Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Help with C is Euler's constant and $\Gamma(0)=\infty$ in paper I am referring to a paper by S. Nadarajah & S. Kotz.
The notation is simple enough to understand, however i having trouble with $C$ is Euler’s constant and $\Gamma(0)=\infty$
by equation (2.3) I have
$$F(z)= \displaystyle\lambda\int_{0}^{\infty}y^{-1-1}ex... | Notes:
*
*The paper referenced is unclear about the parameters $p$ and how $z$ is defined. It would seem that somehow $p \sim \Gamma(0)$. This would work if another parameter is defined, say $\alpha$, such that $p = \alpha \, \Gamma(0) \to \beta$ where $\beta$ is a finite quantity.
*It is of interest to note that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Calculating limit-sequences So i have this limit to calculate:
$\lim_{n\to\infty}\frac{[x] + [2x]+ ... +[nx]}{n^2}$
And i tried to make some boundaries and got this two limits:
$\lim_{n\to\infty}\frac{[x] + [x]+ ... +[x]}{n^2}$
$\lim_{n\to\infty}\frac{[nx] + [nx]+ ... +[nx]}{n^2}$
But i am not sure if it's correct,whil... | $$x-1\leqslant \left \lfloor x \right \rfloor< x\\2x-1\leqslant \left \lfloor 2x \right \rfloor< 2x\\3x-1\leqslant \left \lfloor 3x \right \rfloor< 3x\\...\\\\nx-1\leqslant \left \lfloor nx \right \rfloor< nx$$
now sum of them
$$(x+2x+3x+..+nx)-(1+1+1..1)\leq \left \lfloor x \right \rfloor+\left \lfloor 2x \right \rf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How does this algebraic trick regarding partial fraction works? Suppose I have to evaluate the integral $$\int \frac{x}{(x-1)(2x+1)(x+3)} \, dx $$
I write it as $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$
where $a_1$, $a_2$, $a_3$ are constants. once I have foud relationaships between them I solve for the $... | Notice, the partial fractions $$\frac{x}{(x-1)(2x+1)(x+3)}=\frac{A}{x-1}+\frac{B}{2x+1}+\frac{C}{x+3}$$ On solving for $A, B, C$, we get $A=\frac{1}{12}$, $B=\frac{2}{15}$ & $C=-\frac{3}{20}$
$$=\frac{1}{12(x-1)}+\frac{2}{15(2x+1)}-\frac{3}{20(x+3)}$$
Hence, we have $$\int \frac{x}{(x-1)(2x+1)(x+3)}=\int\frac{1}{12(x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding $\iint_S {z \:ds}$ for some $S$ $$\iint_S {z \:ds}$$
In this double integral above, $S$ is the part of a sphere, $x^2+y^2+z^2=1$, which lies above the cone, $z=\sqrt{x^2+y^2}$. How can I calculate the above double integral.
Can someone help me to solve this? I got $\frac{\pi}{3\sqrt{2}}$ as my answer. Can someo... | This problem is actually best done in cylindrical coordinates. Here when we let $x = r\cos\theta$ and $y = r\sin\theta$ we can rewrite our $z$-limits as $z = r$ and $r^2 +z^2 = 1$. Note that when $z =1$ we have $r = 0$, and similarly when $z = r$ we get that $r^2 + r^2 = 1$ implying that $r = \frac{1}{\sqrt{2}}$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sqrt{6+4\sqrt{2}}-\sqrt{2}$ is rational using the rational zeros theorem Let $r=\sqrt{6+4\sqrt{2}}-\sqrt{2}$, then $r+\sqrt{2}=\sqrt{6+4\sqrt{2}}$.
Squaring both sides, we get $$r^2+2r\sqrt{2}+2=6+4\sqrt{2}$$ which is the same as $r^2-4=2\sqrt{2}$.
Squaring both sides again, we get $r^4-8r^2+16=8$ or $$r^4-... | \begin{align*}r=\sqrt{6+4\sqrt 2}-\sqrt 2&\Rightarrow r^2+2r\sqrt 2+2=6+4\sqrt2\iff r^2-4=2\sqrt 2(2-r)\\
&\Rightarrow(r^2-4)^2=8(r-2)^2\iff(r-2)^2((r+2)^2-8)=0\\
\end{align*}
The only rational root of this polynomial is $r=2$. We have to explain why $r$ can't be one of the irrational roots $\;-2(1\pm\sqrt2)$.
Anyway i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solution to equation $4n+1 = (2k+1)^2-(2u)^2$ over the natural numbers(!) Let $n \in \mathbb{N}$ be given, then I want to know if there are $k ,u \in \mathbb{N}$ (with $k \neq n, u \neq n$) such that $4n+1 = (2k+1)^2-(2u)^2$ holds.
What makes it difficult for me is the fact that I am looking for solutions over the nat... | Given
$$
4 n + 1 = \Big( 2 k + 1 \Big)^2 - \Big( 2 u \Big)^2.
$$
Set
$$
k = u + a,
$$
then we get
$$
n = a^2 + a + \big( 2 a + 1 ) u.
$$
The case $a=0$ is the solution
$$
n = u = k,
$$
which is to be excluded.
For every $a > 0$ we have a linear relation between $n$ and $u$.
$$
\begin{array}{l|ll}
a & n\\
\hlin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understa... | No induction is necessary: it's a matter of a telescoping sum, if you write:
$$\frac1{k(k+2)}=\frac12\Bigl(\frac 1k-\frac1{k+2}\Bigr)$$
Apply this decomposition to the above sum:
\begin{align*}
\sum_{k=1}^n\frac1{k(k+2)}&=\frac12\sum_{k=1}^n\Bigl(\frac 1k-\frac1{k+2}\Bigr)\\
&=\frac12\Bigl(1\color{red}{-\frac13}+\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Rewriting approximated terms The following data are inferred from a presentation slide, so I do not much info.
Using linear approximation and log rules $\sqrt x $ can be rewritten as $\frac{x+1}{2}$, where $(1 \leq x \lt 2) $ ... (1)
I understood the Eq. (1).
Now the slide says, in the range $(2 \leq x \lt 4)$,
$ \... | Probably there is a typo in
$ \sqrt x \approx 2^{\frac{(1+x-1)}{2}}$
because it is not a correct approximate in the range $2 \leq x < 4$
Case of $x=2 $ then $ \sqrt 2 \approx 2^{\frac{(1+2-1)}{2}}=2$ is false.
Case of $x=4 $ then $ \sqrt 4 \approx 2^{\frac{(1+4-1)}{2}}=4$ is false.
If we use the Taylor's series :
$$f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x^3+y^3=72$ and $xy=8$ then find the value of $x-y$. I recently came across a question,
If $x^3+y^3=72$ and $xy=8$ then find the value of $(x-y)$.
By trial and error I found that $x=4$ and $y=2$ satisfies both the conditions. But in general how can I solve it analytically? I tried using $a^3+b^3=(a+b)(a^2+b^2-ab)... | Given $$x^3+y^3=72$$
$$xy=8$$
Notice,
$$(x+y)^3=x^3+y^3+3xy(x+y)$$ $$(x+y)^3=72+3(8)(x+y)$$
$$(x+y)^3-24(x+y)-72=0$$
Above is the cubic equation in terms of $(x+y)$ which has one real root $6$ Hence, we get $$x+y=6$$
Now, $$(x-y)^2=(x+y)^2-4xy$$
$$(x-y)=\pm\sqrt{(x+y)^2-4xy}$$
$$x-y=\pm\sqrt{(6)^2-4\times8 }$$
$$=\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Solutions for differential equation The motion of a particle moving along the x-axis obeys the differential equation:
$ \ddot x - 4 \dot x + 4x =-t^2 $
Find the solution for $ x(t)$, given $ x(0)=0 $ and $ \dot x (0) = 0 $.
Can this be solved by treating it as a second-order non-homogenous differential equation? If so... | Using constant coefficient method we solve for two solutions $x_h$ and $x_p$ where the general solution is $x_g=x_h+x_p$
Solving for $x_h$ implies that we solve the homogeneous system $x'' -4x'+ 4x=0$ which has the characteristic equation
$$\lambda^2 -4\lambda +4 = 0 \implies ( \lambda-2)^2=0 \implies \lambda_{1,2}=2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Why is this sum zero? I have been looking at the following sum (for any positive integer $n$)
$$\left(1-\frac{1^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3^2}{n}\right) + \ldots $$
Note that the $i$th term in the sum has $i$... | This might get you started:
$$\begin{align}
&\left(1-{1^2\over n}\right)+\left(1-{1\over n}\right)\left(1-{2^2\over n}\right)+\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(1-{3^2\over n}\right)+\cdots\\
&\quad=\left(1-{1\over n}\right)\left(1+\left(1-{2^2\over n}\right)+\left(1-{2\over n}\right)\left(1-{3^2\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Proof of the identity: $c\sin \frac{A-B}{2} \equiv (a-b) \cos \frac{C}{2}$ Trigs is not my strongest apparently...
I need to prove $c\sin \frac{A-B}{2} = (a-b) \cos \frac{C}{2}$ for a general triangle $ABC$.
Here is what I do, or rather, here is how I fail at proving it:
$\cos \frac{C}{2} \equiv \sin \frac{A+B}{2}$, s... | Using sine law, $$\dfrac{a-b}c=\dfrac{\sin A-\sin B}{\sin C}$$
Using Prosthaphaeresis & Double Angle Formula,
$$\dfrac{\sin A-\sin B}{\sin C}=\dfrac{2\sin\dfrac{A-B}2\cos\dfrac{A+B}2}{2\sin\dfrac C2\cos\dfrac C2}$$
Now $\dfrac{A+B}2=\dfrac\pi2-\dfrac C2\implies\cos\dfrac{A+B}2=?$
Finally, if $\sin\dfrac C2=0,\dfrac C2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\sum_{n=1}^{\infty}\ln^2 \!\left(1+\frac1{2n}\right) \!\ln^2\!\left(1+\frac1{2n+1}\right)$ Based upon Oloa's question here Evaluating $\sum_{n \geq 1}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)$ I was thinking if we possibly can get a nice way to evaluate the series
$$\sum_{n=1}^{\i... | It can be shown that:
\begin{align}
\sum_{n=1}^{\infty} \ln^{2}\left(1+\frac{1}{2n}\right) \, \ln^{2}\left(1 + \frac{1}{2n+1}\right) = - \frac{\ln^{4} 2}{2} + 2 \, \sum_{n=1}^{\infty} \ln^{2}\left(1 + \frac{1}{n}\right) \, \ln\left(1+\frac{1}{2n}\right) \, \ln\left(1 + \frac{1}{2n+1}\right).
\end{align}
The remaining s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Derivation of Series Expansions
I have the series expansion for
$$I=\frac{1}{\sqrt{1-x^2}}$$
$$|x|\lt1$$
which is
$$1+\sum_{k=1}^\infty\frac{1.3.5.7......(2k-1)x^{2k}}{2^kk!}$$
or so I am assured. Given the above what is the series expansion for
$$\int_0^x\frac{dt}{\sqrt{1-t^2}}$$
can I simply substitut... | You need to use the following theorem:
A power series can be integrated term by term in the interior of region of convergence.
If we restrict to real variables then a power series of the from $$f(x) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots = \sum_{n = 0}^{\infty}a_{n}x^{n}\tag{1}$$ has region of convergence as an interva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove the trigonometric identity $\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1 = \sec x \csc x$ I am doing some practice questions for a Math class and I was told that similar questions would be in the exam. So I need to learn this but I have no idea where to even start with this question:
$$\frac{\c... | Start from the left-hand side:
\begin{align}
\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1
&=\frac{\dfrac{\cos x}{\sin x}}{1-\dfrac{\sin x}{\cos x}}
+\frac{\dfrac{\sin x}{\cos x}}{1-\dfrac{\cos x}{\sin x}}-1\\
&=\frac{\cos^2x}{\sin x(\cos x-\sin x)}
-\frac{\sin^2x}{\cos x(\cos x-\sin x)}-1\\
&=\frac{\cos^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find inverse of 15 modulo 88. Here the question: Find an inverse $a$ for $15$ modulo $88$ so that $0 \le a \le 87$; that is, find an integer $a \in \{0, 1, ..., 87\}$ so that $15a \equiv1$ (mod 88).
Here is my attempt to answer:
Find using the Euclidean Algorithm, we need to find $\gcd(88, 15)$, that must equal to $1$ ... | $x$ is an inverse of $15 \pmod{88}$ if and only if $x=88n-41$ for some (any) integer $n$, because it is necessary and sufficient that $15(-41+x)$ is divisible by $88$, but this is equivalent to $(-41+x)$ being divisible by $88$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Deducing $\sum_{r=1}^{n}r$ from sine summation formula We know the famous formula
$$\sum_{r=1}^{n}\sin r\theta=\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\sin\frac{(n+1)\theta}{2}\ .$$
I have come across a question that use the above result to find $\sum_{r=1}^{n}r$. I thought but could not deduce how this can be done.... | $\sum_{r=1}^{n}\sin r\theta
=\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\sin\frac{(n+1)\theta}{2}
$
My first thought on seeing this
is to somehow use
$\lim_{x \to 0} \frac{\sin x}{x}
= 1
$.
Let's divide both sides by
$\theta$:
$\sum_{r=1}^{n}\frac{\sin r\theta}{\theta}
=\frac{\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluation of trignometric limit I want to find the following limit without L'Hospital.
$ \lim_{x \to \frac{3π}{4}} \frac{1+(\tan x)^{\frac13}}{1-2(\cos x)^2}$
Maybe I should try to get rid of the radical.
| As the OP suggested openness to asymptotic analysis, we proceed accordingly.
First, we note the expansion of the numerator is given by
$$\begin{align}
1+\tan^{1/3} x&=1+\left(-1+2(x-3\pi/4)+O(x-3\pi/4)^2\right)^{1/3}\\\\
&=\frac23 (x-3\pi/4)+O(x-3\pi/4)^2 \tag 1
\end{align}$$
whereas the expansion of the denominator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $\frac{dy}{dx}\frac{dx}{dy} = 1$, does $\frac{d^2 y}{dx^2} \frac{d^2 x}{dy^2} = 1$? I know $\frac{dy}{dx}\frac{dx}{dy} = 1$ because the chain rule says $1 = \frac{dy}{dy} = \frac{dy}{dx}\frac{dx}{dy}$. But does $\frac{d^2 y}{dx^2} \frac{d^2 x}{dy^2} = 1$? Or would that be too good to be true?
| What is true is that (assuming always that $y$ is a one-to-one function of $x$ on some interval)
$$ \eqalign{\dfrac{d^2 x}{dy^2} &= \dfrac{d}{dy} \dfrac{dx}{dy} = \dfrac{dx}{dy} \dfrac{d}{dx} \left(\dfrac{dy}{dx}\right)^{-1}\cr
&= - \left(\dfrac{dy}{dx}\right)^{-3} \dfrac{d^2 y}{dx^2} }$$
In order to have $\dfrac{d^2 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve $\cos \frac{4x}{3}=\cos x+1$
Solve the equation \begin{equation} \cos \frac{4x}{3}=\cos x+1\tag 1\end{equation}
I had tried by taking $\cos\dfrac x3=t$ and from this we have $\displaystyle\cos\frac{4x}3=2\left(2t^2-1\right)^2-1; \cos x=4t^3-3t$
$(1) \iff t\left(8t^3-4t^2-8t+3\right)=0$
But I can't solve $8t^3-4... | HINT:
Notice, we have $$\cos \frac{4x}{3}=\cos x+1$$ Let, $x=3t$ , we get
$$\cos 4t=\cos 3t+1$$ $$2\cos^2 2t-1=4\cos^3 t-3\cos t+1$$
$$2(2\cos^2 t-1)^2-4\cos^3 t+3\cos t-2=0$$
$$8\cos^4 t-8\cos^2t+2-4\cos^3 t+3\cos t-2=0$$
$$\cos t(8\cos^3 t-4\cos^2t+2-8\cos t+3)=0$$
$$\iff \cos t=0\iff t=\frac{\pi}{2}\iff x=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the number of natural number solutions of $a+2b+c=100$
Find the number of natural number solutions of $a+2b+c=100$
I remember something like stars and bars if the equation I change to $a+b_{1}+b_{2}+c=100$
then i get $\dbinom{99}{3}$ ways.
If the equation is like $a_{1}+a_{2}+a_{3}+a_{4}=m$ I can use $\dbinom{m-... | I will assume that by the natural numbers that you mean the positive integers (as opposed to the nonnegative integers).
Observe that $a + c = 100 - 2b$ is an even number. Thus, $a$ and $c$ must have the same parity.
Case 1: The numbers $a$ and $c$ are both even. Let $a = 2u$; let $c = 2v$. Then $u, v \in \mathbb{N}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show using logarithms that the first equation can be transformed into the second. Show using logarithms that if $y^k = (1-k)zx^k(a)^{-1}$ then $y = (1-k)^{(1/k)}z^{(1/k)}x(a)^{(-1/k)}$.
| Take logarithm of both sides of the equation $\,y^k = \left(1-k\right)\,z\,x^k\,\left(a\right)^{-1},\,$ get
$$
\begin{aligned}
y^k = \left(1-k\right)\,z\,x^k\,a^{-1}
&\implies
\ln\left( y^k \right) = \ln\left(\left(1-k\right)\,z\,x^k\,\left(a\right)^{-1} \right)
\\
&\implies
k\ln y = \ln\left(1-k \right) + \ln z ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve $\log_9 (a) + \log_{12} (b) = \log_{16} (a+b)$ for $a/b$ The question:
$$\log_9 (a) + \log_{12} (b) = \log_{16} (a+b)$$
solve for $a/b$.
It gives hints:
put it all in terms of x.
$$9^x=a$$
$$12^x=b$$
$$16^x=a+b$$
Now prove that:
$b^2=a(a+b)$ I did and it does equal.
Then it tells me to divide both sides of $b^2=a... | Hint : You have the following equality
$$1 = \frac{a}{b} \times \frac{a+b}{b}, $$
and observe that $\frac{a+b}{b} = \frac{a}{b}+\frac{b}{b} = \frac{a}{b} + 1$, now we define $Y=\frac{a}{b}$, to get that $Y$ is a solution of
$$1 = Y\times(Y+1)\qquad \Leftrightarrow\qquad Y^2+Y-1=0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator
Prove that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator.
I did in the following way. Are there other ways?
Proof : Let $f(x)=e\pi\frac{\ln x}{x}$. Then,
$$e^{\pi}-{\pi}^e=e^{f(e)}-{e}^{f(\pi)}\tag1$$
Now,
$$f'(x)=\frac{e\pi(1-\ln x)}{x^2},\quad f''... | Consider $f(x,y) = x^y - y^x$ where $x, y \approx 3$. Increasing $x$ should bring $f$ down and increasing $y$ should bring $f$ up. In general, without any real knowledge of $f$:
$$ 1 > f(x+ \epsilon_1, y + \epsilon_2) \approx
f(x, y ) + \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\parti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
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What is the probability that a natural number is a sum of two squares? Some natural numbers can be expressed as a sum of two squares:
$$2=1^2+1^2$$
$$25=3^2+4^2$$
$$50=7^2+1^2$$
If one chooses a random natural number, what would be the probability that that number is a sum of two squares? Is it zero?
I read about Lagra... | Consider the function $r: \mathbb{N} \to \mathbb{C}$:
$$
r(n) = \begin{cases}
1 &n \text{ is the sum of two squares (0 is a square)} \\
0 &\text{otherwise}.
\end{cases}
$$
One way to define the "probability that a random integer is the sum of two squares" would be to consider the distribution on the integers where $n$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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If $a,b,c,d,e,f$ are non negative real numbers such that $a+b+c+d+e+f=1$, then find maximum value of $ab+bc+cd+de+ef$ $(a+b+c+d+e+f)^2=$ sum of square of each number (X)+ $2($ sum of product of two numbers (Y) $)$
$ab+bc+cd+de+ef \le Y$ since all are positive.
Therefore $1\ge X+(ab+bc+cd+de+ef)$
Edit: From AM GM inequa... | I will give you an example with 4 variables to show how to solve such problems:
$a+b+c+d=1,b=1-(a+c)-d, ab+bc+cd=b(a+c)+cd=(1-(a+c))(a+c)-d(a+c)+cd=(1-(a+c))(a+c)-ad \le \left(\dfrac{(1-(a+c))+(a+c)}{2}\right)^2-ad \le \dfrac{1}{4}$
when $d=0,a+c=\dfrac{1}{2},b=\dfrac{1}{2}$ you will get maximum value.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why doesn't quadratic formula lead to a the correct factored form of the original equation? Applying the quadratic formula to $2x^2-3x+1$ we have
\begin{eqnarray*}
a&=&2 \\
b&=&-3 \\
c&=&1
\end{eqnarray*}
which gives me two roots:
\begin{eqnarray*}
x_1&=&1 \\
x_2&=&\tfrac{1}{2}
\end{eqnarray*}
Therefore you can re-wr... | How can $x=\frac{1}{2}$ and $x=1$ both be solutions to $x^2 - \tfrac{3}{2}x+\tfrac{1}{2} = 0$ and $2x^2-3x+1=0$?
The simple fact is that $2x^2-3x+1 \equiv 2\!\left(x^2 - \tfrac{3}{2}x+\tfrac{1}{2}\right)$, and so
$2x^2-3x+1 = 0$ if, and only if, $x^2 - \tfrac{3}{2}x+\tfrac{1}{2}=0$. They have the same solutions.
By a s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I solve $x^4-3x^2+2=0$? How do I solve $x^4-3x^2+2=0$ ?
I would appreciate some kind of hint here. I have no clue how to start this problem.
| We can complete squares as follows:
\begin{align*}
x^4-3x^2+2&=0\\
x^4-3x^2&=-2\\
x^4-3x^2+\color{green}{\frac{9}{4}}&=-2+\color{green}{\frac{9}{4}}\\
\left(x^2-\frac{3}{2}\right)^2&=\frac{1}{4}\\
\left(x^2-\frac{3}{2}\right)^2-\frac{1}{4}&=0\\
\end{align*}
Now we can factor this difference of squares:
\begin{align*}
\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Integrate $\tan^2(\frac{\pi}{12} \cdot y)$ Integrate $\tan^2(\frac{\pi}{12} \cdot y)$
Wolfram gives the answer:
$$\frac{12 \tan(\frac{\pi y}{12})}{\pi} -y + \text{constant}$$
But why is the value of $\tan^2$ not getting differentiated?
According to this rule, the answer should be:
$$\frac{\tan(\frac{\pi}{12} \cdot y)}{... | Let $\frac{\pi}{12}\cdot y=t\implies \frac{\pi}{12}dy=dt\iff dy=\frac{12}{\pi}dt$
We have $$\int \tan^2\left(\frac{\pi}{12}\cdot y\right)dy=\int \tan^2 (t) \frac{12}{\pi}dt$$ $$=\frac{12}{\pi}\int \tan^2(t)dt$$
$$=\frac{12}{\pi}\int (\sec^2(t)-1)dt$$
$$=\frac{12}{\pi}\int\sec^2(t)dt-\frac{12}{\pi}\int dt$$
$$=\frac{12}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a matrix that satisfy a givven inner product
Let $V=M_2(\mathbb{R})$
with $\left\langle A,B\right\rangle:=\text{tr}\left(AB^*\right)$ and let $A=\left(\begin{array}{cc} 1 & -3 \\ -2 & 2 \end{array}\right)$ find $B\in M_2(\mathbb{R})$ such that $B$ is orthonormal and orthogonal to $A$
So we need the both ... | $||B||=1\implies a^2+b^2+c^2+d^2=1$. $A\perp B=0\implies a-3b-2c+2d=0$. Pick $a,b,c,d$ in such a way that $a=\sqrt{0.5}\cos\phi,b=\sqrt{0.5}\sin\phi,c=\sqrt{0.5}\cos\theta,d=\sqrt{0.5}\sin\theta$ so that the $1$-st equality is taken care of. If we pick $\theta,\phi$ in such a way that $a=3b$ and $c=d$ we're done. $\phi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the area of the circle that falls between the circle $x^2+y^2=5$ and the lines $x^2-4y^2+6x+9=0$
Find the area of the circle that falls between the circle $x^2+y^2=5$ and the lines $x^2-4y^2+6x+9=0$.
I tried to solve this question. The lines are $x-2y+3=0$ and $x+2y+3=0$ which intersect at $(-3,0)$ and the circl... | The line of equation $x-2y+3=0$ intersects the circle of equation $x^2+y^2=5$ at
$A(-\frac{11}{5},-\frac{2}{5})$ and $B(1,2)$. It follows that
$AB=\frac{8}{\sqrt{5}}$.
$\qquad\qquad\qquad$
Using the notation of the figure we have $\sin\theta=\frac{AB}{2OA}=\frac{4}{5}$. That is $\theta=\arcsin\frac{4}{5}$. Moreover, $O... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find roots of equation $(x^2+1)\cdot \arccos\left(\frac{2x}{1+x^2}\right)+2x\cdot \mathrm{sgn}(x^2-1)=0$
Find roots of equation $(x^2+1)\cdot \arccos\left(\frac{2x}{1+x^2}\right)+2x\cdot \mathrm{sgn}(x^2-1)=0$
One root is $x=1$ (checking functions $\arccos$ and $\mathrm{sgn}$).
Second root is $x=0.442$.
How to find t... | Taking $x=\tan \theta$ and noting that $2x/(1+x^2)=\sin 2\theta$ will simplify the equation to $$\pi/2-2\theta+\sin 2\theta\cdot\mathrm{sgn}(\tan^2\theta-1)=0\\\implies f(\theta)+g(\theta)=0$$ where $$f(\theta)=\pi/2-2\theta,\ g(\theta)=\sin 2\theta\cdot \mathrm{sgn}(\tan^2\theta-1)=\left\{\begin{array}
&
\sin 2\the... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible? Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible?
What is the best/simplest/elementary way to approach this? Any solutions or hints are greatly appreciated.
| Because the modulus is so small, we can do this the hard way.
All arithmetic will be done in $\mathbb Z_7.$
$p(x)=6x^4+3x^3+6x^2+2x+5$
$-p(x) = x^4 + 4x^3 + x^2 + 5x + 2$
$-p(x-1) = x^4 + 2x^2 + 4x + 2$
Now we suppose that
\begin{align}
-p(x-1)
&= (x^2 + ax + b)(x^2 + cx + d)\\
&= x^4 + (a+c)x^3 + (ac + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving an integral with substitution method I want a hint for the following integral:
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}}$$
where $a$ is a real constant.
In fact,
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}} = -\frac{\sqrt{1- a^{2}+x^{2}}}{x(1- a^{2})}$$
why?
| Here's one approach:
$$ \frac{1}{x^2 \sqrt{1-a^2 + x^2}} = \frac{1}{x^3\sqrt{\frac{x^2-a^2+1}{x^2}}}$$
$$ =\frac{1}{x^3\sqrt{\frac{1-a^2}{x^2} + 1}} $$
Let $\theta = \frac{1-a^2}{x^2} + 1$
$$ \frac{d\theta}{dx} = \frac{-2(1-a^2)}{x^3} \ \implies \ dx = \frac{-x^3}{2(1-a^2)} \ d\theta $$
The integral then reduces to:
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many ways can a selection be done of $5$ letters?
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$.
$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $
Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way
Number of... | If Order is Unimportant
The number of ways to choose $5$ letters (if their order is unimportant) is the coefficient of $x^5$ in
$$
\begin{align}
&\small\overbrace{(1+x)\vphantom{x^2}}^{1\text{ E}}
\overbrace{\left(1+x+x^2\right)}^{2\text{ D's}}
\overbrace{\left(1+x+x^2+x^3\right)}^{3\text{ C's}}
\overbrace{\left(1+x+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
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How to solve this pde equation: $(p^2 + q^2)y = qz$ My attempt at solution
$p^2 y = qz - q^2 = a... (I)$
This equation is of the form $f_1(x,p) = f_2(y,q)$.
Its solution is given by $dz=pdx + qdy$, upon integrating this we get value of $z$.
From (I)
$-yq_2 + zq - a = 0$, solving the quadratic equation for $q$, we get
... | $let\ f(x,y,z,p,q)=(p^2+q^2)y-qz=0 \hspace{1cm} \to (1)$
Now,
$f_x=0$
$f_y=p^2+q^2$
$f_z=-q$
$f_p=2py$
$f_q=2qy-z$
The auxillary equations of the given DE are
$\dfrac{dp}{f_x+pf_z}=\dfrac{dq}{f_y+q.f_z}=\dfrac{dz}{-p.f_p-q.f_q}=\dfrac{dx}{-f_p}=\dfrac{dy}{-f_q}$
$\dfrac{dp}{-pq}=\dfrac{dq}{p^2+q^2-q^2}=\dfrac{dz}{-p(2p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Simplifying $\tan100^{\circ}+4\sin100^{\circ}$ The answer is $-\sqrt3$.
I was wondering if this is just a coincidence?
Also, is there a relation between $$\tan(100^{\circ}+20^{\circ})=\frac{\tan100^{\circ}+\tan20^{\circ}}{1-\tan100^{\circ}.\tan20^{\circ}}=-\sqrt3$$ and the given expression? Or is there a more elegant... | Notice, $$\tan 100^\circ+4\sin 100^\circ$$
$$=\frac{\sin 100^\circ}{\cos 100^\circ}+4\sin 100^\circ$$
$$=\frac{\sin 100^\circ+4\sin 100^\circ\cos 100^\circ}{\cos 100^\circ}$$
$$=\frac{\sin 100^\circ+2\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{(\sin 200^\circ+\sin100^\circ)+\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Coefficient of the generating function $G(z)=\frac{1}{1-z-z^2-z^3-z^4}$ I am seeking the coefficient $a_n$ of the generating function
$$G(z)=\sum_{k\geq 0} a_k z^k = \frac{1}{1-z-z^2-z^3-z^4}$$
The combinatorial background of this question is to solve the recurrence
$$a_n=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4},\qquad (a_0,a_... | Let $\left\{z_i\right\}_{i=1}^4$ be the roots of the polynomial in the denominator. Note that:
\begin{eqnarray}
\frac{1}{1-z-z^2-z^3-z^4} = \frac{-1}{\prod\limits_{i=1}^4 (z-z_i)} =
\sum\limits_{p=1}^4 \frac{-1}{z-z_p} \cdot \left(\prod\limits_{q\neq p} \frac{1}{z_p-z_q}\right)
\end{eqnarray}
Now clearly :
\begin{equa... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this system of exponential equations? Solve the following system of equations ($x,y \in \Bbb R$):
$$\begin{cases}
3^{x+3y-2} + 6\cdot 3^{y^2+4x-2} &=5^{5y-3x} + 2\cdot 3^{y^2-2y+1}\\
1+2\sqrt{x+y-1} &=3\sqrt[3]{3y-2x}.
\end{cases}$$
I think about it but I still have no solution... :(
Since the second equ... | The curves corresponding to the functions $y(x)$ , computed by numerical calculus, are drawn on the figure below.
Since they dosn't intersect, the system of equations has no real solution.
I guess that there is a mistake in the wording of the question or in copying the equations.
The more $x, y$ are large, the more ... | {
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$a_1 =2$ and $a_{n+1}= \frac{2a_n +3}{a_n +2}$ the recursive sequence , converges?, and if yes, show it to which converges I do not know what to do, I try showing the first elements to see if they were behaving in some way, but no
| Another less known way of solving this particular form of recurrence relations.
The key is transform the non-linear recurrence to a linear one.
Notice $$\begin{bmatrix}a_{n+1}\\1\end{bmatrix}
\propto \begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}
\begin{bmatrix}a_{n}\\1\end{bmatrix}, \forall n > 1
\quad\text{ and }\quad
\beg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the series $\sum\frac{(-1)^n cos(3n)}{n^2+n}$ converge absolutely? $\sum|\frac{(-1)^n \cos(3n)}{n^2+n}| \le \sum\frac{1}{n^2+n}$
since $-1 \le cos(3n) \le 1$
$\sum\frac{1}{n^2+n} = \sum\frac{1}{n(n+1)} = \sum(\frac{1}{n} - \frac{1}{n+1})$
$\int(\frac{1}{x} - \frac{1}{x+1}) dx = \log|x| + \log|x+1| + C$
which diver... | In fact, it can be simple. If $n \geq 1$ then
$$
\bigg| \frac{(-1)^{n}\cos 3n}{n^{2}+n} \bigg| \leq \frac{1}{n^{2}+n} < \frac{1}{n^{2}}.
$$
But $\sum_{n} \frac{1}{n^{2}}$ converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem with multivariable calculus: $\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$ Anyone can help me with this limit?
$$\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$
I'm having trouble with proving that this limit really goes to $0$
thank you
| $$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$
Method 1: Using the two path test
Along the path of $y=0$, we have
$$\lim\limits_{(x,0)\to (0,0)} \frac{x^3 + 0^3}{x^2 + 0}=\lim\limits_{x\to 0} \frac{x^3}{x^2}=\lim\limits_{x\to 0} x=0$$
Along the path of $y=p(x)-x^2$, where $p(x)$ is a polynomial that passes ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \... | It should work if we start factoring consecutive integers out of the expression. You have it boiled down to this $$ 6 \cdot 6^3\cdot5^2 \cdot 4^2 \cdot 7 =$$
$$2\cdot 3 \cdot (6^3 \cdot 5^2 \cdot 4^2 \cdot 7)=2 \cdot 3 \cdot 4\cdot 5\cdot 6\cdot 7\cdot (6^2 \cdot 5\cdot 4)$$
This is a good start, but now we need an 8.... | {
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"timestamp": "2023-03-29T00:00:00",
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Breaking a determinant into eight pieces From Strang, Introduction to Linear Algebra, 4th ed. p. 250. This is an example whose answer is given in the text but I'm confused. I've paraphrased and corrected what I think is a typo in the first matrix (row 3 + row 3 changed to row 3 + row 2).
How is the determinant of M rel... | Let's apply the rule you quoted once:
$$
\det M=\left|
\begin{matrix}
\text{row 1}+\text{row 3}\\
\text{row 2}+\text{row 1}\\
\text{row 3}+\text{row 2}\\
\end{matrix}
\right|
=
\left|
\begin{matrix}
\text{row 1}\\
\text{row 2}+\text{row 1}\\
\text{row 3}+\text{row 2}\\
\end{matrix}
\right|+
\left|
\begin{matrix}
\text{... | {
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"url": "https://math.stackexchange.com/questions/1440674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the limit $\lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6}$ exist $$
\lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6}
$$
Does this limit exist?
I've tried substituting y=x^0.5 and y=x^(2/3) which both goes to 0.
| We have $$ \frac{|x|^3y^2}{ x^4 + y^6 } \leq c \sqrt{ |y|}$$
Indeed it suffices to show that
$$ |x|^3y^2\leq c \sqrt{ |y|} (x^4 + y^6) $$
Which we see it holds from AM - GM on $$ x^4/3 +x^4/3 +x^4/3+y^6 \geq C|x|^3\sqrt{|y|^3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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$[\cos x+\sin x]=[\cos x]+[\sin x]$,where [.] is the greatest integer function. Solve the equation in interval $[0,\pi]:[\cos x+\sin x]=[\cos x]+[\sin x]$,where [.] is the greatest integer function.
How should i start this question,breaking it into intervals is difficult.Please guide me.
| Clearly the equality holds true if one of $\sin x,\cos x=0$
This $\implies x=0,\dfrac\pi2,\pi$
$[\sin x]=\begin{cases} 1 &\mbox{if } x=\dfrac\pi2 \\
0 & \text{otherwise} \end{cases} $
$[\cos x]=\begin{cases} 1 &\mbox{if } x=0 \\
0 & \mbox{if } 0<x<\dfrac\pi2\\
-1 & \mbox{if } \dfrac\pi2<x\le\pi\\
\end{cases} $
Fo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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figuring out an integer function $f(1) = 1\\
f(2) = 2\\
f(3) = 6\\
f(4) = 20\\
f(5) = 70\\
f(6) = 252\\
f(7) = 924\\
f(8) = 3432\\
f(9) = 12870$
Then what is $f(n)$ (where $n > 0$)?
I though about many many possibilities but still cannot figure out the expression.
| What about this?
$$\frac{2^{n-1} (2 n-3)\text{!!}}{(n-1)!}$$
which gives
$$
\begin{array}{ccl}
f(1) & = & 1 \\
f(2) & = & 2 \\
f(3) & = & 6 \\
f(4) & = & 20 \\
f(5) & = & 70 \\
f(6) & = & 252 \\
f(7) & = & 924 \\
f(8) & = & 3432 \\
f(9) & = & 12870 \\
f(10) & = & 48620 \\
f(11) & = & 184756 \\
f(12) & = & 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Can i factor out an $x^2$ while completing the square? Original equation $$\int\frac{x}{\sqrt{56+10x^2-x^4}}$$
This equation must be integrated by completing the square. My question is, when completing the square am i able to factor out $x^2$ like this $$-x^4+10x^2+56$$ $$-x^2(x^2-10)+56$$ would that be fine? and if i... | $$\int\frac{x}{\sqrt{56+10x^2-x^4}}dx$$
Using the substitution $x^2 = t$, $\,\,dt = 2xdx$
$$\frac{1}{2}\int\frac{dt}{\sqrt{-(t^2 -10t - 56)}}$$
$$\frac{1}{2}\int\frac{dt}{\sqrt{81-(t-5)^2}}$$
Substitute $s = t-5$, $\,\,ds = dt$
$$\frac{1}{2}\int\frac{ds}{\sqrt{81-s^2}}$$
$$\frac{1}{2}\int\frac{ds}{9\sqrt{1-\frac{s^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving $\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$ without L'Hopital. I am trying to solve the limit
$$\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$
Without using L'Hopital.
Evaluating yields $\frac{0}{0}$. When I am presented with roots, I usually do this:
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} \cdot \frac{3+\... | I would rewrite the quotient as the quotient of two variation rates. Set $x=4+h\enspace(h\to 0)$:
\begin{align*}
\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}&=\frac{\sqrt{9}-\sqrt{9+h}}{\sqrt{1}-\sqrt{1-h}}\\
&=\frac{\sqrt{9+h}-\sqrt{9}}h\cdot\frac h{\sqrt{1-h}-\sqrt{1}}\to\bigl(\sqrt{x}\bigr)'_{x=9}\cdot-\frac1{\bigl(\sqrt{x}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the determinant of $n\times n$ matrix Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc}
-x & a_2&a_3&a_4&\cdots &a_n\\
a_{1} & -x & a_3&a_4&\cdots &a_n\\
a_1&a_{2} & -x &a_4&\cdots &a_n\\
\vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
a_1&a_{2} & a_3&a_4&\cdots & -x\\
\end{array}... | Hint: $\det(M)$ is a polynomial in $x$ and $M$ is clearly singular if $x=-a_k$ for some $k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is there anything wrong with this question? Find an equation of the plane that passes thru the points $p(1,0,1)$ and $q(2,1,0)$ which is parallel to the intersection of the two planes $x+y+z=5$ and $3x-y=4$?
I plotted the two points and the intersection of the two planes using a 3-d grapher and the points aren't parall... | It's the plane that needs to be parallel, not the points.
The points just need to lie in the plane.
Let the equation of the plane be $ax+by+cz=d$. Without loss of generality we can let $d=1$.
Using the values of $(x,y,z)$ from the given points, we find that $a+c=d=1$ and that $2a+b=d=1$.
The line of intersection has di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Problem with definite integral $\int _0^{\frac{\pi }{2}}\sin \left(\arctan \left(x\right)+x\right)dx$ Need to calculate this definite integral. It's seems very strange for me
$$\int _0^{\frac{\pi }{2}}\sin \left(\arctan\left(x\right)+x\right)dx$$
I dont see any reasonable way to calculate this integral. For instance, a... | $\int_0^\frac{\pi}{2}\sin(\tan^{-1}x+x)~dx$
$=\int_0^\frac{\pi}{2}\sin\tan^{-1}x\cos x~dx+\int_0^\frac{\pi}{2}\cos\tan^{-1}x\sin x~dx$
$=\int_0^\frac{\pi}{2}\dfrac{x\cos x}{\sqrt{x^2+1}}~dx+\int_0^\frac{\pi}{2}\dfrac{\sin x}{\sqrt{x^2+1}}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{2(2n)!\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Is $\frac{x^2+4x+7}{x^2+x+1}$ a one-to-one function or not? Is $\frac{x^2+4x+7}{x^2+x+1}$ a one-to-one function or not?
Leaving aside the graphical method is there any other way in which I can do this question?
| An algebraic approach:
The function is one-one if the equation in $x$
$$
k=\dfrac{x^2+4x+7}{x^2+x+1}
$$
has only one solution for any $k$ that gives real solutions.
Since $x^2+x+1 \ne 0$ we have:
$$
x^2(k-1)+x(k-4)+k-7=0
$$
and we can easely see that the discriminant of this equation can be positive, so there is some $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot... | If L'Hopitals' Rule gives you $\frac{f'}{g'}\rightarrow\frac{0}{0}$, apply L'Hopital's Rule to it. :-)
$$\frac{f(x)}{g(x)}=\frac{1-\sqrt{cos(x)}}{x^{2}}
$$
$$f''(x)=-\frac{1}{2}\sqrt{cos(x)}-\frac{sin^{2}(x)}{4cos^{\frac{3}{2}}(x)}
$$
$$\frac{f''(x)}{g''(x)}\rightarrow-\frac{1}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How to solve $\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$ Solved I stuck in this limit, I tried to solve it and gets 1/2 as result. Yet, I was wrong because I forgot a square. Please need help!
$\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$
Note: it's $+\infty$
Tha... | If ${\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} = 0}$ and you are taking a limit of the form
$$\lim_{x \rightarrow \infty} \sqrt{x + f(x)} - \sqrt{x}$$
Then you can proceed by writing the limit as
$$\lim_{x \rightarrow \infty} \sqrt{x} \bigg(\sqrt{1 + {f(x) \over x}} - 1\bigg)$$
Since $\sqrt{1 + \epsilon}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
find all values of a for which the system has nontrivial solutions So i was given this question.
In each of the following, find all values of a for which the system has nontrivial solutions, and determine all solutions in each case.
a) x - 2y + z = 0
x + ay - 3z = 0
-x + 6y - 5z = 0
Here is my attempt:
$$\left[\begin{... | You just need to go a little further with the row reduction.
$$\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & a+6 & -8 & 0 \\ 0 & 2 & -2 & 0 \end{array}\right]\sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & a+6 & -8 & 0 \\ 0 & 1 & -1 & 0 \end{array}\right]\sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
compute the sum of a series $$
\sum_{n=0}^\infty \frac{x^n}{n(n+1)}
$$
what is the sum of this?
I suspect this has something to do with $e^x = \sum x^n/n! $ but I don't know how to go from there
| Your sum has an error in it because for n=0 you are dividing by 0. However, if you start instead with n=1, there is no error, so I will assume you meant that.
Notice that $$\int \int x^n dx dx = \frac{x^{n+2}}{(n+1)(n+2)} +Cx+D =x \cdot \frac{x^{n+1}}{(n+1)(n+2)} +Cx+D$$
Therefore $$\int \int \sum_{n=0}^{\infty} x^n dx... | {
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"url": "https://math.stackexchange.com/questions/1457451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find the limit of $x +\sqrt{x^2 + 8x}$ as $x\to-\infty$ $$\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}$$
I multiplied it by the conjugate:
$\frac{-8x}{x - \sqrt{{x^2} + 8x}}$
I can simplify further and get:
$\frac{-8}{1-\sqrt{1+\frac{8}{x}}}$
I think there is an error with my math, because the denominator should probably be ... | We have
\begin{align}
\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}&=\lim_{x\to -\infty}\frac{(x +\sqrt{x^2 + 8x})(x -\sqrt{x^2 + 8x})}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{x^2 -(x^2 + 8x)}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{-8x}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{\frac{-8x}{|x|}}{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities? $\sin^{4}x+\cos^{4}x$
I should rewrite this expression into a new form to plot the function.
\begin{align}
& = (\sin^2x)(\sin^2x) - (\cos^2x)(\cos^2x) \\
& = (\sin^2x)^2 - (\cos^2x)^2 \\
& = (\sin^2x - \cos^2x)(\sin^2x + \cos^2x) \\
& = (\sin^2x - \c... | Expand in terms of complex exponentials.
$$\sin^4 x + \cos^4 x = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^4 + \left( \frac{e^{ix} + e^{-ix}}{2} \right)^4$$
Notice that $i^4 = +1$, so we get
$$\sin^4 x + \cos^4 x = \frac{1}{16} \left( 2e^{4ix} + 2 e^{-4ix} + 12 \right)$$
where we use the relation $(a+b)^4 = a^4 + 4 a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Convert $r^2= 9 \cos 2 \theta$ into a Cartesian equation This is how I tried so far...
$r^2= 9 \cos 2 ( \theta)$
$\cos (2 \theta) = \cos ^2 (\theta) - \sin^2 (\theta)$ and
$r^2= x^2 + y^2$
so, it will become
$x^2 + y^2 = 9 [\cos^2 (\theta) - \sin^2 (\theta) ]$
| Starting with
$$r^2=9\cos2\theta$$
Applying the double angle identity
$$r^2=9\cos^2\theta-9\sin^2\theta$$
Now multiply both sides by $r^2$, apply $x=r\cos\theta,y=r\sin\theta$ and $x^2+y^2=r^2$ to obtain
$$(x^2+y^2)^2=9x^2-9y^2$$
Which, when expanded gives
$${x}^{4}+2\,{x}^{2}{y}^{2}+{y}^{4}-9\,{x}^{2}+9\,{y}^{2}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
system of three equations with Cramer's rule First of - sorry for my English.
My math teacher gave me three tasks to complete, but I can't complete the second one.
Here is the system:
$$\begin{cases} x+y-z=3 \\ x-3y+2z=1 \\ 7x-4y+z=7\end{cases}$$
We have to complete them using Cramer's rule. Determinant result is 43 fo... | The coefficient matrix for the system $$\begin{cases} x+y-z=3 \\ x-3y+2z=1 \\ 7x-4y+z=7\end{cases}$$ is $$\pmatrix{1 & 1 & -1 \\ 1 & -3 & 2 \\ 7 & -4 & 1}$$
Let's find the determinant:
$$\begin{align}\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -3 & 2 \\ 7 & -4 & 1\end{array}\right| &= \left|\begin{array}{ccc} 1 & 1 & -1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Plotting $f(x) = \sin(x)+\cos(x)$ by converting it to another form Are there any trigonometric identity that can make $f(x) = \sin(x)+\cos(x)$ easier to plot? I have no idea how it becomes a sin graph shape in the end.
| Special Case
It's not that hard. You should just use the summation formula for sines:
$$\sin (x + y) = \sin (x)\cos (y) + \cos (x)\sin (y)$$
This is how it works
$$\eqalign{
\sin (x) + \cos (x) &= \sqrt 2 \left( {{1 \over {\sqrt 2 }}\cos (x) + {1 \over {\sqrt 2 }}\sin (x)} \right) \cr
&= \sqrt 2 \left( {\sin ({... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
What is the probability that the number $3^a+7^b$ has a digit equal to $8$ at the units place? The number $a$ is randomly selected from the set $\left\{0,1,2,3,....,98,99\right\}$.The number $b$ is selected from the same set.What is the probability that the number $3^a+7^b$ has a digit equal to $8$ at the units place?
... | Use the Chinese theorem. $\equiv 8\pmod{10}$ means even and $\equiv 3\pmod{5}$. Now both $3$ and $7\equiv 2$ are generators for $\mathbb{F}_5^*$, hence the working cases are $a\equiv 2\pmod{4}$ and $b\equiv 2\pmod{4}$, $a\equiv 3\pmod{4}$ and $b\equiv 0\pmod{4}$, $a\equiv 0\pmod{4}$ and $b\equiv 1\pmod{4}$. $3^a$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Implicit equation of Semicircle and ellipse May I know what is the Implicit equation to define the upper bound of a circle of radius 1?
Is it $ y^2 - \sqrt{1-x^2} = 0$?
and for the lower bound, $ y^2 + \sqrt{1-x^2} = 0$?
What is the implicit equation of the upper bound of an ellipse of radius 0.3 then?
| The implicit equation of a circle of radius $1$ centered at the origin is $x^2+y^2=1$, so you can find: $y^2=1-x^2$ and the upper esmicircle ($y\ge 0$) is given by: $y=\sqrt{1-x^2}$ ( that you can write as $ y-\sqrt{1-x^2}=0$ if you want). The other semicircle ($y\le 0$) is $y=-\sqrt{1-x^2}$.
For an ellipse with center... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor series of arctan(x) (Spivak) At p. 388 of Calculus, Spivak gives a formula:
$$\frac{1}{1+t^2} = 1 - t^2 + t^4 - ... + (-1)^nt^{2n} + \frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$$
Which can be integrated to find $\arctan(x)$.
I don't understand where this formula comes from, but I found it up to $(-1)^nt^{2n}$ by considerin... | Consider
$$
1+x+x^2+\dots+x^n=\frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}-\frac{x^{n+1}}{1-x}
$$
that can also be written
$$
\frac{1}{1-x}=1+x+x^2+\dots+x^n+\frac{x^{n+1}}{1-x}
$$
Now substitute $x=-t^2$, that gives
$$
\frac{1}{1+t^2}=1+(-t^2)+(-t^2)^2+\dots+(-t^2)^n+\frac{(-t^2)^{n+1}}{1+t^2}
$$
and not it's just a matter of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Help in proof involving matrices
Let $$ P=\begin{pmatrix}11&-6\\18&-10\end{pmatrix}. $$ Show that
there is no $2\times2$ matrix $Q$ such that $Q^2=P$.
I don't know how to approach this question. Is brute force the only way to go? That is, take a general matrix
$$
Q=\begin{pmatrix}a&b\\c&d\end{pmatrix}
$$
and get ... | The Eigen values of $P$ are $2$ and $-1$. Now there exists a Matrix $Q$ such that $Q^2=P$. So $Det(Q^2)=Det(P)$ $\implies$ $Det(Q)=i\sqrt{2}$ or $-i\sqrt{2}$.
Now by Cayley hamilton theorem the characteristic equation of $Q$ is of the form
$$Q^2-Tr(Q)Q+Det(Q)I=0$$ $\implies$ Since $Q^2=P$
$$Tr(Q)Q= P+Det(Q)I=\begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$. I have a homework as follow:
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$.
Please help to prove it.
EDIT: MY ATTEMPT
Suppose that $5\mid a^2-2b^2$, then $a^2-2b^2=5n$,where $n\in Z$,
then $a^2-2b^2=(a+\sqrt2b)(a-\sqrt2b)=5n$,
Since 5 is a prime number, we ge... | You just have to make a table for the function $a^2-5b^2$ in $\mathbf Z/5\mathbf Z$ above the horizontal line, the possible values for $a^2$; to the left of the vertical line, those for $b^2$):
$$\begin{array}{r|rrr}
&0&1&-1\\
\hline
0&0&1&-1\\1&-2&-1&2\\-1&2&-2&1
\end{array}$$
which shows the only case with $a^2-5b^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be equal to
$(A)15^\circ\hspace{1cm}(B)75^\circ\hspace{1cm}(C)115^\circ\hspace{... | here is way to compute the angle $\angle BAD$ using the rule of sines where $D$ is mid point of $BC.$
let me use $t = \angle BAD.$
suing the rule of sin on $\Delta ABD, \Delta ADC$ we have $$\frac{BD}{\sin t} = \frac{AD}{\sin(t + 45^\circ)}, \frac{CD}{\sin(15^\circ)} = \frac{AD}{\sin 30^\circ} $$
from the two equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$? How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$?
I found the derivative to be $f'(x)= (2\sin 4\theta)(1+\cos 2\theta) + (\sin 2\theta)^2... | I’d start by rewriting $f(x)$, using the identities $\sin2\theta=2\sin\theta\cos\theta$ and $\cos^2\theta=\frac12(1+\cos2\theta)$:
$$f(x)=\sin^22\theta(1+\cos2\theta)=8\sin^2\theta\cos^4\theta\;.$$
In fact, let’s go a step further and get rid of the sine:
$$f(x)=8\sin^2\theta\cos^4\theta=8(1-\cos^2\theta)\cos^4\theta=8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solution of quadratic diophantine equations Is there any algorithm so that solution to the following equation can be found?
$(x+a)^2-y^2=c$ where $c$ and $a$ is a constant.
It is similar to Pells eqution with a variation where $D=1$.
I am new to this number theory problem. So can anyone please help me?
| I.e. you want to solve in integers, where $c$ is given (we can let $x$ be $x+a$):
$$x^2-y^2=c$$
The typical solution is factoring $(x-y)(x+y)=c$, noticing the factors of $c$ and checking cases, when $x-y=d$ and $x+y=\frac{c}{d}$. Note $\begin{cases}x-y=d\\x+y=\frac{c}{d}\end{cases}$ is solvable in integers if and only ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $0<|z|<1$, show that $\frac{1}{4}|z|<|1-e^z|<\frac{7}{4}|z|$ My question:
If $0<|z|<1$, show that $\frac{1}{4}|z|<|1-e^z|<\frac{7}{4}|z|$ ($z$ is complex)
what I have tried:
I tried to expand the middle term in its Taylor series but I can't get the appropriate bound.
| The Maclaurin expansion of $e^z - 1$ is
$$z + \frac{z^2}{2!} + \cdots + \frac{z^N}{N!} +\cdots \tag{*}$$
Since $0 < |z| < 1$ and $k! \ge 2\cdot 3^{k-2}$ for all $k \ge 2$, then by the triangle inequality, (*) is strictly bounded above in modulus by
$$\left[1 + \frac{1}{2} + \frac{1}{2}\left(\frac{1}{3}\right) + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Use the PMI to prove the following for all natural numbers $3^n≥1+2^n$ Use the PMI to prove the following for all natural numbers
$3^n≥1+2^n$
Base Case: $n=1$
$3^1≥1+2^1$
$3 ≥ 3$, which is true
Inductive Case:
Assume $3^k ≥ 1+2^k$
[Need to Show for k+1]
$3^{(k+1)} \ge 1+2^{(k+1)}$
Now from here i always get stuck tryi... | Assume $$3^k\ge 1+2^k$$
Add $2^k$ to both sides:
$$3^k+2^k\ge 1+2^k+2^k=1+2^{k+1}$$
Now notice that $3^{k+1}=3\cdot 3^k=3^k+2\cdot 3^k\ge 3^k+2^k$
So, we combine $3^k+2^k\ge 1+2^{k+1}$ with $3^{k+1}\ge 3^k+2^k$ and get
$$3^{k+1}\ge 1+2^{k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I poste... | $x = \underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = \underbrace{\frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \ldots + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 3
} |
A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
What are the steps to solve this equation for $ \theta $?
Because, I am always unable to deal with the product $\sin \theta \cos \theta$.
| $3\sin^2{\theta}-2\cos^2{\theta}+5\sin{\theta}\cos{\theta}=0$
Transfer the $\sin{\theta}\cos{\theta}$ term to the RHS.
$3\sin^2{\theta}-2\cos^2{\theta}=-5\sin{\theta}\cos{\theta} $
Then square both sides.
$(3\sin^2{\theta}-2\cos^2{\theta})^2=(-5\sin{\theta}\cos{\theta})^2$
$9\sin^4{\theta}-12\sin^2{\theta}\cos^2{\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to prove this inequality about $xyz=1$ Let $x,y,z>0$,and such $xyz=1$, show that
$$\dfrac{x+y}{x^3+x}+\dfrac{y+z}{y^3+y}+\dfrac{z+x}{z^3+z}\ge 3$$
I tried use $AM-GM$ inequality
$$\dfrac{x+y}{x^3+x}+\dfrac{y+z}{y^3+y}+\dfrac{z+x}{z^3+z}\ge 3\sqrt{\dfrac{(x+y)(y+z)(z+x)}{(x^3+x)(y^3+y)(z^3+z)}}$$
which shows that $... | firstly we can rewrite the terms like : $$\frac{x+y}{x^3+x}=\frac{1}{1+x^2}+\frac{y}{x+x^3}=1-\frac{x^2}{1+x^2}+\frac{y}{x}-\frac{yx^2}{x+x^3}$$ then we can apply $AM\geq GM$ for the denominators and we get $$\frac{x+y}{x^3+x}\geq 1+\frac{y}{x}-\frac{x^2}{2x}-\frac{yx^2}{2x^2}=1+\frac{y}{x}-\frac{x}{2}-\frac{y}{2}$$ s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
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When is $2^n +3^n + 6^n$ a perfect square? Find all $n$ for which $2^n+3^n+6^n$ is a perfect square.
I do not have a specific idea how to solve this one
| If $n$ is odd then $2^n+3^n+6^n\equiv -1\pmod{3}$, so $2^n+3^n+6^n$ cannot be a square.
For similar reasons $\!\!\pmod{5}$, $n$ has to be a number of the form $4k+2$.
For similar reasons $\!\!\pmod{7}$, $n$ has to be a number of the form $12k\pm 2$, so $2^n+3^n+6^n$ is a multiple of $7$. $n=2$ gives the solution:
$$ 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How do I express 108 as the product of powers of its prime factors? How do I express $108$ as the product of powers of its prime factors?
I've got $2^2 \times 3^3$ is this correct?
| First notice that $108$ is even so $2$ is a factor:
$$\begin{matrix} & & 108 \\ & \large\swarrow & & \large\searrow \\ 2 & & & &54\end{matrix}$$
$54$ is also even:
$$\begin{matrix} & & 54 \\ & \large\swarrow & & \large\searrow \\ 2 & & & &27\end{matrix}$$
$27$ is not even so let's check the next prime: $3$. We can s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Limits without L'Hopitals Rule ( as I calculate it?) Prove that:
$\lim z \to \infty \left (z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}\right )=\frac{-1}{4}$
| $$\begin{align}
z^2+\sqrt{z^4+2z^3}-2\sqrt{z^4+z^3}
&=(\sqrt{z^4+2z^3}-\sqrt{z^4+z^3})+(z^2-\sqrt{z^4+z^3})\\
&={z^3\over\sqrt{z^4+2z^3}+\sqrt{z^4+z^3}}-{z^3\over z^2+\sqrt{z^4+z^3}}\\
&={z^3(z^2-\sqrt{z^4+2z^3})\over(\sqrt{z^4+2z^3}+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+z^3})}\\
&={z^3(-2z^3)\over(\sqrt{z^4+2z^3}+\sqrt{z^4+z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Prove , there exists $\theta , \phi \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta) = 0$ and $f'(\phi)\neq 0$ Let the function
$$f(\theta) = \begin{vmatrix} \sin\theta & \cos\theta & \tan\theta \\ \sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) & \tan(\frac{\pi}{6}) & \\ \sin(\frac{\pi}{3}) & \cos(\frac{\pi}{3}) & ... | Without explicitly expanding the determinant we see that
$$f(\pi/6) = f(\pi/3) = 0$$
since two of the row are equal in these cases giving us a zero determinant. The mean value theorem then tells us that there exists a $\theta\in(\pi/6,\pi/3)$ where $f'(\theta) = 0$.
If II is false then $f’(\theta) \equiv 0 \implies f(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1488824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factorial modular arithemetic
Find $50!/(10^{12}) \pmod{10}$.
Let $N$ represent the number then,
$N \equiv 0 \pmod{2}$
For mod 5 it is harder.
$N^{4} \equiv 1 \pmod{5}$.
How to solve this?
| \begin{align}
\lfloor 50/2 \rfloor &= 25\\
\lfloor 25/2 \rfloor &= 12\\
\lfloor 12/2 \rfloor &= 6\\
\lfloor 6/2 \rfloor &= 3\\
\lfloor 3/2 \rfloor &= 1\\
\lfloor 1/2 \rfloor &= 0\\
25 + 12 + 6 + 3 + 1 &= 47
\end{align}
So $2^{47} \mid 50!$. It follows that $\dfrac{50!}{2^{12}} \equiv 0 \pmod 2$
\be... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A sum including binomial coefficients I would like to prove the following equality:
$$\sum_k (-1)^{n-1}(-2)^k\binom{n}{k+1}\binom{n+k-1}{k}=\sum_{k=0}^{n-2}\binom{2n-k-2}{n-1}\binom{n-2}{k}$$
but the power over two and the switch on the number of sums bothers me.
Any help would be welcome.
(the equality it part of Not... | The following answer is purely algebraic. We transform both sides of OPs expression to finally obtain the same representation. We also use the coefficient of operator $[z^n]$ to denote the coefficient $a_n$ of $z^n$ of a series $\sum_{k=0}^{\infty}a_kz^k$.
At first we transform the right-hand side. It's the easier on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $ then find $a+b$ $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $
if I use this $$\lim _{x\rightarrow -\infty }\sqrt {ax^{2}+bx+c}=\left| x+\dfrac {b} {2a}\right| $$
I find $a=1,b=4$ but if I try to multiple by its conjugate
$$\lim _{x\rightarrow -\... | Here is a slightly different approach.
First note that
\begin{align}
\lim_{x\to \infty}\frac{\sqrt {x^{2}-6x+3}-ax+b}{x^1}&=\lim_{x\to \infty}\sqrt {1-\frac6x+\frac3{x^2}}-a+\frac bx\\
&=1-a
\end{align}
and
\begin{align}
\lim_{x\to \infty}\frac{\sqrt {x^{2}-6x+3}-ax+b-(1-a)x}{x^0}&=\lim_{x\to \infty}\sqrt {x^{2}-6x+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verifying partial order relation I have the following question where i have to verify if the relation is partial order:
$A=\{1,2,3,\ldots,100\}$, relation $x\mathrel{R}y \leftrightarrow \frac{y}x=2^k$, where $k\ge 0$ is an integer.
This how i am thinking :
Reflexive: $x\mathrel{R}x\implies\frac{x}x=1$, therefore the ... | Your argument for reflexivity is correct: if $x\in A$, then $\frac{x}x=1=2^0$, so $x\mathrel{R}x$. Your example showing that $R$ is not symmetric is correct, though you’ve not explained it adequately: $\frac42=2=2^1$, so $2\mathrel{R}4$, but $\frac24=\frac12$ is not of the form $2^k$ for any integer $k\ge 0$, so $4\not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A limit problem about $a_{n+1}=a_n+\frac{n}{a_n}$ Let $a_{n+1}=a_n+\frac{n}{a_n}$ and $a_1>0$. Prove $\lim\limits_{n\to \infty} n(a_n-n)$ exists.
In my view, maybe we can use
$${a_{n + 1}} = {a_n} + \frac{n}{{{a_n}}} \Rightarrow {a_{n + 1}} - \left( {n + 1} \right) = \left( {{a_n} - n} \right)\left( {1 - \frac{1}{{{a_... | We have:
$$ a_n(a_{n+1}-a_n) = n $$
so:
$$ a_{n+1}^2-a_{n}^2 = a_{n+1}(a_{n+1}-a_n) + n = n\left(1+\frac{a_{n+1}}{a_n}\right)=2n+\frac{n^2}{a_n^2}$$
and:
$$ a_{N+1}^2-a_1^2 = N(N+1)+\sum_{n=1}^{N}\frac{n^2}{a_n^2} $$
from which $a_{N+1}\geq \sqrt{N(N+1)}$ and $a_n\geq \sqrt{(n-1)n}$.
If we plug this inequality back i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ I've seen this identity on examsolutions, but I'm unsure on how to prove it.
$$\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
| Notice,
$$RHS=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$$
$$=2\cos \left(\frac{A}{2}+\frac{B}{2}\right)\cos \left(\frac{A}{2}-\frac{B}{2}\right)$$
$$=2 \left(\cos\frac{A}{2}\cos\frac{B}{2}-\sin\frac{A}{2}\sin\frac{B}{2}\right)\left(\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{A}{2}\sin\frac{B}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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multiplication over gf(16) Can some one show me how to do multiplication over gf(16) step by step
I found this example online, http://userpages.umbc.edu/~rcampbel/Math413Spr05/Notes/12-13_Finite_Fields.html#An_Example.
An Example:
$x^4 = x+1$
$x^5 = x^2+x$
$(x^2+x+1) (x^3+x^2+1)$
$= x^5 + 2x^4 + 2x^3 + 2x^2 + x + 1$
$=... | A better way to do the multiplication (not necessarily more efficient, but a better way to think about it) is by using remainders of polynomial division. So for example, $x^{5} = x(x^{4}+x+1) + x + 1$, and since $x^{4}+x+1 = 0$, $x^{5} = x+1$. You really are looking at polynomials $\bmod{(x^{4}+x+1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Probability to identify faulty machines.
There are four machines and it is known that exactly
two of them are faulty. They are tested, one by one,
in a random order till both the faulty machines are
identified. Find the probability that exactly $3$ tests will
be required to identify the $2$ faulty machines.... | The probability for faulty-not faulty-faulty is
$\frac{1}{2}\times \frac{2}{3} \times \frac{1}{2}=\frac{1}{6}$
The probability for not faulty-faulty-faulty is
$\frac{1}{2} \times \frac{2}{3} \times \frac{1}{2}=\frac{1}{6}$
The sum is $\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Integration with substitution of trig: $\int\frac1{(x^2-4)^{3/2}}dx$ $$\int\frac1{(x^2-4)^{3/2}}dx $$
I'm unsure how to go about this integral after subbing $x=2\cosh(\theta)$; to the power of $3/2$ is confusing me when trying to simplify.
Thanks.
| Let $u^2x^2=x^2-4$. Then $x=\pm\sqrt{\frac{-4}{u^2-1}}$, $dx=\pm(-4)\sqrt{\frac{u^2-1}{-16}}\left(\frac{-2u}{\left(u^2-1\right)^2}\right)\,du$.
$$\int\frac{1}{\left(x^2-4\right)^{3/2}}\, dx=\int\frac{\pm(-4)\sqrt{\frac{u^2-1}{-16}}\left(\frac{-2u}{\left(u^2-1\right)^2}\right)\,du}{u^3\cdot \pm\left(\frac{-4}{u^2-1}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$-3\cdot7^x+2\cdot6^x+2\cdot5^x-2\cdot4^x+3^x>0$, for $-1\le x<0$. Show that $-3\cdot7^x+2\cdot6^x+2\cdot5^x-2\cdot4^x+3^x>0$, for $-1\le x<0$.
Here its plotting.
By first derivative it involves natural log and expression become more difficult.
| Note that $a^x$ is a decreasing convex function of $a>0$ for $x<0$. Hence
$$-3\cdot7^x+2\cdot 6^x+5^x>0$$
$$5^x-2\cdot 4^x+3^x>0$$
Add these up to get the needed inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots$. Show that: $$(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots = \dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}.$$
By ... | I would do the following: consider a finite product
$$\mathcal{P}_N:=(1-z)\left(1+\frac z{2}\right)\ldots \left(1-\frac{z}{2N-1}\right)\left(1+\frac{z}{2N}\right)$$
and rewrite it as
\begin{align*}\mathcal{P}_N&=\frac{\prod_{k=0}^{N-1}\left(2k+1-z\right)\prod_{k=0}^{N-1}\left(2k+2+z\right)}{(2N)!}=\\
&=\frac{2^{2N}}{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Show that $\lim \limits_{x \to \infty} (x + x^2 \log \frac{x}{x+1}) = \frac{1}{2}$ I am self studying real analysis and I found this problem in a book.
We have to show that if $x > 1$ then
$$\lim_{x \to \infty} \left[ x + x^2 \log \frac{x}{x+1} \right] = \frac{1}{2}$$.
I can prove that the sequence is bounded.
It is bo... | Set $1/x=h$
to get $$\lim_{h\to0}\dfrac{h-\ln(1+h)}{h^2}$$
Now use $\ln(1+h)=h-\dfrac{h^2}2+O(h^3)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding first three terms of a GP whose sum and sum of squares is given? We have to find three numbers in G.P. such that their sum is $\frac{13}{3}$ and the sum of their squares is $\frac{91}{9}$.
Here's what I have tried to do:-
$$a+ar+ar^2=\frac{13}{3}$$
$$a^2+a^2r^2+a^2r^4=\frac{91}{9}$$
$$\frac{a^2+a^2r^2+a^2r^4}{a... | $a+ar+ar^2=\frac{13}{3}$ [equation 1]
$a^2+a^2r^2+a^2r^4=\frac{91}{9}$ [equation 2]
Square the first equation:
$a^2(1+r+r^2)^2 = \frac{169}{9}$
$a^2(1 + r^2 + r^4 + 2r + 2r^2 + 2r^3) = \frac{169}{9}$ [equation 3]
Now take equation 3 minus equation 2:
$2a^2r(1 + r + r^2) = \frac{26}{3}$
Using equation 1 again,
$2ar \fra... | {
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"url": "https://math.stackexchange.com/questions/1502913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the Integral: $\int^{0.6}_0\frac{x^2}{\sqrt{9-25x^2}}dx$ Evaluate the Integral: $\int^{0.6}_0\frac{x^2}{\sqrt{9-25x^2}}dx$
I believe my work is correct until I try to change the bounds according to theta, $\theta$.
How do I do that? Also, please comment freely on my work, that is, is my method correct and cog... | Notice, you let $$x=\frac{3}{5}\sin \theta\implies \theta=\sin^{-1}\left(\frac{5x}{3}\right)$$ then
upper limit at $x=0.6$ : $\theta=\sin^{-1}\left(\frac{5\times 0.6}{3}\right)=\sin^{-1}\left(1\right)=\frac{\pi}{2}$
lower limit at $x=0$ : $\theta=\sin^{-1}\left(\frac{5\times 0}{3}\right)=\sin^{-1}\left(0\right)=0$
aft... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve for $x$: $x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$ Solve for $x$: $$x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$$
My attempt: I have changed this equation into the fractional part function. so that we know $0 \leq \{x\}<1$. I have final equation $x^2 +7x-x\{x+2\}-\{2x-2\}=14$. How to pro... | Since
$$\lfloor x+2\rfloor=\lfloor x\rfloor+2,\quad \lfloor 2x-2\rfloor=\lfloor 2x\rfloor-2$$
the equation can be written as
$$x(\lfloor x\rfloor+2)+\lfloor 2x\rfloor-2+3x=12,$$
i.e.
$$x\lfloor x\rfloor+5x+\lfloor 2x\rfloor=14\tag 1$$
Now let us separate it into cases :
Case 1 : $x=m+\alpha$ where $m\in\mathbb Z$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Solving the equation $z^7=-1$
Solve the equation $z^7=-1$
My attempt:
$$z=x+yi$$
$$(x+yi)^7+1=0$$
$$(x^2+2yi-y^2)^3(x+yi)+1=0$$
but now it's start to look ugly.
I'm sure that there is a simple way
| $$z^7=-1\Longleftrightarrow$$
$$z^7=|-1|e^{\arg(-1)i}\Longleftrightarrow$$
$$z^7=1\cdot e^{\pi i}\Longleftrightarrow$$
$$z^7=e^{\pi i}\Longleftrightarrow$$
$$z=\left(e^{\left(\pi+2\pi k\right)i}\right)^{\frac{1}{7}}\Longleftrightarrow$$
$$z=e^{\frac{1}{7}\left(\pi+2\pi k\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-6$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Linear Algebra Matrix and Inverse I have 4x4 matrices:
$$A\begin{bmatrix}
3&1&3&-4\\
6&4&8&10\\
3&2&5&-1\\
-9&5&-2&-4
\end{bmatrix} =
\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}
$$
and
$$\begin{bmatrix}
3&1&3&-4\\
6&4&8&10\\
3&2&5&-1\\
-9&5&-2&-4
\end{bmatrix}
B = \begin{bmatrix}
0&1&0&0\\
0... | The right hand matrix is known as permutation matrix and its inverse is its transpose, i.e.
$$
P^{-1}=\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}^T=\begin{bmatrix}
0&0&0&1\\
1&0&0&0\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}=[e_2\: e_4\: e_3 \:e_1]=\begin{bmatrix}e_4'\\e_1'\\e_3'\\e_2'\end{bmatrix}
$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Pythagorean Triple Inequality When finding the Pythagorean triple where $a+b+c=1000$,
Wolfram alpha shows me that $a< -500\left(\sqrt{2} - 2\right)$
When I input $a^2+b^2=c^2, a<b<c$ and $a+b+c=1000$
How does wolfram arrive at that inequality:
$a< -500\left(\sqrt{2} - 2\right)$
Here is the link: Wolfram
| For the inequality, we observe that since $b > a$, we must have
$$
c^2 = a^2+b^2 > 2a^2
$$
or
$$
c > a\sqrt{2}
$$
Thus
$$
1000 = a+b+c > a+a+a\sqrt{2} = a(2+\sqrt{2})
$$
or
$$
a < \frac{1000}{2+\sqrt{2}} = \frac{1000(2-\sqrt{2})}{2^2-(\sqrt{2})^2}
= 500(2-\sqrt{2})
$$
As regards the original problem: Primitive Pythag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $ \int_0^1 x^2 \psi(x) \, dx = \ln\left(\frac{A^2}{\sqrt{2\pi}} \right) $ Basically what the title says:
Prove that $ \displaystyle \int_0^1 x^2 \psi(x) \, dx = \ln\left(\dfrac{A^2}{\sqrt{2\pi}} \right). $
where $A\approx 1.2824$ denotes the Glaisher–Kinkelin constant and $\psi(x) $ denote the digamma fun... | A little late, but here is yet another solution not shown previously.
I will first show that
$$\int_0^1 x \ln\left(\Gamma(x)\right)\,dx=\frac{1}{4} \ln \frac{2 \pi}{A^4} \tag{1}$$
And then from $(1)$ derive the desired result.
Recall Kummer´s fourier expansion for LogGamma $0<x<1$
$$\ln\left(\Gamma(x)\right)=\frac{\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Suppose ${3^x} + {4^x} = {5^x}$. What is zero of this equation? Suppose ${3^x} + {4^x} = {5^x}$.
What is zero of this equation?
| $x=2$
Since
${\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} = 1 = {(\sin x)^2} + {(\cos x)^2} \Rightarrow x = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate a 4th power of sine and cosine? I'm having some trouble figuring out the right substitutions to make to integrate
$$\int \sin^4(\theta)d\theta$$
and
$$\int \cos^4(\theta)d\theta$$
Any hints or suggestions are welcome.
Thanks,
| HINT (using partial integration):
$$\int\sin^4(x)\space\space\text{d}x=$$
$$-\frac{1}{4}\sin^3(x)\cos(x)+\frac{3}{4}\int\sin^2(x)\space\space\text{d}x=$$
$$-\frac{1}{4}\sin^3(x)\cos(x)+\frac{3}{4}\int\left(\frac{1}{2}-\frac{1}{2}\cos(2x)\right)\space\space\text{d}x=$$
$$-\frac{1}{4}\sin^3(x)\cos(x)-\frac{3}{8}\int\cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.