Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers? $\textbf{Question:}$Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers?
that i... | Well, let's see. So we have
$$\begin{cases}
a^2 = x(b+c) \\
b^2 = y(a+c) \\
c^2 = z(a+b)
\end{cases}$$
where all quantities are integers. What does this mean?
For one thing, each of $a,b,c$ is strictly greater than 1.
For another, if $a,b,c$ are all pairwise coprime (as per the problem statement), then so are $b+c, a+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How does $e^x\cdot e^X$ equal $e^{x+X}$? I know that they equal each other, but when I'm trying to prove it, something doesn't match. Please mind the difference between the two equations, one is a lowercase $x$ and the other is an uppercase $x.$ I know that the formula to get $e^x$ is $\frac{x^n}{n!}$. So I apply on $e... | Perhaps you're confused about the arrangement. We should get $e^xe^x=e^{2x}$, which can be written $$e^{2x} = 1 +(2x) + \tfrac{(2x)^2}{2!} + \tfrac{(2x)^3}{3!}+\cdots$$
$$=1 + 2x + \tfrac{4x^2}{2!} + \tfrac{8x^3}{3!} + \cdots$$
So let's multiply series:
$$e^x\cdot e^x = \left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using partial information to factor $x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ I wish to find exact expressions for all roots of $p(x)=x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ By observing that for the roots $x_0 \pm iy_0, x_0 \approx -0.15883609808599033632, y_0 \approx 0.27511219196092896700,$ we have that $x_0$ is the unique real ro... | The hint.
Use the following:
$$x^6+3x^5+5x^4+10x^3+13x^2+4x+1=\left(x^3+\frac{3}{2}x^2-2x-1\right)^2+\frac{3}{4}x^2(3x+4)^2.$$
Now, you can get all roots of the polynomial. Just solve two cubic equations.
| {
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"url": "https://math.stackexchange.com/questions/3761730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Find $w^2$ and compute a quadratic equation satisfied by $w$ I'm given $$\zeta_7 = e^{\frac{2\pi i}{7}} = \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$$
and I know $$\sum_{k=0}^{6}\zeta_7^k=0 $$
I'm then given $$w=\zeta_7+\zeta_7^2+\zeta_7^4$$
and that $\zeta_7^7 = 1$.
I'm asked to find $w^2$ and I found that that's $\zeta^2... | Here's a slightly simpler approach:
$$w^2=(\zeta_7+\zeta_7^2+\zeta_7^4)^2=\zeta_7^2+\zeta_7^4+\zeta_7^8+2\zeta_7^3+2\zeta_7^5+2\zeta_7^6,$$
so
$$w^2+w=2\zeta_7^2+2\zeta_7^4+2\zeta_7^1+2\zeta_7^3+2\zeta_7^5+2\zeta_7^6=2(0-\zeta_7^0)=2(-1),$$
hence
$w^2+w+2=0$.
| {
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Proof that $\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0$ How should I prove
$$\forall n\in\mathbb{N}:\, \frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0?$$
My attempt:
$$\begin{align}\frac{e^{\frac{\pi i}{4}... | I tried what @Joe suggested on the problem statement itself. I'm getting the right answer for even $n$ but the answer for odd $n$ is equal upto a sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x^2+3y = u^2$ and $y^2+3x=v^2$ in positive integers. The question is from the pg - 59 from ' An Introduction to Diophantine Equations ' by Titu Andreescu , Dorin Andrica , Ion Cucurezeanu.
Example 1 : Solve in positive system of equations in positive integers
$$\begin{cases} x^2+3y = u^2 \\ y^2 + 3x = v^2 \end... | You are indeed correct; it is not hard to verify that your solution(s) are indeed solutions. The proof that you quote overlooks the case $k=5$, which yields your two additional solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Roze... | The second problem.
Let $a=\frac{x}{y}$ and $b=\frac{y}{z},$ where $x$, $y$ and $z$ are positives.
Thus, by C-S and Vasc we obtain: $$\sum_{cyx}\frac{1}{b(a+b)}=\sum_{cyc}\frac{1}{\frac{y}{z}\left(\frac{x}{y}+\frac{y}{z}\right)}=\sum_{cyc}\frac{z^2}{y^2+xz}=$$
$$=\sum_{cyc}\frac{z^4}{y^2z^2+xz^3}\geq\frac{(x^2+y^2+z^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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2010 USAMO #5:Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for integers $m$ and $n$, then $m - n$ is divisible by $p$. Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let $S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q(q+1)(q+2)}
$
Prove that if $\frac{1}{p}-2S_q = \frac{... | (Original) Hint: You are almost there with the simplification. Note that you are summing over $\frac1n$ from $n=2$ to $\frac{3p-1}2$ in the first. So
$$
2S_q+1=\sum_{n=(p+1)/2}^{(3p-1)/2}\frac1n
$$
If you tweak the RHS slightly, you would be summing over $\frac1n$ as $n$ runs through representative of each of the non... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Finding a polynomial $f(x)$ of degree 5 such that $f(x)$ is divisible by $x^3$ and $f(x)+2$ is divisible by $(x+1)^3.$ There is some polynomial $f(x)$ of degree $5$ such that both of these properties hold:
$f(x)$ is divisible by $x^3$.
$f(x)+2$ is divisible by $(x+1)^3.$
Find that polynomial.
I know that because $f(x)$... | Consider the equality
$$(x - (x+1) )^5 = -1$$
and expand getting
$$x^5 - 5 x^4 (x+1) +10 x^3 (x+1)^2- 10 x^2 (x+1)^3 +5 x (x+1)^4 - (x+1)^5 = -1$$
or
$$x^5 - 5 x^4 (x+1) + 10 x^3 (x+1)^2 = -1 + (x+1)^3 Q(x)$$
Therefore the polynomial is
$$2 x^3( x^2 - 5 x(x+1) + 10 (x+1)^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the eccentricity of the conic $4x^2+y^2+ax+by+c=0$, if it tangent to the $x$ axis at the origin and passes through $(-1,2)$ Solving this would require three equations
(1) Tangent to x axis at origin
Substituting zeroes in all $x$ and $y$ gives $c=0$
(2) Passes through (-1,2)
$$4(1)+4-a+2b=0$$
$$-a+2b=-8$$
How do I... | As you mentioned, one of the points on the ellipse is (0,0). Equation of the ellipse is
$\frac{(x+\frac{a}{8})^2}{1^2}+\frac{(y+\frac{b}{2})^2}{2^2} = {(\frac{a}{8})}^2 + {(\frac{b}{4})}^2$
For ellipse equation $\frac{(x \pm h)^2}{A^2}+\frac{(y \pm k)^2}{B^2} = 1$, eccentricity of ellipse
= $\frac{\sqrt{B^2-A^2}}{B}$ (... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\binom{n}{0}\binom{n+1}{n} +\binom{n}{1}\binom{n}{n-1} +\binom{n}{2}\binom{n-1}{n-2} +\cdots +\binom{n}{n}\binom{1}{0} = 2^{n-1}(n+2)$
Prove the below:
$$\binom{n}{0}\cdot\binom{n+1}{n} +\binom{n}{1}\cdot\binom{n}{n-1} +\binom{n}{2}\cdot\binom{n-1}{n-2} +\cdots +\binom{n}{n}\cdot\binom{1}{0} = 2^{n-1}\cdot... | We can easily use the fact that you already observed: $$\binom{n} {0}(1+x)^{n}+\binom{n} {1}x(1+x)^{n-1} +\binom{n} {2}x^2(1+x)^{n-2}+ \cdots + \binom {n} {n} x^n=(x+(1+x))^n$$ Hence You just need the coefficient of $x^n$ in the binomial $(x+1)(2x+1)^{n},$ which is $$2^{n-1}\binom{n} {n-1}+2^n\binom{n} {n}=2^{n-1}(n+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not s... | By C-S $$x^2+y^2=2(x-y)+2\leq2\sqrt{(1^2+(-1)^2)(x^2+y^2)}+2,$$ which gives
$$x^2+y^2-2\sqrt2\sqrt{x^2+y^2}+2\leq4$$ or
$$\left(\sqrt{x^2+y^2}-\sqrt2\right)^2\leq4,$$ which gives
$$\sqrt{x^2+y^2}\leq2+\sqrt2.$$
Id est, $$x^2+y^2+\sqrt{32}\leq(2+\sqrt2)^2+4\sqrt2=6+8\sqrt2.$$ The equality occurs for $x=1+\sqrt2$ and $y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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sum of terms of series
If $$F(t)=\displaystyle\sum_{n=1}^t\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$$ find $F(60)$.
I tried manipulating the general term(of sequence) in the form $V(n)-V(n-1)$ to calculate the sum by cancellation but went nowhere. I also tried using the fact that $$2n+\sqrt{4n^2-1}=\frac{1}{2... | With more details, we have that
$$\frac{2n+\sqrt{(4n^2-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)$$
then
$$\frac{4n+\sqrt{(4n^2-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)+\frac{2n}{\sqrt{2n+1}+\sqrt{2n-1}}=\\=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find all solutions for : $a^3 \equiv b^3 \pmod{7^3}$, knowing that $7 \nmid ab$.
Find all integers $a$ and $b$ such that $$a^3 \equiv b^3 \pmod{7^3}\,,$$ knowing that $7 \nmid ab$.
As a try, I noticed that, since $\gcd(b, 7)=1$, there exists $x \in \mathbb{N}$ such that $b\cdot x \equiv 1 \pmod{7} \Rightarrow ... | Given $7\nmid ab$, $a^3\equiv b^3\bmod 7^3\iff (a/b)^3\equiv1\bmod 7^3$.
Let $x\equiv a/b\bmod 7^3$. We are looking for $x$ such that $7^3|x^3-1=(x-1)(x^2+x+1)$.
Now if $7|x-1$ and $x^2+x+1$, then $7|x^2+x+1-(x+2)(x-1)=3$, a contradiction.
So $7^3|x-1$ (i.e., $x\equiv1\bmod7^3$) or $7^3|x^2+x+1$.
Now we are looking fo... | {
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"url": "https://math.stackexchange.com/questions/3776108",
"timestamp": "2023-03-29T00:00:00",
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Singular values of matrices which preserve the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$ Let $0<a<b$, $ab=1$, and let
$$
D_{a,b}=\biggl\{(x,y) \,\biggm | \, \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \biggr\}
$$
be the ellipse with diameters $a,b$.
Let $A \in \operatorname{SL}_2(\mathbb R) \setminus \operatorname{... | That's not true in general: the singular values of $A$ are by definition the square roots of the eigenvalues of $A^*A$. Set
\begin{align*}A(\theta) & = \begin{pmatrix} a& 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1/a& 0 \\ 0 & 1/b \end{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\int_0^{\infty} \frac{\ln{(1+x)}\arctan{(\sqrt{x})}}{4+x^2} \, \mathrm{d}x = \frac{\pi}{2} \arctan{\left(\frac{1}{2}\right)} \ln{5}$ Prove: $$\int_0^{\infty} \frac{\ln{(1+x)}\arctan{(\sqrt{x})}}{4+x^2} \, \mathrm{d}x = \frac{\pi}{2} \arctan{\left(\frac{1}{2}\right)} \ln{5}$$
This might be a repeat question (I c... | Let us try to collect the most interesting ideas in one simple solution.
At first,
$$I=\int\limits_0^\infty \dfrac{\ln(1+x)\arctan\sqrt x}{x^2+4}\text{ d}x
=\int\limits_0^\infty \dfrac{\ln(1+y^2)\arctan y}{y^4+4}\,2y\text{ d}y.$$
At the second, by pisco,
$$\ln(1+y^2) = \ln(1+iy) + \ln(1-iy),\\
\arctan y = \dfrac i2(\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 1
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Write $\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} - \sqrt{5}} - \frac{1}{2}$ with a rational denominator. Write $\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} - \sqrt{5}} - \frac{1}{2}$ with a rational denominator.
How should I solve this question?
| Only the first term
We have $\frac{\sqrt 2}{\sqrt 2 + \sqrt 3 - \sqrt 5}.$ Do :
$$
\frac{\sqrt 2}{\sqrt 2 + \sqrt 3 - \sqrt 5}= \frac{\sqrt 2(\sqrt 2 + \sqrt 3 + \sqrt 5)}{(\sqrt 2 + \sqrt 3)^2 - 5} = \frac{2 + \sqrt 6 + \sqrt {10}}{2\sqrt 6} \\
= \frac{2\sqrt 6 + 6 + \sqrt {60}}{2\sqrt 6 \sqrt 6} = \frac{6 + 2 \sqrt ... | {
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"url": "https://math.stackexchange.com/questions/3780558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to prove this inequlity $\ln(\frac{p}{q}) \le \frac{p-q}{\sqrt{pq}} (0 < q \le p)$ by means of integral inequalities I can solve it by introducing a function, but I need to prove it by applying integral inequalities such as the holder inequality or the schwarz inequality.
| You probably can recognize this is the equivalent to prove
$$
f(x) = \ln(x)\le \sqrt{x} - \frac{1}{\sqrt{x}} = g(x) \quad \text{when} x \ge 1
$$
Easy to see $f(1) = g(1)= 0$
We also have
$$
g'(x) = \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{x}^3} = \frac{1}{2}\left( \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}^3}\right) \ge \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If roots of equation $ax^2+bx+c=0$ is $\alpha,\beta$ find the roots of equation $bcx^2+a(bc+a^2)x+ac^2=0$ in terms of $\alpha,\beta$ If roots of equation $ax^2+bx+c=0$ is $\alpha,\beta$ find the roots of equation $bcx^2+a(bc+a^2)x+ac^2=0$ in terms of $\alpha,\beta$
My Try:
We know that
$$\alpha+\beta=-\frac ba \quad \... | The given quadratic equation has the roots $\alpha$ and $\beta$. Then we have that
$$ \alpha + \beta = -\frac{b}{a} \quad \text{ and } \quad \alpha \cdot \beta = \frac{c}{a} $$
The new quadratic $$ bc \cdot x^2 + a(bc + a^2)x + ac^2 = 0 $$
$\blacksquare~$What I have so far:
Let the roots of the new quadratic be $m, n$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Systems of polynomial equations involving sums of equal powers Given the following system of polynomial equations:
$$
\left\{\begin{array}{lclclcr}
x & + & y & + & z & = & 1
\\
x^{2} & + & y^{2} & + & z^{2} & = & 14
\\
x^{3} & + & y^{3} & + & z^{3} & = & 36
\end{array}\right.
$$
What is $x^{5} + y^{5} + z^{5}\ {\large ... | The requested exponent does not matter that much. Each of the three is a root of the same
$$ 6 t^3 - 6 t^2 - 39 t - 31, $$
one real and two complex conjugates, but we don't need them. Each also obeys
$$ 6 t^{n+3} - 6 t^{n+2} - 39 t^{n+1} - 31 t^n, $$ so that $a_n=x^n + y^n + z^n$ is a sequence with linear recurren... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The number of three digit numbers $abc$, which satisfy $a≤b>c$ is My approach in this question is as follows,
⇒When $a=1$
$-> 1-1-(0) / 1-2-(0,1) / 1-3-(0,1,2) ... / 1-9-(0,1...,8)$ ⇒Total numbers = $1+2+3+...+9$
⇒When $a=2$
$-> 2-2-(0,1) / 2-3-(0,1,2) ... / 2-9-(0,1...,8)$ ⇒Total numbers = $2+3+...+9$
⇒When $a=3$
$->... | The given condition should be extended to
$$1\leq a\leq b>c\geq0\ .$$
Given $b\in\{1,2,\ldots,9\}$ we therefore have $b$ choices for $a$ and $b$ choices for $c$. The number of admissible three digit numbers $abc$ therefore is
$$\sum_{b=1}^9b^2={n\cdot(n+1)(2n+1)\over6}\biggr|_{n=9}={9\cdot10\cdot 19\over 6}=285\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimizing a function by finding its critical points Let $f_n(x)$ equal:
$$(2^n+2) \left(2x+1-\sqrt{2x^2+2x} \right)^n-x^n-(x+1)^n-\left(3x+1-2\sqrt{2x^2+2x}\right)^n-\left(3x+2-2\sqrt{2x^2+2x}\right)^n$$
Mathematica suggests that this function has two critical points on $(0,\infty)$, namely $x_1=1$ and $x_2=1/\sqrt{2}... | The fact that $f_1(x),f_2(x),f_3(x)$ are identically equal to zero is simple (even if a bit tedious) to prove; just expand the expression.
Concerning $f_4(x)$ it reduces to
$$f_4(x)=4 x (x+1) \left(17 x^2+17 x+2\right)-24 \sqrt{2}\, x^{3/2} (x+1)^{3/2} (2 x+1)$$
$$f'_4(x)=8 (2 x+1) \left(17 x^2+17 x+1\right)-12 \sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Obtain sum of a sequence from sum of its odd terms. I would like to compute the sum
$$
\sum_{k=1}^\infty \frac{1}{k^4}
$$
by using the Fourier series of $f(x)=|x|$ over $(-\pi,\pi)$. Coefficients $b_k$ are all $0$ because $f$ is even. Doing the integration stuff, I obtained:
$$
a_0 = \pi
$$
and
$$
a_k = \frac{2}{k^2}\... | You essentially have
$${\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ... = \frac{\pi^4}{96}}$$
You want to find
$${\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ... = ?}$$
in other words, you want to add on
$${\frac{1}{2^4} + \frac{1}{4^4} + ...}$$
Factoring out a ${\frac{1}{2^4}}$ on the above yields
$${\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Better proof of a numerical inequality of $e^x$ The inequality is
$$ e^z \leq 1+z+\frac{z^2/2}{1-|z|/3} \text{ for } |z|<3$$
I proved it by splitting it into 3 cases: $-3<z<0$, $z=0$ and $0<z<3$.
For $z=0$, both sides equals.
The other 2 cases are done with calculus. Define $f(x)=e^x-1-x-\frac{x^2/2}{1-|x|/3}$ and th... | Note that, if $|z|<3$,\begin{align}e^z-1-z&=\frac{z^2}2+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots\\&=\frac{z^2}2\left(1+\frac z3+\frac{z^2}{3\times4}+\frac{z^3}{3\times4\times5}+\cdots\right)\\&\leqslant\frac{z^2}2\left(1+\frac{|z|}3+\frac{|z|^2}{3\times4}+\frac{|z|^3}{3\times4\times5}+\cdots\right)\\&\leqslant\frac{z^2}2\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded, monotone, and find its limit Prove that $x_1 = 0, x_2 = 0, x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded and monotonic. Then find its limit.
My attempt at boundedness:
(Using induction) For the base case we have $0 \leq x_... | No, your argument is not valid. You show that
$$x_{k+1}\le 2\implies x_{k+2}\le 4.$$
If you apply induction, this leads to
$$x_{k+m}\le 2^{m+1}$$ which is not bounded.
But you can use
$$x_k,x_{k+1}\le2\implies x_{k+2}=\frac{x_k}{3}+\frac{x_{k+1}}6+1\le\frac23+\frac26+1=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793084",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Coefficient of $x^7y^6$ in $(xy+x+3y+3)^8$
Find the coefficient of $x^7y^6$ in $(xy+x+3y+3)^8$.
My solution:
Factor $(xy+x+3y+3)^8$ into $(x+3)^8(y+1)^8$. To get an $x^7y^6$ term, we need to find the coefficient of $x^7$ in the first factor and $y^6$ in the second factor. Using the binomial theorem, we get the coeffi... | The factor for $x^7$ is $24$ and for $y^6$ is $28$, which their multiplication is the correct answer. Probably you're just applying the binomial theorem incorrectly.
As an add-on, your method is completely correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Checking the analyticity of a complex function. This was a question in an assignment that demands the use of Cauchy's Integral Formula in the questions. The question goes like this:
Integrate $\displaystyle{g(z)=\frac{e^z}{ze^z-2iz}}$ over $\displaystyle{C:\ \vert z\vert=0.5}$.
What I tried:
First, I noted that $\dis... | If you consider $f$ defined in $U=D(0,r)$ where $\frac{1}{2}<r<\text{ln}(2)$ then $f(z)=\frac{e^z}{e^z-2i}$ is well-defined. This is true because $e^z-2i=0 \Rightarrow |e^z|=|2i| \iff e^{Re(z)}=2$ which is impossible in $U$, therefore $f$ is analytic once is the quotient of two analytic functions. Notice that $\frac{1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Change this integral $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$. It is given that $f(a+b+1 -x) = f(x)$ where $a$ and $b$ are positive real numbers then $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$ is equal to
*
*$\int_{a-1}^{b-1} f(x) dx$
*$\int_{a+1}^{b+1} f(x+1) dx$
*$\int_... | Your question is,
$$\dfrac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$$
Now, in second term i.e. $f(x+1)$, put $x+1$ as $u$ and use $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ and $f(a+b+1 -x) = f(x)$.
You will get $$\dfrac{1}{a+b} \int_{a}^{b} (a+b) \left[ f(x) \right] dx$$. Now substitute $x\rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find this $\prod_{k=-\infty}^{+\infty}\frac{x^2+(4k+1-y)^2}{x^2+(4k+3-y)^2}$ let $x,y$ be real numbers,show that
$$\prod_{k=-\infty}^{+\infty}\dfrac{x^2+(4k+1-y)^2}{x^2+(4k+3-y)^2}=\dfrac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}$$
This is a quest... | Firstly, define the complex number $z=x+yi$. Then, we are asked to prove that
$$
\prod_{k\in\mathbb{Z}}\left|\frac{z-(4k+1)i}{z-(4k+3)i}\right|^2=\frac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}.
$$
Now, let's rewrite the right-hand side of this equality:... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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solutions to $\frac{1}a + \frac{1}b + \frac{1}c = \frac{1}{2018}$
Find, with proof, all ordered triplets of positive integers $(a,b,c)$ so that $\dfrac{1}a + \dfrac{1}b + \dfrac{1}c = \dfrac{1}{2018}.$
In general, if $d$ is a positive integer, then $(a,b,c) = (3d,3d,3d),(d,2d,6d),(d,6d,2d), (2d,d,6d),(2d,6d,d),(6d,d,... | Partial answer. $\,$ Let $d=\text{gcd}(a,b,c)$ and, thus, $a=dx, b=dy, c=dz$ - where $\text{gcd}(x,y,z)=1$. If we assume that $\text{gcd}(x,y)=\text{gcd}(y,z)=\text{gcd}(z,x)=1$ then $\text{gcd}(xyz, xy+yz+zx)=1$ $$\frac1a+\frac1b+\frac1c=\frac1{2018}\iff \frac{abc}{ab+bc+ca}=\frac{dxyz}{xy+yz+zx}=2018$$ This implies t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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obtaining a simplified expression for the coefficient of $x^n$
I was trying to find the coefficient of $x^n$ in the expansion of $(1+x)^{-2}(1-2x)^{-2},$ denoted $[x^n]\{(1+x)^{-2}(1-2x)^{-2}\}$. Using the negative binomial theorem, I know that it is equal to
$$
\begin{split}
\sum_{j=0}^n &([x^j](1+x)^{-2})([x^{n-j}](... | As suggested by @AnginaSeng, you can apply partial fraction decomposition:
\begin{align}
\frac{1}{(1+x)^2(1-2x)^2}
&=\frac{1/9}{(1+x)^2}+\frac{4/27}{1+x}+\frac{4/9}{(1-2x)^2}+\frac{8/27}{1-2x}\\
&=\frac{1}{9}\sum_{n \ge 0}\binom{n+1}{1}(-x)^n+\frac{4}{27}\sum_{n\ge 0} (-x)^n+\frac{4}{9}\sum_{n \ge 0} \binom{n+1}{1}(2x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete? So, for any angle $\alpha$ :
$$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= ... | Your comment is correct. You can only get the final equality by proving that $\tan\left(\frac{\alpha}{2}\right)$ and $\frac{1-\cos \alpha}{\sin\alpha}$ have the same sign.
But this is not complicated to prove. $\tan\left(\frac{\alpha}{2}\right)$ is positive if and only if $\frac{\alpha}{2} \in (k\pi, k\pi +\frac{\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800605",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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The limit and asymptotic analysis of $a_n^2 - n$ from $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$ I came up with the following question which is the follow up of
How to prove that for $a_{n+1}=\frac{a_n}{n} + \frac{n}{a_n}$ , we have $\lfloor a_n^2 \rfloor = n$?
Problem: Let $a_1 = 1,\quad a_{n+1} = \frac{a_n}{n} + \fra... | Since $a_1=1>0$ it is clear that $a_n>0$ for all $n$. Squaring gives $$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2$$ and defining $a_n^2=nb_n$, this recurrence becomes $$b_{n+1}=\frac{b_n}{n(n+1)} + \frac{n}{(n+1)b_n} + \frac{2}{n+1}$$
with $b_1=1$. Now suppose $$1\leq b_n \leq 1+\frac{1}{n}+ \frac{2}{n^2}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Proving $(a^2 + 1)(b ^2 + 1)(c ^2 + 1) ≥ 2(ab + bc + ca)$ where $a,b,c$ are real numbers. The inequality above seems very compelling for the pqr-method.
So this was my attempt-
$$ LHS = (a^2 + 1)(b^2 + 1)(c^2 + 1) = 1 + a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 + a^2b^2c^2 $$
Now substituting $p = a+b+c$ , $q = ab+bc+... | Because $2(ab + bc + ca) \leqslant 2(|a||b| + |b||c| + |c||a|)$ and
$$(a^2 + 1)(b ^2 + 1)(c ^2 + 1) = (|a|^2 + 1)(|b| ^2 + 1)(|c| ^2 + 1),$$
so we need to prove the inequality when $ a,\,b,\,c \geqslant 0.$
Indeed, easy to check $3t^2 \geqslant 3t-1.$ Now, using the AM-GM we have
$$(a^2 + 1)(b ^2 + 1)(c ^2 + 1) \geqsla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all integers such that polynomial $x^4+n$ is reducible in $\mathbb{Z}[x]$ In my studies of abstract algebra and polynomials, I have the following question:
We are asked to find all integers $n$ such that the polynomial $x^4+n$ is reducible in $\mathbb{Z}[x]$.
The only thing I can think of is Eisenstein's criteri... | Let's assume the polynomial is reducible. Then we can see that it must have a quadratic factor - for if it had a linear factor, it must be in form $x-p$, but then $p$ is a root and so is the $-p$ (since the polynomial has only even exponents), and so $(x-p)(x+p)=x^2-p^2$ is a quadratic factor.
So assume a generic facto... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to show the concavity of a function with an undefined point? Look at this function: $f(x)=\left\{\begin{array}{ll}
\frac{x(1-x^2)}{1-x^3}\\
2/3
\end{array}\right.$. Here $2/3=\lim_{x\to1}\frac{x(1-x^2)}{1-x^3}$. I can show that the second derivative of $\frac{x(1-x^2)}{1-x^3}$ is non-positive for $0\leq x<1$ and $x... | If $x\neq1$, then
$$f(x)=\frac{x(1-x^2)}{1-x^3}=\frac{x(1-x)(1+x)}{(1-x)(1+x+x^2)}=\frac{x(1+x)}{1+x+x^2}.$$
And plugging in $x=1$ into $\dfrac{x(1+x)}{1+x+x^2}$ also yields $2/3$.
This means that the given function is in fact $f(x)=\dfrac{x(1+x)}{1+x+x^2}$ for all real numbers. Now you can the second derivative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sq... | Since the limit is defined for $x\to 2^+$, let $y^2=x-2\to 0$ then
$$\lim _{x\to 2}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)=\lim _{y\to 0}\left(\frac{\sqrt{y^2(y^2+4)}+\sqrt{y^2+2}-\sqrt{2}}{y}\right)=$$
$$\lim _{y\to 0}\left(\frac{\sqrt{y^2(y^2+4)}+\sqrt{y^2+2}-\sqrt{2}}{y}\right)=\lim _{y\to 0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum of series without differentiation Given a series $\sum_{i > 0}\frac{i^2}{z^i}$, and $\sum_{i > 0}\frac{i}{z^i} = \frac{z}{(z - 1)^2}$ I need to find the sum
My method does not require differentiation but there is a difficulty.
Let $S = \frac{1^2}{z} + \frac{2^2}{z^2} + \frac{3^2}{z^3} + ... + \frac{i^2}{z^... | You have this
$(z - 1)S
= 1 + \frac{3}{z} + \frac{5}{z^2} + ... + \frac{2i - 1}{z^{i - 1}} - \frac{i^2}{z^i}
$
or, in summation notation,
$(z - 1)S
= \sum_{k=0}^{i-1} \dfrac{2k+1}{z^k}- \frac{i^2}{z^i}
$.
We can now split this into sums
we already know:
$\begin{array}\\
(z - 1)S
&= \sum_{k=0}^{i-1} \dfrac{2k+1}{z^k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that $x_{n+1}=x_n(2-ax_n)$ converges and find the limit
Let $a>0$ and $x_0\in I=\left (\frac{1}{2a}, \frac{3}{2a}\right )$. Show that the sequence $(x_n)$, $n\geq 0$, $$x_{n+1}=x_n(2-ax_n), \quad n \geq 0$$ converges. Which is the limit? Hint: Consider $\phi (x)=x(2-ax)$ and show that $\phi (I)\subset \left [\fra... | Complete the square on the right side to find
$$
1-ax_{n+1}=(1-ax_n)^2
$$
so that the convergence of $y_n=1-ax_n$ is very easy to discuss (subsequence of a geometric sequence).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding $\cos ( 2 \sin^{-1}( \frac{5}{ 13} )) $ The following problem is from the $8$th edition of the book Calculus, by James Stewart. It is problem number $9$ in section $6.6$.
Problem:
Find an exact value for the expression:
$$ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } $$
Answer:
\begin{align*}
... | In the second line there is a mistake!
$$\sin^22\alpha=4\sin^2\alpha\cos^2\alpha.$$
Now, $$\cos2\arcsin\frac{5}{13}=\sqrt{1-4\left(\frac{5}{13}\right)^2\left(\frac{12}{13}\right)^2}=\frac{119}{169}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$ For $a,b,c\geqslant 0.$ Prove: $$\text{Q}=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3},$$
where $k={\frac {27}{8}}+\frac{9\sqrt... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality it's a linear inequality of $w^3$, which says that it's enough to prove our inequality for an extreme value of $w^3$, which by $uvw$ happens in the following cases.
*
*$w^3=0$.
Let $c=0$, $b=1$ and $a^2+1=2ua$.
Thus, $u\geq1$ and we need to prove h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find numerical matrices to remove the same coefficients on the main diameter $M=\left(
\begin{array}{cccc}
\frac{1}{4}m_{11} & m_{12} & ... & m_{1n} \\
m_{12} & \frac{1}{4}m_{22} & ... & m_{2n} \\
\colon & \colon & & \colon\\
m_{1n} & ... & & \frac{1}{4}m_{nn}
\end{array}
\right)$
$M$ is a Symmetric matrix That is, ... | Consider
$$M=\left(
\begin{array}{cccc}
\frac{1}{4} & \frac{1}{4} \\
\frac{1}{4} & \frac{1}{4}
\end{array}
\right)$$
For this matrix, you have
$$M_1=\left(
\begin{array}{cccc}
1 & \frac{1}{4} \\
\frac{1}{4} & 1
\end{array}
\right)$$
If there existed $P$, $Q$ such that $PMQ=M_1$, then taking the determinant, you would ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is the probability of rolling 2 before a second odd? Consider a game of dice:
*
*You win if you roll $2$.
*You lose if you roll two odds ( need not be consecutive ).
*If you roll a $4$ or $6$, you keep playing as you have neither lost nor won.
Eg: $1$, $3$ is a loss. $1$, $6$, $4$, $3$ is a loss.
What's the ... | Your approach is right, but you haven´t calculated correctly the probability of rolling evens and one odd before rolling 2, since you´re supposing you always get the odd in the first roll.
If you get a 2 in the second roll you had (odd, 2) with probability of $\frac{1}{2}*\frac{1}{6}$
Getting 2 in the third roll might ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Stuck on proof of $\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$ for $n \geq 3$ using induction I'm relatively familiar with induction, I'm just stuck on this step. I am currently taking Introduction to Abstract Math, and have taken Calculus I and II.
$P(n)$ is
$$\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7... | You know that
$$\sum_{k=1}^n\frac1{k^2}+\frac1{(n+1)^2}\le\frac74-\frac1n+\frac1{(n+1)^2}\,,\tag{1}$$
and you want to show that lefthand side of $(1)$ is at most $\frac74-\frac1{n+1}$; the most straightforward way to do this is to show that
$$\frac74-\frac1n+\frac1{(n+1)^2}\le\frac74-\frac1{n+1}\,,$$
which amounts to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Let $a_k$ be the coefficient of $x^k$ in the expansion of :- $(x+1) + (x+1)^2 + (x+1)^3 + (x+1)^4 + ... + (x+1)^{99}$ . Find $[\frac{a_4}{a_3}]$ Let $a_k$ be the coefficient of $x^k$ in the expansion of :- $(x+1) + (x+1)^2 + (x+1)^3 + (x+1)^4 + ... + (x+1)^{99}$ . Find $[\frac{a_4}{a_3}].$
What I Tried :- Honestly ther... | You can use the formula for the sum of a geometric progression and then the binomial theorem to find that your sum is
$$
\frac{{1 - (x + 1)^{100} }}{{1 - (x + 1)}} - 1 = \frac{{(x + 1)^{100} - 1}}{x} - 1 = \left( {\sum\limits_{k = 1}^{100} {\binom{100}{k}x^{k - 1} } } \right) - 1 \\ = 99 + \sum\limits_{k = 2}^{100} {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$; find my mistake solve :$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$
My try:
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}\\\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\\left(t\frac{1-x}{1+x}\right)'=\left(x-1\right)\frac{1-x}{1+x}\\\left(t\frac... | You made a mistake for the integrating factor. Since $\int-\frac{2}{x^2-1}dx=\ln(\frac{x+1}{1-x})+C$, your $4$th line should be:
$$(t\frac{1+x}{1-x})'=(x-1)\frac{1+x}{1-x}=-(1+x)$$
$$(t\frac{1+x}{1-x})=-x-\frac{x^2}{2}+c$$
$$t=\frac{(1-x)(c-x-\frac{x^2}{2})}{1+x}=\frac{(x-1)(c_{1}+x(x+2))}{2(1+x)}$$
where $c_{1}=-2c.$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why isn't $x^2+x+1$ a factor of $x^{12}+x^6+1$? When we solve the equation
$$x^{12}+x^6+1=0$$ we obtain $2$ solutions that also satisfy $$x^2+x+1=0$$
namely $-\frac{1}{2}\pm i\frac{\sqrt3}{2}$.
Shouldn't this imply that $x^2+x+1$ is a factor of $x^{12}+x^6+1$? However, the fully factorsied form of
$x^{12}+x^6+1$ is $$(... | I just want to note a technique based on my comment to the original post where a factor of $x^2+x+1$ can be shown longhand for $x^{2n}+x^n+1$ where $n$ is not a multiple of $3$.
Suppose $n=3m+1$ then $$x^{6m+2}+x^{3m+1}+1=x^{6m+2}-x^2+x^{3m+1}-x+x^2+x+1=$$$$=x^2\left(x^{6m}-1\right)+x\left(x^{3m}-1\right)+x^2+x+1$$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$ I just came across the following question:
Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$
I solved it as follows:
$x=y=0$ obvious solution. If $xy\neq0$, let $d=gcd(x, y)$ and we write $x=da$, $y=db$, $a, b\in... | The following is neither intuitive nor simple, but it does give a different approach to the proof.
If $xy\not=0$, let $p$ be an odd prime and write $x=p^ru$ and $y=p^sv$ with $p\not\mid uv$. From $p^{5r}u^5-p^{5s}v^5=16uvp^{r+s}$, we see we cannot have $r=s\not=0$, so we either have $5r=r+s$ or $5s=r+s$. This means th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove $(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$ is an algebraic integer Let $b = (4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$, then $1 = b(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$, and $1 - 4b = b(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$.
Therefore $1 - 12b + 48b^2 - 64b^3 = (1 - 4b)^3 = b^3(3 \cdot 3^{1/3} + 2 \cdot 3... | I think, there is a mistake in your computations.
We obtain:
$$4b-1+3b\sqrt[3]3+2b\sqrt[3]9=0$$ or
$$(4b-1)^3+81b^3+72b^3-3(4b-1)\cdot18b^2=0$$ or $$b^3+6b^2+12b-1=0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Application of AM-GM inequality to specific contest problem Suppose that $x,y\in [0,1]$. Prove that $\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\leq \frac{2}{\sqrt{1+xy}}.$
I suppose that this problem can be solved by some application of AM-GM inequality. I was trying to do the following: since $xy\leq \frac{x^2+y^2}... | We have
$$\left(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\right)^2 = \frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{2}{\sqrt{(1+x^2)(1+y^2)}}$$
Using the AM-GM we have
$$\frac{2}{\sqrt{(1+x^2)(1+y^2)}} \leqslant \frac{1}{1+x^2}+\frac{1}{1+y^2}.$$
Therefore, we need to prove
$$\frac{1}{1+x^2}+\frac{1}{1+y^2}\leqslant \frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A function inequality about $e^x$ and $\ln x$ If for any $x \in (1,+\infty)$，there is the inequality:
$$x^{-3} e^{x}-a \ln x \geq x+1$$
Find the value range of $a$ .
And I tried constructing the function $f(x)=x^{-3} e^{x}-a \ln x - x-1$ and deriving it, but to no avail.
| $$\max{a}=\min_{x>1}\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}.$$
Let $g(x)=\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}.$
Thus, for $e^x=x^3$ or $x=3\ln{x}$ we obtain: $$g'(x)=\frac{x^4+x^3+e^x(x\ln{x}-3\ln{x}-1)-x^4\ln{x}}{x^4\ln^2x}=$$
$$=\frac{x^4+x^3+x^3(x\ln{x}-3\ln{x}-1)-x^4\ln{x}}{x^4\ln^2x}=\frac{x^4-3x^3\ln{x}}{x^4\ln^2x}=0.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3826668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $83$ divides $p$ .
Let $p$ and $q$ be the positive integers such that $1 - \frac{1}{2} + \frac{1}{3} - ... - \frac{1}{54} + \frac{1}{55} = \frac{p}{q}$ . Prove that $83$ divides $p$ .
What I Tried :- To me it does not actually seem like an easy problem for me. (I guess I couldn't figure out the idea)
This ... | Given expression is
$$
1 - \frac{1}{2} + \frac{1}{3} - ... - \frac{1}{54} + \frac{1}{55} \\
= 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{55} - 2(\frac{1}{2} + \frac{1}{4} + ... + \frac{1}{54} ) \\
= \frac{1}{28} + \frac{1}{29} ... + \frac{1}{55} \\
= (\frac{1}{28} + \frac{1}{55}) + (\frac{1}{29} + \frac{1}{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that for any three integers $a,b,c$ there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer Prove that for any three integers $a,b,c$ there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer.
In order to solve this problem I have tried looking at the ... | John Omielan has already provided a nice answer using $\text{mod}\ 4$.
Here is another approach using $\text{mod}\ 4$.
Let $f(n):=n^3+an^2+bn+c$.
Let us prove that at least one of $f(1),f(2),f(3),f(4)$ is not a square number.
Proof :
Let us consider in $\text{mod}\ 4$.
Suppose that $f(1),f(2),f(3),f(4)$ are square numb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum value of $f(x,y,z) = x^z + y^z - (xy)^{\frac{z}{4}}, x > 0, y > 0, z > 0$ $$f(x,y,z) = x^{z}+y^{z}-(xy)^{\frac{z}{4}}$$
for all real positive numbers x, y, z
Does anyone have a clue to find the minimum value of $f(x,y,z)$?
I honestly don't know where to start the solution, I just come up with $AM \geq GM$
$\fra... | For a minimum $f_x=zx^{z-1}-(z/4)y^{z/4}x^{z/4-1}=0$ so $x^{3z/4}-y^{z/4}/4=0$.
As $f(x,y,z)=f(y,x,z)$ we also have $y^{3z/4}-x^{z/4}/4=0$ and equating yields $(4x^{3z/4})^3-x^{z/4}/4=0$. Thus $256x^{2z}-1=0$ which gives $x^z=y^z=1/16$ as $x,y,z>0$.
Hence the minimum value is $1/16+1/16-(1/16\cdot1/16)^{1/4}=-1/8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Roots and point of inflections Let $b$ and $c$ be the roots of a four degree polynomial. Also $x=b$ and $x=c$ are the real points of inflection of this four degree polynomial. If the other two roots of the polynomial be $a$ and $d$ where $a<b<c<d$ then prove that $$\int_{a}^{d}f(x)dx=0$$
My Attempt
I framed the equatio... | If $f$ has degree $4$, $f''$ has degree $2$, so if $b$ and $c$ are inflection points (and thus roots of $f''$), we must have $f''(x) = k (x-b)(x-c)$ for some constant $k$. By integrating twice, we get $f$ as a polynomial of degree $4$ with coefficients of $x^1$ and $x^0$ arbitrary. But those coefficients will be det... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $(1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$ Can someone help me simplify $(1+\sqrt{3})\times\sqrt{2-\sqrt{3}}$?
The end result is $\sqrt{2}$, however, I honestly do not know how to get there using my current skills.
I asked a teacher/tutor and he proposed setting the expression equal to X and working backwards, squ... | In general, $\sqrt{2 - \sqrt3}$ can be denested using the equation $$\sqrt{X\pm Y} = \sqrt{\frac{X + \sqrt{X^2-Y^2}}{2}} \pm \sqrt{\frac{X - \sqrt{X^2-Y^2}}{2}}.$$ (source)
Setting $X = 2$ and $Y = \sqrt3$ gives $X^2-Y^2=1$ and therefore $$\sqrt{2-\sqrt3} = \sqrt{\frac{2+1}{2}} - \sqrt{\frac{2-1}{2}} = \frac{\sqrt3 - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$ z,w\in\mathbb{C},|z|=|w|=R\gt0$. Show that $\left(\frac{z+w}{R^2+zw}\right)^2+\left(\vcenter{\frac{z-w}{R^2-zw}}\right)^2\ge\frac1{R^2}$ Let $ z, w \in \mathbb{C} $ be such that $ |z| = |w| = R > 0 $. Show that
$ \left(\frac{z + w}{R^2 + zw}\right)^2 + \left(\frac{z - w}{R^2 - zw}\right)^2 \geq \frac{1}{R^2} $
Wel... | We must assume that $zw \ne \pm R^2$ because the left-hand side is undefined otherwise. Then we can assume that $R=1$ because of the homogeneity of the inequality, this simplifies the calculation a bit.
You already noticed that the terms on the left-hand side are real numbers, so that
$$
\left(\frac{z + w}{1 + zw}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842864",
"timestamp": "2023-03-29T00:00:00",
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Convergence of $\sum_{k=1}^n\frac{(-1)^k}{\sqrt{k}}$ Does $\sum_{k=1}^n\frac{(-1)^k}{\sqrt{k}}$ converge?
My attempt:
$$\begin{aligned}
\sum_{k=1}^{2n}\frac{(-1)^k}{\sqrt{k}}&= \sum_{k=1, 3, ..., 2n-1}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\\
&=\sum_{k=1, 3,..., 2n-1}\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt... | Your idea is fine but we need to consider
$$
\sum_{k=1}^{2n+1}\frac{(-1)^k}{\sqrt{k}} =-1+ \sum_{k=2}^{2n+1}\frac{(-1)^k}{\sqrt{k}}=-1+\sum_{k=1}^n\left(\frac{1}{\sqrt{2k}}-\frac{1}{\sqrt{2k+1}}\right)
$$
and since
$$\frac{1}{\sqrt{2k}}-\frac{1}{\sqrt{2k+1}}=\frac{\sqrt{2k+1}-\sqrt{2k}}{\sqrt{2k}\sqrt{2k+1}}=\frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the recursive $(g(n) = 2g(n−1) + 1, g(0) = 1)$ is equal to $ g(n) = 2^{n + 1} - 1$ How do I show that the recursive definition below is equivalent to the explicit one?
$$g(n) = 2g(n−1) + 1, g(0)=1$$
$$g(n) = 2^{n + 1} - 1$$
| By back substitution for the recurrence relation:
$$g(n)=2g(n-1)+1$$
$$g(n-1)=2g(n-2)+1$$
$$g(n-2)=2g(n-3)+1$$
$$\cdot\cdot\cdot$$
$$g(1)=2g(0)+1=2+1=3$$
So we have $$g(n)=2g(n-1)+1=2[2g(n-2)+1]+1$$
$$=2^2g(n-2)+2+1$$
$$=2^2[2g(n-3)+1]+2+1$$
$$=2^3g(n-3)+2^2+2+1$$
$$=\cdot\cdot\cdot $$
$$=2^{k}g(n-k)+2^{k-1}+2^{k-2}+..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving inequality: $\sum_{i=1}^n \left(a_i^7+a_i^5\right) \geq 2(\sum_{i=1}^n a_i^3)^2$
Let $a_i$ be distinct positive integers; prove that $$(a_1^7+a_2^7+\cdots + a_n^7)+(a_1^5+a_2^5+\cdots +a_n^5)\ge 2(a_1^3+a_2^3+\cdots + a_n^3)^2$$
I tried using some well known inequalities; obviously, since non homogenous and n... | We can use induction here.
For $n=1$ it's true by AM-GM.
Now, let $$\sum_{k=1}^n(a_k^7+a_k^5)\geq\left(\sum_{k=1}^na_k^3\right)^2.$$
We'll prove that:
$$\sum_{k=1}^{n+1}(a_k^7+a_k^5)\geq\left(\sum_{k=1}^{n+1}a_k^3\right)^2.$$
Indeed, let $a_{n+1}=a=\max\limits_k\{a_k\}$.
Thus, $$\sum_{k=1}^{n+1}(a_k^7+a_k^5)\geq \left(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Examples of vector fields directed towards origin I'm asked to state a vector field equation directed radially in towards origin, would $-\langle x,y\rangle $ suffice, or do I need to divide it by its magnitude r i.e. $-\left\langle \frac{x}{\sqrt {x^2+y^2}},\frac{y}{\sqrt {x^2+y^2}}\right\rangle $ or should I divide i... | Any vector field of the form
$$
F(\mathbf r) = -f(|\mathbf r|)\, \mathbf r = -f(r)\, \mathbf r \ ; \qquad f(r) > 0
$$
is directed towards the origin, where $f$ is function of one argument, and its argument here is the distance $r$ from the origin, which is the length of $|\mathbf r|$. For example, in $\mathbb R^2$ we h... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Dummit and Foote 12.2.16: Determining all $2 \times 2$ matrices with entries from $\mathbb F _{19}$ of order $2$ This is exercise 12.2.16 of Abstract Algebra by Dummit and Foote.
Show that $x^5-1 =(x-1)(x^2-4x+1)(x^2+5x+1)$ in $\mathbb F_{19}[x]$. Use this to determine, up to similarity, all $2 \times 2$ matrices with... | We looks at the divisors of $x^5-1$ and use the decomposition as given. Notice for a $2 \times 2$ matrix it is only possible to get $2$ blocks of size $1$ or $1$ block of size $2\times 2$.
Recall that for a polynomial of the form $b_0 +b_1x +x^2$, we get the following companion matrix according to page 475:
$$ \begin{p... | {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}$. How to find the value?
Find the meaning of the expression
$$\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}$$
What I tried:
$$\tag{1} \frac{-5}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2} - \frac{(\sqrt[3]{3})(\sqrt[3]{3^2}) + (\sqrt[3]{3})(... | Note that in your computation from (1) to (2), it should be
$$(\sqrt[3]{3})(2\sqrt[3]{3})=2\sqrt[3]{3^2}\qquad (\text{it is not $6$)}.$$
It easier to follow the algebraic manipulations if we let $a=\sqrt[3]{3}$:
$$\begin{align}\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}&=\frac{a^3-8}{a^2+2a+4}-a=\frac{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let a, b, c be ints. $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an int, show that each of $\frac{ab}{c}, \frac{bc}{a}, \frac{ac}{b}$ is an int. Let a, b, c $\in \mathbb{Z}$ . If $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an integer, prove that each of $\frac{ab}{c}, \ \frac{bc}{a}, \ \frac{ac}{b}$ is an integ... | You can use divisibility as I show here. First, let
$$\frac{ab}{c} = \frac{d_1}{e_1} \tag{1}\label{eq1A}$$
$$\frac{bc}{a} = \frac{d_2}{e_2} \tag{2}\label{eq2A}$$
$$\frac{ac}{b} = \frac{d_3}{e_3} \tag{3}\label{eq3A}$$
where each fraction $\frac{d_i}{e_i}$ for $1 \le i \le 3$ is in lowest terms, i.e., $\gcd(d_i, e_i) = 1... | {
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Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answ... | If $a<b$ then
$$(a+x)(b-x)$$
is increasing in x for $0\leq x \leq \frac{b-a}{2}$.
Using this $(2+1/2)*(5+1/5)$ is larger than $(2+1/5)*(5+1/2)$ and $(3+1/3)*(4+1/4)$ is larger than $(3+1/4)*(4+1/3)$.
Intuitively, the square maximises the area over all rectangles with the same circumference. To maximise a product where ... | {
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Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, $x < 0, a = 1$ The textbook answer is $\frac {1}{3}$, I went through all the steps, but couldn't interpret it.
Below were my steps.
Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, x < 0, a = 1
First I tried to find the inverse, rearrange the equation to $$x = y ... | The domain of $f$ must be the range of $f^{-1}$, i.e. $f^{-1}$ must only output negative values for any value of $x$. So, you should reject the solution $y=\frac{x+\sqrt{x^2+8}}{2}$ and notice $$\frac{x-\sqrt{x^2+8}}{2} \lt \frac{x-\sqrt{x^2}}{2} \le 0 $$ which satisfies our requirement and the corresponding answer is ... | {
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Intersection of median and segment between two sides of a triangle
In triangle $ABC,$ $M$ is the midpoint of $\overline{BC},$ $AB=12,$ and $AC=16.$ Points $E$ and $F$ are taken on $\overline{AC}$ and $\overline{AB},$ respectively, and $\overline{EF}$ and $\overline{AM}$ intersect at $G.$ If $AE=2AF,$ then what is $EG/... | The key step is the ratio lemma. Let $b$ and $c$ be the measures of angles $BAM$ and $MAC$, respectively, and say $BM=m=MC$. Also, say $AF=\ell$ and $AE=2\ell$. Then, applying the ratio lemma to triangle $ABC$ at $A$, we have $\frac{\sin b}{\sin c}=\frac{m/12}{m/16}=\frac{4}{3}$. Then, we apply the ratio lemma to trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
| \begin{align*}
\frac{3+\cos4x}{4}&=\frac{3+2\cos^22x-1}{4}=\frac{(\cos^2x-\sin^2x)^2+1}{2}\\
&=\dfrac{\sin^4x+\cos^4x+(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}{2}\\
&=\sin^4x+\cos^4x
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove a bijection between $\mathbb{N}^2$ and $\mathbb{N}$.
Prove that the function
$$f(m,n)=\frac{1}{2}\left(m^2+2 m n+n^2+m+3 n\right)$$
is a bijection between $\mathbb{N}^2$ and $\mathbb{N}$.
The problem arose in a series problem. I have to show that for each couple $(m,n)$ we get a different natural number and tha... | Here is a rigorous proof, but first we rewrite $f$:
$$f(m,n) = \frac12(m^2+2mn+n^2 + m + 3n) = \frac12((m+n)(m+n+1)+2n)$$
$\Large \textbf{Injectivity}$
Suppose we have $f(m,n) = f(a,b)$. Then $(m+n)(m+n+1)+2n = (a+b)(a+b+1)+2b$.
First, suppose $m+n\ne a+b$. WLOG suppose $m+n > a+b$. Then:
\begin{align}(m+n)(m+n+1)+2n &... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Best way to evaluate $\lim_{n \rightarrow \infty} |\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}|$ Best way to evaluate $\lim_{n \rightarrow \infty} |\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}|$
Can you all show me some different ways to evaluate this limit? I was thinking of multiply... | We need to factor out the leading terms to obtain
$$\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}
=\frac{(n+1)\cdot 4^{n+1} \cdot 5^n}{5^{n+1} \cdot n \cdot 4^n}\frac{\left(3+\frac4{n+1}\right)\left(1+\frac5{4^{n+1}}\right)\left(1+\frac3{5^n}\right)}{\left(1+\frac3{5^{n+1}}\right)\left(3+\frac4n\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where is the mistake in my evaluation of $\int\frac{x}{x^2+2x+3}\,dx$? Here is how I did it:
First, write $$\int\frac{x}{x^2+2x+3}\,dx=\int\frac{2x+2-x-2}{x^2+2x+3}\,dx=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx.$$
Now consider the integral in the minuend. Letting $u=x^2+2x+3$, one finds $du=(2x+2)\,dx$... | Answer :
$\int_{}^{} \frac{x}{x^2 +2x+3}dx=\frac{1}{2}\int_{}^{} \frac{2x+2-2}{x^2 +2x+3} dx =\frac{1}{2}([ln(|x^2 +2x+3|] +c_1) - \frac{1}{2}\int_{}^{} \frac{2}{x^2 +2x+1 +2} dx= \frac{1}{2}([ln(|x^2 +2x+3|] +c_1) - \frac{1}{2}\int_{}^{} \frac{2}{(x+1)^2 +2 } dx =\frac{1}{2}([ln(|x^2 +2x+3|]+c_1) -\frac{1}{2}\int_{}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $f$ and $g$ by trial and error and a rigorous proof for showing that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}]$ Here is the question I am trying to solve:
Find polynomials $f(x), g(x) \in \mathbb{Q}[x]$ such that $\sqrt{2} = f(\sqrt{2} + \sqrt{3})$ and $\sqrt{3} = g(\sqrt{2} + \sqrt{3}).$ De... | $(\sqrt 2 + \sqrt 3)^k$ will be a linear combination of $\sqrt 2$, $\sqrt 3$, and $\sqrt 6$ so any polynomial $f(\sqrt 2 + \sqrt 3)$ will yield result of $a\sqrt 6 + b \sqrt 2 +c \sqrt 3 + d$ so we need a polynomial where the yielded values are $a=c=d=0$ and $b = 1$ (and for $g$, $a=b=d=0; c=1$). And, for simplicity, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$n\times n$ real matrix with characteristic polynomial $p$, where $p$ is an irreducible polynomial in $\mathbb{R}[X]$. Is there a $2\times 2$ real matrix which has $x^2+x+1$ as its characteristic polynomial?
Edit: For any irreducible polynomial of degree n in $\mathbb{R}[X]$, is there a matrix $A$ in $M_n(\mathbb{R})$ ... | The polynomial $x^2+x+1$ has the zeros
$$
\frac{-1 \pm \sqrt{ 1^2 - 4(1)(1) } }{2 (1) } = \frac{-1 \pm \sqrt{-3 } }{2} = \frac{-1 \pm \iota \sqrt{3} }{2}.
$$
So we can write
$$
x^2 + x + 1 = \left( x - \frac{-1 + \iota \sqrt{3} }{2} \right) \left( x - \frac{-1 - \iota \sqrt{3} }{2} \right) = \left( \frac{-1 + \iota \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute the limit as $x\to 3$ of a $\textit{complicated}$ product and quotient of trigonometric functions $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}$$
I substituted $x-3$ for $u$ and got as far as
$$\frac{1}{6} \lim_ {u\to 0} \frac{\sin(9 \sin u)}{\sin((u+3)^(3)... | You can use the following trick:
As $\sin(x)\sim x$ for $x\to0$, in a product you can replace the sine (tangent) of an argument that tends to zero by the argument itself. In your case,
$$\frac{ \tan\left(\dfrac{x-3}{x+3}\right)\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}\to \frac{ \dfrac{x-3}{x+3}\sin(9(x-3)) }{ (x-3)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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How to find steady-state probabilities in a queuing system? Queuing system set
$$\begin{matrix}\text{intake intensity} & \lambda=26 \\
\text{service channels} & m=4\\
\text{service intensity} & \mu=8\\
\text{maximum queue size} &n=18\end{matrix}$$
It is required to
Draw up a graph of a Markov process, write down the K... | As I understand it, the values of the matrix are derived from the queuing system graph.
Namely we have $m+n+1$ rows and $m+n+1$ columns since we also take cell with the zero state
$$ \Lambda = \begin{pmatrix} 0 & \lambda & 0 & 0 & \dots & 0 & 0\\
\mu & 0 & \lambda & 0 & \dots & 0 & 0 \\
0 & 2\mu & 0 & \lambda & \dots &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3868772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Possible ways of choosing 7 courses out of 20 with the constraint that at least 1 course has to be a stat course To fulfill the requirements for a certain degree, a student can choose to take any $7$ out of a list of $20$ courses, with the constraint that at least $1$ of the $7$ courses must be a statistics course. Sup... | The answers are equal and both correct.
The relationship between the answers is explained by Vandermonde's Identity
$$\binom{m+n}{r}=\sum\limits_{k=0}^r\binom{m}{k}\binom{n}{r-k}$$
In your case, that is $$\binom{20}{7}=\binom{5}{0}\binom{15}{7}+\binom{5}{1}\binom{15}{6}+\binom{5}{2}\binom{15}{5}+\dots+\binom{5}{5}\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem to find exact solution with steepest descent I have a problem that is formulated as this:
$$\begin{matrix}\min\\x \in \mathbb{R}^2\end{matrix} f(\mathbf{x}) := (2 x_1^2 - x_2^2)^2 + 3x_1^2-x_2$$
The task is: Perform one iteration using the steepest descent algorithm when $\mathbf{x}_0 = (1/2, 5/4)^T$.
And I get... | Let $x_1=x$ and $x_2=y$ and $y^2\leq\frac{3}{4}$.
Thus, by AM-GM
$$(2x^2-y^2)^2+3x^2-y=4x^4-4x^2y^2+3x^2+y^4-y=$$
$$\geq x^2(4x^2+3-4y^2)+y^4-y\geq y^4-y=$$
$$=y^4+\frac{3}{4\sqrt[3]4}-y-\frac{3}{4\sqrt[3]4}\geq4\sqrt[4]{y^4\left(\frac{1}{4\sqrt[3]4}\right)^3}-y-\frac{3}{4\sqrt[3]4}=$$
$$=|y|-y-\frac{3}{4\sqrt[3]4}\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Asymptotic expansion of an integral as $x \to 0^{+}$ Find the asymptotic expansion of $$F(x) := \int_{x}^{1} \frac{1}{t \sqrt{1+t^2}} \ dt, \text{ as } x \to 0^{+}$$
I tried expanding $\frac{1}{ \sqrt{1+t^2}} = 1 - \frac{t^2}{2} + \frac{3 t^{4}}{8} + \cdot \cdot \cdot$
The integral is then :$$F(x) := \int_{x}^{1} \fra... | HINT
I would start by making the substitution $t = \sinh(u)$. Thus we get
\begin{align*}
\int\frac{\mathrm{d}t}{t\sqrt{1+t^{2}}} = \int\frac{\cosh(u)}{\sinh(u)\cosh(u)}\mathrm{d}u = \int\frac{\mathrm{d}u}{\sinh(u)} = \int\frac{\sinh(u)}{\sinh^{2}(u)}\mathrm{d}u = \int\frac{\sinh(u)}{\cosh^{2}(u)-1}\mathrm{d}u
\end{alig... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where did I go wrong in applying the factor theorem?
Given that $x + 1$ and $x - 3$ are two of the four factors of the expression $x^4 + px^3 + 5x^2 + 5x + q$, find the values of $p$ and $q$.
I tried to answer this question using the factor theorem but got the answer wrong:
$$ \text{Let } f(x) = x^4 + px^3 + 5x^2 + 5... | Here is a way to see where you went wrong by comparison. Let
\begin{align*}
f(x) & = x^4+px^3 +5x^2+5x+q \\
g(x) & = (x+1)(x-3)(x-a)(x-b)
\end{align*}
Then write down the polynomial $f-g=0$. The coefficients must be all zero, i.e., we have
$$
a + b + p + 2=0,\; ab + 2a + 2b - 8=0,\; - 2ab + 3a + 3b - 5=0, q-3ab=0.
$$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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how would I find integer solutions for $y^2 = mx + b$? For example; $$y^2 = 30x + 1$$
For which one answer is; $$11^2 = 30(4) + 1 = 121$$
WA kindly gave me 4 answers;
$$x = 2 (15 n^2 - 29 n + 14), y = 29 - 30 n, n \in Z$$
$$x = 2 (15 n^2 - 19 n + 6), y = 19 - 30 n, n \in Z$$
$$x = 2 (15 n^2 - 11 n + 2), y = 11 - 30 n,... | $y^2 = mx +b$ so $x=\frac {y^2-b}m$ needs to be an integer.
I don't know of any general way to do it but you could solve $y^2 = b\pmod m$ (which may or may not have tricks; but brute force we can test $m$ values). $\alpha$ is a such a solution then $\alpha + mk$ will be solutions.
For example for $y^2 = 6m + 7$ we mus... | {
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Why is $r^3+4t^3+2s^3-6rts$ non-zero (unless $r=s=t=0$)? When solving How to find the multiplicative inverse of a polynomial? I created a $3 \times 3$ linear system with determinant $r^3+4t^3+2s^3-6rts, \text{ where } r, t, s \in \Bbb Q$. Basic field theory tells me this determinant must be non-zero (unless $r=s=t=0$)... | Let $x = r$, $y =\sqrt[3]{4}t$ and $z = \sqrt[3]{2}s$
$$r^3 + 4t^3 +2s^3 - 6rts = \left(r + t\sqrt[3]{4} + s\sqrt[3]{2}\right)\left(x^2 + y^2 + z^2 - xy -yz - zx\right)$$
$r + t\sqrt[3]{4} + s\sqrt[3]{2} = 0 \iff r = s = t = 0$
$$x^2 + y^2 + z^2 - xy - yz - zx = \frac{1}{2}(x-y)^2 + \frac{1}{2}(y-z)^2 + \frac{1}{2}(z-x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| We have
$$3 = b^2+bc+c^2 = \frac{3(b+c)^2+(b-c)^2}{4} \geqslant \frac{3}{4}(b+c)^2.$$
Therefore $(b+c)^2 \leqslant 4,$ so $b+c \leqslant 2.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I calculate the definite integral $\int_{-1}^1\frac{\sqrt{1-x^2}}{1+x^2}dx$ using complex variables? I have tried solving the integral $\oint_{C}\frac{\sqrt{1-z^2}}{1+z^2}dz$ using the upper semi-circle contour; I am getting the poles $z=\pm i$.
Only $z=i$ exists within the contour and I have evaluated the resid... | Let $I$ be given by
$$I=\int_{-1}^1\frac{\sqrt{1-x^2}}{1+x^2}\,dx$$
Next, let $C$ be the classical dog bone contour around $[-1,1]$ in the complex plane. It is straightforward to show that
$$\oint_C \frac{\sqrt{1-z^2}}{1+z^2}\,dz=-2I$$
since $C$ is traversed in the counter-clockwise direction.
In the following analys... | {
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Prove by Induction. For $n \in \mathbb{N}$, $10|(9^{n+1}+7^{2n})$. So far, this is what I have, but I'm confused as to how to 1. remove the 7 from inside the brackets to be able to substitute 10k and 2. make the whole thing divisible by 10 so that I can prove it.
Basic Step: Let $n = 1$. Therefore,
$$
9^{1+1} + 7^{2 \c... | Can you show that $$9^{k+2}+7^{2k+2}=9\cdot9^{k+1}+49\cdot7^{2k}=9(9^{k+1}+7^{2k})+40\cdot7^{2k}$$ is divisible by $10$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find maximize of $P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$ Let $x,y,z\in \mathbb{R^+}$ such that $6x+3y+2z=xyz$. Find maximize of $$P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$$
We will prove $$P\le \sqrt{\frac{16}{27}}$$
$$\Leftrightarrow \fr... | Also, $uvw$ helps.
Indeed, in my previous solution it's enough to prove that:
$$64(2a+b)^3(2a+c)^3(b+c)^2\geq729(2a+b+c)^4a^2bc.$$
Now, let $b+c=2u$ and $bc=v^2$, where $v>0$.
Thus, by AM-GM $u\geq v$ and we need to prove that $f(u)\geq0,$ where
$$f(u)=16u^2(4a^2+4ua+v^2)^3-729(a+u)^4a^2v^2.$$
But, $$f'(u)=32u(4a^2+4u... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying the derivative of $\sin(\cos^2 x)\cos(\sin^2 x)$ Differentiating the term
$$\sin(\cos^2 x)\cos(\sin^2 x)$$
leads me through the chain and product rule to
$$-\sin(2x)\cos(\cos^2 x)\cos(\sin^2 x)+(-\sin(\sin^2 x)\sin(2x)\sin(\cos^2 x))$$
where the derivative of $\sin^2 x$ equals to
$$\frac{d}{dx} \sin^2 x = 2... | Denote:
$$f(x) = \sin(\cos^2 x)\cos(\sin^2 x)$$
$$g(x) = \sin(\cos^2 x)$$
$$h(x) = \cos(\sin^2 x)$$
Thus:
$$f(x) = g(x)\cdot h(x)$$
$$g'(x) = \cos(\cos^2{x}) \cdot 2 \cos{x} \cdot (-\sin{x})$$
$$h'(x) = -\sin(\sin^2{x}) \cdot 2 \sin{x} \cdot \cos{x}$$
Now we get:
$$f'(x) = g(x)\cdot h'(x) + g'(x)\cdot h(x) $$
$$f'(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Jordan normal form of $\left(\begin{smallmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{smallmatrix}\right)$ I want to find the Jordan normal form of $A=\begin{pmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{pmatrix}$, but somewhere I think that I make a mistake (I am quite new to the computation of the Jordan ... | After you get the Jordan form (as psidaga's answer did)
\begin{align*}
J = \begin{pmatrix}
3 & 1 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix},
\end{align*}
in order to determine the similarity matrix $P$ such that $P^{-1}AP = J$, it is is equivalent to solve the system
\begin{align*}
& A\alpha_1 = 3\alpha_1, \tag{1} \\
... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$ $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$
Can we apply Stolz-Cesaro?
$\lim\limits_{n \to \infty}\frac {\frac{1}{\sqrt{n+1}} - \frac{... | Option:
$n/√n < 1/√1+1/√2....+1/√n;$
$\dfrac{√n}{2 \log (√n)} < \dfrac{1/√1+1/√2+...1/√n}{\log n};$
$(1/2) \lim_{n \rightarrow \infty} \dfrac{√n}{\log √n} = $?
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof Does anyone know hot to prove this inequality?
Having: $a, b, c \gt 0$
$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$
I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I ... | We can prove it without expending by SOS and the Tangent Line method.
Indeed, $$\sum_{cyc}\frac{b^2-a^2}{a+c}=\frac{1}{2}\sum_{cyc}\frac{2b^2-2a^2}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{2b^2-2a^2}{a+c}+a-c\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{b^2-c^2-(a^2-b^2)}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{a^2-b^2}{c+b}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$ A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.
By taking $\cos... | When you used the replacement $\cos \alpha = x$, the expression in the bracket became
$$\frac{\cos \alpha}{2} + \frac{\sqrt{3} \sin \alpha}{2} = \cos \frac{\pi}{3} \, \cos \alpha + \sin \frac{\pi}{3} \, \sin \alpha$$
but note that this can be written as both
$$\cos \left( \frac{\pi}{3} - \alpha \right) \quad \text{or} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of a Bessel function times a sine function times a polynomial I have recently faced the following expression (while trying to compute the Fourier transform of a RKKY-like potential):
\begin{align}
\int_0^{\infty} \text{d} r\; J_0 (k r) \frac{\sin \left(\alpha \sqrt{1+r^2} \right) }{(1+r^2)^2} \; r = -\frac... | According to this reference, Table 8.2 , entry (7):
$$\int_0^\infty \dfrac{1}{\left(1+r^2\right)^\frac{3}{2}}J_0(kr)r \; dr = e^{-k}$$
and according to Table 8.2, entry (41):
$$\int_0^\infty \dfrac{\sin\left[\alpha\left(1+r^2\right)^\frac{1}{2}\right]}{\left(1+r^2\right)^\frac{1}{2}}J_0(kr)r \; dr = \dfrac{\cos\left[\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)... | From the beginning, move one of the absolute values to the other side. It will simplify the calculations:
$$|z+i|=2-|z-i|$$
Square this and you get $$|z+i|^2=(2-|z-i|)^2=4-4|z-i|+|z-i|^2$$
$$x^2+(y+1)^2=4-4|z-i|+x^2+(y-1)^2$$
$$x^2+y^2+2y+1=4-4|z-i|+y^2-2y+1$$
$$4y-4=-4|z-i|$$
$$y-1=-|z-1|$$
$$(y-1)^2=x^2+(y-1)^2$$
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Is there a simple, but tight lower bound for the error made when $\sum_{n=1}^{k}\frac{1}{n^2}$ is used to approximate $\frac{\pi^2}{6}$? As a math-for-fun exercise, I've recently been seeking bounds for the error $R_k$ made when using $\sum_{n=1}^{k}1/n^2$ to estimate its beautiful sum $\pi^2/6$. Applying the Compariso... | Consider $$g(n) = \frac{1}{n-1/2} - \frac{1}{n+1/2}$$
Then
$$ g(n) - \frac{1}{n^2} = \frac{1}{4n^4 - n^2} > 0 \ \text{for}\ n \ge 1$$
Now $1/(4n^4 - n^2)$ is a decreasing function of $n$ for $n > 1$, so
$$\eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &= \frac{1}{N+1/2} - \sum_{n=N+1}^\infty \frac{1}{4n^4-n^2}\cr & > \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ Find the group A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?
Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$
but how i can fro... | Hint: Since $2x+1$ is odd for all integers $x$, we have that $\frac{x^3-3x+2}{2x+1}\in\mathbb Z$ if and only if $\frac{2(x^3-3x+2)}{2x+1}\in\mathbb Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3896747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Sketching a set of $Re(z^2(1+i))$ on the Complex plane. So I am trying to sketch a set of complex numbers that is characterized by inequality: $ Re((1+i)z^2) < 0 $.
I know that if I take $ z = a + bi $ , then:
$$Re(z^2) \implies (a + bi)^2 = a^2 + 2abi - b^2 \implies Re(z^2) = a^2 - b^2$$
Similarly:
$$Re(z^2(1+i)) \im... | When dealing with complex number, polar form is almost always more useful than breaking it into real and imaginary part.
$$
Re((1+i)z^{2})<0\implies \frac{\pi}{2}<\arg{((1+i)z^{2})}<\frac{3\pi}{2}
$$
Now we use the following
$$
\begin{align}
\arg{((1+i)z^{2})}&=\arg{(1+i)}+ 2\arg{(z)}\\
&=\frac{\pi}{4}+2\arg{(z)}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Domain of function $\frac{1}{x + \sqrt{x+1}}$ I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$
I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
... | $$x\ge-1$$ is obvious.
Then the denominator is zero when
$$x+\sqrt{x+1}=0$$ or by squaring $$x^2=x+1$$
and there is a single valid root
$$x=1-\phi\ge-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Minimum value of $(\sin x + \cos x)^3 + \frac{1}{\sin^2x\cos^2x}$ Let $\sin x + \cos x = t\qquad (\vert t \vert \leq \sqrt2)$
and solve it as $t^3 + \dfrac{4}{(t^2 - 1)^2}$
Is there any easier solution to solve this $(\sin x + \cos x)^3 + \dfrac{1}{\sin^2x\cos^2x}$?
Is there any available inequality can solve it?
Pleas... | There may not be a specific inequality, but sure there is a faster way.
$$(\sin x + \cos x)^3 + \dfrac{1}{\sin^2x\cos^2x}=2\sqrt{2}\left(\sin\left(x+\frac{\pi}{4}\right)\right)^{3}+4\csc^{2}\left(2x\right)$$
Observe that only the expression having an odd power can contribute to reducing the value of the expression. Tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3899922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the remainder when $1^n + 2^n + 3^n + \ldots + 99^n$ is divided by $1 + 2 + 3 + \ldots + 99$? My first idea on how to approach was to create a polynomial expansion for the first few terms and then try to find a pattern for the rest but this became cumbersome and I don't think that this is a right approach as th... | An ad-hoc solution:
$n$ odd one can group $1^n+99^n, 2^n+98^n,...50^n$ to get $S_n$ divisible by $50$ and similarly $1^n+98^n,..49^n+50^n$ to get $S_n$ divisible by $99$ so as noted above the remainder is zero.
For $n$ even we notice that modulo $3$ each group $3k-2,3k-1, k=1,..33$ gives a sum that is $2$ modulo $3$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
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Quadratic equation where constant term is an integral function Consider a quadratic equation ${x^2} + 2x = k + \int\limits_0^1 {\left| {t + k} \right|dt}$. Then choose the correct option(s),
(A) Roots are Real
(B) Roots are Imaginary
(C) Roots are Distinct
(D) Roots are complex number
My approach is as follow
Let $-c=k... | The discriminant $b^2 - 4ac$ is actually
$$d = 4 + 4(k + \int_{0}^{1}\left| t + k \right| dt) \tag{1}\label{eq1A}$$
Regarding calculating $k + \int_{0}^{1}\left| t + k \right| dt$, there are $3$ cases to consider:
Case $1$: $\; k \lt -1$
Since $t + k \lt 0$ for $t$ from $0$ to $1$, this gives
$$\begin{equation}\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1... | Hint: notice that $$\sum(2k)^2=\sum4k^2=4\sum k^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.