Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Could someone please explain the steps in the Maclauren series?
Find the limit of $$\lim_{x\to 0}{{1-(\cos x)({\cos2x})^{1/2}(\cos3x)^{1/3}}\over{x^2}}$$
Steps to solving this problem, from the book:
Step 1: $$\lim_{x\to 0}{{1-(1-\frac{x^2}{2})({1-2x^2})^{1/2}(1-\frac{9x^2}{2})^{1/3}}\over{x^2}}$$
Step 2: $$\lim_{x\t... | You should probably get rid of the book. Most calculus texts are like this. The right approach is to use the little / big o notation.
The key here is to use Taylor / Maclaurin expansions as $x\to 0$ $$ \cos x=1-\frac{x^2}{2}+o(x^2)\tag{1}$$ and $$(1+x)^n=1+nx+o(x)\tag{2}$$ where $o(f(x)) $ represents a function $g(x) $... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. Th... | Since $G$ is a group then if $x\in G$ , there exists $x^{-1}\in G$ such that $$x\cdot x^{-1}=x^{-1}\cdot x=e\in G$$having this result and exploiting the other properties of group we obtain $$x^{-1}\cdot (xyz)=(x^{-1}\cdot x)\cdot yz=e\cdot yz=yz=x^{-1}\cdot e=x^{-1}$$therefore $$yz\cdot x=yzx=x^{-1}\cdot x=1$$and the ... | {
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"url": "https://math.stackexchange.com/questions/3069921",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer. Prove that $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer using mathematical induction.
I tried using mathematical induction but using binomial formula also it becomes little bit complicated.
Please show me ... | @I like Serena has a great answer but since the OP asked for a proof by induction, I'll show what that would look like. Define
$$f(k)=\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}=\frac{15k^7 + 21k^5+70k^3-k}{105}$$
For our base case, let $k=1$. Then we have
$$f(1)=\frac{15+21+70-1}{105}=1$$
which is an inte... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$
Prove that for $a>1$,$b>1$ and $c>1$ where $a,b,c\in \mathbb{R}$
$$\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$$
My attempt: it is not so clear why is $a>1$, $b>1$ and $c>1$, but I factor left side, and all put ond side to prove $\geq 0$, but it is... | Well here is a start. Set $A=a+1$ and $B=b-1$ and $C=c$ and the equation becomes $$\frac 13(A+B+C)^2\le A^2+B^2+C^2$$
Now consider $$(A-B)^2+(B-C)^2+(C-A)^2\ge 0$$ so that $$2A^2+2B^2+2C^2-2AB-2BC-2AC\ge 0$$ so that $$AB+BC+AC\le A^2+B^2+C^2$$
Now $$(A+B+C)^2=A^2+B^2+C^2+2AB+2BC+2AC\le 3(A^2+B^2+C^2)$$
| {
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Proving uniform continuity of $f(x) = \sqrt{1-x^2}$ on $[-1,1]$ I want to prove that $f(x) = \sqrt{1-x^2}$ is uniform continuous on the interval $[-1,1]$. Let $f(x) = \sqrt{1 - x^2}$. Then I need to show:
$\forall \epsilon > 0 \enspace \exists \delta > 0 \enspace \forall x,y \in [-1,1] \enspace |x-y|<\delta \implies |f... | Hint:
$$|\sqrt{1-x^2} - \sqrt{1-y^2}|^2 \leqslant |\sqrt{1-x^2} - \sqrt{1-y^2}||\sqrt{1-x^2} + \sqrt{1-y^2}| = |x^2 - y^2|$$
| {
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Distinct prime number solution of an equation I am trying to solve the following question:
$$\textrm{Find all distinct prime numbers } p, q \textrm{ and } r \textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p \geq 5, q = 3, r \geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
... | $$3p^4=5q^4+4r^2+26$$
If $s\ne 5$ is an odd prime then $s^2$ ends with $1$ or $9$, and $s^4$ ends with $1.$
If $r>5$ or $r=3$ then the right side ends with $0$ or $7$ (which is impossible) so $5\mid 3p^4\implies p=5$. So the left side is odd and thus right also. But this can be only if $q=2$. Then $$ 3\cdot 5^4 -5\cdo... | {
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Help with complex numbers Algebraic form of the $z^3 = (3 + i)^6$. Can someone help me to solve this? My answer is $z = 8 + 6i$, but I'm not sure that this is true.
| Whenever you have $z^3 = k$ there will be $3$ solutions.
If $z^3 = (3+i)^6 = ((3+i)^2)^3$ then one solution is obviously $z = (3+i)^2 = 9 + 6i + i^2 = 9+6i -1 = 8+6i$.
But that is only one solution.
However $w^3 = 1$ has three solutions.
If we let $w$ be one of the solutions and we let $z = w(8+6i) = w(3+i)^2$ then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079959",
"timestamp": "2023-03-29T00:00:00",
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Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I s... | $$ \begin{align}L&=\lim_\limits{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)\\&
=\lim_\limits{n\to\infty}n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\\&=\lim_\limits{n\to\infty}\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length. Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length.
Attempt:
let $x = \sin^{3}t, y = \cos^3({t})$
$r(t) = (\sin^{3}(t), \c... | Now $||r'(t)||=\sqrt{(3 \sin^2(t) \cos (t))^2+(-3 \cos^2(t) \sin (t))^2}=\sqrt{9\sin^4(t)\cos^2(t)+9\cos^4(t)\sin^2(t)}=\sqrt{9\sin^2(t)\cos^2(t)(\sin^2(t)+\cos^2(t)}=\sqrt{9\sin^2(t)\cos^2(t)}=|3\sin(t)\cos(t)|=|\frac{3}{2}\sin(2t)|$
$$L=\int_0^{2\pi}\frac{3}{2} |\sin(2t)| \ dt=\frac{3}{2}\int_0^{2\pi}|\sin(2t)| \ dt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081247",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
At... | Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign.
Let
$$I(a) = \int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (a \tan x) \, dx, \qquad a > 1.$$
We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later ... | {
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"url": "https://math.stackexchange.com/questions/3082875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Mixed Fractions and Multiplication (with Variables) I stumbled over this expression: $3 \frac{1}{x^3}$.
How should you interpret something like that?
While you could see that as implicitit multiplication ($3 * \frac{1}{x^3}$),
you could also argue that $3 \frac{1}{x^3}$ is a mixed fraction ($3 + \frac{1}{x^3}$).
I thin... | This can be written as $$\frac{3}{1}\cdot \frac{1}{x^3}=\frac{3\cdot 1}{1\cdot x^3}=\frac{3}{x^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ I'm wondering if one could derive a closed form expression for the series
$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$
$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k... | Here is my way of evaluating this sum also offering a different way to calculate that polylogarithmic integral.
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}$$
First let's consider the following central binomial coefficient generating function.
$$\sum _{k=1}^{\infty }\frac{x^{2k}}{k\:4^k}\binom{2k}{k}=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+2}<2$ I wanted to prove this by mathematical induction. I have proved that it is less than two and that the sum is greater than 0 but I am stuck on how to prove it to be greater than 1. Please help!!!
I would like a conclusive proof for this. Thank you so much!
... | This is a matter of comparison testing.
First render $n+2>n+1>0, n+3>n+1>0, \text{ etc}$. Thereby
$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+2}<\dfrac{1}{n+1}+\dfrac{1}{n+1}+\dfrac{1}{n+1}+...+\dfrac{1}{n+1}$
where the right side has $2n+2=2(n+1)$ identical terms. So
$\color{blue}{\dfrac{1}{n+1}+\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Maximizing $ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$, such that $a^2+b^2+c^2=1 $
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$
I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
| We need to find a minimal value of $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2\leq k(a^2+b^2+c^2)$$ or
$$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30ac\geq0$$ or
$$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2\geq0,$$ for which we need
$$k>41$$ and
$$9(4b... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $? How to calculate
$\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $?
Since $\lim\limits_{n\to \infty} \frac {\cos(1/n)-\sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $\frac{1}{e}$, but since the form $(\to 1)^{\to ... | You may consider
*
*$\left(\frac{\cos x - \sin x}{1-x}\right)^{\frac{1}{x}}$ for $x\to 0$ and use
*$\lim_{t\to 0} (1+t)^{\frac{1}{t}}= e$
\begin{eqnarray*} \left(\frac{\cos x - \sin x}{1-x}\right)^{\frac{1}{x}}
& = & \left( \underbrace{\left(1 + \frac{\cos x - \sin x -1 +x}{1-x}\right)^{\frac{1-x}{\cos x - \sin x ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the constant term of $P(x-1)$ if the remainder of the division of $P(x)$ by $x-3$ is $18$ and $P(x+1) = (x^2 -4)Q(x)+3ax+6$
Assume that $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$
and that the remainder of the division of polynomial $P(x)$ by $x-3$ is $18$. Evaluate the constant term of polynomial $P(x-1)$.
All I cou... | We are given that $P(x+1) = (x^2 -4)Q(x)+3ax+6$ $(\ast)$. Plugging $x=2$ into $(\ast)$ yields $P(3)=6a+6$. On the other hand, for every $b$, $P(b)$ is the remainder of the division of polynomial $P(x)$ by $x-b$ hence $P(3)=18$, which shows that $a=2$.
Finally, plugging $x=-2$ into $(\ast)$ yields $P(-1)=-6a+6=-6$ henc... | {
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How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mes... | Hint:
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)} = \frac{k(k+2)+1}{(k+1)(k+2)} = \frac{(k+1)(k+1)}{(k+1)(k+2)}.$$
Regarding your calculations, note that
$$ \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)} = \frac{k^2+2k+1-k^2-2k}{(k+1)(k+1+1)}=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \time... | In you increase $n$ by $1$, all terms increase by $2k+1$ and there comes an extra term $2n+1$. Then $k^2+2k+1=(k+1)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Power series representation of $f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \left( \frac{1+x}{1-x} \right)$
We consider the power series:
$$ f(x) = \sum_{n=1}^{+ \infty} \frac{1}{n(n+1)(2n+1)}.x^{2n+2} $$
Prove that:
$$ f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \bigg( \frac{1+x}{1-x} \biggr) $$
Starting from the... | Partial fraction decomposition of the coefficient yields to
$$\frac1{n(n+1)(2n+1)}=\frac1n+\frac1{n+1}-\frac4{2n+1}$$
Therefore we may write the original series as
$$\sum_{n=1}^\infty\left[\frac1{n(n+1)(2n+1)}\right]x^{2n+2}=\sum_{n=1}^\infty\left[\frac1n+\frac1{n+1}-\frac4{2n+1}\right]x^{2n+2}$$
The easiest way from h... | {
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A sum of Fibonacci numbers Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$\sum_{n=0}^{\infty}(-1)^n(n+1)^2\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}}=F_{k}\tag1$$
which simplifies to:
$$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2... | Using Wolfram for these sums, we get approximate values
\begin{align*}
C \approx 0.86082, \qquad B \approx -0.11649, \qquad A \approx -0.07897
\end{align*}
From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define
\... | {
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Sum of an infinite series of fractions involving multiple terms in the denominator This is the series in question:
$$S = \frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4} \ldots$$
The general term seems to be:
$$T_n=\frac{n}{1+n^2+n^4}$$
In the original question, which is from Resonance DLPD Algebra, the valu... | We are looking for: $S_n=\sum\limits_{n=1}^\infty\frac{n}{1+n^2+n^4}$
Let's realize that $\frac{n}{1+n^2+n^4}=\frac{1}{2}\big(\frac{1}{1-n+n^2}-\frac{1}{1+n+n^2}\Big)=\frac{1}{2}\big(\frac{1}{(n-\frac{1}{2})^2+\frac{3}{4}}-\frac{1}{(n+\frac{1}{2})^2+\frac{3}{4}}\Big)$
Which provide telescopic sum so $S=\frac{1}{2}$
| {
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How do I simplify $\sqrt {4(2- \sqrt{3})}$ into $\sqrt{6} - \sqrt{2}$ This might be a stupid question, but how do I get from $$\sqrt {4(2- \sqrt{3})}$$ to $$\sqrt{6} - \sqrt{2}$$ It is obvious if you squared both, they both equal $8 - 4 \sqrt{3}$, but I'm wondering how you can find the answer from the original expressi... | Suppose $8-4\sqrt 3= (a-b\sqrt 3)^2$
This gives $a^2+3b^2=8$ and $2ab=4$ so that $ab=2$
This means that $a^2b^2+3b^4=8b^2$ or $4+3b^4=8b^2$ or $$3b^4-8b^2+4=0=(3b^2-2)(b^2-2)$$
So either $b^2=2$ or $b^2=\frac 23$.
We have also $a^2b^2=4$ so that $a^2=2$ or $a^2=6$
This should suffice to identify the solution which corr... | {
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Solving $\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$
Solve for $x$ such that
$$\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$$
First, I use the co-function formula:
$$\sin 80^\circ = \cos 10^\circ \tag{1}$$
Also,
$$\sin 20^\circ = 2\sin 10^\circ \cos 10^\circ \tag{2}$$
From th... |
A geometric solution:
First draw triangle $ABC$ where $\angle BAC=20$ and $AB=AC$. Easy to find out $\angle ABC=\angle ACB=80$.
Now add point $D$ on line $AC$ such that $\angle CBD = x$, so $\angle ABD=80-x$ and $\angle BDC = 100-x$.
${sin(80-x)\over sin20}={sin(100-x)\over sin80} \implies {AD\over BD}={BC\over BD}\im... | {
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} |
How can I prove the following matrix equality? Let $n\in\mathbb{N}$ be arbitrary. Why is the following equality true?
$$
\left(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}
+ 3 \begin{pmatrix}4 & 6\\-2 & -3\end{pmatrix}\right)^n
= \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}
+ (4^n-1)\begin{pmatrix}4 & 6\\-2 & -3\end{pmatrix}
.... | Just use binomial formula and the fact that $A^2 = A \Rightarrow A^k = A$
$$(I + 3A)^n = \sum_{k=0}^n \binom{n}{k}3^kA^k =I + A\sum_{k=1}^n \binom{n}{k}3^k = I + A ((1+3)^n-1) = I +(4^n-1)A$$
Note that you can apply the binomial formula since $IA = AI$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum point of $x^2+y^2$ given that $x+y=10$ How do you I approach the following question:
Find the smallest possible value of $x^2 + y^2$ given that $x + y = 10$.
I can use my common sense and deduce that the minimum value is $5^2 + 5^2 = 50$. But how do you approach this mathematically?
Thanks in advance.
| $\large \text{How about this approach:}$
$
x + y = 10, \text{ so } y = 10 - x \\
x^2 + y^2 = k, y = \sqrt{k - x^2}\\
\rightarrow 10 - x = \sqrt{k - x^2}\\
\rightarrow 100 - 20x + x^2 = k - x^2\\
\rightarrow 2x^2 - 20x + 100 = k\\
f\prime = 4x - 20\\
\text{When x is 5, } f\prime = 0\\~\\
\text{Testing values on the righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 9,
"answer_id": 1
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What are the solutions of $x=\cot x$? Need to find intervals in which the function $y=\frac{x}{2}\cdot \cos x$ is increasing and decreasing. I tried to solve it on the way below but don't know how to continue.
$ \\
y=\frac{x}{2}\cdot \cos x,\ x\in (0,2\pi)\\
y=\frac{1}{2}\cdot x\cdot \cos x \\
{y}'=({\frac{1}{2}\cdot ... | It is a transcendental equation and you will have to find the solution numerically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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What is the value of $x+y$ if x and y are co-primes and $PR=\dfrac{x}{y}$ in the diagram?
$ST$ is the perpendicular bisector of $PR$ and $SP$ is the angle
bisector of $\angle QPR$. If $QS=9cm$ and $SR=7cm$ then
$PR=\dfrac{x}{y}$ where x, y are co-primes. $x+y$=?
Source: Bangladesh Math Olympiad 2015 junior categ... |
As shown, by the angle bisector theorem,
$${(\sqrt{7^2-a^2} + \sqrt{9^2-a^2})\over 9}={(\sqrt{7^2-a^2} + \sqrt{7^2-a^2})\over 7}$$
Simplify,
$$7\sqrt{9^2-a^2}=11\sqrt{7^2-a^2}\implies 7^29^2-49a^2=7^211^2-121a^2\implies 72a^2=$$
$$7^2(11^2-9^2)=7^2\times40\implies a^2={245\over 9}$$
Therefore $PR=2\sqrt{7^2-a^2}=2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to convert the following sum to a geometric series?
Find $$
\sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n}
$$
There are many ways to find that the limit is divergent, but the question explicitly states the sum must be interpreted as a geometric series. I know how to render the exponential terms to $(n-1)$; it'... | Expanding of Ross Millikan's comment, you could try
\begin{align}
\sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n}
& = \sum_{n = 1}^{\infty} \frac{6}{3^n} - \sum_{n = 1}^{\infty} \frac{2^{2n - 1}}{3^n}
= 2 \cdot \sum_{n = 1}^{\infty} \left( \frac{1}{3} \right)^{n - 1} - \frac{1}{2} \sum_{n = 1}^{\infty} \left( \frac{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How can $\frac{4}{3} \times 3=4$ if $ \frac{4}{3}$ is $1.3$? Ok use your closest calculator, and type $\frac{4}{3}$, which is $1.3333333333$,and then multiply it with $3$ which is $3.9999999999$ but then type $\frac{4}{3} \times 3=4$ how?. How can it be $4$ if $\frac{4}{3}$ is $1.3333333333$ and when you multiply it wi... | Rational numbers are numbers that can be expressed as $\frac{p}{q}$, where p and q are relatively prime integers. Notice that the definition does not mention the decimal expansion. The decimal representation of $\frac{4}{3}$ is defined as $1+\sum_{i=1}^{\infty}\frac{3}{10^i}$. Multiplying by 3, the decimal expansion fo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find the difference of ceiling functions For $k \geq 1, 1 \leq r \leq k, t \geq 1, x \geq 1$, is there a lower bound or upper bound on:
$$\left\lceil \dfrac{k(t+x)}{r} \right\rceil - \left\lceil \dfrac{k(t)}{r} \right\rceil$$
edit: all the variables $k,r,t,x$ are integers
| The remainder theorem says that for any two positive integers, $q$ and $n$ you can divide $n$ by $q$ and get two unique integers $d$ and $s$ so that $n = dq + s$ and $0 \le s <q$.
So let $kt = Mr +s$ and let $kx = Nr + w$.
Then $\lceil \frac {k(t+x)}r \rceil = \lceil M + N + \frac {s+w}r\rceil= M + N + \lceil \frac {s+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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$15 [\tan 2\theta + \sin 2\theta] + 8 = 0$ if ...
QUESTION: $15[\tan 2\theta + \sin 2\theta] + 8 = 0$ if...
(a) $\tan\theta = \frac{1}{2}$
(b) $\sin\theta = \frac{1}{4}$
(c) $\tan\theta = 2$
(d) $\cos\theta = \frac{1}{5}$
What I have tried:
$$\tan2\theta + \sin2\theta = -\frac{8}{15}$$
$$\frac{2\tan\theta... | From the final equation you gave, one gets
$$\frac{4\tan\theta}{1-\tan^4\theta} = -\frac{8}{15} \quad\Rightarrow\quad
8\tan^4\theta - 60\tan\theta -8=0.$$
This quartic factors as $4(\tan\theta-2)(2\tan^3\theta + 4\tan^2\theta + 8\tan\theta + 1)$.
This shows that (c) is a solution while (a) is not. To see that (b) and (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$
If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!
| Swapping numerator and denominator,
$$
\frac{\sin^2 y+\ln(1+x^2)}{x^2+y^2}=\frac{y^2}{x^2+y^2}\frac{\sin^2 y}{y^2}+\frac{x^2}{x^2+y^2}\frac{\ln(1+x^2)}{x^2}.
$$ It is equal to
$$
\frac{y^2}{x^2+y^2}\left(\frac{\sin^2 y}{y^2}-1\right)+\frac{x^2}{x^2+y^2}\left(\frac{\ln(1+x^2)}{x^2}-1\right)+1
$$ The first term tends to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Maximum value of expression $a+b+c$
If $a,b,c$ are non negative integers such that $$2(a^3+b^3+c^3)=3(a+b+c)^2.$$
Then maximum value of $a+b+c$ is ?
My Try: Using Jensen Inequality
Let $f(x)=x^3$. Then $f''(x)>0$ for $x>0$ is convex function
So $$\frac{f(a)+f(b)+f(c)}{3}\geq f\bigg(\frac{a+b+c}{3}\bigg)$$
$$\frac{a^... | You can solve it also like this. By Cauchy inequality we have $$(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2$$
Further: $$a^2+b^2+c^2\geq {1\over 3}(a+b+c)^2$$
so $${3\over 2}(a+b+c)^3\geq {1\over 9}(a+b+c)^4$$
and thus $$a+b+c\leq {27\over 2}$$
Since $2\mid a+b+c$ we have $a+b+c\leq 12$.
Also, since $x^3\equiv x\pmod 3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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find equation of tangent for function $f(x)$ parallel to a line I dont have a clue how to solve this kind of problem.
I dont know how to find out slope of $\ln x$.
$f(x) = \ln x$
line $p: 2x-y-3=0$
I have searched for slope of $\ln x$, I know how find out slope for one point, for example:
first I need to derivate the $... | Equation of tangent to $y = f(x) = \ln(x)$ at $(x_1, y_1)$ is of form
$ (y - y_1) = f'(x_1)(x - x_1) $ where
$ f'(x) = \frac{d}{dx} \ln(x) = \frac{1}{x} $
$
\Rightarrow (y - y_1) = \frac{1}{x_1}(x - x_1)
$
Since Slope of $p: 2x-y-3=0$ is 2, hence,
Slope of || tangent line = Slope of $p$ = 2
$
\Rightarrow \frac{1}{x_1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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$x+x^2+x^3=x^4+x^5+x^6$ implies $x^4=x$ in a ring Let $(A,+,.)$ be a ring s.t. $x+x^2+x^3=x^4+x^5+x^6$ for all $x \in A$. Prove that $x^4=x$ for all $x$ in $A$. Can somebody give me some tips, please?
| Here is a variant of the solution by @Servaes that seems simpler to me:
Plugging $x=-1$ gives $2=0$ in $A$.
Therefore, $x+x^2+x^3=x^4+x^5+x^6$ is the same as $x+x^2+x^3+x^4+x^5+x^6=0$.
Multiplying that by $x-1$ gives $x^7-x=0$ and so $x^8-x^2=0$.
Now $x^8-x^2=(x^4-x)^2$.
Let $z=x^4-x$. Then $z^2=0$ and $z+z^2+z^3=z^4+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Using induction to prove an inequality for a sequence of numbers We have the sequence $d_n = \begin{cases} 1 &\text{ if } n=0 \\
\frac{n}{d_{n-1}} &\text{ if } n>0 \end{cases}$
for all natural numbers $n$.
($d_{n-1}$ is the previous number of the sequence.)
examples: $d_0 = 1$, $d_1 = 1$, $d_2 = 2$, $d_3 = \frac{3}{2}$... | Your observation $$d_{2n-1} = d_{2n-3} \, \frac{2n-1}{2n-2}$$ is homologous with $$d_{2k+1} = d_{2k-1} \, \frac{2k+1}{2k}.$$
\begin{align}
d_{2k-1} \, \frac{2k+1}{2k} &\le \sqrt{2k-1} \, \frac{\bbox[yellow, 2px]{2k+1}}{2k} \tag{induction hypothesis}\\
&\le \color{blue}{\sqrt{2k-1}} \, \frac{\bbox[yellow, 2px]{\color{bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the rectangular function a convolution of $L^1$ functions? Do there exist functions $f,g$ in $L^1(\mathbf{R})$ such that the convolution $f \star g$ is (almost everywhere) equal to the indicator function of the interval $[0,1]$ ?
| Following the idea of Christian Remling, I have two explicit functions that have the property that $f*g = \mathbf{1}_{[-1,1]}$:
\begin{eqnarray}
f(x) &=& \frac{K_\frac{1}{4}\left(|x|\right)}{2^{1/4}\pi\Gamma\left(\frac{1}{4}\right)|x|^{1/4}}\\\\
g(x) &=& \frac{\mathrm{sgn}(1-x)}{\sqrt{|1-x|}}\,_1F_2\left(\begin{matrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 3,
"answer_id": 1
} |
maximum value of $\sum ab-2abc$ If $a+b+c=1$ and $a,b,c\in(0,1)$, then what is the maximum value of $(ab+bc+ca-2abc)$?
What I've tried:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 4(ab+bc+ca)$
$(a-b)^2=a^2+b^2-2ab\geq 0$
$a^2+b^2\geq 2ab,b^2+c^2\geq 2bc,c^2+a^2\geq 2ca$
$ab+bc+ca\leq\frac14$
How do I solve it help me please... | Σ(ab) = Σ ( a²b+b²a ) + abc and by using AM-GM to the terms under the modified sigma we get that it is always greater than 6abc and using AM - GM on a,b,c we get abc<1/27 hence the maximal value must be 7/27
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Reduction of hypergeometric function for integer parameters Consider integers $a,b,c,d>0$ and the hypergeometric function
$${}_3F_2(1-d,b,a+b;a+b+c,b+1;1)$$
I don't know much about hypergeometric functions but I understand that when the parameters are integer valued we get some simplification. Unfortunately, I can't se... | Wolfram functions reference is a good place to find identities like this. Here's the one for $\;_3F_2$. For this, you probably want. Unfortunately I'm not finding much in values at $z = 1$, but there's a lot to go through.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite
If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite.
My attempt so far:
$$a^2+b^2+c^2+d^2$$
$$=a^2+d^2+2ad+b^2+c^2+2bc-2ad-2bc$$
$$=(a+d)^2+(b+c)^2-2(b^2+bc+c^2)-2bc$$
$$=(a+d)^2+(b+c)... | Assume that $a - b - c + d = 1$. Then you'll get from your result that
$$a^2 + b^2 + c^2 + d^2 = (a+b+c+d)(a-b-c+d) = a+b+c+d \tag{1}\label{eq1}$$
Now, since $a,b,c,d\in\mathbb{Z^+}$, note that $a^2 \gt a$ if $a \gt 1$, and likewise for $b, c, d$. Thus, \eqref{eq1} can only be true is $a = b = c = d = 1$. In that case,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Picture proof that the area of a right triangle is $xy$ I stumbled on the following result by accident:
Let $A, B, C$ be the vertices of a right triangle, with opposite side lengths $a, b, c$ respectively, where $\angle C = 90^\circ$ and $a^2 + b^2 = c^2$.
Draw the incircle, and let $x, y, z$ be the length of the tang... | Let A be the area of the original triangle cut up in the obvious way as on the picture. Then the area of the rectangle is $A_r = 2 A = ab = xy + A$. Thus $xy = A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Find the Length of $QP$ Given two circles with radii $8$ and $6$ units with centers $A$ and $B$ such that $AB=12$ If $P$ is mid point of $QR$
Find Length of $QP$
My try:
I assumed $A(0,0)$ and $B(12,0)$
So the equations of circles are:
$$x^2+y^2=64$$
$$(x-12)^2+y^2=36$$
Solving above equations we get:
$$P\left(\frac{... |
You don't need any trigonometry or analytical geometry. The following equations are quite obvious (Pythagora):
$$p^2+\frac{x^2}{4}=8^2=64\tag{1}$$
$$q^2+\frac{x^2}{4}=6^2=36\tag{2}$$
$$(p-q)^2+x^2=12^2=144\tag{3}$$
Introduce substitution: $y=\frac{x^2}{4}$ and you get:
$$p=\sqrt{64-y}\tag{4}$$
$$q=\sqrt{36-y}\tag{5}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum value of $PA+PB$ is If $P(x,y,z)$ lie on line $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}$ and $A(5,3,4)$ and $B(1,-1,2)$ . Then minimum value of $PA+PB$ is
what i try
let $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}=\lambda$
Then $P(2\lambda-2,2\lambda-7,\lambda+2)$
$PA+PB=\sqrt{(2\... | By Minkowski we obtain:
$$\sqrt{9\lambda^2-72\lambda+153}+\sqrt{
9\lambda^2-36\lambda+45}=$$
$$=3\left(\sqrt{(4-\lambda)^2+1}+\sqrt{(\lambda-2)^2+1}\right)\geq$$
$$\geq3\sqrt{(4-2)^2+(1+1)^2}=6\sqrt2.$$
The equality occurs for $\lambda=3,$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solution to the definite integral $\int_0^{\infty}x^{-\alpha} e^{-\beta/x}\,dx$ Is there a explicit solution to this definite integral, with $\alpha>0$ and $\beta>0$ :
$$\int_0^{\infty}x^{-\alpha} e^{-\beta/x}\,dx$$
If yes, what is it?
| The method prescribed by egreg is far superior to what I'm going to put forward.
Here we will address your integral:
\begin{equation}
I\left(\alpha, \beta \right) = \int_0^\infty x^{-\alpha} e^{-\beta/x}\:dx\nonumber
\end{equation}
We now employ Fubini's Theorem and take the Laplace Transform with respect to $\beta$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find digits $a,b$ such that $7ab + 4ba = 1a21$. I have to find all the digits $a$ and $b$ such that $7ab + 4ba = 1a21$.
Note: there is no multiplication, those are three decimal numbers.
I put this equality this way: $$7\cdot100 + a \cdot 10 + b + 4 \cdot 100 + b \cdot 10 + a = 1 \cdot 1000 + a \cdot 100 + 2 \cdot 10 +... | Another approach to solve such equations is to write it as an algebraic equation, instead of with base-10 digits. We get
$$
700 + 10a + b + 400 + 10b + a = 1000 + 100a + 21
$$
which simplifies to:
$$
11(a + b) + 1100 = 1021 + 100a
$$
or
$$
11b + 79 = 89a.
$$
Now we use the fact that $a$ and $b$ are between $0$ and $9$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3145603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Number of possible unique patterns of $5$ points on an $8\times 8$ grid. I'm after a specific answer and workings/formula for the number of possible unique patterns that $5$ points can have on an $8\times 8$ grid. The pattern must be unique and not match another pattern when it's rotated or mirrored. Can the workings b... | We solve the case of the points going into the slots between the grid,
so there are $64$ slots. We put at most one point into one slot. We
use the Polya Enumeration Theorem (PET), which requires the cycle
index of the action of the symmetries on the slots. Supposing the grid
is $n\times n$ with $n$ even we g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all polynomials with coefficients from set $\{-1,1\}$ and which have all their roots real.
Find all polynomials with coefficients from
set $\{-1,1\}$ and which have all their roots real.
What I have tried:
Assume polynomial is $a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots +a_{0}=0$ and $a_{i}\in \{-1,1\}$ for... | WLOG, let $a_n = 1$. Also, let $r_1,\ldots,r_n$ be the roots of the polynomial.
As you noted, $\displaystyle\sum_{k = 1}^{n}r_k = -a_{n-1}$, $\displaystyle\sum_{1 \le k < \ell \le n}r_kr_{\ell} = a_{n-2}$, and $\displaystyle\prod_{k = 1}^{n}r_k = (-1)^na_0$. Hence,
$$\displaystyle\sum_{k = 1}^{n}r_k^2 = \left(\sum_{k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$
$a$, $b$ and $c$ are three sides of a triangle such that $a + b + c = 2$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$
Every task asking for finding the minimum value of an expression containing the product of all of t... | I do not know if you had to use AM-GM but the problem is quite simple using pure algebra.
Considering
$$ P = 4(a^3 + b^3 + c^3) + 15abc \qquad \text{with} \qquad a+b+c=2$$ eliminate $c$ from the constaint to get
$$P=3 a^2 (8-9 b)-3 a (b-2) (9 b-8)+8 (3 (b-2) b+4)$$ Now
$$\frac{\partial P}{\partial a}=6 a (8-9 b)-3 (b-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Parametrizing the square spiral Related to this question concerning number spirals I have another one, more specific.
While it is rather easy to arrange the natural numbers along an Archimedean spiral by
$$x(n) = \sqrt{n}\cos(2\pi\sqrt{n})$$
$$y(n) = \sqrt{n}\sin(2\pi\sqrt{n})$$
it's much more difficult to arrange them... | [This answer is inspired by user SMM's answer. Thanks for it.]
Consider "piecewise linear approximations" of the sine and cosine function, periodically defined on the unit interval, i.e. $x \in [0,1]$.
Let
$$\boxed{\cos_\bigcirc(x) = \cos(2\pi x)\\\sin_\bigcirc(x) = \sin(2\pi x)}$$
and compare this to
$$\boxed{\cos_\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Solve the following system of equations - (3)
Solve the following system of equations:
$$\large
\left\{
\begin{align*}
3x^2 + xy - 4x + 2y - 2 = 0\\
x(x + 1) + y(y + 1) = 4
\end{align*}
\right.
$$
I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.
| Solving the first equation for $y$ we get $$y=\frac{-3x^2+4x+2}{2+x}$$ for $$x\neq -2$$
plugging this in the second equation we get after simplifications
$$(5x+4)(x-1)^3=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
$x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
Substituting $y=x+45^{\circ}$, we get
$$3\tan{(y-60^{\circ})}=\tan{(y-30^{\circ})}$$
$$3\frac{\tan y - \sqrt3}{1+\sqrt3\tan y}=\frac{\tan y - 1/\sqrt3}{1+1/\sqrt3\cdot\tan y}$$ ... | You got $$0\cdot\tan^2{y}+9=1.$$ Id est, you got that $\tan{y}$ does not exist.
Thus, $$y=\frac{\pi}{2}+\pi n,$$ where $n\in\mathbb{Z},$ or
$$x+45^{\circ}=\frac{\pi}{2}+\pi n,$$ which gives $$x=\pi n+\frac{\pi}{4}.$$
We did not get a contradiction in Math!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation. I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the functio... | Consider $z^2+z-2=(z+2)(z-1)=-2(1+\frac{1}{2}z)(1-z)$. Its inverse (as a formal power series) is
$$
-\frac{1}{2}(1+z+z^2+\dotsb)\left(1-\frac{1}{2}z+\frac{1}{4}z^2+\dotsb\right)
=-\dfrac{1}{2}-\dfrac{1}{4}z-\dfrac{3}{8}z^2+\dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$
Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x -... | After squaring one times we obtain
$$-2\sqrt{x+2}\sqrt{3-x}=(x^2-6x+9)^2-5$$
squaring again we obtain
$$ \left( x-2 \right) \left( {x}^{7}-22\,{x}^{6}+208\,{x}^{5}-1096\,{x}
^{4}+3468\,{x}^{3}-6552\,{x}^{2}+6772\,x-2876 \right)
=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Dudeney’s solutions to haberdasher's problem exact measures of sections What is the IG length if the side of the square is 1? I wonder if it is half of the square side. The triangle below represents the haberdasher's problem.
version 2
version 1 (added after edit, here the question is about JL)
The images are taken ... | Denote by $a$ the side of the triangle. Since the areas of the triangle and the square are the same, we have $\frac{a^2\sqrt 3}{4}=1$, so $a=\frac{2}{\sqrt[4]3}$.
Set the coordinate system as follows: $A=(0,0)$, $C=(a,0)$ and $B=(\frac{a}{2},\frac{a\sqrt 3}{2})$. Then $D=\frac{A+B}{2}=(\frac{a}{4},\frac{a\sqrt 3}{4})$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Examining an inequality involving exponential functions and hyperbolic cosine Let $a,b$ be real numbers with $0 < a < b$.
Problem: I would like to prove/disprove that $$ \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b} \leq \cosh(x \log{2})
$$ is true for all real $x \geq 0$.
Approach:
*
*I defined a function $f: \mathbb{R... | Using
$$
\frac{a}{a+b} = \frac 12 \left( 1 - \frac{b-a}{a+b}\right) \, , \quad
\frac{b}{a+b} = \frac 12 \left( 1 + \frac{b-a}{a+b}\right)
$$
we have
$$
\begin{aligned}
\frac{a \cdot 2^x + b \cdot 2^{-x}}{a+b} &= \left(\frac{2^x + 2^{-x}}{2}\right) - \frac{b-a}{a+b} \left(\frac{2^x - 2^{-x}}{2}\right) \\
&= \cosh(x \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Number of ways to express sum.
Consider three sets of cards colored Blue, Red and Yellow. Each set has cards numbered $1-10$. The $4$ remaining cards are all indistinguishable cards numbered $0$.
Card numbered $i$ has the value of $2^i$. How many ways are there to choose a group of cards that sums up to $2016$.
I'm h... | We can represent the number of selecting zero up to four indistinguishible cards which have value $2^0=1$ as
\begin{align*}
1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}
\end{align*}
We can represent the number of selecting zero up to three distinguishible cards numbered with value $2^i$ as
\begin{align*}
\left(1+x^{2^i}\right)^3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+...+x^{11}$ Find the remainder when the polynomial $$1+x^2+x^4+x^6+....+x^{22}$$ is divided by $$1+x+x^2+x^3+...+x^{11}$$
$1+x^2+x^4+x^6+....+x^{22}=\frac{x^{24}-1}{x^2-1}$
$1+x+x^2+x^3+...+x^{11}=\frac{x^{12}-1}{x-1}$
Now$$\... | Assume $\deg Q(x)=11$ and $\deg R(x)\le10$, with
\begin{align*}
(1+x^2+\cdots+x^{22})=(1+x+\cdots+x^{11})Q(x)+R(x).\tag 1
\end{align*}
Multiply $(x^2-1)$ on both sides,
\begin{align*}
\color{blue}{(x^{12}-1)}(x^{12}+1)&=\color{blue}{(x^{12}-1)}(x+1)Q(x)+(x^2-1)R(x)
\end{align*}
implies $\color{blue}{(x^{12}-1)}\mid(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solutions of sin(x) = cos(x) I know that the solutions to the equation $\sin(x) = \cos(x)$ are :
$ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°)
However when I try to solve it algebraically I get the following :
$$ \sin x = \cos x$$
$$ \sin^2 x = \cos^2 x$$
$$ \sin^2 = 1 - \sin^2 x$$
$$ 2\sin^2 x = 1$$
$$... | A better alternative would be to do the following: $$\begin{aligned}\sin x&=\cos x\\\tan x&=1\\x&=\arctan 1\\ x&=\dfrac{\pi}{4}+\pi n, n\in \mathbb{Z}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without L'Hospital's Rule. I need to solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without using L'Hospital's Rule. Using that rule I found the equation becomes $\lim_{x \to 0}(\frac{1}{\sqrt{1+2x}} - \frac{2}{\sqrt{1-4x}}) = \frac{-1}{\sqrt{1}}$.... | $$\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x} =\frac{1+x + o(x)-1 + 2x + o(x)}{x} = 3 + o(1) $$
So the limit is $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find length of the arc of $y^2=x$ I am embarrassingly stuck on this example. My textbook provides the answer (picture below) and the steps, but I am unable to follow the math. I have been stuck for more then an hour.
Find the length of the arc of the parabola $y^2=x$ from (0,0) to (1,1).
Since $x=y^2$, we have $\frac{... | If you have a sum under the square root sign ($\sqrt{a^2+x^2}=\sqrt{\frac{1}{a^2}\left(1+a^2x^2\right)}=\frac{1}{a}\sqrt{1+a^2x^2},\ a>0$), your trigonometric substitution would be $x=\frac{1}{a}\tan{\theta}$:
$$
\int\sqrt{1+4y^2}\,dy=\int\sqrt{1+2^2y^2}\,dy\rightarrow y=
\frac{1}{2}\tan{\theta}, dy=\frac{dy}{d\theta}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation... | How about using Cramer's Rule? Define $\Delta_x=\left[\begin{matrix}36 & 2 \\ 64 & 4\end{matrix}\right]$, $\Delta_y=\left[\begin{matrix}3 & 36\\ 5 & 64\end{matrix}\right]$
and $\Delta_0=\left[\begin{matrix}3 & 2\\ 5 &4\end{matrix}\right]$.
Now computation is trivial as you have: $x=\dfrac{\det\Delta_x}{\det\Delta_0}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 7
} |
What is $\int_{0}^{\pi} (a^2+1+2a \cos\theta)^{\frac{p-1}{2}}d\theta$? I have tried the problem using $\theta\rightarrow\theta/2$, then taking $z=e^{i\theta}$ we get $\frac{1}{2i}\int_{|z|=1}\frac{(a^2+1+a\sqrt{z}+\frac{a}{\sqrt{z}})^{\frac{p-1}{2}}}{z}dz$.
| With the change of variables $t = 2 \arccos \sqrt u$,
$$\int_0^\pi (a^2 + 2 a \cos t + 1)^{(p - 1)/2} dt = \\
|a - 1|^{p - 1} \int_0^1 u^{-1/2} (1 - u)^{-1/2}
\left( 1 + \frac {4 a} {(a - 1)^2} u \right)^{(p - 1)/2} du = \\
\pi |a - 1|^{p - 1} {_2 \hspace{-1px} F_{\hspace{-1px} 1}} {\left(
\frac 1 2, \frac {1 - p} 2;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the v... | Let's start with the basic symmetric expressions: $ab+bc+ca$ and $abc$. You can refer to giannispapav's answer for details, which shows that
$$ab+bc+ca = -1/2, abc = 1/6.$$
With that, Vieta's formulas implies that $a,b,c$ satisfy:
$$ x^3 -x^2 - x/2 -1/6=0,\tag{1}$$
Or
$$x^3 = x^2 + x/2 + 1/6.$$
That means, for $x$ equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Solve for $x^2 + 7x +1 = 3n(x^2 + x +1), n \in \Bbb{Z}$ Solve for $x$ and $n$ in the equation
$$ \dfrac{x^2 + 7x +1}{ x^2 + x +1}= 3n,\qquad n \in \Bbb{Z}.$$
The original problem was a trigonometric equation but solved it till I got stuck here. I am a high school student who is self studying and preparing for college ... | $3n
=\dfrac{x^2 + 7x +1}{ x^2 + x +1}
$
so
$3nx^2+3nx+3n
=x^2+7x+1
$
so
$(3n-1)x^2+(3n-7)x+3n-1
=0
$
or
$\begin{array}\\
x
&=\dfrac{-3n+7\pm \sqrt{(3n-7)^2-4(3n-1)^2}}{2(3n-1)}\\
&=\dfrac{-3n+7\pm \sqrt{(3n-7-2(3n-1))(3n-7+2(3n-1))}}{2(3n-1)}\\
&=\dfrac{-3n+7\pm \sqrt{(-3n-5)(9n-9)}}{2(3n-1)}\\
&=\dfrac{-3n+7\pm 3\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Very indeterminate form: $\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x \longrightarrow (\infty-\infty)^{\infty}$ Here is problem:
$$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$
The solution I presented in the picture below was made by a Mathematics Teacher
I tried to solve t... | The proof can be accelerated, using binomial Maclaurin series in the form of
$$\sqrt{x^2+2x+3} = (x+1)\sqrt{1+\dfrac2{(x+1)^2}} = (x+1)\left(1 + \dfrac1{(x+1)^2}+O\left(x^{-4}\right)\right)$$
$$ = x+1+\dfrac1x+O\left(x^{-2}\right),$$
$$\sqrt{x^2+3} = x\left(1+\dfrac3{2x^2}+O(x^{-4})\right) = x + \dfrac3{2x}+O(x^{-3}).$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 5
} |
Representing a function as a power series and finding the coefficients of the terms? My prblem asks me to find the coefficients of the first five terms of the power series representation of the function:
$$f(x)=\frac{10}{17+x}$$
$$f(x)=10*\frac{\frac{1}{17}}{1-\frac{-x}{17}}$$
$$\sum_{n=0}^{\infty}10*\frac{1}{17}*(\fra... | You allready found $$f(x)=\frac{10}{17} + \frac{10}{17}\frac{-x}{17} + \frac{10}{17}\frac{x^2}{17^2} + \frac{10}{17}*\frac{-x^3}{17^3} + \frac{10}{17}\frac{x^4}{17^4}+...$$
$$=\frac{10}{17} - \frac{10}{17^2}x + \frac{10}{17^3}x^2 - \frac{10}{17^4}x^3 + \frac{10}{17^5}x^5 \mp ...$$
The general coefficient $a_n$ is given... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$f: \mathbb{R} \to \mathbb{R},\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space$find $f(3)$ in terms of $f(0)$.
$f: \mathbb{R} \to \mathbb{R},\space\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space\space\space$
Find $f(3)$ in terms of $f(0)$.
My approach:
$$f(x)-2f(\frac{... | I'll go ahead and assume that $f$ is continuous at $0$.
For any $x \in \mathbb{R}$, let $P(x)$ denote the property that $f(x) - 2f(\tfrac{x}{2}) + f(\tfrac{x}{4}) = x^2$. Then, for any $x$, we have the following:
\begin{align}
P(x)&: \quad f(x) - 2f(\tfrac{x}{2}) + f(\tfrac{x}{4}) = x^2
\\
P(\tfrac{x}{2})&: \quad f(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The smallest value of the expression: $(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$ lies in which of the following intervals? Question:
Let $a_1,a_2,a_3,a_4\in\Bbb R$, such that $a_1+a_2+a_3+a_4 = 0$ and $a_1^2+a_2^2+a_3^2+a_4^2 = 1$.
Then the smallest value of the expression, $$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2... | Hint: Your sum simplifies to $$2-2(a_1a_3+a_2a_3+a_3a_4+a_4a_1)=2-2(a_2(a_1+a_3)+a_4(a_1+a_3))=2-2(-(a_2+a_4)^2)$$ and $$2+2(a_2+a_4)^2\geq 2$$. The minimum is equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\left(\sqrt{\sqrt{2}-4-x}\right)+x^{\frac{1}{4}}=2^{\frac{-1}{4}}$ Solve the Equation in real/Complex numbers:
Solve $$\left(\sqrt{\sqrt{2}-4-x}\right)+x^{\frac{1}{4}}=2^{\frac{-1}{4}}$$
My try:
Letting $x=t^4$ we get
We get
$$\left(\sqrt{\sqrt{2}-4-t^4}\right)+|t|=2^{\frac{-1}{4}}$$
Then:
$$\left(\sqrt{\sqrt{2... | As said in comments, I seriously wonder it the problem is correctly stated.
This equation shows two complex roots. Using Newton method, we have the awful iterates
$$\left(
\begin{array}{cc}
n & x_n \\
0 & +1.00000+1.00000\, i \\
1 & -3.58944+1.23135\, i \\
2 & -0.75253+0.37882\, i \\
3 & -1.89384+0.00982\, i \\
4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Real roots of the equation $\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$
Find the set of real roots of the equation$$\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$$
My Attempt
$$
2x+3>0, 5x+4>0, 2x+3,5x+4\neq1\implies x>-4/5\;\&\;x\neq -1\;\&\;x\neq -3/5
$$
$$
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)(2x... | Since $10x^2+23x+12=(2x+3)(5x+4)$, my suggestion is to set $t=\log_{(2x+3)}(5x+4)$, so that
$$
\log_{(5x+4)}(2x+3)=\frac{1}{t}
$$
and the equation becomes
$$
\frac{3}{t}-1-t=1
$$
so $t^2+2t-3=0$ and $t=1$ or $t=-3$.
The first case yields $5x+4=2x+3$, hence $x=-1/3$.
The second case yields $5x+4=(2x+3)^{-3}$. If we cons... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Calculate $\sum_{n=1}^{\infty}\left(\frac{1}{3}\frac{2}{5}\cdots\frac{n}{2n+1}\frac{1}{n+1}\right)$
Denote $a_n=\frac{1}{3}\cdot\frac{2}{5}\cdots\frac{n}{2n+1}\cdot\frac{1}{n+1}$.
Please prove
$$\sum_{n=1}^{\infty}a_n=\frac{\pi^2}{8}-1$$
This is the answer given by my friend. He also used the integral $\int_{-\pi}^{\... | Using the Factorial aswell as the Double Factorial we may rewrite $a_n$ as
$$a_n=\frac13\frac25\cdots\frac n{2n+1}\frac1{n+1}=\frac{n!}{(2n+1)!!}\frac1{n+1}$$
Expressing the utilized Double Factorial in terms of the Factorial this can be further reduced
$$a_n=\frac1{(2n+1)!!}\frac{n!}{n+1}=\frac{2^{n+1}(n+1)!}{(2n+2)!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Writing $3.8473221018630726$ in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$. I attempted the following question on Brilliant which has to do with finding roots of a cubic polynomial. I was successful in finding what the only real root is but I am facing a problem rewriting the root in the sought expression.
The equ... |
Would the fact that $x=\frac{2}{\sqrt[3]{4}}+\sqrt[3]{4}+1=\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1\approx3.8473221018630726$ help somewhere in determining $a, b$ and $c$?
Yes, it would very much. If $\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1$ is the root you're after, then
$$
\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1 = \sqrt[3]{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How do i integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$? Integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$
I put $x^2+3x+2=t,$ and notice that $2x+3 dx=dt$, but the $dx$ is above! Please help me!
| Let $ t = \sqrt{x^2+3x+2}$, $dt = \frac{2x+3}{2\sqrt{x^2+ 3x +2}}dx$, $dx = \frac{2\sqrt{x^2+ 3x +2}}{2x+3}dt$
$I = \int{ \frac{2\sqrt{x^2+ 3x +2}}{2x+3}. \frac{1}{{(2x+3)}.{\sqrt{x^2+ 3x +2}}}dt} =\int{ \frac{2}{2x+3}. \frac{1}{{2x+3}}dt} =\int{ \frac{2}{4x^2+6x+9}dt} = 2\int{\frac{1}{4t^2 + 1}} $(on solving x in te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$1+\sqrt{2}$ seems special: its powers quickly approach integers $ \alpha = 1+\sqrt{2}$ has some interesting properties. One of which is that if you look at powers
\begin{align*}
\alpha^2 &= 3+2\sqrt{2} \approx 5.8284 \\
\alpha^3 & \approx 14.0710 \\
\alpha^4 & \approx 33.9705 \\
& \vdots \\
\alpha^{10} & \approx 6725.... | You could observe the pattern if you would calculate the powers:
$$\begin{align}\alpha^1&=(1+\sqrt{2})^1=1+1\sqrt{2}\\
a^2&=(1+\sqrt{2})^2=3+2\sqrt{2}\\
a^3&=(1+\sqrt{2})^3=7+5\sqrt{2}\\
a^4&=(1+\sqrt{2})^4=17+12\sqrt{2}\\
\vdots
\end{align}$$
So, basically, the second terms are important. The coefficients $1,2,5,12,..... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
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When recreating the quadratic formula by completing the square of $ax^2+bx+c=0$ I cannot shorten the right hand side I am attempting to derive the quadratic formula by completing the square on the generic generic expression:
$$ax^2+bc+c=0$$
I'm struggling with the right hand side of the equation which, for the step I... | Note that
$$
\frac{c}{a}=\frac{4ac}{4a^2}
$$
whence
$$
\frac{b^2}{4a^2}-\frac{c}{a}=\frac{b^2-4ac}{4a^2}
$$
In your approach
$$
\frac{b^2a-4a^2c}{4a^3}=\frac{b^2-4ac}{4a^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integrate $S = \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy$ $S = \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy$
$$
\begin{align}
S &= \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy \\
&= 2\pi \int_{0}^{1} (y+y^3) \sqrt{1+(1+3y^2)^2}dy \\
&= 2\pi \int_{0}^{2} \sqrt{1+u^2}du ~~~~~~\textrm{substitution with $u = y ... | We see that
$$S/2\pi=\int_0^1\sqrt{1+(1+3x^2)^2}\,xdx+\int_0^1x^2\sqrt{1+(1+3x^2)^2}\,xdx$$
Then set $t=1+3x^2\Rightarrow xdx=\frac16dt$ in both integrals to get
$$S/2\pi=\frac16\int_1^4 \frac{t-1}{3}\sqrt{1+t^2}\,dt+\frac16\int_1^4\sqrt{1+t^2}\,dt\\
=\frac1{18}\int_1^4t\sqrt{1+t^2}\,dt+\frac19\int_1^4\sqrt{1+t^2}\,dt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to solve Shonk Sequences? A Shonk sequence is a sequence of positive integers in which
*
*each term after the first is greater than the previous term, and
*the product of all the terms is a perfect square
For example: 2, 6, 27 is a Shonk sequence since 6>2 and27>6 and 2*6*27 = 324 or 18^2
a. If 12, x, 24 is a... | Although you don't show any work in your question text, I assume you have at least tried to answer them yourself. If you haven't, I suggest you do this first before looking at the solutions hidden below.
For question (a):
For $12, x, 24$ to be a Shonk sequence, note that $12 \times 24 = 2^5 \times 3^2$. Thus, $x$ mus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$ Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$
After few calculations I get that if I take $\frac{3}{2}x$ then after integral I get $\frac{3}{4}x^2+ C$ and
$$ \int_{\pi/6}^{\pi/2} \frac{3}{4}x^2+ C = \pi^2 / 6$$ so I shou... | Call the given integral as I, then $$I=\int_{\pi/6}^{\pi/2} \frac{x}{\sin x} dx, \mbox{use}~ x=2y~ ~ \mbox{ and get} ~ I= \int_{\pi/12}^{\pi/4} ~\frac{2y}{\sin 2y} 2 dy= 2 \int_{\pi/12}^{\pi/4} \frac{y}{\sin y \cos y} dy.$$Next, let $y=\tan^{-1} z$. then $$I=2 \int_{2-\sqrt{3}}^{1} \frac{\tan^{-1}z}{z} dz < 2(\sqrt{3}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How many Taylor series terms are needed to accurately approximate $\sqrt{a+x}-\sqrt{a}$? Naive evaluation of $\sqrt{a + x} - \sqrt{a}$ when $|a| >> |x|$ suffers from catastrophic cancellation and loss of significance.
WolframAlpha gives the Taylor series for $\sqrt{a+x}-\sqrt{a}$ as:
$$\frac{x}{2 \sqrt{a}} - \frac{x^2}... | To avoid cancellation error the first thing to do is to write:
$$
\sqrt{a+x}-\sqrt{a}=\frac{x}{\sqrt{a+x}+\sqrt{a}}=\sqrt{a}\frac{x}{a}
\frac{1}{1+\sqrt{1+\frac{x}{a}}}
$$
then with $y=\frac{x}{a}$ you must approximate this
$$
\sqrt{a}\frac{y}{1+\sqrt{1+y}}
$$
fonction for $y\in[10^{-300},1]$. This function has nothin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $a$ such that $-3\sin^{2}(x)-4\sin(x)+3-a=0$ has solution I have the function $f:[0,2\pi]\rightarrow \mathbb{R},f(x)=3\cos^{2}(x)-4\sin(x)$
I need to find the values of $a$, a real parameter such that $f(x)=a$ has solution.
My try:I got $-3\sin^{2}(x)-4\sin(x)+3=a$ so $-3\sin^{2}(x)-4\sin(x)+3-a=0$
I noted $\sin(x... | Consider the function
$$
f(x)=3\cos^2x-4\sin x=-3\sin^2x-4\sin x+3
$$
Then $f'(x)=-6\sin x\cos x-4\cos x=-2\cos x(3\sin x+2)$. The derivative vanishes for $x=\pi/2$, $x=3\pi/2$, $x=\arcsin(-2/3)$ and $x=\pi-\arcsin(-2/3)$.
The second derivative is
$$
f''(x)=12\sin^2x+4\sin x-6
$$
and $f''(\pi/2)=10>0$, $f(3\pi/2)=2>0$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of positive definite sequence with terms $a_{n} = \frac{a_{n - 1} + 1}{2}$ or $a_{n} = sin(a_{n - 1})$ In the sequence all terms are positive.
Each term can be either $a_{n} = \frac{a_{n - 1} + 1}{2}$ or $a_{n} = \sin{a_{n - 1}}$.
Can this sequence have a limit in the interval $(0, 1)$?
| No, it's not possible. Let us use an indirect prrof and assume that we have such a sequence, convergent to $g\in(0,1)$. Then by the definition of convergence $$ \forall \epsilon >0 \,\exists N \,\forall n>N :|a_n-g|<\epsilon$$
Let us take $\epsilon < \min\{\frac{1-g}{4},\frac{g-\sin g}{2}\}$. For $n$ big enough we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that the number $3^{3^n} + 1$ has at least $2n + 1$ prime factors. For any natural $n,$ prove that $3^{3^n} + 1$ has at least $2n + 1$ prime factors.
My idea was to use induction:
*
*for $n = 1$:
$$f(1) = 3^3 + 1 = 28 = 7*2^2$$
*let it be true for $n = k$, then for $n = k + 1$:
$$f(k + 1) = 3^{3^{k + 1}} + 1 ... | We have:
$$3^{2\cdot3^k} - 3^{3^k} + 1=(3^{3^k}+3^{\frac{3^k+1}{2}}+1)(3^{3^k}-3^{\frac{3^k+1}{2}}+1)$$
And for $k>0$ both factors are greater than one. This factorization can be deduced from the fact that $f(2)=387400807=19441\cdot19927$ and that both factors are close to each other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Proof of the proposition which says that the column rank of $A$ is equal to the row rank of $A$. (Gilbert Strang's new lecture) I am watching this new lecture by Gilbert Strang.
I have the following question.
Let $A = \begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}$.
Prof. Strang showed that the column rank of $... | We have $A=C R$ as in the lecture. the row space of $A$ is the column space of $A^T$ and $A^T=R^T C^T$ so the colums space of $A^T$ is generated by the columns of $C^T$ (using coefficients from $R^T$) and those columns are just the rows of $C$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can't solve $\int_{0}^{\pi} \frac{x}{1 + \cos^2x} dx$ I tried this :-
Let $$I =\int_{0}^{\pi}\frac{x}{1 + \cos^2x}dx\tag{1}$$
then $$I = \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2(\pi-x)}dx= \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2x}dx\tag{2}$$
Adding (1) and (2), we get
$$
2I = \int_{0}^\pi\frac{\pi}{1 + \cos^2x}dx\\
= \pi\in... | As $\tan x$ is discontinuous at $\dfrac\pi2,$ let's fold the integral in the first quadrant
$$I=\int_0^{2a}f(x)\ dx=\int_0^af(x)\ dx+\int_a^{2a}f(x)\ dx$$
Now set $2a-x=y$ in $\displaystyle J=\int_a^{2a}f(x)\ dx$
to find $\displaystyle J=\int_a^0f(2a-y)\ (-dy)=\int_0^af(2a-y)\ dy=\int_0^af(2a-x)\ dx$
$$\displaystyle I=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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How do I show that $w = \frac{1}{(z-a)^n}$ is an analytic function using the Cauchy Riemann equations,where z = x + iy? I know I need to use the Cauchy Riemann equations, to test for the analyticity, but how do I break the function $w = \frac{1}{(z-a)^n}$ into real and imaginary parts?
| With
$z = x + iy, \tag 1$
$a = \sigma + i\omega, \tag 2$
$z - a = (x - \sigma) + i(y - \omega); \tag 3$
$\dfrac{1}{z - a} = \dfrac{1}{(x - \sigma) + i(y - \omega)} = \dfrac{(x - \sigma) - i(y - \omega)}{(x - \sigma)^2 + (y - \omega)^2}$
$= \dfrac{x - \sigma}{(x - \sigma)^2 + (y - \omega)^2} - i\dfrac{y - \omega}{(x - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Long division of $\frac{3x^3-x^2-13x-13}{x^2-x-6}$ I'm self-studying from Stroud & Booth's amazing textbook "Engineering Mathematics", and am on the "Partial Fractions" chapter. As part of an exercise I need to do long division of two polynomial equations.
The problem is, long division of polynomials was never explaine... | First, you see which monomial you must multiply $x^2-x-6$ by in order to get something close to $3x^2-x^2-13x-13$. Clearly, that would be $3x$: $3x(x^2-x-6)=3x^3-3x^2-18x$. Now, you subtract this from $3x^2-x^2-13x-13$:$$3x^2-x^2-13x-13-(3x^3-3x^2-18x)=2x^2+5x-13.$$Now, you start all over again. What is the monomial yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Evaluate the Limit Without L'Hopital Rule Let $\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=L$, find the value of $\lim_{x\to a}\frac{x(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}$ for $a\neq0$.
Using L'Hopital rule I found that the answer is $a^4-aL$. My question is how to solve this pr... | Hint: The numerator can be written as $$-x \left( \left| x \right| -x \right) \left( x \left| x \right| +2\,
{a}^{2}-{x}^{2} \right)
$$ and now distinguish the cases $$x\geq 0$$ or $$x<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove that hat if $a,b \ge 2$ then $ab \ge a+b$ Prove that if $a,b \ge 2$ then $a+b \le ab$
so if $a \ge 2$ and $b \ge 2$ then $a-1 \ge 1$ and $b-1 \ge 1$
$(a-1)(b-1) \ge b-1$
$(a-1)(b-1) \ge 1$
$(a-1)(b-1) - 1 \ge 0$
$ab -a -b\ge 0$
$ab \ge a + b$
Thanks in advance, is this valid?
| $$ ab\ge 2a$$
$$ab\ge 2b$$
Add both sides
$$2ab \ge 2a+2b$$
$$ ab\ge a+b $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$ if $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$
My try
Since this is a factor I tired finding A, B, C & D such that,
$(x^2+kx+1)(Ax^3+Bx^2+cx+D)$ =$px^5+qx^2+r$
if I can find k in terms of p, q & r ... | Let $px^5+qx^2+r=(x^2+kx+1)(px^3+ax^2+bx+r)$
Let's compare the coefficients of $x$
$0=kr+b\iff b=-kr$
That of $x^4$
$0=a+kp\iff a=-kp$
That of $x^3$
$0=b+ka+p=-kr+k(-kp)+p$
which is on rearrangement, a quadratic equation in $k$
Compare the values of $a$ to find $k$
Coefficients of $x^2$
$q=r+kb+a=r+(-kr)(k-1)$
Replace ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation: $x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$ Here is the equation:
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$$
The equation looks very simple. But, it's a little misleading (for me).
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}\Longrightarrow x(x^2-b^2)-a^2x(x^2-b^2)+b^2+a(x^... | Your solution is almost right, except that 2) is not a solution because that yields $0=2$ in your equation.
Here’s a simpler method.
Adding $\frac{b^2}{b^2-x^2}$ to both sides, note that the RHS now becomes $1$. So we now need to solve $x(1-a^2)=1-a$, which factorises to $(1-a)(x(a+1)-1)=0$ (where $x \neq \pm b$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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$f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$ $f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$, where $f^{(9)}$ is the $9$-th derivative of $f$.
Domain of $f$ is $\mathbb{R} - \{-1, 2\}$. I've got that $f(x) = \frac{1}{1 - (-x)} + \frac{1}{1 - \frac{1}{2}x} = \sum_{n=0}^\infty ((-1)^n + 2^{-n})x^... | I've solved this using Julian Mejia's suggestion by making substitution $t = x -1$. We have $g(t) = \frac{1}{2} \cdot \left(\frac{4}{1 - t} + \frac{1}{1 + \frac{1}{2}t}\right) = \sum_{n=0}^\infty (2-(-\frac{1}{2})^{n+1})t^n$ which converges for $t \in (-1, 1)$. Now $9$-th derivative is just $(2-(-\frac{1}{2})^{10}) \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Undetermined coefficients in a perturbative expansion In order to familiarize myself with perturbation methods, I've been trying to derive the Lorentz transformations, given by
\begin{align*}
x \rightarrow \frac{x + vt}{\sqrt{1 - v^2}} & = (x + vt)(1 + \frac{1}{2}v^2 + \frac{3}{8}v^4 + ...) \\
& = (x + vt + \frac{1}{2}... | What you're trying to do here cannot possibly work. The parameter $v$ in the Lorentz transformation is an arbitrary parameter that you can redefine at will. Your approach has just as much reason to yield the coefficients in the expansion in $2v$ or $\sin v$ as it does to yield the coefficients in the expansion in $v$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Combinatorial proof of $\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$ and then use result to find a formula for $1^2 + 2^2 + \ldots + n^2$ a) Give a combinatorial proof that for every $n \geq r \geq 1$ that:
$$\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$$
And use (a) to concoct a formula for $1^2 + 2^2... | Maybe not combinatorial, but maybe it helps somebody else for a proper idea.
\begin{align}
\binom{n+1}{r+1} &= \frac{n+1}{r+1}\binom{n}{r} \\
&=\binom{n}{r} + \frac{n-r}{r+1} \binom{n}{r} =\binom{n}{r} + \frac{n}{r+1} \binom{n-1}{r} =\binom{n}{r} + \binom{n}{r+1} \\
&=\binom{n}{r} + \binom{n-1}{r} + \binom{n-1}{r+1} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Angle formed by orthocenter, incenter and circumcenter of a triangle $>135^\circ$? If $H$ is the orthocenter, $I$ the incenter and $O$ the circumcenter of a triangle , the I need to show that the angle $HIO>135^\circ$
With the assumptions of $OI^2=R^2-2Rr$, $OH^2=9R^2-(a^2+b^2+c^2)$, $HI^2=2r^2-4R^2\cos A\cos B \cos C... | Too long for a comment.
Put $X=4R^2\cos A\cos B \cos C$.
We have to show that
$$-\frac 1{\sqrt{2}}>\cos (HIO)=\frac{HI^2+OI^2-OH^2}{2\cdot HI\cdot OI}.$$
Since it is assmed that $2\cdot HI\cdot OI>0$, we have to show that
$$OH^2- HI^2-OI^2>\sqrt{2}\cdot HI\cdot OI$$
$$2Rr-2r^2-X>\sqrt{2(2r^2-X)(R^2-2Rr)}$$
Since $H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the ratio of twelfth terms if the sum of first n terms of two AP's are $3n+8,7n+15$
The sum of first n terms of two AP's are $3n+8,7n+15$. Then the ratio of their twelfth term is:
$$
\frac{a_{12}}{A_{12}}=\frac{a+11d}{A+11D}=\frac{2a+22d}{2A+22D}=\frac{s_{23}}{S_{23}}=\frac{3(23)+8}{7(23)+15}=\frac{77}{176}=\fra... | The question should have been
The ratio of the sum of first $n$ terms of two AP's are $3n+8:7n+15$
instead of
The sum of first $n$ terms of two AP's are $3n+8,7n+15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For every integer $n$, the quantity $n^2 + 2n \equiv 0\pmod 4$ or $n^2 + 2n \equiv 3\pmod 4$ I'm trying to prove this question using induction
So far I have
Base Case
Let $n = 1$, $(1^2 + 2)\equiv 3 \pmod 4 $
Claim holds for base case
Induction
Assume $n = k$ holds, that is $k^2 + 2k \equiv 0\pmod 4$ or $k^2 + 2k \equ... | Separate cases for $n$ even and $n$ odd.
If $n$ is even, than it is $n=2m$.
We get: $(2m)^2+2(2m)=4m^2+4m=4(m^2+m)\equiv 0\mod 4$
If $n$ is odd, then it is $n=2m+1$.
We get: $(2m+1)^2+2(2m+1)=4m^2+4m+1+4m+2=4(m^2+2m)+3\equiv 3\mod 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Checking a proof that the equality $ax^2+bx+c=0$ is never true for integer $a , b$ and $c$ if $ x = 2^{\frac{1}{3}} $ The solution of $ax^2+bx+c=0$ is $x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ . Since $ x^3=2$ then after raising both sides to the third power we get $$ 2=-\frac{b^3}{8a^3}+\frac{\left(b^2-4ac\right)^{\... | There is a fast way to finish. Once you have $2^\frac{4}{3} = \frac{m-b}{a}$, note that the LHS is $2 \times 2^\frac{1}{3}$. Dividing both sides by $2$ we obtain that $2^\frac{1}{3}$ is rational; this yields a contradiction in a similar method to the proof that $\sqrt 2$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that $n$ is Divisible by $6$ when the Quadratics have integer roots If $m,n \in \mathbb{Z}$ and Both the Quadratics $x^2+mx-n=0$ and $x^2-mx+n=0$ has integer roots, Prove that $n$ is a Multiple of $6$
My try:
Let $n$ is not a multiple of $6$
We have
$$n=6q\pm1$$
OR
$$n=6q\pm 2$$
OR
$$n=6q\pm 3$$
Now since the Dis... | You mention in your question that it already "has an existing thread". You should specify where this is to help avoid people duplicating efforts. However, if it's Prove that $n$ is divisible by $6$, note my solution below is somewhat different and, as you asked, I try to continue your solution.
You started with the two... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\frac{a^8 + b^8}{a^8 - b^8} + \frac{a^8 - b^8}{a^8 + b^8}$ based on $\frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2}$.
Given that $a$ and $b$ are distinct reals,
calculate $\dfrac{a^8 + b^8}{a^8 - b^8} + \dfrac{a^8 - b^8}{a^8 + b^8}$ based on $\dfrac{a^2 + b^2}{a^2 - b^2} + \dfrac{a^2 - b^2}{a^2 ... | Let
$$x=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}=2\frac{a^{4}+b^{4}}{a^{4}-b^{4}}.$$
Then
$$\frac{a^{4}+b^{4}}{a^{4}-b^{4}}+\frac{a^{4}-b^{4}}{a^{4}+b^{4}}=\frac{x}{2}+\frac{2}{x}.$$
We can repeat this to get
$$\frac{a^{8}+b^{8}}{a^{8}-b^{8}}+\frac{a^{8}-b^{8}}{a^{8}+b^{8}}=\frac{\frac{x}{2}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the maximum value of $\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$ where $x, y, z > 0$ and $x^2 + y^2 + z^2 = 3xyz$.
$x$, $y$ and $z$ are positives such that $x^2 + y^2 + z^2 = 3xyz$. Calculate the maximum value of $$\large \frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 +... | Also, by AM-GM and Muirhead we obtain:
$$\sum_{cyc}\frac{x^2}{x^4+yz}=\sum_{cyc}\frac{\frac{9x^4y^2z^2}{(x^2+y^2+z^2)^2}}{x^4+\frac{9x^2y^3z^3}{(x^2+y^2+z^2)^2}}=\sum_{cyc}\frac{9x^2y^2z^2}{x^2(x^2+y^2+z^2)^2+9y^3z^3}\leq$$
$$\leq\sum_{cyc}\frac{9x^2y^2z^2}{6x(x^2+y^2+z^2)\sqrt{y^3z^3}}=\frac{3\sum\limits_{cyc}x\sqrt{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.