Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.
What is $\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n}$ for $x > 4$.
Here is what I got so far (using Cauchy's integral formula) :
$$\sum\limits_{n=0}^\infty \binom{2n}{n}\frac{1}{x^n} ~=~ \sum\limits_{n=0}^\infty \left(\frac{1}{2\pi i}\... | You just need to keep going: factor the quadratic $z^2-(x-2)z+1=(z-r_1)(z-r_2)$, apply partial fractions to get
$${1\over(z-r_1)(z-r_s)}={1\over r_1-r_2}\left({1\over z-r_1}-{1\over z-r_2} \right)$$
and then use the fact that one of the roots is inside the unit circle and the other is outside when applying the Residue... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Partial Derivative with product & chain rule I cannot for the life of me work out the answer to this partial derivative.
$$\frac{\delta}{\delta x}\left(\frac{x^2}{(x+y)^2(x+z)^2}\right) $$
My first thought was:
*
*Split into two equations:
$$\frac{\delta}{\delta x}\left(\frac{x}{(x+y)^2}\cdot \frac{x}{(x+z)^2}\ri... | $$\left\lbrack\left(\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}\right)\cdot \frac{x}{(x+{\bf{z}})^2}\right\rbrack + \left\lbrack\left(\frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3}\right)\cdot \frac{x}{(x+{\bf{y}})^2}\right\rbrack = \\
\frac{x(y-x)}{(x+y)^3(x+z)^2}+\frac{x(z-x)}{(x+z)^3(x+y)^2} = \\
\frac{x(y-x)(x+z)+x(z-x)(x+y)}{(x+y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with t... | This answer uses similar ideas to those in Jack D'Aurizio's solution:
First we let $\displaystyle c_n=\frac{n^2-n+2}{2}$ and let $\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}$ for each n,
and we consider the related series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$ obtained by grouping terms with the same s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
How to diagonalize a matrix with several eigenvalues of zero? I'm looking to diagonalize a matrix A seen below. (Find a $P$ and a $D$ such that $AP = PD$).
$$
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 \\
... | We need to find the eigenvectors $\mathbf{x}$ such that:
$A\mathbf{x} =\mathbf{0}.$ Because the system is homogeneous, we can use the row echelon form of $A$, thus we need to solve the sytem:
$$ \begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$
Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equ... | We only need to prove that
$$(5a+1)^8 + (5a+2)^8 + (5a+3)^8 +(5a+4)^8 \equiv_{25} (5a+1)^4+(5a+2)^4+(5a+3)^4+ (5a+4)^4$$
In this equations the elements of the polynomials that only matter are the elements of the form $\dbinom{8}{i}(5a)^ik^{8-i}$ with $i=0,1$, because for $i>1$, the number is $\equiv_{25} 0$. This is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$\epsilon-\delta$ proof of $\lim_{ x\to 5} \frac{1}{x-4}=1$ $$\lim_{x\rightarrow 5} \frac{1}{x-4}=1$$
So far I have started the proof
For every $\epsilon > 0$ there exists a $\delta>0$ such that $\left|\frac{1}{x-4}-1\right|< \epsilon$, whenever $0 < |x-5| < \delta.$
I am having trouble figuring out what I need to fact... | We see $$\left \lvert \frac{1}{x-4} - 1\right \rvert = \left \lvert \frac{1}{x-4} - \frac{x-4}{x-4}\right \rvert = \left \lvert \frac{5-x}{x-4} \right \rvert.$$ For fixed $\epsilon > 0$, take $\delta = \min\{1/2, \epsilon / 2 \}$. Then for $\lvert x - 5 \rvert < \delta$ we have $\lvert x - 4 \rvert > 1/2$ so $\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving the fractional equation: $\{2^{n-1}\sqrt{3}\}=0.b_nb_{n+1}\ldots_{(2)}$
Prove that $$\{2^{n-1}\sqrt{3}\}=0.b_nb_{n+1}\ldots_{(2)}$$ where $\sqrt 3 = 1.b_1b_2b_3 \dots _{(2)}$.
(Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)
I am not sure how to prove this result because of the f... | From
$$
\sqrt 3 = 1.b_1b_2b_3 \cdots _{\,(2)} \tag1
$$ one gets, using $b_n \in \{0,1\}$,
$$
\begin{align}
\left\{2^{n-1}\sqrt 3\right\} &=\left\{2^{n-1}\left( 1+\frac{b_1}{2}+\frac{b_2}{2^2}+\cdots+\frac{b_{n-1}}{2^{n-1}}+\frac{b_n}{2^{n}}+\frac{b_{n+1}}{2^{n+1}}+\cdots \right)\right\}
\\\\&=\left\{2^{n-1}\left( 1+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplify $\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$ Simplify $$\frac{1}{\sqrt{4+2\sqrt{3}} - \sqrt{4-2\sqrt{3}}}$$
I know there is another easier method except the one I answered. I cannot find it. Can you please help? Thanks in advance.
| Here is another way, a few lines shorter,$$\sqrt\frac{1}{(\sqrt{4+2\sqrt3}-\sqrt{4-2\sqrt{3}})^2}=\frac{1}{\sqrt{4}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Prove $1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$ $n \ge 1$
$F_n$; Fibonacci numbers
$\frac{1+\sqrt5}{2}=\phi$
Prove
$$1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$$
I can't go any further can anyone hel... | notice that $A=B^2C$, and since $B>0$, $\sqrt A = B\sqrt C$.
Since $\sqrt A + B > 0$, $1 + 1/(\sqrt A + B) = \sqrt C \iff \sqrt A + B + 1 = \sqrt C(\sqrt A + B)$.
But $\sqrt A+B+1 = B\sqrt C + B + 1$ and $\sqrt C(\sqrt A+B) = BC + B\sqrt C$, so $ \sqrt A + B + 1 = \sqrt C(\sqrt A + B) \iff B+1 = BC \iff \phi^{2n+1}F_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Need help solving $x^4-3x^3-11x^2+3x+10=0$ Solve $x^4-3x^3-11x^2+3x+10=0$
I have tried to solve this equation using 'general formula from roots' from https://en.wikipedia.org/wiki/Quartic_function.
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x_{1,2}=-\frac b{4a}-S \pm 0.5\sqrt{-4S^2-2P+ \frac q S}$$
$$x_{3,4}=-\frac b{4a} + S \pm 0.5\... | We have
$$x^4-3x^3-11x^2+3x+10$$
Using rational root test:
$$(x-1)(x^3-2x^2-13x-10)$$
And rational root test again:
$$(x-1)(x+1)(x^2-3x-10)$$
Factoring quadratic:
$$(x-1)(x+1)(x-5)(x+2)$$
Thus, roots are $x=\pm1,\; x=-2,\; x=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Least value of $b$ for which inequality is true.
find the least natural number $b$ for which $x+bx^{-2}>2,\forall \space x\in (0,\infty)$
What I got confused with
In my book they have done this: rearranged the function as $$f(x)=x^3-2x^2+b$$ an $f(x)>0$
They have told to find least value of $b$, $f(x)$ should be mini... | $$
g(x) = x + b/x^2 > 2 \quad (x > 0) \quad (*) \iff \\
x^3 + b > 2 x^2 \quad (x > 0) \iff \\
f(x) = x^3 - 2 x^2 + b > 0 \quad (x > 0) \quad (**)
$$
A $b$ that will keep $(*)$ true, will also keep $(**)$ true and vice versa.
Thus we are are talking the same set
$$
B = \{ b \mid g(x) > 2, x > 0 \} = \{ b \mid f(x) > 0, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solution of $x^y=y^x$ and $x^2=y^3$ Solve the given set of equations:
$x^y=y^x$ and $x^2=y^3$ where $x,y \in \mathbb{R}$
Would any other solution exist other that $x=y=1$ because I think $x^2=y^3$
will only be true for $x=y=1$ or $x=y=0$
| Write $x=y^{3/2}$ and substitute in the first equation to get $y^{3y/2}=y^{y^{3/2}}$; if $y \neq 0,1$, then the exponents must be equal so that $3y/2=y^{3/2}$, which gives $y=\dfrac{9}{4}$ and $x=\dfrac{27}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
$8^a=3$ and $3^b=5$ and $10^c=5$ then find $c$ in terms of $a$ and $b$. if $8^a=3$ and $3^b=5$ and $10^c=5$ then find $c$ using $a$ and $b$.
My Attempt:
if $8^a=3$ and $3^b=5$ then we can say that $8^{ab}=5$ and then we have $2^{3ab}=10^c$ but i cant solve this equation.
| We have $$2^{3a}=3\implies 2^{3ab}=5\implies 2\times 2^{3ab}=10\implies 2^{3ab+1}=10$$
Thus $$10^c=2^{c(3ab+1)}$$ But we know that $2^{3ab}=5$ so we are asked to solve $$c(3ab+1)=3ab\implies c=\frac {3ab}{3ab+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the condition such that the roots of the polynomial are in AP
$f(x)=x^3+3px^2+3qx+r$ has roots in AP.Find the relation between $p,q$ and $r$.
[Answer:$-2p^2-3pq+r=0$]
My attempt:-
Taking $d$ as the common difference of the roots in AP we have $f(x)=(x-(a-d))(x-a)(x-(a+d))=x^3-3x^2a+(3a^2-d^2)x-a(a^2-d^2)$.
Compa... | $(p=-a),q=(\frac{3a^2-d^2}{3}),r=-a(a^2-d^2)\\
r=p(a^2-d^2)\\
\frac rp=(a^2-d^2)\\
3q=(3a^2-d^2)\\
3q=2p^2 + \frac rp\\
0=2p^3 - 3pq + r$
For the answer given, it is not correct.
Check it against $(x+1)(x+2)(x+3) = x^3 + 6x^2 + 11x + 6$
$-2p^2 - 3pq + r = -24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Factorization of $x^5+x$ I need to make the decomposition in $\mathbb{R}$ of:
$$x^5+x$$
Here my steps:
$$x(x^4+1)$$
$$x\big[ (x^2)^2+1 \big]$$
$$x\big[(x^2+1)^2-2x^2\big]$$
how should I proceed?
| Another (less ingenious but more systematic?) way is to first work over complex numbers:
$$\begin{align}x^4+1&=(x^2)^2-i^2=(x^2-i)(x^2+i)\\
&=(x+\frac{1+i}{\sqrt2})(x-\frac{1+i}{\sqrt2})\cdot(x+\frac{-1+i}{\sqrt2})(x-\frac{-1+i}{\sqrt2})\\
&=(x-\frac{1+i}{\sqrt2})(x+\frac{-1+i}{\sqrt2})\cdot(x-\frac{-1+i}{\sqrt2})(x+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Most natural way to prove $\sum_{n=1}^{\infty}\frac{1}{n+2}$ diverges I don't know how my teacher wants me to prove that
$$\sum_{n=1}^{\infty}\frac{1}{n+2}$$
diverges. All I know is that I have to use the $a_n>b_n$ criteria and prove that $b_n$ diverges.
I tried this:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1... | For all $n$, $n + 2 \leq n + 2n$, so
$$
\sum_{n=1}^\infty \frac{1}{3n} \leq \sum_{n=1}^\infty\frac{1}{n + 2},
$$
and $\sum_{n=1}^\infty \frac{1}{3n}$ diverges since it is a constant multiple of a divergent series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Permutations conjugated
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
and
$\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such th... | $\;\gamma^{-1}\;$ will be the permutation mapping each cycle of $\;\alpha\;$ to a corresponding cycle of $\;\beta\;$ of the same type, thus for example:
$$\gamma^{-1}:=\begin{cases}1\to2\\2\to3\\5\to4\\.......\\4\to1\\6\to5\\.......\\3\to6\end{cases}\implies\gamma=(123654)$$
and we can now check that
$$\gamma^{-1}\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric s... | $$\sum\limits_{k = 1}^n {\left( {4k - 3} \right)} = 4\sum\limits_{k = 1}^n k - 3 \cdot \sum\limits_{k = 1}^n 1 = 4\frac{{n\left( {n + 1} \right)}}{2} - 3 \cdot n = 2n^2 - n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 5
} |
Is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$ Is there a simple intuitive way to show that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
I sense it could be done more simple than this:
1 - take the derivative
$f'(x)=1-\frac{x}{\sqrt{x^2-1}}<0$ if $x>1$ so the slope of $f(x)$ in $(1,\infty)$ is negat... | $f(x)=x-\sqrt{x^2-1}= \frac{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}{x+\sqrt{x^2-1}}=\frac{x^2-x^2+1}{x+\sqrt{x^2-1}}=\frac{1}{x+\sqrt{x^2-1}}$
From this, it should be easy to deduct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
} |
Showing that $\sum_{n=0}^{\infty}\frac{2^{n+2}}{{2n\choose n}}\cdot\frac{n-1}{n+1}=(\pi-2)(\pi-4)$ Showing that
(1)
$$\sum_{n=0}^{\infty}\frac{2^{n+2}}{{2n\choose n}}\cdot\frac{n-1}{n+1}=(\pi-2)(\pi-4)$$
see here
(2)
$$(\arcsin(x))^2=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
Look very much sim... | Hint. By using the Euler beta function, one has
$$
\frac{2^{n+2}(n-1)}{{2n \choose n}(n+1)}=\frac{4(n-1)(2n+1)}{n+1}\int_0^1(2x(1-x))^ndx,\quad n \geq 0. \tag1
$$ Then, one may obtain
$$\begin{align}
&\sum_0^\infty\frac{2^{n+2}(n-1)}{{2n \choose n}(n+1)}
\\\\&=\sum_0^\infty\frac{4(n-1)(2n+1)}{n+1}\int_0^1(2x(1-x))^ndx\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}$ I would like to evaluate the following integral:
$$
\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}
$$
I tried various methods but without success.
| Hint. One may write
$$
\begin{align}
\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} &=\int_{-1}^0 \frac{dx}{(e^x+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)}
\\\\&=\int_0^1 \frac{dx}{(e^{-x}+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)}
\\\\&=\int_0^1 \frac{e^x\:dx}{(e^{x}+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove $\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$ $$\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$$
$u=x^4$ $\rightarrow$ $du=4x^3dx$
$x \rightarrow \infty$, $u\rightarrow \infty$
$x\rightarrow 0$, $u\rightarrow 0$
$$=\int_{0}^{\infty}\frac{2x}{u^2+2u+1}\cdot\frac{du}{4x^3}$$
$$=\frac{1}{2}\int_{0... | By subbing $x=\sqrt{z}$, then $\frac{1}{z^2+1}=u$
$$\int_{0}^{+\infty}\frac{2x}{(x^4+1)^2}\,dx = \int_{0}^{+\infty}\frac{dz}{(z^2+1)^2}=\frac{1}{2}\int_{0}^{1}u^{1/2}(1-u)^{-1/2}\,du$$
that by Euler's beta function and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ equals:
$$ \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
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how many strings of length 21 from letters {D,L,O} with repetition that have DOLL appearing only once The word DOLL has length $4$. There will be $18$ possible positions where 'DOLL' can appear, each with $3^{17}$ choices. So there will $18 \cdot 3^{17}$ cases where it can appear at least once. But I also know that the... | (This is simply a supplement to the two previous answers.)
As shown in N.F. Taussig's answer, the number of words with DOLL appearing at least once is given by
$\hspace{.5 in}\displaystyle\binom{18}{1}3^{17} - \binom{15}{2}3^{13} + \binom{12}{3}3^9 - \binom{9}{4}3^5 + \binom{6}{5}3=2,161,418,679$
Similarly, the number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$.
Show that if $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a$, $b$ or $c$ is divisible by $7$.
Here it seems Fermat's theorem has no use. We could consider many different cases of remainders of $a,b,c$ modulo $7$ but that's ... | Suppose that none of $a,b,c$ is divisible by $7$.
Write $a^3+b^3 \equiv -c^3$, square both sides, and use Fermat:
$ a^6 + 2a^3b^3 + b^6 \equiv c^6 $
$ 1+2a^3b^3 + 1\equiv 1 $
$ (ab)^3 \equiv 3$
But $3$ is not a cube mod $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\int x^8\sqrt{x^2+1}\ dx$
Evaluation of $\displaystyle \int x^8\sqrt{x^2+1}\ dx$
$\bf{My\: Try::}$ Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d \theta$
So $$I = \int \sec ^{3}\theta \cdot \tan^{8}\theta d \theta=\int \frac{\sin^{8}\theta}{\cos^{5}\theta }d\theta$$
Now How can I solve it after that... | HINT:
Integrating by parts,
for $m+1\ne0,$
$$I_m=\int x^m\sqrt{x^2+1}\ dx=\sqrt{x^2+1}\int x^m\ dx-\int\left(\dfrac{d\sqrt{x^2+1}}{dx}\int x^m\ dx\right)dx$$
$$(m+1)I_m=x^{m+1}\sqrt{x^2+1}-\int\dfrac{x^{m+2}}{\sqrt{x^2+1}}dx$$
$$=x^{m+1}\sqrt{x^2+1}-\int\dfrac{x^m(1+x^2)-x^m}{\sqrt{x^2+1}}dx$$
$$(m+2)I_m=x^{m+1}\sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 2
} |
Exponential Equations with Fractions I have had some issues with the following two equations:
$$ \frac{3^{n-2}}{9^{1-n}}=9$$
$$\frac{5^{3n-3}}{25^{n-3}}=125$$
If anyone could work them out step by step that would be awesome. I keep arriving at the wrong answer and believe it could be related to the addition of expo... | $\frac{3^{n-2}}{9^{1-n}}=9$
$\implies\frac{3^{n-2}}{3^{2(1-n)}}=9$
$\implies\frac{3^{n-2}}{3^{2-2n}}=9$
$\implies3^{(n-2)-(2-2n)}=3^{2} [a^{(m-n)}=\frac{a^m}{b^m}]$
$\implies3^{3n-4}=3^2$
$\implies{3n-4}=2$
$\implies n=2$
$\frac{5^{3n-3}}{25^{n-3}}$
$=\frac{5^{6-2}}{25^{2-3}} [put(n=2)]$
$=\frac{5^4}{5^{-1}}$
$=5^{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $f(1-x) + 2f(x) = 3x$, what is $f(0)?$ I did the following:
$$f(1-0) + 2f(0) = 3\cdot 0$$
$$f(1) + 2f(0) = 0$$
This reminds me of the equation of the straight line in the plane, then:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{f(1)}\\
{f(0)}
\end{pmatrix} \right> =0$$
$(1,2)$ is a normal v... | Just in case if you want to solve for the actual function next time:Substituting $x$ with $1-x$
$$f(1-(1-x))+2f(1-x)=3(1-x)$$$$f(x)+2f(1-x)=3-3x\tag{1}$$Now multiply the original equation by 2:$$2f(1-x)+4f(x)=6x\tag{2}$$
Subtract: $(2)-(1)$ We get
$$3f(x)=9x-3$$Therefore$$f(x)=3x-1$$$$f(0)=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Five white balls, five black balls There are 5 white balls and 5 black balls in a box. A visually-challenged man takes 5 balls from the box... What's the probability that all the balls which this individual picked are white if we know for sure that there 3 white balls among those 5 balls he took?
Using the formula $P(A... | Your answer is correct if you are doing "three white balls among five chosen balls", but is incorrect if the problem is looking for "at least three white balls".Stricktly according to your formula, $P(A | B) = \frac{P(A \cap B)}{P(B)}$
$B=$ at least three white balls among five chosen balls, $P(B)=\frac{\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove $\sum\limits_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}} \geq \frac{(\sum\limits_{i=1}^{n}a_i)^2}{\sum\limits_{i=1}^{n}b_i}$ So I have the following problem, which I'm having trouble solving:
Let $a_1$ , $a_2$ , ... , $a_n$ be real numbers. Let $b_1$ , $b_2$ , ... , $b_n$ be positive real numbers. Prove
$$ \frac{a_{1}^{2}}... | We will use induction.
First, suppose that $n=2$, the given inequality is
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}\geq \frac{(a_1+a_2)^2}{b_1+b_2}$$
This can be proven directly by subtracting, then factoring the equation into a complete square.
Suppose the given inequality holds for $n=k$
$$\frac{a_1^2}{b_1}+\frac{a_2^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\int_c\frac{1}{\sin\frac{1}{z}}\text{d}z$ over $C= \{z\big\vert|z|=\frac{1}{5}\}$ Evaluate
$$\int\limits_{|z|=\frac{1}{5}} \frac{1}{\sin\frac{1}{z}}\text{d}z$$
My attempt: I know that this function has non isolated singularity at $0$, and simple poles at $\frac{1}{n \pi}$. Which ever methods I know are a... | $$f(z) = \frac1{\sin{\left ( \frac1{\zeta+\frac1{n \pi}} \right )}} $$
where $\zeta = z-\frac1{n \pi}$. Then
$$\begin{align}f(z) &= \frac1{\sin{\left ( \frac{n \pi}{1+ n \pi \zeta} \right )}} \\ &= \frac1{\sin{\left [ n \pi \left (1-n \pi \zeta + n^2 \pi^2 \zeta^2-\cdots \right )\right ]}}\\ &= \frac1{\sin{n \pi} \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that for all odd integer $n, n^{4}=1\pmod {16}$ My answer: $n=2k+1$
$n^{4}=(2k+1)^{4}$=$16k^{4}+32k^{3}+8k^{2}+24k+1$.
I do not know how to conclude; really needed help here.
| Alternatively, you can simply consider the following cases:
*
*$n\equiv 1\pmod{16} \implies n^4\equiv 1^4\equiv 1\equiv1\pmod{16}$
*$n\equiv 3\pmod{16} \implies n^4\equiv 3^4\equiv 81\equiv1\pmod{16}$
*$n\equiv 5\pmod{16} \implies n^4\equiv 5^4\equiv 625\equiv1\pmod{16}$
*$n\equiv 7\pmod{16} \implies n^4\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 13,
"answer_id": 8
} |
How can I solve this hard system of equations? Solve the system below
\begin{align}
&\sqrt {3x} \left( 1+\frac {1}{x+y} \right) =2\\
&\sqrt {7y} \left( 1-\frac{1}{x+y} \right) =4\sqrt{2}
\end{align}
Frankly I am disappointed, because I spent around 2 hours in solving this equation, but, finally I didn't do it so, I... | Squaring 1st equation you have $3x(x+y+1)^2=4(x+y)^2$, squaring second you have $7y(x+y-1)^2=32(x+y)^2$. Put $s=x+y,y=s-x$ and we get $3s^2x+6sx+3x-4s^2=0,7s^3-7s^2x+14sx-46s^2+7s=0$. The first of these gives $x=\frac{4s^2}{3s^2+6s+3}$.
Substituting in the second we get $$s(21s^4-124s^3-178s^2-124s+21)=0$$ or $$s(3s^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can thi... | If $7+4\sqrt{3}$ is a perfect square this means $7=a^2+3b^2$ is the sum of two squares in $\Bbb{Z}[\sqrt{3}]$ and $4\sqrt{3}=2ab\sqrt{3}$. So the possibilities are
$$a=\pm 1,b=\pm 2\\a=\pm 2,b=\pm 1$$
Only the latter gives $a^2+3b^2=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 2
} |
Evaluate $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$ $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
My Attempt: I tried to use the regular way.
$A=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
$A^2=1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}$
But then I saw that nothing happened twice and I... | Setting the Scene
Evaluate $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }}$
Let $$f_n(0)=\sqrt{1+n}$$also that $$f_n(k)=\sqrt{1+(n-k)f_n(k-1)}.$$
Then clearly $0<f_n(0)<n+1$ when $n>0$. Now if we assume that $f_n(k)<n+1-k$ and by inducti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all natural roots of: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=d$ given that: $a
Find all natural roots of: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=d$ given that: $a<b<c$
Rearranging the equation gives: $$ab+bc+ac=abcd$$
What can we do with this?
| First observe that$$d \le \frac{1}{1}+\frac{1}{2}+\frac{1}{3}<2$$Thus $d=1$. And notice that if $a \ge 3$ then$$d \le \frac{1}{3}+\frac{1}{4}+\frac{1}{5}<1$$ So we have $a \le 2$. Its easy to see that $a \neq 1$, so $a=2$. Equation reduces to$$\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$$Again if $b \ge 4$ then$$\frac{1}{b}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the Fourier Sine Transform of $\frac{1}{x(a^2+x^2)}$? (Please check my solution) My textbook says that the answer is $\frac{\pi}{2a^2}(1-e^{-as})$ while my answer is $\frac{1}{a^2}\sqrt{\frac{\pi}{2}}(1-\frac{e^{-as}}{2}+\frac{e^{as}}{2})$. Here is my solution:
\begin{align}
\hat{f_s} & = \sqrt{\frac{2}{\pi}} \... | If you differentiate,
\begin{align}
\left(-\frac{d^2}{ds^2}+a^2\right)\int_{0}^{\infty}\frac{1}{t(t^2+a^2)}\sin(st)dt
&= \int_{0}^{\infty}\frac{\sin(st)}{t}dt \\
&=\int_{0}^{\infty}\frac{\sin(st)}{(st)}d(st) \\
&=\int_{0}^{\infty}\frac{\sin(u)}{u}du=\frac{\pi}{2}.
\end{align}
Therefore, $f(s)=\int_{0}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $\frac{d^ny}{dx^n}$ if $y = \frac{7}{1-x}$ I am wondering what is $\frac{d^n}{dx^n}$ if $y = \frac{7}{1-x}$
Basically, I understand that this asks for a formula to calculate any derivative of f(x) (correct me if I'm wrong). Is that related to Talylor's theory? How do I end up with such a formula?
Thanks!
| Let $f(x)= \sum_{n\geq 0} a_n(x-c)^n$
Then $f^{(n)}(c)=n!a_n $,
\begin{align*}
\frac{7}{1-x} &= \frac{7}{1-c-(x-c)} \\
&= \frac{7}{1-c}\frac{1}{1-\frac{x-c}{1-c}}\\
&= \frac{7}{1-c}\sum_{n\geq 0 } \frac{1}{(1-c)^n} (x-c)^n\\
&=\sum_{n\geq 0 } \frac{7}{(1-c)^{n+1}} (x-c)^n
\end{align*}
Therefore $$f(c)= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Compute antiderivative of $x \frac{3}{4} \sqrt{ -x}$ I need to compute the area under $x \times\frac{3}{4} \times \sqrt{|x|}$ from x=-1 to x=1. Therefore I need to compute the antiderivative of $x \times \frac{3}{4} \times \sqrt{x} $ for $0 \le x \le 1$, which is straight forward, $ \frac{3}{10} \times x^{2.5}$. And fo... | $x\frac34\sqrt{|x|}$ is an odd function - so the integral from -1 to 1 is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $(a+b-c)/(a+b+c)$
Given $$2a^2 +17b^2 + 8c^2 -6ab -20bc =0 \\ abc \neq 0.$$
Find the value of $(a+b-c)/(a+b+c)$.
I tried factorizing the equation, but still can't find a solution
| $$0=2\cdot LHS=(2a-3b)^2+(5b-4c)^2$$
and so $2a=3b$ and $5b=4c$. This gives $$\frac{a+b-c}{a+b+c}=\frac{3b/2+b+5b/4}{3b/2+b-5b/4}=1/3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A common term for $a_n=\begin{cases} 2a_{n-1} & \text{if } n\ \text{ is even, }\\ 2a_{n-1}+1 & \text{if } n\ \text{ is odd. } \end{cases}$ When I was answering a question here, I found a sequence as a recursive one as given below.
$a_1=1$, and for $n>1$,
$$a_n=\begin{cases}
2a_{n-1} & \text{if } n\ \text{ is even, }\... | It can often be helpful to just write out the first couple of terms:
$$\begin{align}
a_1 &= 1 \\
a_2 &= 2 \\
a_3 &= 5 = 2^2 + 1\\
a_4 &= 10 = 2^3 + 2 \\
a_5 &= 21 = 2^4 + 5 = 2^4 + 2^2 + 1\\
a_6 &= 42 = 2^5 + 10 = 2^5 + 2^3 + 2\\
a_7 &= 85 = 2^6 + 21 = 2^6 + 2^4 + 5 = 2^6 + 2^4 + 2^2 + 1\\
a_8 &= 170 = 2^7 + 42 = 2^7 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Seeking closed form for infinite sum $\sum \limits_{ n=1 }^{ \infty }{ \frac { { \left(n! \right) }^{ 2 } }{ { n }^{ 3 }(2n)! } }$ $\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { \left(n! \right) }^{ 2 } }{ { n }^{ 3 }(2n)! } }$ is approximately $.5229461921333351$ but I've been assured that there is a closed form... | The series can be expressed in the form
\begin{align}
\sum_{n=1}^{\infty} \frac{(n!)^2}{n^3 \, (2n)!} &= \frac{1}{2} \, {}_{4}F_{3}\left(1,1,1,1; 2,2,\frac{3}{2}; \frac{1}{4} \right) \\
&= \frac{1}{2} \, \left[ - \frac{8}{3} \, \zeta(3) + \frac{\pi}{4 \sqrt{3}} \, \psi^{(1)}\left(\frac{1}{3}\right) - \frac{\pi}{12 \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $\arcsin \frac{x-1}{x+1}$ I was looking at a question that asks for the derivative of $\arcsin (\frac {x+1}{x-1}) $.
The solution starts by saying $y = \frac{x+1}{x-1}$, so $1-y^2= \frac{4x}{(x+1)^2}$ and $\frac{1}{\sqrt{1-y^2}}$ and thus $\frac{x+1}{2\sqrt{x}} $
However, I don't see why $1-y^2= \frac... | $1-y^2=1-\left(\dfrac{x-1}{x+1}\right)^2=1-\dfrac{x^2-2x+1}{x^2+2x+1}=\dfrac{(x^2+2x+1)-(x^2-2x+1)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}=\dfrac{4x}{(x+1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displa... | Using $c=-a-b$,
$\displaystyle \hspace{.2 in}\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right) \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\displaystyle=\left(\frac{a}{a+2b}-\frac{b}{2a+b}-\frac{a+b}{a-b}\right)\left(\frac{a+2b}{a}-\frac{2a+b}{b}-\frac{a-b}{a+b}\right)$
$\displaystyle=\left(\frac{2(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
'Strange' trigonometric roots of $x^5-4x^4+2x^3+5x^2-2x-1$ - could someone explain? This quintic equation has $5$ real roots:
$$x^5-4x^4+2x^3+5x^2-2x-1=0 \tag{1}$$
The roots are, from left to right:
$$x_1=\frac{\cos \frac{19}{22} \pi}{\cos \frac{1}{22} \pi}$$
$$x_2=\frac{\cos \frac{9}{22} \pi}{\cos \frac{19}{22} \pi}$$... | Using the equality $\cos(\pi \pm x)=-\cos(x)$ you get
$$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\
x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\
x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\
x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\
x_5=-\frac{\cos \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to prove that given a binary operator $*$, $a * b = a^2 - ab + b^2$ is associative? No matter how hard I try I cannot seem to prove that, given the binary operator $*$, the operation $a * b = a^2 - ab + b^2$ is associative.
This is what I have tried:
$(a * b) * c = a * (b * c) [Associativity];
$(a^2 - ab + b^2) * c... | Just do it:
$(a*b)*c = (a^2 - ab + b^2)*c = (a^2 - ab + b^2)^2-a^2c +abc-b^2c+c^2=a^4 -2a^3b + 3a^2b^2 -2ab^3+b^4-a^2c+abc-b^2c+c^2$
$a*(b*c) = a^2-a (b^2-bc+c^2) +(b^2 - bc + c^2)^2$
It's pretty clear these need not be equal. One is "dominated" by $a^4$ and $b^4$ while the other is dominated by $c^4$ and $b^4$.
Let $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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hard integral problems to solve I'm practicing harder integration using techniques of solving with special functions
I have difficulties with these two hard integrals; don't even know how to start,
$$\int_0 ^\infty x^p e^{-\frac{\theta}{x}+Bx}dx$$
where $\theta,B>0$
$$pv\int_0 ^\infty \frac{x^p e^{\cos{\theta x}} \cos... | Let $\theta=b$ and $B = -a \lt 0$. Then the first integral is a generalization of this integral. Using the same substitutions:
$u=a x+b/x$. Then $a x^2-u x+b = 0$, and therefore
$$x = \frac{u}{2 a} \pm \frac{\sqrt{u^2-4 a b}}{2 a}$$
$$dx = \frac1{2 a}\left (1 \pm \frac{u}{\sqrt{u^2-4 a b}}\right ) du $$
Now, it sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why can't we use the law of cosines to prove Fermat's Last Theorem? In investigating approaches to Fermat's Last Theorem I came across the following and I can't figure out where I am going wrong. Any input would be greatly appreciated.
We want to show that $a^n + b^n = c^n$ cannot hold for odd $n>1$ and pairwise relat... | How do you get to the conclusion that $ab\vert s$? I honestly can't see it. The way I see it you have:
$$\dfrac{r}{s}=\dfrac{(a+b)^2-c^2}{2ab}. $$
Now $(a+b)^2-c^2$ is even. You can check this case by case, when $a,b$ are odd then $c$ has to be even and so forth. So at least one of them is even but by your assumption m... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Evaluate $\int_0^1\frac{1}{\lfloor1-\log_{2}(1-x)\rfloor}dx$ Problem:
$$\int_0^1\frac{1}{1-\log_{2}(1-x)}dx$$
Which I simplified down to this:
$$2\log(2)\int_0^1\frac{1}{2^xx} dx$$
Now I am stuck as the answer is supposed to be computed without the use of a calculator, meaning non standard use of Ei(x) would not be app... | The answer given by @clark is correct for the originally given integral. Things change quite a bit when we bring in the floor function.
In a comment on the question you asked what the floor function means in this instance. It means the same thing it always does: $\lfloor y \rfloor$ is the greatest integer less than ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Eliminate $\theta$
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$
My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
| using shorthand
$$ s = \sin \theta; c = \cos \theta; s2 = \sin 2 \theta; c2 = \cos 2 \theta; $$
Square and add
$$ s^2 +1/s^2 + c^2 + 1/c^2 = m^2+n^2$$
or
$$ sc= \frac{1}{\sqrt{m^2n^2-1}}, \, s2= \frac{2}{\sqrt{m^2+n^2-1}} $$
$$c2 = \sqrt{1-s2^2}=\sqrt{ \frac{ m^2+n^2-5}{m^2+n^2-1}}$$
$$s = \sqrt { \frac{ 1-c2}{2}}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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What is the actual meaning of power which is less than 1? $3^4 = 3 \cdot 3 \cdot 3 \cdot 3$
$3^2 = 3 \cdot 3$
$3^1 = 3$
But my problem is for below 1:
Example:
$3^{0.7} = \ ?$
Can you please explain it to clear my doubt?
| Exponents less than 1 are better understood in terms of fractions rather than decimals:
$3^4=3^{2+2}=3^2•3^2$
Therefore:
$3^1=3^{\frac{1}{2}+\frac{1}{2}}=3^\frac{1}{2}•3^\frac{1}{2}=(3^\frac{1}{2})^2$
What value, when multiplied by itself, is equal to 3? The only possibility is ±$\sqrt{3}$. This is why we understand an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solving $\cos^{-1}x+\cos^{-1}2x=-\pi$ gives the invalid answer $x=0$ To solve the inverse trigonometric equation
$$\cos^{-1}x+\cos^{-1}2x=-\pi,$$
I use the normal cosine addition
\begin{align}
\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&=-\pi\\
2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=\cos(-\pi)\\
2x^2 -\sqrt {1-x^2} \sqrt{... |
\begin{align}
\cos^{-1}x+\cos^{-1}2x=-\pi\tag{*}\\
\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&=-\pi\tag1\\
2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=-1\tag2\\
(2x^2+1)^2&=(1-x^2)(1-4x^2)\tag3\\
4x^4+1+4x^2&=1-4x^2-x^2+4x^4\tag4\\
x&=0\tag5.
\end{align}
Note that $$\cos^{-1}x+\cos^{-1}2x=\begin{cases}2\pi-&\cos^{-1}(2x^2 -... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Exact value of $\cos^2(\frac{\pi}{8})+\sin^2(\frac{15\pi}{8})?$ I tried separating it into $\cos^2(\frac{\pi}{8})+\sin^2(\frac{\pi}{8}+\frac{14\pi}{8})?$ and using the angle sum identity but it didn't help.
| Notice that $\frac{15\pi}{8} = 2\pi - \frac{\pi}{8}$. So $\sin \frac{15\pi}{8} = -\sin \frac{\pi}{8}$. Your expression then reduces/evaluates down to:
$$\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
point of maximal curvature on $y=\ln x$ While attempting the problem below, I got the cross product of the function's derivative and second derivative to equal 0. If this happens, what does this mean for the curvature formula?
My work:
$$y = f(x)$$
$$\rm{curvature}(x) = {f''(x)\over(1+(f'(x)^2)^{3/2}}$$
$$f'(x) = {1\ov... | For $x>0$,
\begin{align*}
y &= \ln x \\
y' &= \frac{1}{x} \\
y'' &= -\frac{1}{x^2} \\
\kappa &= \frac{|y''|}{(1+y'^2)^{3/2}} \\
&= \frac{\frac{1}{x^2}}{\left( 1+\frac{1}{x^2} \right)^{3/2}} \\
&= \frac{x}{(1+x^2)^{3/2}} \\
\frac{d\kappa}{dx} &=
\frac{1}{(1+x^2)^{3/2}}-
\frac{x\times 2x \times \frac{3}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive
$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$
$$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt... | Use Titus lemma:
$$\frac{x^2}{a}+\frac{y^2}b\ge \frac{(x+y)^2}{a+b}$$
Then $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}=\frac{\sqrt{x}^2}{\sqrt{y}}+\frac{\sqrt{y}^2}{\sqrt{x}}\ge \frac{(\sqrt{x}+\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})}=\sqrt{x}+\sqrt{y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Find all $a, b, c, d$ Find all $a, b, c, d$ satisfying
$$\frac{x^4+ax^3+bx^2-8x+4}{(x^2+cx+d)^2} = 1$$
My answers are:
$$\begin{cases}
d=2\\
d=-2
\end{cases}
\quad \begin{cases}
a=4\\
a=-4
\end{cases} \quad
\begin{cases}
c=2\\
c=-2
\end{cases} \quad
\begin{cases}
b=8\\
b=0
\end{cases}$$
Is it right?
| For another perspective:
Let $r_1, r_2, r_3, r_4$ be the zeroes of
$$f(x) = x^4 + ax^3 + bx^2 - 8x + 4.$$
Since
$$x^4 + ax^3 + bx^2 - 8x + 4 = (x^2 + cx + d)^2,$$
$r_1, r_2, r_3, r_4$ are also the zeroes of
$$g(x) = (x^2 + cx + d)^2.$$
Using the quadratic formula, we get (without loss of generality) that
$$r_1 = r_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving the Inequality $\frac{14x}{x+1}<\frac{9x-30}{x-4}$ The question says to find all the integral values of x for which the inequality holds.
the question is
$$\frac{14x}{x+1}<\frac{9x-30}{x-4}$$
My Solution
\begin{align}
& \frac{14x}{x+1} < \frac{9x-30}{x-4} \\[6pt]
& \frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}<0 \\... | Correcting your initial mistake will lead to the inequality
\begin{equation}
\dfrac{(x+1)(x-6)}{(x+1)(x-4)}<0
\end{equation}
which is positive on the interval $(6,\infty)$ since the product of $x$ coefficients is positive, and the expression changes sign at each zero and pole as you move left along the $x$-axis since e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$
What is the coefficient of coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$?
Using summation of G.P., this is equivalent to finding the coefficient of $x^{41}$ in
$$\left(x^5 \left(\frac{1-x^5}{1-x}\right)\right)^5$$
and thus finding coefficient... | It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
[x^{16}]\left(\frac{1-x^5}{1-x}\right)^5
&=[x^{16}](1-x^5)^5\sum_{k=0}^\infty \binom{-5}{k}(-x)^k\tag{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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where is my mistake with this equation ot hyperbolic problem? Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,
but we use
$$\begin{cases}
y^2=4x\\
x^2-y^2=1... | I know it's rude to give complex solutions to real problems, but I think we can easily see where the second solution $x=2-\sqrt{5}$ came from.
We want to find solutions for: $$\begin{cases} y^2=4x \\ x^2-y^2=1 \end{cases}$$ Let's try find them in $\mathbb{C}$.
We got, as you said before, the equation $x^2-4x-1=0$ that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the inverse matrix I have these matrices:
Find the inverse matrices:
\begin{bmatrix}
1 & 1 & 0& 0&\dots & 0& 0\\0 & 1 & 1& 0&\dots & 0& 0 \\0 & 0 & 1& 1&\dots & 0& 0 \\\dots & \dots & \dots& \dots&\dots & \dots& \dots\\0 & 0 & 0& 0&\dots & 1& 1 \\0 & 0 & 0& 0&\dots & 0& 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & a... | Using the usual algorithm, you can calculate that $$\begin{pmatrix}
1&1&0&0&\mid&1&0&0&0\\
0&1&1&0&\mid&0&1&0&0\\
0&0&1&1&\mid&0&0&1&0\\
0&0&0&1&\mid&0&0&0&1
\end{pmatrix}\xrightarrow{R_3\rightarrow R_3-R_4}\begin{pmatrix}
1&1&0&0&\mid&1&0&0&0\\
0&1&1&0&\mid&0&1&0&0\\
0&0&1&0&\mid&0&0&1&-1\\
0&0&0&1&\mid&0&0&0&1
\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$
Prove that
$$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$
My attempt
\begin{align}
\text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\
&=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac... | Recall the identity $$\sin \alpha \cos \beta = \frac{1}{2} \left( \sin(\alpha+\beta) + \sin(\alpha - \beta) \right).$$ Then
$$\begin{align*} \sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right)
&= \frac{1}{2} \left( \sin \frac{4\pi}{7} + \sin \frac{6\pi}{7} + \sin \frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\cos 36^\circ - \cos 72^\circ$ without the aid of a calculator I have a quick question about a difficult trigonometric functions problem that I have been assigned. The problem is as follows: Evaluate $$\cos36° - \cos72°$$ without the aid of a calculator. In terms of my attempts at the problem, I have converte... | First write it as $$\cos 36°-\cos 72°=2\sin { \frac { 72°-36° }{ 2 } } \sin { \frac { 36°+72° }{ 2 } } =2\sin { 18°\sin { 54° } } =2\sin { 18° } \cos { 36° } $$
then let calculate $\sin { 18° } $
$$\cos { 36° } =\sin { 54° } \\ \cos { 2\cdot 18°= } \sin { 3\cdot 18° } \\ \theta =18°\\ \cos { 2\theta } =\sin { 3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find $\tan x $ from $(a+1)\cos x + (a-1)\sin x=2a+1$? How do I find $\tan x$ from this equation?
$$(a+1)\cos x + (a-1)\sin x=2a+1$$
Thanks for any help!!
| If $a=1$, we obtain $2\cos x=3$, which is invalid (unless you want to consider complex values). Set $X=\cos x$ and $Y=\sin x$ to get
$$
\begin{cases} (a+1)X+(a-1)Y=2a+1 \\[6px] X^2+Y^2=1 \end{cases}
$$
Therefore $(a-1)Y=2a+1-(a+1)X$ and therefore
$$
(a-1)^2X^2+(2a+1-(a+1)X)^2=(a-1)^2
$$
With standard steps we find
$$
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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how to partial fraction $\frac{1}{(x+1)^2}$ I need to integrate $\frac{1}{(x^2+2x+1)}$, so I need to use partial fraction as the polynomial can be factored as $\frac{1}{(x+1)^2}$. This is what I've tried:
$$\frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
$$A\cdot(x+1)^2 + B\cdot(x+1)$$
$$Ax^2+2Ax+A+Bx+B$$
$$Ax^2+(2A+B)x+(A+B)$$
... | You can integred it directly without partial fraction as $$\int { \frac { dx }{ { \left( x+1 \right) }^{ 2 } } =\int { \frac { d\left( x+1 \right) }{ { \left( x+1 \right) }^{ 2 } } } =-\frac { 1 }{ x+1 } +C } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integrate $\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$ Integrate using integrating by parts :
$$\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
My attempt :
$$I=\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
$$I=\int \frac{e^x}{(1+x^2)^2}\cdot(1-x)^2dx$$
$$I=\frac{e^x}{(1+x^2)^2}\cdot\frac{(1-x)^3}{-3}-\int \frac{(1-x)^... | $$\int \frac { (1-x)^{ 2 }\cdot e^{ x } }{ (1+x^{ 2 })^{ 2 } } dx=\int { \frac { { e }^{ x } }{ (1+x^{ 2 }) } dx-2\int { \frac { x{ e }^{ x } }{ (1+x^{ 2 })^{ 2 } } dx= } } \\ =\int { \frac { d{ e }^{ x } }{ 1+x^{ 2 } } } -2\int { \frac { x{ e }^{ x } }{ (1+x^{ 2 })^{ 2 } } dx= } \left( \frac { { e }^{ x } }{ 1+x^{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Geometry problem related to right angled triangles.
In the given figure $AC = 12 cm, AE = 6 cm$ and $CD = 8 cm$. CD is perpendicular to $AD$ and $BE$ is perpendicular to $AC$. How can we find the value of $BE$?
| This is a method without using similar triangles:
First of all, we can find $\text{AD}$ by using the Pythagorean Theorem:
$$\text{AD}=\sqrt{\text{AC}^2-\text{CD}^2}\to\text{AD}=\sqrt{12^2-8^2}=4\sqrt{5}\space\text{cm}$$
So, for $\text{DE}$:
$$\text{DE}=\text{AD}-\text{AE}\to\text{DE}=\left(4\sqrt{5}-6\right)\space\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Factors of -2 that sum to I'm trying to simplify this equation: $x^2+x-2$. I put it on WolframAlpha and it gives me this:
The factors of -2 that sum to are $2$ and $-1$. So, $x^2+x-2 = (x+2)(x-1)$
Can someone please explain to me what does this mean? How is this technique called?
Thanks!
| Alternative way to factor
$$x^{ 2 }+x-2=x^{ 2 }+x+\frac { 1 }{ 4 } -\frac { 1 }{ 4 } -2={ \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }-\frac { 9 }{ 4 } =\left( x+\frac { 1 }{ 2 } -\frac { 3 }{ 2 } \right) \left( x+\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \right) \\ =\left( x-1 \right) \left( x+2 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\f... | We start from $\left( 1 - \frac{n}{(n+1)^2}\right)\left(\frac{n+2}{n+1}\right)$:
$ \left( 1 - \frac{n}{(n+1)^2}\right)\left(\frac{n+2}{n+1}\right) = \frac{((n+1)^2 - n)(n+2)}{(n+1)^3}$
We then set up the inequality and simplify in order to see if we can get a trivial inequality.
$ \frac{((n+1)^2 - n)(n+2)}{(n+1)^3} \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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A pair of sequences defined by mutual addition/multiplication Define sequences $\{a_n\},\,\{b_n\}$ by mutual recurrence relations:
$$a_0=b_0=1,\quad a_{n+1}=a_n+b_n,\quad b_{n+1}=a_n\cdot b_n.\tag1$$
The sequence $\{a_n\}$ begins:
$$1,\,2,\,3,\,5,\,11,\,41,\,371,\,13901,\,5033531,\,69782910161,\,...\tag2$$
It appears i... | Given a sequence $\{A_n\}$ with the Fibonacci recurrence relation
$$A_{n+2} = A_{n+1} + A_n$$
we get the well-known asymptotic
$$A_n \sim C\phi^n$$
where $\phi$ is the golden ratio and $C$ depends on the initial terms $A_1$ and $A_2$. For example if $A_1 = A_2 = 1$ we get the Fibonacci sequence and $C = 1/\sqrt{5}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
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Neglected constant curvature difference surfaces What are some surfaces where $ \kappa_1-\kappa_2$ is constant? On a sphere where all are umbilical points.. is a special case.
For the $ \kappa_1+\kappa_2$ = constant case we have DeLaunay and Minimal surfaces.
$ \kappa_1,\kappa_2$ are the principal curvatures.
EDIT1:
... | Another trivial case is a cylinder with $\kappa_{1}=\dfrac{1}{r}$ and $\kappa_{2}=0$.
How about another variation with $2\kappa_{1}-\kappa_{2}=0$?
Mylar balloon
\begin{align*}
\left(
\begin{array}{c}
x \\
y \\
z \\
\end{array}
\right)
&=
r\left(
\begin{array}{c}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the following using AM-GM inequality The least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \ge 1 $for all $x \gt 0 $, is
Using AM-GM inequality
$$\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{a}$$
$$4ax^2 + \frac{1}{x} \ge 3\sqrt[3]{a}$$
Now my question start from here .
Can I do that for le... | No, I think. Because does not exist $a>0$, for which $3\sqrt[3]a>3\sqrt[3]a$.
For the original problem we see that since for $4ax^2=\frac{1}{2x}$ we get an equality and we obtain $a\geq\frac{1}{27}$, so the answer is $\frac{1}{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim_{x\rightarrow c}x^2=c^2$ What is the common proof for $\lim_{x\rightarrow c}x^2=c^2$.
I tried like this
Let $\varepsilon>0$, want to find a $\delta>0$ such that $\forall x\in\mathbb{R},0<|x-c|<\delta\Rightarrow |x^2-c^2|<\varepsilon$.
Therefore $$-\varepsilon<x^2-c^2<\varepsilon$$
$$c^2-\varepsilon<x^2... | $$lim_{x \to c} x^2 = c^2$$
Let $\epsilon>0$ , we know that
$$x^2-c^2=(x-c)(x+c)$$
$$|x^2-c^2|=|x-c||x+c|$$
If
$$|x-c|<2 \implies -2<|x|-|c|<2 (by\ \ triangular \ \ property)$$
$$|c|-2<|x|<2+|c|$$
By triangular property, we have
$$|x+c|\le |x|+|c|<2+|c|+|c|$$
$$|x+c|<2+2|c|$$
So,
$$|x^2-c^2|<2.(2+2|c|)$$
$$|x^2-c^2|<(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\tan(A+B)$ using $\cos(A-B)$ and $\sin(A-B)$ Use $\cos(A-B)$ and $\sin(A-B)$ to prove $$\tan(A+B)=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$$
It seems like we cannot simply change $A+B$ to $A+(-B)$ to prove it? Any ideas?
| $$\begin{align*}\tan (A+B) = \frac{\sin (A+B)}{\cos (A+B)} &= \frac{\sin A\cos B + \sin B\cos A}{\cos A\cos B - \sin A\sin B} \div \frac{\cos A\cos B}{\cos A\cos B} \\ & = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A\sin B}{\cos A\cos B}} \\ & = \frac{\tan A + \tan B}{1 - \tan A \tan B}\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Using power series representation find an approximation to $\int_{0}^{0.1} \arctan(2x)dx$ Using power series representation find an approximation to
$$\int_{0}^{0.1} \arctan(2x)dx$$
This is my solution:
$$\frac{d}{dx}\arctan(2x)=\ \frac{2}{1+4x^2} $$
$$\int_{0}^{0.1}\frac{d}{dx}\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4... | We have:
$$ I=\int_{0}^{\frac{1}{10}}\arctan(2x)\,dx = \frac{1}{2}\int_{0}^{\frac{1}{5}}\arctan(x)\,dx \stackrel{IBP}{=} \frac{\arctan\frac{1}{5}}{10}-\frac{\log\left(1+\frac{1}{25}\right)}{4}\tag{1}$$
where, for any $x\in(0,1)$,
$$ \arctan(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)},\qquad \log(1+x^2)=\sum_{n\geq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find the derivative of $f(x) = \frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}$ Question:
Find the derivative of:
$$f(x) = \frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}$$
Attempted Solution
The most productive approach seems to be logarithmic differentiation:
$$\ln |f... | Take the natural log to get $$\ln f = \ln e^{x^2} + 2\ln \arcsin x + \ln x + \frac{1}{2} \ln \cos x - 6\ln \ln x - 2\ln \sin x$$
And then simplifying that a bit, i.e: $\ln e^{x^2} = x^2$ and differentiating gives $$f' = f\left(2x + \cdots - 2 \frac{\cos x}{\sin x}\right)$$
So that gives you the correct first and last... | {
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"url": "https://math.stackexchange.com/questions/1874198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum of f(x,y,z) let $x,y,z> -1$,then how can we find the minimum of this function
$$f(x,y,z)=\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}$$
I think that we can use a famous inquality but I can not find it.
So thanks
| For every $t\in\mathbb{R}$, we have
$$t\le\frac{1+t^2}{2}$$
thus
$$\frac{1+x^2}{1+y+z^2}\ge\frac{1+x^2}{1+z^2+\frac{1+y^2}{2}}$$
$$\frac{1+y^2}{1+z+x^2}\ge\frac{1+y^2}{1+x^2+\frac{1+z^2}{2}}$$
$$\frac{1+z^2}{1+x+y^2}\ge\frac{1+z^2}{1+y^2+\frac{1+x^2}{2}}$$
Set $a=1+x^2$ , $b=1+y^2$ and $c=1+z^2$, so
$$f(x,y,z)\ge \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Landau symbol and taylor series $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{1}{2}$$
I saw the calculation $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=$$...$$=\sqrt x (\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1)= $$ $$\sqrt x(1+\frac{1}{2}\sqrt{1/x+\sqrt{1/x^3}}+ \color{red}{\mathca... | Use formula $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$ in order to get:
$$L=\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}.$$
Now divide numerator and denumerator with $\sqrt{x+\sqrt{x}}$ and calculate the limit:
$$L=\lim_{x\rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Closed form for the integral $\int_0^\infty \frac{dx}{\sqrt{(x+a)(x+b)(x+c)(x+d)}}$ Let's consider the function defined by the integral:
$$R(a,b,c,d)=\int_0^\infty \frac{dx}{\sqrt{(x+a)(x+b)(x+c)(x+d)}}$$
I'm interested in the case $a,b,c,d \in \mathbb{R}^+$. Obviously, the function is symmetric in all four parameters... |
Hint:
Assume that $0<a<b<c<d$. Using the magical linear-fractional transformation,
$$\frac{\left(d-b\right)\left(x+a\right)}{\left(d-a\right)\left(x+b\right)}=t,$$
we obtain:
$$\begin{align}
R{\left(a,b,c,d\right)}
&=\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{\left(x+a\right)\left(x+b\right)\left(x+c\right)\left(x+d\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Sequence and Series bounded Problem: Let $ x_n $ be a sequence defined by $x_n = \displaystyle {\sum_{k=n+1}^{ 2n}\frac{1}{k}}$. Show that $ x_n$ converge.
I can bounded $ x_n <\frac{ n}{ n+1}$, but I can't show that $ x_n$ is increasing, thus would use that es increasing and bounded then is convergent.
| Simply notice that, since $\frac{1}{2n+1} > \frac{1}{2n+2}$, $$x_{n+1} - x_n = \sum_{k=n+2}^{2n+2} \frac{1}{k} - \sum_{k=n+1}^{2n} \frac{1}{k} = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} > \frac{2}{2n+2} - \frac{1}{n+1} = 0.$$
Hence the sequence is increasing. Then apply your bound $x_n \le \frac{n}{n+1} < 1$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}... | $$\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots$$
When we use the LOG function get (when $n$ is positive):
$$\ln\left(\left(2^n\right)^{\frac{1}{2^n}}\right)=\frac{n\ln(2)}{2^n}$$
And using, when... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
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Prove the inequalitiy
If $a, b, c$ are positive reals such that:
$\frac a {b + c + 1} + \frac b {a + c + 1} + \frac c {b + a + 1} \le
1$ then:
$\frac 1 {b + c + 1} + \frac 1 {a + c + 1} + \frac 1 {b + a + 1} \ge
1$
I tried using inequality of arithmetic and harmonic means, to no avail.
Any help is appreciated.
| .WLOG, $a <b <c$, then $\frac{1}{b+c} < \frac{1}{a+c} < \frac{1}{a+b}$
Note that $$\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} = (a+b+c+1)\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) - 3$$
Hence,
$$
(a+b+c+1)\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) \leq 4
$$
Further, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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solve $\tan{x} = \tan{3x}$ I'm asked to solve $\tan{x} = \tan{3x}$
Here's my attempt:
$$\tan{x} = \tan{3x}$$
$$\tan{x} = \tan{(x + 2x)}$$
$$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$
Recall the identity:
$$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$
So then we have:
$$\tan{x} = \frac{\tan{x} + \frac{2\tan{... | When you arrive at
$$
-\tan^2x \frac{2\tan x}{1-\tan^2x} = \frac{2\tan x}{1-\tan^2x}
$$
you can't multiply both sides by $\frac{1-\tan^2x}{2\tan x}$, because this operation is allowed only when the multiplier is nonzero.
You should rather move everything to the right-hand side and collect terms, getting
$$
0=\frac{2\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Evaluate $\int_\frac12 ^2 \frac1x\tan(x-\frac1x)dx $ $$\int\limits_\frac{1}{2} ^2 \frac{1}{x}\tan\left(x-\frac{1}{x}\right)\mathrm{d}x$$
I have tried substitution and by parts and it seems failed at all. Can anyone give me some hints?
| $$I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\tan\left(x - \frac{1}{x}\right)dx = \int_{2}^{\frac{1}{2}}u\cdot \frac{-1}{u^2}\tan\left(\frac{1}{u}-u\right)du = -\int_{\frac{1}{2}}^{2}\frac{1}{u}\tan\left(u - \frac{1}{u}\right)du = -I$$
Where the second step was obtained by letting $\displaystyle x = \frac{1}{u}, dx = \frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Elementary matrix operations I am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.
I am asked to find both the nullspace and the general solution of the following systems $A\tilde{x}=\tilde{b}$.
The first is
$$
A=
\begin{pmatrix}
1 & 1 & 0 & 2\\
2 & 1 & 1 &... | Your solution is correct. You can obtain the professor's solution from yours via the substitution $\beta=\alpha+2$. Both are valid answers, and unless there was some given format your answer was supposed to be in, I don't see why yours could be considered incorrect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help simplify using quadratic formula $\dfrac{4\pm\sqrt{28}}{2}=2\pm\sqrt7$
My Question is how did $\dfrac{4\pm\sqrt{28}}{2}$ become simplified as $2\pm\sqrt7$
Can you help me by explaining the steps clearly
:) Many Thanks
| Notice that $\sqrt{28}=\sqrt{4\times7}=\sqrt{4}\times\sqrt{7}=2\sqrt{7}$.
Hence $\dfrac{4\pm\sqrt{28}}{2}=\dfrac{4\pm2\sqrt{7}}{2}=\dfrac{2(2\pm\sqrt{7})}{2}=2\pm\sqrt{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to break $\frac{1}{z^2}$ into real and imaginary parts? $$ \frac{1}{(x+iy)^2}=\frac{1}{x^2+i2xy-y^2}=\frac{x^2}{(x^2+y^2)}-\frac{2ixy}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
So I thought I could just say:
$$ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
and
$$ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)}... | \begin{eqnarray}
f(z)=1/(z^2)=1/((x+iy)^2)&=&1/(x^2-y^2+2ixy) \newline
&=&(1/(x^2-y^2+2ixy))\cdot((x^2-y^2-2ixy)/(x^2- y^2-2ixy)) \newline
&=&((x^2-y^2-2ixy)/((x^2-y^2)^2+4x^2y^2)) \newline
&=&(x^2-y^2-2ixy)/(x^2+y^2)^2
\end{eqnar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How does $\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$? How does $$\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$$
| There's a trigonometric identity for $\arctan$: $\arctan(\alpha) \pm \arctan(\beta) = \arctan(\frac{\alpha \pm \beta}{1 \mp \alpha\beta})$.
In our case:
$$
\pi-\arctan(\frac{8}{x})-\arctan(\frac{13}{20-x})=\arctan(\frac{x}{8})+\arctan(\frac{20-x}{13})\\
\pi-\arctan(\frac{13}{20-x})-\arctan(\frac{20-x}{13})=\arctan(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution of inequality Question:
$x$ and $y$ are different integers and both $x$ and $y$ are greater than $0$. If $x^2-y^2<8$ and $x+y>3$, what is the maximum value of $x$?
My solution:
I used plugged in number method, and found for $x=4$ and $y=3$, $4^2-3^2=7$ and $4+3=7$, satisfies both equations. but, if $x=5$ and ... | As $x^2 - y^2 = (x-y)(x+y) < 8 $, and $x+y > 3$ we know $x-y<2$.
So, if there is a additional condititon $x \neq y$ (I will asume that, also that $x>y$), then $x=1+y$.
So $(y+1)^2 - y^2 < 8$, so $2y+1<8 $, and maximum of $y$ is $3$.
Without the condition $x \neq y$ the value of $y$ is unbounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Shortest distance between z axis and the line $x+y+2z=3, 2x+3y+4z+4=0$ z axis is : $x=0=y$
$x + B(y) = 0$ --> (1)
$x+y+2z-3+ A (2x+3y+4z+4 ) = 0$ --> (2)
These are two planes constructed through the two lines. We find the parallel planes so that it is easier to calculate the distance ... | Solving the given equations for $x$ and $y$ in function of $z$ gives $x=13-2z$, $y=-10$. It follows that $$\sqrt{x^2+y^2}=\sqrt{(13-2z)^2+100}$$ is minimal when $z={13\over2}$, and the minimal value is $10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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When is ${\large\int}\frac{dx}{\left(1+x^a\right)^a}$ an elementary function? Consider the following indefinite integral:
$$F_a(x)=\int\frac{dx}{\left(1+x^a\right)^a}.$$
Here $a\in\mathbb R$ is a parameter, and $x>0$ is a variable. For what values of the parameter $a$ the function $F_a(x)$ is an elementary function of ... | Using integration by parts leads to \begin{align}\int\frac{dx}{(1+x^a)^a}&=\frac x{(1+x^a)^a}+\int\frac{a^2x^adx}{(1+x^a)^{a+1}}\\
&=\frac x{(1+x^a)^a}+\int a^2\left\{\frac1{(1+x^a)^a}-\frac1{(1+x^a)^{a+1}}\right\}dx\\
&=\frac x{(1-a^2)(1+x^a)^a}-\frac{a^2}{1-a^2}\int\frac{dx}{(1+x^a)^{a+1}}
\end{align}
so you can also... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
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normalization of the nodal cubic Consider the nodal cubic $R = k[x,y]/(y^2 - x^3 + x^2)$, so $x,y$ satisfy $y^2 = x^3 - x^2$.
The idea behind normalization is to adjoin "$\frac{y}{x}$", since then in the fraction field $K$ of $R$, one has $\left(\frac{y}{x}\right)^2 = x-1$.
It's straightforward to show that the subring... | If $xt=y$ and $t^2-x+1=0$, then multiplying the second equation by $x^2$ and using the first one shows that $y^2-x^3+x^2=0$. This means that you forget about this last equation, and $S$ is $k[x,t,y]/(xt-y,t^2-x+1)$.
Now in this quotient, we have $x=t^2+1$ so we can remove the generator $x$, and it is isomorphic to $k... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Show that: $nf(\frac{x_1+\cdots+x_n}{n})+n\left(\frac{f(M)+f(m)}{2}-f(\frac{M+m}{2})\right) \ge f(x_1)+\cdots+f(x_n)$ I am looking for a proof of the problem as following:
Let $f(x)$ is a real continuous function that is strictly convex ($f''>0$) on $[m, M]$, let $m \le x_i \le M$, for $i=1,2,\ldots,n$ then show that:
... | The inequality does not hold in general. For a simple counterexample, consider the convex function:
$$
f(x) =
\begin{cases}
0 & \text{if $x \lt \frac{2}{3}$} \\
x - \frac{2}{3} & \text{if $x \ge \frac{2}{3}$}
\end{cases}
$$
and $m=0, M=1, n=3, x_1=0, x_2=x_3=1$.
Since $f(0)=f(\frac{1}{2})=f(\frac{2}{3})=0, f(1)=\fra... | {
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"source": "stackexchange",
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"answer_id": 0
} |
Loxodromics (curves with fixed angle with meridians) of a revolution surface I am trying to find the curves with a fixed angle $\phi$ with meridians for the revolution surface given by the revolution of the graph of $z=\frac{1}{x}$ around the $z$ axis. The surface is easily parametrized as
$$(x,\theta)\mapsto(x \cos \t... | After Futurologist's answer, you just need $$I=\int \frac{\sqrt{x^4+1}}{x^3}dx=\int \frac{\sqrt{x^4+1}}{x^4}x\,dx$$ So, define $$x^2=y\implies x=\sqrt y\implies dx=\frac{dy}{2 \sqrt{y}}$$ which make $$I=\frac 12\int \frac{\sqrt{y^2+1}}{ y^2}\,dy=\frac{1}{2} \left(\sinh ^{-1}(y)-\frac{\sqrt{y^2+1}}{y}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Question about Spivak's chapter 5 problem 3 part 3 Question:
Find a $\delta$ such that $|f(x)-L|<\epsilon$ for all x satisfying $0<|x-a|<\delta$
Relating equations to the problem;
$$f(x)=x^4+\frac{1}{x} \\a=1, L=2 $$
rewriting the equation as the definition of the limits
$$|x^4+\frac{1}{x}-2|<\epsilon$$ if and only if ... | There are many ways to solve this. We can do a bit of simplification at the beginning using the triangle inequality, and end up with simpler factors, as follows:
$$\begin{aligned}
\left|x^4 + \frac{1}{x} - 2\right| &= \left|(x^4 - 1) + \left(\frac{1}{x} - 1\right)\right| \\
&\leq |x^4 - 1| + \left|\frac{1}{x} - 1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $A$ and $B$ are two matrices such that $AB=B$ and $BA=A$ then how to show that $A^2+B^2$ equals $A+B$?
If $A$ and $B$ are two matrices such that $AB=B$ and $BA=A$ then $A^2+B^2$ equals ?
(a) $2AB$
(b) $2BA$
(c) $A+B$
(d) $AB$
I tried
$(A+B)^2=A^2+B^2+AB+BA$
or,$A^2+B^2=(A+B)^2-AB-BA$
$=(A+B)^2-A-B$
$
=(A+B)^2-(A... | More precisely, there is a positive integer $p$ s.t. $A,B$ are simultaneously similar to $A'=\begin{pmatrix}I_p&0\\0&0_{n-p}\end{pmatrix},B'=\begin{pmatrix}I_p&Q\\0&0_{n-p}\end{pmatrix}$ where $Q$ is an arbitrary $(p\times n-p)$ matrix.
Proof. Since $A^2=A$, there is $p$ s.t. $A$ is similar to $A'=\begin{pmatrix}I_p&0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to simplify this expression of square roots? If one simplifies the infinite nested radical
$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}$$
Does one get $$\sqrt x\frac{1+\sqrt5}2$$
Any help would be appreciated!
| You have
$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}$$
$$ = \sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
$$ = \sqrt{x}\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
The continued square root $\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$ is a well known form of $\phi = \frac{\sqrt{5}+1}{2}$ (see here) and so you end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Spivak's Calculus 4th ed: Absolute value definition p. 11 After defining the absolute value $|a|$ of $a$ as
$$
|a| =
\begin{cases}
a, & a\geq 0\\
-a, & a\leq 0,
\end{cases}
$$
the text provides some examples. While $|-3| = 3$ and $|7| = 7$ are simple enough (by direct application of the above definition), I do not und... | For the first absolute value, it is enough to compare the squares: $(1+\sqrt2)^2=3+2\sqrt 2>(\sqrt 3)^2$.
For the second absolute value, $3+2\sqrt 2<3+4<10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a Differential equation using eigenvectors I want to solve the following differential equation using eigenvectors:
$x'=y$
$y' = 2x-5y$
Or using matrices:
$\begin{pmatrix}
x' \\
y'
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
2 & -5
\end{pmatrix} \begin{pmatrix}
x \\
y
\end{pmatrix}$
With the initial conditio... | You have done most of the work already. $A's$ eigenvalues are:
$$\lambda_{1, 2} = \dfrac{-5\pm\sqrt{33}}{2}$$
We now have to find the eigenvectors and solve $[A - \lambda_i I]v_i = 0$. So for $\lambda_{1,2}$, we have:
$$[A - \lambda_{1,2} I]v_{1,2} = [A - \dfrac{1}{2} (-5 \pm \sqrt{33}) I]v_{1,2} = \begin{pmatrix}
- ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The square of any odd integer is odd. Suppose that $n$ is an odd integer. Then $n = 2k + 1$ for some integer $k$.
Hence $n^2=(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1$.
Since $k$ is an integer, $2k^2 + 2k$ is an integer. Thus $n^2 = 2k' + 1$ for some integer $k'$.
Therefore $n^2$ is odd.
Is this correct?
| you can write more generally that the product of 2 odd integers is odd
let $a = 2k+1$ and $b = 2p+1$
then $ab = (2k+1)(2p+1) = 2(2kp+k+p)+1 $ which is the general format of odd numbers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Prove this inequality $\sum\limits_\text{cyc}\sqrt{1-xy}\ge 2$ Let $x,y,z\ge 0$, and $x+y+z=2$, show that
$$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2.$$
Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$
or
$$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$
| WLOG, assume that $x \ge y \ge z$.
We have
$$\sqrt{1 - xy} \ge \sqrt{1 - (x + y)^2/4}.$$
Also, we have
$$\sqrt{1 - yz} + \sqrt{1 - zx} \ge 1 + \sqrt{1 - yz - zx}.$$
(Note: Squaring both sides, it is just
$2\sqrt{1 - yz - zx + z^2xy} \ge 2\sqrt{1 - yz - zx}$.)
Thus, it suffices to prove that
$$\sqrt{1 - (x + y)^2/4} + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Finding the maximum value of the sum of the product of sets of 3 variables given their sum Given $x_{1}+x_{2}+...+x_{2014}=2014$, what is the maximum value of: $x_{1}.x_{2}.x_{3}+x_{2}.x_{3}.x_{4}+x_{3}.x_{4}.x_{5}+...+x_{2012}.x_{2013}.x_{2014}$.
It looks like an application of AM-GM inequality on terms obtained after... | $$\begin{align}
\text{Objective} &= x_{1}.x_{2}.x_{3} + x_{2}.x_{3}.x_{4}+x_{3}.x_{4}.x_{5}+...+x_{2012}.x_{2013}.x_{2014} \\
&\leqslant (x_1 + x_4 + \dots+x_{2014})(x_2 + x_5 + \dots+x_{2012})(x_3 + x_6 + \dots+x_{2013}) \\
&\leqslant \left(\frac{x_1 + x_2 + x_3 + \dots + x_{2014}}{3} \right)^3 =\left( \frac{2014}3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"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.