Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of so... | We have three expressions:
$$S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$
$$M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}$$
$$N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}$$
Obviously, we have $M+N=3$ According to AM-GM,we have:
$$M+S=\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}\geqslant 3$$
$$N+S=\frac{a+c}{b+c}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1... | alternatively, using abbreviations $c=\cos x$ and $s=\sin x$ we have
$$
s(1+c+s)=s(1+c) + s^2 = s(1+c) + 1-c^2=s(1+c)+(1-c)(1+c)=(1+s-c)(1+c)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Prove that $xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge 2x^2+2y^2+2z^2$ for every $x,y,z$ strictly positive I checked the inequality for $y=z$. In this particular case, after some simplifications, the inequality becomes:
$$
2x^3+y^3\ge 3x^2y,
$$
which is true, according to the arithmetic-geomet... | simplifying and factorizing the term $$xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}-2(x^2+y^2+z^2)$$ we get
$${\frac { \left( xy+zx+yz \right) \left( {y}^{3}{x}^{3}-{y}^{2}z{x}^{3
}-y{z}^{2}{x}^{3}+{z}^{3}{x}^{3}-{y}^{3}z{x}^{2}+3\,{z}^{2}{x}^{2}{y}^
{2}-y{z}^{3}{x}^{2}-{y}^{3}{z}^{2}x-{z}^{3}{y}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving $\lim \limits _{x \to \infty} (\sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}-x)$ $$\lim \limits _{x \to \infty}\bigg(\sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}-x\bigg)$$
We can see the limit is of type $\infty-\infty$. I don't see anything I could do here. I can only see the geometric mean which is the $n$-th root term.... | It's more complicated to write in LaTeX, than to solve. Remember that
$$A-B = \frac {A^n - B^n} {A^{n-1} + A^{n-2} B + \dots + A B^{n-2} + B^{n-1}} .$$
Choosing $A = \sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}$ and $B = \sqrt[n] {x^n}$, note that the largest power of $x$ in $A^n - B^n$ is $x^{n-1}$, and it is multiplied by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find critical numbers in awkward function I have problem to find critical numbers in this awkward function with Euler's constants.
$$f(x, y) = e^{2 x+3 y} \left(8 x^2-6 x y+3 y^2\right)$$
Task: Find critical numbers and determine if it's maximum or minimum
| We find $f_x$ and $f_y$ set them equal to zero and find the critical points.
$$f_x = 2 e^{2 x+3 y} \left(8 x^2-6 x y+3 y^2\right)+e^{2 x+3 y} (16 x-6 y)=0$$
$$f_y = 3 e^{2 x+3 y} \left(8 x^2-6 x y+3 y^2\right)+e^{2 x+3 y} (6 y-6 x)=0$$
Clearly the exponential terms do not give zeros (divide those out), so we are left w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve the integral $\int\frac{x-1}{\sqrt{ x^2-2x}}dx $ How to calculate $$\int\frac{x-1}{\sqrt{ x^2-2x}}dx $$
I have no idea how to calculate it. Please help.
| Substitute $u=x^2-2x,du=2(x-1)dx$. Then:
\begin{align*}
\int \frac{1}{2\sqrt{u}}du &=\frac{1}{2}\int u^{-0.5}du
\end{align*}
User the power rule:
$$\int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$$
$$\Rightarrow \frac{1}{2}\int u^{-0.5}du =\frac{1}{2}\frac{u^{-0.5+1}}{-0.5+1}$$
Back substitute and simplify:
$$\Rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
If the inequality $\log_a(x^2-x-2)>\log_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{9}{4}$ in the interval $(x_1,x_2)$ If the inequality $\log_a(x^2-x-2)>\log_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{9}{4}$ in the interval $(x_1,x_2)$,then find the product $x_1x_2$.
Here $a$ is not specified .I know... | You have done interval for $x<0$ plug in $-\infty$ you will see why. So the correct interval is $(2,5/2)$ so the product is $5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show $\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}$. Let $p$ be prime and $d \ge 2$. I want to show that
$$
\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}.
$$
I have a proof, but I think it is complicated, and the statement appears in a book as if it is very easy to see. So is there... | You need to prove that $p\mid\frac{(p^{d-1}-1)(p^d-1)}{(p-1)(p^2-1)}-1$
It is not difficult to see that $p\mid (p^{d-1}-1)(p^d-1)-(p-1)(p^2-1)$. Furthermore, we have that $(p-1)(p^2-1)\mid (p^{d-1}-1)(p^d-1)$ because either $d$ or $d-1$ is even and so either $p^{d}-1$ or $p^{d-1}-1$ is divisible by $p^2-1$ and the othe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all real numbers $x,y > 1$ such that $\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$
Find all real numbers $x,y > 1$ such that $$\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$$
Attempt
We can rewrite this as $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$. Then I get a multivariate cubic, which I find hard to find all solutions to.
| user236182 was on the right track by using Cauchy-Schwarz. We have from Cauchy-Schwarz that $$((y-1)+(x-1)) \left (\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1} \right) \geq \left (\sqrt{y-1}\sqrt{\dfrac{x^2}{y-1}}+\sqrt{x-1}\sqrt{\dfrac{y^2}{x-1}} \right)^2$$ and letting $a = x+y$ we get $8 \geq \dfrac{a^2}{a-2} \implies a < 2$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? 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?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
| $$\lim_{ x\to +\infty } \frac { (x+2)!+4^{ x } }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\lim_{ x\to +\infty } \frac { x!\left( x+1 \right) (x+2)+4^{ x } }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\lim_{ x\to +\infty } \frac { x!\left( \left( x+1 \right) (x+2)+\frac { { 4 }^{ x } }{ x! } \right) }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Finding the limit of the sequence $T_1=0,T_2=1, T_n=\frac{T_{n-1}+T_{n-2}}{2}$ Given that $T_1=0$, $T_2=1$ and $T_n=\frac{T_{n-1}+T_{n-2}}{2}$, show that the sequence converges to $\frac{2}{3}$.
| Start from the matrix representation of the problem
$$\begin{bmatrix}
T_{n} & T_{n-1}
\end{bmatrix}=\begin{bmatrix}
T_{n-1} & T_{n-2}
\end{bmatrix}
\begin{bmatrix}
\frac{1}{2}& 1\\
\frac{1}{2}& 0
\end{bmatrix}$$
$$\begin{bmatrix}
T_{2} & T_{1}
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}$$
Chaining it all y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Decomposing $x^4-5x^2+6$ over some fields My book asks me to decompose
$$x^4-5x^2+6$$
over:
$K = \mathbb{Q},\\ K = \mathbb{Q[\sqrt{2}]},\\ K = \mathbb{R}$
For $K = \mathbb{Q}$, I substituted $x² = a$ to get:
$$a²-5a+6 = (a-3)(a-2)$$
So Getting back to $a = x²$ we get:
$$x^4-5x^2+6 = (x²-3)(x²-2)$$
Also, for $\mathbb{R}... | Your work is correct; for $\mathbb{Q}[\sqrt{2}]$, you surely have
$$
x^4-5x^2+6=(x-\sqrt{2})(x+\sqrt{2})(x^2-3)
$$
Is $x^2-3$ reducible over $\mathbb{Q}[\sqrt{2}]$? This would mean…
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Denoting the sum of $n$ odd numbers $1+3+5+\cdots+(2n-1) = 1+3+5+\cdots+(2n+1)$? Let's say we are denoting the sum of $n$ odd numbers.
Then in symbols $1+3+5+\cdots+(2n-1)$.
If we substitute $(k+1)$ for $n$. $2n-1=2k+1$
So $1+3+5+...+2k+1$
Then can we use $1+3+5+\cdots+(2n+1)$ instead of $1+3+5+\cdots+(2n-1)$?
Logic... | Suppose your inductive hypothesis is $$1+3+5+...+(2n-1) = n^2.$$
You initially show the inductive hypothesis is true when $n=1$, e.g. by saying $1=1^2$ is indeed correct
Then you assume the inductive hypothesis is true for $n=k$ so $1+3+5+...+(2k-1) = k^2.$ You add $(2k+1)$ to both sides so the left hand side becomes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\pi=2\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}}$ How to prove that $$\pi=2\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}}$$
Where $F_{n}$ is the Fibonacci Number.
| The goal is to write $\arctan\left(\dfrac1{F_{2n+1}}\right)$ as $\arctan(a_{n+1}) - \arctan(a_{n})$. This means we need
$$\dfrac{a_{n+1}-a_n}{1+a_na_{n+1}} = \dfrac1{F_{2n+1}}$$
Recall that from Cassini/Catalan identity we have
$$F_{2n+1}^2 = 1+F_{2n+2}F_{2n}$$
Hence, let $a_n = F_{2n}$. We then have
$$\dfrac{a_{n+1}-a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint to move everything to the left hand side of the inequality and then factor an... | Multiplying out, this is equivalent to $a^7b^2 + a^2b^7 \geq a^5b^4 + a^4b^5$, or $a^5 + b^5 \geq a^3 b^2 + a^2b^3$ (the case $ab=0$ is easy).
We can prove this using weighted AM-GM:
$$\frac{3}{5}a^5 + \frac{2}{5}b^5 \geq a^3 b^2$$
$$\frac{2}{5}a^5 + \frac{3}{5}b^5 \geq a^2 b^3$$
(Sorry, I didn't notice that you specif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(v^2)\geq0$, where $f(v^2)=3u^2-4v^2+w^2$.
Thus, $f$ is a linear function, which says that $f$ get's a minimal value for an extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a=x^3$ and $c=1$.
Hence, we need to p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then... | Hint:
If $a+ b +c=0, a^3+b^3+c^3=3abc$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Show $\lim_{a \rightarrow \infty} \int_0^1 {f(x)x\sin(ax^2)}=0$.
Suppose $f$ is integrable on $(0,1)$, then show $$\lim_{a \rightarrow \infty} \int_0^1 {f(x)x\sin(ax^2)}=0.$$
I tried to write $$(0,1) = \bigcup _{k=0}^{{a-1}} \left(\sqrt{\frac{k}{a}},\sqrt{\frac{k+1}{a}}\right),$$ but cannot make the integral converge... | Suggestion:
Split $[0, 1]$
into the intervals where
$ax^2 = 2\pi n$,
$ax^2 = \pi(2 n+1)$,
$ax^2 = \pi(2 n+2)$.
These are
$I_{2n}
=[\sqrt{\frac{2\pi n}{a}}, \sqrt{\frac{\pi(2 n+1)}{a}})
$
and
$I_{2n+1}
=[\sqrt{\frac{\pi(2 n+1)}{a}}, \sqrt{\frac{\pi(2 n+2)}{a}})
$.
Since
$\sin(ax^2) > 0$
in
$I_{2n}$
and
$\sin(ax^2) < 0$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve equation on mathematical physics Show that
$$\gamma_+ - \gamma_-=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}$$
where
$$\gamma_+=(1-\beta^2_+)^{-\frac{1}{2}} \ \mbox{and} \ \beta_+=\frac{\beta_0+\beta}{1+\beta_0\beta}$$
$$\gamma_-=(1-\beta^2_-)^{-\frac{1}{2}} \ \mbox{and} \ \beta_-=\frac{\beta_0-\beta}{1... | I was moved $\gamma_-$ to other side equation:
$$\gamma_+=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}+\gamma_-$$
So
$$\gamma_+=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}+\frac{1}{\sqrt{1-\beta^2_-}}$$
Therefore
$$\gamma_-=\frac{2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}}{\sqrt{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Find the value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Now the trivial method is to put $X=5+2\sqrt{-4}$ in the polynomial and calculate but this is for $2$ marks only and that takes a hell lot of time for $2$! So I was thinking may be there is some tri... | It's not the that much work.
1) Just calculate $x, x^2, x^4 = (x^2)^2, x^3 = x^2 * x$ to the side first. and
2) consider every number as having two "parts"; a "normal" number part and a "square root of negative one part"
$x = -5 + \sqrt{-4} = -5 + 2\sqrt{-1}$
$x^2 = 25 + 4(-1) + 20\sqrt{-1} = 21 + 20\sqrt{-1}$
$x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sqrt{(a+b+1)(c+2)}+\sqrt{(b+c+1)(a+2)}+\sqrt{(c+a+1)(b+2)}\ge{9}$. Let $a,b,c$ be non-negative real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$. Prove that
$\sqrt{(a+b+1)(c+2)}+\sqrt{(b+c+1)(a+2)}+\sqrt{(c+a+1)(b+2)}\ge{9}$.
Here equality holds when $a=b=c=1$ so by using AM-GM on $c+1+1$ we get
$\sq... | Cauchy Schwarz actually works:
$$
\sum_{cyc}\sqrt{(a+b+1)(1+1+c)}≥\sum_{cyc}\sqrt{\left(\sqrt a+\sqrt b+\sqrt c\right)^2}=3(\sqrt a+\sqrt b+\sqrt c)=9
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all solutions to the diophantine equation $(x+2)(y+2)(z+2)=(x+y+z+2)^2$ Solve in postive integer the equation
$$(x+2)(y+2)(z+2)=(x+y+z+2)^2$$
It is rather easy to find several parametric solutions, (such $(a,b,c)=(2,1,1),(2,2,2)$).but it seems harder to find a complete enumeration of all the solutions.
and I have ... | For the equation.
$$(x+2)(y+2)(z+2)=(x+y+z+2)^2$$
If we use the solutions of Pell equations.
$$p^2-(z^2-4)s^2=1$$
Using my replacement.
$$a=p^2-2(z+2)ps+(z^2-4)s^2$$
$$b=p^2-2zps+(z^2-4)s^2$$
Decisions will be.
$$x=-2a^2+2(z+2)ab-z(z+2)b^2$$
$$y=-za^2+2(z+2)ab-2(z+2)b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Value of $a^2 + b^2$ given $a^3 - 3ab^2 = 44$ and $b^3 - 3a^2 b = 8$ If $a$ and $b$ are real numbers such that $a^3-3ab^2=44$ and $b^3-3a^2b=8$ what is value of $a^2+b^2$? I have tried by adding and subtracting these equations, but can't find anything.
| Notice $$(a+ib)^3 = (a^3-3ab^2) + (3a^2b-b^3)i = 44-8i$$
Multiply both sides by their complex conjugates, one find
$$(a^2+b^2)^3 = 44^2 + 8^2 = 2000\quad\implies\quad
a^2 + b^2 = \sqrt[3]{2000} = 10\sqrt[3]{2}$$
Update
If one really want to hide the use of complex numbers, one can
take the squares of both conditions, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that the angles satisfy $x+y=z$ How can I show that $x+y=z$ in the figure without using trigonometry? I have tried to solve it with analytic geometry, but it doesn't work out for me.
| Imagine that all the the squares are 1 by 1 and so the rectangle has base 3 and a height of $1$.
There are three right-angled triangles in the diagram. The one with angle $x$ has base 3, and height 1. The triangle with angle $y$ has base 2 and height 1. The triangle with angle $z$ has base 1 and height 1.
Using the st... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:
Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from... | Why not just
$$ \begin{align} 2^{2n+1} &=\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}+\binom{2n+1}{n+1}+\cdots+\binom{2n+1}{2n+1} \\
&=\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}+\binom{2n+1}{n}+\cdots+\binom{2n+1}{0} \\
&=2\left[\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}\right] \end{align} $$
Then divide each extremity by 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
An inequality on three constrained positive numbers Assume $a,b,c$ are all positive numbers, and $2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2$.
Prove that:
$$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2\ge(ab+bc+ca)^2$$
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition gives $\sum\limits_{cyc}(a^3b+a^3c-a^2b^2)=\sum\limits_{cyc}(a^3c-a^3b)$
or $9u^2v^2-9v^4+uw^3=u\prod\limits_{cyc}(a-b)$
or $(9u^2v^2-9v^4+uw^3)^2=27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$
or $28u^2w^6+18u(6u^4-8u^2v^2-v^4)w^3-54u^2v^6+81v^8=0$, which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Coupon collector's problem using inclusion-exclusion Coupon collector's problem asks:
Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once?
The well-known solution is $E(T)=n \cdot H_n$, where T is the time to collect all n coupons(proof).
I a... | By way of enrichment here is a proof using Stirling numbers of the
second kind which encapsulates inclusion-exclusion in the generating
function of these numbers.
First let us verify that we indeed have a probability distribution
here. We have for the number $T$ of coupons being $m$ draws that
$$P[T=m] = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{a+ab}+\frac{1}{b+bc}+\frac{1}{c+ca} \geq \frac{3}{2}.$
Let $a,b,$ and $c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1}{a+ab}+\dfrac{1}{b+bc}+\dfrac{1}{c+ca} \geq \dfrac{3}{2}.$$
I thought about substituting in $abc = 1$ to get $$\dfrac{1}{a+\dfrac{1}{c}}+\dfrac{1}{b+\dfra... | Let $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
$$\sum_{\text{cyc}}\frac{1}{\frac{x}{y}+\frac{x}{y}\frac{y}{z}}=\sum_{\text{cyc}}\frac{yz}{xz+xy}$$
Let $yz=c$, $xz=d$, $xy=e$. Then your inequality follows from Nesbitt's inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try:
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because
$ax+by+cz\le{\frac{a+b+c}... | Let $a+b+c=k$.
Hence, $(a-kx)^2+(b-ky)^2+(c-kz)^2\geq0$ gives
$(a-kx)^2+(b-ky)^2+(c-kz)^2\geq(a-kx+b-ky+c-kz)^2$ or
$(xy+xz+yz)k^2-(ay+bx+az+cx+bz+cy)k+ab+ac+bc\geq0$, which gives
$4(ab+ac+bc)(xy+xz+yz)\leq(ay+bx+az+cx+bz+cy)^2$ or
$ax+by+cz+2\sqrt{(ab+ac+bc)(xy+xz+yz)}\leq a+b+c$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Factore $(x+1)(x+2)(x+3)(x+4)-35$ I need to factor
$$(x+1)(x+2)(x+3)(x+4)-35$$
I know that the answer will be
$$(x^2+5x+11)(x^2+5x-1)$$
I go out only
$$(x^2+5x)(x^2+5x+10)-11$$
Help me.
| Put $t=x^2+5x$
Then the given expression will be $t(t+10)-11$
$\Rightarrow t^2+10t-11=t^2-t+11t-11=(t+1)(t-11)$
Now put the value of $t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find $P$ and $Q$. A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find the coordinates of $P$ and $Q$.
I found the lengths of the ... | Your initial equation setup is incorrect. Note that $P$ and $Q$ do not have the same $x$-coordinate! Because of this faulty assumption, anything you do after that will not lead to the right answer.
A good way to fix this is to give the coordinate different variable names. Let $P\left(a, \frac{3a-5}{2}\right)$ and $Q\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show this fraction is equal to 1/2? I have the fraction:
$$\frac{\left(2 \left(\frac {a}{\sqrt{2}}\right) + a \right) a} {2(1 + \sqrt{2})a^2}$$
Using Mathematica, I've found that this simplifies to $\frac{1}{2}$, but how did it achieve the result? How can I simplify that fraction to $\frac12$?
| Assume $a\neq 0$, we have \begin{align*}
\frac{\left(2\left(\frac {a}{\sqrt{2}}\right)+a\right)a} {2(1+\sqrt{2})a^2}&=\frac{\left(\frac{2a}{\sqrt 2}+a\right)}{2(1+\sqrt 2)a}\tag 1\\
&=\frac{\left(\frac{2}{\sqrt2}+1\right)a}{2(1+\sqrt 2)a}\tag2\\
&=\frac{\left(\frac{2}{\sqrt2}+1\right)}{(2+2\sqrt 2)}\\
&=\frac{\sqrt 2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Probability of getting the same number three times. If I have a set of numbers $\{1 \ldots n\}$ where $n \ge 1$ and I pick $3$ numbers from the set independently and uniformly. Whats the probability I'll get all $2$'s, the probability I get all the same numbers and the probability that my first two numbers are the same... | *
*You're right, the probability of all 2s is
$$P(222) = P(2)P(2)P(2) = \frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6} = \left(\frac{1}{6}\right)^3= \frac{1}{216},$$
where use the product by independence.
*When calculating probability that they are all the same is, you have to account for all the choices you have for a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the complex roots of $y^3-\frac{1}{3}y+\frac{25}{27}$ I've been trying to solve this for hours and all found was the real solution by Cardano"s formula.
I vaguely remember that if $\alpha$ is a root of a complex number, the other roots are $\omega \alpha$ and $\omega ^2 \alpha$ where $\omega = - \frac{1}{2... | By $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+\omega b+\omega^{2} c)(a+\omega^{2} b+\omega c)$;
and replacing $a, b, c$ by $x, -u, -v$ respectively,
$x^{3}-3uvx-(u^{3}+v^{3})=0 \implies x_{k}=u\, \omega^{k}+ v\, \omega^{2k}$ for $k=0,1,2$.
$u, v$ are known as resolvents.
Let your reduced cubic be $y^{3}-3py-2q=0$.
(Will substit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
an upper bound for number of primes in the interval $[n^2+n,n^2+2n]$ What is an upper bound for the number of primes in an interval of $n$ consecutive
numbers?
What is an upper bound for the number of primes in the interval $[n^2+n,n^2+2n]$?
| According to the Prime-counting function
$$\text{number of primes below natural }n = \pi(n) \approx \frac{n}{\ln n}$$
so you just want an approximation of $\pi(n^2+2n) - \pi(n^2+n)$:
$$\begin{align}
\pi(n^2+2n) - \pi(n^2+n) &\approx \frac{n^2+2n}{\ln (n^2+2n)} - \frac{n^2+n}{\ln (n^2+n)}\\
&\approx \lim_{n \to \infty} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
To evaluate the limits $\lim\limits_{n \to \infty} \{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\}$ To me it seems like that we need to manipulate the given sum into the Riemann sum of some function. First writing in the standard summation form;
$$\{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3... | Note that
$$\frac{k^2}{k^3+n^3}=\dfrac1n\cdot\dfrac{\left( \frac{k}{n} \right)^2}{\left( \frac{k}{n} \right)^3+1}$$
so you can rewrite the summation to be
$$\begin{align}\lim_{n\to \infty} \Big(\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\Big) &= \lim_{n\to \infty} \sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\omega$ is an imaginary fifth root of unity, then $\log_2 \begin{vmatrix} 1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\ \end{vmatrix}$ =
If $\omega$ is an imaginary fifth root of unity, then $$\log_2 \begin{vmatrix}
1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\
\end{vmatrix} =$$
My approach :
$$\omega^5 = ... | Notice:
$$\omega^5=1\Longleftrightarrow$$
$$\omega^5=e^{0i}\Longleftrightarrow$$
$$\omega=\left(e^{2\pi ki}\right)^{\frac{1}{5}}\Longleftrightarrow$$
$$\omega=e^{\frac{2\pi ki}{5}}$$
With $k\in\mathbb{Z}$ and $k:0-4$
So, the solutions are:
$$\omega_0=e^{\frac{2\pi\cdot0i}{5}}=e^{\frac{0}{5}}=e^0=1$$
$$\omega_1=e^{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sum_{i=0} ^{\log n} i2^i$
Solve (or simplify): $$\sum_{i=0} ^{\log n} i2^i$$ (without integrals)
Trying to change the parameter: $j=i2^i$, so since $ 0\le i \le \log n$, then the maximum value for $j$ is when $j=n\log n$. So we get $ \displaystyle \sum_{i=0} ^{\log n} i2^i = \sum_{j=0} ^{n\log n} j$
Now I ... | Note that
\begin{align}
\sum_{i=1}^{n}i \cdot 2^{i} &= 2+(4+4)+(8+8+8)+\ldots+(\underbrace{ 2^{n}+2^{n}+\ldots+2^{n}}_{n\text{ times}})\\
&=(2+4+8+\ldots+2^{n-1}+2^{n})+(4+8+\ldots+2^{n-1}+2^{n})+(8+\ldots\\
&\quad+2^{n-1}+2^{n})+\ldots+(2^{n-2}+2^{n-1}+2^{n})+(2^{n-1}+2^{n})+2^{n}\\
&=2(1+2+4+\ldots+2^{n-2}+2^{n-1})+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \... | Hint: $$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$$$$= \sec^2 \theta + \csc^2 \theta + \sec^2 \theta + \csc^2 \theta -1 = 2sec^2 \theta + 2\csc^2 \theta -1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Rolling a dice $5$ times
Rolling a dice $5$ times find the probability that we will get the continuum:
*
*A. $12345$
*B. $1234$
*C. $434$
My attempt:
*
*A: for each throwing there are $5$ possibilities so
$P(1)=P(2)=P(3)=P(4)=P(5)=\frac 1 6\Longrightarrow$ we are looking for $P(1)\cdot P(2)\cdot P(3)\cdot P... | Assuming it's a fair die and it has actually 6 faces then
*
*$P(1,2,3,4,5) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \left( \frac{1}{6} \right) ^ 5 = \frac{1}{6^5}$
*$P(1,2,3,4) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = 2 \cdot \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit problem $\ln(x)$ and $1^\infty$ Can anyone help me with this limit problem without L'Hopital rule and Taylor series?
$$\lim_{x\rightarrow\ 1}\left(\frac{2^x+2}{3^x+1}\right)^{1/\ln(x)}$$
| $$\frac{2^x+2}{3^x+1}=\left(1+(\frac{2^x+2}{3^x+1}-1)\right)=1+\frac{2^x-3^3+1}{3^x+1}$$ Puting $\frac{2^x-3^3+1}{3^x+1}=h$ one gets $$\left(\frac{2^x+2}{3^x+1}\right)^{\frac{1}{\ln x}}=(1+h)^{\frac{1}{\ln x}}=(1+h)^{{\frac{h}{h\ln x}}}$$
$$\lim_{x\rightarrow\ 1}\left(\frac{2^x+2}{3^x+1}\right)^{1/\ln(x)}=\lim_{h\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Integration by parts - hint I'm stuck on a passage on my textbook:
$$ \int \frac{1}{(1+t^2)^3} dt = \frac{t}{4(t^2+1)^2}+\frac{3}{4} \int \frac{1}{(t^2+1)^2} dt$$
I know that it should be easy but I just can't figure out what is the product of functions considered in this integration by parts..
can you help me?
thanks ... | Notice, in general, use integration by parts as follows $$\int\frac{1}{(1+x^2)^{n-1}}\ dx=\int \underbrace{\frac{1}{(1+x^2)^{n-1}}}_{I}\cdot \underbrace{1}_{II}\ dx$$
$$=\frac{1}{(1+x^2)^{n-1}}\int 1\ dx-\int\left(\frac{d}{dx}\frac{1}{(1+x^2)^{n-1}}\cdot \int 1\ dx\right)\ dx$$
$$=\frac{x}{(1+x^2)^{n-1}}-\int\left(-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Proving Trig Identities (Complex Numbers)
Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n + {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)
I learnt to prove the first part in another post l... | I don't think that the hint is really that useful. It would be better to do
$$
\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}
$$
that gives you
\begin{align}
64\cos^6\theta
&=(e^{i\theta}+e^{-i\theta})^6 \\[3px]
&=e^{6i\theta}+6e^{4i\theta}+15e^{2i\theta}
+20+15e^{-2i\theta}+6e^{-4i\theta}+e^{-6i\theta} \\[3px]
&=(e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding $\lim_{(x,y)\to (0,0)} \frac{3x^2\sin^2y}{2x^4+2\sin y^4}$ I had following limit of two variables as a problem on my calculus test. How does one show whether the limit below exists or does not exist? I think it does not exist but I was not able to show that rigorously. There was a hint reminding that $\lim_{t\t... | $$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{x^2sin^2(y)}{x^4+siny^4})$$
If Dividing by $y^4$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{\frac{x^2sin(y)sin(y)}{y^2yy}}{\frac{x^4}{y^4}+\frac {sin(y^4)}{y^4}})$$
Know:$\lim_{t\to0}\frac{sint}{t}=1$
$(y^4=t)$,$(y=t)$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{\frac{x^2}{y^2}}{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A particle moves along the x-axis find t when acceleration of the particle equals 0 A particle moves along the x-axis, its position at time t is given by
$x(t)= \frac{3t}{6+8t^2}$, $t≥0$,
where t is measured in seconds and x is in meters.
Find time at which acceleration equals 0.
I got the answer 0.5 and 0 but apparent... | The acceleration is expressed by the second derivative.
We have $$x(t)=\frac{3t}{6+8t^2}$$ $$x'(t)=\frac{3(6+8t^2)-3t\cdot 16t}{(6+8t^2)^2}=\frac{18+24t^2-48t^2}{(6+8t^2)^2}=\frac{18-24t^2}{(6+8t^2)^2}$$ $$ x''(t)=\frac{-48t(6+8t^2)^2-2(18-24t^2)(6+8t^2)(16t)}{(6+8t^2)^4}=\frac{(6+8t^2) \left [-48t(6+8t^2)-2(18-24t^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solving modulo equations with one variable Given the following equation:
$$10 = 4^x \pmod {18}$$
How can one know what are the correct values for $x$ ?
| First, $x>0$, because $x=0$ is not a solution, and $4$ has no inverse mod $18$.
Also, $2$ is a primitive root mod $9$, because by Euler's Theorem (since $\phi(9)=6$ and $\gcd(2,9)=1$) we get $2^6\equiv 1\pmod{9}$, also $2^{6/2}\equiv 8\not\equiv 1\pmod{9}$ and $2^{6/3}\equiv 4\not\equiv 1\pmod{9}$.
$$4^x\equiv 10\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solution to the equation $\sqrt{x^2 - 2x + 1} - 5 = 0$ I had this equation on my exam :
$$\sqrt{x^2-2x+1} - 5 = 0$$
My friends said the the solution could be :
$$|x-1| = -5$$
So the solution is nothing!
But I say the solution is:
$$x^2-2x+1 = 25 $$
so $$x = 6\ |\ x = -4$$
My Question here is which solution is right... | Your solution is definitely correct:
\begin{align}
&\sqrt{x^2 - 2x +1} - 5 = 0 \\
\Rightarrow \; &x^2 -2x+1 = 25 \\
\Rightarrow \; &(x-6)(x+4) = 0 \\
\Rightarrow \; &x = 6 \text{ or } x=-4
\end{align}
Your friend's approach is also correct, but he (or she) has a typo. After correcting this, you get:
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Series expansion of $\frac{1}{\sqrt{x^3-1}}$ near $x \to 1^{+}$ How can I arrive at a series expansion for $$\frac{1}{\sqrt{x^3-1}}$$ at $x \to 1^{+}$? Experimentation with WolframAlpha shows that all expansions of things like $$\frac{1}{\sqrt{x^y - 1}}$$ have $$\frac{1}{\sqrt{y}\sqrt{x-1}}$$ as the first term, which I... | Set $x:=1+\epsilon$, with $\epsilon \to 0^+.$ Then, by the binomial theorem,
$$
x^3=(1+\epsilon)^3=1+3\epsilon+3\epsilon^2+\epsilon^3
$$ giving
$$
\sqrt{x^3-1}=\sqrt{3\epsilon+3\epsilon^2+\epsilon^3}=\sqrt{3}\:\sqrt{\epsilon}\:\sqrt{1+\epsilon+O(\epsilon^2)} \tag1
$$ Observe that, by the Taylor expansion, as $\epsilon ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $a(x+y+z) = x(a+b+c)$ If $(a^2+b^2 +c^2)(x^2+y^2 +z^2) = (ax+by+cz)^2$
Then prove that $a(x+y+z) = x(a+b+c)$
I did expansion on both sides and got:
$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $
but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?
| By C-S inequality, $(ax+by+cz)^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)$ with equality iff $(x,y,z)=\lambda(a,b,c)$ for some $\lambda$ or $(a,b,c)=(0,0,0)$. But, if $(a,b,c)=(0,0,0)$, the problem is trivially true. If it not the case, then $x=\lambda a$, $y=\lambda b$ and $z=\lambda c$.
Then $x+y+z=\lambda(a+b+c)$. Multiplying... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Number of integer solutions of $\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}$ How can we find number of integer solutions of
$\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}$
I want to ask what approach in general should be followed in such types of question?
| $$\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}\Longleftrightarrow$$
$$\frac{y}{xy}+\frac{x}{xy}=\frac{1}{2016}\Longleftrightarrow$$
$$\frac{x+y}{xy}=\frac{1}{2016}\Longleftrightarrow$$
$$xy=2016(x+y)\Longleftrightarrow$$
$$xy-2016x=2016y\Longleftrightarrow$$
$$x\left(y-2016\right)=2016y\Longleftrightarrow$$
$$x=\frac{2016y}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to simplify $\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}$ The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
| By direct calculation we have the identity
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac).$
Then set $a=\sqrt[3]{7+5\sqrt{2}},b=\sqrt[3]{7-5\sqrt{2}}$ and $c=-2$. Note that $ab=-1$ and $a^3+b^3=14$, so $a^3+b^3+c^3=3abc$.
Since $(a-b)^2+(b-c)^2+(b-c)^2>0\Longleftrightarrow a^2+b^2+c^2-ab-bc-ac>0$ holds for any three d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Trigonometric Equation Simplification $$3\sin x + 4\cos x = 2$$
To solve an equation like the one above, we were taught to use the double angle identity formula to get two equations in the form of $R\cos\alpha = y$ where $R$ is a coefficient and $\alpha$ is the second angle being added to $x$ when using the double angl... | A different approach:
You can trade the trigonometric functions for a rational expression with
$$3\sin x+4\cos x=3\frac{2t}{1+t^2}+4\frac{1-t^2}{1+t^2}=2.$$
Thus, the quadratic equation
$$6t^2-6t-2=0,$$ which you can readily solve.
Then
$$\tan x=\frac{2t}{1-t^2}$$
and
$$x=\arctan\frac{-12\pm2\sqrt{21}}5+k\pi.$$
To de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\sqrt{x^2+y^2+2x-4y+5} + \sqrt{x^2+y^2-6x+8y+25}$, if $3x+2y-1=0$ As the title says, given $x,y \in \mathbb{R}$ where $3x+2y-1=0$ and $x \in [-1, 3]$, calculate $A = \sqrt{x^2+y^2+2x-4y+5} + \sqrt{x^2+y^2-6x+8y+25}$.
I tried using the given condition to reduce the complexity of the roots, but couldn't get r... | $-1\le x\le3\iff-2\le x-1\le2$
WLOG $x-1=2\cos u\iff y=-1-3\cos u$
$x^2+y^2-6x+8y+25=(x-3)^2+(y+4)^2=(2^2+3^2)(1-\cos u)^2$
$\implies\sqrt{x^2+y^2-6x+8y+25}=\sqrt{13}(1-\cos u)$ as $1-\cos u\ge0$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability of getting $5$ heads on $10$ (fair) coin flips? Even before attempting the problem, I immediately defaulted to an answer: $\frac{1}{2}$.
I thought that this was a possible answer since the probability of flipping a head on one flip is definitely $\frac{1}{2}$.
I then worked through the problem:
Let E be the... | Let's imagine a case with $4$ rolls where we desire $2$ heads (MathJax diagrams only go so far...)
$$\newcommand{\mychoose}[2]{\bigl({{#1}\atop#2}\bigr)}$$
$$\begin{array}{ccccccccccc} & & & & & & & H & & & & & \\
& & & & & & \swarrow & & \searrow & \\
& & & & & H & & & & T & \\
& & & & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Evaluate this limit of inverse trigonometric and radical functions without l'Hospital How can I solve this using only 'simple' algebraic tricks and asymptotic equivalences? No l'Hospital.
$$\lim_{x \rightarrow0}
\frac
{\sqrt[3]{1+\arctan{3x}} - \sqrt[3]{1-\arcsin{3x}}}
{\sqrt{1-\arctan{2x}} - \sqrt{1+\arcsin{2x}}}
$$
R... | We will use the following standard limits $$\lim_{x \to 0}\frac{\arctan x}{x} = 1 = \lim_{x \to 0}\frac{\arcsin x}{x},\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ We have
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - \sqrt[3]{1 - \arcsin 3x}}
{\sqrt{1 - \arctan 2x} - \sqrt{1 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) wi... | $$ x+y= p\tag1$$
Square, since $( x^2+y^2=1 )$
$$ 1+ 2 x\;y = p^2, \; x y= \dfrac{p^2-1}{2} \tag2$$
From (1) and (2)
$$ \dfrac{1}{x}+ \dfrac{1}{y} = \dfrac{x+y}{x y}= \dfrac{2p}{p^2-1} $$
$$ = \dfrac{35}{12},\;$$
if
$$\;p= \dfrac{7}{5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 9
} |
Minimizing distance between line and point So I am given the problem where a line of the form $ax+by+c=0$ and point $(x_0,y_0)$ are given and I have to find the minimum distance between these two. I was able to do it with the projection of a vector normal to the line and found the answer to be $$\frac{|ax_0+by_0+c|}{\s... | To find the minimum distance between a point $P(x_0,y_0)$ and a line $ax + by + c = 0$, we will use the distance formula $$f(x) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
to create a function that represents the distance between $P(x_0,y_0)$ and the line $ax + by + c = 0$ as a function of $x$.
First, let's rewrite $ax + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving for an inverse diophantine equation $\frac{1}{x_1} +\frac{1}{x_2} + ... + \frac{1}{x_n} +\frac{1}{x_1 x_2 ... x_n} = 1$ How can I prove that the Diophantine equation $$\frac{1}{x_1} +\frac{1}{x_2} + ... + \frac{1}{x_n} +\frac{1}{x_1 x_2 ... x_n} = 1$$ has at most one solution? All $x_i$ and $n$ are natural n... | Assuming that you want the number of solutions (you have referred to as root so it is a bit confusing).
This is NOT true, for example if $n=5$, then there are $3$ solutions.
\begin{align*}
1 & = \frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{1807}+\frac{1}{2 \cdot 3 \cdot 7 \cdot 43 \cdot 1807}\\
& = \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I generalize this? I'm not sure what to tag this question as or whether its a bit nonsensical, but I'm a bit curious. I asked a question on a pretty (turned out to be) easy question about the Collatz Sequence here:
Collatz $4n+1$ rule?
My question is observe the following sequences (next term is $4n+1$):
$$3\ 1... | You could notice that the formula describing each of those sequences is of the form $\frac{1}{3}(-1+4^n+3\cdot 4^n\cdot(2j-1))$, where $2j-1$ is the initial odd value.
Then the numbers $2k-1$ which have already been covered in a sequence must be of the form $2k-1=\frac{1}{3}(-1+4^n+3\cdot 4^n\cdot(2j-1))$ for some $n,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: L... | $$\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt k+1} > \sqrt{k} + \frac{1}{\sqrt {k+1}} =
\frac{\sqrt{k(k+1)} + 1}{\sqrt{k+1}} > \frac{\sqrt{k^2} + 1}{\sqrt{k+1}} = \sqrt{k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $x,y,z$ are positive real number number, Then minimum value of $\frac{x^4+y^4+z^2}{xyz}$
If $x,y,z$ are positive real number number, Then minimum value of $\displaystyle \frac{x^4+y^4+z^2}{xyz}$
$\bf{My\; Try::}$ Given $x,y,z>0.$ So Using $\bf{A.M\geq G.M\;,}$ We get
$$\displaystyle x^4+y^4\geq 2x^2y^2$$ and then ... | $$x^4+y^4+z^2=x^4+x^4+(\sqrt{2}x^2)^2=4x^4\\
xyz=xx\sqrt{2}x^2=\sqrt{2}x^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?
please explain why my procedure is wrong i am not able to find out??
I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions)... | The basic limits needed are
$\frac{e^x-1}{x}
\to 1
$,
$\frac{\sin x}{x}
\to 1
$,
and
$\cos x
\to 1
$ as
$x \to 0$.
First,
as $x \to 0$,
$\begin{array}\\
\tan x -\sin x
&=\frac{\sin x}{\cos x}-\sin x\\
&=\sin x(\frac{1}{\cos x}-1)\\
&=\sin x(\frac{1-\cos x}{\cos x})\\
&=\sin x(\frac{2\sin^2(x/2)}{\cos x})\\
&\to x(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
Probability of combinations so that only $2$ of the $3$ types remains. $15$ telephones have just been received at an authorized service center. $5$ of these telephones are cellular, $5$ are cordless, and the other $5$ are corded phones. Suppose these components are randomly allocated the number $1,2,3, \dots$.
What is... | To start, lets fix a telephone type, solve the problem for that type and then multiply with $3$ the obtained probability to obtain the correct answer (due to symmetry). We need to service all $5$ phones of this type and at least $1$ phone of each of the other $2$ types (in order to exhaust exactly this type and not $2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Infinity Series to Approximate a fraction I have observed that the following series is a good approximation for
$\frac{1}{10}$.
$\frac{1}{8}- \frac{1}{16} + \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \frac{1}{512} + \frac{1}{1024} - \frac{1}{2048} -
\frac{1}{(2\cdot2048)} + \frac{1}{(4\cdot2048)} + ... | $$\frac{1}{8}-\frac{1}{16}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+1}}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+2}}=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
State the range of this How do I state the range of the following equation:
$$8x-4x^2$$ using $$c-\frac{b^2}{4a}$$
I'm again unsure of what $a$, $b$, and $c$ are.
| Generally, $f(x)=ax^2+bx+c$ has its extreme value at $f'(x)=0$, i.e.
$$2ax+b=0\implies x=-\frac{b}{2a}.$$
Substituting $x=-\frac{b}{2a}$ into $f(x)$ gives:
$$f\left(-\frac{b}{2a}\right)=a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c=\frac{ab^2}{4a^2}-\frac{b^2}{2a}+\frac{2ac}{2a}$$
$$=\frac{ab^2}{4a^2}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $f(x)=8$ for all natural numbers $x\ge{8}$ A function $f$ is such that $$f(a+b)=f(ab)$$ for all natural numbers $a,b\ge{4}$ and $f(8)=8$. Prove that $f(x)=8$ for all natural numbers $x\ge{8}$
| $\>$$\>$$\>$$\>$$\>$SS_C4's answer has two problems; the untrue claim that $2y>y+6$, for $y\geq 6$, and not demonstrating that $f(x)=f(8)$, for $9\leq x<17$, but SS_C4's proof can be saved as follows. The basic ideas belong to SS_C4. The value of $f(8)$ is immaterial. Lower-case Latin letters, except $f$, denote positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
How can I solve $7^{77}\mod 221$ How is it possible to solve this without calculator etc.: $$7^{77} \mod 221$$
I started with:
\begin{align}
7&\equiv 7 \mod 221 \\
7^2 &\equiv 49 \mod 221 \\
7^4 &\equiv \ ? \mod 221
\end{align}
Sure i can calculate this by my own, but is there a trick to calclulate this with any tools?... | Without the Chinese Remainder Theorem :
$$
\eqalign{
7^5 &\equiv 11 \pmod {221} \cr
7^{75} &\equiv 11^{15} \pmod {221} \cr
7^{77} &\equiv 7^2 \cdot 11^{15} \pmod {221} \cr
}
$$
Also
$$
\eqalign{
11^3 &\equiv 5 \pmod {221} \cr
11^{15} &\equiv 5^{5} \pmod {221} \cr
}
$$
To conclude
$$
7^{77} \equiv 7^2 \cdot 11^{15} \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let $a_{99}=K\times10^4$ .The value of $K$ is______ .
My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+... | $$a_n-a_{n-1}=6n^2+2n$$
Now sum both sides from $n=2$ to $x$. You end up with a telescoping sum on the left hand side which will drastically make things easier and help you find the closed form, assuming you know some summation formulas.
$$\sum_{n=2}^x (a_n-a_{n-1})= \sum_{n=2}^x (6n^2+2n)$$
$$a_x-a_1=\sum_{n=2}^x (6n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+C... | I'd try the following: substitute $\;t^2=x\implies 2t\,dt=dx\;$, and your integral becomes
$$I=\int\frac{2t^2dt}{(t^2+1)^2}$$
and already here integrate by parts:
$$\begin{cases}u=t&u'=1\\{}\\v'=\frac{2t}{(t^2+1)^2}&v=-\frac1{t^2+1}=\end{cases}\;\;\;\implies$$
$$I=-\frac t{t^2+1}+\int\frac1{1+t^2}dt=-\frac t{1+t^2}+\ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Solving $3^x \cdot 4^{2x+1}=6^{x+2}$ Find the exact value of $x$ for the equation $(3^x)(4^{2x+1})=6^{x+2}$
Give your answer in the form $\frac{\ln a}{\ln b}$ where a and b are integers.
I have tried using a substitution method, i.e. putting $2^2$ to be $y$, but I have ended up with a complicated equation that hasn't p... | $$\color{violet}{(3^x)(4^{2x+1})=6^{x+2}}$$
$$\color{indigo}{3^x\cdot 2^{2(2x+1)}=2^{x+2}\cdot 3^{x+2}}$$
$$\color{blue}{2^{2(2x+1)}\cdot 3^x=2^{x+2}\cdot 3^{x+2}}$$
$$\color{green}{\frac{2^{2(2x+1)}}{2^{x+2}}=\frac{3^{x+2}}{3^x}}$$
$$\color{brown}{2^{2(2x+1)-(x+2)}=3^{(x+2)-x}}$$
$$\color{orange}{2^{3x}=3^2}$$
$$\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How to prove the trigonometric identity $\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$ $$\frac{1-\sin\varphi}{1+\sin\varphi}$$
I have no idea how to start this, please help. This problem is essentially supposed to help me solve the proof of
$$\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi -... | $$(\sec\varphi - \tan\varphi)^2 = \left(\frac{1}{\cos \varphi}-\frac{\sin \varphi}{\cos \varphi}\right)^2 = \frac{(1 - \sin \varphi)^2}{\cos^2 \varphi}$$
$$ = \frac{(1 - \sin \varphi)^2}{ 1 - \sin^2 \varphi} = \frac{(1 - \sin \varphi)^2}{(1 - \sin \varphi)(1+\sin \varphi)} = \frac{1-\sin \varphi}{1+\sin\varphi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is $\,a\bmod 63\,$ if $\,a\,$ is both a square and a cube? Let there be integers $a,m,n$ such that $a=m^2=n^3$. Show that $a\equiv 0,1,28,36\pmod{63}$.
I have established that $a$ must be a sixth power of an integer:
$$a=\frac{a^3}{a^2} = \frac{m^6}{n^6} =\left (\frac{m}{n}\right )^6$$
$n^6\mid m^6$ implies $n\mid... | You must show $x^6\equiv 0,1,28,36\pmod{63}$ For $ 0 \leq x \leq 62$. By Fermat's little Theorem we conclude $x^6 \equiv 1 \pmod{7}$ if $x\neq 7k$ and it's easy to check $x^6 \equiv 1 \pmod{9}$ if $x\neq 3k$.
Now because of theorem :
$a\equiv b \pmod {m_1}$ and $a\equiv b \pmod {m_2}$ then $a\equiv b \pmod {lcm (m_1,m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and... | Hint: Try $u=x^7+1$. $\,\,\,\,\,$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Antiderivative problem What is the antiderivative of $(2x+7)^{1/2}$?
My understanding is that it would be $\frac 23 \times(2x+7)^{3/2}$ but according to the source I am working from the answer is $\frac 13 \times (2x+7)^{3/2}$.
| Notice:
$$\int x^n\space\text{d}x=\frac{x^{1+n}}{1+n}+\text{C}$$
$$\int\sqrt{2x+7}\space\text{d}x=$$
Substitute $u=2x+7$ and $\text{d}u=2\space\text{d}x$:
$$\frac{1}{2}\int\sqrt{u}\space\text{d}u=\frac{1}{2}\int u^{\frac{1}{2}}\space\text{d}u=\frac{1}{2}\cdot\frac{u^{1+\frac{1}{2}}}{1+\frac{1}{2}}+\text{C}=\frac{u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ If $x$ is real, the maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ is?
Is it necessary that this function will attain maximum when the denominator will be minimum?
| You need to solve:
$$\frac{\text{d}}{\text{d}x}\left[\frac{3x^2+9x+17}{x^2+2x+9}\right]=0\Longleftrightarrow$$
$$\frac{47+x(20-3x)}{(9+x(2+x))^2}=0\Longleftrightarrow$$
$$47+x(20-3x)=0\Longleftrightarrow$$
$$x(20-3x)=-47\Longleftrightarrow$$
$$-3x^2+20x=-47\Longleftrightarrow$$
$$3x^2-20x=47$$
If you look to it graphic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Understanding a solution to $m^3-n^2=2$ I am reading through a proof that the only integer solutions to $m^3-n^2=2$ are $(m,n)=(3,\pm 5)$.
Working in $\mathbf{Z}[\sqrt{-2}]:$ we have $m^3=n^2+2=(n+\sqrt{-2})(n-\sqrt{-2})$. Also: $$\text{gcd}\,\big(n+\sqrt{-2},n-\sqrt{-2}\big)\,\big|\,2\sqrt{-2}$$
That's all fine. The n... | First: $2$ can't divide $n+\sqrt{-2}$, and thus neither can $2\sqrt{-2}$. So $\gcd(n+\sqrt{-2},n-\sqrt{-2})\mid \sqrt{-2}$.
If two numbers are relatively prime and their product is a cube in a unique factorization domain, then both numbers are cubes times a unit. Since the only units here are also cubes, this means if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Nested radical $\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ I am studying the $f(x) = \sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ for $x \in (0,\infty)$ and I am trying to get closed form formula for this, or at least some useful series/expansion. Any ideas how to get there?
So far I've got only trivial values, which... | You can obtain an expansion in negative powers of $x$, valid for large $x$, by trying to find $$\frac{f(x)}{\sqrt{2x}}$$. If you pull the $\frac{1}{\sqrt{2x}}$ into the square roots (note that it gets raised to progressively higher powers as it is pulled into more inner square roots) you get an expansion of the form $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
For what values of $k$ does the integral $\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$ converge? For what values of $k$ does the integral $$\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$$ converge?
I believe it does not converge for any $k$.
| $\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}=\int_{0^+}^{1} \frac{dx}{x^k \sqrt{x+x^2}}+\int_{1}^{2} \frac{dx}{x^k \sqrt{x+x^2}}+\int_{2}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$
If $0<x<1$, $\sqrt{x+x^2}\ge \sqrt{x}$ hence $\int_{0^+}^{1} \frac{dx}{x^k \sqrt{x+x^2}}\le \int_{0^+}^{1} x^{-k-\frac{1}{2}} dx<\infty$ if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding Maximum Area of a Rectangle in an Ellipse
Question: A rectangle and an ellipse are both centred at $(0,0)$.
The vertices of the rectangle are concurrent with the ellipse as shown
Prove that the maximum possible area of the rectangle occurs when the x coordinate of
point $P$ is $x = \frac{a}{\sqrt{2... | Your mistake is here
$$
A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) +\color{red}{\frac12} \times4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-2b^2x}{a^2} \right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.
| Not as elegant as Roman83's solution but here you go:
Solve the quadratic equation in $x$:
$$ x = \frac{-B\pm \sqrt{B^2 - 4AC}}{2A}$$
$A = 1, \ B = 2\cos b\cos c, \ C =\cos^2 b +\cos^2 c - 1 $
$$ x = \frac{-2\cos b\cos c\pm \sqrt{(2\cos b\cos c)^2 - 4(\cos^2 b +\cos^2 c - 1)}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to find the limit of $\lim_{x\to 0} \frac{1-\cos^n x}{x^2}$ How can I show that
$$
\lim_{x\to 0} \frac{1-\cos^n x}{x^2} = \frac{n}{2}
$$
without using Taylor series $\cos^n x = 1 - \frac{n}{2} x^2 + \cdots\,$?
| I have seen several limit problems on MSE where people don't use the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ whereas frequent use is made of other standard limits like $$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Prove that largest root of $Q_k(x)$ is greater than that of $Q_j(x)$ for $k>j$. Consider the recurrence relation:
$P_0=1$
$P_1=x$
$P_n(x)=xP_{n-1}-P_{n-2}$;
1). What is closed form of $P_n$?
2). Let $k,j.\in\{1,2,...,n+1\}$ and $Q_k(x)=xP_{2n+1}-P_{k-1}.P_{2n+1-k}$. Then prove that largest root of $Q_k$ is greater than... | The expression for $P_n$ in terms of $n$ is pretty ugly,
$$P_n=xP_{n-1}-P_{n-2}$$
Exact form for $P_n$ can be solved by using characteristic equation:
It comes from guessing $P_n=ar^n$ might be an answer for $P_n$, you can search "recurrence relation" online to see how it works.
$ar^n=x(ar^{n-1})-ar^{n-2}$
$r^2=xr-1$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $ , What is the exact value of $\sin(3x)$?
Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $
What is the exact value of $\sin(3x)$?
What I have done:
Given $\tan(x) = 2\sqrt2 $ , I drew a right angled triangle and found the hypotenuse to be $3$... | $$\sin^2x=\frac1{1+\cot^2x}=\frac1{1+\frac1{8}}=\frac89\implies \sin x=-\frac{2\sqrt2}3\,,\,\,\text{since}\;\;x\in[\pi,\,3\pi/2]\implies$$
$$\sin3x=3\sin x-4\sin^3x=-2\sqrt2+\frac{64\sqrt2}{27}=\frac{10\sqrt2}{27}$$
since
$$\pi\le x\le\frac{3\pi}2\implies3\pi\le 3x\le4\pi+\frac\pi2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$
Prove that $$\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$$
My try: Using $$\displaystyle \sin x<x$$ and $$\frac{\sin x-0}{x-0}>\frac{1-0}{\frac{\pi}{2}-0}=\fra... | This is not an answer but it is too long for a comment.
I cannot resist the pleasure of reporting again the magnificent approximation proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ (see here).
So, as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Limits and continuity: $\lim_{ x\rightarrow0} {\frac{ax+bx^3}{x^2}} = 2$. Find the values of $a$ and $b$ Please help
$$\lim_{ x\rightarrow0} {\frac{ax+bx^3}{x^2}} = 2$$
Find the values of $a$ and $b$.
| \begin{align}
\lim_{ x\rightarrow0} \frac{ax+bx^3}{x^2}
&=\lim_{ x\rightarrow0} \frac{x^2(\frac{a}{x}+bx)}{x^2}\\[6px]
&=\lim_{ x\rightarrow0} \left(\frac{a}{x}+bx\right)\\[6px]
&=\begin{cases}
\pm\infty&\text{if $a\neq 0$ (*)}\\[6px]
0 &\text{if $a=0$}
\end{cases}
\end{align}
(*) Depending on the sign of $a$ and i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality involving four numbers Show that, if $a$, $b$, $c$ and $d$ are four positive numbers with sum $1$, then $$\frac 3 {1-a} + \frac 3 {1-b} + \frac 3 {1-c} + \frac 3 {1-d} \ge \frac 5 {1+a} + \frac 5 {1+b} + \frac 5 {1+c} + \frac 5 {1+d}.$$ I tried to subtract the fractions, but I didn't get to any result.
| I’ll show that the function $$g(a,b,c,d)=\frac 3 {1-a} + \frac 3 {1-b} + \frac 3 {1-c} + \frac 3 {1-d} - \left(\frac 5 {1+a} + \frac 5 {1+b} + \frac 5 {1+c} + \frac 5 {1+d}\right)$$ is minimized over positive $a,b,c,d$ with sum $1$ when $a=b=c=d=\frac{1}{4}$. Its minimum is then $g(\frac14,\frac14,\frac14,\frac14)=0$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the correct u-substitution for this integral? I'm wondering how I can solve the following indefinite integral:
$$\int\frac{x}{x^2-2x+2}dx$$
The $u$-substitution I did was $x^2-2x+2$, but I've gotten a little stuck. I've shown my steps below for clarification on the problem.
$$u=x^2-2x+2$$
$$du = (2x-2)dx$$
$$du... | $$
\int\frac{x}{x^2-2x+2}dx = \frac{1}{2}\int\frac{2x-2+2}{x^2-2x+2}dx\\
= \frac{1}{2}\int\frac{2x-2}{x^2-2x+2}dx + \int\frac{1}{(x-1)^2+1}dx\\
= \frac{1}{2} \log(x^2-2x+2) + \tan^{-1}(x-1) + c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$ Problem : Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$
My Approach : I would assume that we can prove by induction.
Base case $n=2$.
$$=2{2 \choose 2}+ {2 \choose 1}= (2\cdot 1)+2 =4$$
$$n^2 =2^2 = 4.$$
Assume for $n\ge 2$, $n\ge 2$, $2{n \c... | It's much simpler to give a direct proof:
$$
2 \binom{n}{2} + \binom{n}{1}
=
2 \cdot \frac{n(n-1)}{2} + n
=
n^{2} - n + n
=
n^{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How can we find a condition on $a$ so that there exists a regular matrix $P$ s.t. $B=P^{-1}AP$ $$ A=\begin{pmatrix} 4 & 1 & 1 \\ -1 & 1 & 0 \\ -2 & -1 & 0 \\ \end{pmatrix}, \, \, B=\begin{pmatrix} 0 & 1 & 0 \\ -3 & 4 & 0 \\ a & 2 & 1 \\ \end{pmatrix} $$
How can we find a condition on $a$ so that there exists a regula... | $(A-I)(A-3I)\not=0,(A-I)^2(A-3I)=0$ and $(B-I)(B-3I)=(a+2)\begin{pmatrix}0&0&0\\0&0&0\\-3&1&0\end{pmatrix},(B-I)^2(B-3I)=0$.
Thus $A,B$ are similar iff $a\not=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that the determinant is a multiple of $17$ without developing it Let, matrix is given as :
$$D=\begin{bmatrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{bmatrix}$$
Prove that the determinant is a multiple of $17$ without developing it?
I saw a resolution by the Jacobi method , but could not apply the methodology... | $$
|D|
=
\begin{vmatrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 9 \\
0 & 7 & -2 \\
0 & 4 & -6\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 9 \\
0 & 7 & -2 \\
0 & 0 & -\dfrac{34}{7}\end{vmatrix}
=
1 \times 7 \times -\dfrac{34}{7}
=
-34
=-2 \times 17
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
divergence/convergence $\sum_{k=1}^{\infty} \frac{1}{5k+2}$
$$\sum_{k=1}^{\infty} \frac{1}{5k+2}$$
Can I say that $$\sum_{k=1}^{\infty} \frac{1}{5k+2}=\sum_{k=1}\frac{1}{5k}+\sum_{k=1}^{\infty} \frac{1}{2}=\frac{1}{5}\sum_{k=1}^{\infty}\frac{1}{k}+\frac{1}{2}$$
and we know that $\\sum_{k=1}^{\infty} \frac{1}{k}$ dive... | I would say that
$$ \sum_{k=1}^\infty\frac{1}{10k}<\sum_{k=1}^\infty \frac{1}{5k+2}$$
which is true because
$$ \frac{1}{10k}<\frac{1}{5k+2}\:\:\forall k\in \mathbb{N}.$$
$$ \sum_{k=1}^\infty \frac{1}{10k}=\frac{1}{10}\sum_{k=1}^\infty \frac{1}{k}$$
which implies that it diverges. So, because
$$ \sum_{k=1}^\infty\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum length of a line segment with endpoints on the coordinate axes that passes through the point $(1, 1)$ A line passes through the point $P=(1,1)$ and through the $x$ and $y$-axes at points $A$ and $B$ respectively. Find the minimum length of the line segment $AB$.
| $\frac{y-1}{x-1}=m$ defines the line.
$A$ is the $x$ intercept which is $(1-\frac{1}{m}, 0)$ and $B$ is the $y$ intercept which is $(0,1-m)$.
By minimizing the distance between these two points we can calculate $m$ which will then give us the distance. To make the algebra simple we will actually minimize the distance s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to calculate this integral without any integration techniques?
Differentiate $f(x) = (5x+2)\ln(2x+1)$ with respect to $x$. Hence, find $\int \ln(2x+1)^3dx$.
Because of the word "Hence" I'm assuming that the question doesn't allow integration techniques such as integration by parts or substitution.
The first part... | Alright so from my first experiment we have the derivatie $3\ln(2x+1) + \dfrac{2(3x+2)}{2x+1}$. To get this to just $3\ln(2x+1)$ we subtract $3$ and $\dfrac{1}{2x+1}$. This gives us the expression $$\bigg(3\ln(2x+1) + \dfrac{6x}{2x+1} + \dfrac{4}{2x+1} \bigg) - 3 - \dfrac{1}{2x+1}$$ Integrating gives us $(3x+2)\ln(2x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\iint_{D}\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\text{d}x\text d y$
Evaluate using polar coordinates:$$\iint_{D}\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\text{d}x \, \text d y\qquad D=\bigg\{\frac{x^2}{a^2}+\frac{y^2}{b^2}\leqslant 1\bigg\}$$
What I did:
I found that the Jacobian matrix equal to $a\cdot b\c... | If the change is $x = a r\cos\theta$, $y = b r\sin\theta$:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = r^2\cos^2\theta + r^2\sin^2\theta = r^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $(1)\;\;\left(\frac n3\right)^n
Proving $$(1)\;\;\;\;\left(\frac n3\right)^n<n!<\left(\frac n2\right)^n\forall\,\,n\ge 6,n\in\mathbb{N}$$
and $$(2)\;\; \sqrt{n^n}\le n!\,\,\forall n\in\mathbb{N}$$ without induction
$\bf{My\; Try::}$ For $\bf{R.H.S}$ Inequality.
Using $\bf{A.M\geq G.M\;,}$ We get $$\frac{2+3+4+5... | For (2), note that if $n$ is even we have:
$$\begin{align*}\dfrac{n^n}{(n!)^2}&=\dfrac{n\cdot n\cdot n\cdot n\cdots n\cdot n}{n\cdot n\cdot (n-1)\cdot (n-1)\cdots2\cdot 2\cdot 1\cdot 1}\\&=\dfrac{n\cdot n}{n\cdot n}\cdot \dfrac{n\cdot n}{(n-1)(n-1)}\cdots \dfrac{n\cdot n}{(n/2+1)(n/2+1)}\cdot \dfrac{1}{(n/2)(n/2)}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11) If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11)
Not sure if I'm approaching this problem correctly, any help is appreciated.
So far I have:
$7x^2 - 4x + 1 \equiv 0 $ (mod 11)
$21x^2 - 12x + 3 \equiv 0 $ (mod 11)
$-x^2 - x + 3 \equiv 0 $ (mod 11)
$-x^2 - x \eq... | In $\mathbb F_{11}$, one has $$7x^2 - 4x + 1= 0\iff-4x^2-4x+1=0\iff x(x+1)=\frac {1}{2^2}=2^8=3$$
For all $x\in\mathbb F_{11}$ we have $x(x+1)\ne 3$ (successively one has $0,2,6,1,9,8,9,1,6,2,0$)
Hence there is not solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$ Let $x,y,z>0$ and $x+y+z=1$, then find the least value of
$${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$
I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am n... | Here we use the AM-HM (harmonic mean) inequality.
\begin{align*}
{{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}
&=-3+2\left(\frac{1}{2-x}+\frac{1}{2-y}+\frac{1}{2-z}\right)\\
&\geq
-3+2\left[\frac{3^2}{(2-x)+(2-y)+(2-z)}\right]\\
&=-3+\frac{18}{5}=\frac{3}{5}.
\end{align*}
Also, when $x=y=z=\frac{1}{3}$, the above ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1708395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Finding the error in an evaluation of the limit $\lim_{x\to0} \frac{e^x-x-1}{x^2} $ \begin{align}
\lim_{x\to0} \frac{e^x-x-1}{x^2}
&= \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} \\
&= \lim_{x\to0} \frac{e^x-1}{x}\lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\ &= \lim_{x\to0} \frac{1}{x} - \lim_{x\t... | The first claimed equality,
$$\lim_{x\to0} \frac{e^x-x-1}{x^2} \color{red}{\stackrel{\textrm{(false)}}{=}} \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} ,$$ already doesn't hold: Neither of the limits on its r.h.s. exist.
To evaluate the limit:
Hint Apply l'Hopital's Rule (twice) or expand $e^x$ in a Maclau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
To prove $X^2$ and $Y^2$ are independent for a given joint pdf $f(x,y)=\dfrac{1}{4}(1+xy)$ To prove $X^2$ and $Y^2$ are independent for a given joint pdf $f(x,y)=\dfrac{1}{4}(1+xy)$
$f(x,y) =
\begin{cases}
\dfrac{1}{4}(1+xy), & |x|<1,|y|<1 \\
0, & \text{otherwise}
\end{cases}$
The given solution starts with trying to ... | The answer is rather simple. Just verify by a simple calculation that
$\frac{1}{4} \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} uv \, du dv = 0$
and you're done.
Edit: here all steps:
$\int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} f(u,v) \, du dv = \int\limits_{-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Some uncommon improper integral. Convergence. I bet the integral $\int_1^3\frac{dx}{\sqrt{\tan(x^3-7x^2+15x-9)}}$ converges, but have no idea about proof, except expansion of tan in a series near zeros of its argument (limits of the integration).
| Factoring the polynomial in the function gives us
$$\sqrt{\tan(x^3-7x^2+15x-9)}=\sqrt{\tan((x-1)(x-3)^2)}$$
The tangent function can be bounded above as the following: $$\tan(y)=\frac{\sin(y)}{\cos(y)} < \frac{y}{\cos(y)} < 2y$$ when $y$ is sufficiently close to zero. Therefore $$\sqrt{\tan((x-1)(x-3)^2)} < \sqrt{2(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.