Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Derivative of $\sec^{-1}x$ and integral of $\frac{1}{x\sqrt{x^2-1}}$ My attempt is as follows:-
*
*Derivative of $\sec^{-1}x$
Let $\theta=\sec^{-1}x,$ where $\theta\in [0,\pi]-{\dfrac{\pi}{2}}.$
$$\sec\theta=x.$$
Differentiating both sides with respect to $x:$
$$\sec\theta\cdot\tan\theta\cdot\dfrac{\mathrm d\theta}{\... | To avoid $\pm$ signs, proceed as follows:
*
*Derivative of $\sec^{-1}x$. Let $\theta=\sec^{-1}x,$ with $\theta\in [0,\pi]-{\dfrac{\pi}{2}}$. Then, $\sin \theta >0$
$$\cos\theta=\frac1{\sec\theta}=\frac1x\implies \sin\theta\cdot\dfrac{\mathrm d\theta}{\mathrm dx}=\frac1{x^2}\\
\dfrac{\mathrm d\theta}{\mathrm dx}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How many five-digit numbers with distinct digits can be formed from the digits $0, 1, 2, 3, 4, 5, 6, 7$ under the given conditions?
You can use the digits $0,1,2,3,4,5,6,7$.
You have to make a number of $5$ digits, above $9999$ (no $0$ in the start).
All digits should be different from each other, the number must have... | You have to take the conditions into account both when you count the five-digit strings and when you count the five-digit strings that begin with $0$.
The answer in the text is incorrect.
Method 1: We use your method.
Five-digit numbers of the form _ _ _ $25$ that use three even and two odd digits, all of which are dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Tricky complex contour integration Show that $$\int_{0}^{2\pi} \frac{\cos^2{(\theta)}}{13-5\cos{(2\theta)}} \; d\theta = \frac{\pi}{10}$$ using complex contour integration.
I let $z = e^{i\theta}$ which means $dz = ie^{i\theta} \; d\theta = iz \; d\theta$ (meaning we know integrate over the unit circle $C$). Hence,
$$\... | On the last step, you should have
$$I=\frac i2\oint_C\frac{(z^2+1)^2}{\color{red}z(z^{\color{red}2}-5)(5z^{\color{red}2}-1)}~\mathrm dz$$
Now use the two roots in the unit circle from $5z^2-1=0\Rightarrow z=\pm\sqrt5/5$ and $z=0$, which are first order poles, to get
$$I=\frac\pi{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int\frac{dx}{(1+\sqrt{x})(x-x^2)}$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}
$$
Set $\sqrt{x}=\cos2a\implies\dfrac{dx}{2\sqrt{x}}=-2\sin2a.da$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}=\int\frac{-4\sin2a\cos2a.da}{2\cos^2a.\cos^22a.\sin^22a}\\
=\int\frac{-2.\sec^2a.da}{\sin2a\cos2... | Note that there is an inconsistency. The solution you listed is not for the integral posted. The integral instead should be
$$
I=\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}
$$
If so, first use the substitution $t=\sqrt x$ to rewrite it as
$$I=\int\frac{2dx}{(1+t)\sqrt{1-t^2}}$$
Then, let $u=\frac{1-t}{1+t}$ and the integr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int \frac{\cos^2x}{1+\tan x}dx$ Evaluate $\int \dfrac{\cos^2x}{1+\tan x}dx$
Here are my various unsuccessful attempts:-
Attempt $1$:
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
$$\ln(1+t)=y$$
$$\dfrac{1}{1+t}=\dfrac{dy}{dt}$$
$$\int \dfrac{dy}{\left(1+(e^y-1)^2\right)^2}$$
$$... | You may continue and finish the second approach as follows,
$$I=\dfrac{1}{4}\int\dfrac{3\cos x+\cos3x}{\cos x+\sin x}dx$$
Rewrite the term $\cos3x $ in the integrand as
$\cos3x = (\cos2x-\sin2x)(\sin x+ \cos x)+\sin x$ and the integral becomes
$$I=\dfrac{1}{4}\int\left(2+\cos2x-\sin2x+\dfrac{\cos x-\sin x}{\cos x+\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How many ways can we get a number by addition if each part of the addition has to be smaller or equal to a set value? For example, if we need to get 5 with the largest number we can use being 3, we can use:
*
*3 + 2
*3 + 1 + 1
*2 + 2 + 1
*2 + 1 + 1 + 1
*1 + 1 + 1 + 1 + 1
Is there any way to find out the soluti... | If you wish to find the number of ways to express a positive integer $n$ as the sum of $m$ positive integers ($m\le n$) and each of the positive integers must be less than or equal to a certain value $\alpha$ then the number of such integers will be the number of positive integral solutions to these equations:
$x_1+x_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)? If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?
$$5x^2+ax+b=0?$$
I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confu... | Since the roots are $4/5\pm\sqrt{31}/5$, we have that $x^2+(a/5)x+b/5=(x-(4/5+\sqrt{31}/5))(x-(4/5-\sqrt{31}/5))$.
So, $x^2+(a/5)x+b/5=x^2-(8/5)x+(16/25-31/25)=x^2-(8/5)x-3/5$.
Hence $a=-8, b=-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Calculate $2^{5104} \bmod 10$ using mental arithmetic I am practising interview for Jane Street's trader internship and I found the following question.
Question: Calculate $2^{5104} \bmod 10$ using mental arithmetic.
I know that $2^5 \bmod 10 \equiv 2 \bmod 10.$
So,
\begin{align*}
2^{5104} & = (2^5)^{1020} 2^4 \\
... | The Chinese Remainder Theorem says:
$$2^{5104}\equiv x \pmod{2\cdot 5} \iff \begin{cases}2^{5104}\equiv x\pmod 2 \\ 2^{5104}\equiv x\pmod 5\end{cases}$$
It follows that $x\equiv 0\pmod 2$ and $x\equiv 1\pmod 5$.
Consequently $x\equiv 6\pmod{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
Factoring $1 - y-x^2-y^2-yx^2+y^3$ I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?
$$1 - y-x^2-y^2-yx^2+y^3$$
Answer: $$ \ ( 1+y)[(1-y)^2... | Here is how I factored it:
$$1-y-x^2-y^2-yx^2+y^3=$$
$$(1-y^2)-(y-y^3)-(x^2+yx^2)=$$
$$(1-y^2)-y(1-y^2)-x^2(1+y)=$$
$$(1-y^2)(1-y)-x^2(1+y)=$$
$$(1-y)(1+y)(1-y)-x^2(1+y)=$$
$$(1+y)(1-y)^2-(1+y)x^2=$$
$$\boxed{(1+y)[(1-y)^2-x^2]}$$
You can factorize it further:
$$(1+y)[(1-y)^2-x^2]=$$
$$(1+y)[(1-y-x)(1-y+x)]=$$
$$\boxed... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Solve the irrational radical equation Solve equation:$$\frac{x + \sqrt{3}}{\sqrt{x} + \sqrt{x + \sqrt{3}}} + \frac{x - \sqrt{3}}{\sqrt{x} - \sqrt{x - \sqrt{3}}} = \sqrt{x}$$
answer: $S=\{2\}$
Attemp:
$$\frac{x+\sqrt{3}}{\sqrt{x}+\sqrt{x+\sqrt{3}}}+\frac{x-\sqrt{3}}{\sqrt{x}-\sqrt{x-\sqrt{3}}}=\sqrt{x}\Leftrightarrow \... | When you inserted factors to obtain a common denominator, you made the error
$$\left( \sqrt{x}+\sqrt{x+\sqrt{3}} \right) \left( \sqrt{x}-\sqrt{x-\sqrt{3}} \right) \neq -\sqrt{3} \text{.} $$
That you did not get the same denominator in both fractions should have highlighted a problem.
Since we require $x - \sqrt{3} \geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Vector, Matrix multiplication notation Given a vector $v=\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix}$ and matrix $M=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$ is there a standard notation for scaling each row of $M$ by the corresponding element of $v$. e.g.
$$\begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix} ... | You could do it by
$$\begin{bmatrix} 1&0&0 \\0& 2&0 \\0&0& 3\end{bmatrix}
\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix} = \begin{bmatrix} 1 & 2 & 3\\ 8 & 10 & 12 \\ 21 & 24 & 27\end{bmatrix}$$
easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$
If real numbers $a$, $b$, $c$, $d$ satisfy
$$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$
then find $(a,b,c,d)$.
What I try:
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, pl... | You can observe this as quadratic equation on $a$ :
$$8a^2-8a(1+b) +(8b^2+8c^2 -8bc+7)=0$$ which has to have a solution so the discriminant must be nonnegative:
$$64(1+b)^2-32(8b^2+8c^2 -8bc+7)\geq 0$$
so $$6b^2-4b(1+2c)+8c^2+5\leq 0$$
Now again, the discriminat must $\geq 0$ since that inequality has a solution...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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evaluate the integral $\int x^3(5x^2+3)^5 dx$ How to evaluate for the integral:
$$\int x^3(5x^2+3)^5 dx$$
What I've tried:
For example,
(1) $u = x^2$ whence $du = 2x$ $dx$
(2)$u = ax^2+b$ whence $du = 2ax$ $dx$
Therefore,
(3) $u = 5x^2+3$ whence $du = 10x$ $dx$
$$I = \int \frac{du}{10}u^5 = \frac{1}{10} \int \frac{u^6... | $$\int x^3(5x^2+3)^5 dx=\frac{1}{5}\int(5x^3+3x-3x)(5x^2+3)^5 dx=$$
$$=\frac{(5x^2+3)^7}{5\cdot7\cdot10}-\frac{3(5x^2+3)^6}{5\cdot6\cdot10}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Lagrange multipliers... Three variables. Let $f:\mathbb{R}^3\mapsto \mathbb{R}$ be defined by
\begin{equation*}
f(x,y,z) := x+3y+5z.
\end{equation*}
Compute the minimum of $f$ subject to the constraint that
\begin{equation*}
g(x,y,z) = x^2+y^2+z^2-1 = 0.
\end{equation*}
The method of Lagrange multipliers is as follows:... | Yes, your computations and your result are correct. Using the Cauchy-Schwarz inequality, we can also obtain the same result very quickly:
Note that (where $\langle\cdot,\cdot\rangle$ is the euclidean inner product and $\|\cdot\|$ is the euclidean norm) $$f(x,y,z)=\left\langle(x,y,z),(1,3,5)\right\rangle \text{ and that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Nested radical of Ramanujan and cyclic inequality Inspired by a nested radical of Ramanujan I propose this :
Let $a,b,c\in (2\sqrt[4\,]{5},6-2\sqrt[4\,]{5})$ such that $a+b+c=9$ then we have :
$$\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{b+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\fr... | Remark: For applications of the Symmetric Function Theorem, also see
Prove that $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+ab+b^2}}\geq\frac{2}{\sqrt{ab+ac+bc}}+\sqrt{\frac{a+b+c}{3(a^3+b^3+c^3)}}$
How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$
Proof: We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Is this solution correct for equation $3x^2-4y^2=13$? Prove equation $3x^2-4y^2=13$ has no integer solution.
Solution: Suppose (y, 3)=1, We have:
$y≡( 1, 2) \mod (3)$
⇒ $4y^2≡( 1, 2) \ mod(3)$
$3x^2≡0 \mod (3)$
The common remainder between $3x^2$ and $4y^2$ is 0 , so we may write:
⇒ $3x^2-4y^2≡ 0 \ mod(3)$
But, $13≡1 \... | Here is another method:
$3x^2-4y^2=13$
Clearly x must be odd; let $x=2k+1$, then:
$3x^2-4y^2=12k^2+12k+3-4y^2=4(3k^2+3k-y^2)=10$
⇒$2(3k^2+3k-y^2)=5$
LHS is even but RHS is odd, so the equation can not have integer solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$
I should solve the following system: $$\begin{cases} x^3-y^3=19(x-y) \\
x^3+y^3=7(x+y) \end{cases}$$
by reducing the system to a system of second degree.
We can factor:
$$\begin{cases} (x-y)(x^2+xy+y^2)=19(x-y) \\
(x+y)(x^2-xy+y^2)=7(x+y) \end{cases}... | You can divide by variables if you ensure they are not zero. Here, you can consider the cases $x=y$ and $x=-y$ first. If $x=y$ the first equation is trivial and the second becomes $2x^3=14x$ or $x=0,\pm \sqrt 7$. You can do the same for $x=-y$ and find a pair of solutions.
Then decree that $x+y \neq 0, x-y \neq 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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number of real solution in exp equation
Find number of real values of $x$ in
$$3^x (3^x-1)=|3^x-1|+|3^x-2|$$
What i try
Put $\displaystyle 3^{x}-\frac{1}{2}=t.$ Then
$$\bigg(t-\frac{1}{2}\bigg)\cdot \bigg(t+\frac{1}{2}\bigg)= \bigg|t-\frac{1}{2}\bigg|+\bigg|t-\frac{3}{2}\bigg|$$
So $$t^2-\frac{1}{4}=|t-0.5|+|t-1.... | If $3^x > 0$ and $|A| + |B| \ge 0$ so $3^x(3^x-1) \ge 0$ so $3^x-1 \ge 0$. So $x \ge 0$.
If $x=0$ then LHS $= 0$ and RHS $=0 + 1= 1$. So $x = 0$ is not a solution.
Case 1: If $0 < x < \log_3 2$ then $3^x -1 > 0$ but $3^x - 2 < 0$ so
$3^x(3^x-1) = (3^x -1) + (2 - 3^x)$. Let $a = 3^x$ and solve for $a(a-1)=(a-1)+(2-a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\frac{\sqrt{\sqrt[4]8-\sqrt{\sqrt2+1}\;}}{\sqrt{\sqrt[4]8+\sqrt{\sqrt2-1}\;} -\sqrt{\sqrt[4]8-\sqrt{\sqrt2-1}\;}}=\frac1{\sqrt2}$ Days ago, I tried to demonstrate this equality, reducing radicals, multiplying by the conjugate of the denominator, etc. But, I did not reach anything similar to the right side.
$$
\fr... | Apply the denest formula $\sqrt{a-\sqrt c}=\sqrt{\frac{a+\sqrt{a^2-c}}2}-\sqrt{\frac{a-\sqrt{a^2-c}}2} $ to the numerator
$$N=\sqrt{\sqrt[4]8-\sqrt{\sqrt2+1}}=\sqrt{\frac{\sqrt[4]8+\sqrt{\sqrt2-1}}2}-\sqrt{\frac{\sqrt[4]8-\sqrt{\sqrt2-1}}2}=\frac1{\sqrt2}D$$
where $D$ is the denominator.
Alternatively, evaluate
$$D^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that the following recursive series is bounded and monotonic I need to show that the following recursive series $$a_1=1, \quad a_{n+1}=\frac{a_n+3}{5}$$ is bounded and monotonic.
At first, I want to show that the series is bounded from below with $\frac{3}{4}$. By induction:
For $n=1$ we have $a_1=1>\frac{3}{4}$. ... | $$
\begin{align}
a_{n+1}-\frac34
&=\frac{a_n+3}5-\frac34\\
&=\frac{a_n-\frac34}5
\end{align}
$$
Thus, if $a_n\ge\frac34$, then $a_{n+1}\ge\frac34$. Therefore, since $a_1=1$, $a_n\ge\frac34$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation. A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation.
My attempt is as fol... | A different method (but not much shorter):
The centre of the circle also lies on the line $3x+4y=k$ for some $k\in\mathbb{R}$. Solving with $x+y=2$, the centre is $(8-k,k-6)$. The radius is $\dfrac{|4(8-k)-3(k-6)+4|}5=\dfrac{|54-7k|}5$.
Therefore, $\displaystyle (8-k)^2+(k-6-1)^2=\frac{(54-7k)^2}{25}$.
$25(2k^2-30k+113... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Finding the minimum of a discrete function from a continuous version In another post is asked to minimize
$$n+\left\lceil\frac dn\right\rceil$$ where $n$ is a positive integer ($d$ is also an integer, but we can relax this condition).
We obviously have
$$n+\dfrac dn\le n+\left\lceil\frac dn\right\rceil< n+\dfrac dn+1$$... | Let $f(n,d)=n+\lceil\frac{d}{n}\rceil$.
Let $m(d)$ denote the minimum value of $f(n,d)$ for positive integers $n$. The function $x+\frac{k^2}{x}$ has minimum $2k$ over the reals and $x+\lceil\frac{k^2}{x}\rceil\ge x+\frac{k^2}{x}$, so $m(k^2)=2k$.
Note that $f(n,d+1)=f(n,d)$ or $f(n,d)+1$, so $m(d+1)=m(d)$ or $m(d)+1\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$? How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$?
\begin{align*}
2\cos x(\sin x)^2+(\cos x)^2 & = \frac{1}{2}(\cos x- \cos 9x) +(\cos 4x)^2\\
4\cos x(1-(\cos x)^2)+2(\cos x)^2 & = \cos x + \cos 9x +(2(\cos 4x)^2-1)+1\\
\cos x(... | I put this to Wolfram Alpha with $x=\cos\theta$ and got an output of
$$(2 x - 1) (2 x + 1) (x - 1) (x + 1) (32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6 x - 1)$$
It it returned roots of
$$x\in\{-0.841254,-0.415415,0.142315,0.654861,0.959493\}$$
for the last factor, so you need to solve a quintic to get all the roots it seems.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there som... | Since $y=\frac{1}{x}$ is convex we have:-
$\dfrac15+\dfrac16+\dfrac17>\dfrac36=\dfrac12$
$\dfrac18+\dfrac19+\dfrac1{10}+\dfrac1{11}+\dfrac1{12}>\dfrac5{10}=\dfrac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Rewriting inequality given by a fraction so I got the following solution for a problem:
$u(x,)=\begin{cases}
1 & x\leq y \\
\sqrt{\frac{x-1}{y-1}} & 1< x < y\\
0 & x\geq 1
\end{cases} \tag{1}$
and the master solution is
$u(x,)=\begin{cases}
1 & x\leq y \\
\sqrt{\frac{1-x}{1-y}} & y< x<1\\
0 & x\geq 1
\end{cases} \tag{2... | No, it's true:$$\sqrt{\frac{1-x}{1-y}}=\sqrt{\frac{-(1-x)}{-(1-y)}}=\sqrt{\frac{x-1}{y-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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The value of the series: $\sum_{p=1}^{n} \sin(\frac{p}{n^2})$
Find the limit as $n\rightarrow \infty$ of:
$$\sum_{p=1}^{n} \sin\left(\frac{p}{n^2}\right)$$
This question seems odd to me because it was included in my differentiability problems set.
My Attempt:
We have: $\sin(x) \le x $, Thus clearly:
$$\sum_{p=1}^{n}... | Maple says
$$
\sum_{p=1}^n\sin\left(\frac{p}{n^2}\right) =
{\frac {\sin \left( \frac{1}{n} \right) \cos \left( \frac{1}{n^2} \right) +
\left( \cos \left( \frac{1}{n} \right) -1 \right) \sin \left( \frac{1}{n^2}
\right) -\sin \left( \frac{1}{n} \right) }{2\,\cos \left( \frac{1}{n^2}
\right) -2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Compute $\sigma$, given $\sigma^{11}$ in $S_{10}.$
Let $\sigma \in S_{10}$ with $\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)$. How to compute $\sigma$?
I tried:
$\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 9 & 1 & 3 & 2 & 8 & 10& 7 & 6 & 5\end{pmatrix}$
Now I tried to fin... | The first thing to note is that $11$ is $-1$ modulo $3$ and $4$ modulo $7$. In particular, we note that $(1,3,4)^{11} = (3,1,4)$, so if we can find an element $\tau$ of the symmetric group on $\{2,5,6,7,8,9,10\}$ whose eleventh power is $(5,2,9,6,8,7,10)$, we're done. Now, we note that the $n$th power of a $k$ cycle is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore... | Binomial theorem:
$\begin{align*}
\frac{1}{x^2}
&= (2 + (x - 2))^{-2} \\
&= \frac{1}{4}
\cdot \left(
1 + \frac{1}{2}(x - 2)
\right)^{-2} \\
&= \frac{1}{4}
\cdot \sum_{k \ge 0}
\binom{-2}{k}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to solve the ordinary differential equation $ x^2 y'' - 2 x y' + 2y = x^4 \mathrm{e}^x $ Please tell me how to solve this differential equation.
$$ x^2 y'' - 2 x y' + 2y = x^4 \mathrm{e}^x $$
I tried to solve it but finally I stuck tell me how to go further more if anyone gives me a hint(I want to know how to fin... | Operators and first and second integrals are two methods to find a solution. This solution uses the series solution.
Let
$$y(x) = \sum_{n=0}^{\infty} a_{n} \, x^n$$
which easily leads to
\begin{align}
x^2 \, y'' - 2 x y' + 2 y &= x^4 \, e^{x} \\
\sum_{n=0}^{\infty} a_{n} \, (n(n-1) - 2n + 2) \, x^n &= \sum_{n=0}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Combination of problem having identical objects without using generating function approach
A person goes in for an examination in which there are $4$ papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is
My Attempt
Number of ways of partitioning $n$ identical object... | $\#$ of ways $n$ identical objects can be divided into $r$ distinct groups if empty groups are allowed is:${}^{n+r-1}C_{r-1}$
Req. # of ways=$${}^{2m+4-1}C_{4-1}-\Big[\text{#{> m for paper 1}+#{> m for paper 2}+#{> m for paper 3}+#{> m for paper 4}}\Big]\\+\Big[\text{#{strictly m for paper 1}+#{strictly m for paper 2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding limit for given function
If $f(x)=\lim\limits_{n \to \infty} \frac{x}{n} \left(\frac{1}{1+e^{-x/n}}+\frac{1}{1+e^{-2x/n}}+\dots+\frac{1}{1+e^{-x}} \right)$
then what is $\lim\limits_{x\to 0} \left(\frac{2f(x)-x}{x^2}\right)$?
My solution is :
$f(x)=0$ since $\lim\limits_{n \to \infty} \frac{x}{n}$ part wi... | Your answer is not correct because $f$ is not identically zero.
Note that the definition of $f$ is the limit of a Riemann sum:
$$\begin{align}f(x)&=\lim\limits_{n \to \infty} \frac{x}{n} \left(\frac{1}{1+e^{-x/n}}+\frac{1}{1+e^{-2x/n}}+\dots+\frac{1}{1+e^{-x}} \right)\\&=x\int_0^1\frac{dt}{1+e^{-tx}}=\ln\left(\frac{e^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let $X,Y$ be independent normally distributed random variables. Find the density of $\frac{X^2}{Y^2+X^2}$ Let $X,Y$ be independent standard normally distributed random variables and $X,Y\neq 0$. Find the density of $\frac{X^2}{Y^2+X^2}$
I was given the tip of first calculating the density of $(X^2,Y^2)$ and then calcul... | Observe that the distribution of the vector $(X,Y)$ is rotationally invariant. That is, if we rotate it at any angle, the resulting vector has the same distribution as $(X,Y)$.
Why? Rotation $\alpha$ radians counterclockwise gives the vector $(X\cos \alpha-Y\sin \alpha,X\sin \alpha+Y \cos \alpha)$, which is centered ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Help with proof by induction, please! I may not be understanding how to work with factorials.
Prove that for all integers $n \ge 1, 1\cdot1! + 2\cdot2! + 3\cdot3! + \cdots + n\cdot n! = (n+1)! - 1$.
OK, so I verified base case $n = 1: 2! - 1 = 1$
I assumed that for all $k \ge n, P(k) = (k+1)! - 1$
I am unsure how to ... | Hint: If $f(n) = 1*1! + 2*2! + ..... + n*n!$ then $f(n+1) = 1*1! + 2*2! + ..... + n*n! + (n+1)(n+1)! = f(n) + (n+1)(n+1)!$.
Hint 2: $m!(m+1) = (m+1)!$ and $m!*m + m! {=\over{\text{factor}}} m!(m+1) = m!(m+1)= m!$.
So if we assume $f(k) = 1*1! + 2*2! + ..... + k*k!= (k+1)! - 1$.
.
Then $f(k+1) = f(k) + (k+1)(k+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Solving the inequality $1+1.2 + 1.2^2 + 1.2 ^3 + \cdots + 1.2^n < N/16?$ Given the inequality $$1+1.2 + 1.2^2 + 1.2 ^3 +\cdots + 1.2^n < \frac{N}{16}?$$
I need the value of $n$, or just an approximation. $N$ is known.
| $1+1.2+1.2^2+\ldots +1.2^n = \frac{1.2^{n+1}-1}{1.2-1}$ $= 5(1.2^{n+1}-1)$.
So you want to find the values of $n$ such that the inequality
$$5(1.2^{n+1}-1) < \frac{N}{16}$$
This gives
$$1.2^{n+1} < \frac{N}{80} - 1$$
Taking logs this gives
$$n \log 1.2 < \log \left(\frac{N}{80}-1\right)$$
which as $\log 1.2 < .2$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?
| The standard method uses partial fractions decomposition.
Here is how it begins for the first integral: factorout the denominator into irreducible factors with high-school identities:
$$x^3+1=(x+1)(x^2-x+1)$$
Newt proceed to decompose into partial fractions:
$$\frac{1}{(x^3+1)^2}=\frac{1}{(x+1)^2(x^2-x+1)^2}=\frac A{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find all integer values of $m$ such that the equation $\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$ has exactly four distinct real roots.
Find all integer values of $m$ such that the equation $$\large \sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$$ has exactly four distinct real roots.
$\left(x \in [0, 9], m \in ... | $\sqrt{9-x}=\sqrt{3m-x^2+9x}-\sqrt x$
Equations $\sqrt{9-x}=\sqrt x$ has no solution. For the moment we assume that equation $\sqrt{3m-x^2+9x}=\sqrt x$ has also no solution. Therefore the roots of equation $\sqrt{9-x}=\sqrt{3m-x^2+9x}$ must be satisfied by $\sqrt x$; considering $x∈[0, 9]$, we may write for integer x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How can we get all positive integer solution for these two equations While solving a problem i was able to reduce the question to finding positive integer solutions that satisfy the equations:-
$x+y+z = 2020$ and $x^2 + y^2 + z^2 = xyz+4$.
I found
$$(x,y,z)= (2,1009,1009),\ (1009,2,1009) \text{ and }(1009,1009,2)$$
as... | You can write second equation as $(x+y+z)^2-2(xy+yz+xz)=xyz+4$ that is $2020^2-4= xyz+ 2(xy+yz+xz)$. Now it could be useful to write RHS as a product. You can calculate $(2+x)(2+y)(2+z)=8+4(x+y+z)+ 2(xy+yz+xz)+xyz=8088+ 2(xy+yz+xz)+xyz$.
So we can write $(2+x)(2+y)(2+z)=8088+2020^2-4=2020^2+4\cdot 2020+4=2022^2$. Are y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How many integer solutions are there to $x+y+z=8$
How many integer solutions are there to $x+y+z=8$ When $x,y,z>0$? When $x,y,z\geq -3$?
So I know there is a formula for computing the number of nonnegative solutions
${8+3-1 \choose 3-1}={10\choose 2}$
So I then just subtracted cases where one or two integers are $0$.... | Start by looking at the number of solutions of
\begin{eqnarray*}
X+Y+Z=17
\end{eqnarray*}
which has $\binom{17+3-1}{3-1}$ solutions.
This is the same as the number of solutions of
\begin{eqnarray*}
x+y+z=8
\end{eqnarray*}
where $x=X-3$,$y=Y-3$ and $z=Z-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$. Let $A$be a $5 \times 5$ matrix whose characteristic polynomial is given by $(\lambda -2)^3(\lambda+2)^2.$ If $A$ is diagonalizable,find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$
Solution:
Characteristic polynomial of $A$,$Ch_{A}(\lambda)=(\lambda -2)^3(... | Here's another way of solving your problem:
We're given that the characteristic polynomial of $A$ is $c_A(x)=(x-2)^3(x+2)^2$. Since $A$ is diagonalizable, that implies that the minimal polynomial must be $m_A(x)=(x-2)(x+2)$. Hence we have that
$$\begin{align*}
0 &= m_A(A)\\
&= (A-2I)(A+2I)\\
&= A^2-4I
\end{align*}$$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$ Let $a,$ $b,$ and $c$ be positive real numbers such that $a + b^2 + c^3 = \frac{325}{9}.$ Find the minimum value of
$a^2 + b^3 + c^4.$
| Note that for the surface $a^2 + b^3 + c^4=k$ to have the minimum value $k$, it is tangential to the surface $a + b^2 + c^3 = \frac{325}{9}$. Calculate their normal vectors
$(1,2b,3c^2)$ and $(2a,3b^2,4c^3)$, and match them to get
$$\frac{2a}1=\frac{3b}2=\frac{4c}3$$
Then, plug $b= \frac43a$ and $c=\frac32a$ into $a +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Solving $15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1$ I need help solving this logarithm exercise:
$$15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1$$
What I've done is re-writing the equation
$$\Rightarrow \qquad 5^{\log_5(3)}\cdot 3^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1 \tag{1}$$
Then applying logarithms on both sides
$$\Rightarro... | $15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1\\
x^{\log_5(9x)+1}=15^{-\log_5(3)}$
Take the $\log_5$ of both sides
$(\log_5(9x)+1)\log_5 x =(-\log_5(3))(\log_5 15)\\
(\log_5 x + \log_5 9+1)\log_5 x =(-\log_5(3))(\log_5 15)\\
(\log_5 x)^2 + (\log_5 9+1)\log_5 x + \log_5(3)(\log_5 3 + 1) = 0\\
$
let $u = \log_5 x, b = 2\log_5 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Let $x∈R$ and suppose that$ |x−2| < 1/3$. Prove that $|x^2+x−6| < 2$. I need help studying for a midterm coming up and am doing this problem for practice:
Let $x∈R$ and suppose that $|x−2| < 1/3$. Prove that $|x^2+x−6| < 2$.
If $|x-2| < 1/3, x < 7/3$ and $x > 5/3$.
$|x^2 + x - 6| = |x-2||x+3| < 2$.
I'm really not sur... | Let $x \in \mathbb R$ and suppose that $|x−2| < \dfrac 13$. Prove that $|x^2+x−6| < 2$.
\begin{align}
|x-2| < \dfrac 13
&\implies \dfrac 53 < x < \dfrac 73 \\
&\implies 3+\dfrac 53 < x+3 < 3 + \dfrac 73 \\
&\implies -\dfrac{16}{3} < x+3 < \dfrac{16}{3} \\
&\implies |x+3| < \dfrac{16}{3} \\
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find : $\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$ Find :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
My attempt : i don't know is correct or no!
I use this rule :
$$\lim\limit... | Assuming that you would like to get more than the limit itself, starting with
$$a_n=\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$ take logarithms
$$\log(a_n)=n^4\log\left(1+\frac{1}{n^3}\right)-(n+1)^4\log\left(1+\frac{1}{(n+1)^3}\right)$$ Use twice the expansion
$$\log(1+x)=x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 3
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Find $\lim\limits_{x\rightarrow 0^+}\frac{1}{\ln x}\sum\limits_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}$
Find $$\lim\limits_{x\rightarrow 0^+}\dfrac{1}{\ln x}\sum_{n=1}^{\infty}\dfrac{x}{(1+x)^n-(1-x)^n}.$$
Consider $$f(x):=(1+x)^n,$$
By Lagrange MVT, we can obtain
$$\frac{2x}{f(x)-f(-x)}=\frac{1}{f'(\xi)}, -x\gtrles... | If $x>0$, then by the mean value argument
$$
\frac{x}{{(1 + x)^n - (1 - x)^n }} > \frac{1}{{2n(1 + x)^{n - 1} }}.
$$
Assume now that $0<x<1$. Then
$$
\frac{1}{{\log x}}\sum\limits_{n = 1}^\infty {\frac{x}{{(1 + x)^n - (1 - x)^n }}} < \frac{1}{{\log x}}\sum\limits_{n = 1}^\infty {\frac{1}{{2n(1 + x)^{n - 1} }}}
\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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proof that there doesn't exist integers $a, b, c>0$, that satisfy $a^2bc+2$, $ab^2c+2$ and $abc^2+2$ are all square numbers. please proof that there doesn't exist a,b,c which are all positive integers, that satisfy a^2bc+2, ab^2c+2 and abc^2+2 are all square numbers.
I know squares remain 1 or 0 when divided by 4, so ... | You idea will work. We just need to be careful.
Suppose all $3$ $a^2bc + 2$ and $ab^2c + 2$ and $abc^2+2$ are square numbers.
Then all $a^2bc, ab^2c, abc^2$ will be $\equiv 2,3 \pmod 4$.
Now suppose $a$ is even. Then $a^2 \equiv 0\pmod 4$. And $a^2bc \equiv 0 \pmod 4$ . That is a contradiction. So $a$ is odd.
The s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$... | Call the function you're minimizing $f(a,b)$. Note that the first three terms are homogeneous of order $2$ in $a$ and $b$, while the fourth is homogeneous of order $1$.
Thus $$f(ta, tb) = t^2 \left(a^2 + b^2 + \frac{a^2 b^2}{(a+b)^2}\right) - 2 t \frac{a^2+ab+b^2}{a+b} = t^2 g(a,b) + t h(a,b)$$
where $g \ge 0$, and th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$ I want to solve a limit without l'Hospital, just with algebraic manipulation:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}$$
I started with:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{(3-\sqrt{6+x})(3+\... | Clearly, it is enough to compute$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac x6}{x-3}\tag1$$and, in order to compute that, it will be enough to compute$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}\text{ and }\lim_{x\to3}\frac{\frac x6-\frac12}{x-3},\tag2$$since $(1)$ is equal to$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Stability Function system Hi I have this question
For the system
$$\begin{align}
\dot p_1 &= -p_1 + 6p_2 \\
\dot p_2 &= -7p_2+(p_1+p_2)\cos p_1
\end{align}
$$
Use the function $L(p)=\frac{1}{2}(p_1^2 + p_2^2)$ ,and show that the origin is locally asymptotically stable.
What I have done so far:
*
*Differentiated $... | Since
$$x_1^2 + x_2^2 \ge 2x_1x_2 \implies x_1x_2 \le \frac{1}{2}x_1^2+\frac{1}{2}x_2^2$$
for all real numbers $x_1,x_2$, we have
\begin{align}\dot V(x)&=-x_1^2+x_1x_2+(x_1x_2+x_2^2)\sin x_1-3x_2^2\\&\le
-x_1^2+\frac{1}{2}x_1^2+\frac{1}{2}x_2^2+\left(\frac{1}{2}x_1^2+\frac{1}{2}x_2^2+x_2^2\right)|\sin x_1|-3x_2^2\\&=
-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Interesting thing about sum of squares of prime factors of $27$ and $16$. Let
$$n=p_1×p_2×p_3×\dots×p_r$$
where $p_i$ are prime factors and
$f$ is the functions
$$f(n)=p_1^2+p_2^2+\dots+p_r^2$$
If we put $n=27,16$ and $27=3×3×3$, $16=2×2×2×2$ then
$$\begin{split}f(27)&=3^2+3^2+3^2=27\\f(16)&=2^2+2^2+2^2+2^2=16.\end{sp... | For two factors $$f(pq)=p^2+q^2\gt pq$$ so $pq$ is not a solution.
For three factors: If $3$ is a factor then $3^2+p^2+q^2$ is only a multiple of $3$ if $p=q=3$ as well. If $3$ is not a factor then $p^2=q^2=r^2=1\pmod3$, so the sum is a multiple of $3$, and $pqr$ is not a solution. So $27$ is the only solution with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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Complex numbers is an algebraic extension of Reals so what real Polynomial has root x+iy If I have an arbitrary complex number $x+iy$ then how do I work out a polynomial in the ring of polynomials over $\mathbb{R}$ for which it is a root?
| $z= (x+yi)$
$z-x = yi$
$(z-x)^2 = -y^2$
$z^2 -2xz + (x^2 + y^2)$ is a polynomial that has $x+yi$ as a root.
..... or .....
$P(z) = az^2 + bz + c$ then the root of $P$ is $\frac {-b\pm\sqrt{b^2-4ac}}{2a}$ so we let $2a=1$ need $x=-b$ and $b^2-4ac=b^2+2c= -y^2$
So let $b=-x$, $a=\frac 12$ then $c=\frac {y^2 + x^2}2$.
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int \frac{1}{1+x+x^4}dx$
Find $$\int \frac{{\rm d}x}{1+x+x^4}.$$
WA gives a result, which introduces complex numbers. But according to the algebraic fundamental theorem, any rational fraction can be integrated over the real number field.
How to do this?
| Why did you have to choose this rational function... ? Well there is a factorization like this:
$$1+x+x^4=\left(x^2-\sqrt{\frac{\sqrt[3]{\frac{1}{2} \left(9+i
\sqrt{687}\right)}}{3^{2/3}}+\frac{4}{\sqrt[3]{\frac{3}{2} \left(9+i \sqrt{687}\right)}}}
x+\frac{1}{2} \left(\frac{\sqrt[3]{\frac{1}{2} \left(9+i
\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim _{x\to \infty }\left(x\left(\arctan2x\:-\arccos\left(\frac{1}{x}\right)\right)\right)$ Find $$\lim _{x\to \infty }\left(x\left(\arctan(2x)-\arccos\left(\frac{1}{x}\right)\right)\right)$$
My idea was to let $t=\frac{1}{x}$ then I get
$$\lim _{t\to 0 }...$$
but i did not know what to do with $\arctan(2x)$
an... | The parantheses goes to $0$, so using $\lim\limits_{x\to 0} \dfrac{\sin x}{x} = 1$, the limit equals:
$$\lim _{x\to \infty }\left(x\left(\arctan 2x-\arccos \frac{1}{x}\right)\right)=\lim_{x\to \infty}x\left(\sin\left(\arctan 2x-\arccos\frac{1}{x}\right)\right)$$
and using the trigonometric identities:
$$\sin(a-b)=\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cd... | You can apply the Ratio Test and use $\displaystyle{\lim_{k\to\infty}\left(\frac{k+1}{k}\right)^k=e}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Value of $\arg(\sin \theta +i\cos \theta)$
If $\displaystyle \frac{3+i\sin \theta}{4-i\cos \theta}$ is purely real number , where $\theta \in [0,2\pi].$ Then what is $\arg(\sin \theta +i\cos \theta)$?
What I tried:
\begin{align*}
\frac{3+i\sin \theta}{4-i\cos \theta} & =\frac{(3+i\sin \theta)(4+i\cos \theta)}{(4-i\co... | So your work is good until the third from last line. You forgot the arctan.
So your solution should be $\arctan(\frac{-4}{3})$ now remember arctan is an odd function so $\arctan(\frac{-4}{3})=-\arctan(\frac{4}{3})$ which will be a negative number and you want $\theta \in [0,2 \pi]$ So we add $\pi$ to get what we want... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A geometric inequality for acute triangle $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ I am trying to prove $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ for an acute triangle. There is:
$$\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt{3\sum\frac{b^2+c^2-a^2}{a^2+2bc}}$$
So we have to prove: $\sum\fra... | We can end the beautiful Atticus's idea also by the following C-S:
$$\sum_{cyc}\frac{x}{\sqrt{\frac{y^2+z^2}{2}+x^2+yz}}=\sum_{cyc}\frac{x\sqrt6}{\sqrt{(1+2)(2x^2+(y+z)^2}}\leq\sum_{cyc}\frac{x\sqrt6}{\sqrt2x+\sqrt2(y+z)}=\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdot \ldots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$
Evaluate $$\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$$
Let $$y=x + \frac{1 \cdot 3 \cdot}{2!} x^2 + \frac{1 \cdot 3 \cdot 5}{3!} x^3+\ldots$$ be the given expression.(replaci... | $$S=\sum_{r=1}^{\infty} \frac{1.3.5....(2r-1)}{r!}\left(\frac{2}{5}\right)^r$$
$$\implies S=\sum_{r=1}^{\infty} \frac{(2r)!}{r!~ r!} 5^{-r} =\sum_{r=1}^{\infty} {2r \choose r} 5^{-r}=\frac{1}{\sqrt{1-4/5}}-1=\sqrt{5}-1.$$
Here we have used $$\sum_{k=0}^{\infty} {2k \choose k} x^k=\frac{1}{\sqrt{1-4x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3581695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Range of $f(x) = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $ Here's what I did :
$$\text{Let }\quad y = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $$
$$\text{Let } \sin^2x=t$$
$$\Rightarrow\ (y-1)t^2-(y+1)t+2y+1=0 $$
$$\text{Since } t= \sin x\text{ is real,} $$
$$\text{Discriminant} \geqslant 0 $$
$... | Note that to find the maximum value of the range you can just simply reason as follows:
Max($\sin^2(x)) = 1$ = Max($\sin(x)$). Hence the max is
$$y_{\text{max}} = \frac{1+1-1}{1-1+2} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Hopf map $S^3 \to S^2$ stereographic projection I'm reading up on this topic and I've seen lots of literature suggest that the complex number representation of the stereographic projection of $S^2$ can be expressed as
$\frac{\zeta^1 + i\zeta^2}{1-\zeta^3} = \frac{x^1 + ix^2}{x^3 +ix^4}$
where
$\zeta^1 = 2(x^1x^3 + x^2... | Write points of $S^3$ as things in $\Bbb C^2$, i.e., pairs $(z, w)$ where $|z|^2 + |w|^2 = 1$. Then the Hopf-map (to $\Bbb C \cup \{ \infty \}$) is defined by
$$
(z, w) \mapsto z/w
$$
Now the codomain (the extended complex plane) isn't exactly $S^2$, so you need to stereographically project a point $p + q i$ up to the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use Taylor series to compute $\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $ Use Taylor series to solve $$\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $$
This equals $$\lim\limits_{x \to 0} \left( \sum_{n=0}^{\infty}\frac{(2n+1)!}{x^{2n+1}(-1)^{n+1}} - \sum_{n=0}^{\infty}\f... | I'm not quite sure how you got your formula. Using just the first $2$ terms of the Taylor series for $\sin(x)$ (you actually only need just $1$ term, as shown in some of the other answers, but I thought using $2$ may help to better see what is going on), such as given in Trigonometric functions, and as suggested in cop... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solving for $x$ in a (modified) quadratic I've come across this question which can be simplified (with a generous amount of manipulation) to a quadratic. Here it is:
$$\left(\sqrt{49+20\sqrt6}\right)^{\sqrt {a \sqrt{a\sqrt{a\cdots \infty}}}}+(5-2\sqrt{6})^{x^2+x-3-\sqrt{x\sqrt{x\sqrt{x}\cdots \infty}}}=10 $$
$ \text{ ... | There is nothing wrong but if $x=\pm\sqrt 2$ then $a=-1<0$ so $\sqrt{a\sqrt{a\dots}}$ is not well-defined. Hence, $x=\pm\sqrt 2$ must be "disqualified" as a solution.
Usually it is always good to check all solutions you got with the original equation since we usually only prove the direction that if $x$ satisfies then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{x^2}{(7x^3+2)^2} dx$ using u substitution Section 5.2
Can somebody verify my solution? Thanks!!
Evaluate $\int \frac{x^2}{(7x^3+2)^2} dx$ using u substitution
Let $u=7x^3+2$. Then $\frac{du}{dx}=21x^2$ and so $\frac{du}{21x^2}=dx$.
Thus we have:
\begin{align}
& \int \frac{x^2}{(7x^3+2)^2} \, dx \... | I worked it out and I got the same answer you did.
One thing to point out: When you have $$\dfrac {du}{dx} = 21 x^2$$ you can write the subsitution as either $$\dfrac {du}{21 dx} = x^2$$ if you prefer to use $\dfrac {du}{dx}$ notation or $$\dfrac {1}{21}du = x^2 dx$$ if you prefer differential notation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Line Integral with Change of Variables I'm a bit rusty on my computational math, and genuinely can't solve this question which is frustrating me.
QUESTION:
$$\int_Csin(y)dx+xcos(y)dy$$ where C is the ellipse defined as follows: $x^2+xy+y^2=1$
MY ATTEMPT:
Define variable $u= \sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y}$
Defi... | Hint: Use Green's theorem instead to ease out calculations:
$\int_C Pdx+Qdy=\int\int_R (\partial Q/\partial x-\partial P/\partial y) dx dy$
where $R$ is the region bounded by closed curve $C$ on $xy-$ plane oriented counterclockwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Roots of the equation $(1-4x)^4+32x^4=\frac{1}{27}$ find all real roots of the equation $(1-4x)^4+32x^4=\dfrac{1}{27}$
i try to use binomial expansion and am-gm inequality but i don't know how to do next.
| Clearly there are no solutions if $ x > \frac{1}{4}$ or $ x < 0$. So, we may apply the power mean inequality to $ 1-4x, 2x, 2x$.
$$\sqrt[4]{\frac{( 1- 4x)^ 4 + (2x)^4 + (2x)^4 }{3}}\geq \frac{ (1-4x) + 2x + 2x } { 3} = \frac{1}{3} $$
This tells us that $( 1- 4x)^ 4 + 32x^4 \geq \frac{1}{27}$.
Equality holds iff $ 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Compute: $\lim_{n\to\infty}\sum_{i=1}^{2n^4+n^2}\frac{5n^2+1}{n^4+i}$ The question is to find $\lim_{n\to\infty} \left ( \frac{5n^2+1}{n^4+1}+\frac{5n^2+1}{n^4+2}+\frac{5n^2+1}{n^4+3}+...\frac{5n^2+1}{3n^4+n^2} \right )$
The answer is 5, according to the booklet.
I tried to bring the sum to the form $ \sum \frac{1}{n}f... | Let $a_n=\frac{5n^2+1}{n^4+1}+\frac{5n^2+1}{n^4+2}+\frac{5n^2+1}{n^4+3}+...+\frac{5n^2+1}{n^4+n^2}$, then notice
$$
a_n \geq n^2\cdot \frac{5n^2+1}{n^4+n^2}=\frac{5+\frac{1}{n^2}}{1+\frac{1}{n^2}}
$$
and similarly
$$
a_n \leq n^2 \cdot \frac{5n^2+1}{n^4+1} = \frac{5+\frac{1}{n^2}}{1+\frac{1}{n^4}}.
$$
Now just use the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Diophantine equation from "Solving mathematical problems" by Terence Tao
Find all integers $n$ such that the equation $\frac{1}{a} + \frac{1}{b} = \frac{n}{a+b}$ is satisfied for some non-zero values of $a$ and $b$ (with $a + b \neq 0$).
I'm reading "Solving mathematical problems" by Terence Tao and I'm a bit stuck ... | Regarding your first equation, with the formula
$$a^2+ab(2-n)+b^2 \tag{1}\label{eq1A}$$
consider that $n$ and $b$ are constants, with only $a$ being a variable. In that case, it's a quadratic polynomial in $a$, of the form
$$a^2 + ca + d \tag{2}\label{eq2A}$$
where $c = b(2-n)$ and $d = b^2$. Thus, using the quadratic ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How do I get from $x^4+2x^3y+3x^2y^2+2xy^3+y^4$ to $(x^2+xy+y^2)^2$? I was doing an example $$(x+y)^4+x^4+y^4$$ and I need to factor it. I've tried and couldn't really do much, so I checked if there was anything to help, and I came across a post asking about the same thing. But my question is how do I know that $$x^4+2... | The given quartic form can be written as follows
$$q (x,y) := x^4+2x^3y+3x^2y^2+2xy^3+y^4 = \begin{bmatrix} x^2\\ x y\\ y^2\end{bmatrix}^\top \begin{bmatrix} 1 & 1 & t\\ 1 & 3-2t & 1\\ t & 1 & 1\end{bmatrix} \begin{bmatrix} x^2\\ x y\\ y^2\end{bmatrix}$$
Choosing $t = 1$, we obtain a symmetric, positive semidefinite, r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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How to find the common factor I am currently stomped at this problem $(2 x^2-(y+z) (y+z-x))/(2 y^2-(z+x) (z+x-y))$ I am supposed to factor it and I can't find a way how to get to the answer.
It says that
$(2 x^2+x y-y^2+x z-2 y z-z^2)/(2 y^2-x^2+x y-2 x z+y z-z^2)$ has a common factor of x+y+z. Can anyone elaborate o... | From the numerator, you can pick up a $2x-y-z$, obtaining:
$$ x^2+x y-y^2+x z-2 y z-z^2=(2x-y-z)\cdot(x+y+z)$$
From the denominator, you can pick up $2y-x-z$, arriving at:
$$2 y^2-x^2+x y-2 x z+y z-z^2=(2y-x-z)\cdot(x+y+z)$$
Now, you can rewrite your fraction as:
$$\frac{(2 x^2+x y-y^2+x z-2 y z-z^2)}{(2 y^2-x^2+x y-2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integral involving elliptic integral functions Recently I came across this identity:
$$\int _0^1\:\frac{K\left(x\right)}{1+x}dx=\frac{\pi ^2}{8}$$
Where:
$$K\left(x\right)=\int _0^{\frac{\pi }{2}}\:\frac{1}{\sqrt{1-\left(x\sin \left(\theta \right)\right)^2}}d\theta $$
Any hints to how to prove this identity?
| Well, we are trying to find the following integral:
$$\mathcal{I}:=\int_0^1\frac{1}{1+x}\cdot\left(\int_0^\frac{\pi}{2}\frac{1}{\sqrt{1-\left(x\sin\left(\theta\right)\right)^2}}\space\text{d}\theta\right)\space\text{d}x\tag1$$
Now, the inner integral is known as the complete elliptic integral of the first kind. And can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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If $x$ and $y$ are rational numbers such that : $(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$
If $x$ and $y$ are rational numbers such that : $$(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$$ then which of the following is true?
A) $\;x=1$, $y=1$
B) $\;x=2$, $y=1$
C) $\;x=5$, $y=1$
D) $\;x$ and $y$ can take infinitely ... | Since
$\sqrt{2}$
and
$\sqrt{6}$
are linearly independent, their coefficients must be zero.
Therefore
$0 = x-2y = x-y-1$
so $x=2, y=1$.
They threw in
$x+y=2x-y$
for the fun of it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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if $x_{m}$be prime number, show that $2m+1$ must prime number! give the integer $a>1$, and define
$$x_{0}=1,x_{1}=4a+1,x_{n+1}=(4a+2)x_{n}-x_{n-1},n\ge 1$$
if $x_{m}$be prime number, show that $2m+1$ must prime number!
this problem seem interesting,Now I post follow my try :
since Characteristic equation
$$r^2-2(2a+... | As far as I can tell, everything you've shown is correct. As such, as you stated near the end of your question text, with
$$\alpha = \sqrt{a + 1} + \sqrt{a}, \; \; \beta = \sqrt{a + 1} - \sqrt{a}, \; \; \alpha\beta = 1 \tag{1}\label{eq1A}$$
you get
$$x_{n} = \frac{\alpha^{2n + 1} + \beta^{2n + 1}}{\alpha + \beta} \tag{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What are the asymptotics of the solutions to $a_n\ ^{2} = (1 - a_n)a_{n-1} \ ^2$ as $n \to \infty$? In
Prove $a_t \leq \frac{c}{t}$
I showed that
the solutions to
$a_0 \in (0, 1]$,
$a_n^2 = (1 - a_n)a_{n-1}^2$
satisfy
$a_n \le \dfrac{2}{n}$.
Numerical experiments
seem to show that
$na_n \to 2$
for any initial
$a_0 \in ... | I will answer (1) and (2) and try to explain a potential strategy to get the finer results of (3).
First of all let $$f(x) = \frac{2}{1 + \sqrt{1 + \frac{4}{x^2}}},$$ then the sequence $(a_n)$ is defined by $$a_0 > 0, \forall n \geq 0, a_{n+1} = f(a_n).$$
Now you have already shown that $\lim_{n \rightarrow \infty} a... | {
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"url": "https://math.stackexchange.com/questions/3608878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $P(x)$ a degree-$n$ polynomial such that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,\ldots,n$, find $P(n+1)$. The problem, from USAMO $1975$, reads:
If $P(x)$ denotes a polynomial of degree $n$ such that $P(k) = \frac{k}{k+1}$ for $k = 0,1,2, \ldots, n$. Find $P(n+1)$.
My attempt:
We can represent $p(x)$ as another polyno... | So $$Q(x) = (x+1)P(x)-x$$ is of degree $n+1$ and has zeroes $0,1,2,...,n$, thus we can write it $$Q(x) = ax(x-1)(x-2)\cdots (x-n)$$ so $$(x+1)P(x) =ax(x-1)(x-2)\cdots (x-n)+x$$ which is valid for all $x$ and in particulary for $x=-1$: $$0 =a(-1)^{n+1}(n+1)!-1$$ and for $x=n+1$ we get: $$(n+2)P(n+1) =a(n+1)!+n+1$$ so $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the triangle with the maximum area with a given perimeter
Question: Of all triangles with a given perimeter, find the triangle with the maximum area. Justify your answer.
My approach: Let us have any $\Delta ABC$ such that the sides opposite to $A$ is of length $a$, the side opposite to $B$ is of length b an... | Use the Heron's formula
$$A =\sqrt{s(s-a)(s-b)(s-c)}, \>\>\>\>\> s=\frac p2$$
and apply the AM-GM inequality to get
$$A =\sqrt{s(s-a)(s-b)(s-c)} \le \left[ s \left( \frac{3s-(a+b+c)}{3} \right) ^3\right]^{1/2}
= \frac{s^2}{3\sqrt3}=\frac{p^2}{12\sqrt3}$$
where the equality, or the maximum area, occurs at $a=b=c=\frac p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $ without using L'Hopital's rule. Let $ n $ be a positive integer greater than $ 1 $.
Find : $$ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $$
Without L'Hopital's rule or series expansion.
Here is ... | By definition of the derivative, the desired limit is equal to $-f'(0)$ where
$$f(x) = \sqrt{1+x} \sqrt[3]{1-x} \sqrt[4]{1+x} \cdots \sqrt[2n+1]{1-x}.$$
We can now use logarithmic differentiation to find
$$f'(x) = f(x) \sum_{k=2}^{2n+1} \frac{(-1)^k}{k} (1 + (-1)^k x)^{-1}.$$
Plugging in $x = 0$ and $f(0) = 1$ will giv... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$ Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$
Listing out a few terms and simplifying the denominators gets me $$-1+\sqrt{3}+2-\sqrt{2}-\sqrt{3}+\sqrt{5}\cdots-\sqrt{97}+3\sqrt{11}-7\sqrt{2}+10-3\sqrt{11}+\sqrt{101}.$$
A lot of these term... | We can rewrite the sum as:
$$\sum_{n=1}^{99} \frac{2}{\sqrt{n} + \sqrt{n+2}}\frac{\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}}=
\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}$$
It's a telescopic sum, where most of the terms are cancelled.
We can expand it:
$$\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}=
(\sqrt{3}-\sqrt{1})+(\sqrt{4}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim_{x\to1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-\sqrt{x^4+1}}.$ I don't know how to solve these kind of limit problems. I was wondering if someone could help me about it to learn them. $$\lim_{x\to 1^+}\frac{\sqrt{x+1}+\sqrt{x^2 -1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2 +1}-... | We can rewrite the numerator as $$N=\sqrt{x-1}\sqrt{x+1}+\frac{x(1-x)(1+x)}{\sqrt{1+x}+\sqrt{1+x^3}}=\sqrt{x-1}\left(\sqrt{1+x}-\frac{x(1+x)\sqrt{x-1}}{\sqrt{1+x}+\sqrt {1+x^3}}\right)$$ and hence $N/\sqrt {x-1}\to\sqrt{2}$. Similarly one can show that if $D$ is the denominator then $D/\sqrt{x-1}\to 1$. And therefore $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve the following radical equation?$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$
How to solve the following radical equation?
$$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$$
I'm looking for an easy way to solve it.
| As hinted
For real $x,\sqrt{2-x}\ge0\implies x\le2$
But $x=2$ doesn't satisfy the given equation
and $x=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}>0$
So, $0<x,<2$
WLOG let $x=2\cos32t, 0<32t<\dfrac\pi2$
Use $\cos2x=2\cos^2x-1=2-2\sin^2x$
$$\sqrt{2-x}=+2\sin16t$$
Now $\sqrt{2+\sqrt{2-x}}=\sqrt{2(1+\sin16t)}=+\sqrt2(\sin8t+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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For positive a,b,c,d, if $a^2+b^2+c^2+d^2+abcd=5$, show $a+b+c+d\leq 4$. One can use Lagrange multiplier, but I am looking for a more elementary proof.
I try to find the maximum of $a+b+c+d$ and follow the standard approach.
Construct $a+b+c+d+\lambda (a^2+b^2+c^2+d^2+abcd-5)$.
One can obtain $1+2a\lambda +bcd =0$ and... | Alternative solution:
Fact 1: For $a, b, c, d \ge 0$, it holds that
$$a^2 + b^2 + c^2 + d^2 + abcd - 5 - 3(a+b+c+d - 4) \ge 0.\tag{1}$$
From Fact 1, the desired result follows.
Edit 2022/02/14: A nice proof of Fact 1 is given by mudok@AoPS.
See: #4 in https://artofproblemsolving.com/community/c6t243f6h2780412_hard_ineq... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $
I proved this relationship by incident. I tried to directly prove thi... | Remark: My previous solution is ugly. I give another solution.
Clearly, we only need to prove the case when $x > 0$.
For $n=1$, clearly the inequality is true.
For $n=2$, the inequality is equivalent to $2\cos x - 2 + x^2\ge 0$ which is true.
For $n\ge 3$ and $x\in [\frac{\pi}{2}, \infty)$:
It is easy to prove that
$-n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Intersection of a circle $x^2+(y-1)^2=1$ and a parabola $ax^2=y$, bounds for a I have a circle $x^2+(y-1)^2=1$ and a parabola $x^2=\frac{y}{a}$ and it is required to find the bounds for $a$ if they intersect at point other than origin.
So I proceeded as follow, substituted $x^2=y/a$ into $x^2+(y-1)^2=1$
which led me to... | We can compute very easily the intersections of the two conics, or:
$$\frac{y}{a}+(y-1)^2=1 \leftrightarrow ay^2+y(1-2a)=0 \leftrightarrow y(ay+1-2a)=0$$
Now, a solution is $y=0$ that holds for evry $a$. The other solution is $y=\frac{2a-1}{a}$ and so $x=\frac{1}{a}\sqrt{2a-1}$. Now, taking from here, it's very simple.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Conditional probability of continuous throwing a dice A person will keep throwing a dice until he gets $6 .$ If he gets 6 then the experiment will be stopped.
The probability that experiment ends at 6 th trial, given that prime number does not appear is
(A) $\left(\frac{2}{5}\right)^{5}\left(\frac{1}{6}\right)$
(B) $\l... | Context is a bit confusingly worded to me... isn't it just simply $(B)$ because if no prime numbers appear then $2,3$ and $5$ won't appear, reducing the dice to only 3 possibilities of which only 1 of those 3 is a 6 so isn't it simply just $({2 \over 3})^5*(1/3)$? I could be wrong however...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Use mathematical induction to prove that for all integers $n \geq 3,\, 2n + 1 < 2^{n}$ This is what I've got so far.
Let $P(n)$ be the statement that $2n + 1 < 2^n$
Basis:
Let $n = 3$. Show that $P(3)$ is true.
$2(3) + 1 = 7$ and $2^3 = 8$.
Since $7 < 8$, $P(3)$ is true.
Inductive Hypothesis:
Suppose that $P(k)$ is tr... | Continuing from your step $2(k+1)+1<2^{k}+2$.
Since $k\geq 3$, then $2<2^{k}$.
So, $2(k+1)+1<2^{k}+2<2^{k}+2^{k}=2^{k+1}$
$\therefore 2(k+1)+1<2^{k+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove such $|\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n}|\le\frac{1}{a}$ let $a$ is give postive real number,and $a_{1},a_{2},\cdots,a_{n}$ be postive real numbers,and such
$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}=1,a_{1}+a_{2}+\cdots+a_{n}=a$$
show that: there exist $\mu_{i}\in\{-1,1\},i=1,2,\cdots,n$ such
$$|\mu_{1}... | Let us show the following more general proposition.
Let $a_{1},a_{2},\cdots,a_{n}$ be positive real numbers. Then there exist $\mu_{i}\in\{-1,1\},i=1,2,\cdots,n$ such
$$0\le(\mu_{1}a_{1}+\mu_{2}a_{2}+\cdots+\mu_{n}a_{n})(a_{1}+a_{2}+\cdots+a_{n})\le a_{1}^2+a_{2}^2+\cdots+a_{n}^2\tag{1}$$
Proof by induction on $n$.
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3632481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove this has real roots $(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$ Prove this has real roots
$(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$
My Work
\begin{align*}
\Delta&=(a+b)^2(a-c)^2-4a(b-c)(a^2-bc) \\
&=a^4+2a^3c+a^2c^2-2a^3b+b^2a^2-4a^2bc-2abc^2+2ab^2c+b^2c^2.
\end{align*}
How do I show that this is positive? Simplification doesn'... | Hint: remarkably, that expression for $\Delta$ is a perfect square...! Can you factor the expression knowing that?
| {
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Solving $ \sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x} $ I am reviewing my high-school math (year 10-ish) but I am hitting a wall with this nasty inequality:
$$
\sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x}
$$
I find (steps below) the solution to be
$$
1 \le x \le 3 \text{ with } x \ne 2
$$
but the text-book and W... | $$\begin{cases}\sqrt{3-x}-\sqrt{x-1}>\sqrt{4-x}-\sqrt x\\ 1\le x\le 3\end{cases}$$
If you want to square both current sides, then you must first evaluate their signs, namely using $A>B\iff \begin{cases}A\ge 0\\ B\ge 0\ \\ A^2>B^2\end{cases}\lor \begin{cases}A\ge 0\\ B<0\end{cases}\lor\begin{cases}A< 0\\ B< 0\ \\ A^2<B^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how to find positive integer $m,n , (m^2 + n^2 )(n^2-m)= 2mn(m+n^2) $ how to find positive integer $m,n $
$(m^2 + n^2 )(n^2-m)= 2mn(m+n^2) $
answer is $(36,18), (36, 12) $
$n$-th try........
$ gcd (m,n)= p$, then $m=ap, n= bp$
$(a^2 p^2 + b^2 p^2 )(b^2p^2 -ap)= abcp^2(ap+ b^2p^2)$
$(a^2+ b^2 )(b^2p-a) = 2ab (a+ b^2... | $(36, 18)$ and $(36, 12)$ are the only positive integer solutions.
Expand the equation and move all the terms to one side:
$$m^3+mn^2+2m^2n-m^2n^2-n^4+2mn^3=0\tag{1}$$
Let $p=gcd(m,n)$ and $m=ap, n=bp$. Then by rewriting as a cubic equation in $a$, we get
$$a^3-(b^2p-2b)a^2+(2b^3p+b^2)a-b^4p=0$$
By the Rational Root Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find volume of region bounded by plane, bounded by $y^2 = 1 - x$ and $x = -1$, and parabolic cylinder, $z = 1 - x^2$ Question:
Let $R$ denote the finite plane region on the $xy$-plane bounded by the line $x = -1$ and the parabola $y^2 = 1 - x$. Let $D$ denote the solid region under the surface of the parabolic cylinder... | Note: you don't have to find the area of region $R$ first. As $D$ is the volume under the surface of the form $z=f(x,y)$, we first set the limits of $z$ and then find the projection of the given surface on $xy$-plane that is our region $R$ and then we set the limits of $x, y$.
HINT:
The volume of the solid region $D$ c... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove $(\frac{n+1}{n})^{n+1}$ is decreasing I managed to prove a similar fact: the following sequence is increasing: $\left( \dfrac{n+1}{n}\right)^n$, which means $\left( \dfrac{n+1}{n}\right)^n < \left( \dfrac{(n+1)+1}{(n+1)}\right)^{n+1}$.
All this took was some simple algebra and bernoulli's inequality. For the one ... | Easy approach with A.M$\gt $G.M
Let $u_n=(\frac{n+1}{n})^{n+1}=\left(1+\frac{1}{n}\right)^{n+1}$
Let us consider n+2 positive numbers $(1-\frac{1}{n+1}),(1-\frac{1}{n+1}),(1-\frac{1}{n+1}),••••,(1-\frac{1}{n+1}),[(n+1)times] $and $1$.
Applying A.M$\gt $G.M.,we have,
$$\frac{(n+1)(1-\frac{1}{n+1})+1}{n+2}\gt (1-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to calculate the sum of the first n natural numbers? I already know one way to prove the sum of the n first natural number is equal to $\frac{n(n+1)}{2}$, but I found another way which involves calculating $(k+1)^2 - k^2, k\in \mathbb{N},\ k\geqslant 1$ . Here is the whole demonstration:
$
\begin{gather}
(k+1)^2 -k... | Let's try writing some terms out:
\begin{align*}
\sum\limits_{k=1}^n ((k+1)^2 - k^2) & = ((1+1)^2 - 1^2) + ((2+1)^2 - 2^2) + ((3+1)^2 - 3^2) + \ldots + ((n+1)^2 - n^2) \\
& = (2^2 - 1^2) + (3^2 - 2^2) + (4^2 - 3^2) + \ldots + ((n+1)^2 - n^2) \\
& = (2^2 - 2^2) + (3^2 - 3^2) + \ldots + (n^2 - n^2) + (n+1)^2 - 1^2 \\
& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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MOP 2011 inequality If $a,b,c$ are positive integers prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$
My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +... | Another way.
By AM-GM $$\sum_{cyc}\sqrt{a^2-ab+b^2}\leq\sum_{cyc}\left(\frac{a^2-ab+b^2}{a+b}+\frac{a+b}{4}\right)=$$
$$=\sum_{cyc}\left(\frac{a^2+2ab+b^2-3ab}{a+b}+\frac{a+b}{4}\right)=\frac{5}{2}(a+b+c)-3\sum_{cyc}\frac{ab}{a+b}=$$
$$=4(a+b+c)-3\sum_{cyc}\left(\frac{ab}{a+b}+\frac{a+b}{4}\right)\leq$$
$$\leq 4(a+b+c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dividing polynomial $f(x^3)$ by $x^2+x+1$ Given two polynomials with real coefficients $g(x)$ and $h(x)$.
Additionally, $x^2+x+1$ is a factor of $f(x^3)= h(x^3)+xg(x^3)$.
Prove that $x-1|h(x)$ and $x-1|g(x)$
I have tried to solve it by doing this;
$(x-1)(x^2+x+1)|(x-1)(h(x^3)+xg(x^3))$
then;
$(x^3-1)|(x)(h(x^3)-g(x^3)... | Suppose
$$
\left.x^2+x+1\,\,\middle|\,\,h\!\left(x^3\right)+x\,g\!\left(x^3\right)\right.\tag1
$$
Let ${\bar h}(x)=h(x+1)$ and ${\bar g}(x)=g(x+1)$, then $(1)$ becomes
$$
\left.x^2+x+1\,\,\middle|\,\,{\bar h}\!\left(x^3-1\right)+x\,{\bar g}\!\left(x^3-1\right)\right.\tag2
$$
Write $\bar h$ and $\bar g$ as
$$
\bar h(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which integration is correct? Or are both correct? What's the correct method to integrate the function $\frac{1}{2(3x+1)}$?
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \int\frac{1}{6x+2}dx\\
&= \frac{1}{6}ln(6x+2)+c
\end{align}
$$or$$
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \frac{1}{2}\int\frac{1}{3x+1}dx \\
&= (\frac{... | Yes, you are right
Both the feasible answers are correct & the only difference between two is constant of integration
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Rewriting elliptic curve with point of order 3 I have a question regarding a change of coordinates with elliptic curves. I tried to show that an elliptic curve $E: y^2 = x^3 + ax^2 + bx +c$ with coefficients in $\mathbb{F}_q[t]$ and where $P$ is a rational point of order 3 can be rewritten as:
\begin{equation*}\label{}... | Let us write $z=f/g$ with polynomials $f,g\in\Bbb F_p[t]$. We multiply with $g^6$ the last equation $y^2=x^3+a(x+z)^2$ getting:
$$
(yg^3)^2 = (xg^2)^3 + ag^2(xg^2+fg)^2\ ,
$$
and use the new "coordinates" $X = xg^2$, $Y=yg^3$.
Note: $y^2=x^3+C^2$ is not of the wanted form. It has the torsion points $T_\pm=(0,\pm C)$ of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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How many integers solutions does $x^2 + xy - y^2 = 1$ have? Also, if we know if a binary form represents an integer $n$, is there an algorithm to find all the solutions?
| Multiply equation with 4: $$4x^2+4xy-4y^2=4$$ so
$$(2x+y)^2-5y^2=4$$
$$(2x+y-2)(2x+y+2)=5y^2$$ Clearly $\gcd(x,y)=1$. Let prime $p\mid \gcd(2x+y-2,2x+y+2)$
then $p\mid 4$ (so $p\mid y$). Let $y=2z$, so $$(x+z-1)(x+z+1)=5z^2$$
Let prime $q\mid \gcd(x+z-1,x+z+1)$, then $q\mid 2$ and thus $2\mid z$ and $2\mid x-1$. Now w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$? I am trying to solve the following question involving floor/greatest integer functions.
$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$ with the notations $\lfloor x \rfloor$ denoting the greatest integer less than or equal to $x$ and $\{x\}$ t... | Write $\{x\}=x-\lfloor x\rfloor$. Then we have
$$
5\lfloor x\rfloor - \lfloor x^2\rfloor = 2x
$$Since the LHS is an integer, the RHS must be as well. There are two cases: $x$ is an integer, or $x$ is a half-integer.
*
*$x$ an integer. Drop the brackets:
$$
5x-x^2=2x;\qquad x=0,3
$$
*$x$ is a half-integer. Write $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find k'th integer not divisible by n? Although this was a programming question I want the mathematical intuition behind it. So we were given two numbers $n$ and $k$. We were asked to find out $k{-th}$ number **not divisible** by $n$. For example, $n=3$ and
$k=7$.
If we see numbers $1$ to $12$. $$1, 2, 3... | We need the $k^{th}$ integer not divisible by $n$.
At first, we list all the integers not divisible by $n$. We don't know the values of these integers yet.
$$1^{st}, 2^{nd}, 3^{rd}, ..., k^{th}$$
This list omits the integers divisible by $n$. We need to know exactly how many such integers are omitted to find the value ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Why $c$ closed to $-2\times10^n$ in $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$? I have tried many times to evaluate $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$ as polynomial for some values of integer $n$ which are greater than $1$ for the solution of the titled equation for looking why exa... | Your expression is
$$(1 - c^2)^3 + (c^3 + 10^n c^2 - 1)^3 + (10^n c^2 - 1)^3 = n \tag{1}\label{eq1A}$$
for $n \gt 1$. For some real $d$, let
$$c = d \times 10^n \tag{2}\label{eq2A}$$
Note that within each of the $3$ terms on the left side of \eqref{eq1A}, there's a term of $1$ or $-1$, but the other terms are all at le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove the Condition for Construction of a Triangle Let $a,b,c>0$ be real numbers. Show that you can construct the triangle with length sides $a,b,c$ if and only if $pa^2+qb^2>pqc^2$, for any $p,q$ such that $p+q=1$.
| As Exodd said, the statement is false. Let $f_{a,b,c}(p)=pa^2+2b^2-p(1-p)c^2=c^2p^2+(a^2-c^2)p+2b^2$. Then your statement says: $a,b,c>0$ are the sides of a triangle if, and only if, $f_{a,b,c}>0$.
Take $a=c=1$, $b=3$. Then $f_{a,b,c}(p)=p^2+6$, which is always positive, but $(1,1,3)$ cannot be the three sides of a tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $4\sin (x)\,\big(\sin(3x)+\sin (7x)+\sin(11x)+\sin(15x)\big)$ when $x=\frac{\pi}{34}$.
If $x=\frac{\pi}{34}$, then find the value of
$$S=4\sin (x)\,\big(\sin(3x)+\sin (7x)+\sin(11x)+\sin(15x)\big).$$
Source: Joszef Wildt International Math Competition Problem
My attempt:
$$\sin(3x)+\sin(15x)=2\sin(9x)\cos(6x)$... | as You simplified and as @WE Tutorial School pointed out,
$$S=16\cos\frac{\pi}{17}\cos\frac{2\pi}{17}\cos\frac{4\pi}{17}\cos\frac{8\pi}{17}.$$
Applying $\sin 2x = 2 \sin x \cos x$ 4 times,
$$ S \cdot \sin \frac \pi{17} = \sin \frac {16\pi}{17}$$
$$ S = \frac {\sin \left( \pi - \frac \pi{17} \right)}{\sin \frac \pi{17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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contour integration of $\int_0^\infty \frac{\ln(x)}{x^2-1}dx$ I asked for a problem for few days ago, regarding integration of
$$I=\int_0^\infty \frac{\ln(x)}{x^2-1}dx$$
I know that I could do the substitution $x=it$ and do the integral where the denominater is $x^2+1$ and so on... But I tried anyway to create a conto... | \begin{eqnarray*}
\int_0^\infty\int_0^\infty\frac{dxdy}{(1+y)(1+x^2y)}
&=&\int_0^\infty\frac{dy}{1+y}\int_0^\infty\frac{dx}{1+x^2y}\\
&=&\int_0^\infty\frac{dy}{1+y}\int_0^\infty\frac{1}{y}\frac{dx}{\frac{1}{y}+x^2}\\
&=&\int_0^\infty\frac{dy}{1+y}\int_0^\infty\frac{1}{y}\frac{1}{\frac{1}{\sqrt{y}}}arctan \frac{x}{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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