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How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am not sure how this is useful.
|
Like Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$,
as $2$ is a primitive root $\pmod{37}$ as
$2^5\equiv-5\pmod{37}\implies2^{10}\equiv(-5)^2\equiv-12\implies2^{12}\equiv-12\cdot2^2\equiv26\not\equiv1$
and $2^{18}\equiv(-10)^3\equiv-1$
using Discrete Logarithm , $3$ind$_2x\equiv0\pmod{36}\iff$ind$_2x\equiv0\pmod{12}$
$\implies x\equiv2^{12k}\pmod{37}$ where $k\equiv0,1,2\pmod3$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find limit of $\frac{2x+7}{\sqrt{x^2+2x-1}}$ (check my steps please..)
Compute the limit $$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}$$
Here are my steps:
$$\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}=\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}\cdot\frac{-1/x}{-1/x}=\frac{\displaystyle \lim_{x\to-\infty}-2-\frac{7}{x}}{\displaystyle \lim_{x\to-\infty}\sqrt{\frac{x^2+2x-1}{x^2}}}=\frac{-2}{1}=-2$$
But I am not confident about $\displaystyle \lim_{x\to-\infty}{\sqrt\frac{x^2+2x-1}{x^2}}$. I just suppose that $x$ can be 'large enough' to ignore $+2x-1$.
|
To avoid confution the best way when we deal with limit $\to -\infty $ is to change variable and set $x=-y$ with $y\to +\infty$ then
$$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}=\lim_{y\to+\infty}{\frac{-2y+7}{\sqrt{y^2-2y-1}}}=\lim_{y\to+\infty}{\frac{y}{y}\frac{-2+7/y}{\sqrt{1-2/y-1/y^2}}}=-2$$
|
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How prove this inequality $H(a_1)+H(a_2)+\cdots+H(a_m)\leq C\sqrt{\sum_{i=1}^{m}i a_i}$
Prove that: There exists a constant $C>0$ such that
$$H(a_1)+H(a_2)+\cdots+H(a_m)\leq C\sqrt{\sum_{i=1}^{m}i a_i}$$
holds for arbitrary positive integer $m$ and any $m$ positive integers $a_1,a_2,\cdots,a_m$,
where $H(n)=\sum_{k=1}^{n}\frac{1}{k}.$
This is 2018 China TST 3 Day 1 Q3. Maybe it is from some paper. The reason is that in the past, the Chinese training team selected most of the questions from articles.
|
In view of the rearrangement inequality, it suffices to check the inequality when $(a_i)$ is decreasing, i.e., $a_1 \geq a_2 \geq \cdots \geq a_m$. Also, it is no harm to introduce $a_{m+1} = 0$. Then
$$ \sum_{i=1}^{m} H(a_i)
= \sum_{i=1}^{m} \sum_{k=i}^{m} \left( H(a_k) - H(a_{k+1}) \right)
= \sum_{k=1}^{m} k \left( H(a_k) - H(a_{k+1}) \right) $$
and likewise
$$ \sum_{i=1}^{m} i a_i
= \sum_{i=1}^{m} \sum_{k=i}^{m} i (a_k - a_{k+1})
= \sum_{k=1}^{m} \frac{k(k+1)}{2} (a_k - a_{k+1}). $$
Now we invoke the following simple lemma:
Lemma. If $0 \leq a < b$ are integers, then
$$ \frac{H(b) - H(a)}{\sqrt{b-a}}
\leq \sqrt{\frac{1}{a+\frac{1}{2}} - \frac{1}{b+\frac{1}{2}}} $$
Before proving this lemma, let us see how this implies the desired inequality. Applying the Cauchy-Schwarz inequality and the lemma above, we obtain
\begin{align*}
\sum_{i=1}^{m} H(a_i)
&\leq \left( \sum_{k=1}^{m} \frac{2\left( H(a_k) - H(a_{k+1}) \right)^2}{a_k - a_{k+1}} \mathbf{1}_{\{a_k > a_{k+1} \}} \right)^{1/2} \left( \sum_{k=1}^{m} \frac{k^2}{2} (a_k - a_{k+1}) \right)^{1/2} \\
&\leq \left( 2 \sum_{k=1}^{m-1} \left( \frac{1}{a_{k+1}+\frac{1}{2}} - \frac{1}{a_k+\frac{1}{2}} \right) \right)^{1/2} \left( \sum_{i=1}^{m} i a_i \right)^{1/2} \\
&\leq 2 \left( \sum_{i=1}^{m} i a_i \right)^{1/2}.
\end{align*}
Therefore the claim is true with $C = 2$.
Proof of Lemma. Notice that for $x \geq 1$,
$$
\int_{x-\frac{1}{2}}^{x+\frac{1}{2}} \frac{dt}{t}
= \int_{x}^{\infty} \left( \frac{1}{t-\frac{1}{2}} - \frac{1}{t+\frac{1}{2}} \right) \, dt
= \int_{x}^{\infty} \frac{4 dt}{4t^2-1}
\geq \int_{x}^{\infty} \frac{dt}{t^2}
= \frac{1}{x}.
$$
(Alternatively, this is the result of the convexity of $\frac{1}{x}$. Indeed, the tangent line $\frac{1}{x} - \frac{1}{x^2}(t-x)$ at $x$ lies below $\frac{1}{t}$, and integrating both sides from $x-\frac{1}{2}$ to $x+\frac{1}{2}$ gives the inequality above.) Then by Cauchy-Schwarz inequality,
\begin{align*}
H(b) - H(a)
&= \sum_{k=a+1}^{b} \frac{1}{k}
\leq \int_{a+\frac{1}{2}}^{b+\frac{1}{2}} \frac{dx}{x} \\
&\leq \left( \int_{a+\frac{1}{2}}^{b+\frac{1}{2}} \frac{dx}{x^2} \right)^{1/2}\left( \int_{a+\frac{1}{2}}^{b+\frac{1}{2}} dx \right)^{1/2} \\
&= \Bigg( \frac{1}{a+\frac{1}{2}}-\frac{1}{b+\frac{1}{2}} \Bigg)^{1/2} \sqrt{b-a}.
\end{align*}
|
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solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\frac{1}{2} d\ln(1+x^2)=\frac{1}{2}d\ln(1+y^2)(1+x^2)=0$$
Let $c=(1+y^2)(1+x^2)$, so our equation becomes:
$$
d\ln c=0
$$
So what should I do here, should I integrate, or should I divide by $dx$?
If I divide by dx I get the expression $2x+2yy'+2xy^2+2x^2yy'=0$ which has $x$, $y$ and $y'$ and doesn't help me get anywhere.
Thanks in advance.
|
The DE is
$$\frac12d(x^2)+\frac12d(y^2)+\frac12d(x^2y^2)=0$$
Then, the solution is
$$\boxed{\frac12x^2+\frac12y^2+\frac12x^2y^2=c}$$
|
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|
Use Ramanujan’s method to denest $\sqrt[3]{7\sqrt[3]{20}-1}$ and $\sqrt[3]{7\sqrt[3]{20}-19}$ A possible way to denest $(2^{1/3}-1)^{1/3}$ is by first setting $x=\sqrt[3]{2}$ so$$x^3-1=1\implies x-1=\frac 1{1+x+x^2}=\frac 3{1+3x+3x^2+x^3}$$Multiply both sides by $9$ so$$9(x-1)=\left(\frac 3{1+x}\right)^3\implies\sqrt[3]{9(\sqrt[3]2-1)}=1-\sqrt[3]2+\sqrt[3]4$$However, I'm having trouble adapting this method to denest
both
$$\begin{align*}\sqrt[3]{7\sqrt[3]{20}-1} & =\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}\tag1\\\sqrt[3]{7\sqrt[3]{20}-19} & =\sqrt[3]{\frac 49}-\sqrt[3]{\frac {80}9}+\sqrt[3]{\frac {25}9}\tag2\end{align*}$$
I've only started on (1) so far. Here's my work.
My work: Let $x^3=6860$ so $x^3-1=19^3$. Hence$$x-1=\frac {19^3}{1+x+x^2}=\frac {19^3\cdot3}{3+3x+3x^2}$$However, I'm not quite sure what to do after that. I don't see an easy relationship between three and $x$.
Perhaps you guys can help?
|
HINT:
A method that proves the first equality but is not able to find a denesting.
One checks that both $\sqrt[3]{7\sqrt[3]{20}-1}$ and $\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$ are roots of the polynomial $x^9 + 3 x^6 + 3 x^3 - 6859$, a polynomial with a unique real root.
ADDED: Note that the LHS is a root of the equation $(x^3+1)^3= 7^3 \cdot 20=6860$. The polynomial function $(x^3+1)^3$ is strictly increasing so the equation has a unique solution.
Say we want to right LHS as a combination of cubic roots. Such a combination in general has a minimal polynomial of degree $3\cdot 3 \cdot 3=27$, unless there is some multiplicative combination of these roots that gives a rational number. For our examples, one notices that the product of these radicals is a rational number
$$\sqrt[3]{\frac {16}9}\cdot \sqrt[3]{\frac 59}\cdot \sqrt[3]{\frac {100}9}=\frac{20}{9}$$
Therefore, in general we are looking for a RHS of form
$$\sqrt[3]{a}+\sqrt[3]{b}-\frac{d}{\sqrt[3]{ab}}$$
with $a$, $b$, $d$ (positive) rational. Now we want the RHS to satisfy an equation of form $(x^3+1)^3 = 6860$ ( in general, if we tried to denest an expression of form $\sqrt[3]{\sqrt[3]{\alpha}-\beta}$ the equation should be
$(x^3+\beta)^3=\alpha$). Note that in fact the RHS is of the form
$$\sqrt[3]{u^2}+\sqrt[3]{v^2}- p \sqrt[3]{u v}$$ with $u$,$v$, $p$ are rational(positive) (there should be a reason for that, not entirely clear at this point). I'll leave it here for now.
ADDED: We might as well find an equation of degree $9$ with root $\sqrt[3]{a}+\sqrt[3]{b}-\frac{d}{\sqrt[3]{ab}}$. It is a polynomial with rational coefficients expressed as a product
$$\prod_{k,l=0,1,2}(x-(\sqrt[3]{a}\omega^k+\sqrt[3]{b} \omega^l-\frac{d}{\sqrt[3]{ab}\omega^{k+l}}))$$
where $\omega= \exp(2\pi i/3)$.
I don't have a CAS system at the moment so I leave it like this. One should place conditions on $a$, $b$, $d$ so that it is a polynomial of form $(x^3+\beta)^3-\alpha$. The problem seems doable now. It may also appear necessary that $a$,$b$ are squares of rational numbers.
|
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The convergence of a recursive sequence
The sequence $(a_n)$ is defined by $a_1 = 1$ and
$a_{n+1}=a_{n}+\sqrt{1+a_{n}^{2}}$. A sequence $(b_n)$ is defined by
$b_n = \dfrac{a_n}{2^n}$. It is easy to see $(b_n)$ is monotone but is
it convergent?
Suppose $(b_n)$ be bounded. Then $(b_n)$ is convergent by Monotone Convergence Theorem. Say, $b_n$ converges to a limit $b$, then $(b_{n+1}) \to b$.
We have $a_{n+1}=a_{n}+\sqrt{1+a_{n}^{2}}$. Dividing $2^{n+1}$ in both sides of the equation,
$\Rightarrow \dfrac{a_{n+1}}{2^{n+1}}=\dfrac{a_{n}+\sqrt{1+a_{n}^{2}}}{2^{n+1}}$
$\Rightarrow b_{n+1} = \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}$
$\Rightarrow \lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} \dfrac{b_n}{2} + \lim_{n \to \infty} \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}$
$\Rightarrow b = \dfrac{b}{2} + \sqrt{0 + \dfrac{b^2}{4}}$
$\Rightarrow b = b$.
No decisive conclusions could be made.
However numerical evidences suggest that $ b_n \approx 0.63$ whenever $100\le n \le 200$. Does it diverge to infinity but sure, it diverges pretty slow? I tried to prove otherwise that $b_n$ is bounded above by $1$ but it didn't work but neither can find an counterexample to it. How far do I need to go in the sequence to surpass $1$?
|
$$
b_{n+1} = \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}
$$
where $b_1 = 1/2$. If you try to bound the relation, you get
$$
\dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}} \le b_n + x_n
$$$$
\dfrac{1}{2^{2n+2}} \le b_nx_n + x_n^2
$$
so $x_n = 2^{-n}$ is enough. This proves that
$$b_{n+1}\le b_n + 2^{-n}\le b_1 + 1$$
so the sequence is bounded, and the monotone sequence converges.
|
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Find the value of the constant c such that $\sum_{n=2}^\infty(1+c)^{-n} = 2$
Find the value of the constant c such that $$\sum_{n=2}^\infty(1+c)^{-n} = 2 $$
For this question,I'm not sure if I'm doing it right. If I am doing it right, I'm not sure how to get further. Here is what I have so far. Can anyone please help me out?
Comparing the series to the conventional form $a + a r + a r^2 + \cdots = a/(1-r)$, we have
$$r = 1+c \qquad\text{and}\qquad a = (1+c)^{-2} \qquad\text{and}\qquad 2 = \frac{a}{1-r}$$
So ...
$$\begin{align}
2 &= \frac{(1+c)^{-2}}{1-(1+c)} \\
2 &= \frac{(1+c)^{-2}}{-c} \\
-2c &= (1+c)^{-2} \\
-2c &= \frac{1}{(1+c)^2} \\
-2c(1+c)^2 &= 1 \\
-2c(1+2c+c^2) &= 1
\end{align}$$
|
When the series converges,
$$\sum_{n=2}^\infty(1+c)^{-n}=\frac{(1+c)^{-2}}{1-(1+c)^{-1}}=\frac1{(1+c)\,c}=2.$$
Solve $$c^2+c-\frac12=0.$$
After solving, check the convergence condition, $|1+c|>1$.
|
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If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$? Suppose $f$: $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ and $f(x)^2 = x + (x+1)f(x+2)$, what is $f(1)$? Or more in general, what is $f(x)$?
The motivation behind this problem is that I want to find what the number of this nested radical $\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8...}}}}$. This can be written more generally as $f(x)=\sqrt{x+(x+1)f(x+2)}$ where $x=1$. This is where the problem arises from. If anybody can find an expression for the nested radical or find $f(x)$ I would be very happy!
|
Here is a suggested alternate approach.
Consider the recursive sequence $a_k\in\mathbb{R}$
\begin{eqnarray}
a_0&=&\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8\sqrt{9+...}}}}}\\
a_{k+1}&=&\frac{a_k^2+1}{2(k+1)}-1 \tag{1}
\end{eqnarray}
This gives the increasing unbounded sequence
\begin{eqnarray}
a_1&=&\sqrt{3+4\sqrt{5+6\sqrt{7+8\sqrt{9+10\sqrt{11+...}}}}}\\
a_2&=&\sqrt{5+6\sqrt{7+8\sqrt{9+10\sqrt{11+12\sqrt{13+...}}}}}\\
a_3&=&\sqrt{7+8\sqrt{9+10\sqrt{11+12\sqrt{13+14\sqrt{15+...}}}}}\\
&\vdots&
\end{eqnarray}
This converts the problem of finding $a_0$ into the problem of finding a generating function
\begin{equation}
G(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots
\end{equation}
and a formula for the general term $a_n$ of the coefficient sequence.
|
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How to quickly find the $x^{24}$ term in this expansion? Is there a swift way to find the $x^{24}$ coefficient in the expansion of
$$ \left(1-x^6\right)^{-2} \left(1-x^3\right)^{-1} \left(1-x\right)^{-1} $$
The general term of each bracket is $(r+1)x^{6r}$, $x^{3r}$ and $x^{r}$ respectively.
|
For convenience, in the index notation, I only sum over nonnegative integers.
\begin{align}
\sum_{6r_1+3r_2+r_3=24} (1+r_1) &= \sum_{2r_1+r_2+k=8} (r_1+1) \text{, we let } r_3 = 3k \\
&= \sum_{2r_1+2w=8} (2w+1)(r_1+1) \text{, we let } r_2+k = 2w \\
&= \sum_{r_1+w=4} (2w+1)(r_1+1)\\
&=\sum_{w=0}^4(2w+1)(5-w) \\
&= -2 \sum_{w=0}^4w^2 + 9 \sum_{w=0}^4w+5 \sum_{w=0}^4 1 \\
&= -2 \cdot \frac{4(5)(9)}{6} + 9\cdot \frac{4(5)}{2}+ 5 \cdot 5 \\
&= - 60 +90+ 25 \\
&= 55
\end{align}
|
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Prove the relation is an equivalence relation. Problem
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \vert(a^2 + b) $. Prove that $R$ is an equivalence relation.
This is what I have so far.
Claim:
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \mid(a^2 + b) $. The relation $R$ is an equivalence relation.
Proof:
Part 1 (Reflixivity):
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that $b \in \Bbb{N}$.
Then, for some integer $k$:
$$\require{enclose}
\enclose{downdiagonalstrike}{\begin{align}
2 \mid b^2 +b \iff 2k &= b^2 +b \\
& = b^2 + b - (b^2 + b) + (b^2 + b) \\
& = 2b^2 + 2b - (b^2 + b) \\
& = -2b^2 - 2b + (b^2 + b)
\end{align}}$$
Therefore,
$$\enclose{downdiagonalstrike}{\begin{align}
2 \mid b^2 +b \iff b^2 + b &= 2k -2b^2 - 2b \\
& = 2(k - b^2 - b)\\
&
\end{align}}$$
$\enclose{horizontalstrike}{\text{Thus, $2 \mid b^2 +b$ for some integer $(k - b^2 - b)$. Which implies that $R$ is Reflexive.}} $
EDIT: Thanks to some positive feed back I have been let known that this is not showing Reflexivity.
Part 2 (Symmetry)
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that, for any $a,b \in \Bbb{N}$, $ a\mathbf{R}b \leftrightarrow b\mathbf{R}a.$
Then, for some integer $k$:
$$\enclose{updiagonalstrike}{\begin{align}
2 \mid a^2 + b & \iff 2k = a^2 + b \\
& \iff 2k + (a + b^2) = (a^2 + b) + (a + b^2)\\
& \iff (a + b^2) = (a + b) + (a^2 + b^2) - 2k \\
&
\end{align}}$$
$\enclose{horizontalstrike}{\text{Since, the relation $R$ is proven to be Reflexive, let the integer $ m = a = b $ and let the integer $ n = a^2 = b^2 $. Then, }}$
EDIT: This is not a valid way to show Symmetry, since Part 1 (Reflexivity) has not been proven.
$$\enclose{updiagonalstrike}{\begin{align}
a + b^2 = (a + b) + (a^2 + b^2) - 2k & \iff (a + b^2) = 2m + 2n - 2k \\
& \iff (a + b^2) = 2(m + n -k)\\
&
\end{align}}$$
$\enclose{horizontalstrike}{\text{Thus, $2 \mid b^2 + a$ which implies that $R$ is Symmetric since $ a\mathbf{R}b \leftrightarrow b\mathbf{R}a $.}}$
Part 3 (Transitivity)
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that, for any $a,b,c \in \Bbb{N}$, $ a\mathbf{R}b \text{ and } b\mathbf{R}c.$ Then, let $k$ and $h$ be some integers:
\begin{align}
2k = a^2 + b \text{ and } 2h = b^2 + c & \implies 2(k + h) = (a^2 + b) + (b^2 + c) \\
& \\
& \\
&
\end{align}
Comment: And, this is where I get stuck. I am struggling to find a way to show that $ (2 \mid a^2 + b) \land (2 \mid b^2 + c) \implies 2 \mid a^2 + c$. I've also tried eliminating $b$ like so, $2h = (2k - a^2)^2 + c$, but this doesn't seem to get me any where. I feel like I'm running in circles here.
My Question
Can you argue that $R$ is transitive since it has already been shown that $2 \mid b^2 + b$?
This would imply something like "$2(k - b^2 - b) + 2h = (a^2 + c) +(b^2 + b)$ is logically equivalent to $(2 \mid b^2 + b) \land (2 \mid a^2 + c)$." And, this simplifies to just $2 \mid a^2 + c$ by the inference rule of simplification [$(p \land q) \to p$]. Which ultimatily I believe gets me to my goal, but I'm not sure if it is two far of a leep to go from $2(k - b^2 - b) + 2h = (a^2 + c) +(b^2 + b) \implies (2 \mid b^2 + b) \land (2 \mid a^2 + c)$.
I hope my question was specific enough. Otherwise, I would much appreciate some guidance on showing how this relation is transitive if anyone is feeling generous. Thanks!
|
Your approach is too technical.
Note that $$2| (a^2 + b)$$ if and only if both $a$ and $b$ are odd or both are even.
Reflexivity: $a$ and $a^2$ are either both even or both odd.
Symmetry: If both $a$ and $b$ are odd then both $b$ and $a$ are also odd. Similarly for the even case.
Transitivity: If $a$ and $b$ are related and $b$ and $c$ are related then all three of them are odd or all three are even .
In any case a and c are both odd or both even.
|
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|
Evaluating trigonometric limit $\csc^2(2x) - \frac{1}{4x^2}$
$$\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right]$$
I've tried to use l'Hôpital's rules but still can't find the answer. Here's my approach:
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2(\sin^2(2x))} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-4\sin(2x)\cos(2x)}{8x\sin^2(2x)+16x^2\sin(2x)\cos(2x)} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-2\sin(4x)}{8x\sin^2(2x)+8x^2\sin(4x)} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x-\sin(4x)}{4x\sin^2(2x)+4x^2\sin(4x)} \right] \\
\end{aligned}
$$
Am I using the correct way? How to solve it correctly?
P. S. I tried to use calculator and it outputs one third $1/3$.
|
hint: replace the denominator on the third line of your proof $\sin^2(2x)$ by $(2x)^2$ and apply L'hopitale rule three times to the expression: $\dfrac{4x^2-\sin^2(2x)}{16x^4}$. I did it and it works. Try it. Note that "replace" here means you write: $\sin^2 (2x) = 4x^2\cdot \left(\dfrac{\sin(2x)}{2x}\right)^2$, and the second factor approaches $1$ when $x \to 0$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\frac{y}{\sqrt{1-y^2}}\bigg]=a+\frac{x}{\sqrt{1-x^2}}\\
\frac{dy}{dx}=\frac{a\sqrt{1-x^2}+x}{\sqrt{1-x^2}}.\frac{\sqrt{1-y^2}}{a\sqrt{1-y^2}+x}=\sqrt{\frac{1-y^2}{1-x^2}}.\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}
$$
How do I poceed further and find the derivative ?
|
After my comment you will get $$a\sqrt{1-x^2}+x=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and
$$a\sqrt{1-y^2}-y=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and you will get the desired result!
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the sum of series $1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$ I am trying to solve this series question which is
$$1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$$
Till now I've only been able to write the general form of this series which is
$$\sum_{n=1}^\infty\frac1{(2n-1)4^{n-1}}$$
but I don't know how to proceed.
|
Hint Reindexing your expression for the series for convenience gives
$$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k .$$
Now, the factors $\frac{1}{2 k + 1}$ suggest considering the power series
$$\operatorname{artanh} x \sim \sum_{k = 0}^\infty \frac{1}{2 k + 1} x^{2 k + 1} .$$
Rewrite the given series to get $$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k = 2 \sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\color{red}{\frac{1}{2}}\right)^{2 k + 1} = 2 \operatorname{artanh} \color{red}{\frac{1}{2}} = 2 \cdot \frac{1}{2}\log \frac{1 + \color{red}{\frac{1}{2}}}{1 - \color{red}{\frac{1}{2}}} = \boxed{\log 3} .$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2723875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius
My Attempt
From sine law,
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$b=2R \sin B$$
$$c=2R \sin C$$
Then,
$$8R^2=a^2+b^2+c^2$$
$$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$
$$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$
$$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$
|
By the law of cosines,
\begin{align}
a^2+b^2+c^2
&=
2ab\cos\gamma+
2bc\cos\alpha+
2ca\cos\beta
\tag{1}\label{1}
\end{align}
Using expressions for the area $S$ of triangle
\begin{align}
S&=\tfrac12ab\sin\gamma=
\tfrac12bc\sin\alpha=
\tfrac12ca\sin\beta
,\\
S&=2R^2\sin\alpha\sin\beta\sin\gamma
,
\end{align}
we have
\begin{align}
a^2+b^2+c^2
&=
2ab\sin\gamma\,\cot\gamma+
2bc\sin\alpha\,\cot\alpha+
2ca\sin\beta\,\cot\beta
\\
&=4S(\cot\alpha+\cot\beta+\cot\gamma)
\\
&=8R^2
\sin\alpha\sin\beta\sin\gamma
(\cot\alpha+\cot\beta+\cot\gamma)
=8R^2
,
\end{align}
\begin{align}
\cos\alpha\sin\beta\sin\gamma+
\sin\alpha\cos\beta\sin\gamma+
\sin\alpha\sin\beta\cos\gamma
&=1
,\\
(\cos\alpha\sin\beta+\sin\alpha\cos\beta)\sin\gamma+
\sin\alpha\sin\beta\cos\gamma
&=1
,\\
\sin(\alpha+\beta)\sin(\alpha+\beta)
+
\sin\alpha\sin\beta\cos\gamma
&=1
,\\
\sin^2(\alpha+\beta)
+
\sin\alpha\sin\beta\cos\gamma
&=1
,\\
1-\cos^2(\alpha+\beta)
-
\sin\alpha\sin\beta\cos(\alpha+\beta)
&=1
,\\
-\cos(\alpha+\beta)(\sin\alpha\sin\beta+\cos(\alpha+\beta))
&=0
,\\
-\cos(\alpha+\beta)(\sin\alpha\sin\beta+\cos\alpha\cos\beta-\sin\alpha\sin\beta)
&=0
,\\
-\cos(\alpha+\beta)\cos\alpha\cos\beta&=0
,\\
\cos\alpha\cos\beta\cos\gamma&=0
.
\end{align}
|
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"url": "https://math.stackexchange.com/questions/2724255",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
If $\alpha$ $\in$ $(0,2)$ study the convergence of the sequence given with recurrence relation $X_{n+1}=\alpha X_{n}-(1-\alpha)X_{n-1}$ If $\alpha$ $\in$ $(0,2)$ study the convergence of the sequence given with recurrence relation $X_{n+1}=\alpha X_{n}-(1-\alpha)X_{n-1}$.
Find the limit of the sequence.
Can somebody help me with this problem?
I only studied sequences with recurrence relation give with firsts conditions. I don't know how to deal with these ones.
I would appreciate some help
|
$X_n=Aa^n+Bb^n$ for some constants $A$ and $B$, where $a$ and $b$ are the roots of the equation $x^2-\alpha x+(1-\alpha)=0$. The sequence is convergent if both $|a|\le 1$ and $|b|\le 1$.
When $\alpha^2-4(1-\alpha)<0$, i.e. when $0<\alpha<2\sqrt{2}-2$, the quadratic equation has unreal roots, $|a|=|b|$ and $|a||b|=|ab|=1-\alpha<1$. The sequence is convergent.
When $\alpha^2-4(1-\alpha)\ge0$, i.e. when $2\sqrt{2}-2\le \alpha<2$, the quadratic equation has real roots. Let $a\le b$ Then $\displaystyle a=\frac{\alpha-\sqrt{\alpha^2+4\alpha-4}}{2}=\frac{2(1-\alpha)}{\alpha+\sqrt{\alpha^2+4\alpha-4}}$ and $\displaystyle b=\frac{\alpha+\sqrt{\alpha^2+4\alpha-4}}{2}$.
So, when $2\sqrt{2}-2\le \alpha\le 1$, $0\le a\le b$. We have to consider $|b|$ only.
\begin{align*}
|b|=\frac{\alpha+\sqrt{\alpha^2+4(\alpha-1)}}{2}\le \frac{\alpha+\sqrt{\alpha^2+4(\alpha-\alpha)}}{2}=\alpha\le1
\end{align*}
The sequence is convergent.
If $1<\alpha<2$, then
\begin{align*}
|b|=\frac{\alpha+\sqrt{\alpha^2+4(\alpha-1)}}{2}\ge \frac{\alpha+\sqrt{\alpha^2}}{2}=\alpha>1
\end{align*}
The sequence is divergent.
The sequence is convergent when $0<\alpha\le1$.
When $0<\alpha<1$, $\displaystyle \lim_{n\to\infty}X_n=0$.
When $\alpha=1$, $X_{n+1}=X_n$ and the seqeunce is a constant sequence.
As Youem pointed out in the comment, there is a special case that the sequence is convergent even when $1\le\alpha<2$.
If $1\le\alpha<2$, then
\begin{align*}
0\ge\frac{2(1-\alpha)}{\alpha+\sqrt{\alpha^2+4\alpha-4}}\ge\frac{2(1-\alpha)}{\alpha+\sqrt{\alpha^2+4\alpha-4\alpha}}=\frac{1}{\alpha}-1\ge-1
\end{align*}
and hence $|a|\le1$.
$X_n=Aa^n+Bb^n$ is convergent (when $1<\alpha<2$) if $B=0$, i.e. $X_n=Aa^n$
So, when $1\le\alpha<2$, The sequence is convergent to $0$ when $\displaystyle X_2=\left(\frac{\alpha-\sqrt{\alpha^2+4\alpha-4}}{2}\right)X_1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of equation regarding ratio of gamma function I get the result from Wolfram that
$$\lim_{a\to \infty}a-\frac{\Gamma(a+1/2)^2}{\Gamma(a)^2}=\frac{1}{4}.$$
I am trying to prove it. It seems the Stirling's formula cannot be used here.
Can anyone help me on this?
Thank you very much!
I tried to use Stirling's approximation as:
\begin{equation}
\begin{aligned}
a-\frac{\Gamma(a+1/2)^2}{\Gamma(a)^2}
&\to a-\frac{(a-1/2)^{2a}}{(a-1)^{2a-1}}e^{-1}\\
&=a-\big(\frac{a-1+1/2}{(a-1)}\big)^{2(a-1)}e^{-1}\frac{a^2-a+1/4}{a-1}\\
&\to a-a-\frac{1}{4(a-1)}
\end{aligned}
\end{equation}
|
$\Gamma$ is log-convex by the Bohr-Mollerup theorem/characterization, hence it is enough to prove
$$ \lim_{n\to +\infty} n-\left(\frac{n\sqrt{\pi}}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{4}.\tag{1} $$
On the other hand
$$ \frac{1}{4^n}\binom{2n}{n} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),\qquad \left(\frac{1}{4^n}\binom{2n}{n}\right)^2=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{2}$$
and
$$\pi\left(\frac{n}{4^n}\binom{2n}{n}\right)^2=n\prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)^{-1}\tag{3}$$
by Wallis' product. We also have
$$\prod_{k>n}\left(1+\frac{1}{4k(k-1)}\right)^{-1}=\exp\sum_{k>n}\left[-\frac{1}{4k(k-1)}+O\left(\frac{1}{k^4}\right)\right]=\exp\left(-\frac{1}{4n}\right)\left(1+O\left(\frac{1}{n^3}\right)\right) $$
and the RHS of the last line equals
$$\left(1-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\right)\left(1+O\left(\frac{1}{n^3}\right)\right)=1-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\tag{4}$$
so the claim is proved without resorting to the full power of Stirling's approximation/inequality.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Equation in 3d-Space Consider we have 3 fixed points A,B and C also,We have this Equation :
$MA^2+MB^2+MC^2=30$.
What does this Equation represent in 3d-Space,( where M is such a Non-fixed point in that space)?
Suppose $A(a1,b1,b1),B(a2,b2,c2),C(a3,b3,c3),M(x,y,z)$
any help will be appreciated.
|
Given $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3),M(x,y,z)$ and $r>0$.
What subset of $\mathbb{R}^3$ is specified by the equation
\begin{equation} |MA|^2+|MB|^2+|MC|^2=r^2 \tag{1}\end{equation}
\begin{eqnarray}
|MA|^2&=&(x-x_1)^2+(y-y_1)^2+(z-z_1)^2\\
|MB|^2&=&(x-x_2)^2+(y-y_2)^2+(z-z_2)^2\\
|MC|^2&=&(x-x_3)^2+(y-y_3)^2+(z-z_3)^2
\end{eqnarray}
So we can re-write equation $(1)$ as
\begin{eqnarray}
&&3x^2-2x(x_1+x_2+x_3)+(x_1^2+x_2^2+x_3^2)\\
&+&3y^2-2y(y_1+y_2+y_3)+(y_1^2+y_2^2+y_3^2)\\
&+&3z^2-2z(z_1+z_2+z_3)+(z_1^2+z_2^2+z_3^2)\\
&=&r^2
\end{eqnarray}
Define
\begin{eqnarray}
\bar{x}&=&\frac{x_1+x_2+x_3}{3}\\
\bar{y}&=&\frac{y_1+y_2+y_3}{3}\\
\bar{z}&=&\frac{z_1+z_2+z_3}{3}\\
\end{eqnarray}
Then the equation can be written in the form
\begin{eqnarray}
(x-\bar{x})^2+(y-\bar{y})^2+(z-\bar{z})^2=R^2
\end{eqnarray}
where $R$ is a constant defined in terms of $r$ and the coordinates of $A,\,B$ and $C$.
This is the equation of a sphere.
Note that OP has amended the question to ask what conditions on $A,\,B$ and $C$ guarantee that the solution will be a sphere.
First let
$$D=\left(\bar{x},\bar{y},\bar{z}\right)$$
The equation can also be written in the form
\begin{equation}
x^2-2x\bar{x}+y^2-2y\bar{y}+z^2-2z\bar{z}+\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)=\frac{1}{3}r^2
\end{equation}
which can be written
\begin{eqnarray}
(x-\bar{x})^2+(y-\bar{x})^2+(y-\bar{y})^2&=&\frac{1}{3}r^2+\bar{x}^2+\bar{y}^2+\bar{z}^2-\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)\\
&=&\frac{1}{3}r^2+\vert D\vert^2 -\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)\\
&>&0
\end{eqnarray}
So the condition is
$$ \vert A\vert^2+\vert B\vert^2+\vert C\vert^2-3\vert D\vert^2<r^2 $$
Thus, if we have $A(2,2,2),\,B(3,3,3),\,C(4,4,4)$ and $r^2=30$, then we have $D(3,3,3)$. So
\begin{eqnarray}
\vert A\vert^2+\vert B\vert^2+\vert C\vert^2-3\vert D\vert^2&=&12+27+48-3(27)\\
&=&6\\
&<&30
\end{eqnarray}
so there will be a sphere for that collection of three points.
Using the previous result
\begin{equation}
R^2=\frac{1}{3}r^2+\vert D\vert^2 -\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)=8
\end{equation}
we get
$$ (x-3)^2+(y-3)^2+(z-3)^2=8 $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Divisibility property for sequence $a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}$ Let $(a_n)$ be the sequence uniquely defined by $a_1=0,a_2=1$ and
$$
a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}
$$
Can anybody show (or provide a counterexample) that $p|a_{p-2}$ and
$p|a_{p-1}$ for any prime $p\geq 5$ ? I have checked this fact for
$p\leq 200$.
|
Define $$a_1=0,\; a_2=1,\;\text{and}\quad a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}.\tag{0}$$
Define $$\;R(x) := \sqrt{1-4x} = 1-2\sum_{n=1}^\infty C_{n-1}x^n = \sum_{n=0}^\infty {2n \choose n}\frac{x^n}{1-2n}\;$$ which is the generating function of the OEIS integer sequence A002420.
Define $$\;B(x) := \frac{ (1+4x+8x^2)R(-x-2x^2)-(1+6x+18x^2+20x^3)}{2x^2}$$
$$ = 3x^2 - 6x^3 + 10x^4 - 12x^5 + 3x^6 + \dots
=: \sum_{n=0}^\infty b_nx^n$$
which is the generating function for $b_n.\;$ It is easy to check that $b_n$ is also an integer sequence and $\;b_1=0, b_2=3.\;$ Using derivatives we get
the differential equation
$$\;(8x^2-2x-2)B(x)+12x^2=x(1+4x+8x^2)\frac{d}{dx}B(x)\;$$
and using some algebra on power series coefficients we get the equation
$$\;(2+n)b_{n}=(2-4n)b_{n-1}+(24-8n)b_{n-2}\quad\forall\;n>2.$$
Redefine $$\;a_n:=\frac{(n+2)!}{2^{n+1} 9}b_n.\tag{1}$$
It is easy to check that this satisfies the same initial values and recursion as given in $(0)$, but is it an integer sequence? It will be if
$\;2^{n+1}\;|\;(n+2)!\;b_n\;$ for all $n>1.$
The key results to prove is that if
$\;C_n\;$ is the Catalan numbers and if $\;v_2(n)\;$ is the 2-adic valuation of $n$, then
$$v_2(b_n) = v_2(C_{n+1}) = v_2\Big(\frac{2^{n+1}}{(n+2)!}\Big),\; v_2(a_n)=0\quad \forall n>1. \tag{2}$$
Now the original equation $(0)$ already implies that $\;a_n\;$ is an integer sequence using induction, $\;a_n\;$ is odd if $n>1$, and the rest of equation $(2)$ is true.
However, there is an alternative approach using the exponential generating function of $\;a_n,\;$ namely $\;A(x):=\sum_{n=0}^\infty a_nx^n/n!.\;$
Using equation $(0)$ for $\;a_n\;$ we get the differential equation
$$ 1 = -6A(x) +(3+6x)\frac{d}{dx}A(x) + (1+2x+2x^2)\frac{d^2}{dx^2}A(x).$$
Taking the derivative of this equation gives another differential equation
$$ 0 = 5(1+2x)\frac{d^2}{dx^2}A(x) +(1+2x+2x^2)\frac{d^3}{dx^3}A(x).$$
Now define the sequence $\;y_n(x)\;$ by
$\,y_0(x) := 1+O(x),\,$ and recursion
$$ y_{n+1}(x) := 1-\int_0^x 5(1+2x)/(1+2x+2x^2)\;y_n(x)\; dx. \tag{3}$$
The limit as $\;n\to\infty$ of $\;y_n(x) =\frac{d^2}{dx^2}A(x).\;$ It is easy to check that the recursion $(3)$ preserves exponential generating functions of integer sequences. Thus, again $\;a_n\;$ is an integer sequence.
If $n>0$ is odd, then $(n+2)\;|\;a_n$ and if $n>2$ is even, then $(n+1)\;|\;a_n.\;$ This since one of the $(n\!+\!1)(n\!+\!2)$ factors in the numerator of $(1)$ are not cancelled by the $2^{n+1}$ in the denominator.
|
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|
Solve $(1+x)y’=y$ by power series Solve $(1+x)y’=y$ by power series. $$$$Start with $y$ and $y’$:
$y=a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n$
$y’=a_1+2a_2x+3a_3x^2+...+na_nx^{n-1}$
Then $(1+x)y’=a_1(1+x)+2a_2(1+x)x+3a_3(1+x)x^2+...+na_n(1+x)x^{n-1}$
Then $(1+x)y’-y=[a_1(1+x)+2a_2(1+x)x+3a_3(1+x)x^2+...+na_n(1+x)x^{n-1}]-[a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n]=0$ $$$$$\implies [a_1+a_1x+2a_2x+2a_2x^2+3a_3x^2+3a_3x^3+...+(na_n+na_nx^n)]-[a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n]=0$$$$$$(a_1-a_0)+(a_1+2a+2-a_1)x+(2a_2+3a_3-a_2)x^2+(3a_3+a_3)x^3+...+=0$
I’m missing the trick when equating terms. Any help would be appreciated!
|
Are you required to "use series"? Here's a very quick solution. Rewrite it as
$$ \frac{y'}{y} = \frac{1}{1+x}. $$
Note that
$$ \frac{y'}{y} = \frac{d\phantom{x}}{dx}\ln(y) $$
and
$$ \frac{1}{1+x} = \frac{d\phantom{x}}{dx}\ln(1+x). $$
Integrating gives
$$ \ln(y) = \ln(1+x) + C. $$
Therefore $y=c(1+x)$.
|
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|
$1-\cos (x) \leq \frac{x^2}{2} + \frac{x^3}{6}$, for $x > 0$, using Taylor expansion I want to solve this problem using Taylor expansions.
I tried
\begin{align*}
1 - \cos (x) = 1 - \left( \sum_{k=0}^{n} (-1)^k \frac{x^{2k}}{(2k)!} +R_{2n}(x) \right) \;,
\end{align*}
where $R_{2n}(x)$ is the remainder function.
For $n=2$, this yields
\begin{align*}
1 - \cos (x) = 1 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} + R_4(x) \right) = \frac{x^2}{2} - \frac{x^4}{24} - o(x^4) \;.
\end{align*}
Is it now possible to just state
\begin{align*}
|o(x^4)| \leq \frac{x^4}{24} \; ,
\end{align*}
and hence
\begin{align*}
\frac{x^2}{2}-\frac{x^4}{24}-o(x^4)\leq \frac{x^2}{2} \leq \frac{x^2}{2} + \frac{x^3}{6} \; ?
\end{align*}
|
I think I found an appropriate method.
The formula for the remainder term of an $n$-degree Taylor polynomial around $a$ is given by
\begin{align*}
R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} \; ,
\end{align*}
for some constant $c$ in the open interval between $a$ and $x$, so $c\in(a,x)$ or $c\in(x,a)$.
The remainder term for the second-order Taylor polynomial $T_2(x)$ of $\cos x$ is given by\begin{align*}
R_2(x) = \frac{-\cos c}{6} x^3 \; ,
\end{align*}
which can be bounded: $-x^3/6 \leq R_2(x) \leq x^3/6$.
From this, the claim now immediately follows
\begin{align*}
1 - \cos x = 1 - (T_2(x) + R_2(x)) = 1 - (1 - \frac{x^2}{2} + R_2(x)) = \frac{x^2}{2} - R_2(x) \leq \frac{x^2}{2} + \frac{x^3}{6} \;.
\end{align*}
|
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|
Conceptual problem in Theory Of Equations While doing some self-study from the book: Higher Algebra by Hall & Knight, I encountered some articles that I could not understand properly. Those articles were Art. 562 & Art. 563.
Art 562: The result of the preceding article(given below) enables us very easily to find the sum of an assigned power of the roots of an equation.
Result of preceding article: $f'(x)=\frac{f(x)}{(x-a)}+\frac{f(x)}{(x-b)}+....+\frac{f(x)}{(x-k)}$,
where a,b,....,k are the roots of the equation $f(x)=0$
Art 563: When the coefficients are numerical we may also proceed as in the following example.
Here is where the problem lies
Example: Find the sum of the fourth powers of the roots of
$$x^3-2x^2+x-1=0$$
Here $f(x)=x^3-2x^2+x-1$, $f'(x)=3x^2-4x+1$. $$\frac{f'(x)}{f(x)}=\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}$$
$$=\sum{\Bigl(\frac{1}{x}}+\frac{a}{x^2}+\frac{a^2}{x^3}+....\Bigl)\qquad\qquad...(1)$$
$$=\frac{3}{x}+\frac{S_1}{x^2}+\frac{S_2}{x^3}+\frac{S_3}{x^4}+.....;\qquad\qquad...(2)$$
hence $S_4$ is equal to the coefficient of $\frac{1}{x^5}$ in the quotient of $f'(x)$ by $f(x)$, which is very conveniently obtained by the method of synthetic division as follows:
Hence the quotient is $\frac{3}{x}+\frac{2}{x^2}+\frac{2}{x^3}+\frac{5}{x^4}+\frac{10}{x^5}+.....;$ thus $S_4=10$.
I could not understand the method of synthetic division and the steps (1) and (2).
Any help will be appreciated.
|
For all $x$ such that $|x|>\max (|a|,|b|,|c|)$ and for each $d\in \{a,b,c\}$ we have $$\frac {1}{x-d}=(1/x)\cdot\frac {1}{1-d/x}=(1/x)\sum_{n=0}^{\infty}(d/x)^n=$$ $$=(1/x)+(d/x^2)+(d^2/x^3)+...$$ If we sum this over each of the 3 series with $d=a, d=b$ and $d=c,$ the resulting co-efficient of $x^{-5}$ is $a^4+b^4+c^4.$
Imagine that $f(x$) and $f'(x)$ are representations of numbers in number-base $x,$ although we allow the digits (the co-efficients) to be any numbers, not just integers from $0$ to $x-1.$ Synthetic division follows the same rules of long division that you would use for computing all of the digits (co-efficients) of all integer-powers of of the base, including especially the negative powers.
|
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|
Prove: $(a+b) \sqrt {ab}+(a+c) \sqrt {ac}+(b+c)\sqrt {bc}≥ \frac {(a+b+c)^2}{2}$
EDİTED:
I tried to solve the last question.I can not get the state of equality and I can not find my mistake. Can you show me my mistake(s)?
4.
Let, $a,b,c$ be the lengths of sides of a triangle. Prove the inequality:
$$(a+b) \sqrt {ab}+(a+c) \sqrt {ac}+(b+c)\sqrt {bc}≥ \frac {(a+b+c)^2}{2}$$
I will use:
$(1)$
$\begin{cases} a+b≥2\sqrt{ab} &\\ b+c≥2\sqrt{bc}&\\ a+c≥2\sqrt{ac} \end{cases}\Longrightarrow \begin{cases} (a+b)\sqrt{ab}≥2ab &\\ (b+c)\sqrt{bc}≥2bc&\\ (a+c)\sqrt{ac}≥2ac \end{cases} \Longrightarrow (a+b)\sqrt{ab}+(b+c)\sqrt{bc}+(a+c)\sqrt{ac}≥2(ab+bc+ac)$
$(2)$
$\begin{cases} b(a+c)>b^2 &\\ c(a+b)>c^2&\\ a(b+c)>a^2 \end{cases}\Longrightarrow 2(ab+bc+ac)>a^2+b^2+c^2 \Longrightarrow \frac{a^2+b^2+c^2}{2}<ab+bc+ac$
Applying $(1)$ and $(2)$ we have
$(a+b) \sqrt {ab}+(a+c) \sqrt {ac}+(b+c)\sqrt {bc}- \frac {(a+b+c)^2}{2}≥2ab+2bc+2ac- \frac {(a+b+c)^2}{2}≥ab+bc+ac-\frac{a^2+b^2+c^2}{2}>0≠≥0$
|
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$\sum_{cyc}(2x+y+z)\sqrt{(x+y)(x+z)}\geq2(x+y+z)^2$$ and since by C-S $$\sqrt{(x+y)(x+z)}\geq x+\sqrt{yz},$$ it's enough to prove that
$$\sum_{cyc}(2x+y+z)(x+\sqrt{yz})\geq2(x+y+z)^2$$ or
$$\sum_{cyc}\left(\sqrt{x^3y}+\sqrt{x^3z}-2xy+2x\sqrt{yz}\right)\geq0$$ or
$$\sum_{cyc}\left(\sqrt{xy}(\sqrt{x}-\sqrt{y})^2+2x\sqrt{yz}\right)\geq0.$$
Done!
The equality does not occur because $\sum\limits_{cyc}x\sqrt{yz}>0.$
|
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|
Conditions on coefficients of positive semidefinite matrix with certain symmetries I have a real matrix, with certain symmetries, defined as
$
A = \left( {\begin{array}{*{20}{c}}
1-x&a&b&c\\
a&x&d&b\\
b&d&x&a\\
c&b&a&1-x
\end{array}} \right),
$
with $x,a,b,c,d \in \mathbb{R},{\rm{~ }}0 \le x \le 1$.
I want to obtain conditions on coefficients $x,a,b,c,d$ for the matrix to be positive semidefinite.
For the particular case $x=0$, I obtain with the Mathematica Reduce command the following conditions for the eigenvalues to be nonnegative:
$a = b = 0,{\rm{~}} -1 \le c \le 1,{\rm{ ~}}d = 0$.
In[1]:= A = {{(1 - x), a, b, c}, {a, x, d, b}, {b, d, x, a}, {c, b,
a, (1 - x)}}
Out[1]= {{1 - x, a, b, c}, {a, x, d, b}, {b, d, x, a}, {c, b, a,
1 - x}}
In[2]:= FullSimplify[Eigenvalues[A /. x -> 0]]
Out[2]= {1/2 (1 - c - d - Sqrt[4 (a - b)^2 + (1 - c + d)^2]),
1/2 (1 - c - d + Sqrt[4 (a - b)^2 + (1 - c + d)^2]),
1/2 (1 + c - Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d),
1/2 (1 + c + Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d)}
In[3]:= Reduce[
1/2 (1 - c - d - Sqrt[4 (a - b)^2 + (1 - c + d)^2]) >= 0 &&
1/2 (1 - c - d + Sqrt[4 (a - b)^2 + (1 - c + d)^2]) >= 0 &&
1/2 (1 + c - Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d) >= 0 &&
1/2 (1 + c + Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d) >= 0]
Out[3]= b == 0 && a == 0 && d == 0 && -1 <= c <= 1
Similarly, for the particular case $x=1~$ I get
$~a = b = 0,{\rm{~}} -1 \le d \le 1,{\rm{ ~}}c = 0$.
However, for a general $~0\le x \le1$, Mathematica takes weeks without giving an answer. I suspect that for $~0< x <1~$ the condition $a = b = 0~$ must be satisfied.
Do you think there is a way to obtain some conditions for the general case $~0\le x \le1$? The general eigenvalues of matrix $A$ have the following expressions:
$
\frac{1}{2} \left(-\sqrt{4 (a-b)^2+(-c+d-2 x+1)^2}-c-d+1\right),\frac{1}{2} \left(\sqrt{4 (a-b)^2+(-c+d-2 x+1)^2}-c-d+1\right),\frac{1}{2} \left(-\sqrt{4 (a+b)^2+(c-d-2 x+1)^2}+c+d+1\right),\frac{1}{2} \left(\sqrt{4 (a+b)^2+(c-d-2 x+1)^2}+c+d+1\right).
$
|
You have both $1-(c+d)-\sqrt{4(a-b)^2+(1-2x+d-c)^2}$ and $1-(c+d)+\sqrt{4(a-b)^2+(1-2x+d-c)^2}$ non negative iff:
$$1-(c+d)\geq 0$$
$$(1-(c+d))^2 \geq 4(a-b)^2 +(1-2x+d-c)^2$$
the second condition can be simplified as:
$$((1-x)-c)(x-d) \geq (a-b)^2$$
In particular $(1-x)-c$ and $x-d$ have the same sign.
As the sum $(1-x)-c+(x-d)=1-(c+d)$ both terms are positive.
Finally $1-(c+d) \pm \sqrt{4(a-b)^2+(1-2x+d-c)^2} \geq 0$ iff $c\leq 1-x$, $d<x$ and $((1-x)-c)(x-d) \geq (a-b)^2$.
The same computation on the two other eigenvalues give the final conditions:
$$-(1-x) \leq c \leq (1-x)$$
$$-x \leq d \leq x$$
$$((1-x)-c)(x-d) \geq (a-b)^2$$
$$((1-x)+c)(x+d) \geq (a+b)^2$$
If $x=1$ or $x=0$ these conditions are exactly the ones you described, if $0 \leq x \leq 1$ there is maybe a simpler way to express them but it is worth noticing that this is no longer necessary to have $a=b=0$.
|
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|
How do I determine this integral? $\int_{0}^{+\infty}\sin^2(1/x)\frac{dx}{(4+x^2)^2}$ $$\int_{0}^{\infty}\mathrm dx{\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}$$
$${\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1-\cos^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1\over 2(4a^2+x^2)^2}-{\cos\left({2a\over x}\right)\over 2(4a^2+x^2)^2}$$
$${1\over 2}\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$
$${\pi\over 8(2a)^3}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$
$$\int \mathrm dx{1\over (b^2+x^2)^2}={x\over 2b^2(b^2+x^2)}+{1\over 2b^3}\arctan\left({x\over b}\right)+K$$
$$\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$
Enforcing a substitution of $u=\dfrac{2a}{x}$
$${1\over (2a)^3}\int_{0}^{\infty}\mathrm du {u^2\cos(u)\over (1+u^2)^2}$$
Now this integral is more harder than the original due to the extra $u^2$ at the numerator.
This is an even function, so can be expressed as
$${1\over 2(2a)^3}\int_{-\infty}^{\infty}\mathrm du {u^2\cos(u)\over (1+u^2)^2}$$
Decomposition of fraction
$${u^2\over (1+u^2)^2}={Au+B\over 1+u^2}+{Cu+B\over (1+u^2)^2}$$
This look like a nightmare, so how do I determine this integral?
|
The value of $a$ is irrelevant, the $a$-parameter can be removed through a suitable substitution, and by replacing $x$ with $\frac{1}{x}$ the problems boils down to computing
$$ \int_{0}^{+\infty}\frac{\sin^2(x)}{(4x+1/x)^2}\,dx=\frac{1}{16}\int_{0}^{+\infty}\frac{1-\cos(x)}{(x+1/x)^2}\,dx =\frac{1}{16}\left[\frac{\pi}{4}-\int_{0}^{+\infty}\frac{\cos(x)}{(x+1/x)^2}\,dx\right]$$
or
$$ \text{Re}\int_{-\infty}^{+\infty}\frac{x^2 e^{ix}}{(x^2+1)^2}\,dx\stackrel{\text{Residues}}{=}\color{red}{0} $$
which leads to $\int_{0}^{+\infty}\frac{\sin^2(x)}{(4x+1/x)^2}\,dx = \color{red}{\frac{\pi}{64}}$.
|
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|
For how many primes $p$ is $p^2 + 2$ is also prime? For how many primes $p$ is $p^2 + 2$ also prime?
I'm not sure but I think you can consider every prime number as $(3k+1)$ or $(3k+2)$
|
We will not consider $p=2$ because $2^2+2=6$ which is obviouly not prime.
Most definitely $p\neq n^2 + 1$ for all odd $p$ because if we assume otherwise, we want $$(n^2+1)^2+2=n^4+2n^2+3$$ to be prime. Therefore $3\nmid n$. (The only prime divisible by $3$ is $3$ itself, so $n=0$ is a solution, but note that $0^2+1=1$ which is neither prime nor composite.) Every prime is of the form $3k+1$ or $3k+2$, so if $p=n^2+1$ then $p\neq 3k+1$ and thus $p= 3k+2$. $$\begin{align}\therefore n^2+1&=3k+2 \\ \Leftrightarrow 3k&=n^2-1 \\ &=(n+1)(n-1).\end{align}$$ Since $$(n+1)-(n-1)=2<3$$ then $3\mid n+1$ or $3\mid n-1$ but not both. $$\therefore n = 3l+1\text{ or } 3l-1$$ which when we substitute, we get that $3\mid p^2+2$ for all odd $p = n^2+1$.
It is also provable by induction that $3\mid n^4 + 2n^2$ for all integers $n$. The same result is thus implied.
|
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|
How does Tom Apostol deduce the inequality $\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}$ in section I 1.3 of his proof by induction example? In his book Calculus, Vol. 1, Tom Apostol writes the following proof for proving the following predicate by induction:
$$A(n): 1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3}.$$
I have omitted the Base Case due to lack of specific relevance. An excerpt from the book's page:
Assume the assertion has been proved for a specific value of $n$, say $n = k$. That is, assume we have proved
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 \lt \frac{k^3}{3}$$
for a fixed $k \ge 1$. Now using this, we shall deduce the corresponding result for $k + 1:$
$$A(k + 1): 1^2 + 2^2 + \cdots + k^2 \lt \frac{(k + 1)^3}{3}.$$
Start with $A(k)$ and add $k^2$ to both sides. This gives the inequality
$$1^2 + 2^2 + \cdots + k^2 \lt \frac{k^3}{3} + k^2.$$
To obtain $A(k + 1)$ as a consequence of this, it suffices to show that
$$\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}.$$
But this follows at once from the equation
$$\frac{(k + 1)^3}{3} = \frac{k^2 + 3k^2 + 3k + 1}{3} = \frac{k^3}{3} + k^2 + k + \frac13.$$
Therefore, we have shown that $A(k + 1)$ directly follows from $A(k)$.
I can not understand the last two steps. Why do the expressions seem to flip on the inequality in step 4 such that $\frac{k^3}{3} + k^2$ is now on the other side of the less than symbol, and how does the final equation prove the assertion?
Thank you.
|
There's possibly a better way to see what's going on; instead of manipulating what you want to show, proceed from the left-hand side:
\begin{align}
1^2+2^2+\dots+(k-1)^2+k^2
&<\frac{k^3}{3}+k^2 && \text{(induction hypothesis)}\\[4px]
&=\frac{1}{3}(k^3+3k^2) && \text{(because we want $1/3$)}\\[4px]
&=\frac{1}{3}(k^3+3k^2+3k+1-3k-1) && \text{(complete the cube)}\\[4px]
&=\frac{1}{3}(k+1)^3-\frac{1}{3}(3k+1) && \text{(separate the cube)}\\[4px]
&<\frac{1}{3}(k+1)^3
\end{align}
|
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|
Lagrange Multiplier: Distance to the Origin Find the points on the curve $x^2+xy+y^2=2$ that are closest to the origin.
Is there a way to use Lagrange multipliers to answer this question?
|
Without using calculus,
$$8=(2x+y)^2+3y^2$$
WLOG $\sqrt3y=2\sqrt2\sin t, 2x+y=2\sqrt2\cos t\iff x=?$
So, we need to minimize $$\left(\dfrac{2\sqrt2\sin t}{\sqrt3}\right)^2+\left(\dfrac{2\sqrt2\cos t-\dfrac{2\sqrt2\sin t}{\sqrt3}}{2}\right)^2$$
$$=\dfrac{8\sin^2t}3+\dfrac{8(\sqrt3\cos t-\sin t)^2}{12}$$
$$=\dfrac{8\sin^2t+3\cos^2t+\sin^2t-2\sqrt3\sin t\cos t}3$$
$$=\dfrac{9(1-\cos2t)+3(1+\cos2t)-2\sqrt3\sin2t}6$$
$$=\dfrac{12-2\sqrt3(\sqrt3\cos2t+\sin2t)}6$$
using $\cos2t=1-2\sin^2t=2\cos^2t-1$
and use $$-\sqrt{a^2+b^2}\le a\cos2t+b\sin2t\le\sqrt{a^2+b^2}$$
|
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|
Find the limit $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$ and prove it. Find $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$.
I claim that $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$. To prove this, for given $\varepsilon >0$, I have to find $M\in N$ such that $|\frac {2n^2+10n+5}{n^2}-2|<\varepsilon$ for $n \ge M$.
By Archimedean property, we can find $M \in N$ such that $\frac {15}M<\varepsilon$, and note that $n\ge M \rightarrow \frac 1n \le \frac 1M \rightarrow \frac {15}n \le \frac {15}M$.
Then, for $n \ge M$, we have that $|\frac {2n^2+10n+5}{n^2}-2|=|\frac {10n+5}{n^2}| < |\frac {15n}{n^2}|$ (since $n \ge M \in N$) $<\frac {15}n\le \frac {15}M<\varepsilon$.
Therefore, by definition of convergence, $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$.
Should I say $M\in Z^+$ (because I am worried about the case where $M=0$)? Can you find any mistakes in this proof?
Thank you in advance.
|
Another possibility: apply two times Cesàro-Stolz.
$$
\lim_{n\to\infty}\frac{2n^2 + 10n + 5}{n^2} =
\lim_{n\to\infty}\frac{((2(n+1)^2 + 10(n+1) + 5) - (2n^2 + 10n + 5)}{(n+1)^2 - n^2} =
\lim_{n\to\infty}\frac{4n + 12}{2n + 1} =
\lim_{n\to\infty}\frac{(4(n+1) + 12) -(4n + 12)}{(2(n+1) + 1) - (2n + 1)} =
\lim_{n\to\infty}\frac42 = 2.
$$
|
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|
Finding the number of multiple of $3$ in the coefficients of $x(x+1)(x+2)\cdots (x+239)$ Suppose
$$
x(x+1)(x+2)\cdots(x+239)=\sum_{n=1}^{240}a_nx^n
$$
What's the total number of $a_n$ which is exactly the multiple of $3$?
I've calculated using Mathematica and got the answer is $160$, but I don't know how to solve it using Number Theory.
This is my Mathematica code:
Tr[Divisible[CoefficientList[Product[(x + i), {i, 1, 240}], x][[2 ;; 240]], 3]]
and get
80 False + 159 True
|
Some variant based on Robert Z's answer not requiring evaluating $\binom{80}{k}\pmod 3$, even though it is the same at final.
Note that $(x-1)^2(x+1)^2=(x^2-1)^2=x^4-2x^2+1\equiv 1+x^2+x^4\pmod 3$
Then we also have a pattern for the cube of such expressions:
*
*$(1+x^2+x^4)^3\equiv 1+x^6+x^{12}\pmod 3$
*$(1+x^6+x^{12})^3\equiv 1+x^{18}+x^{36}\pmod 3$
*$(1+x^{18}+x^{36})^3\equiv 1+x^{54}+x^{108}\pmod 3$
Since $80=2+6+18+54$
The product is in fact $$\quad x^{80}(1+x^2+x^4)(1+x^6+x^{12})(1+x^{18}+x^{36})(1+x^{54}+x^{108})\pmod 3$$
And we can see it develops to $\displaystyle \sum\limits_{k=40}^{120} x^{2k}$ because all powers are different.
There are $120-40+1=81$ terms.
|
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|
Area of $\triangle ABC$, given $AB=7$, $AC=15$, and median $AM=10$ I've been working on this interesting problem for a while already, and here it is:
In $\triangle ABC$, $AB = 7$, $AC = 15$, and median $AM = 10$. Find the area of $\triangle ABC$.
I have figured out that $BM$ and $CM$ are both $4\sqrt2$ using Stewart's Theorem. Now, I tried to use Heron's Formula to calculate the area, which was a mess.
Any help is appreciated. Thanks.
|
Let $|AB|=7=c$,
$|AC|=15=b$,
$|AM|=10=m_a$,
$|BM|=|MC|=\tfrac{a}2=x$.
Then by Stewart’s theorem for $\triangle ABC$
\begin{align}
b^2x+c^2x&=2x(m_a^2+x^2)
,\\
x&=\sqrt{\tfrac12(b^2+c^2)-m_a^2}
=\sqrt{\tfrac12(49+225)-100}
=\sqrt{37}
,\\
a&=2x=2\sqrt{37}
.
\end{align}
And the area
\begin{align}
S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
\\
&=\tfrac14\sqrt{4\cdot4\cdot37 \cdot 225
-(4\cdot37+225-49)^2}
.
\end{align}
So, the answer is$\dots$ still 42.
|
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|
inequality of series, showing their difference is less than a number given it is monotonically increasing Given $a_1=1$ $$a_{n+1} =
\frac{2+2a_n}{2+a_n} \text{is bounded: } 1\leq a_n \leq 2, \quad a_{n+1}-a_n=\frac{2(a_n-a_{n-1})}{(2+a_n)(2+a_{n-1})}$$
Prove $$|a_{n+1}-a_n|\leq \frac{1}3\left(\frac{2}9\right)^{n-1}$$
So I can show that $$\lvert a_{n+1}-a_n \rvert \leq \frac{2\lvert a_n-a_{n-1}\rvert }9,$$ but I then end up with $2/9\ldots$
Any hints? How should I be thinking of this? Like where does the $^{n-1}$ come from?
|
You have done most of the work already. Let $b_n=|a_{n+1}-a_n|$ for $n\in \mathbb{N} ^*$.
Then, you proved that $\forall n\in \mathbb{N}^*, b_n\leqslant \frac{2} {9} b_{n-1}$. You can prove $b_n \leqslant b_1 \left( \frac{2} {9} \right)^{n-1}$ using induction since $b_n$ is always positive.
Goal:
Prove that, $\forall n\in \mathbb{N}^*, b_n \leqslant b_1\left( \frac{2}{9} \right)^{n-1}$
Initialization:
At $n=1$, $b_1 = b_1 \times \left( \frac{2}{9} \right)^{1-1}$. Thus the property is initialized
Heredity:
If $b_{n-1} \leqslant b_1\left( \frac{2}{9} \right)^{n-2}$, then, since $b_n\leqslant \frac{2}{9} b_{n-1}$ then, $b_n \leqslant \frac{2}{9} \times b_1\left( \frac{2}{9} \right)^{n-2}=b_1\left( \frac{2}{9} \right)^{n-1}$
We thus proved the result for all $n>0$
As you noticed, $b_1=\frac{1}{3}$ hence the final result.
That is pretty much the proof for the general term of a geometric sequence.
|
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|
Short way for upper triangularization
We are given a matrix $$A =
\begin{bmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{bmatrix}$$
and we are asked to find a matrix $P$ such that $P^{-1}AP$ is upper triangular.
Here, we first find one eigenvalue as $\lambda= 4$. Then the matrix $$
A-4I =
\begin{bmatrix}
-1 & 0 & 1 \\
-1 & 0 & -3 \\
-1 & 0 & 1 \\
\end{bmatrix}$$
has basis formed from $f_1 = (-1,-1,-1)^T$, $f_2 = (1,-3,1)^T$. We extend this to a basis of the whole space by adjoining $f_3 = (1,0,0)^T$, and so we have base-change matrix
$$ P =
\begin{bmatrix}
-1 & 1 & 1 \\
-1 & -3 & 0 \\
-1 & 1 & 0 \\
\end{bmatrix}.$$
Then, by using some computational tools, we find that
$$ P^{-1}AP =
\begin{bmatrix}
3 & 1 & 1 \\
-1 & 5 & 0 \\
0 & 0 & 4 \\
\end{bmatrix}$$
Now, we need to look at $$B = \begin{bmatrix}
3 & 1 \\
-1 & 5 \\
\end{bmatrix},$$ which has eigenvalue $\lambda = 4$. So we have $$B-4I = \begin{bmatrix}
-1 & 1 \\
-1 & 1 \\
\end{bmatrix}.$$
So basis for the image of this is $(1,1)^T$. We extend this to the basis $(1,1)^T$, $(1,0)^T$ of $\mathbb{R}^2$. Now, going back to $\mathbb{R}^3$, we have the matrix $$Q = \begin{bmatrix}
1 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}.$$
Then, by using calculation tools, we get
$$Q^{-1}P^{-1}APQ = \begin{bmatrix}
4 & -1 & 0 \\
0 & 4 & 1 \\
0 & 0 & 4 \\
\end{bmatrix},$$
which is in upper triangular form.
Now, what I wanted to ask is that is there a way to directly find the matrix $R = PQ$ such that $R^{-1}AR$ is upper triangular, without going through these steps?
|
There is a misprint in the announcement. The $(1,3)$ element of $A$ should be $+1$ rather than $-1$ in order for $A$ to be consistent with the calculations that follow in your posted question.
So let $A=\begin{bmatrix} 3&0&1\\-1&4&-3\\-1&0&5 \end{bmatrix}$ and set
$N=A-4I= \begin{bmatrix} -1&0&1\\-1&0&-3\\-1&0&1 \end{bmatrix}$.
Then $N^2= \begin{bmatrix} 0&0&0\\4&0&-4\\0&0&0 \end{bmatrix}$ and $N^3=0$. So $N$ is nilpotent and $4$ is a triple eigenvalue of $A$ having a `Jordan chain' of length 3. A simple way to triangularize $A$ is then to calculate 3 vectors that generate this chain. For this you may note that $\ker N =\langle v_1\rangle $ is generated by $v_1=\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}^T$ and $\ker N^2 =\langle v_1,v_2\rangle$ with e.g. $v_2=\begin{bmatrix} -1 & 1 & -1\end{bmatrix}^T$. Finally, set e.g. $v_3=\begin{bmatrix} 2 & 0 & -2\end{bmatrix}^T$ so that $\ker(N^3)={\Bbb R}^3=\langle v_1,v_2,v_3\rangle$.
Now set $R= [v_1 \; v_2\; v_3] = \begin{bmatrix} 0&-1&2\\1&1&0\\0&-1&-2\end{bmatrix}$ and we get: $A R = R \begin{bmatrix} 4 & 4 & 0\\ 0&4&4\\0&0&4\end{bmatrix}$ as wanted.
You may, of course, replace $v_2$ by $a v_2+b v_1$ with $a\neq 0$ which will change e.g. the $(1,2)$ element of the Jordan form (and similarly for $v_3$). My choice was just in order to get easy numbers and a nice end-result.
|
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|
sequences above 0,1,2 start&end with 0 with no 2 same successive digits find the number of sequences of length n above {0,1,2} that start and end with 0 and without 2 successive digits that are equal (00,11,22).
Find and solve a recursice equation.
I would be thankful for a clue or an approach to this.
|
We define a recurrence relation as follows. Let
*
*$a_k$ denote the number of valid strings with $k$-th symbol $0$.
*$b_k$ denote the number of valid strings with $k$-th symbol $1$.
*$c_k$ denote the number of valid strings with $k$-th symbol $2$.
Since valid strings start with $0$ we have
\begin{align*}
a_1=1\qquad b_1=0\qquad c_1=0\tag{1}
\end{align*}
Valid strings do not have equal consecutive characters. The recurrence relation for $2\leq k\leq n-1$ is
\begin{align*}
a_k&=b_{k-1}+c_{k-1}\\
b_k&=a_{k-1}+c_{k-1}\tag{2}\\
c_k&=a_{k-1}+b_{k-1}
\end{align*}
and since the $n$-th character of a valid string is $0$, we have for $n\geq 2$:
\begin{align*}
a_n=b_{n-1}+c_{n-1}\tag{3}
\end{align*}
We are looking for $a_n, n\geq 0$.
Due to the symmetry of $b_n$ and $c_n$ in the recurrence relations (1) - (3) we observe $b_n=c_n, n\geq 1$ and we obtain a simplified recurrence relation
\begin{align*}
a_1&=1, b_1=0\\
a_k&=2b_{k-1}\qquad\qquad 2\leq k\leq n-1\tag{4}\\
b_k&=a_{k-1}+b_{k-1}\\
a_n&=2b_{n-1}\tag{5}
\end{align*}
We obtain from (4) a recurrence relation for $b_k, k\geq 1$.
\begin{align*}
b_1&=0, b_2=1\\
b_k&=b_{k-1}+2b_{k-2}\qquad\qquad k\geq 3\tag{6}
\end{align*}
Setting $B(x)=\sum_{k=0}^\infty b_kx^k$ we obtain from (6)
\begin{align*}
\sum_{k=3}^\infty b_kx^k&=\sum_{k=3}^\infty b_{k-1}x^k+2\sum_{k=3}b_{k-2}x^k\\
&=x\sum_{k=2}^\infty b_{k}x^k+2x^2\sum_{k=1}b_{k}x^k\\
B(x)-x^3&=xB(x)+2x^2B(x)\\
\color{blue}{B(x)}&\color{blue}{=\frac{x^3}{1-x-2x^2}}
\end{align*}
Since $a_n=2b_{n-1}$ according to (5), we finally obtain for $n\geq 2$
\begin{align*}
a_n&=2[x^{n-1}]B(x)\\
&=2[x^n]xB(x)\\
&=[x^n]\frac{2x^3}{1-x-2x^2}\\
&=[x^n]\left(2 x^3 + 2 x^4 + 6 x^5 + \color{blue}{10} x^6 + 22 x^7 + 42 x^8 + \cdots\right)
\end{align*}
The last line was calculated with some help of Wolfram Alpha.
The coefficient $\color{blue}{10}$ of $x^6$ for example tells us we have $10$ valid strings
\begin{align*}
&010120,010210,012010,012020,012120,\\
&020120,020210,021010,021020,021210
\end{align*}
|
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|
$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$. Find its 10th term. Given this sequence, find its 10th term and its exact limit.
$$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$$
I've been stuck in this question forever. I can't find any relation between them. The answer for the 10th term is $\frac{5333}{3771}$
|
$$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},...$$
$$\frac{1}{2},\frac{5}{(2+1)},\frac{11}{(2+1+5)},\frac{27}{(2+1+5+11)},...$$
$$\frac{1}{2},\frac{(2+3)}{3},\frac{(3+8)}{8},\frac{(8+19)}{19},...$$
In this way I would define the series as follows:
$$\text{ for } a_1=\frac{x_1}{y_1} = \frac{1}{2}$$
$$a_n := \frac{x_n}{y_n} = \frac{y_{(n-1)} + y_n}{x_{(n-1)}+y_{(n-1)}} $$
so
$$a_n := \frac{x_n}{y_n} = \frac{y_{(n-1)} + y_n}{x_{(n-1)}+y_{(n-1)}} = \frac{y_{(n-1)} + \left(x_{(n-1)}+y_{(n-1)}\right)}{x_{(n-1)}+y_{(n-1)}} = \frac{2y_{(n-1)} + x_{(n-1)}}{x_{(n-1)}+y_{(n-1)}} =$$
$$ a_n = \left(1 + \frac{y_{(n-1)}}{y_{(n-1)}+x_{(n-1)}}\right)$$
the series until the 10th term:
$$\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},\frac{65}{46}, \frac{157}{111}, \frac{379}{368}, \frac{915}{647}, \frac{2209}{1562}, \frac{5333}{3771}$$
|
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|
Question about functional equations
Let $F(x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equation $$F(x) + F\left(\frac{x-1}x\right) = 1+x$$ Find $F(x)$ satisfying these conditions.
Write $F(x)$ as a rational function with expanded polynomials in the numerator and denominator.
I substituted $x = \frac{x-1}x, -\frac{1}{x-1}$. This gave me the system of equations
$F(x) + F\left(\frac{x-1}x\right) = 1+x,
F( \frac{x-1}x ) + F(- \frac{1}{x-1} ) = 1+ \frac{x-1}x,
F(- \frac{1}{x-1} ) + f(x) = 1- \frac{1}{x-1}.$
However, solving this system of equations gave me $F(x) = \frac{x^3-x^2-1}{2(x^2-2)}$, which is incorrect. Where did I go wrong?
|
It's just a simplification error. Let $g(x)=\frac{x-1}{x}$. Then, note
$$
g(x)=1-\frac{1}{x},\quad g(g(x))=\frac{1}{1-x},\quad g(g(g(x)))=x.
$$
And so you have the system:
\begin{align*}
1+x&=F(x)+F(g(x)),\\
1+g(x)&=F(g(x))+F(g(g(x))),\\
1+g(g(x))&=F(g(g(x)))+F(x)
\end{align*}
from which
\begin{align*}
F(x)&=[F(x)+F(g(x))+F(g(g(x)))]-[F(g(x))+F(g(g(x)))]\\
&=\frac{1}{2}[1+x+1+g(x)+1+g(g(x))]-(1+g(x))\\
&=\frac{x^3-x^2-1}{2(x^2-\color{red}{x})}.
\end{align*}
|
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|
How many different strings of length $9$ containing only the letters a, b, and c have exactly two a's or exactly three b's? The question as stated in the title is how many different strings of length $9$ containing only the letters a, b, and c have exactly two a's or exactly three b's?
I came up with the idea that there is $C(9,2)$ ways to choose two a's and then the rest of the $7$ positions is $2^7$, and I added that to $C(9,3) \cdot 2^6$ with similar logic. However, I cannot seem to get the right numeric answer. Any help is appreciated.
|
This would be a good case for inclusion/exclusion.
Number of ways with exactly $2$ $A$ will $2^7*{9\choose 2}$. (This includes those with $0,1,2,3,4,5$ $B$s.... so this includes those with $2$ $A$ and $3$ $B$s).
Number of ways with exactly $3$ $B$s will be $2^6*{9\choose 3}$ (This includes those with $0,1,2,3,4$ $A$s.... so this includes those with $2$ $A$ and $3$ $B$s.).
And numbers of ways with exactly $2$ $A$s and $3$ $B$s would be $1^4*{9\choose 2}*{7\choose 3}$.
So exactly $2$ $A$ or exactly $3$ $B$s will be: $2^7*{9\choose 2} + 2^6{9\choose 3}-1^4{9\choose 2}*{7\choose 3}$.
|
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|
Elementary proof that the MacLaurin series of $\sin x$ converges to $\sin x$ for all $x$ In my book it is given:
$\sin x = x- \dfrac {x^3}{3!}+\dfrac{x^5}{5!}- \dfrac{x^7}{7!}...$
I googled around for a proof but couldn't understand any of them. I would like to know if there's any elementary high school level proof the series
|
I faced same problem here. We know that series approximated around $ x= 0$ and it converged but how do we know that it converge to $\sin x $ ?
As limited knowledge in math. I prefered to start with Euler's formular see more in wikipedia
$$e^{i\theta} =\cos \theta + i\sin \theta $$
Which can be proof as shown in wiki's page by
$$ f(\theta) = e^{-i\theta}(\cos \theta + i \sin \theta)$$
$$ f'(\theta) = e^{-i\theta}(-\sin \theta + i \cos \theta) + -ie^{-i\theta}(\cos \theta + i \sin \theta) = 0$$
And
$$ f(0) = e^{-0i}(\cos 0 + i \sin 0) = 1$$
As $f'(\theta) = 0$ then $f(\theta)$ is constant. Then
$$ f(\theta) = 1$$
$$ e^{-i\theta}(\cos \theta + i \sin \theta) = 1$$
$$ \cos \theta + i \sin \theta = e^{i\theta}$$
Now, we look at power factor of $e^x$
$$ e^x = \sum_{n=0}^{\infty} \frac {x^n} {n!}$$
Then
$$ e^{ix} = 1 + ix + \frac {(ix)^2} {2!} + \frac {(ix)^3} {3!} + \frac {(ix)^4} {4!} + \frac {(ix)^5} {5!}+...+ \frac {(ix)^n} {n!}$$
$$ e^{ix} = 1 + ix - \frac {(x)^2} {2!} - \frac {(ix)^3} {3!} + \frac {(x)^4} {4!} + \frac {(ix)^5} {5!}-...\pm \frac {(ix)^n} {n!}$$
$$ e^{ix} = (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!}-...) + i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}-...) $$
Just for sin part, you see that
$$sinx = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}-...$$
Now the only thing you need to proof is power factor of $e$ which easier to find the proof and again, with my limited knowledge of math I will go wih this.
$$ \frac d {dx} e^x = e^x$$
And.
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty} \frac d {dx} \frac {x^n} {n!}$$
$$ \frac d {dx} e^x = 0 + \sum_{n=0}^{\infty} \frac d {dx} \frac {x^n} {n!}$$
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty} \frac {nx^{n-1}} {n!}$$
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty} \frac {x^{n-1}} {(n-1)!} = \sum_{n=0}^{\infty} \frac {x^n} {n!}$$
As derivative of series equal to itself for all $x$ and For $x = 0,\sum_{n=0}^{\infty} \frac {x^n} {n!} = 1$
which means power series of $e^x$ is valid for all of x
|
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|
Solution of system of equations involving $x_{1},x_{2},x_{3}$
Solve for $x_{1},x_{2},x_{3}$, given
$ax^2_{1}+bx_{1}+c=x_{2}$
$ax^2_{2}+bx_{2}+c=x_{3}$
$ax^2_{3}+bx_{3}+c=x_{1}$
Try: from $(1)$ and $(2)$
$a(x^2_{1}-x^2_{2})+b(x_{1}-x_{2})=(x_{2} - x_{3})$
And from $(2)$ and $(3)$
$a(x^2_{2}-x^2_{3})+b(x_{2}-x_{3})=(x_{3}-x_{1})$
Could some help me to find $x_{1},x_{2},x_{3}$, Thanks
|
option 1:
$x_1 = x_2 = x_3$
$ax^2 + bx + c = x\\
ax^2 + (b-1)x + c = 0\\
x= \frac {-(b-1) \pm \sqrt {(b-1)^2 - 4ac}}{2a}\\
$
option 2:
$x_1 \ne x_2 \ne x_3$
Find a polynomial that intersects:
$(x_1,x_2),(x_2,x_3),(x_3,x_1)\\
p(x) = x_2 + (x-x_1)\frac {x_3-x_2}{x_2-x_1} + (x-x_1)(x-x_2)\left(\frac {x_1 - x_2}{(x_3-x_2)(x_3-x_1)} - \frac {x_2-x_1}{(x_3-x_2)(x_2-x_1)}\right)$
|
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|
Weird infinite sum Evaluate
$\sum_{n=0}^{\infty}{\frac{(-1)^n}{3n+1}=1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+...}
$
This looks a lot like the series expansion for $\ln(1+x)$ when $x=1$, but I cannot find the relationship.
|
The log series hints at finding the function
$$ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^nx^{3n+1}}{3n+1} $$
when $x=1$. Taking the derivative, we find
$$ f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{3n} = \sum_{n=0}^{\infty} (-x^3)^n = \frac{1}{1+x^3} $$
Therefore
$$ f(1) = \int_0^1 \frac{1}{1+t^3} dt $$
You can solve the integral using partial fractions. WolframAlpha gives the solution as well as the anti-derivative here
$$ f(x) = \frac13 \ln (x+1) - \frac16 \ln(x^2-x+1) + \frac{1}{\sqrt{3}}\arctan \left(\frac{2x-1}{\sqrt{3}}\right) $$
$$ f(1) = \frac13 \ln 2 + \frac{\pi}{3\sqrt{3}} $$
|
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|
$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
|
$\sqrt[5]{2^4\sqrt[3]{16}} =$
$\sqrt[5\cdot 3]{2^{4\cdot 3}\color{blue}{16}}=$
$\sqrt[15]{2^{12}2^{\color{blue}{4}}}=$
$\sqrt[15]{2^{12}2^{\color{blue}{4}}} = \sqrt[15]{2^{\color{blue}{16}}} = 2^{\frac{\not 1\not 5\color{red}{16}}{\not{\color{blue}{1\not 6}}\color{red}{15}}}$
But in my opinion your method seems a little scatter-shot and undirected. In particular $\sqrt[k]{b} = \sqrt[mk]{b^m}$ rubs me the wrong way. It's not wrong per se, but is seems that we are going in the wrong direction and making things complicated rather than simpler. And I fear extraneous roots and sign errors. (Ex: $\sqrt[5]{(-1)^3} \ne \sqrt [5*2]{(-1)^{3*2}}$).
Although there is no universal right or wrong way to do things try to develop a more systematic approach. I'd personal reduce to a common base, convert radicals to fractional exponents, and then just do the math:
$\sqrt[5]{2^4\sqrt[3]{16}}=$
$\sqrt[5]{2^4\sqrt[3]{2^4}} =$
$(2^4(2^4)^{\frac 13})^{\frac 15}=$
$2^{\frac 15(4 + 4*\frac 13)} =$
$2^{\frac{16}{15}}$.
Also, I suppose I should point out that:
$2^{\frac {16}{15}} = 2^{1 \frac 1{15}} = 2\sqrt[15]{2}$
which could be an acceptable answer. As is $2\times 2^{\frac 1{15}}$.
Which of these three answers $2^{\frac {16}{15}}, 2\sqrt[15]{2}, 2\times 2^{\frac 1{15}}$ it the correct one? Well, none are, or they all are.
I personally prefer $2\sqrt[15]{2}$ as it.... well, I get a better sense of what makes the number. I would prefer $2\times 2^{\frac 1{15}}$ as I'd general like to conform radicals and exponents, but in this case requiring the $\times$ sign bugs me.
But this is purely subjective.
|
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|
A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probability that the ball you toss lands in any one of the bins. What is the expected number of tosses?
Answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1}) \\
p_3 &=& 1 - (\frac{n-1}{n})(\frac{n-1-1}{n-1}) \\
p_3 &=& 1 - (\frac{n-2}{n}) = \frac{2}{n} \\
p_4 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1})(1 - \frac{1}{n-2}) \\
p_4 &=& 1 - ( \frac{n-1}{n} )( \frac{n-2}{n-1} )( \frac{n - 2 -1}{n - 2} ) \\
p_4 &=& 1 - \frac{n-3}{n} = \frac{3}{n} \\
\end{eqnarray*}
Now for $1 <= i <= n$ we have: $p_i = \frac{i-1}{n}$.
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
E &=& \sum_{i = 1}^{n} \frac{i(i+1)}{n} = \frac{1}{2n} \sum_{i=1}^{n} i^2 + i \\
E &=& \frac{1}{2n}(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} ) \\
E &=& \frac{n+1}{4n} ( \frac{2n+1}{3} + 1 ) \\
\end{eqnarray*}
Here is an update to my answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\newline
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (\frac{n-1}{n})( \frac{n-2}{n}) \\
p_3 &=& 1 - \frac{(n-1)(n-2)}{n^2} = \frac{n^2 - (n^2 - 3n + 2)}{n^2} \\
p_3 &=& \frac{3n-2}{n^2} \\
p_4 &=& 1 - (\frac{n-1}{n})(\frac{n-2}{n})(\frac{n-3}{n}) \\
p_4 &=& 1 - \frac{(n^2-3n+2)(n-3)}{n^3}\\
p_4 &=& 1 - \frac{n^3-3n^2+2n - 3n^2 +9n - 6}{n^3}\\
p_4 &=& \frac{3n^2-2n + 3n^2 - 9n + 6}{n^3}\\
p_4 &=& \frac{3n^2 + 3n^2 - 11n + 6}{n^3}\\
\end{eqnarray*}
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
\end{eqnarray*}
Now, am on the right track? That is, is what I have so far correct?
Thanks,
Bob
|
Let $p_i$ denote the number of ways we can obtain $i$ tosses.Obviously $2\le i\le (n+1)$
To find $p_i$ -
1.First find the number of ways we can put $(i-1)$ balls in the bins such that no two go into the same bin
Number of ways = $^np_{i-1}$
2.Put the next ball in one of the bins that already has a ball
Number of ways = $i-1$
$\therefore p_i=(i-1)^np_{i-1}$
$\therefore$ The total number of ways=$$\sum^{n+1}_{i=2}(i-1)^np_{i-1}$$
Putting the values in each case, we can obtain the required expected value for the number of tosses as $$\sum^{n+1}_{i=2}\frac{i(i-1)^np_{i-1}}{\sum^{n+1}_{i=2}(i-1)^np_{i-1}}=\frac{\sum^{n+1}_{i=2}i(i-1)^np_{i-1}}{\sum^{n+1}_{i=2}(i-1)^np_{i-1}}$$
|
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|
What's the Probability of a drunk man open a door with $n$ possible keys? I have this following problem in my problem set and I would like to check if my work is in the right directio.
A drunk man with $n$ keys wants to open his door and tries the keys at random. Exactly one key will open the door. Find the mean number of trials if
a) unsuccessful keys are not eliminated from further selections;
b) unsuccsesful keys are eliminated
It seems clear to me that in the first case we have a Geometric distribution with parameter $1/n$, so the expected number of trials is just $n$.
For the second case, my reasoning follows. Let $X$ denote the number of trials until he opens the door.
$P(X = 1) = 1/n$
$P(X = 2) =(n-1)/n \cdot 1/n-1 = 1/n$
$P(X = 3) = (n-1)/n \cdot (n-2)/(n-1) \cdot 1/(n-2) = 1/n$, and so on.
I'm inclined to say that, in the second case, the probabilities over the possible $n$ values of $X$ are uniformly distributed. If this is case, then the average number of trails should be $(n+1)/2$.
Does it seem correct? If not, how to do it? Thanks in advance!!
|
Yes, you are correct. Let $X$ be the number of trials. Expectation value $E[X]$ is given by the standard formula:
$$E[X] = \sum\limits_{k=1}^{\infty} k \cdot P(X=k),$$
where $P(X=k)$ is the probability that in the $k$-th trial the guy opened the door.
In the first case, we are returning the key back. If $k$-th trial was successful, then previous $k-1$ should not be. We can write:
$$P(X=k) = \left(\frac{n-1}{n} \right)^{k-1}\frac{1}{n},$$
and the expectation value is
$$E[X] = \frac{1}{n}\sum\limits_{k=1}^{\infty} k \left(\frac{n-1}{n} \right)^{k-1} = \frac{1}{n} \frac{1}{[1 - (n-1)/n]^2} = n.$$
In the second case, wrong keys are thrown away. The probability in this case is equal to
$$
\begin{align}
P(X=1) & = \frac{1}{n}, \\
P(X=2) & = \frac{n-1}{n} \cdot \frac{1}{n-1} = \frac{1}{n}, \\
P(X=3) & = \frac{n-1}{n} \cdot \frac{n-2}{n-1} \cdot \frac{1}{n-2} = \frac{1}{n}, \\
P(X=k) & = \frac{n-1}{n} \cdot \frac{n-2}{n-1} \cdot \ldots \cdot \frac{n-(k-1)}{n-(k-2)} \cdot \frac{1}{n-(k-1)} = \frac{1}{n}, \\
P(X \geqslant n+1) & = 0.
\end{align}
$$
The probability that the guy successfully opens the door after any number of trials is the same. Expectation value equals:
$$E[X] = \frac{1}{n}\sum\limits_{k=1}^{n}k = \frac{1}{n} \frac{1+n}{2}n = \frac{1+n}{2}.$$
|
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|
Finding parameters of an ellipse in terms of Semi-Latus Rectum and Directrix. I am trying to solve for all the parameters of an ellipse in terms of the Semi-Latus Rectum, $\ell$, and Directrix, $x$. This is for some equation tables I am making so I am looking for the most simplistic expression. In nearly all the cases it comes down to solving a cubic function.
For example in the case of the linear eccentricity, $c$, I know that:
$$
\ell=\frac{\sqrt{c}(c-x)}{\sqrt{x}}
$$
Which gets rearranged into:
$$
c^3-2c^2 x+c x^2-x \ell^2 = 0
$$
Solving for c using the normal general solution to the cubic equation gives me something nasty. Mathematica also doesn't help. Using a trigonometric solution helps however it gives me a piecewise solution:
$$
c =\begin{cases}
& \dfrac{2x}{3} \left (1 + sin \left (\frac{arcsin\left (1 - \frac{27 \ell^2}{2 x^2}\right )}{3} \right ) \right )\\
& \dfrac{4x}{3} sin^2 \left (\frac{arccos\left (1 - \frac{27 \ell^2}{2 x^2}\right )}{6} \right )
\end{cases}
$$
I feel that this can be simplified further through either assumptions on $x$ and $\ell$ or some trig identity I am unaware of.
If anyone has a reference where they solved for the parameters of an ellipse in terms of just the semi-latus rectum and directrix that would be wonderful. Otherwise any insights on how to reduce the answer further or other methods to attack this problem would be appreciated.
|
There's one bug in the first equation though it's unaffected after squaring both sides.
\begin{align}
x &= \frac{a^2}{c} \\
a &= \sqrt{cx} \\
\ell &= \frac{b^2}{a} \\
&= \frac{a^2-c^2}{a} \\
&= \frac{cx-c^2}{\sqrt{cx}} \\
\ell \sqrt{x} &= \sqrt{c} \, \color{red}{(x-c)} \\
0 &= c\sqrt{c}-x\sqrt{c}+\ell \sqrt{x} \\
\sqrt{c} &= 2\sqrt{\frac{x}{3}} \sin
\left(
\frac{1}{3} \sin^{-1} \frac{3\sqrt{3}\, \ell}{2x}+\frac{2k\pi}{3}
\right) \\
c &= \frac{4x}{3} \sin^2
\left(
\frac{1}{3} \sin^{-1} \frac{3\sqrt{3}\, \ell}{2x}+\frac{2k\pi}{3}
\right)
\end{align}
It is quite standard for a cubic equation. The roots are irreducible and also inconstructible by compasses and ruler.
Further points to be noticed:
*
*$k=0,1 \implies\sqrt{c}>0 \implies \text{two ellipses}$
*$k=2 \quad \implies\sqrt{c}<0 \implies \text{a hyperbola}$
*$\dfrac{X^2}{\frac{4x^2}{3} \sin^2
\left(
\frac{1}{3} \sin^{-1} \frac{3\sqrt{3}\, \ell}{2x}+\frac{2k\pi}{3}
\right)}+
\dfrac{Y^2}{\frac{2\ell x}{\sqrt{3}} \sin
\left(
\frac{1}{3} \sin^{-1} \frac{3\sqrt{3}\, \ell}{2x}+\frac{2k\pi}{3}
\right)}=1$
*$e=\dfrac{2}{\sqrt{3}}
\left|
\sin \left(
\dfrac{1}{3} \sin^{-1} \dfrac{3\sqrt{3}\, \ell}{2x}+
\dfrac{2k\pi}{3}
\right)
\right|$
|
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|
The inverse image of an interval Someone help me to find $f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilon[)$ where $$f(x)=\begin{cases} 0,~\text{if}~ x<0\\ x^2+1,~\text{if}~ x\geq0\end{cases}$$
I know that
$f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilon[)=\{y\in \mathbb{R}, f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\varepsilon[\}$
let $y<0$ then $f(y)=0$ so $f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\varepsilon[ $ if $x^2+1-\varepsilon <0$ that is $ \varepsilon >x^2+1$
so if $\varepsilon >x^2+1$ $f^{-1}(y)=]-\infty,0[$ right ?
someone have an idea on how to do this ?
|
Hint:Based on the graph of $f(x)$
$${{f}^{-1}}\left( \left] a,b \right[ \right)=\left\{ \begin{align}
& \phi \ \quad \quad \quad \quad \quad \quad \quad \quad ,if\ b\le 0\ \\
& \left] -\infty ,0 \right[\quad \quad \quad \quad \quad \quad ,if\ b\le 1\ AND\ a<0 \\
& \phi \quad \quad \quad \quad \quad \quad \quad \quad ,if\ b\le 1AND\ a\ge 0 \\
& \left] \sqrt{a-1},\sqrt{b-1} \right[\quad \ \quad ,if\ a\ge 1 \\
& \left] 0,\sqrt{b-1} \right[\quad \quad \quad \ \quad ,if\ b>1\ AND\ a\ge 0 \\
& \left[ 0,\sqrt{b-1} \right[\quad \quad \quad \ \quad ,if\ b>1\ AND\ a<0 \\
\end{align} \right\}
$$
|
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|
Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?
Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?
The correct answer is $\dfrac13$ while mine is $\dfrac{445}{2003}$
My attempt:
For $x^3+y^3$ to be divisible by $3$, EITHER both $x$ and $y$ should be a multiple of $3$ OR one of them should leave remainder $1$ when divided by $3$ and the other should leave remainder $2$.
Therefore, $$\text{no. of ways} = \frac{668 \times 667 + 668\times 668}{2004\times 2003} = \frac{445}{2003}$$
|
With replacement
As you basically note $x^3+y^3 \equiv x+y \pmod 3$, which follows from Fermat's Little Theorem.
We note that $2004 \equiv 0 \pmod 3$, so there's exactly $2004/3=668$ numbers equivalent to $i \pmod 3$, for all $i \in \{0,1,2\}$.
So, no matter what $x$ value is randomly chosen, there are $2004/3$ out of $2004$ (i.e., $1/3$ probability) of randomly choosing an $y$ value for which $y \equiv -x \pmod 3$.
We can do this like the method in the question:
$$
\frac{\overbrace{668 \times 668}^{x \equiv 0, y \equiv 0} + \overbrace{668 \times 668}^{x \equiv 1, y \equiv 2} + \overbrace{668 \times 668}^{x \equiv 2, y \equiv 1}}{2004^2}=\frac{1}{3}.
$$
Without replacement
If $x$ and $y$ are drawn without replacement, i.e., we assume $x \neq y$, then there's two distinctions: (a) when $x \equiv 0 \pmod 3$, there are $2004/3-1$ distinct $y$ values for which $x+y \equiv 0 \pmod 3$, and (b) there are $2004 \times 2003$ ordered pairs $(x,y)$, which also gives the probability:
$$
\frac{\overbrace{668 \times 667}^{x \equiv 0, y \equiv 0} + \overbrace{668 \times 668}^{x \equiv 1, y \equiv 2} + \overbrace{668 \times 668}^{x \equiv 2, y \equiv 1}}{2004 \times 2003}=\frac{1}{3}.
$$
To interpret this using the first method I mention, we have $1/3$ probability of any given value of $x \pmod 3$, and given a value of $x \pmod 3$, we have probability of either $667/2003$ (when $x \equiv 0 \pmod 3$) or $668/2003$ (when $x \not\equiv 0 \pmod 3$) of randomly choosing $y \equiv -x \pmod 3$. Since these are mutually exclusive events, we get the probability
$$
\overbrace{\frac{1}{3} \times \frac{667}{2003}}^{x \equiv 0, y \equiv 0}+\overbrace{\frac{1}{3} \times \frac{668}{2003}}^{x \equiv 1, y \equiv 2}+\overbrace{\frac{1}{3} \times \frac{668}{2003}}^{x \equiv 2, y \equiv 1}=\frac{1}{3}.
$$
The given answer seems to assume that $x \neq y$, which is this second case. However, the main problem is that it doesn't account for both cases $(x,y) \equiv (-1,1) \pmod 3$ and $(x,y) \equiv (1,-1) \pmod 3$.
|
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|
Minimum of $xy+yz+xz=k$ Given that the sum of x,y,z is 3 find the minimum of xy from the relation$$xy+yz+xz=k$$ Is there anything wrong with my solution since someone said the correct answer differs? $$xy=k-z(x+y)<=>xy=k-z(3-z)=k+z^2-3z=k+\left(z-\frac{3}{2}\right)^2-\frac{9}{4}$$ So $xy\ge k-\frac{9}{4}$ Can anyone give a different solution to confirm?
|
With Lagrange multipliers the problem can be formulated as
$$
L(x,y,z,\lambda_1,\lambda_2) = x y + \lambda_1(x+y+z-3)+\lambda_2(xy+yz+xz-k)
$$
The stationary points are the solutions for
$$
\lambda_1 + y + \lambda_2 (y + z) = 0\\
\lambda_1 + x + \lambda_2 (x + z) = 0\\
\lambda_1 + \lambda_2 (x + y) = 0\\
x + y + z = 3\\
x y + x z + y z = k
$$
The solution gives
$$
x y = k - \frac{9}{4}
$$
NOTE
$$
x y = \frac{1}{9}(3\pm \sqrt{9-3k})^2
$$
are solutions also but $k - \frac{9}{4}\le \frac{1}{9}(3\pm \sqrt{9-3k})^2 $
The full set of solutions is
$$
\begin{array}{ccccc}
x & y & z & \lambda_1 & \lambda_2\\
\frac{1}{4} \left(3-\sqrt{45-16 k}\right)&\frac{1}{4} \left(\sqrt{45-16 k}+3\right)& \frac{3}{2}& \frac{3}{2}& -1\\
\frac{1}{4} \left(\sqrt{45-16 k}+3\right)& \frac{1}{4} \left(3-\sqrt{45-16k}\right)& \frac{3}{2}& \frac{3}{2}& -1\\
\sqrt{1-\frac{k}{3}}+1&\sqrt{1-\frac{k}{3}}+1& 1-2 \sqrt{1-\frac{k}{3}}& -\frac{2 \left(\left(\sqrt{9-3 k}+6\right) k-6 \left(\sqrt{9-3k}+3\right)\right)}{9 (k-3)}& \frac{k-\sqrt{9-3 k}-3}{3 (k-3)}\\
1-\sqrt{1-\frac{k}{3}}&1-\sqrt{1-\frac{k}{3}}& 2 \sqrt{1-\frac{k}{3}}+1& \frac{2 \left(\left(\sqrt{9-3 k}-6\right) k-6 \sqrt{9-3 k}+18\right)}{9
(k-3)}& \frac{k+\sqrt{9-3 k}-3}{3 (k-3)}
\end{array}
$$
The minimum solutions are the first and second in the set.
|
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|
What do polynomials solve for? Roots? I had a question in which I’ve been hung up over on. I understand if we had a graph x-y plane and we drew points that intercept the $x$-axis at $2$ and $3$, we would write a quadratic equation that satisfies our condition as $(x+3)(x+2)$, or $x^2+5x+6$, and set this equal to $0$ using the null factor law. Then, we would solve it by factoring and would eventually get back to $(x+3)(x+2)=0$.
And we would get $x=-2,-3$ but how come these are the roots(points that intercept the $x$-axis), when we first created this quadratic equation using the points $2$ and $3$ on $x$-axis? I want to understand in short a logical explanation what we are solving for when we are factoring polynomials?
|
The graph of a function $y=f(x)$ intersects the $x$-axis when $y=0$, that is when $f(x)=0$.
Now if a factor $(x-a)$ appears in your polynomial, then the polynomial is zero when $x-a = 0$, which occurs precisely when $x=a$. Therefore
$$\begin{align*}
(x-a)(x-b) = 0 & \Leftrightarrow x-a=0 \text{ or } x-b = 0\\
& \Leftrightarrow x=a \text{ or } x=b
\end{align*}$$
So the graph $y=(x-a)(x-b)$ intersects the $x$-axis when $x=a$ or $x=b$.
This explains the $-$ signs.
So the quadratic $(x+3)(x+2)$ has roots $-3$ and $-2$, whereas the quadratic $(x-3)(x-2)$ has roots $3$ and $2$.
|
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|
Application of AM-GM Inequality
If $a,b,c >0$ so that $a+b+c=27$ then what is the maximum value of $(a^2)(b^3)(c^4)$?
I tried the AM-GM inequality but the product term has different powers of $a,b,c$. So how to go about it?
|
Hint:
$$a+b+c=27 \iff \frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}=27.$$
|
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|
Find the derivative at (1,2)
$$f(x) = x^2 \sqrt{5 - x^2}$$
Find the derivative at $(1, 2)$.
\begin{align}
\frac{d}{dx} \left[ x^2 \sqrt{5 - x^2} \right] & = \frac{d}{dx} \left[ x^2 (5 - x^2)^{1/2} \right] \\
& = x^2 \frac12 (5 - x^2)^{-1/2}(-2x) + (5 - x^2)^{1/2}(2x)
\end{align}
The equation is formed using the product rule and the chain rule. The author explained that $(-2x)$ was inserted as "multiply as the inside derivative", and $(2x)$ as the derivative of the first term.
I assume the following:
$(-2x)$ is the derivative of the inside derivative which is $(5-x^2)$.
$(2x)$ is the derivative of $x^2$.
My question is, why isn't $(-2x)$ multiplied for the 2nd part of the equation (after the $+$ sign)?
|
The derivative actually reads $f'(x) = u'(x)v(x) + u(x)v'(x)$ with $u(x)=x^2$ and $v(x)=\sqrt{5-x^2}$.
For $u(x)$, we have $u'(x)=2x$.
For $v(x)$, we can rewrite $v(x)=f(g(x))$ where $f(t)=\sqrt{t}$ and $g(x)=5-x^2$, so that $v'(x)=g'(x)f'(g(x))=-\frac{2x}{2\sqrt{5-x^2}}$.
Hence $f'(x)=2x\sqrt{5-x^2}-(x^2)(-\frac{2x}{2\sqrt{5-x^2}})=2x\sqrt{5-x^2}+x^2\frac{2x}{2\sqrt{5-x^2}}$, which is the equation shown.
|
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|
How would I obtain the square root of this multinomial? I was doing some problems from a book I found on finding the square root of a polynomial expression. I came across this problem:
$$\frac{a^4}{64}+\frac{a^3}{8}-a+1$$
I utilised the method outlined here, and obtained the following result
$$\frac{a^4}{64}+\frac{a^3}{8}-a+1)\frac{a^2}{8} + \frac{a}{2}$$
$$\frac{a^2}{4} + \frac{a}{2} )\frac{a^3}{8} - a + 1$$
$$\frac{a^2}{4} + a + 1) -\frac{a^2}{4} - a + 1$$
$$\frac{a^2}{4} + a + 1 ) -2$$
I know that I didn't format it well, but, basically, when I used the method they suggested, I had a remainder at the end. I don't know whether I did something wrong or whether the polynomial is a perfect swuare.
|
We can start small and improve a bit at a time. We can begin with $\frac{a^2}{8}$ which does give the top term when squared. Next,
$$ \left( \frac{a^2}{8} + B a \right)^2 = \frac{a^4}{64} + B \frac{a^3}{4} + B^2 a^2. $$ To get the $a^3/8$ we take $B = 1/2.$ So far, we have
$$ \left( \frac{a^2}{8} + \frac{a}{2} \right)^2 = \frac{a^4}{64} + \frac{a^3}{8} + \frac{ a^2}{4}. $$ Not bad. Next,
$$ \left( \frac{a^2}{8} + \frac{a}{2} +C\right)^2 = \frac{a^4}{64} + \frac{a^3}{8} + \frac{ a^2}{4} + C\frac{a^2}{4} + C a + C^2. $$
To get rid of the $a^2$ term, we need only take $C = -1$ and get
$$ \left( \frac{a^2}{8} + \frac{a}{2} -1\right)^2 = \frac{a^4}{64} + \frac{a^3}{8} - a + 1. $$
|
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|
When does $\frac{a+b}{2}$ and $\sqrt{ab}$ have inversed tens digits and ones digits? Let $a$ and $b$ be natural numbers, and
$$A = \frac{a+b}{2}$$
$$B = \sqrt{ab}$$
It's given that $A$ and $B$ are two-digit numbers such that the tens digit of $A$ is the same as the ones digit of $B$, and the tens digit of $B$ is the same as the ones digit of $A$.
So $A = 10x + y\;\,$and$\;B = 10y + x$.
Also given is $A\ne B$.
What is $a$ and $b$?
My teacher gave us the answer without explaining it as: $a = 98$ and $b = 32$, which makes $A = 65$ and $B=56$.
My question is: How do you prove this? I know $98 = 2\cdot 7^2$ and $32 = 2^5$, but I don't understand how to find this specific answer.
|
Solve for $a$ in the second equation. Plug that into the first. Then solve for $b$. You get:
$b=A\pm \sqrt{A^2-B^2}$.
Plug in for $A=10x+y$ and $B=10y+x$ and expand. You get:
$b=10x+y\pm 3\sqrt{11(x^2-y^2)}$
The only multiples of 11 that are perfect squares are even powers of 11 times even powers of other primes. So, either $x^2=y^2$, which would give $A=B$, or $x^2-y^2=11$. Since $x,y \in \{1,2,3,4,5,6,7,8,9\}$, it is a simple matter of trial and error to find $6^2-5^2 = 36-25=11$.
Again with simple trial and error, you can verify that you cannot achieve $x^2-y^2$ to be 11 times some product of primes to even powers.
Edit: Check out this table of possible results:
http://www.wolframalpha.com/input/?i=Table%5BTable%5B(x,y,(x%5E2-y%5E2)%2F11),%7Bx,y%2B1,9%7D%5D,%7By,1,8%7D%5D
|
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|
A question about exponential function/equation. I'm solving the following exponential equation
$$4^{x}-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}$$
(My attempt is below)
\begin{align}4^{x}-3^{x-\frac{1}{2}}&=3^{x+\frac{1}{2}}-2^{2x-1}\\\\
4^{x}+2^{2x-1}&=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}\\\\
2^{2x}+2^{2x}\cdot2^{-1}&=3^{x}\cdot3^{\frac{1}{2}}+3^{x}\cdot3^{-\frac{1}{2}}\\\\
2^{2x}(1+\frac{1}{2})&=3^{x}(\sqrt{3}+\frac{1}{\sqrt{3}})\\\\
\frac{3}{2}\cdot2^{2x}&=\frac{4\sqrt{3}}{3}\cdot3^{x}\\\\
3\cdot2^{2x-1}&=4\sqrt{3}\cdot3^{x-1}\\\\
2^{2x-3}&=3^{x-\frac{3}{2}}\end{align}
It is clear that $$x=\frac{3}{2}$$ is a solution because all exponential functions are equal to 1 when the exponent is 0. But how do I prove it's the only solution? My apologies if it's trivial.
|
Exponential functions are strictly increasing. At $x<3/2$ we have that $3^{x-\frac32} < 2^{2x-3}$ and at $x>3/2$ we have that $3^{x-\frac32} > 2^{2x-3}$. Using these observations We can say that $3/2$ is the only solution.
|
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|
Is there a fast way to prove a tridiagonal matrix is positive definite? I' m trying to prove that
$$A=\begin{pmatrix}
4 & 2 & 0 & 0 & 0 \\
2 & 5 & 2 & 0 & 0 \\
0 & 2 & 5 & 2 & 0 \\
0 & 0 & 2 & 5 & 2 \\
0 & 0 & 0 & 2 & 5 \\
\end{pmatrix}$$
admits a Cholesky decomposition.
$A$ is symmetric, so it admits a Cholesky decomposition iff it is positive definite. The only methods I know for checking this are:
*
*$X^tAX > 0, \quad \forall X \in \mathbb{K}^n- \{0\}$.
*If $\lambda$ is an eigenvalue of $A$, then $\lambda>0.$
I have failed to prove it using 1 and 2 is taking me so much time. Is there any easier way to do this, given that $A$ is tridiagonal?
|
I have failed to prove it using 1
Note:
$$\begin{pmatrix}a&b&c&d&e\end{pmatrix}\begin{pmatrix}
4 & 2 & 0 & 0 & 0 \\
2 & 5 & 2 & 0 & 0 \\
0 & 2 & 5 & 2 & 0 \\
0 & 0 & 2 & 5 & 2 \\
0 & 0 & 0 & 2 & 5 \\
\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}=\\
\begin{pmatrix}4a+2b&2a+5b+2c&2b+5c+2d&2c+5d+2t&2d+5e\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}=\\
4a^2+4ab+5b^2+4bc+5c^2+4cd+5d^2+4de+5e^2=\\
(2a+b)^2+(2b+c)^2+(2c+d)^2+(2d+e)^2+4e^2.$$
|
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|
Simplifying $\operatorname{tanh}(\operatorname{arsinh}(x))$ So I am trying to simplify $\tanh(\operatorname{arsinh}(x))$ to $\frac{x}{\sqrt{1+x^2}}$
In general,
$$\tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}$$
and $$\operatorname{arsinh}(x)= \ln(x+\sqrt{x^2-1})$$
therefore
\begin{align}\tanh(\operatorname{arsinh}(x)) & =\frac{e^{\ln(x+\sqrt{x^2-1})}-e^{\ln(\frac{1}{x+\sqrt{x^2-1}})}}{e^{\ln(x+\sqrt{x^2-1})}+e^{\ln(\frac{1}{x+\sqrt{x^2-1}})}}\\
&=\frac{x+\sqrt{x^2-1}-\frac{1}{x+\sqrt{x^2-1}}}{x+\sqrt{x^2-1}+\frac{1}{x+\sqrt{x^2-1}}} \\
&=\frac{(x+\sqrt{x^2-1})^2-1}{(x+\sqrt{x^2-1})^2+1} \\
&=\frac{x^2+2x\sqrt{x^2-1}+x^2-1-1}{x^2+2x\sqrt{x^2-1}+x^2-1+1} \\
&=\frac{x^2+x\sqrt{x^2-1}-1}{x^2+x\sqrt{x^2-1}} \\
&=\frac{x^2+x\sqrt{x^2-1}}{x^2+x\sqrt{x^2-1}}-\frac{1}{x^2+x\sqrt{x^2-1}} \\
&=1-\frac{1}{x(x+\sqrt{x^2-1})} \\
&=1-\frac{x-\sqrt{x^2-1}}{x(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})} \\
&=1-\frac{x-\sqrt{x^2-1}}{x(x^2-x^2+1)} \\
&=1-\frac{x-\sqrt{x^2-1}}{x} \\
&=\frac{\sqrt{x^2-1}}{x} \\
&\ne \frac{x}{\sqrt{1+x^2}} \end{align}
I'm struggling to get my answer into the required form.
|
Let arcsinh$(x)=y\implies2x=2\sinh(y)=e^y-e^{-y}$
$\implies(e^y+e^{-y})^2=(e^y-e^{-y})^2+4e^y\cdot e^{-y}=4(x^2+1)$
and tanh$(y)=\dfrac{e^y-e^{-y}}{e^y+e^{-y}}=?$
|
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|
Find the volume of the region common to the interiors of the cylinders First cylinder $x^2+y^2=4$
Second cyclinder $x^2+z^2=4$
My progress so far
$$V=8\int_0^2\int_0^\sqrt{4-y^2}\int_0^\sqrt{4-x^2}dzdxdy$$
I know I can substitute $x$ with $2\sin(\theta)$ in $\int\sqrt{2-x^2}$ but if I do that then I came across $sin(4\sin\theta)$ after substituting $y$ with $2\sin\theta$ in $\int_0^\sqrt{4-y^2}\cos^2\theta$
|
You may use a triple integral to find the volume.
$$ V= 8\int _0^2 \int _0^{\sqrt {4-x^2}}\int _0^{\sqrt {4-x^2}}dzdydx$$
|
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|
Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots )\\
1& = \Big ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big ) \cdot \Big ( z^3 \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \Big )\\
\end{align*}
After some more thought...
Is it true to say that because f(z) has a pole of order 4 at $z=0$ that our $b_n$'s only go out to the 4th term? Meaning there are no $b_5$, $b_6$, etc like how wrote previously. That is,
$\frac{1}{z^3 \sin{(z)}}
= \Big ( \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big )\\$
followed by
\begin{align*}
1
&= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots\Big )\\
\end{align*}
Which then when multiplying out $b_1z^2$ with each term from sin(z)'s Laurent expansion will never yield a $\frac{1}{z}$ term, concluding that the coefficient $b_1 = 0$?
|
Hint. The function
$$
z \mapsto \frac{1}{\sin(z)}-\frac{1}{z}
$$ is regular and odd near $0$ thus the residue of
$$
f(z) = \frac{1}{z^3 \sin(z)}
$$is equal to zero.
|
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|
Finding every solution for equation of complex numbers I need to find every solution for:
$\ z^{3} + 3i \overline z = 0 $
So I tried was just to compare imaginary and complex part of $\ z^{3} $ and $\ 3i\overline z$
Ill spare you the alegbra, here is the result:
$$\ a^{3} - 3ab^{2} + i(3a^{2}b-b^{3}) = -3b -3ai \\a^{3} - 3ab^{2} = -3b \\3a^{2}b - b^{3} = -3a$$ and so $$\ a^{3} -3ab^{2} + 3b = 0 \\ b^{3} -3a^{2}b-3a=0
$$
but I'm pretty stuck here. not sure what do next. I also tried using eulers rule so
$$\ z^{3} = -3i\overline z \\ r^{3}e^{{i\theta}^{3}} = -3i \times re^{-i\theta} \\ r^{3}e^{{i\theta}^{3}} = -3i \times re^{2\pi-i\theta} \\ 3\theta = 2\pi - \theta +2\pi k \\ 4\theta = 2\pi + 2\pi k \\ \theta = \frac{\pi}{2} + \frac{\pi k }{2}$$
and
$$\ r^{3} = -3ir \\ r^{2} = -3i$$
|
Write your equation in the form: $$z(z^2+3i)=0$$
and then you will get: $$x^2-y^2+i(2xy+3)=0$$ or $$z=0$$
|
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|
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \cdot2x-\int\frac{2x}{1+x^2}\,dx=2x\tan^{-1}x-\log(1+x^2)+C
$$
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|\leq{1}\\
\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x<0
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x\text{ if }|x|\leq{1}\\
\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x<0
\end{cases}
$$
$$
\int\sin^{-1}\frac{2x}{1+x^2}\,dx=\begin{cases}\int2\tan^{-1}x\,dx&\text{ if } |x|\leq{1}\\\int\pi\, dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x>0\\-\int\pi \,dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}=\begin{cases}\color{red}{2x\tan^{-1}x-\log(1+x^2)+C\text{ if } |x|\leq{1}}\\\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x>0\\-\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}$$
So don't we have two more cases for our solution rather than that is given in my reference, right ?
|
You're right; but it depends on what the full question is; if one is asked to compute
$$
\int_{0}^{1}\arcsin\frac{2x}{1+x^2}\,dx
$$
then just the antiderivative $2\arctan x$ is sufficient. Not if one wants to compute an integral involving points outside the interval $[-1,1]$.
Here's a shorter way to get at your result.
Consider
$$
f(x)=\arcsin\frac{2x}{1+x^2}
$$
Then
$$
f'(x)=\frac{1}{\sqrt{1-\dfrac{4x^2}{(1+x^2)^2}}}\frac{2(1+x^2)-4x^2}{(1+x^2)^2}
=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}
$$
hence
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} & x\in(-1,1) \\[4px]
-\dfrac{2}{1+x^2} & x\in(-\infty,-1)\cup(1,\infty)
\end{cases}
$$
Therefore, knowing that $f(0)=0$,
$$
f(x)=\begin{cases}
-\pi-2\arctan x & x<-1 \\[4px]
2\arctan x & -1\le x\le 1 \\[4px]
\pi-2\arctan x & x>1
\end{cases}
$$
In order to find an antiderivative we can consider
$$
\int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx
=x\arctan x-\frac{1}{2}\log(1+x^2)
$$
Thus an antiderivative of $f$ has the form
$$
F(x)=\begin{cases}
c_- -\pi x-2x\arctan x+\log(1+x^2) &\qquad x<-1 \\[4px]
2x\arctan x-\log(1+x^2) &\qquad -1\le x\le 1 \\[4px]
c_+ +\pi x-2x\arctan x+\log(1+x^2) &\qquad x>1
\end{cases}
$$
and you just need to determine $c_-$ and $c_+$ to ensure continuity at $-1$ and $1$.
The other antiderivatives differ from $F$ by a constant.
|
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|
If $x^2-bx+c=0$ has real roots, then prove that both are greater than $1$ when $c+1>b>2$.
If $ x^2-bx+c=0$ has real roots, prove that both roots are greater than $1$, when $c+1>b>2$.
Working
I tried to prove the given inequality by taking roots greater than $1$.
Let $\alpha$, $\beta$ be the roots of the quadratic equation. So $$\alpha+\beta =b$$
$$\alpha\cdot\beta =c$$
Since $\alpha>1$ and $\beta>1$, it can be deduced that,
$$\alpha+\beta >2\implies b>2$$ $$\alpha\cdot\beta >1\implies c>1\implies c+1>2$$
to combine these two inequalities I need another link between $c$ and $b$. How to proceed? Thanks.
|
The roots of
$x^2-bx+c=0$
are
$x
=\dfrac{b\pm\sqrt{b^2-4c}}{2}
$.
If
$c+1 > b > 2$,
the smallest root is
$\dfrac{b-\sqrt{b^2-4c}}{2}
$
and
$b^2 > 4c$
so
we want
$b-\sqrt{b^2-4c}
\gt 2$
or
$(b-2)^2
\gt b^2-4c
$
or
$b^2-4b+4
\gt b^2-4c
$
or
$c+1 > b$
which we are given.
|
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|
Confused about finding LCM of three numbers I need to find the LCM of three numbers:
*
*$x^2-36$
*$2x^2-12x$
*$x^2-6x$
I factor:
*
*$x^2-36=(x+6)(x-6)$
*$2x^2-12x=2x(x-6)$
*$x^2-6x=x(x-6)$
Then I multiply all factors. Every factor that appears in all three expressions I multiply only once. $(2x)(x)(x-6)(x+6)$. And I get $2x^2(x+6)(x-6)$. But the solution is $2x(x+6)(x-6)$. Where I'm wrong?
|
Note that $x$ is already contained in $2x$ thus $2\cdot x\cdot (x+6)\cdot (x-6)$ is the correct answer.
Indeed the different factors which appear are
*
*$2$
*$x$
*$(x-6)$
*$(x+6)$
|
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|
which of the following statement is TRue ?..... let $f : R \rightarrow R$ be a continious and nonnegative function.
which of the following statement is TRue ?
a) if there exist $ c \in (0,1)$ such that $f(c) = 100$ then $\int_{0}^{1} f(x) dx \ge \frac {1}{2}.$
b)$\int_{0}^{1} f(x) dx > \frac {1}{2}$.then $f(c) > \frac{1}{2}$ for some $c \in (0,1).$
c)$\int_{0}^{1} f(x) dx = \frac {1}{2}$ then there exist $c\in (0,1)$ such that $f(c) = \frac {1}{2}$
d) None of these
My answer : option b) and C) is true.. by intermediate theorem
For option a) if i take $f(x) = 200x$ now put $x = \frac{1}{2}$..then $\int_{0}^{1} f(x) dx = \frac {200 x^2}{2} |_0^1$...we will not get $\int_{0}^{1} f(x) dx = \frac {1}{2}.$..so option a) is false
Is its right or wrong ?? Pliz tell me
Any hints/ solution
|
In the same vein as GNU Supporter's answer, but not as elementary (and neat :D)
Consider another "peak function", namely
$$f(x) = 100\ e^{-a(x-\frac{1}{2})^2}\ .$$
Clearly $f(\frac{1}{2}) = 100.$ We can compute
\begin{align*}
\int_0^1 100\ e^{-a(x-1/2)^2} \ dx
&= 100 \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-a x^2} \ dx \\
&= \frac{100}{\sqrt{a}} \int_{-\frac{\sqrt{a}}{2}}^{\frac{\sqrt{a}}{2}} e^{-x^2} \ dx \\
&= \frac{100 }{\sqrt{a}} \sqrt{\pi}\ \mathsf{erf}\left(\frac{\sqrt{a}}{2}\right)
\end{align*}
where $\mathsf{erf}$ is the error function. Now, all we have to do is find for what values of $a$
\begin{align*}
\frac{100 }{\sqrt{a}} \sqrt{\pi}\ \mathsf{erf}\left(\frac{\sqrt{a}}{2}\right) < \frac{1}{2}\ .
\end{align*}
This can be solved with WolframAlpha, and the answer is roughly $a\geq 125\ 664$.
So an example would be with $a = 100^3$, and we get
\begin{align*}
\int_0^1 100\ e^{-100^3 (x-1/2)^2} \ dx \approx 0.177 < \frac{1}{2}
\end{align*}
|
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|
Integrate $\int x\sin^2 (x) dx$ Integrate $\int x\sin^2 (x) dx$
My attempt:
$$=\int x\sin^2 (x) dx\\
=x^2\sin^2 (x) - \int 2\sin (x)\cos (x)x^2 dx\\
=x^2\sin^2 (x) - \int \sin (2x) x^2 dx.$$
|
Here is a method using tabular integration, where we use $\sin^2 x = \frac{1}{2} (1 - \cos 2x)$:
$$\begin{array}{c|c} D & I \\ \hline \color{red}{x} & \frac{1}{2} - \frac{1}{2} \cos(2x) \\ \hline \color{blue}{1} & \color{red}{\frac{1}{2}x - \frac{1}{4} \sin(2x)} \\ \hline 0 & \color{blue}{\frac{1}{4}x^2 + \frac{1}{8} \cos(2x)} \end{array}$$
hence:
$$\int x \sin^2 (x) \ dx = \color{red}{+}x \left(\frac{1}{2}x - \frac{1}{4} \sin(2x) \right) \color{blue}{-} \left(\frac{1}{4}x^2 + \frac{1}{8} \cos(2x) \right) +C$$
$$= \frac{1}{4}x^2 - \frac{1}{4} x \sin(2x) - \frac{1}{8} \cos(2x) +C.$$
|
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|
solving differential equation $\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$ How would you solve this third order differential equation:
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
My first thought was to take a double integral:
$$\iint\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}dxdx=\iint{x^2+2x+2}dxdx$$
so:
$$y+\frac{dy}{dx}=\frac{x^4}{12}+\frac{x^3}{3}+x^2+c_1x+c_2$$
This is what I got to but I am unsure beyond here
|
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
Substitute $z=y''$
$$z'+z=x^2+2x+2$$
$$(z-x^2-2)'+(z-x^2-2)=0$$
$$v'=-v$$
Where $v=z-x^2-2$
$$\implies \ln|v|=-x+K \implies v=K_1e^{-x} $$
$$\implies z=x^2+2+K_1e^{-x}$$
Integrate twice to get y
$$\boxed{y=K_1e^{-x}+K_2x+K_3+x^2+\frac {x^4}{12}}$$
|
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"url": "https://math.stackexchange.com/questions/2791300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Integral $\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$ I am tring to evaluate
$$I=\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$$
The first thing I did was to notice that
$$\frac{1}{x^2+2x+2}=\frac{1}{(x+1)^2+1}=\frac{d}{dx}\arctan(x+1)$$
So I integrated by parts in order to get
$$I=\arctan 2\arctan 3-\int_0^2\frac{\arctan(x+1)}{1+x^2}dx$$
I let $x=u+1$ but when I do that I get
$$I=\arctan 2\arctan 3+\int_{-1}^1\frac{\arctan(u)}{1+(1+u)^2}du
=\arctan 2\arctan 3$$
Now this is not close to the approximation given by wolfram. What have I done wrong and how to solve this?
|
There is no need to evaluate the integral to answer the original question.
The original question is a multiple choice question so ruling out every option but the right one leads to the right answer of course. One sees that the integrand is positive almost everywhere. Moreover the arctangent function is increasing, so one has:
\begin{align}
0 < \int^2_ 0 \frac{\arctan(x)}{x^2+2x+2}\,dx &\leq \arctan(2) \int^2 _0 \frac{1}{x^2+2x+2}\,dx\\&= \arctan(2)\left( \arctan(3)-\arctan(1)\right)
\end{align}
By the addition formula for arctangent function one sees that:
\begin{align}
\arctan(3)-\arctan(1) = \arctan\left( \frac 1 2 \right)
\end{align}
Now define for $x>0$ the function:
$$f(x): = \arctan(x)\arctan\left( \frac 1 x\right)$$
This function is strictly positive. Moreover it tends to zero as $x\to 0^+$ and as $x\to\infty$ and it is differentiable with only one stationary point $x=1$ (check this!) which clearly corresponds with the maximum. So:
\begin{align*}
f(x) \leq f(1) = \arctan(1)^2 = \frac{\pi^2}{4^2} < \frac{4^2}{4^2}=1
\end{align*}
Hence:
\begin{align}
0 < \int^2_ 0 \frac{\arctan(x)}{x^2+2x+2}\,dx < 1 < \pi < 2\pi
\end{align}
So what is the only option that can be the right answer?
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find the Eigenvalues and its Eigen vectors. Consider the $ \ n \times n \ $ matrix $$ \begin{pmatrix} a & -1 & & & & \\ -1 & a & -1 & & & \\ & -1 & a & -1 & & & \\ & & -1 & a & -1 & \\ & & ..... & .... & .... \\ & & && a & -1 \\ & & & &-1 & a \end{pmatrix} $$
Find the Eigenvalues and its Eigen vectors.
Also determine the values of $ \ a \ $ for which the matrix is positive definite.
Answer:
For our convenience consider the $ \ 3 \times 3 \ $ matrix as follows
$$ A=\begin{pmatrix} a & -1 & \\ -1 & a & -1 \\ & -1 & a \end{pmatrix} $$
The blank positions must be filled with $ \ 0 \ $. Thus,
$$ A=\begin{pmatrix} a & -1 & 0 \\ -1 & a & -1 \\ 0& -1 & a \end{pmatrix} $$
Let $ \ \lambda \ $ be the Eigen value of matrix $ \ A \ $ , then
$ |A-\lambda I |=0 \\ \Rightarrow \begin{vmatrix} a-\lambda & -1 &0 \\ -1 & a-\lambda & 0 \\ 0 & -1 & a-\lambda \end{vmatrix} =0 \\ \Rightarrow (a-\lambda)^3-(a-\lambda)=0 \\ \Rightarrow (a-\lambda) [(a-\lambda)^2-1)=0 \\ \Rightarrow \lambda=a, \ a-1, \ a+1 $
For $ \ 2 \times 2 \ $ such matrix we have
$$ A'=\begin{pmatrix} a & -1 \\ -1 & a \end{pmatrix} $$
The Eigen values of $ \ A' \ $ are $ \ a-1 , \ a+1 \ $
For $ \ 4 \times 4 \ $ such that
the eigen values are $ \ a-1, a-1 , a+1, a+1 \ $
Thus in general the eigen vlaues are ($ \ if \ n=even \ $ )
$ a-1, a-1, ...... \frac{n}{2} \ times \ \\ a+1 , a+1, ........ \frac{n}{2} \ times \ $
If $ n=odd \ $ , then the eigen values of the $ \ n \times n \ $ matrix are
$ a-1 , a-1, ................. \frac{n-1}{2} \ times \\ a+1,a+1,..............\frac{n-1}{2} \ times \\ and \ \ a \ \ $
Am I right ?
But how to find the eigen vectors ?
|
First question: No, it appears you've made a computational error. I'm getting eigenvalues that follow in the example.
Given a matrix $A,$ then one computes eigenvectors with eigenvalue $\lambda$ by computing a basis for $$\ker(A-\lambda I_n).$$
Example:
$$A=\begin{pmatrix} a & -1 &0 \\ -1 & a & -1 \\0& -1 & a \end{pmatrix}$$
then for the eigenvalue $a$
$$\ker(A-aI_3)=\text{span}\left\{\begin{pmatrix}-1\\0\\-1\end{pmatrix}\right\}$$
for the eigenvalue $-\sqrt{2}+a$
$$\ker(A-(-\sqrt{2}+a)I_3)=\text{span}\left\{\begin{pmatrix}1\\\sqrt{2}\\1
\end{pmatrix}\right\}$$
for the eigenvalue $\sqrt{2}+a$
$$\ker(A-(\sqrt{2}+a)I_3)=\text{span}\left\{\begin{pmatrix}1\\-\sqrt{2}\\1
\end{pmatrix}\right\}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
${3^n\choose k}$ is divisible by $3$? How can I prove that ${3^n\choose k}$ is divisible by $3$ for all positive integer values of $n$? (where $k$ is any positive integer smaller than $3^n$) Can you use induction? Thanks.
|
Looking at the fractional expansion of the binomial coefficient, you can cancel out factors of 3 in the $(3n+1)^\text{th}$ value in the numerator with the $(3n)^\text{th}$ value in the denominator, for example:
$${3^3 \choose k} = \begin{array} {c}
27 & 26 & 25 & \color{red}{24} & 23 & 22 & \color{blue}{21} & 20 & 19 & \color{green}{18} & 17 & 16 & \color{orange}{15} & \dots & 27 - (k - 1) \\ \hline
1 & 2 & \color{red}{3} & 4 & 5 & \color{blue}{6} & 7 & 8 & \color{green}{9} & 10 & 11 & \color{orange}{12} & 13 & \dots & k
\end{array}$$
So you get $\pi_3(\text{Numerator}) - \pi_3(\text{Denominator}) = \pi_3(3^n) - \pi_3(k) > 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2804314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possible other method?
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The first thing that screams at me is $3|x^3 + x^3$ so
$x^3 \equiv -y^3 \mod 3$ so $x\equiv -y \mod 3$
Let $x \equiv i \mod 3$ and $y \equiv -i \mod 3$. If $x = 3k + i$ and $y = 3j -i$ then $\frac {x^3 + y^3}{3} = 9(k^3 + j^k) + 9(k^2i-j^2i)+3(ki -ji) + \frac {i^3 - i^3}3 \equiv 0 \mod 3$
So $xy + \frac {x^3 + y^3}3 \equiv i^2 \mod 3$ and $2007\equiv 0 \mod 3$ so $i = 0$ and $3|x$ and $3|y$.
Let $x = 3a$ and $y = 3b$ and we get
$9ab + 9(a^3 + b^3) = 2007$ so
$ab + a^3 + b^3 = 223$
Hmmm, still trial and error but $b^3, a^3 < \sqrt[3]223 \approx 6$ so not too many to test.
One of $a$ or $b$ must be odd. We have $ab + a^3 + b^3\equiv ab + a +b \equiv 1 \mod 3$: That means:
$(a,b)\equiv (0, k) \to 0*k + 0 + k\equiv k \mod 3$
So $a\equiv 0$ (wolog) and $b \equiv 1$ is possible.
$(a,b) \equiv (1,1) \to 1 + 1 + 1 \equiv 0 \mod 3$.
$(a,b) \equiv (1,-1) \to -1 + 1 + -1 \equiv -1 \mod 3$.
$(a,b) \equiv (-1,-1) \to 1 -1 -1 \equiv -1 \mod 3$.
So $a \equiv 0$ and $b \equiv 1$ and at least one is odd.
So
$[0,1],[3,1],[3,4],[6,1]$ are the only four options.
Obviously if the average of $ab, a^3, b^3$ is $\frac {223}3 \approx 70 > 4^3$, the first three can't possibly work. Don't even need to test them.
So $[6,1]$ is only option.
An indeed $6 + 6^3 + 1 = 223$.
So, if my reasoning is right (which.... I really don't see any error or faulty assumptions... but ... a self-editor is always blind....). $\{x,y\} = \{18,3\}$ are the only solutions.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding equation with roots $\cos(2k+1)\pi/9$ and using Vieta's formula's Question
From the equation who's roots are $\cos\frac{\pi}9,\cos\frac{3\pi}9,\cos\frac{5\pi}9,\cos\frac{7\pi}9$ and hence prove
a) $8\cos\frac{\pi}9\cos\frac{5\pi}9\cos\frac{7\pi}9=1=8\cos\frac{\pi}9\cos\frac{2\pi}9\cos\frac{4\pi}9$
b) $\sec^4\frac{\pi}9+\sec^4\frac{2\pi}9+\sec^4\frac{4\pi}9=1104$
My Attempt:
Say $y=e^{i\pi(2k+1)/9}$
Thus $y^9+1=0$ has solutions $e^{i\pi(2k+1)/9}$ for $k\in\{0,1,\dots,8\}$
Let $y+1/y=2x\implies y=x\pm\sqrt{x^2-1}$
Thus $(x\pm\sqrt{x^2-1})^9+1=0$ has solutions $\cos\frac{(2k+1)\pi}{9}$ for $k\in\{0,1\dots8\}$
Thus $(x\pm\sqrt{x^2-1})^9+1=\prod_{k=0}^8(x-\cos\frac{(2k+1)\pi}{9})$
But if we put $x=0$, LHS is complex while RHS is Real
What am i doing wrong?
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$$\implies x^4-3x^2+1=x(x^2-2)$$
$$\implies(x^4-3x^2+1)^2=x^2(x^2-2)^2$$
Set $\sec^2\dfrac{(2k+1)\pi}9=\dfrac4{x^2}=v$(say)
On replacement & simplification
$$ v^4-v^3(24+16)+v^2(144+32+64)+v(\cdots)+1=0$$
whose roots are $\sec^2\dfrac{(2k+1)\pi}9,k=0,1,2,3$
$$\sum_{k=1}^4\left(\sec^2\dfrac{(2k+1)\pi}9\right)^2$$
$$=\left(\sum_{k=1}^4\sec^2\dfrac{(2k+1)\pi}9\right)^2-2\sum_{k_1,k_2=0,1,2,3, k_1>k_2}\sec^2\dfrac{(2k_1+1)\pi}9\cdot\sec^2\dfrac{(2k_2+1)\pi}9$$
$$=40^2-2(144+32+64)=?$$
Now for $k=1,\sec^2\dfrac{(2k+1)\pi}9=?$
Finally $\sec(\pi-t)=-\sec t,\sec^2(\pi-t)=\sec^2t$
Set $t=\dfrac{2\pi}9,\dfrac{4\pi}9$
|
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"timestamp": "2023-03-29T00:00:00",
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|
differentiation under the integral sign $\int_{0}^{\pi}\ln\left(1-2r\cos x +r^2 \right)dx$ Using differentiation with respect to the parameter,show that for $|r|<1$
$$\mathbf{F}(r)=\int_{0}^{\pi}\ln\left(1-2r\cos x +r^2 \right)dx =0$$
my attempt is
$$\mathbf{F}'(r)=\int_{0}^{\pi}{-2\cos x + 2r \over1-2r\cos x +r^2}dx$$
let $u=\tan{x \over2}$ then $\cos x = {1-u^2 \over 1+u^2} $ and $dx={2du \over 1+u^2}$
Now
$$\mathbf{F}'(r)=\int_{0}^{\infty}{-2{1-u^2 \over 1+u^2} + 2r \over1-2r{1-u^2 \over 1+u^2} +r^2}{2du \over 1+u^2}$$
$$\mathbf{F}'(r)= 4\int_{0}^{\infty}{r+ru^2+u^2-1 \over 1+u^2-2r+2ru^2+r^2}{du \over 1+u^2}$$
but I do not know how to continue,
I tried with partial fractions but it becomes too tedious.
|
If you continue $$A={r+ru^2+u^2-1 \over 1+u^2-2r+2ru^2+r^2}{1 \over 1+u^2}=\frac{(r+1) u^2+(r-1) } {(1+u^2)((r+1)^2 u^2+(r-1)^2)} $$ For the time being, let $t=u^2$ to get
$$A=\frac{(r+1) t+(r-1) } {(1+t)((r+1)^2 t+(r-1)^2)} $$
being and partial fraction decomposition leads to
$$A=\frac{1}{2 r}\frac{1}{
\left(t+1\right)}+\frac{r^2-1}{2 r} \frac{1}{ (r+1)^2 t+(r-1)^2}$$ and , bach to $u$
$$A=\frac{1}{2 r}\frac{1}{
\left(u^2+1\right)}+\frac{r^2-1}{2 r} \frac{1}{ (r+1)^2 u^2+(r-1)^2}$$ does not seems too bad.
|
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|
Prove: $\frac{x+ 2\,y}{x^{2}+ 3\,y+ 5}+ \frac{y+ 2\,x}{y^{2}+ 3\,x+ 5}+ \frac{1}{4\left ( x+ y- 1 \right )}\geqq \frac{7}{8}$ Prove: $$\frac{x+ 2\,y}{x^{2}+ 3\,y+ 5}+ \frac{y+ 2\,x}{y^{2}+ 3\,x+ 5}+ \frac{1}{4\left ( x+ y- 1 \right )}\geqq \frac{7}{8}$$
for $x,\,y\in \left [ 1,\,2 \right ]$
My unsuccessful try (with my method like ABC method, but I can't continue):
We have: $$x\in \left [ 1,\,2 \right ]\rightarrow x- 1\in \left [ 0,\,1 \right ]\rightarrow \frac{1}{x- 1}\in\left [ 1, +\infty \right )\rightarrow \underbrace{\frac{1}{x- 1}-1}_{u}\in\left [ 0, +\infty \right )\rightarrow \underbrace{x= \frac{u+ 2}{u+ 1}}_{u>0}$$
Similarly, $\underbrace{y= \frac{v+ 2}{v+ 1}}_{v>0}$, we have a new expression:
https://www.wolframalpha.com/input/?i=(x%2B2y)%2F(x%5E2%2B3y%2B5)%2B(y%2B2x)%2F(y%5E2%2B3x%2B5)%2B1%2F(4(x%2By-1))+with+x%3D(u%2B2)%2F(u%2B1),y%3D(v%2B2)%2F(v%2B1)
That's too large & too big! I need to the help & your opinions about my method! Thanks!
|
An alternate way to your substitution: $x\in [1, 2] \implies (x-1)(x-2)\leqslant 0 \implies x^2\leqslant 3x-2$. Similarly $y^2\leqslant 3y-2$. Using these in the LHS, we get
$$LHS \geqslant \frac{x+2y}{3(x+y+1)}+\frac{y+2x}{3(x+y+1)}+\frac1{4(x+y-1)}$$
Simplifying and letting $x+y=t$, we are trying to find the minimum of the univariate
$$f(t) = \frac{t}{t+1}+\frac1{4(t-1)}$$
which is easily found setting $f'(t)=0$ as $f(t) \geqslant f(3) = \frac78$ for $t \in [2, 4]$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$.
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it is divisible by $2\cdot5=10$.
Is it correct?
|
Another way to see the solution is by using
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1})\tag{1}$$
leading to
$$1^n-3^n-6^n+8^n=(8^n-3^n)-(6^n-1^n)=\\
(8-3)(8^{n-1}+8^{n-2}\cdot3+...+8\cdot3^{n-2}+3^{n-1})-\\
(6-1)(6^{n-1}+6^{n-2}\cdot1+...+6\cdot1^{n-2}+1^{n-1})=...$$
we can see that $8^{n-1}+8^{n-2}\cdot3+...+8\cdot3^{n-2}$ is always even and so it $6^{n-1}+6^{n-2}\cdot1+...+6\cdot1^{n-2}$ then
$$...=5(2m+3^{n-1})-5(2n+1)=5\left(2(m-n)+3^{n-1}-1\right)$$
and $3^{n-1}-1$ is even from the same application of $(1)$, thus we can "extract" a 2 from $\left(2(m-n)+3^{n-1}-1\right)$ and the result follows.
|
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|
Finding limit for infinite quantities.
Find $$\lim_{x\rightarrow 0}{\color{red}{x}} \cdot\bigg(\dfrac{1}{1+x^4}+\dfrac{1}{1+(2x)^4}+\dfrac{1}{1+(3x)^4}\cdots\bigg)$$
As terms are written infinitely my intuitions doesn't let to give answer as $0$. So tried calculated that weird sum something like,
$T_n=\dfrac{1}{1+(nx)^4}=\dfrac{1}{(n^2x^2+\sqrt{2}nx+1)\cdot(n^2x^2-\sqrt{2}nx+1)}$ now this not results in telescopic sum, what should I do, is answer $0$, if so then why should we assume that $\color{red}x$ will stay in numerator no matter whatever happens.
Please help.
|
Making $n = \frac{1}{x}$
$$
\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^4}\equiv\int_0^1\frac{d\xi}{1+\xi^4} = \frac{\pi +2 \coth ^{-1}\left(\sqrt{2}\right)}{4 \sqrt{2}}
$$
NOTE
For integration purposes
$$
\frac{1}{1+\xi^4} = \frac{a_1\xi+b_1}{\xi^2+\sqrt 2 \xi + 1}+\frac{a_2 \xi+ b_2}{\xi^2-\sqrt 2 \xi + 1}
$$
so that
$$
\int\frac{d\xi}{1+\xi^4} = \frac{1}{2} \left(-2 \left(a_1+\sqrt{2} b_1\right) \tan ^{-1}\left(1-\sqrt{2} \xi \right)+a_1 \log \left(\xi ^2-\sqrt{2} \xi
+1\right)-2 \left(a_2-\sqrt{2} b_2\right) \tan ^{-1}\left(\sqrt{2} \xi +1\right)+a_2 \log \left(\xi ^2+\sqrt{2} \xi
+1\right)\right)+C
$$
|
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|
Proof involving generating function The following is part of a proof that the number of ways of associating a product with $n$ terms (different ways of inserting parentheses) is $$
a_1 = 1,\ a_n = \frac{1}{n} \binom{2n-2}{n-1},$$
and the relationship$$
a_{n+1} = a_1a_n + a_2a_{n-1} + a_3a_{n-2} + \cdots + a_na_1$$
is already established.
The proof starts with the generating function $$f_A(x) = \sum_{n=1}^\infty a_nx^n,$$
then by the relationship, $$f_A(x) = x + \sum_{n=2}^\infty (a_1a_{n-1} + \cdots + a_{n-1}a_1)x^n.$$
I am unclear about the next step:
$$f_A(x) = x + \left(\sum_{n=1}^\infty a_nx^n\right)\left(\sum_{n=1}^\infty a_nx^n\right).$$
I tried using an upper bound of three in each expression. For the first expression I obtained $$a_1^2x^2 + 2a_1a_2x^3.$$
For the second expression I obtained $$a_1^2x^2 + 2a_1a_2x^3 + a_2^2x^4 + 2a_2a_3x^5 + a_3^2x^6.$$
As it appears the first two terms are the same. Can we ignore the extra terms as the upper bound approaches infinity?
|
Let
$$f_n = \frac{1}{n} \binom{2n-2}{n-1} = \frac{1}{2n - 1} \binom{2n-1}{n}$$
\begin{align}
f(x) &= \sum_{n=1}^\infty f_nx^n\tag{1}\\
&= \sum_{n=1}^\infty \frac{1}{2n - 1} \binom{2n-1}{n}x^n\\
&= \sum_{n=1}^\infty \frac{(2n-2)!}{n!(n-1)!}x^n\\
&= \sum_{n=1}^\infty \frac{1}{n} \binom{2n-2}{n-1}x^n\tag{2}\\
&= \frac{1}{2} - \frac{1}{2}\sqrt{1 - 4x}\\
\end{align}
$$x + \left(\sum_{n=1}^\infty f_nx^n\right)\left(\sum_{n=1}^\infty f_nx^n\right) =x + [f(x)]^2 = f(x)$$
But from equation (1),
$$[f(x)]^2 = \sum_{n=2}^\infty\left(\sum_{i=1}^{n-1}f_if_{n-1}\right)x^n$$
Therefore,
$$x + \sum_{n=2}^\infty\left(\sum_{i=1}^{n-1}f_if_{n-1}\right)x^n = \sum_{n=1}^\infty f_nx^n$$
An observation $f_n = \frac{1}{n} \binom{2n-2}{n-1} = \frac{(2n-2)!}{n!(n-1)!} = C_{n-1}$. Where $C_n$ is the Catalan Number
|
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|
Integral $\int_0^{\pi} \frac{\cos(2018x)}{5-4\cos{x}}dx$ I wish to evaluate $$I(2018)=\int_{0}^{\pi}\frac{\cos(2018x)}{5-4\cos x} dx$$ Considering $$X=I(k)+iJ(k)=\int_{-\pi}^{\pi}\frac{\cos{kx}}{5-4\cos x} dx +i\int_{-\pi}^{\pi}\frac{\sin{kx}}{5-4\cos x} dx=\int_{-\pi}^{\pi}\frac{e^{ikx}}{5-4\cos x} dx$$ let us substitute $$e^{ix}=z\rightarrow dx=\frac{dz}{iz} \, ,|z|=1$$ Due to Euler's formula we can rewrite $$\cos x=\frac{z^2+1}{2z}$$ $$X=\oint_{|z|=1} \frac{z^k}{5-4\frac{z^2+1}{2z}}\frac{dz}{iz}=\frac{1}{i}\oint_{|z|=1} \frac{z^k}{-2z^2+5z-2}dz$$ $$-2z^2+5z-2=-\frac{1}{2}((2z)^2-5(2z)+4)=-\frac{1}{2}(2z-4)(2z-1)=-2(z-2)(z-\frac{1}{2})$$ Now let us notice that in our contour $|z|=1\,$ only the pole $z_2=\frac{1}{2}$ is found. Thus the integral we seek to evaluate is $$\frac{1}{i} \cdot 2\pi i \, \text{Res} (f(z),z_2)$$ where $f(z)=\frac{z^k}{-2(z-2)(z-\frac{1}{2})}$ $$X=2\pi \lim_{z\to z_2} (z-z_2)\frac{z^k}{-2(z-2)(z-z_2)}=\frac{2}{3}\pi \frac{1}{2^k}$$ therefore $$I(k)=\Re (X) =\frac{2\pi}{3}\frac{1}{2^k}$$ And $$\int_{0}^{\pi}\frac{\cos(2018 x)}{5-4\cos x} dx=\frac{\pi}{3}\cdot\frac{1}{2^{2018}}.$$ Now someone told me that the answer is $0$ and I am wrong (also wolfram gives $0$ as an answer). Could you please clarify? Or maybe give another solution to this integral if it's $0$ or another answer?
|
$$
\frac{1}{5-4\cos\left(x\right)}=\frac{1}{5-2e^{ix}-2e^{-ix}}=\frac{e^{ix}}{5e^{ix}-2e^{2ix}-2}
$$
Let $X=e^{ix}$ then
$$
-2X^2+5X-2=-\left(2X-1\right)\left(X-2\right)
$$
Now we do a partial decomposition
$$-\frac{1}{\left(2X-1\right)\left(X-2\right)}=\frac{2}{3}\frac{1}{2X-1}-\frac{1}{3}\frac{1}{X-2}
$$
So far we have
$$
\frac{1}{5-4\cos\left(x\right)}=\frac{1}{3}\frac{1}{2e^{ix}-1}-\frac{2}{3}\frac{1}{e^{ix}-2}
$$
We'll express this as a series, so we transform it into an adapted form
$$
\frac{1}{5-4\cos\left(x\right)}=\frac{1}{3}\left(\frac{1}{2e^{ix}-1}-\frac{2}{e^{ix}-2}\right)=\frac{1}{3}\left(\frac{\frac{1}{2}e^{-ix}}{\displaystyle {1-\frac{1}{2}e^{-ix}}}+\frac{1}{1-\frac{1}{2}e^{ix}}\right)$$
Hence
$$
\frac{1}{5-4\cos\left(x\right)}=\frac{1}{3}\left(\sum_{n=1}^{+\infty}\left(\frac{1}{2}\right)^ne^{-inx}+\sum_{n=0}^{+\infty}\left(\frac{1}{2}\right)^ne^{inx}\right)
$$
which finally gave us
$$
\frac{1}{5-4\cos\left(x\right)}=\frac{1}{3}\left(1+\sum_{n=1}^{+\infty}\left(\frac{1}{2}\right)^{n-1}\cos\left(nx\right)\right)$$
Hence using normal convergence
$$
\int_{0}^{\pi}\frac{\cos\left(2018x\right)}{5-4\cos\left(x\right)}\text{d}x=\int_{0}^{\pi}\frac{\cos\left(2018 x\right)}{3}\text{d}x+\sum_{n=1}^{+\infty}\left(\frac{1}{2}\right)^{n-1}\int_{0}^{\pi}\cos\left(nx\right)\cos\left(2018x\right)\text{d}x
$$
Using that
$$
\int_{0}^{\pi}\cos\left(Kx\right)\text{d}x=0
$$
for all $K \in \mathbb{Z}$, we have
$$
\int_{0}^{\pi}\frac{\cos\left(2018x\right)}{5-4\cos\left(x\right)}\text{d}x=\frac{\pi}{3}\left(\frac{1}{2}\right)^{2018}
$$
|
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|
If for a real $x, x +\frac1x$ is an integer, prove $x^{2017}+\frac1{x^{2017}}$ is also. For some real $x$, let $y=x +\frac1x$, with $y\in \mathbb{Z}$.
$y=x +\frac1x\implies x^2 -xy +1 =0\implies x= \frac{y\pm \sqrt{y^2 -4}}2$, so $x+\frac1x = \frac{y\pm \sqrt{y^2 -4}}2+ \frac2{y\pm \sqrt{y^2 -4}}$
This implies: $ y = \frac{(y\pm \sqrt{y^2 -4})^2+4}{2\cdot (y\pm \sqrt{y^2 -4})}$
Not sure of proper logic (request guidance on this part), but hope that can take positive & negative signs for $ \sqrt{y^2 -4}$ alternately, in both numerator & denominator simultaneously.
Case (a) : positive $ \sqrt{y^2 -4}$ :
$ y = \frac{(y+ \sqrt{y^2 -4})^2+4}{2\cdot (y+ \sqrt{y^2 -4})}=\frac{y^2 + y\sqrt{y^2-4}}{y+\sqrt{y^2-4}}\implies y$
Case (b) : negative $ \sqrt{y^2 -4}$ :
$ y = \frac{(y- \sqrt{y^2 -4})^2+4}{2\cdot (y-\sqrt{y^2 -4})}=\frac{y^2 - y\sqrt{y^2-4}}{y-\sqrt{y^2-4}}\implies y$
There can be formed no conclusion with above, except vindicating that the roots are correct.
Now to show further that $y'= x^{2017}+\frac1{x^{2017}}$ is also an integer, there seems no way 'algebraically (i.e., direct multiplication) ' except possibly by modulus arithmetic. The direct route is not clear to me, as modulus is to be different (higher with each step) powers of $x^i+\frac1{x^i}, i\in \mathbb{Z+}$.
A variant of the above idea can be with strong induction, which considers all powers of $x, \frac1x$ till $2017$, or any positive integral power $i$ as follows:
Step 1: Base case of $n=1$ holds true, i.e. for some real $x, y = x+\frac1x$ is an integer.
Step 2: Suppose hypothesis holds for all $i\le n$ for the induction hypothesis step, i.e. :$x^n+\frac1{x^n}$ is an integer too.
Step 3: Need prove that for $i = n+1, x^i +\frac 1{x^i}$ is also an integer.
$$(x^n+\frac1{x^n})(x+\frac1x)=x^{n+1}+\frac1{x^{n-1}}+x^{n-1}+\frac1{x^{n+1}}=(x^{n+1}+\frac1{x^{n+1}})+ (\frac1{x^{n-1}}+x^{n-1})$$
so
$$x^{n+1}+\frac1{x^{n+1}}=(x^n+\frac1{x^n})(x+\frac1x)-(x^{n-1}+1/x^{n-1})$$
with the r.h.s. being an integer, hence proved.
Have two doubts:
1. Is it possible to prove by finding the complex roots of $x^i +\frac1{x^i}$ and proving using finding roots, by changing / extending domain of $x$ to be in $\mathbb{C}$.
2. No idea for negative integer values of powers of $x, \frac1x$ by the strong induction approach.
|
Let $A_n = (x + 1/x)^n$.
Then $A_n$ is a palindromic polynomial with integer coefficients.
Let $Y_k = x^{k} + 1/x^{k}$.
Then $Y_{k+1} = A_{k+1} - \sum_{j = 1}^k c_k Y_k$ where $c$ is the list of integer coefficients necessary to make the construction.
So if $x$ is such that $A_1$ (and thus $Y_1$) is an integer, then all $A_n$ must be integers and so all $Y_n$ must also be integers.
|
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|
Finding $C_n=4C_{n-1}+6n-1$ 2.$$\begin{cases}
C_n=4C_{n-1}+6n-1\\
C_0=2\\
\end{cases} $$
$\begin{cases}
C_n=D_n+dn+e\\
D_n=\lambda D_{n-1}\end{cases}$
$D_n+dn+e=4C_{n-1}+6n-1$
But $C_{n-1}=D_{n-1}+d(n-1)+e$
So $D_n+dn+e=4(D_{n-1}+d(n-1)+e)+6n-1$
$D_n+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$
But for $\lambda=4$ we get $4D_{n-1}+dn+e=4D_{n-1}+4dn-4d+4e+6n-1$ and
$0=-dn-e+4dn-4d+4e+6n-1$ or $0=(3d+6)n+3e-4d-1$
So $d=-2,e=1.5$
$C_n=D_n+dn+e$ So $C_n=D_n-2n+1.5$ And $C_0=D_0+1.5$ So $2=D_0+1.5$ And $D_0=-0.5$
$D_n=4D_{n-1}$
So
$C_n=D_n+dn+e=4^n-2n+1.5$ which is $O(4^n)$
Is it correct?
|
No, it's not correct.
$3d+6=0$ and $3e-4d-1=0$ gives $d=-2$ and $e=-\frac73$.
From $D_n=4D_{n-1}$, we get $D_n=D_0 4^n$ with $D_0 = C_0 - 0d - e = 2 + \frac73 = \frac{13}{3}$, that is, $D_n=\frac{13}{3} 4^n$.
This gives $C_n = D_n -2n -\frac73 = \frac{13}{3} 4^n -2n -\frac73 = \frac13 (13 \cdot 4^n - 6 n - 7)$.
Your other mistake was concluding that $D_n=4^n$ instead of $D_n=D_0 4^n$.
|
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|
Arithmetic doubt while studying limits of sequences While studying limits of sequences, I came across these expressions.
$\left|\frac{-5}{n+2}\right|<\delta \iff\frac{5}{n+2}<\delta \iff n+2>\frac{1}{\delta }$
$n\in \mathbb N$
$\delta\in \mathbb R$ and $\delta >0$
I'm struggling to understand how it went from the second expression ($\frac{5}{n+2}<\delta$) to the third ($n+2>\frac{1}{\delta }$)
|
To treat that as an equal sign just makes it dead wrong. What they mean is implies the next. (Actually; we must assume $\delta > 0$ and that $n + 2 > 0$. I presume those are conditions stated in the source that you omitted? That $n\in \mathbb N$?)
(Note: these implications only work one way.)
$\left|\frac{-5}{n+2}\right|<\delta \implies\frac{5}{n+2}<\delta \implies n+2>\frac{1}{\delta }$
$|\frac {-5}{n+2}| < \delta \implies \frac {5}{n+2} < \delta$.
That should be clear $|\frac {-5}{n+2}|= \frac {5}{n+2}$ (if $n+2 > 0$). We could make the stronger. $|\frac {-5}{n+2}| < \delta \iff \frac {5}{n+2} < \delta$.
Now...
$\frac{5}{n+2}<\delta \implies n+2>\frac{1}{\delta }$
This is presumably the part you have trouble with.
$\frac{5}{n+2}<\delta \iff$
$\frac{5}{n+2}\frac{n+2}{\delta} < \delta \frac{n+2}{\delta}\iff$ (assuming $n+2 > 0; \delta > 0$)
$\frac 5{\delta} < n+2$
So $\left|\frac{-5}{n+2}\right|<\delta \iff\frac{5}{n+2}<\delta \iff n+2>\frac{5}{\delta }$
Now $\frac 1{\delta} < \frac 5{\delta}$ so
$\frac 1{\delta} < \frac 5{\delta}< n+2$.
So $\frac {5}{n+2} \le \delta \implies n+2 > \frac 1{\delta}$
But obviously that is only one way. $n+2 > \frac 1{\delta} \implies (\frac 1{n+2} < \delta; \frac 1{\delta} < \frac 5{\delta}) \not \implies \frac {5}{n+2} < \delta$.
|
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|
integration by parts $\int \frac{x^{2}+4x}{x+2}\,dx$ $$\int \frac{x^{2}+4x}{x+2}\,dx$$
I have written it in this form:
$$\int \frac{(x+2)^{2}-4}{x+2}\,dx$$: on this stage I try to do integration by parts, which gets me to :
$$\frac{x^{2}}{2}+2x-4\ln\left | {x+2} \right |$$
but it's wrong for some reason. The right answer is : $$\frac{x^{2}+4x+4}{2}-4\ln\left | x+2 \right |$$
Where am I wrong?
|
Note that we have $$\frac{x^2+4x+4}2-4\ln|x+2|+C=\frac{x^2}2+2x\color{red}{+2}-4\ln|x+2|+C$$ and this is the same as your expression if we let $C_1=C+2$!
A quicker way: $$\int \frac{(x+2)^{2}-4}{x+2}\,dx=\int\left[x+2-\frac4{x+2}\right]\,dx=\frac{x^2}2+2x-4\ln|x+2|\color{blue}{+C}.$$
|
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|
Is element $5 + 2i$ irreducible in the ring $\mathbb{Z}[i]$? I want to prove that $5 + 2i$ is an irreducible in the ring $\mathbb{Z}[i]$. I was wondering how to do so without the use of the norm defined on this ring.
Let's suppose that it is reducible, i.e. $5 + 2i = (a + bi)(c + di)$ for some $a, b, c, d \in \mathbb{Z}$ and $a + bi, c + di$ are not units. Thus
$$\begin{cases} 5 = ac - bd, \\ 2 = ad + bc.\end{cases}$$
How can I prove now that this system of equation does not have an integer solution?
EDIT
Extracting $c$ from the first equation, I get $c = \frac{5 + bd}{a}$, which substituted to the second one gives me $2a - 5b = d(a^2 + b^2)$ and therefore $d$ divides both $2a$ and $5b$. If $d \neq 1$, we've got $d |a$ and $d|b$, so the element $a + bi$ is invertible, which is contrary to the hypothesis?
|
Assume $5+2i = (a+ib)(c+id) = (ac - bd) + i(ad + bc)$.
Notice that
$$(a-ib)(c-id) = (ac - bd) - i(ad + bc) = 5-2i$$
Multiply these two equalities:
$$29 = 5^2 + 2^2 = (5+2i)(5-2i) = (a+ib)(c+id)(a-ib)(c-id) = (a^2+b^2)(c^2+d^2)$$
Since $29$ is prime, this implies $a^2+b^2 = 1$ or $c^2+d^2 = 1$. In the first case we have
$$(a+ib)(a-ib) = a^2 + b^2 = 1 \implies (a+ib)^{-1} = a-ib$$
and in the second case we have
$$(c+id)(c-id) = c^2 + d^2 = 1 \implies (c+id)^{-1} = c-id$$
Hence, one of $a+ib$ and $c + id$ is invertible.
Therefore, $5+2i$ is irreducible.
|
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|
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
|
The polynomial $x^4+6x^3+11x^2+6x+1$ has symmetric coefficents - more precisely, it's called a palindromic polynomial:
$$p(x)=x^4+ax^3+bx^2+ax+1$$
The goal is to factor $p(x)$ (and show that it factors to a square of some expression). Let's start by dividing by $x^2$ and refactoring:
$$\frac{p(x)}{x^2} = q(x) = x^2+\frac{1}{x^2}+a\left(x+\frac{1}{x}\right)+b$$
It's tempting to make a substitution $y=x+\frac{1}{x}$:
$$q(x) \rightarrow q(y) = y^2+ay+(b-2)$$
This: $q(y)=0$, being a quadratic equation, is something that can be automatically solved:
$$y_{1,2}=\frac{-a\pm \sqrt{a^2-4(b-2)}}{2}$$
and inserting the values $a=6$ and $b=11$, one obtains (note the expression under the square root is equal to zero):
$$y_1=y_2=-3$$
so that $q(y)=(y+3)^2$ - at this point one sees it's a perfect square, and due to $q(y)=\frac{p(x)}{x^2}$, basically we're done with the proof at this point.
Just to take things to their end:
Because $q(y)=0 \Leftrightarrow q(x)=0 \Leftrightarrow p(x)=0 $:
$$q(y) = (y+3)^2 = \left(x+\frac{1}{x}+3\right)^2 = \frac{(x^2+3x+1)^2}{x^2} = \frac{p(x)}{x^2}$$
hence $x^4+6x^3+11x^2+6x+1 = (x^2+3x+1)^2$ - a perfect square indeed.
|
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|
The value of x satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ The value of x satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$
(A) $2 \cos 10$
(B) $2 \cos 20$
(C) $2 \cos 40$
(D) $2 \cos 80$
I was dumb enough to square the expression to reach $x^8-4x^6+4x^4+2-x=0$ which is clearly a dead end ;-;
|
Using the trigonometric identity, $$\cos2x=2\cos^2x-1=1-2\sin^2x$$
$$2\cos2x+2=4\cos^2x$$
We are looking for an angle that allows for the relationship $\sin x=\cos(\frac{\pi}{2}-x)$ in the final step. This is because, working from the inside out, the first gives a cosine, the second gives a sine, so in order for the last one to be a cosine, we must change the sine into a cosine through the above relation.
A simple check/use of intuition finds that C is the answer.
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
If $x=2\cos40$,
$$\sqrt{2+\sqrt{2-\sqrt{2+2\cos40}}}=\sqrt{2+\sqrt{2-\sqrt{4\cos^220}}}$$
$$=\sqrt{2+\sqrt{2-2\cos20}}=\sqrt{2+\sqrt{4\sin^210}}=\sqrt{2+2\sin10}$$
$$=\sqrt{2+2\cos80}=2\cos40=x$$
|
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|
Evaluating trivial integral with linear algebra $$\int_{-\infty}^{\infty} \frac{\text{d}x}{ax^2+bx+c} = \frac{\pi}{\sqrt{\det(A)}}$$
where $A = \begin{bmatrix}a&\frac{b}{2}\\\frac{b}{2}&c\end{bmatrix}$
The connection to matrix quadratic form is given via the equivalence:
$$ax^2 + bx + c \equiv \begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}a&\frac{b}{2}\\\frac{b}{2}&c\end{bmatrix} \begin{bmatrix}x\\1\end{bmatrix} $$
What standard results from Linear Algebra make this integral a linear transformation of the standard integral?
$$\int_{-\infty}^{\infty} \frac{\text{d}x}{x^2 + 1} = \pi$$
|
Note: I am relatively new here, so if I didn't interpret your actual question correctly, sorry in advance.
The integral in question is
$$I=\int_{-\infty}^{\infty} \frac{\text{d}x}{ax^2+bx+c}.$$
Completing the square for the denominator yields
$$I=\frac{1}{a}\int_{-\infty}^{\infty} \frac{\text{d}x}{(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}}.$$
Factoring out $\frac{4ac-b^2}{4a^2}$ from the denominator gives
$$I=\frac{1}{a} \frac{1}{\frac{4ac-b^2}{4a^2}}\int_{-\infty}^{\infty} \frac{\text{d}x}{\frac{4a^2}{4ac-b^2}(x+\frac{b}{2a})^2+1}=\frac{4a}{4ac-b^2} \int_{-\infty}^{\infty} \frac{\text{d}x}{\frac{4a^2}{4ac-b^2}(x+\frac{b}{2a})^2+1}.$$
Let $\frac{4a^2}{4ac-b^2}(x+\frac{b}{2a})^2=\tan^2(u)\rightarrow x=\frac{\sqrt{4ac-b^2}}{2a}\tan(u)-\frac{b}{2a}$ such that $\text{d}x=\frac{\sqrt{4ac-b^2}}{2a}(\tan^2(u)+1)\text{d}u$, where $u\in(-\frac{\pi}{2},\frac{\pi}{2})$:
$$I=\frac{4a}{4ac-b^2}\frac{\sqrt{4ac-b^2}}{2a}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \text{d}u=\frac{2\pi}{\sqrt{4ac-b^2}}.$$
The determinant of A is given by
$$\det(A)=ac-(\frac{b}{2})^2=\frac{4ac-b^2}{4}.$$
Thus, it follows that
$$\frac{\pi}{\sqrt{\det(A)}}=\frac{2\pi}{\sqrt{4ac-b^2}}.$$
This is equal to the previously derived result.
|
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|
Solving a congruence of the form $a^x = b \pmod m$ without indices or primitive roots Consider $9^x\equiv 7 \mod 19$. So $9^x\equiv 26 \equiv 45$, $9^{x-1} \equiv 5 \equiv 24 \equiv 43 \equiv 62 \equiv 81$, so $x=3$, and $19 \mid 722$.
But what I really want to solve is $12^x\equiv 17 \mod 25$. Using the same method, $$2^{2x}\cdot3^x \equiv 17 \equiv 42,$$ $$ 2^{2x- 1}3^{x-1}\equiv 7 \equiv 32,$$ $$2^{2x-6}3^{x-1}\equiv 1\equiv 26, $$ $$ 2^{2x-7}3^{x-1}\equiv 13 \equiv38,$$ $$ 2^{2x-8}3^{x-1} \equiv 19\equiv 44, $$ $$2^{2x-10}3^{x-1}\equiv 11 \equiv 36 \equiv (2^2)(3^2),$$ so $2x-10=2$, hence $x=6$, and $x-1=2$, so $x=3$.
Why doesn't this work? Is it because $12$ is not the power of a prime and $9$ is? Any help is appreciated!
|
There are several possible answers to your question.
*
*You don't do the "same thing", since you are breaking the left hand side in two powers. In fact, it is not true that $a^xb^y\equiv a^zb^t \pmod n$ implies that $x=z$ and $y=t$, even not modulo the respective order $\pmod n$, or even if they are both "primes" (which has no meaning modulo n) or "coprimes" (again).
*What you do works: you continue until you get the same $x$ from both powers: that $x$ will work.
*Why do you repeated 6 times the procedure? In the third step you already got
$$2^{2x−6}3^{x−1}\equiv 1\equiv 2^03^0$$
so $2x-6=0$ and $x-1=0$, hence $x=3$ and $x=1$!
*What you do it is not really well defined. For example you could change $9$ in the first case to $9=28=2^27$, and try it again. In fact $2^{2x}7^x\equiv 7$, so $2x=0$ and $x=1$!!! Or you can change $12$ in the second case by $12=37$, which is prime, and do the same procedure.
|
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"url": "https://math.stackexchange.com/questions/2839062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\binom{n}{k}$ for $n>k^{100}$ This paper I'm reading says this
"We used the fact that $4k^2n^{k-1}<4(k+1)!\binom{n-1}{k-1}$ for $n>k^{100}$." The only condition here is that $k\geq3$ and $n$ can be sufficiently large. However, this seems incorrect to me.
|
Note that
$$\begin{align}\frac{(k-1)!{n-1\choose k-1}}{n^{k-1}}
&=\frac{(n-1)!}{n^{k-1}(n-k)!}\\
&>\frac{(n-k+1)^{k-1}}{n^{k-1}}\\
&=\left(1-\frac{k-1}{n}\right)^{k-1}\\
&>1-\frac{(k-1)^2}n \end{align}$$(where we first use that $\frac{(n-1)!}{(n-k)!}$ consists of $k-1$ factors $\ge n-k+1$ and in the last step use Bernoulli's inequality).
As we are given that $k\ge 1$ and $n\ge k^3$, we have
$$ n\ge k^3>(k^2-1)(k-1)=(k+1)(k-1)^2,$$
so that
$$ 1-\frac{(k-1)^2}n>1-\frac1{k+1}$$
and ultimately
$$\begin{align}4(k+1)!{n-1\choose k-1}&=4(k+1)k\cdot \frac{(k-1)!{n-1\choose k-1}}{n^{k-1}}\cdot n^{k-1}\\&>4(k+1)k\left(1-\frac1{k+1}\right)n^{k-1}\\&=4k^2 n^{k-1}.\end{align}$$
|
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|
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$5^x+7^x+11^x=(5+1)^x+(7+1)^x+(11-2)^x$$
and tried to think of some AM-GM mean inequality or to divide everything by $11^x$, but those don't seem like the way to go. Any hints?
|
Consider the function for $x>1,n>1$
$$f(n)=n^x-(n-1)^x$$
$$f'(n)=xn^{x-1}-x(n-1)^{x-1}=x(n-1)^{x-1}[(1+\frac{1}{n-1})^{x-1}-1]>0$$
So $f(n)$ is increasing for $x>1,n>1$.
Now rewrite the equation
$$5^x+7^x+11^x=6^x+8^x+9^x$$
$$\color{red}{(11^x-10^x)}+\color{blue}{(10^x-9^x)}=\color{red}{(8^x-7^x)}+\color{blue}{(6^x-5^x)}$$
Comparing the red and blue parts, LHS is larger than RHS due to increasing $f(n)$.
So the equation holds only if $0\le x\le 1$ for natural number solutions, i.e. $x=0$ or $x=1$.
|
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|
Solving $a! + b! = 2^n$
$a! + b! = 2^n$, find $a,b,n$
My obseravtions thus far:
*
*If $a>4$, $b<4$, otherwise $a! + b! = 0 \mod 10$
*If $a=1$, $b=1$
If we try all combination with the numbers 2 to 4, we get following solutions:
(1,1,1), (2,2,2), (2,3,3),(3,2,3)
Are there more solutions and how to find them?
|
I guess I would do: wolog $b < a$ so $a! + b! = b!([a-b]! + 1) = 2^n$.
If $b \ge 3$ then $3|b!([a-b]! + 1)=2^n$ but that's impossible. So $b \le 2$.
Also the only factor of $2^n$ are powers of $2$ so $[a-b]! + 1 = 2^m$ so $[a-b]! = 2^m -1$ which is an odd number if $m > 0$. But if $[a-b] \ge 2$ then $[a-b]!$ is even.
So . . . Putting that together.
$b = 0,1,2$.
If $b = 0$ or $1$ then $b!([a-b]! + 1) = (a! + 1)=2^n$ so $a! = 2^n-1$. If $n =0$ then $a!=0$ which is impossible. If $n \ge 1$ then $a!$ is odd so $a=0$ or $1$.
And for $a=b=0$ or $b=0; a=1$ or $a=b=1$ we have $a! + b! = 2$ so $n=1$.
If $b=2$ then $b!([a-b]! + 1)=2([a-2]! + 1) = 2^n$ so $[a-2]! =2^{n-1} -1$. And be the same reasoning as above $a-2 = 0$ or $1$ so $a = 2$ or $a = 3$.
If $a = 3$ then $a! + b! = 8$ so $n =3$.
And if $a = 2$ and $b=2$ we have $a! + b! = 4$ and $n =2$.
So the solutions are
$(a,b,c) = (0,0,1), (0,1,1),(1,0,1),(1,1,1),(2,2,2),(2,3,2),(3,2,3)$.
|
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|
Find the minimum value of $\frac{a+b+c}{b-a}$ Let $f(x)=ax^2+bx+c$ where $(a<b)$ and $f(x)\geq 0$ $\forall x\in R$.
Find the minimum value of $$\frac{a+b+c}{b-a}$$
If $f(x)\geq 0$ $\forall x\in R$ then $b>a>0$ and $b^2-4ac\leq 0$ implying that $c>0$. After this not able to find way out.
|
Given $f(x) = ax^2+bx+c\geq 0\forall x \in \mathbb{R}$
Now put $x=-2,$ We get $f(-2)\geq 4a-2b+c\geq 0\Rightarrow 2a+c\geq 2(b-a)$
So $\displaystyle \frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}\geq 1+2=3.$
|
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|
For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction:
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, thanks!
|
$$\frac{4x^2+8x+13}{6(x+1)} = \frac{4(x+1)^2+9}{6(x+1)} \ge \frac{2\sqrt{4(x+1)^2\cdot 9}}{6(x+1)} = \frac{2\cdot2\cdot3}{6} =\frac{12}{6} = 2$$
|
{
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|
Using De Moivre's to find roots of a polynomial a) Use De Moivre's theorem to express $\frac{\sin 8\theta}{\sin\theta \cos\theta}$ as a polynomial in$ s$, where $s=\sin\theta$
b) Hence solve the equation $x^6-6x^4+10x^2-4=0$
I've been able to do the first question and I worked out the answer to be $8(1-10s^2+24s^4-16s^6)$ but I can't seem to be able to do part b) because I can't see a relationship between the two polynomials.
Any help is appreciated thanks :)
|
Solution
$$x^6-6x^4+10x^2-4=(x^2-2)(x^4-4x^2+2)=(x^2-2)[(x^2-2)^2-2]=0$$
Thus, let $x^2-2=0$ or $x^4-4x^2+2=0$. We obtain $$x=\pm\sqrt{2},$$or$$x=\pm\sqrt{2\pm\sqrt{2}}.$$
|
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|
How can these two equations be solved by elimination? In this question, the following two equations were solved using elimination. With a google crash course I sort of get how elimination works, but it seems like these are much too complex to add the left sides together in a way that cancels out x or y.
$$
\frac{3x(3x^2+9)}{2y}
-
\frac{(3x^2+9)}{8y^3}^3
-
y
=
6 \pmod {23}
$$
$$
\frac{(3x^2+9)^2}{(2y)^2}
-
2x
=
12 \pmod {23}
$$
Ultimately I'm trying to figure out any way (the easiest preferably) to reduce these such that x=18, y=10 or x=19, y = 3 in order to convert it to a function in a scripting language.
If elimination is the way to go, what were the steps involved in arriving at the following equation?
$$x^4-48 x^3-18 x^2+13968 x=-86481 \pmod {23}$$
|
If you would like to see how to get the equation by hand:
Introduce a third variable $$z = \frac{3x^2+9}{2y}$$ We have three equations
$$2yz - 3x^2 = 9\\3xz - z^3 - y = 6\\z^2 - 2x = 12$$
Multiplying the second equation by $2z$ and adding the first eliminates $y$:
$$6xz^2 - 2z^4 - 3x^2 = 12z + 9$$
Substituting for $z^2$ from the final equation gives $$6x(2x + 12) - 2(2x +12)^2- 3x^2 = 12z + 9\\x^2 - 24x - 297 = 12z$$
Squaring both sides, then substituting for $z^2$ again
$$(x^2 - 24x - 297)^2 = 144(2x + 12)\\x^4 - 48x^3 - 18x^2 +13968x + 86481 = 0$$
Modulo $23$, this reduces to
$$x^4 - 2x^3 + 5x^2 + 7x + 1 \equiv 0 \mod 23$$
|
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|
Number of solutions of $\left\{x\right\}+\left\{\frac{1}{x}\right\}=1$ Find the number of solutions of $$\left\{x\right\}+\left\{\frac{1}{x}\right\}=1,$$ where $\left\{\cdot\right\}$ denotes Fractional part of real number $x$.
My try:
When $x \gt 1$ we get
$$\left\{x\right\}+\frac{1}{x}=1$$ $\implies$
$$\left\{x\right\}=1-\frac{1}{x}.$$
Letting $x=n+f$, where $n \in \mathbb{Z^+}$ and $ 0 \lt f \lt 1$, we get
$$f=1-\frac{1}{n+f}.$$
By Hint given by $J.G$, i am continuing the solution:
we have
$$f^2+(n-1)f+1-n=0$$ solving we get
$$f=\frac{-(n-1)+\sqrt{(n+3)(n-1)}}{2}$$ $\implies$
$$f=\frac{\left(\sqrt{n+3}-\sqrt{n-1}\right)\sqrt{n-1}}{2}$$
Now obviously $n \ne 1$ for if we get $f=0$
So $n=2,3,4,5...$ gives values of $f$ as
$\frac{\sqrt{5}-1}{2}$, $\sqrt{3}-1$, so on which gives infinite solutions.
|
Let $x:=n+f$. The equation is
$$f+\frac1{n+f}=1,$$
giving the solutions in $f$
$$f=\frac{\pm\sqrt{(n+1)^2-4}-n+1}2.$$
The negative sign cannot work, nor the negative $n$. Then $n\ge1$ is required, but $n=1$ yields $x=1$, which is wrong. Finally,
$$f=\frac{\sqrt{(n+1)^2-4}-n+1}2, \forall n>1.$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate $\int \frac{\cos^2 x}{1 + \sin^2 x}dx$ My approach was to use a double-arc formula for cosine and sine, but I did not succeed.
$\sin^2 x = \frac{1 - \cos(2x)}{2}$ and $\cos^2x = \frac{1 + \cos(2x)}{2}$
|
We reduce the degree of the numerator with
$$\int\frac{\cos^2x}{1+\sin^2x}dx=\int\frac{2-(1+\sin^2x)}{1+\sin^2x}dx=2\int\frac{dx}{1+\sin^2x}-x$$
and we introduct a tangent to rationalize,
$$\int\frac{dx}{1+\sin^2x}=\int\frac{dx}{\cos^2x+2\sin^2x}=\int\frac{dx}{\cos^2x(1+2\tan^2x)}=\int\frac{dt}{1+2t^2}
\\=\frac1{\sqrt2}\int\frac{du}{1+u^2}=\frac1{\sqrt2}\arctan u=\frac1{\sqrt2}\arctan\sqrt2t=\frac1{\sqrt2}\arctan(\sqrt2\tan x).$$
|
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|
Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$
I solved it as follows but I seek better or quicker way:
$\text{ Assume }x, y \in \mathbb Z^+\\ 1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x}=1+\frac{y+1}{y}+4 \\
\Rightarrow \left(1+\frac{x}{y}\right)\left(1+\frac{y+1}{x}\right)=6+\frac{1}{y}\\
\Rightarrow (x+y)(x+y+1)=x(6y+1)\\
\Rightarrow x\mid (x+y) \;\text{ or }\; x\mid (x+y+1)\\
\Rightarrow x\mid y \;\text{ or }\; x\mid (y+1)\\
\text{Put}\; y=nx ,n \in \mathbb Z^+\;\Rightarrow\; \frac{x}{nx} +\frac{nx+1}{x}=4 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x}=4-n \;\rightarrow\;(1)\\
\text{But}\; \frac{1}{n} +\frac{1}{x} \gt 0 \;\Rightarrow\; 4-n \gt 0 \;\Rightarrow\; n \lt 4\\
\text{Also}\; \frac{1}{n},\frac{1}{x}\le 1 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x} \le 2 \;\Rightarrow\; 4-n \le 2 \;\Rightarrow\; n \ge 2\\
\;\Rightarrow\; n=2 \;\text{ or }\; 3, \;\text{substituting in eq. (1), we find no integral values for } x.\\ \text{The same for the other case.}\\
$
So is there any other better or intelligent way to get this result?
|
Rewrite the equation: $x^2 - 4yx + y^2+y = 0\implies\triangle'=(-2y)^2-1(y^2+y)= 4y^2-y^2-y = 3y^2-y=k^2 \implies y(3y-1)=k^2$ . Observe that $\text{gcd}(y,3y-1) = 1$ since if $d = \text{gcd} \implies d \mid y, d \mid 3y-1 \implies y = md, 3y-1 = nd\implies 1 = 3y -nd= 3md - nd = d(3m-n)\implies d = 1\implies y = u^2, 3y-1 = v^2, uv = k\implies 3u^2-v^2=1$ . This is a Pell equation and it either has infinitely many solutions or no solutions at all. Please check its status via google.
|
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|
Angles on a point inside a triangle Let $ABC$ be an isosceles triangle with $AB=AC$ and $∠BAC = 100$. A point $P$
inside the triangle $ABC$ satisfies that $∠CBP=35$ and $∠PCB= 30$. Find the
measure, in degrees, of angle $∠BAP$. Attached is the figure of the triangle
I tried to Angle Chase but it seemed true for all values of $BAP$. I then tried using the sine law. In the triangle $PBC$, We have
$$\frac{PB \sin(35)}{\sin(30)}= PC$$
Trying it with triangles $APB$ ($x= BAP$) and the fact that they are isosceles
$$\frac{PC\sin(x+70)}{\sin(100-x)}=\frac{PB\sin(175-x)}{\sin (x)}$$
Which becomes,
$$\frac{\sin(35)\sin(x+70)}{\sin(30)\sin(100-x)} = \frac{\sin(175-x)}{\sin(x)}$$
Where in I don't know how to solve it. Other methods are welcome.
|
Let $M$ be the midpoint of $BC.$ Let $Q$ be the intersection of $CP$ with $AM.$
Note that $\angle ABC = \frac12(180^\circ - 100^\circ) = 40^\circ$
and therefore
$$ \angle PBA = \angle ABC - \angle PBC = 40^\circ - 35^\circ = 5^\circ.$$
Since $Q$ is on the perpendicular bisector of $BC,$ triangle $\triangle BQC$
is isoceles with $BQ = CQ.$
Moreover,
\begin{align}
\angle QBC &= \angle QCB = 30^\circ,\\
\angle PBQ &= \angle PBC - \angle QBC = 35^\circ - 30^\circ = 5^\circ,\\
\angle PQB &= \angle QCB + \angle QBC = 60^\circ,\\
\angle PQA &= \angle MQC = 90^\circ - \angle QCB = 60^\circ.\\
\end{align}
In summary, $\angle PBA = \angle PBQ$ and $\angle PQA = \angle PQB.$
That is, the rays $BP$ and $QP$ are the angle bisectors of angles
$\angle ABQ$ and $\angle AQB$ of triangle $\triangle ABQ.$
The angle bisectors of a triangle are concurrent,
hence $AP$ is an angle bisector of $\angle BAQ.$
But $\angle BAQ = 50^\circ,$
so $\angle BAP = \frac12\angle BAQ = 25^\circ.$
|
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|
For prime $p$ do we have $p^3+p^2+p+1=n^2$ infinitely often? This is a question to ponder about the occurrence of prime $p$ giving
$p^3 + p^2 + p +1=n^2$ as is true with $7$ giving $400=20^2$. Do you
think this will ever happen again?
|
There are no more positive prime solution other than $p = 7$.
First, we known $p = 2$ doesn't work, we can assume $p$ is an odd prime.
Let's say $p$ is an odd prime such that
$$p^3 + p^2 + p + 1 = n^2\quad\iff\quad(p^2+1)(p+1) = n^2$$
Since $\gcd(p+1,p^2+1) = \gcd(p+1,2) = 2$, we find for some $k, m > 0$,
$$\begin{cases}
p^2 + 1 &= 2m^2\\
p + 1 &= 2k^2\\
\end{cases} \quad\iff\quad \begin{cases} m = \sqrt{\frac{p^2+1}{2}}\\k = \sqrt{\frac{p+1}{2}}\end{cases}$$
Subtracting the two equations on the left, we obtain
$$p(p-1) = 2(m+k)(m-k)$$
Since $0 < m - k < m = \sqrt{\frac{p^2+1}{2}} < \sqrt{\frac{p^2+p^2}{2}} = p$, both $2$ and $m-k$ are coprime to $p$.
Since $p$ is a prime, we obtain $p | (m+k)$.
Since $m+k < 2m < 2p$, this forces $p = m + k$.
As a result,
$$\begin{align}
& 2(p^4 + m^4 + k^4) - (p^2 + m^2 + k^2)^2 = (p+m+k)(p+m-k)(p-m+k)(p-m-k) = 0\\
\iff &
p^4 - 6p^3 - 7p^2 = (p-7)(p+1)p^2 = 0\\
\implies & p = 7\end{align}
$$
In short, the equation
$$p^3 + p^2 + p + 1 = n^2$$
has one and only one positive prime solution. Namely, $p = 7$.
|
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|
Evaluating $\lim_{x\to \sqrt 2}\frac{\sqrt{3+2x} - (\sqrt 2+1)}{x^2-2}$ I used rationalization trick where you multiply the top and bottom with opposites of given functions and thus got $\frac{2(x^2+2)}{(x^2-2)({\sqrt {3+2x} + (\sqrt2 + 1)})}$ but by then it has become too complex. The book says answer should be $\frac{1}{2(2+\sqrt 2)}$.
|
You made a mistake in the numerator.
\begin{align}
\frac{\sqrt{3+2x} - (\sqrt 2+1)}{x^2-2}\cdot \frac{\sqrt{3+2x} + (\sqrt 2+1)}{\sqrt{3+2x} + (\sqrt 2+1)} &= \frac{3+2x - (\sqrt{2}+1)^2}{(x^2-2)(\sqrt{3+2x} + (\sqrt 2+1))}\\
&= \frac{2(x-\sqrt{2})}{(x^2-2)(\sqrt{3+2x} + (\sqrt 2+1))}\\
&= \frac{2}{(x+\sqrt{2})(\sqrt{3+2x} + (\sqrt 2+1))}\\
&\xrightarrow{x\to\sqrt{2}} \frac{2}{(\sqrt{2}+\sqrt{2})(2\cdot (\sqrt{2}+1))}\\
&= \frac{1}{2(2+\sqrt2)}
\end{align}
|
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|
Using inverse Laplace transform to solve differential equation The differential equation is as follows-
$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions-
$X(s) = \frac{1}{(s-1)(s+3)(s+2)} = \frac{A}{s-1} + \frac{B}{s+3} + \frac{C}{s+2} $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $
I am stuck from here on how to carry on this partial fraction
Can I sub all s values to be 0 ?
For example
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $
|
So, if we have that
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
We can expand everything to get
$$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$
$$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$
So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.
Or alternatively:
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
If we substitute $s=1$ we get that
$$1=A(1+3)(1+2)$$
With $s=-2$ we get that
$$1=C(-2-1)(-2+3)$$
And with $s=-3$ we get that
$$1=B(-3-1)(-3+2)$$
And here is a video about the partial fraction decomposition, you might find it helpful.
|
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|
How to determine two matrices are conjugate. Which of the following statements are true?
*
*The matrices $
A=\left[ {\begin{array}{cc}
1 & 1 \\
0 & 1\\
\end{array} } \right]$ and $
B=\left[ {\begin{array}{cc}
1 & 0 \\
1 & 1\\
\end{array} } \right]$ are conjugate in $GL_2(\mathbb{R})$
*The matrices $
A=\left[ {\begin{array}{cc}
1 & 1 \\
0 & 1\\
\end{array} } \right]$ and $
B=\left[ {\begin{array}{cc}
1 & 0 \\
1 & 1\\
\end{array} } \right]$ are conjugate in $SL_2(\mathbb{R})$
*The matrices $
C=\left[ {\begin{array}{cc}
1 & 0 \\
0 & 2\\
\end{array} } \right]$ and $
D=\left[ {\begin{array}{cc}
1 & 3 \\
0 & 2\\
\end{array} } \right]$ are conjugate in $GL_2(\mathbb{R})$
I know the conjugate matrices have the same eigenvalues. But all of this matrices have the same eigenvalues. I am not sure how to determine that which matrices are conjugate to each other.
|
Quiver has already told you (and I am sure it is in you textbook) that two matrices, A and B, are conjugate if and only if there exist an invertible matrix, P, such that $A= P^{-1}BP$. That is equivalent to $PA= BP$.
In the first problem, $A= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ and $B= \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$. We can show they are conjugate by finding an appropriate P!
Since A and B are 2 by 2 P must be also and we can write it $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ so we must have
$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix}$.
$\begin{bmatrix}a & a+ b\\ c & c+ d\end{bmatrix}= \begin{bmatrix}a * b \\ a+ c & c+ d\end{bmatrix}$.
We must have a= a, a+ b= b, c= a+ c, and c+ d= c+ d.
Both a+ b= b and c= a+ c reduce to a= 0 while c+ c= c+ d is always true.
Any matrix of the form $\begin{bmatrix}0 & b \\ c & d\end{bmatrix}$ will do.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2860140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
Maximum of $\sqrt{\frac{1}{4}\cdot \sin^2(t)+\sin^2(t+\frac{\pi}{3})}$ Maximum of $\sqrt{\frac{1}{4}\cdot\sin^2(t)+\sin^2(t+\frac{\pi}{3})}$
In my opinion, the maximum of the sinus is 1, so I calculated $\sqrt{\frac{1}{4}\cdot1+1}$
This is wrong, why?
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Other people have pointed out the OP's error. Now, I am supplying a solution.
Note that
$$\sin\left(t+\frac{\pi}{3}\right)=\frac{1}{2}\sin(t)+\frac{\sqrt{3}}{2}\cos(t)\,.$$
That is,
$$\begin{align}\frac{1}{4}\sin^2(t)+\sin^2\left(t+\frac{\pi}{3}\right)&=\frac{\sin^2(t)+\big(\sin(t)+\sqrt{3}\cos(t)\big)^2}{4}\\&=\frac{5}{8}+\frac{2\sqrt{3}\sin(2t)+\cos(2t)}{8}\,.\end{align}$$
We then apply the Cauchy-Schwarz Inequality to get
$$-\frac{\sqrt{13}}{8}\leq \frac{2\sqrt{3}\sin(2t)+\cos(2t)}{8}\leq +\frac{\sqrt{13}}{8}\,.$$
Consequently,
$$\sqrt{\frac{5-\sqrt{13}}{8}}\leq \sqrt{\frac{1}{4}\sin^2(t)+\sin^2\left(t+\frac{\pi}{3}\right)}\leq \sqrt{\frac{5+\sqrt{13}}{8}}\,.$$
It is not difficult to see that both the inequality on the left-hand side and the inequality on the right-hand side are sharp.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2861289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Determine all real $x$ that satisfy $\det A=0$ I want to find all real $x$ that satisfy
$$
\textrm{det } X=
\begin{vmatrix}
x &2 &2 &2\\
2 &x &2 &2\\
2 &2 &x &2\\
2 &2 &2 &x
\end{vmatrix}\\
$$
My teacher does this by adding the three bottom rows to the top row
$$
\textrm{det } X=
\begin{vmatrix}
x+6 &x+6 &x+6 &x+6\\
2 &x &2 &2\\
2 &2 &x &2\\
2 &2 &2 &x
\end{vmatrix}\\
$$
and then subtracting a row of $2$'s from the bottom three rows
$$
\textrm{det } X=
(x+6)
\begin{vmatrix}
1 &1 &1 &1\\
0 &x-2 &0 &0\\
0 &0 &x-2 &0\\
0 &0 &0 &x-2
\end{vmatrix}.
$$
The answer is
$$
x\in \{-6,2\}.
$$
I think I understand the operations (although subtracting an arbitrary row of numbers from a matrix/determinant row is something I've never seen before, but I don't see why that wouldn't be allowed. Just like you can subtract arbitrary coefficients on both sides of an equation, right?), my main issue is why they are performed.
*
*Why can't I just in the same way subtract a row of $2$'s from the three bottom rows in the first determinant? If I do that I get a different answer.
*I know I want a column of all zeroes except one column-element, but why do I need to perform the first operation beforehand? Is it somehow necessary that all the top row elements to be the same, $(x+6)$?
|
"Subtracting a row of 2s" is NOT a valid matrix operation. But subtracting one row from another is. Perhaps what your teacher did (or meant to do) is, first subtract the third row from the second to get
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 2 & 2 & x & 2 \\ 2 & 2 & 2 & x \end{array}\right|$.
Now, subtract the fourth row from the third to get
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 0 & 0 & x- 2 & 2-x \\ 2 & 2 & 2 & x \end{array}\right|$.
Finally, you can subtract the first row from the fourth to get
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 0 & 0 & x- 2 & 2-x \\ 2- x & 0 & 0 & x- 2 \end{array}\right|$.
That's almost an "upper triangular" matrix. We can calculate the determinant reasonably easily by "expanding on the first column:
$x\left|\begin{array}{cc} x- 2 & 2- x & 0 \\ 0 & x- 2 & 2- x \\ 0 & 0 & x- 2 \end{array}\right|$$- 2\left|\begin{array}{cc} 2 & 2 & 2 \\ x- 2 & 2- x & 0 \\ 0 & x- 2 & 2- x \end{array}\right|$.
That first determinant is $x(x- 2)^3$. For the second we can further expand that three by three determinant on the first column to get $-2\left(2\left|\begin{array}{cc}2- x & 0 \\ x- 2 & 2- x\end{array}\right|- (x- 2)\left|\begin{array}{cc}2 & 2 \\ x- 2 & 2- x\end{array}\right|\right)= -4(2- x)^2- (x- 2)(4- 2x- 2x+ 4)= -4(2- x)^2- (x- 2)(8- 4x)= -4(2- x)^2+ 4(2- x)^2= 0$.
So the determinant of the given matrix is $x(x- 2)^3$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2861589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.