Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$ I'm having trouble computing the integral:
$$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$
I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.
An... | $$= \frac{1}{2} \cdot \int \frac{\sin(x) + \sin(x)}{\sin(x) + \cos(x)} dx = \frac{1}{2} \int \frac{\sin(x) + \cos(x) + \sin(x) - \cos(x) }{\sin(x) + \cos(x)} dx $$ $$= \frac{x}{2} - \frac{1}{2} \int \frac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)} dx$$ Let $u = \sin(x) + \cos(x)$ $$ \implies \frac{x}{2} - \frac{1}{2} \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "88",
"answer_count": 8,
"answer_id": 7
} |
Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Possible Duplicate:
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity?
If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg ... | Since $C = \pi - A - B$, what needs to be shown is
$$\sin(A) + \sin(B) + \sin(\pi - A - B) = 4\cos({A \over 2})\cos({B \over 2})\cos({\pi \over 2} - {A + B \over 2})$$
Equivalently, you need that
$$\sin(A) + \sin(B) + \sin( A + B) = 4\cos({A \over 2})\cos({B \over 2})\sin( {A + B \over 2})$$
Using the trig identity $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How do I transform the equation based on this condition? If a and b are the roots of the equation $$2x^2-px+7=0$$ Then a-b is a root of ?
| So, $a+b=\frac{p}{2}$ and $ab=\frac{7}{2}$
If $y=a-b=>y^2=(a+b)^2-4ab$
$=>y^2=(\frac{p}{2})^2-4 \frac{7}{2}$
So, a-b is a root of $4y^2-p^2+56=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $x$ in the displayed figure Find $x$ in the following figure.
$AB,AC,AD,BC,BE,CD$ are straight lines.
$AE=x$, $BE=CD=x-3$, $BC=10$, $AD=x+4$
$\angle BEC=90^{\circ}$, $\angle ADC=90^{\circ}$
NOTE: figure not to scale.
| $∆BCE$ is right angle triangle.
Hence $BC^2 = BE^2 + EC^2$
$EC = \sqrt{(BC^2 - BE^2})= \sqrt{(100 - (x-3)^2)}$
$∆ACD$ is right angle triangle.
Hence $AC^2 = CD^2 + AD^2$
$(AE + EC)^2 = CD^2 + AD^2$
Substitute the values,
$(\sqrt{(100 - (x-3)^2)} + x)^2 = (x-3)^2 + (x+4)^2$
Then you can solve this equation easil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/183088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Properties of Mediants If $\frac{a}{c} > \frac{b}{d}$, then the mediant of these two fractions is defined as $\frac{a+b}{c+d}$ and can be shown to lie striclty between the two fractions.
My question is can you prove the following property of mediants: if $|\frac{a}{c} - x| > |x - \frac{b}{d}|$ then $|b/d - mediant| < ... | $|\frac{a}{c}-x|>|x-\frac{b}{d}|$
$=>(\frac{a}{c}-x)^2>(x-\frac{b}{d})^2$
$=>(\frac{a}{c}-\frac{b}{d})(\frac{a}{c}+\frac{b}{d}-2x)>0$
$=>2x<\frac{a}{c}+\frac{b}{d}=>-x>\frac{-1}{2}(\frac{a}{c}+\frac{b}{d})$
$\frac{b}{d}-mediant=\frac{b}{d}-\frac{a+b}{c+d}=\frac{bc-ad}{d(c+d)}=\frac{-c}{c+d}(\frac{a}{c}-\frac{b}{d})$
$m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving: $x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$ What is the proof that:
$x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$
if:
$0< x;y;z< 1$
| André’s solution is much nicer, but the problem can also be solved in a more routine fashion. Let $a=x+y$; then
$$\begin{align*}
x(1-y)+y(1-z)+z(1-x)&=x(1-a+x)+(a-x)(1-z)+z(1-x)\\
&=x^2-ax+a+(1-a)z\\
&=\left(x-\frac{a}2\right)^2+a+(1-a)z\;.\tag{1}
\end{align*}$$
If $a\le 1$, the third term is bounded above by $(1-a)\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Manipulating Exponents I'm doing my homework and there are a couple of things that I am having trouble grasping. All my homework asks is that I simplify the exponents. For example: 6^5 * 6^3 = 6^8
There are 2 problems I am unsure on what to do. They have to do with multiplying fractions:
(7^3)^5/6 = 7^5/2 is the answe... | For any $x$, $a$, and $b$.
$(x^a)^b = x^{ab}$.
Hence for the first one, letting $x = 7$, $a = 3$, $b = \frac{5}{6}$, you have
$(7^3)^{\frac{5}{6}} = 7^{3 \cdot \frac{5}{6}} = 7^{\frac{5}{2}}$
For the second letting $x = 7$, $a = \frac{3}{5}$ and $b = \frac{5}{6}$, you have
$(7^{\frac{3}{5}})^{\frac{5}{6}} = 7^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\... | $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots }}}}}}$
$x^2 =2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots }}}}}$
$x^2 =2+x$
$x^2-x-2=0$
$x^2-2x+x-2=0$
$(x-2)(x+1)=0$
we have, $x=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \... | you have to resolve this inequality depending $x$ : $$\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$$
that is equivalent with: $\displaystyle \frac{x(2x-13)-15(2x+3)}{x(2x+3)} \lt 0$ so:
$\displaystyle \frac{2x^2-13x-30x-45}{2x^2+3x} \lt 0$ and now the simple form is :
$$\frac{2x^2-43x-45}{2x^2+3x} \lt 0$$
Now we have two e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Integrating $ \int \frac{1}{1-\tan(x)}dx $ $$ \int \frac{dx}{1-\tan(x)} $$
Please help me to solve this problem as I'm trying this since last 1 day...
| An alternative is to use the substitution $t=\tan x$ and then expand into partial fractions.
$$\begin{equation*}
dt=\left( 1+\tan ^{2}x\right) dx=\left( 1+t^{2}\right) dx
\end{equation*}$$
$$\begin{equation*}
I=\int \frac{1}{1-\tan x}dx=\int \frac{1}{\left( 1-t\right) \left(
1+t^{2}\right) }\,dt.
\end{equation*}$$
Sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\e... | Homogenize the given problem into,
$$\frac{\sqrt{ab}}{a+3b}+\frac{\sqrt{ab}}{b+3a}\leq \frac 12.$$
Now note that, using the AM-GM inequality we have $a+3b\geq 2\sqrt{2b(a+b)},$ so that $$\dfrac{\sqrt{ab}}{a+3b}\leq \frac{\sqrt{a}}{2\sqrt{2(a+b)}}.$$
Hence it suffices to check that $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{2(a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Simplify $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.
Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.
I have tried completing square by several method but all failed. Can anyone help me please? Thank you.
p.s. I'm a poor question-tagger.
| There is a formula. I'm not too sure how to prove it, but I know that there is a formula where you can denest $$\sqrt{\sqrt[3]{\alpha}+\sqrt[3]{\beta}}$$Into$$\pm\frac {1}{\sqrt{f}}\left(-\frac {s^2\sqrt[3]{\alpha^2}}{2}+s\sqrt[3]{\alpha\beta}+\sqrt[3]{\beta^2}\right)$$ where $$f=\beta-s^3\alpha$$ and $s$ is a real num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 1
} |
Inequality with unusual constraint $a,b,c\in (0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$ Suppose as in the title that $a,b,c$ are three real positive numbers in $(0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$. Then I was asked to prove that $$a+b+c\leq \frac 32.$$ I am not very good at inequalities, so can anybo... | It is easy to see that the equality is achieved at a=b=c=1/2. Therefore, let's expand around that point, i.e. define $x,y,z$ by $a=x+1/2$, $b=y+1/2$, $c=z+1/2$. Then, the first equality becomes
$$1=2abc+a^2+b^2+c^2\\
\Longleftrightarrow 0=2xyz+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)$$
Next, Jensen's inequality gives yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}$ Please help me for solving $$\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}$$
| $\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}$=$\lim_{x\to\infty}\frac{\ln x^2(1-\frac{1}{x}+\frac{1}{x^2})}{\ln x^{10} (1+\frac{1}{x^9}+\frac{1}{x^{10}})}$=$\lim_{x\to\infty}\frac{\ln x^2+\ln(1-\frac{1}{x}+\frac{1}{x^2})}{\ln x^{10}+\ln(1+\frac{1}{x^9}+\frac{1}{x^{10}})}$=$\lim_{x\to\infty}\frac{\ln x^2+\ln 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
triangles and trigonometry A triangle has sides $a,b,c$ and angles $\alpha,\beta,\gamma$ such that:
$$ a \,\cos\beta + b \, \cos\gamma+ c \, \cos\alpha = \frac{a+b+c}{2}$$
Prove that the triangle is isosceles.
I tried writing $\cos$ in terms of the sides (using the cosine theorem),
for example $ \cos\alpha= \frac{a^2-b... | To continue your proof, write $$a^3\left(b-c\right)+b^3\left(c-b+b-a\right)+c^3\left(a-b\right)=0$$ which implies $$\left(a^3-b^3\right)\left(b-c\right)=\left(b^3-c^3\right)\left(a-b\right).$$
If $b\neq a$ and $b\neq c$ then we can simplify the equation to $$a^2+ab+b^2=b^2+bc+c^2$$ which is equivalent to $$\left(a-c\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequality. $a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}$ Could you help me please with the following inequality
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3.$ Prove that:
$$a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}.$$
| $$a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2}=\sqrt{a}\sqrt{2ba+c^2a}+\sqrt{b}\sqrt{2bc+a^2b}+\sqrt{c}\sqrt{2ac+b^2c}$$
Using Cauchy-Schwarz we obtain :
$$\sqrt{a}\sqrt{2ba+c^2a}+\sqrt{b}\sqrt{2bc+a^2b}+\sqrt{c}\sqrt{2ac+b^2c} \leq \sqrt{(a+b+c)\left(a^2b+b^2c+c^2a+2ab+2bc+2ca\right)}=\sqrt{3\left(\sum{2ab}+\sum{c^2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$(1+i)$ to the power $n$
Possible Duplicate:
Complex number: calculate $(1 + i)^n$.
I came across a difficult problem which I would like to ask you about:
Compute
$ (1+i)^n $ for $ n \in \mathbb{Z}$
My ideas so far were to write out what this expression gives for $n=1,2,\ldots,8$, but I see no pattern such that I ca... | You got to this expression:
$$(1+i)^n=\left(\sqrt{2}\right)^n \left(\cos\left(n\frac{\pi}{4}\right)+i\sin\left(n\frac{\pi}{4}\right)\right)$$
Now, you should note that this expression (without $\left(\sqrt{2}\right)^n$) has periodicity of $8$, i.e. for $n=k$ and $n=k+8$ the value will be the same. So you can just look... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Multiplicative inverse of a quadratic algebraic number $\,a+b\sqrt 2$ Find the multiplicative inverse of $1+ 3\sqrt{2}$ in the ring $\mathbb{Q}(\sqrt{2})$ and use it to solve the equation $(1+3\sqrt{2})x=1-5\sqrt{2}$.
I think that the inverse is the conjugate, so it would be $1-3\sqrt{2}$, but then I don't know where t... | Let $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ be the inverse of $1+3\sqrt{2}$, i.e. $(a+b\sqrt{2})(1+3\sqrt{2})=1$. Then
$$
1=a+6b+(3a+b)\sqrt{2},
$$
i.e.
$$
3a+b=0,\ a+6b=1.
$$
It follows that
$$
a=-\frac{1}{17},\ b=\frac{3}{17}.
$$
Now
$$
(1+3\sqrt{2})x=1-5\sqrt{2} \iff x=(1+3\sqrt{2})^{-1}(1-5\sqrt{2}),
$$
i.e.
$$
x=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
A inequality proposed at Zhautykov Olympiad 2008 An inequality proposed at Zhautykov Olympiad 2008.
Let be $a,b,c >0$ with $abc=1$. Prove that:
$$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$
Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
Our inequality becomes:
$$\sum_{\mathrm{cyc}}{\frac{z^2... | Since $\mathrm{LHS}$ of last inequality is homogeneous we can assume $x^2 + y^2 + z^2 = 1$. Then it becomes
$$
\mathrm{LHS} = 2\sum_{cyc} \frac {x^2} {1 + z^2} =:2I
$$
Now using Cauchy-Schwarz inequality we get
$$
1 = (x^2 + y^2 + z^2)^2 = \left(\sum_{cyc} x\sqrt{1 + z^2} \cdot \frac x {\sqrt{1 + z^2}}\right)^2 \leq\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
A problem about multiples.
For any positive integers $a$, $ b$, if $ab+1$ is a multiple of $16$, then $a+b$ must be a multiple of $p$. Find the largest possible value of $p$.
I have no idea how to solve this. Please help. Thank you.
| We can consider the problem in $\mathbb{Z}_{16}$.
Then $ab+1$ is a multiple of 16 means that $\overline{ab+1}=\overline{0}$, then $\overline{a}\overline{b}=\overline{ab}=\overline{15}$, in $\mathbb{Z}_{16}$, $\overline{15}$ only has four decompositions, means $\overline{15}=\overline{1}\cdot\overline{15}=\overline{3}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/206509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Correctness of Fermat Factorization Proof I have asked similar questions regarding this proof. But now I would like to know if my reformulation (after perseverance and different thinking) is correct.
Prove: An odd integer $n \in \mathbb{N}$ is composite $\iff$ $n$ can be written as $n = x^2 - y^2 s.t. y+1 < x$
Proof: $... | You need to specify that $0\le b\le a$ and that $b$ is not $1$. If $n$ is composite, we can certainly find such $a$ and $b$. Then we have
$$x-y=\frac{a+b}{2}-\frac{a-b}{2}=b\gt 1.$$
(If you want the inequality $y+1\gt x$, interchange the roles of $x$ and $y$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/206844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\sum_{k=1}^n\lfloor \sqrt{k} \rfloor$ I know that $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$.
How can I use this fact to evaluate $\sum_{k=1}^n\lfloor \sqrt{k} \rfloor$ ?
| For $1\le r<\lfloor\sqrt n\rfloor$, there are exactly $2r+1$ summands of size $r$, namely for $r^2\le k<(r+1)^2$. The remaining $n+1-\lfloor\sqrt n\rfloor^2$ summands are $\lfloor\sqrt n\rfloor$ each. Therefore, with $m:=\lfloor\sqrt n\rfloor$,
$$\begin{align}\sum_{k=1}^n\lfloor\sqrt k\rfloor &= 2\sum_{r=1}^{m-1}r^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
$$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$
by Mathematical induction，n$\ge $1
| Here's an elementary proof, that requires no lengthy computations. Let
$$S_n=\sum_{k=n+1}^{2n}\frac1k$$
We show by induction that $S_n \le \frac{25}{36} - \frac{1}{4n+1}$ for all $n \ge 2$. To start with, $S_2 = \frac13+\frac14=\frac{25}{36} - \frac19$, so the hypothesis is true for $n=2$. Now suppose it is true for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 1
} |
Finding the $n$-th derivative of $f(x) =e^x \sin x$, solving the recurrence relation I am trying to find a closed solution for the nth derivative of the function:
$f(x) = e^x \sin x$
So far I have been able to obtain the derivative as:
$f^{(n)}(x) = e^x S_n \sin x + e^x C_n \cos x$
The sequences S and C are defined as ... | The recurrence $S_n=S_{n-1}-2S_{n-3}$ can be solved mechanically. Its auxiliary equation is $x^3-x^2+2=0$. By inspection $-1$ is a solution, so $x+1$ is a factor of the cubic: $$x^3-x^2+2=(x+1)(x^2-2x+2)\;.$$ The other roots are $$\frac{2\pm\sqrt{-4}}2=1\pm i\;,$$ so the solution is of the form $S_n=A(-1)^n+B(1+i)^n+C(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ How to show the following equality?
$$\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$$
| It is well known that
$$\sum_{n=-\infty}^\infty f(n)= -\sum_{j=1}^k \operatorname*{Res}_{z=j}\pi \cot (\pi z)f(z) $$
Assume $a \neq 0$.
To find the residues of $g(z) := \pi \cot (\pi z)\frac{1}{a^2+n^2}$, we see
$$\frac{1}{a^2+n^2} = \frac{1}{(n+ia)(n-ia)}$$
so $g$ has poles at $z_1 = ia$ and $z_2 = -ia$. Their respe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 5,
"answer_id": 3
} |
How to solve the recurrence relation of $f_n(x)=A_n(a,b)x+B_n(a,b)f(x)$ $$f(f(x))=ax+bf(x)$$
$$f(f(f(x)))=f_3(x)=af(x)+b(ax+bf(x))=abx+(a+b^2)f(x)$$
$$f(f(f(f(x))))=f_4(x)=abf(x)+(a+b^2)(ax+bf(x))=(a^2+ab^2)x+(2ab+b^3)f(x)$$
$$f_n(x)=A_n(a,b)x+B_n(a,b)f(x)$$
$$f_n(f(x))=A_n(a,b)f(x)+B_n(a,b)(ax+bf(x))$$
$$f_n(f(x))=aB_... | I would like to share my solution for the problem
$$f(f(x))=ax+bf(x)$$
$f(x)=rx$ is a solution of the equation.
$$r^2x=(a+br)x$$
$$r^2-br-a=0$$
$$r_1=\frac{b+\sqrt{b^2+4a}}{2}$$
$$r_2=\frac{b-\sqrt{b^2+4a}}{2}$$
$$f_n(x)=r^nx$$
$$f_n(x)=A_n(a,b)x+B_n(a,b)f(x)=(A_n(a,b)+rB_n(a,b))x$$
$$A_n(a,b)+(\frac{b+\sqrt{b^2+4a}}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/210401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sequence of solutions to $x\sin x=1$
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.
Consider a sequence $x_n, n\ge1$ formed by positive solutions to $x \sin{x}=1$.
How can we find
$$\lim _{n\rightarrow \infty}(n(x_{2n+1}-2\pi n))= L$$... | Let $y$ be a variable tending to zero. Put
$$
\begin{array}{l}
a=y-\frac{5}{6}y^3, \\
b=y-\frac{5}{6}y^3+\frac{169}{120}y^5
\end{array}
$$
Using Taylor expansions, one finds that
$$
\begin{array}{l}
\sin(a)\bigg(\frac{1}{y}+a\bigg)=1-\frac{169}{120}y^4+O(y^5), \\
\sin(b)\bigg(\frac{1}{y}+b\bigg)=1+\frac{5021}{1680}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/211333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Limit of a Recurrence Sequence $a_0=c$ where $c$ is positive, with $a_n=\log{(1+a_{n-1})}$,Find
\begin{align}\lim_{n\to\infty}\frac{n(na_n-2)}{\log{n}}\end{align}
I'have tried Taylor expansion, but I can't find the way to crack this limit. Thanks alot for your attention!
| You may find an asymptotic formula for $a_n$ by improving the accuracy in an adaptive manner.
Step 1. Since
$$0 < a_{n+1} = \log (1+a_n) < a_n,$$
it is a monotone decreasing sequence which is bounded. Thus it must converge to some limit, say $\alpha$. Then $\alpha = \log(1 + \alpha)$, which is true precisely when $\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/217961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
The limit of a nested radical Let $f(a,x)=\sqrt{a(a-1)+x}$, $f^{(n)}(a,x)$ denotes the $n^{th}$ iteration of $x$, where
\begin{align}f^{(1)}(a,x)=f(a,x),f^{(n)}(a,x)=f^{(n-1)}(a,f(a,x))\end{align}
Find
\begin{align}\lim_{n\to\infty}(2a)^{\frac{n}{2}}\sqrt{a-f^{(n)}(a,0)}\end{align}
This is a problem from a discussion... | Steps for solution ... With $a=2$ write $f(2,x) = f(x)$.
For the appropriate values of $\theta$ ...
$$\begin{align}
&2\cos\theta = \sqrt{2+2\cos(2\theta)}
\\
&f^{(n)}(2\cos\theta) = 2\cos\frac{\theta}{2^n}
\\
&f^{(n)}(0) = 2\cos\frac{\pi}{2^{n+1}}
\\
&\sqrt{2-2\cos\theta} = 2 \sin \frac{\theta}{2}
\\
&\sqrt{2-f^{(n)}(0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/219457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Order of elements modulo p Let $p$ be prime. Suppose that $x\in Z$ has order 6 mod p. Prove that $(1-x)$ has order 6 mod p as well.
I know that I need to show that the order can't be 2 or 3 (4 and 5 are trivial cases), but I'm even having difficulty showing that $(1-x)^6$ is 1. I've expanded $x^{6}-1$ to get a nice con... | If $ord_px=6\implies p\mid(x^6-1)$ ,i.e, $p\mid(x^3-1)(x^3+1)$
If $p\mid(x^3-1), x^3\equiv 1\pmod p\implies ord_px=3\ne 6$
So, $p\mid(x^3+1)\implies p\mid (x+1)(x^2-x+1)$ and $x^3\equiv -1\pmod p$
If $p\mid(x+1), x\equiv -1\pmod p\implies x^2\equiv 1\pmod p\implies ord_px=2\ne 6$
So, $p\mid(x^2-x+1)$
(i) $ x^2\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/220493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$
| Using the chain rule where $\cfrac {df}{dx} = \cfrac {dg}{du} \cfrac {du}{dx} $ if $f(x) = g(u(x))$ and the product rule where $\cfrac {d}{dx} (uv) = v\cfrac {du}{dx} + u\cfrac {dv}{dx} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/220858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding the equations of two lines that are tangent to the curve and parallel to a line. I have no idea how to do this as i have been absent from class for medical reasons.
the curve $y=3+x^2$
and the line $3x-y=-6$
The answer is a possibility of the following:
a) $ y=3x, y=3x+4$
b) $ y=12x-15, y=12x+17$
c) $ y=-3x+9, ... | The equation of any line parallel to $3x-y=-6$ can be written as $3x-y=k$
Now, let this line touches the curve $y=3+x^2$ at $(a,b).$
So, $3a-b=k$ or $b=3a-k$ and $b=3+a^2$
Comparing the values of $b$ we get, $3a-k=3+a^2,$ or $a^2-3a+k+3=0$
Now, this is a quadratic equation in $a,$ for the tangency both the root should... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/221496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Partial Fractions - Calculus Evaluate the integral: $$\int \dfrac{9x^2+13x-6}{(x-1)(x+1)^2} dx$$
For some reason I cannot get the right answer. I split up the equation into three partial fractions but I cannot seem to find A, B, or C from the three subsequent equations. Thanks!
| The motivation is to write $\dfrac{9x^2 + 13x - 6}{(x-1)(x+1)^2}$ as $$\dfrac{A}{(x+1)^2} + \dfrac{B}{x+1} + \dfrac{C}{x-1}$$
This gives us
\begin{align}
9x^2 + 13x - 6 & = A(x-1) + B(x^2-1) + C(x+1)^2\\
& = (B+C)x^2 + (2C+A)x + (C-A-B)
\end{align}
This gives us $$B+C = 9\\ 2C+A = 13\\ C - A - B = -6$$
Adding all the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
| Let $$M=\dfrac{a^3}{a+b}+\dfrac{b^3}{b+c}+\dfrac{c^3}{c+a}$$ be the original expression, we introduce its "conjugate"
$$N=\dfrac{ab^2}{a+b}+\dfrac{bc^2}{b+c}+\dfrac{ca^2}{c+a}$$
Direct computing, and using $a^2+b^2+c^2\geq ab+bc+ca$, yields
$$M-N =\frac{a(a^2-b^2)}{a+b}+\frac{b(b^2-c^2)}{b+c}+\frac{c(c^2-a^2)}{c+a}\\=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Analyzing a function I am having a problem with the following function:
$f(x)=\sin^2(x)-\cos(3x)$
I need to examine the sign of $f'(x)$
I noticed that f is $2\pi$-periodic, therefore we need to analyze f(x) on $[0;2\pi]$
In addition to that $f(x)=-4\cos^3(x)-\cos^2(x)+3\cos(x)+1$
Hence $\forall x \in ]0;2\pi[, f'(x)=(-... | $$
\begin{align}
f'(x)&=2 \cos x\sin x +3 \sin (3x) = 2 \cos x \sin x +3 (3 \sin x - 4 \sin^3 x) \\
&= \sin x \left( 2 \cos x + 9 -12 \sin^2 x \right) = \sin x \left( 2 \cos x + 9 -12 (1-\cos^2 x) \right)
\end{align}
$$
so that you have $\sin x=0$ and $12 \cos^2 +2 \cos x -3=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$(2^m -1)(2^n-1)$ divides $(2^{mn} -1)$ if and only if $\gcd(m,n) = 1$. If $\gcd(m,n) = 1$ then $(2^m-1)(2^n-1)$ divides $2^{mn} - 1$ because each of $2^m-1$, $2^n-1$ divide $2^{mn}-1$ and $\gcd(2^m-1, 2^n-1) = 2^{\gcd(m,n)}-1 = 1$. How about the converse? If $\gcd(m,n) > 1$, can one show that $(2^m-1)(2^n-1)$ does not... | Here's another proof I found. First we need the following:
Lemma If $m,n$ are positive integers with $m > n$, then $(2^m - 1)(2^n-1) \mid (2^{mn}-1)$ implies $(2^{m-n}-1)(2^n-1) \mid (2^{(m-n)n} - 1)$.
Proof of lemma. $(2^m-1)(2^n-1) \mid (2^{mn}-1)$ is equivalent to $(2^{mn} - 1)/(2^m-1) \equiv 0 \pmod{2^n-1}$. Using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/223818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
How to factor a polynomial in real numbers? Which of the following polynomials can't be factored in real numbers?
Multiple choice question from an old test.
Got the following polynomials:
$x^8$
$(x-3)(x^2+x+1)$
$(x-2)^3(x^2-1)$
$x^8(x^2-1)$
$(x^2-1)^3$
$x(x-1)(x+1)$
Solution:
Solving $x^2+x+1$ isnt possible with real n... | The only polynomial in that list that doesn't split into linear factors is the one with $x^2 + x + 1$ because its discriminant is $-3$. For the other ones,
\begin{gather}
\begin{aligned}
x^8 & = x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\
(x-2)^3 (x^2-1) & = (x-2)(x-2)(x-2)(x-1)(x+1) \\
(x^2-1)^3 &= (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/225924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Area fractal pentagrams I When I saw this image I was a little curious.
How can I find the area of this fractal?
| Each segment of the pentagram is the initiator of the fractal. Take its length to be 1. Now the generator consists of 2 line segments each of length $\frac{1}{3}$.
Hence on each iteration $n$ the area can be expressed as follows:
$$A_n=10\sum_{k=0}^{n}2^kS_k +S_{p}$$
Where $S_p$ is the area of the regular pentagon and:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Area of triangle ABC inside circle Consider the following diagram:
$AB+AD=DE$, $\angle BAD= 60$, and $AE$ is $6$. How do we find the area of the triangle $ABC$?
|
From the above picture,
$x+y=z$ , $ y+z=6$ and $\frac x {y+z} = \cos 60^\circ = \frac 1 2$
After calculation, $x=3$, $y=\frac 3 2$ and $z = \frac 9 2$
$\angle ABE = 90^\circ$ and $\angle BAD = 60^\circ$
So, $\angle AEB = 30^\circ = \angle ACB$ (properties of a circle)
Now, $\cos 60^\circ = \frac {x^2 + y^2 - w^2}{2xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/231944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the area of a triangle using analytic geometry Given are the points $P (1,0)$ and $Q (3,2)$. The points $P$ and $Q$ have the same distance to a certain line $l$, which intersects the positive x-axis in the point $A$ and the positive y-axis in the point $B$. The area of the triangle $ABO$ is minimal. Get the equati... |
Let the equation of the line $l(AB)$ be $\frac x a+\frac y b=1$
So, $A(a,0)$ and $B(0,b)$ and $a>0$ and $b>0$.
The area of $\triangle ABO=\frac{ab}2$
The distance of $l:b x+ay-ab=0$ from $P(1,0)$ is $\frac{\mid b-ab\mid}{\sqrt{a^2+b^2}}$ and that of from $Q(3,2)$ is $\frac{\mid 3b+2a-ab\mid}{\sqrt{a^2+b^2}}$
So, $(b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
absolute value integrated For example I have the function $f(x) = |x^2-1| = \sqrt{(x^2-1)^2}$
$\int \sqrt{(x^2-1)^2}dx = \frac{x(x^2-3)\sqrt{(x^2-1)^2)}}{3 (x^2-1)}+constant$
But plotted it looks like this: Plot 1. There are values < 0.
I have to take the absolute value again to get the correct function.
Plot 2 Why do ... | To evaluate $\int_a^b |x^2-1|dx$ there are 6 possibilities depending on where $a,b$ lie. I am assuming $a\leq b$ in the following, if $a>b$ then use $\int_a^b = -\int_b^a$.
If $a,b \in (-\infty,-1]$ or $a,b \in [1,\infty)$, then $\int_a^b |x^2-1|dx= \frac{1}{3} (b^3-3b-a^3+3a)$.
If $a,b \in (-1,1)$, then $\int_a^b |x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integral table contradiction? In a table of integrals, I see the following two formulas:
$\int \frac{dx}{(a+x)(b+x)} = \frac{1}{b-a}\ln\frac{a+x}{b+x}$, and
$\int \frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}$.
How can these both be true? It seems like if we expand $(a+x)(b+x)$ o... | You take the square root of a negative number $\left(-(a-b)^2\right)$, so the argument of arctan is complex. Using the following formula you should be able to turn one solution into the other:
$$\arctan(z) = {\ln(1 - iz) - \ln(1 + iz) \over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/233725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Express each power of the root $\alpha$ of $\frac{\mathbb Z_2[x]}{\langle x^3+x^2+1\rangle}$ as linear combinations of $1, \alpha$ and $\alpha^2$ There are $8$ elements in $\frac{\mathbb Z_2[x]}{\langle x^3+x^2+1\rangle} = GF(8)$
and this generates the set $\{0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1\}$
We're required to expr... | If $\alpha$ is a root of $x^3+x^2+1$, then $\alpha^3+\alpha^2+1=0$, so $\alpha^3=\alpha^2+1$ (since $1=-1$ over $\mathbb{Z}_2$).
From here:
$$\begin{array}{l} \alpha^4=\alpha\alpha^3=\alpha(\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1\\
\alpha^5=\alpha\alpha^4=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/234093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum of the sum of the sum of the first $n$ natural numbers I have here another problem of mine, which I couldn't manage to solve.
Given that: $$x_n = 1 + 2 + \dots + n \\ y_n = x_1 + x_2 + \dots + x_n
\\ z_n = y_1 + y_2 + \dots + y_n $$
Find $z_{20}$.
I know the answer but I'm having a hard time reaching it. I recog... | $$2y_n=2\sum_{1\le r\le n}x_1=\sum_{1\le r\le n}(r^2+r)=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2=\frac{n(n+1)(n+2)}3$$
$\implies 6y_n=n^3+3n^2+2n$
$$6z_n=6\sum_{1\le r\le n}y_r=\sum_{1\le r\le n}(r^3+3r^2+2r)$$
$$=\left(\frac{n(n+1)}2\right)^2+3\frac{n(n+1)(2n+1)}6+2\frac{n(n+1)}2$$
$$=\frac{n(n+1)\{3n(n+1)+6(2n+1)+12\}}{12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/234304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 6
} |
Rolling a fair die You and your friend play a game in which you and your friend take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. Play continues until either player wins if, after the player rolls, the number
on the running tally is a multiple of 7. Should yo... | I thought I'd answer this question as another opportunity to learn some math that's way out of my league in hopes that some of my usual favorites will evaluate it and see if I'm on the right track.
The first thing I did was consider a few trials:
*
*Player 1 has a 0/6 chance of winning
*Player 2 has a 1/6 chance of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/238359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Possible values of $N$ Find the number of values of $N$ such that the below expression is an integer:
$(n+1)^2\over n+7$ is an integer
| $(n+1)^2=n^2+2n+1=(n+7)(n-5)+36$
So, $\frac{(n+1)^2}{n+7}=n-5+\frac{36}{n+7}$
Assuming $n$ to be an integer, $(n+7)\mid36 \iff (n+7)\mid(n+1)^2$
So, $n+7$ can be any divisor of $36,$ namely $\pm1,\pm2,\pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36$
If we constrain $n$ to be non-negative i.e., if $n+7\ge 7,$ then $n+7$ can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/238703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
| Here's a generating-function approach to both derive and prove the desired formula.
Write $T(n)$ as $T_n$, and let $f(z) = \sum_{n=1}^\infty T_n z^n$ be the ordinary generating function. Now use the recurrence relation and initial condition to obtain
$$T_1 z^1 + \sum_{n=2}^\infty (T_n-2T_{n-1}) z^n = z + \sum_{n=2}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 8
} |
double absolute values I am having a little bit of problem with an inequality with nested absolute values:
$$|z^2-1| \ge |z+|1-z^2||$$
I've tried solving it by making three cases, $z\ge1$, $z\le-1$ and $z$ between $1$ and $-1$ and thus getting rid of absolute values for $z^2-1$ and $1-z^1$, and I am only left with 1 ab... | $|z^2-1| \ge |z+|1-z^2||$
Case 1: Suppose $z \ge 1$. Then $|z^2 - 1| = z^2 - 1$ and $|1 - z^2| = z^2 - 1$:
$z^2-1 \ge |z+(z^2-1)||$
Also $z + (z^2 - 1) > 0$ so:
$z^2-1 \ge z+(z^2-1)$
$0 \ge z$
This is a contradiction.
Case 2: Suppose $z \le -1$. Then $|z^2 - 1| = z^2 - 1$ and $|1 - z^2| = z^2 - 1$:
$z^2-1 \ge |z+(z^2-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/244718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & ... | $$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
$$
P_1A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
$$
P_2P_1A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 1 & 2
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
If $x$, $y$, $x+y$, and $x-y$ are prime numbers, what is their sum?
Suppose that $x$, $y$, $x−y$, and $x+y$ are all positive prime numbers. What is the sum of the four numbers?
Well, I just guessed some values and I got the answer.
$x=5$, $y=2$, $x-y=3$, $x+y=7$. All the numbers are prime and the answer is $17$.
Supp... | Note that one of $x$ and $y$ has to be even, as if $x$ and $y$ are both odd, $x+y$ and $x-y$ are even, and there is only one even positive prime. As $x - y > 0$, we have $x > y$ and hence, as $2$ is the only even prime and the smallest prime, we have $y = 2$. No $x-2$, $x$ and $x+2$ are prime. But one of them is divisi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/250584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "100",
"answer_count": 5,
"answer_id": 1
} |
Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$? As much as it embarasses me to say it, but I always had a hard time understanding the following equality:
$$
\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}
$$
I always thought that the left-hand side of the above equation was equivalent to
$$
\frac{a}{\frac{b}{x}} ... | $$
\frac{a}{\large\frac{b}{x}} = \large\frac {x}{\not x} \frac{a}{\frac{b}{\not x}} = x\cdot \frac{a}{b}
$$
Multiplying by $\dfrac xx = 1$ does not change the expression; but by multiplying numerator and denominator by $x$, the numerator becomes $ax$ and the denominator becomes $x\cdot \dfrac{b}{x} = b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/251317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
} |
Problem related with the similarity of matrices I came across this problem which says:
Let $A$ be a $2 \times 2$ matrix such that only $A$ is similar to itself.Then show that A is a scalar matrix, that is $A$ is of the form \begin{pmatrix}
a &0 \\
0 & a
\end{pmatrix}
?
My attempts:
Since $A$ is similar to itself,th... | Alternatively, that $A$ is only similar to itself means $PA=AP$ for all invertible matrix $P$. Therefore
$$
\begin{pmatrix}a+cx&b+dx\\ cy&dy\end{pmatrix}=
\underbrace{\begin{pmatrix}1&x\\0&y\end{pmatrix}}_{P}
\underbrace{\begin{pmatrix}a&b\\c&d\end{pmatrix}}_{A}=
\begin{pmatrix}a&b\\c&d\end{pmatrix}
\begin{pmatrix}1&x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/251470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$ If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?
Please provide a solution without the use of calculus. I know the solution in that way.
| Here is an attempt. I believe it has to be reconsidered for negative numbers, though.
$0\leq (x+y+z-1)^2$
$0\leq x^2+y^2+z^2+1-2(x+y+z)+2xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$2(x+y+z)-2xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\leq x^2+y^2+z^2+1$
EDIT: As pointed out my bound is the wrong way :( But I hope the first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/252178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
If $(a,b)=1$ then prove $(a+b, ab)=1$.
Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.
$(a,b)=1$ means $a$ and $b$ have no prime factors in common
$ab$ is simply the product of factors of $a$ and factors of $b$.
Let's say $k\mid a+b$ where $k$ is some factor of... | Here is another proof using Bezout's Identity:
$$
\begin{align}
ma+nb&=1\tag{1}\\
(n-m)b&=1-m(a+b)\tag{2}\\
(m-n)a&=1-n(a+b)\tag{3}\\
-(m-n)^2ab&=1-(m+n)(a+b)+mn(a+b)^2\tag{4}\\
1&=((m+n)-mn(a+b))\color{#C00000}{(a+b)}-(m-n)^2\color{#C00000}{ab}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $(a,b)=1$ and Bezout's Identity
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/257434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 16,
"answer_id": 1
} |
An inequality from the handbook of mathematical functions (by Abramowitz and Stegun) Prove that
$$\frac{1}{x+\sqrt{x^2+2}}<e^{x^2}\int\limits_x^{\infty}e^{-t^2} \, \text dt \le\frac{1}{x+\sqrt{x^2+\displaystyle\tfrac{4}{\pi}}}, \space (x\ge 0)$$
| (The following argument is adapted from Dümbgen, ''Bounding Standard Gaussian Tail Probabilities.'')
Approximating $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$
Suppose we want to approximate $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ with a function of the form $\dfrac{e^{-x^2}}{h(x)}.$ Let
$$\Delta(x) = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/259356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
How to get from $\frac{x}{x+1}\;$ to $\;1 - \frac{1}{x+1}$? Please show me how to manipulate $\dfrac{x}{x+1}\;\;$ to get $\;\;1 - \dfrac{1}{x+1}$
| Sometimes when working with polynomial division, it helps to recall how we handle division of plain old integers:
Consider the fraction $\dfrac{17}{20}$. Note that $\dfrac{17}{20} = \dfrac{20 - 3}{20} = \dfrac{20}{20} - \dfrac{3}{20}$.
We can do the same for $\dfrac{x}{x+1}$:
$$\dfrac{x}{(x+1)}\; = \;\dfrac{(x+1) - 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/259498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do we deal with recurrence relation characteristic equations that are not quadratic or have imaginary roots? Suppose we have $$H(n) = H(n-1)-H(n-2) \rightarrow x^2-x+1 \rightarrow r_1 = \frac{1+\sqrt{-3}}{2}, r_2 = \frac{1-\sqrt{-3}}{2}$$
or
$$H(n) = H(n-1)+H(n-2)+H(n-3) \rightarrow x^3-x^2-x-1=0$$
In either case, ... | The first one repeats
$$ a,b,b-a,-a,-b,a-b, a,b,b-a,-a,-b,a-b, a,b,b-a,-a,-b,a-b, a,b,c,-a,-b,a-b,\ldots $$
This does follow, eventually, from the description
$$ H(n) = A \, r_1^n + B \, r_2^n $$
for some complex constants $A,B.$
Indeed,
$$
\left( \begin{array}{r}
A \\
B
\end{array}
\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/259727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the solutions of Boolean equations It's given 4 Boolean equations. I need to find the number of solutions of each.
$a)\ x_{1}x_{2}\oplus x_{2}x_{3}\oplus\ ...\ \oplus\ x_{n-1}x_{n}=1$
$b)\ x_{1}x_{2}\vee x_{2}x_{3}\vee\ ...\ \vee\ x_{n-1}x_{n}=1$
$c)\ x_{1}x_{2}\oplus x_{3}x_{4}\oplus\ ...\ \oplus\ x_{2n-1}x_{2... | Part d)
There are $4^n$ possible values for $(x_1, x_2, \ldots, x_{2n-1}, x_{2n})$ which
we can divide into $n$ $2$-bit vectors $(x_1,x_2), (x_3,x_4),\ldots, (x_{2n-1},x_{2n})$
each of which can take on $4$ values. Then,
$$x_{1}x_{2}\vee x_{3}x_{4}\vee\ ...\ \vee\ x_{2n-1}x_{2n}= 0$$
if and only if $(x_1,x_2) \neq (1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/261515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can I systematically find the roots of $ x^4 + 1?$ Is there some algorithm?
Possible Duplicate:
How to find the root of $x^4 +1$
What algorithms can be used for finding all roots of the given polynomial:
\begin{equation}
x^4 + 1 = 0
\end{equation}
| $$x^4 + 1 = (x^2 + 1)^2 - 2x^2 = \\
(x^2+1+\sqrt2 x) (x^2+1-\sqrt2 x) =\\
(x^2 - x\sqrt 2+1)(x^2 + x\sqrt2+1)=0$$
*
*$x^2 - x\sqrt 2+1=0$
$$x_{1}={\sqrt2+\sqrt{-2}\over2}={\sqrt2\over2}(1+i)$$
$$x_{2}={\sqrt2-\sqrt{-2}\over2}={\sqrt2\over2}(1-i)$$
*$x^2 + x\sqrt2+1=0$
$$x_{3}={-\sqrt2+\sqrt{-2}\over2}=-{\sqrt2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $ Are there any solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $,
for $i>3$ and $j>2$ ?
Thanks! $:)$
| Gottfried Helms has dealt with $2^i=3^j+1$. For completeness we deal with $2^i =3^j-1$.
This has the solutions $i=j=1$ and $i=3$, $j=2$. We show there are no others.
First we deal with odd $j$. Note that
$$3^j-1=(3-1)\left(3^{j-1}+3^{j-2}+\cdots +1\right).$$
The term $3^{j-1}+3^{j-2}+\cdots +1$ is the sum of an odd nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the value of D here? Number $S$ is obtained by squaring the sum of digits of a two digit number $D$. If the difference between $S$ and $D$ is $27$, then the two digit number $D$ is?
My thoughts:
Let the two digit number $D$ be $AB$.
And so $S=(A+B)^2$
If $\,S-D=27,\,$ then $\,(A+B)^2 -AB=27$
$$A^2 + 2AB + B^2 ... | I believe I see 2 solutions for this problem. First let's redefine $D$ algebraicly as $D=10A+B$. So our equation is
$$(A+B)^2-10A-B=27$$
I don't know if you know any modular arithmetic. You may at least be aware of the divisibility test for $9$. The sum of the digits of a number is closely related to its remainder ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $d$ divides $n$, then $2^d -1$ divides $2^n -1$ Prove that if $d$ divides $n$, then $2^d -1$ divides $2^n -1$.
Use the identity $x^k -1 = (x-1)*(x^{k-1} + x^{k-2} + \cdots + x +1)$
| EDIT: Here's a visual proof. In binary base, $2^{d}-1 = \underbrace{1\cdots 1}_{d \text{ times}}$, and
\begin{align*}
2^n-1 &=\underbrace{1\cdots 1}_{n \text{ times}} = \underbrace{\underbrace{1\cdots 1}_{d \text{ times}} \cdots \underbrace{1\cdots 1}_{d \text{ times}}}_{n/d \text{ times}} \\
&=\und... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots... | Let $M_n$ be the $n \times n$ matrix. Calculate the determinant by expanding along the first row and then by the second column, we get $ Det(M_n) = 5 Det(M_{n-1} ) - 4 Det(M_{n-2})$.
Let $Det(M_n) = D_n$, so $D_n$ satisfies the recurrence relation $D_n - 5 D_{n-1} + 4 D_{n-2} = 0$, with initial values $D_0 = 1, D_1 = 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 4
} |
Why do we choose $3$ to be positive after $\sqrt{9 - x^2}$ in the following substitution? The integral $$\int \frac{\sqrt{9 - x^2}}{x^2}dx$$ is solved in my book by letting $x = 3\sin\theta$ where $-\frac {\pi}{2} \le \theta \le \frac {\pi}{2}$.
Then, $dx = 3\cos\theta\,d\theta$
and, $$\sqrt{9-x^2} = 3|\cos\theta| = ... | The core of your problem seems to be a confusion about the meaning of $\left| \cdot \right|$. $\left| x \right|$ is not the same as $\pm x$. Instead:
*
*$\left| x \right| = -x$ if $x \le 0$
*$\left| x \right| = x$ if $x \ge 0$
Now, because $\sqrt{\cdot}$ is a real-valued function, it can only return one of the qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the limit of the expression $2^{n+1}\sqrt{2-t_n}$ as $n\to\infty$, where $t_n=\sqrt{2+t_{n-1}}$
Possible Duplicate:
Why is this series of square root of twos equal $\pi$?
Find the limit of the expression $2^{n+1}\sqrt{2-t_n}$ as $n\rightarrow\infty$, where $t_1=\sqrt{2}$, $t_2=\sqrt{2+\sqrt{2}}$, $t_3=\sqrt{2+\... | $L_n = 2^{n+1} \sqrt{2-t_n}$. Then $L_n = 2^{n+1} \sqrt{2 - \sqrt{2+t_{n-1}}}$. Now $$2 - \sqrt{2+t_{n-1}} = \dfrac{4 - (2+t_{n-1})}{2 + \sqrt{2+t_{n-1}}} = \dfrac{2-t_{n-1}}{2 + t_n} = \dfrac{2-t_{n-1}}{t_{n+1}^2}$$
Hence, $$L_n = 2^{n+1} \sqrt{\dfrac{2-t_{n-1}}{t_{n+1}^2}} = \dfrac{2L_{n-1}}{t_{n+1}}$$
We have $L_0 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\tan(x) = x$. Find the values of $x$ How can I find the possible values of $x$ for:
$\tan(x)=x$
mathematically?
| $$\frac{\sin(x)}{\cos(x)}=x$$
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$
$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}...$$
Your question is equivalent to solving the equation
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=x-\frac{x^3}{2!}+\frac{x^5}{4!}-\cdots$$
$$x^3\left(\frac{1}{3!}-\frac{1}{2!}\right)-x^5\left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Find all polynomials $P$ such that $P(x^2+1)=P(x)^2+1$ Find all polynomials $P$ such that
$$P(x^2+1)=P(x)^2+1$$
| Let $P(y)=\sum_{0\le r\le n}a_ry^r$
So, $$P(1+x^2)=\sum_{0\le r\le n}a_r(1+x^2)^r=a_0+a_1(1+x^2)+a_2(1+\binom 21x^2+x^4)+\cdots
+a_{n-1}(1+\binom {n-1}1x^2+\binom {n-1}2x^4+\cdots+\binom {n-1}{n-2}x^{2(n-2)}+\binom {n-1}{n-1}x^{2(n-1)})
+a_n(1+\binom n1x^2+\binom n2x^4+\cdots+\binom n{n-1}x^{2(n-1)}+\binom n nx^{2n})$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/271337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 6,
"answer_id": 0
} |
Show that $\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0$ I am asked to prove this statement $^{*}$. I am trying now, but it is getting to small and tiny steps that I even loose my way. my steps are as follows:
$$\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0^{*}$$
$\lim_{n\rightarrow... | Hint:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
Apply for $a=\sqrt[3]{n+\sqrt n}$ and $b=\sqrt[3] n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/271480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving:$(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$ How to prove that :
$$(1+x)(1+x^2)(1+x^3)\cdots (1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$$
| It's wrong of course! Try $n=3$ and $x=-1$.
For non-negative $x$ it's just Holder:
$$(1+x)(1+x^2)...(1+x^n)\geq\left(1+\sqrt[n]{x\cdot x^2\cdot...\cdot x^n}\right)^n=\left(1+x^{\frac{n+1}{2}}\right)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/273383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve an equation involving the sine and the inverse tangent The equation is
$$ \sin\left(\frac{x}{x-1}\right) + 2 \tan^{-1}\left(\frac{1}{x+1}\right)=\frac{\pi}{2} $$
The answer is $0$, but I do not know how they got that.
| $$2\tan^{-1}\left(\frac1{x+1}\right)=\cos^{-1}\left(\frac{1-\left(\frac1{x+1}\right)^2}{1+\left(\frac1{x+1}\right)^2}\right)=\cos^{-1}\frac{x^2+2x}{x^2+2x+2}$$
as $$2\tan^{-1}y=\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)$$ (Proof below)
$$\implies \sin^{-1}\left(\frac x{x-1}\right)=\frac\pi2-\cos^{-1}\left(\frac{x^2+2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I solve $\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$? The limit is: $$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$
The limit is the kind of $1^\infty$, since:
$$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x=\left(\frac{2\times\frac{\pi}{2}}{\pi}\right)^\infty=\left(\frac{\pi... | There is a way to complete your argument, although the work is a little tedious. Observe that
$$
x \cdot \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) =
\frac{\ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right)}{\left( \dfrac{1}{x} \right)}.
$$
As
$$
\lim_{x \to \infty} \ln \left( \frac{2 \cdot {\tan^{-1}}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Prove quotient of factorials is integral If $n$ is an integer $\gt 0$, prove
$$\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$$
is also an integer. I understand that a general approach is proving that the power of any prime factor is greater in the numerator than it is in the denominator, but I haven't been able to formulate this... | This is a gross brute force answer.
We will show that for any positive integer $D$:
$$0\leq\left\lfloor\frac{30n}{D}\right\rfloor + \left\lfloor\frac{n}{D}\right\rfloor - \left\lfloor\frac{15n}{D}\right\rfloor - \left\lfloor\frac{10n}{D}\right\rfloor - \left\lfloor\frac{6n}{D}\right\rfloor$$
This is enough to show your... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Use Taylor Series Expansion in Calculating Integral I want to approximate this integral:
$$
I = \int_0^\infty {{e^{ - bx}}\ln \left( {{a_1}{e^{ - {b_1}x}} + {a_2}{e^{ - {b_2}x}}} \right)dx}
$$
where $b,{a_1},{a_2},{b_1},{b_2} > 0$ and $b_2>b_1$.
Here is my answer:
We can observe that there exists a constant $c$ so t... | Substitution and repeated integration by parts yields
$$
\int_0^\infty x^ne^{-kx}\mathrm{d}x=\frac{n!}{k^{n+1}}
$$
Assume $b_1\le b_2$ and $|a_2|\le|a_1|$, but not both equal, then
$$
\begin{align}
&\int_0^\infty e^{-bx}\log\left(a_1e^{-b_1x}+a_2e^{-b_2x}\right)\mathrm{d}x\\
&=\int_0^\infty e^{-bx}\left(\log(a_1)-b_1x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the integer solutions Find all integer solutions of
$$(a + b^2)(a^2 + b) = (a − b)^3.$$
Obviously $b = 0$ is one. But how to get other solutions?
| On simplification, $$2b^3+(a^2-3a)b^2+b(a+3a^2)=0$$
If $$b\ne 0, 2b^2+(a^2-3a)b+(a+3a^2)=0 $$ which is a quadratic equation in $b$
As $b$ is integer, the discriminant must be perfect square.
So, $$(a^2-3a)^2-4\cdot2\cdot(a+3a^2)=a^4-6a^3-15a^2-8a=f(a)$$(say)
Clearly, $f(0)=0=f(-1)$ so, $a^4-6a^3-15a^2-8a=a(a+1)(a^2-7a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Summing $r(r+3)$ using induction We want to prove the following summation by induction:
$$\sum_{r=1}^{n}r(r+3)=\frac{1}{3}n(n+1)(n+5)$$
The problem is posted for a friend, but others can look at the solution if they want/need.
| Let us denote the sum by $S_{n}$. We first establish the base case, that $S_{1}=1(1+3)=4$ is equal to $\frac{1}{3}\cdot 1\cdot 2\cdot 6=4$, which holds.
Then, assume that for some $k$, that $S_{k}=\frac{1}{3}k(k+1)(k+5)$. Obviously $S_{k+1}=S_{k}+(k+1)(k+4).$ Tidying up, we have
$$S_{k+1}=\frac{1}{3}(k+1)\left[k^{2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Rational function partial fraction help? So I have to integrate
$$
\frac{3x+2}{(x)(x+1)^3}
$$ so I have the terms
$$
\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}
$$
... and then I have
$$
3x+2=A(x+1)^3 + B(x)(x+1)^2 +C(x)(x+1)^2+ D(x)(x+1)^2
$$
... here I tried to equal the quotients and I have... | You’re starting off on the wrong foot. When you put
$$\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}\tag{1}$$
back together into a single fraction, that fraction must have the same denominator as the original one. $(1)$ becomes
$$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}\;.$$
and you want to cho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Working with exponent on series Hi have this sequence:
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}$$
I understand that this is a Geometric series so this is what I've made to get the sum.
$$\sum\limits_{n=1}^\infty (-1)^n\frac{3^{n}\cdot 3^{-2}}{4^n}$$
$$\sum\limits_{n=1}^\infty (-1)^n\cdot 3^{-2}{(\frac{3}{4}... | The objective here is to transform your sum into a sum of the form:
$$\sum_{n=1}^\infty ar^{n-1}$$
$$\text{Transformation: }\quad\quad\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-3)^{n-1}}{4^{n-1}} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\left(\frac{-3}{4}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Weird question pertaining to HCF I encountered this question which seems weird/incomplete to me :
Q: H.C.F. of 3240, 3600 and a third number is 36, and their L.C.M. is
$2^4 \cdot 3^5 \cdot 5^2 \cdot 7^2$ . The third number is?
Can anyone please teach me concept wise how to solve it?
| Find the prime power factorizations of the two given numbers. We get
$3240=2^3\cdot 3^4\cdot 5^1$ and
$3600=2^4\cdot 3^2\cdot 5^2$.
Let our unknown number be $n$. Because the LCM of $3240$, $3600$, and $n$ only involves the primes $2$, $3$, $5$, and $7$, we know that the prime power factorization of $n$ can involve no ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/288142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I find a number $a$ such that this limit is 1 I want to find the number $a$ and $b$ such that $\lim_{x\to 0} \frac{\sqrt{ax+b}-2}{x}=1$.
First of all, I know that $b$ has to be 4, because the limit of the numerator has to be zero because the denominator is zero when we take its limit.
My problem is with the num... | When $a=b=4$. We can calculate the binomial series of $\sqrt{ax+b}$ and we find that
$$\sqrt{ax+b} \sim \sqrt{b} + \frac{a}{2\sqrt{b}}x + \cdots$$
Using this approximation, we see that
$$\frac{\sqrt{ax+b}-2}{x} \sim \frac{\sqrt{b}-2}{x} + \frac{a}{2\sqrt{b}} + \cdots $$
where the tail "$+\cdots$" consists of terms divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How do I solve this exponential equation? $5^{x}-4^{x}=3^{x}-2^{x}$ How do I solve this exponential equation?
$$5^{x}-4^{x}=3^{x}-2^{x}$$
| $$5^x - 4^x = \int_4^5 x y^{x-1} \,dy$$
$$3^x - 2^x = \int_2^3 x y^{x-1} \,dy$$
$$= \int_4^5 x (y-2)^{x-1} \,dy$$
So the difference between $5^x - 4^x$ and $3^x - 2^x$ is
$$ \int_4^5 x (y^{x-1} - (y-2)^{x-1})\,dy$$
In the integrand here, since $y \rightarrow y^{x-1}$ is monotone whenever $x \neq 1$, the
expression $(y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Evaluate:: $ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 +\cdots + \frac 1n\right) $ How to evaluate the series:
$$ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$
According to Mathematica, this converges to $ (\log 2)^2 $.
| This is a special case of a more general result derived here.
$$S = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n+1} \sum_{k=1}^n \dfrac1k$$
Recall that $\dfrac1k = \displaystyle \int_0^1 x^{k-1} dx$ and $\dfrac1{n+1} = \displaystyle \int_0^1 y^n dy$.
Now use the following fact.
$$\sum_{k=0}^{\infty} \int_0^1 (-z)^k dz = \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 6,
"answer_id": 5
} |
Show $\mathbb{Q}[\sqrt[3]{2}]$ is a field by rationalizing I need to rationalize $\displaystyle\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}$
I'm given what I need to rationalize it, namely $\displaystyle\frac{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}$
But I fail to... | You rationalize by using Euclid's algorithm as used to prove Bézout's identity.
Let $\alpha = \sqrt[3]{2}$. We have obviously to assume $a+b\sqrt[3]2 + c(\sqrt[3]2)^2 = a+b\alpha + c \alpha^2 \ne 0$.
Consider the polynomial $0 \ne a + b x + c x^2 \in \mathbb{Q}[x]$. Since $x^3 - 2$ is irreducible in $\mathbb{Q}[x]$, f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/294993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Discussion on even and odd perfect numbers. First of all thank you so much for answering my previous post. These are few interesting problems drawn from Prof. Gandhi lecture notes. kindly discuss:
1) If $n$ is even perfect number then $(8n +1)$ is always a perfect square.
2) Every odd perfect number has at least three ... | $1.$ By a theorem of Euler, any even perfect number is of the shape $2^{p-1}(2^p-1)$ where $p$ is prime. We don't need the primality part. Multiply by $8$, add $1$. We get $2^{2p+2}-2(2^{p+1})+1$, which is the square of $2^{p+1}-1$.
$2.$ We show that an odd perfect number cannot have only $2$ distinct prime factors (we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Number theory fun problem Say $a,b > 2 $ are integers. Then we have that $2^a + 1$ is not divisible by $2^b - 1$.
Any thoughts on how to tackle this problem???
| Let $a = bq + r$ where $0 \leq r < b$. Hence, $$2^a+1 = 2^{bq} \cdot 2^r +1 = (2^{bq}-1)\cdot 2^r + 2^r+1$$
But $2^b-1$ divides $2^{bq}-1$ since $(x-1)$ divides $x^m-1$ for all $m \in \mathbb{N}$.
Hence, if $2^b-1$ divides $2^a+1$, then $2^b-1$ has to divide $2^r+1$. But $0 \leq r < b$. Hence, we need $$2^b-1 \leq 2^r ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/296710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
$n$-digits numbers made of 1, 2, 3, such that none of two consecutive digits differ by more than one
We call a number to be good if none of two consecutive digits differ by more than one. How many good $n$-digits numbers made from digits $1$, $2$ and $3$ are there?
For example, $12232$ is good, but $12\textbf{31}2$ i... | Let the $n$ digit number be $a_1 a_2 \ldots a_n$. Let $N_1(n)$ be the number of $n$ digit numbers that are good starting with $1$, $N_2(n)$ be the number of $n$ digit numbers that are good starting with $2$ and , $N_3(n)$ be the number of $n$ digit numbers that are good starting with $3$.
We have
\begin{align}
N_1(n) &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/296766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The number of paths on a graph of a fixed length w/o repeatings Sorry for bad English.
Consider a graph $G$ with the adjacency matrix $A$. I know that the number of paths of the length $n$ is the sum of elements $A^n$.
But what if we can't walk through a vertex more than one times?
| [This is a counter-example to wece's answer, posted only as an answer because it's too long for a comment.]
wece's method is only correct for $A_2$. Counterexample: Let $A=\left[\begin{array}{ccccc}0&1&0&0&0\\1&0&1&0&0\\0&1&0&1&0\\0&0&1&0&1\\0&0&0&1&0\end{array}\right]$ representing a chain of 5 vertices connected by 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/299300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
$\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices I want to prove that for $X\in M_2(\mathbb{R})$ the formula $\det(\exp X)=e^{\mathrm{Tr}\, X}$ holds, writing $X$ in normal form gives $X=PJP^{-1}$, where $J$ is the Jordan matrix, now $\exp (PJP^{-1})=P(\exp J)P^{-1}$ and $\det P(\exp J)P^{-1}=\det \exp J$.... | Your calculation of $\exp J$ is incorrect. $$\exp J = \exp \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \exp \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} e^a & 0 \\ 0 & e^a \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e^a & e^a\\ 0 & e^a \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rolling three dice...am I doing this correctly? Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
F... | As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3\cdot6\cdot5=90$ outcomes for two of three the same. There are a total of $6^3=21... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
How to prove the Mantel's theorem of graph theory 's bound is best possible? The theorem state that every graph of order $n$ and size greater than floor function $\lfloor \frac{n^2}{4} \rfloor$ contain a triangle.
I already know a proof of the number of the edge of graph $\leq\frac{n^2}{4}$ if it does not contains a tr... | We can't use a smaller bound because we can show that for each natural number $n$, there exists a graph of order $n$ with exactly $\lfloor \frac{n^2}{4} \rfloor$ edges and no triangles. In response to your edit, this gives us an infinite number of examples.
So let $n$ be a natural number, and we will find a graph of or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/305021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A series involving the harmonic numbers : $\sum_{n=1}^{\infty}\frac{H_n}{n^3}$ Let $H_{n}$ be the nth harmonic number defined by $ H_{n} := \sum_{k=1}^{n} \frac{1}{k}$.
How would you prove that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}?$$
Simply replacing $H_{n}$ with $\sum_{k=1}^{n} \frac{1}{k}$ does not... | I will try to reduce the sum to an integral:
$$
\sum_{n=1}^\infty \frac{H_n}{n^3} = \sum_{n=1}^\infty H_n \frac{1}{\Gamma(3)} \int_0^\infty x^2 \mathrm{e}^{-n x} \mathrm{d} x = \frac{1}{2} \int_0^\infty x^2 \sum_{n=1}^\infty H_n \mathrm{e}^{-n x} \mathrm{d} x \tag{1}
$$
We now make use of the following generating func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/305476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 6,
"answer_id": 1
} |
Geometry Problem and Isosceles Triangle $ABC$ is an isosceles triangle with $AB=AC$, $\angle BAC=96^\circ$. $D$ is a point such that $\angle DCA=48^\circ$, $AD=BC$ and angle $DAC$ is obtuse. What is the measure (in degrees) of $\angle DAC$?
| First, draw. Then calculate the angles you might already know and name the unknown ones. So, using that $ABC$ is isosceles, we have $\angle BAC=42^\circ$. Since $42+48=90$, we have $\cos 48^\circ=\sin 42^\circ$. Let $\delta:=\angle ADC$, which is $<90^\circ$ since $\angle DAC$ is obtuse. After finding $\delta$, we will... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) = 3(2)^{k+1} + 2(5)^{k+1}$ $7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) = 3(2)^{k+1} + 2(5)^{k+1}$
The problem is part of a proof. If you could also talk me through your thought process for solving this problem, I would greatly appreciate it. I've played ... | Assuming the missing right parenthesis goes at the end:
$$7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) =\\21\cdot 2^k+14\cdot 5^k-30\cdot 2^{k-1} -20\cdot 5^{k-1}=\\ 21\cdot 2^k +14\cdot 5^k-15\cdot 2^k-4\cdot 5^k=\\ 6\cdot 2^k+10\cdot 5^k=\\3(2)^{k+1} + 2(5)^{k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is it possible to have a rule which generates: 2, 4, 6, 8, 10, 12, 14, 16, -23? This is on Lagrange Interpolations . . .
Is it possible to have a rule which generates the sequence: 2, 4, 6, 8, 10, 12, 14, 16, -23?
The hint that he gave us is to use Summation Products, the only thing I can come up with is $\sum_{i=1}^{8... | Thomas Andrews's solution is the best one, but you may also consider this simplistic but longer and bulky solution:
$$
\begin{array}{lccr} \\
f(x) & = & & a_1\times(-1)^{1-1}\times\frac{1}{9!}(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \\
& & + & a_2\times(-1)^{2-1}\times\frac{2}{9!}(x-1)(x-3)(x-4)(x-5)(x-6)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Solving an equation with fractional powers I was trying to find the maximum value for a function. I took the first derivative and arrived at this horrible expression:
$$ (x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0$$
How can I find the extrema by hand?
| After some simplifying steps,
$$(x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0, \\
(x^2 + y^2)^\frac{3}{2} - 3y^2(x^2 + y^2)^{\frac{1}{2}} = 0, \\
(x^2 + y^2)^{\frac{1}{2}}\left(x^2+y^2-3y^2 \right)=0, \\
x^2+y^2=0 \Rightarrow x=0,\;y=0, \\
\text{or}\\
x^2-2y^2=0 \Rightarrow |x|=\sqrt{2}|y|.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0... | OK let's tackle the one giving you grief. Just add the fractions and massage it:
$$
\begin{align*}
\frac{2^{k + 1} + (-1)^k}{3 \cdot 2^k} + \left(- \frac{1}{2}\right)^{k + 1}
&= \frac{2^{k + 1} + (-1)^k}{3 \cdot 2^k} + \frac{(-1)^{k + 1}}{2^{k + 1}} \\
&= \frac{2^{k + 2} + 2 \cdot (-1)^k + 3 \cdot (-1)^{k + 1}}{3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
What is $q(x)$ and $r(x)$ when $(x^2-6x+9)q(x) + r(x) = x^3 -27$? I just failed this question on a test, so I would please like to get some feedback on where my thinking was wrong.
I need help with determining $q(x)$ and $r(x)$ when: $$(x^2-6x+9)q(x) + r(x) = x^3 -27$$
I know that: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
... | Long division!
$$\require{enclose}
\begin{array}{rll}
x\;+6\qquad\qquad\quad\\[-3pt]
x^2-6x+9\enclose{longdiv}{x^3\qquad\qquad\;\;-27}\\
\underline{x^3-6x^2+9x\phantom{\qquad\,}}\\[-3pt]
6x^2-\phantom{0}9x-27\\[-3pt]
\underline{6x^2-36x+54\;}\\[-3pt]
27x-81
\end{array}$$
So $q(x) = x + 6$ and $r(x) = 27 x - 81$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
CDF of $X+Y$,$X−Y$,$XY$ for $(X,Y)$ Chosen Uniformly Inside Triangle Let $(X,Y)$ be chosen uniformly on the triangle $\{(x,y)\in\mathbb R^2:x+y\leq1,x\geq0,y\geq0\}$. What is the joint density function of $(X,Y)$? Find the CDFs of $X+Y$, $X-Y$, and $XY$.
What I've tried:
$\displaystyle \frac{1}{Area\hspace{1mm} of\hs... | The definition of the cumulative distribution function of a random variable $X$ is the function given by $\displaystyle F_X(x)=P(X \leq x)$.
If $Z=X+Y$,
$\displaystyle F_{Z}(z)=P\{X + Y \leq z\} = \int_{0}^{z}\int_{0}^{z-x}2\,\mathrm dy\,\mathrm dx=2\int_{0}^{z}(z-x)\mathrm dx=z^2$
(Rather than integrating, it is easy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
positive integer value of $n$ for which $n^2-19n+99$ is a perfect square. Calculate positive integer value of $n$ for which $n^2-19n+99$ is a perfect square.
My Try:: Let $n^2-19n+99 = k^2$ where $k\in \mathbb{Z}$
$4n^2-76n+396 = 4k^2 $ or $(2n-19)^2-35 = (2k)^2$
$(2n-19)^2-(2k)^2 = 35$
now we have to take two perfect ... | $(2n-19)^2-(2k)^2 = -35$
$\Rightarrow (2n-19-2k)(2n-19+2k)=35$
$35=5\times 7=35\times 1=-5\times -7=-35\times -1$
Check the possible pairs.
Another way: $n^2-19n+99 = k^2$ is a quadratic in n so for it to have integral(perhaps rational) roots its determinant is a square.
$\Rightarrow 19^2-4(99-k^2)=y^2,y\in \mathbb{N}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the remainder. Let $a,b$ be positive integers such that $7$ divide $a^2+b^2$ .How to find the remainder when we divide $ab-1$ by $7$
| Note that $7$ divides $a^2+b^2$ if and only if $7$ divides both $a$ and $b$.
You can prove this in two ways. One way is to calculate $a^2+b^2$ modulo $7$ for all possibilities. This is not as tedious as it sounds, since a square is congruent to $0$, $1$, $4$, or $2$ modulo $7$. No sum of these is congruent to $0$ modul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Diagonalise a matrix and show the formula I have diagonlised P to get
$$P=\left(\begin{matrix}
-1 &0 &0\\
0 &0 &0\\
0 &0 &1
\end{matrix}\right)$$
however am unsure on how to proceed, would appreciate any help!
By diagonalising P by a transformation of similarity, show that
$$ e^{Pt} = (I_3 - P^2) + P \sinh( t) + P^2... | Given:
$$P=\left(\begin{matrix}
-1 &0 &0\\
0 &0 &0\\
0 &0 &1
\end{matrix}\right)$$
By diagonalising P by a transformation of similarity, show that
$$ e^{Pt} = (I_3 - P^2) + P \sinh( t) + P^2 \cosh(t)$$
If we diagonalize the matrix $P$, we arrive at:
$$P = S J S^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 0 & 1\\ 1 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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.