Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Check proof that $\prod_{k=1}^n(1+{1\over a_k})$ is bounded if $a_{n+1} = (n+1)(a_n + 1)$ and $a_1 = 1$, $n\in \mathbb N$
Let $n \in \mathbb N$ and:
$$
\begin{cases}
a_1 = 1 \\
a_{n+1} = (n+1)(a_n + 1)
\end{cases}
$$
Prove that
$$
x_n = \prod_{k=1}^n\left(1+{1\over a_k}\right)
$$
is a bounded sequence.
Obviou... | Observe that $b_{n+1} = \dfrac{a_{n+1}}{(n+1)!}= \dfrac{a_n}{n!} + \dfrac{1}{n!}= b_n+\dfrac{1}{n!}\implies b_n = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2})+\cdots+(b_2 - b_1)+ b_1 = \dfrac{1}{(n-1)!}+\dfrac{1}{(n-2)!}+\cdots + \dfrac{1}{1!} + 1\implies a_n
= n!b_n = n + n(n-1) + n(n-1)(n-2)\cdots + n(n-1)(n-2)\cdot\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Partial Fraction of $\int \frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)}$ If
$$\int\frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)} = \int \frac{A \ \mathrm{d}x}{(\sin x)} + B \int\frac{\sin x \ \mathrm{d}x}{ 1 + \sin^2 x} + C \int \frac{\mathrm{d}x}{1 + \s... | The integrand can be written
$$
\frac{\cos x+\sin2x}{(1+\sin^2x)\sin x}
$$
Let's try and determine $A$, $B$ and $C$ so that
$$
\frac{A}{\sin x}+\frac{B\sin x}{1+\sin^2x}+\frac{C}{1+\sin^2x}
$$
is the same. Removing the denominators this forces
$$
A(1+\sin^2x)+B\sin^2x+C\sin x=\cos x+\sin2x
$$
If we set $x=0$, we get $A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$ A while ago I encountered this integral $$I=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$$ To be fair I spent some time with it and solved it in a heuristic way, I want to avoid that way so I won't show that approach, but the result I got is $\frac{\pi}{6} \ln(2+... | Integrate by parts
\begin{align}
I&=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx
= \int_0^1 \frac{1}{1+x} \cot^{-1}\frac x{1-x^2}dx
\end{align}
Let $J(a) =\int_0^1 \frac{1}{1+x} \cot^{-1}\frac {2x\sin a}{1-x^2}\ dx$
\begin{align}
J’(a) &= \int_0^1 \frac{2\cos a \ (x^2-x)}{(x^2+1)^2-(2x\cos a)^2}dx
=\frac12\left( a \ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Is this induction proof mathematically correct? Proof that $\frac{n^3}{3} < 3n-3$ is true for $n=2$ but false for every other $ n \in \mathbb{N}$.
Idea is to proof that $\frac{n^3}{3} \geq 3n-3$.
Let $n$ be $n=3$
$\frac{27}{3} \geq 9-3 $.
That means $\exists n \in \mathbb{N}$ such that $\frac{n^3}{3} \geq 3n-3$ and t... | Dont write $n=n+1$ you can write $n\mapsto n+1$.
The typical way to estimate from $n^2+2n-\frac23$ is to use, that $n\geq 3$.
This helps dealing with the quadratic term. The rest is simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$ Prove that $$\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$$
Hence, show that $$(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^5+i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^5=0.$$
In the first ... | Replace $\theta$ with $90^\circ-2y$
$$F=\dfrac{1+\cos2y+i\sin2y}{1+\cos2y-i\sin2y}$$
Using Double angle formula,
$$F=\dfrac{2\cos^2y+2i\sin y\cos y}{2\cos^2y-2i\sin y\cos y}$$
If $\cos y\ne0$
$$F=\dfrac{\cos y+i\sin y}{\cos y-i\sin y}=(\cos y+i\sin y)^2=\cos2y+i\sin2y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help with this proof by induction with inequalities. Show that mathematical induction can be used to
prove the stronger inequality $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n + 1}}$ for all integers greater than 1, which, together with a verification for the case where n = 1, establishes the weaker i... | You need to show that $$\frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\
< \frac{1}{\sqrt{3n+4}}\\$$Square both sides and cross multiply to get $$ (2n+1)^2 (3n+4)<(3n+1)(2n+2)^2$$
Multiply and cancel equal terms to get $$ 12n^2+12n+3 <12n^2 +13n+3 $$
Which is true for all positive integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving Sums or Products by Induction I am currently teaching Extension 1 Mathematics and seem to be coming unstuck on a few induction questions.
$$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$$
First step is to prove the equation works for "1" - which it does. The second step is to prove the function works for $... | Fro the induction step we need to prove that
$$\sum_{k=1}^{n+1}\frac{1}{(2k-1)(2k+1)}=\frac{1}{(2n+1)(2n+3)}+\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}\stackrel{Ind. Hyp.}=\frac{1}{(2n+1)(2n+3)}+\frac{n}{2n+1}\stackrel{?}=\frac{n+1}{2n+3}$$
that is
$$\frac{1}{(2n+1)(2n+3)}+\frac{n}{2n+1}\stackrel{?}=\frac{n+1}{2n+3}$$
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding complex roots of $(z-i)^n-(z+i)^n =0$ I have a problem. First part of it was to find $z$ such that $$\left|\frac{z-i}{z+i}\right| = 1$$ and I've quickly figured out, that $z \in \Bbb R$.
How ever, now I have to (using previous part of the problem) find all roots of the following equation: $$(z-i)^n - (z+i)^n =... | $$(z-1)^n-(z+1)^n=0\ (n\ \epsilon\ \mathbb{Z}^+)$$
$$\Rightarrow (z-1)^n=(z+1)^n \Rightarrow \left(\frac{z-1}{z+1}\right)^n=1=1\angle 2\pi k\ (k=0,1,2,...n-1)$$
$$\Rightarrow \frac{z-1}{z+1}=1\angle \frac{2\pi k}{n}\Rightarrow 1 -\frac{2}{z+1}=1\angle \frac{2\pi k}{n}$$
$$\frac{2\pi k}{n}=\theta \Rightarrow 1 -\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivati... | Let $x = \sec \theta, \ 0 \le \theta \le \pi, \ \theta \ne \frac{\pi}{2}$
Then $\sqrt {\dfrac {x+1}{x-1}} =\sqrt {\dfrac {\sec \theta+1}{\sec \theta-1}} = \sqrt {\dfrac {1+\cos \theta}{1-\cos \theta}} = \tan \dfrac{\theta}{2} $ (positive root as $\dfrac{\theta}{2}$ is in the 1st Quadrant.
$\therefore y = \tan^{-1} \tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $ $$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= ... | Let the fractions be $\frac{a}{k}$, $\frac{b}{k+1}$, and $\frac{c}{k+2}$.
$\frac{a}{k}+\frac{b}{k+1}+\frac{c}{k+2}=\frac{a(k+1)(k+2)+bk(k+2)+ck(k+1)}{k(k+1)(k+2)}=\frac{2k-1}{k(k+1)(k+2)}$
We want the following
$a+b+c=0$
$3a+2b+c=2$
$2a=-1$
Solve, $a=-\frac{1}{2}$, $b=3$, and $c=-\frac{5}{2}$.
The rest is standard.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Probability that tossing $10$ coins has $4$ consecutive heads I'd like to compute the probability of having a string of four consecutive heads after $10$ coin tosses.
I tried to solve this problem by considering $10$ blank spaces. To have $4$ consecutive heads, we can consider $7$ blank spaces, where the $7^{\text{th}... | The total number of possible outcomes is $2^{10}$
Using the counting method for the numerator .......
The number of outcomes with a single exact maximum $HHHH$ is
$HHHHT----- \Rightarrow 2^5 - 3$
$THHHHT---- \Rightarrow 2^4 - 1$
$-THHHHT--- \Rightarrow 2^4$
$--THHHHT-- \Rightarrow 2^4$
$---THHHHT- \Rightarrow 2^4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Power Series (Cambridge Tripos 1900) If
$$a/(a+bz+cz^2)=1+p_1z+p_2z^2
+\dots$$
then
$$1+p_1^2z+p_2^2z^2+\dots=\frac{a+cz}{a-cz}\frac{a^2}{
a^2-(b^2-2ac)z+c^2z^2}$$
A tricky problem from G.H.Hardy's "A Course in Pure Mathematics", or maybe I'm missing the obvious.Any help will be greatly appreciated.
| If we start with the original function, we can reformulate is conveniently as
$$
\frac{a}{a+bz+cz^2} = \frac{1}{1+(\frac{b}{a})z+(\frac{c}{a})z^2} = \frac{1}{(1-\frac{z}{z_+})(1-\frac{z}{z_-})} = \frac{z_+}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_-})} - \frac{z_-}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_+})}
$$
with $z_\pm$ the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_1^2\frac{\arctan(x+1)}{x}\,dx$
Evaluate the following integral
$$\int_1^2\frac{\arctan(x+1)}{x}\,dx$$
with $0\leq\arctan(x)<\pi/2$ for $0\leq x<\infty.$
I proceeded the following way
$$\begin{aligned}
&\int_1^2\frac{\arctan(x+1)}{x}\,dx\to {\small{\begin{bmatrix}&u=x+1&\\&du=dx&\end{bmatrix}}}
\t... | Related Integral
First, let me tackle another integral which will come in useful later:
$$J=\int_{\arctan 2}^{\arctan 3}\ln(\tan u-1)du$$
$$\tag 1=\int_{\frac{3\pi}4-\arctan 2}^{\frac{3\pi}4-\arctan 3}\ln\Biggl(\tan\left(\frac{3\pi}4-u\right)-1\Biggr)(-du)$$
$$\tag 2=-\int_{\arctan 3}^{\arctan 2}\ln\left(\frac{\tan \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
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Solution to equation using permutation I am hoping I am on the right track for my homework in Discrete Math.
How many solutions does the equation $$a + b + c + d + e + f = 30$$ have if each variable must be a non-negative integer and $a ≤ 3, b ≤ 7$ and $d ≥ 8$?
I started my solution using the formula: ${m+n-1 \choos... | The number of all 6-couples in our universe, which is set of all solutions of the equation $a+b+c+d'+e+f =22$ where $d'=d-8\geq 0$ is $${27\choose 5}$$
Let $A$ be a set of all 6-couples where $a\geq 4$ and $B$ be a set of all 6-couples where $b\geq 8$.
So $$|A| = {23\choose 5}\;\;\;{\rm and}\;\;\;|B| = {19\choose 5}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $y=\frac25+\frac{1\cdot3}{2!}\left(\frac25\right)^2 +\frac{1\cdot3\cdot5}{3!}\left(\frac25\right)^3+\ldots$ then what is the value of $y^2+2y$? If
$$y=\frac25+\frac{1\cdot3}{2!}\left(\frac25\right)^2 +\frac{1\cdot3\cdot5}{3!}\left(\frac25\right)^3+\ldots$$
then what is the value of $y^2+2y$?
This is a question from... | Recall the binomial series $(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}+\dots$
Let $x=\frac{-4}{5}$ and $\alpha=\frac{-1}{2}$. Then
$$
\sqrt{5} = \left(1-\frac{4}{5}\right)^{\frac{-1}{2}}
= 1 +\frac{2}{5} + \frac{1\cdot3}{2!}\cdot\left(\frac{-1}{2}\right)^2\cdot\left(\frac{-4}{5}\right)^2+\dots =y+1
$$
so $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove convergence of sequence given by $a_{1}=1$ and $a_{n+1}= \frac{1}{a_1+a_2+\ldots+a_n}$ For sequence given by $a_{1}=1$ and $a_{n+1}= \frac{1}{a_1+a_2+\ldots+a_n}$ I have to prove that it converges to some number and find this number.
I tried toshow that it's monotonic by calculating
$$
\frac{a_{n+1}}{a_{n}} = \f... | Let $s_n = \sum\limits_{k=1}^n a_k$. We can rewrite the recurrence relation as
$$s_{n+1} - s_n = a_{n+1} = \frac{1}{s_n} \quad\implies s_{n+1} = s_n + \frac{1}{s_n}$$
This implies
$$s_{n+1}^2 = s_n^2 + 2 + \frac{1}{s_n^2} \ge s_n^2 + 2$$
So for all $n > 1$, we have
$$s_n^2 = s_1^2 + \sum_{k=1}^{n-1} (s_{k+1}^2 - s_k^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Combinatorial proof of $\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$?
Give a combinatorial proof of the following identity: $$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3.$$
I've been working on this proof for hours, however I'm not able to show LHS = RHS-
I completely understand binomial theorem and few c... | Let $A,B,C$ be the $3$ groups of $n$ students each with grades (marks) $A,B,C$, respectively. You need to select $3$ students.
The LHS is simply the combination of $3n$ students chosen $3$ at a time.
The RHS is to choose $3,2,1$ or $0$ students from $A,B$ and $C$, respectively:
$${n\choose 3}{n\choose 0}{n\choose 0}+{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ Find the value of
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.
I tried to use the formula which is... | For simplicity, let me use $\alpha=a, \beta=b,\gamma=c$.
We can find:
$$\begin{cases}a+b+c=-2 \\ ab+bc+ca=3\\ abc=-3\end{cases} \Rightarrow \\
\begin{cases}\color{red}{a^2+b^2+c^2}=4-2(ab+bc+ca)=\color{red}{-2} \\ \color{blue}{a^2b^2+b^2c^2+c^2a^2}=9-2abc(a+b+c)=\color{blue}{-3}\end{cases}$$
We can express:
$$x^3+2x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) Fo... | Consider the function
$$f(x)=x(1-ax) \implies f'(x)=1-2ax\implies f''(x)=-2a$$ The first derivative cancels at $x_*=\frac 1 {2a}$ and $$f\left(\frac{1}{2 a}\right)=\frac{1}{4 a}$$ The point $x_*$ is a maximum if $a>0$ by the second derivative test and a minimum otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Proving that for $n \equiv 0 \pmod{2}$, we get ${n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n\choose n/2}$ etc. How can one prove that for $n \equiv 0$ mod $2$ we have
$${n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n \choose n/2}>{n\choose n/2+1}>\ldots>{n\choose n-1}>{n\choose n}\,?$$
Can I... | Let's try using induction on $m$ where $n=2m$. For $m=0$, there is nothing to be proven, for $m=1$, then ${2\choose 0} = {2\choose 2} = 1 < 2 = {2\choose 1}$.
To get the induction step, observe that ${n+2\choose k} = {n+1\choose k-1} + {n+1 \choose k} = {n\choose k-2} + 2 {n\choose k-1} + {n\choose k}$. Suppose that $k... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If $a=b+c$ prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer
If $a=b+c$, and $a$,$b$,$c\in \Bbb N$, prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer.
Source: a list of problems used in the preparation to math contests.
My attempt:
By making the substitution $a=b+c$ in $S$ and de... | $(b+c)^4=b^4+4b^3c+6b^2c^2+4bc^3+c^4=(b^4+c^4)+4bc(b^2+c^2)+6b^2c^2=>$
$$
S=a^4+b^4+c^4=a^4+a^4-4bc(b^2+c^2)-6b^2c^2=2a^4-4bc(b+c)^2+2b^2c^2=2a^4-4bca^2+2b^2c^2=2(a^4-2bca^2+b^2c^2)=2(a^2-bc)^2
$$
$=>S=2(a^2-bc)^2$
$$
Q.E.D
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Simultaneous real solution of $x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$ I am trying to solve the following system of non-linear equations in real numbers:
$x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$, with $x,y$ real.
I can only see that $xy\ne 0$.
I have no clue whether a solution exists or not and how to find any solution. I cannot... | The given set of equations has three real solutions:
$$(x,y) =
\left(\frac{1}{y_2},y_1\right),
\left(\frac{1}{y_3},y_2\right),
\left(\frac{1}{y_1},y_3\right)
$$
where $y_1 = 2\cos\frac{\pi}{9}$, $y_2 = 2\cos\frac{7\pi}{9}$ and $y_3 = 2\cos\frac{13\pi}{9}$.
To derive them, we will exploit a symmetry hidden underneath t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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If $11z^8+20iz^7+10iz-22=0$,then show that $1<|z|<2$ If $$11z^8+20iz^7+10iz-22=0$$then show that $$1<|z|<2$$
My Attempt:
If $z=x+iy$; then$$z^7=\frac{22z-10iz}{11z+20i}$$
$$|z|^7=\sqrt{\frac{400+440y+100(x^2+y^2)+84}{400+440y+100(x^2+y^2)+21(x^2+y^2)}}$$
If $x^2+y^2=4$ i.e. $|z|=2$
then $|z|^7=1$
which is clearly a con... | Let $p(z) = 11z^8+20iz^7 + 10iz - 22$.
The part for $|z|<2$ is already in the other answer, but I include it for the sake of completeness. When $|z|=2$,
$$|20iz^7 + 10iz - 22| < 20(2^7)+10(2)+22 \\
< 21(2^7) < 22(2^7) = 11(2^8) = |11z^8|.$$
$11z^8$ has eight zeros (counted as many times as its multiplicity) in the ope... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What value does $\frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots(2n)}$ tend to? I need to find where this sequence tends to:
$$\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}$$
My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $\dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence... | $$ a_n
= \frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots (2n)}
= \sqrt[n]{1(1+\frac{1}{n})(1+\frac{2}{n})\cdots (1+\frac{n}{n})}.
$$
$$
\log a_n
= \frac{1}{n} \{ \log 1 + \log (1+\frac{1}{n}) + \log (1+\frac{2}{n})
+ \cdots + \log (1+\frac{n}{n}) \}.
$$
We know the above equations. Hence,
$$
\lim\limits_{n \rightarrow \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Polynomial division with remainder
If the polynomial
$$x^4-x^3+ax^2+bx+c$$
divided by the polynomial
$$x^3+2x^2-3x+1$$
gives the remainder $$3x^2-2x+1$$
then how much is (a+b)c?
So what I know, and how I solved these problems before, I can write this down like this:
$x^4-x^3+ax^2+bx+c=Q(x)(x^3+2x^2-3x+1)+3x... | Let (original) polynomial $P(x) = x^4 - x^3 + ax^2 + bx + c$ and divisor polynomial $D(x) = x^3 + 2x^2 - 3x + 1$. The remainder polynomial $R(x) = 3x^2 - 2x + 1$. We can also define quotient polynomial $Q(x) = \alpha x + \beta$.
We have that $P(x) = Q(x)\cdot D(x) + R(x)$.
By inspection of the coefficient of $x^4$ on b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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$M$ is a point in an equaliateral $ABC$ of area $S$. $S'$ is the area of the triangle with sides $MA,MB,MC$. Prove that $S'\leq \frac{1}{3}S$.
$M$ is a point in an equilateral triangle $ABC$ with the area $S$. Prove that $MA, MB, MC$ are the lengths of three sides of a triangle which has area $$S'\leq \frac{1}{3}S$$
| Choose a coordinate system so that triangle $ABC$ is lying on the unit circle centered at origin and $A$ on the $x$-axis. Let $a = AM$, $b = BM$, $c = CM$ and $S'$ be the area of a triangle with sides $a,b,c$. In this coordinate system, $S = \frac{3\sqrt{3}}{4}$, we want to show $S' \le \frac{\sqrt{3}}{4}$. Using Heron... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. I have no idea how to do this question.
I'm given $\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\frac{\pi}{\sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tr... | $$I(r)=\int_{-\infty}^{\infty} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}=\int_{-\frac 12}^{\infty} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}+\int_{-\infty}^{-\frac 12} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}$$
Substitute $x+\frac 12=u$
$$I(r)=\int_0^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Is there a $\gamma>1$ such that $\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+\cdots+\binom{4p}{p-1}}>\gamma^{p}$ In a proof of the Larman-Rogers conjecture (there is $\gamma>1$ such that $\chi(\mathbb{R}^{d})>\gamma^d) $ they used that there is a $\gamma>1$ such that $\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\bino... | By induction on $n$, using Pascal's rule:
$$
\binom{n}{r} = \binom{n - 1}{r} + \binom{n - 1}{r - 1} \quad (n \geqslant r \geqslant 1),
$$
we get, for $n > r \geqslant 0$:
\begin{equation}
\tag{$*$}\label{eq:id}
\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{r} = \binom{n - 1}{r} + 2\binom{n - 2}{r - 1} + \cdots + 2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$? If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?
I tried with Tchebyshev inequality on sets $\{a, b, c\}$ and $\{a^2, b^2 , c^2\... | For $a=b=c=3$ we obtain a value $81$.
We'll prove that it's a minimal value.
Indeed, we need to prove that $$a^3+b^3+c^3\geq81,$$ which is true because
$$a^3+b^3+c^3-81=\sum_{cyc}(a^3-27)=$$
$$=\sum_{cyc}\left(a^3-27-\frac{9}{2}(a^2-9)\right)=\frac{1}{2}\sum_{cyc}(a-3)^2(2a+3)\geq0.$$
Also, Holder helps:
$$a^3+b^3+c^3=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Evaluating $\int_0^1\frac{\ln(1+x-x^2)}xdx$ without using polylogarithms.
Evaluate $$I=\int_0^1\frac{\ln(1+x-x^2)}xdx$$ without using polylogarithm functions.
This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.
The motivation of writing this post i... | Put
\begin{equation*}
I=\int_{0}^1\dfrac{\ln(1+x-x^2)}{x}\,\mathrm{d}x = \int_{0}^1\dfrac{\ln(1+x(1-x))}{x}\mathrm{d}x.
\end{equation*}
If we change $x$ to $ 1-x $ we get
\begin{equation*}
I=\int_{0}^1\dfrac{\ln(1+x-x^2)}{1-x}\,\mathrm{d}x.
\end{equation*}
Consequently
\begin{equation*}
2I = \int_{0}^1\ln(1+x-x^2)\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How do I simplify this expression? How do I simplify:
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
If simplification is possible, it should be possible with elementary algebra, but I'm completely lost as to how to go about it.
What I've done so far:
$$d=\frac{-(x+\alpha x)^... | $$\frac{-\alpha^2x^2+\alpha^2y^2-2\alpha x^2+2\alpha y^2}{\sqrt{\alpha^2x^2+\alpha^2y^2}}=\frac{\alpha^2(y^2-x^2)+2\alpha(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}$$
$$\frac{\alpha^2(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}+\frac{2\alpha(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}=\alpha\frac{(y^2-x^2)}{\sqrt{x^2+y^2}}+2\frac{y^2-x^2}{\sqrt{x^2+y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Binomial Theorem with Three Terms $(x^2 + 2 + \frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
| The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$\left((x^2+2)+\frac{1}{x}\right)^7=\sum _{k=0}^7 \binom{7}{k} \left(x^2+2\right)^k x^{k-7}$$
Second step
$$\left(x^2+2\right)^k=\sum _{m=0}^k 2^{k-m} x^{2 m} \binom{k}{m}$$
Hence you get a double sum in which the power of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Find $\alpha$,$\beta$ if $\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x -\beta] = 0$ Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = \frac {-b±\sqrt {b^2-ac}}{a} = \frac {-b±\sqrt D}{a}$$
Hence;
$$\sqrt {ax^2+2bx+c} = \sqrt {(x+\frac {b-√D}{a})(x+\frac {b+√D}{a})}$$
$$=x\sqrt {(1+\frac {b-√D}{ax})(1+\f... | Another approach could be Taylor series
$$\sqrt{a x^2+2 b x+c}=x \sqrt{a+\frac{2 b}{x}+\frac{c}{x^2} }$$ So, for large $x$,
$$\sqrt{a x^2+2 b x+c}=\sqrt{a} x+\frac{b}{\sqrt{a}}+\frac{a c-b^2}{2 a^{3/2}
x}+O\left(\frac{1}{x^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Number of integral solutions $x_1 + x_2 + x_3 = 10$ $x_1 + x_2 + x_3 = 10, \ \ 0 \leq x_1 \leq 10 , \ 0 \leq x_2 \leq 6 , \ 0 \leq x_3 \leq 2 $
$[x^{10}] (1 + x^1 + x^2 + ... + x^{10})(1 + x^1 + x^2 + ... + x^{6})(1 + x + x^2)$
$[x^{10}] \Large (\frac{1 - x^{11}}{1-x})(\frac{1 - x^{7}}{1-x})(\frac{1 - x^{3}}{1-x})$
... | Using
$$(1-x)^{-n}=1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+\dotsb+\frac{n(n+1)\dotsb(n+k-1)}{k!}x^k+\dotsb$$
We get
$$(1-x)^{-3}=1+3x+\frac{12}{2}x^2+\frac{60}{6}x^3+\dotsb +\frac{3\cdot 4\cdot \dotsb (3+k-1)}{k!}x^k+\dotsb$$
Now you want coefficient of $x^{10}$ in $(1 - x^3 - x^7 + x^{10})\color{blue}{(1-x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the function equation $f(x+f(y)+yf(x))=y+f(x)+xf(y)$ find all functions $f:\Bbb R\to \Bbb R$ and such for any $x,y\in\Bbb R$,and such
$$f(x+f(y)+yf(x))=y+f(x)+xf(y)$$
I have proven that $$f(f(x))=x$$
proof:Let $y=0$,we have
$$f(x+f(0))=f(x)+xf(0)$$
let $x=0,y\to x$,we have
$$f(f(x)+xf(0))=x+f(0)$$
so we have
$$... | This is only a partial result (EDIT: full result is given below), but I'll post it anyway. Let us write $(x,y)$ if $y = f(x)$ holds. Since $f(f(x)) = x$, $(x,y)$ is equivalent to $(y,x)$ and $(x,y)\wedge (x,y')$ implies $y= y'$. What we can directly observe is that $0= f(f(0)) = f(0)$ by letting $x=y=0$. What is less t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Contour intergal of a rational trigonometric function, can't find my mistake. This is the integral :
$$I = \int_{0}^{2\pi} \frac{dx}{(5+4\cos x)^2}\ $$
Which, according to wolfram alpha, should evaluate to $\frac{10\pi}{27} $,
but the value i find is $\frac{20\pi}{27} $.
These are my calculations :
Using complex form ... | You made an error in calculating the factorization of the denominator. You should get $$5t+2t^2+2=(2+t)(1+2t)=2(t+2)(t+1/2),$$ where you missed to transfer the leading factor $2$ to the factorized expression. The correction leads to $I=-\frac i4 I_a$ and the additional division by $4$ corrects the residuum that is comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Condition in terms of b and a if $ax^2+bx+c=0$ has two consecutive odd positive integers as roots
The roots of the equation$$ax^2+bx+c=0$$, where $a \geq 0$, are two consecutive odd positive integers, then
(A) $|b|\leq 4a$
(B) $|b|\geq 4a$
(C) $|b|=2a$
(D) None of these
My attempt
Let p and q be the roots then if th... | Since we are given that there are two distinct roots and that $a \ge 0$, we must have
$a > 0, \tag 0$
since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.
Let
$n \ge 0, \tag 1$
$r = 2n + 1, \tag 2$
$s = 2n + 3; \tag 3$
suppose for the moment
$a = 1; \tag 4$
then
$(x - r)(x - s) = (x - (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$ If $x \ge 5$, is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$
I believe the answer is yes.
Here is my thinking:
(1) $\log_2{5} > 2.32 > 2.284 > 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
(2) Assume up to $x$ that $\... | We have that
$$1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x \iff 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \frac{\ln x}{\ln 2}$$
$$\ln 2\left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} \right)< \ln x$$
then recall that by harmonic series
$$\sum_{k=1}^x \frac1k \sim\ln x+\gamma+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I simplify $\frac{\sin(2x)}{1-\cos (2x)}$? $$\dfrac{\sin(2x)}{1-\cos (2x)}$$
How do I simplify given expression?
My attempt:
We know the double-angle identity for $\sin(2x)$ and $\cos(2x)$ as shown below
$$\sin(2x) = 2\sin (x)\cos(x)$$
$$\cos(2x) = 2\cos(x)-1$$
So we have that
$$\dfrac{2\sin (x)\cos(x)}{1-2\cos... | You got one of the signs mixed up.
$$\frac{\sin(2x)}{1-\cos (2x)} = \dfrac{\sin(2x)}{1-(2\cos^2x-1)} = \frac{\sin(2x)}{1-2\cos^2x+1}$$
Instead, you accidentally used
$$\frac{\sin(2x)}{1-\cos (2x)} \color{red}{\neq \frac{\sin(2x)}{1-2\cos^2x-1}}$$
As for the simplification itself,
$$\frac{\sin(2x)}{1-\cos (2x)} = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Recurrence k-th pattern I am trying to solve this recurrence $T(n) = 6 T(\frac{n}{3}) + n$.
1st recurrence: $6^2T(\frac{n}{3^2}) + \frac{6n}{3} + n$
2nd: $6^3T(\frac{n}{3^3}) + \frac{6^3n}{3^2} + n$
3rd: $6^4T(\frac{n}{3^4}) + \frac{6^6n}{3^3} + n$
I am having trouble describing the general pattern after the k-th itera... | If you are trying to solve the recurrence, try finding the initial pattern and the number of terms - this will help solve the equation in most of the cases.
In the first step, we have $n$ work to do, we do $n$ work and have 6 problems of the same type but of size $\frac{n}{3}$ this time.
In the second step, we have $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of $\{n\sqrt{2}\}$ I would like to prove (rigorously, not intuitively) that
$$\sum_{n=1}^N \{n\sqrt{2}\}=\frac{N}{2}+\mathcal{O}(\sqrt{N})$$
where $\{\}$ is the "fractional part" function. I understand intuitively why this is true, and that's how I came up with this claim - $\{n\sqrt{2}\}$ behaves like a random var... | Writing
\begin{align}
&\quad \sum_{n=1}^N \left( n\sqrt{2} - \lfloor n\sqrt{2} \rfloor \right) \\
&=\int_1^N \left( n\sqrt{2} - \lfloor n\sqrt{2} \rfloor \right) {\rm d}n + {\cal O}(1) \tag{1} \\
&=\frac{1}{\sqrt{2}} \int_\sqrt{2}^{\sqrt{2}N} \left(x- \lfloor x \rfloor \right) {\rm d}x + {\cal O}(1) \\
&=\frac{1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Finding a mean of biased sample An RV $X$ follows a uniform distribution over $[0,1]$.
Suppose that we cannot observe the realization $x$ of $X$.
Instead, we can observe a value $y$, meaning that $x\in\{y,\frac{1}{2}+\frac{y}{2},\frac{3}{4}+\frac{y}{2}\}$. That is, the true $x$ is one of the three value.
If so, what... | First off, the problem is far less confusing if you state given the value $x$, what values of $y$ can you observe. For example, if $y=1$, then as you put it, $x$ is 1, 1, or 5/4. The last is impossible, and what does two equivalent possibilities mean?
Next, the problem is underdetermined unless you know the probabiliti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3018353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Inequality for $\sin(20°)$ Prove that $$\frac{1}{3} < \sin{20°} < \frac{7}{20}$$
Attempt
$$\sin60°=3\sin20°-4\sin^{3}(20°)$$
Taking $\sin20°$=x
I got the the equation as
$$8x^3-6x+\sqrt{3} =0$$
But from here I am not able to do anything. Any suggestions?
Thanks!
Edit-graph of p(x)
| Mostly for fun, here's a way to show that
$$6\left(1\over3\right)-8\left(1\over3\right)^3\lt\sqrt3\lt6\left(7\over20\right)-8\left(7\over20\right)^3$$
with only a small amount of multi-digit arithmetic:
$$6\left(1\over3\right)-8\left(1\over3\right)^3=2-{8\over27}\lt2-{8\over28}=2-{2\over7}={12\over7}$$
and $12^2=144\lt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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identities and binomial coefficients
I'm having some problems proving this identity. I tried using some formulas I found on the internet so I can turn that $2$ base number into something else but i'm not really sure how to do that. I would be really thankful if someone could help!
| We have by inspection that
$$\sum_{k=0}^n {m+k\choose k} 2^{n-k}
= 2^n [z^n] \frac{1}{1-z} \frac{1}{(1-z/2)^{m+1}}.$$
This is
$$2^n \times \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1-z} \frac{1}{(1-z/2)^{m+1}}
\\ = - 2^n \times \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{2^{m+1}}{(2-z)^{m+1}}
\\ = 2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding determinant of $3\times3$ matrix $$A = \left(\begin{matrix}
\lambda - 1 & -1 & -1 \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{matrix}\right)$$
$$\det A = \begin{vmatrix}A\end{vmatrix} =
(\lambda - 1) \begin{vmatrix}
\lambda - 3 & 1 \\
1 & \lambda + 1 \\
\end{vmatrix} + 1\begin{vmatrix}
1 & 1 \\
-3 &... | HINT
We can simplify a little bit summing the third to the first row
$$\begin{vmatrix}
\lambda - 1 & -1 & -1 \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{vmatrix}= \begin{vmatrix}
\lambda -4 & 0 & \lambda \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{vmatrix}$$
and using Laplace expansion in the firs... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that for all $n$ $\gcd(n, x_n)=1$, given $x_{n+1}=2(x_n)^2-1$ and $x_1=2$ I have a sequence $x_{n+1} = 2(x_n)^2-1$; first values are $2, 7, 97, 18817,\dots$
I noticed that if prime $p$ divides $x_n$, then $x_{n+1} \equiv -1\pmod p$ and for all $k>n+1$, $x_k\equiv 1\pmod p$. But I have no idea what to do next.
| I'll do my usual playing around
and see what happens.
tl;dr - Nothing but dead ends.
If
$x_{n+1}
= 2(x_n)^2-1
$
then
$x_{n+1} -1
= 2(x_n)^2-2
= 2(x_n^2-1)
= 2(x_n-1)(x_n+1)
$.
Therefore,
if $y_n =x_n-1$ then
$y_{n+1}
=2y_n(y_n+2)
$.
Since
$x_1 = 2$,
$y_1 = 1$.
Therefore
$y_2 = 2\cdot 1\cdot 3 = 6$,
$y_3 = 2\cdot 6\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Conjecture: if $a$, $b$ and $c$ have no common factors, dividing each of them by their sum yields at least one irreducible fraction
Let $a$, $b$ and $c$ be $3$ integers with no common factors.
I conjecture that at least one of the three fractions:
$$\frac{a}{a+b+c},\quad\frac{b}{a+b+c},\quad\frac{c}{a+b+c}$$
is irredu... | Counterexample: $5,7,58$.
\begin{align*}\frac{5}{5+7+58} &= \frac{5}{70} = \frac1{14}\\
\frac{7}{5+7+58} &= \frac{7}{70} = \frac1{10}\\
\frac{58}{5+7+58} &= \frac{58}{70} = \frac{29}{35}\end{align*}
How did I find it? Choose $a$ and $b$ odd and relatively prime, then let $c=2ab-a-b$. This $c$ must be even, which makes ... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove: $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2} < \infty$ without L'Hôpital's Given $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}$, prove that it converges.
I tried to use the Ratio test.
I got a terrible algebraic expression: $$\lim_{n \to \infty} \frac{a_{n+1}}... | Try the root criterion.
Prove that
$$\sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}}=\left(1-\frac{n}{n^2+n+1}\right)^n$$
and that this tends to $L<1$ as $n\to \infty$. Then, the criterion implies that the sum converges.
To find the limit, remember that if $a_n\to 0$ then
$$(1+a_n)^{\frac1{a_n}}\to e.$$
So
$$\left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do we write $f(x+2)$ in terms of $f(x)$? $$f : R^+ \rightarrow R^+$$
$$f(x) = \dfrac{x}{x+1} $$
How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.
Regards
| A more general method:
$$f(x) = \frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=\frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=\frac{x+2}{x+3}=\frac{\frac{f(x)}{1-f(x)}+2}{\frac{f(x)}{1-f(x)}+3}=\frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=\frac{2-f(x)}{3-2f(x)}$$
Note: I usually think that this can easily go wrong, so you m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is right the inequality: $|a|+|b| \leq 2|a+ib|$
I want know if is right this inequality: $\forall a,b \in \mathbb{R}$,
$$|a|+|b| \leq 2|a+ib|$$
| For real numbers $a$ and $b$ we have
\begin{align*}
&&a^2&\le a^2+b^2\\
&\implies&\sqrt{a^2}&\le\sqrt{a^2+b^2}\\
&&|a|&\le\sqrt{a^2+b^2}
\end{align*}
In the same way we can prove $|b|\le\sqrt{a^2+b^2}$. Since $\sqrt{a^2+b^2}=|a+ib|$, the asked inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Compute the limit $\lim_{n\to\infty} I_n(a)$ where $ I_n(a) :=\int_0^a \frac{x^n}{x^n+1}\,\mathrm{d}x, n\in N$. For $a>0$ we define
$$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$
*
*Prove that $0\le I_n(1) \le \frac{1}{n+1}$
*Compute $\lim_{n\to\infty} I_n(a)$
My attempt:
*
*I regard $I_n(1)... | Lets consider the interval $(1, a)$. We have
$|\frac{x^n}{1 + x^n} - 1| = |\frac{1}{1+x^n}|$
By Bernoulli's Inequality, $x^n = (1 + (x - 1))^n \geq 1 + n(x - 1)$. This implies that
$|\frac{x^n}{1 + x^n} - 1| \leq \frac{1}{2 + n(x-1)}$
Now, fix $\epsilon > 0$ small enough and choose $n$ big enough such that $\frac{1}{2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proof verification: For $a$, $b$, $c$ positive with $abc=1$, show $\sum_{\text{cyc}}\frac{1}{a^3(b+c)}\geq \frac32$ I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've use... | You make a little mistake i think: $$f''(x)=\frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
integration question $\int_{-1}^{1}\left(\frac{1-x}{1+x}\right)^a\frac{dx}{(x-b)^2}$ How to integrate this integral $\displaystyle\int_{-1}^{1}\bigg(\frac{1-x}{1+x}\bigg)^a\frac{dx}{(x-b)^2}$ where $0<a<1$ and $b>1$
Answer. I put this into online integral calculator, but it said it cann't do this integration.
| First, substitute $t = (1-x)/(1+x)$.
$$\begin{aligned} I = \int_{-1}^{1}\left(\frac{1-x}{1+x}\right)^{a}\frac{\mathrm{d}x}{(x-b)^{2}} &= \frac{1}{2}\int_{0}^{\infty}t^{a}\left(\frac{x+1}{x-b}\right)^{2}\mathrm{d}t \\
&= 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{(1-t-b(1+t))^{2}} \\
&= 2\int_{0}^{\infty}\frac{t^{a}\,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given c... | Since $$|z| = |z^2-(-1)|\geq |z|^2-|-1|$$
Write $r=|z|$ and we have $$r^2-r-1\leq 0\implies r\in [0,r_2]$$ where $$r_{1,2} ={1\pm \sqrt{5}\over 2} $$
So $r\leq \displaystyle{1+ \sqrt{5}\over 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^... | If $x+3y=0$ so from the second equation we obtain $y=0$, which gives $x=0$, which is impossible.
Thus, $x+3y\neq0$ and since $$\frac{(x^2+y^2)y}{x+3y}=1,$$ from the first equation we obtain:
$$x+\frac{3x-y}{x^2+y^2}\cdot\frac{(x^2+y^2)y}{x+3y}=3\sqrt{\frac{(x^2+y^2)y}{x+3y}}$$ or
$$x^2+6xy-y^2=3\sqrt{y(x^2+y^2)(x+3y)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Prove $x^3+3y^3+9z^3-9xyz=1$ has infinity integers solutions! A question from Alibaba Global Mathematics Competition (number theory)
Prove $\displaystyle x^{3} +3y^{3} +9z^{3} -9xyz=1$ has infinitely many integer solutions.
The hint for the question is to transform the left side to a complex polynomial.
I found that:
... | this is my anser
first suppose we have an answer x,y,z , for simplicity,
set $\displaystyle a=x,b=3^{1/3} y\ ,c=3^{2/3} z$
and take both sides pow3 ,we get
$\displaystyle ( a+b+c)^{3}\left( a+\lambda b+\lambda ^{2} c\right)^{3}\left( b+\lambda ^{2} b+\lambda c\right)^{3}$=1
⟹
$\displaystyle \begin{array}{{>{\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
The volume of solid of revolution rotated about the line $y=x$ Find the volume of solid of revolution of region between curves $y=\sqrt x$ and $y=x^2$ in $xy-$plane about the line $y=x$. I know the answer, $\pi/30\sqrt 2$, but how we can obtain it? Should we rotate axis?
|
$x + w/\sqrt{2} = \sqrt{x - w/\sqrt{2}}$
means $w = \frac{1}{2} \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)$.
Volume:
$$\int\limits_{x=0}^1 \sqrt{2} \pi \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)^2\ dx = {\pi \over 30 \sqrt{2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find all integers $a$ that satisfy $c \equiv a \pmod{ab+1}$ Let $a,b,c$ positive integers with $b|c$ I am looking for triplets that satisfy $c \equiv a \pmod{ab+1}$.
What I found by now is that given a solution, we can write $c=q(ab+1)+a=a(qb+1)+q$
So we get $c \equiv q \pmod{qb+1}$ where q is the whole part of $\frac{... | First, we have $b \mid c$, which means that we can replace $c = bk$. Now, we have $a \equiv bk \pmod{ab+1}$. As you noted, we can see that $\gcd(b,ab+1)=1$. Thus, we can instead write $k \equiv \frac{a}{b} \pmod{ab+1}$.
We can also note that- $$ab \equiv -1 \pmod{ab+1} \implies \frac{1}{b} \equiv -a \pmod{ab+1} \implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\fra... | Following Thomas Shelby and Dr. Sonnhard Graubner, I provide you a full proof:
\begin{align}\sin(a)+\sin(b)+\sin(a+b)&=2\cos\left(\frac{a-b}{2}\right)\sin\left(\frac{a+b}{2}\right)+2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a+b}{2}\right)\\&=2\sin\left(\frac{a+b}{2}\right)\left[\cos\left(\frac{a-b}{2}\right)+\cos\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Proof of polynomial divisibility without using complex numbers? My question is the same as polynomial of degree n and its divisor except I want a solution that does not make use of complex numbers
Problem: Find all positive integers $n$ such that $x^2+x+1\mid (x+1)^n+x^n+1$
Using wolframalpha, I can see that a numbe... | You can use a form of polynomial modular arithmetic here, working modulo $x^2+x+1$
We have $x^2+x+1\equiv 0$ so that $x+1\equiv -x^2$
also $x^3+x^2+x\equiv 0$ so that $x^3\equiv -x^2-x\equiv 1$
So $p_n(x)=(x+1)^n+x^n+1\equiv (-1)^nx^{2n}+x^n+1$
Now since $x^3\equiv 1$ and you have a $(-1)^n$ there you can work modulo $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
find the smallest $N$ that $\varphi(n)\ge5$ for every $n\ge N$ I know that the solution is with Euler function. I could not understand how to show this. thanks.
| The smallest $N$ where $\phi(N) \ge 5$ for every $n \ge N$ would be $M + 1$ where $M$ is the largest $M$ where $\phi(M) \le 4$.
Now if $\gcd(n,k) = 1$ then $\phi(nk) = \phi(n)\phi(k)$ and for any $p^m$ for a prime $p$, then $\phi (p^m) = (p-1)p^{m-1} \ge p-1$. So if $\phi(M) \le 4$ then $p \le 5$ so the only possible ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}\cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$ Problem
Evaluate the infinite product
$$\frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}\cdot
\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$$
Attempt
For convenience，let's rewrite the limit. Denote$$x_1=\sqr... | Use $\cos2A=2\cos^2A-1,$
$$\dfrac2{\sqrt2}=\dfrac2{2\cos\dfrac\pi4}\text{ and }2+\sqrt2=2\left(1+\cos\dfrac\pi4\right)=4\cos^2\dfrac\pi8$$
$$\implies\dfrac2{\sqrt{2+\sqrt2}}=\dfrac2{2\cos\dfrac\pi8}$$
$\dfrac1{\cos\dfrac\pi4}\dfrac1{\cos\dfrac\pi8}=\dfrac{2\sin\dfrac\pi8}{\cos\dfrac\pi4\sin\dfrac\pi4}=\dfrac{4\sin\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Generalizing $\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2$ I was looking at this paper on section [17],
$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2\tag1$$
Let generalize $(1)$
$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)\cdots [2n-(2k+1)]}\tag2$$
Where $k\ge... | Using Lemma $(13)$ in this paper provided by the OP
$$\sum_{n=1}^{\infty}H_n{2n \choose n}x^n=\frac{2}{\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right)$$
Relpace $x$ with $\displaystyle \frac{x^2}{4}$, we get
$$\sum_{n=1}^{\infty}\frac{H_n}{4^n}{2n \choose n}x^{2n}=\frac{2}{\sqrt{1-x^2}}\ln\left(\frac{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Find the $\frac mn$ if $T=\sin 5°+\sin10°+\sin 15°+\cdots+\sin175°=\tan \frac mn$ It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.
$$T=\sin(5^\circ) +\sin(10^\circ) + \sin(15^\circ) + \cdots +\sin(175^\circ) =\tan \frac mn$$
Find the $\frac mn=?$, where $m$ and $n$ are p... | Not exactly the most elementary approach, but with complex numbers we could write
$$
T = \operatorname{Im}\left(1 + e^{\frac{\pi}{36}i} + e^{2\frac{\pi}{36}i} + \cdots + e^{35\frac{\pi}{36}i}\right) \\
=\operatorname{Im}\left(1 + z + z^2 + \cdots + z^{35}\right) \quad \left(z = e^{\frac{\pi}{36}i}\right)\\
= \operatorn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find every $z$s that fit $\cos(z) = -2$ I couldn't find any, I tried to write $\cos(z)$ as $\cos(x)\cos(iy)-\sin(x)\sin(iy)$ which then gave me
$\cos(x)\cosh(y) - i\sin(x)\sinh(y) = -2$
$\sin(x)=0$ so that imaginary part become $0$
now we have to find $\cosh(y) = -2$ which is not true for no $y$.
is it right or i made... | Be:
$$
\cos z=\frac{e^{iz}+e^{-iz}}{2}=-2
$$
Then:
\begin{eqnarray}
\frac{e^{iz}+e^{-iz}}{2} &=& -2 \\
e^{iz}+e^{-iz} &=& -4 \\
e^{2iz} + 1 &=& -4e^{iz} \\
\left(e^{iz}\right)^2 + 4\left(e^{iz}\right) + 1 &=& 0 \\
e^{iz} &=& \frac{-4 \pm \sqrt{16-4}}{2} \\
e^{iz} &=& -2 \pm \sqrt{3} \\
\end{eqnarray}
Then si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$ Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$
I have thought about doing the follo... | 1) $(x+y)^2- xy= 3;$
$xy = (x+y)^2 -3;$
$f(x,y)=3xy = 3(x+y)^2-9.$
$f_{min}=-9$, at $x=-y$;
$-3x^2= -9; x=\pm √3; y=\mp √3$;
2) $(x-y)^2+3xy= 3;$
$f(x,y)= 3xy= 3-(x-y)^2$;
$f_{max}= 3$, at $x=y$.
$x^2=1$; $x =\pm 1$, $y=\pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
In $\Delta ABC$, find $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ if $b+c=3a$
If in a triangle ABC, $b+c=3a$, then $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ is equal to ?
My reference gives the solution $2$, but I have no clue of where to start ?
My Attempt
$$
\cot\dfrac{B}{2}.\cot\dfrac{C}{2}=\frac{\cos\frac{B}{2}.\cos\frac{C}{2}... | Hint:
$$b+c=3a\implies\sin B+\sin C=3\sin A$$
$$\iff2\sin\dfrac{B+C}2\cos\dfrac{B-C}2=6\sin\dfrac A2\cos\dfrac A2$$
Now use $\dfrac{B+C}2=\dfrac\pi2-\dfrac A2,\cos\dfrac{B+C}2=?$
As $0<A<\pi,\sin\sin\dfrac A2\ne0$
$\implies\cos\dfrac{B-C}2=3\sin\dfrac A2=3\cos\dfrac{B+C}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac... | $$\dfrac1{\tan\dfrac B2}=\cot B/2=\tan(A/2+C/2)=?$$
Use
$$\sin2x=\dfrac{2\tan x}{1+\tan^2x}=?$$ for $2x=A,B,C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Finding the Jordan Form of a matrix... I know that this type of question has been asked on here before but I am still having a hard understanding what is going on. The text that I am learning from is "Linear Algebra Done Right by Sheldon Axler" and I don't think it covers this topic very well. Does anyone have any sugg... | The characteristic polynomial is
$$
p(\lambda)=\lambda^3-5\lambda^2+8\lambda-4=(\lambda-1)(\lambda-2)^2.
$$
The vectors involved in the Jordan blocks associated with eigenvalue $2$ is the column space of
$$
A-I = \begin{pmatrix}-1 & -1 & -1 \\
-3 & -2 & -2 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all parameters $a,b,c,d \in \mathbb R$ for which the function $f: \mathbb R \rightarrow \mathbb R$ a) has a limit at point 0 b) is continuous My function $f: \mathbb R \rightarrow \mathbb R$ it is given a pattern:
$$f(x)=\begin{cases} \frac{2^{7^{x}-1}-a}{\ln(1-x)},& x<0\\ b,& x=0 \\ \frac{\sin\left(c \sqrt{x^{2... | The first thing to note is that as $x\nearrow 0,$ the denominator $\ln(1-x)$ vanishes, so the only way to avoid $\bigl|f(x)\bigr|\to+\infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-a\to2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.
If you'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Super hard system of equations
Solve the system of equation for real numbers
\begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}
First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)... | Not really complete, but an interesting result using simple algebraic manipulations.
Write:
$$\begin{align}
a^2+b^2&=(a+b)^2-2ab\\
a^3+b^3&=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)\\
a^4+b^4&=((a^2)^2+(b^2)^2)=\cdots=(a+b)^4+2a^2b^2-4ab(a+b)^2\\
\vdots
\end{align}$$
From $(2)$, we have:
$$\begin{align}
(a^2+b^2)(c^2+d^2)&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 0
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What is $\int_0^{\pi/2}\sin^7(\theta)\cos^5(\theta)d\theta$ I have to integrate the following:
$\int_0^\limits\frac{\pi}{2}\sin^7(\theta)\cos^5(\theta)d\theta$
I decided to use a $u$ substitution of $u=\sin^2(\theta)$, and $\frac{du}{2}=\sin(\theta)\cos(\theta)$
and arrived at this integral
$\int_\limits{0}^{1}u^3(1-u)... | To lower the exponents a bit, notice that substitution $\theta \mapsto \frac\pi2-\theta$ yields
$$\int_0^{\frac\pi2} \cos^7\theta\sin^5\theta\,d\theta = \int_0^{\frac\pi2} \sin^7\theta\cos^5\theta\,d\theta$$
so we have
$$2I = \int_0^{\frac\pi2}\sin^5\theta\cos^5\theta(\cos^2\theta+\sin^2\theta)\,d\theta = \int_0^{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Integral over recurrence relationship I'm interested in evaluating the following definite integral
\begin{equation}
I_n = \int_0^{\gamma} F_n(x)\:dx
\end{equation}
Where $\gamma \gt 0$ and $F_n(x)$ is based on the recurrence relationship:
\begin{equation}
F_{n + 1}(x) = \frac{1}{1 + F_n(x)}
\end{equation}
Here $F_0(x)... | Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $F_0=f$. By Mathematica, it is easy to find the general expression should be of the form
$$
F_n=\frac{a_n+b_nf}{a_n+b_n+a_nf}
$$
for $n\ge 1$, where $a_n$ and $b_n$ are constants. By the recurrence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Maximizing $f$ in $\mathbb{R}^3$
Find the domain and the maximum value that the function
$$f(x,y,z)=\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$
may attain in its domain.
I have found the domain of the function to be $\mathbb{R^3\backslash\mathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=\frac{-... | Consider the vectors $\,\vec{u}=(x,y,z)\,$ and $\,\vec{v}=(1,2,3)$, then we can write
$$\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}=\frac{\vec{u}\boldsymbol{\cdot}\vec{v}}{\Vert\vec{u}\Vert}=\frac{\Vert\vec{u}\Vert\Vert\vec{v}\Vert\cos(\alpha)}{\Vert\vec{u}\Vert}=\Vert\vec{v}\Vert\cos(\alpha)=\sqrt{1^2+2^2+3^2}\cos(\alpha
)=\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the limit using Taylor seris with Big-O notation I have a limit
$$
\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2}
$$
I've tried to solve it like this:
\begin{align}
&\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2} =\\
&\lim_{x \to 0} \frac{x^2 - \sin x}{x^2\sin x} = \\
&\lim_{x \to 0} \frac{ x^2 - x + \frac{x^3}... | Intuitively. You understand that it is the indeterminate form $\infty-\infty$. To avoid it take $\frac1x$ out of brackets:
$$\frac1{\sin x}-\frac1{x^2}=\frac1x\left(\frac x{\sin x}-\frac1x\right).$$
When $x\to 0^+$, the limit is $(+\infty)(-\infty)=-\infty$. When $x\to 0^-$, the limit is $(-\infty)(+\infty)=-\infty$. B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find all integer solutions to $x^2+xy+y^2=((x+y)/3 +1)^3$
Find all ordered pairs of integers $(x,y)$, that satisfy the following relation:
$$x^2+xy+y^2=((x+y)/3 +1)^3$$
I tried bashing at first. Then I assumed $x+y = 3k$ for some integer $k$ so that $y=3k-x$, substituted in the given equation and got a cubic polynomi... | Bashing seems like a good approach. Setting $x+y=3k$ as you suggest, the equation becomes $$x^2-3kx+9k^2=(k+1)^3.$$ For a fixed integer $k$, we can consider this as a quadratic in $x$, and it has integer roots iff the discriminant $$(-3k)^2-4(9k^2-(k+1)^3))=-27k^2+4(k+1)^3$$ is a perfect square. Serendipitously, thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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What is the maximum value of $(a+ b+c)$ if $(a^n + b^n + c^n)$ is divisible by $(a+ b+c)$ where the remainder is 0?
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$\ge b\geq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisib... | The answer is indeed $6$. Here is a complete solution.
First, take a prime $p$ such that, $p\mid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.
*
*If $p\nmid a,b,c$, then by taking $n=p-1$, we h... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a matrix and analytic formula for a linear map $f: \mathbb R^{4} \rightarrow \mathbb R^{3}$ This linear map $f: \mathbb R^{4} \rightarrow \mathbb R^{3}$ meets the conditions:
\begin{align}\newcommand{\ran}{\operatorname{ran}}
f(1,2,3,1) &=(1,3,1) \\
(1,5,4,1) &\in \ker f\\
(1,1,2) &\in \mathrm{im} f \\
(7,5,0) &\i... | $\newcommand{\ran}{\operatorname{ran}}\newcommand{\R}{\mathbb{R}}$I guess you want a matrix which correspond with the linear map with respect to the standard basis.
First of all, let me tell you that you cannot find a unique solution to this exercise. So at some point you have to choose some properties of your solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Using the R method for finding all solutions to $\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$. My solution differs from official answer.
How many solutions does
$$\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$$
have between $-90^\circ$ and $90^\circ$?
I used the R method and got
$$2a-45^\circ = \arcsin\left(\frac{\sqrt... | Hint:
$$2a-45^\circ=180^\circ n+(-1)^n\arcsin\dfrac{\sqrt3}2$$
If $n$ is odd$=2m+1$(say)
$$2a-45=360m+180-60$$
But $-180-45\le2a-45\le180-45$
$-225\le360m+120\le135$
$?\le m\le?$
What if $n$ is even $=2m$(say)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim\limits_{x \to 1}\dfrac{x-x^x}{1-x+\ln x}$. Problem
Evaluate $$\lim\limits_{x \to 1}\frac{x-x^x}{1-x+\ln x}$$.
Solution
Consider using Taylor's formula. Expand $x^x $ and $\ln x$ at $x=1$. We obtain
$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$\ln x=(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
\begin{align*}... | $1-x=h\implies$
$$\lim_{h\to0}(1-h)\cdot\dfrac{1-(1-h)^{-h}}{h+\ln(1-h)}=\lim_{h\to0}\dfrac{1-\left(1+(-h)(-h)+\dfrac{(-h)(-h-1)}2\cdot h^2+O(h^3)\right)}{h-\left(h+\dfrac{h^2}2+O(h^3)\right)}=\dfrac1{\dfrac12}$$
So, you are correct
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Limit distribution of a reducible Markov chain I had a simple question yesterday when I was trying to solve an exercise on a reducible,aperiodic Markov Chain. The state spase S was $$S=\{1,...,7\}$$ and we could partition it into two closed classes $\{5,6,7\} \bigcup \{3,4\}$ but the class $\{1,2\}$ was open (thus tran... | As dan_fulea noticed above, it suffices to calculate the case when $ P[X_0=1]=1 $ and $ P[X_0=2]=1 $ . The first is already treated in the notes: it gives the limit vector $$ \begin{pmatrix} 0, & 0, & \frac{11}{51}, & \frac{22}{51}, & \frac{18}{221}, & \frac{24}{221}, & \frac{36}{221} \end{pmatrix} $$
For the case $ P... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$Ax^2+Bxy+Cy^2=1$ with $A=C=1$ and $B=2$ should be a parabola (because $B^2 = 4 AC$). Instead, it represents parallel lines. What went wrong? My calculus book says
The equation $Ax^2+Bxy+Cy^2=1$ produces a hyperbola if $B^2>4AC$ and
an ellipse if $B^2<4AC$. A parabola has $B^2=4AC$.
When I set $B=2,A=C=1$, the equa... | You are right: a nondegenerate parabola doesn't admit an equation of the form $Ax^2+2Bxy+Cy^2=1$.
Indeed, if $B^2=AC\ne0$, we can multiply all coefficients by $A$, finding
$$
A^2x^2+2ABxy+B^2y^2=B
$$
that becomes $(Ax+By)^2=B$. If $B>0$ this factors as
$$
(Ax+By-\sqrt{B})(Ax+By+\sqrt{B})=0
$$
which is a pair of paralle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Is there always exactly one solution to $a \cos\left(\frac{x}{2}\right)- b \sin\left(\frac{x}{2}\right) = 0$ in the interval $0\leq x\leq 2\pi$? I have a probably simple question.
If I have an equation like
\begin{align}
a \cos \left(\frac{x}{2}\right) - b \sin \left(\frac{x}{2}\right) &= 0\\
\frac ab \cos \left(\frac{... | Simply note the definition of tangent: $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$.
$$a\cos\left(\frac{x}{2}\right)-b\sin\left(\frac{x}{2}\right) = 0 \iff a\cos\left(\frac{x}{2}\right) = b\sin\left(\frac{x}{2}\right) \iff \\ \frac{a}{b} = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \iff \frac{a}{b} = \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I s... | Like Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $,
$$F=\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4}+1}}-\sqrt{2}n\right)\right)
=\lim_{y\to0^+}\dfrac{\sqrt{1+\sqrt{1+y}}-\sqrt2}y$$
Set $\sqrt{1+\sqrt{1+y}}-\sqrt2=u\implies u\to0$
and $1+\sqrt{1+y}=(u+\sqrt2)^2=2+2\sqrt2u+u^2$
$\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
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$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I ... | Let $\alpha=a+b=c+d$, $\beta=a^3+b^3=c^3+d^3$.
Then $$ \alpha^3=(a+b)^3=a^3+3a^2b+3ab^2+b^3=\beta+\alpha ab.$$
It follows that $a,b$ (and likewise $c,d$) are the two solutions of
$$ X^2-\alpha X+\frac{\alpha^3-\beta}{\alpha}=0,$$
i.e., $\{a,b\}=\{c,d\}$ and hence $$a^n+b^n=c^n+d^n $$
for arbitrary integers $n$. this a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
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Differential Equations - Is the solution to $(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$, $y=\ln(-5(e^x+1)^5+c)$? This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because som... | The wrong step is here :
$$-\frac{1}{e^y+1}=\frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
The correct steps are :
Multiply both left-hand and right-hand terms by $(e^y+1)(5(e^x+1)^5)$
$$\left(-\frac{1}{e^y+1}\right)(e^y+1)\big(5(e^x+1)^5\big)=\left(\frac{1}{5(e^x+1)^5}+c_3\right)(e^y+1)\big(5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
At... | I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let
$$ J(t) = \int_{0}^{\infty} \frac{\operatorname{arsinh}(tx)}{1+x^2} \, \mathrm{d}x. $$
Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=\sqrt{u^{-2}-1}$, we obtain
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 3
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How to Find the integral $\int_{S}−(xy^2)~dydz + (2x ^2 y)~dzdx − (zy^2)~dxdy$ where is the portion of the sphere $x^2 + y^2 + z^2 = 1$
Find the integral
$$
\int_{S}−(xy^2)~dydz + (2x^2y)~dzdx − (zy^2)~dxdy
$$
where $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z=\frac{1}{2}$. Choose th... | Let $$
\Omega: \frac{1}{2}\le z\le \sqrt{1-x^2-y^2}
$$ be the region enclosed by $S\bigcup S_1$ where $S_1 :z=\frac{1}{2}, x^2+y^2\le\frac{3}{4}$. We obtain
$$
\iiint_\Omega 2(x^2-y^2)\ \mathrm{d}x\mathrm{d}y\mathrm{d}z =\iint_{S\bigcup S_1}\vec{F}\cdot \vec{n}\ \mathrm{d}S
$$ by divergence theorem. Note that
$$
\iiint... | {
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"url": "https://math.stackexchange.com/questions/3083747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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pdf of transformed variable Given pdf $f_X(x)=\frac{x+2}{18}$ where $-2 < x < 4$, I wanted to find another r.v. $Y = \frac{12}{|X|}$. I think the support of $Y$ would be $3 < y < \infty$ but I wasn't super sure. I found the cdf of $X$, which was $F_X(x) = \int_{-2}^x\frac{x+2}{18}dx=\frac{x^2}{36}+\frac{x}{9}+\frac{1}{... | Take care with the absolute signage and the supports for the functions.
Note that $12/\lvert X\rvert$ folds both the domains $(-2;0)$ and $(0;2)$ onto $(6;\infty)$, and the domain $[2;4)$ onto $(3;6]$.
More clearly, only when $Y>6$ it is mapped to by a positive and negative value for $X$. When $3<Y\leq 6$ then $Y$ is ... | {
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"url": "https://math.stackexchange.com/questions/3086587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Transforming integral to polar coordinates By transforming to polar coordinates, show that
$$\int_{0}^{1} \int_{0}^{x}\frac{1}{(1+x^2)(1+y^2)} \,dy\,dx$$
Is equal to
$$ \int_{0}^{\pi/4}\frac{\log(\sqrt{2}\cos(\theta))}{\cos(2\theta)} d\theta$$
I have tried the standard $x=r\cos(\theta)$ etc. And can easily see that the... | You have correctly converted the region of integration and thus we have:
\begin{align}
I &= \int_0^{\frac{\pi}{4}}\int_1^{\sec(\theta)} \frac{1}{\left(1 + r^2\cos^2(\theta)\right)\left(1 + r^2\sin^2(\theta)\right)} \cdot r\:dr\:d\theta \\
\end{align}
Applying a partial fraction decomposition we arrive at:
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
The Integral $\int \frac {dx}{(x^2-2ax+b)^n}$ Recently I came across this general integral,
$$\int \frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±\sqrt {a^2-b} = a±\sqrt {∆}$$
Hence the integrand can be written as,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\frac {1}{(x-a-\sqrt ∆)^n(x-a+\sqrt ∆)^n}
$$
Resolving ... | Let $a$, $b$ and $x$ be real and $n$ be a positive integer. Let $\Delta:= a^2-b$.
then the following formula holds:
\begin{eqnarray}
\frac{1}{(x^2-2 a x+b)^n} &=&
\sum\limits_{l_1=1}^n \binom{2 n-1-l_1}{n-1} \frac{(-1)^n}{(x-a-\sqrt{\Delta})^{l_1}} \cdot \frac{1}{(-2 \sqrt{\Delta})^{2 n-l_1}} + \\
&& \sum\limits_{l_1=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to convert $\frac{1}{(1-x)(1-x^3)}$ into a sum of multiple fractions? What I mean is, that one can convert $\frac{1}{(1-x)^2(1-x^2)}$ into the following sum:
$\frac{1}{8}(\frac{1}{1+x}+\frac{1}{1-x} +\frac{2}{(1-x)^2} + \frac{4}{(1-x^3)})$
But I can't seem to do the same here, because when I try to simplify $1-x^3$... | Since you know that$$\frac1{(x-1)(x^3-1)}=\frac1{(x-1)^2(x^2+x+1)},$$use this to deduce that$$\frac1{(x-1)(x^3-1)}=\frac{x+1}{3\left(x^2+x+1\right)}-\frac1{3(x-1)}+\frac1{3 (x-1)^2},$$by partial fraction decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Extracting two balls with the same color vs distinct color I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(\text{distinct color})=P(... | Your inequality
\begin{align*}
(a-b)^2< a + b
\end{align*}
can be expressed as
\begin{align*}
a - \sqrt{2a + \frac{1}{4}}+\frac{1}{2} < b < a + \sqrt{2a + \frac{1}{4}}+\frac{1}{2}
\end{align*}
or
\begin{align*}
\left|b - a - \frac{1}{2}\right| < \sqrt{2a + \frac{1}{4}}
\end{align*}
This domain isn't pretty in any way, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$ I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$
is irrational, I assumed
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$
I cubed both sides and got
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$
I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhe... | Let $a=\sqrt[3]{3}$ and $b=\sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$
So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $? How to calculate
$\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $?
Since $\lim\limits_{n\to \infty} \frac {\cos(1/n)-\sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $\frac{1}{e}$, but since the form $(\to 1)^{\to ... | We have :
$\begin{align*}
\lim\limits_{n\to \infty} \left( \cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right) \right) ^n&=\lim\limits_{n\to \infty}\left(1+\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)-1\right)^n\\&=\lim\limits_{n\to \infty}\left[\bigg(1+\cos(\frac{1}{n})-\sin(\frac{1}{n})-1\bigg)^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Evaluate $\sum_{n=1}^{\infty}\arctan(\frac{2}{n^2})$? $$\begin{align}
\sum_{n=1}^{\infty}\arctan\left(\frac{2}{n^2}\right)&=\sum_{n=1}^{\infty}\arctan\left(\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}\right)\\
&=\sum_{n=1}^{\infty}\arctan(n+1)-\arctan(n-1)\\
&= \frac\pi4
\end{align}$$
Is this correct?
| The series telescopes. The partial sum $S_N$ is equal to
\begin{align}
S_N &= [\arctan (2) - \arctan (0)] + [\arctan (3) - \arctan (1)] + \cdots\\
& \qquad \cdots + [\arctan (N) - \arctan (N - 2)] + [\arctan(N + 1) - \arctan (N - 1)]\\
&= - \arctan (0) - \arctan (1) + \arctan (N) + \arctan (N + 1)\\
&= -\frac{\pi}{4} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inverse of Roto-Translation Matrix I'm trying to implement two JavaScript functions to calculate
*
*A 2D coordinate transformation from the reference frame $O$ to $O'$;
*The inverse transformation from $O'$ to $O$;
The first step it's very simple for me and it works. My mathematical process is this:
$$
\begin{p... | Pay attention to what transformation you want to represent, whether active or passive.
Anyway, if $R$ is the matrix that represents a rotation of angle $\theta$ in a plane, its inverse is the matrix that represents a rotation of the opposite angle $- \theta$, around the same origin, and it is easy to see that it is re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do I show $\sum\limits_{j=0}^{\infty}e^{-wj}=\frac{1}{w}+\frac{1}{2}+\frac{w}{12}-\frac{w^3}{720}+\frac{w^5}{30240}+\cdots$? Since $$\sum_{k=0}^{\infty}x^k=\frac{1}{1-x},$$we have
$$\sum_{j=0}^{\infty}e^{-wj}=\sum_{j=0}^{\infty} (e^{-w})^j=\frac{1}{1-e^{-w}}=\frac{e^w}{e^w-1}$$
Also, since $$e^x=\sum_{n=0}^{\infty}... | The Bernoulli numbers are defined by
$$\frac{t}{e^t-1}=\sum_{n=0}^{\infty} B_n \frac{t^n}{n!}$$
We have
$$\begin{split}
\sum_{j=0}^{\infty}e^{-wj} &=\frac{1}{1-e^{-w}}\\
&=\frac 1 w \frac{(-w)}{e^{-w}-1}\\
&= \frac 1 w \sum_{n=0}^{\infty} B_n \frac{(-w)^n}{n!}\\
&= \frac 1 w + \sum_{n=0}^{\infty} \frac{(-1)^{n+1}B_{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
$$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.
What did I do wrong?
| A good way of checking your working is to put some numbers in. Let's take $x=4,y=2$. Then what you have said is that $$\frac{64-4}{16+4\times2+4+2}=\frac{4}{4\times 2+4+2}$$
but we can see that the first fraction is: \begin{align}\frac{64-4}{16+4\times2+4+2}&=\frac{60}{30}\\
&= 2\end{align} and the second fraction is \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
value of $x$ in Trigonometric equation Find real $x(0<x<180^\circ)$ in
$\tan(x+100^\circ)=\tan(x-50^\circ)+\tan(x)+\tan(x+50^\circ)$
what i try
$\displaystyle \tan(x+100^\circ)-\tan(x)=\tan(x+50^\circ)+\tan(x-50^\circ)$
$\displaystyle \frac{\sin(100^\circ)}{\cos(x+100^\circ)\cos x}=\frac{\sin(2x)}{\cos(x+50^\circ)\co... | Let $\cos100^\circ=a$ and $\sin100^\circ=b$; let $\cos2x=X$ and $\sin2x=Y$. The final equation can be written
$$
\frac{b}{aX-bY+a}=\frac{Y}{X+a}
$$
that is,
$$
bX+ab=aXY-bY^2+aY
$$
together with $X^2+Y^2=1$. This is the intersection between a hyperbola and a circle, so generally a degree 4 equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"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.