\pi}{2}+2n\pi && \text{(from $\cos x=0$)} \\[4px]
x&=-\frac{\pi}{6}+2n\pi &\text{or}&& x&=\frac{7\pi}{6}+2n\pi && \text{(from $\sin x=-1/2$)}
\end{align}
(you grouped together the families of solutions of $\cos x$ and of $\sin x=-1/2$, but I prefer to keep them separated). Good job.
Alternative method: $\cos x=-\sin2x=\cos(\frac{\pi}{2}+2x)$. Therefore either
$$
x=\frac{\pi}{2}+2x+2n\pi \to x=-\frac{\pi}{2}-2n\pi
$$
or
$$
x=-\frac{\pi}{2}-2x+2n\pi \to 3x=-\frac{\pi}{2}+2n\pi \to x=\frac{2n\pi}{3}-\frac{\pi}{6}
$$
How do you recover the previous sets of solutions?
One set is already present. For the other three, consider the cases when $n=3k$, $n=3k+1$ or $n=3k+2$, with integer $k$. Then
\begin{align}
n&=3k & x&=\frac{6k\pi}{3}-\frac{\pi}{6}=2k\pi-\frac{\pi}{6} \\[4px]
n&=3k+1 & x&=\frac{(6k+2)\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{2\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{\pi}{2}\\[4px]
n&=3k+2 & x&=\frac{(6k+4)\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{4\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{7\pi}{6}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of the expression $\lim_{x\to 0}\left(\frac{\sin x}{\arcsin x}\right)^{1/\ln(1+x^2)}$ Limit: $\lim_{x\to 0}\left(\dfrac{\sin x}{\arcsin x}\right)^{1/\ln(1+x^2)}$ I have tried to do this: it is equal to $e^{\lim\frac{\log{\frac{\sin x}{\arcsin x}}}{\log(1+x^2)}}$, but I can't calculate this with the help of l'Hopital rule or using Taylor series, because there is very complex and big derivatives, so I wish to find more easier way.
$$\lim_{x\rightarrow 0}{\frac{\log{\frac{\sin x}{\arcsin x}}}{\log(1+x^2)}} = \lim_{x\rightarrow 0}\frac{\log1 + \frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)}}{\log(1+x^2)} = \lim_{x\rightarrow0}\frac{\frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)} + o(\frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)})}{x^2+o(x^2)}$$ using Taylor series. Now I think that it's not clear for me how to simplify $o\left(\frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)}\right)$.
| HINT
By Taylor's series
$$\frac{\sin x}{\arcsin x}=\frac{x-\frac16x^3+o(x^3)}{x+\frac16x^3+o(x^3)}=\frac{1-\frac16x^2+o(x^2)}{1+\frac16x^2+o(x^2)}=$$$$=\left(1-\frac16x^2+o(x^2)\right)\left(1+\frac16x^2+o(x^2)\right)^{-1}$$
Can you continue form here using binomial series for the last term?
| {
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Showing that the distance from any point on an ellipse to the foci points is constant Let $a$ be the largest x-value of an ellipse centered at the origin. Similarly, let $b$ be the largest y-value of the ellipse. Assume that $a>b$. We know that the foci points will be at $c_1=(-\sqrt {a^2-b^2},0)$ and $c_2=(+\sqrt {a^2-b^2},0)$ respectively. Then select some point on the ellipse $p_1$ with x-value $x_1$ between $-a$ and $a$. If we only consider the positive y-values of the ellipse, then we know from the equation of an ellipse that the y-value of $p_1$, will be $y_1=\sqrt{b^2(1-\frac{x_1^2}{a^2}})$ or better yet $y_1=\frac{b\sqrt{a^2-x_1^2}}{a}$. So for a generic x-value, the point on the ellipse will be $p_1=\left(x_1, \frac{b\sqrt{a^2-x_1^2}}{a}\right)$.
I'd like to show that the sum distance of $p_1$ to $c_1$ and $p_1$ to $c_2$ will be a constant that does not depend on the value of $x_1$. To do so, I planned to use Euclid's Metric, and have cancellation of the $x_1$ terms algebraically.
In other words, I would like to show a formula in terms of only $a$ and $b$ for the distance of a point on the ellipse to each foci. I am looking for how to manipulate Euclid's metric to show this cancellation.
For thoroughness, I'll give the original equation for the generic distance from $p_1$ to $c_1$: $\sqrt{(x_1-\sqrt{a^2-b^2})^2+\left(\frac{b\sqrt{a^2-x_1^2}}{a}\right)^2}$.
| Every point of elipse $(bx)^2 +(ay)^2 =(ab)^2 $ can be represent as $P=(a\cos t , b\sin t)$ hence $$c_1 P +c_2 P =\sqrt{(\sqrt{a^2 -b^2} +a\cos t )^2 + b^2\sin^2 t } +\sqrt{(\sqrt{a^2 -b^2} -a\cos t )^2 + b^2\sin^2 t }=\sqrt{(a+\sqrt{a^2 -b^2}\cos t)^2 } +\sqrt{(a-\sqrt{a^2 -b^2}\cos t)^2 } =2a$$
| {
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Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| \leq 3$ Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| \leq 3$.
I have been trying a simpler case $4a^2 - b^{b+1} = 0$. I found that when $b$ is odd then $b^{b+1}$ have to be in form $b^{b+1} = c^2$ where $c$ is integer. So we have $(2a + c)(2a - c)=0 \implies 2a = c$ (otherwise $a$ would be negative). However, $c$ is odd because $b$ is also odd (odd number times odd number is odd). But this means that $a$ (since $2a = c$) isn't integer and that is contradiction. Therefore in this simpler case $b$ can't be odd.
But above applies only for specific simpler case. For example for $a = b = 1$ original inequality holds.
Can someone help me with this please?
| For the sake of simplicity, I will consider positive integers $a,b$.
First, let's consider the case of $4a^2-b^{b+1}=0$. You've already noticed that $b$ cannot be odd, and so we shall consider only the case of even $b$. If $b$ is even, then $b+1$ is odd and so $b^{b+1}$ is a perfect square iff $b$ is a perfect square, so we have that $b$ is a perfect square. Since it is an even perfect square, we may write $b=4c^2$ and $b^{b+1}=4\cdot 4^{4c^2}\cdot c^{8c^2+2}$ and we may let $a=4^{2c^2}\cdot c^{4c^2+1}$ for a solution. Thus, we have found one possible solution set:
$$(a,b)=(4^{2c^2}\cdot c^{4c^2+1},4c^2)$$
Now suppose that $4a^2-b^{b+1}=1$. Then $(2a+1)(2a-1)=b^{b+1}$, and so $b$ must be odd, meaning that $b+1$ is even and $b^{b+1}$ is a perfect square. But $4a^2=b^{b+1}+1$ is also a perfect square, giving a contradiction. So $4a^2-b^{b+1}\ne 1$.
Similarly, suppose that $4a^2-b^{b+1}=-1$, or $4a^2=b^{b+1}-1$. We have again that $b$ must be odd, and so $b+1$ is even and $b^{b+1}$ is a perfect square, which is a contradiction since $b^{b+1}-1$ is a perfect square.
Now suppose that $4a^2-b^{b+1}=2$, or $4a^2=b^{b+1}+2$. Then $b$ must be even, and we have that the $2$-adic valuation of $b^{b+1}+2$ is equal to $1$. But the $2$-adic valuation of $4a^2$ is at least $2$, giving a contradiction.
Use the same reasoning for the case of $4a^2-b^{b+1}=-2$.
Suppose that $4a^2-b^{b+1}=3$, or $4a^2=b^{b+1}+3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}+3$ is also a perfect square, which is only possible if $b^{b+1}=1$. This gives the solution pair
$$(a,b)=(1,1)$$
and no others.
Finally, suppose that $4a^2-b^{b+1}=-3$ or $4a^2=b^{b+1}-3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}-3$ is also a perfect square, which can only occur if $b^{b+1}=4$, but this never happens for positive integers $b$.
We are done! We have only the solutions $(1,1)$ and
$$(4^{2c^2}\cdot c^{4c^2+1},4c^2)$$
for $c\in\mathbb N$.
| {
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} |
Find the maximize value of $\sqrt{\frac{a}{a+c}}+\sqrt{\frac{b}{b+c}}-\frac{9\sqrt{c^2+1}}{8c}$ Let $a;b;c>0$ such that $ab+bc+ca=1$. Find the maximize value of $$K=\sqrt{\frac{a}{a+c}}+\sqrt{\frac{b}{b+c}}-\frac{9\sqrt{c^2+1}}{8c}$$
I can see: $$a=b=\frac{1}{\sqrt{7}};c=\frac{3}{\sqrt{7}}\rightarrow K=\frac{-1}{2}$$ so i will prove $K=\frac{-1}{2}$ is maximize value of $K$.
Indeed, i will prove $\sqrt{\frac{a}{a+c}}+\sqrt{\frac{b}{b+c}}-\frac{9\sqrt{c^2+1}}{8c}\le -\frac{1}{2}$
Or $\sqrt{\frac{a}{a+c}}-\frac{1}{2}+\sqrt{\frac{b}{b+c}}-\frac{1}{2}-\left(\frac{9\sqrt{c^2+ab+bc+ca}}{8c}-\frac{3}{2}\right)\le 0$
Or $$\frac{\frac{3a-c}{4a+c}}{\sqrt{\frac{a}{a+c}}+\frac{1}{2}}+\frac{\frac{3b-c}{b+c}}{\sqrt{\frac{b}{b+c}}+\frac{1}{2}}-\frac{3}{2}\cdot \frac{\frac{-7c^2+9ab+9bc+9ca}{3\sqrt{c^2+ab+bc+ca}+4c}}{4c}\le 0$$
Or $$\frac{\frac{3a-c}{4a+c}}{\sqrt{\frac{a}{a+c}}+\frac{1}{2}}+\frac{\frac{3b-c}{b+c}}{\sqrt{\frac{b}{b+c}}+\frac{1}{2}}-\frac{3}{2}\cdot \frac{\frac{9ab-c^2+3c\left(3a-c\right)+3c\left(3b-c\right)}{3\sqrt{c^2+ab+bc+ca}+4c}}{4c}\le 0$$
Then we have: $9ab\le \frac{\left(a+b\right)^2}{4}\Leftrightarrow 9ab-c^2\le \frac{9\left(a+b\right)^2}{4}-c^2=\left(\frac{3a-c+3b-c}{2}\right)\left(\frac{3a+3b}{2}+c\right)$
Or $$\frac{\frac{3a-c}{4\left(a+c\right)}}{\sqrt{\frac{a}{a+c}}+\frac{1}{2}}+\frac{\frac{3b-c}{4\left(b+c\right)}}{\sqrt{\frac{b}{b+c}}+\frac{1}{2}}-\frac{3}{2}\frac{\frac{\left(3a-c\right)\cdot \frac{1}{2}\cdot \left(\frac{3a+3b}{2}+c\right)+\left(3b-c\right)\cdot \frac{1}{2}\cdot \left(\frac{3a+3b}{2}+c\right)+3c\left(3b-c\right)+3c\left(3a-c\right)}{3\sqrt{c^2+ab+bc+ca}+4c}}{4c}\le 0$$
It's easy to recognize the equality occurs at $3a=3b=c$ so i am trying to take it becomes $(3a-c)^2\cdot f(a;b;c)+(3b-c)^2\cdot f(a;b;c)\ge 0$ but i do not know how to do it. Help me !!
P/s:This problem was extended by Prove that $2\sqrt{\frac{a}{a+c}}-\frac{9\sqrt{c^2+1}}{8c}<1$ with $a;c>0$
| By your work we need to prove that
$$\sqrt{\frac{a}{a+c}}+\sqrt{\frac{b}{b+c}}-\frac{9\sqrt{(a+c)(b+c)}}{8c}\leq-\frac{1}{2}.$$
Now, let $\frac{a}{a+c}=\frac{x^2}{4}$ and $\frac{b}{b+c}=\frac{y^2}{4},$ where $x$ and $y$ be positives.
Thus, $0<x<2$ and $0<y<2$ and since
$$\frac{a+c}{c}=\frac{1}{\frac{c}{a+c}}=\frac{1}{1-\frac{a}{a+c}}=\frac{1}{1-\frac{x^2}{4}}=\frac{4}{4-x^2},$$
we need to prove that
$$\frac{x}{2}+\frac{y}{2}-\frac{9}{2\sqrt{(4-x^2)(4-y^2)}}\leq-\frac{1}{2}$$ or
$$(x+y+1)\sqrt{(4-x^2)(4-y^2)}\leq9$$ or
$$\left(\frac{x+y+1}{3}\right)^2\cdot\frac{2+x}{3}\cdot\frac{2+y}{3}\cdot(2-x)\cdot(2-y)\leq1,$$ which is true by AM-GM:
$$\left(\frac{x+y+1}{3}\right)^2\cdot\frac{2+x}{3}\cdot\frac{2+y}{3}\cdot(2-x)\cdot(2-y)\leq$$
$$\leq\left(\frac{2\cdot\frac{x+y+1}{3}+\frac{2+x}{3}+\frac{2+y}{3}+2-x+2-y}{6}\right)^6=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the coefficient of $x^{70}$ in $(x^1-1)(x^2-2)(x^3-3)\cdots(x^{12}-12)$.
Find the coefficient of $x^{70}$ in $(x^1-1)(x^2-2)(x^3-3)\cdots (x^{12}-12)$.
I tried to solve this problem using theory of equation the coefficient of $x^{70}$ will be the sum of products taking two at a time. But this very very exhaustive I want to know some another method as it will be proficient in higher powers.
| For any formal Laurent series $f(z) = \sum_{k=-\infty}^\infty a_k z^k$ in any variable $z$, let $[z^k]f(z)$ be a short hand for the coefficient $a_k$.
Let $y = x^{-1}$, we can simplify the coefficient at hand as
$$\begin{align}\mathcal{C} \stackrel{def}{=} [x^{70}]\prod_{k=1}^{12}(x^k-k)
&= [x^{70}]x^{78} \prod_{k=1}^{12}(1-kx^{-k}) \\
&= [y^8]\prod_{k=1}^{12}(1-ky^k)
= [y^8]\prod_{k=1}^8(1-ky^k)\end{align}$$
We can split last product into two pieces
$$\begin{align}\prod_{k=1}^3 (1-ky^k) &= (1-y)(1-2y^2)(1-3y^3)
= (1-y-2y^2+2y^3)(1-3y^3)\\
&= 1-y-2y^2-y^3 + 3y^4 +6y^5 - 6y^6\\
\prod_{k=4}^8 (1-ky^k) &=
1 - 4y^4 - 5y^5 - 6y^6 - 7y^7 - 8y^8 + O(y^9)
\end{align}$$
Throwing away terms which doesn't contribute to $[y^8]$, we obtain
$$\begin{align}\mathcal{C} = &\; [y^8]\big((1-y-2y^2-y^3+3y^4)(1-4y^4-5y^5-6y^6-7y^7-8y^8)\big)\\
= &\; (1)(-8) + (-1)(-7) + (-2)(-6)+(-1)(-5)+(3)(-4)\\
= &\; -8 + 7 + 12 + 5 - 12\\
= &\; 4\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Pair up $\{1..2n\}$ that the sums of each pair are different primes. Pair up $\{1..2n\}$ that the sums of each pair are different primes.
I found 9 examples:
$\{(1,2)\},$
$\{(1,2),(3,4)\},$
$\{(1,2),(3,4),(5,6)\},$
$\{(1,4),(2,5),(3,8),(6,7)\},\{(2,3),(1,6),(4,7),(5,8)\},$
$\{(1,4),(2,5),(3,8),(6,7),(9,10)\},\{(2,3),(1,6),(4,7),(5,8),(9,10)\},$
$\{(1,4),(2,5),(3,8),(6,7),(9,10),(11,12)\},\{(2,3),(1,6),(4,7),(5,8),(9,10),(11,12)\}.$
Is there another example like this?
| There are no more. There are none for $n=7$. We would need the sums to be seven distinct odd primes below $27$. The most the sum of these sums can be is $5+7+11+13+17+19+23=95$ but the sum of all the numbers up to $14$ is $105$.
For $n=8$ we would need the sum of eight primes below $31$ to be $120$. We can only do this with $3,5,11,13,17,19,23,29$. We need $1+2$ to get $3$ but cannot get $5$.
For $n=9$ we need all the primes up to $31$ except $2,5$, but then $1+2=3,3+4=7,5+5=11$ and we are stuck for $13$.
For $n=10$ we need all the primes up to $37$ except $2,5$ and the $n=9$ proof works.
For $n=11$ we need all the primes up to $43$ except $2,7,11$ or $2,5,13$. We can do $21+22=43$ but cannot get $41$.
For $n=12$ the sum of all the primes up to $43$ is $271$ while the sum of the numbers up to $24$ is $276$
Now going from $n$ to $n+1$ the sum of the numbers increases by $4n+3$ while the sum of primes increases by $4n+1, 4n+3,$ or $8n+4$. As less than $\frac {2\cdot 3 \cdot 5}{3 \cdot 5 \cdot 7} \lt \frac 12$ are primes, the sum of primes will never catch up. $n=13$ doesn't add any new primes because the odd numbers added are $49,51$ and again we miss at $n=19$. Each intervening $n$ has only added one prime.
| {
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"timestamp": "2023-03-29T00:00:00",
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Use the sine rule to prove trig identity Using the sine rule:
$$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$$
prove, for triangle ABC:
$$\sin\left(\frac{B-C}{2}\right) = \frac{b-c}{a} \cos\left(\frac A2\right)$$
Using the sine rule it's easy to translate the RHS into:
$$\sin\left(\frac{B-C}{2}\right) = \frac{\sin(B)-\sin(C)}{\sin(A)} \cos\left(\frac A2\right)$$
Yes, I can expand out the LHS, and use the difference of 2 sines in the RHS, but neither makes an obvious equality, especially with terms in a and A in the RHS.
A nudge in the right direction would really be appreciated. Thanks.
| Since $A+B+C=\pi$, the difference formula for $\sin$ gives \begin{align}\frac{\sin B-\sin C}{\sin A} \cos\frac A2&=\frac{2\cos\frac{B+C}2\sin\frac{B-C}2}{\sin(\pi-B-C)}\cos\frac{\pi-B-C}2\\&=\frac{2\sqrt{\frac{1+\cos(B+C)}2}\sin\frac{B-C}2}{\sin(B+C)}\sqrt{\frac{1+\cos(\pi-B-C)}2}\\&=\frac{2\sqrt{\frac{1+\cos(B+C)}2}\sin\frac{B-C}2}{\sin(B+C)}\sqrt{\frac{1-\cos(B+C)}2}\\&=\frac{\sin\frac{B-C}2}{\sin(B+C)}\sqrt{1-\cos^2(B+C)}\\&=\frac{\sin\frac{B-C}2}{\sin(B+C)}\sin(B+C)=\sin\frac{B-C}2\end{align} as required.
| {
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Proof Checking a "Logic Equation" Puzzle Did a "logic equation" puzzle that took me several hours. Feel like it should have been faster, and that there surely is a more elegant path to the solution.
Sharing the proof to spot redundancies, omissions, faulty logic and areas to simplify. If there are best practices I should follow in terms of formatting and expression, that would also be a nice contribution.
You can try your hand at the puzzle here before you look at my proof, if you like.
Given that all variables are unique integers, and given the following
statements:
$B \ne 1; B \ne 2; B \ne 5$
$D \ne 4; D \ne 6$
$D+E=C+F$
$B+C=A+F$
Find the value of all variables.
$$ \begin{array}{c|lcr} & \text{1} & \text{2} & \text{3} & \text{4} &
\text{5} & \text{6} \\ \hline A & & & & & & \\ B & x & x & & & x & \\
C \\ D & & & & x & & x\\ E \\ F \\ \end{array} $$
Proof:
Given that the sum of all variables (A-F) is 21 and that $B + C = A + F$,
*
*$21 - (D+E) = A + B + C + F$
*$D+E$ must produce an odd-numbered sum so that $21 - (D+3)$ produces an even-numbered difference.
*$D$ and $E$ must have opposite parity.
Given that the sum of all variables (A-F) is 21 and that $A+F \ne C+F$,
*
*A sum of 7 is impossible for $D+E$, $B+C$ or $A+F$ as it would create a paradox $(A+F=C+F)$
Given that $D+E = C+F$ and that $D$ and $E$ have opposite parity,
*
*$C$ and $F$ must have opposite parity.
Given that $D+E=C+F$ and $B+C=A+F$,
*
*Sums of 3, 10 and 11 are not possible for $D+E$, $C+F$ or $A+F$ because they cannot be produced two or more ways in this data-set.
Given that $(21-(D+E))/2=B+C$, that $D+E$ must be odd, that $D+E>4$, that $D+E \ne 7$, that $D \ne 4$ and that $D \ne 6$,
*
*$E \ne 1$
*If $D+E$ produced a sum equal to 5, both $B+C$ and $A+F$ would produce a sum of 8.
*If $D+E$ produced a sum equal to 9, both $B+C$ and $A+F$ would produce a sum of 6.
Testing the only eligible odd number for B (3) given that $B+C$ must produce a sum of 6 or 8, the only possible option for C would be 5. This leads to a paradox when entering values for $D+E=C+F$ ($5=7$).
Given that both $B+C$ and $A+F$ must both produce equal- and even-numbered sums; this signals common parity for B and C, and common parity for A and F.
Given that the remaining options for B are 4 and 6; this signals even parity for B and C, and odd parity for A and F.
As $B+C$ must equal 8 given that $B$ cannot be 1, 2, 3, or 5, $B=6$ and $C=2$.
Given that $C=2$ and that $C+F \ne 7$, $F=3$.
Given that $D+E=5$ and that $E \ne 1$, $D=1$ and $E=4$.
$A+3=8$, so $A=5$.
Result:
$$ \begin{array}{c|lcr} & \text{1} & \text{2} & \text{3} & \text{4} &
\text{5} & \text{6} \\ \hline A & & & & & ✓ & \\ B & x & x & & & x & ✓ \\
C & & ✓ \\ D & ✓ & & & x & & x\\ E & & & & ✓\\ F & & & ✓\\ \end{array} $$
| This goes a little faster:
Start is same as yours:
Since $B+C=A+F$, we have $B+C+A+F$ is even, and since $A$ through $F$ add up to $21$, $D+E$ must be odd
But we cannot have $D+E=7$, for then $B+C=A+F=7$, and also $C+F=7$, and so $A+F=C+F$ which is impossible.
OK, so far it's the same.
But at this point in your proof you note that $D+E$ cannot be $3$,$10$, or $11$ ... but note for same reason you state it also cannot be $4$ (that is: $D+E=C+F=4$ does not work)
So: $D+E$ is odd, greater than $4$, and smaller than $10$, but is not equal to $7$, so there are only two options left: $D+E=5$ or $D+E=9$
But the latter doesn't work, for then $D+E=C+F=9$, so $A+B=3$, and $B$ is neither $1$ or $2$
So, $D+E=C+F=5$ and $A+B=11$, and since $B$ is not $5$, we have $B=6$ and $A=5$
Since $B+C=A+F$ that means $F=C+1$, and since $D+E=C+F$ and since you only have numbers $1,2,3,4$ left, that means $F=3$ and $C=2$.
So $D$ and $E$ are $1$ and $4$, and since $D$ is not $4$, we have $D=1$ and $E=4$
Note that I never used the clue that $D$ is not $6$ ....
| {
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Taylor series of $\ln\frac{1+x}{1-x}$
Let $f(x)=\ln\frac{1+x}{1-x}$ for $x$ in $(-1,1)$. Calculate the Taylor series of $f$ at $x_0=0$
I determined some derivatives:
$f'(x)=\frac{2}{1-x^2}$; $f''(x)=\frac{4x}{(1-x^2)^2}$; $f^{(3)}(x)=\frac{4(3x^2+1)}{(1-x^2)^3}$; $f^{(4)}(x)=\frac{48x(x^2+1)}{(1-x^2)^4}$; $f^{(5)}(x)=\frac{48(5x^2+10x^2+1)}{(1-x^2)^5}$
and their values at $x_0=0$:
$f(0)=0$; $f'(0)=2$; $f''(0)=0$; $f^{(3)}(0)=4=2^2$
$f^{(4)}(0)=0$; $f^{(5)}(0)=48=2^4.3$; $f^{(7)}(0)=1440=2^5.3^2.5$
I can just see that for $n$ even, $f^{(n)}(0)=0$, but how can I generalize the entire series?
| Beginning with
$$\frac{1}{1 - x} = \sum_{k = 0}^{\infty} x^{k}, $$
we can integrate to get $$\log(1 - x) = -\sum_{k = 1}^{\infty} \frac{x^{k}}{k}.$$
Also, if we plug in $(-x)$ for $x$ into the equation for $\frac{1}{1 - x}$, we get
$$\frac{1}{1 + x} = \sum_{k = 0}^{\infty}(-x)^{k}.$$
Integrating, we get
$$\log(1 + x) = \sum_{k = 1}^{\infty} \frac{(-x)^{k}}{k}.$$
Now, by properties of $\log$, $\frac{\log(1 + x)}{\log(1 - x)} = \log(1 + x) - \log(1 - x)$. So,
$$\frac{\log(1 + x)}{\log(1 - x)} = \sum_{k = 1}^{\infty} \frac{(-x)^{k}}{k} + \sum_{k = 1}^{\infty} \frac{x^{k}}{k}$$
$$= \sum_{k = 1}^{\infty} \frac{(-x)^{k} + x^{k}}{k}.$$
Note that when $k$ is odd, the terms vanish.
| {
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"timestamp": "2023-03-29T00:00:00",
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Series convergence without sigma notation Consider the following series:
$$\frac{1}{1} + \frac{10}{2} + \frac{100}{3} - \frac{37}{4} - \frac{37}{5} - \frac{37}{6} + \frac{1}{7} + \frac{10}{8} + \frac{100}{9} - \frac{37}{10} - \frac{37}{11} - \frac{37}{12} + \dots$$
This series seems to converge but I am not able to prove it. It seems impossible to write thing whole thing into summation notation. I tried to group by modulo 6, but then for example sum of $\frac{1}{6n+1}$ diverges already so that doesn't seem right. Also tried to change the 37s into powers of 10 to maybe use the comparison test but also failed.
Any suggestions?
| I presume the pattern $(1, 10, 100, -37, -37, -37)$ continues forever, so this is
$$ \sum_{n=1}^\infty \frac{a_n}{n}$$
where $$ a_n = \cases{1 & if $n \equiv 1 \mod 6$\cr
10 & if $n \equiv 2 \mod 6$\cr
100 & if $n \equiv 3 \mod 6$\cr
-37 & if $n \equiv 4,5$ or $0 \mod 6$\cr}$$
Note that $1+10+100 - 3\times 37 = 0$. Then the $6m$
-th partial sum
$$ \eqalign{S_m &= \sum_{n=1}^{6m} \frac{a_n}{n}\cr
& = \sum_{j=0}^m \frac{1}{6j+1} +
10 \sum_{j=0}^m \frac{1}{6j+2} + 100 \sum_{j=0}^m \frac{1}{6j+3}
- 37 \sum_{k=4}^6 \sum_{j=0}^m \frac{1}{6j+k}\cr
&= \frac{1}{6} \left(\Psi(m+1+1/6) - \Psi(1/6) + 10 (\Psi(m+1+2/6) - \Psi(2/6)) + 100 (\Psi(m+1+3/6) - \Psi(3/6)) - 37 (\Psi(m+1+4/6)+\Psi(m+1+5/6)+\Psi(m+1+6/6)-\Psi(4/6)-\Psi(5/6)-\Psi(6/6))\right)\cr}$$
and since $\Psi(x) = \ln(x) + O(1/x)$ as $x \to \infty$, the sum converges.
In fact, the limit is
$$ \frac{64}{3} \ln(2) - \frac{63}{4} \ln(3) +\frac{161}{36} \pi \sqrt{3}$$
| {
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Proving fraction is irreducible Example: The fraction $\frac{4n+7}{3n+5}$ is irreducible for all $n \in \mathbb{N}$, because $3(4n+7) - 4(3n+5) = 1$
and if $d$ is divisor of $4n+7$ and $3n+5$, it divides $1$, so $d=1$.
I want to know if there is some general method of finding $x, y \in \mathbb{Z}$,
so that $$x(an+b) + y(cn+d) = 1$$ when $(an+b, cn+d) = 1$, instead of trial and error,
or some quicker and easier way (for not so pretty fractions) for determining whether it is irreducible.
|
In general, for $a,b,c,d
\in\Bbb N$, the following statements are equivalent:
$(i)$ there are integers $x,y$ s.t. $x(an+b)+y(cn+d)=1$ for all $n\in \Bbb N$;
$(ii)$ $ad-bc$ divides $\gcd(a,c)$;
$(iii)$ $|ad-bc| =\gcd(a,c)$.
Note also that any of the statements $(i)$, $(ii)$, and $(iii)$ implies that $(iv)$ the rational number $\frac{an+b}{cn+d}$ is in the lowest form for all $n\in \Bbb N$.
Obviously $(ii)\iff (iii)$ because $\gcd(a,c)$ always divide $ad-bc$. In the case $ad-bc\mid \gcd(a,c)$, we can take $x=-\frac{c}{ad-bc}$ and $y=\frac{a}{ad-bc}$. So $(ii)\implies (i)$. We now prove that $(i)\implies (ii)$.
Suppose that such $x$ and $y$ exist. Then,
$$ax+cy=0\wedge bx+dy=1.$$
That is, $(x,y)$ is an integer solution to
$$\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\1\end{pmatrix}.$$
Observe that the determinant $ad-bc$ of $\begin{pmatrix}a&c\\b&d\end{pmatrix}$ cannot be $0$ (otherwise $(a,b)$ and $(c,d)$ are proportional, and so $an+b$ and $cn+d$ are also proportional). That is, the matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is invertible and
$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}a&c\\b&d\end{pmatrix}^{-1}\begin{pmatrix}0\\1\end{pmatrix}=\frac{1}{ad-bc}\begin{pmatrix}d&-c\\-b&a\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}.$$
So $(x,y)=\frac{1}{ad-bc}(-c,a)$. That is, $ad-bc\mid c$ and $ad-bc\mid a$. So $ad-bc\mid \gcd(a,c)$.
In your example, $a=4$, $b=7$, $c=3$, and $d=5$. So, $ad-bc=-1 \mid \gcd(a,c)$, and we can take $x=-\frac{c}{ad-bc}=3$ and $y=\frac{a}{ad-bc}=-4$.
I should like to mention that $(iv)$ is not equivalent to any of the statements $(i)$, $(ii)$, and $(iii)$. The rational numbers of the form $\frac{2n+1}{2n+3}$ is reduced for any $n\in \Bbb N$, but it does not meet $(i)$, $(ii)$, or $(iii)$ (i.e., $(a,b,c,d)=(2,1,2,3)$, so $\gcd(a,c)=2$, but $ad-bc=4\nmid\gcd(a,c)$). However, $(iv)$ is equivalent to the condition that for any prime divisor $p$ of $ad-bc$, there does not exist $n\in\Bbb N$ such that $p$ divides both $an+b$ and $cn+d$.
| {
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Evaluate $\int\frac{x{\rm d}x}{(1+x)\sqrt{1-x-x^2}}$ My solution
Make a Euler's substitution. Let $\sqrt{1-x-x^2}=xz-1$. Then
\begin{align*}z&=\frac{1+\sqrt{1-x-x^2}}{x},\\x&=\frac{2z-1}{z^2+1},\\
{\rm d}x&=\frac{2(1+z-z^2)}{(z^2+1)^2}{\rm d}z,\\(1+x)\sqrt{1-x-x^2}&=\frac{z(z+2)(1+z-z^2)}{(z^2+1)^2}.
\end{align*}
Thus
\begin{align*}
\int\frac{x{\rm d}x}{(1+x)\sqrt{1-x-x^2}}&=2\int\frac{2z-1}{z(z^2+1)(z+2)}{\rm d}z\\
&=2\int\left(\frac{1}{z^2+1}+\frac{1}{2(z+2)}-\frac{1}{2z}\right){\rm d}z\\
&=2\arctan z+\ln|z+2|-\ln|z|+C\\
&=2\arctan z+\ln\left|\frac{z+2}{z}\right|+C.
\end{align*}
The result is right? Who can verify it for me? Thanks in advance.
| If you want an alternative method,
as $1-x-x^2=\dfrac{5-(2x+1)^2}4,$ let $2x+1=\sqrt5\sin t$
$$\int\dfrac x{(1+x)\sqrt{1-x-x^2}}\ dx=\int\dfrac{\sqrt5\sin t-1}{\sqrt5\sin t+1}\ dt$$
$\dfrac{\sqrt5\sin t-1}{\sqrt5\sin t+1}=1-\dfrac2{\sqrt5\sin t+1}$
For the second part, use Weierstrass substitution
| {
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"url": "https://math.stackexchange.com/questions/3039444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $g(k)$ be the greatest odd divisor of $k$ show that $ 0< \sum_{k=1}^n \frac {g(k)}{k} - \frac {2n}{3} \lt \frac 23$
Prove that for all positive intergers $n$,
$$ 0< \sum_{k=1}^n \frac {g(k)}{k} - \frac {2n}{3} < \frac {2}{3}$$
Where $g(k)$ denotes the greatest odd divisor of $k$.
Here's my try:
All numbers $k$ can be written as $k= 2^ts$, for nonnegative $t$ and odd $s$, therefore if $k= 2^ts$, then $\frac {g(k)}{k} = \frac {1}{2^k}$ i.e $\frac {g(k)}{k}$ is equal to $1$ divided by the highest power of $2$ dividing $k$. I first thought of proving the inequality for $k= 2^n (n>1)$. Let $Q= \sum_{k=1}^{2^n} \frac {g(k)}{k}$, then:
$$Q = \frac {1}{2}q_1 +\frac {1}{2^2}q_2 + \cdots + \frac {1}{2^{n -1}} q_{2^{n -1}}+ \frac {1}{2^n} q_{2^n}+ 2^{n-1} $$.
$q_i$ is the number of times $\frac {1}{2^i}$ is added in the summation. It's easy to notice that $q_{2^n} =1$. $q_i$ for $0< i < 2^n$ would be equal to $2^{n-1-i}$ (comes from this $(2^i)(2(2^{n-1-i})-1)$). Then:
$$Q= 2^{n-1}+ \frac {1}{2^n} + \sum_{i=1}^{n-1} \frac{1}{2^i} \cdot 2^{n-1-i} $$
$$Q = 2^{n-1}+ \frac {1}{2^n} + 2^{n-1} \sum_{i=1}^{n-1} (\frac{1}{4})^i $$
EDITED
With some algebra we get that:
$$Q - \frac {2}{3} \cdot 2^n= - \frac {4}{3} \cdot 2^{n-1} + 2^{n-1}+ \frac {1}{2^n} + 2^{n-1} \cdot \frac {4^{n-1}-1}{4^{n-1}} \cdot \frac {1}{3} < \frac {1}{2^n} < \frac {2}{3} $$
Also:
$$ Q - \frac {2}{3} \cdot 2^n = \frac {1}{2^n} - \frac {2^{n-1}}{4^{n-1}} \cdot \frac {1}{3}= \frac {1}{2^n} - \frac {1}{2^{n-1}} \cdot \frac {1}{3}> \frac {1}{2^n} - \frac {1}{2^{n-1}} \cdot \frac {1}{2}= 0$$
Then I would get:
$$ 0< Q - \frac {2}{3} \cdot 2^n < \frac {2}{3}$$
Well, I thought proving the inequality for $k=2^n$ because I had some idea about how many times the powers of $2$ appeared. I then thought that I could go backwards with induction but I'm stuck. I would like to see some other approaches but I would also like to know if it's possible to solve the problem from the point where I am.
I'll appreciate any hints or help, thanks in advance.
| Let $v(k)$ be the largest integer $m$ such that $2^m$ divides $k$. As you noted, $\dfrac{g(k)}{k} = 2^{-v(k)}$.
So, let $S_n = \displaystyle\sum_{k = 1}^{n}\dfrac{g(k)}{k} = \sum_{k = 1}^{n}2^{-v(k)}$.
Then, we have:
\begin{align*}
S_{2n} &= \displaystyle\sum_{k = 1}^{2n}2^{-v(k)}
\\
&= \sum_{k = 1}^{n}2^{-v(2k-1)} + \sum_{k = 1}^{n}2^{-v(2k)}
\\
&= \sum_{k = 1}^{n}2^{-0} + \sum_{k = 1}^{n}2^{-(1+v(k))}
\\
&= \sum_{k = 1}^{n}1 + \dfrac{1}{2}\sum_{k = 1}^{n}2^{-v(k)}
\\
&= n + \dfrac{1}{2}S_n
\end{align*}
Also, $S_{2n+1} = \displaystyle\sum_{k = 1}^{2n+1}2^{-v(k)} = 2^{-v(2n+1)} + \sum_{k = 1}^{2n}2^{-v(k)} = 2^{0} + S_{2n} = S_{2n} + 1$.
We can now proceed by induction. Trivially, $S_1 = 1$, so $S_1 > \dfrac{2}{3}$ and $S_1 < \dfrac{4}{3}$.
Now, suppose that for some integer $n$, we have $\dfrac{2n}{3} < S_n < \dfrac{2n}{3} + \dfrac{2}{3}$.
Then, we have:
$$S_{2n} = \dfrac{1}{2}S_n + n > \dfrac{1}{2}\left(\dfrac{2n}{3}\right) + n = \dfrac{4n}{3} = \dfrac{2 \cdot 2n}{3}$$
$$S_{2n} = \dfrac{1}{2}S_n + n < \dfrac{1}{2}\left(\dfrac{2n}{3}+\dfrac{2}{3}\right) + n = \dfrac{4n}{3} + \dfrac{1}{3} < \dfrac{2 \cdot 2n}{3} + \dfrac{2}{3}$$
$$S_{2n+1} = S_{2n}+1 > \left(\dfrac{4n}{3}\right)+1 > \dfrac{4n}{3}+\dfrac{2}{3} = \dfrac{2(2n+1)}{3}$$
$$S_{2n+1} = S_{2n}+1 < \left(\dfrac{4n}{3}+\dfrac{1}{3}\right)+1 = \dfrac{2(2n+1)}{3} + \dfrac{2}{3}.$$
Hence, $\dfrac{2 \cdot 2n}{3} < S_{2n} < \dfrac{2 \cdot 2n}{3} + \dfrac{2}{3}$ and $\dfrac{2(2n+1)}{3} < S_{2n+1} < \dfrac{2(2n+1)}{3} + \dfrac{2}{3}$.
So by induction, we have that $\dfrac{2n}{3} < S_n < \dfrac{2n}{3} + \dfrac{2}{3}$ for all $n$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Simplification of rational expression gone wrong(high school math) Currently doing high school math and can't get this one right. I think I'm using entirely incorrect practices and am trying to pinpoint what it is. Could someone tell me where exactly I went wrong?
Original expression:
$$\left(\frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab}\right)\times \left(\frac{a^2}{b^3-a^2b}+\frac{1}{a+b}\right)^{-1}$$
When I simplify the equations in parentheses I get:
$$\frac{b^2-ab+a^2}{ab(a+b)}\times\frac{1}{a^{2}}$$
Which should result in:
$$\frac{b^2+a(a-b)}{a^3b(a+b)}=\frac{b^2-a}{a^3b}$$
That is of course not right. The correct answer according to the book is $\frac{b-a}{a}$
| Go slowly. Do it in blocks.
First block:
\begin{align}
\frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab}
&=\frac{b}{a(a+b)}-\frac{b-a}{b(a+b)} \\[4px]
&=\frac{b^2-a(b-a)}{ab(a+b)} \\[4px]
&=\frac{a^2-ab+b^2}{ab(a+b)}
\end{align}
Second block:
\begin{align}
\frac{a^2}{b^3-a^2b}+\frac{1}{a+b}
&=\frac{a^2}{b(b^2-a^2)}+\frac{1}{a+b} \\[4px]
&=\frac{a^2}{b(b-a)(b+a)}+\frac{1}{a+b} \\[4px]
&=\frac{a^2+b(b-a)}{b(b-a)(b+a)} \\[4px]
&=\frac{a^2-ab+b^2}{b(b-a)(b+a)}
\end{align}
Putting the pieces together:
$$
\left(\frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab}\right)\times \left(\frac{a^2}{b^3-a^2b}+\frac{1}{a+b}\right)^{-1}=
\frac{a^2-ab+b^2}{ab(a+b)}\frac{b(b-a)(b+a)}{a^2-ab+b^2}=
\frac{b-a}{a}
$$
Now you see that you managed the first block correctly, but failed in the second one. Check carefully your steps and you'll see where you went astray.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Find all the solutions of $z^2-(1+3i)z-8-i=0$ I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.
I want to find all the roots of:
$$z^2-(1+3i)z-8-i=0$$
There are two ways I tried to approach this.
*
*Quadratic formula:
$$\begin{align} z_1,z_2 &=-\frac{p}{2}\pm\sqrt{\left( \frac{p}{2}\right)^2-q} \\ & \implies z_1,z_2=\frac{1+3i}{2}\pm\sqrt{\left( \frac{(-1-3i)}{2}\right)^2-(-8-i)}\\ &= \frac{1+3i}{2} \pm\sqrt{\frac{24+10i}{4} }=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{align}$$
*Completing the square:
$$\begin{aligned} z^2-(1+3i)z-8-i &=0 \\ & \iff \left(z-\left( \frac{1+3i}{2}\right) \right)^2-8-i=\left( \frac{1+3i}{2}\right)^2 \\ & \iff u^2=\frac{24+10i}{4} \\ & \iff u=\pm\frac{\sqrt{24+10i}}{2} \\ & \iff z_{1,2}=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{aligned} $$
Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:
$$z_1=-2+i \\ z_2=3+2i$$
I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?
| Hint: Use the fact: $$ 24+10i = (5+i)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do we solve for all values of $x$? Suppose
$$(x^2 - 5)^{8} (x+1)^{-16} = 1$$
How do we solve for all values of $x$?
I think we can equate the bases to $1$ in order to get $$1 \times 1 = 1$$
$$x^2-5 = 1, x^2-5=-1$$
or
$$x + 1 = 1, x + 1 = -1$$
Could you assist me with this?
| $(x^2 - 5)^{8} (x+1)^{-16} = 1$ means
$(x^2 - 5)^{8} = (x+1)^{16}$.
Assuming you only want real and not complex answers then
$ \sqrt[8]{(x^2 - 5)^{8}} = \sqrt[8]{(x+1)^{16}}$
$|x^2 - 5| = (x+1)^2$
or either A) $x^2 - 5 = (x+1)^2$ or B) $5- x^2 = (x+1)^2$
A) $x^2 - 5 = x^2 + 2x + 1$
$2x = -6$
$x = -3$.
Verification: $(x^2 - 5)^8(x+1)^{-16}=\frac{(9-5)^8}{(-2)^{16}}= \frac {4^8}{2^{16}}=\frac {2^{16}}{2^{16}} = 1$.
B)$5- x^2 = (x+1)^2$
$5-x^2 = x^2 + 2x + 1$
$2x^2 + 2x -4 = 0$
$x^2 + x -2 = 0$
$(x +2)(x-1) = 0$ so either C) $x+2 = 0$ and $x=-2$ or $x-1=0$ and $x = 1$.
So solutions are $x = -3; x = -2;$ or $x = 1$.
Verification:
$(x^2 - 5)^8(x+1)^{-16} = \frac {(4-5)^8}{(-2 + 1)^{16}}= \frac {(-1)^8}{(-1)^{16}} = \frac 11 = 1$. so $x = -2$ is a solution.
$(x^2 - 5)^8(x+1)^{-16} = \frac {(1-5)^8}{(1 + 1)^{16}}= \frac {(-4)^8}{(2)^{16}} = \frac {4^8}{4^8} = 1$. so $x = 1$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
| \begin{align}
\sqrt[4]{28+16 \sqrt 3}=&\ \sqrt[4]{4(7+2 \sqrt {3\cdot4})}
= \sqrt[4]{4(\sqrt4+\sqrt {3})^2}\\
= &\ \sqrt{4+2\sqrt{1\cdot3}}=\sqrt{(1+\sqrt3)^2}=1+\sqrt3
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
Proving that $\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$ The following integral was proposed by Cornel Ioan Valean and appeared as Problem $12054$ in the American Mathematical Monthly earlier this year.
Prove $$\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$$
I had small tries for it, such as writting:
$$I=\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx\overset{ x\to \tan \frac{x}{2}}=-\frac12 {\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx}$$
And with Feynman's trick we obtain:
$$J(t)=\int_0^\frac{\pi}{2} \frac{x\ln(1-t\sin x)}{\sin x}dx\Rightarrow J'(t)=\int_0^\frac{\pi}{2} \frac{x}{1-t\sin x}dx$$ But I don't see a way to obtain a closed from for the above integral.
Also from here we have the following relation:
$$\int_0^1 \frac{\arctan x \ln(1+x^2)}{x} dx =\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx$$
Thus we can rewrite the integral as:
$$I=\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx -2\int_0^1 \frac{\arctan x \ln(1-x)}{x}dx$$
Another option might be to rewrite:
$$\ln\left(\frac{1+x^2}{(1-x)^2}\right)= \ln\left(\frac{1+x}{1-x}\right)+\ln\left(\frac{1+x^2}{1-x^2}\right)$$
$$\Rightarrow I= \int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x}{1-x}\right)dx+\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{1-x^2}\right)dx$$
And now to use the power expansion of the log functions to obtain:
$$\small I=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1 \frac{\arctan x}{x} \, \left(x^{2n+1}+x^{4n+2}\right)dx=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1\int_0^1 \frac{\left(x^{2n+1}+x^{4n+2}\right)}{1+y^2x^2}dydx$$
This seems like an awesome integral and I would like to learn more so I am searching for more approaches.
Would any of you who also already solved it and submitted the answer to the AMM or know how to solve this integral kindly share the solution here?
Edit:
In the meantime I found a nice solution by Roberto Tauraso here and another impressive approach due to Yaghoub Sharifi here.
| By the identity
$$
\Im~{\log^2\left(\frac{1-x}{1+\text i x}\right)}=\arctan x\log\left(\frac{1+x^2}{(1-x)^2}\right)
$$
one easily has
$$
\begin{align}
&\int_0^1\frac{\arctan x}x\log\left(\frac{1+x^2}{(1-x)^2}\right)\text d x
\\=&\Im{\int_0^1\log^2\left(\frac{1-x}{1+\text ix}\right)\frac{\text d x}{x}}
\\=&\Im{\int_0^1\log^2y\left(\frac{1+y}{(1-y) (1+y^2)}+\text i~\frac{1}{1+y^2}\right)\text d y}
\\=&\int_0^1\frac{\log^2y}{1+y^2}\text d y
\\=&\frac{\pi^3}{16}
\end{align}
$$
with the obvious substitution $y=\dfrac{1-x}{1+\text ix}$. The last equality comes directly from series expansion.
| {
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"url": "https://math.stackexchange.com/questions/3050696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
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What is the best way to factor $25x^2-121$? What is the best way to factor $25x^2-121$ ?
The methods I have been taught doesn't work for this problem. Is there an another way to solve this problem? If so, how?
| The hard way:
We want to factor, IF it can be done, as
$25x^2 -121 = (ax + b)(cx + d) = acx^2 + (bc+ad)x + bd$ where
$ac = 25; bd= -121; bc+ad= 0$.
IF this can be factored to $ac =25$ means $a,c$ can be factored as $1,25$ or $5,5$.
And $bd =121$ means $b,d$ can be factored to $\pm 1;\mp 121$ or $\pm 11, \mp 11$.
But we have $ad = -bc$ of the options, if $a=25$ then $b=1$ and we can't have $25d = -c$ as $c$ can't have anything to do with $5$.
The same issues occur if $b = 121$.
To get $ad = -bc$ it helps if $|a| = |c| = 5$ and $|b|=|d| = 11$. Then everything is nicely balanced and should cancel out.
And indeed: $(5x + 11)(5x - 11) = 5x\times 5x + 11\times 5x + 5x\times(-11) + 11\times(-11) = 5x^2 - 121$.
The easy way:
Remember the "difference of squares" rule: $a^2 - b^2 = (a+b)(a-b)$.
So $25x^2 = (5x)^2$ and $121 = 11^2$ so $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Different ways of proving that $|\sum^{\infty}_{k=1}(1-\cos(1/k))|\leq 2 $ I've found two ways of proving that
\begin{align} \left|\sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]\right|&\leq 2 \end{align}
Are there any other ways out there, for proving this?
METHOD 1
Let $k\in \Bbb{N}$, then
\begin{align} f:[ 0&,1]\longrightarrow \Bbb{R}\\&x \mapsto
\cos\left(\dfrac{x}{k}\right) \end{align}
is continuous. Then, by Mean Value Theorem, there exists $c\in [ 0,x]$ such that
\begin{align} \cos\left(\dfrac{x}{k}\right)-\cos\left(0\right) =-\dfrac{1}{k}\sin\left(\dfrac{c}{k}\right)\,(x-0), \end{align}
which implies \begin{align} \left|\sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]\right| &=\left|\sum^{\infty}_{k=1}\dfrac{x}{k}\sin\left(\dfrac{c}{k}\right)\right| \leq \sum^{\infty}_{k=1}\dfrac{\left|x\right|}{k}\left|\sin\left(\dfrac{c}{k}\right)\right|\leq \sum^{\infty}_{k=1}\dfrac{\left|x\right|}{k}\dfrac{\left|c\right|}{k}\\&\leq \sum^{\infty}_{k=1}\left(\dfrac{\left|x\right|}{k}\right)^2\leq \sum^{\infty}_{k=1}\dfrac{1}{k^2}=1+ \sum^{\infty}_{k=2}\dfrac{1}{k^2}\\&\leq 1+ \sum^{\infty}_{k=2}\dfrac{1}{k(k-1)}\\&= 1+ \lim\limits_{n\to\infty}\sum^{n}_{k=2}\left(\dfrac{1}{k-1}-\dfrac{1}{k}\right)\\&=1+ \lim\limits_{n\to\infty}\left(1-\dfrac{1}{n}\right)\\&=2, \end{align}
METHOD 2
Let $x\in [0,1]$ be fixed, then
\begin{align} \sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]&=\sum^{\infty}_{k=1}\dfrac{1}{k}\left[-k\cos\left(\dfrac{x}{k}\right)\right]^{1}_{0}=\sum^{\infty}_{k=1}\dfrac{1}{k}\int^{1}_{0}\sin\left(\dfrac{x}{k}\right)dx \\&=\sum^{\infty}_{k=1}\int^{1}_{0}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)dx \end{align}
The series $\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)$ converges uniformly on $[0,1]$, by Weierstrass-M test, since
\begin{align} \left|\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right) \right|\leq \sum^{\infty}_{k=1}\dfrac{1}{k}\left|\sin\left(\dfrac{x}{k}\right) \right|\leq \sum^{\infty}_{k=1}\dfrac{1}{k^2}. \end{align}
Hence,
\begin{align} \sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]&=\sum^{\infty}_{k=1}\int^{1}_{0}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)dx=\int^{1}_{0}\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)dx, \end{align}
and
\begin{align} \left|\sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]\right|&=\left|\int^{1}_{0}\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)dx\right|\leq\int^{1}_{0}\sum^{\infty}_{k=1}\dfrac{1}{k}\left|\sin\left(\dfrac{x}{k}\right)\right|dx \\&\leq\int^{1}_{0}\sum^{\infty}_{k=1}\dfrac{1}{k^2}dx \\&\leq 2 \end{align}
| Another method which gives a better bound than $2$ is the following: the inequality $1-\cos\, x \leq \frac {x^{2}} 2$ holds for all real $x$ and $\sum \frac 1 {k^{2}} =\frac {\pi^{2}} 6$. Use the fact that $\frac {\pi^{2}} {12}<2$. [ $1-\cos\, x - \frac {x^{2}} 2$ vanishes at $0$ and its derivative is $\sin\, x -x <0$. This gives the inequality above]. Note that LHS $<1$ in fact!
| {
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"url": "https://math.stackexchange.com/questions/3057268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Rationalizing denominator of $\frac{7}{2+\sqrt{3}}$. Cannot match textbook solution I am given this expression and asked to simplify by rationalizing the denominator:
$$\frac{7}{2+\sqrt{3}}$$
The solution is provided:
$14 - 7\sqrt{3}$
I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:
$$\frac{7}{2+\sqrt{3}} * \frac{2-\sqrt{3}}{2-\sqrt{3}}$$
=
$$\frac{14 - 7\sqrt{3}}{4}$$
Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?
| Trick to remember forever and use again and again.
$(a + b)(a-b) = a(a-b) + b(a-b)=$
$a^2 - ab + ab - b^2 = a^2 - b^2$.
So
1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$
and
2) If you ever need to get rid of a radical sign in $a+\sqrt b$ you can always multiple by $(a + \sqrt b)(a - \sqrt b) = a^2 - \sqrt b^2 = a^2 -b$.
So
3) to deradicalize a $\frac m{\sqrt a + \sqrt b} = \frac {m(\sqrt a - \sqrt b)}{(\sqrt a - \sqrt b)} = \frac {m(\sqrt a - \sqrt b)}{a - b}$.
So:
$\frac {7}{2 + \sqrt 3} = $
$\frac {7(2 - \sqrt 3)}{(2+\sqrt 3)(2 - \sqrt 3)} =$
$\frac {7(2-\sqrt 3)}{2^2 - \sqrt 3^2} =$
$\frac {7(2-\sqrt 3)}{4-3} =$
$\frac {7(2-\sqrt 3)}{1} =$
$7(2-\sqrt 3)$.
Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.
You will be using it for the REST OF YOUR LIFE!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does one evaluate the multiplication $f(2)\cdot f(3)\cdot f(4)\cdots f(15)$ by formulating? Suppose
$$f : \mathbb{Z}^+ \rightarrow \mathbb{R}$$
$$f(x) = 1-\dfrac{1}{x^2}$$
How does one evaluate the multiplication $\prod_{i=2}^{15} f(i)=f(2)\cdot f(3)\cdot f(4)\cdots f(15)$?
Here I have to see the trick that directly yields the calculation.
How come that we write this using $\Pi$ (product) notation? I'll be glad to hear your dear thoughts.
| Hint: $f(x) = \frac{x^2-1}{x^2} = \frac{(x-1)(x+1)}{x^2}$
and take product of $f(2),f(3),...,f(n)$
Look:
$$
\frac{(1)(3)}{2^2} \cdot \frac{(2)(4)}{3^2} \cdot \frac{(3)(5)}{4^2} \cdot ... \frac{(n-3)(n-1)}{(n-2)^2} \cdot \frac{(n-2)n}{(n-1)^2} \cdot \frac{(n-1)(n+1)}{n^2} = \frac{1}{2} \cdot \frac{n+1}{n} $$
and take this for your task
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! \approx \sqrt{2\pi\cdot 15}
\cdot \left(\frac{15}{e}\right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
| Another way to reason is to note that $15!$ is divisible by $2\cdot4\cdot2\cdot8\cdot2\cdot4\cdot2=2^{11}$, which means $1\square0767436\square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8\mid1000$ and $8\mid360$, the final $\square$ must be either $0$ or $8$. But it can't be $0$, since $15!$ has only three powers of $5$ (from $5$, $10$, and $15$), and those were already accounted for in the final three $0$'s of the number. Thus the final $\square$ is an $8$. Casting out $9$'s now reveals that the first $\square$ is a $3$.
Remark: It wasn't strictly necessary to determine the exact power of $2$ (namely $2^{11}$) that divides $15!$, merely that $2^6$ divides it, but it wasn't that hard to do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Verification of the approach on finding the limit of $x_n$ as $n\to\infty$ given $x_{n+1} = \sqrt[k]{5x_n}$ where $x_1 = \sqrt[k]{5}, k\in\Bbb N$
Let $x_n$ denote a sequence, $n\in\Bbb N$:
$$
\begin{cases}
x_{n+1} = \sqrt[k]{5x_n}\\
x_1 = \sqrt[k]{5}\\
k\in\Bbb N
\end{cases}
$$
Find:
$$
\lim_{n\to\infty} \sqrt[k]{x_n}
$$
I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
$$
x_1 = \sqrt[k]{5}\\
x_2 = \sqrt[k]{5\sqrt[k]{5}}\\
x_3 = \sqrt[k]{5\sqrt[k]{5\sqrt[k]{5}}}\\
\dots
$$
Rewrite as:
$$
\begin{align}
x_1 &= 5^{1\over k}\\
x_2 &= \left(5\cdot 5^{1\over k}\right)^{1\over k} = 5^{\frac{k+1}{k^2}}\\
&\cdots\\
x_n &= 5^{\frac{k^{n-1} + k^{n-2} + \cdots + 1}{k^n}}
\end{align}
$$
This is hardly readable using powers in mathjax. Rewrite:
$$
\log_5 x_n = \frac{k^{n-1} + k^{n-2} + \cdots + 1}{k^n}
$$
Nominator is a regular geometric series:
$$
1 + k + \cdots + k^{n-2} + k^{n-1} = \frac{k^n - 1}{k - 1}
$$
Thus:
$$
\frac{k^n - 1}{k - 1} \cdot \frac{1}{k^n} = \frac{k^n}{(k - 1)k^n} - \frac{1}{(k-1)k^n}\tag 1
$$
Since $a^x$ is continuous we may consider the limit of $(1)$:
$$
\lim_{n\to\infty} \left(\frac{k^n}{(k - 1)k^n} - \frac{1}{(k-1)k^n}\right) = \frac{1}{k-1}
$$
Which implies:
$$
\lim_{n\to\infty}x_n = \sqrt[k-1]{5}
$$
Is it the right way to approach this problem? Could it be simplified? Thank you!
| We have $x_2 = \sqrt[k]{5\sqrt[k]{5}} \ge \sqrt[k]{5} = x_1$. If we assume that $x_{n} \ge x_{n-1}$, it follows
$\sqrt[k]{5x_{n}} \ge \sqrt[k]{5x_{n-1}}$, or $x_{n+1} \ge x_n$. Induction implies that $(x_n)_n$ is increasing.
We have $x_1 = \sqrt[k]{5} \le \sqrt[k-1]{5}$. If we assume $x_n \le \sqrt[k-1]{5}$, we get $$x_{n+1} = \sqrt[k]{5x_n} \le \sqrt[k]{5\sqrt[k-1]{5}} = \left(5\cdot 5^{\frac1{k-1}}\right)^{1/k} = \sqrt[k-1]{5}$$
Induction implies that $x_n \le \sqrt[k-1]{5}, \forall n\in\mathbb{N}$.
The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = \lim_{n\to\infty} x_n$.
Letting $n\to\infty$ in the relation $x_{n+1} = \sqrt[k]{5x_n}$ gives
$$L = \sqrt[k]{5L} \implies L(L^{k-1}-5) = 0$$
It cannot be $L = 0$ because $x_n \ge x_1 =\sqrt[k]{5}, \forall n\in\mathbb{N}$. Therefore $L = \sqrt[k-1]{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that a sequence is Cauchy. Show that $\left( x_{n}\right) $ is a Cauchy sequence, where
$$
x_{n}=\frac{\sin1}{2}+\frac{\sin2}{2^{2}}+\ldots+\frac{\sin n}{2^{n}}.
$$
We try to evaluate $\left\vert x_{n+p}-x_{n}\right\vert $ and to show that
this one is arbitrary small. So
\begin{align*}
\left\vert a_{n+p}-a_{n}\right\vert & =\left\vert \frac{\sin\left(
n+1\right) }{2^{n+1}}+\frac{\sin\left( n+2\right) }{2^{n+2}}+\ldots
+\frac{\sin\left( n+p\right) }{2^{n+p}}\right\vert \\
& \leq\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\ldots+\frac{1}{2^{n+p}}.
\end{align*}
Now, $\left( \frac{1}{2^{n+j}}\right) $ converge all to zero. Hence
$\left\vert a_{n+p}-a_{n}\right\vert \rightarrow0$.
Does it work this argument according to the definition of Cauchy sequence
$$
\forall\varepsilon>0,
\quad
\exists n_{\varepsilon}\in
\mathbb{N}
,
\quad
\forall n\in
\mathbb{N}
,
\quad
\forall p\in
\mathbb{N}
:n,p\geq n_{\varepsilon}\Rightarrow\left\vert a_{n+p}-a_{n}\right\vert
<\varepsilon?
$$
My question was born from the fact that for the sequence
$$
a_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n},
$$
we have that
\begin{align*}
\left\vert a_{n+p}-a_{n}\right\vert & =\left\vert \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+p}\right\vert \\
& \leq\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+p}
\end{align*}
and also all sequences $\left( \frac{1}{n+j}\right) _{n}$ converge to zero,
but this time $\left( a_{n}\right) $ is not a Cauchy sequence.
| It is not sufficient that all $\left( \frac{1}{2^{n+j}}\right)$ converge to zero. A correct argument would be that
$$
\left\vert a_{n+p}-a_{n}\right\vert \le
\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\ldots+\frac{1}{2^{n+p}}
= \frac{1}{2^{n}} \left( \frac 12 + \frac 14 + \ldots + \frac{1}{2^{p}}\right) < \frac{1}{2^{n}}
$$
becomes arbitrary small for large $n$ and arbitrary $p$.
That won't work for the harmonic series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $ Show that if $a,b,c > 0$, such that $ab + bc + ca = 1$, then the following inequality holds:
$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $$
What I have tried is firstly using the inequality:
$\frac{x^2}{a} + \frac{y^2}{b} \geq \frac{(x + y)^2}{a + b}$, for any $x, y$ and any $a,b > 0$.
Using this inequality we obtain $\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} \geq \frac{a + b + c}{2}$, and then we have:
$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq \frac{a + b + c}{2} + \frac{36}{a + b + c} = \frac{(a + b + c)^2 + 72)}{2(a + b + c)} $$.
Using $ab + bc + ca = 1$, we would then have to prove that:
$$a^2 + b^2 + c^2 + 74 - 40(a + b + c) \geq 0 $$ and then I tried replacing in this inequality $c = \frac{1 - ab}{a + b}$, but I didn't get anything nice.
I also tried rewriting the lhs:
$$\frac{a^2}{b + c} = \frac{a^2(ab + bc + ca)}{b + c} = a^3 + \frac{a^2bc}{b + c}$$
And this would result in: $a^3 + b^3 + c^3 + abc(\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}) + \frac{36}{a + b + c} \geq 20$, but I didn't know how to continue from here.
Do you have any suggestions for this inequality?
| Let $p=a+b+c,\,q=ab+bc+ca,$ we write your inequality as
$$ \sum \frac{a^2}{b+c} + \frac{36(ab+bc+ca)} {a+b+c}\geqslant 20\sqrt{ab+bc+ca},$$
or
$$(a+b+c) \sum \frac{a}{b+c} - (a+b+c) + \frac{36(ab+bc+ca)} {a+b+c}\geqslant 20\sqrt{ab+bc+ca}.$$
Use
$$ \sum \frac{a}{b+c} \ge \frac{a^2+b^2+c^2}{ab+bc+ca} = \frac{p^2}{q}-2.$$
We need to prove
$$p\left(\frac{p^2}{q}-2\right) - p + \frac{36q} {p}\geqslant 20\sqrt{q},$$
equivalent to
$$\frac{(p^4-3p^2q+36q^2)^2}{p^2q^2} \geqslant 400q,$$
or
$$\frac{(p^4+2p^2q+81q^2)(p^2-4q)^2}{p^2q^2} \geqslant 0.$$
Equality occur when $a=b=1,\,c=0.$ The proof is completed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
A question about the number 541456 I found that the concatenation in base 10 of $2^{541456}-1$ and $2^{541455}-1$, gives a probable prime.
I also found about number $541456$ that:
$$5\cdot(5^2+4^2+1+4^2+5^2+6^2)=(5^3+4^3+1+4^3+5^3+6^3)$$
the equation $a\cdot(2\cdot a^2+2\cdot b^2+c^2+d^2)=(2\cdot a^3+2\cdot b^3+c^3+d^3)$ besides $a=5, b=4, c=6, d=1$ has other non trivial solutions with a,b,c,d positive integers?
| Your equation $a*(2*a^2+2*b^2+c^2+d)=(2*a^3+2*b^3+c^3+d)$ is equivalent to
$$2ab^2+ac^2+ad=2b^3+c^3+d$$
I find the below solutions assuming all of the variables are in the range $1$ to $9$. I just did a search.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Least Squares solution for a symmetric singular matrix I want to solve this system by Least Squares method:$$\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\5\\-2\end{pmatrix} $$ This symmetric matrix is singular with one eigenvalue $\lambda1 = 0$, so $\ A^t\cdot A$ is also singular and for this reason I cannot use the normal equation: $\hat x = (A^t\cdot A)^{-1}\cdot A^t\cdot b $.
So I performed Gauss-Jordan to the extended matrix to come with $$\begin{pmatrix}1 & 2 & 3\\\ 0 & 1 & 2 \\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\3\\-1\end{pmatrix} $$
Finally I solved the $\ 2x2$ system: $$\begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}1\\3\end{pmatrix} $$ taking into account that the best $\ \hat b\ $ is $\begin{pmatrix}1\\3\\0\end{pmatrix}$
The solution is then $\ \hat x = \begin{pmatrix}-5\\3\\0\end{pmatrix}$
Is this approach correct ?
EDIT
Based on the book 'Lianear Algebra and its applications' from David Lay, I also include the Least Squares method he proposes: $(A^tA)\hat x=A^t b $
$$A^t b =\begin{pmatrix}5\\9\\13\end{pmatrix}, A^tA = \begin{pmatrix}14 & 20 & 26 \\
20 & 29 & 38 \\
26 & 38 & 50\end{pmatrix}$$
The reduced echelon from the augmented is:
$$ \begin{pmatrix}14 & 20 & 26 & 5 \\
20 & 29 & 38 & 9 \\
26 & 38 & 50 & 13 \end{pmatrix} \sim \begin{pmatrix}1 & 0 & -1 & -\frac{35}{6} \\
0 & 1 & 2 & \frac{13}{3} \\
0 & 0 & 0 & 0
\end{pmatrix} \Rightarrow \hat x = \begin{pmatrix}-\frac{35}{6} \\ \frac{13}{3} \\ 0 \end{pmatrix}$$ for the independent variable case that $z=\alpha , \alpha=0 $
| Since the matrix has the eigenvector $\pmatrix{1&-2&1\cr}^t$ with eigenvalue $0$, one has
$$\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x+t\\y-2t\\z+t\end{pmatrix}$$
for all $t$, so there is a least squares solution with $z=0$, which makes it a least squares solution for
$$\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x\\y\\0\end{pmatrix} \approx\begin{pmatrix}1\\5\\-2\end{pmatrix} \text{ or, equivalently, } \begin{pmatrix}1 & 2 \\\ 2 & 3 \\\ 3 & 4 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} \approx\begin{pmatrix}1\\5\\-2\end{pmatrix}$$
That makes it a regular least squares problem with solution $\pmatrix{x&y\cr}^t=\pmatrix{-35/6&13/3\cr}^t$, so the solutions for the orig1nal problem are $$\begin{pmatrix}-\frac{35}6 +t\\\frac{13}3-2t\\t\end{pmatrix}$$
| {
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"url": "https://math.stackexchange.com/questions/3062701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my working:
$\frac{\sqrt{8}}{1-\sqrt{3x}}$ = $\frac{\sqrt{8}}{1-\sqrt{3x}}$ * $\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$ = $\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$ = $\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$ = $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Is $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$?
| The second equality is where you mess up - note that
$$\sqrt{8}\cdot(1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\cdot\sqrt{3x}$$
by the distributive property of real numbers
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 5
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Simplify $\sqrt[4]{\frac{162x^6}{16x^4}}$ is $\frac{3\sqrt[4]{2x^2}}{2}$ (I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)
I'm asked to simplify $\sqrt[4]{\frac{162x^6}{16x^4}}$ and am provided the text book solution $\frac{3\sqrt[4]{2x^2}}{2}$.
I arrived at $\frac{3\sqrt[4]{2x^6}}{2x^4}$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.
Here is my working:
$\sqrt[4]{\frac{162x^6}{16x^4}}$ = $\frac{\sqrt[4]{162x^6}}{\sqrt[4]{16x^4}}$
Denominator: $\sqrt[4]{16x^4}$ I think can be simplified to $2x^4$ since $2^4$ = 16
Numerator: $\sqrt[4]{162x^6}$ I was able to simplify (or over complicate) to $3\sqrt[4]{2}\sqrt[4]{x^6}$ since:
$\sqrt[4]{162x^6}$ = $\sqrt[4]{81}$ * $\sqrt[4]{2}$ * $\sqrt[4]{x^6}$ = $3 * \sqrt[4]{2} * \sqrt[4]{x^6}$
Thus I got:
$\frac{3\sqrt[4]{2}\sqrt[4]{x^6}}{2x^4}$ which I think is equal to $\frac{3\sqrt[4]{2x^6}}{2x^4}$ (product of the radicals in the numerator).
How ca I arrive at the provided solution $\frac{3\sqrt[4]{2x^2}}{2}$?
| You made an error:
$$\sqrt[4]{16x^4} = 2\vert x\vert$$
because $\sqrt[4]{16x^4} = \sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get
$$\frac{3\sqrt[4]{2x^6}}{2\vert x\vert} = \frac{3\sqrt[4]{2x^4x^2}}{2x} = \frac{3\vert x\vert\sqrt[4]{2x^2}}{2\vert x\vert} = \frac{3\sqrt[4]{2x^2}}{2}$$
As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).
$$\sqrt[4]{\frac{162x^6}{16x^4}} = \sqrt[4]{\frac{2\cdot3^4x^2}{2^4}} = \frac{3\sqrt[4]{2x^2}}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Logarithm rules i.e. $-1/2 \cdot \log_2(2/9)$ to $2/9\cdot \log_2(2)$? I'm struggling to understand the flow of calculation as shown in the picture below.
It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.
Thank you! :)
| I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.
\begin{align}
&1 - \frac{-\frac{1}{3}\log_2\left(\frac{1}{3}\right) - \frac{1}{2}\log_2\left(\frac{2}{9}\right)}{-\sum\limits_{s = 1}^9 \frac{1}{9}\log_2\left(\frac{1}{9}\right)}\\\\
&\textrm{Denominator: You're just adding the same thing 9 times}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(\left(\frac{1}{3}\right)^{-1}\right) + \frac{1}{2}\log_2\left(\left(\frac{2}{9}\right)^{-1}\right)}{-9 \cdot \frac{1}{9}\log_2\left(\frac{1}{9}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{-\log_2\left(\frac{1}{9}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{\log_2\left(\left(\frac{1}{9}\right)^{-1}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{\log_2\left(9\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\left(\log_2\left(9\right) - \log_2\left(2\right)\right)}{\log_2\left(9\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(9\right) - \frac{1}{2}}{\log_2\left(9\right)}\\\\
\end{align}
For this to be equal to the second step, you'd need $\frac{1}{2}\log_2\left(9\right) - \frac{1}{2} = \frac{2}{9}\log_2\left(2\right) = \frac{2}{9}$ to be true. It's not: $\frac{1}{2}\log_2\left(9\right) - \frac{1}{2} \approx 1.1$ and $\frac{2}{9} \approx 0.2$.
| {
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Prove Fibonacci Sequence Property: $x^2_n + x^2_{n+1}=x_{2n+1}$ For Fibonacci Sequence, We know that the recursive difference equation is:
$$
x_{n+2} = x_n + x_{n+1}\ \ \ \ n\ \geq 0
$$
And that the closed form solution is:
$$
x_n = \frac{1}{\sqrt{5}}\left[ \left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n} \right]\ \ \ \ n \geq 0
$$
But how to prove that this fibonacci property is true using this information?:
$$x^2_n + x^2_{n+1}=x_{2n+1}$$
Hint: substitute the closed form expression for the fibonacci sequence into difference equation and verify that its true.
well...I've tried this at least 3 different ways of doing exactly as hinted without much luck...the equation always blows up on me. Any ideas how to prove it?
| The 'standard proof' uses strong induction, and I'll leave you to figure that out.
Let $\varphi$ and $\psi$ represent the two roots of $x^2-x-1=0$ where $\varphi > \psi$ (i.e. $\varphi,\psi=\frac{1\pm \sqrt{5}}{2}$). The given closed form solution is $$F_n=\frac{\varphi ^n-\psi ^n}{\sqrt{5}}$$ so we know $$F_n^2=\frac{\varphi ^{2n}-2(\varphi \psi)^n+\psi ^{2n} }{5}$$ and $$F_{n+1}^2=\frac{\varphi ^{2n+2}-2(\varphi \psi )^{n+1}+\psi ^{2n+2}}{5}$$ so $$F_n^2+F_{n+1}^2=\frac{\varphi ^{2n}-2(\varphi \psi)^n+\psi ^{2n}+\varphi ^{2n+2}-2(\varphi \psi )^{n+1}+\psi ^{2n+2} }{5}$$But we know that $\varphi \psi =-1$, so either $(\varphi\psi)^n=1$ and $(\varphi\psi)^{n+1}=-1$ or $(\varphi\psi)^n=-1$ and $\varphi\psi)^{n+1}=1$. Either way, the two terms cancel each other, so$$F_n^2+F_{n+1}^2=\frac{\varphi^{2n}(1+\varphi ^2)+\psi ^{2n}(1+\psi^2)}{5}$$It can be easily seen that $\frac{1+\varphi^2}{\sqrt{5}}=\varphi$ and $\frac{1+\psi^2}{\sqrt{5}}=-\psi$ so the above becomes \begin{align*}F_n^2+F_{n+1}^2&=\frac{\varphi^{2n}(1+\varphi ^2)+\psi ^{2n}(1+\psi^2)}{5}\\ &=\frac{\varphi^{2n}\cdot \varphi\sqrt{5} + \psi^{2n}\cdot (-\psi\sqrt{5})}{5} \\ &=\frac{\sqrt{5}(\varphi^{2n+1}-\psi^{2n+1})}{5} \\ &=\frac{\varphi^{2n+1}-\psi^{2n+1}}{\sqrt{5}} \\ &= F_{2n+1}\end{align*}completing the proof.
| {
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Simplify $\sqrt{\dfrac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}$ into $\dfrac{\sqrt{3}}{3}$ I am on the final question of a textbook chapter on radicals and this question feels more challenging, perhaps that's the idea. If you view my post history I typically make a effort to provide some working to simplify the expression to an extent, but here I am very confused about where to go or my first steps.
I am to simplify:
$$\sqrt{\frac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}$$
The solution is $\displaystyle \frac{\sqrt{3}}{3}$
The expression makes me "feel" like there is a rule when dividing radicals with the same radicand but different index' with different index'. Is that true? In this case, how to divide $\frac{\sqrt[3]{64}}{\sqrt{64}}$? I know that the 3rd root and sq roots are 4 and 8 which would leave me with 1/2. using a calculator I can see that the 4th root of $256$ is 4 but I think I'm to arrive at the solution without a calculator.
Is there a prescribed approach or order of operations to simplifying an expression like this?
How can I arrive at $\dfrac{\sqrt{3}}{3}$
| $ 256 $ is a perfect square because $ 16 \cdot 16 = 16^2 = 256 $. Thus $ \sqrt{256} = 16 $. The expression $$ \sqrt{\frac{\sqrt[3]{64} + \sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}} $$ evaluates to $$ \sqrt{\frac{4 + 4}{8 + 16}} $$ which simplifies to $$ \sqrt{\frac{8}{24}} $$ Reduce the fraction inside the radical by dividing both the numerator and denominator by the greatest common divisor that divides both $ 8 $ and $ 24 $, which is $ 8 $. $$ \sqrt{\frac{8/8}{24/8}} = \sqrt{\frac{1}{3}} $$
Now from one of the basic properties involving radicals, $ \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} $. Using this property,
$$ \sqrt{\frac{1}{3}} = \frac{\sqrt{1}}{\sqrt{3}} = \frac{1}{\sqrt{3}} $$
"Rationalize" the denominator by multiplying the numerator and denominator by $ \sqrt{3} $ to achieve the desired result.
$$ \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3} $$
| {
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Calculating $\int \frac{dx}{(x^2+1)^3}$ integral Would anyone help me calculate the following integral? $\int \frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $\int \frac{dx}{(x^2+1)^2}$ like that:
$\int \frac{dx}{(x^2+1)^2} = \int \frac{x^2+1-x^2}{(x^2+1)^2}dx = \int \frac{1}{x^2+1}dx - \int \frac{x^2}{(x^2+1)^2}dx = $
$= \Biggr\rvert \begin{equation}
\begin{split}
& u = x \quad v' =\frac{x}{(x^2+1)^2} =\frac{1(x^2+1)'}{2(x^2+1)^2}\\
& u' = 1 \quad v = -\frac{1}{2} \frac{1}{x^2+1}
\end{split}
\end{equation} \Biggr\rvert$
$= \arctan x - (-x\frac{1}{2}\frac{1}{x^2+1} + \frac{1}{2} \int \frac{dx}{x^2+1})$
$= \arctan x + \frac{x}{2(x^2+1)} - \frac{1}{2}\arctan x + C = \frac{1}{2}\arctan x + \frac{x}{2(x^2+1)} + C$
Thank you.
| I used:
$$p=x^2+1\to dx =\frac{1}{2\sqrt{p-1}}$$
which leads to:
$$\int{\frac{dx}{(x^2+1)^3}=\frac12 \int{\frac{dp}{p\sqrt{p-1}}}}$$
Then the substitution $$q=\sqrt{p-1}\to dp=2q\space dq$$
Leads to $$\frac12\int\frac{dp}{p\sqrt{p-1}}=\frac12\int\frac{2q}{q(q^2+1)}dq=\int\frac{dq}{q^2+1}$$
Which is resolved simply.
| {
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Minimum value of a linear program $x$ and $y$ are real numbers such that $ x\ge 0$ and $y \ge 0$.
If $$ x + y \le 5$$
$$ x + 2y \ge 8$$
Then what is the minimum value of $5x+y$?
There is something wrong in my approach.
I write the first inequality as $$ 5 \ge x + y$$
Adding with the second inequality yields, $$5 + x + 2y \ge x + y + 8$$
Or $$ y \ge 3$$ Thus $$ 9y \ge 27$$
Multiplying the second inequality by $5$ we obtain $$ 5x + 10y \ge 40$$ $$9y \ge 27$$
Hence $5x+y \ge 13$
| Our conditions give an interior of $\Delta ABC$,where $A(2,3)$, $B(0,5)$ and $C(0,4).$
The system $x=0$ and $x+2y=8$ gives $C(0,4)$;
The system $x=0$ and $x+y=5$ gives $B(0,5)$
and the system $x+y=5$ and $x+2y=8$ gives $A(2,3).$
Let $f(x,y)=5x+y.$
Thus, $$\min_{x\geq0,y\geq0,x+y\leq5,x+2y\geq8}f=\min\{f(0,5),f(0,4),f(2,3)\}=f(0,4)=4.$$
| {
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Why do we divide by $ \sqrt{A^2 + B^2} $ to convert a function from the standard form to the normal form? To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ \cos \omega \ x + \sin \omega \ y = p $, we first divide by $ \sqrt{A^2 + B^2} $. Why?
For example, to convert the equation $ \sqrt 3 \ x + y - 8 = 0 $, we divide by $ \sqrt{A^2 + B^2} = 2 $ to get $ \frac{\sqrt 3}{2} \ x + \frac 12 \ y = 4 $ and then solve for $ \cos \omega = \frac{\sqrt 3}{2} $ and $ \sin \omega = \frac 12 $to find the value of $ \omega $.
I understand that we can't possibly convert this directly, since there is no $ \omega $ for which $ \cos \omega = \sqrt 3 $--but why do we specifically divide by $ \sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ \cos \omega $ and $ \sin \omega $ between 0 and 1 and then solve it?
| Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/\sqrt{A^2+B^2}$ and $b=B/\sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.
| {
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Find closed form of sum of fraction of binomial coefficients can somebody give me a hint for this exercise, where I have to find the specific closed form?
$\sum_{k=0}^m \frac{\binom{m}{k}}{\binom{n}{k}}, m,n\in\mathbb{N}$ and $m\leq n$
What I have done so far:
$\sum_{k=0}^m \frac{\binom{m}{k}}{\binom{n}{k}} = \sum_{k=0}^m \frac{\frac{m!}{(m-k)!\cdot k!}}{\frac{n!}{(n-k)!\cdot k!}} = \sum_{k=0}^m \frac{m!}{n!}\cdot \frac{(n-k)!}{(m-k)!} = \frac{m!}{n!}\cdot \sum_{k=0}^m \frac{(n-k)!}{(m-k)!}$
Look at example 1: $n=8, m=5$
$\frac{5!}{8!}\cdot(\frac{8!}{5!} + \frac{7!}{4!} +\frac{6!}{3!} +\frac{5!}{2!} + \frac{4!}{1!} +\frac{3!}{0!}) = \\\frac{5!}{8!} \cdot (8\cdot7\cdot6+7\cdot6\cdot5+6\cdot5\cdot4+5\cdot4\cdot3+4\cdot3\cdot2+3\cdot2\cdot1) = \\
\frac{5!}{8!} \cdot (336+210+120+60+24+6) = \frac{5!}{8!}\cdot 756 = 2.25$
I can't find a pattern.
Edit 1: Maybe there is an approach with recursion.
Edit 2: Okay I found the solution.
$\sum_{k=0}^m \frac{m!}{n!}\cdot \sum_{k=0}^m \frac{(n-k)!}{(m-k)!}= \frac{m!}{n!}\cdot\frac{1}{4}\cdot t\cdot(t+1)\cdot(t+2)\cdot(t+3)$ with $t=(n-m)!$
Edit 3: This formula works well for example 1, but fails for example 2: $n=9, m=3$
$\frac{3!}{9!}\cdot(\frac{9!}{3!} + \frac{8!}{2!} +\frac{7!}{1!} +\frac{6!}{0!}) = \\\frac{3!}{9!} \cdot (9\cdot8\cdot7\cdot6\cdot5\cdot4+8\cdot7\cdot6\cdot5\cdot4\cdot3+7\cdot6\cdot5\cdot4\cdot3\cdot2+6\cdot5\cdot4\cdot3\cdot2\cdot1) = \\
\frac{3!}{9!} \cdot (60480+20160+5040+720) = \frac{3!}{9!}\cdot 86400 = 1.428$
So I have still no general solution. Can someone help?
| Let
$$S(m,n):=\sum_{k=0}^m \frac{\binom{m}{k}}{\binom{n}{k}}$$
We are going to prove:
$$S(m,n)=\frac{n+1}{n+1-m}.\tag{1}$$
Obviously (1) is valid for $m=0$ and arbitray $n\ge 0$. Further, if (1) is valid for $(m-1,n-1)$ it is valid for $(m,n)$ as well:
$$
S(m,n):=1+\sum_{k=1}^m \frac{\frac mk\binom{m-1}{k-1}}{\frac nk\binom{n-1}{k-1}}
=1+\frac{m}{n} S(m-1,n-1)\\
\stackrel{I.H.}{=}1+\frac{m}{n}\frac{n}{n+1-m}=\frac{n+1}{n+1-m}.
$$
Thus by induction (1) is valid for arbitrary $0\le m \le n$.
| {
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If $z=\cos\theta + i\sin \theta$ prove $\frac{z^2-1}{z^2+1}=i\tan\theta$
If $z=\cos\theta + i\sin \theta$ prove $$\frac{z^2-1}{z^2+1}=i\tan\theta$$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$\frac{(\cos\theta + i\sin \theta)^2-1}{(\cos\theta + i\sin \theta)^2+1}$$
$$\frac{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)-1}{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)+1}$$
$$\frac{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta) -1}{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta)+1}$$
$$\frac{\cos2\theta + i\sin 2\theta -1}{\cos2\theta + i\sin 2\theta +1}$$
I understand how I can do it with using $z=e^{i \theta}$, however I want to solve it using double angle identities.
| $\cos \theta +i\sin \theta =e ^{i \theta}$. Then
$$\begin{aligned}
\frac{z^2-1}{z^2+1}&=\frac{z-1/z}{z+1/z}\\
&=\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}\\
&=\frac{2 i \sin \theta}{2 \cos \theta}\\
&=i\tan\theta
\end{aligned}$$
| {
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Precalc Trig Identity, verify: $1 + \cos(x) + \cos(2x) = \frac 12 + \frac{\sin(5x/2)}{2\sin(x/2)}$ Working with LHS:
I've tried using the sum to product trig ID to get:
$1 + 2\cos(3x/2)\cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 - x)$ and using sum identity, but this just makes things even messier.
I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum.
| For $\sin\frac{x}{2}\neq0$ we obtain:
$$1+\cos{x}+\cos2x=\frac{2\sin\frac{x}{2}+2\sin\frac{x}{2}\cos{x}+2\sin\frac{x}{2}\cos2x}{2\sin\frac{x}{2}}=$$
$$=\frac{2\sin\frac{x}{2}+\sin\frac{3x}{2}-\sin\frac{x}{2}+\sin\frac{5x}{2}-\sin\frac{3x}{2}}{2\sin\frac{x}{2}}=\frac{1}{2}+\frac{\sin\frac{5x}{2}}{2\sin\frac{x}{2}}.$$
I used the following formula.
$$\sin\alpha\cos\beta=\frac{1}{2}(\sin(\alpha+\beta)+\sin(\alpha-\beta)).$$
For example,$$2\sin\frac{x}{2}\cos{x}=2\cdot\frac{1}{2}\left(\sin\left(\frac{x}{2}+x\right)+\sin\left(\frac{x}{2}-x\right)\right)=\sin\frac{3x}{2}-\sin\frac{x}{2}.$$
| {
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If $x$ and $y$ are acute, and $\sin y = 3 \cos (x+y) \sin x$, then find the maximum value of $\tan y$
Given $x,y$ are acute angles such that
$$\sin y = 3 \cos(x+y)\sin x$$
Find the maximum value of $\tan y$.
Attempt:
We have
$$\begin{aligned} 3(\cos x \cos y - \sin x \sin y) \sin x & = \sin y \\ 3 \cos x \sin x - 3 \sin^2 x \tan y & = \tan y \\ 3 \cos x \sin x & = \tan y(1 +3 \sin^2 x) \\ \tan y & = \dfrac{3 \sin x \cos x} {1+3 \sin^2 x} \end{aligned}$$
Now, how about the next step? Or maybe I did some mistakes?
| $$\tan y=\dfrac{3\tan x}{1+4\tan^2x}$$
$$\iff(4\tan y)\tan ^2x-3\tan x+\tan y=0$$ which is a Quadratic Equation $\tan x$
As $\tan x$ is real, the discriminant must be $\ge0$
i.e., $$3^2-4(4\tan y)\ge0\tan y\iff\tan^2y\le\dfrac9{16}\iff-\dfrac34\le\tan y\le\dfrac34$$
| {
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Geometry problem related to circle, triangles. Given acute triangle $\triangle ABC$ satisfying $|\overline{AB}| \ne |\overline{AC}|$. Let $D,E$, respectively, be the midpoints of $\overline{AB}, \overline {AC}$. Let $Q, P$ be the intersections of $(\triangle ADE)$ and $(\triangle BCD)$, $(\triangle ADE)$ and $(\triangle BCE)$, respectively. Prove that $|\overline{AP}| = |\overline{AQ}|$.
I have already tried using radical axis but still cannot figure out the solution. Please help me with this. Thanks.
| We shall use Barycentric Coordinates with reference triangle $ABC$, that is $P=(x,y,z) \stackrel{def}{\iff} \vec P=x\vec A+y\vec B +z \vec C$ and $x+y+z=1$. Also, $(x:y:z)=(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z})$. Let $a=BC,\ b=CA,\ c=AB.$
The general equation of a circle(G.E.C.) is $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0.$
Let $T=PB\cap QC$. $AP=AQ\iff\angle AEP=\angle ADQ \iff \angle PBC= \angle QCB \iff K$ lies on perpendicular bisector of $BC$.
$A=(1:0:0),\ D=(1:1:0),\ E=(1:0:1)$. Plugging into G.E.C., we obtain $(ADE): -a^2yz-b^2zx-c^2xy+\frac{(x+y+z)}{2}(c^2y+b^2z)=0$. Also, $B=(0:1:0)$ and $C=(0:0:1)$. Therefore, $(BEC): -a^2yz-b^2zx-c^2xy+\frac{(x+y+z)}{2}(b^2x)=0$.
Without loss of generality, let $P=(1:y_1:z_1)$. Plugging $P$ into $(ADE)$ and $(BCE)$ and solving for $z_1$, $z_1=\frac{b^2-c^2}{2a^2}$. By symmetry, if $Q=(1:y_2:z_2)$, $y_2=\frac{c^2-b^2}{2a^2}$.
Therefore, $T=PB\cap QC=(1:\frac{c^2-b^2}{2a^2}:\frac{b^2-c^2}{2a^2})=(2a^2:c^2-b^2:b^2-c^2)$. Clearly, $T$ satisfies the equation of perpendicular bisector of $BC$ :
$a^2(y-z) +x(b^2-c^2)=0$
$\blacksquare$
| {
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Counting the ways to form 4 different teams We have 40 players and we need to form 4 teams. For each team is necessary to indicate explicitly 6 players and 4 reserves. How many ways can teams be formed?
My attempt: first we can start to calculate in how many ways we can choose 6 players and 4 reserves. Since a distinction is made between players and reserves, I think we should consider it in the calculation (in the sense that, otherwise, it would have been simply 11 players). I think that we can calculate this number by multiplicate $\binom{40}{6}$ ways to choose the players and $\binom{34}{4}$ ways to choose the reserves. Now, the multiplication produces a very large number and makes me believe that I'm wrong (because then I need to calculate in how many ways we can assign one of the combinations to a team). Am I wrong? Is this a correct reasoning?
| You're on the right track.
The number of ways to pick Team 1 is equal to $\binom{40}6\cdot\binom{34}4$. After that, the number of ways to pick out Team 2 is $\binom{30}6\cdot \binom{24}4$. And so on.
Ultimately, the number of ways to pick out the full $4$ teams is
$$
\binom{40}6\cdot\binom{34}4\cdot\binom{30}6\cdot\binom{24}4\cdot\binom{20}6\cdot\binom{14}4\cdot\binom{10}6\cdot\binom{4}4\\
= \frac{40!}{(6!)^4\cdot (4!)^4}
$$
(Also known as the multinomial coefficient $\binom{40}{6, 4, 6, 4, 6, 4, 6, 4}$.)
However, we don't care which team is Team 1 and which team is Team 4. We just care which four teams are picked. Since the same four teams can be picked out in $4!$ ways, the total number of different team compositions is
$$
\frac{40!}{(6!)^4\cdot (4!)^5}
$$
| {
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Find the convergence of $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ I want to find what the series $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ converges to exactly or show that it diverges.
By taking the partial sum of the series $S_N$ = $ \sum_{n=1}^{N} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ then $S_N = 2^\frac{1}{3} - 1 + 3^\frac{1}{3} - 2^\frac{1}{3} +4^\frac{1}{3}-3^\frac{1}{3} + ... + (N+1)^\frac{1}{3} - N^\frac{1}{3}$
And at the end I'm left with $S_N = -1 + (N+1)^\frac{1}{3}$ and $\lim_{N \to \infty} S_N = -1 + \infty= \infty $
So $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3} = \infty$
Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?
Thank you in prior.
| Another idea:
$$
1=(n+1)-n = ((n+1)^{1/3}-n^{1/3})((n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}).
$$
Hence, roughly speaking, $(n+1)^{1/3}-n^{1/3}$ grows like $n^{-2/3}$. Hence, asymptotically, I'd expect $\sum_n (n+1)^{1/3}-n^{1/3} \sim \sum_n n^{-2/3}$, which is well-known to be diverging.
| {
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Let $a;b;c\in R+$ such that $a+\frac{b}{16}+\frac{c}{81}\le \:3;\frac{b}{16}+\frac{c}{81}\le 2;c\le 81$. Find maxima of $A$ Let $a;b;c\in R+$ such that $a+\frac{b}{16}+\frac{c}{81}\le \:3;\frac{b}{16}+\frac{c}{81}\le 2;c\le 81$. Find the maxima of function $$A=\sqrt[4]{a}+\sqrt[4]{b}+\sqrt[4]{c}$$
Wlog $a\le b\le c$
$f''(a)=-3/16a^{7/4}$ then $f(a)=a^{\frac{1}{4}}$ is concave function ,$(a;b;c)\succ (1;16;81)$
By Karamata :$A=\sqrt[4]{a}+\sqrt[4]{b}+\sqrt[4]{c}\le 1+2+3=6$
It is one of the most difficult problem on my exam. Then this is my try, check it for me
| Let $a=x$, $\frac{b}{16}=y$ and $\frac{c}{81}=z.$
Thus, $$x+y+z\leq3,$$ $$y+z\leq2$$ and $$z\leq1$$ and since $f(x)=\sqrt[4]{x}$ is a concave function, by Jensen we obtain:
$$\sqrt[4]a+\sqrt[4]b+\sqrt[4]c=\sqrt[4]x+2\sqrt[4]y+3\sqrt[4]z=$$
$$=\sqrt[4]x+\sqrt[4]y+\sqrt[4]z+\sqrt[4]y+\sqrt[4]z+\sqrt[4]z\leq$$
$$\leq3\sqrt[4]{\frac{x+y+z}{3}}+2\sqrt[4]{\frac{y+z}{2}}+\sqrt[4]z\leq3+2+1=6.$$
The equality occurs for $x=y=z=1,$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the smallest prime number $p$ such that $p\, | \,n^2-n-2023$ for some integer $n$. Find the smallest prime number $p$ such that $p\, | \,n^2-n-2023$ for some integer $n$.
since $n^2-n =n(n-1)$ is the product of two consecutive integers they must be even so the difference between an even and odd number is always odd so $n^2-n-2023$ is always odd which implies $p$ is not even and the only even prime is $2$ so $p\neq 2$ but after this I do not know what to do please help.
| Note that $2023 = 7 \cdot 17^2$
$p=3$
\begin{align}
n^2-n-2023 \equiv 0 \pmod 3 \\
n^2+2n-1 \equiv 0 \pmod 3 \\
(n+1)^2 \equiv 2 \pmod 3
\end{align}
\begin{array}{|c|cc|}
\hline
n \mod 3 & 0 & 1,2 \\
n^2 \mod 3 & 0 & 1 \\
\hline
\end{array}
The perfect squares modulo $3$ are $0$ and $1$. So there is no solution.
$p=5$
\begin{align}
n^2-n-2023 &\equiv 0 \pmod 5 \\
n^2+4n+2 &\equiv 0 \pmod 5 \\
(n+2)^2+3 &\equiv 0 \pmod 5 \\
(n+2)^2 &\equiv 2 \pmod 5
\end{align}
\begin{array}{|c|ccc|}
\hline
n \mod 5 & 0 & 1,4 & 2,3 \\
n^2 \mod 5 & 0 & 1 & 4 \\
\hline
\end{array}
The perfect squares modulo $5$ are $0, 1, 4$. So there is no solution.
$p=7$
\begin{align}
n^2-n-2023 &\equiv 0 \pmod 7 \\
n^2-n &\equiv 0 \pmod 7 \\
n(n-1) &\equiv 0 \pmod 7 \\
\end{align}
Clearly $n \equiv 0 \pmod 7$ and $n \equiv 1 \pmod 7$ are solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Different ways to express $\sqrt{a}+\sqrt{b}$ I had been thinking about this for a long time. I can’t express $\sqrt{a}+\sqrt{b}$ in ways that are useful. (I’m not wanting a formula, but just some ways to express this.) I can only think of $\sqrt{a+2\sqrt{ab}+b}$.
| You can use the conjugate:
\begin{align*}
\sqrt{a}+\sqrt{b} &= \frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\\\
&=\frac{a^2-b^2}{\sqrt{a}-\sqrt{b}}
\end{align*}
However, this does leave a difference of square roots in the denominator. In most cases, having a sum or difference of square roots in the numerator is less of a hassle than in the denominator.
Most of the time, you will use this technique of multiplying by the conjugate to remove pesky expressions in the denominator of a fraction.
It also works for complex numbers, for which multiplying by the conjugate can remove the imaginary part from a denominator:
\begin{align*}
\frac{a+bi}{c+di}&=\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\\\
&=\frac{(a+bi)(c-di)}{c^2+d^2}
\end{align*}
Multiplying the entire expression by $\sqrt{ab}/\sqrt{ab}$ is interesting as well:
\begin{align*}
\sqrt{a}+\sqrt{b} &= \frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{ab}\right)}{\sqrt{ab}}\\\\
&=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}-\sqrt{b}}
\end{align*}
If for some reason you were using a log table, this identity could be useful:
$$\sqrt{a}+\sqrt{b} = a^{1/2} + b^{1/2} = e^{(\ln a)/2} + e^{(\ln b)/2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is my proof of $x^n-y^n = \ldots$ correct? I am solving problems from Spivak's calculus book's 1st chapter. Basically Spivak wants me to prove this:
$$x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$$
Question 1: Is my proof correct?
Question 2: How to make it perfect? Any advise?
My proof:
I guess Spivak wants to say:
$$
x^n-y^n = (x-y)\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
$$
Let:
$$
f(x,y,n) = \left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
$$
Then we can say (is this sub-proof perfect?):
$$\begin{split}
x^{n+1}-y^{n+1} &= (x-y)\left(\sum_{i=1}^{i=n+1} x^{(n+1)-i}y^{i-1}\right)\\
&= (x-y)\left(
\begin{split}
&x^{(n+1)-(n+1)} y^{(n+1)-1}\\
&+ \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
\end{split}
\right)\\
&= (x-y)\left(
x^{0} y^{n}
+ \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
\right)\\
&= (x-y)\left(
y^{n}
+ \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
\right)\\
&= (x-y)\left(
y^{n}
+ x\sum_{i=1}^{i=n} x^{n-i}y^{i-1}
\right)\\
&= (x-y)\left(
y^{n}
+ xf(x,y,n)
\right)\\
\end{split}$$
We can also rewrite what Spivak wants into:
$$
x^n-y^n = (x-y)\Big(y^{n-1} + xf(x,y,n-1)\Big)
$$
We've already proven in [spivak_calc_probs.1.1.2]:
$$\begin{split}
x^2-y^2 &= (x-y)(x+y)\\
&= (x-y)\left(\sum_{i=1}^{i=2} x^{2-i}y^{i-1}\right)\\
&= (x-y)\Big(y^{2-1} + xf(x,y,2-1)\Big)\\
\end{split}$$
Then, by induction, we prove that:
$$\begin{split}
x^2-y^2 &= (x-y)\Big(y^{2-1} + xf(x,y,2-1)\Big)\\
x^3-y^3 &= (x-y)\Big(y^{3-1} + xf(x,y,3-1)\Big)\\
\vdots\\
x^n-y^n &= (x-y)\Big(y^{n-1} + xf(x,y,n-1)\Big)\\
\end{split}$$
And since $(x-y)(y^{n-1} + xf(x,y,n-1))$ is only a rewrite of what Spivak
wants, i.e. $(x-y)(x^{n-1}$ $+$ $x^{n-2}y$ $+$ $\ldots$ $+$ $xy^{n-2})$,
therefore Q.E.D already.
alternative proof:
Using the distributive axiom:
$$\begin{split}
& (x-y)\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-y\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-y\left(x^{n-n}y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-y\left(x^{0}y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-y\left(y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x\left(x^{n-1}y^{1-1} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x\left(x^{n-1}y^{0} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x\left(x^{n-1} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= \left(x^{n} + \sum_{i=2}^{i=n} x^{n-i+1}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-(i+1)+1}y^{(i+1)-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-i-1+1}y^{i+1-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- y^{n} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\\
\end{split}$$
Then using additive associative axiom:
$$\begin{split}
& x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- y^{n} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\\
&= x^{n}
- y^{n}
+ \left(\sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x^{n}
- y^{n}
+ \left(0\right)\\
&= x^{n}
- y^{n}
\end{split}$$
| More simply, in your first proof, for brevity let $S(n)$ be the sentence $(x-y)f(x,y,n)=x^n-y^n.$ Then, using your calculation that $f(x,y,n+1)=y^n+xf(x,y,n),$ we have $$S(n)\implies (x-y)f(x,y,n+1)=$$ $$=(x-y)(y^n+xf(x,y,n))=$$ $$=xy^n-y^{n+1}+(x)(( x-y)f(x,y,n))=$$ $$=xy^n-y^{n+1}+(x)(x^n-y^n)=$$ $$=x^{n+1}-y^{n+1}.$$ That is, we have $$S(n)\implies S(n+1).$$ And $S(1)$ is true because $f(x,y,1)=1.$ So by induction on $n$ we have $S(n)$ for all $n\in \Bbb N.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Compute $\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$ Compute $$I=\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$$
My attempt:
$$u=x^{1/3}\implies I = 3\int_0^{\infty}\frac {u}{u^6+1}du=\frac 32\int_0^\infty \frac {1}{u^3+1}du=\frac 32 \int_0^\infty \frac {1}{(x+1)(x^2+x+1)}dx\\\implies I=\frac 32 \int_0^{\infty} \frac 1{x+1}-\frac x{x^2+x+1}dx$$ And I'm stuck here, what can I do from here?
| As an alternative approach:$$I=\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx\,\overset{\large x^{2/3}=u}=\,\frac32 \int_0^\infty \frac{1}{u^3+1}du$$
We will substitute $\displaystyle{u=\frac{1-t}{1+t}\Rightarrow du=-\frac{2}{(1+t)^2}dt}$.
The reason behind it is that $(1-t)^3+(1+t)^3=2(3t^2+1)$, thus we get rid of the third powers.
$$\Rightarrow I=\frac32 \int_{-1}^1 \frac{t+1}{3t^2+1}dt=\frac32 \cdot 2\int_0^1 \frac{dt}{3t^2+1}dt=3\cdot \frac{1}{\sqrt 3}\arctan(\sqrt 3 t)\bigg|_0^1=\frac{\pi}{\sqrt 3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
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Fractions in Questions and Answers
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