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Find $\sum_{n=1}^{\infty} x^{\left\lfloor {n \over 2}\right\rfloor} y^{\left\lfloor {n + 1 \over 2}\right\rfloor}$ Let $x,y > 0, xy <1$. Find the sum
$$\sum_{n=1}^{\infty} x^{\left\lfloor {n \over 2}\right\rfloor} y^{\left\lfloor {n + 1 \over 2}\right\rfloor}$$
While I have some ideas how to test convergence, I don't quite know how to get started on the actual sum.
Edit:
If we break the sum down to a sum of two infinite series:
$$(y + xy^2 + x^2y^3 + \dots) + (xy + x^2y^2 + \dots)$$
and use the formula for the sum of the first n terms of a geometric series twice, we get:
$$y {1 - (xy)^n \over1 - xy} + xy {1 - (xy)^n \over1 - xy} \to {y \over1 - xy} + {xy \over1 - xy} = {y(1+x) \over 1 - xy}$$
Is this reasoning correct?
| Note that for $n \in \mathbf N$ we have
$$ \def\fl#1{\left\lfloor#1\right\rfloor}\fl{\frac n2} + 1
= \fl{\frac{n+2}2}$$
and
$$ \fl{\frac{n+1}2}+1 = \fl{\frac{n+3}2} $$
So, if we call the sum $s := \sum_{n=1}^\infty x^{\fl{n/2}}y^{\fl{(n+1)/2}}$ and assume it converges we have,
\begin{align*}
xys &= \sum_{n=1}^\infty x^{\fl{(n+2)/2}}y^{\fl{(n+3)/2}}\\
&= \sum_{k=3}^\infty x^{\fl{k/2}}y^{\fl{(k+1)/2}}\\
&= s - x^0y^1 - x^1y^1\\
&= s - y(1+x)
\end{align*}
Now solve for $s$, we have
$$ xys = s - y(1+x) \iff s(1-xy) = y(1+x) \iff s = \frac{y(1+x)}{1-xy} $$
If you do not want to assume convergence, you can do the following: Note that
\begin{align*}
\sum_{n=1}^\infty x^{\fl{n/2}}y^{\fl{(n+1)/2}}
&= \sum_{k=1}^\infty x^{\fl{(2k-1)/2}}y^{\fl{2k/2}}
+ \sum_{k=1}^\infty x^{\fl{2k/2}}y^{\fl{(2k+1)/2}}\\
&= \sum_{k=1}^\infty x^{k-1}y^k + \sum_{k=1}^\infty x^k y^k\\
&= y\sum_{k=0}^\infty (xy)^k + xy \sum_{k=0}^\infty (xy)^k\\
&= (y+xy) \sum_{k=0}^\infty (xy)^k\\
&= \frac{y(1+x)}{1-xy}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/2034276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
| Let $x=(2a,b,1)$ and $y=(1,2a,b)$. Then by, Cauchy-Schwarz,
$2ab+2a+b= <x,y> \le ||x||*||y||=\sqrt{4a^2+b^2+1}*\sqrt{4a^2+b^2+1}=4a^2+b^2+1$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$. Need help with proof. If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$.
Working:
$\frac{1}{2}(1-a)\leq 0<\frac{1}{2}(b-a)$ and $(1-\sqrt{a})\leq 0<\sqrt{b}-\sqrt{a}$
Trying to show $\frac{1}{2}(b-a)-(\sqrt{b}-\sqrt{a})$ is positive but don't know what to do next.
| One may observe that, for $b>a\ge1$,
$$
\begin{align}
(b-a)-2(\sqrt{b}-\sqrt{a})&=(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})-2(\sqrt{b}-\sqrt{a})
\\\\&=(\sqrt{b}-\sqrt{a})\left[\sqrt{b}+\sqrt{a}-2 \right]
\\\\&=(\sqrt{b}-\sqrt{a})\left[(\sqrt{b}-1)+(\sqrt{a}-1) \right]
\\\\&>0
\end{align}
$$ by using that $x \mapsto \sqrt{x}$ is strictly increasing over $[1,\infty)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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About fractions whose sum is a natural number Some days ago I found an old problem of an olympiad that I always found interesting. It asks to replace each $\boxed{}$ with the numbers $1,2\ldots 30$ without repeating any number, such that their sum is an integer number.
$$\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed{}}$$
I got a solution by trial-and-error and it's: $$\frac{14}{1}+\frac{23}{2}+\frac{11}{22}+\frac{19}{3}+\frac{10}{15}+\frac{29}{4}+\frac{9}{12}+\frac{17}{6}+\frac{5}{30}+\frac{25}{8}+\frac{21}{24}+\frac{16}{7}+\frac{20}{28}+\frac{27}{18}+\frac{13}{26}=53.$$
I was wondering if there is another method different than trial-and-error. Thanks in advance for your answers.
| Here's what one can play around with (and supposedly, Ian Miller started similarly):
We can build as many integer fractions as possible, namely
$$\tag1\frac{30}{15}+\frac{28}{14}+ \frac{26}{13}+\frac{24}{12}+ \frac{22}{11}+
\frac{20}{10}+ \frac{18}{9}+ \frac{16}{8},$$
each summand equalling $2$.
As $14$ is already in use, we continue with $$\tag2\frac{21}7.$$
All multiples of $6$ are used, so we postpone that for a moment and continue with
$$\tag3\frac{25}5. $$
Now we are left with $29,27,23,19,17,6,4,3,2,1$ to form five fractions. As barak manos observed, we must have the high primes $29,23,19,17$ in numerators (or otherwise, $\frac{20!}p$ times our sum would not be an integer - even if we had not started with the above strategy).
Because of what we did so far, we also cannot have $27$ in the denominator (or otherwise $12$ times our sum would not be an integer) - but that might be different if we had started differently.
Now we try to exploit that $\frac16+\frac13+\frac12$ is an integer.
We observe that $29\equiv -1\pmod 6$, $23\equiv -1\pmod 3$, and e.g. $27\equiv -1\pmod 2$, so that $$\tag4\frac{29}6+\frac{23}3 +\frac{27}2$$ is an integer.
We are now left with $19,17,4,1$, and definitely end with something having a $4$ in the denominator, namely $$\tag5\frac{19}4+\frac{17}1\quad\text{or}\quad\frac{17}4+\frac{19}1.$$
We play around with the previous fractions to mend this. For example, we can dissect our $\frac{28}{14}$ and $\frac{21}{7}$ to form $\frac 7{28}+\frac{21}{14}=\frac74$ from its parts. This adjusts our quarter-integer result to either an integer result (and we are done) or to a half-integer result. Even though we can verify that we can guarantee an integer here by making the right choice in $(5)$, it should be noted that by flipping one of the fractions in $(1)$, we could turn a half-integer sum into an integer if necessary.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of $\sum_{n=1}^{\infty}\left(\sqrt{n^2+1} - \sqrt[3]{n^3+1}\right)$ I am trying to test the convergence of
$$\sum_{n=1}^{\infty}\left(\sqrt{n^2+1} - \sqrt[3]{n^3+1}\right)$$
The ratio test yields and inconclusive $1$. I am able to show $\sqrt{n^2+1} - \sqrt[3]{n^3+1} < 1$ but this is not enough for convergence. One idea would be to rearrange the expression so that it is of the form ${1 \over f(n)}$ and then show that $\text{deg}f > 1$. With the fact that $\sum{1 \over n^s}$ converges if $s>1$ this would give me convergence. So, we could use the formula for $a^3-b^3$ and then the formula for $a^2-b^2$. Is there a quicker way?
| $$ \sqrt{n^2-1}-\sqrt[3]{n^3-1} = \left(n-\sqrt[3]{n^3-1}\right)-\left(n-\sqrt{n^2-1}\right)\tag{1} $$
so, due to $\frac{a^2-b^2}{a-b}=(a+b)$ and $\frac{a^3-b^3}{a-b}=(a^2+ab+b^2)$ we have:
$$\left(n-\sqrt[3]{n^3-1}\right)\leq \frac{1}{n^2},\qquad \left(n-\sqrt{n^2-1}\right)\geq \frac{1}{2n}\tag{2} $$
and these inequalities ensure that the given series is divergent.
| {
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"url": "https://math.stackexchange.com/questions/2037637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a_{1}=1$ and $b_{1}=2$ show that $b_{5}-a_{5}<1/2^{45}$. Need help interpreting solution. If $a_{1}=1$ and $b_{1}=2$ show that $b_{5}-a_{5}<1/2^{45}$.
$a_{n+1}=\sqrt{a_{n}b_{n}}$ , and $b_{n+1}=\frac{1}{2}(a_{n}+b_{n})$.
I've managed to work out that $b_{n+1}-a_{n+1}<\frac{1}{8}(b_{n}-a_{n})^{2}$. It was the solution in the back of the book. I do not however understand how this relates to the inequality which is to be shown.
| From
$$
b_{n+1}-a_{n+1}<\frac{1}{8}(b_{n}-a_{n})^{2},\quad n\ge1,
$$ one gets, with $n=1,2,3,4$,
$$
\begin{align}
b_{5}-a_{5}&<\frac{1}{8}(b_{4}-a_{4})^{2}
\\\\&<\frac{1}{8^3}(b_{3}-a_{3})^{2}
\\\\&<\frac{1}{8^7}(b_{2}-a_{2})^{2}
\\\\&<\frac{1}{8^{15}}(b_{1}-a_{1})^{2}.
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $G_4(i)\neq 0$, and $G_6(\rho)\neq 0$, $\rho=e^{2\pi i /3}$ Let $G_k$ denote the Eisenstein series of weight $k$. I know that $G_k(i)=0$ if $k \not\equiv 0 \ (mod \ 4)$ and $G_k(\rho)=0$ if $k \not\equiv 0 \ (mod \ 6)$. However, I want to know how to show, that $G_4(i)\neq0$ and $G_6(\rho)\neq0$, without using the $\frac{k}{12}$-formula.
| Here's a direct proof that $G_4(i) \neq 0$ that uses only the series.
The point is simply to estimate the tail $|a|,|b| \geq N$ and then calculate the first few terms for $|a|,|b| < N$. In fact, using very simply bounds we can take $N=2$.
First note that $|a|+|b| \leq \sqrt{3}|a+bi|$ which follows from
$$ 2|a||b| \leq 2 (a^2+b^2) \implies a^2+b^2 + 2|a||b| \leq 3(a^2+b^2) \implies (|a|+|b|)^2 \leq 3(a^2+b^2). $$
This with integral comparison, easily gives \begin{align*} \left|\sum_{|a|,|b| \geq N} \frac{1}{(a+bi)^4} \right| &\leq \sum_{|a|,|b| \geq N} \frac{1}{|a+bi|^4} \\\\ &\leq \sqrt{3}^4 \sum_{|a|,|b| \geq N} \frac{1}{(|a|+|b|)^4} \\\\ &= 9 \cdot 4 \sum_{a,b \geq N} \frac{1}{(a+b)^4} \\\\ &= 36 \sum_{a \geq N} \sum_{b \geq N} \frac{1}{(a+b)^4} \\\\ &= 36 \sum_{a \geq N} \sum_{b \geq N+a} \frac{1}{b^4} \\\\ &\leq 36 \sum_{a \geq N} \int_{N+a-1}^{\infty} \frac{1}{x^4}dx \\\\&= 12 \sum_{a \geq N} \frac{1}{(N+a-1)^3} \\\\&= 12 \sum_{a \geq 2N-1} \frac{1}{a^3} \\\\ &\leq 12 \int_{2N-2}^{\infty} \frac{1}{x^3} dx \\\\ &= \frac{12}{(2N-2)^2} \\\\ &= \frac{3}{(N-1)^2}. \end{align*}
Thus, when $N=2$, the tail is bounded by $\Large{\color{red}{3}}$.
Now, since conjugation permutes $\mathbb{Z} + i\mathbb{Z}$, we know that $G_4(i)$ is real. Thus, we need only worry about the real part of the sum. We have $$\text{Re} \left(\sum_{\substack{|a|,|b| < N \\ (a,b) \neq (0,0) }} \frac{1}{(a+bi)^4} \right) = \sum_{\substack{|a|,|b| < N \\ (a,b) \neq (0,0) }} \frac{a^4+b^4-6a^2b^2}{(a^2+b^2)^4}.$$ With $N=2$, a very short calculation gives $\Large{\color{red}{5}}$, a positive number. Note that there is no way that the tail will make it negative. Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of occurrences of k consecutive 1's in a binary string of length n (containing only 1's and 0's) Say a sequence $\{X_1, X_2,\ldots ,X_n\}$ is given, where $X_p$ is either one or zero ($0 < p < n$). How can I determine the number of strings, which do contain at least one occurrence of consequent $1$'s of length $k$ ($0 < k < n$).
For example, a string $\{1, 0, 1, 1, 1, 0\}$ is such a string for $n = 6$ and $k = 3$.
Here I have found an answer for arbitrary $n$ and $k = 2$, ($k = 1$ is trivial), but I need a more general answer for any natural number $k$ smaller than $n$.
| Here is an approach based upon generating functions. We start considering words with no consecutive equal characters at all.
These words are called Smirnov words or Carlitz words. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)
A generating function for the number of Smirnov words over a binary alphabet is given by
\begin{align*}
\left(1-\frac{2z}{1+z}\right)^{-1}\tag{1}
\end{align*}
Generating function: $G_k(z)$
In order to set up for the binary strings we are looking for, we replace occurrences of $1$ in a Smirnov word by runs of $1$ with length up to $k-1$ assuming $k\geq 2$. This corresponds to a substitution
\begin{align*}
z&\longrightarrow z+z^2+\cdots+z^{k-1}=\frac{z\left(1-z^{k-1}\right)}{1-z}\tag{2}\\
\end{align*}
Since there are no restrictions to $0$ at all, we replace occurrences of $0$ in a Smirnov word by any runs of $0$'s with length $\geq 1$.
\begin{align*}
z&\longrightarrow z+z^2+z^3+\cdots=\frac{z}{1-z}\tag{3}\\
\end{align*}
We obtain by substituting (2) and (3) in (1) a generating function $A_k(z)$
\begin{align*}
A_k(z)&=\left(1-\frac{\frac{z}{1-z}}{1+\frac{z}{1-zt}}-\frac{\frac{z\left(1-z^{k-1}\right)}{1-z}}{1+\frac{z\left(1-z^{k-1}\right)}{1-z}}\right)^{-1}\\
&=\frac{1-z^k}{1-(t+1)z+tz^{k+1}}\tag{4}\\
\end{align*}
counting all binary words having no 1-runs of length $k$. To obtain a generating function which counts all binary words having at least one 1-run of length $k$, we take the generating function
\begin{align*}
\frac{1}{1-2z}=1+2z+4z^2+8z^3+\cdots
\end{align*}
which counts all binary words and subtract $A_k(z)$ from it.
We conclude from (4) a generating function counting all binary words having at least one 1-run of length $k$ is $G_k(z)$ with
\begin{align*}
\color{blue}{G_k(z)}&\color{blue}{=\frac{1}{1-2z}-\frac{1-z^k}{1-2z+z^{k+1}}}\\
&\color{blue}{=\frac{(1-z)z^k}{(1-2z)(1-2z+z^{k+1]})}}
\end{align*}
Explicit formula:
We derive from $G_k(z)$ an explicit formula of the wanted numbers. Denoting with $[z^n]$ the coefficient of $z^n$ in a series we obtain using the geometric series expansion
\begin{align*}
[z^n]\frac{1}{1-2z+z^{k+1}}&=[z^n]\sum_{j=0}^\infty z^j(2-z^k)^j\tag{5}\\
&=\sum_{j=0}^n [z^{n-j}](2-z^k)^j\tag{6}\\
&=\sum_{j=0}^{\left\lfloor\frac{n}{k}\right\rfloor}[z^{kj}](2-z^k)^{n-kj}\tag{7}\\
&=\sum_{j=0}^{\left\lfloor\frac{n}{k}\right\rfloor}[z^{kj}]
\sum_{l=0}^{n-kj}\binom{n-kj}{l}(-z^k)^l2^{n-kj-l}\\
&=\sum_{j=0}^{\left\lfloor\frac{n}{k}\right\rfloor}\binom{n-kj}{j}(-1)^j2^{n-(k+1)j}\tag{8}
\end{align*}
Comment:
*
*In (5) we use the binomial series expansion.
*In (6) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]F(z)=[z^p]z^qF(z)$. We also set the upper limit of the sum to $n$ since the exponent of $z^{n-j}$ is non-negative.
*In (7) we change the order of summation $j\rightarrow n-j$ and respect that only multiples of $k$ give a contribution to the sum.
*In (8) we select the coefficient of $z^{kj}$.
The sum in (8) gives the first part of the formula we need. We can use it to derive the second part
\begin{align*}
[z^n]\frac{z^k}{1-2z+z^{k+1}}&=[z^{n-k}]\frac{1}{1-2z+z^{k+1}}\\
&=\sum_{j=0}^{\left\lfloor\frac{n-k}{k}\right\rfloor}\binom{n-k(j+1)}{j}(-1)^j2^{n-k-(k+1)j}
\end{align*}
Finally putting all together we conclude the number of binary words of length $n$ containing 1-runs of length $k\geq 2$ is
\begin{align*}
\color{blue}{[z^n]G_k(z)=2^n-\sum_{j=0}^{\left\lfloor\frac{n}{k}\right\rfloor}
\left(\binom{n-kj}{j}-\frac{1}{2^k}\binom{n-k(j+1)}{j}\right)(-1)^j2^{n-(k+1)j}}
\end{align*}
We can also use $G_k(z)$ to derive a recurrence relation for the coefficients $[z^n]G_k(z)$. Due to the specific structure of
\begin{align*}
G_k(z)&=\frac{1}{1-2z}-A_k(z)\\
\end{align*}
it seems to be more convenient to derive a recurrence relation for the coefficients $a_n=[z^n]A_k(z)$ and subtract them from $2^n$.
Recurrence relation:
We obtain by coefficient comparison for $k\geq 2$
\begin{align*}
A_k(z)&=\frac{1-z^k}{1-2z+z^{k+1}}\\
\left(1-2z+z^{k+1}\right)A_k(z)&=1-z^k\\
\color{blue}{a_n-2a_{n-1}+a_{n-k-1}}&\color{blue}{=}
\color{blue}{\begin{cases}
1&\qquad n=0\\
-1&\qquad n=k\\
0&\qquad n \neq 0,k
\end{cases}}
\end{align*}
where we set $a_n=0$ if $n<0$.
Example: $k=2$
In case of $k=2$ we obtain
\begin{align*}
A_2(z)&=\frac{1-z^2}{1-2z+z^3}\\
&=1+2z+3z^2+5z^3+8z^4+13z^5+21z^6+\cdots
\end{align*}
and finally
\begin{align*}
G_2(z)&=\frac{1}{1-2z}-\frac{1-z^2}{1-2z+z^3}\\
&=z^2+3z^3+8z^4+19z^5+43z^6+94z^7+\cdots
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Non-linear recurrence with square root: $a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$ How should I approach this problem:
$a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$
where $a_0 = 2, a_1=8$
| An alternative approach to the one given by Oliver Oloa:
$$a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$$
$$\implies a^2_{n+2}={a_{n+1}\cdot a_{n}}$$
$$\implies \frac{a_{n+2}}{a_{n+1}}=\frac{a_{n}}{a_{n+2}} \tag1$$
Similarly we can write that
$$\frac{a_{n+1}}{a_{n}}=\frac{a_{n-1}}{a_{n+1}} \tag2$$
$$\frac{a_{n}}{a_{n-1}}=\frac{a_{n-2}}{a_{n}} \tag3$$
$$\ldots$$
$$\frac{a_{2}}{a_{1}}=\frac{a_{0}}{a_{2}} \tag{n+1}$$
Multiplying these $n+1$ relations, we get
$$\frac{a_{n+2}}{a_{n+1}}\cdot\frac{a_{n+1}}{a_{n}}\cdot\frac{a_{n}}{a_{n-1}}\ldots \frac{a_{2}}{a_{1}}=\frac{a_{n}}{a_{n+2}}\cdot\frac{a_{n-1}}{a_{n+1}}\cdot\frac{a_{n-2}}{a_{n}}\ldots\frac{a_{0}}{a_{2}}$$
$$\implies \frac{a_{n+2}}{a_{1}}=\frac{a_{1}\cdot a_{0}}{a_{n+2}\cdot a_{n+1}}$$
$$\implies \frac{a_{n+2}}{8}=\frac{16}{a_{n+2}\cdot a_{n+1}}$$
$$\implies a_{n+2}=\frac{12}{\sqrt{a_{n+1}}}$$
$$\implies a_{n+2}=\sqrt{12}\cdot \sqrt[4]{a_{n}}$$
$$\implies \boxed{a_{n}=\sqrt{12}\cdot \sqrt[4]{a_{n-2}}}$$
So we can write that $$a_{n}=12^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots + \frac{(-1)^{r-1}}{2^{r-1}}}\cdot (a_{n-r})^{\frac{(-1)^r}{2^r}} \,\,\,\,\,\,\, \text{where} \,\,\, r \ge 1$$
For $r=n$, we get
$$a_{n}=12^{\frac{1-(-\frac{1}{2})^n}{1-(-\frac{1}{2})}}\cdot (a_0)^{\frac{(-1)^n}{2^n}}$$
$$\boxed{a_{n}=12^{\frac{1}{3}\cdot\frac{2^n-(-1)^n}{2^{n-1}}}\cdot (2)^{\frac{(-1)^n}{2^n}}} \,\,\,\,\,\,\,\,\, \forall \,\,\,\,\,\, n\ge0$$
This is the closed form.
Hope this helps you.
| {
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How do I express x in terms of y for the following expression ? $y=2x+\frac{8}{x^2}-5$ How do I express x in terms of y for the following expression ?
$$y=2x+\frac{8}{x^2}-5$$
The reason I want to do this is to calculate the area between this curve, the y-axis, y=5 and y=1.
I found the answer to be 5 using a roundabout method.
But my first attempt was to try and express x in terms of y.
Basic algebra gave me an expression with an $x^3$.
| Basic algebra is correct.
From
$y=2x+\frac{8}{x^2}-5
$,
multiplying by $x^2$
we get
$yx^2=2x^3+8-5x^2
$,
or
$2x^3-(5-y)x^2+8
= 0
$.
This is,
unfortunately,
a cubic
which can be solved by the
traditionally messy formula.
Its derivative is
$6x^2-2(5-y)x
= 0
$
which has roots
$x=0$
and
$x=\dfrac{5-y}{3}
$.
Note that
at $x=0$
the function is $8$
and at
$\dfrac{5-y}{3}
$
the value is
$2(\frac{5-y}{3})^3-(5-y)(\frac{5-y}{3})^2+8
=(\frac{5-y}{3})^3(2-3)+8
=-(\frac{5-y}{3})^3+8
=(\frac{y-5}{3})^3+8
$.
From this,
we can determine which
values of $y$
give an equation with
$1$ or $3$ real roots.
| {
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"url": "https://math.stackexchange.com/questions/2057584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you find the real solutions to these simultaneous equations? I am looking for all real $(a,b,c)$ that satisfy the following
\begin{equation}
\left\{
\begin{array}{l}2a + a^2b = b\\
2b + b^2c = c\\
2c + c^2a = a\\
\end{array}
\right.
\end{equation}
I know that $a=b=c = 0$ is the only real solution to the problem I know of but I don't know how to prove it.
I was also given the hint, substitute $a = \tan(x)$.
| $b = \frac {2a}{1-a^2}$
$a = \tan t$
$b = \tan 2t\\
c = \tan 4t\\
a = \tan 8t$
$tan t = tan 8t\\
t + n\pi = 8t\\
7t = n\pi\\
t = \frac {n}{7} \pi$
$a,b,c = \tan \frac {n\pi}{7}, \tan \frac {2n\pi}{7},\tan \frac {4n\pi}{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$
Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned}
u_x&= y *(x^2+y^2)^{-2}*-1*2x
\\ &= -y*2x(x^2+y^2)^{-2}=-2xy(x^2+y^2)^{-2}
\\ u_{xx} &= -2xy*(x^2+y^2)^{-3}*-2*2x+-2y*(x^2+y^2)^{-2}
\\&= \frac{-2xy}{(x^2+y^2)^3}*-4x +\frac{-2y}{(x^2+y^2)^2}
\\&=\frac{8x^2y}{(x^2+y^2)^3}+\frac{-2y}{(x^2+y^2)^2}
\\
\end{aligned} $$
Partial Derivatives with y
$$\begin{aligned}
u_y&=1(x^2+y^2)^{-1}+y*(x^2+y^2)^{-2}*-1*2y
\\ &=(x^2+y^2)^{-1}-2y^2(x^2+y^2)^{-2}
\\ u_{yy}&=-1(x^2y^2)^{-2}*2y -2*2y(x^2+y^2)^{-2}-2y^2*-2(x^2+y^2)^{-3}*2y
\\ &=-2y(x^2+y^2)^{-2}-4y(x^2+y^2)^{-2}+8y^3(x^2+y^2)^{-3}
\\ &=\frac{-6y}{(x^2+y^2)^2} + 8y^3(x^2+y^2)^{-3}
\end{aligned} $$
From here need to show that $u_{xx}+u_{yy}=0$ and technically say why the other partials are continous right?? This was a test question whith 3 lines of paper by the way
| $$
f(z) = \frac 1z = \frac{1}{x+iy} = \frac{x}{x^2+y^2} + i\frac{-y}{x^2+y^2}
$$
is holomorphic in $\Bbb C \setminus \{ 0 \}$. It follows that
$$
-\operatorname{Im} f(z) = \frac{y}{x^2+y^2}
$$
is harmonic in the same domain.
(It is a direct consequence of the Cauchy-Riemann
differential equations that real and imaginary part of a holomorphic
function are harmonic.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = \sqrt{2}$, and $s_{n+1} = \sqrt{2 + \sqrt{s_n}}$, what is the limit of this sequence? Here's Prob. 3, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
If $s_1 = \sqrt{2}$, and $$s_{n+1} = \sqrt{2 + \sqrt{s_n}} \ \ (n = 1, 2, 3, \ldots),$$ prove that $\left\{ s_n \right\}$ converges, and that $s_n < 2$ for $n = 1, 2, 3, \ldots$.
My effort:
We can show that $\sqrt{2} \leq s_n \leq 2$ for all $n = 1, 2, 3, \ldots$. [Am I right?]
Then we can also show that $s_n < s_{n+1}$ for all $n = 1, 2, 3, \ldots$. [Am I right?]
But how to calculate the exact value of the limit? Where does this sequence occur in applications?
| For the first part of your question, not answered above:
If $s_1=\sqrt{2}$ and
$$
s_{n+1}=\sqrt{2+\sqrt{s_n}}
$$
prove 1) that $\{ s_n \}$ converges and 2) that $s_n<2$ for any $n\in \mathbb{N}$..
First we show $\{ s_n\}$ is increasing by induction.
Base case:
$$
2>0\Rightarrow 2+\sqrt{2}>\sqrt{2}\Rightarrow \sqrt{2+\sqrt{2}}>\sqrt{2}
\Rightarrow s_2>s_1
$$
Inductive hypothesis: Suppose $s_n>s_{n-1}$, then
$$
\sqrt{s_n}>\sqrt{s_{n-1}}\Rightarrow
2+\sqrt{s_{n}}>2+\sqrt{s_{n-1}}\Rightarrow
\sqrt{2+\sqrt{s_{n}}}>\sqrt{2+\sqrt{s_{n-1}}}
\Rightarrow s_{n+1}>s_{n}
$$
Now, showing part 2 to be true will allow us to conclude part 1, as increasing
sequences bounded above converge.
Base case: $\sqrt{2}<2$.
Inductive hypothesis: Suppose $s_n<2$. Then
$$
\sqrt{2+\sqrt{s_{n-1}}}<2\Rightarrow 2+\sqrt{2+\sqrt{s_{n-1}}}<4\Rightarrow
\sqrt{2+\sqrt{2+\sqrt{s_{n-1}}}}<2\Rightarrow \sqrt{2+s_n}<2\\
\stackrel{\text{since $s_n>1\Rightarrow s_n>\sqrt{s_{n}}>0$}}{\Rightarrow}\sqrt{2+
\sqrt{s_n}}<\sqrt{2+s_n}<2\Rightarrow s_{n+1}<2
$$
and we can conclude.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$
I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$
$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$
gathering everything together,
$(n+1)^{n-1}(n+2)^n=\frac{(n+1)^n(n+2)^{n+1}}{(n+1)(n+2)}>\frac{(n+1)(n!)^22^{2n+1}}{(n+1)(n+2)}$
$\iff(n+1)^{n-1}(n+2)^n>(n!)^2\times\frac{2^{2n+1}}{(n+2)}$
but $\frac{2^{2n+1}}{(n+2)}>3^n$ is not true for $n=2$
can you suggest another approach ?
| Hint
Induction for the step $n+1$:
$$(n+2)^{n}(n+3)^{n+1}=\frac{(n+3)^{n+1}}{(n+1)^{n-1}}(n+1)^{n-1}(n+2)^{n}>3^n(n!)^2\frac{(n+3)^{n+1}}{(n+1)^{n-1}}$$
We may expect
$$\frac{(n+3)^{n+1}}{(n+1)^{n-1}}>3(n+1)^2 \quad (1)$$
in order to finish the induction.
Backing to $(1)$ we have an equivalent expression:
$$\left(\frac{n+3}{n+1}\right)^{n+1}>3 \Leftrightarrow \left(1+\frac{2}{n+1}\right)^{n+1}>3$$
Using the Bernoulli inequality $(1+x)^m \ge1+mx$ for $x>-1$.
Taking $m=n+1$ and $x=\frac{2}{n+1}$ we get what we want.
| {
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How to find the largest integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive integers I need to find the largest integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive integers. Example: $7! = 7 \cdot 8 \cdot 9 \cdot 10 $
| $n=23$ has an answer.
Given that $\binom{K}{n-3} = \frac{K(K-1)\cdots(K-(n-2))}{(n-3)!}$ this means you want a $K$:
$$\binom{K}{n-3} = n(n-1)(n-2)$$
Now, $\binom{K}{n-3}$ is an increasing function as $K$ increases.
And we have:
$\binom{n}{n-3} = \frac{n(n-1)(n-2)}{6} < n(n-1)(n-2)$ and $\binom{n+1}{n-3} = \frac{(n+1)n(n-1)(n-2)}{24}> n(n-1)(n-2)$ when $n+1>24$. So you only have to check the cases $n\leq 23$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Natural number which can be expressed as sum of two perfect squares in two different ways? Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two different ways. One example I got is $50$ which is $49+1$ and $25+25$. But here second pair contains same numbers. Does any one have other examples ?
| Any Pythagorean triple $(A^2+B^2=C^2)$ provides a candidate where the $C^2$ can have $2$-or-more combinations of $A^2$ and $B^2$ that add up to it. These can be found by testing natural numbers of the form $(4n+1)$ with a range of $m$ values defined as shown below to see which, if any yield integers. We begin with Euclid's formula and solve the $C$-function for $n$ in terms of $C$ and $m$.
$$A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$$
$$C=m^2+n^2\implies n=\sqrt{C-m^2}\\
\text{where}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \big\lfloor\sqrt{C-1}\space\big\rfloor$$
The lower limit ensures $m>n$ and the upper limit ensures $n\in\mathbb{N}$.
$$C=65\implies\\
\bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \big\lfloor\sqrt{65-1}\big\rfloor=8\\
\text{ and we find} \quad m\in\{7,8\}\Rightarrow n\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\quad F(8,1)=(63,16,65) $$
Here, we have $\space 33^2+56^2=65^2\space$ and $\space 63^2+16^2=65^2$
There infinitely many of these pairs of triples; the next that meet the criteria are
$$(13^2,84^2,85^2),\qquad (77^2,36^2,85^2)\\ (75^2,100^2,125^2),\qquad (117^2,44^2,125^2)\\(17^2,144^2,145^2),\qquad (143^2,24^2,145^2)\\ ...$$
$\textbf{Edit:}$ If $\space n \space$ is the number of distinct prime fators of $\space C,\space$ there are $2^{n-1}$ primitive triples with that same
$C$-value. This means e.g. that for $1105=5\times13\times17,\space$ there are $2^{3-1}=4$ primitives with $C=1105.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 13,
"answer_id": 12
} |
What is the minimal polynomial of $\sqrt{3} + \sqrt[3] {2}$ over $\mathbb{Q}$? What is the minimal polynomial of $\sqrt{3} + \sqrt[3]{2}$ over $\mathbb{Q}$?
I know the basic idea of what a minimal polynomial is--it is the lowest degree monic polynomial in $\mathbb{Q}[x]$ that has the above as a root. But how do you go about calculating it? How do you know intuitively if it does or doesn't exist?
| Start with $x=\sqrt{3}+\sqrt[3]{2}$. Then
\begin{gather*}
x-\sqrt{3} = \sqrt[3]{2} \\
(x-\sqrt{3})^3 = 2 \\
x^3 - 3\sqrt{3}x^2 + 9x - 3\sqrt{3} = 2 \\
x^3 + 9x - 2 = 3\sqrt{3}(x^2+1) \\
(x^3+9x-2)^2 = 27(x^2+1)^2 \\
x^6 + 18x^4 - 4x^3 + 81x^2 - 36x + 4 = 27x^4 + 54x^2 + 27 \\
x^6 - 9x^4 - 4x^3 + 27x^2 - 36x-23 = 0.
\end{gather*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Use the sum identity and double identity for sine to find $\sin 3x$. Q. Use the sum identity and double identity for sine to find $\sin 3x$.
$$
\begin{align}
\sin 3x &= \sin (2x + x)\\
&=\sin 2x \cos x + \cos 2x \sin x \\
&= (2\sin x \cos x) \cos x + (1 - 2\sin^2 x) \sin x\\
&=2\sin x \cos^2 x + \sin x - 2\sin^3 x \\
&=2\sin x (1 - \sin^2 x) + \sin x - 2\sin^3 x\\
&= " 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x \\
&=3\sin x - 4\sin^3 x"
\end{align}
$$
The part of the problem I'm having trouble with is in quotations.
My question:
is how does $\sin x - 2\sin^3 x = 4\sin^3 x$?
I see it as this $\sin x - 2\sin^3 x = 2\sin^{3-1} x = 4\sin x$.
| You seem to have forgotten the first two terms inside the quotation marks.
$$2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x = 3 \sin x - 4 \sin^3 x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2-1}}{2^{2n}}
\end{align}
I could not move on. Please help me to continue.
| $\frac {2^{n(1-n)}.2^{n-1}.4^n}{2.2^n.2^{(n-1)}}$
First of all both N & D contains $2^{n-1}$ Eliminate them. Now you have,
$\frac {2^{n(1-n)}.4^n}{2.2^n}$
$\frac {2^n.2^{-n^2}.2^{2n}}{2.2^n}$
Now both N and D have $2^n$ eliminate it.
$\frac {2^{-n^2}.2^{2n}}{2}$
= $\frac {2^{2n}}{2.2^{n^2}}$
Edit -
According to your answer -
=$\frac {2^{4n-n^2-1}}{2^{2n}}$
=$2^{4n-n^2-1-2n}$
=$2^{2n-n^2-1}$
=$2^{-(n^2-2n+1)}$
=$2^{-(n-1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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How to use the 'modulus' operator? This is a problem from BdMO $2012$ Dhaka region Question Paper:
The product of a number with itself is called its square. For example,
$2$ multiplied by $2$ is $4$, so $4$ is the square of $2$. If you take a square
number and multiply it with itself, what will be the largest possible
remainder if the product is divided by $10$?
I came up with this: $$x^4 \mod {10}$$
I know that the modulus (%) operator calculates the remainder of a division. And that it can be used to see, suppose, whether $N$ is a multiple of $M$ or not. Nothing more than that. I am much familiar with mod because of my programming experience with mid-level languages like C and C++. It was not until later that I came to know that modulus is used in mathematics as well.
Now, how to use the 'modulus' operator? How can I use this to go further into solving this problem?
| So you can simply try out all $10$ possible options:
*
*$x\equiv0\pmod{10} \implies x^4\equiv0^4\equiv 0\equiv0\pmod{10}$
*$x\equiv1\pmod{10} \implies x^4\equiv1^4\equiv 1\equiv1\pmod{10}$
*$x\equiv2\pmod{10} \implies x^4\equiv2^4\equiv 16\equiv6\pmod{10}$
*$x\equiv3\pmod{10} \implies x^4\equiv3^4\equiv 81\equiv1\pmod{10}$
*$x\equiv4\pmod{10} \implies x^4\equiv4^4\equiv 256\equiv6\pmod{10}$
*$x\equiv5\pmod{10} \implies x^4\equiv5^4\equiv 625\equiv5\pmod{10}$
*$x\equiv6\pmod{10} \implies x^4\equiv6^4\equiv1296\equiv6\pmod{10}$
*$x\equiv7\pmod{10} \implies x^4\equiv7^4\equiv2401\equiv1\pmod{10}$
*$x\equiv8\pmod{10} \implies x^4\equiv8^4\equiv4096\equiv6\pmod{10}$
*$x\equiv9\pmod{10} \implies x^4\equiv9^4\equiv6561\equiv1\pmod{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and should be added to fraction, while the book treats it as part of fraction itself, thus multiplying it with $n^2+5$.
So, I just want to understand which convention is correct.
This is from problem 6 in exercise 9.1 on page 180 of the book Sequences and Series.
Here is the answer sheet from the book (answer 6, 3rd element):
*
*$3,8,15,24,35$
*$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6}$
*$2, 4, 8, 16 \text{ and } 32$
*$-\dfrac{1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6},\dfrac{7}{6}$
*$25,-125,625,-3125,15625$
*$\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$
*$65, 93$
*$\dfrac{49}{128}$
*$729$
*$\dfrac{360}{23}$
*$3, 11, 35, 107, 323$; $3+11+35+107+323+...$
*$-1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24},\dfrac{-1}{120}$; $-1+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(d\frac{-1}{120})+...$
*$2, 2, 1, 0, -1$; $2+2+1+0+(-1)+...$
*$1,2,\dfrac{3}{5},\dfrac{8}{5}$
| $3 \times \frac{3^2+5}{4} = 3 \times \frac{9+5}{4} = 3\times \frac{14}{4} = 3\times \frac{7}{2} = \frac{21}{2}$
Normally (even though in calculator this is often not true) the convention is that a number on the side of a fraction is multiplying that fraction. There should be a "times" either a cross or a dot, however often you can omit it as a shortcut (in veritas, it is almost always omitted apart from very specific cases when someone wants to emphasise the steps as in my answer above)
| {
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"question_score": "22",
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"answer_id": 5
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Find the maximum power of $24$ in $(48!)^2$? Find the maximum power of $24$ in $(48!)^2$ ?
How to approach for such questions ?
| $24=\color\red{2^3}\cdot\color\green{3^1}$
The multiplicity of $\color\red{2}$ in $48!$ is $\sum\limits_{n=1}^{\log_{\color\red{2}}48}\Big\lfloor\frac{48}{\color\red{2}^n}\Big\rfloor=24+12+6+3+1=\color\red{46}$
The multiplicity of $\color\green{3}$ in $48!$ is $\sum\limits_{n=1}^{\log_{\color\green{3}}48}\Big\lfloor\frac{48}{\color\green{3}^n}\Big\rfloor=16+5+1=\color\green{22}$
Therefore:
*
*The maximum value of $n$ such that $(\color\red{2^3})^n$ divides $48!$ is $\Big\lfloor\frac{\color\red{46}}{\color\red{3}}\Big\rfloor=\color\red{15}$
*The maximum value of $n$ such that $(\color\green{3^1})^n$ divides $48!$ is $\Big\lfloor\frac{\color\green{22}}{\color\green{1}}\Big\rfloor=\color\green{22}$
Therefore:
*
*The maximum value of $n$ such that $(\color\red{2^3})^n$ divides $(48!)^2$ is $\Big\lfloor\frac{\color\red{46}\cdot2}{\color\red{3}}\Big\rfloor=\color\red{30}$
*The maximum value of $n$ such that $(\color\green{3^1})^n$ divides $(48!)^2$ is $\Big\lfloor\frac{\color\green{22}\cdot2}{\color\green{1}}\Big\rfloor=\color\green{44}$
Therefore, the maximum value of $n$ such that $24$ divides $(48!)^2$ is $\min(\color\red{30},\color\green{44})=30$.
| {
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"url": "https://math.stackexchange.com/questions/2075577",
"timestamp": "2023-03-29T00:00:00",
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If $ f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval If $\displaystyle f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval
assume $\sin x= t$ where $|\sin x|\leq 1$
let $\displaystyle y = \frac{t^2+t-1}{t^2-t+1}$
$\displaystyle yt^2-yt+y=t^2+t-1$
$(y-1)t^2-(y+1)t+(y-1)=0$
for real roots $(y+1)^2-4(y-1)^2\geq 0$ or $(y+1)^2-(2y-2)^2\geq0$
$(3y-3)(3y-1)\leq 0$ or $\displaystyle \frac{1}{3}\leq y\leq 1$
but walframalpha shows different answer
https://www.wolframalpha.com/input/?i=range+of+f(x)+%3D+%5Cfrac%7B%5Csin%5E2+x%2B%5Csin+x-1%7D%7B%5Csin%5E2+x-%5Csin+x%2B2%7D
could some help me with this, thanks
| This is an answer without using derivative.
As you did, let $t=\sin x$.
First of all, we have $y\not=1$ since for $y=1$ we get $t=\frac 32\gt 1$.
Since $y\not=1$, we have
$$\begin{align}y=\frac{t^2+t-1}{t^2-t+2}&\iff (y-1)t^2+(-y-1)t+2y+1=0\\&\iff t^2+\frac{-y-1}{y-1}t+\frac{2y+1}{y-1}=0\\&\iff \left(t-\frac{y+1}{2y-2}\right)^2+\frac{7y^2-6y-5}{4(y-1)^2}=0\tag1\end{align}$$
Let $g(t)$ be the LHS of $(1)$.
Note that we want to find the condition on $y$ such that $g(t)=0$ has at least one real solution satisfying $-1\le t\le 1$.
Here, $Y=g(t)$ is a parabola whose vertex is $(\alpha,\beta)$ where $\alpha=\frac{y+1}{2y-2},\beta=\frac{7y^2-6y-5}{4(y-1)^2}$.
Case 1 : When $\alpha\lt -1$, we have to have $\beta\lt 0$ and $g(-1)\le 0$ and $g(1)\ge 0$
In this case, we have $\frac 13\lt y\le \frac 12$.
Case 2 : When $\alpha=-1$, we have to have $\beta\le 0$ and $g(1)\ge 0$
In this case, we have $y=\frac 13$.
Case 3 : When $-1\lt\alpha\lt 1$, we have to have $\beta\le 0$ and "$g(-1)\ge 0$ or $g(1)\ge 0$"
In this case, we have $\frac{3-2\sqrt{11}}{7}\le y\lt\frac 13$.
Case 4 : When $\alpha=1$, we have to have $\beta\le 0$ and $g(-1)\ge 0$
There are no such $y$.
Case 5 : When $\alpha\gt 1$, we have to have $\beta\lt 0$ and $g(-1)\ge 0$ and $g(1)\le 0$
There are no such $y$.
Hence, the answer is
$$\color{red}{\frac{3-2\sqrt{11}}{7}\le y\le\frac 12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Use Mean value theorem to prove the following inequality A) Use the Mean value theorem to prove that
\begin{equation}
\sqrt{1+x} < 1 + \frac{1}{2}x \text{ if } x>0
\end{equation}
B) Use result in A) to prove that
\begin{equation}
\sqrt{1+x}>1+\frac{1}{2}x-\frac{1}{8}x^2 \text{ if } x>0
\end{equation}
Can someone give an answer for part B) ?
| OK, I'll show how I'd solve a:
We have
$$
f(x) = 1+ \frac{x}{2} - \sqrt{1+x}
$$
Clearly $f(0) = 0$. Let's look at the derivative of $f$. If we show that the derivative is strictly positive for $x>0$, the function is strictly increasing, hence positive (in combination with $f(0)$) for $x>0$. We have
$$
f'(x) = \frac{1}{2} - \frac{1}{2\sqrt{1+x}} = \frac{1}{2}\bigg(\frac{x}{\sqrt{1+x}} \bigg)
$$
Since $x>0$, the term in the brackets is always positive, hence the derivative is strictly positive and the function is increasing. Hence,
$$
f(x)>0 \to 1+\frac{x}{2} > \sqrt{1+x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| An easy way is to simplify the nested radical, and then combine like terms.
An easy way to denest $\sqrt[m]{A+B\sqrt[n]{C}}$ is to assume the form $a+b\sqrt[n]{C}$ and expand both sides with the binomial theorem. $\sqrt[3]{2+\sqrt5}$ becomes$$2+\sqrt5=(a^3+15ab^2)+(3a^2b+5b^3)\sqrt5\tag1$$
From which we get a system of equations\begin{align*} & a^3+15ab^2=2\tag2\\ & 3a^2b+5b^3=1\tag3\end{align*}
Cross multiplying, we get$$a^3+15ab^2=6a^2b+10b^3\\a^3-6a^2b+15ab^2-10b^3=0\tag4$$
Dividing the last equation of $(4)$ by $b^3$ and substituting $x=a/b$ gives the cubic$$x^3-6x^2+15x-10=0\implies x=1$$
Therefore, $a=b$ and plugging that back into the system, we get$$a^3+15ab^2=2\implies 16b^3=2\implies a=b=\dfrac 12$$
Hence, $\sqrt[3]{2+\sqrt5}=\dfrac 12+\dfrac {\sqrt5} 2$ and similarly, $\sqrt[3]{2-\sqrt5}=\dfrac 12-\dfrac {\sqrt5}2$.
Therefore, we have$$\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}=\dfrac 12+\dfrac 12+\dfrac {\sqrt5}2-\dfrac {\sqrt5}2=\boxed{1}\tag5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Identifying $\mathbb{C} D_8$ as a product of matrix rings Let $G = D_8$ be the dihedral group of order $8$, i.e.,
\begin{align*}
G = \langle a,b \mid a^4 = 1 = b^2, ab = ba^{-1} \rangle.
\end{align*}
By standard results from ring/represention theory, the group algebra $\mathbb{C} G$ decomposes as a product of matrix rings as follows:
\begin{align*}
\mathbb{C}G \cong \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}).
\end{align*}
I would like to explicitly find elements of $\mathbb{C} G$ which behave like the elements $(1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (0,0,0,1,0), (0,0,e_{ij})$, where $e_ij$ is the matrix with a $1$ in the $(i,j)$th position, and zeros elsewhere.
I've found the following pairwise orthogonal idempotents which hopefully correspond to $1$ in each copy of $\mathbb{C}$:
\begin{align*}
e_1 = \tfrac{1}{8}(1+a+a^2+a^3+b+ab+a^2b+a^3b)\\
e_2 = \tfrac{1}{8}(1+a+a^2+a^3-b-ab-a^2b-a^3b)\\
e_3 = \tfrac{1}{8}(1-a+a^2-a^3+b-ab+a^2b-a^3b)\\
e_4 = \tfrac{1}{8}(1-a+a^2-a^3-b+ab-a^2b+a^3b)
\end{align*}
The element
\begin{align*}
e_{11} = \tfrac{1}{4}(1+ia-a^2-ia^3+b+iab-a^2b-ia^3b)\\
\end{align*}
is also an idempotent and is orthogonal to the $e_i$, and so it makes sense to set $e_{22} = 1-e_1-e_2-e_3-e_4-e_{11}$. However, I can't seem to find suitable elements for $e_{12}$ and $e_{21}$. Am I going in the right direction at the moment?
(Also, my method has been largely ad hoc; is there a uniform method for approaching this sort of problem for an arbitrary finite group $G$?)
| I managed to work this out, so I'll provide my solution, which carries on from my work in the question.
The elements
\begin{align*}
e_{11} = \tfrac{1}{4}(1+ia-a^2-ia^3) \\
e_{22} = \tfrac{1}{4}(1-ia-a^2+ia^3)
\end{align*}
are orthogonal idempotents which are also orthogonal to each of the $e_i$. Observe that $e_{11}b = be_{22}$ and $be_{11} = e_{22}b$. These identities ensure that if we set
\begin{align*}
e_{12} = e_{11}b = \tfrac{1}{4}(b+iab-a^2b-ia^3b) \\
e_{21} = e_{22}b = \tfrac{1}{4}(b-iab-a^2b+ia^3b),
\end{align*}
then the correct relations hold between these matrices. Additionally, one can check that $e_1+e_2+e_3+e_4+e_{11}+e_{22} = 1$. Finally, the matrix
\begin{align*}
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 & i & -i & 0 & 0 \\
1 & 1 & 1 & 1 & -1 & -1 & 0 & 0 \\
1 & 1 & -1 & -1 & -i & i & 0 & 0 \\
1 & -1 & 1 & -1 & 0 & 0 & 1 & 1 \\
1 & -1 & -1 & 1 & 0 & 0 & i & -i \\
1 & -1 & 1 & -1 & 0 & 0 & -1 & -1 \\
1 & -1 & -1 & 1 & 0 & 0 & -i & i
\end{pmatrix}
\end{align*}
has nonzero determinant, so the eight elements we have found are linearly independent. It follows that the map
\begin{gather*}
\phi: \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}) \to \mathbb{C}D_8 \\
e_1 \mapsto \tfrac{1}{8}(1+a+a^2+a^3+b+ab+a^2b+a^3b)\\
e_2 \mapsto \tfrac{1}{8}(1+a+a^2+a^3-b-ab-a^2b-a^3b)\\
e_3 \mapsto \tfrac{1}{8}(1-a+a^2-a^3+b-ab+a^2b-a^3b)\\
e_4 \mapsto \tfrac{1}{8}(1-a+a^2-a^3-b+ab-a^2b+a^3b)\\
e_{11} \mapsto \tfrac{1}{4}(1+ia-a^2-ia^3) \\
e_{22} \mapsto \tfrac{1}{4}(1-ia-a^2+ia^3) \\
e_{12} \mapsto \tfrac{1}{4}(b+iab-a^2b-ia^3b) \\
e_{21} \mapsto \tfrac{1}{4}(b-iab-a^2b+ia^3b)
\end{gather*}
is a surjective ring homomorphism between $8$-dimensional spaces, and hence an isomorphism.
| {
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"url": "https://math.stackexchange.com/questions/2083462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find recurrence relation in a counting problem In the book Galois Theory of Ian Stewart, problem 1.11 states: "Let $P(n)$ be the number of ways to arrange $n$ zeroes and ones in a row, given that ones occur in groups of three or more. Show that $P(n)=2P(n-1)-P(n-2)+P(n-4)$".
From the recurrence relation, I guess we have to divide the problem. Here's what I did
*
*If the first position is $0$, we have $P(n-1)$ ways.
*If the first position is $1$, we have $P(n-1)$ ways subtracting some situations
And this is where I'm stuck. Any help would be appreciated. Thank you
| $$
\begin{align}
P(n)
&=\overbrace{P(n-1)}^{\substack{\text{# of arrangements}\\\text{of $n-1$ digits}\\\text{prepending a $0$}}}
+\overbrace{P(n-1)}^{\substack{\text{# of arrangements}\\\text{of $n-1$ digits}\\\text{prepending a $1$}}}
-\overbrace{P(n-2)}^{\substack{\text{# of arrangements}\\\text{of $n-2$ digits}\\\text{prepending a $10$}}}
+\overbrace{P(n-4)}^{\substack{\text{# of arrangements}\\\text{of $n-4$ digits}\\\text{prepending a $1110$}}}\\
\end{align}
$$
The "# of arrangements of $n-1$ terms prepending a $1$" counts the arrangements starting with $4$ or more ones as well as those starting with "$10$", so we need to subtract the latter. We then need to add those arrangements starting with only $3$ ones.
Using Generating Functions
$$
\begin{align}
&\sum_{k=0}^\infty\overbrace{\vphantom{\left(\frac{x^3}{1-x}\right)^k}\left(1+\frac x{1-x}\right)}^{\text{$0$ or more zeros}}\overbrace{\left(\frac{x^3}{1-x}\frac x{1-x}\right)^k}^{\substack{\text{$3$ or more ones and}\\\text{$1$ or more zeros}}}\overbrace{\vphantom{\left(\frac{x^3}{1-x}\right)^k}\left(1+\frac{x^3}{1-x}\right)}^{\substack{\text{$0$ or $3$ or more ones}}}\\
&=\frac{1-x+x^3}{(1-x)^2}\sum_{k=0}^\infty\left(\frac{x^3}{1-x}\frac x{1-x}\right)^k\\
&=\frac{1-x+x^3}{(1-x)^2}\frac1{1-\frac{x^4}{(1-x)^2}}\\
&=\frac{1-x+x^3}{1-2x+x^2-x^4}
\end{align}
$$
The denominator of the generating function requires the recursion
$$
P(n)=2P(n-1)-P(n-2)+P(n-4)
$$
as mentioned in the text.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The left-hand side is $a\cos x+b\sin x=\frac{ab}{c}+\frac{ab}{c}=2\frac{ab}{c}$. The right-hand side is $\sqrt{a^2+b^2}\cos(x-\alpha)=c\left(\cos{x}\cos{\alpha}+\sin{x}\sin{\alpha}\right)=c\left(\frac{ab}{c^2}+\frac{ab}{c^2}\right)=2\frac{ab}{c}$.
Is my proof valid? Is there a more general way to prove it?
For the second part of the question, I think it should be $\alpha=\arccos\frac{a}{\sqrt{a^2+b^2}}=\arcsin\frac{b}{\sqrt{a^2+b^2}}$. Is this correct?
| You are largely correct. However you could prove the first part in a very simple way:
We can write $$P= a\cos x+b\sin x =\sqrt {a^2+b^2} \left[\frac {a}{\sqrt {a^2+b^2}}\cos x+\frac{b}{\sqrt {a^2+b^2}}\sin x \right] $$ Now we can take $\frac{a}{\sqrt {a^2+b^2}} $ as $\cos \alpha $ giving us $$P=\sqrt {a^2+b^2}[\cos \alpha \cos x+\sin \alpha \sin x]=\sqrt{a^2+b^2}\cos (x-\alpha) $$ And also $$\alpha =\arccos \frac {a}{\sqrt {a^2+b^2}} $$ which I think maybe a small typo in your calculation.
You need not prove the first part with the help of a triangle but your approach is fine. Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Provide a different method of proving:$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}$ Accidentally founded this particular integral producing a rational number
I can't be for sure it is correct, so can one provide a proof of it.
$$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}\tag
1$$
I found related to $(1)$ is this
Let enforce a substitution of $u=x+e^{\pi}$ then $du=dx$
$$\int_{-\infty}^{\infty}{1\over [\pi{u}^2+\pi^{1/3}]^2}du={1\over 2}\tag2$$
To avoid confusing with too much $\pi$ symbol, we write a general
$$\int_{-\infty}^{\infty}{1\over [A{u}^2+B]^2}du={1\over 2}\tag3$$
We could apply partial decomposition
$${au+b\over Au^2+B}+{cu+d\over (Au^2+B)^2}=1\tag4$$ then find a,b,c and d.
I found a general integral of
$$\int{dx\over(x^2+a^2)^2}={x\over2a^2(x^2+a^2)}+{1\over 2a^3}\tan^{-1}{\left(x\over a\right)}\tag5$$
I am sure this is enough to prove $(1)$
Question: what are other methods can we apply to prove $(1)?$
| Starting with,
$$\int_{-\infty}^{\infty}{1\over [A{u}^2+B]^2}du$$ you could let $u=\sqrt{\frac{B}{A}} \tan(\theta)$ so $du=\sqrt{\frac{B}{A}} \sec^2(\theta)d \theta.$ Using $1+\tan^2(x)=\sec^2(x)$, this gives, $$\sqrt{\frac{B}{A}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2(\theta)}{(B\tan^2(\theta)+B)^2}d\theta=\sqrt{\frac{1}{AB^3}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta)d\theta.$$ Now using $cos^2(x)=\frac{1}{2}(cos(2x)+1)$, we have, $$\frac{1}{2\sqrt{AB^3}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(2\theta)+1 d\theta=\frac{1}{2\sqrt{AB^3}}(\frac{\sin(2\theta)}{2} +\theta)|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2\sqrt{AB^3}}.$$ Plugging in $A=\pi$ and $B=\pi^{\frac{1}{3}}$ gives the desired result.
| {
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"url": "https://math.stackexchange.com/questions/2085788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)}.$$
Let's prove $\frac{n+2}{3n+1}$ is a decreasing sequence.
$$a_n>a_{n+1} \Leftrightarrow \frac{n+2}{3n+1}>\frac{n+3}{3n+4}\Leftrightarrow(n+2)(3n+4)>(n+3)(3n+1)\Leftrightarrow3n^2+10n+8>3n^2+10n+3\Leftrightarrow 8>3$$
So $\frac{n+2}{3n+1}$ is a decreasing sequence and we know that $\frac{1}{n(3n+1)}$ is also decreasing so our given sequence is a decreasing sequence as a sum of $2$ decreasing sequences.
| You're solution is fine. Here is another, perhaps more efficient way forward.
We start with the decomposition in the OP as expressed by
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)} \tag 1$$
Then, we simply note that the first term on the right-hand side of $(1)$ can be written as
$$\frac{n+2}{3n+1}=\frac13 \frac{3n+6}{3n+1}=\frac13 \left(1+ \frac{5}{3n+1}\right) \tag 2$$
from which we see by inspection that $\frac{n+2}{3n+1}$ is decreasing. And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the $\gcd[x+y+z; x^2+xy+z^2; y^2+yz+z^2; z^2+zx+x^2]$ What I have done:
There exists a non-zero integer $t$ such:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\dfrac{v-w}{y-z}=\dfrac{w-u}{z-x}=k$$
$\implies$
$$x+y+z=kt $$
$$k(x-y)=u-v$$
$$k(y-z)=v-w $$
$$k(z-x)=w-u $$
$\implies$
$$u=w+k(x-z) $$
$$v=w+k(y-z)$$
$$w=w $$
$\implies$
$$x+y+z=kt$$
$$x^2+xy+y^2=[w+k(x-z)]t$$
$$y^2+yz+z^2=[w+k(y-z)t]t$$
$$z^2+zx+x^2=wt$$
$\implies$
| I first noticed that $(x,y,z)$ are not interchangeable, and found only $gcd=1$ and $gdg=3$, with $(x,y,z)$ in the range $1$ to $999$ and co-prime in pairs.
I’ll use $(f_1,f_2,f_3,f_4)$ for $(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)$
*
*When $x=3a+1,y=3b+1,z=3c+1$
$$f_1=3*(a+b+c+1)$$
$$f_2=3*(3a^2+3ab+3a+b+3c^2+2c+1)$$
$$f_3=3*(3b^2+3bc+3b+3c^2+3c+1)$$
$$f_4=3*(3a^2+3ac+3a+3c^2+3c+1)$$
*When $x=3a+2,y=3b+2,z=3c+2$
$$f_1=3*(a+b+c+2)$$
$$f_2=3*(3a^2+3ab+6a+2b+3c^2+4c+4)$$
$$f_3=3*(3b^2+3bc+6b+3c^2+6c+4)$$
$$f_4=3*(3a^2+3ac+6a+3c^2+6c+4)$$
*Otherwise, in the range tested,
$$gcd(f_1,f_2,f_3,f_4)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Sum of $n$ terms of the given series. Find the sum of $n$ terms of the series:
$$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\frac{4x^3}{(x+1)(x+2)(x+3)(x+4)}+.......$$
Could someone give me slight hint to proceed in this question?
| Hint. One may observe that
$$
\frac{nx^{n-1}}{\prod _{k=1}^n (x+k)}=\left(1-\frac{x^n \Gamma(x+1)}{\Gamma(x+n+1)}\right)-\left(1-\frac{x^{n-1} \Gamma(x+1)}{\Gamma(x+n)}\right)
$$ giving
$$
\sum_{n=1}^N\frac{nx^{n-1}}{\prod _{k=1}^n (x+k)}=1-\frac{x^N \Gamma(x+1)}{\Gamma(x+N+1)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Exponent rules- basic algebra Can some one point out where I've gone wrong. The correct answer, aside from the 9, is on the left page. I tried it again, changing up what I used as the denominator and it looks even more incorrect. Was what I did on the right page wrong?
| We have: $\dfrac{\dfrac{x^{2}y^{-3}}{3z^{2}}-\dfrac{z^{-3}y^{-3}}{3x^{2}}}{\dfrac{x^{-4}y^{2}}{3z^{-2}}}$
$=\bigg(\dfrac{x^{2}y^{-3}}{3z^{2}}-\dfrac{z^{-3}y^{-3}}{3x^{2}}\bigg)\cdot\dfrac{3z^{-2}}{x^{-4}y^{2}}$
$=\bigg(\dfrac{(3x^{2})(x^{2}y^{-3})-(3z^{2})(z^{-3}y^{-3})}{(3z^{2})(3x^{2})}\bigg)\cdot\dfrac{3z^{-2}}{x^{-4}y^{2}}$
$=\bigg(\dfrac{3x^{4}y^{-3}-3y^{-3}z^{-1}}{9x^{2}z^{2}}\bigg)\cdot\dfrac{\dfrac{3}{z^{2}}}{\dfrac{y^{2}}{x^{4}}}$
$=\left(\dfrac{\dfrac{3x^{4}}{y^{3}}-\dfrac{3}{y^{3}z}}{9x^{2}z^{2}}\right)\cdot\dfrac{3}{z^{2}}\cdot\dfrac{x^{4}}{y^{2}}$
$=\left(\dfrac{\dfrac{(y^{3}z)(3x^{4})-(y^{3})(3)}{(y^{3})(y^{3}z)}}{9x^{2}z^{2}}\right)\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$
$=\left(\dfrac{\dfrac{3x^{4}y^{3}z-3y^{3}}{y^{6}z}}{9x^{2}z^{2}}\right)\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$
$=\dfrac{3x^{4}y^{3}z-3y^{3}}{9x^{2}y^{6}z^{3}}\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$
$=\dfrac{9x^{8}y^{3}z-9x^{4}y^{3}}{9x^{2}y^{8}z^{5}}$
$=\dfrac{9x^{4}y^{3}\big(x^{4}z-1\big)}{9x^{2}y^{8}z^{5}}$
$=\dfrac{x^{2}\big(x^{4}z-1\big)}{y^{5}z^{5}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Drawing $\lvert z-2\rvert + \lvert z+2\rvert=5$ in complex plane If we substitute $z=x+iy$ to $\lvert z-2\rvert + \lvert z+2\rvert=5$ and solve for $iy$ we will get
$$ iy=-\frac{1}{2}(2x-5) $$
$$ iy=-\frac{1}{2}(2x+5) $$
Then we can draw like in the $\mathbb{R}^2$ plane. This way we get two parallel lines.
Do you think this approach is correct for the complex plane? Just want to make sure.
| $1)$ If you want go through a geometric approach: by definition of ellipse one can conclude that $z$ is on an ellipse with foci $-2$ and $2$ and bigger axis equal to $5$.
$2)$ If you want go through an analytic approach you can write $z=x+iy$ and then
$$\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=5$$
$$\sqrt{(x-2)^2+y^2}=5-\sqrt{(x+2)^2+y^2}$$
Square both sides:
$$(x-2)^2+y^2=25-10\sqrt{(x+2)^2+y^2}+(x+2)^2+y^2$$
$$10\sqrt{(x+2)^2+y^2}=25+8x \rightarrow 100[(x+2)^2+y^2]=(25+8x)^2$$
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
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$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$?
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$.
I think this is a simple exercise, but I get this:
$(n+1)^2+1=n^2+2n+2$.
$n^2+2n+2 = (n^2+1)+(2n+1)$
then $\gcd(n^2+1, (n+1)^2+1)=\gcd(n^2+1, 2n+1)$
and $\displaystyle n^2+1 = \frac{n(2n+1)}{2}+\left(-\frac{n}{2}+1\right)$
then $\gcd(n^2+1, 2n+1)=\gcd(2n+1, \frac{n}{2}-1)$.
But gcd is on integer numbers and $\dfrac{n}{2}-1$ is not always an integer number, so, I need some help?
| Note that $$(2n+3)(n^2+1) - (2n-1)((n+1)^2+1) = 5.$$
To get this equation, set $(n+a)(n^2+1)-(n+b)(n^2+2n+2) = c$ for some constants $a,b,c$, and equate coefficients. This gives $b = -\tfrac{1}{2}$, $a = \tfrac{3}{2}$, and $c = \tfrac{5}{2}$. Then, multiply both sides by $2$.
Hence, $\text{gcd}(n^2+1,(n+1)^2+1)$ divides $5$ (which is prime).
Then, since $n^2+1 \equiv 0 \pmod{5}$ iff $n \equiv 2,3 \pmod{5}$ and $(n+1)^2+1 \equiv 0 \pmod{5}$ iff $n \equiv 1,2 \pmod{5}$ we have that $$\text{gcd}(n^2+1,(n+1)^2+1) = \begin{cases}5 & \text{if} \ n \equiv 2 \pmod{5} \\ 1 & \text{otherwise}\end{cases}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1$ Let $x_1,x_2,x_3,\cdots ,x_n (n\ge2)$ be real numbers greater than $1.$ Suppose that $|x_i-x_{i+1}|<1$ for $i=1,2,3,\cdots,(n-1)$.
Prove that $$\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1$$
Please help!!!
| Proof: I get this proof for two case
(1): if for all $k=1,2,\cdots,n-1$,have $a_{k}\le a_{k+1}$,then we have
$$a_{k}\le a_{k+1}<a_{k}+1$$
so we have
$$a_{i}<a_{i-1}+1<a_{i-2}+2<\cdots<a_{1}+(i-1)$$
then we have
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n}}{a_{1}}<(n-1)+\dfrac{a_{1}+(n-1)}{a_{1}}<(n-1)+1+(n-1)=2n-1$$
(2):define set $A=\{k|a_{k}>a_{k+1}\}$,Assmue that $|A|=p$,for $k\in A$,we have
$$a_{k+1}<a_{k}<a_{k+1}+1,\dfrac{a_{k}}{a_{k+1}}<\dfrac{a_{k+1}+1}{a_{k+1}}
<1+\dfrac{1}{a_{k+1}}<2$$
then we have
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}<2p+(n-1-p)=n-1+p$$
and
$$\dfrac{a_{n}}{a_{1}}=\dfrac{(a_{n}-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots+(a_{2}-a_{1})+a_{1}}{a_{1}}<\dfrac{(n-1-p)+a_{1}}{a_{1}}<(n-1-p)+1=n-p$$
so for this case also have
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n}}{a_{1}}<n-1+p+n-p=2n-1$$
| {
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Find the Frenet Frame of a curve
Find the Frenet Frame of $\alpha(t)=(2t,t^2,\frac{t^3}{3})$ for $t\in\mathbb{R}$
First finding $\alpha'=(2,2t,t^2)$ and $\alpha''=(0,2,2t)$
$||\alpha'||=\sqrt{4+4t^2+t^4}=\sqrt{(t^2+2)^2}=t^2+2 \neq 1$ so not arc length parameterized
$T=\frac{\alpha'}{||\alpha'||}=(\frac{2}{t^2+2}, \frac{2t}{t^2+2}, \frac{t^2}{t^2+2})$
$N=\frac{T'}{||T'||}$
However I am struggling from here:
$T'=(\frac{-4t}{(t^2+2)^2}, \frac{4-2t^2}{(t^2+2)^2}, \frac{4t+2t^3-2t^4}{(t^2+2)^2})$
$$||T'||=\sqrt{\frac{16t^2+16-16t^2+4t^3+16t^2+8t^4-8t^5+8t^4+4t^6-4t^7-8t^5-4t^7+4t^8}{(t^2+2)^4}}=\sqrt{\frac{16+20t^2+16t^4-16t^5+4t^6-8t^7+4t^8}{(t^2+2)^4}}$$
Providing my calculation is correct, how do I simplify this crazy expression?
| So using $B=\frac{\alpha' \times \alpha''}{||\alpha' \times \alpha''||}$
$\alpha' \times \alpha'' = (2t^2, -4t, 4)$
$||\alpha' \times \alpha''||=\sqrt{4t^4+16t^2+16}=2\sqrt{(t^2+2)^2}=2(t^2+2)$
$B=(\frac{t^2}{t^2+2}, \frac{-2t}{t^2+2}, \frac{2}{t^2+2})$
$N=-(T \times B)=(\frac{-2t^3-4t}{(t^2+2)^2}, \frac{-t^4+4}{(t^2+2)^2}, \frac{2t^3+4t}{(t^2+2)^2})$
| {
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"url": "https://math.stackexchange.com/questions/2091839",
"timestamp": "2023-03-29T00:00:00",
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then maximum and minimum value of $x+y$ if $x,y\in R$ and $x^3+y^3=2\;,$ then maximum and minimum value of $x+y$
using $\displaystyle \frac{x^3+y^3}{2}\geq \left(\frac{x+y}{2}\right)^3$
So $(x+y)^3\leq 2^3$ so $x+y\leq 2$
could some help me to find minimum value, thanks
| If $x,y$ are real numbers then there is no absolute minimum. In fact the value of $x+y$ can be made as close to zero as we like. We have $y = \sqrt[3]{2 - x^3}$
Obviously for very large $x$ we have that $y \approx -x$. Additionally the value must be positive, as $y > -x$, from the condition.
On the other side if we add the condition that $x,y$ are positive, then we have:
$$(x+y)^3 \ge x^3 + y^3 = 2 \implies x+y \ge \sqrt[3]{2}$$
And the minimum is obtained for $(x,y) = \{(\sqrt[3]{2},0), (0,\sqrt[3]{2})\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Irreducibles in $\mathbb{Q}$ and Inverses of Quadratics in $\mathbb{Q}(\alpha)$ Consider $f(x) = x^3 - 9x + 3 \in \mathbb{Q}[x],$ and let $\alpha$ be a root of $f(x).$ Prove that $f(x)$ is irreducible over $\mathbb{Q}.$ Furthermore, in the field $\mathbb{Q}(a),$ find $(3 \alpha^2 + 2 \alpha + 1)^{-1}$ in terms of the basis $1, \alpha, \alpha^2.$
$\textit{Proof.}$ Observe that $f$ is 3-Eisenstein, hence $f$ is irreducible in $\mathbb{Q}[x].$
$\textit{Solution.}$ Because $f$ is irreducible, it follows that $f$ is relatively prime to $g(x) = 3x^2 + 2x + 1.$ By definition of relatively prime, there exist polynomials $u$ and $v$ in $\mathbb{Q}[x]$ such that $$(3x^2 + 2x + 1)u(x) + (x^3 - 9x + 3)v(x) = 1.$$
By the Division (Euclidean) Algorithm, I know that we can explicitly find our $u(x)$ and $v(x),$ but I am not sure how to proceed with the algorithm because it looks a little funky to multiply by these unknown polynomials. Ultimately, if I can find $u(x),$ I know that $u(\alpha) = (3 \alpha^2 + 2 \alpha + 1)^{-1}.$ Can anyone point me in the right direction with this. Thanks very much for your consideration.
| I really like Lubin's approach. It is something I haven't seen before. The Euclidean Algorithm is tried and true, but as you will see, it is not fun in this case.
Using the division algorithm, we divide $x^3-9x+3$ by $3x^2 + 2x + 1$ to see that:
$$ x^3 - 9x + 3 = \left( \frac{1}{3}x-\frac{2}{9}\right) (3x^2+2x+1) + \left(-\frac{80}{9}x + \frac{29}{9}\right).$$
Now divide $3x^2 + 2x + 1$ by $\left(-\frac{80}{9}x + \frac{29}{9}\right)$ to see that:
$$ 3x^2+2x+1 = \left(-\frac{80}{9}x + \frac{29}{9}\right) \left(-\frac{27}{80}x - \frac{2223}{6400} \right) + \frac{13563}{6400}$$
Therefore:
\begin{align*}
1 &= \frac{6400}{13563}(3x^2+2x+1) - \frac{6400}{13563}\left(-\frac{80}{9}x + \frac{29}{9}\right)\left(-\frac{27}{80}x - \frac{2223}{6400} \right)\\
&= \frac{6400}{13563}(3x^2+2x+1) - \frac{6400}{13563}\left((x^3-9x+3) - \left( \frac{1}{3}x-\frac{2}{9}\right) (3x^2+2x+1) \right)\left(-\frac{27}{80}x - \frac{2223}{6400} \right)_.
\end{align*}
From here it is just some gross simplification, then modding out by $x^3-9x+3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the general solution to this Differential Equation. Given that $z=f(x,y)$ and $a\in \mathbb{R}$ is a constant, I have to solve the following differential equation:
$$ \frac{z \,dz+y \,dy}{y^2+z^2}=\frac{dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$
I have not seen anything like this before so any ideas/hints would be much appreciated.
| Let rewrite equation:
$$\frac{z\,dz+y\,dy}{y^2+z^2}=\frac{dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$
$$\frac{d(y^2+z^2)}{y^2+z^2}=\frac{2dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$
With changing $y^2+z^2=t^2$ and $x-a=s$:
$$\frac{2tdt}{t^2}=\frac{2ds}{\sqrt{s^2+t^2}+s}.$$
or
$$\frac{\sqrt{s^2+t^2}+s}{t}dt=ds$$
This is a homogeneous equation. Another substitution $s=ut$ gives us
$$\frac{dt}{t}=\frac{du}{\sqrt{u^2+1}}$$
and we conclude
$\ln(t)=\sinh^{-1}(u)+C$. Now we return main variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\quad\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}$ find limit :
$$\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=\;\; ? \;\quad \text {given }\,n \in \mathbb{N}, n>2 ,\text{odd}$$
I tried :
$$\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=\frac{x+x-2x }{x+x-2x}=\frac{0}{0}$$
| If we apply binomial expansion, we get
$$L=\lim_{x\to\pm\infty}\frac{\frac9{nx^{1-1/n}}+\mathcal O\left(\frac1{x^{2-1/n}}\right)}{-\frac9{nx^{1-1/n}}+\mathcal O\left(\frac1{x^{2-1/n}}\right)}\to-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the values of $k$ when $kx^2+x+k$, has equal zeroes? Show that the quadratic equation $kx^2 + 2(x+1)=k$ has real roots for all the values of $k\in \mathbb{R}$
what i did
$kx^2+2x+2-k=0$
$4-4(2-k)(k)>0$
$4-8k+4k^2>0$
$(64±64)÷ 4(2)>0$
$128÷ 8>0$
$16>0$
please help me check
| We can just compute the roots using factorization:
$$
kx^2+2(x+1) = k \\
kx^2 +2x + 2 - k = 0 \\
kx^2 -x^2 + x^2 +2x + 1 + 1-k = 0 \\
(k-1)x^2 + (x+1)^2 + 1-k = 0 \\
(k-1)(x^2 - 1) + (x+1)^2 = 0 \\
(k-1)(x+1)(x - 1) + (x+1)^2 = 0 \\
(x+1)((k-1)(x - 1) + (x+1)) = 0 \\
(x+1)(kx - k - x + 1 + x+1) = 0 \\
(x+1)(kx - k + 2) = 0 \\
$$
Hence the roots are $-1$ and, if $k$ is non-zero, $1-2/k$ .
| {
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Power Series proofs $$\alpha(x) =\sum_{j=0}^\infty \frac{x^{3j}}{(3j)!}$$
$$\beta(x) = \sum_{j=0}^\infty \frac{x^{3j+2}}{(3j+2)!}$$
$$\gamma(x) = \sum_{j=0}^\infty \frac{x^{3j+1}}{(3j+1)!}$$
Show that $\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)$ for every $x, y \in\mathbb R$.
Show that $\alpha(x)^3 + \beta(x)^3 + \gamma(x)^3 − 3\alpha(x)\beta(x)\gamma(x) = 1$ for every $x\in\mathbb R$.
My work:
Ive noticed that $$\sinh(x)=\sum_{j=0}^\infty \frac{x^{2j+1}}{(2j+1)!}$$
and $$\cosh(x)=\sum_{j=0}^\infty \frac{x^{2j}}{(2j)!}$$
Can hyperbolic trig identies be used to solve these and if so how?
| For the latter, notice that \begin{align*} \alpha' &= \beta, \\ \beta' &= \gamma, \\ \gamma' &= \alpha. \end{align*} Then we see \begin{align*}\frac d {dx} (\alpha^3 + \beta^3 + \gamma^3 - 3\alpha \beta \gamma ) &= 3\alpha^2 \alpha'+ 3\beta^2 \beta' + 3\gamma^2 \gamma' - 3\alpha' \beta \gamma - 3\alpha \beta' \gamma - 3\alpha \beta \gamma' \\
&= 3\alpha^2 \beta + 3\beta^2 \gamma + 3\gamma^2\alpha - 3\beta^2 \gamma - 3\gamma^2 \alpha - 3\alpha^2\beta = 0.
\end{align*} Thus $\alpha^3 + \beta^3 + \gamma^3 - 3\alpha \beta \gamma$ is constant. Then $\alpha(0) = 1, \beta(0) = \gamma(0) = 0$ shows that $$\alpha(x)^3 + \beta(x)^3 + \gamma(x)^3 - 3\alpha(x) \beta(x) \gamma(x) = 1, \,\,\,\, \forall x \in \mathbb R.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$ The sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$
Attempt Assume $\displaystyle S = \sum^{50}_{r=1}((2r)^2-1)((2r+4)^2-1) = \sum^{50}_{r=1}(4r^2-1)(4r^2+16r+15)$
$\displaystyle S = \sum^{50}_{r=1}(16r^4+64r^3+56r^2-16r-15)$
could some help me how to solve it, thanks
| You can find the sum using the idea of telescoping sum.
Since
$$((2r)^2-1)((2r+4)^2-1)=(2r-1)(2r+1)(2r+3)(2r+5)$$
we get
$$\small\begin{align}&\sum_{r=1}^{50}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}\color{red}{10}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}\color{red}{((2r+7)-(2r-3))}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}((2r+7)(2r+5)(2r+3)(2r+1)(2r-1)-(2r+5)(2r+3)(2r+1)(2r-1)(2r-3))\\\\&=\frac{1}{10}(107\cdot 105\cdot 103\cdot 101\cdot 99-7\cdot 5\cdot 3\cdot 1\cdot (-1))\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of given infinite series: $\frac14+\frac2{4 \cdot 7}+\frac3{4 \cdot 7 \cdot 10}+\frac4{4 \cdot 7 \cdot 10 \cdot 13 }+....$ Find the sum of infinite series
$$\frac{1}{4}+\frac{2}{4 \cdot 7}+\frac{3}{4 \cdot 7 \cdot 10}+\frac{4}{4 \cdot 7 \cdot 10 \cdot 13 }+....$$
Generally I do these questions by finding sum of $n$ terms and then putting $ \lim{n \to \infty}$ but here I am not able to find sum of $n$ terms. Could some suggest as how to proceed?
| Notice that
$$\frac k{\prod_{m=1}^k(3m+1)}=\frac1{3\prod_{m = 1}^{k-1} (3m+1)}-\frac{1}{3\prod_{m = 1}^k (3m+1)}$$
Which gives us a telescoping series:$$S_N=\frac{1}{3} - \frac{1}{3\prod_{m = 1}^N (3m+1)}$$
which tends to $1/3$ as suspected.
| {
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$\int\sin^2(x) \, dx$ - Which one is right? If I enter $\int\sin^2(x)\,dx$ into an integral calculator it returns:
$$\int\sin^2(x)\,dx=-\frac{\sin(2x)-2x}{4}+c$$
If I calculate it with Angular multiples I come to:
$$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{\sin(2x)-2x} 4 +c$$
Here is my calculation
If I calculate it with partial integration and substitution I come to:
$$\int\sin^2(x)\,dx=-\frac{\sin(x)\cos(x)+x} 2 +c$$
Here is my second calculation
If I plot all of the functions they look also different...
So which one is right? Or maybe are all of them right due to the "+c"?
EDIT:
I have added my calculations, maybe you can help me to spot the wrong signs - thanks!
| With angular multiples,
$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{x}{2} - \frac{\sin(2x)}4 + c$
Now $\frac{x}{2} - \frac{\sin(2x)}4 + c$
$\frac{x}{2} - \frac{2\sin(x)\cos(x)}4 + c$
$\frac{x}{2} - \frac{\sin(x)\cos(x)}2 + c$
$\frac{- \sin(x)\cos(x) + x}2 + c$
That is with partial integration and substitution.
Again $\frac{x}{2} - \frac{\sin(2x)}4 + c$
$ \frac{- \sin(2x)}4 + \frac{x}{2} + c$
$ - \frac{\sin(2x) - 2x}4 + c$
That is with integral calculator.
So all are same. I think you have either typing mistake in your third solution or mistake in solving.
| {
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Evaluating some limits $$f\left(x\right)\:=\:\frac{1}{1-e^{\frac{1}{x}}}, x \in (0, \infty )$$
As far as i know, $\lim _{x\to \infty }f\left(x\right)$ is $\frac{1}{+0}=\infty$. But, $\lim _{x\to \infty }\left(f\left(x\right)\:+\:x\right)=\frac{1}{2}$. How so ?
| As mentioned, $\frac{1}{1-e^{\frac{1}{x}}} \xrightarrow[x \to \infty]{} -\infty$. As for the limit of $f(x) + x$,
$\frac{1}{1-e^{\frac{1}{x}}} + x = \frac{1}{1-(1+\frac{1}{x} +\frac{1}{2x^2} +\mathcal{O}(\frac{1}{x^3}))} + x$
$= \frac{-x}{1+\frac{1}{2x} + \mathcal{O}(\frac{1}{x^2})} + x = -x(1-\frac{1}{2x} + \mathcal{O}(\frac{1}{x^2})) + x $
$= \frac{1}{2} + \mathcal{O}(\frac{1}{x}) \xrightarrow[x \to \infty]{} \frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
Calculate volume between two geometric figures I have a figure C that is defined as the intersection between the sphere $x^2+y^2+z^2 \le 1 $ and the cyllinder $x^2+y^2 \le \frac{1}{4}$.
How should i calculate the volume of this figure?
| Here is a sketch of the sphere and part of the infinite length cylinder:
(Large version)
It suggests to split the intersection volume into three volumes:
*
*the fully enclosed cylinder part $V_m$ in the middle and
*the two spherical caps, one on top $V_t$, one at the bottom $V_t$, where by symmetry $V_t = V_b$
At the cap base circles we have both
$$
x^2 + y^2 + z^2 = 1 \\
x^2 + y^2 = 1/4
$$
Subtracting gives
$$
z^2 = 3/4 \iff \\
z = \pm \sqrt{3}/2
$$
This gives
$$
V_m= \pi r^2 h = \pi (1/2)^2 \cdot 2 \cdot \sqrt{3}/2 = \pi \sqrt{3}/4
$$
For the top cap we stack up disc shaped volume elements
$$
dV = A(z) \, dz
$$
where $A(z)$ is the area of a disc at height $z$ and get
\begin{align}
V_t
&= \int\limits_{\sqrt{3}/2}^1 A(z) \, dz \\
&= \int\limits_{\sqrt{3}/2}^1 \pi r(z)^2 \, dz \\
&= \int\limits_{\sqrt{3}/2}^1 \pi (1-z^2) \, dz \\
&= \pi \, \left[z - z^3/3 \right]_{\sqrt{3}/2}^1 \\
&= \pi \left( 1-1/3 - (\sqrt{3}/2 - (\sqrt{3}/2)^3/3) \right) \\
&= \pi \left( 2/3 - \sqrt{3}/2 + \sqrt{3}/8 \right) \\
&= \pi \left( 2/3 - 3\sqrt{3}/8 \right) \\
\end{align}
This gives
\begin{align}
V &= V_t + V_m + V_b \\
&= \pi \sqrt{3}/4 + 2 \pi \left( 2/3 - 3 \sqrt{3} / 8 \right) \\
&= \pi \left( 4/3 - 2 \sqrt{3}/4 \right) \\
&= \pi \left( 4/3 - \sqrt{3}/2 \right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$. If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$.
My first shot would be to assume the perfect square is $2^{2018}$, but how would I prove that? Even if it is, what is $n$? All help is appreciated.
| We are going to solve a more general question: For any given natural number $k$, find the integer $n$ satisfying that
$$
2^{2k}+2^{2k+3}+2^n=m^2,
$$
where $m$ is a natural number.
Simplifying the equation to $$9\cdot 2^{2k}+2^n=m^2 \Leftrightarrow 2^n=(m+p)(m-p), $$
where $p=3\cdot 2^{k}.$
Splitting the power of $2$ yields
$$
\left\{\begin{array}{l}
m+p=2^s \cdots (1)\\
m-p=2^{n-s} \cdots (2)
\end{array}\right.
$$
for some integer $s>n-s.$
$(1)-(2)$ gives $$
3 \cdot 2^{k}=p=2^{n-s-1}\left(2^{2s-n}-1\right)
$$
$$
\Rightarrow \left\{\begin{array}{l}
n-s-1=k \\
2 s-n=2
\end{array}\right. \Rightarrow \boxed{n=2k+4},
$$
which is the unique solution of the equation,
Back to our equation, when $2k=2014, n=2018$ is its unique solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Patterns in $\frac{80}{81}$ and $\frac{10}{81}$. The decimal form of $\frac{80}{81}$ is $0.987654320\ldots$ notice the expected $1$ is missing. The decimal form of $\frac{10}{81}$ is $0.12345679\ldots$ notice the expected $8$ is missing. Can someone expansion why the decimal form is the way they are? I think it has something to do with $(10-1)^2=81$.
| This is because
$\dfrac1{(1-x)^2}
=\sum_{n=0}^{\infty} (n+1)x^n
$.
Putting $x=.1$,
this is
$\dfrac1{.9^2}
=\dfrac1{.81}
=\dfrac{100}{81}
=1+2/10+3/100+4/1000 + ...
$
so
$\dfrac{10}{81}
=1/10+2/100+3/1000+4/10000 + ...
=0.1234567...
$.
The next terms are
$8/10^8+9/10^9 +10/10^{10}+...
$,
but we get a carry here
(from the $10/10^{10}$)
and these terms have a value of
$9/10^8+0/10^9 +0/10^{10}+...
$
which explains the
$....6790...$.
The other is just $1-x$
where $x$ is a decimal:
$\dfrac{80}{81}
=1-\dfrac{1}{81}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Given $AO:OA' + BO:OB' + CO:OC' = 92$ find the value of $AO:OA' \times BO:OB' \times CO:OC'$.
In $\triangle ABC$, points $A',B',C'$ are on sides $BC,AC,AB$ respectively. $AA', BB', CC'$ are concurrent at point $O$.
Given $AO:OA' + BO:OB' + CO:OC' = 92$ find the value of $AO:OA' \times BO:OB' \times CO:OC'$.
My work
I can find these relations -
$AO:OA' =[AOB]:[BOA'] = [AOC]:[COA']\\
BO:OB' =[AOB]:[AOB'] = [COB]:[COB']\\
CO:OC' =[AOC]:[AOC'] = [BOC]:[BOC']$
Now how to continue?
Source: BdMO 2015 national secondary.
| Another way to solve this problem is to use the method of Mass Points. Assign the masses $a,b,c$ to the points $A, B, C$. This implies that we should assign the masses $(a+b), (b+c), (c+a), (a+b+c)$ to $C', A', B', O$. Then we have:
$$\frac{AO}{A'O} = \frac{b+c}{a} \quad \frac{BO}{B'O} = \frac{c+a}{b} \quad \frac{CO}{C'O} = \frac{a+b}{c}$$
Then using this we have:
$$92 = \frac{AO}{A'O} + \frac{BO}{B'O} + \frac{CO}{C'O} = \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} = \frac{bc(b+c) + ac(a+c) + ab(a+b)}{abc} = \frac{(a+b)(b+c)(c+a) - 2abc}{abc} = \frac{AO}{A'O} \times \frac{BO}{B'O} \times \frac{CO}{C'O} - 2$$
Therefore: $\frac{AO}{A'O} \times \frac{BO}{B'O} \times \frac{CO}{C'O} = 94$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $x^6+(x^2+y)^3$ is a perfect square, then $y$ is a multiple of $x^2$
Let $x$ and $y$ be integers with $x \neq 0$. Prove that if $x^6+(x^2+y)^3$ is a perfect square, then $y \equiv 0 \pmod{x^2}$.
We can expand the given expression as $$x^6+(x^2+y)^3 = 2x^6+3x^4y+3x^2y^2+y^3 = (2x^2+y)(x^4+x^2y+y^2)=k^2,$$ for some integer $k$. Now assume $x \neq 0$. We proceed by contradiction. Let $y = mx^2+d$, where $0 < d \leq x^2-1$ for some $m$. Then the original equation is equivalent to $$((2+m)x^2+d)(x^4+mx^4+dx^2+(m^2x^4+2dmx^2+d^2))$$ and to $$((2+m)x^2+d)((1+m+m^2)x^4+(d+2dm)x^2+d^2).$$ This seemed like a computational way of solving this, but could we still solve it like this?
| As far as I am aware, this is how one can prove it.
Let $a,b$ be integers that satisfy your condition.
So $$a^6+(a^2+b)^3=t^2$$
Then, dividing by $a^6$, we have $$1+\left(1+\frac{b}{a^2} \right)^3=\left(\frac{t}{a^3} \right)^2$$
However, the elliptic curve $1+x^3=y^2$ is an elliptic curve with rank $0$, , so it only has finitely many rational points, all of which are “torsion”, or of finite order. As it has rational points of finite order it follows from the the Nagell-Lutz theorem,
that all rational points are integer points.
As $\left(1+\frac{b}{a^2}, \frac{t}{a^3} \right)$
is a rational point on $1+x^3=y^2$, it follows that $1+\frac{b}{a^2}$ is an integer, or $$b \equiv 0 \pmod{a^2}$$
As desired. We are done. (Also see here).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Ask about beautiful properties of $e$ One of students asked me about "some beautiful properties (or relation) of $e$". Then I list like below
\begin{align}
& e \equiv \lim_{x \to \infty} \left(1+\frac{1}{x} \right)^x\\[10pt]
& e = \sum_{k=0}^\infty \frac{1}{k!}\\[10pt]
& \frac{d}{dx} (e^x) = e^x\\[10pt]
& e^{ix} = \cos x + i \sin x \quad \text{(Euler)}\\[10pt]
& e^{i \pi} + 1 = 0
\end{align} After this, he asked me for more relation or properties. I said
I'll think and answer ...
Now I want help to add some relation, properties, or visual things (like proof without words)
Please help me to add something more. Thanks in advance.
***The class was math. 1. engineering
| Here is the (simple) continued fraction for $e-1.$ The pattern continues forever, 1,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,...
To get $e$ itself add one, this is the same as changing the first $1$ to a $2$
$$
\begin{array}{ccccccccccccccccccccccc}
& & 1 & & 1 & & 2 & & 1 & & 1 & & 4 & & 1 & & 1 & & 6 & \\
\frac{0}{1} & \frac{1}{0} & & \frac{1}{1} & & \frac{2}{1} & & \frac{5}{3} & & \frac{7}{4} & & \frac{12}{7} & & \frac{55}{32} & & \frac{67}{39} & & \frac{122}{71} & & \frac{799}{465} \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\frac{(n+2)^{n+1}}{(n+1)^n}-\frac{(n+1)^n}{n^{n-1}}For all $n\in\mathbb N$ prove that:
$$\frac{(n+2)^{n+1}}{(n+1)^n}-\frac{(n+1)^n}{n^{n-1}}<e$$
We can rewrite it in the following form
$$(n+2)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^{n-1}<e$$ or
$$(n+1)\left(1+\frac{1}{n+1}\right)^{n+1}-n\left(1+\frac{1}{n}\right)^{n}<e$$
and how should I proceed?
| I do not know if this is a satisfactory answer for you.
Consider $$A=\frac{(n+2)^{n+1}}{(n+1)^n}\qquad , \qquad B=\frac{(n+1)^n}{n^{n-1}}$$ Take logarithms and expand as Taylor series for large values of $n$. This would lead to $$\log(A)=1+\log \left(n\right)+\frac{1}{2 n}+\frac{1}{3
n^2}-\frac{13}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ $$\log(B)=1+\log \left(n\right)-\frac{1}{2 n}+\frac{1}{3
n^2}-\frac{1}{4 n^3}+O\left(\frac{1}{n^4}\right)$$ Taylor again $$A=e^{\log(A)}=e n+\frac{e}{2}+\frac{11 e}{24 n}-\frac{43 e}{48
n^2}+O\left(\frac{1}{n^3}\right)$$
$$B=e^{\log(B)}=e n-\frac{e}{2}+\frac{11 e}{24 n}-\frac{7 e}{16
n^2}+O\left(\frac{1}{n^3}\right)$$ $$A-B=e-\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculation error the length of an angle bisector How to find the length of an angle bisector ($BK$) in a triangle $A(1;4),~B(7;8),~C(9;2)$.
I use this formula:
$\frac{A_{1} \cdot x+B_{1} \cdot y+C_{1}}{\sqrt{A_{1}^{2}+ B_{1}^{2}}}=\frac{A_{2} \cdot x+B_{2} \cdot y+C_{2}}{\sqrt{A_{2}^{2}+ B_{2}^{2}}}$.
And, my result: $\frac{2x−3y+10}{\sqrt{4+9}}=\frac{3x+y−29}{\sqrt{9+1}}$, but it is not equation of angle bisector.
What's my mistake?
Update: $\frac{x-7}{\frac{1+\frac{\sqrt{13}}{\sqrt{10}} \cdot 9}{1+\frac{\sqrt{13}}{\sqrt{10}}} - 7} - \frac{y-8}{\frac{4+\frac{\sqrt{13}}{\sqrt{10}} \cdot 2}{1+\frac{\sqrt{13}}{\sqrt{10}}} - 8}=0$ - is it bad?
| Maybe you might find interesting this approach:
Once you have all tree points you have all sides ($AB,AC,BC$) and then you can use the angle bisector theorem:
$$\frac{AB}{AK}=\frac{BC}{KC}$$
and using that $AK+KC=AC$ you can find $AK$ and $KC$. After that you can use Stewart's theorem in order to find $BK$:
$$\frac{AB^2}{AK\cdot AC}-\frac{BK^2}{AK\cdot KC}+\frac{BC^2}{KC\cdot AC}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integration by parts: the variance of a standard normal I am calculating the variance of a standard normal, but I stuck with the following part (the answer is different from what I know). What is wrong with my calculation?
$$
\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} y^2 e^{- y^2 / 2} = \left[ y^2 \cdot \left( - \frac{1}{y} \right) \cdot e^{- y^2 / 2} \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} 2y \cdot \left( - \frac{1}{y} \right) \cdot e^{- y^2 / 2} dy = 2?({\rm Should \ be\ 1})
$$
I used integration by parts:
$$
\int_{a}^{b} f(x) g'(x) dx = \left[f(x)g(x) \right]_{a}^{b} - \int_{a}^{b} f'(x) g(x) dx
$$
I thought $f(y) = y^2$, so $f'(y) = 2y$, and $g'(y) = e^{- y^2 / 2}$, so $g(y)= (- 1/y) \cdot e^{- y^2 / 2}$
I also used the result of Gaussian integral:
$$ \int_{-\infty}^{\infty} e^{-a x^2} dx = \sqrt{\frac{\pi}{a}}$$
| \begin{align}
\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} y^2 e^{- y^2 / 2}dy
&=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} y^2 e^{- y^2 / 2} dy\\
&\text{put $\frac{y^2}{2}=t$,we have $ydy=dt$}\\
&=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- t} \sqrt{2t}dt\\
&=\frac{2\sqrt 2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- t} t^{\frac12}dt\\
&=\frac{2\sqrt 2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- t} t^{\frac32-1}dt\\
&=\frac{2\sqrt 2}{\sqrt {2\pi}}\Gamma(\frac32)\\
&=\frac{2\sqrt 2}{\sqrt {2\pi}}\times\frac{\sqrt \pi}{2}\\
&=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find length of intersection between 2 points and a sphere I have a sphere and 2 points. The points have (x,y,z) coordinates and the sphere is defined by its centre (0,0,0) and radius R. I am trying to find the length between the 2 points which intersects the sphere. How can I obtain the equation to describe this length?
See below, my objective is Length, L:
| The sphere is: $ x^2 + y^2 + z^2 =R^2 $ . The line is: $ x=x \\ y=x\cdot\frac{y_2-y_1}{x_2-x_1} -\frac{x_1(y_2-y_1)}{x_2-x_1} +y_1 =:r_yx+y_0 \\ z=x\cdot\frac{ z_2-z_1 }{x_2-x_1} -\frac{x_1(z_2-z_1)}{x_2-x_1} +z_1 =:r_zx+z_0$.
Sphere and line intersect when:
$ x^2 + (r_yx+y_0)^2 + (r_zx+z_0)^2 =R^2 \implies \\
x^2\cdot(1+r_y^2+ r_z^2) + x\cdot(2r_yy_0+2r_zz_0)+(y_0^2+z_0^2-R^2)=0\\
\\
$
Solving for $x$ :
$$
x=\frac{ -(2r_yy_0+2r_zz_0) \pm \sqrt{(2r_yy_0+2r_zz_0)^2-4(1+r_y^2+ r_z^2)(y_0^2+z_0^2-R^2)}}{2(1+r_y^2+ r_z^2)}
$$
So for the intersection:
$$
(\Delta x)^2 = \frac{(2r_yy_0+2r_zz_0)^2-4(1+r_y^2+ r_z^2)(y_0^2+z_0^2-R^2)}{(1+r_y^2+ r_z^2)^2} \\
(\Delta y)^2 =(\Delta x)^2(\frac{y_2-y_1}{x_2-x_1})^2\\
(\Delta z)^2 =(\Delta x)^2(\frac{z_2-z_1}{x_2-x_1})^2\\
$$
Length of the line segment is:
$$
\Delta x \cdot \sqrt{1 +(\frac{y_2-y_1}{x_2-x_1})^2 + (\frac{z_2-z_1}{x_2-x_1})^2} = \\
\sqrt{ \frac{(2r_yy_0+2r_zz_0)^2-4(1+r_y^2+ r_z^2)(y_0^2+z_0^2-R^2)}{(1+r_y^2+ r_z^2)^2}}\cdot\sqrt{1 +r_y^2 + r_z^2}=
$$
$$
2\cdot\sqrt{ \frac{(r_yy_0+r_zz_0)^2-(1+r_y^2+ r_z^2)(y_0^2+z_0^2-R^2)}{(1+r_y^2+ r_z^2)}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$ if : $$abc=8 :a,b,c\in \mathbb{R}_{> 0}$$
then :
$$(a+1)(b+1)(c+1)\ge 27.$$
My try :
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+abc$$
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+8, $$
then?
| By Holder $$\prod\limits_{cyc}(a+1)\geq\left(\sqrt[3]{abc}+1\right)^3=27$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Help Determinant Binary Matrix I was messing around with some matrices and found the following result.
Let $A_n$ be the $(2n) \times (2n)$ matrix consisting of elements $$a_{ij} = \begin{cases} 1 & \text{if } (i,j) \leq (n,n) \text{ and } i \neq j \\ 1 & \text{if } (i,j) > (n,n) \text{ and } i \neq j \\ 0 & \text{otherwise}. \end{cases} $$
Then, the determinant of $A_n$ is given by $$\text{det}(A_n) = (n-1)^2.$$
Example: $$A_2 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}, A_3 = \begin{pmatrix} 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \end{pmatrix},$$ with det$(A_2)$ and det$(A_3)$ being $1$ and $4$ respectively. I was wondering if anybody could prove this statement for me.
| As I commented : $detA = detB\cdot detc$, where $B = C$ with zero elements $a_{i,j}$ when $i = j$ and $1$ otherwise.
Now it's easy to see that $detB = n-1$, using definition of determinant. We have only one zero position in every row and only one $\sigma \in S_n$, which give us non-zero addition. So we get $det(B) = 1+\dots +1 = n - 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this proportion Okay I think I'm just having a major brain block, but I need help solving this proportion for my physics class.
$$\frac {6.0\times 10^{-6}}{ x^2} = \frac {2.0\times 10^{-6}}{ (x-20)^2}$$
What's confusing me is the solution manual to this problem lists writing the proportion as,
$$\frac {(x-20)^2} { x^2} = \frac {2.0\times 10^{-6}}{ 6.0\times 10^{-6}}$$
and then proceeds to solve the problem from there... but that doesn't seem right to me. Usually you would cross multiply a proportion and solve, but they seemed to do some illegal math or something. Could you guys work me through how to solve this? This answer is 47 by the way.
| The proportion they gave is correct. Simply divide both sides of the equation you started with by $6\cdot10^{-6}$ and multiply both sides by $(x-20)^2$.
So we have
$$
\frac{(x-20)^2}{x^2} =
\frac{2.0\cdot10^{-6}}{6.0\cdot10^{-6}}=\frac{1}{3}.$$
The left-hand side becomes
$$\frac{(x-20)^2}{x^2}=\frac{x^2-40x+400}{x^2}=\frac{1}{3}.$$
Multiplying both sides by $x^2$, we get
$$x^2-40x+400=\frac{x^2}{3} \quad\Rightarrow\quad\frac{2}{3}x^2-40x+400=0.$$
From here you use the quadratic equation to give you $x=10(3±\sqrt{3})$. It's physics, so you probably want the positive one, so $x=10(3+\sqrt{3})\approx47$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question of computing a limit which tends to infinity $$f(x) = (x^3 + 2x^2 - 3x)/ (3x^2 + 3x - 6)$$
Question 1:
It is known that $\lim_{ x \rightarrow \infty} (f(x) - (mx +c))=0$
Find the values of $m$ and $c$,
Using the difference and sum rule and splitting out the values, simplifying $f(x)$,by dividing it with $x^2$, I would be able to prove that $f(x) = \infty$.
Same goes for $mx$ where the answer would be $\infty$ and $c$ being a constant would remain a $c$.
Hence my equation being
$\infty - \infty + c = 0$,
Am I going about this the right way or is there anything wrong in my method of computation?
| $\lim_{x \to \infty}((\frac {x^3+2x^2-3x}{3x^2+3x-6})-(mx+c))=0 \Rightarrow \lim_{x \to \infty} \frac {(x^3+2x^2-3x)-(mx+c)(3x^2+3x-6)}{3x^2+3x-6}=0 \Rightarrow \lim_{x \to \infty} \frac {x^3+2x^2-3x-[3mx^3+(3m+3c)x^2+(-6m+3c)x-6c]}{3x^2+3x-6}=0 \Rightarrow \lim_{x \to \infty} \frac {(1-3m)x^3+(2-3m-3c)x^2+(-3+6m-3c)x+6c}{3x^2+3x-6}=0.$
We want the coefficient of $x^3$ to be zero in the numerator. So $m=\frac 13$.
Now $\lim_{x \to \infty} \frac {(2-3m-3c)+\frac {(-3+6m-3c)}{(x)}+\frac {6c}{x^2}}{3+\frac 3x - \frac {6}{x^2}}=0 \Rightarrow 2-3m-3c=0 \Rightarrow 1-3c =0 \Rightarrow c=\frac 13.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{4a+2b+3}}\leq1$. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{4a+2b+3}}+\sqrt{\frac{b}{4b+2c+3}}+\sqrt{\frac{c}{4c+2a+3}}\leq1$$
The equality "occurs" also for $a\rightarrow+\infty$, $b\rightarrow+\infty$ and $a>>b$.
I tried AM-GM, C-S and more, but without any success.
| We begin with a theorem :
Theorem :
Let $a,b,c,d,e,f$ be positive real number , with $a\geq b \geq c$ , $d\geq e \geq f $ under the three following conditions :
$a\geq d$ , $ab\geq de$ , $abc\geq def$ so we have :
$$a+b+c\geq d+e+f$$
Here we suppose that we have :
$a\geq b \geq 1 \geq c $
So to get the majorization we prove this :
$\sqrt{\frac{a}{4a+2b+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$
Wich is equivalent to :
$\frac{a}{4a+2b+3}\geq \frac{c}{4c+2a+3}$
Or :
$a(4c+2a+3)\geq c(4a+2b+3)$
Wich is obvious under the previous conditions.
With the same reasoning we can prove that we have :
$\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$
Now we study the case :
$\sqrt{\frac{a}{4a+2b+3}}\geq\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem
And
$0.5-\frac{1}{8.2a}\geq 0.5-\frac{1}{8.2b}\geq \frac{1}{8.2a}+\frac{1}{8.2b} $
wich corresponding to $d\geq e \geq f $ in the initial theorem
It's clear that we have :
$\sqrt{\frac{a}{4a+2b+3}}\leq \sqrt{\frac{a}{4a+3}}\leq 0.5-\frac{1}{8.2a}$
And
$\sqrt{\frac{b}{4b+2c+3}}\leq \sqrt{\frac{b}{4b+3}}\leq 0.5-\frac{1}{8.2b}$
So we have :
$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})$
And
$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\sqrt{\frac{c}{4c+2a+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})(\frac{1}{8.2a}+\frac{1}{8.2b})$
Wich is true because we have with the condition $abc=1$
$$27\leq \prod_{cyc}\sqrt{4a+2b+3}$$
So now you just have to apply the theorem with this majorization .
The case $\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{a}{4a+2b+3}} \geq \sqrt{\frac{c}{4c+2a+3}}$ is the same.
And for the case $a\geq 1 \geq b \geq c$ you just have to make the following substitution $B=\frac{1}{b}$ to find the previous case $a\geq b \geq 1 \geq c $
Edit :
With the previous substitution the original inequality becomes with $a\geq b \geq 1 \geq c$ and $ac=b$:
$$\sqrt{\frac{ab}{4ab+2+3b}}+\sqrt{\frac{1}{4+2cb+3b}}+\sqrt{\frac{c}{4c+2a+3}}$$
We can briefly prove that we have :
$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}$
And
$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$
Now we study the case :
$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem
And
$0.5-\frac{1}{11(ab)^2}\geq \frac{1}{3} \geq 1-\frac{1}{3}-(0.5-\frac{1}{11(ab)^2})$
wich corresponding to $d\geq e \geq f $ in the initial theorem
Now you just have to apply the theorem with this majorization .
The case $\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}\geq \sqrt{\frac{1}{4+2cb+3b}}$ works this the same majorization.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Work Proof... Need help! I'm having trouble figuring out a proof at work, and to spare the business jargon I put it in terms of variables. Could someone help?
Prove:
$$\frac{A-B+Y}{1-C} \cdot C +\frac{A-B+X}{1-D} \cdot D = \frac{A-B}{1-C-D} \cdot(C+D)$$
where:
$$X=\frac{A-B+Y}{1-C} \cdot C$$
$$Y=\frac{A-B+X}{1-D} \cdot D$$
| From the condition we get
$(1-C)X-CY=(A-B)C$ and $-DX+(1-D)Y=(A-B)D$ or
$$(1-C)(1-D)X-C(1-D)Y=(A-B)(1-D)C$$ and $$-DCX+C(1-D)Y=(A-B)CD$$ and after adding we get $((1-C)(1-D)-DC)X=(A-B)C$, which gives
$X=\frac{(A-B)C}{1-C-D}$ and $Y=\frac{(A-B)D}{1-C-D}$.
Thus, $$\frac{A-B+Y}{1-C} \cdot C +\frac{A-B+X}{1-D} \cdot D=\frac{A-B+\frac{(A-B)D}{1-C-D}}{1-C} \cdot C +\frac{A-B+\frac{(A-B)C}{1-C-D}}{1-D} \cdot D=$$
$$=(A-B)\left(\frac{1+\frac{D}{1-C-D}}{1-C} \cdot C +\frac{1+\frac{C}{1-C-D}}{1-D} \cdot D\right)=\frac{(A-B)(C+D)}{1-C-D}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$ If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$
i have converted tan to sin and cos and reached to $\sin^2\theta(a^2-c^2+2ac) + c^2-b^2-ac=0$. how do i proceed
thanks
| We have $a\tan\theta +b \sec\theta=c$. So,
\begin{align*}
c-a\tan \theta & = b \sec \theta\\
(c-a\tan \theta)^2 & = (b\sec \theta)^2\\
(a^2-b^2)\tan^2 \theta -2ac \tan \theta+(c^2-b^2) & = 0.
\end{align*}
This is a quadratic in $\tan \theta$. It has two solutions $\tan \alpha$ and $\tan \beta$. So
\begin{align*}
\tan \alpha + \tan \beta & = \frac{2ac}{a^2-b^2}\\
\tan \alpha \cdot \tan \beta & = \frac{c^2-b^2}{a^2-b^2}
\end{align*}
Now,
$$\tan(\alpha+\beta)=\frac{\tan \alpha +\tan\beta}{1-\tan \alpha \tan \beta}$$
$$\tan(\alpha+\beta)=\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}$$
$$\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify derivatives
The math problem asks to find the derivative of the function
$$y=(x+1)^4(x+5)^2$$
I get to the part $$(x+1)^4 \cdot 2(x+5) + (x+5)^2 \cdot 4(x+1)^3$$
How do they arrive at the answer
$$2(x+1)^3(x+5)(3x+11) ?$$
| $$(x+1)^4 \cdot 2(x+5) + (x+5)^2 \cdot 4(x+1)^3$$
$$(x+1)^3((x+1) \cdot 2(x+5) + (x+5)^2 \cdot 4)$$
$$(x+1)^3(x+5)( (x+1)\cdot 2 + (x+5) \cdot 4)$$
$$(x+1)^3(x+5)( 2)((x+1)+ (x+5)2)$$
$$2(x+1)^3(x+5)(x+1+2x+10)$$
$$2(x+1)^3(x+5)(3x + 11)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2\cos(\frac {2\pi}{9})-1]$$.
What should I do further?
| $\cos 3x=4\cos^3x-3\cos x \Rightarrow 2\cos 3x=8\cos^3x-6\cos x$.
You can now see that if you put $x=\frac{\pi}{9}$, you get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
use of triple scalar product to prove an equality $a,b,c$ are unit vectors mutually inclined at an angle $\theta$.
Given that $$(a\times b) + (b\times c)=pa+qb+rc$$ where $p,q,r$ are constants and $a,b,c$ represent the above mentioned vectors. Prove that
$$r^2 + p^2 + \frac{q^2}{\cos\theta}=2.$$
| Let $\, \mathbf{v}=\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c} =
p\mathbf{a}+q\mathbf{b}+r\mathbf{c}$
$\mathbf{a} \cdot \mathbf{v}$ gives
$$\mathbf{a} \cdot \mathbf{b} \times \mathbf{c} =
p+(q+r)\cos \theta \tag{1}$$
$\mathbf{b} \cdot \mathbf{v}$ gives
$$0 = q+(r+p)\cos \theta \tag{2}$$
$\mathbf{c} \cdot \mathbf{v}$ gives
$$ \mathbf{c} \cdot \mathbf{a} \times \mathbf{b} =
r+(p+q)\cos \theta \tag{3}$$
$(1)-(3)$,
$$0=p(1-\cos \theta)+r(\cos \theta-1)$$
For $\cos \theta \ne 1$,
$$p=r \tag{4}$$
Substitute $(4)$ in $(2)$,
$$q=-2p\cos \theta \tag{5}$$
$\mathbf{v} \cdot \mathbf{v}$ gives
$$
(\mathbf{a}\times \mathbf{b})^2+(\mathbf{b}\times \mathbf{c})^2+2
\begin{vmatrix}
\mathbf{a} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{b} \\
\mathbf{a} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{c}
\end{vmatrix} =
p^2+q^2+r^2+2(pq+qr+rp)\cos \theta $$
\begin{align*}
2\sin^2 \theta+2(\cos^2 \theta-\cos \theta) &=
(p^2+r^2)+q^2+2rp\cos \theta+2q(r+p)\cos \theta \\
2(1-\cos \theta) &=
2p^2+4p^2\cos^2 \theta+2p^2\cos \theta-8p^2\cos^2 \theta \\
&= 2p^2(1+\cos \theta-2\cos^2 \theta) \\
&= 2p^2(1-\cos \theta)(1+2\cos \theta) \\
p^2 &= \frac{1}{1+2\cos \theta} \\
p^2+r^2+\frac{q^2}{\cos \theta} &=
\frac{2}{1+2\cos \theta}+\frac{\cos \theta}{1+2\cos \theta} \\
&= 2
\end{align*}
provided $\, -\dfrac{1}{2} < \cos \theta <1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ABCD$ is a rectangle. I am looking for a synthetic proof Let $ABCD$ a parallelogram and $ BE \perp AC, E \in (AC)$, $M$ is the middle of $[AE]$ and $N$ is the middle of $[CD]$. If $ BM \perp NM$ then $ABCD$ is a rectangle. I try to show that $MNCB$ is inscriptible. I am looking for a synthetic proof.
| $$\overrightarrow{BM}\cdot \overrightarrow{MN} = 0$$
$$ \Leftrightarrow (\overrightarrow{BA}\cdot \overrightarrow{BE})\cdot(\overrightarrow{AD} \cdot \overrightarrow{EC}) = 0$$
$$ \Leftrightarrow \color\red{\overrightarrow{BA}\cdot \overrightarrow{AD}} + \color\RoyalBlue{\overrightarrow{BA} \cdot \overrightarrow{EC}} + \color\green{\overrightarrow{BE} \cdot \overrightarrow{AD}} + \overrightarrow{BE}\cdot \overrightarrow{EC}= 0$$
*
*$\color\red{\overrightarrow{BA} \cdot \overrightarrow{AD} = \overrightarrow{BA} \cdot \overrightarrow{BC} = (\overrightarrow{BE} + \overrightarrow{EA})\cdot(\overrightarrow{BE}+\overrightarrow{EC}) = \overrightarrow{BE}^2 + \overrightarrow{EA}\cdot {EC}}$
*$\color\RoyalBlue{\overrightarrow{BA} \cdot \overrightarrow{EC} = (\overrightarrow{BE}+\overrightarrow{EA})\cdot \overrightarrow{EC} = \overrightarrow{EA} \cdot \overrightarrow{EC}}$
*$\color\green{\overrightarrow{BE}\cdot \overrightarrow{AD} = \overrightarrow{BE} \cdot \overrightarrow{BC} = \overrightarrow{BE}\cdot (\overrightarrow{BE} + \overrightarrow{EC}) = \overrightarrow{BE}^2}$
$$\Longrightarrow 2(\overrightarrow{BE}^2 + \overrightarrow{EA}\cdot \overrightarrow{EC}) = 0$$
$$\Rightarrow BE^2 = EA\times EC$$
$$\Rightarrow \angle{ABC} = 90^\circ.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is $\cos(4x)+\sin(2x)$ periodic and how do I calculate the primitive period? My first attempt is under this, i can work out the primitive period of both of the $\cos(4x)$ and $\sin(2x)$ but how do I calculate the primitive period of $\cos(4x)+\sin(2x)$?
My attempt:
Let $u=4x$ then $x=\frac{u}{4}$ and $\cos(u)$ is $2\pi$ periodic thus $T=\frac{2\pi}{4}$ hence $\cos(4x)$ is periodic with primitive period $T=\frac{\pi}{4}$.
Now Let $H=2x$ and thus $x=\frac{H}{2}$ and $\sin(H)$ is also $2\pi$ periodic Thus $\sin(2x)$ is periodic with primitive period $T=\frac{2\pi}{2}=\pi$ but i dont know how to combine these results to calculate the primitive period of the sum of both $\cos(4x)$ and $\sin(2x)$
| Your attempt seems fine. Choose $T = \pi$, because that is a multiple of $T_1 = \pi /4$ and $T_2 = \pi$. Then, \begin{align} f(x + T) & = \cos (4(x+T)) + \sin (2(x+T)) \\ & = \cos (4x + 4T) + \sin (2x + 2T) \\ & = \cos(4x+4\pi) + \sin (2x + 2 \pi) \\ & = \cos (4x) + \sin (2x) \\ & = f(x). \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Probability of the second heart card If you pull a card and it is a heart, you turn it back and you pull another card. If it is not heart, you pull another card without returning the first card back. What is the probability that the second card is a heart.
| Case 1 -
First card is heart = $\frac {13}{52} = \frac {1}{4}$
Second card is heart = $\frac {13}{52} = \frac {1}{4}$
Probability = $\frac {1}{4} × \frac {1}{4} = \frac {1}{16}$
Case 2 -
First card not heart = $\frac {39}{52} = \frac {3}{4}$
Second card is heart = $\frac {13}{51}$
Probability = $\frac {3}{4} × \frac {13}{51} = \frac {13}{68}$
Total probability = $\frac {1}{16} + \frac {13}{68} = \frac {21}{272}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}$$
I think I should be using recurrence relations here but I'm not entirely sure how that method works. I tried this:
Multiplying the first row by $(-1)$ and adding it to all rows:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
Applying Laplace's method to the first column
$$D=5\begin{vmatrix}
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}+6\begin{vmatrix}
5 &5 &\ldots &5 &5 &-1\\
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
I can see that this one is $D$ but of order $99$...Is this leading anywhere? How would you solve this?
| Add all the rows to the last row and in the last row all entries will be $99 \cdot 5 - 1 = 494$. Now you can take $494$ outside of the determinant and now start subtracting $-5$ times the last row from each other row. Then the entries in the second diagonal will be $-6$, the last row will be $1$'s , while all other entries will be $0$. Then you can start changing the first row with the last one, the second with the second to last and so on. As there will be $50$ changes (even number), the value of the determinant will not change. Eventually you will get a triangular number and hence the determinant is:
$$494 \cdot (-6)^{99}$$
| {
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"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
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An expression for $f(2x)$ as a rational function of $f(x)$ A real-valued function $f$ defined on the $\mathbb{R}\setminus\{1\}$ by
\begin{equation*}
f(x) = \frac{3x + 1}{x - 1} = 3 + \frac{4}{x - 1}
\end{equation*}
is invertible, and
\begin{equation*}
f^{-1}(x) = \frac{x + 1}{x - 3} = 1 + \frac{4}{x - 3} .
\end{equation*}
Evaluate constants $A$, $B$, and $C$ such that
\begin{equation*}
f(2x) = \frac{Af(x) + B}{f(x) + C}
\end{equation*}
for every real number $x$ distinct from $1/2$ and 1.
This is a problem from a past high school competition given in a county - Monroe County - in New York. (I will post a solution to the problem.) Does every rational function that is equal to the quotient of two linear functions have such a property? Can $f(3x)$ be expressed analogously? How about $f(kx)$ for any positive integer $k$? Is there any advantage to such an expression over substituting "$2x$" in for "$x$"?
| Here is the intended solution to the problem. This is a lot of work for a problem on a ten-minute competition. (There are six topics at each competition, and in each topic, there are three problems. The problem in the post comes from the topic called "Functions." I doubt anyone solved this problem during the competition.)
Solution
If $x > 1$, for example, there is exactly one real number $y > 3$ such that $f^{-1}(y) = x$.
\begin{equation*}
f(2x) = \frac{Ay + B}{y + C} .
\end{equation*}
$2x > 1$, $f(2x) > 3$, and $f(2x)$ is in the domain of $f^{-1}$.
\begin{equation*}
\frac{2y + 2}{y - 3} = 2x = f^{-1}\bigl(f(2x)\bigr)
= f^{-1}\left(\frac{Ay + B}{y + C}\right)
= \frac{\dfrac{Ay + B}{y + C} + 1}{\dfrac{Ay + B}{y + C} - 3}
= \frac{(A + 1)y + B + C}{(A - 3)y + B - 3C} .
\end{equation*}
\begin{equation*}
(2y + 2)\bigl[ (A - 3)y + B - 3C \bigr] = (2A - 6)y^{2} + (2A + 2B - 6C - 6)y + 2B - 6C ,
\end{equation*}
and
\begin{equation*}
(y - 3)((A + 1)y + B + C) = (A + 1)y^{2} + [-3A + B + C - 3]y - 3B - 3C .
\end{equation*}
For every real number $y > 3$,
\begin{equation*}
(2A - 6)y^{2} + (2A + 2B - 6C - 6)y + 2B - 6C = (A + 1)y^{2} + (-3A + B + C - 3)y - 3B - 3C .
\end{equation*}
So, $(A, \, B, \, C)$ is the solution to the system of linear equations
\begin{equation*}
\begin{cases}
2x - 6 = x + 1 \\
2x + 2y - 6z - 6 = -3x + y + z - 3 \\
2y - 6z = -3y - 3z
\end{cases}
\end{equation*}
in the variables $x$, $y$, and $z$. The only solution to this system of equations is $(7, \, 3, \, 5)$. So, $A = 7$, $B = 3$, and $C = 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Finding function when dx is integrated $\int (5-x^2+\frac{18}{x^4})dx$ I have a question on how to evaluate this integral:
$\int (5-x^2+\frac{18}{x^4})dx$
Is this correct?
$\int (5-x^2+\frac{18}{x^4})dx$
$ \int 5 dx = 5x + C$
$ \int -x^2 = \frac {-x^3}{3}= -\frac{1}{3}x^3 +C $
$ \int \frac {18}{x^4} = \int 18x^{-4} dx = \frac {18x^{-3}}{3}=6x^{-3} + C$
So, $\int (5-x^2+\frac{18}{x^4})dx= 5x-\frac{1}{3}x^3+6x^{-3}+C$
| It is not quite correct. Note that
$$\int 18x^{-4}\,dx=18\,\left(\frac{x^{(-4+1)}}{(-4+1)}\right)+C=-6x^{-3}+C$$
On a side note, it is not good form to write "$5=5x$" as short hand notation for "$\int 5\,dx=5x$."
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate: $\lim_{x\to 0}{\frac{\tan x-\sin x}{x^3}}$
Calculate: $\lim{\frac{\tan x-\sin x}{x^3}}$ as $x\to 0$
I have solved it using a way that gives a wrong answer but I can't figure out why:
$$\lim{\frac{\tan x-\sin x}{x^3}}\\
=\lim{\frac{\tan x}{x^3}-\lim\frac{\sin x}{x^3}}\\
=\lim{\frac{1}{x^2}}\cdot\lim{\frac{\tan x}{x}-\frac{1}{x^2}}\cdot\lim{\frac{\sin x}{x}}\\
=\lim{\frac{1}{x^2}}-\lim{\frac{1}{x^2}}
=0$$
The answer using the standard method gives $\frac{1}{2}$.
It would be great if someone could explain to me why this method is wrong.
| $\frac{\tan{x}-\sin{x}}{x^3}=\frac{2\sin{x}\sin^2\frac{x}{2}}{x^3\cos{x}}\rightarrow2\cdot\left(\frac{1}{2}\right)^2=\frac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Cross-product properties from abstract definition Given two $3$D vectors $\mathbf{u}$ and $\mathbf{v}$ their cross-product $\mathbf{u} \times \mathbf{v}$ can be defined by the property that, for any vector $\mathbf{x}$ one has $\langle \mathbf{x} ; \mathbf{u} \times \mathbf{v} \rangle = {\rm det}(\mathbf{x}, \mathbf{u},\mathbf{v})$.
From this a number of properties of the cross product can be obtained quite easily. It is less obvious that, for instance $|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 - \langle \mathbf{u} ; \mathbf{v} \rangle ^2$, from which the norm of the cross-product can be deduced.
Is it possible to obtain these properties nicely (i.e. without dealing with coordinates), but with elementary linear algebra only (i.e. without the exterior algebra stuff, only properties of determinants and matrix / vector multiplication).
Thanks in advance!
| I believe I've found an elegant proof.
*
*Assume that $ \mathbf{u}\times \mathbf{v} \ne 0$. For $\mathbf{x} = \mathbf{u}\times \mathbf{v}$ we have
\begin{align}
\|\mathbf{u}\times \mathbf{v}\|^4 &= \langle \mathbf{u}\times \mathbf{v},\mathbf{u}\times \mathbf{v}\rangle^2 \\
&=\det(\mathbf{u}\times \mathbf{v},\mathbf{u},\mathbf{v})^2\\
&= \det\left(\begin{bmatrix} \mathbf{u}\times \mathbf{v} \\ \mathbf{u} \\ \mathbf{v}\end{bmatrix}\begin{bmatrix} \mathbf{u}\times \mathbf{v} ,\mathbf{u} ,\mathbf{v}\end{bmatrix}\right)\\
&= \begin{vmatrix} \|\mathbf{u}\times \mathbf{v}\|^2 & \langle \mathbf{u}\times \mathbf{v}, \mathbf{u}\rangle & \langle \mathbf{u}\times \mathbf{v},\mathbf{v}\rangle \\ \langle \mathbf{u},\mathbf{u}\times \mathbf{v}\rangle & \langle \mathbf{u},\mathbf{u}\rangle & \langle \mathbf{u},\mathbf{v}\rangle \\ \langle \mathbf{v},\mathbf{u}\times \mathbf{v}\rangle & \langle \mathbf{v},\mathbf{u}\rangle & \langle \mathbf{v},\mathbf{v}\rangle\end{vmatrix}\\
&= \begin{vmatrix} \|\mathbf{u}\times \mathbf{v}\|^2 & 0 & 0 \\ 0 & \|\mathbf{u}\|^2 & \langle \mathbf{u},\mathbf{v}\rangle \\ 0 & \langle \mathbf{u},\mathbf{v}\rangle & \|\mathbf{v}\|^2\end{vmatrix}\\
&= \|\mathbf{u}\times \mathbf{v}\|^2(\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \langle \mathbf{u},\mathbf{v}\rangle^2).
\end{align}
Dividing by $\|\mathbf{u}\times \mathbf{v}\|^2$ gives the result.
*Assume that $ \mathbf{u}\times \mathbf{v} = 0$. Then the $\det(\mathbf{x},\mathbf{u},\mathbf{v}) = 0$ for all vectors $\mathbf{x}\in\Bbb{R}^3$ so $\{\mathbf{u},\mathbf{v}\}$ are linearly dependent. Equality condition in Cauchy-Schwartz gives
$$\|\mathbf{u}\times \mathbf{v}\|^2 = 0 = \|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \langle \mathbf{u},\mathbf{v}\rangle^2.$$
| {
"language": "en",
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"question_score": "5",
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How many 3-tuples satisfy $x_{1} + x_{2} + x_{3} = 11;$ $(x_{1} ,x_{2} ,x_{3}$ are nonnegative integers?) I know that the total number of choosing without constraint is
$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13·12}{2} =78$
Then with x1 ≥ 1, x2 ≥ 2, and x3 ≥ 3.
the textbook has the following solution
$\binom{3+5−1}{5}=\binom{7}{5}=21$ I can't figure out where is the 5 coming from?
The reason to choose 5 is because the constraint adds up to 6? so 11 -6 =5?
| For the given equation: $$x_1+x_2+x_3 = 11, \;\text{ with } x_1, x_2, x_3 \;\text{ non-negative },$$ your solution is correct.
$$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13!}{2!11!} = \frac{13·12}{2} =78$$
Your final answer is correct, (Now corrected: (but you failed to show that you need to divide $13\times 12$ by $2$ to obtain $78$).
The other solution (from the text) would be the solution to $x_1+x_2+x_3 = 5$, with $x_1, x_2, x_3$ non-negative. In this case, there are $$\binom{3+5-1}{5} = \frac{7!}{2!5!} = \frac{7\cdot 6}{2} = 21$$
Edit after another question-update: We now are solving $$x_1 + x_2 + x_3 = 11, \text{ with }\; x_1\geq 0+1,\,x_2 \geq 0 + 2,\,x_3 \geq 0 + 3.$$
We can solve in the same manner by writing $(x_1+1)+ (x_2 + 2) + (x_3+3) = 11-1-2-3 = 5$.
Then we may simply ascribe $y = x_1 + 1, y_2 = x_2+2, y_3 = x_3 + 3$ to get $$y_1+y_2 +y_3 = 5$$
From here, the text's solution (addressed above) solves the number of the required solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the system of non-homogeneous differential equations using the method of variation of parameters My exam question reads as follows:
Given the system:
$y'_1=y_1-y_2+\frac{1}{\cos \left(x\right)}$
$y'_2=2y_1-y_2$
Solve it using the method of variation of parameters, before that describe that method for solving systems of non-h. differential equations.
I'm not sure on how to solve this system, that is only remember learning about homogeneous systems so not sure even on how to describe the general case. Either way after some googling I find out that as always the solutions is given as $Y=Y_h+Y_p$ and finding the solution to the homogeneous system is something I know how to do. But then how can I find the particular solution?
| I will map it out and you can fill in the details. We are given
$$y'= Ay + g = \begin{pmatrix}1 & -1 \\ 2 & -1\end{pmatrix}y+\begin{pmatrix}\dfrac{1}{\cos x} \\ 0\end{pmatrix}$$
Using eigenvalues / eigenvectors (or other approaches), we find the homogeneous solution
$$Y_h(x) = c_1\begin{pmatrix}\sin x + \cos x \\ 2\sin x\end{pmatrix} + c_2 \begin{pmatrix}-\sin x \\ \cos x - \sin x\end{pmatrix}$$
For Variation of Parameters (there are other approaches too), we will follow Example 2
$$Y = \begin{pmatrix}\sin x + \cos x & -\sin x \\ 2\sin x & \cos x - \sin x \end{pmatrix} \implies Y^{-1} =
\begin{pmatrix}
\cos x-\sin x & \sin x \\-2 \sin x & \sin x + \cos x
\end{pmatrix}$$
We now form
$$Y^{-1} g = \begin{pmatrix}
\cos x -\sin x & \sin x \\-2 \sin x & \sin x + \cos x
\end{pmatrix} \begin{pmatrix}\dfrac{1}{\cos x} \\ 0\end{pmatrix} =
\begin{pmatrix}
\sec x ~(\cos x - \sin x) \\
-2 \tan x
\end{pmatrix}
$$
Next we integrate the previous result
$$\displaystyle \int Y^{-1}g ~dx = \begin{pmatrix}
x+\ln (\cos x) \\
2 \ln (\cos x) \\
\end{pmatrix}$$
We can now write the particular solution $Y_p(x) = Y \displaystyle \int Y^{-1}g~ dx$
$$Y_p(x) = \begin{pmatrix} \cos x (x+\ln (\cos x))+\sin x (x-\ln (\cos x)) \\2 (x \sin x +\cos x \ln (\cos x))\end{pmatrix}$$
Now write
$$Y(x) = Y_h(x) + Y_p(x)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To evaluate $\frac{\cot25+\cot55}{\tan25+\tan55}+ \frac{\cot55+\cot100}{\tan55+\tan100}+\frac{\cot100+\cot25}{\tan100+\tan25}$ To evaluate $$\frac{\cot25^{\circ}+\cot55^{\circ}}{\tan25^{\circ}+\tan55^{\circ}}+ \frac{\cot55^{\circ}+\cot100^{\circ}}{\tan55^{\circ}+\tan100^{\circ}}+\frac{\cot100^{\circ}+\cot25^{\circ}}{\tan100^{\circ}+\tan25^{\circ}}$$
i took lcm and after usual trigonometric identities i have reduced above expression to $$\cot25^{\circ}\cot55^{\circ}+\cot55^{\circ}\cot100^{\circ}+\cot100^{\circ}\cot25^{\circ}$$
How do i proceed from here?
Thanks
| We know that $$\cot A + \cot B = \frac {\cot A \cot B -1}{\cot (A+B)} $$
Now using this identity, we get, $$P =\cot 55^\circ [\cot 25^\circ + \cot 100^\circ] + \cot 25^\circ \cot 100^\circ $$ $$P =\cot 55^\circ [\frac {\cot 25^\circ \cot 100^\circ -1}{\cot 125^\circ}] + \cot 25^\circ \cot 100^\circ $$ $$P = \cot 25^\circ \cot 100^\circ [1 + \frac {\cot 55^\circ}{\cot 125^\circ}] - \frac {\cot 55^\circ}{\cot 125^\circ} $$
Now notice $\cot (180 - \alpha) = -\cot \alpha $. When $\alpha =55^\circ $, then $\cot 125^\circ = - \cot 55^\circ $.
Thus, $$P = \cot 25^\circ \cot 100^\circ [1-1] - \frac {\cot 55^\circ}{-\cot 55^\circ} =1$$ Hope it helps.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I find the point equidistant from three points $(x, y, z)$ and belonging to the plane $x-y+3z=0?$ I struggle to find the point ${P}$ equidistant from the points ${A(1,1,1), B(2,0,1), C(0,0,2)}$ and belonging to the plane ${x-y+3z=0}$.
Any tips?
| In $\Delta ABC$,
\begin{align*}
a &= BC \\
&= \sqrt{5} \\
b &= CA \\
&= \sqrt{3} \\
c &= AB \\
&= \sqrt{2} \\
a^2 &= b^2+c^2 \\
\angle A &= 90^{\circ} \\
O &= \frac{B+C}{2} \tag{circumcentre of $\Delta ABC$} \\
&= \left( 1,0,\frac{3}{2} \right) \\
\vec{AB} \times \vec{AC} &=
(1, -1, 0) \times (-1, -1, 1) \\
&= (-1,-1,-2)
\end{align*}
Equation of axis of enveloping cone for circular section $ABC$ is
$$\mathbf{r}=\left( 1,0,\frac{3}{2} \right)+t(-1,-1,-2)$$
Substitute into $x-y+3z=0$,
\begin{align*}
(1-t)-(-t)+3\left( \frac{3}{2}-2t \right) &= 0 \\
t &= \frac{11}{12} \\
(x,y,z) &= \left( \frac{1}{12},-\frac{11}{12},-\frac{1}{3} \right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convert rectangular equation $(x^2+y^2)^2 - 4(x^2-y^2) = 0$ to polar form So I am not sure how to do this problem : "Convert rectangular equation $(x^2+y^2)^2 - 4(x^2-y^2) = 0$ to polar form". Any help would be appreciated.
| $$0=(x^2+y^2)^2 - 4(x^2-y^2) $$
Recall that $x^2+y^2=r^2$ and $x=r\cos(\theta)$ and $y=r\sin(\theta)$
$$ 0=(r^2)^2-4(r^2 \cos^2(\theta)-r^2\sin^2(\theta))$$
$$ 0=r^4-4(r^2(\cos^2(\theta)-\sin^2(\theta))$$
$$ 0 = r^4-4r^2\cos(2\theta)$$
$$ 0 = r^2(r^2-4\cos(2\theta))$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
I will keep it short and take only an extract (most important part) of
the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$
What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /
If anyone wants see, here is my long solution which I'm not happy with:
$$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$
| Oh goodness, you needn't so much work. All you need to do is this:
$$\begin{align}\frac{n(n+1)(2n+1)}6+(n+1)^2&=\frac{n(n+1)(2n+1)}6+\frac{6(n+1)^2}6\\&\tag 1=c\bigg(n(2n+1)+6(n+1)\bigg)\\&\tag2=c\bigg(2n^2+7n+6\bigg)\\&\tag3=c\bigg((n+2)(2n+3)\bigg)\end{align}$$
$(1)$ factor out $(n+1)/6$ and call it $c$.
$(2)$ expand the remaining stuff
$(3)$ factor the remaining stuff.
| {
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"url": "https://math.stackexchange.com/questions/2137292",
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"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
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Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$
My Attempt:
$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$
$$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$
$$=\dfrac {\cos^2 40°.\cos^2 80°+\cos^2 20°.\cos^2 80°+\cos^2 20°.\cos^2 40°}{\cos^2 20°.\cos^2 40°.\cos^2 80°}$$.
I got paused here. Please help to prove this..
| Starting like dxiv,
let $a=\cos20^\circ,b=-\cos40^\circ,c=-\cos80^\circ,$
As $\cos(3\cdot20^\circ)=\dfrac12$
$\cos(3\cdot40^\circ)=-\dfrac12$
$\cos(3\cdot80^\circ)=-\dfrac12$
As $\cos3x=\cos60^\circ, 3x=360^\circ m\pm60^\circ$ where $m$ is any integer.
$\implies x=120^\circ m+20^\circ$ where $m\equiv-1,0,1\pmod3$
Now as $\cos3x=4\cos^3x-3\cos x$
The roots of $4t^3-3t-\dfrac12=0$ are $a,b,c$
and we need to find $\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}=\dfrac{a^2b^2+b^2c^2+c^2a^2}{(abc)^2}$
$=\dfrac{(ab+bc+ca)^2-2abc(a+b+c)}{(abc)^2}$
By Vieta's formula
$a+b+c=\dfrac04$
$ab+bc+ca=-\dfrac34$
$abc=-\dfrac1{2\cdot4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137609",
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$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$ on 2015 MIT Integration Bee So one of the question on the MIT Integration Bee has baffled me all day today $$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$$ I have tried a variety of things to do this, starting with
Integration By Parts Part 1
$$\frac{\tan x-x}{1+x^2}\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\frac{-2x(\tan x -x)}{\left (1+x^2 \right )^2}\text{d}x$$ which that second integral is not promising, so then we try
Integration By Parts Part 2
$$\tan^{-1} x\tan^2 x\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}2\tan^{-1} x\tan x\sec^2 x\text{d}x$$ which also does not seem promising
Trig Substitution
$x=\tan\theta$ which results $$\int_{0}^{\tan^{-1}\frac{\pi}{4}}\tan^2 \left (\tan\theta\right )\text{d}\theta$$ which I think too simple to do anything with (which may or may not be a valid reason for stopping here) I had some ideas following this like power reducing $\tan^2 x=\frac{1-\cos 2x}{1+\cos 2x}$ which didn't spawn any new ideas. Then I thought maybe something could be done with differentiation under the integral but I could not figure out how to incorporate that. I also considered something with symmetry somehow which availed no results. I'm also fairly certain no indefinite integral exists. Now the answer MIT gave was $\frac{1}{3}$ but wolfram alpha gave $\approx$ .156503. Note
The integral I gave was a simplified version of the original here is the original in case someone can do something with it $$\int_{0}^{\frac{\pi}{4}}\frac{1-x^2+x^4-x^6...}{\cos^2 x+\cos^4 x+\cos^6 x...}\text{d}x$$ My simplification is verifiably correct, I'd prefer no complex analysis and this is from this Youtube Video close to the end.
| This is not the whole answer to this question but it is something new that wasn't mentioned here so I thought I would post it. Let us denote the unknown integral by $J$. Then substituting for $y= \tan(x)$ we get:
\begin{equation}
J = \int\limits_0^1 \frac{y^2}{1+\arctan(y)^2} \cdot \frac{1}{1+y^2} d y =
\int\limits_0^1 \frac{1}{1+\arctan(y)^2} d y - \int\limits_0^1 \frac{1}{1+\arctan(y)^2} \cdot \frac{dy}{1+y^2}=
\left(\sum\limits_{n=0}^\infty (-1)^n \int\limits_0^1 \arctan(y)^{2 n} d y\right) - \arctan(\frac{\pi}{4})
\end{equation}
Now, clearly the series converges which follows from the estimate $\arctan(y) < y$ for $y\in(0,1)$. Integrating by parts we have:
\begin{eqnarray}
\int\limits_0^1 \arctan(y)^2 d y &=& (\frac{\pi}{4})^2+ \frac{\pi}{4} \log(2) - G \\
\int\limits_0^1 \arctan(y)^{2 n} d y &=& \cdots
\end{eqnarray}
where $G$ is the Catalan constant. Now, of course the real challenge is to calculate the last integral above in closed form. This requires more work. However I have a feeling that this integral can be reduced to poly-logarithms for arbitrary values of $n$ and as such does have a "closed form".
We use the following identities:
\begin{eqnarray}
[\arctan(y)]^2 &=& \frac{1}{2} \sum\limits_{n=0}^\infty (-1)^n \frac{(y^2)^{n+1}}{n+1} \cdot \left( \Psi(-1/2) + \Psi(n+3/2)\right) \\
&=& -\frac{1}{4} \log ^2\left(\frac{y+i}{y-i}\right)+\frac{1}{2} i \pi \log \left(\frac{y+i}{y-i}\right)+\frac{\pi ^2}{4}
\end{eqnarray}
Now, raising the identity above to the $n$th power and integrating we readily see that the only non-trivial integral we are dealing is, is the following:
\begin{eqnarray}
{\mathcal A}_n &:=& \int\limits_0^1 [\log(\frac{x+\imath}{x-\imath})]^n dx \\
&=& (-2 \imath) \int\limits_{-1}^{\imath} \frac{ [\log(u)]^n}{(1-u)^2} d u \\
&=&(-\imath)(\imath \pi)^n + (1+\imath) (\imath \frac{\pi}{2})^n + \\
&&(-\imath)n! (-1)^n \left((\log(2)+\imath \frac{\pi}{2}) 1_{n=1} + 2(S_{1,n-1}(2) - S_{1,n-1}(1-\imath)) 1_{n > 1}\right)
\end{eqnarray}
where the path in middle integral is a quarter of a unit circle starting at $-1$ and ending at $\imath$. The quantities $S_{1,n-1}()$ in the bottom formula are the Nielsen generalized poly-logarithms. Here $n\ge 1$. Combining the two identities above we obtain the following identity :
\begin{eqnarray}
&&\int\limits_0^1 \arctan(y)^{2 n} dy = \\
&&
\left(\frac{\pi^2}{16}\right)^n + \\
&& Re\sum\limits_{0\le p_1 \le p_2 \le n}
\frac{n! (2(p_2-p_1)+n-p_2)!}{p_1! (p_2-p_1)!(n-p_2)!}
\left(\frac{\pi^2}{4}\right)^{p_1} \left(\frac{1}{4}\right)^{p_2-p_1} \left(\frac{\pi}{2}\right)^{n-p_2} \cdot \\
&&
\left(
0 \cdot 1_{n+p_2-2 p_1=0} +
(\log(2)) \cdot 1_{n+p_2-2 p_1=1} +
2 (-\imath)^{n+p_2-2 p_1+1}
(S_{1,n+p_2-2 p_1-1}(2) - S_{1,n+p_2-2 p_1-1}(1-\imath)) \cdot
1_{n+p_2-2 p_1>1}
\right)
\end{eqnarray}
valid for $n=1,2,\cdots$. This concludes the calculation.
| {
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\sqrt{3}abc$$
, which is horribly weak !
I can not use equality constraint whichever method I use. Thanks to Wolfram|Alpha, I know the answer is $125$ when $(a,b,c) \equiv (1,1,1).$
Any help would be appreciated. :)
| By Lagrange multipliers, $(a,b,c)=(1,1,1)$ is the only stationary point inside the closed region $(a,b,c)\geq 0, a+b+c=3$. To prove it is a minimum is is enough to study the given function under the assumptions $c=0$ and $b=3-a$. The one variable function
$$ f(a) = 3(3+2a^2)(3+2(3-a)^2) $$
has absolute minima over the interval $[0,3]$ at $a=\frac{3\pm\sqrt{3}}{2}$. It follows that the minimum value of the original function over $(a,b,c)\geq 0, a+b+c=3$ is $162>125$.
An algebraic alternative: by looking at the LHS as a norm in $\mathbb{R}[i]$, we get that is equals
$$ 3(3-2ab-2ac-2bc)^2 +2 (3a+3b+3c-2abc)^2\\ = 3(a^2+b^2+c^2-6)^2+2(9-2abc)^2$$
and by GM-AM-QM is is trivial that the minimum is achieved only at $a=b=c=1$.
| {
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"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\prod \limits_{cyc}(a^3+a+1 ) \leq 27$
Let $a,b,c > 0$ such that $a^2 + b^2 + c^2 =3$. Prove that
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$$
My attempt :
Let $\lambda = \prod \limits_{cyc} a^3+a+1$ .
Applying AM-GM on the set $\{(a^3+a+1),(b^3+b+1),(c^3+c+1)\}$ :
$$\lambda^{\frac{1}{3}}\leq\dfrac{\sum \limits_{cyc} a^3+a^2+a}{3} \leq \dfrac{\sum \limits_{cyc} a^3 +2a^2 }{3} = 2+\dfrac{a^3+b^3+c^3}{3}$$
I am struck now and do not how to proceed. The question seems easy but I can not figure out the right strategy. Also, C-S does not seem to help.
Thanks in Advance !
| From the expansion of $a^4-4a^3+6a^2-4a+1=(a-1)^{4} \ge 0$, we can simply the equation to get the following: $$4a^3+4a+4 \le a^4+6a^2+5 $$
So we have that
$$\prod_{cyc}(a^3+a+1)\leq\frac{1}{4^3}\prod_{cyc}(a^2+1)\prod_{cyc}(a^2+5) \le 27$$
From $\text{AM-GM}$ Inequality. We have the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving functions are linearly independent I'm currently going through Harvard's Abstract Algebra using Michael Artin's book, and have no real way of verifying my proofs, and was hoping to make sure that my proof was right.
The question reads:
Let $V$ be the vector space of functions on the interval $[0, 1]$. Prove that the functions $x^{3}$, $\sin(x)$, and $\cos(x)$ are linearly independent.
My proof goes as follows:
For these to be linearly dependent there must exist an $a_{i} \neq0$, where $ i = 1, 2, 3$ such that
$$a_{1}x^{3} + a_{2}\sin(x) + a_{3}\cos(x) = 0. $$
So, we'll do this in 3 cases:
Case 1: $x = 0$
In this case, $x^{3} = 0$, $\sin(x) = 0$ but $\cos(x) = 1$.
So, we have
$$0\times a_{1} + 0\times a_{2} + 1\times a_{3} = 0.$$
So, $a_{1}$ and $a_{2}$ could be anything but $a_{3}$ must be 0.
Case 2: $x \in (0,1)$
In this case, $x^{3} \neq 0$, $\sin(x) \neq 0$ and $\cos(x) \neq 0$.
So, for this to be true, $a_{1}$, $a_{2}$ and $a_{3}$ all must be $0$.
Case 3: $x = 1$
In this case, $x^{3} = 1$, $\sin(x) = .8...$ and $\cos(x) = .5...$.
So, we have
$$1\times a_{1} +.8\times a_{2} + .5\times a_{3} = 0.$$
So, $a_{3}$ could be any value, while $a_{1}$ and $a_{2}$ must be $0$.
So, if $a_{1} \neq 0$ then we have a problem in Case 3. If $a_{2} \neq 0$ we have a problem in Case 3. If $a_{3} \neq 0$ we have a problem in Case 1. So, we know that all of the $a$ values must be $0$ and we complete the proof.
| Why not assume that $f(x) = a_1x^3+a_2 \sin x + a_3 \cos x$ is the zero function. Then note that $0 = f(0) = a_3$, so we may rewrite $f(x) = a_1x^3 + a_2 \sin x$. Evaluate this rewritten expression for $f$ at the point $\pi$ to show that $0 = f(\pi) = a_1 \pi^3 \Longrightarrow a_1 = 0$. Thus we can again rewrite $f(x) = a_2 \sin x$. Finally, evaluate this latest rewritten expression for $f$ at the point $\pi/2$ to find $0 = f(\pi/2) = a_2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Solutions to a Cubic Root I'm looking for a sanity check to my work.
We are supposed to find the solutions of the cubic polynomial, $x^3+3x+5=0$.
Let $x=w-\frac{1}{w}$ and substitute.
Then $(w-\frac{1}{w})^3+3(w-\frac{1}{w})+5=0$.
Then $w^3+5w^3-1=0$.
Finally $w^3=\frac{-5+\sqrt{29}}{2}$ and $\frac{-5-\sqrt{29}}{2}$.
Solution 1: $\sqrt[3]\frac{-5+\sqrt{29}}{2}-\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
Solution 2: $\left(\frac{-1+\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5+\sqrt{29}}{2}-\left(\frac{-1-\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
Solution 3: $\left(\frac{-1-\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5+\sqrt{29}}{2}-\left(\frac{-1+\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
| Here's a sanity check that shows something is wrong:
$$\sqrt[3]{-5+\sqrt{29}\over2}-\sqrt[3]{-5-\sqrt{29}\over2}=\sqrt[3]{-5+\sqrt{29}\over2}+\sqrt[3]{5+\sqrt{29}\over2}\gt0$$
since $\sqrt{29}\gt5$ (so that the first cube root is positive). But $x^3+3x+5\gt0$ for all $x\gt0$, so the real roots of $x^3+3x+5$ must be negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}$ I tried first without L'Hôpital's rule:
$$\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}} =
\frac{\sqrt{x}}{\sqrt[3]{x}} \cdot \frac{1+\frac{x}{\sqrt x}}{1+ \sqrt[3] x} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\frac{\sqrt x+x}{\sqrt x}}{\frac{\sqrt[3]{x}+x}{\sqrt[3]{x}}} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\sqrt[3]{x}(\sqrt x +x)}{\sqrt x (\sqrt[3]{x}+x)} = \frac{\sqrt x \sqrt[3]{x}(\sqrt x +x)}{\sqrt[3] x \sqrt x (\sqrt[3]x +x )} = \frac{\sqrt x +x }{\sqrt[3] x +x }$$
That didn't work, so then I tried L'Hôpital's:
$$\frac{\frac{1}{2}(x)^{-\frac{1}{2}}+1}{1+\frac{1}{3}(x)^{-\frac{1}{3}}} = \frac{\frac{1}{2\sqrt x }+1}{1+\frac{1}{{3\sqrt[3]{x}}}} = \frac{\frac{1+2\sqrt x}{2\sqrt x}}{\frac{1+3\sqrt[3]{x}}{3\sqrt[3]{x}}} = \frac{3\sqrt[3]{x}(1+2\sqrt x)}{2\sqrt x (1+3\sqrt[3]{x})} = \frac{3\sqrt[3]{x}+3\sqrt[3]{x}2\sqrt{x}}{2\sqrt{x} + 2\sqrt{x}3\sqrt[3]{x}} = ??? $$
How do I solve this? If you can solve this with and without L'Hôpital's, please do so.
| In such cases, it is easier to perform a suitable transformation to obtain integer-valued powers: this suggests letting $x = u^6$, hence we have the equivalent limit $$\lim_{x \to \infty} \frac{x^{1/2} + x}{x+x^{1/3}} = \lim_{u \to \infty} \frac{u^3 + u^6}{u^6 + u^2} = \lim_{u \to \infty} \frac{u^{-3} + 1}{1 + u^{-4}}$$ and the rest is trivial.
| {
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Geometric interpretation of complex number.
Let $z \in \mathbb{C}, z^2, z^3$ be the verticies of a right triangle. Find the geometric images of $z$.
I did not understand the question but I guess it want me to find the figure formed by $z$ under these constraints.
Let $z = x + iy$, then $z^2 = x^2 - y^2 +2xyi$ and $z^3 = x^3 - iy^3 + 3x^2iy - 3xy^2$
It forms a triangle, So, $|z - z^2|^2 + |z^3 - z^2|^2 = |z - z^3|^2$ by Pythagoras theorem.
But when I put the values in I get polynomial in two degree with 6th degree with many extra terms, in short a complete mess.
How can I do this question without actually explanding the Pythogoras theorem ?
| If the vertex at the right angle is $z^2$, then $\dfrac{z^3-z^2}{z-z^2}$ must be purely imaginary. Thus $\dfrac{z^2-z}{1-z} = -z$ is purely imaginary. Thus $z = ri$ for some real $r$.
Let $z = re^{i\theta}$ be the vertex at the right angle. From the figure we have
\begin{align*}
AB^2 &= r^2 + r^4 -2r^3 \cos \theta \\
BC^2 &= r^4 + r^6 - 2r^5 \cos \theta\\
AC^2 &= r^2+r^6 - 2r^4 \cos 2\theta
\end{align*}
Using $BC^2 = AB^2 + AC^2$, we gave
$$ (r^2 + r^4 -2r^3 \cos \theta)+ (r^2+r^6 - 2r^4 \cos 2\theta) = r^4 + r^6 - 2r^5 \cos \theta$$
Simplifying we get
$$1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta = 0$$
When $\theta = 90^\circ$, we have
$1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta = 1+r^2 > 0$ and when $\theta = 180^\circ$, we have $1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta = (1+r)(1-r^2)$ When $r > 1$, this quantity is negative and hence there is a $\theta$ that $\dfrac{\pi}{2} < \theta < \pi $ for which the above equation is satisfied. Hence there is a right angled triangle with $z$ as the vertex at the right angle. An example is shown below.
Similar situation happens when the vertex at right angle is $z^3$.
| {
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Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$ $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$
I want to prove this by Induction
$$n=1 \checkmark\\
n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}\geq \sqrt k\\
n=k+1 \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt {k+1}}\geq \sqrt {k+1}$$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}+\dfrac{1}{\sqrt {k+1}}\geq \sqrt k+\dfrac{1}{\sqrt {k+1}}$$now we prove that $$\sqrt k+\dfrac{1}{\sqrt {k+1}} >\sqrt{k+1} \\\sqrt{k(k+1)}+1 \geq k+1 \\ k(k+1) \geq k^2 \\k+1 \geq k \checkmark$$ and the second method like below ,
and I want to know is there more Idia to show this proof ? forexample combinatorics proofs , or using integrals ,or fourier series ,....
Is there a close form for this summation ?
any help will be appreciated .
| Combining AM-HM $$\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\geq n^2$$
Thus
$$\left(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\right)\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\right)\geq n^2$$
and
$$n\sqrt{n}\geq\left(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\right)$$
so
$$n\sqrt{n}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\right)\geq n^2$$
and
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\geq \sqrt{n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A relation between two elements Let $G $ be a group and $ a, b\in G $ such that $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$.
How can we show that $a=b$$\,$?
| $\bbox[5px,border:2px solid]{\begin{array}{c}b^2a=ab^3\\a^2b=ba^3\end{array}}\ $ these relations are equivalent to OP's since elements in a group are invertible.
If $G$ is commutative or if $a,b$ commute
We have $ab=ba$.
$b^3=a^{-1}b^2a=a^{-1}ab^2=b^2\Rightarrow b=1$
$a^3=b^{-1}a^2b=b^{-1}ba^2=a^2\Rightarrow a=1$
We get $a=b=1$.
G any group, but either $a$ or $b$ is of order $\le$ 4
For instance $a=1$
$b^3=a^{-1}b^2a=b^2\Rightarrow b=1$
For instance $a^2=1$
$a=a^3=b^{-1}a^2b=b^{-1}b=1$ and we conclude like previously
For instance $a^3=1$
$b=ba^3=a^2b\Rightarrow a^2=1$ and we conclude like previously
For instance $a^4=1$
$(ba)a^2=ba^3=a^2b=a^2ba^4=a^2(ba^3)a=a^2(a^2b)a=a^4(ba)=ba\Rightarrow a^2=1$ and we conclude like previously
Everything being symetrical in $a,b$, the same conclusion arises if we make the hypothesis on $b$.
So here also we get $a=b=1$
G any group, but either $a$ or $b$ order is finite
Let assume a^n=1
$a^2(ba)=(a^2b)a=(ba^3)a=(ba)a^3\Rightarrow \bbox[5px,border:2px solid]{\forall k\in\mathbb N,\ a^{2k}(ba)=(ba)a^{3k}}$
If n even then n=2p
$a^{2p}(ba)=a^n(ba)=(ba)=(ba)a^{3p}=(ba)a^na^{p}=(ba)a^{p}\Rightarrow a^{p}=1$ and since $p<n$ we can conclude by recurrence.
If n odd then n=2p+1
$a^{2p}(ba)=a^{n-1}(ba)=a^{-1}(ba)=(ba)a^{3p}=(ba)a^na^{p-1}=(ba)a^{p-1}\Rightarrow (ba)=(ab)a^p$
$b=(ab)a^{p-1}$
$(ab)b^2=ab^3=b^2a=(ab)a^{p-1}\,(ab)a^{p-1}\ a=(ab)a^{p-2}(a^2b)a^p=(ab)a^{p-2}(ba^3)a^p$
$b^2=a^{p-2}ba^{p+3}=a^{p-4}(a^2b)a^{p+3}=a^{p-4}(ba^3)a^{p+3}=a^{p-4}ba^{p+6}$
$\bbox[5px,border:2px solid]{\forall k\in\mathbb N,\ b^2=a^{p-2k}\,b\,a^{p+3k}}$
If p even
For $p=2k$ we have $b^2=ba^{5k}\Rightarrow b=a^{5k}$ then $a,b$ commute and we can conclude.
If p odd
For $p=2k+1$ we have $b^2=aba^{5k+1}\Rightarrow (ab)b^2=ab^3=b^2a=aba^{5k+2}=(ab)a^{5k+2}$
$b^2=a^{5k+2}\Rightarrow ab^3=b^2a=a^{5k+3}\Rightarrow b^3=a^{5k+2}=b^2\Rightarrow b=1$ and we can also conclude.
Everything being symetrical in $a,b$ the same conclusion arises if we make the hypothesis $b^n=1$.
Thus we have proved it in all cases either directly if $n$ odd, either by recurrence if $n$ is even.
G infinite non-commutative group, order of $a,b$ infinite
I have no idea... :-(
| {
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$\lim \limits_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$ find the limit :
$$\lim_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$$
my try :
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\tan 2x}}{3^{\sin 2x}-3^{2\tan x}}=$$
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\frac{\sin 2x}{\cos2x}}}{3^{\sin 2x}-3^{\frac{2\sin x}{cosx}}}=$$
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\sin 2x}+2^{\cos 2x}}{3^{\sin 2x}-3^{2\sin x}+3^{\cos x}}=?$$
now ?
| $$\lim_{x\to0}\frac {2^{\tan 2x}\left(2^{(2\sin x-\tan 2x)}-1\right)}{3^{(2\tan x} \left(3^{(\sin 2x-2\tan x)}-1\right)}$$
$$=\lim_{x\to0}\frac {2^{\tan 2x}}{3^{2\tan x}}\cdot\underbrace{\lim_{x\to0}\dfrac{2^{(2\sin x-\tan 2x)}-1}{2\sin x-\tan 2x}}_{(1)}\cdot\lim_{x\to0}\dfrac1{\underbrace{\dfrac{3^{(\sin 2x-2\tan x)}-1}{\sin 2x-2\tan x}}_{(2)}}\cdot\lim_{x\to0}\dfrac{2\sin x-\tan 2x}{\sin 2x-2\tan x}$$
For $(1),(2)$ use $\lim_{h\to0}\dfrac{a^h-1}h=\ln a$
Now $\lim_{x\to0}\dfrac{2\sin x-\tan 2x}{\sin 2x-2\tan x}$
$=\lim_{x\to0}\dfrac{\cos x}{\cos2x}\cdot\lim_{x\to0}\dfrac{2\sin x\cos2x-\sin2x}{\sin2x\cos x-2\sin x}$
$=\lim_{x\to0}\dfrac{2\sin x(\cos2x-\cos x)}{2\sin x(\cos^2x-1)}$
$=2\cdot\dfrac32\cdot\dfrac12\lim_{x\to0}\dfrac{\sin\dfrac{3x}2}{\dfrac{3x}2}\cdot\lim_{x\to0}\dfrac{\sin\dfrac x2}{\dfrac x2}\cdot\dfrac1{\left(\lim_{x\to0}\dfrac{\sin x}x\right)^2}=?$
| {
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"timestamp": "2023-03-29T00:00:00",
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Letters are arranged in boxes
Ways =(selecting one box from upper and lower rows)(selecting four boxes out of remaining)(number of ways arranging six letters in these boxes)
$$={3\choose 1}{3\choose 1}{6 \choose 4}\cdot6!$$
| [\begin{align}
& There\text{ are 3 cases to be considered : }\left( 1,1,1,3 \right)\text{ ,}\left( 2,1,1,2 \right)\text{ }and\text{ }\left( 3,1,1,1 \right) \\
& \underline{case\text{ 1}}\text{ : }top\text{ }row\text{ }1:6C1*3 \\
& \text{ }second\text{ }row\text{ }1:5C1 \\
& ~\text{ }third\text{ }row\text{ }1:\text{ }4C1 \\
& ~\text{ }fourth\text{ }row\text{ }3\text{ : }3C3*3! \\
& Soln~\text{ : }6C1*3*5C1*4C1*\text{ }3C3*3!=2160 \\
& \text{-------------------------------------------------------------------------------------} \\
& \underline{case\text{ 2}}\text{ : }top\text{ }row\text{ }1:6C2*3P2 \\
& \text{ }second\text{ }row\text{ }1:4C1 \\
& ~\text{ }third\text{ }row\text{ }1:\text{ }3C1 \\
& ~\text{ }fourth\text{ }row\text{ }3\text{ : }3C3*3P2 \\
& Soln~\text{ : }6C2*3P2*4C1*3C1*3C3*3P2=6480 \\
& ------------------------------------------ \\
& \underline{case\text{ 3}}\text{ : }top\text{ }row\text{ }1:6C3*3! \\
& \text{ }second\text{ }row\text{ }1:3C1 \\
& ~\text{ }third\text{ }row\text{ }1:\text{ }2C1 \\
& ~\text{ }fourth\text{ }row\text{ }3\text{ : }1C1*3 \\
& Soln~\text{ : }6C3*3!*3C1*2C1*\text{ }1C1*3=2160 \\
& ------------------------ \\
& Total\text{ }Soln=case1+\text{ }case2+case3=2160+4680+2160=10800 \\
\end{align}]
| {
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Maximum radius of intersection of a plane and an ellipsoid The question is to use Lagrange multipliers to show that the maximum value of $r = (x^2 + y^2 + x^2)^{1/2}$ subject to the following conditions
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \quad\quad\quad lx + my + nz = 0$$ satisfies $$\frac{l^2a^2}{a^2-r^2} + \frac{m^2b^2}{b^2-r^2} + \frac{n^2c^2}{c^2-r^2} = 0$$ where $a$, $b$, $c$, $l$, $m$, $n$ are arbitrary constants.
I have defined the Lagrangian function $L(x,y,z,\lambda,\mu)$ with two multipliers $\lambda$ and $\mu$, one for each of the constraints. After setting $\nabla L = 0$, I get the following system of equations:
$$\frac{x}{r} + \frac{2x\lambda}{a^2} + \mu l = 0$$
$$\frac{y}{r} + \frac{2y\lambda}{b^2} + \mu m = 0$$
$$\frac{z}{r} + \frac{2z\lambda}{c^2} + \mu n = 0$$
All of my attempts to eliminate $\lambda$ and $\mu$ and solve this system of equations have ended in either an algebraic mess or inconsistencies. I have a suspicion that there is an elegant solution there somewhere... can anyone help me find it?
| Multiply your first equation with $x$, second with $y$, third with $z$ and add them together. You will get $$(x^2+y^2+z^2)\frac{1}{r}+2\lambda\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=r+2\lambda=0$$
Now use $2\lambda=-r$ to plug into your equations, find $x,y,z$ and plug it into the equation of the plane:
$$x\left(\frac{1}{r}-\frac{r}{a^2}\right)+\mu l=0$$
$$x=-\frac{\mu l a^2 r}{a^2-r^2}$$ and similar expressions for $y$ and $z$. When you add plug them into the equation of the plane you get
$$-\mu r \left(\frac{l^2a^2}{a^2-r^2}+\frac{m^2b^2}{b^2-r^2}+\frac{n^2c^2}{c^2-r^2}\right)=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Natural parametrization of curves I want to find curvatures and torsions for the following curves but get stuck with their natural parametrizations ($s$ is natural if $|\dot{\gamma}(s)| = 1$). Can anyone help me?
(a) $e^t(\cos t,\sin t,1)$
(b) $(t^3+t,t^3-t,\sqrt{3}t^2)$
(c) $3x^2+15y^2=1, z=xy$
Update:
Here are my attempts on solving (a):
$\dot{\gamma}(t) = (\dot{t}e^t \cos t - \dot{t} e^t \sin t, \dot{t} e^t \sin t + \dot{t}e^t \cos t, \dot{t} e^t)$ which gives $|\dot{\gamma}(t)| = \sqrt{2}\dot{t}e^t=1$ and the solution for this ODE is $t = \ln\frac{\tau}{2}$.
But if I substitute $t$ with $t=\ln\frac{\tau}{2}$ the result will be $\dot{\gamma}(\tau) = (\frac{1}{\sqrt{2}}\cos\ln\frac{\tau}{2} - \frac{1}{\sqrt{2}}\sin\ln\frac{\tau}{2},\frac{1}{\sqrt{2}}\sin\ln\frac{\tau}{2}+\frac{1}{\sqrt{2}}\cos\ln\frac{\tau}{2},\frac{1}{2})$ and $|\dot{\gamma}(\tau)| = \sqrt{\frac{3}{2}}$. So for $|\dot{\gamma}(\tau)| = 1$ we should take $t = \ln\frac{\tau}{3}$. Where is my mistake?
Any help and hints will be very appreciative.
| $(a)$
\begin{align*}
\mathbf{\dot{r}}(t) &= e^{t}(\cos t-\sin t, \sin t+\cos t,1) \\
|\mathbf{\dot{r}}(t)| &= e^{t}\sqrt{(\cos t-\sin t)^2+(\sin t+\cos t)^2+1} \\
&= e^{t}\sqrt{3} \\
s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\
&= \sqrt{3}(e^{t}-1) \\
t &= \ln \left( 1+\frac{s}{\sqrt{3}} \right)
\end{align*}
$(b)$
\begin{align*}
\mathbf{\dot{r}}(t) &= (3t^2+1, 3t^2-1,2t\sqrt{3}) \\
|\mathbf{\dot{r}}(t)| &= \sqrt{(3t^2+1)^2+(3t^2-1)^2+12t^2} \\
&= \sqrt{2(9t^4+6t^2+1)} \\
&= \sqrt{2}(3t^2+1) \\
s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\
&= \sqrt{2}(t^3+t) \\
t &= \sqrt[3]{\sqrt{\frac{s^2}{8}+\frac{1}{27}}+\frac{s}{2\sqrt{2}}}-
\sqrt[3]{\sqrt{\frac{s^2}{8}+\frac{1}{27}}-\frac{s}{2\sqrt{2}}} \\
\end{align*}
$(c)$
\begin{align*}
\mathbf{r}(t) &=
\left(
\frac{\cos t}{\sqrt{3}},
\frac{\sin t}{\sqrt{15}},
\frac{\cos t \sin t}{3\sqrt{5}}
\right) \\
\mathbf{\dot{r}}(t) &=
\left(
-\frac{\sin t}{\sqrt{3}},
\frac{\cos t}{\sqrt{15}},
\frac{\cos^2 t-\sin^2 t}{3\sqrt{5}}
\right) \\
|\mathbf{\dot{r}}(t)| &=
\sqrt{\frac{15\sin^2 t+3\cos^2 t+(\cos^2 t-\sin^2 t)^2}{45}} \\
&= \frac{3-\cos 2t}{3\sqrt{5}} \\
s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\
&= \frac{6t-\sin 2t}{6\sqrt{5}}
\end{align*}
there is no close form for $t$ in terms of $s$.
You still can find $\kappa$ and $\tau$ by keeping $t$ parametrization
\begin{align*}
\kappa &=
\frac{|\mathbf{\dot{r}} \times \mathbf{\ddot{r}}|}
{|\mathbf{\dot{r}}|^3} \\
\tau &=
\frac{\mathbf{\dot{r}} \cdot \mathbf{\ddot{r}} \times \mathbf{\dddot{r}}}
{|\mathbf{\dot{r}} \times \mathbf{\ddot{r}}|^2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $ I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)
Here is what I've done
$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\theta} + \frac{\sin 3\theta}{\cos 3\theta}= \frac {\sin \theta}{\cos \theta} \frac {\sin 2\theta} {\cos 2\theta} \frac{\sin 3\theta}{\cos 3\theta}$$
$$ \frac {\sin \theta \cos 2\theta \cos 3\theta + \cos \theta \sin 2\theta \cos 3\theta + \cos \theta \cos 2\theta \sin 3\theta - \sin \theta \sin 2\theta \sin 3\theta }{\cos\theta \cos 2\theta \cos 3\theta} = 0 $$
$$\cos 2\theta \{\sin\theta \cos 3\theta + cos \theta \sin 3\theta \} + \sin 2\theta \{\cos \theta \cos 3\theta - \sin \theta \sin 3\theta \} = 0 $$
$$\cos 2\theta \sin(3\theta + \theta) +\sin2\theta \cos(3\theta + \theta) = 0 $$
$$ \cos 2\theta \sin 4\theta + sin 2\theta cos 4\theta = 0$$
$$ \sin (2\theta + 4\theta) = 0$$
$$\sin 6\theta = 0 $$
$$ \theta = \frac {n\pi}{6}, n \in Z$$
I understand from this question that whatever mistake I am making is in the third step, where I remove $\cos \theta \cos 2\theta \cos 3\theta $ from the denominator. However, despite reading through the aforementioned post, I couldn't really get the intuition behind why this is wrong.
I'd like :
*
*To understand the intuition behind why removing $\cos \theta \cos 2\theta cos 3\theta $ is a mistake.
*To know how to solve this question correctly
*How do I avoid making these types of mistakes when solving trigonometric equations
| You could have solved the problem using the tangent multiple angle formulae.
Using $t=\tan(\theta)$ $$\tan(2\theta)=\frac{2t}{1-t^2}\qquad \tan(3\theta)=\frac{3t-t^3}{1-3t^2}$$ This makes, after some minor simplifications
$$\tan \theta + \tan 2\theta + \tan 3\theta - \tan \theta \tan 2\theta \tan 3\theta=\frac{2 t \left(3-t^2\right)}{1-t^2}=0$$ and then the solutions assuming that ${1-3t^2}\neq 0$ and ${1-t^2}\neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{9a^2+b^2}}+\frac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \frac32.$
Prove that $$\dfrac{a}{\sqrt{a^2+b^2}}+\dfrac{b}{\sqrt{9a^2+b^2}}+\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \dfrac{3}{2}.$$
When is equality attained ?
My Attempt :
I could not think of anything suitable for the entire LHS. The last term $\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2}} $ suggested C-S
. So I applied C-S on the values $\left(\dfrac{2a}{\sqrt{a^2+b^2}}\right)$ and $\left(\dfrac{b}{\sqrt{9a^2+b^2}}\right)$ to get :
$$\left(\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2}}\right)^2\leq \left(\dfrac{4a^2}{a^2+b^2}\right)\left(\dfrac{b^2}{9a^2+b^2}\right)\leq\left(\dfrac{4a^2}{b^2}\right)\times \left(\dfrac{b^2}{9a^2}\right)=\dfrac{4}{9}.\,\,\,\,(♦)$$
For the first term $\dfrac{a}{\sqrt{a^2+b^2}}$, I applied C-S on the terms $\left(a\right)$ and $\left(\dfrac{1}{\sqrt{a^2+b^2}}\right)$ to get :
$$\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)\leq \left(\dfrac{a^2}{a^2+b^2}\right)^{1/2}\leq 1.\,\,\,\,(♣)$$
Using the same logic for the second term, I get :
$$\dfrac{b}{\sqrt{9a^2+b^2}}\leq 1 \,\,\,\,(♠)$$
Adding all the inequalities, I get a very "weak" inequality when compared to the problem.
What is the best way to prove this inequality ?
| It's enough to prove our inequality for positives $a$ and $b$.
Let $a^2=\frac{x}{3}$, $b^2=y$ and $x^2+y^2=2kxy$.
Hence, $k\geq1$ and we need to prove that
$$\sqrt{\frac{x}{x+3y}}+\sqrt{\frac{y}{3x+y}}+2\sqrt{\frac{xy}{(3x+y)(x+3y)}}\leq\frac{3}{2}$$ or
$$\frac{\sqrt{x(3x+y)}+\sqrt{y(x+3y)}}{\sqrt{(3x+y)(x+3y)}}\leq\frac{3}{2}-2\sqrt{\frac{xy}{(3x+y)(x+3y)}}$$
and since $$2\sqrt{\frac{xy}{(3x+y)(x+3y)}}=2\sqrt{\frac{xy}{3(x^2+y^2)+10xy}}\leq2\sqrt{\frac{xy}{6xy+10xy}}=\frac{1}{2}<\frac{3}{2},$$
we need to prove that
$$\frac{3x^2+3y^2+2xy+2\sqrt{xy(3x+y)(x+3y)}}{3(x^2+y^2)+10xy}\leq\left(\frac{3}{2}-2\sqrt{\frac{xy}{(3x+y)(x+3y)}}\right)^2$$ or
$$\frac{6k+2}{6k+10}+\frac{2}{\sqrt{6k+10}}\leq\left(\frac{3}{2}-\frac{2}{\sqrt{6k+10}}\right)^2$$ or
$$15k+49\geq16\sqrt{6k+10},$$
which is AM-GM:
$$15k+49=12k+20+3k+29\geq12k+20+32\geq2\sqrt{(12k+20)\cdot32}=16\sqrt{6k+10}$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x} = \\
\frac{e^{x}(1+e^{-2x}-\frac{2}{e^x})}{x(x-2)} = \frac{e^x(1+e^{-2x})-2}{x(x-2)} = \frac{e^x(1+e^{-2x})}{x(x-2)} - \frac{2}{x(x-2)} = \\
\frac{1+e^{-2x}}{x} \cdot \frac{e^x}{x-2} - \frac{2}{x(x-2)} = ???$$
What do I do next?
| Simply define $f(x) = e^x + e^{-x}.$ Then the expression equals
$$\frac{f(x) - f(0)}{x-0}\cdot \frac{x}{x^2+2x}.$$
As $x\to0,$ the first fraction $\to f'(0)$ by the definition of the derivative. The second fraction $\to 1/2.$ Since $f'(0)=0,$ the limit in question is $0\cdot (1/2) = 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $f(n)$ be the integer closest to $\sqrt{n}.$ then value of $ \sum^{2016}_{k=1}\frac{1}{f(k)}$ If $f(n)$ be the integer closest to $\sqrt{n}.$ then value of $\displaystyle \sum^{2016}_{k=1}\frac{1}{f(k)}$
could some help me with ths, thanks
| Taking a cue from sequence A000194 in the OEIS, we have $$S = \sum_{k=1 }^{2016} \frac{1}{f (k)}$$ $$ = \frac {2}{1} + \frac {4}{2} + \frac {6}{3} + \cdots + \frac {2 \lfloor \sqrt {2016} \rfloor}{\lfloor \sqrt {2016} \rfloor} + \frac {36}{\lceil \sqrt {2016} \rceil}\,\,(\text{ why? })$$ $$=2+2+2\cdots +2 +\frac {36}{45} $$ where $2$ is added $44$ times giving an answer of $\boxed {88.8} $.
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Continued Fraction Identity Problem
Question: What went wrong in my work to$$\ln(1-x)=-\cfrac x{1-\cfrac x{2+x-\cfrac {2x}{3+2x-\cfrac {3x}{4+3x-\ddots}}}}\tag{1}$$
I started with the expansion$$\begin{align*}\ln(1-x) & =-x-\dfrac {x^2}2-\dfrac {x^3}3-\dfrac {x^4}4-\&\text{c}.\\ & =-x\left\{1+\left(\dfrac x2\right)+\left(\dfrac x2\right)\left(\dfrac x{3/2}\right)+\&\text{c}.\right\}\tag{2}\end{align*}$$
And by Euler's Continued Fraction, $(2)$ can be rewritten into$$\begin{align*} & -x\left\{1+\left(\dfrac x2\right)+\left(\dfrac x2\right)\left(\dfrac x{3/2}\right)+\left(\dfrac x{2}\right)\left(\dfrac x{3/2}\right)\left(\dfrac x{4/3}\right)+\&\text c.\right\}\\ & =-\cfrac {x}{1-\cfrac x{2+x-\cfrac {2x}{3+2x-\ddots}}}\end{align*}\tag{3}$$
However, if we set $x=2$, then we have the LHS as $\ln(1-2)=\ln -1=\pi i$. The RHS becomes
$$-\cfrac 2{1-\cfrac 2{4-\cfrac 4{7-\cfrac 6{10-\ddots}}}}=\pi i\tag{4}$$
But the LHS is real while the RHS is imaginary. What went wrong?
| You went wrong with conversion from the series to the Euler CF. It was almost right, but the numeric numerator factor should be squared. That is,
$$ \ln (1-x) = -\cfrac{x}{1 - \cfrac{1^2x}{2 + x - \cfrac{2^2x}{3 + 2x - \cfrac{3^2x}{\ddots}}}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
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