Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ in the field $\mathbb{Q}(\sqrt{5},\sqrt{7})$.
I know that all the elements of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ are of the form:
$a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35}$, where $a,b,c,d \in \mathbb{Q} $
So I could simply solve: $(2+\sqr... | Hint: Consider the product $(2+\sqrt{5}+2\sqrt{7})(2-\sqrt{5}+2\sqrt{7})(2+\sqrt{5}-2\sqrt{7})(2-\sqrt{5}-2\sqrt{7})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find multivariable limit $\frac{x^2y}{x^2+y^3}$ Find multivariable limit of: $$\lim_{ \left( x,y\right) \rightarrow \left(0,0 \right)}\frac{x^2y}{x^2+y^3}$$
How to find that limit? I was trying to do the following, but i am not able to find a proper inequality:
$$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$... | Note that for $x \neq 0$ and $y \neq 0$ we have
$$ \left | \frac{x^2y}{x^2 + y^3} \right| = \left| \frac{y}{1 + \frac{y^3}{x^2}} \right| = \left| \frac{1}{\frac{1}{y} + \frac{y^2}{x^2}} \right| \leq \left| \frac{1}{\frac{1}{y}} \right| = \left| y \right|. $$
If $x = 0$ or $y = 0$ (but not $x = y = 0$), then we also hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Proving $ \frac{1}{c} = \frac{1}{a} + \frac{1}{b}$ in a geometric context
Prove or disprove
$$
\frac{1}{c} = \frac{1}{a} + \frac{1}{b}.
$$
I have no idea where to start, but it must be a simple proof.
Trivia. This fact was used for determination of resistance of two parallel resistors in some circumstances long time a... | I'm not sure what the protocol is for adding a separate answer when it uses a different approach. If there is an issue with doing this, I can move this to my existing entry.
Here is a method using analytic geometry. I have changed the labeling of two points, since I wish to place the origin of the coordinate system a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Compute $\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$ Given
$$\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$$
I couldn't evaluate this integral. My only idea here was evaluating this as integration by parts.
\begin{align}
\int\frac{x \log(x)}{(1+x^2)^2}dx & = \frac{1}{2} \int\frac{ \log(x)}{(x^2+1)^2}d(x^2+1)\\
& =... | Here is a general technique. Let's consider the integral
$$ I = \int_{0}^{\infty} \frac{x^{s-1}}{(1+x^2)^2} dx = -\frac{1}{4}\,{\frac { \left( s-2 \right) \pi }{\sin \left( \pi \,s/2
\right) }},$$
which is the Mellin transform of $\frac{1}{(1+x^2)^2}$. Your integral can be evaluate using $I$ as
$$ \int_{0}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
computing an integration with a floor function I am trying to compute
$$\int_0^1 \left(\frac{1}{x} - \biggl\lfloor \frac{1}{x}\biggr\rfloor\right) dx$$
with no success.
Any hints?
| $$\int_0^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx=\lim_{k\to +\infty}\int_{\frac{1}{k}}^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx$$
Now to compute the other integral use the decomposition:
$$\int_{\frac{1}{k}}^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Primitive of the function $(\sin x)/x$ I know that for some functions, for instance $f(x) = e^{-x^2}$, there does not exist a primitive.
Does there is a primitive for the function $f(x) = \frac{\operatorname{sin}(x)}{x}$?
| No It is not In terms of elementary functions, But we can write in Infinite Series form.
Using $$\displaystyle \sin x= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+..........$$
So $$\displaystyle \frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{4!}-\frac{x^6}{6!}.........$$
So $$\displaystyle \int \frac{\sin x}{x}dx = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding roots of $z^3 = 8$ I am having trouble finding the cubed roots of $8$ as a complex number.
$$z^3 = 8+0i$$
$$z^3 = r^3 e^{3\theta i}=8e^{2i\pi k},\quad k\in \Bbb Z$$
$$\implies r=2,3\theta = 2\pi k\implies \theta = \frac{2\pi k}{3}$$
$$z=2e^{\frac{2i\pi k}{3}}$$
I would think I have then $$z=2,2e^{\frac{2i\pi}{3... | $z^3-2^3=(z-2)(z^2+2z+4)=0$
So first term has solution $z=2$.
Now-
$z^2+2z+4=0$
$\frac{-2 \pm \sqrt{4-16}}{2}=z$
So, $z=-\sqrt{3}i-1 \ and\ z=\ \sqrt{3}i-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Algebraic Manipulation What is the best method to get the LHS equal to RHS?
$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$
Thank you.
| Firstly, take $(n+1)(n+2)$ out as a factor:
$((n+1)(n+2))[\frac n3 +1] = \frac{n+3}{3} (n+1)(n+2)$
Then divide through by $(n+1)(n+2)$:
$(\frac n3 +1) = \frac{n+3}{3}$
Multiply both sides by $3$:
$n+3 = n+3$
Therefore LHS = RHS
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove equality of two numbers written in complex polar form. Show that these two numbers are equal:
$$
z_1=\frac{e^{\tfrac{2\pi i}{9}}-e^{\tfrac{5\pi i}{9}}}{1-e^{\tfrac{7\pi i}{9}}}
$$
and
$$z_2=\frac{e^{\tfrac{\pi i}{9}}-e^{\tfrac{3\pi i}{9}}}{1-e^{\tfrac{4\pi i}{9}}}=\frac{1}{e^{\tfrac{\pi i}{9}}+e^{\tfrac{-\pi i}{... | Let $w = \exp(i\pi/9)$. Then
$$
z_1 = \frac{w^2-w^5}{1-w^7} = \frac{w^2(1-w^3)}{1-w^7}
$$
and
$$
z_2 = \frac{w-w^3}{1-w^4} = \frac{w(1-w^2)}{(1-w^2)(1+w^2)} = \frac{w}{1+w^2}.
$$
Hence
$$
z_1 - z_2 = \frac{w^2(1-w^3)(1+w^2)-w(1-w^7)}{(1-w^7)(1+w^2)} =
\frac{w(1-w)(w^6-w^3+1)}{(1-w^7)(1+w^2)}.
$$
It remains to show tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$ The task is to evaluate $$\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$$
My best approach has been substitution of $u=x+\frac{1}{2}$, and from there onto (some terrible) trig sub - finally arriving at a messy answer.
Is there a slicker way of doing this?
| We have
\begin{align}
\dfrac{x^2}{\sqrt{1+x+x^2}} & = \dfrac{x^2+x+1}{\sqrt{x^2+x+1}} - \dfrac{x+1}{\sqrt{x^2+x+1}}\\
& = \sqrt{x^2+x+1} - \dfrac12 \dfrac{2x+1}{\sqrt{x^2+x+1}} - \dfrac12 \dfrac1{\sqrt{x^2+x+1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Use de Moivre's theorem to obtain an expression for $\sin^6x$ as a sum of terms in the form $\cos ax$ I'm not exactly sure if I'm on the right lines but I've started with a binomial expansion:
$(\cos x+i\sin x)^6=\cos 6x +i \sin 6x= \cos^6 x + i(6\cos^5x \sin x)-15\cos^4x \sin^2x-i(20\cos^3x \sin^3 x)+15 \cos^2x \sin^4... | Let $z = e^{ix}$. $\sin x = \frac1{2i} (z-z^{-1})$ and
$$\sin^6 x = -\frac1{64} (z - z^{-1})^6 = -\frac1{64} (z^6 - 6z^4 + 15z^2 - 20 + 15 z^{-2} - 6 z^{-4} + z^{-6})$$
Now remember that $\cos ax = \frac12 (z^a + z^{-a})$ to get
$$\sin^6 x = -\frac1{32} \cos 6x + \frac3{16} \cos 4x - \frac{15}{32} \cos 2x + \frac5{16}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\bigg(\frac{-2}{p}\bigg)= \begin{cases} 1 & \text{ if $p\equiv 1$ or $3 \mod 8$} \\ -1 & \text{ if $p\equiv 5$ or $7 \mod 8$} \\\end{cases}$ Show that
$$\bigg(\frac{-2}{p}\bigg)=
\begin{cases}
1 & \text{ if $p\equiv 1$ or $3 \mod 8$} \\
-1 & \text{ if $p\equiv 5$ or $7 \mod 8$} \\\end{cases}$$
$\textbf{Proof:}$
\begi... | 1) Recall that if $p$ is an odd prime, then $2$ is a QR of $p$ if $p\equiv 1$ or $7 \pmod{8}$, and $2$ is an NR of $p$ if $p\equiv 3$ or $5\pmod{8}$.
2) Also, $-1$ is a QR of $p$ if $p\equiv 1\pmod{4}$ and is an NR of $p$ if $p\equiv 3\pmod{4}$. This can be restated as $-1$ is a QR of $p$ if $p\equiv 1$ or $5 \pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x)$
If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x).$
My Solution:: Let $$\displaystyle y = \sin^4 x+\cos^2 x \leq \sin^2 x+\cos^2 x=1$$
And for Minimu... | We can also solve this problem as follows.
Let $f(x)=\sin^4x+\cos^2x$. This simplifies $f(x)=\frac{1}{8}(7+\cos 4x).$ Then $f_{min}=\frac{3}{4}$ which occurs when $\cos 4x=-1$ and $f_{max}=1$ which occurs when $\cos 4x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$
I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\fr... | Since $\arctan x >- \frac{\pi}{2}$ and the function is increasing, $xy<1$ would yield $$\arctan x + \arctan y < \arctan x + \arctan \frac{1}{x}=\frac{\pi}{2}< \pi + \arctan \frac{x+y}{1-xy}$$ if none of $x$ and $y$ is negative, or WLOG only $y$. If they are both negative we would have a negative LHS, while the RHS rema... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Why is $n^2+4$ never divisible by $3$? Can somebody please explain why $n^2+4$ is never divisible by $3$? I know there is an example with $n^2+1$, however a $4$ can be broken down to $3+1$, and factor out a three, which would be divisible by $4$.
| Well, You have two cases
1st case $n$ is odd then $n=2k+1$ for some integer $k$
Then $n^2 + 4 = (2k+1)^2 + 4 = 4k^2 + 4k + 5 = 4k(k+1) +5$
Here we have either $k$ or $k+1$ is even and the other is odd or vice versa and we know that even $\times$ odd = even and so $4k(k+1) + 5 = 4(2m)+5$ for some integer $m$ and so have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
Summation rule index In the simplification shown below, The index are changed but when i follow the index change rule, i cant find the same final answer.
$$\sum^{n-1}_{i=0} \frac{n}{n-i}=n\sum^{n}_{i=1} \frac{1}{i}$$
Thanks
| Write it as
$$\sum^{n-1}_{i=0} \frac{n}{n-i}= \frac{n}{n} + \frac{n}{n-1} + \dots + \frac{n}{1}$$
Read from the end to the beginning.
$$\frac{n}{1} + \frac{n}{2} + \dots + \frac{n}{n} = n \Big(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n}\Big) = n\sum^{n}_{i=1} \frac{1}{i}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Indefinite Integral of a function $$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$
I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.
| \begin{align}
\int \frac{1}{5}(x^3)-2x+\frac{3}{x}+e^x dx & = \int \frac{1}{5}x^3 dx- \int 2x dx+ \int \frac{3}{x} dx+ \int e^x dx \\ &= \frac{1}{20}x^4 - x^2 + 3 \ln(x) + e^x + C
\end{align}
Important rules:
•Derivative of $x^n$ is $\frac{1}{n+1}x^{n+1}$ for $n \neq -1$
•Derivative of $x^{-1}$ is $\ln(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2, $or $6$
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2,$ or $6$.
I'm having a hard time with the $0,2,6$ and part but here is what I have so far.
Since $n$ is the product of two consec... | When you multiply two numbers, the units place of the product is determined only by the units' places of the two numbers. So it is sufficient to check the cases one by one.
$$
0 \times 1 \text{ gives } 0 \\
1 \times 2 \text{ gives } 2\\
2 \times 3 \text{ gives } 6\\
3 \times 4 \text{ gives } 2\\
4 \times 5 \text{ gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$? Let $a,b,c>0: (a+b)(b+c)(c+a)=ab+bc+ca$. How to prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$$
| By Cauchy-Schwarz $\left(\sum\limits_{cyc}\frac{a}{\sqrt{a+b}}\right)^2\leq\sum\limits_{cyc}\frac{a}{(a+b)(a+b+2c)}\sum\limits_{cyc}a(a+b+2c)$.
Hence, it remains to prove that $\frac{27(a+b)(a+c)(b+c)}{16(ab+ac+bc)\sum\limits_{cyc}(a^2+3ab)}\geq\sum\limits_{cyc}\frac{a}{(a+b)(a+b+2c)}$, which after full expanding gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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Computing $\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx$ I wish to compute
$$\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx, \quad a>0$$
but have no contour to work with. Does anyone have ideas on how to compute this integral?
| This integral can be evaluated with calculus only:
\begin{align*}
\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3} \, dx
&= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{x^2 + 1}{x^6 + 1} \, dx \tag{1} \\
&= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{1}{(x - \frac{1}{x})^2 + 1} \, \frac{dx}{x^2} \\
&= \frac{1}{2a^{3/2}} \int_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove the sum of any $n$ consecutive numbers is divisible by $n$ (when $n$ is odd).
Let $n \in \mathbb N$ be odd. Prove that the sum of any $n$ consecutive numbers is divisible by $n$.
I started out with $s = x + (x + 1) + (x + 2) + … + (x + n) = kx + n.$ What I am interested in is if that's a right way to sum $x + (... | $(x+1)+(x+2)+\cdots+(x+n)=nx+(1+2+\cdots+n)$
$=nx+\frac{n(n+1)}{2}=n\left(x+\frac{n+1}{2}\right)$. Since $n$ is odd, $x+\frac{n+1}{2}$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Finding the number of real roots of an unusual(!) equation How many real roots does the below equation have?
\begin{equation*}
\frac{x^{2000}}{2001}+2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0
\end{equation*}
A) 0 B) 11 C) 12 D) 1 E) None of these
I could not come up with anything.
(Turkish Math Olympiads 2001)
| Consider the discriminant of $f(X) = 2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0$:
$$(-2\sqrt{5})^2 - 4(2\sqrt{3})\sqrt{3} = 20 - 24 < 0.$$
Therefore $f(x)$ has no real roots.
But $f(0) = \sqrt{3} > 0$, so $f(x) > 0$ everywhere.
Now combine this with $$\frac{x^{2000}}{2001} \geq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Dice outcomes probability
Two dice are rolled. What is the probability that the sum of the
numbers on the dice is at least 10
Let $Z$ denote the set of successful outcomes:
$Z=\{(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\}\\
\text{Sample space: }
S=6^2$
So the answer gives the probability as, $P=\frac{6}{36}$.
My question... | Your sample space $S=\{1,2,3,4,5,6\}^2$ has exactly 36 unique outcomes, with each having probability $\frac{1}{36}$. $(4,6)$ and $(6,4)$ are separate outcomes, whereas $(5,5)$ and $(5,5)$ are the same.
Intuitively, to get an outcome involving one $4$ and one $6$, I could roll either a $4$ or a $6$ for the first die, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square. I successfully thought of bounding our expression examining consecutive squares that attain values close to it, and this led to the solution I'll post as an answer, which was the one reported. However, before that, I had briefly ... | There are only two additional non-positives that also give perfect squares.
Case 1: $\boldsymbol{n}$ odd
$$
\begin{align}
\left(n^2+\frac{n-1}2\right)^2
&=n^4+n^2(n-1)+\frac{n^2-2n+1}4\\
&=n^4+n^3-\frac34n^2-\frac12n+\frac14
\end{align}
$$
smaller than $n^4+n^3+n^2+n+1$ if $\frac74n^2+\frac32n+\frac34\gt0$
$$
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a)... | Yes, this is enough. Basically, if you are trying to solve:
$$\begin{pmatrix}d_1&d_2&d_3\end{pmatrix}A = \begin{pmatrix}y_1&y_2&y_3\end{pmatrix}$$ you are looking for a polynomial $f(x)=d_1+d_2x+d_3x^2$ such that $f(a)=y_1$, $f(d_2)=y_2$, and $f(d_3)=y_3$.
This we can find by simple interpolation of polynomials - give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
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How do you add two series together How do you add the series
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
?
is this right?
$$\begin{aligned}
&\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}}{(z-3)^{n+1}} + \frac{(z-3)^{n}}{4^{n+1}}\right)\\
=\... | $$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
$$=\frac{1}{2(z-3)}\sum_{n=0}^{\infty} \left( \frac{2}{z-3}\right) ^n +
\frac18 \sum_{n=0}^{\infty} \left( \frac{z-3}{4}\right) ^n $$
evaluate the two sums using the formula for geometric series... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Points of intersection of $\sin x$ and $\cos x$ I'm trying to find the points of intersection of $\sin x$ and $\cos x$ between $0$ and $2\pi$. I've tried but I keep getting 4 solutions... Would someone please be able to take me through the process?
I squared $\sin x$ and $\cos x$ and then got $\sin x = \frac{\pm 1}{\sq... | $\sin x = \cos x \implies \sin^2x = \cos^2 x $ together with the proviso that $\sin x$ and $\cos x$ must share the same sign.
So the only two solutions are
$\sin x=\frac{1}{\sqrt 2}$, $\cos x=\frac{1}{\sqrt 2}$ giving $x=\frac{\pi}{4}$
and
$\sin x=\frac{-1}{\sqrt 2}$, $\cos x=\frac{-1}{\sqrt 2}$ giving $x=\frac{5\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The value of ${\sum_{k=0}^{20}}(-1)^k\binom{30}{k}\binom{30}{k+10}$ $\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\c}[1]{{}^{30}{\mathbb C}_{#1}}
\newcommand{\r}[1]{\frac1{x^{#1}}}$
The value of $$\sum_{k=0}^{20}(-1)^k\binom{30}{k}\binom{30}{k+10}$$
It is also the coefficient of $x^{10}$ in:
$$\b{\c0\r0-\c1\r1+\c2\r... | Suppose we are interested in the value of
$$S(n,m) =
\sum_{k=0}^n (-1)^k {n+m\choose k} {n+m\choose k+m}.$$
Introduce
$${n+m\choose k+m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m}}{z^{k+m+1}} \; dz.$$
This integral controls the range being zero when $k>n$ so we can
extend the summation to infinity ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Determinant properties Prove without expanding:
\begin{equation}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3 & c^3\end{vmatrix} = (ab + ac + bc)(b - a)(c - a)(c - b)\end{equation}
*I tried to zero some elements and expand until I reach the Right hand side.
*Also tried C1-C3, C2-C3 then decompose the determinant into t... | First note that the determinant is cyclic. Hence, it is of the form $f(a,b,c)$, where $f$ is a polynomial of degree $5$. Further, we have
$f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$, which means $(a-b)$, $(b-c)$ and $(c-a)$ are factors, i.e., the determinant is $g(a,b,c)(a-b)(b-c)(c-a)$, where $g(a,b,c)$ is a cyclic polynomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can anyone explain why this series converges? Can anyone explain why this series converges?
$1+\frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...$
The answer is given, but i do not understand it:
$$**|S_{6n}-S_{3n}|= \frac{1}{3n+1}+\frac{1}{3n+2}-\frac{1}{3n+3}+...+\frac{1}{6n-2}+\frac{1}{6n-1}-\fr... | This inequality holds because each of the elements $\frac{1}{3n+1}.....\frac{1}{6n-2}$ is greater or equal to the last term. ie $\frac{1}{3n+1} \gt \frac{1}{6n-2}$ etc for all the terms. So the sum $\frac{1}{3n+1}+\frac{1}{3n+4}...$ (n terms) is definitrly less than $\frac{1}{6n-2}+\frac{1}{6n-2}$...(n terms)..$\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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number of subsets of even and odd Let $A$ be a finite set. Prove or disprove: the number of subsets of $A$ whose size is even is equal to the number of subsets of $A$ whose size is odd.
Example: $A = {1,2}$. The subsets of $A$ are {},{1},{2}, and {1,2}. Since there are two subsets of odd size ({1} and {2}) and two su... | The number of subsets of size $k$ in a set with $n$ elements is $\binom{n}{k}$.
The Binomial Theorem states that
$$(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k}a^{n - k}b^k$$
In particular, if we let $a = b = 1$, we obtain
$$2^n = (1 + 1)^n = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}1^k = \sum_{k = 0}^{n} \binom{n}{k}$$
Ther... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find numbers $a, b, c$ given that $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$
Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$
Suppose $a, b, c$ are roots of $P(x)$.
$$P(x) = k(x - a)(x - b)(x - c)$$
But then I get $(k = 1)$
$$P(x) = x^3 - 12x^2 + x(ab + ac + bc) - abc$$
Cant go further.... | $$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$
So $ab+ac+bc=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=47$
Similarly
$$a^3+b^3+c^3=(a+b+c)^3+3abc-3(a+b+c)(ab+ac+bc)$$
So $abc=\frac{a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)-(a+b+c)^3}{3}=44$
And $a,b,c$ are solutions of the cubic
$$X^3-12X^2+47X-44=0$$
No simple solution except Cardan formulaes
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int \frac{x^5 dx}{\sqrt{1+x^3}}$ I took $1+x^3$ as $t^2$ . I also split $x^5$ as $x^2 .x^3$ . Then I subsituted the differentiated value in in $x^2$ . I put $x^3$ as $1- t^2$ . I am getting the last step as $2/9[\sqrt{1+x^3}x^3 ]$ but this is the wrong answer , i should get $2/9[\sqrt{1+x^3}(x^3 +2)]$.
Plea... | This is an integration by parts problem, but first, set $u=x^3$. Then $ du = 3x^2 dx $, so the integral becomes
$$ \frac{1}{3}\int \frac{u}{\sqrt{1+u}} \, du, $$
which you can see how to do parts on:
$$ \int \frac{u}{\sqrt{1+u}} \, du = 2u\sqrt{1+u} - 2\int \sqrt{1+u} \, du = 2u\sqrt{1+u} - \frac{4}{3}(1+u)^{3/2} +C, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Quadratic solutions puzzle The equation $x^2+ax+b=0$, where $a\neq b$, has solutions $x=a$ and $x=b$. How many such equations are there?
I'm getting $1$ equation as I can only find $a=b=0$ as an equation, which is not allowed.
$$x=\frac{±\sqrt{a^2-4 b}-a}2$$
$x=a$ or $b$ so these are the equations
$$a=\frac{\sqrt{a^2-4... | if $x=a$ is a solution to $x^2+ax+b=0$ then $b=-2a^2$
if $x=b$ is a solution $b^2+ab+b=0$ so either $b=0$ or $a+b+1=0$
the second solution gives $=-2a^2+a+1=0$
this equation has solutions $a=1$ and $a=-\frac12$
so it looks like we can have $x^2+x-2$ with solutions of 1 and -2
and $x^2 -\frac12 x -\frac 12$ which has ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do we find eigenvalues from given eigenvectors of a given matrix? For instance let
$$A=\begin{pmatrix}
3 & -1 & -1 \\
2 & 1 &-2 \\
0 & -1 & 2 \\
\end{pmatrix}$$
be a matrix and
$$u_1=\begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix},$$
$$u_2=\begin{pmat... | If $u$ is an eigenvector of $A$ and $\lambda$ is the corresponding eigenvalue, you know the following:
$$Au = \lambda u$$
So in your example, you can do the following (I’ll take the second one):
$$Au_2 = \begin{pmatrix}3 & -1 & -1 \\ 2 & 1 & -2 \\ 0 & -1 & 2\end{pmatrix}\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} = \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$
Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $$a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$$
The inequality can be written in the condensed form
$$\sum\limits_{Sym}a\ge\sum\limits_{Sym}a^3+\sum\limits_{Sym}ab$$
I was... | I got some hints from Crux Problem 3059, click here for more details=)
Based on the hints (which tries to relate the inequality with a constrained optimization problem), I worked out a proof as follows:
We have
\begin{align}
\frac{1}{2}(a+b+c+d)^2=&\frac{1}{2}(a^2+b^2+c^2+d^2)+ab+ac+ad+bc+bd+cd\\
=&\frac{1}{2}+ab+ac+ad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Mathematically Expressing the Sum of... I noticed that $$\large{2x-1 = \frac{x}{2} + \frac{\frac{x}{2}}{2} + \frac{\frac{\frac x2}2}2} + \cdots$$ until the output of one of the steps in the pattern equals $1$. Or in other words $2x-1$ is equal to the sum of $\large{\frac{x}{2}}$ plus that result divided by two plus tha... | The partial sum of this geometric series is $\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}$, if you start at k=0. Thus in your case it is $\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}-1$, because $2^0=1$.
$=\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}-\frac{1-\frac{1}{2}}{1-\frac{1}{2}}$=$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find the solutions to the equation $4\sin^2\theta + 1 = 6\sin\theta$ in the interval $0^\circ \leq \theta < 360^\circ$ Find the solutions to the following equation for $0^\circ \leq θ < 360^\circ$:
$$4\sin^2 θ + 1 = 6\sin θ$$
My work:
$$4\sin^2\theta - 6\sin\theta + 1 = 0$$
Factor
$$\sin\theta= \frac{1}{4}(3+ \sqrt{5})... | what do you mean? $\frac{3 -\sqrt 5}4 = 0.19098$ and $\theta = 11.01^\circ, 168.989^\circ.
$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the 2x2 matrix given 2 equations. Find the $2 \times 2$ matrix ${A}$ such that ${A}^2 = {A}$ and
${A} \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$
I have tried to express A as a matrix with variables a,b,c, and d, but it gets too messy with the equations. Any help is appreciated... | First, $A^2-A=0$ means $A$ must have eigenvalues 1 or 0. $A$ is not a zero matrix, so the two eigenvalues of $A$ cannot be all 0. The two eigenvalues cannot be all 1 either for otherwise $A$ would either be (1) the identity matrix which is certainly the case (2) a nondiagonalizable matrix but judging from the polynomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285881",
"timestamp": "2023-03-29T00:00:00",
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Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv... | just apply Chinese remainder theorem (See Burton)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 5
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Can $\sin(\pi/25)$ be expressed in radicals I suspect that sin(pi/25) is not expressible in elementary forms in radicals because it is the root of some quintic (or rather cos(pi/25) is). Can anyone prove that that particular quintic has no solution in radicals?
I know that (from chebyshev polynomials) if $$u=\cos{\frac... | As requested, R. Israel's Maple answer can be simplified. That complicated expression,
$$u = -\frac{-3125\sqrt{-10}+12500\sqrt{5}\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}+15625\sqrt{-2}-25000\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}}{2048(\sqrt{5}-1)^2\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}}$$
is just equal to,
$$u=\big(\tfrac{5}{4}\big)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$1989 \mid n^{n^{n^{n}}} - n^{n^{n}}$ for integer $n \ge 3$ Before anyone comments, yes this is kind of a duplicate of Prove that $1989\mid n^{n^{n^{n}}} - n^{n^{n}}$ . The problem that I'm having I don't see the $n=5$ as a counterexample. Also if anyone wants to know where I got this problem from here.
I'm looking at ... |
Lemma 1
$$ \forall n \in \mathbb{N}_{>0} \hspace{1.5em} n^{n^n} \equiv n^n \pmod3 $$
This is trivial.
Lemma 2
$$\forall n \in \mathbb{N} \hspace{1.5em} n\ge 3 \implies n^{n^n} \equiv n^n \pmod{16} $$
Proof. If $n$ is even, then $n\ge4$ and thus $16 \mid n^n, n^{n^n}$. If $n$ is congruent to $\pm 1$ modulo $16$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
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Find the limit of: $\lim_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$
Find the limit of: $\lim\limits_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$
I tried multiplying by the conjugate of $\sqrt{2x-1}-1$ ,but I obtain
$\frac{2x-2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{2x-1}+1\right)}$,
which is again z... | \begin{align}
L & = \lim_{x \to 1} \dfrac{\sqrt{2x-1}-1}{x^2-1} = \lim_{x \to 1} \dfrac{\sqrt{2x-1}-1}{x^2-1} \cdot \dfrac{\sqrt{2x-1}+1}{\sqrt{2x-1}+1}\\
& = \lim_{x \to 1} \dfrac{2x-1-1}{(x^2-1)(\sqrt{2x-1}+1)} = \lim_{x \to 1} \dfrac{2(x-1)}{(x^2-1)(\sqrt{2x-1}+1)}\\
& = \lim_{x \to 1} \dfrac2{(x+1)(\sqrt{2x-1}+1)} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Describe all odd primes p for which 7 is a quadratic residue I need to describe all odd primes $p$ for which $7$ is a quadratic residue.
Now let $\left(\frac{a}{b}\right)$ be the Legendre Symbol.
Then if $7$ is a quadratic residue $p$ we must have:
$$1=\left(\frac{7}{p}\right)=(-1)^{\frac{3(p-1)}{2}} \left(\frac{p}{7}\... | $$(-1)^n\equiv\begin{cases}
\hphantom{-}1 & \text{if }n\equiv 0\bmod 2\\
-1 & \text{if }n\equiv 1\bmod 2
\end{cases}$$
Since $p$ is odd, we know that $p-1$ is even, so $\frac{p-1}{2}$ is an integer.
By definition, $\frac{p-1}{2}\equiv 0\bmod 2$ if and only if $\frac{p-1}{2}=2x$ for some integer $x$, which is equivalent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given that $1 = 27 \times 11 - 74 \times 4$, solve the following equations in modulo $74$: $ 3x - y = 1$; $2x + 3y = 0$ Given that $1 = 27 \times 11 - 74 \times 4$, solve the following equations in modulo $74$:
$3x - y = 1$;
$2x + 3y = 0$.
Thank you.
| In mod $74$, we have
$$27\times 11\equiv 1\iff 3\times 27\times 11\equiv 3.$$
Then,
$$\begin{align}y\equiv 3x-1&\Rightarrow 3y\equiv 9x-3\\&\Rightarrow -2x\equiv 9x-3\\&\Rightarrow 11x\equiv 3\\&\Rightarrow 11x\equiv 3\times 27\times 11\\&\Rightarrow x\equiv 3\times 27\equiv 81\equiv \color{red}{7}\end{align}$$
Hence... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $7^{8^9}\mod 100$ I'm preparing myself for discrete math exam and here's one of the preparation problems:
Evaluate $$7^{8^9}\mod 100$$
Here's my solution:
$7^2\equiv49 \mod 100\implies (7^2)^2\equiv49^2=2401\equiv 1\mod 100$
So, as it turns out $7^4\equiv 1\mod 100$. It will be useful later. Now let's examine ... | Carmichael's function $\lambda(100) = \text{lcm}(\lambda(25),\lambda(4)) = \text{lcm}(20,2) = 20$.
$7$ is coprime to $100$, so $7^{8^9} \equiv 7^{8^9 \bmod 20} \bmod 100$
$\lambda(20) = \text{lcm}(\lambda(5),\lambda(4)) = \text{lcm}(4,2) = 4$
and $9 \equiv 1 \bmod 4$
$8$ has a higher exponent than $20$ in $2$, their o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to solve $12-\sin(\theta)=\cos(2\theta)$? $$12-\sin(\theta)=\cos(2\theta)$$
What's the correct answer on the $[0,2\pi]$?
I started with $12-\sin(\theta)=1-2\sin^2(\theta)$ and then i cant get anything sensible as i end up with $12=\sin(\theta)+2\sin^2(\theta)$
| Note that
\begin{eqnarray*}
12\sin \theta &=&\cos (2\theta ) \\
&=&1-2\sin ^{2}\theta
\end{eqnarray*}
then
\begin{equation*}
2\sin ^{2}\theta +12\sin \theta -1=0.
\end{equation*}
Let $X=\sin \theta ,$ then
\begin{equation*}
2X^{2}+12X-1=0
\end{equation*}
This quadratic equation have two roots: $X_{1}=\frac{1}{2}\sqrt... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 3
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Let $X$ be a set of primes $p$ so that $5^{p^2}+1 \equiv 0 \pmod {p^2}$ Which of these sets is $X$ equal to? $5^{p^2}+1\equiv 0\pmod {p^2}$
$1.$ $\emptyset $
$2.$ {$3$}
$3.$ All primes of the form $4k+3$
$4.$ All primes except $2$ and $5$
$5.$ All primes
This one is pretty easy to get right through the process of elimi... | In the congruance holds you also have : $5^{p^2}+1\equiv 0 [\rm mod p]$.
On the other hand $5^p\equiv 5 [\rm mod p]$ ($equation : x^p \equiv x [\rm mod p]$) so we should have $5+1\equiv 0 [\rm mod p]$ (apply equation twice).
So the only possible primes are 2 or 3. 2 is not a solution and 3 is a solution indeed:
$$5^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $a_1,a_2,a_3$ are roots $x^3+7x^2-8x+3,$ find the polynomial with roots $a_1^2,a_2^2,a_3^2$ If $a_1,a_2,a_3$ are the roots of the cubic $x^3+7x^2-8x +3,$ find the cubic polynomial whose roots are:
$a_1^2,a_2^2,a_3^2$ and the polynomial whose roots are $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}.$ Not really sure wh... | HINT:
For the first let $y=x^2$
$\implies x^2(x+7)=8x-3\implies y=x^2=\dfrac{8x-3}{x+7}$
Express $x$ in terms of $y$ and put the value of $x$ in the given cubic equation and simplify.
For the second, let $z=\dfrac1x\implies x=\dfrac1z$
and put the value of $x$ in the given cubic equation and simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$, partial fraction braindead decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$
the way my teacher wants us to solve is by substitution values for x,
I set it up like this:
(after setting the variables to the common denominator and getting rid of the denominator in the original equat... | If you substitute $x=2i$, this gives
$\;\;\;-1-4i=((2Ci+D)(2i-1)^2=(-3-4i)(2Ci+D)=(-3D+8C)-(6C+4D)i$
Therefore $8C-3D=-1\;\;$ and $\;\;6C+4D=4,\;$ so
$\hspace{.6 in}16C-6D=-2\;\;$ and $\;\;9C+6D=6\implies 25C=4\implies C=\frac{4}{25}$.
Then $D=1-\frac{3}{2}C=1-\frac{6}{26}=\frac{19}{25},\;\;$ and $\;\;A=-\frac{4}{25}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Trigonometry Identity proving If $\sin(x-y) =\cos y$ prove that $\tan y = \frac{1+ \sin y}{\cos y} $.
Is there an error with the question? I don't seem to be able to get the answer. Should it be $\tan x$ instead of $\tan y = \frac{1+\sin y}{\cos y}$ ?
$$\tan x\times \cos y = 1 + \sin y $$
$$\frac{\sin x}{\cos x} \time... | Since $\sin(x-y)= \cos(y)$, expanding it gives
\begin{equation*}
\sin (x) \cos(y)-\cos(x) \sin(y)=\cos(y).
\end{equation*}
Dividing both sides by $\cos(y)$ we have
\begin{equation*}
\sin(x)-\cos(x)\tan(y)=1.
\end{equation*}
So from here, we get
\begin{equation*}
\tan(y)=\dfrac{\sin(x)-1}{\cos(x)}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
First order differential equation: did i solve this equation right So i'm trying to solve:
$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
$$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$
I think i've solved it and got
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i... | $$x^2\frac{dy}{dx} + 2xy = y^3$$
Divide both sides by $x^2$
$$\frac{dy}{dx} + \frac{2}{x}y = x^{-2} y^3$$
Consider
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
We know that
*
*n = 3
*1- n = 1-3 = -2
*p(x) = $ \frac{2}{x}$
*q(x) = $x^{-2}$
*u = $y^{1-3} = y^{-2}$
Subbing these in...
$$
\frac{du}{dx} + (-2)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solution to $y'=y^2-4$ I recognize this as a separable differential equation and receive the expression:
$\frac{dy}{y^2-4}=dx$
The issue comes about when evaluating the left hand side integral:
$\frac{dy}{y^2-4}$
I attempt to do this integral through partial fraction decomposition using the following logic:
$\frac{1}{(... | Saturday morning: I do like robjohn's point about $\log |x|$ giving the student some incorrect expectations. What I wrote here is correct and careful, but the same conclusions come from dropping the absolute value signs and doing the calculations three times, $y < -2,$ $-2 < y < 2,$ $y > 2.$ At some point the vertical... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
decimal to fractions When being asked how to solve the Arithmetic Means of $8, 7, 7, 5, 3, 2,$ and $2$, I understand that adding these numbers then dividing by $7$ (the amount of numbers) gives me the decimal $4.85714...$ But when being asked to change this number into a fraction, I do not understand where the $\frac{6... |
Arithmetic mean $= \frac{8+7+7+5+3+2+2}{7} = \frac{34}{7}$
Now, we calculate $\frac{34}{7}$ by using long division and soon we will get a remainder of $6$, the quotient as $4$, and the divisor as $7$ ($x\frac{y}{z}$ ---- $y$ as the remainder, $x$ as the quotient, and $z$ as the divisor). As the result, we will get:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the Standard Matrix for Linear Transformation We are asked to find the standard matrix $A$ for $T$:
Consider the transformation $T : \mathbb{R^3} \rightarrow \mathbb{R^4}$ given by
$$T(x_1, x_2, x_3) = (x_1 + x_2 + x_3, x_2 + x_3, 3x_1 + x_2, 2x_2 + x_3)$$
for every $$(x_1,x_2,x_3) \in \mathbb{R^3}$$
I am confu... | We are asked to find the standard matrix $A$ for $T$:
Consider the transformation $T : \mathbb{R^3} \rightarrow \mathbb{R^4}$ given by
$$T(x_1, x_2, x_3) = (x_1 + x_2 + x_3, x_2 + x_3, 3x_1 + x_2, 2x_2 + x_3)$$
for every $$(x_1,x_2,x_3) \in \mathbb{R^3}$$
Your update is correct $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Basis of eigenspace For eigenvalue $1$ I did $(A-I)x=0$ which is
$$\left(\begin{array}{ccc}
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
2 & 2 & 1 & 0\\
-1 & 1 & -1 & 1
\end{array}\right).$$
I got row reduced echelon form
$$\left(\begin{array}{ccc}
1 & 0 & 0 & 1/3\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & -2/3\\
0 & 0 & 0 & 0
\end{array}\rig... | Your reduced row echelon matrix is not correct. It should be
$$RREF(A-I) = \left[\begin{array}{cccc}1 & 0 & 0 & 1\\0 & 1 & 0 & 0\\0 & 0 & 1 & -2\\0 & 0 & 0 & 0\end{array}\right]$$
Now to see how to find the eigenspace from this let's rewrite this matrix as a set of linear equations:
$$\left[\begin{array}{cccc}
1 & 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Bisector of two lines in the euclidean space $\mathbb{E}_3$ Let $$r: \begin{cases} x + z = 0 \\ y + z + 1 = 0\end{cases}$$ and $$s: \begin{cases} x - y - 1 = 0 \\ 2x - z -1 = 0\end{cases}$$
be two lines in the euclidean space $\mathbb{E}_3$. It is easily seen that their intersection is one point.
How do I find th... | The unit direction vectors of the first line and the second line are
$$\overline a=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$$
and
$$\overline b=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}\right),$$
respectively. (The scalar product of these two vectors is $0$. That is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove that $\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c$ How to prove that
\begin{equation*}\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c,\ where \ a,b,c>0\end{equation*}
I tried the following:
\begin{equation*}abc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge a+b+c\end{equation*}
Using Chebyshev's inequality
... | The inequality which you want to prove is symmetric, so we can assume without loss of generality that $a\geq b\geq c$.
Then $ab\geq ac\geq bc$ and $\frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a}$.
Thus, from the rearrangement inequality, we get that
$$\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq\frac{ab}{b}+\frac{ac}{a}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 1... | $\left(2x^2+x\right)^2=\left(2x^2+x\right)\left(2x^2+x\right)=4x^4+4x^3+x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
| The simplest, as we have a homographic function, is to write it in canonical form:
\begin{align*}
2<\frac{x}{x-1}\leq 3&\iff 2 < 1+\frac1{x-1}\leq 3 \iff 1< \frac1{x-1}\leq 2\\
&\iff \frac12\le x-1 <1 \iff \frac32 \le x <2
\end{align*}
The third equivalence is valid because all numbers at the end of the first line have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | Below is a simple way to implement the (Euclidean) denominator descent implicit in Ivo Terek's answer (which was John Conway's favorite way to present this proof).
Theorem $\quad \rm r = \sqrt{n}\:$ is an $\rm\color{#c00}{integer}$ if rational,$\:$ for $\:\rm n\in\mathbb{N}$
Proof $\ \ \ $ Put $\ \ \displaystyle\rm r ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 18
} |
Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.
What are all integral solutions of this equation?
I can only find $x=y=z=1$.
| From the identity
$$(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x)$$
we obtain $8 = (x+y)(y+z)(z+x)$. It follows that $(3−x)(3−y)(3−
z) = 8$. On the other hand,
$$(3−x)+(3−y)+(3−z)−3(x+y+z) = 6,$$
implying that either $|3−x|, |3−y|, |3−z|$ are all even, or exactly one of
them is even. In the first case, we g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Power Series Coefficients Find the sum of the coefficients of $x^{20}$ and $x^{21}$ in the power series expansion of $\frac 1{(1-x^3)^4}$.
I don't know a lot on power series at the moment, and I was wondering how do I find the coefficients?
Thanks
| Recall that
$$\frac1{1-s} = \sum_{n=0}^\infty s^n. $$
Differentiating we have
$$\frac{\mathsf d^3}{\mathsf ds^3}\left[\frac1{1-s}\right] = \frac6{(1-s)^4},
$$
and
$$\begin{align*}
\frac{\mathsf d^3}{\mathsf ds^3}\left[\sum_{n=0}^\infty s^n\right] &= \sum_{n=0}^\infty \frac{\mathsf d^3}{\mathsf ds^3}[s^n]\\
&= \sum_{n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Evaluate $\lim\limits_{n\to\infty}\prod\limits_{k=2}^{n}\frac{k^2+k-2}{k^2+k}$ I can't find the product of a sequence.
We have
$$\frac{(2+2)(2-1)}{2(2+1)}\frac{(3+2)(3-1)}{3(3+1)}...\frac{(k+2)(k-1)}{k(k+1)}$$
I am stuck with $$P=\frac{2(n+2)}{n^2(n-1)}$$ but that isn't correct. Can the squeeze theorem be used?
| $$\prod_{k=2}^{K}\frac{k^2+k-2}{k^2+k}=\prod_{k=2}^{K}\frac{k-1}{k}\prod_{k=2}^{K}\frac{k+2}{k+1}=\frac{1}{K}\cdot\frac{K+2}{3}$$
hence the limit equals $\color{red}{\displaystyle\frac{1}{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find the asymptotic behavior of solutions of the equation Find the asymptotic behavior of solutions $y$ of the equation
$$x^5 + x^2y^2=y^6,$$
which tends to $0$ when $x$ tends to $0$.
My solution: if $y=Ax^n$, then
$$x^5 + A^2x^{2+2n}=A^6x^{6n}.$$
If $2+2n\ge 5$, we should have $5=6n$, but $2+2\cdot5/6 \not> 5$.
If $2+... | Now lets use a correction term
$$y=\sqrt x + f$$
Insert and use only the first two terms in the binomial expansion
$$x^5+x^2(x+2x^{0.5} f)=x^3+6x^{2.5}f$$
$$x^5=4x^{2.5}f$$
$$f=\frac{x^{2.5}}{4}$$
Now, you may want to insert $y=\sqrt x +\frac{x^{2.5}}{4}+f$
Adding (or multiplying) a correction term and inserting into t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1314681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$\frac{1}{{9\choose r}} -\frac{1}{{10\choose r}} = \frac{11}{6{11\choose r}}$. Is there a way to find $r$ without using algebra?
$$\frac{1}{\dbinom 9r} -\frac{1}{{\dbinom{10}r}} = \frac{11}{6\times \dbinom{11}r}$$
I guess directly applying algebra for this problem would be enough. But are there any simpler and pretti... | This method is very simple :
$\frac{1}{{9\choose r}} -\frac{1}{{10\choose r}} =\frac{{10\choose r}-{9\choose r}}{{10\choose r}{9\choose r}}=\frac{\frac{{10\choose r}-{9\choose r}}{{9\choose r}}}{{10\choose r}}=\frac{(\frac{{10\choose r}-{9\choose r}}{{9\choose r}})\frac{11*6}{11-r}}{{10\choose r}\frac{11*6}{11-r}}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$
Solving for $n$ in the equation $$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$$
Can anyone show me a numerical method step-by-st... | Classic problem for Newton's Method.
Note that $F(n) = \dfrac{1}{2^n} + \dfrac{1}{4^n} + \dfrac{3^n}{4^n}-1$ is continuously differentiable. Furthermore, $F(0) = 2$ and $\lim_{n \rightarrow \infty} F(n) =-1$ so by the Intermediate Value Theorem, there is definitely a zero.
I used Maple to calculate the root to a toler... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
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Integrate $f(y)=\int_{0}^{\frac{\pi}{2}} \ln(y^2 \cos^2x+ \sin^2x) .dx$ This Integral came up while attempting another question:
$$f(y)=\int_{0}^{\frac{\pi}{2}} \ln(y^2 \cos^2x+ \sin^2x) .dx$$
The suggested solution was as follows:
$$f'(y) = 2y \int_{0}^{\pi/2}\frac{cos^{2}x}{sin^{2}x + y^{2}cos^{2}x}dx$$
$$= 2y \in... | Your calculation of $2y\int_0^{\pi/2}\sec^2 x/(\tan^2 x + y^2)\, dx$ is correct so far, and the value is $\pi$. Now
\begin{align}2y\int_0^{\pi/2} \frac{\tan^2 x}{\tan^2 x + y^2}\, dx &= 2y\int_0^{\pi/2} \left(1 - \frac{y^2}{\tan^2 x + y^2}\right)\, dx \\
&= 2y\cdot\frac{\pi}{2} - y^2\cdot 2y\int_0^{\pi/2} \frac{1}{\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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a limit problem? $$\lim_{x\to 0}{e^x+ \ln{1-x\over e }\over \tan x-x} $$
i tried doing l'hospital. but couldn't do ! could anyone help ?
| This solution uses the standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}} &=&\frac{1}{6}
\\
\lim_{x\rightarrow 0}\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}} &=&-\frac{1}{%
3} \\
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3}.
\end{eqnarray*}
Re-write the original ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} -... | Your extraction of $x$ is incorrect: what you need is
$$ \sqrt{x^2+3x} = \sqrt{x^2(1+3/x)} = x\sqrt{1+3/x}. $$
Then you have
$$ \lim_{x \to \infty} \frac{3x-1}{x(\sqrt{1+3/x}+\sqrt{1+1/x^2})} = \lim_{x \to \infty} \frac{3}{\sqrt{1+3/x}+\sqrt{1+1/x^2}} - \frac{1}{x(\sqrt{1+3/x}+\sqrt{1+1/x^2})} $$
The square roots in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$ How does one calculate:
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$
Is the best way to just take the first term times the following two, and the second two times the next two to see the pattern?
First ter... | Instead of working forward, work backward:
Calculate the $z^2$ term of the result, then the $z^3$ term, and so on.
Clearly the coefficients of the $z^0$ and $z^1$ terms are zero. The $z^2$ term must be the result of squaring the $\frac z{2!}$ term, so the result is $\frac{z^2}{2!^2}$.
Now let's consider the $z^3$ ter... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to retrieve the expression of $a^3+b^3+c^3$ in terms of symmetric polynomials? I recently had to find without any resources the expression of $a^3+b^3+c^3$ in terms of $a+b+c$, $ab+ac+bc$ and $abc$.
Although it's easy to see that $a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$, I couldn't come up myself with $a^3+b^3+c^3=(a+b+... | In terms of the products of the three polynomials, the only possible way of getting an $a^3$ term is from $$(a+b+c)^3=a^3+b^3+c^3+3\sum a^2b+6abc$$
Where the sum is of all the expressions of the same kind. To obtain $a^2b$ from the symmetric polynomials, you need $a\cdot ab$ which comes from $$(a+b+c)\cdot(ab+bc+ca)=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$
Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$
Values $a,b,c$ are all positive reals. I tried Muirhead and a few AM $\geq$ GM.
This problem is equivalent to proving $a^4b^3 + b^4c^3 + c^4a^3 \geq c^2a^5 + b^2c^5+a^2b^5$.
| The inequality does not change if $(a, b, c)$ are permuted
cyclically, therefore we can assume
that $a \le b \le c$ or $a \ge b \ge c$.
In the first case $a \le b \le c$ we can
apply the Rearrangement inequality to
$$
x_1 = a^4, x_2 = b^4, x_3 = c^4 \\ y_1 = a^3, y_2 = b^3, y_3 = c^3 \, .
$$
The rearrangement inequalit... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is it true that two $3 \times 3$ matrices in $R$ are similar $\iff$ they have same determinant and same trace? Is it easy to cite some example of $3 \times 3 $ non-similar matrices of real entries with equal determinant and trace?
| Even simpler examples than Chappers' one would be
$$ \begin{pmatrix} 0&0&0\\0&0&0\\0&0&0 \end{pmatrix} \quad\text{and}\quad
\begin{pmatrix} 0&0&1\\0&0&0\\0&0&0 \end{pmatrix} $$
(determinant and trace both $0$, but the zero matrix is similar only to itself) or
$$ \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1 \end{pmatrix} \quad\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Points of intersection of two functions Through the following steps I found the x-coordinates of the intersection points of two functions:
$(x)= -x^{2}+3x+1\: and\: g(x)=3/x $
The numbers I found are x=(1, 2, 3)
But on the graph, one of the points has a negative x value, could you guys point me to anything I have mi... | $x(-x^{2}+3x+1)=3\quad$ doesn't imply $x=3$ or $-x^{2}+3x+1=3$. Instead of that we have
$$x(-x^{2}+3x+1)=3\iff -x^3+3x^2+x-3=0\iff-(x+1)(x-1)(x-3)=0$$
Therefore the graphs of the functions mets at $x=-1$, $x=1$ and $x=3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equa... | $$1^3 +2^3 +\ldots +n^3 =\left(\frac{n(n+1)}{2}\right)^2.$$
We will prove by induction on $n$, that
$$\sum_{k=1}^n k^3 = 1^3 +2^3 +\ldots +n^3 =\frac{n^2 (n+1)^2}{4}.$$
For $n=1$, we have $1^3 = \dfrac{1^2 2^2}{4} = 1$.
We shall prove that
$$\sum_{k=1}^{n+1} k^3 = 1^3 +2^3 +\ldots +n^3 +(n+1)^3 =\frac{(n+1)^2 (n+2)^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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List the elements of the field $K = \mathbb{Z}_2[x]/f(x)$ where $f(x)=x^5+x^4+1$ and is irreducible Since $\dim_{\mathbb{Z}_2} K = \deg f(x)=5$, $K$ has $2^5=32$ elements. So constructing the field $K$, I get:
\begin{array}{|c|c|c|}
\hline \text{polynomial} & \text{power of $x$} & \text{logarithm} \\\hline
0 & 0 & -\... | Elaborating more on the answer of Burde and generalizing:
Let $F$ be any finite field with $q$ elements, and $f(x)$ an irreducible polynomial from $F[X]$, say of degree $m$. Then the quotient ring by the principal ideal generated by $f(x)$ is in fact a field, has $q^m$ elements and the elements of this field is the set... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the definite integral: $\int_0^1 x\sqrt{px + 1}\,dx$ with $p>0$?
How to integrate $$I = \int_0^1 x\sqrt{px + 1}\,dx$$ , $p>0, p\in\mathbb{R}$ ?
I tried this:
$$ t = \sqrt{px + 1} \implies x = \frac{t^2 - 1}{p}$$
$$ x = 0 \implies t = 1$$
$$ x = 1 \implies t = \sqrt{p+1}$$
so I have:
$$ I = \int_1^\sqrt{p+1} \... | $$I(p)=\int_{0}^{1}x\sqrt{px+1}\,dx = \frac{1}{p^2}\int_{0}^{p}x\sqrt{x+1}\,dx $$
but integration by parts gives:
$$ \int x\sqrt{x+1}\,dx = \frac{2}{15}(3x-2)(1+x)^{3/2} $$
hence:
$$ I(p) = \frac{4}{15p^2}+\frac{2}{15}\sqrt{1+p}\left(3+\frac{1}{p}-\frac{2}{p^2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\prod\limits _{r=3 }^{\infty }{ \frac { { r }^{ 3 }-{ 8 } }{ { r }^{ 3 }+{ 8 } } } $ How should I go about evaluating this product? I have not been able to figure out.
$$\lim_{n\to\infty}\prod _{r=3 }^{n }{ \frac { { r }^{ 3 }-{ 8 } }{ { r }^{ 3 }+{ 8 } } } $$
| We have
$$\begin{align}
\prod_{n=3}^{N}\frac{n^3-8}{n^3+8}&=\prod_{n=3}^{N}\frac{(n-2)(n^2+2n+4)}{(n+2)(n^2-2n+4)}\\\\
&=\prod_{n=3}^{N}\frac{(n-2)}{(n+2)}\prod_{n=3}^{N}\frac{(n+1)^2+3}{(n-1)^2+3}\tag 1
\end{align}$$
Now, analyzing the individual partial products we see that the first partial product reduces to
$$\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$ then quotient $\frac xy$ is equal to?
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$, then quotient $\frac xy$ is equal to?
Other conditions ... | $$(2x+i)(1-i\sin 3\alpha)=(y+i)(1+i\sin \alpha) \Rightarrow (2x+\sin 3\alpha)+i(1-2x\sin 3\alpha) = (y-\sin\alpha)+i(1+y\sin \alpha)$$
Equate the imaginary parts,
$$\Rightarrow \frac{x}{y}=-\frac{\sin \alpha}{2\sin 3\alpha}=-\frac{\sin \alpha}{2(\sin \alpha \cos 2\alpha+\cos \alpha \sin 2\alpha)}$$
Use $\sin2\alpha =2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $ without using L'hopital's rule Find $$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $$ without using L'hopital's rule. Using the rule, I got it as 1/12. What's a way to do it without L'hopital?
| Integration by parts is a reasonable choice. Since:
$$ \int \frac{t}{t^4+4}\,dt = \frac{1}{4}\arctan\left(\frac{t^2}{2}\right)\tag{1}$$
we have:
$$ \int_{0}^{x}\frac{t\log(1+t)}{t^4+4}=\frac{\log(1+x)}{4}\arctan\left(\frac{x^2}{2}\right)-\frac{1}{4}\int_{0}^{x}\frac{\arctan\frac{t^2}{2}}{1+t}\,dt \tag{2}$$
and by consi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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How to proceed with evaluating $\int\frac{dx}{\sqrt{9+4x^2}}$ and $\int\tan^2(3x)dx$
*
*$\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}$
*$\displaystyle\int\tan^2(3x)dx$
For the first one i'm not sure if I did it correctly, here is what I did:
Let $2x=3\tan(t)$, so $x=\frac{3}{2}\tan(t)$ and $dx=\frac{3}{2}\sec^2(t)dt... | Put $3x=u
$ to get $$\frac{1}{3}\int\tan^{2}\left(u\right)du=\frac{1}{3}\int\left(\sec^{2}\left(u\right)-1\right)du=\frac{1}{3}\tan\left(u\right)-\frac{u}{3}+C=\frac{1}{3}\tan\left(3x\right)-x+C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Quadrature on segment Is there a quadrature formula on the segment [0,1], such that on [0,1/2] the points and weights are symetric with respect to 1/4, on [1/2,1] they are symetric with respect to 3/4, and such that the formula exactly integrates polynomials of degree 5. If not, what maximal order can be expected ?
| A simple example of such quadrature is
$$
x = \frac{1}{2} + \frac{1}{4}\begin{pmatrix}
-1 - \sqrt\frac{3}{5}&
-1&
-1 + \sqrt\frac{3}{5}&
1 - \sqrt\frac{3}{5}&
1&
1 + \sqrt\frac{3}{5}
\end{pmatrix}\\
w = \frac{1}{36}\begin{pmatrix}5 & 8 & 5 & 5 & 8 & 5\end{pmatrix}
$$
which is just a combination of two three-point Gauss... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $f\circ f$ if $f(t)=\dfrac{t}{(1+t^2)^{1/2}},\ \ t\in \mathbb{R}$
Find $f\circ f$ if $f(t)=\dfrac{t}{(1+t^2)^{1/2}},\ \ t\in \mathbb{R}$
$ a.)\ \dfrac{1}{(1+2t^2)^{1/2}} \\
\color{green}{ b.)\ \dfrac{t}{(1+2t^2)^{1/2}}}\\~\\
c.)\ (1+2t^2)\\~\\
d.)\ \text{none of these} \\$
I tried to put the value $t=1$ and ... | Our formula for $f$ is
$$f(\Box)=\frac{\Box}{\sqrt{1+\Box^2}}$$
With this in mind we have
\begin{align*}
f\bigl(f(t)\bigr)
&=f\left(\fbox{$f(t)$}\right) \\
&=\frac{\fbox{$f(t)$}}{\sqrt{1+\fbox{$f(t)$}^2}} \\
&=\frac{\frac{t}{\sqrt{1+t^2}}}{\sqrt{1+\left(\frac{t}{\sqrt{1+t^2}}\right)^2}} \\
&=\frac{t}{\sqrt{1+\frac{t^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given primitive solution to $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, show $a+b$ is a perfect square If $a,b,c$ are positive integers and
$\gcd(a,b,c)$ is $1$. Given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then prove that $a+b$ is a perfect square.
I was trying to get something useful from the information given in... | Let $d=(a,b)$ be the g.c.d. of $a$ and $b$. We will show that $a+b=d^2$. Write $a=a_1d$ and $b=b_1d$. You also have $(a_1,b_1)=1$. We thus have $$\frac{1}{a}+\frac{1}{b}=\frac{1}{d}\left(\frac{1}{a_1}+\frac{1}{b_1}\right)=\frac{1}{c}$$$$\frac{a_1+b_1}{a_1b_1}=\frac{d}{c}$$ Observing that $(d,c)=1$ and $(a_1+b_1,a_1b_1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding an isomorphism between polyomial quotient rings Let $F_1 = \mathbb{Z}_5[x]/(x^2+x+1)$ and $F_2 = \mathbb{Z}_5[x]/(x^2+3)$. Note neither $x^2+x+1$ nor $x^2+3$ has a root in $\mathbb{Z}_5$, so that each of the above are fields of order 25, and hence they are isomorphic from elementary vector space theory, howeve... | Hint $\ $ Complete the square $\ 4(x^2\!+\!x\!+\!1) = (2x\!+\!1)^2\!+\!3 = X^2\!+\!3,\ $ for $\,X = 2x\!+\!1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ t... | Another approach:
$$\text{Euler's formula: }\quad a^{\phi(n)}\equiv 1 \pmod{n} \text{ when} \gcd(a,n)=1$$
$$\phi(100)=\phi(2^2)\phi(5^2)=2\cdot 20=40$$
$$\gcd(100,33)=1$$
$$33^{40}\equiv 1 \pmod {100}$$
$$33^{100}\equiv (33^{40})^{2}3^{20}\equiv 3^{20}\pmod {100}$$
$$\equiv (3^5)^4\equiv43^4\equiv 49^2\equiv 01\pmod {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
| $$(1 - \omega + \omega^2)(1 + \omega - \omega^2)$$
$$=((1 + \omega^2) - \omega)(1 + \omega - \omega^2)$$
$$=(-\omega - \omega)(-\omega^2 - \omega^2)$$
$$=(-2\omega)(-2\omega^2)$$
$$=4\omega^3=4$$
Where $1+\omega+\omega^2=0$ and $\omega^3=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Trigonometry - log/ln and absolute sign in equations Will this equation still hold if the absolute sign is being used at different places
For example,
This trigonometry identity;
$\frac{-1}{3}log|\frac{cos3x+1}{cos3x-1}|=\frac{2}{3}ln|(\frac{sin\frac{3x}{2}}{\cos\frac{3x}{2}})|$.
My question is
Does $\frac{-1}{3}log\... | $$1+cos(x)=2cos^(\frac{x}{2})\\1+cos(x)=2sin^2(\frac{x}{2})$$ so $$\frac{|cos(3x)+1|}{|cos(3x)-1|} =|\frac{cos(3x)+1}{cos(3x)-1}|=\frac{|2cos^2(\frac{3x}{2})|}{|-2sin^2(\frac{3x}{2})|}$$ as we know $|ab|=|a||b| \rightarrow |-a|=|-1||a|=|a|$
$$\frac{|2cos^2(\frac{3x}{2})|}{|-2sin^2(\frac{3x}{2})|}=\frac{|2cos^2(\frac{3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $(z^1+z^2+z^3+z^4)^3$ using Pascals Triangle? In an exercise it seems I must use Pascal's triangle to solve this $(z^1+z^2+z^3+z^4)^3$. The result would be $z^3 + 3z^4 + 6z^5 + 10z^ 6 + 12z^ 7 + 12z^ 8 + 10z^ 9 + 6z^ {10} + 3z^ {11} + z^{12}$. But how do I use the triangle to get to that result? Personally... | The expression factors as
$$z^3(1+z)^3(1+z^2)^3=z^3(1+3z+3z^2+z^3)(1+3z^2+3z^4+z^6).$$
Then I see no better way than to perform the multiply (though there is a symmetry)
$$\begin{align}
&1+3z+&3z^2+&z^3\\
&&3z^2+&9z^3+&9z^4+&3z^5\\
&&&&3z^4+&9z^5+&9z^6+&3z^7\\
&&&&&&z^6+&3z^7+&3z^8+z^9\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Quadratic polynomials describe the diagonal lines in the Ulam-Spiral I'm trying to understand why is it possible to describe every diagonal line in the Ulam-Spiral with an quadratic polynomial $$2n\cdot(2n+b)+a = 4n^2 + 2nb +a$$ for $a, b \in \mathbb{N}$ and $n \in 0,1,\ldots$.
It seems to be true but why?
Wikipedia s... | Reading georgmierau comment, I've been trying to reach this "arbitrary" 8, so here what I've done:
Let $A(n)$ be the cardinal number of the $n^{th}$ ring in Ulam's spiral.
$$A(1)=1\\A(2)=8=(2\cdot2-1)^2-1^2\\A(3)=16=(2\cdot3-1)^2-3^2\\ \vdots \\ A(n)=(2n-1)^2-(2n-3)^2\Rightarrow A(n)=8n-8, n\neq1$$
So, $A(n+1)-A(n)=8$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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What the difference between the smallest two numbers from these numbers? There are infinitely many integers $n$ bigger than $1$, such that if we divide $n$ by any integer $k$ where $2\leq k\leq 11$, the remainder is equal to $1$.
What the difference between the smallest two such integers?
Help guys please. this is my ... | What you wan can be set up as:
$x\equiv1\bmod 2$
$x\equiv 1\bmod 3$
$x\equiv1\bmod 4$
$x\equiv 1\bmod 5$
$x\equiv1\bmod 6$
$x\equiv 1\bmod 7$
$x\equiv1\bmod 8$
$x\equiv 1\bmod 9$
$x\equiv1\bmod 10$
$x\equiv 1\bmod 11$
which is equivalent to
$x\equiv 1 \bmod 2^3$
$x\equiv 1 \bmod 3^2$
$x\equiv 1 \bmod 5$
$x\equiv 1 \bmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simultaneous equation with fractional solutions. How do you get to find $x$ when $y$ is a fraction ? Anyone mind to explain it step by step for the clearest explanation.=)
$$-7x +2y = 2$$
$$14x + 3y = -5$$
Answer: $x=?, y=-1/7$
| The correct solution to your system of equations is $y = -\frac{1}{7}$ and you want to find $x$. To do so, substitute this value of $y$ into the first equation to get $$-7x - 2\cdot \frac{1}{7} = 2 \iff -7x = \frac{16}{7} \iff x = -\frac{16}{49}$$ by re-arranging and solving for $x$.
Alternatively, you could substitute... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding $\sum\limits_{k=0}^n k^2$ using summation by parts Sorry to bother you guys again, but I still have some doubts. I do think I'm making some progress, though.
So, again, the formula that I'm using for summation by parts is
$\sum\limits_{k=o}^n f(k)g(k) = g(n)(\sum\limits_{k=0}^n f(k)) - \sum\limits_{k=0}^{n-1} ... | As you probably know:
$$\sum_{k=0}^{n-1} k = \frac{(n-1)^2+(n-1)}{2} = \frac{n^2-2n+1+n-1}{2} = \frac{n^2-n}{2}$$
You also know:
$$\sum_{k=0}^{n}k^2=n^2+\sum_{k=0}^{n-1}k^2 \rightarrow \sum_{k=0}^{n-1} k^2=\sum_{k=0}^{n} k^2 - n^2$$
So now you can substitute:
$$2\sum_{k=0}^n k^2=n^3+n^2-\left(\sum_{k=0}^n k^2-n^2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Let $a$, $b$, and $c$ be positive real numbers. Let $a$, $b$, and $c$ be positive real numbers. Prove that
$$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$
Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?
| An Algebraic Solution:
Note that $$\sqrt{a^2-ab+b^2}=\sqrt{\left(a-\frac{b}{2}\right)^2+\left(\frac{\sqrt{3}b}{2}\right)^2}$$ and $$\sqrt{a^2-ac+c^2}=\sqrt{\left(\frac{c}{2}-a\right)^2+\left(\frac{\sqrt{3}c}{2}\right)^2}\,.$$ Using the Triangle Inequality in the form $\sqrt{\sum_{i=1}^nx_i^2}+\sqrt{\sum_{i=1}^ny_i^2}\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What are the ways to solve trig equations of the form $\sin(f(x)) = \cos(g(x))$? if I have the following trig equation:
$$\sin(10x) = \cos(2x)$$
I take the following steps to solve it:
*
*I rewrite $\cos(2x)$ as $\sin\left(\frac{\pi}{2} + 2x\right)$ or as $\sin\left(\frac{\pi}{2} - 2x\right)$ cause $\sin\left(\frac{... | You may prefer to transform the sine into cosine:
$$
\cos\left(\frac{\pi}{2}-10x\right)=\cos(2x)
$$
This splits into two:
$$
\frac{\pi}{2}-10x=2x+2k\pi
$$
or
$$
\frac{\pi}{2}-10x=-2x+2k\pi
$$
The trick is that $\cos\alpha=\cos\beta$ if and only if $\alpha=\beta+2k\pi$ or $\alpha=-\beta+2k\pi$ (with integer $k$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How do I evaluate this integral :$\int \frac{1}{ \sqrt{1-x^2} } \arctan( \frac{\sqrt{1-x^2}}{2})dx$? Is there somone who can show me how do I evaluate this integral :$$\int \frac{1}{ \sqrt{1-x^2} } \arctan\left(\frac{\sqrt{1-x^2}}{2}\right)dx$$
Note: I took : $x=\cos t$ , but it didn't work
Thank you for any help .
| For the integral
\begin{align}
I = \int \frac{1}{ \sqrt{1-x^2} } \, \arctan\left(\frac{\sqrt{1-x^2}}{2}\right) \, dx
\end{align}
let $u^{2} = 1 - x^{2}$ to obtain
\begin{align}
I = - \, \int \tan^{-1}\left(\frac{u}{2}\right) \, \frac{du}{\sqrt{1-u^{2}}}.
\end{align}
Using $2 \tan^{-1}(x) = \ln(1 - ix) - \ln(1+ix)$ then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
Let $0\le a\le b\le c,abc=1$, then show that
$$a+b^2+c^3\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}$$
Things I have tried so far:
$$\dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}=\dfrac{b^2c^3+ac^3+ab^2}{bc^2}$$
Since $abc=1$, it suffices to ... | we can write the inequality as
$$\left(a-\dfrac{1}{c^3}\right)+\left(c^3-\dfrac{1}{a}\right)\ge\dfrac{1}{b^2}-b^2$$
or
$$(c^3a-1)\left(\dfrac{1}{a}+\dfrac{1}{c^3}\right)\ge\dfrac{1-b^4}{b^2}\tag{1}$$
since $c\ge 1,bc\ge 1$,then
$$c^3a\ge c\cdot abc=c\ge 1$$
so it only prove $b\le 1$case.
$1$.case if $b^2\ge a$，then we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $A$, $B$ are $3\times 3$ matrices, and all elements are different from each other and greater in their absolute value than 3, then is $AB \ne 0$?
Let $A$ and $B$ be two $3\times 3$ matrices. All entries of $A$ are distinct and all entries of $B$ are distinct. All entries in both matrices are greater in their absolu... | $$\begin{pmatrix} 4 & 8 & 12 \\ 5 & 10 & 15 \\ 7 & 14 & 21 \end{pmatrix}
\begin{pmatrix} -28 & -35 & -49 \\ 8 & 10 & 14 \\ 4 & 5 & 7 \end{pmatrix}
=
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
The idea was to start with
$$\begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Way to make this homogoneous ODE seperable? Is my algebra correct, turning this:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4y-3x}{2x-y}$$
Into this:
I split into the difference of the two fractions,
then factored x out of the left fraction, and factored y
out of the right fraction, getting:
$$\frac{4\frac{y}{x}}{x(1-\... | The usual way of doing such a question is to divide the top and bottom of the fraction by $x$ yielding $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4\frac{y}{x} - 3}{2 - \frac{y}{x}}$$ and then use the substitition $v = \frac{y}{x} \implies y = vx$ so that $\frac{\mathrm{d}y}{\mathrm{d}x} = v + x\frac{\mathrm{d}v}{\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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