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proving the limit of a function using the $\epsilon , \delta $ definition Use the $\epsilon , \delta $ definition of a limit to verify that $$\lim \limits_{x \to 1} \frac{x^2 +x}{3x-2}=2$$
Let, $$\lim \limits_{x \to 1} \frac{x^2 +x}{3x-2}-2 = \frac{x^2+x -6x+4}{3x-2}= \frac{x^2-5x+4}{3x-2} = \frac{(x-4)(x-1)}{3x-2}$$
but $x \ne \frac{2}{3}$ so we need to find $\lvert x-1 \rvert< $ some number betwen 1 and $\frac{2}{3}$ say $\frac{5}{6}$
So $\lvert x-1 \rvert < \frac{5}{6}$ or $\frac{1}{6}< x< \frac{11}{6}$
We want the upper bound so we use $x< \frac{11}{6}$ and the upper bound is $\frac{(\frac{11}{6}-4)(\frac{11}{6}-1)}{3(\frac{11}{6})-2} $ but that doesn't make any sense cause then the upper bound would be negative. Should I have used the lower bound?
| Continuing your analysis: If $|x-1|<{5\over 6}$ then
$$
\left|{x^2+x\over 3x-2}-2\right|=\left|{(x-4)(x-1)\over 3x-2}\right|\le {23/6\over 3/2}\cdot|x-1|={23\over 9}|x-1|,
$$
because $|x-4|<23/6$ and $|3x-2|>3/2$ for $x\in(1/6,11/6)$. Now make your choice of $\delta>0$ (for a given $\epsilon>0$) taking care that it is no larger than $5/6$ so that the above inequality is in force.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving some identities in the set of natural numbers without using induction... I'm not sure how to prove some of the identities without using induction, for example:
$$1+2+3+...+n=\frac{n(n+1)}{2}$$
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
$$1^3+2^3+...+n^3=(\frac{n(n-1)}{2})^2$$
What my teacher suggested and did for the first example is, take two successive members and sum their squares, then using some transformations get $\frac{n(n+1)}{2}$, here's what the teacher did:
$$(k+1)^2-k^2=2k+1$$
we sum $k$-s from $1$ to $n$:
$$\sum_{k=1}^{n}((k+1)^2-k^2)=2\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$$
$$2^2-1^2+3^2-2^2+4^2-3^2+...+(n+1)^2-n^2=2\sum_{k=1}^{n}k+n$$
$$(n+1)^2-1=2\sum_{k=1}^{n}k+n$$
$$n^2+2n-n=2\sum_{k=1}^{n}k$$
$$2\sum_{k=1}^{n}k=n^2+n$$
$$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$
What is the method my teacher used here? The teacher also suggested that, for example, if we have the sum of squares of successive integers (like in the second example), we should take two successive numbers and sum their cubes, or if we have the sum of cubes, then we take two successive members of the sum and sum their 4th degrees. Is there a name for this method of solving so I could google it and examine it a bit more?
| For the first one there is a nice way to find it is sum let
$S = 1 + 2 + 3 \cdots + n $ then
$S = n + (n-1) + (n-2) + \cdots + 2 + 1 $ add them by terms (first term with the first second with the second we get
$2S = (n+1) + (n+1) + (n+1) \cdots + (n+1) $ but we have $n$ of $(n+1)$ so
$2S = n(n+1) \Rightarrow S = \frac{n(n+1)}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490653",
"timestamp": "2023-03-29T00:00:00",
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Continuity of $\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ at (0, 0) I am having trouble proving that $\dfrac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ is continuous at $(0, 0)$ if we set the value at $(0, 0)$ to be $0$.
I don't see a way to prove this as I cannot factor this into partial fractions.
| As an another approach:
$$\bigg|\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}\bigg|=|x|\bigg|\frac{x^4-4x^2y^2-y^4}{(x^2+y^2)^2}\bigg|=|x|\bigg|\frac{(x^2+y^2)^2-6x^2y^2-2y^4}{(x^2+y^2)^2}\bigg| \\ \le |x|\bigg(1+\bigg|\frac{6x^2y^2+2y^4}{(x^2+y^2)^2}\bigg|\bigg)\le |x|\bigg(1+\bigg|\frac{6x^2y^2+6y^4}{(x^2+y^2)^2}\bigg|\bigg) \\ =|x|\bigg(1+\bigg|\frac{6y^2}{x^2+y^2}\bigg|\bigg)\le |x|(1+6)=7|x|.$$
Therefore, the limit is zero as $x$ approaches zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verify $\int\sec x\ dx=\frac12 \ln \left\lvert\frac{1+\sin x}{1-\sin x}\right\rvert + C$ Question says it all, how can I verify the following?
$$\int\sec x\ dx=\frac12 \ln \left|\frac{1+\sin x}{1-\sin x}\right| + C$$
| We have
$$\begin{array}{lll}
\sec^2 x&=& \tan^2 x + 1\\
\sec^2 - \tan^2x&=& 1\\
(\sec x + \tan x)(\sec x - \tan x)&=&1\\
(\sec x + \tan x)^2(\sec x - \tan x)&=&\sec x + \tan x\\
(\sec x + \tan x)^2&=&\frac{\sec x + \tan x}{\sec x - \tan x}\\
\mid\sec x + \tan x\mid&=&\bigg(\frac{\sec x + \tan x}{\sec x - \tan x}\bigg)^\frac{1}{2}\\
\ln\mid\sec x + \tan x\mid&=&\ln\bigg(\frac{\sec x + \tan x}{\sec x - \tan x}\bigg)^\frac{1}{2}\\
\ln\mid\sec x + \tan x\mid&=&\frac{1}{2}\ln\bigg(\frac{\sec x + \tan x}{\sec x - \tan x}\bigg)\\
\end{array}$$
Now all that needs to be done is to show that
$$\int\sec x dx = \ln\mid\sec x + \tan x\mid+C$$
Hint
$$\frac{\sec x + \tan x}{\sec x + \tan x}=1\implies \sec x = \frac{\sec^2 x+\sec x\tan x}{\sec x+\tan x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A polynomial problem related to lx^2 + nx + n If the roots of $lx^2 + nx + n = 0$ are in the ratio $p:q$, find the value of $\sqrt{\frac{p}{q}}$ + $\sqrt{\frac{q}{p}}$ + $\sqrt{\frac{n}{l}}$.
How to go about this problem?
| Notice,
let $kp$ & $kq$ be the roots of $lx^2+nx+n=0$ hence satisfying the equation we get $$lp^2k^2+npk+n=0\tag 1$$
$$lq^2k^2+nqk+n=0\tag 2$$
solving (1) & (2) for $k$, as follows $$\frac{k^2}{n^2p-n^2q}=\frac{k}{nlq^2-nlp^2}=\frac{1}{nlp^2q-nlpq^2}$$
$$\implies k^2=\frac{n^2p-n^2q}{nlp^2q-nlpq^2}=\frac{n^2(p-q)}{nlpq(p-q)}=\frac{n}{lpq}\tag 3$$
$$\implies k=\frac{nlq^2-nlp^2}{nlp^2q-nlpq^2}=\frac{nl(q+p)(q-p)}{nlpq(p-q)}=\frac{-(p+q)}{pq}\tag 4$$ from (3) & (4), we get
$$\left(\frac{-(p+q)}{pq}\right)^2=\frac{n}{lpq}$$
$$\frac{p^2+q^2+2pq}{pq}=\frac{n}{l}$$
$$\frac{p}{q}+\frac{q}{p}+2=\frac{n}{l}$$
$$\left(\sqrt{\frac{p}{q}}\right)^2+\left(\sqrt{\frac{q}{p}}\right)^2+2\left(\sqrt{\frac{p}{q}}\right)\left(\sqrt{\frac{q}{p}}\right)=\frac{n}{l}$$
$$\left(\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}\right)^2=\frac{n}{l}$$
$$\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}=\pm\sqrt{\frac{n}{l}}$$
taking negative sign, we get
$$\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}=-\sqrt{\frac{n}{l}}$$
$$\color{blue}{\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}}=\color{red}{0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to find this limit : lim x to infinity how can I find this limit which is become infinite
$\lim _{x\to \infty }\left(x(\sqrt{x^2+1}-x)\right)$
can I use conjugate method ?
that what I'm doing until now
$= x\left(\frac{\sqrt{x^2+1}-x}{1}\cdot \:\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)\:$
$= x\left(\frac{1}{\sqrt{x^2+1}+x}\right)$
$= \left(\frac{x}{\left(\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}+\frac{x}{x}\right)}\right)\:$
$= \:\left(\frac{x}{\left(\sqrt{1+\frac{1}{x^2}}+1\right)}\right)\:$
| Yes, you can use conjugates for $\left(\sqrt{x^2+1}-x\right)$ as follows
Notice,
$$\lim_{x\to \infty}x\left(\sqrt{x^2+1}-x\right)$$
$$=\lim_{x\to \infty}\frac{x\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\left(\sqrt{x^2+1}+x\right)}$$
$$=\lim_{x\to \infty}\frac{x(x^2+1-x^2)}{\sqrt{x^2+1}+x}$$
$$=\lim_{x\to \infty}\frac{x}{\sqrt{x^2+1}+x}$$
$$=\lim_{x\to \infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}$$
$$=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find all the solutions of diophantine eq: $x^3-2xy^2+y^3-s^2=0$ Given $x,y,s$ are natural numbers:
$$x^3-2xy^2+y^3-s^2=0$$
I found the solutions using wolfram alpha
$$(x,y,s) = (1,2,1), (6,10,4), (4,8,8)$$
But how do I prove these are the only solutions? Any tips or reference to papers that study this diophantine equation would be much appreciated.
| There is a beautiful parameterization by Lagrange and Legendre:
$$x^3 + a x^2y + b x y^2 + c y^3 = z^2\tag1$$
where,
$$x = u^4 - 2b u^2 v^2 - 8c u v^3 + (b^2 - 4a c)v^4$$
$$y = 4v(u^3 + a u^2v + b u v^2 + c v^3)$$
and $z$ is a $6$-deg polynomial.
In your case, with $a,b,c = 0,-2,1$, we have,
$$x^3 - 2x y^2 + y^3 = z^2$$
where,
$$x = u^4 + 4 u^2 v^2 - 8 u v^3 + 4 v^4$$
$$y = 4 v (u^3 - 2 u v^2 + v^3)$$
$$z = u(u^5 - 10 u^3 v^2 + 20 u^2 v^3 - 20 u v^4 + 8 v^5)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\frac xy + \frac yx \ge 2$ I am practicing some homework and I'm stumped.
The question asks you to prove that
$x \in Z^+, y \in Z^+$
$\frac xy + \frac yx \ge 2$
So I started by proving that this is true when x and y have the same parity, but I'm not sure how to proceed when x and y have opposite partiy
This is my proof so far for opposite parity
$x,y \in Z^+ $ $|$ $x \gt 0,$ $y \gt 0$. Let x be even $(x=2a, $ $ a \in Z^+)$ and y be odd $(y=2b+1, $ $b \in Z^+)$. Then,
$\frac xy + \frac yx \ge 2$
$\frac {2a}{2b+1} + \frac {2b+1}{2a} \ge 2$
$\frac {2b^2 + 4a^2 + 4a + 1}{2b(2a+1)} \ge 2$
$4b^2 + 4a^2 +4a + 1 \ge 4b(2a+1)$
$4b^2 + 4a^2 + 4a +1 \ge 8ab + 4b$
$4b^2 - 4b + (2a + 1)^2 \ge 8ab$
$(2b-1)^2 + 1 + (2a+1)^2 \ge 8ab$
I feel like this is the not the correct way to go about proving it, but I can't think of a better way to do it. Does anyone have any suggestions? Just hints please, not a solution.
| If $x=y,$ then it is true.
Suppose $x>y$ then there is exists positive integer $n$ such that $x=y+n.$
This implies $\frac{x}{y}=1+\frac{n}{y}$ and $\frac{y}{x}=1-\frac{n}{x}$.
Adding these equations, we obtain
\begin{align}
\frac{x}{y}+ \frac{y}{x}=2+n\left(\frac{1}{y}-\frac{1}{x}\right)=2+n\frac{x-y}{xy}> 2\quad as\quad x>y.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
| Reducing to the same denominator, we get
$$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}\iff -\frac{4x+3}{(x+2)(x-1)(x-3)}>0 $$
Now the sign of $-\dfrac{4x+3}{(x+2)(x-1)(x-3)}$, when defined, is the sign of the product $p(x)=-(4x+3)(x+2)(x-1)(x-3)$, which by Bolzano's theorem can change sign only at $-2, -\dfrac34,1$ and $3$. As $p(0)=-\frac12<0$, we only have to alternate signs to obtain the following table:
$$\begin{array}{*{9}{c}}
&-2&&-\frac34&&1&&2& \\
\hline
-&\Vert& + &0&\color{red}{\mathbf-}&\Vert&+&\Vert&-\end{array}, $$
whence the solutions:
$$(-2,-\frac34)\cup(1,2).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve using AM GM Inequality if possible Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to?
I don't really know how to solve this. any methods would be welcome. I was solving a couple of AM GM inequality questions and I'm assuming this should be solved in a similar way. Correct me.
| This method doesn't use the $AM-GM$ inequality.
$a^3 +b^3 +c^3 = (a+b+c)^3 -3(a+b+c)(ab+ac+bc)+3abc$
Applying it to your question, we have
$\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$
= $(\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x})^3 -3(\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x})(\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z}) +3$
=$(7^3)-3(7)(9) +3$
=$157$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495357",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Distributing multiplication of rational functions I am having trouble distributing with fractions. This
$$ \left(\frac{1}{(x + 3)} + \frac{(x + 3)}{(x - 3)}\right)\, (9 - x^2) = -\frac{(x^2-3)}{9-x^2}(9-x^2) $$
has the answer $\left\{\left[x = - \frac{9}{7}\right]\right\}$.
I start solving by cancelling $(9-x^2)$ on the right hand side, and multiplying $(9-x^2)$ with the terms on the left hand side by foiling (multiplying term by term). So the first step is
$$9\times\frac{1}{x+3}+\frac{9 (x+3)}{x-3}-x^2\times\frac{1}{x+3}-\frac{x^2 (x+3) }{x-3}=-x^2-3$$
This is apparently wrong because it has another answer, $\left\{\left[x = - \frac{3}{7}\right]\right\}$ What am I doing wrong?
| $\left(\frac{1}{x+3} + \frac{x+3}{x-3} \right) ( 9 -x^2) -x^2 -3$
It will be
$\frac{9-x^2}{x+3} + \frac{(x+3)(9-x^2)}{x-3} = -x^2 - 3 $
after that
$\frac{(9-x^2)(x-3)}{(x+3)(x-3)} + \frac{(x+3)(x+3)(9-x^2)}{(x-3)(x+3)} = -x^2 -3 $
Note that $9-x^2 = -(x^2 - 9) = -(x+3)(x-3)$, i.e., you can cancel it with the denominator
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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What is the number of solution to $x^2\equiv 1\pmod{p^k}$? We learned about quadratic reciprocity and how it relates to the number of solution to $x^2 \equiv 1 \pmod{p}$.
I was wondering, is there a general formula for the
number of solutions modulo $p^k$ to $x^2 \equiv 1 \pmod{p^k}$?
Thank you very much!
| If $p$ is odd, then $x^2\equiv 1\pmod{p}$ has exactly two solutions, because if $p\mid x^2-1=(x+1)(x-1)$, then either $p\mid x+1$ or $p\mid x-1$ (by Euclid's Lemma). The solutions are $x\equiv \pm 1\pmod{p}$.
$x^2\equiv 1\pmod{p^k},\, k\ge 2$ for an odd prime $p$ too has exactly two solutions $x\equiv \pm 1\pmod{p^k}$. This is because if $p^k\mid x^2-1=(x+1)(x-1)$, then assume for contradiction $p\mid x+1$ and $p\mid x-1$. But then $p\mid (x+1)-(x-1)=2$, so $p=2$; contradiction. We get either $p^k\mid x+1$ or $p^k\mid x-1$, i.e. $x\equiv \pm 1\pmod{p^k}$.
If $p=2$, then $x^2\equiv 1\pmod{2}$ has exactly one solution, namely $x\equiv 1\pmod{2}$. Also $x^2\equiv 1\pmod{2^k},\, k\ge 2$ has all the solutions given by $x\equiv \pm 1\pmod{2^{k-1}}$. This is because if $2^k\mid x^2-1=(x+1)(x-1)$, then $x+1,x-1$ are both even and $\gcd(x+1,x-1)=2$, so either $2^{k-1}\mid x+1$ or $2^{k-1}\mid x-1$.
$x\equiv \pm 1\pmod{2^{k-1}}$ gives $2$ solutions when $k=2$ (namely $x\equiv 1,3\pmod{4}$) and $4$ solutions given by $x\equiv 2^{k-1}l\pm 1\pmod{2^k},\, l\in\{0,1\}$ when $k\ge 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Expanding brackets problem: $(z - 1)(1 + z + z^2 + z^3)$ I have:
$(z - 1)(1 + z + z^2 + z^3)$
As, I have tried my own methods and enlisted the help on online software, but as well as them not all arriving at the same solution, I can't follow their reasoning.
I tried to gather all the like terms:
$(z - 1)(z^6+1)$
And then thought it may offer a route to a difference or two cubes if I shifted powers across the brackets, but that's $z^7$ in total, isn't it? I can't just do $z^7-1$, can I?
If I try to expand, I see:
$(z - 1)(z^6 + 1) = z^7+z-z^6-1$
Well...
$z^7-z^6=z$
Add the other z leaves:
$z^2-1$?
I expect this is all wrong.
Will you help please?
On a side note, I have difficulty with understanding how I'm expected to tag this post accurately when I can't used "expand", "multiply", or use any of the subjects it covers as tags?
| I like to do factorizations like this by writing
$$
z^7-1=(z-1)(\cdots)
$$
and try to figure out what should replace the dots. First, we'd like to have a factor of $z^7$ in the end result, so the $z$ on the right side should be multiplied by $z^6$ to get this. Then, we write
$$
z^7-1=(z-1)(z^6+\cdots).
$$
At this point, by the distributive law, the right side will multiply out to $z^7-z^6+\cdots$. Since we don't want the $z^6$ in the final product, it must cancel with something. We can cancel the $-z^6$ by multiplying $z$ by $z^5$ to get
$$
z^7-1=(z-1)(z^6+z^5+\cdots)
$$
because after distributing, the product is $z^7-z^6+z^6-z^5+\cdots$ and the $z^6$'s cancel. Now, we're left with $z^7-z^5+\cdots$, so the $-z^5$ must cancel with another term, since $z^5=z\cdot z^4$, we have
$$
z^7-1=(z-1)(z^6+z^5+z^4+\cdots).
$$
At this point, the product is $z^7-z^4+\cdots$. In order to cancel the $-z^4$, we introduce a $z^3$, which results in an extra $-z^3$. This extra term is cancelled with a $z^2$, which results in an extra $-z^2$ in the product. This $-z^2$ is cancelled with a $z$, but that gives an extra $-z$ in the product. The extra $-z$ is cancelled with a $-1$. Therefore, we have
$$
z^7=(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1+\cdots).
$$
But, after multiplying everything out, we find that the left side already matches the right side and there is nothing more in the dots! Therefore,
$$
z^7=(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ How can one solve the equation :
$[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ ?
Such that $[x]$ is the integer part of $x$.
By definition :
$[\frac{x^2-x+1}{2}]<=\frac{x^2-x+1}{2}<[\frac{x^2-x+1}{2}]+1$.
So :
$\frac{x-1}{3}<=\frac{x^2-x+1}{2}<\frac{x-1}{3}+1$.
1)
$\frac{x-1}{3}<=\frac{x^2-x+1}{2}$ <=> $3x^2-5x+5>0$.
2)
$\frac{x^2-x+1}{2}<\frac{x-3}{2}+1$ <=> $3x^2-5x-1<0$.
How can I solve the system ?
| Write the original equation as:
$$\left\lfloor\frac{x^2-x+1}{2}\right\rfloor=\frac{x-1}{3} \tag{1}$$
By definition of the floor function, both sides of the equation must be integers, i.e. $x\in\mathbb{Z}$. So then $x^2-x+1$ is an odd integer, whether $x$ is even or odd. Therefore
$$\left\lfloor\frac{x^2-x+1}{2}\right\rfloor=\frac{x^2-x+1}{2}-\frac{1}{2}=\frac{x^2-x}{2}$$
Substituting back into (1):
$$\frac{x^2-x}{2}=\frac{x-1}{3} \implies \frac{x(x-1)}{2}=\frac{x-1}{3} \implies x=\tfrac{2}{3}\text{ or }x=1$$
Of these $x=1$ is the only integer potential solution, and checking against (1) shows that it is a solution to the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the determinant of this $5 \times 5$ matrix? How can I find the determinant of this matrix?
I know in matrix $3 \times 3$
$$A= 1(5\cdot 9-8\cdot 6)-2 (4\cdot 9-7\cdot 6)+3(4\cdot 8-7\cdot 5) $$
but how to work with a $5\times 5$ matrix?
| Once the matrix starts getting large, it can be easier to use row- or column-reduction to find the determinant, especially if there aren’t many sparse rows or columns to take advantage of in iterated Laplace expansions. This technique makes use of the facts that swapping two rows/columns changes the sign of the determinant, multiplying a row/column by a scalar multiplies the determinant by the same amount, and adding a scalar multiple of a row/column to another leaves the determinant unchanged.
For your matrix, we can start by adding $3$ times the first row to the fourth: $$\begin{vmatrix}
1 & 2 & 3 & 4 & 1 \\
0 & -1 & 2 & 4 & 2 \\
0 & 0 & 4 & 0 & 0 \\
0 & 0 & 0 & 0 & 7 \\
0 & 0 & 1 & 1 & 1 \notag
\end{vmatrix}$$
Clear the first and second rows:$$\begin{vmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 4 & 0 & 0 \\
0 & 0 & 0 & 0 & 7 \\
0 & 0 & 1 & 1 & 1 \notag
\end{vmatrix}$$
Clear the third and last columns:$$\begin{vmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 4 & 0 & 0 \\
0 & 0 & 0 & 0 & 7 \\
0 & 0 & 0 & 1 & 0 \notag
\end{vmatrix}$$
Swap the fourth and fifth rows:$$-1\cdot\begin{vmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 4 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 7 \notag
\end{vmatrix}$$
At this point, we can stop and multiply the diagonal elements together to find the determinant, which is 28.
Update: I should note that I did much more work than necessary above. You don’t need to perform a complete row-reduction—it’s enough to get the matrix into upper-triangular form since the determinant of such a matrix is also the product of its main diagonal elements. After the first step above, we can skip directly to adding $-\frac14$ times the third row to the last: $$\begin{vmatrix}
1 & 2 & 3 & 4 & 1 \\
0 & -1 & 2 & 4 & 2 \\
0 & 0 & 4 & 0 & 0 \\
0 & 0 & 0 & 0 & 7 \\
0 & 0 & 0 & 1 & 1 \notag
\end{vmatrix}$$ and then swap the last two rows$$-1\cdot\begin{vmatrix}
1 & 2 & 3 & 4 & 1 \\
0 & -1 & 2 & 4 & 2 \\
0 & 0 & 4 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 7 \notag
\end{vmatrix}.$$ This obviously has the same determinant as the result of the full row-reduction above.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$A+B+C+7$ is perfect square where $A,B,C$ numbers with repeating digits. Let $A=\underbrace{444...4}_{2m},$ $B=\underbrace{222...2}_{m+1}$ and $C=\underbrace{888...8}_{m}$.
Prove that $A+B+C+7$ is perfect square.
How to prove that problem?
I have checked it for small values of $m$ and it's true but I Can't prove it in general case.
| Proof:
$A=\underbrace{444...4}_{2m}=\dfrac{4}{9}\cdot \underbrace{999...9}_{2m}=\dfrac{4}{9}(10^{2m}-1).$
$B=\underbrace{222...2}_{m+1}=\dfrac{2}{9}\cdot \underbrace{999...9}_{m+1}=\dfrac{2}{9}(10^{m+1}-1).$
$C=\underbrace{888...8}_{m}=\dfrac{8}{9}\cdot \underbrace{999...9}_{m}=\dfrac{8}{9}(10^{m}-1).$
Hence $$A+B+C+7=\dfrac{4}{9}(10^{2m}-1)+\dfrac{2}{9}(10^{m+1}-1)+\dfrac{8}{9}(10^{m}-1)+7=$$$$=\dfrac{1}{9}(4\cdot 10^{2m}+28\cdot 10^m+49)=\dfrac{1}{9}(2\cdot 10^{m}+7)^2=\left(\dfrac{2\cdot 10^m+7}{3}\right)^2$$ but $$2\cdot 10^m+7\equiv 0 \pmod{3}$$ since $10^m\equiv 1 \pmod{3} \Rightarrow 2\cdot 10^m\equiv 2 \pmod{3} \Rightarrow 2\cdot 10^m+7\equiv 2+7\equiv 0 \pmod{3}.$
Q.E.D.
P.S. By the way very nice problem.
| {
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"source": "stackexchange",
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In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?
I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!
| Hint:
Notice that $$x+3-4\sqrt{x-1}=x-1-4\sqrt{x-1}+4=(\sqrt{x-1}-2)^2$$
and
$$x+8-6\sqrt{x-1}=x-1-6\sqrt{x-1}+9=(\sqrt{x-1}-3)^2=(3-\sqrt{x-1})^2$$
After, you can try by cases.
| {
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"source": "stackexchange",
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Can a pre-calculus student prove this?
a and b are rational numbers satisfying the equation $a^3 + 4a^2b = 4a^2 + b^4$
Prove $\sqrt a - 1$ is a rational square
So I saw this posted online somewhere, and I kind of understand what the question is saying. I'm interesting in doing higher order mathematics but don't quite have the mathematical skills yet to grasp them (hence, precalculus) but is this problem way more complex than it seems or are the tools within the reach of a precalculus student?
| As randomgirl and Michael point out, there are counterexamples, such as $a=b=0$. However, it is true when $a \ge 1$. In principle it can be proved with just pre-calculus mathematics, but only an exceptional pre-calculus student could prove it.
Suppose $a^3 + 4a^2b = 4a^2+b^4$, where $a$ and $b$ are rational numbers with $a \ge 1$. We would like to show that $\sqrt{a}-1$ is the square of a rational number. Let $t = \sqrt{\sqrt{a}-1}$. Then $t$ is a real number because $a \ge 1$, and we would like to show that $t$ is rational. In terms of $t$, we have $a = (1+t^2)^2$, and a first question is whether we can also solve for $b$ in terms of $t$.
Substituting $a = (1+t^2)^2$ into $a^3 + 4a^2b - (4a^2+b^4)$ and factoring, we find that either $b = t^3+t^2+t+1$, $b = -t^3+t^2-t+1$, or $b^2 + 2(1+t^2)b + t^6 + 5t^4 + 7t^2 + 3 = 0$. The latter is a quadratic equation in $b$ with discriminant $-4(t^2+2)(t^2+1)^2$, which is negative, so it has no real root. Thus, $b$ must be either $t^3+t^2+t+1$ or $-t^3+t^2-t+1$, and without loss of generality we can assume $b = t^3+t^2+t+1$ by switching the sign of $t$ if necessary, because replacing $t$ with $-t$ does not affect the equation $a=(1+t^2)^2$.
Thus, the problem reduces to showing that if $a = (1+t^2)^2$ and $b = t^3+t^2+t+1$ with $a$ and $b$ rational, then $t$ must be rational as well. We will prove this by solving for $t$ in terms of $a$ and $b$. To do so, first note that $b = (t+1)(t^2+1)$, from which it follows that $b^2/a = (t+1)^2$. Then $b^2/a+b-2 = t(t^2+2t+3)$, while $b^2/a + 2 = t^2+2t+3$ (which is strictly positive because it equals $(t+1)^2+2$). Putting this together, we find that $t = (b^2/a+b-2)/(b^2/a+2)$, which is rational because $a$ and $b$ are rational.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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$\epsilon-N$ for $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$. I want to prove using $\epsilon-N$ that $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Firstly we will find a sufficently large $n\in \Bbb N$:
\begin{align*}
\epsilon\quad\gt&\quad\left|\frac{3n+1}{2n+5} - \frac32 \right|\\
=&\quad \left|\frac{3n+1 -3n-7.5}{2n+5}\right|\\
=&\quad \left|\frac{-13}{4n+10}\right|\\
\implies&\quad \epsilon \gt \frac{13}{4n+10}\\
\implies&\quad \frac1\epsilon \lt \frac{4n+10}{13}\\
\implies&\quad \frac14\left(\frac{13}\epsilon-10\right) \lt n\\
\end{align*}
So $\forall \epsilon\gt 0, \exists N(\epsilon)>\frac14\left(\frac{13}\epsilon-10\right)$ such that $\forall n\in N, n>N$, $a_n=|\frac{3n+1}{2n+5}-\frac32|\lt \epsilon$
Proof: $\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt} \left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|\lt\epsilon$
Assuming this is even right. Everything feels good except the $\color{red}{\lt}$ in red, and then when I check, since we have
\begin{align*}
&\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt}\left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|,\quad n\gt N\\
&\implies \left|\frac{3n+1}{2n+5} - \frac32\right| - \left|\frac{3(n+1)+1}{2(n+1)+5} - \frac{3}{2}\right|\gt 0
\end{align*}
But this fails for all $n$? What did I do wrong?
Also, since $N\in\Bbb N$, we have to choose our $N$ so it is a natural number simultaneous to being greater than $\frac14\left(\frac{13}\epsilon-10\right)$, how do I ensure that subbing in $\frac14\left(\frac{13}\epsilon-10\right)$ is greater than subbing in the natural number that is the ceiling of this?
| What is the problem?
$\large\mid{3n+1\over2n+5}-{3\over2}\mid={13\over4n+10}$
$\large\mid{3N+1\over2N+5}-{3\over2}\mid={13\over4N+10}$
$\large\mid{3(n+1)+1\over2(n+1)+5}-{3\over2}\mid={13\over4(n+1)+10}$
So for $n>N$ the second one is obviously larger than the first one and the first one is obviously larger than the third one.
| {
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Sum of $(a+\frac{1}{a})^2$ and $(b+\frac{1}{b})^2$ Prove that:
$$
\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}
$$
if $a,b$ are positive real numbers such that $a+b=1$.
I have tried expanding the squares and rewriting them such that $a+b$ is a term/part of a term but what I get is completely contradictory to what is asked to prove
| Without loss of generality we can choose $a=sin^2x$ and $b=cos^2x$ for $x \in \left(0 \: \frac{\pi}{2}\right)$
Now $$a^2+b^2=sin^4x+cos^4x=1-2sin^2xcos^2x=1-\frac{sin^22x}{2}=\frac{3+cos4x}{4} \ge \frac{1}{2}$$ with Equality at $x=\frac{\pi}{4}$
Also $$2sinxcosx \le 1$$ $\implies$
$$\frac{1}{sin^4xcos^4x} \ge 16$$
Now by $CS$ Inequality $$\left(\frac{1}{a^2}+\frac{1}{b^2}\right)(1^2+1^2) \ge\left(\frac{1}{a}+\frac{1}{b}\right)^2=(sec^2x+cosec^2x)^2=(sec^2xcosec^2x)^2=sec^4xcosec^4x=\frac{1}{sin^4xcos^4x} \ge 16$$ Thus
$$\left(\frac{1}{a^2}+\frac{1}{b^2}\right)(1^2+1^2) \ge 16$$
$$\left(\frac{1}{a^2}+\frac{1}{b^2}\right) \ge 8$$
Thus $$\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2 \ge \frac{1}{2}+8+4=\frac{25}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Complex numbers lies on ellipse $ z $ is a variable complex number such that $ |z|=2$. Show that the point $ z+ \frac{1}{z} $ lies on an ellipse of eccentricity $\frac{4}{5}$ in the complex plane.
| As you know $|z|=2$, you can write $z$ as a function of its argument $$z = 2\cos\varphi + 2i\sin\varphi $$
and then $$\begin{align}\frac 1z & = \frac 12\cos(-\varphi) + \frac i2\sin(-\varphi) \\ & = \frac 12\cos\varphi - \frac i2\sin\varphi\end{align}$$
so $$\begin{align}z+\frac 1z & = \left(2+\frac 12\right)\cos\varphi + i\left(2-\frac 12\right)\sin\varphi \\ & = \frac 52\cos\varphi + i\frac 32\sin\varphi \end{align}$$
Now you need only to show that the locus of $(x,y)=(\frac 52\cos\varphi, \frac 32\sin\varphi)$ is an ellipse of a given eccentricity.
| {
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a conjecture of two equivalent q-continued fractions related to the reciprocal of the Göllnitz-Gordon continued fraction A111374-OEIS Given the square of the nome $q=e^{2i\pi\tau}$ and ramanujan theta function $f(a,b)=\sum_{k=-\infty}^{\infty}a^{k(k+1)/2}b^{k(k-1)/2}$ with $|q|\lt1$, define,
$$\begin{aligned}M(q)=\cfrac{1-q^3}{1-q^2+\cfrac{q^2(1-q^{-1})(1-q^5)}{1-q^6+\cfrac{q^4(1-q)(1-q^7)}{1-q^{10}+\cfrac{q^6(1-q^3)(1-q^{9})}{1-q^{14}+\ddots}}}}\overset{\color{blue}{?}}=\prod_{n=1}^\infty\frac{(1-q^{8n-3})(1-q^{8n-5})}{(1-q^{8n-1})(1-q^{8n-7})}\\[1.5mm]&\end{aligned}$$
$$=\frac{f(-q^3,-q^5)}{f(-q,-q^7)}$$
and
$$\begin{aligned}N(q)=\cfrac{1-q^3}{1+q^2-\cfrac{q^2(1+q^{-1})(1+q^5)}{1+q^6+\cfrac{q^4(1-q)(1-q^{7})}{1+q^{10}-\cfrac{q^6(1+q^3)(1+q^{9})}{1+q^{14}+\ddots}}}}\overset{\color{blue}{?}}=\prod_{n=1}^\infty\frac{(1-q^{8n-3})(1-q^{8n-5})}{(1-q^{8n-1})(1-q^{8n-7})}\\[1.5mm]&\end{aligned}$$
$$=\frac{f(-q^3,-q^5)}{f(-q,-q^7)}$$
Q: How do we prove that the two q-continued fractions are equal to the q-products such that the continued fractions are equivalent $M(q)=N(q)$?
| For $n\in\mathbb{N}$, let $q_n=\exp\frac{2\pi\mathrm{i}\tau}{n}$, so $q_n^n=q$.
Use formula $(***)$ from
that post,
but with $q$ replaced with $q^2$, so it reads
$$\small\cfrac{1}{1-q^2+\cfrac{(a+bq^2)(aq^2+b)}
{1-q^6+\cfrac{q^2(a+bq^4)(aq^4+b)}
{1-q^{10}+\cfrac{q^4(a+bq^6)(aq^6+b)}{1-q^{14}+\cdots}}}}
= \frac{(-a^2q^6;q^8)_\infty\,(-b^2q^6;q^8)_\infty}
{(-a^2q^2;q^8)_\infty\,(-b^2q^2;q^8)_\infty}
\qquad(ab=q^2)\tag{1}$$
Now set $a=-\mathrm{i}q_2^5$, $b=\mathrm{i}q_2^{-1}$.
This implies $ab=q^2$, $a/b=-q^3$, $-a^2=q^5$, $-b^2=q^{-1}$, hence
$$\underbrace{\cfrac{1}{1-q^2+\cfrac{q^2(1-q^{-1})(1-q^5)}
{1-q^6+\cfrac{q^4(1-q)(1-q^7)}
{1-q^{10}+\cfrac{q^6(1-q^3)(1-q^9)}{1-q^{14}+\cdots}}}}}_{\frac{M(q)}{1-q^3}}
= \frac{(q^{11};q^8)_\infty\,(q^5;q^8)_\infty}
{(q^7;q^8)_\infty\,(q;q^8)_\infty}$$
which implies your statement for $M(q)$.
In $(1)$, replace $q$ with $\mathrm{i}q$. The formula requires $ab=-q^2$ then,
and odd powers of $q^2$ in $(1)$ change sign.
Now set $a=-q_2^5$, $b=q_2^{-1}$.
This implies $ab=-q^2$, $a/b=-q^3$, $-a^2=-q^5$, $-b^2=-q^{-1}$, hence
$$\underbrace{\cfrac{1}{1-q^2-\cfrac{q^2(1+q^{-1})(1+q^5)}
{1+q^6+\cfrac{q^4(1-q)(1-q^7)}
{1+q^{10}-\cfrac{q^6(1+q^3)(1+q^9)}{1+q^{14}+\cdots}}}}}_{\frac{N(q)}{1-q^3}}
= \frac{(q^{11};q^8)_\infty\,(q^5;q^8)_\infty}
{(q^7;q^8)_\infty\,(q;q^8)_\infty}$$
which implies your statement for $N(q)$.
From a bird's view, this works because the right-hand side in (1) is preserved
if the signs of $q^2$, $a^2$ and $b^2$ are changed simultaneously.
The corresponding left-hand sides must then be equivalent too.
| {
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Coefficient of $x^{5}$ in expansion of determinant How to find the coefficient of $x^{5}$ in expansion of determinant $\begin{vmatrix}
7 & 2 & 1 & 3 & x & 7 & 2\\
2 & 8 & 7 & x & 5 & 2 & 8\\
1 & 7 & x & 8 & 4 & 1 & 7\\
3 & x & 8 & 10 & 6 & 3 & 9\\
x & 5 & 4 & 6 & 2 & 10 & x\\
7 & 2 & 1 & 3 & 10 & x & 2\\
2 & 8 & 7 & 9 & x & 2 & 8
\end{vmatrix}$?
The only way I see is to write all possible combinations with $x^{5}$ like $-(2*2)*x^{5} + (7 * 2)*x^{5} ...$
| Here is one approach which should work:
Let
$$A(X) =\begin{vmatrix}
7 & 2 & 1 & 3 & x & 7 & 2\\
2 & 8 & 7 & x & 5 & 2 & 8\\
1 & 7 & x & 8 & 4 & 1 & 7\\
3 & x & 8 & 10 & 6 & 3 & 9\\
x & 5 & 4 & 6 & 2 & 10 & x\\
7 & 2 & 1 & 3 & 10 & x & 2\\
2 & 8 & 7 & 9 & x & 2 & 8
\end{vmatrix}$$
Then, you are looking for $\frac{A^{5}(0)}{5!}$.
Since determinants are defined by products, the rule of differentiating a determinant is easy to find and prove: to differentiate a determinant, you differentiate a column, and keep the rest unchanged, and add all matrices together.
In this case, differentiating a column twice will give you zero. Because of this you get
$$A^{(5)}(x) = \sum_{K} A_K$$
where $K$ runs over all $5$ elements subset of $\{ 1,2,3,4,5,6,7 \}$ and
$A_K$ is the determinant obtained from $A$ by differentiating the columns corresponding to $K$.
For example, if $K= \{ 1,2 ,4,5,7\}$ we have
$$A_K= \begin{vmatrix}
0 & 0 & 1 & 0 & 1 & 7 & 0\\
0 & 0 & 7 & 1 & 0 & 2 & 0\\
0 & 0 & x & 0 & 0 & 1 & 0\\
0 & 1 & 8 & 0 & 0 & 3 & 0\\
1 & 0 & 4 & 0 & 0 & 10 & 1\\
0 & 0 & 1 & 0 & 0 & x & 0\\
0 & 0 & 7 & 0 & 1 & 2 & 0
\end{vmatrix}$$
As at least $4$ of the five columns will have six zeroes, and the last one has at least 5 zeroes, your determinants will be easy to calculate. Esspecially if you plug in $x=0$ before calculating each $A_K$.
Unfortunately there are 21 of them, so the solution will be very vert long but simple.
| {
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Number of ordered positive rationals (x,y,z) satisfying following conditions. How many ordered triples $(x,y,z)$ of positive rational numbers satisfy the conditions: $x+y+z$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, and $xyz$ are all integers.
| Then $yz+xz+xy$ is also an integer. It follows from the Vieta relations that $x$, $y$, and $z$ are the roots of a monic cubic with integer coefficients, so they are integers.
Thus we want all triples $(x,y,z)$ of positive integers such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is an integer.
Now it is a short search, since the smallest of the integers must be $\le 3$, and indeed is $3$ only in the case $x=y=z=3$.
Smallest is $2$ yields $(2,3,6)$ and $(2,4,4)$ (and permutations). And smallest is $1$ again yields essentially two solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the value of integers of a logarithmic function. Expressing $\frac{2x^2-9x-6}{x(x^2-x-6)}$ in partial fraction would give you: $\frac{16}{5(x+2)}$-$\frac{1}{5(x-3)}$-$\frac{1}{x}$
Given that $\int_{4}^6\frac{2x^2-9x-6}{x(x^2-x-6)}dx=ln\frac{m}{n},$ determine the values for the integers m and n.
| It turns out the partial fraction decomposition is incorrect.
It can be verified that instead,
$$\begin{align}\int_4^6 \frac{2x^2 - 9x - 6}{x(x^2 - x - 6)} dx &= \int_4^6\frac{1}{x} + \frac{2}{x + 2} - \frac{1}{x-3}dx\\
&=\left[\ln x + 2 \ln (x + 2) - \ln (x - 3)\right]_4^6\\
&= \ln\frac{6}{4} + 2\ln \frac{8}{6} - \ln\frac{3}{1}\\
&= \ln \frac{3}{2} + \ln\frac{16}{9} - \ln 3\\
&= \ln \frac{8}{9}\end{align}$$
| {
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how to solve the following mordell equation:$ y^2 = x^3 - 3$ i just started solving mordell's equations and get a little bit stuck. For example:
$ y^2 = x^3 - 3$.
I know that $x$ must be odd, for if $x$ is even $y^2 \equiv 5 \pmod{8}$. So $x \equiv \{1,3\} \pmod{4}$. Now note the following if we add 4 at both sides:
$y^2 + 4 = (x+1)(x^2 -x +1)$. The right hand side is the same as $(x-1)^2 + x$
I now know $(x^2 -x +1)$ is divisible by a prime divisor $3 \pmod{4}$ if $x \equiv 3 \pmod {4}$ and divisible by $1 \pmod{4}$ if $x \equiv 1 \pmod{4}$.
So we get $y^2 + 4 \equiv 0\pmod{p}$, which is solvable if $\left(\frac{-2}{p}\right)$ = 1. (I don't think this step is valid). this is only true if the following occurs:
$p \equiv 1 \pmod{4}$ and $p \equiv +- 1\pmod{8}$.
or
$p \equiv 3 \pmod{4}$ and $p \equiv +- 3\pmod{8}$.
so $p \equiv \{1,3\}\pmod{8}$
but $y^2 + 4$ is either $4 \pmod{8}$ or $5 \pmod {8}$, so there are no solutions?
Am i solving this the correct way or is there a shorter path?
Kees
| Yes, I think this is the best way. Keith Conrad has answered this with many examples in his article Examples of Mordell's equation; e.g., compare with Theorem $2.2$ and $2.3$, which is very similar, for $y^2=x^3-5$ and $y^2=x^3-6$.
| {
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"url": "https://math.stackexchange.com/questions/1520668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate the Integral: $\int^\pi_0\cos^6\theta\ d\theta$ $\int^\pi_0\cos^6\theta\ d\theta$
So I split the trig value into:
$\int^\pi_o\cos^5\theta\ cos\theta\ d\theta$
Then I utilized the Pythagorean theorem for $cos^5\theta$
$\int^\pi_o(1-sin^5\theta)\ cos\theta$
I utilized u-substitution:
$u=sin\ \theta$
$du=cos\ \theta$
Thus:
$\int^{x=\pi}_{x=0}\ (1-u^5)\ d\theta$
I intergated
$(\frac{1}{6}u^6)+(\frac{1}{6}u^6)$
$-(\frac{\pi^6}{6})+(0)$
$-(\frac{\pi^6}{6})$
Is my answer right?
| Answer (using partial integration):
$$\int\cos^6(x)\space\space\text{d}x=$$
$$\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{6}\int\cos^4(x)\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{8}\int\cos^2(x)\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{8}\int\left(\frac{1}{2}\cos(2x)+\frac{1}{2}\right)\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{32}\int\cos(2x)\space\space\text{d}x+\frac{5}{16}\int 1\space\space\text{d}x=$$
Substitute $u=2x$ and $\text{d}u=2\space\space\text{d}x$:
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{32}\int\cos(u)\space\space\text{d}u+\frac{5}{16}\int 1\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5\sin\left(u\right)}{32}+\frac{5x}{16}+\text{C}=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5\sin\left(2x\right)}{32}+\frac{5x}{16}+\text{C}=$$
$$\frac{1}{192}\left(60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right)+\text{C}$$
With boundaries:
$$\left[\frac{1}{192}\left(60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right)\right]_{0}^{\pi}=$$
$$\frac{1}{192}\left[60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right]_{0}^{\pi}=$$
$$\frac{1}{192}\left(60\pi+45\sin(2\pi)+9\sin(4\pi)+\sin(6\pi)\right)=$$
$$\frac{1}{192}\left(60\pi+0+0+0\right)=\frac{60\pi}{192}=\frac{5\pi}{16}$$
| {
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"url": "https://math.stackexchange.com/questions/1521105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &\neq \sqrt{x+2}\\
\end{align}
*
*Is my proof correct?
*I'm interested in more ways of proving it.
| Hint #1:
Assume that $\sqrt{x + 2} - \sqrt{x + 1} = \sqrt{x + 1} - \sqrt{x}$ for some $x > 0$.
Hint #2:
Derive a contradiction.
Hint #3:
This proof (by contradiction) results to some changes in the notation you used in your proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove that $\lim\limits_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$ using the definition
Prove $$\lim_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$$ using the definition.
Let there be $\varepsilon>0$ we need to find $N<n$ such that $\Big|\frac{n^2+n-1}{3n^2+1}-\frac{1}{3}$$
\Big|<\varepsilon$
$$\left|\frac{n^2+n-1}{3n^2+1}-\frac{1}{3}
\right|=\left|\frac{3n^2+3n-3-3n^2-1}{9n^2+3}
\right|=\left|\frac{3n-4}{9n^2+3}\right|$$
What should I do next, can I say that because $n\to \infty$ we can neglect $-4$ and $3$ from the expression?
| $|\frac{3n-4}{9n^2+3}|<\frac{3n}{9n^2}=\frac{1}{3n}\leq\frac{1}{3N}<\epsilon$ whenever $n\geq N>\frac{1}{3\epsilon}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Associative Property What I learn as a basic theory of associative is that $$(a \times b) \times c = a \times (b \times c) \text{ and } (a+b)+c = a+(b+c).$$
However when doing my exercises on this topic, I came across questions with only 2 variables. Eg. $a^2+b^2$. Is this associative and why?
The binary operation is given as $a * b = a^2 + b^2$.
| For $a\ast b$ to be associative it should be
$$(a\ast b)\ast c=a\ast (b\ast c),\text{ for all }a,b,c\in \Bbb R,\tag{1}$$
since I assume we are discussing a sum of squares of real numbers. This is, it should be
$$(a^2+b^2)^2+c^2=a^2+(b^2+c^2)^2,\ \forall a,b,c\in \Bbb R.$$
But now it's not difficult to find a counterexample to the above equality:
$$(0^2+1^2)^2+2^2=(0+1)^2+2^2=1^2+2^2=1+4=5$$
while
$$0^2+(1^2+2^2)^2=0+(1+4)^2=0+5^2=0+25=25.$$
Since $(0\ast 1)\ast 2\ne 0\ast (1\ast 2)$, $(1)$ does not hold, this is
$$a\ast b:=a^2+b^2$$
is not associative over the real numbers (and it is not over any subset of reals containing $0$, $1$ and $2$, like $\Bbb Q$ or $\Bbb N$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of the series $\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$? To analyze the convergence of the
$$\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$$
series I used the criterion of integral $$\displaystyle\int_4^\infty {\frac{{x + 1}}{{(x + 5)(x + 4)(x - 3)}}dx},$$ but calculate this improper integral is a very laborious task.
Is there a shorter way? What criteria of convergence would be most effective or simple?
| Assume that the series begins at $n=4$. Then, we have
$$n+5\ge n$$
$$n+4\ge n$$
$$n-3\ge \frac14 n$$
$$n+1\le 2n$$
Therefore, have
$$\frac{n+1}{(n+5)(n+4)(n-3)}\le \frac{2n}{\frac14 n^3}=8\frac1{n^2}$$
Finally, using the result $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ reveals
$$\sum_{n=4}^{\infty}\frac{n+1}{(n+5)(n+4)(n-3)}\le 8\sum_{n=4}^{\infty}\frac1{n^2}=4\left(\frac{\pi^2}{3}-3\right)$$
and the series converges by the comparison test.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding limit of $\left(\frac{n^2 + n}{n^2 + n + 2}\right)^n$ Please, help me to find limit of this sequence:
$\lim_{n\to \infty} \left(\frac{n^2 + n}{n^2 + n + 2}\right)^n$
| $$\lim_{n\to\infty}\left(\frac{n^2+n}{n^2+n+2}\right)^n=$$
$$\lim_{n\to\infty}\exp\left(\ln\left(\left(\frac{n^2+n}{n^2+n+2}\right)^n\right)\right)=$$
$$\lim_{n\to\infty}\exp\left(n\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}n\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\ln\left(\frac{n^2+n}{n^2+n+2}\right)}{\frac{1}{n}}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right)}{\frac{\text{d}}{\text{d}n}\left(\frac{1}{n}\right)}\right)=$$
$$\exp\left(\lim_{n\to\infty}-\frac{2n(2n+1)}{(n+1)(n^2+n+2)}\right)=$$
Since $-2n(2n+1)$ grows asymptotically slower than $(n+1)(n^2+n+2)$ as $n$ approaches $\infty$:
$$\exp\left(0\right)=e^0=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\frac{1}{(x^2+1)^2}$ How do I integrate $\frac{1}{(x^2+1)^2}$?
I've tried to use the fact that $\int \frac{1}{(x^2+1)}=Arctan(x)$ but I don't know how to. I think it's by parts. Tried using $u'=(x^2+1)^{-1}$ and $v=(x^2+1)^{-1}$ in the formula $\int u'v=vu-\int uv'$, but it doesn't seem to help...
| First consider
$$I = \int \frac{dx}{(x^2+1)^{2}}$$
and the substitution $x=\tan\theta$. This leads to $dx = sec^{2}\theta \, d\theta$ and $x^2 + 1 = sec^{2}\theta$ and
\begin{align}
I &= \int \frac{sec^{2}\theta}{sec^{4}\theta} \, d\theta \\
&= \int \frac{d\theta}{sec^{2}\theta} \\
&= \int \cos^{2}\theta \, d\theta \\
&= \frac{1}{2} \, \int \left( 1 + \cos(2\theta) \right) \, d\theta\\
&= \frac{1}{2} \, \left[ \theta + \frac{1}{2} \, \sin(2\theta) \right] + c_{0}
\end{align}
Upon backward substitution this becomes
\begin{align}
I = \frac{1}{2} \, \tan^{-1}(x) + \frac{1}{4} \, \sin(2 \, \tan^{-1}(x)) + c_{0}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding points of a function's graph that are closest to a given point A question from my calculus book states,
Which points on the graph $y=4-x^2$ are the closest to the point (0,2)?
Using some of my notes, I have a formula as follows (not sure what it's actually called):
$$d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$
So I plugged in $(0,2)$ as my $x_1/y_1$ values:
$$d=\sqrt{(x_2-0)^2 + (y_2-2)^2}$$
$$d^2 = (x-0)^2 + (y-2)^2$$
I know $y=4-x^2$ so I plug it in:
$$d^2 = (x-0)^2 + ((4-x^2)-2)^2$$
$$F(x) = (x-0)^2 + (2-x^2)^2$$
Finding the derivative using the chain rule I get:
$$F'(x) = 2(x-0)(1) + 2(2-x^2)(-1)$$
$$0 = 2(x-0)- 2(2-x^2)$$
$$0 = 2x-0-4+2x^2$$
$$0 = 2x^2+2x-4$$
Thus I have a polynomial, so I tried using the quadratic equation to solve for $[x]$ and ended up with $$x=(-2 + 6)/(4) | x=(-2 - 6)/(4)$$, so $x=1/x=-2$?
Can someone help me out with what I'm missing or doing wrong in trying to find the points of the graph to point $(0,2)$?
| This is a classic optimization problem where you are trying to minimize the distance from the point $(0, 2)$. Given the function $$f(x) = 4 - x^2$$, we can represent all of the points on that curve as, $(x, 4 - x^2)$. Now, we can use the distance formula: $$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$. When we plug in $(x, 4 - x^2)$ and $(0, 2)$ for $(x_1, y_1)$ and $(x_2, y_2)$, respectively, we obtain, $$d = \sqrt{x^4-3x^2+4}$$ We want to minimize the distance function, so we should find when the derivative is equal to $0$. Since the square root will always increase with the inside function, we can disregard the square root and call $$f(x)=x^4-3x^2+4$$. Taking the derivative yields $$f'(x)=4x^3-6x$$ Which when factored is $$f'(x)=2x(2x^2-3)$$. We can see that the critical numbers of the function $f(x)$ are $x = 0$ and $x = \pm\sqrt{\frac{3}{2}}$ Applying the first derivative test shows that $x = \pm\sqrt{\frac{3}{2}}$ are minimums. Therefore the points that are a minimum distance from $(0, 2)$ and are on $y=4-x^2$ are $(\sqrt{\frac{3}{2}}, \frac{5}{2})$ and $(-\sqrt{\frac{3}{2}}, \frac{5}{2})$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so far: $\int_{0}^{a}\sqrt{1+\left(\left(\sqrt{x}\right)'\right)^{2}}dx=\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\left[\begin{array}{cc}
t^{2}=1+\frac{1}{4x} & \sqrt{1+\frac{1}{4N}},\sqrt{1+\frac{1}{4a}}\\
2tdt=-\frac{1}{8x^{2}}dx & x=\frac{1}{4\left(t^{2}-1\right)}
\end{array}\right]=\\\lim_{N\to0}-\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\frac{t^{2}dt}{\left(t^{2}-1\right)^{2}}=\lim_{N\to0}\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\left(\frac{1}{4\left(t+1\right)}-\frac{1}{4\left(t+1\right)^{2}}-\frac{1}{4\left(t-1\right)}-\frac{1}{4\left(t-1\right)^{2}}\right)dt=\\=\lim_{N\to0}\frac{1}{4}\left[\ln\left(t+1\right)+\frac{1}{t+1}-\ln\left(t-1\right)+\frac{1}{t-1}\right]_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4a}}}{1+\frac{1}{4a}-1}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4N}}}{1+\frac{1}{4N}-1}\right]=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-8N\sqrt{1+\frac{1}{4N}}\right]=\\=\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}-0+0=\ln\left(4a\left(\sqrt{\frac{1}{a}+4}+2\right)+1\right)-8a\sqrt{1+\frac{1}{4a}}$
But it doesn't seem right... any Ideas what went wrong?
| You need to make a first substitution:
let u= 1+1/4x
yhen you will have to integrate: sqrt(1+u^2)du which is easy.
at the end you will get a formula with ln, tan inverse, and sec.
Hope it works
| {
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Solving the recurrence $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$ using generating functions
Solve the following recurrence using generating functions: $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$.
My partial solution:
We can rewrite $a_{n+2} = 3a_{n+1} - 2a_n$, as $a_{n+2} - 3a_{n+1} + 2a_n = 0$, and we let $A(z) = \sum a_n z^n$. The goal is to compute $A(z)$ as this can be done as follows: $$A(z) - a_0 - a_1z - 3z(A(z) - a_0) + 2z^2A(z) = 0$$
$$(1-3z+2z^2)A(z) = a_0 + a_1z -3a_0z$$
$$A(z) = \frac{a_0 + (a_1 - 3a_0)z}{1-3z+2z^2}$$
$$\quad \quad = \frac{a_0 + (a_1 - 3a_0)z}{(1-z)(1-2z)}$$
$$\quad \quad = \frac{C}{(1-z)}+\frac{D}{(1-2z)}$$
And, I don't know how to continue, I cannot figure out the remaining. I'm pretty sure it is obvious, but I just cannot see it. If someone can help me I would be glad.
| \begin{align}
\frac{C}{1-z}+\frac{D}{1-2z} &= \frac{C(1-2z)}{(1-z)(1-2z)} + \frac{(1-z)D}{(1-z)(1-2z)} \\\\ &= \frac{(C+D)+(-2C-D)z}{(1-z)(1-2z)} \\\\ &= \frac{(a_0)+(a_1-3a_0)z}{(1-z)(1-2z)}
\end{align}
You now have 2 linear equations and easily solve for $C$ and $D$ in terms of $a_0$ and $a_1$.
What is the power series for $\dfrac{1}{1-z}$? It's a simple one, try long hand division, and you'll see the coefficient pattern after a few terms. Note that $\dfrac{1}{1-2z}$ is just $\dfrac{1}{1-y}$ with $y=2z$.
Let $\displaystyle F(z) = \dfrac{1}{1-z} = \sum_{n=0}^{\infty}f_n z^n$ be that power series.
Then $\displaystyle A(z) = CF(z)+DF(2z) = \sum_{n=0}^{\infty} (C+D2^n)f_n z^n$. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Construct generator matrix given generator polynomial? How would I take a generator polynomial and construct a generator matrix out of it for a cyclic code?
For example, I have a cyclic code in:
$R_{15}=GF(2)[x] / \langle x^{15} + 1\rangle$
This is given by the generator polynomial:
$g(x) = x^8 + x^4 + x^2 + x + 1$
So, the length is 15 and the dimension is 15 - 8 = 7.
How would I go about constructing a generator matrix from that?
Should be a k x n matrix which is 7 x 15 correct?
$G = \begin{bmatrix}
1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \end{bmatrix}$
That is what I come up with, just wondering if this is the correct way to do it?
| This is correct. The rows of $G$ correspond to the generator polynomial (expressed as a binary $n-$tuple) with all its (non cyclical) $k$ shifts, which correspond to the "canonical" input messages $u_0=(1,0,0,0 \cdots) \equiv 1$, $u_1=(0,1,0,0 \cdots) \equiv x $, $u_2=(0,0,1,0 \cdots) \equiv x^2 $, and so on.
This is the generator for the standard (non-systematic) cyclic code. If you instead want to build a systematic cyclic code, that's a different story.
| {
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Given 3 equally likely outcomes - H, T, and B) - why is P(exactly 1 H in 3 tries) = 0.4444.... and not 0.381? I have a three-sided dice, with three letters: $H$, $T$ and $B$. I want to figure out the probability that, out of 3 rolls, I get exactly 1 H. I figured I'd start by finding the probability that the 3 rolls do not all result in the same outcome (I don't get three H's, three T's, or three B's.) That probability, by my calculation, = $1 - 3/27 = 0.8889$.
I figured that in $3$ rolls of the dice, given that they do not result in $3$ $H$'s, $3$ $T$'s or $3$ $B$'s, then these are the only possible outcomes:
$2$ $H$'s and a $T$
$2$ $H$'s and a $B$
$2$ $T$'s and a $B$
$2$ $T$'s and an $H$ -
$2$ $B$'s and a $T$
$2$ $B$'s and an $H$ -
$1$ $H$, $1$ $B$, and $1$ $T$ -
So, there are seven outcomes and only three ways to then get only $1$ $H$
Since there are only two ways to get $1$ $H$, therefore $P(1H$|not $3H$, $3T$, or $3B) = 3/7$.
Multiplying the two together, we get $0.381$, when my binomial calculator says the real answer is $0.4444....$ I feel as though I'm missing something here.
| These are the probabilities:
*
*$P_0=P(H=\color\red0,T=\color\green0,B=\color\orange3)=\dfrac{3!}{\color\red0!\cdot\color\green0!\cdot\color\orange3!}\cdot\dfrac{1}{3^3}=\dfrac{1}{27}$
*$P_1=P(H=\color\red0,T=\color\green1,B=\color\orange2)=\dfrac{3!}{\color\red0!\cdot\color\green1!\cdot\color\orange2!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$
*$P_2=P(H=\color\red0,T=\color\green2,B=\color\orange1)=\dfrac{3!}{\color\red0!\cdot\color\green2!\cdot\color\orange1!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$
*$P_3=P(H=\color\red0,T=\color\green3,B=\color\orange0)=\dfrac{3!}{\color\red0!\cdot\color\green3!\cdot\color\orange0!}\cdot\dfrac{1}{3^3}=\dfrac{1}{27}$
*$P_4=P(H=\color\red1,T=\color\green0,B=\color\orange2)=\dfrac{3!}{\color\red1!\cdot\color\green0!\cdot\color\orange2!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$
*$P_5=P(H=\color\red1,T=\color\green1,B=\color\orange1)=\dfrac{3!}{\color\red1!\cdot\color\green1!\cdot\color\orange1!}\cdot\dfrac{1}{3^3}=\dfrac{6}{27}$
*$P_6=P(H=\color\red1,T=\color\green2,B=\color\orange0)=\dfrac{3!}{\color\red1!\cdot\color\green2!\cdot\color\orange0!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$
*$P_7=P(H=\color\red2,T=\color\green0,B=\color\orange1)=\dfrac{3!}{\color\red2!\cdot\color\green0!\cdot\color\orange1!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$
*$P_8=P(H=\color\red2,T=\color\green1,B=\color\orange0)=\dfrac{3!}{\color\red2!\cdot\color\green1!\cdot\color\orange0!}\cdot\dfrac{1}{3^3}=\dfrac{3}{27}$
*$P_9=P(H=\color\red3,T=\color\green0,B=\color\orange0)=\dfrac{3!}{\color\red3!\cdot\color\green0!\cdot\color\orange0!}\cdot\dfrac{1}{3^3}=\dfrac{1}{27}$
Hence the requested probability is $\dfrac{P_4+P_5+P_6}{1-(P_0+P_3+P_9)}=\dfrac{12}{24}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
if $a ≡ b\pmod {2n}$ then prove $a^2 ≡ b^2 \pmod {4n}$
Let $n$ be positive number, if $a \equiv b \pmod{2n}$, prove that
$a^2 \equiv b^2 \pmod{4n}$.
By the congruence in hypothesis, we have $a-b = 2nk$ where $k$ is an integer.
Then $a = b+2nk$ and $a^2 = b^2+4n^2k^2+4knb$. From this we get $a^2-b^2 = 4kn(kn+b)$.
Now I have a question here. On the right hand side, we only know that $k$ and $4$ are integers because question says nothing about $n$ and $b$. So can we go and say that $\frac{a^2-b^2}{4n} = k(kn+b)$ is an integer?
| Suppose
$a \equiv b \bmod 2kn
$.
($k=1$ in this case.)
Then
$a
=b+2jkn
$
for some $j$,
so
$a^2
=b^2+4bkjn+4j^2k^2n^2
=b^2+4kn(bj+j^2kn)
$
so
$a^2
\equiv b^2 \bmod 4kn
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{x(1 - 0.5\cos x) - 0.5\sin x}}{{{x^3}}}$ For evaluating the limit $\lim\limits_{x \to 0} \frac{x(1 - 0.5\cos x) - 0.5\sin x}{x^3}$, I proceeded as follows:
$$\lim_{x \to 0} \left(\frac{x(1 - 0.5\cos x)}{x^3} - \frac{0.5}{x^2}\left(\frac{\sin x}{x}\right)\right)$$
Using the fact that $\lim\limits_{x \to 0} \frac{\sin x}{x}=1$ and $\lim\limits_{x \to a} \{ f(x) - g(x)\} = \lim\limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$, I got
$$\lim\limits_{x \to 0} \frac{1 - 0.5\cos x}{x^2} - \lim\limits_{x \to 0} \frac{0.5}{x^2}$$
$$ = \lim_{x \to 0} \left(\frac{1 - 0.5\cos x - 0.5}{x^2} \right)$$
Now, after applying L'Hopital's Rule I got the final answer as $0.25$. However, on evaluating the original limit using Mathematica, I got the answer as $\frac{1}{3}$.
Can someone please tell me where am I going wrong.
Somehow, I believe that you cannot use the fact that $\lim\limits_{x \to a} \{ f(x) - g(x)\} = \lim \limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$ in case of indeterminate forms.
Thanks in advance!
| Use Taylor's development at order $3$:
*
*$\cos x= 1-\dfrac{x^2}2+o(x^2)$, hence $\;x\bigl(1-\frac12\cos x\big)=\dfrac x2+\dfrac{x^3}4+o(x^3)$,
*$\sin x=x-\dfrac{x^3}6+o(x^3)$.
Thus the numerator is
$$\dfrac x2+\dfrac{x^3}4-\frac x2+\dfrac{x^3}{12}+o(x^3)=\dfrac{x^3}3+o(x^3)$$
and finally
$$\frac{x\bigl(1-\frac12\cos x\big)-\frac12\sin x}{x^3}=\frac{\dfrac{x^3}3+o(x^3)}{x^3}=\frac13+o(1)\to\frac13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$
I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.
| $$x^2=1-x$$
$$x^4=(1-x)^2=x^2-2x+1=1-x-2x+1=2-3x$$
$$x^6=(1-x)(2-3x)=3x^2-5x+2=3(1-x)-5x+2=5-8x$$
$$x^8=(2-3x)^2=9x^2-12x+4=9(1-x)-12x+4=13-21x$$
$$x^{10}=(1-x)(13-21x)=21x^2-34x+13=21(1-x)-34x+13=34-55x$$
Therefore,
$$x^{10}+x^8+x^2+1=34-55x+13-21x+1-x+1=49-77x$$
$$x^{10}+x^6+x^4+1=34-55x+5-8x+2-3x+1=42-66x$$
Your equation
=$$\frac{49-77x}{42-66x}=\frac{7}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
sum of solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$ What is the sum of non-negative integer solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$?
I know that $\{(x,y,z)\mid x+y+z=k\}$ has $\binom{k+3-1}{3-1}=\binom{k+2}{2}$ non-negative integer solutions. Thus, the answer to the question above is $$\sum_{k=1}^N \binom{k+2}{2}.$$
Can we further simplify the summation?
| The set of solutions of the set of equations
$$x + y + z = k \qquad 1 \leq k \leq N \tag{1}$$
in the non-negative integers is the set of solutions of the inequality
$$x + y + z \leq N \tag{2}$$
in the non-negative integers with the exception of $(0, 0, 0)$. The number of solutions of the inequality in the non-negative integers is the number of solutions of the equation
$$w + x + y + z = N \tag{3}$$
where $w = N - x - y - z$. The number of solutions of equation 3 in the non-negative integers is equal to the number of ways three addition signs can be placed in a row of $N$ ones, which is
$$\binom{N + 3}{3}$$
since we must select which $3$ of the $N + 3$ symbols (three addition signs and $N$ ones) will be addition signs. Hence, the number of solutions of the set of equations $x + y + z = k$ if $1 \leq k \leq N$ in the non-negative integers is
$$\binom{N + 3}{3} - 1$$
whence
$$\sum_{k = 1}^{N} \binom{k + 2}{2} = \binom{N + 3}{3} - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
My work:
We consider the congruences $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$, $x \equiv 5 \pmod 6$. We can reduce this further to $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$. We have
$N_1 = 4 \cdot 5 = 20 \implies 20 x_1 \equiv 1 \pmod{3} \implies 2x_1 \equiv 1 \pmod{3} \implies x_1 \equiv 2 \pmod {3}$
$N_2 = 3 \cdot 5 = 15 \implies 15x_2 \equiv 1 \pmod{4} \implies -x_2 \equiv 1 \pmod{4} \implies x_2 \equiv 3 \pmod {4}$
$N_3 = 3 \cdot 4 = 12 \implies 12 x_3 \equiv 1 \pmod{5} \implies 2x_3 \equiv 1 \pmod{5} \implies x_3 \equiv 3 \pmod {5}$
Now,
\begin{align*}
\bar x &= a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 \\
&= 3 \cdot 20 \cdot 2 + 4 \cdot 15 \cdot 3 + 5 \cdot 12 \cdot 3 \\
&= 480 \equiv 0 \pmod {3 \cdot 4 \cdot 5}
\end{align*}
Is this correct, or is something wrong in my work? I don't like how I have $0$ remainder.
| I too found the answer to be: x ≡ -1 (mod 60).
My solution:
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding expected value of random variable You and a friend go to a wine bar. The wine bar offers the following wine styles for ordering:
*
*2 of the wines offered are a Rosé style wine
*3 of the wines offered are a White style wine
*and 5 of the wines offered are a Red style wine
All in all there are 10 unique wines to choose from.
You and your friend order a uniformly random subset of seven wines (no duplicates).
Y = the number of white wines in this order
What is the expected value of Y?
Here is what I think:
So for this I first break down the types of events and what the values are
*
*Y = 0 => { $XXXXXXX$ } (X meaning not a rosé style wine)
*Y = 1 => { $XXXXXXR, XXXXXRX, XXXXRXX, XXXRXXX, XXRXXXX, XRXXXXX, RXXXXXX$ } => $1 \choose 7$
*Y = 2 => { $XXXXXRR,
XXXXRXR,
...
RXRXXXX,
RRXXXXX$ } => $2 \choose 7$
*Y = 3 => { $XXXXRRR,
XXXRXRR,
XXRXXRR,
..
RRRXXXX$ } => $3 \choose 7$
So now to get the expected value do I add up all the probabilities of each ie. Pr(Y=1) = $|Y=1|/|S|$ = $1 \choose 7$$/$$ 3 \choose 7$, so I would then do something like $Pr(Y = 0) + Pr(Y = 1) + Pr(Y = 2) + Pr(Y = 3)$
Thanks
| We first do it more or less in the way you attempted. There are $\binom{10}{7}$ equally likely ways to choose $7$ wines.
First we find the number of ways to choose $0$ Pinks. There is only one way, so $\Pr(Y=0)=\frac{1}{\binom{10}{7}}$. For the future, note that this is $\frac{\binom{3}{0}\binom{7}{7}}{\binom{10}{7}}$. So now we know $\Pr(Y=0)$.
Next we find the number of ways to choose $1$ Pink, and therefore $6$ non-Pinks. There are $\binom{3}{1}$ ways to choose the Pink, and for each of these there are $\binom{7}{6}$ ways to choose the non-Pinks. It follows that $\Pr(Y=1)=\frac{\binom{3}{1}\binom{7}{6}}{\binom{10}{7}}$.
A similar analysis shows that $\Pr(Y=2)=\frac{\binom{3}{2}\binom{7}{5}}{\binom{10}{7}}$, and $\Pr(Y=3)=\frac{\binom{3}{3}\binom{7}{4}}{\binom{10}{7}}$.
Finally, $E(Y)=0\cdot\Pr(Y=0)+1\cdot \Pr(Y=1)+2\cdot \Pr(Y=2)+3\cdot \Pr(Y=3)$. Calculate.
Another way: Imagine we pick the wines one at a time. Let random variable $X_i$ be equal to $1$ if on the $i$-th pick we choose Pink, and let $X_i=0$ otherwise. Then $Y=X_1+\cdots +X_7$, and by the linearity of expectation $E(Y)=E(X_1)+\cdots+E(X_7)$. For any $i$, $\Pr(X_i=1)=\frac{3}{10}$, so $E(X_i)=\frac{3}{10}$. It follows that $E(X_i)=7\cdot \frac{3}{10}$. Easier!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem.
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx = \int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx
$$
We can also replace $1-x$ in the denominator with $\sqrt{(1-x)^2}$
$$
\int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx = \int \frac{\sqrt{(1-x)^2}}{\sqrt{(1-x)(2x+1)}}\,dx
$$
If we simplify this fraction we get:
$$
\int \frac{\sqrt{1-x}}{\sqrt{2x+1}}\,dx
$$
Next we apply the following substitutions
$$
u = -x
$$
so : $-du = dx$
We can rewrite the integral as following:
$$-\int \frac{\sqrt{1+u}}{\sqrt{1-2u}}\,du$$
Then we apply another substitution:
$\sqrt{1+u} = t $ so $ \frac{1}{2\sqrt{1+u}} = dt $
We rewrite: $ \sqrt{1+u} $ to $\frac{1}{2}t^2 \,dt $
We can also replace $\sqrt{1-2u} $ as following:
$$\sqrt{-2t^2+3}=\sqrt{-2(1+u)+3}=\sqrt{1-2u}$$
With al these substitutions the integral has now the following form:
$$-\frac{1}{2}\int \frac{t^2}{\sqrt{-2t^2+3}}\,dt$$
Next we try to ''clean'' up the numerator:
$$-\frac{1}{2} \int \frac{t^2}{\sqrt{\frac{1}{2}(6-t^2)}} \, dt$$
$$-\frac{\sqrt{2}}{2} \int \frac{t^2}{\sqrt{6-t^2}} \, dt$$
And that's where I got stuck. I can clearly see that an arcsin is showing up in the integral but don't know how to get rid of the $t^2$.
| Hint: When you arrived at $~\displaystyle\int\sqrt{\frac{1-x}{2x+1}}~dx,~$ you should have immediately substituted
$\dfrac{1-x}{2x+1}=u^2.~$ Then the entire integrand would have been reduced to a rational function.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Find the value of $a+b+c$ $a=\sqrt{57+40\sqrt2}-\sqrt{57-40\sqrt2}$ and $b=\sqrt{25^{\frac{1}{\log_85}}+49^{\frac{1}{\log_67}}}$ and $c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Find the value of $a+b+c$.
I tried to solve and simplify this problem but no luck.Please help me.Thanks.
| (1) $a=(5+4\sqrt{2})-(4\sqrt{2}-5)=10$
(2) $b=\sqrt{25^{\log_5{8}}+49^{\log_7{6}}}=\sqrt{8^2+6^2}=10$
(3) $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}=(\sqrt{2}+1)-\frac{1}{\sqrt{2}+1}=(\sqrt{2}+1)-(\sqrt{2}-1)=2$, $c=8+6-14=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\tanh ^{-1} (\sin \theta)=\cosh^{-1} (\sec \theta)$ As the question says
How to prove
$$\tanh ^{-1} (\sin \theta)=\cosh^{-1} (\sec \theta)$$
I have tried to solve it
The end result that got for RHS
$$=\log \frac{1+\tan\frac{\theta}{2}}{1-\tan \frac{\theta}{2}}$$
I am stuck here Please help
| We know by formula,$$\tanh^{−1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$$ and $$\cosh^{-1} x=\log (x+\sqrt{x^2-1})$$
Now putting $x=\sin \theta$ in the formula for $\tanh ^{-1}x$, we have that
$$\tanh^{−1}(\sin \theta)=\frac{1}{2}\log\left(\frac{1+\sin \theta}{1-\sin \theta}\right)$$
$$=\frac{1}{2}\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})^2}{(\cos \frac{\theta}{2}-\sin \frac{\theta}{2})^2}\right)$$
$$=\frac{1}{2}\log\left(\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\right)^2$$
$$=\frac{1}{2}\cdot 2\log\left(\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\right)$$
$$=\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}{(\cos \frac{\theta}{2}-\sin \frac{\theta}{2})}\cdot \frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}\right)$$
$$=\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})^2}{\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}}\right)$$
$$=\log (\frac{1+\sin \theta}{\cos \theta})$$
$$=\log (\tan \theta+ \sec \theta)$$
$$=\log (\sec \theta+\sqrt{(\tan \theta)^2})$$
$$=\log (\sec \theta+\sqrt{(\sec \theta)^2-1})$$
$$=\cosh^{-1} (\sec \theta)$$
Hence proved, assuming $0 < \theta < \frac{\pi}{2}$
EDIT: The proof is valid for $0 < \theta < \frac{\pi}{2}$ but not in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552782",
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"source": "stackexchange",
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Question on indefinite integrals I have to integrate:
$$I_2 = \int \frac{e^{2x} - e^{x} + 1}{(e^x\cos(x) - \sin(x))\cdot \left(e^x\sin(x) + \cos(x)\right)} \text{d}x$$
I simply can't understand from where to begin with. Please help me in solving this problem.
| I have a better solution to your problem
Here it goes:
\begin{aligned}
& \int \frac{e^{2 x}-e^{x}+1}{\left(e^{x} \sin x+\cos x\right)\left(e^{x} \cos x-\sin x\right)} dx \\
=& \int \frac{e^{2 x}-e^{x}+1}{\left(e^{2 x}+1\right) \sin \left(x+\tan ^{-1} \frac{1}{e^{x}}\right) \cdot \cos \left(x+\tan ^{-1} \frac{1}{e^{x}}\right)} d x \\
=&2 \int \frac{e^{2 x}-e^{x}+1}{\left(e^{2 x}+1\right) \sin \left(2 x+2 \operatorname{ta}^{-1}\left(\frac{1}{e^{x}}\right)\right)} d x \\
=&2 \int \frac{e^{2 x}-e^{x}+1}{\left(e^{2 x}+1\right) \sin \left(2 x+2 \cot ^{-1}\left(e^{x}\right)\right)} dx \\=&\int \operatorname{cosec}t~ dt=\ln \left|\tan \frac{t}{2}\right|+c \end{aligned}
where $t=2 x+2 \cot ^{-1}\left(e^{x}\right)$. Note that
$$\frac{d t}{d x}=2-\frac{2}{1+e^{2 x}} \cdot e^{x}=\frac{2\left(e^{2 x}-e^{x}+1\right)}{e^{2 x}+1}$$
Also, $\frac{t}{2}=x+\tan ^{-1}\left(\frac{1}{e^{x}}\right)$ so we have $$
\tan \frac{t}{2}=\frac{\tan x+\frac{1}{e^{x}}}{1-\frac{t x}{e^{x}}} =\frac{e^{x} \sin x+\cos x}{e^{x} \cos x-\sin x}$$
Thus the integral is $$ \boxed{\ln \left|\frac{e^{x} \sin x+\cos x}{e^{x} \cos x-\sin x}\right|+c} $$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How do I determine if a matrix is in a span of two other matrices? If \begin{align}
A =\begin{bmatrix}
1 & 1 \\
-1 & 1 \\
\end{bmatrix}
\end{align}
and
\begin{align}B = \begin{bmatrix}
1 & -1 \\
1 & 0 \\
\end{bmatrix}
\end{align}
So, how can I check if $C$ is in span $(A,B)$ when
\begin{align}C=
\begin{bmatrix}
1 & 2 \\
3 & 4 \\
\end{bmatrix}
\end{align}
| Hint: If you can find $κ, λ$ such that $$κ\begin{pmatrix}1 & 1\\-1&1 \end{pmatrix}+λ\begin{pmatrix}1 & -1\\1&0 \end{pmatrix}=\begin{pmatrix}1 & 2\\3&4 \end{pmatrix}$$ then, the answer is yes. You have $4$ equations with $2$ unknowns, the first of the equations being $κ\cdot1+λ\cdot1=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554664",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
| The problem is that you assume $\frac00$ is = some $x$ belong to the real number group.$\frac00$ is just not defined for you to assume it to be $=x$.
Basically its one of the seven indeterminate forms of mathematics.Look here for the complete list-https://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "52",
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Find the area of the region enclosed by the inner loop of the curve $r = 4 + 8 \sin \theta$ The loop is generated for $\theta \in \left[\frac76\pi, \frac{11}{6}\pi\right]$
(this is from setting $r = 0$).
So, $$A = \int_{\frac76\pi}^{\frac{11}{6}\pi} \frac12(4 + 8 \sin \theta)^2 \,d\theta=\\
= \int_{\frac76\pi}^{\frac{11}{6}\pi} 8 + 32 \sin \theta + 32 \sin^2\theta\, d\theta =\\
= \int_{\frac76\pi}^{\frac{11}{6}\pi} 8 + 32 \sin \theta + 16(1 - \cos(2\theta))\, d\theta=\\
= \int_{7\pi/6}^{11\pi/6} 24 + 32 \sin \theta - 16 \cos(2\theta)\, d\theta =\\
= \bigg[24\theta - 32 \cos \theta - 8 \sin(2\theta)\bigg]_{\frac76\pi}^{\frac{11}{6}\pi} =\\
= (44\pi - 16\sqrt{3} + 4\sqrt{3}) - (28\pi + 16\sqrt{3} - 4\sqrt{3}) =\\
= 16\pi - 24\sqrt{3}$$
Is my answer right?
| As much as Wolfram|Alpha is impressive, here is the answer again using the SageMath free/libre software:
sage: integral((1/2)*(4 + 8 * sin(x))^2, x, 7*pi/6, 11*pi/6).simplify()
16*pi - 24*sqrt(3)
| {
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Combinatorics : Choosing r elements from a multiset of n elements, given some elements in multiset are identical to each other Given a multiset = {a, a, a, b, b, b, b, c, c}. The multiset has:
number of a's = 3
number of b's = 4
number of c's = 2
In how many ways we can select any 3 elements from this multiset?
My solution to this question:
to get the required number of ways in which we can select three elements from that multiset we need to find the coefficient of $x^3$ in $\{\left( 1+x+x^2+x^3 \right) \left( 1+x+x^2+x^3 +x^4 \right) \left( 1+x+x^2 \right)\}$
Now, I'm stuck. How to proceed with this? Do I have to manually sum the number of $x^3$ we can obtain after multiplication, to get the coefficient.
Can this be done using Generating functions? If yes, then how?
Give references to book/study material online link, if you want to escape writing down complete solution.
| Here the numbers are so small that it’s not hard simply to write out the terms of degree $3$ and count them:
$$\begin{align*}
&x^3\cdot1\cdot1\\
&x^2\cdot1\cdot x\\
&x^2\cdot x\cdot 1\\
&x\cdot 1\cdot x^2\\
&x\cdot x\cdot x\\
&x\cdot x^2\cdot 1\\
&1\cdot x\cdot x^2\\
&1\cdot x^2\cdot x\\
&1\cdot x^3\cdot 1
\end{align*}$$
However, counting multisets of a given cardinality isn’t too hard a problem theoretically. If we had at least $3$ of each letter available, the answer would be simply
$$\left(\!\!\binom{3}{\color{red}3}\!\!\right)\;,$$
the number of multisets of cardinality $3$ taken from a set of $3$ available types of object. By a standard stars and bars argument this is equal to
$$\binom{3+\color{red}3-1}{\color{red}3}=\binom53=10\;.$$
However, we have only $2$ $c$s available, so we have to throw away any multiset with more than $2$ $c$s. Since we’re looking at multisets of size $3$, there is only one, and we get $10-1=9$ as the final count.
You can also work directly with your generating function. Note that
$$1+x+x^2+x^3=\frac{1-x^4}{1-x}\;,$$
and similarly for the other two factors, so your generating function can be rewritten as
$$\frac{(1-x^3)(1-x^4)(1-x^5)}{(1-x)^3}\;.\tag{1}$$
Now
$$\begin{align*}
(1-x^3)(1-x^4)(1-x^5)&=(1-x^3-x^4+x^7)(1-x^5)\\
&=1-x^3-x^4-x^5+x^7+x^8+x^9-x^{12}\;,
\end{align*}$$
so $(1)$ is a sum of terms of the form $\dfrac{x^m}{(1-x)^3}$. A useful generating function to know is
$$\frac1{(1-x)^3}=\sum_{n\ge 0}\binom{n+2}nx^n=\sum_{n\ge 0}\binom{n+2}2x^n\;,$$
from which we get
$$\frac{x^m}{(1-x)^3}=\sum_{n\ge 0}\binom{n+2}2x^{m+n}\;.\tag{2}$$
The coefficient of $x^3$ in $(2)$ is
$$\binom{(3-m)+2}2=\binom{5-m}2\;,$$
which is $0$ for $m>3$, so we need only deal with $\dfrac1{(1-x)^3}-\dfrac{x^3}{(1-x)^3}$, with $m=0$ and $m=3$, getting
$$\binom52-\binom22=9$$
as before.
| {
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Proving of this trigonometric identity $$\frac{\cot \beta}{\csc \beta - 1} + \frac{\cot \beta}{\csc \beta + 1} = 2 \sec \beta$$
What I've done:
$$\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} +1} +
\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} -1}=\\
=\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1-\sin\beta}{\sin \beta}} +
\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1+\sin\beta}{\sin \beta}}=\\
=\cos \beta\frac{(1 - \sin\beta)}{\sin^2\beta} + \cos \beta\frac{1 + \sin\beta}{\sin^2\beta}=\\
=\frac{ \cos \beta - \cos\beta \sin\beta + \cos\beta + \cos\beta \sin\beta}{\sin^2 \beta}=\\
=\frac{2\cos\beta}{\sin^2\beta}=\\
=\frac{2\cos\beta}{1-\cos^2\beta}$$
Right side:
$$2 \sec\beta=\frac{1}{2\cos\beta}$$
Could you tell me where I went wrong? I've tried using a proof program online (symbolab) though those steps are a bit hard for me to follow.
Note: I only want to use what I have on the left side to solve the left side.
Thank you very much.
(Might be some errors first time using math jax..)
| The mistake is in the third line of your derivation: the denominators should be $(1-\sin\beta)(1+\sin\beta)$, or $1-\sin^2\beta$, instead of $\sin^2\beta$.
| {
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If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\;,$ We get
$$k\sin C\left[k\sin A+k\sin B\right]\cdot \cos \frac{B}{2} = k\sin B\left[k\sin A+k\sin C\right]\cdot \cos \frac{C}{2} $$
So we get $$\sin C\left[\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\sin \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
Now Using $A+B+C=\pi\;,$ We get $\displaystyle \frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$ and $\displaystyle \frac{A+C}{2}=\frac{\pi}{2}-\frac{B}{2}$
So we get $$\sin C\left[\cos \frac{C}{2}\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\cos \frac{B}{2}\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
So we get $$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
Now if we put $B=C\;,$ Then these two are equal.
My question is how can we prove it.
Help me, Thanks
| $$\sin C\cos\dfrac{A-B}2=\sin\dfrac C2\cdot2\sin\dfrac{A+B}2\cos\dfrac{A-B}2 =\sin\dfrac C2\left(\sin\dfrac A2+\sin\dfrac B2\right)$$
So,
$$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
$$\iff\sin\dfrac C2\left(\sin\dfrac A2+\sin\dfrac B2\right)=\sin\dfrac B2\left(\sin\dfrac A2+\sin\dfrac C2\right)$$
As $\sin\dfrac A2\ne0,\sin\dfrac C2=\sin\dfrac B2$
Can you take it from here?
| {
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On the complete solution to $x^2+y^2=z^k$ for odd $k$? While trying to answer this question, I was looking at a computer output of solutions to $x^2+y^2 = z^k$ for odd $k$ and noticed certain patterns. For example, for $k=5$ we have $x,y,z$,
$$10, 55, 5\\25, 50, 5\\38, 41, 5\\117, 598, 13\\122, 597, 13\\338, 507, 13\\799, 884, 17$$
Question: Is it true that all integer solutions to $x^2+y^2=z^k$ for odd $k>1$ are given by just two formulas, namely,
*
*Primitives $\gcd(x,y)=1$:
$$A^2+B^2 = (a^2+b^2)^k\tag1$$
where $A,B$ is the expansion of $(a+bi)^k = A+Bi$. Example, $$(a+bi)^3 =(a^3 - 3 a b^2)+ (3 a^2 b - b^3)i$$ hence, $$(a^3 - 3 a b^2)^2+ (3 a^2 b - b^3)^2 = (a^2+b^2)^3$$ and so on for other $k$.
*Non-primitives $\gcd(x,y)\neq1$:
$$a^2(a^2+b^2)^{k-1}+b^2(a^2+b^2)^{k-1} = (a^2+b^2)^k\tag2$$
where, for both Forms $1$ and $2$, we use some rational $a,b$?
Example. The first three solutions for $k=5$ above use:
$$a,b = 2/5,\;11/5\quad \text{Form 2}$$
$$a,b = 1,\;2\quad \text{Form 2}$$
$$a,b = -2,\;1\quad \text{Form 1}$$
| If $x^2+y^2=z^k$, then $(x+iy)(x-iy)=z^k$. The Gaussians are a UFD, so if $\gcd(x+iy,x-iy)=1$ then this forces $x+iy=(r+si)^k$ for some integers $r$ and $s$. I think this is your form 1.
If there exists $d$ dividing both $x+iy$ and $x-iy$, then $d$ must divide their sum, $2x$, and their difference, $2iy$. So if $x$ and $y$ are relatively prime, then $d$ is nontrivial if and only if $x$ and $y$ are both odd, and $d=1+i$. But if $x$ and $y$ are odd, then $x^2+y^2\equiv2\bmod4$, while $z^k\equiv0\bmod4$, contradiction.
So we're left with the case where there is an integer $d>1$ dividing both $x$ and $y$. To be continued....
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all incongruent solutions of $x^8\equiv3\pmod{13}$. Find all incongruent solutions of $x^8\equiv3\pmod{13}$.
I know that $2$ is a primitive root of $13$ and that $2^4\equiv3\pmod{13}$, so we want to solve $x^8\equiv2^4\pmod{13}$.
Now, $\gcd(8,\phi(13))=\gcd(8,12)=4$ divides the exponent of $2$, which is $4$, so $x^8\equiv3\pmod{13}$ has exactly $4$ incongruent solutions modulo $13$.
I was able to find on my calculator (using brute force) that these solutions are $4,6,7,$ and $9$ (i.e. $\pm4,\pm6$), but how would I go about finding them without using a calculator?
| We have that $Z/13Z$ it's a field. In $Z/13Z$ the inverse of $3$ is $9$ so if you multiply by $9$ will have $9x^8$ is congruent by 1 modulo 9. In Z/13Z, $x^{12}$ is congruent by 1 modulo 13. So $9x^8-x^{12}$ is congruent by 0 modulo 13. So $x^8(x^4-9)$ is congruent with 0 modulo 13. Result that x is congruent by 0 mod 13(is not solution) or $x^4-9=(x^2-3)(x^2+3)$ is congruent by 0 modulo 13 and now it is more clear.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of the following $n \times n$ determinantes Find the value of the following $n \times n$ determinantes
*
*$$\begin{vmatrix}
a_1+x & x & x & \ldots & x \\
x & a_2+x & x & \ldots & x \\
x & x & a_3+x & \ldots & x \\
\vdots & \vdots& &\ddots& \vdots\\
x & x & x & \ldots & a_n+x \\
\end{vmatrix}$$
*$$\begin{vmatrix}
a_1+x & a_2 & a_3 & \ldots & a_n \\
a_1 & a_2+x & a_3 & \ldots & a_n \\
a_1 & a_2 & a_3+x & \ldots & a_n \\
\vdots & \vdots& &\ddots& \vdots\\
a_1 & a_2 & a_3 & \ldots & a_n+x \\
\end{vmatrix}$$
Both seem to be equally complicated to solve, I reckon that it's needed to subtract the $x$ from the diagonal in each term. I tried by subtracting the $(k-1)$-th row from the $k$-th row, however that doesn't really lead me to anything more comfortable whatsoever. So help is greatly appreciated, also perhaps a link to some methods on solving these kind of problems, would also be helpful.
| Let
\begin{align}
T_n &=
\begin{vmatrix}
a_1+x & x & x & \cdots & x \\
x & a_2+x & x & \cdots & x \\
x & x & a_3+x & \cdots & x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x & x & x & \cdots & a_n+x
\end{vmatrix} \\
&=
\begin{vmatrix}
a_1+x & x & x & \cdots & x \\
-a_1 & a_2 & 0 & \cdots & 0 \\
-a_1 & 0 & a_3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-a_1 & 0 & 0 & \cdots & a_n
\end{vmatrix}
\end{align}
Expand $T_n$ with the last column, and we can get
$$
T_n = x a_1 a_2 \cdots a_{n-1} + a_n T_{n-1}
$$
It's obvious that
$$
T_1 = a_1 + x
$$
By induction, we can get
$$
T_n = x \sum^n_{j=1} \prod^n_{\scriptstyle i=1 \atop \scriptstyle i\ne j} a_i + \prod^n_{i=1} a_i
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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prove inequality $2(x+z)^3<(3x+z)(x+3z)$ Let $0<x<1,0<z<1$. Then
$$
2(x+z)^{3}<(3x+z)(x+3z)
$$
This checks out numerically, but I don't know why.
| In the proof I'll use the fact that : $x^3<x^2$, $z^3<z^2$ ,$x^2<x$, $z^2<z$ and $2xz<x^2+z^2$. We have
\begin{align}
2(z+x)^3=2x^3+2z^3+6x^2z+6z^2x&< 2x^2+2z^2+12xz\\ &=2x^2+2z^2+10xz+2xz\\
&<2x^2+2z^2+10xz+z^2+x^2\\
&=(3x+z)(x+3z)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
| Assume $1\le a\le b\le c$. Then $2^n=a!(1+b\cdots(a+1)+c\cdots(a+1)$.
It follows that either $a=2$ or $a=1$.
Consider the case $a=2$. Then one of the numbers $b\cdots(a+1)$ and $c\cdots(a+1)$ must be even, and the other odd, and since $b\cdots(a+1)$ divides $c\cdots(a+1)$ it follows that $b\cdots(a+1)$ must be odd, and $c\cdots(a+1)$ must be even. Hence $b=2$ or $b=3$, and $c\ge 4$. So we are looking for $c$ such that either $\displaystyle 2+\frac{c!}2$ is a power of $2$, or $\displaystyle 4+\frac{c!}2$ is a power of $2$.
These are only ideas how to possibly start (seems right), I will think if I could add more details and a more definite answer. Edit: I added all details below.
Consider the case $a=1$. Then $2^n=1+b!+c!$. It follows that $b=1$ and $\displaystyle 2^n=2(1+\frac{c!}2)$, so $\displaystyle 2^{n-1}=1+\frac{c!}2$. This works if (and only if) $c=2$ or $c=3$. So we get solutions $(a,b,c)=(1,1,2)$ with $1+1+2=4=2^2$, and $(a,b,c)=(1,1,3)$ with $1+1+6=8=2^3$. It looks like you missed these small integers solutions.
Back to the case $a=2$. First let $b=2$. Then $\displaystyle 2^n=2(1+1+\frac{c!}2)$, so $\displaystyle 2^{n-2}=1+\frac{c!}4$. No solutions in this case.
Finally, let $a=2$, $b=3$. Then $\displaystyle 2^n=2(1+3+\frac{c!}2)$, so $\displaystyle 2^{n-3}=1+\frac{c!}8$. Works if $c=4$ or $c=5$. Solutions $(a,b,c)=(2,3,4)$ when $2+6+24=32=2^5$, and $(a,b,c)=(2,3,5)$ when $2+6+120=128=2^7$ (these are the two solutions that you found). This finishes all possible cases, no other solutions.
| {
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Find the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$ Find the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$
Since the domain of definition of $\sec^{-1} x$ is $x\leq-1$ or $x\geq 1$.
So $(\frac{2+x}{1-x})^{\frac{1}{x}}\leq -1$ or $(\frac{2+x}{1-x})^{\frac{1}{x}}\geq 1$
The domain of $(\frac{2+x}{1-x})^{\frac{1}{x}}$ is known to be $[-2,0)\cup(0,1)$.
I am stuck here.Please help.
| $\Longrightarrow$The Range of $(\frac{2+x}{1-x})^{\frac{1}{x}}$ =
[0,$\infty$)
The domain of Sec$^{-1}x$ = $\mathbb{R}$- (-1,1)
$\Longrightarrow$1$\leq$$\left(\frac{2+x}{1-x}\right)^{\frac{1}{x}}$$\Longrightarrow$
x$\geq$ -$\frac{1}{2}$$\Longrightarrow$ the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$
is [$\frac{-1}{2}$,1) -$\left\{ 0\right\} $
| {
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Infinite trigonometric summation Can the following summation be written in a finite number of terms:
$$\sum_{r=1}^{\infty}\frac{\tan(\theta/2^r)}{2^{r-1}\cos(\theta/2^{r-1})}$$
I tried to simplify the expression using trigonometric identities and then converting the infinite summation into a definite integral. I couldn't think of any way to get it into that since the variable $r$ occurs as an exponent.
So is there any other possible simplification for the expression?
| Let $a_n = \dfrac{\tan(\frac{x}{2^n})}{2^{n-1}\cos(x/2^{n-1})}$. We want to find $\sum_{n=1}^{\infty} a_n$.
Let's write out the first term:
$$a_1 = \frac{\tan(x/2)}{\cos(x)}$$
and invoke the identity $\tan(\frac{x}{2}) = \csc(x)-\cot(x)$:
$$a_1 = \frac{\csc(x)-\cot(x)}{\cos(x)} \\ = \frac{1}{\sin(x)\cos(x)} - \frac{\cos(x)}{\sin(x)\cos(x)} \\ = \frac{2}{2\sin(x)\cos(x)} - \frac{1}{\sin(x)} \\ = \frac{2}{\sin(2x)} - \csc(x) \\ a_1 = 2\csc(2x) - \csc(x).$$
By performing the same procedure for the next couple terms ($a_2$ and $a_3$), we have the following:
$$a_2 = \csc(x) - \frac{1}{2}\csc(x/2) \\
a_3 = \frac{1}{2}\csc(x/2) - \frac{1}{4}\csc(x/4) \\ \vdots \\ \vdots \\ a_n = \frac{1}{2^{n-2}} \csc(x/2^{n-2}) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}).$$
Now it's time to consider partial sums. Let $s_n = a_1 + a_2 + \cdots + a_n$. A pattern emerges quickly because of the telescoping nature of the terms:
$$s_1 = a_1 = 2\csc(2x) - \csc(x) \\ \ \\ \ \\ s_2 = s_1 + a_2 = [2\csc(2x) - \csc(x)] + [\csc(x) - \frac{1}{2}\csc(x/2)] \\ s_2 = 2\csc(2x) - \frac{1}{2}\csc(x/2) \\ \ \\ \ \\ s_3 = s_2 + a_3 = [2\csc(2x) - \frac{1}{2}\csc(x/2)] + [\frac{1}{2}\csc(x/2) - \frac{1}{4}\csc(x/4)] \\ s_3 = 2\csc(2x) - \frac{1}{4}\csc(x/4) \\ \vdots \\ \vdots \\ s_n = 2\csc(2x) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}).$$
Now that we have an expression for the $n$-th partial sum, we can attempt considering the limiting case since $\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} s_n$. That is,
$$\lim_{n \to \infty} s_n \\
= \lim_{n \to \infty} \left[ 2\csc(2x) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}) \right] \\
= 2\csc(2x) - \lim_{n \to \infty} \left[\frac{1}{2^{n-1}}\csc(x/2^{n-1}) \right] \\
= 2\csc(2x) - \lim_{n \to \infty} \left[\frac{1}{2^{n-1}\sin(x/2^{n-1})}\right]
$$
Unfortunately, this limit is not defined. However, for $x\neq 0$, it can be shown that we're left with
$$\lim_{n \to \infty} s_n = 2\csc(2x) - \frac{1}{x}.$$
So altogether, and replacing $n$ with $r$ and replacing $x$ with $\theta$, we have $$ \sum_{r=1}^{\infty} \frac{\tan(\frac{\theta}{2^r})}{2^{r-1}\cos(\theta/2^{r-1})} = 2\csc(2\theta) - \frac{1}{\theta}.$$
I'm not sure how to find out the restrictions on $x$ this late at night. Fro example, since $x=0$ gives zero for the sum but not this answer, perhaps someone else could take over at this point. However, I hope the derivation is of some use to you!
| {
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"url": "https://math.stackexchange.com/questions/1573701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Existence of primitive Pythagorean triples Let $(a,b,c)$ be a Pythagorean triple, which means $c^2=a^2+b^2$.
If $c$ is odd and $a$ & $b$ are relatively prime, then there exist integers $m$ and $n$ such that $c=m^2+n^2, ~a=m^2-n^2, ~b=2mn$.
One can easily check the above by proving $\gcd(\frac{a+c}{2},\frac{a-c}{2})=1$.
My question is whether the converse of the above also holds or not.
;Whenever the odd number $c$ is a sum of 2 squares, then there exist integers $a$ and $b$ satisfying $c^2=a^2+b^2$ and $\gcd(a,b)=1$.
Any help will be appreciated.
| $$\text{For any C, if } n=\frac{-1\pm \sqrt{2C-1}}{2}\text { yields an integer for }n, \text{then you have a primitive triple.}$$
$$A=2n^2+1\quad B=2n^2+2n\quad C=2n^2+2n+1 \text{ where }C-B=1$$
$$\text{If }n=\sqrt{\frac{C-1}{4}}\text{ yields an integer for n, you have a primitive triple.}$$
$$A=4n\quad B=4n^2-1\quad C=4n^2+1\quad \text{where }C-B=2$$
Otherwise
$$C=m^2+n^2\Rightarrow n=\sqrt{C-m^2}\text{ where } \biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor$$
For any non-integer $n$, that value of $C$ is not part of any primitive. For any integer value of $n$ then $C$ is either part of a primitive or a multiple where the multiple is either $2$ or a perfect square. The only way to be sure is to generate them with Euclid's formula and test the GCD.
For example; let $C=25$ then $m_{min}=\lceil\sqrt{12.5}\space\rceil=4$ and $m_{max}=\lfloor\sqrt{25}\rfloor=5.$ We can see that $\sqrt{25-16}=3$ and $f(4,3)=(7,24,25)$. We can also see by inspection, however that $m=5$ leads to a trivial solution. This only happens when $C$ itself is a perfect square.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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About equality of nested radicals. Allow me, please, reformulate this problem.
The equal numbers $$a=\sqrt{13}+\sqrt{10+2\sqrt{13}}$$
$$b=\sqrt{5+2\sqrt3}+\sqrt{18-2\sqrt3+2\sqrt{65-26\sqrt3}}$$ have the same minimal polynomial (over $\mathbb Z$) $$f(x)=x^4-46x^2-104x-43$$ in which it appears just one sign change so the equation $f(x)=0$ has at most a positive root (Descartes' Sign Rule). Thus, since $a>0$ and $b>0$ we verify that $a=b$.
This elegant way is not possible to be apply to the equal numbers $$\alpha=\sqrt {29}+\sqrt{14+2\sqrt {29}}$$ $$\beta=\sqrt{7+2\sqrt 5}+\sqrt{36-2\sqrt 5+2\sqrt{203-58\sqrt 5}}$$ whose common minimal polynomial is $$f(x)=x^4-86x^2-232x+109$$
How to verify using $f(x)$ that $\alpha=\beta$ ?
It is not hard but, is there a method something elegant (like the first given example) to do it?
| The two examples are of the following form:
Take $a,b,c\in \mathbb{R}$ such that $a=b^2-4c$, and suppose that all the square roots below are well defined. Then if
$$\alpha=\sqrt{a}+\sqrt{2b+2\sqrt{a}}$$
and
$$\beta=\sqrt{b+2\sqrt{c}}+\sqrt{a+b-2\sqrt{c}+2\sqrt{ab-2a\sqrt{c}}}$$
we have $\alpha=\beta$.
We have only to see that $$a+b-2\sqrt{c}+2\sqrt{ab-2a\sqrt{c}}=\big (\sqrt{a}+\sqrt{b-2\sqrt{c}} \big)^2$$
to simplify by $\sqrt{a}$, and then to take $\sqrt{b-2\sqrt{c}}+ \sqrt{b+2\sqrt{c}}$ and $\sqrt{2b+2\sqrt{a}}$ to the power two.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the points on the graph of the function $\frac{1}{3}x^3 + x^2 - x - 1$ at which the slope is $-1$; $2$ Please help me. Find the points on the graph of $\frac{1}{3}x^3 + x^2 - x - 1$ at which the slope is
(a) $-1$; (b) $2$.
I don't know where to begin.
| The slope of the graph is equal to the derivative $$\frac{d}{dx}\left(\frac13x^3+x^2-x-1\right)=x^2+2x-1$$ Now solve for (a) $$x^2+2x-1=-1 \iff x^2+2x=0 \iff x(x+2)=0$$ and for (b) $$x^2+2x-1=2 \iff x^2+2x-3=0 \iff (x-1)(x+3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576515",
"timestamp": "2023-03-29T00:00:00",
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Can you simplify this term? $$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}$$
| $X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^{T}}-1}$
$====================$
$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c-r}{r}+\frac{1-c}{(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r^2(1+r)^{T+1}}}{\frac{(c-r)(1+r)^{T}+(1-c)r}{r(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r(1+r)}}{\frac{(c-r)(1+r)^{T}+(1-c)r}1}$
$====================$
$X=\frac{(c(1+r)^{T+1}+(1-c)r^2)1}{r(1+r)((c-r)(1+r)^{T}+(1-c)r)}$
$====================$
$X=\frac{c(1+r)^{T+1}+(1-c)r^2}{r(1+r)((c-r)(1+r)^{T}+(1-c)r)}$
| {
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"url": "https://math.stackexchange.com/questions/1578116",
"timestamp": "2023-03-29T00:00:00",
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Circle and a line that passes through it Given a line with equation: $y=ax-3$ that passes through a circle with equation $(x-1)^2+(y-1)^2= 1$. Find the range of values of $a$.
I tried graphing and got: $0<x<2$ and $0<y<2$.
I also tried finding $a$ by substituting $x$ and $y$ into $y=ax-3$
which really confuses me.
Could you help me in solving this problem?
| we substitute
$y=ax-3$
in
$(x-1)^2+(y-1)^2= 1$
$(x-1)^2+(ax-3-1)^2= 1$
$(x-1)^2+(ax-4)^2= 1$
$x^2-2x+1+a^2x^2-8ax+16=1$
we have this second degree equation
$x^2(1+a^2)+x(-2-8a)+16=0$
with
$\Delta =(-2-8a)^2-4\cdot(1+a^2)\cdot 16$
if $\Delta <0$
the equation have no solution and the line not intersect the circle
if $\Delta =0$
the equation have one solution and the line is tangent to the circle
if $\Delta >0$
the equation have two solutions and the line intersect the circle in 2 points
$\Delta =4+32a+64a^2-64-64a^2$
$\Delta =32a-60$
so if
$32a-60>0$
$a>\frac{15}{8}$
the line intersect the circle in 2 points
and if
$32a-60=0$
$a=\frac{15}{8}$
the line is tangent to the circle
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{x}{\sqrt{1+x^2}} \lt \arctan x$ for every $x \gt 0$
Prove $$\frac{x}{\sqrt{1+x^2}} \lt \arctan x$$ for every $x \gt 0$.
I proved half of it with lagrange rule but that I got stuck. Any ideas?
I can upload my work if you want.
| let be
$\displaystyle f:[0,\infty) \rightarrow \mathbb{R} $
$f(x)=\frac{x}{\sqrt{1+x^2}} - \arctan x$
$\displaystyle f'(x)=\frac{x' \cdot \sqrt{1+x^2} - x \cdot (\sqrt{1+x^2})'}{(\sqrt{1+x^2})^2} -(\arctan x)'$
$\displaystyle f'(x)=\frac{1 \cdot \sqrt{1+x^2} - x \cdot \frac{1}{2 \sqrt{1+x^2}}\cdot 2 \cdot x}{(\sqrt{1+x^2})^2} - \frac{1}{1+x^2}$
$\displaystyle f'(x)=\frac{ \sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}} }{{1+x^2}} - \frac{1}{1+x^2}$
$\displaystyle f'(x)=\frac{ \frac{1+x^2}{\sqrt{1+x^2}} - \frac{x^2}{\sqrt{1+x^2}} -1}{{1+x^2}} $
$\displaystyle f'(x)=\frac{ 1+x^2-x^2 -\sqrt{1+x^2} }{{(1+x^2) \cdot \sqrt{1+x^2}}} $
$f'(x)=\frac{1-\sqrt{1+x^2}}{(1+x^2)\sqrt{1+x^2}} <0$
because $f'(x) <0$ we have that $f$ is decreasing on $ [0,\infty)$
and thus
$f(0)>f(x)$ for any $x$ in $ (0,\infty)$
but $f(0)=0$, so
$0>f(x)$
$0>\frac{x}{\sqrt{1+x^2}} - \arctan x$
$\arctan x>\frac{x}{\sqrt{1+x^2}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can one show the inequality Let $a,b,n$ be natural numbers (in $\mathbb{N}^*$) such that
$a>b$ and $n^2+1=ab$
How can one show that $a-b\geq\sqrt{4n-3}$, and for what values of $n$ equality holds?
I tried this:
We suppose that $a-b\geq\sqrt{4n-3}$ so
$$a^2-2ab+b^2\geq 4n-3\,.$$
And we have $ab=n^2+1$.
So $a^2-2(n^2+1)+b^2-4n+3\geq 0$.
So $a^2-2n^2+b^2-4n+1\geq 0$.
I'm stuck here! How do I proceed?
| Let $x:=a-n$ and $y:=b+x-n$ (that is, $a=n+x$ and $b=n-x+y$). It can be easily seen that $x$ and $y$ are positive integers. We want to find $a-b=2x-y$. Note that $n^2+1=ab$ implies that $x^2-xy=ny-1$, or equivalently, $(2x-y)^2=y^2+4(ny-1)$. As $y\geq 1$, we have $$(a-b)^2=(2x-y)^2\geq 1+4(n-1)=4n-3\,$$ Therefore, $a-b\geq \sqrt{4n-3}$. The equality holds iff $y=1$, which leads to $$(n,a,b)=\left(x^2-x+1,x^2+1,x^2-2x+2\right)$$ for a positive integer $x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I would like to calculate $\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$ I want to calculate the following limit: $$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$$
or prove that it does not exist. Now I know the result is $-3$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
| Notice, $$\lim_{x\to \pi/6}\frac{2\sin^2 x+\sin x-1}{2\sin^2 x-3\sin x+1}$$
$$=\lim_{x\to \pi/6}\frac{\sin x-(1-2\sin^2 x)}{2-3\sin x-(1-2\sin^2 x)}$$
$$=\lim_{x\to \pi/6}\frac{\sin x-\cos 2x}{2-3\sin x-\cos 2x}$$
Apply L'hospital's rule for $\frac00$ form
$$=\lim_{x\to \pi/6}\frac{\cos x+2\sin 2x}{-3\cos x+2\sin2x}$$
$$=\frac{\frac{\sqrt 3}{2}+\sqrt 3}{-\frac{3\sqrt 3}{2}+\sqrt 3}=\color{red}{-3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$ If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$
$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\pi-\arcsin z$
$x\sqrt{1-y^2}+y\sqrt{1-x^2}=z$
Similarly,$y\sqrt{1-z^2}+z\sqrt{1-y^2}=x$
Similarly,$x\sqrt{1-z^2}+z\sqrt{1-x^2}=y$
Adding the three equations,we get
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=\frac{x+y+z}{2}$
I am stuck here,please help me.Thanks.
| Let $\sin^{-1}(x) = A\Rightarrow x=\sin A$ and $\sin^{-1}(y)=B\Rightarrow y=\sin (B)$
and $\sin^{-1}(z)=C\Rightarrow z=\sin C$
So above we have given $A+B+C = \pi$
Now we have to prove $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$
So for $\bf{L.H.S}$
$$\sin 2A+\sin 2B+\sin 2C = 2\sin (A+B)\cdot \sin (A-B)+2\sin C\cos C$$
$$=2\sin C\cdot \sin (A-B)+2\sin C\cos C$$
$$=2\sin C\left[\cos(A-B)+\cos(\pi-(A+B))\right]$$
$$=4\sin C\cdot \left[\cos(A-B)-\cos(A+B)\right]$$
$$=4\sin A\sin B\sin C$$
| {
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If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$? The common multiplication rules are
$$(-1) \cdot (-1) = +1 \\
(+1) \cdot (+1) = +1$$
But these rules seem asymmetric. Because of these rules it is not possible e.g. to solve the equation
$$x^2 = -1 $$
without introducing complex numbers. There would be two "more symmetric" alternatives:
$$(-1) \cdot (-1) = -1 \\
(+1) \cdot (+1) = +1$$
or
$$(-1) \cdot (-1) = +1 \\
(+1) \cdot (+1) = -1$$
Both alternatives would allow to solve the equations
$$x^2 = +1 \\ x^2 = -1 $$
without the need to introduce complex numbers. However the two more symmetric alternatives are not used. What is the reason for that? Would math break down, if one would adopt one of these different conventions?
| Math would certainly break down, here's why. Assume that every number is equal to itself, addition is associative, multiplication distributes over addition , $1$ and $-1$ are additive inverses, $1$ is the multiplicative identity, and $0$ is the additive identity. Then it follows that:
$$\begin{array}{lll}
\bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1) &=& \bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1)\\
\bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + \bigg(1\cdot (-1) + (-1)\cdot(-1)\bigg)\\
1\bigg(1+(-1)\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + (-1)\bigg(1 + (-1)\bigg)\\
1\bigg(0\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + (-1)\bigg(0\bigg)\\
0 + (-1)\cdot(-1) &=& 1\cdot 1 + 0\\
(-1)\cdot(-1) &=& 1\cdot 1 \\
\end{array}$$
The only loose end not assumed above is $(-1)\cdot 0 = 0$
$$(-1)\cdot 0 = (-1)(0+0)$$
$$(-1)\cdot 0 = (-1)\cdot 0+ (-1)\cdot 0$$
$$\dots$$
| {
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Prove This Inequality ${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$ $${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$$
I see I should use Riemann sum, and that
$$\int_0^{\infty} {dx \over {1+x^2}} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}}$$
But how do I explain this exactly, and how to get the other half of the inequality (with ${\pi \over 2} + 1$)?
| Hint. For each $n=0,1,2,\cdots$, you have, for all $x \in [n,n+1]$,
$$
n^2+1\leq 1+x^2\leq (n+1)^2+1
$$ giving
$$
\frac1{(n+1)^2+1}\leq \frac1{x^2+1}\leq \frac1{n^2+1}
$$ then integrating with respect to $x$ from $n$ to $n+1$,
$$
\int_n^{n+1}\frac1{(n+1)^2+1}\:dx\leq \int_n^{n+1}\frac1{x^2+1}\:dx\leq \int_n^{n+1}\frac1{n^2+1}\:dx,
$$ observe that
$$
\int_n^{n+1}\frac1{(n+1)^2+1}\:dx=\frac1{(n+1)^2+1},\quad \int_n^{n+1}\frac1{n^2+1}\:dx=\frac1{n^2+1}
$$ giving
$$
\frac1{(n+1)^2+1}\leq \int_n^{n+1}\frac1{x^2+1}\:dx\leq \frac1{n^2+1}
$$ then sum from $n=0$ to $+\infty$, and with a change of indice on the left, you may conclude easily using
$$
\int_0^\infty\frac1{x^2+1}\:dx=\lim_{x \to +\infty}\arctan x-\arctan 0=\frac{\pi}2.
$$
| {
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Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series and seems to me of the form:
$-3-7-11-15\ldots $
I feel like its of the closed form:
$\sum(-4i+1)$
So how do I prove that the equality is right?
| Hint: In order to show
\begin{align*}
\sum_{j=0}^k(-1)^{j-1}j^2=(-1)^{k-1}\frac{k(k+1)}{2}\qquad\qquad k\geq 0\tag{1}
\end{align*}
we consider sequences $(a_k)_{k\geq 0}$ and the corresponding generating functions $\sum_{k=0}^{\infty}a_kx^k$ as building blocks to generate the left hand sum in (1). This enables us to calculate the right hand side of (1).
\begin{array}{crl}
(a_k)_{k\geq 0}\qquad &\qquad A(x)=&\sum_{k=0}^{\infty}a_kx^k\\
\hline\\
((-1)^{k-1})_{k\geq 0}\qquad&\qquad -\frac{1}{1+x}=&\sum_{k=0}^{\infty}(-1)^{k-1}x^k\\
((-1)^{k-1} k)_{k\geq 0}\qquad&\qquad -\left(x\frac{d}{dx}\right)\frac{1}{1+x}
=&\frac{x}{(1+x)^2}\\
&=&\sum_{k=0}^{\infty}(-1)^{k-1}kx^k\\
((-1)^{k-1} k^2)_{k\geq 0}\qquad&\qquad -\left(x\frac{d}{dx}\right)^2\frac{1}{1+x}
=&\frac{x(1-x)}{(1+x)^3}\\
&=&\sum_{k=0}^{\infty}(-1)^{k-1}k^2x^k\\
(\sum_{j=0}^{k}(-1)^{j-1} j^2)_{k\geq 0}\qquad&\qquad -\frac{1}{1-x}\left(x\frac{d}{dx}\right)^2\frac{1}{1+x}
=&\frac{x}{(1+x)^3}\\
&=&\sum_{k=0}^{\infty}\left(\sum_{j=0}^{k}(-1)^{j-1}j^2\right)x^k\tag{2}\\
\end{array}
We can see in the small intro above the operator $x\frac{d}{dx}$ transforms $a_k$ to $ka_k$ and the multiplication with $\frac{1}{1-x}$ acts as summation operator.
It is also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a generating series.
We obtain from (2) for $k\geq 1$
\begin{align*}
\sum_{j=0}^{k}(-1)^{j-1}j^2&=[x^k] \frac{-1}{1-x}\left(x\frac{d}{dx}\right)^2\frac{1}{1+x}\\
&=[x^k] \frac{x}{(1+x)^3}\\
&=[x^{k-1}]\sum_{n=0}^{\infty}\binom{-3}{n}x^{n}\tag{3}\\
&=\binom{-3}{k-1}\\
&=(-1)^{k-1}\binom{k+1}{2}\tag{4}\\
&=(-1)^{k-1}\frac{k(k+1)}{2}
\end{align*}
Comment:
*
*In (3) we use the binomial series representation.
*In (4) we use the formula $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$ for binomial coefficients.
| {
"language": "en",
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"question_score": "4",
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Remainder of the numerator of a harmonic sum modulo 13 Let $a$ be the integer determined by
$$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}.$$
Determine the remainder of $a$ when divided by 13.
Can anyone help me with this, or just give me any hint?
| HINT:
Take the equation:
$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}$
Multiply each side by $23!$:
$\frac{23!}{1}+\frac{23!}{2}+...+\frac{23!}{23}=a$
Divide each side by $13$:
$\frac{23!}{1\times13}+\frac{23!}{2\times13}+...+\frac{23!}{23\times13}=\frac{a}{13}$
There is only one non-integer element in the series: $\frac{23!}{13\times13}$.
Hence the remainder of $a$ when divided by $13$ is equal to the remainder of $\frac{23!}{13}$ when divided by $13$.
In other words: $a\equiv\frac{23!}{13}\pmod{13}$.
| {
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} |
Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$
Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$ for any $p,q$ with $p+q = 1$.
Should I prove this using the triangle inequality? Also should I split the proof up into two parts for the if and only if?
| Below is a complete solution. Don't look at it if you want to solve this problem by yourself. Also, yes, the Triangle Inequality is definitely required in this problem.
Assume that $a,b,c>0$ satisfy the condition that $pa^2+qb^2>pqc^2$ for any $p,q \in\mathbb{R}$ such that $p+q=1$. Choosing $p$ and $q$ to be $\frac{b}{a+b}$ and $\frac{a}{a+b}$, respectively, we obtain $c<a+b$. Choosing $p$ and $q$ to be $\frac{b}{b-a}$ and $\frac{a}{a-b}$, respectively, we obtain $c>|a-b|$, which implies $a<b+c$ and $b<c+a$. Ergo, $a$, $b$, and $c$ form a triangle.
Conversely, suppose that $a,b,c>0$ form a triangle. Let $p,q\in\mathbb{R}$ be such that $p+q=1$. If $p=0$ or $q=0$, then the inequality $pa^2+qb^2>pqc^2$ holds trivially. Now, suppose that $p,q>0$. Then, by the Cauchy-Schwarz Inequality, $$\frac{a^2}{q}+\frac{b^2}{p}=\left(\frac{a^2}{q}+\frac{b^2}{p}\right)(q+p)\geq (a+b)^2>c^2\,,$$ whence $pa^2+qb^2>pqc^2$ holds. If $p>0$ and $q<0$, then we note by symmetry that $\frac{1}{1-q}a^2+\frac{-q}{1-q}c^2>\frac{(-q)}{(1-q)^2}b^2$, which gives $$pa^2-pqc^2>\frac{p(-q)}{1-q}b^2=-qb^2\,,$$ as $p=1-q$. The previous inequality is precisely $pa^2+qb^2>pqc^2$. If $p<0$ and $q>0$, we can prove the inequality $pa^2+qb^2>pqc^2$ in a similar manner.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$
My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1}$$
Now put $\left(x-\frac{1}{x}\right)=t,x\ne0$. Then we get $$f(t) = \frac{t}{t^2+1} = \frac{1}{2}\left(\frac{2t}{1+t^2}\right)$$
Now put $t=\tan \theta$. Then $$f(\theta) = \frac{1}{2}\frac{2\tan \theta}{1+\tan^2 \theta} = \frac{1}{2}\sin 2\theta$$
So we get $$-\frac{1}{2}\leq f(\theta)\leq \frac{1}{2}\Rightarrow f(\theta)\in \left[-\frac{1}{2}\;,\frac{1}{2}\right]$$
My question is: Is my solution right? If not, then how can we solve it?
| take you solution,
$$f(t)=\dfrac{t}{t^2+1}$$
since
$$f(t)=-f(-t)$$
so we only consider $t\ge 0$ Use AM-GM inequality we have
$$f(t)=\dfrac{t}{t^2+1}\le\dfrac{t}{2t}=\dfrac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Volume Integral of this set I'm not sure about this exercise. Be:
$$E=\left\{(x,y,z) \in \mathbb{R^3} : x\geq 0, y\geq 0, 0\leq z\leq \frac{1}{\sqrt{x^2+y^2}}-1\right\}$$
Find:
$$\int_{E} z\, \max\{x,y\}\: dx \, dy \, dz$$
My idea is that since $x,y\geq 0$, one of the two variables will be greater than the other in a section of the quadrant divided by the bisector and vice versa. For example, let's suppose that $x\geq y$, the section of the volume would be something like:
$$\left\{(x,y,z)\in \mathbb{R^3} : 0\leq x\leq \infty, 0\leq y\leq x,0\leq z\leq \frac{1}{\sqrt{x^2+y^2}}-1\right\}$$
The volume of this would be:
$$\int_0^\infty\int_0^x\int_0^{\frac{1}{\sqrt{x^2+y^2}}-1} zx\: dz\, dy\, dx$$
Then the volume of $E$ would be this volume plus the volume of the case where $y\geq x$, does this look correct to you?
| Partial answer:
The upper limit of integration for the $x$ variable doesn't actually go to infinity. Note that the bounds for $z$ imply an additional constraint on the range of $x$ and $y$:
$$\begin{align}
0\le z\le\frac{1}{\sqrt{x^2+y^2}}-1
&\implies0\le\frac{1}{\sqrt{x^2+y^2}}-1\\
&\implies1\le\frac{1}{\sqrt{x^2+y^2}}\\
&\implies\sqrt{x^2+y^2}\le1\\
&\implies x^2+y^2\le1\\
&\implies0\le x\le1\land0\le y\le\sqrt{1-x^2}.\\
\end{align}$$
With the correct integration bounds for your triple integral, and strategically splitting up the integration intervals to eliminate the max functions, we find:
$$\begin{align}
I
&=\iiint_{E}\mathrm{d}^{3}v\,\left[z\max{\left(x,y\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&~~~~~+\int_{\frac{1}{\sqrt{2}}}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&~~~~~+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{x}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&~~~~~+\int_{\frac{1}{\sqrt{2}}}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,xz\\
&~~~~~+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{x}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,yz\\
&~~~~~+\int_{\frac{1}{\sqrt{2}}}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,xz.\\
\end{align}$$
Presumably you can take the helm from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Uniform convergence of $ \sum_{n=1}^\infty \frac{x\cdot\sin\sqrt {x/n}} {n + x} $ I need to check on the uniform convergence.
$$ \sum_{n=1}^\infty \frac{x\cdot\sin\sqrt {\frac x n}} {n + x} $$
on the interval:
a) $\; (0, 1) $
b) $\; (1, +\infty) $
I think I need to use Cauchy ratio, but I cannot understand how
| For part $(a)$:
$$\begin{align}
\left|\sum _{n=1}^\infty \frac{x\sin {\sqrt {\frac{x}{n}}}}{n+x}\right| &\le \sum _{n=1}^\infty \left|\frac{x\sin {\sqrt {\frac{x}{n}}}}{n+x}\right|\\\\
&\le \sum_{n=1}^\infty \frac{\left|x\,\sqrt{\frac{x}{n}}\right|}{n+x}\,\,\text{since} |\sin x|\le |x|\\\\
&\le \sum_{n=1}^\infty \left|\frac{x^{\frac{3}{2}}}{n^{\frac{1}{2}}(n+x)}\right|\\\\
&\le \left|\sum_{n=1}^\infty \frac{x^{\frac{3}{2}}}{n^{\frac{3}{2}}}\right|\\\\
&\le \sum_{n=1}^\infty \frac{1}{n^{\frac{3}{2}}} \to 0
\end{align}$$
provided $|x|<1$
By Weistrass M Test we have the series converges uniformly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Calculate $\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}{\mathrm{B}(1 - x\,, \,x)}\,\mathrm{d}x$ How does one calculate
$\displaystyle{\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}
{\mathrm{B}(1 - x\,,\,x)}\,\mathrm{d}x}$ ?.
The observation
$\displaystyle{\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}{\Gamma\left(1 - x\right)\Gamma\left(x\right)}\,\mathrm{d}x =
\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}
{\mathrm{B}(1 - x\,,\,x)}\,\mathrm{d}x}$ seems useless here.
The answer, according to wolfram, is $2$.
| For the proof, we will be using the following fact $$(1-\alpha)^{\left(\frac{1}{\alpha}-1\right)} = \int_0^1 \frac{\sin\left(\pi x \right)}{\pi (1-\alpha x)} \left[x^x\left(1-x\right)^{1-x}\right]^{-1}\ dx.$$
Taking your original sum, we have
\begin{align}
\int_0^1 \frac{x^{-1-x}(1-x)^{x-2}}{\Gamma(x)\Gamma(1-x)} dx &= \int_0^1 \frac{\sin(\pi x)}{\pi x(1-x)} \left[x^x\left(1-x\right)^{1-x}\right]^{-1} \ dx \\
&= \int_0^1 \frac{\sin(\pi x)}{\pi} \left[x^x\left(1-x\right)^{1-x}\right]^{-1} \left( \frac{1}{x}+\frac{1}{1-x}\right)\ dx \\
&=2\int_0^1 \frac{\sin(\pi x)}{\pi (1-x)} \left[x^x\left(1-x\right)^{1-x}\right]^{-1} \ dx \\
&= 2\lim_{\alpha \to 1^-} (1-\alpha)^{\left(\frac{1}{\alpha}-1\right)} =2\times 1 = 2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Finding conditional expectation on the basis of conditional probabilities Knowing that
*
*$P(Y=1\mid X=5)=1/3$,
*$P(Y=5\mid X=5)=2/3$.
Calculate $E(Y\mid X=5)$ and $E(XY^2\mid X=5)$. How to solve this question? I have no idea whatsoever.
| Since the conditional probabilities sum up to $1$ then that is all for $Y\mid X=5$. So $$E[Y\mid X=5]=1\cdot P(Y=1\mid X=5)+5\cdot P(Y=5\mid X=5)=1\cdot\frac13+5\cdot\frac23=\frac{11}3$$ and
\begin{align}E[XY^2\mid X=5]&=E[5Y^2\mid X=5]=5E[Y^2\mid X=5]\\[0.2cm]&=5\left((1)^2\cdot P(Y=1\mid X=5)+(5)^2\cdot P(Y=5\mid X=5)\right)\\[0.2cm]&=5\left(1\cdot\frac13+25\cdot\frac23\right)=\frac{255}3\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Problem in a Pell equation If you take the continued fraction out of $\sqrt7$, you get [2;1, 1, 1, 4...] which yields:
$$2+\frac1{1+\frac1{\frac11+1}}=\frac8{3}$$
and indeed, $$8^2-7⋅3^2=1$$
However, if you take the continued fraction out of $\sqrt13$ you get [3;1, 1, 1, 1, 6...] which yields: $$3+\frac1{1+\frac1{1+\frac1{\frac11+1}}}=\frac{18}5$$
but that gets $$18^2-13⋅5^2=\textbf{-1}$$not $\textbf{1}$ but $\textbf{-1}$! Why? Where is my mistake?
| If you take a prime $p \equiv 1 \pmod 4$ you are guaranteed to get $-1$ first, but get back to $1$ if you repeat the periodic part of the CF.
Note $$18^2 + 13 \cdot 5^2 = 649,$$
$$ 2 \cdot 18 \cdot 5 = 180, $$
$$ 649^2 - 13 \cdot 180^2 = 1. $$
The continued fraction for $\sqrt {13}$ in the display I like. For a convergent $p/q,$ the number directly below is $p^2 - 13 q^2.$ This includes the fake convergent $1/0,$ which needs to be there to start the process.
$$
\small
\begin{array}{cccccccccccccccccccccccccccccc}
& & 3 & & 1 & & 1 & & 1 & & 1 & & 6 & & 1 & & 1 & & 1 & & 1 & & 6 & \\
\frac{0}{1} & \frac{1}{0} & & \frac{3}{1} & & \frac{4}{1} & & \frac{7}{2} & & \frac{11}{3} & & \frac{18}{5} & & \frac{119}{33} & & \frac{137}{38} & & \frac{256}{71} & & \frac{393}{109} & & \frac{649}{180} & & \frac{4287}{1189} \\
\\
& 1 & & -4 & & 3 & & -3 & & 4 & & -1 & & 4 & & -3 & & 3 & & -4 & & 1 & & -4
\end{array}
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ 2^2 - 5 \cdot 1^2 = -1,$$
$$2^2 + 5 \cdot 1^2 = 9,$$
$$ 2 \cdot 2 \cdot 1 = 4, $$
$$ 9^2 - 5 \cdot 4^2 = 1. $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ 70^2 - 29 \cdot 13^2 = -1,$$
$$70^2 + 29 \cdot 13^2 = 9801,$$
$$ 2 \cdot 70 \cdot 13 = 1820, $$
$$ 9801^2 - 29 \cdot 1820^2 = 1. $$
I put a complete picture for $\sqrt {29}$ at Solving Diophantine Equation - odd Periods
| {
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Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
| Hint:
$$\int{\frac{(x^2 + 4)dx}{x^2 + 6x +10}}=\int{\frac{x^2 + 6x + 10-6x-6}{x^2 + 6x +10}}dx=x-6\int{\frac{x+1}{x^2 + 6x +10}}dx$$
Next hint:
$$6\frac{x+1}{x^2 + 6x +10}=3\frac{\color{blue}{2x+6}-4}{x^2 + 6x +10}$$
Next hint:
$$x^2 + 6x +10 = (x+3)^2+1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving functional equation $f(4x)-f(3x)=2x$ Given that $f(4x)-f(3x)=2x$ and that $f:\mathbb{R}\rightarrow\mathbb{R}$ is an increasing function, find $f(x)$. My thoughts so far: subtituting $\frac{3}{4}x$, $\left(\frac{3}{4}\right)^2x$, $\left(\frac{3}{4}\right)^3x$, $\ldots$, we get that:
$$f(4x)-f(3x)=2x$$
$$ f\left(4\cdot\frac{3}{4}x\right)-f\left(3\cdot\frac{3}{4}x\right)=2\cdot\frac{3}{4}x $$
$$ f\left(4\cdot\left(\frac{3}{4}\right)^2x\right)-f\left(3\cdot\left(\frac{3}{4}\right)^2x\right)=2\cdot\left(\frac{3}{4}\right)^2 $$
$$ f\left(4\cdot\left(\frac{3}{4}\right)^3x\right)-f\left(3\cdot\left(\frac{3}{4}\right)^3x\right)=2\cdot\left(\frac{3}{4}\right)^3 $$
$$\ldots$$
Then note that after adding all these equations we get:
$$ f(4x)=\sum_{k=0}^{\infty}\left(\frac{3}{4}\right)^k2x $$
And this series obviously converges to $8x$. Substituting $\frac{1}{4}x$ we get:
$$ f(x)=2x. $$
Is this correct? ;)
| It's almost correct. You are correct that we can deduce that
$$f(4x)=f(3x)+2x$$
and by repeatedly applying this we get
$$f(4x)=f\left(4\left(\frac{3}4\right)^kx\right)+\sum_{n=0}^{k-1}2\left(\frac{3}4\right)^{n}x$$
You make an error on the next step, however. You mean to take a limit as $k$ goes to infinity, but you to this incorrectly. In particular, the correct expression would be:
$$f(4x)=\lim_{k\rightarrow\infty}f\left(4\left(\frac{3}4\right)^kx\right)+8x$$
where we have a term of $\lim_{k\rightarrow\infty}f\left(4\left(\frac{3}4\right)^kx\right)$ that you omitted; in particular, this can be any constant, and the constant can be different for positive and negative numbers. It does, however, exist since $f$ is increasing. Thus, the solutions are of the form:
$$f(x)=\begin{cases}2x+c_1&&\text{if }x>0\\c_2&&\text{if }x=0\\2x+c_3&&\text{if }x<0\end{cases}$$
for some constants $c_1\geq c_2 \geq c_3$.
| {
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Use the identity $\cos 3\theta = 4 \cos^3\theta- 3 \cos \theta$ to solve the cubic equation $t^3 + pt + q = 0$ when $p, q \in \mathbb{R}$. I'm self studying Ian Stewart's Galois Theory and this is Exercise 1.8 from his Third Edition:
Use the identity $\cos 3\theta = 4 \cos^3\theta- 3 \cos \theta$ to
solve the cubic equation $t^3 + pt + q = 0$ when $p, q \in \mathbb{R}$
such that $27 q^2 + 4p^3 < 0$.
I read through many times his method of solving the cubic equation where he didn't use the identity above; yet I'm not sure where the identity can come into play.
His method is sketched below:
First, he substitutes $t = \sqrt[3]{u} + \sqrt[3]{v}$ and express $t^3$ in terms of $u$ and $v$ as well. Then plugging $t$ and $t^3$ in terms of $u$ and $v$ back to the original equation. Finally solving for $u$ and $v$ will immediately give the zeros.
Thanks very much for hints and help!
| Note that, since $27q^2+4p^3<0$ and since $27q^2\geqslant0$, $p<0$. So, it makes sense to define $u=\sqrt{-\frac43p}$. Consider the substitution $t=u\cos\theta$. Then $t^3+pt+q$ becomes $u^3\cos^3\theta+pu\cos\theta+q$ and\begin{align}u^3\cos^3\theta+pu\cos\theta+q=0&\iff\frac{u^3\cos^3\theta+pu\cos\theta+q}{u^3/4}=0\\&\iff4\cos^3\theta+\frac{4p}{u^2}\cos\theta+\frac{4q}{u^3}=0\\&\iff4\cos^3\theta-3\cos\theta=-\frac{4q}{u^3}.\end{align}But$$\left(-\frac{4q}{u^3}\right)^2=\frac{16q^2}{(u^2)^3}=\frac{16q^2}{-\frac{64}{27}p^3}=\frac{27q^2}{-4p^3}$$and\begin{align}27q^2+4p^3<0&\iff27q^2<-4p^3\\&\iff\frac{27q^2}{-4p^3}<1\end{align}and therefore $-\frac{4q}{u^3}\in(-1,1)$. So, there is some $\theta\in\Bbb R$ such that $\cos(3\theta)=-\frac{4q}{u^3}$; just take $\theta=\frac13\arccos\left(-\frac{4q}{u^3}\right)$. Then, since $4\cos^3\theta-3\cos\theta=\cos(3\theta)$, $\cos\theta$ is a root of the cubic $t^3+pt+q=0$. But you also have$$\cos\left(3\left(\theta+\frac{2\pi}3\right)\right)=-\frac{4q}{u^3}\quad\text{and}\quad\cos\left(3\left(\theta+\frac{4\pi}3\right)\right)=-\frac{4q}{u^3},$$and therefore $\cos\left(\theta+\frac{2\pi}3\right)$ and $\cos\left(\theta+\frac{4\pi}3\right)$ are also roots of that cubic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Is there a relationship between Rotors and the Rodrigues' rotation formula I am trying to understand quaternion in general, and it seems like the path to making sense of how they actually work is to first understand rotors and other techniques related to rotations. By looking at the equations and just reading briefly about these topics, it seems like there are relationships between all these topics: rotors, the Rodrigues' rotation formula which involves a scalar and a vector product as well as quaternions. Before getting on to the topic of quaternions all I am trying to do now, is finding out if there is any sort of connection between rotors (or more precisely using rotors to rotate a vector) as the equation to rotate a vector using a rotor also involves scalar and vector product, and the Rodrigues' formula:
$\mathbf{v}_\mathrm{rot} = \mathbf{v} \cos\theta + (\mathbf{k} \times \mathbf{v})\sin\theta + \mathbf{k} (\mathbf{k} \cdot \mathbf{v}) (1 - \cos\theta)$
If so, could someone explain it please?
| We could "derive" the Rodrigues formula by "vectorizing" the quaternion sandwitch product. First, let us recall the sadwitch product:
$v' = Q \ v \ Q^*$
Where $v$ is a pure quaternion (its real part equal to zero) and $Q$ is a unit quaternion and $Q^*$ is its conjugate:
$Q^{*} = q_0 - i q_1 - j q_2 - k q_3$
Aplying the Euler relation:
$v' = Q \ v \ Q^{-1} = (\cos \frac{\theta}{2} + \sin \frac{\theta}{2} \ b) \ v \ (\cos \frac{\theta}{2} - \sin \frac{\theta}{2} \ b)$
$v' = (Q \ v) \ Q^{-1} = (\cos \frac{\theta}{2} \ v + \sin \frac{\theta}{2} \ b \ v) \ (\cos \frac{\theta}{2} - \sin \frac{\theta}{2} \ b)$
$v' = \cos^2 \frac{\theta}{2} \ v - (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ v \ b + (\sin \frac{\theta}{2} \ \cos \frac{\theta}{2}) \ b \ v - \sin^2 \frac{\theta}{2} \ b \ v \ b$
The product $v \ b$ that apperars in the above expressions is the product of two
pure quaternions and it is defined in vector form as:
$v \ b = (i v_0 + j v_1 + k v_2) \ (i b_0 + j b_1 + k b_2) = -v \cdot b + v \times b$,
where $v \cdot b$ is the dot product and $v \times b$ is the cross product.
Notice that the algebraic sum $-v \ b + b \ v$ is not zero, since the cross product is non-commutative. As a consequence $-v \ b + b \ v = (v \cdot b - v \times b) + (-b \cdot v + b \times v) = 0 + 2 \ b \times v = 0 - 2 \ v \times b$.
Replacing that into our last equation we get:
$v' = \cos^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v - \sin^2 \frac{\theta}{2} \ b \ v \ b$
Interestingly the expression $b \ v \ b$ is nothing more than the reflection of $v$ in the plane with normal $b$. Where $b$ and $v$ are pure quaternions. Reflection can be expressed in vector form as:
$b \ v \ b = 2 \ (-v \cdot b) \ b + v$
where the vector $v$ is translated in direction of the negative normal $b$ twice the distance of $v$ projected in $b$. Finally the equation is:
$v' = \cos^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v - \sin^2 \frac{\theta}{2} \ (2 \ (-v \cdot b) \ b + v)$
$v' = \cos^2 \frac{\theta}{2} \ v - \sin^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v + 2 \ \sin^2 \frac{\theta}{2} \ (v \cdot b) \ b$
Applying the following trigonometric identities:
$\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2}$
$\sin \theta = 2 \cos \frac{\theta}{2} \ \sin \frac{\theta}{2}$
$1 - \cos \theta = 2 \ \sin^2 \frac{\theta}{2}$
We get the Rodrigues formula:
$v' = \cos \theta \ v + \sin \theta \ b \times v + (1 - \cos \theta) \ ( v \cdot b ) \ b $
It was discovered by Olinde Rodriguez three years before Hamilton discover quaternions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Obtaining an estimate for the Lagrangian $L=H^*$ from the Hamiltonian $H$ This is from C. Evans' PDE book, page 130. The convex function $H:\mathbb{R}^n\to\mathbb{R}$ is $C^2$ and satisfies $$ H\big(\frac{p_1+p_2}{2}\big) \leq \frac{1}{2}H(p_1) + \frac{1}{2}H(p_2) - \frac{\theta}{8}|p_1-p_2|^2. $$ The lagrangian $L$ is defined as the Legendre transform of $H$: $$ L(v)=(H^*)(v)=\max_{p\in\mathbb{R}^n}p\cdot v - H (p). $$ It's a fact that the statements $p\cdot v = L(v) + H(p), $ $ p=DL(v), $ $v=DH(p)$ are equivalent.
One must show that from the above estimate it follows that $$ \frac{1}{2}L(v_1)+\frac{1}{2}L(v_2)\leq L\big(\frac{v_1+v_2}{2}\big) + \frac{1}{8\theta}|v_1-v_2|^2.$$ I haven't been able to show this. Below is a screenshot of the book's page.
Thank you.
| As your hint says, for every $q_1, q_2 \in \mathbb R^n$ there exist $p_1, p_2 \in \mathbb R^n$ such that $L(q_j) = p_j \cdot q_j - H(p_j)$ for $j \in \{1,2\}$. Combining with the inequality for $H$,
\begin{align}
\frac{1}{2} L(q_1) + \frac{1}{2} L(q_2)
&= \frac{p_1 \cdot q_1 + p_2 \cdot q_2}{2} - \frac{1}{2} H(p_1) - \frac{1}{2} H(p_2) \\
&\leq \frac{p_1 \cdot q_1 + p_2 \cdot q_2}{2} - H\left(\frac{p_1+p_2}{2}\right) - \frac{\theta}{8} \lvert p_1 - p_2\rvert^2.
\end{align}
Note that
\begin{align}
\frac{p_1 \cdot q_1 + p_2 \cdot q_2}{2} - H\left(\frac{p_1+p_2}{2}\right)
&= \frac{p_1 - p_2}{2} \cdot \frac{q_1 - q_2}{2} + \frac{p_1 + p_2}{2} \cdot \frac{q_1 + q_2}{2} - H\left(\frac{p_1+p_2}{2}\right) \\
&\leq \frac{p_1 - p_2}{2} \cdot \frac{q_1 - q_2}{2} + \max_{p \in \mathbb R^n} \left(p \cdot \frac{q_1 + q_2}{2} - H(p)\right) \\
&= \frac{p_1 - p_2}{2} \cdot \frac{q_1 - q_2}{2} + L\left(\frac{q_1 + q_2}{2}\right),
\end{align}
which means
$$
\frac{1}{2} L(q_1) + \frac{1}{2} L(q_2)
\leq L\left(\frac{q_1 + q_2}{2}\right) - \frac{\theta \lvert p_1 - p_2\rvert^2 - 2(p_1 - p_2) \cdot (q_1 - q_2)}{8}.
$$
The desired inequality for $L$ is now immediate from the fact that
$$
\left\lvert \sqrt\theta (p_1 - p_2) - \frac{q_1 - q_2}{\sqrt\theta}\right\rvert^2
\geq 0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction that $5|11^n-6$ for all positive integers $n$
Prove by induction that $5|11^n-6$ for all positive integers $n$
Let $p(n) = 11^n-6.$ We have $p(1) = 5$, thus it holds for $p(1)$. Assume it holds for $p(k)$. We will prove that it's true for $p(k+1)$. We have $p(k+1) = 11^{k+1}-6$. But $6 = 11^k-f(k).$ Thus
$$f(k+1) = 11\cdot 11^k-11^k+f(k) = 10\cdot 11^k+f(k). $$
$5|(10\cdot 11^k+f(k))$ by hypothesis thus the statement is true for $p(k+1)$. Hence it's true for $p(n)$.
Is the above correct?
| Suppose $5\mid 11^n-6\implies 11^n-6=5k$ for some $k\in\mathbb{Z}$.
\begin{align*}
11(11^n-6)&=11\cdot5k \\
11^{n+1}-11\cdot 6+60&=11\cdot 5k+60 \\
11^{n+1}-6&=11\cdot 5k+60 \\
11^{n+1}-6&=5(11k+12)
\end{align*}
$$11^{n+1}-6=5(11k+12)\implies 5\mid 11^{n+1}-6$$
So we have shown
$$5\mid 11^n-6\implies 5\mid 11^{n+1}-6$$
Check $5\mid 11^1-6$, and conclude $\forall n\in\mathbb{N},5\mid 11^n-6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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solving $\int \frac{dx}{\sqrt{-x^2-12x+28}}$
$$\int \frac{dx}{\sqrt{-x^2-12x+28}}$$
First we need to use completing the square $-(x^2+12x-28)=-(x+6)^2+64$
So we have $\int \frac{dx}{\sqrt{-(x+6)^2+64}}$ I know that it is a general form of $\arcsin(\frac{x+6}{8})$ but how can I solve it using substitution?
| Its not really necessary to go for substitution. Use basic differentiation of inverse trig functions.
$\begin{align}
\int\frac{dx}{\sqrt{a^2-x^2}}&=
\sin^{-1}(\frac{x}{a})+C\\
&=\int\frac{dx}{\sqrt{64-(x+6)^2}}\\
&=\sin^{-1}\frac{(x+6)}{8}+C
\end{align}
$
| {
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"timestamp": "2023-03-29T00:00:00",
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Has this equation appeared before? I want to know if the following equation has appeared in mathematical literature before, or if it has any important significance.
$$\sqrt{\frac{a+b+x}{c}}+\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}=\sqrt{\frac{a+b+c}{x}},$$
where $a,b,c$ are any three fixed positive real and $x$ is the unknown variable.
| I edit this answer to explicit the degree of the resulting polynomial having $x$ as a root (I wonder about the exact origin of this strange unknown x. Explanation is tedious and it will be presented abbreviated as possible, just to have the degree of the polynomial without using algebraic number theory which is not so useful without knowing the particular values of $a,b,c$).
First at all one has $$\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}=\sqrt{\frac{a+b+c}{x}}-\sqrt{\frac{a+b+x}{c}}$$
Squared twice and, for convenience, make the following notations:
$$\begin {cases}A=\frac{a+b+x}{c}+\frac{a+c+x}{b}\\B=\frac{a+b+c}{x}+\frac{b+c+x}{a}\\M=(\frac{a+b+x}{c})(\frac{a+c+x}{b})\\N=(\frac{a+b+x}{x})(\frac{b+c+x}{a})\\C=\frac{B-A}{2}\\D=\frac{B^2-A^2+4N-4M}{4}\end{cases}$$
One has the linear system
$$\begin{cases}\sqrt M+\sqrt N=C\\ A\sqrt M+B\sqrt N=D\end{cases}\qquad(*)$$
So, for instance $$\sqrt M=\frac {CB-D}{B-A}\Rightarrow M=\left(\frac{CB-D}{B-A}\right)^2\qquad(**)$$
Thus a rational resultant is
$$\left(\frac{a+b+x}{c}\right)\left(\frac{a+c+x}{b}\right)\left(B-A\right)^2=\left(B^2+A^2+4M-2AB-4N\right)^2\qquad (***) $$
Explaining $(***)$ with values of $a, b, c, x$ gives
$$G(x,a,b,c)-(H(x,a,b,c))^2=0$$ with
$$G(x,a,b,c)=bc(ax)^2(x^2+A_1x+A_2)(A_3+x)^2(abc-A_4x)^2$$
$$H(x,a,b,c)=\{bc(x^2+(b+c)x+A_5)\}^2+\{ax((b+c)x+A_6)\}^2-2abcx\{(b+c)x+A_7)(x^2+(b+c)x+A_8)+2(ax^2+A_9x-A_{10})\}$$
where $$\begin{cases}A_1=2a+b+c\\ A_2=a^2+ab+ac+bc\\ A_3=a+b+c\\ A_4=ab+ac-bc\\ A_5=a(a+b+c)\\ A_6=b(a+b)+c(a+c)\\ A_7=A_6\\ A_8=A_5\\A_9=a^2+ac-bc\\ A_{10}=b^2c+bc^2\end{cases}$$
Therefore one can see at first sight that degree of $G$ is $8$ and so is for the degree of $H^2$ so that the degree of the resultant equation could be in general $8$.
►However it could be maybe for particular values of $a,b,c$ that the coefficients of $x^8$ in $G$ and $H^2$ be equal so the degree were less than $8$; this happen if
$$bca^2(ab+ac-bc)^2=[b^2c^2+4a^2+a^2(b+c)^2(1-4b^2c^2)]^2$$ which seems to be possible for some values of $a,b,c$.
►On the other hand, solving also in $(*)$ for $N$, one gets $$N=\left(\frac{D-AC}{B-A}\right)^2$$ we can, in general, by multiplication and subtraction of the coefficients of $x^8$, eliminate the degree $8$ so this degree can be reduced.
►With the two polynomials giving by the system $(*)$, we can eliminate the unknown $x$ and find a relation $F(a,b,c)=0$ in order both polynomials have common roots (this can be made by Sylvester’ method, for instance, which would give a large zero determinant for $F(a,b,c)$
I stop here. I have wanted to help @pritam, assuming he is really interested in the rare equation proposed (I don’t discard some typo in my answer).
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then I would get $\geq$ instead of $>$.
| Hint, $(z-x)^3=\left((z-y)+(y-x)\right)^3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$
Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$
Attempt
The $ab+bc+ca = 1$ condition reminds of the rearrangement inequality. Thus, I would say that $a^2+b^2+c^2 \geq ab+bc+ca = 1$ then rewrite the given inequality as $4(a+b+c)^2 = 4(a^2+b^2+c^2) + 8(ab+bc+ca) = 4(a^2+b^2+c^2) \leq 9(a^2+1)(b^2+1)(c^2+1)$
I don't know what to do next.
| Set $a=\cot A$ etc.
$\implies\sum\cot A\cot B=1$
$\iff\tan A+\tan B+\tan C=\tan A\tan B\tan C$
$\implies A+B+C=n\pi$ where $n$ is any integer
Now use In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$
| {
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"timestamp": "2023-03-29T00:00:00",
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System and triangle If $a,b$ and $c$ are the lengths of sides of the triangle, find $x,y$ and $z$ such that
$$\begin{array}{l}
x^2y^2+x^2z^2=axyz\\
y^2x^2+y^2z^2=bxyz\\
z^2x^2+z^2y^2=cxyz
\end{array}$$
I see that $x=y=z=0$ is a solution. Moreover, $x=y=0$ and $z \in \mathbb{R}$ is solution, or $x=z=0$ and $y \in \mathbb{R}$ is a solution, $y=z=0$ and $x \in \mathbb{R}$ too is a solution and if $a=b=c$ $x=y=z=a/2$ is solution. Are There other solutions? This problem was proposed in the RPM- Revista do professor de Matemática in Brazil.
| If $a, b, c$ are the side length of a triangle then
$$
u = \frac 12(b+c-a) \, , \quad v = \frac 12(c+a-b) \, ,\quad w = \frac 12(a+b-c)
$$
are positive real numbers such that
$$
a = v + w \, , \quad b = w + u \, ,\quad c = u + v \, .
$$
(This is the so-called "Ravi substitution", see for example
Ravi Substitution explained.)
Then your system of equations is equivalent to
$$
y^2 z^2 = u x y z \\
z^2 x^2 = v x y z \\
x^2 y^2 = w x y z
$$
Now multiply the first two equations. This gives
$x^2 y^2 z^4 = uv (xyz)^2$ and therefore $z^2 = uv$ or $xyz = 0$.
It follows that positive solutions $x, y, z$ must satisfy
$$
x = \sqrt{vw} \, , \quad y = \sqrt{wu} \, , \quad z = \sqrt{uv}
$$
and conversely one can verify that this is indeed a solution.
| {
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How to integrate $\int (\tan x)^{1/ 6} \,\text{d}x$? How do I compute the following integral
$$
I=\int (\tan x)^{1/ 6} \,\text{d}x
$$
| HINT (the integral will become very very big):
$$\int\left(\tan(x)\right)^{\frac{1}{6}}\space\text{d}x=\int\sqrt[6]{\tan(x)}\space\text{d}x=$$
Substitute $u=\tan(x)$ and $\frac{\text{d}u}{\text{d}x}=\sec^2(x)$:
$$\int\frac{\sqrt[6]{u}}{1+u^2}\space\text{d}u=$$
Substitute $s=\sqrt[6]{u}$ and $\frac{\text{d}s}{\text{d}u}=\frac{1}{6\sqrt[6]{u^5}}$:
$$\int\frac{5s^6}{1+s^{12}}\space\text{d}s=5\int\frac{s^6}{1+s^{12}}\space\text{d}s=$$
$$5\int\left[\frac{s^6+s^2}{3(s^8-s^4+1)}-\frac{s^2}{3(s^4+1)}\right]\space\text{d}s=$$
$$5\left[\int\frac{s^6+s^2}{3(s^8-s^4+1)}\space\text{d}s-\int\frac{s^2}{3(s^4+1)}\space\text{d}s\right]=$$
$$5\left[\frac{1}{3}\int\frac{s^6+s^2}{s^8-s^4+1}\space\text{d}s-\frac{1}{3}\int\frac{s^2}{s^4+1}\space\text{d}s\right]=$$
$$5\left[\frac{1}{3}\int\frac{s^6+s^2}{s^8-s^4+1}\space\text{d}s-\frac{1}{3}\int\left[-\frac{s}{2\sqrt{2}\left(-s^2+\sqrt{2}s-1\right)}-\frac{s}{2\sqrt{2}\left(s^2+\sqrt{2}s+1\right)}\right]\space\text{d}s\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$.Find $\frac{a^2+b^2+c^2}{S}$ In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$
Find the value of $\frac{a^2+b^2+c^2}{S}$,where $S$ is area of the triangle $ABC$and find the value of $1+\frac{R}{r}$ where $R$ is the circumradius and $r$ is the inradius of the triangle $ABC$
My attempt:$a^4+b^4+c^4-4abc+1=0$
I expanded $(a+b+c)^4=a^4+4a^3b+4a^3c+6a^2b^2+6a^2c^2+12a^2bc+4ab^3+4ac^3+12ab^2c+12abc^2+4bc^3+4b^3c+6b^2c^2+b^4+c^4$
But this expression has got complicated and not seeming helpful and i do not know any other method to solve this question.
| HINT:
Using AM-GM inequality
$$\dfrac{a^4+b^4+c^4+1}4\ge\sqrt[4]{a^4b^4c^4\cdot1}$$
Can you take it from here?
| {
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Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try:
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because
$ax+by+cz\le{\frac{a+b+c}{3}}$ is not always true.
| WLOG: $$a+ b+ c= 1$$
AM-GM:
$$ax+ by+ cz+ 2\sqrt{\left ( xy+ yz+ zx \right )\left ( ab+ bc+ ca \right )}$$
$$\leq ax+ by+ cz+ xy+ yz+ zx+ ab+ bc+ ca$$
$$\Leftarrow xy+ yz+ zx+ ab+ bc+ ca= \frac{1- x^{2}- y^{2}- z^{2}}{2}+ \frac{1- a^{2}- b^{2}- c^{2}}{2}$$
$$\leq 1- ax- by- cz$$
$$\Leftrightarrow \left ( x- a \right )^{2}+ \left ( y- b \right )^{2}+ \left ( z- c \right )^{2}\geq 0$$
| {
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Finding Laurent's series of a function I am trying express the function $$f(z)=\frac{z^3+2}{(z-1)(z-2)}$$ like a Laurent's series in each ring centering in $0$, but I do not now how could I express it, in first I said that $$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
Ok, now, I see two posibilities: $A\equiv open\ ring=A(C;r,R)$ where $$C\equiv center\\ r<R$$
so, I see:
$a)$ $A(0;1;2)$
$b)$ $D(0,1)\cup\{\mathbb{C}-\overline{D}(0,2)$
It is correct? Ok, then I suposse that $a)$ I think that I have to try express like series of potence with $\frac{1}{z}$, and in the second case, $b)$, in $\overline{D}(0,1)$ like potences of $z$ and in $\mathbb{C}-\overline{D}(0,2)$ like potences of $\frac{1}{z}$ but I can not, I need someone help, please.
All I know is that $\sum_{0}^{\infty} z^n=\frac{1}{1-z}$ when $|z|<1$
I do not know if it is so, I need someone clarify my doubts
| Executing the polynomial division we have
$$
f(z)=\frac{z^2+3}{(z-1)(z-2)}=1+\frac{3z+1}{(z-1)(z-2)},
$$
and with partial fractions
$$
f(z)=\frac{z^2+3}{(z-1)(z-2)}=1-\frac{4}{z-1}+\frac{7}{z-2}.
$$
This can be rewritten as
$$
f(z)=1+4\cdot\frac{1}{1-z}-\frac{7}{2}\cdot\frac{1}{1-\frac{z}{2}}.
$$
Using the geometric series known expansion
$$
\frac{1}{1-z}=\sum_{n=0}^\infty z^n,\quad|z|<1
$$
we have
$$
f(z)=1+4\cdot\sum_{n=0}^\infty z^n-\frac{7}{2}\cdot\sum_{n=0}^\infty \left(\frac{z}{2}\right)^n=1+\sum_{n=0}^\infty\left(4-\frac{7}{2^{n+1}}\right)z^n.
$$
This expansion is only valid in $|z|<1$.
For $1<|z|<2$ we have to use the Laurent series, whose coefficients are given by the formula
$$
a_n=\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z^{n+1}}dz=\sum_k\operatorname{Res}\left[\frac{f(z)}{z^{n+1}},z_k\right],
$$
where the curve can be taken as a circle centered in $z=0$ and with radius $1<r<2$, and the sum is extended to the residues of the function inside the circle.
The same formula, applied to the circle $|z|<1$, where we already have found the coefficients, provides
$$
\operatorname{Res}\left[\frac{f(z)}{z^{n+1}},0\right]=\begin{cases}
\displaystyle{1+4-\frac{7}{2}}, & n=0, \\
\\
\displaystyle{4-\frac{7}{2^{n+1}}}, & n > 0.
\end{cases}
$$
Furthermore
$$
\operatorname{Res}\left[\frac{f(z)}{z^{n+1}},1\right]=\lim_{z\to1}(z-1)\frac{f(z)}{z^{n+1}}=\left.\frac{z^2+3}{z^{n+1}(z-2)}\right|_{z=1}=-4,
$$
So, for $1<|z|<2$
$$
a_n=\begin{cases}
-4, & n < 0,\\
\\
\displaystyle{1-\frac{7}{2}}, & n=0, \\
\\
\displaystyle{-\frac{7}{2^{n+1}}}, & n > 0,
\end{cases}
$$
i.e.
$$
f(z)=-4\sum_{n=-1}^{-\infty}z^n+1-\sum_{n=0}^\infty\frac{7}{2^{n+1}}z^n=-4\sum_{n=1}^{\infty}\frac{1}{z^n}+1-\sum_{n=0}^\infty\frac{7}{2^{n+1}}z^n.
$$
Finally, for $|z|>2$ we have to add the residue in $z=2$, given by
$$
\operatorname{Res}\left[\frac{f(z)}{z^{n+1}},2\right]=\lim_{z\to2}(z-2)\frac{f(z)}{z^{n+1}}=\left.\frac{z^2+3}{z^{n+1}(z-1)}\right|_{z=2}=\frac{7}{2^{n+1}},
$$
so that
$$
f(z)=-\sum_{n=-1}^{-\infty}\left(4-\frac{7}{2^{n+1}}\right)z^n+1=1-\sum_{n=1}^{\infty}\left(4-7\cdot 2^{n-1}\right)\frac{1}{z^n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\lim_{x \to 0} \frac{\sin(\sin x)-x(1-x^2)^\frac{1}{3}}{x^5}$ by Taylor's theorem. I have to calculate : $$\lim_{x \to 0} \frac{\sin(\sin x)-x(1-x^2)^\frac{1}{3}}{x^5}$$ by using Taylor's theorem.
I know that :$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots$$
But I don't know how to continue.
Thanks.
| Expand $\sin(\sin x)$ similar to $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$, i.e. $$\sin(\sin x)=\sin x-\frac{\sin^3x}3!+\frac{\sin^5 x}{5!}-...$$ near $x=0$.
And $(1-x^2)^{1/3}$ similar to $(1+x)^n=1+nx+\frac{(n-1)n}{2!}x^2+\frac{(n-2)(n-1)n}{3!}x^3+...$
Now, expand each $\sin x$ using taylor then forming combination of useful terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to
$(A)\frac{1}{2}\hspace{1cm}(B)8\hspace{1cm}(C)2\hspace{1cm}(D)3$
$(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$
$(x,y)$ satisfies $(x-2)^2+(y+1)^2=2^2$
The expression $x^2+y^2-10x-6y+34$ can be written as $(x-5)^2+(y-3)^2$
But i do not know how to further solve it.Please help.
| As you note, the restriction is
$$(x-2)^2+(y+1)^2=2^2$$
That means the distance from point $P(x,y)$ to the point $A(2,-1)$, $AP$, is $2$.
As you also note, the function is
$$f(x,y)=(x-5)^2+(y-3)^2$$
which is the square of the distance of point $P(x,y)$ to point $B(5,3)$, $PB$. Note that the distance from point $A(2,-1)$ to point $B(5,3)$, $AB$, is
$$AB=\sqrt{(5-2)^2+(3--1)^2}=5$$
To find the limits on that distance we use the triangle inequalities
$$|AB-AP| \le PB \le AB+AP$$
or
$$|5-2| \le PB \le 5+2$$
$$3 \le PB \le 7$$
So the square of distance PB, or function value, ranges between $9$ and $49$. None of the multiple choices are in that range, so none of the given choices can be correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Not getting the same solution when using the rule sin(x)\x=1 on a limit There is a rule in limits that when $x$ approaches zero:
$$\frac{\sin\left(x\right)}{x}=1$$
So I used this rule on the following exercise:
Evaluate
$$
\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin\left(2x\right)-\tan\left(2x\right)}
$$
I substituted $\sin(2x)$ with $2x$ by the following way:
$$\sin\left(2x\right)=\frac{\sin\left(2x\right)}{2x}\cdot 2x=1\cdot 2x=2x \Rightarrow \lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{2x-\tan\left(2x\right)}$$
But according to symbolab:
$$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{2x-\tan\left(2x\right)}=-\frac{1}{16}$$
while
$$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin(2x)-\tan\left(2x\right)}=-\frac{1}{24}$$
Why am I getting this contradiction?
More over if I susbsitute the following I do get the right answer
$$\tan\left(2x\right)=\frac{\sin\left(2x\right)}{\cos\left(2x\right)}=\frac{2x}{\cos\left(2x\right)}\Rightarrow \lim \:_{x\to \:0}\:\frac{x-\sin\left(x\right)}{2x-\frac{2x}{\cos\left(2x\right)}}=-\frac{1}{24}$$
If you want to test yourself: Symbolab with the excersice preloaded
| The problem you encountered was that both $\sin(x)-\tan(x)=O\!\left(x^3\right)$ and $x-\sin(x)=O\!\left(x^3\right)$, this means that substituting $x$ for $\sin(x)$ may substantively change the limit.
To be precise,
$$
\begin{align}
\sin(x)-\tan(x)
&=\left(x-\frac{x^3}6+O\!\left(x^5\right)\right)-\left(x+\frac{x^3}3+O\!\left(x^5\right)\right)\\
&=-\frac{x^3}2+O\!\left(x^5\right)
\end{align}
$$
whereas
$$
\begin{align}
x-\tan(x)
&=x-\left(x+\frac{x^3}3+O\!\left(x^5\right)\right)\\
&=-\frac{x^3}3+O\!\left(x^5\right)
\end{align}
$$
This explains why the limits you got differ by a factor of $\frac32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \ \ \ d.)\ 7 $
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta \\
=\sin^{2} \theta +\dfrac{1}{\sin^{2} \theta }+\cos^{2} \theta+\dfrac{1}{\cos^{2} \theta }+\tan^{2} \theta+\dfrac{1}{\tan^{2} \theta } \\
\color{blue}{\text{By using the AM-GM inequlity}} \\
\color{blue}{x+\dfrac{1}{x} \geq 2} \\
=2+2+2=6 $
Which is not in options.
But I am not sure if I can use that $ AM-GM$ inequality in this case.
I look for a short and simple way .
I have studied maths upnto $12$th grade .
| Just using algebra$$y=\sin^{2} (\theta) +\cos^{2} (\theta)+\sec^{2} (\theta)+\csc^{2} (\theta)+\tan^{2} (\theta)+\cot^{2} (\theta)=2 \csc ^2(\theta )+2 \sec ^2(\theta )-1$$ Taking derivatives and simplifying $$y'=4 \tan (\theta) \sec ^2(\theta)-4 \cot (\theta) \csc ^2(\theta)=-32 \cot (2 \theta) \csc ^2(2 \theta)$$ $$y''=64 \csc ^4(2 \theta)+128 \cot ^2(2 \theta) \csc ^2(2 \theta)=64 (\cos (4 \theta)+2) \csc ^4(2 \theta)$$ The second derivative is always positive, so we can only find minimum values if $y'=0$ and this happens for $\theta=\pm\frac \pi 4$ and $\theta=\pm\frac {3\pi} 4$. For these values, $y=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
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