Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding first three terms of a GP whose sum and sum of squares is given? We have to find three numbers in G.P. such that their sum is $\frac{13}{3}$ and the sum of their squares is $\frac{91}{9}$.
Here's what I have tried to do:-
$$a+ar+ar^2=\frac{13}{3}$$
$$a^2+a^2r^2+a^2r^4=\frac{91}{9}$$
$$\frac{a^2+a^2r^2+a^2r^4}{a... | Take 3 numbers $a/r$ ,a,ar $a/r$+a+ar=$13/3$ and $a^2/r^2$+$a^2$+$a^2r^2$=$91/9$ from this take $a^2$ common so on sim plifying we get $a^2/r^2$(1+$r^2+r^4$)=$(91/9$ Then square 1 st condition by using idw
Identity $(a+b+c)^2$=$a^2+b^2+c^2+2ab+2bc+2ac=$169/9$ from this we get $91/9$+$2a(a/r+a+ar)$=$169/9$ so $91/9$+2a$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $a$ and $b$ be real numbers satisfying $a^3 - 3ab^2 = 47$ and $b^3 - 3a^2 b = 52$. Find $a^2 + b^2$. Let $a$ and $b$ be real numbers satisfying $a^3 - 3ab^2 = 47$ and $b^3 - 3a^2 b = 52$. Find $a^2 + b^2$.
I observed that adding the two has a suspicious similarity to (a+b)^3, but I cannot relate the two. Can someo... | Good try with $(a+b)^3$, but you will find it better to expand
$$\eqalign{(a^2+b^2)^3
&=a^6+3a^4b^2+3a^2b^4+b^6\cr
&=(a^3-3ab^2)^2+(b^3-3a^2b)^2\ .\cr}$$
If you have begun studying complex numbers, you can use the hint from Joey Zou to explain the "miracle" factorisation in the last line. However if you don't know... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve for $x$: $x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$ Solve for $x$: $$x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$$
My attempt: I have changed this equation into the fractional part function. so that we know $0 \leq \{x\}<1$. I have final equation $x^2 +7x-x\{x+2\}-\{2x-2\}=14$. How to pro... | I would let $x=y+z$, with integer $y$ and $0\leq z<1,z=\{x\}$. Then either
$$(y+z)(y+2)+2y-2+3y+3z=12$$
or
$$(y+z)(y+2)+2y-1+3y+3z=12$$
So, for example, $z(y+5)$ is an integer, so only some fractions are allowed.
Also, 12 is between $y(y+2)+2y-2+3y$ and $(y+1)(y+2)+2y-2+3y+3$, for the first case, or something similar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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double factorial series problem In the problem the sum is given as $$\sum\limits_{n=1}^\infty \frac{(2n-1)!!}{(2n)!!}$$
and then when I try to solve it using Gauss's test I get
$$\frac{a_{n}}{a_{n+1}}=\frac{2n+2}{2n+1}$$
but in the solution there is given:
$$\frac{a_{n}}{a_{n+1}}=\frac{2n}{2n-1}$$
my reasoning was:
$$\... | $$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2n)!!^2} = \frac{1}{4^n}\binom{2n}{n} $$
but the RHS behaves like $\frac{1}{\sqrt{\pi n}}$, hence the series is diverging. Also without the exact asymptotics:
$$ \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\prod_{k=1}^{n}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7 For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7.
I'm not sure how to do this proof so any help would be appreciated.
| If $n=1$, then
$$
4^{n} + 10\cdot 9^{2n-2} = 4+10 = 14,
$$
divisible by $7$;
if $n \geq 1$ is an integer such that $4^{n} + 10\cdot 9^{2n-2} = 7k$ for some integer $k \geq 1$, then
$$
4^{n+1} + 10\cdot 9^{2(n+1) - 2} = 4\cdot 4^{n} + 10\cdot 9^{2n} = 4(4^{n} + 10\cdot 9^{2n-2}) = 28k,
$$
divisible by $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Finding Angles Given Two Triangles with Equal Perimeters The two right triangles shown below have equal perimeters
The hypotenuse of
the orange triangle is one leg of the green triangle stacked on top of it. If the smallest angle of the orange triangle is 20 degrees, what are the angles of the green right triangle?
| I think I have a solution but it will be difficult to explain. I will use the following variables:
$$
\begin{split} a&=\text{base of orange triangle} \\ b&=\text{height of orange triangle} \\ c&= \text{"left" side of green triangle} \end{split}
$$
Then the hypotenuse of the orange triangle is $\sqrt{a^2+b^2}$ and the h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is my method for finding the Basis for the Row(A) correct?
$$A=\begin{pmatrix}-1&0&2\\3&2&0\\0&1&3 \end{pmatrix}$$
Find a basis for $\text{Row} (A)$
$$Row(A)=\text{span} \left[\begin{pmatrix}-1\\0\\2\end{pmatrix},\begin{pmatrix}3\\2\\0\end{pmatrix},\begin{pmatrix}0\\1\\3\end{pmatrix}\right]$$
$$\begin{pmatrix}-1&3&0\... | Elementary row operations preserve the row space of $A$ (but change the column space). If you want to find a basis for $\mathrm{row}(A)$, perform elementary row operations on $A$ until you reach the reduced row echelon form. The non-zero rows of the reduced row echelon form will form a basis for $\mathrm{row}(A)$. In f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~... | $$xy+x+y=44$$
$$\implies xy+x+y+1=45=3\times15$$
$$\implies (y+1)(x+1)=3\times15$$
therefore $y=2$ and $x=14$ , also it satisfies the second identity. Hence $x^2+y^2=196+4=200.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Linear Transformation: Orthogonal Projections
Define $\mathbf{u_1} =$ $\begin{align} \begin{bmatrix}
0 \\
0 \\ 1 \\1\\
\end{bmatrix}\end{align}$ and $\mathbf{u_2} =$ $\begin{align} \begin{bmatrix}
1 \\
1 \\ 0 \\0\\
\end{bmatrix}\end{align}$. Def... | I think that your solution is correct, other than the fact that you computed $\mathbf{u}_3$ and $\mathbf{u}_4$. Note that the completion of an orthogonal linearly independent set to an orthogonal base is not unique; however, the rest of the question can be answered even without knowing who the $\mathbf{u}_i$-s are.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What are roots of $x^{3}+3x^{2}+4x+1$? There are no divisor of 1 in this polynomial for which would be satisfied $x^{3}+3x^{2}+4x+1=0$.
How to find roots here?
| Put $x = y - 1$, this yields:
$$y^3 + y - 1 = 0$$
Then compare with the identity:
$$\begin{split}
(a+b)^3 \equiv a^3 + 3 a^2 b + 3 b a^2 + b^3\Longrightarrow \\
(a+b)^3 - 3 a b (a+b) - (a^3 + b^3) \equiv 0
\end{split}
$$
So, if we can find two numbers $a$ and $b$ such that:
$$-3 a b = 1$$
and
$$a^3 + b^3 = 1$$
then $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Is it true that $ \sum_{t = 1}^T \frac{T-t}{ t+ \sqrt{T-t}} \in O(T) $? Is the following true?
$$
\sum_{t = 1}^T \frac{T-t}{ t+ \sqrt{T-t}} \in O(T)
$$
| The idea is similar to A.S.' answer, I just want to make the calculation more explicit and clearer.
In fact we can show that
$$\limsup_{T \to \infty} \frac{1}{T}\sum_{t = 1}^T \frac{T - t}{t + \sqrt{T - t}} = \infty, \tag{$*$}$$
thus the conjecture is false.
We prove $(*)$ by considering a subsequence of $\{1, 2, \ld... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to evaluate the following integral ? What substitution will be helpful? $$\int\frac{cotx }{(1-sinx)(secx+1)}dx $$
we can write this as
$$\int \frac {cosecx.cotx}{(cosecx-1)(secx+1)}dx $$
Now $cosecx=t$ gives $cosecx.cotx=-dt $ which appears in the numerator, what to do about $secx+1$ ?
| $$\frac{\cot x}{(1-\sin x)(1+\sec x)}=\frac{1+\sin x}{\sin x(1+\cos x)}=\frac{1}{\sin x(1+\cos x)}+\frac{1}{1+\cos x}$$
Enforcing the Wiereatrass Substitution $\tan(x/2)=u$ and $dx=\frac{2}{u^2+1}\,du$ reveals
$$\begin{align}
\int \frac{\cot x}{(1-\sin x)(1+\sec x)}\,dx&=\int \left(\frac{(u^2+1)^2}{4u}+\frac{u^2+1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all distinct $a,b,c$ such that $a +1/b = b + 1/c = c+1/a$ Could anyone advise me how to find all values of $a,b, c$ such that $a + \dfrac{1}{b} = b + \dfrac{1}{c} = c + \dfrac{1}{a} \ , $ where $a,b,c $ are distinct real numbers?
I have found that $a, b, c$ can take on either $1,-2, $ or $-\dfrac{1}{2} $, so there... | Easy parameterization:
$$
(a,b,c) = \left(a, \frac{1}{1-a}, 1-\frac{1}{a}\right),
$$
where $a$ is any real, $a\ne 0, a\ne \frac{1}{2}, a\ne 1$.
Then $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How can I find the rank of $A^6$ if $A$ is a matrix of order $3$ and rank of $A^3$ is $2$? I have tried it in the following manner. We know that $$rank(XY)\ge rank(X)+rank(Y)-n$$ where $X$ and $Y$ are two matrices of order $n$ and also $$rank(XY) \le \min{\{rank(X),rank(Y)\}}$$ If we take $X=Y=A^3$ then using the above... | Assume $A \in M_3(\mathbb{C})$. Let $a,b,c$ be the root of characteristic polynomial of $A$ or say eigenvalue of $A$. So $a^3,b^3,c^3$ will be eigenvalue of $A^3$. As kernel of $A^3$ is one dimensional one of the eigen value of $A^3$ will be $0$, say $c^3 = 0$. So $c=0$.
First note $a,b$ simultaneously can't be $0$, in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof: $f(x)=x^2+x-4$ is continuous at $x=2$ Proof: Let $\epsilon>0$ and $\delta =\min\{1,\epsilon/6\}$.Then if $|x-2|<\delta$, $|x-2|<1$, so $|x+3|<6$.
Thus, $$|f(x)-f(c)|= |x^2+x-4-(2^2+2-4)| = |x^2+x-6|=|x-2||x+3|<6\delta \leq \epsilon$$
Doing this problem as practice. Thoughts? comments?
| Late answer but let us see if $\delta= \min \lbrace 1,\frac{\epsilon}{6} \rbrace $ works or not. First case is if $\delta=1.$ If $|x-2|<1,$ then $$|f(x)-f(2)|=|x^2+x-6|=|x-2||x+3|<(1)\left(6\right)<\frac{\epsilon}{6}(6)=\epsilon$$ since $|x-2|<1$ implies both $|x-2| < \frac{\epsilon}{6}$ and $|x+3|<6.$
Now let $\del... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $5$ divides $3^{3n+1}+2^{n+1}$
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{... | Or we could just expand and rearrange$$
\begin{align}
3^{3n+1}+2^{n+1}
&= 3\cdot27^n + 2\cdot2^n \\
&= 3\cdot(25+2)^n + 2\cdot2^n \\
&= 3\left(5k+2^n \right) + 2\cdot 2^n \tag{Using Binomial Theorem}\\
&= 5\cdot k'+3\cdot2^n+2\cdot2^n \\
&= 5\cdot k'+5\cdot 2^n \\
\end{align}
$$
Hence proved
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
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Find polynomial $f(x)$ based on divisibility properties of $f(x)+1$ and $f(x) - 1$ $f(x)$ is a fifth degree polynomial. It is given that $f(x)+1$ is divisible by $(x-1)^3$ and $f(x)-1$ is divisible by $(x+1)^3$. Find $f(x)$.
| If $(x-1)^3$ divides $f(x)+1$, then $(x-1)^2$ divides $f'(x)$.
If $(x+1)^3$ divides $f(x)-1$, then $(x+1)^2$ divides $f'(x)$.
As deg$\,f=5$, then deg$\,f'=4$, and hence $f'(x)=a(x-1)^2(x+1)^2=a(x^4-2x^2+1)$,
for some $a\in\mathbb R$.
Thus
$$
f(x)=\frac{a}{5}x^5-\frac{2a}{3}x^3+ax+b,
$$
for some $b\in\mathbb R$.
Now, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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Evaluating $\lim_\limits{x\to 1 }\bigl( (2^x x + 1)/(3^x x)\bigr)^{\tan(\pi x/2)}$ I have to calculate limit
$$\lim_{x\to 1 } \left(\frac{2^x x + 1}{3^x x}\right)^{\tan(\frac{\pi x}{2})}.$$
I know $\tan(\frac{\pi x}{2})$ is undefined in $x = 1$, but can I just put $x = 1$ into $\frac{x\cdot 2^x + 1}{x\cdot3^x}$ and ge... | For $x$ near $0$, $a^x=1+x\log(a)+O\left(x^2\right)$.
Furthermore, if $\lim\limits_{n\to\infty}\left|b_n\right|=\infty$ and $c=\lim\limits_{n\to\infty}a_nb_n$, then $\lim\limits_{n\to\infty}\left(1+a_n\right)^{b_n}=e^c$.
Therefore,
$$
\begin{align}
\lim_{x\to1}\left(\frac{2^xx+1}{3^xx}\right)^{\tan\left(\frac{\pi x}2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove $ 5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3) $ if $a^2+b^2+c^2=3$ Let $a,b,c\in\mathbb{R}$ such that $a^2+b^2+c^2=3$. Prove that:
$$
5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3)
$$
I tried to homogenize the inequality to get:
$$
5(a^4+b^4+c^4)+(a^2+b^2+c^2)^2≥\frac{8}{\sqrt3}(a^3+b^3+c^3)\sqrt{(a^2+b^2+c^2)}
$$
I hoped that one don't ne... | Hint:
It is enough to show $f(x)=5x^4+3-8x^3+2(x^2-1)=(x-1)^2(5x^2+2x+1)\ge0$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)... | Here is a mildly disguised version of the same idea.
Let $x=e^u$
Then we have $$\cosh u=2$$
And we want $$x^2+\frac {1}{x^2}=2\cosh 2u=2(2\cosh^2 u-1)=2(8-1)=14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Alternate approaches to solve this Integral Evaluate $$I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\:dx$$
I have used parts taking first function as Integrand and second function as $1$ we get
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}-\int \frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right) \times x dx$$
now $... | First rationalize:
$$\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$$
$$=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\times\sqrt{\frac{1-\sqrt{x}}{1-\sqrt{x}}}$$
$$=\frac{\sqrt{1-\sqrt{x}}}{\sqrt{1+\sqrt{x}}}\times\frac{\sqrt{1-\sqrt{x}}}{\sqrt{1-\sqrt{x}}}$$
$$=\frac{1-\sqrt{x}}{\sqrt{1-x}}$$
$$=\frac{1}{\sqrt{1-x}}-\frac{\sqrt{x}}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the Maximum value.
Maximize $xy^2$ on the ellipse $b^2x^2 +a^2y^2= a^2b^2$
The steps I tried to solve:
$$\nabla f = (y^2,2yx)\lambda\qquad g = (2xb^2,2y^2a^2)\lambda$$
$$y^2= 2xb^2\lambda$$
$$2yx= 2y^2a^2\lambda$$
$$
\left.
\begin{array}{l}
\text{}&y^2= 2xb^2\lambda\\
\text{}&
\end{array}
\right\}
*a^2y
$$ ... | Since you started off using Langrange multipliers, let’s continue down that path. Using the correct value for $g_y$, we have $$\begin{align}
y^2 &= 2b^2\lambda x \\
2xy &= 2a^2\lambda y
\end{align}$$ which upon eliminating $\lambda$ gives $$
a^2y^3 = 2b^2x^2y,
$$ so either $y=0$ or $a^2y^2=2b^2x^2$. Substituting this i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Number of Polynomials with Integer Coefficients that are bounded by $x^2$ and $x^4+1$ What is the number of polynomials $p(x)$ with integer coefficients, such that
$x^2≤p(x)≤x^4+1$ for all real numbers $x$?
| Indeed, no polynomial with degree $3$ can satisfy this property because it has to have a negative value at some point.
Also, it is clear that (by taking very large $x$), the degree of this polynomial must be $2,3$ or $4$.
So this polynomial has to have a degree of $2$ or $4$.
Case 1. deg$p$ is $2$.
Let $p(x)=ax^2+bx+c$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Partial sum of $\sum \frac {1} {k^2}$ It is pretty well-known that $\sum_{k = 1}^{\infty} \frac {1} {k^2} = \frac {\pi^2} {6}$. I am interested in evaluating the partial sum $\sum_{k = 1}^{N} \frac {1} {k^2}$. Here is what I have done so far. Since we have
$$\sum_{k = 1}^{N} \frac {1} {k^2} = 1 + \sum_{k = 2}^{N} \frac... | The Euler summation formula gives, for all $x\ge 1$,
$$
\sum_{n\le x}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{1}{x}+O(x^{-2}),
$$
by taking $f(t)=t^{-2},\, f'(t)=-2t^{-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral with inverse trig functions and u substitution I've been trying to find the integral of $\dfrac{dx}{x \sqrt{x^2-4}}$. Currently, following the example problem, I have it as $\dfrac{1}{2} S \dfrac{1\,du}{x \sqrt{4 [(\frac{x}{2})^2-1]}}$. From what I've got it looks like it's going to the formula matching $\sec^... | \begin{align}
& \frac{dx}{x \sqrt{x^2-4}} = \frac{dx}{x\sqrt{4\left(\left(\frac x 2\right)^2 - 1\right)}} = \frac 1 {\sqrt 4} \cdot \frac{dx}{x\sqrt{\left( \frac x 2 \right)^2 - 1}} = \frac 1 2 \cdot \frac{dx/2}{\frac x 2 \sqrt{\left( \frac x 2 \right)^2 - 1}} \\[10pt]
= {} & \frac 1 2 \cdot \frac{du}{u\sqrt{u^2 -1}} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you solve $\frac{|x^2+5x+6|}{|x|-3} = 1$ How do you solve
$ \frac{|x^2+5x+6|}{|x|-3} = 1 $ ?
I have tried rearranging, polynomial division, multiplying both sides by a a fraction to simplify to no avail for the last hour.
| $$ \frac{|x^2+5x+6|}{|x|-3} = 1 \implies |x^2+5x+6|=|x|-3$$
Solving Algebraically:
Case$1$: When, $x <-3$,
$$ x^2+5x+6=-x-3 \implies x^2+6x+9=0 \implies x=-3 \text{ [No solution]. }$$
Case$2$: When, $-3 < x \leq -2$,
$$ -x^2-5x-6=-x-3 \implies x^2+4x+3=0 \implies (x+3)(x+1)=0 \implies x=-3,-1 \text{ [No solution from t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify the expression $(a+1)(a^2+1)(a^4+1)\cdots(a^{32}+1)$ How do I simplify this expression?
$$\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)$$
| Oh this is so cute!
We know that $(x + y)(x - y) = x^2 - y^2$ and therefore for for any $(a^n - 1)(a^n + 1) = (a^{2n} - 1)$.
So if we simply multiply $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)$ by $(a-1)$ we get
$(a -1)\left( ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove $\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+c^3+abc}+\frac{1}{b^3+c^3+abc} \leq \frac{1}{abc}$ I have to apologize in advance that the question is missing some vital information (like, $a,b,c$ are positive? natural? rational?) I took it off some video off youtube.
In the video, an 11 year old was in the process of solvin... | We have $$\sum \frac{1}{a^3+b^3+abc} \le \sum \frac{1}{a^2b+ab^2+abc} = \frac{1}{a+b+c} \sum \frac{1}{ab} = \frac{1}{abc}$$
Note that $$a^3+b^3 \ge a^2b+ab^2 \iff (a-b)^2(a+b) \ge 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The number of integers $n$ such that the quadratic equation $nx^2+(n+1)x+(n+2)=0$ has rational roots is The number of integers $n$ such that the quadratic equation $nx^2+(n+1)x+(n+2)=0$ has rational roots is
$(A)0\hspace{1cm}(B)1\hspace{1cm}(C)2\hspace{1cm}(D)3$
The condition for the rational roots is that the discrim... | Complete the square:
$$-3n^2-6n+1 = -3n^2-6n-3+3+1 = -3(n+1)^2+4.$$
For this to be nonnegative, we must have $(n+1)^2 \le \frac{4}{3}$. So $n=0, -1, -2$ are the only solutions, and $n=0$ gives a linear, not a quadratic, equation (though this special case would probably stop me from using this problem).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
The least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
$\bf{My\; Try::}$ Let $$K = 2x^2+y^2+2xy+2x-3y+8$$
So $$\displaystyle y^2+(2x-3)y+2x^2+2x+8-K=0$$
Now For real values of $y\;,$ We have $\bf{Discriminant\geq 0}$
... | \begin{align}
&2x^2+y^2+2xy+2x-3y+8\\
&=\left(x^2+y^2+\frac94+2xy-3x-3y\right)+x^2+5x+\frac{23}{4}\\
&=(x+y-3/2)^2+x^2+5x+\frac{23}4\\
&=(x+y-3/2)^2+\left(x+\frac52\right)^2-\frac12\\
&\geq-\frac12\\
&(x=-\frac52,\space y=4)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Seemingly Simple Integral $\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$. Evaluate $$\int_0^1 f(x) dx$$ where
$$f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}$$
I started off with the substitution $x=\sin y$, which resulted in the integrand reducing to
$$\sin^2y\cdot \ln (\sin y) dy$$
Then I used the property of definite integrals th... | $$
\begin{aligned}
I &\stackrel{x=\sin \theta}{=} \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta \ln (\sin \theta)}{\cos \theta} \cdot \cos \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \sin ^{2} \theta \ln (\sin \theta) d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d\left(\frac{\theta}{2}-\frac{\sin 2 \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must... | The value of $A$ is right. $A = -1$. When you plug in $x = -2$, you've forgotten to divide it by $x^2$, so you get $C = 12$. Actually, $C = 3$.
\begin{align}
-(x+2) + Bx(x+2) + 3x^2 &= 5x^2+3x-2 \\
3x^2-x-2 + Bx(x+2) &= 5x^2+3x-2 \\
Bx(x+2) &= 2x^2 + 4x = 2x(x+2)\\
B &= 2
\end{align}
\begin{align}
& \int \frac{-1}{x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Let z = 1 + i. Find the real and imaginary parts of z^19 Let z = 1 + i. Find the real and imaginary parts of $z^{19}$
| It is quickly determined that $\frac{z}{\sqrt{2}} = \frac{1+i}{\sqrt{2}} = e^{\pi i/4}$ and
\begin{align}
\left(\frac{z}{\sqrt{2}} \right)^{19} &= e^{19 \pi i/4} = e^{4 \pi i + 3 \pi i/4} = \cos(4\pi) \, \left( \cos\left(\frac{3 \pi}{4}\right) + i \, \sin\left(\frac{3\pi }{4} \right) \right) \\
&= \frac{-1 + i}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find BC/AB if I lies in the circumcircle of AEF Let $ABC$ be a right triangle with angle $B=\frac{\pi}{2}$. Let $E$ and $F$ be the midpoint of $AB$ and $AC$ respectively. If $I$ the in centre of $ABC$ lies on the circumcircle of $AEF$, find the ratio $BC/AB$.
My guess would be that $AB=BC$ but I am unable to prove it.... | Without loss of generality you may choose coordinates
$$A=\begin{pmatrix}0\\0\end{pmatrix}\qquad
B=\begin{pmatrix}1\\0\end{pmatrix}\qquad
C=\begin{pmatrix}1\\\lambda\end{pmatrix}\qquad
E=\begin{pmatrix}1/2\\0\end{pmatrix}\qquad
F=\begin{pmatrix}1/2\\\lambda/2\end{pmatrix}$$
Now the barycentric coordinates of the incent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of the digits of the sum of the digits? Problem
Let $(10^{2016}+5)^2=225N$. If $S$ is the sum of the digits of N, then find the sum of the digits of S
Attempt
Let's look at some smaller cases. We have $\dfrac{(10^{3}+5)^2}{225} = 4489$,$\dfrac{(10^{4}+5)^2}{225} = 444889$, and $\dfrac{(10^{5}+5)^2}{22... | hint
*
*$U_{n+1}=\frac{(10^{n+1}+5)^2}{225}=\frac{(10^{n}+5+10^n*9)^2}{225}$
= $\frac{(10^{n}+5)^2}{225}+\frac{2*(10^{n}+5)*(10^n*9)}{225}+\frac{(10^{n+1}+5)^2}{225}=U_{n}+10^{n-1}(44*10^{n-1}+4)$ after simplification
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How can one show the inequality $\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge a+b+c$ How can one show the inequality
$$\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge a+b+c$$
Where $a,b,c$ are real and $ab+bc+ac$ is no equal t... | Hint
Use Holder inequality:
$$\left(\sum_{cyc}\dfrac{a^2}{\sqrt{b^2-bc+c^2}}\right)^2\left(\sum_{cyc}a^2(b^2-bc+c^2)\right)\ge (a^2+b^2+c^2)^3$$
so we need prove to
$$(a^2+b^2+c^2)^3\ge (a+b+c)^2\sum_{cyc}a^2(b^2-bc+c^2)$$
This is not hard to prove it.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $x^{10}y^6z^{-21}$ in $(3x^2 -5y^{\frac{1}{2}} + z^{-3})^{24}$ with Binomial theorem. Find $x^{10}y^6z^{-21}$ and $x^{16}y^4z^{-18}$ in $(3x^2 -5y^{\frac{1}{2}} + z^{-3})^{24}$ with Binomial theorem.
Well, I believe I found $x^{10}y^6z^{-21}$ and $x^{16}y^4z^{-18}$ .
But, I'm a bit hesitated about the solution, be... | It is also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an algebraic expression.
Instead of using the multinomial version we could also successively apply the binomial theorem. We obtain
\begin{align*}
[x^{10}&y^6z^{-21}](3x^2-5y^{\frac{1}{2}}+z^{-3})^{24}\tag{1}\\
&=[x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ ... | Ніnt:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=- \left( x+8 \right) \left( x+2 \right) \left( x-1 \right) ^{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluating Indefinite Integral $$∫ \frac{1}{(1-x^3)^{1/3}}\, dx$$
I tried substituting $1-x^3$ as $t^3$ but I am not able to calculate it after that.
Thanks!
| For solving
$$
I=∫\frac{1}{\sqrt[3]{1-x^3}}\, \mathrm{d}x
$$
Let $t=\dfrac{x}{\sqrt[3]{1-x^3}}$,
solved
$$
\begin{aligned}
x&=\frac{t}{\sqrt[3]{t^3+1}}\\
\mathrm{d}x&=\frac{1}{(t^3+1)\sqrt[3]{t^3+1}}\mathrm{dt}\\
\end{aligned}
$$
and
$$
\frac{1}{\sqrt[3]{1-x^3}}=\sqrt[3]{t^3+1}
$$
then
$$
I = \int\frac{\sqrt[3]{t^3+1}}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$ Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
Partial integration can't ... | For $x \in [1,\infty)$, you have $$0 \le \frac{1}{x\sqrt[3]{x^2+1}} \le \frac{1}{x^\frac{5}{3}}$$ and $\int_1^\infty \frac{dx}{x^\frac{5}{3}}$ is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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A misunderstanding concerning $\pi$ The very well-known expression
$$\frac {\pi} {4} = 1 - \frac {1} {3} + \frac {1} {5} - \frac {1} {7} + \cdots$$
puts me face to face with a contradictory position. Let
$$s_N = \sum_{k = 0}^{N} \frac {1} {4k + 1} - \sum_{k = 0}^{N} \frac {1} {4k + 3}.$$
Then it is obvious that
$$\frac... | such mistakes are sometimes difficult to track down! however, your idea is an interesting one. it leads to the following expansion:
$$
S=\sum_{k=0}^{\infty} \frac1{4k+1}-\frac1{4k+3} \\
=1 -\sum_{k=1}^{\infty} \frac1{4k-1}-\frac1{4k+1}
$$
hence
$$
1-S =\sum_{k=1}^{\infty} \frac1{4k-1}-\frac1{4k+1} \\
=2\sum_{k=1}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Let $x^2+kx=0;k$ is a real number .The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set. Let $f(x)=x^2+kx;k$ is a real number.The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set.
The equation $x^2+kx=0$ has solutions $x=0,... | Let me try this way
$$f(x)=x^2+kx$$
$$f(f(x))=f(x)^2+kf(x)=f(x)(f(x)+k)$$
From $f(f(x))=f(x)(f(x)+k)$ the zeros of $f(x)$ are the zeros of $f(f(x))$. This makes you looking for other two remaining zeros since $f(f(x))$ is of 4-th degree.
Obviously they are contained in $f(x)+k$. Since you want to have the same real zer... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Hessian of a function Given this equation: $f(x,y,z) = \sqrt{1+x^2+y^2+z^2}$
I tried to calculate the Hessian --> for example $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{1+x^2+y^2+z^2}}$ The second derivativ respect to x is hard to calculate for me: I tried the product rule: $x*(1+x^2+y^2+z^2)^{-1/2}$.
Then i get: ... | What you get it's quite correct, because $$\frac{\partial f}{\partial x^2}=\frac{1}{\sqrt{1+x^2+y^2+z^2}}-\frac{x^2}{(1+x^2+y^2+z^2)^{3/2}}$$
And making some algebra you get $$\frac{\partial f}{\partial x^2}= \frac{1+y^2+z^2}{(1+x^2+y^2+z^2)^{3/2}}$$
I think that the other may be similar to this
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to use Mathematical Induction to prove $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}$? $$\frac{1}{1 \cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}$$
What I have so far in the induction is:
$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + ... | $$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{k+1}{k+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Identity involving $\arcsin$ and $\sin$. $$\sin z = \sin \left(\arcsin \frac{\pi}{6} - \arccos \frac{\pi}{6} \right)$$
Is there a specific expansion of the expression on the right hand side of the equation? To evaluate the value of $z$?
| As pointed out by @Bye_World, this looks like the trigonometric identity:
$$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \tag{1}$$
Since you already have:
$$arcsin \frac{\pi}{6} \Rightarrow \sin \alpha = \frac{\pi}{6} \tag{2}$$
$$arccos \frac{\pi}{6} \Rightarrow \cos \beta = \frac{\pi}{6} \tag... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find a straight line orthogonal to a curve? We want to find the equation of line passing through $(3,6)$ and cutting the curve $y=\sqrt{x}$ orthogonally .
I thought that it means we have to find a normal through that point equation of normal is $y=mx -2am -am^3 $.
Putting $(3,6)$ in, we get $(m+2)(m^2-4m+3)$.
T... | You need this result:
If two curves $f(x)$ and $g(x)$ (or a curve and a line), pass each other orthogonally at $x=c$, then $f(c)=g(c)$ and $f'(c)g'(c)=-1$.
Let the equation of the line be $g(x)=ax+b$ and let $f(x)=\sqrt{x}$. Then we get the following system of equations:
$$3a+b=6$$
$$ac+b=\sqrt{c}$$
$$a \cdot \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimal c satisfying $x+y-(xy)^c \geq 0$ for all $x,y\in [0,1]$ What is the minimal real $c$ satisfying $x+y-(xy)^c \geq 0$ for all $x,y \in [0,1]$?
Experimentally (though my experiments weren't necessarily accurate enough) I reached as low as $c=\tfrac{13}{32}$, where $c=\tfrac{12}{32}$ violates it.
| Your minimal value also does violate it. Consider $x=y=0.02$, then
$$x+y-(xy)^{13/32}=-0.00164730...$$
Consider the implicit function $$x+y=(xy)^c$$
It is easy to show that the plot is symmetric over $y=x$. Changing coordinates to polar and considering function $r(\theta)$ we can show that $\theta=\tfrac\pi4$ is the ma... | {
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"url": "https://math.stackexchange.com/questions/1583533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of $\sum \limits _{n=1}^{\infty} (-1)^{n} \frac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$ $\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$
How to check this? I've tried using Leibniz test, it's easy to prove that this one is monotonous, but its limit is rather no... | You had the right idea to try to use the alternating series test, however you should double check your calculations for the limit of the terms:
$\dfrac{2^n n!}{5 \cdot 7 \cdots (2n+3)}$ $= \dfrac{3 \cdot 2^nn!}{1 \cdot 3 \cdot 5 \cdots (2n+3)}$ $= \dfrac{3 \cdot 2^nn!}{1 \cdot 3 \cdot 5 \cdots (2n+3)} \cdot \dfrac{2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating functions - deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ I would like some help with deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ using generating functions.
I have managed to do this for $1^2 + 2^2 + 3^2 +\cdots$ by putting
$$f_0(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 +\cdots$$
$$f_1(x)... | To answer otherwise, you can use the known fact that the sum of the first $n$ positive odd numbers is equal to $n^2$. So you have
$$\begin{cases}1=1^2\\1+3=2^2\\1+3+5=3^2\\1+3+5+7=4^2\\................\\1+3+5+…….+(2n-1)=n^2 \end{cases}$$ Therefore
$$\sum_{i=1}^{i=n}n^2=n+(n-1)3+(n-2)5+(n-3)7+………+2(2n-3)+(2n-1)$$ Tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584854",
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"source": "stackexchange",
"question_score": "11",
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Proving that an equation has a solution
If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}=0$$
has at least one solution in the interval $(-1,1)$
My solution is as follows, I could not be sure on its correctness. If you have any other solutions, could you please... | Sketch of solution. May be unstable. :-)
Rewrite $f(x)$ as
$$
f(x) = \frac{a(x^3+x-2)+b(x^3+2x^2-1)}{(x^3+2x^2-1)(x^3+x-2)}
$$
and denote the numerator by
$$
g(x) = (a+b)x^3+2bx^2+ax-(2a+b)
$$
and the denominator by
$$
h(x) = (x^3+2x^2-1)(x^3+x-2)
$$
Note that $h(x)$ has a zero (and therefore $f(x)$ has a pole) at $x_0... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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What is $\frac{1}{1+\sqrt[3]{2}}$ in $\mathbb{Q}(\sqrt[3]{2})$? Since $\mathbb{Q}(\sqrt[3]{2})$ is a field, any number $\neq 0$ has a reciprocal.
How then to write $\frac{1}{1+\sqrt[3]{2}}$ as a number $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ with fractions $a,b,c \in \mathbb{Q}$?
In general, how to "rationalize denominato... | Use the expansion $(1+x^3)=\left( x+1 \right) \left( {x}^{2}-x+1 \right)$.
Then
$$
\frac{1}{1+\sqrt[3]{2}}=\frac{1-\sqrt[3]{2}+(\sqrt[3]{2})^2}{(1+\sqrt[3]{2})(1-\sqrt[3]{2}+(\sqrt[3]{2})^2)}=\frac{1-\sqrt[3]{2}+(\sqrt[3]{2})^2}{3}=\frac{1}{3}-\frac{\sqrt[3]{2}}{3}+\frac{(\sqrt[3]{2})^2}{3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Nice way to solve $\int\int \frac{1}{1-(xy)^2} dydx$? This is something I've been thinking about lately;
$$\int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx$$
Solutions I've read involve making the substitutions: $x= \frac{sin(u)}{cos(v)}$ and $y= \frac{sin(v)}{cos(u)}$. This reduces the integral to the area of a right triang... | Here is an alternative derivation. The form of the integrand suggest expanding it a geometrical series
$$\frac{1}{1-(xy)^2} = \sum_{n=0}^\infty x^{2n}y^{2n}$$
Now integrating term by term we get
$$\int_0^1\int_0^1\frac{{\rm d}x{\rm d}y}{1-(xy)^2} = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$
and from the well known result $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Sine of argument with large n approximation I have worked an integral and reduced the integral to
$$\frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}$$
I want to show that for $$n\rightarrow \infty$$ the above equation reduces to
$$\frac{1}{2}$$
Evidently, this means the $... | First notice that
$$\begin{align}
\sin(\frac{3n\pi}{2})&=\sin(\frac{n\pi}{2}+n\pi) \\
&=\sin(\frac{n\pi}{2})\cos(n\pi)+\cos(\frac{n\pi}{2})\sin(n\pi) \\
&=\sin(\frac{n\pi}{2})\cos(n\pi)+0 \\
&=(-1)^n\sin(\frac{n\pi}{2})
\end{align}$$
and hence
$$\sin(\frac{n\pi}{2})-\sin(\frac{3n\pi}{2})=(1-(-1)^n)\sin(\frac{n\pi}{2})$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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How to prove $2\sqrt{2+\sqrt{3}}=\sqrt{2}+\sqrt{6}$? My calculator and I were arguing one day about the cosine of some number.
The calculator said "$\cos(\frac x2)=\sqrt{2}+\sqrt{6}$".
I said "That's absurd because $\cos(\frac x2)=\sqrt{\frac{1+\cos(x)}2}$, which evaluates to $2\sqrt{2+\sqrt{3}}$ for this particular $x... | A way of showing this that has not yet been given by anyone is to obtain a "nice equation" (i.e. with integer coefficients) having $\;\sqrt{2}+\sqrt{6}\;$ as a solution and then solving this equation by standard methods.
Start by setting $\;x = \sqrt{2}+\sqrt{6}.\;$ Subtracting $\sqrt 2$ from both sides and then squari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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Mathematically, how does one find the value of the Ackermann function in terms of n for a given m? Looking at the Wikipedia page, there's the table of values for small function inputs. I understand how the values are calculated by looking at the table, and how it's easy to see that 5,13,29,61,125 is $2^{n+3}-3$, but ho... | $$A(0,n) = n+1 \;\text{(by definition)}$$
$$A(1,n) \rightarrow A(0,A(1,n-1)) \rightarrow A(1,n-1)+1 \rightarrow A(1,n-2)+2\Rightarrow A(1,0)+n$$
$$\rightarrow A(0,1)+n \rightarrow 2+n = \color{red}{2+(n+3)-3}$$
$$A(2,n) \rightarrow A(1,A(2,n-1)) \rightarrow A(2,n-1)+2 \rightarrow A(2,n-2)+4 \Righta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587557",
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"source": "stackexchange",
"question_score": "5",
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How to prove $\sum\left(\frac{a}{b+c}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$ The question is to prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$$
$$a,b,c>0$$
I tried Cauchy, AM-GM, Jensen, etc. but... | $$\sum_{cyc}\frac{a^2}{(b+c)^2}-\frac{3(a^2+b^2+c^2)}{4(ab+ac+bc)}=$$
$$=\sum_{cyc}\left(\frac{a^2}{(b+c)^2}-\frac{3a^2}{4(ab+ac+bc)}\right)=$$
$$=\sum_{cyc}\frac{a^2(4ab+4ac-2bc-3b^2-3c^2)}{4(ab+ac+bc)(b+c)^2}=$$
$$=\sum_{cyc}\frac{a^2((a-b)(3b+c)-(c-a)(3c+b))}{4(ab+ac+bc)(b+c)^2}=$$
$$=\frac{1}{4(ab+ac+bc)}\sum_{cyc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$... | Use AM-HM Inequality for both.
$$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}} \Rightarrow \frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$$
Similarly, you get $$\frac{1}{a}+\frac{1}{c}\ge \frac{4}{a+c}$$ and $$\frac{1}{c}+\frac{1}{b}\ge \frac{4}{c+b}$$
Now add the three and get the left inequality.
For right inequality,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi$ How to compute
$$\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi ?$$
The answer I am provided with is $\dfrac {(-1)^n\pi r^n} {x^{n+1}}$ for $\ x>r$, but I have no id... | Suppose we seek to evaluate
$$\int_0^{2\pi} \frac{(r\cos\phi+x)\cos(n\phi)}{r^2+2xr\cos\phi+x^2}
\; d\phi.$$
Introduce $z=\exp(i\phi)$ so that $dz=iz \; d\phi$ to get
$$\int_{|z|=1}
\frac{(r(z+1/z)/2+x)(z^n+1/z^n)/2}{r^2+xr(z+1/z)+x^2}
\frac{dz}{iz}
\\ = \frac{1}{4i} \int_{|z|=1}
\frac{(r(z+1/z)+2x)(z^n+1/z^n)}{(r^2+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How can I count solutions to $x_1 + \ldots + x_n = N$? I am interested in how many non-negative integer solutions there are to:
$$x_1 + \ldots + x_N = B$$
where at least $K$ of the variables $x_1, \ldots , x_N \geq C$
For example when:
$B = 5, N = 3, K = 2, C = 2$
I want to count the solutions to:
$$x_1 + x_2 + x_3 = 5... | You could move to new variables $x_i' = x_i - C \ge 0$ for the variables with the minimum value $C$. Then the equation transforms to
$$
x'_1 + \ldots x'_K + x_{K+1} + \ldots x_N = B - K \, C
$$
assuming WLOG GRUMBL MUMBL that those variables are the first $K$ ones.
This is a problem of the form
$$
n_1 + \ldots + n_N = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find all non negative integers x,y,z so we get a whole square Find all $x,y,z\in\mathbb{N_0}$ so that there exists a $k\in\mathbb{N}$ so that
$$4^x+4^y+4^z=k^2$$.
We can transform this problem to:
Find all $a,b\in\mathbb{N_0}$ so that there exists a $t\in\mathbb{N}$ so that
$$4^a+4^b+1=t^2$$
where $a=x-z, b=y-z, k=2^z ... | Consider ,
$4^{x}+4^{y}+4^{z}=k^{2}.$
Assume that $x \geq y \geq z$.
It is immediately obvious that $k$ is even.
Let $k=2m$ where $m$ is an integer.
Thus we have,
$4^{x}+4^{y}+4^{z}=4m^{2}$, assuming that $x,y,z \geq 2$, we have,
$4^{x-1}+4^{y-1}+4^{z-1}=m^{2}$, and so we return to our intial equation.
If we repeat t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n... | You only need modular arithmetic here: both $3$ and $5$ have order $2$ modulo $8$, i.e. $3^r\equiv3^{r\mod 2},\enspace 5^r\equiv5^{r\mod 2}\pmod 8$. Now
*
*If $n$ is odd, $5^n\equiv 5$ and $3^{n-1}\equiv 1\mod8$, so
$$5^n + 2*3^{n-1}+ 1\equiv 5+2+1\equiv 0\mod8.$$
*If $n$ is even, $5^n\equiv 1$ and $3^{n-1}\equiv 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $S=\{z\in\mathbb C:z\neq0\text{ and }\left|z+\frac{1}{z}\right|=a\}$ find $\min\limits_{z\in S}|z|$
If $a$ is a positive real number and $S=\{z\in \mathbb C : z\neq 0\ \text{ and } \vert z+\frac{1}{z}\vert = a\}$, Find $\min\limits_{z\in S}\vert z\vert$ and $\min\limits_{z\in S}\vert\frac{1}{z}\vert$
I first tried... | Let $z=re^{i\phi}$.$\,\,\,$ If $|z+z^{-1}|=a$, then
$$r^2+r^{-2}+2\cos(2\phi)=a^2 \tag 1$$
We can write $(1)$ in a more convenient form as
$$\left(r+\frac1r\right)^2=a^2-4\sin^2(\phi) \tag 2$$
Note that $|a|\ge 2$ since the minimum of $r+\frac1r$ is $2$.
Now, differentiating both sides of $(2)$ with respect to $\phi$... | {
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"timestamp": "2023-03-29T00:00:00",
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How to find $P(-1)$ for $\frac{P(2x)}{P(x+1)}=8-\frac{56}{x+7}$ and $P(1)=1$? $P(x)$ is a polynomial such that $P(1)=1$ and $\frac{P(2x)}{P(x+1)}=8-\frac{56}{x+7}$ and $P(-1)$ is rational. How to find $P(-1)$?
| $8-\frac{56}{x+7}=\frac{8x+56-56}{x+7}=\frac{8x}{x+7}=\frac{P(2x)}{P(x+1)}$.
Expanding, we get,
$8xP(x+1)=(x+7)P(2x)$
It can be seen that $2x|P(2x)$, so $P(x)=xQ(x)$ for some polynomial $Q$.
Also, $x+7|P(x+1)$, so $P(x)=(x+6)R(x)$ for some polynomial $R$.
Combining the above results, $P(x)=x(x+6)S(x)$. We attempt to su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\... | For the second inequality multiply by $a$ to get
$$\ln \left(1 + \frac 1a\right)^a < 1$$ or
$$\left(1 + \frac 1a\right)^a < e $$
which is correct since function $f(a) = \left(1 + \frac 1a\right)^a$ is strictly increasing and ha a limit at $+\infty$ equal to $e$.
To prove the first part use the function $g(a) = \left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute $\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used? Thanks
$$\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\ri... | $$
\lim_{x \to 1^+} \left(\frac{\tan\sqrt{x-1} }{\sqrt{x^2-1}}\right) = \lim_{x \to 1^+} \left(\frac{\tan\sqrt{x-1} }{\sqrt{x-1}\sqrt{x+1}}\right) = \lim_{x \to 1^+} \left(\frac{\sin\sqrt{x-1} }{\sqrt{x-1}} \cdot \frac{1}{\cos\sqrt{x-1}} \cdot \frac{1}{\sqrt{x+1}}\right) =$$
$$ \lim_{x \to 1^+} \left(\frac{\sin\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$ Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
I found the general term of the sequen... | Given $$1+\frac{1}{k^2}+\frac{1}{(k+1)^2} = 1+\frac{1}{k^2}+\frac{1}{(k+1)^2}-\frac{2}{k(k+1)}+\frac{2}{k(k+1)}$$
So $$1^2+\left[\frac{1}{k}-\frac{1}{(k+1)}\right]^2+2\left[\frac{1}{k}-\frac{1}{(k+1)}\right]=\left[1+\frac{1}{k}-\frac{1}{k+1}\right]^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Approximation of $\frac{x}{\sqrt{x^2+R^2}}$ How do you prove this statement?
If $x\gg R$ then
$$\frac{x}{\sqrt{x^2+R^2}}\cong 1-\frac{1}{2}\left(\frac{R}{x}\right)^2$$
I have no ideas even how to start.
| Compare these two functions at $x=\infty$. If their quotient $f/g$ tends to $1$, it means $f\sim g$:
$$\lim_{x\to\infty} \frac{\frac{x}{\sqrt{x^2+R^2}}}{1-\frac{1}{2}\left(\frac{R}{x}\right)^2}=
\lim_{x\to\infty} \frac{\frac{x}{\sqrt{x^2+R^2}}}{\frac{2x^2-R^2}{2x^2}}=
\lim_{x\to\infty} \frac{x\cdot 2x^2}{\sqrt{x^2+R^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
| $$x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)=\frac{x\left(\left(\sqrt{x^2-x}\right)^2-\left(\sqrt{x^2-1}\right)^2\right)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
$$=\frac{x(1-x)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
Since we're searching for the limit as $x\to -\infty$, let $x<0$. Then:
$$=\frac{\frac{1}{-x}(x(1-x))}{\sqrt{\frac{x^2}{(-x)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx... | When you have a second degree polynomial in the denominator with $\Delta <0$, you can proceed like this: find the solutions, in this case $\frac{-1}{4} \pm \frac{i\sqrt {23}}{4}$, and then write the polynomial in the usual way
$$2\Big(x+\frac{1}{4} + \frac{i\sqrt {23}}{4}\Big)\Big(x+\frac{1}{4} - \frac{i\sqrt {23}}{4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
| Setting $t=\sqrt[3]{35-x^3}$, we have
$$tx(t+x) = 30$$
and
$$t^3+x^3=35 \implies (t+x)(t^2+x^2-tx)=35$$
This gives us
$$\dfrac{t^2+x^2-tx}{tx} = \dfrac76 \implies 6t^2-13tx+6x^2 = 0 \implies 6t^2 - 9tx - 4tx + 6x^2 = 0$$
This gives us
$$3t(2t-3x)-2x(2t-3x) = 0 \implies (2t-3x)(3t-2x) = 0 \implies t = \dfrac{2x}3, \dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve $ \int {\frac{(x-1)dx}{(x-2)(x+1)^2 x^2 }} $ $$ \int {\frac{(x-1)dx}{(x-2)(x+1)^2 x^2 }} $$
I don't know about decomposition of fractions a lot,but I know that it is method which I have to use in this example. Please, help me.
| $$ \frac{x-1}{(x-2) x^2 (x+1)^2} = \frac A{x-2}+\frac{Bx+C}{x^2} + \frac{D(x+1)+E}{(x+1)^2} $$
$$ \frac{x-1}{ x^2(x+1)^2} = A + \frac{Bx+C}{x^2}(x-2) + \frac{D(x+1)+E}{(x+1)^2}(x-2). $$
$x\to 2: A = \frac1{36}.$
$$ \frac{x-1}{(x-2)(x+1)^2} = \frac A{x-2}x^2+Bx+C + \frac{D(x+1)+E}{(x+1)^2}x^2 $$
$x\to 0: C = \frac12.$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$ Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$
Let $P(x)=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$
By using $AM-GM$ inequality on the first two terms give me
$8(27)^{\log_{6}x}+27(8)^{\log_... | Angelo Mark has provided a good hint, but it seems that the OP needs more.
We can write
$$\begin{align}8\cdot 3^t+27\cdot 2^t-6^t&=8\cdot 3^t+27\cdot 2^t-2^t\cdot 3^t-8\cdot 27+8\cdot 27\\&=8(3^t-27)-2^t(3^t-27)+8\cdot 27\\&=(3^t-27)(8-2^t)+8\cdot 27\end{align}$$
Note here that $(3^t-27)(8-2^t)$ is non-positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Which integers can be represented as the most pair of difference of two squares? Let $f(x)$ be the number of $a,b,x\in \mathbb N$ where $a^2-b^2=x$.
For example, 1971 is not only $986^2-985^2$ but also $50^2-23^2$, $114^2-105^2$, $330^2-327^2$. So, $f(1971)=4$. Is there some sort of limit to $f(x)/x$ or $f(x)/ln(x)$ or... | Partial answer. If $x$ is odd, then $a=(x+1)/2$, $b=(x-1)/2$, satisfy $x=a^2-b^2$.
If $x$ is even, and $a$, $b$ exist, so that $x=a^2-b^2$, then $x=(a-b)(a+b)$, and clearly, both $a+b$, $a-b$ are even, and hence $x$ is divisible by $4$. Then, $a=(x+4)/4$ and $y=(x-4)/4$, satisfy $x=a^2-b^2$.
Note that in the first case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{... | Let $A=\cos x.$ We have $\cos 2x=2A^2-1.\quad$$ \cos 3 x= 4A^3-3A.\quad$$ \cos 4x=2\cos^22x-1=8A^4-8A^2+1.\quad$$\cos 6x=2(\cos 3x)^2-1=32A^6-48A^4+18A^2-1.$ From the expressions for $\cos 2x,\; \cos 4x,\;\cos 6x$ we have $32A^6=\cos 6x+6\cos 4x+15\cos 2x+10$. So $\cos^6x=\sum_{n=0}^{\infty}(-1)^nx ^{2n}2^{-5}(6^{2n}+6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Evaluating the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$
Evaluate $$\int_{0}^{\infty}\frac{dx}{1+x^3}.$$
I tried integration by partial fraction. My work is below:
$$\int_{0}^{\infty}\frac{dx}{1+x^3}=\frac{1}{3} \int_{0}^{\infty}\frac{1}{x+1}+\frac{1}{3} \int_{0}^{\infty}\frac{2-x}{x^2-x+1}.$$
It seems that... | Hint. You may first integrate over the finite set $[0,M]$ then let $M \to \infty$:
$$\int_{0}^M\frac{dx}{1+x^3}=\frac{1}{3} \int_{0}^M\frac{1}{x+1}\:dx+\frac{1}{3} \int_{0}^M\frac{2-x}{x^2-x+1}\:dx,$$ you will see that the $\log$ terms cancel.
Edit. Let $M>0$. We have $$\frac{1}{3} \int_{0}^M\frac{1}{x+1}\:dx=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
how to find hyperbola equation knowing tangent line and point I have a problem.
A hyperbola passes through point $(3,2)$ and $9x+2y-15=0$ is a tangent line. Find the equation of hyperbola.
| Let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be the hyperbola.It passes through $(3,2)$.
So $\frac{9}{a^2}-\frac{4}{b^2}=1................(1)$
Also tangent $9x+2y-15=0$ is given.
Put $y=\frac{-9x+15}{2}$ in the equation of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{x^2}{a^2}-\frac{1}{b^2}\times\frac{81x^2+225-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\lim \frac{x^2-\sin^2{x}}{\tan(3x^4)}$ as $x$ goes to $0$
Calculate $\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)}$
How does one calculate this limit?
Is it valid to say, since $\sin^2{x}$ is approximated by $x^2$ as $x \to 0$, we have:
$\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)} =... | No, it is not valid to approximate $\sin^2 x$ by $x^2$, since there are higher order terms that are still significant. Using the expansion
$$\sin x = x - \frac{x^3}{6} + O(x^5)$$
we have that
$$\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$$ so that
$$\frac{x^2 - \sin^2 x}{\tan(3x^4)} = \frac{\frac 1 3 x^4 + O(x^6)}{\tan(3x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.
That's what I've tried:
Let a Cauchy-Schwarz Inequality be :
\begin{array}
(((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2... | If you insist on C-S, let
$$
u=(a,b,c),\quad v=(b,c,a).
$$
Then,
$$
\text{RHS}=u\cdot v\underbrace{\leq}_{\text{C-S}}||u||\times||v||=\text{LHS}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show the following inequality holds I want to show that $( 1 + x_1 ) (1 + x_2 )... ( 1 + x_n ) \ge ( 1 + (x_1 x_2 ... x_n ) ^\frac {1} {n } ) ^ n $ for all $x_i > 0$.
I started by taking logarithm on both sides and trying to use the concavity of logarithm but this reverses the inequality. I don't really know how to sta... | by induction:
$n=2, (1+x_1)(1+x_2)=1+x_1+x_2+x_1x_2 \ge 1+2\sqrt{x_1x_2}+x_1x_2=(1+(x_1x_2)^{\frac{1}{2}})^2$
when $n=k, (1+x_1)(1+x_2)...(1+x_k) \ge (1+(x_1x_2...x_k)^{\frac{1}{k}})^k$ holds,
$n=k+1, (1+x_1)(1+x_2)...(1+x_k)(1+x_{k+1}) \ge (1+(x_1x_2...x_k)^{\frac{1}{k}})^k(1+x_{k+1}) $
now, we prove
$(1+(x_1x_2...x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Why this system have one solution Let $b\in (1,2),x\in (0,\frac{\pi}{2})$,if such
$$\begin{cases}
2b^2+b-4=2\sqrt{4-b^2}\cos{x}\\
2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}
\end{cases}$$
show that:$$x=\dfrac{\pi}{6}$$
Here is what I already got.
First of all, one should notice equation $x=\d... | Solving this is mainly just fiddly algebra.
$$2b^2+b-4=2\sqrt{4-b^2}\cos{x}$$
$$2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}$$
The second can be rewritten as:
$$2b^2-4=2b\left(\cos x\cos\frac{\pi}{18}-\sin x\sin\frac{\pi}{18}\right)-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}$$
Rearranging the first one ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve $\sin 2x=-\cos x$ I'm working on solving the problem $\sin(2x)=-\cos(x)$ but I got stuck.
I got the following:
$\sin 2x=-\cos x \Leftrightarrow \sin 2x=\cos(x+\pi )\Leftrightarrow \sin 2x=\sin\left(\frac{\pi }{2}-(x+\pi)\right)\Leftrightarrow \sin 2x= \sin\left(-\frac{\pi }{2}-x\right)$
then I did
$2x =-\frac{\... | $$\sin 2x = -\cos x $$
$$2\sin x\cos x = -\cos x$$
$$2\sin x\cos x + cos x= 0$$
$$\cos x (2\sin x + 1) = 0$$
$$\cos x = 0 \text{ or } \sin x = -\frac{1}{2}$$
$$x = (2n + 1)\frac{\pi}{2} \text{ or } x = m\pi + (-1)^{m+1}\frac{\pi}{6}$$
where $m,n\in\Bbb Z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Interval for area bounded by $r = 1 + 3 \sin \theta$ I'm trying to calculate the area of the region bounded by one loop of the graph for the equation
$$
r = 1 + 3 \sin \theta
$$
I first plot the graph as a limaçon with a maximum outer loop at $(4, \frac{\pi}{2})$ and a minimum inner loop at $(-2, -\frac{3 \pi}{2})$. I ... | $$
\begin{align}
\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12r^2\,\mathrm{d}\theta
&=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12(1+3\sin(\theta))^2\,\mathrm{d}\theta\\
&=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12\left(1+6\sin(\theta)+9\sin^2(\theta)\right)\mathrm{d}\theta\\
&=\int_{\arcsin(-1/3)}^{\pi-\arcs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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A $\log \Gamma $ identity: Where does it come from? $$\log \Gamma (n)=n\log n -n +\frac{1}{2} \log \frac{2\pi}{n}+\int_0^\infty \frac{2\arctan (\frac{x}{n})}{e^{2\pi x}-1} \,\mathrm{d}x$$
Is an identity that is derived from using Sterling's approximation. I can't quite figure out how it was used, and was wondering for ... |
Proposition : $$\int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right]$$
Proof : Let $ \displaystyle \text{I} (a) = \int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x$
$\displaystyle = -\sum_{r=1}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How to show $\phantom{d}_d C_0+\phantom{d}_d C_4 + \cdots = 2^{d-2} + 2^{\frac{d}{2}-1} \cos(\frac{d \pi}{4}) $? I want to show following identities
\begin{align}
&\phantom{d}_d C_0+\phantom{d}_d C_4 + \cdots = 2^{d-2} + 2^{\frac{d}{2}-1} \cos(\frac{d \pi}{4}) \\
&\phantom{d}_d C_1+\phantom{d}_d C_5 + \cdots = 2^{d... | HINT:
As the subscript of each subsequent binomial coefficient differ $4,$
If $x^4=1\implies x=\pm1,\pm i$
Set $x=\pm1,\pm i$ one by one in the following identity $$(1+x)^n=\sum_{r=0}^n\binom nr x^r$$
Find $$\dfrac{(1+1)^n\pm(1-1)^n\pm(1+i)^n\pm(1-i)^n}4$$
$1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)$
$(1+i)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $A,B,A+I,A+B$ are idempotent matrices how to show that $AB=BA$?
If $A,B,A+I,A+B$ are idempotent matrices how to show that $AB=BA$ ?
MY ATTEMPT:
$A\cdot A=A$
$B\cdot B=B$
$(A+I)\cdot (A+I)=A+I$ or, $A\cdot A+A\cdot I+I\cdot A+I\cdot I=A+I$ which implies $A\cdot I+I\cdot A=0$ (using above equations)
$(A+B)\cdot (A+B... | Let $a$ and $b$ be elements of a (possibly noncommutative) ring. If $a$, $b$, $a+1$ and $a+b$ are idempotent, then $ab = ba$.
Proof. From $a^2 = a$ and $(a+1)^2 = a+ 1$ one gets $a^2 + a + a + 1 = a + 1$, whence $a^2 + a = 0$. Since $a^2 = a$, it follows that $a + a = 0$ and finally $a = -a$. Now, from $(a+b)^2 = a + b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $\log_{1/3}(x^2-3x+3)≥0$ I want to solve $$\log_{1/3}(x^2-3x+3)≥0$$
Now I know the result is: $x ∈ <1;2>$, but i am not sure how to get it.
My thoughts: $\frac{1}{3}$ to the power of positive number $= (x^2-3x+3)$, now I would solve $x^2-3x+3$ with the help of discriminant to get the points where $x$ is zero, and... | $$\log_{1/3}(x^2-3x+3) \ge 0$$
This occurs when:
$$x^2-3x+3 \le 1$$
$$x^2-3x+2 \le 0$$
Factorizing, we get:
$$(x-1)(x-2) \le 0$$
Solving the inequality, we get:
$$1 \le x \le 2$$
ALTERNATE SOLUTION
Consider the following equation:
$$\log_{1/3}(x^2-3x+3) = 0$$
If we solve, we get:
$$x^2 -3x + 2 = 0$$
$$(x-1)(x-2) = 0$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Limit: $\lim_{n \to \infty} \frac{3^n+5^n+10^n}{-2^{n+1}+5^{n+1}+10^{n+1}}$ Limit: $$\lim_{n \to \infty} \frac{3^n+5^n+10^n}{-2^{n+1}+5^{n+1}+10^{n+1}}$$
I know the limit is $\frac{1}{10}$, but I am not sure how to get to it. Any help would be greatly appreciated.
| $$\lim _{ n\to \infty } \frac { 3^{ n }+5^{ n }+10^{ n } }{ -2^{ n+1 }+5^{ n+1 }+10^{ n+1 } } =\lim _{ n\to \infty } \frac { { 10 }^{ n }\left( \frac { { 3 }^{ n } }{ { 10 }^{ n } } +\frac { { 5 }^{ n } }{ { 10 }^{ n } } +1 \right) }{ { 10 }^{ n+1 }\left( -\frac { { 2 }^{ n+1 } }{ { 10 }^{ n+1 } } +\frac { { 5 }^{ n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find invertible matrices without Jordan form Here is the statement of the problem: Let $ A $ be the following matrix with entries from $ \mathbb{R} $ $$ \begin{pmatrix} 1 & 2 & -4 & 4 \\ 2 & -1 & 4 & -8 \\ 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 3\end{pmatrix} $$
Let $ A^t $ be its transpose. Find an invertible matrix $ P \in M... | Normally, for "small" computational problems, it is usually best to work them in a straightforward manner, even if it is a bit tedious, because that is usually faster than finding a shortcut. In this problem however, you seem to have found the key to the shortcut.
By using the useful identity in your question, we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.
I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps.
On this topic we learned also about Cauchy inequality, but I have... | If you don't like to my first solution, here is a more simple solution.Assume without loss of generality that A,B,C belongs to the first quadrant.And so, it's easy to see that:
\begin{align*}
x+y\geq 2\sqrt{xy} \tag{1}
\end{align*}
\begin{align*}
x+z\geq 2\sqrt{xz} \tag{2}
\end{align*}
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 6
} |
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
$729=9^3$
For any number to be divisible by $9$, the sum of the digits have to be divisible by $9$. The given number is divisible by $9$.
Then I tried dividing th... | According to Wolfy,
or by repeated use of
$x^3-1
=(x-1)(x^2+x+1)
$,
$\frac{x^{81}-1}{x-1}
= (x^2+x+1) (x^6+x^3+1) (x^{18}+x^9+1) (x^{54}+x^{27}+1)
$.
Since
$x^{3n} \equiv 1 \bmod (x^3-1)$,
$x^{6n}+x^{3n}+1
\equiv 3 \bmod (x^3-1)
$.
Therefore
the right 3 factors
are all
$\equiv 3 \bmod (x^3-1)
$.
Therefore the whole pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Find all natural numbers $x,y$ such that $3^x=2y^2+1$. Find all natural numbers $x,y$ such that
$$3^x=2y^2+1$$
solutions are $(1,1)$, $(2,2)$, $(5,11)$. I found that parity of both is same and If $x$ Is odd it is of the form $4k+1$.
| Here's a somewhat simpler proof using Pell equations
(as Eric Towers suggested might be possible).
If $x$ is odd, say $x=2k+1$, then $3z^2 = 2y^2 + 1$ where $z = 3^k$.
Hence $(2y)^2 - 6z^2 = -2$ and we have
$$
2y + 3^k \sqrt{6} = (2+\sqrt{6}) (5 + 2\sqrt{6})^m
$$
for some $m=0,1,2,\ldots$. The first two solutions $m=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}... | Your problem is very simple. :)
Why admin delete my ANS?????????????????????
Here's my Hint:
$$\left | \dfrac{x}{x+4} \right |<4 \overset{x \ne -4}{\rightarrow} -4<\dfrac{x}{x+4}<4 \Leftrightarrow \left\{\begin{matrix}
&-4<\dfrac{x}{x+4} \\
& \dfrac{x}{x+4}<4
\end{matrix}\right.
\Leftrightarrow
\left\{\begin{matrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Is $7^{101}+8^{101}$ divisible by 25? If not, what is $ 7^{101} + 8^{101} \bmod 25$ What I derived is: $$\begin{align}7^{101}+8^{101} &\equiv (5+2)^{101}+ (5+3)^{101} \\
&\equiv 2^{101}+101\cdot5\cdot2^{100}+3^{101}+101\cdot 5\cdot 3^{100} \\
&\equiv 2^{101}+(100+1)\cdot5\cdot 2^{100}+3^{101}+(100+1)\cdot5\cdot3\cdot2^... | By Euler's theorem, for any $n$ coprime with $25$, $n^{\varphi(25)}=n^{20}\equiv1\mod25$, hence $n^k\equiv n^{k\bmod20}\mod25$, so that
$$7^{101}+8^{101}\equiv 7+8=15\mod 25.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $x^{4}-2x^{3}+4x^{2}+6x-21=0$ Solve the equation $$x^{4}-2x^{3}+4x^{2}+6x-21=0$$ given that two of its roots are equal in magnitude but opposite in sign.
I don't know how to solve it. The roots are given as $\pm\sqrt{3},1\pm i\sqrt{6}. $
| You are given the information that some polynomial $x^2-c$ divides $x^4-2x^3+4x^2+6x-21$. Now polynomial division gives:
$x^4-2x^3+4x^2+6x-21 = (x^2-c) \cdot (x^2-2x+c+4) + 2 (3-c) x+ (c^2+4c-21).$
But the remainder is supposed to be zero, so we get:
$2 (3-c) x+ (c^2+4c-21) = 0$
and in particular $c=3$. Thus,
$x^4-2x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then
$f(... | We can solve this one in the same manner as here.
So, it is easy to see that $f(0)=0$ and $f$ is odd.
Let $f(1)=0$. If we put $y=1$ we get $$f(x^2-1) = (x-1)f(x)$$ and if we put $-x$ instead of $x$ we get $$f(x^2-1) = -(x+1)f(-x) = (x+1)f(x)$$ So we have $$(x-1)f(x)= (x+1)f(x)\implies f(x)=0\;\;\;\forall x$$ and we are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to simplify this? What did I do wrong? $$(3-\sqrt3) (2-\sqrt3)-\sqrt3\cdot\sqrt{27}.$$
So I simplified the $\sqrt3\cdot\sqrt{27}$ part into
$$\sqrt 3\cdot\sqrt{9\cdot3} = \sqrt3\cdot3\cdot\sqrt3=4\cdot\sqrt3$$
Then I multiplied the brackets :
$$6+\sqrt3-2\cdot\sqrt3-3\cdot\sqrt3 =6-4\cdot\sqrt3.$$
Then minus the $... | Your mistake was in the equality $$\sqrt 3 \cdot 3 \cdot \sqrt 3 = 4\cdot \sqrt3$$ which is false.
$$\sqrt{3}\sqrt{9\cdot 3} = \sqrt{3}\cdot 3\cdot \sqrt{3} = \sqrt{3}\cdot \sqrt{3} \cdot 3 = 3\cdot 3 = 9 \neq 4\cdot \sqrt 3$$
Also, you made a mistake in multiplying the brackets:
$$(3-\sqrt 3)(2-\sqrt 3) = 6-3\sqrt 3 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| There are only three possibilities for the divisibility of an integer by $3$, which are: no remainder, a remainder of $1$, or a remainder of $2$. But if we multiply $2$ by itself over and over again, the no remainder option is impossible, as that would mean that $2$ is a multiple of $3$, which it is not.
The thing is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 8
} |
Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.
The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequalit... | We have to use the word "triangle" at some point.
$$ a^2+b^2+c^2 \le a(b+c)+b(a+c)+c(a+b)=2(ab+ac+bc)=6$$
hence
$$(a+b+c)^2\le 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Error when approximating $\int_0^{1}(x^{2}+x)dx$ with midpoint rule. Task is to define the exact error when approximating $\int_0^{1}(x^{2}+x)dx$ with midpoint rule using n subintervals. I know the error term is $E(f)=\frac{1}{24}(b-a)f^{''}(\varepsilon)h^{2}$ but im not sure if this is the exact form and if it is how ... | I believe i know the answer now if someone needs it. $M_n=h(f(\frac{1}{2n})+f(\frac{3}{2n})+f(\frac{5}{2n})+,..,+f(\frac{1-\frac{1}{n}+1}{2}))$
$M_n=\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{2k+1}{2n})=\frac{1}{n}\sum_{k=0}^{n-1}(\frac{2k+1}{2n})+\frac{1}{n}\sum_{k=0}^{n-1}(\frac{2k+1}{2n})^{2}$
$M_n=\frac{1}{2n^{2}}\sum_{k=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this limit involving cube root and infinity? $\lim_{x\to\infty} \left((x^3+x^2)^{1/3} - (x^3-x^2)^{1/3})\right)$ How can I solve this limit? I know the answer is $2/3$.
I tried factorisation, but solving the complicated denominator using L'Hopital's Rule returns a wrong answer, $0$.
$$ \lim_{x\to\infty} \l... | You can use the identity $$a^3−b^3=(a−b)(a^2+ab+b^2)$$ with $a=(x^3+x^2)^{1/3}$ and $b=(x^3-x^2)^{1/3}$. So, you want to multiply (numerator and denominator) with $a^2+ab+b^2$. This will give you \begin{align}a-b&=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}=\frac{a^3-b^3}{a^2+ab+b^2}\\[0.2cm]&=\frac{x^3+x^2-(x^3-x^2)}{(x^3+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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