Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the minimun of $MN+\frac{3}{5}MP$, $MN$ and $MP$ is two sides of a quadrilateral. In a quadrilateral $OPMN$ ，$\angle NOP=90^\circ$,$ON=1$,$OP=3$, and $M$ satisfy $\vec{MO}\cdot\vec{MP}=4$, find the minimum of
$MP+\frac{3}{5}MN$
I choose the vertex $O$ of $OPMN$ as the origin of the coordinate system, and ON as th... | from Maple:
$y = -\frac{969+85 \sqrt{273} }{596} = -3.982265922$
Minimum of $MP+\frac{3}{5}MN$ is $\frac{\sqrt{298}}{5} = 3.452535300$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Counting with Recurrence Relations Find the recurrence relation for a(n) - number of ternary strings of length n, containing the number 2 odd times.
Some of these: 012,112,12,02,... .
| We divide all that sequences into $4$ non-intersecting subclasses:
a) Odd count of $2$s, ends with $1$ or $0$,
b) Even count of $2$s, ends with $1$ or $0$,
c) Odd count of $2$s, ends with $2$,
d) Even count of $2$s, ends with $2$,
Let $a_n,\,b_n,\,c_n,\,d_n$ be the cardinality of a corresponding class of sequences of l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Combinations series: $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$
Evaluate $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$ for $n=10$.
Attempt: I'll deal with the case n being even, as we need to evaluate for n=10.
the numerator is
$${n \choose 1}(n-1)^3+{... | Following the given hint, we have that
$$\begin{align}\sum_{r \text{ odd}} {10 \choose r}(10-r)^3&=\frac{1}{2}\left[\left((e^x+1)^{10}-(e^x-1)^{10}\right)'''\right]_{x=0}\\
&=\left[360(e^x+1)^7e^{3x}+135(e^x+1)^8e^{2x}+5(e^x+1)^9e^x\right.\\
&\quad \left.-360(e^x-1)^7e^{3x}-135(e^x-1)^8e^{2x}-5(e^x-1)^9e^x\right]_{x=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Proving That $\sum^{n}_{k=0} \bigl(\frac{4}{5}\bigr)^k < 5$
Using induction, prove that $$\sum_{k=0}^n \biggl(\frac 4 5 \biggr)^k = 1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\bigg(\frac{4}{5}\bigg)^3+\cdots +\bigg(\frac{4}{5}\bigg)^n<5$$ for all natural numbers $n.$
What I have tried is as follows.
Consider the stateme... | $$1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\bigg(\frac{4}{5}\bigg)^3+\cdots +\bigg(\frac{4}{5}\bigg)^n = 5\left(1-\dfrac{4^{n+1}}{5^{n+1}} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Given a tetrahedron, whose sides are $AB=3,AC=4,BC=5,AD=6,BD=7,CD=8$ . Find the volume of the tetrahedral $ABCD$ .
Given a tetrahedron, whose sides are $AB= 3, AC= 4, BC= 5, AD= 6, BD= 7, CD= 8$ . Find the volume of the tetrahedral $ABCD$ .
Assume that the tetrahedral $ABCD$ has its height $DH$ , whose length I will ... | The volume of the pyramid in terms of its side lengths
can be found as
\begin{align}
V&=
\frac1{12}\,
\left(
4\, u^2\, v^2\, w^2+(u^2+v^2-c^2)\, (v^2+w^2-a^2)\, (u^2+w^2-b^2)
\right.
\\
&\phantom{=}
\left.
-u^2\, (v^2+w^2-a^2)^2-v^2\, (u^2+w^2-b^2)^2-w^2\, (u^2+v^2-c^2)^2
\right)^{1/2}
,
\end{align}
where
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find maximum $k \in \mathbb{R}^{+}$ such that $ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $ Find maximum $k \in \mathbb{R}^{+}$ such that $$ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $$
for all $a, b, c$ that are distinct positive real numbers (... | This is a partial answer when $a,b,c$ form a side of a triangle.
Letting $c\to 0^+$, the inequality is in the form of:
$$\dfrac{a^3}{b^2}+\dfrac{b^3}{a^2}-k(a+b)\geq o(c),$$
for $o(c)\geq 0.$
But the left hand side is:
$$(a+b)\left(\dfrac{(a-b)^2(a^2+ab+b^2)}{a^2b^2}\right)+(a+b)(1-k).$$
If $k>1,$ then the whole thing ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
A problem of definite integral inequality? $$
\text { Minimum odd value of $a$ such that }\left.\left|\int_{10}^{19} \frac{\sin x d x}{\left(1+x^{a}\right)}\right|<\frac{1}{9} \text { is (where } a \in N\right)
$$
I proceed this way
As $|\sin x|<1$
Integration $|\frac{\sin x}{1+x^a}|<|\frac{1}{1+x^a}|$
And using trial ... | Continuing from where you stopped we have that $$\left| \frac{1}{1+x^a}\right| \le \left| \frac{1}{1+10^a}\right|$$ on the interval given. Multiplying by $|\sin x|$ tells us that then integral is less than $$\int_{10}^{19}\frac{1}{1+10^a}\mathrm dx=\frac{9}{1+10^a}.$$
For this quantity to be less than $1/9$ we must hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving a second-order recurrence relation with complex characteristic roots in polar form. I am self-studying this topic from a textbook and am stuck with trying work through one example.
Suppose we are solving the recurrence equation, $u_n = 2u_{n-1} - 2u_{n-2}$.
This has the characteristic equation $r^2 - 2r + 2 = 0... | Solving the characteristic polynomial is just a way of providing linearly independent solutions... In fact, both the real and imaginary parts will be solutions to your equation. So, the general solution will be in fact
$$
(\sqrt{2})^n (C \cos \frac{n \pi}{4} + D \sin \frac{n \pi}{4}).
$$
Very simply speaking, you have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Problem with $a\sin(x)+b\cos(x)=\pm\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) $
Consider $f(x)=a\sin(x)+b\cos(x)$
where $a,b$ are some real constants.
Putting $f(x)=R\sin(\alpha+x)$, I got
$$f(x)=\pm\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) \tag{1}$$
According to my Desm... | Denote
$$
\alpha = \arctan \frac{b}{a}.
$$
Then evidently
$$
\tan \alpha = \frac{b}{a}.
$$
Let's start with identity
$$
\sin^2 \alpha + \cos^2 \alpha = 1
$$
and divide each term by $\cos^2 \alpha$ to get
$$
\tan^2 \alpha + 1 = \frac{1}{\cos^2 \alpha} \implies \cos^2 \alpha = \frac{1}{1+\tan^2 \alpha} = \frac{a^2}{a^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the angle $x$ in this triangle
This image was doing the rounds on a popular text messaging application, so I decided to give it a try.
From sine rule in $\triangle ABP$:
$$\frac{AB}{\sin(150^\circ)} = \frac{AP}{\sin(10^\circ} \\ \implies AP = 2AB \sin(10^\circ)$$
Applying sine rule again in $\triangle APC$:
$$\fr... | Let $\angle ACP =z $. By trigonometric form of Ceva's theorem we have:
$$\frac {\sin60^\circ}{\sin20^\circ}\frac {\sin10^\circ}{\sin40^\circ}\frac {\sin(50^\circ-z)}{\sin z}=1.\tag1
$$
Further we have the following property for product of sines:
$$
\prod_{k=1}^{n-1}2\sin\frac{k\pi}n=n.
$$
Particularly for $n=9$ it give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$ Finding this minimum must be only done using ineaqualities.
$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$
Using inequalities of arithemtic and geometric means:
$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}... | We need to find a positive number $a$ such that
$$x^3+{1\over x^2}\ge a^3+{1\over a^2}$$
for all $x\gt0$. The existence of such an $a$ is not in doubt, since $x^3+{1\over x^2}\to\infty$ as $x\to0$ and $x\to\infty$. But
$$\begin{align}
x^3+{1\over x^2}\ge a^3+{1\over a^2}
&\iff(x^3-a^3)-\left({1\over a^2}-{1\over x^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:- Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long... | Since your function is even, your integral is$$\require{cancel}\frac12\int_{-\infty}^\infty\frac{\mathrm dx}{x^8+x^4+1}.$$On the other hand, $x^8+x^4+1=\dfrac{x^{12}-1}{x^4-1}$ and therefore the roots of $x^8+x^4+1$ are the roots of order $12$ of $1$ which are not fourth roots of $1$. Among these, those with imaginary ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Diophantine equation : $6^m+2^n+2=x^2$
Find $m,n,x\in\mathbb{N}$ such that $6^m+2^n+2=x^2$.
My first approach is to show that for $m,n\geq2$, there exist no solution for $x$ by using modulo $4$.
Case $1$ : $m=1$, $x^2=2^n+8$.
As $n\geq1\implies2\mid RHS\implies2\mid x^2\implies4\mid x^2\implies4\mid LHS\implies 4\mi... | Assume that $m$ and $n$ are both greater than $1$. Then, we have:
$$x^2 \equiv 6^m+2^n+2 \equiv 0+0+2 \equiv 2 \pmod{4}$$
which is impossible. Thus, we either have $m=1$ or $n=1$.
Case $1$ : $m=1$
Substituting yields:
$$2^n+8=x^2$$
If $n>3$, then $8 \mid x^2$ but $16 \nmid x^2$ which would be a contradiction. Thus, $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that $a^2bc + ab^2c + abc^2 \le a^3b+ac^3+b^3c$ Note: $a,b$ and $c$ are positive real numbers.
I tried to use excel and I believe that, after going through a bunch of numbers, this preposition is true. However, I do not know how to prove it mathematically. Can someone help me to prove this question.
| You have to show
$$abc(a+b+c)\leq a^3b+ac^3+b^3c$$
or (since $a,b,c > 0)$
$$a+b+c \leq \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a}$$
Now, Cauchy-Schwarz inequality helps as follows
$$(a+b+c)^2 = \left(\sqrt c\frac a{\sqrt c}+\sqrt a \frac b{\sqrt a }+\sqrt b \frac c{\sqrt b } \right)^2\stackrel{C.-S.}{\leq} (c+a+b)\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+... | $$I = \int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx$$
Take $x^4$ factor out from the square root.
$$ = \int\frac{x^2-1}{x^5\sqrt{2-2x^{-2}+x^{-4}}}$$
Now substitute $\sqrt{2-2x^2+x^4}=t$
We have,
$$\,dt = \frac{4(x^{-3}-x^{-5})}{2\sqrt{2-2x^{-2}+x^{-4}}}\,dx$$
$$I = \frac{1}{2}\int dt$$
$$I = \frac{1}{2}\sqrt{2-2x^{-2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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On the joint numerical range of a pair of symmetric matrices Proposition 13.4 of Alexander Barvinok's A Course in Convexity shows the existence of the following result:
Let $n\ge 3$. For two $n\times n$ symmetric matrices $A$ and $B$, and a PSD matrix $X$ with $\mbox{trace}(X) = 1$, there exists a unit vector $x$ such... | Some thoughts
Let $\alpha = \mathrm{Tr}(AX)$ and $\beta = \mathrm{Tr}(BX)$.
We need to find $x$ such that
\begin{align}
x^\mathsf{T} A x &= \alpha, \\
x^\mathsf{T} B x &= \beta, \\
x^\mathsf{T} x &= 1.
\end{align}
We use the Newton-Raphson method to solve the above system of three quadratic equations.
The Newton-Raphso... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Integer solutions to $2^m - 3^n = p \cdot C$ where $m, n , p$ are positive integer variables and $C$ is a positive integer constant. How many solutions are there for equation $2^m - 3^n = p \cdot C$ where $m, n, p$ are positive integer variables and $C$ is an odd positive integer constant greater than $3$?
Can we sa... | There are infinitely many solutions. Let $C=11$, $n=1$ and $m=10l+8,\exists l\in\mathbb{N}$ then $$2^{m}-3^n\equiv 2^{10l+8}-3\equiv 2^8-3\equiv 0\pmod{11}$$
and thus, we can let natural number $p=\dfrac{2^{10l+8}-3}{11}$ so that $(m,n,p,C)=(10l+8,1,1,11)$ are the solutions to the equation $2^m-3^n=11k$ $\Box$
And w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Rate of convegence of $\frac{E^2[X^n]}{ E[X^{n-1}] E[X^{n+1}]}$ as $n \to \infty$ Let $X$ be a non-negative discrete random variable such that $0\le X\le B$ and$$f(n)=\frac{E^2[X^n]}{E[X^{n-1}]E[X^{n+1}]}$$for $n \ge 1$.
I am interested in the rate of growth of $f(n)$ as $n \to \infty$. In other words, is there a way t... | Some idea
Let $m\ge 2$. Let $0 \le a_1 < a_2 < \cdots < a_m$ and $\mathrm{Pr}(a_1) = p_1, \mathrm{Pr}(a_2) = p_2, \cdots, \mathrm{Pr}(a_m) = p_m$ with $p_1+p_2+\cdots + p_m = 1$. We have
$$f(n) = \frac{(1+A)^2}{(1+B)(1+C)}$$
where
\begin{align}
A = \frac{\sum_{j=1}^{m-1} p_j a_j^n}{p_m a_m^n}, \
B = \frac{\sum_{j=1}^{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluate integral: $\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x +b^2\sin^2x)dx$? For $a,b>0$, show that
$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\pi\ln\left(\frac{a+b}{2}\right)$$
I came across to this problem in the book Table of integrals, series and product. So here is my try to prove the closed form.
For... | Let
$$ I(a)=\int_{0}^{\frac{\pi}{2}} \ln(a^2 \cos^2 x+b^2 \sin^2 x) dx $$
Thus
$$ I(b)=\int_{0}^{\frac{\pi}{2}} \ln(b^2 \cos^2 x+b^2 \sin^2 x) dx=\pi \ln b $$
\begin{align}
I'(a)&=\int_{0}^{\frac{\pi}{2}} \frac{2a\cos^2 x}{a^2 \cos^2 x+b^2 \sin^2 x}dx \\
&=2a \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{(a^2+b^2 \tan^2 x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\f... | $$\frac{dy}{dx} -\frac{3}{2x + 1}y = 3x^2$$
For an integrating factor we have
$$\frac{dy}{dx} + Py = Q, \quad I = \exp(\int P \; dx)$$
then $$Iy = \int IQ\;dx$$
For our method
$$I = \exp(\int -\frac{3}{2x + 1} \; dx) = \exp(-\frac{3}{2}ln(2x + 1)) = (2x + 1)^{-\frac{3}{2}}$$
so
$$y = (2x + 1)^\frac{3}{2} \int \frac{3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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$\int \frac{x^3+3x+2}{(x^2+1)^2 (x+1)} \ dx$ without using partial fraction decomposition One way to evaluate the integral $$\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \ dx $$ is to rewrite it as $$ \int \frac{x^3+x+2x+2}{(x^2+1)^2(x+1)} dx \\=\int\frac{x(x^2+1) +2(x+1)}{(x^2+1)^2(x+1)}dx\\=\int\frac{x}{(x^2+1)(x+1)}dx+\int\... | $$\operatorname*{Res}_{z=-1}\frac{z^3+3z+2}{(z^2+1)^2(z+1)}=\lim_{z\to -1}\frac{z^3+3z+2}{(z^2+1)^2}=-\frac{1}{2}$$
so we know in advance that the integrand function plus $\frac{1}{2(x+1)}$ can be written as $\frac{p(x)}{(x^2+1)^2}$:
$$ \frac{x^3+3x+2}{(x^2+1)^2(x+1)}+\frac{1}{2(x+1)} = \frac{x^3+x^2+x+5}{2(x^2+1)^2}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Equations via Lambert's W function I'm studying Lambert's W function and I came across the equation $2^x = 2x$.
Upon inspection it is easy to see that $x = 1$ and $x = 2$ are the real solutions to the equation.
Solving for Lambert's W function, we have:
$$
\begin{split}
&2^x = 2x \ \Rightarrow \ x2^{-x} = \frac{1}{2} \... | $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
When the argument of $\W(u)$ is in the range $(-\tfrac1\e,0)$,
the number $u$ always can be expressed in two equivalent forms:
\begin{align}
u&=\ln\left(a^{\tfrac 1{1-a}} \right)\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recursive squaring algorithm by squaring five numbers From Erickson (PDF):
**(c) Describe a recursive algorithm that squares any $n$-digit number in $O(n^{\log_3{5}})time$, by reducing to squaring only five $\left( n/3+O(1)\right)$-digit numbers. [Hint: What is $(a+b+c)^2+(a−b+c)^2$?]
Solving the hint,
$$\begin{ali... | Apparently the algorithm that the question is looking for simply Toom-Cook, and that the missing last square is $\left(x(-2)^2 + y(-2) + z\right)^2 = (4x-2y+z)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$
Simplify:
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
After the substitution as $\cos(x)=a$ and $\s... | Another way is to express everything in terms of $\cos 2x$: $$\frac{4\cos^22x -3(\cos^2x-\sin^2x)-\cos^2x}{4\cdot\frac{1-\cos2x}{2} -(1-\cos^22x)} \\=\frac{4\cos^22x-3\cos 2x -\frac{1+\cos 2x}{2}}{2(1-\cos 2x)+\cos^22x-1}\\=\frac{8\cos^22x-7\cos 2x-1}{2(1-\cos 2x)^2}\\=\frac{(8\cos 2x+1)(\cos 2x-1)}{2(\cos 2x -1)^2}\\=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The derivative of $f(x)=\frac{3 \sin x}{2+\cos x}$ My solution:
The background
$$\begin{align} &3\frac{d}{dx}\left(\frac{\sin(x)}{2+\cos(x)}\right)\\
&=3\frac{\frac{d}{dx}(\sin(x))(2+\cos(x))-\frac{d}{dx}(2+\cos(x))\sin(x)}{(2+\cos(x))^2}\\
&=3\frac{\cos(x)(2+\cos(x))-(-\sin(x))\sin(x)}{(2+\cos(x))^2}\\
&=\frac{3+6\cos... | Your answer is correct. Use trig. identity $\sin^2(x)+\cos^2(x)=1$
$$\therefore \frac{3(\sin^2(x)+\cos^2(x)+2\cos(x))}{(2+\cos(x))^2}=\frac{3(1+2\cos(x))}{(2+\cos(x))^2}=\frac{3+6\cos(x)}{(2+\cos(x))^2}$$
Above matches your solution
| {
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Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 1... | From $$n\equiv -5 \pmod{n+5} \Rightarrow \\
n^2\equiv 25 \pmod{n+5} \Rightarrow\\
n^2+n\equiv 20 \pmod{n+5} \Rightarrow \\
(n^2+n)^2\equiv 400 \pmod{n+5} \Rightarrow \\
n+5 \mid 4\cdot\left(\frac{n^2(n+1)^2}{4}-100\right)$$
but we want
$$\sum\limits_{k=1}^n k^3=\frac{n^2(n+1)^2}{4}=Q\cdot (n+5)+17$$
or
$$n+5 \mid 4\cdo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$... | The Euler's identity states that for all $x,y,z$ reals we have
$$x^3+y^3+z^3=3xyz+(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
But if $A,B,C$ angles of an acute triangle $ABC$ and $x=\cos A$, $y=\cos B$, $z=\cos C$, we have
$$
x+y+z=1+\frac{\rho}{R},\tag 1
$$
$$
xyz=\frac{\rho_0}{2R}\tag 2
$$
$$
x^2+y^2+z^2=1-\frac{\rho_0}{R}\tag 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Suppose that $z = x + iy$, with $y > 0$. Show that there are positive real numbers $u$ and $v$ with $2u^{2} = |z| + x$ and $2v^{2} = |z| - x$ Suppose that $z = x + iy$, with $y > 0$. Show that there are positive real numbers $u$ and $v$ with $2u^{2} = |z| + x$ and $2v^{2} = |z| - x$. Calculate $(u+iv)^{2}$. Show that $... | I think you accidentally solved for $x$ and $y$ instead of $u$ and $v$. Remember that $z = x+iy$ is fixed, so we're looking for $u$ and $v$ in terms of $x$ and $y$. I think you can probably reuse most of the work you already have for that, though.
All they mean by showing $z$ has two complex roots is that since you s... | {
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If $a,b,c$ are sides of a triangle, then find range of $\frac{ab+bc+ac}{a^2+b^2+c^2}$ $$\frac{ab+bc+ac}{a^2+b^2+c^2}$$
$$=\frac{\frac 12 ((a+b+c)^2-(a^2+b^2+c^2))}{a^2+b^2+c^2}$$
$$=\frac 12 \left(\frac{(a+b+c)^2}{a^2+b^2+c^2}-1\right)$$
For max value, $a=b=c$
Max =$1$
How do I find the minimum value
| In $\Delta ABC$, $a^2=b^2+c^2-2bc \cos A > b^2+c^2-2bc$.
Adding up the similar inequalities gives $ab+bc+ca > \frac{1}{2} (a^2+b^2+c^2)$
| {
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Can someone explain the limit $\lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)$? $$\begin{aligned}
&\text { Find the following limit: } \lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\rig... | $$\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+1}}\leq \frac{1}{\sqrt{n^2}}\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+2}}\leq \frac{1}{\sqrt{n^2}}\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+3}}\leq \frac{1}{\sqrt{n^2}}\\
\vdots\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+n}}\leq \frac{1}{\sqrt{n^2}}.$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741401",
"timestamp": "2023-03-29T00:00:00",
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If $ 1+ \frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}=\frac{A}{B}$ where $A$ and $B$ are coprime positive integers, then $5\nmid A$ and $5\nmid B$.
Let the sum $$1+ \frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}=\frac{A}{B}$$ where $A,B\in \mathbb{N}$ and $\gcd(A,B)=1$. Show that neither $A $ nor $B $ is divisible by $5$.... | Let $S:=\displaystyle\sum_{k=1}^{100}\,\dfrac1{k}$, and write $[n]:=\{1,2,\ldots,n\}$ for each positive integer $n$. Note that $B$ is not divisible by $5$ because
$$\begin{align}S&=\sum_{\substack{k\in[100]\\ 5\nmid k}}\,\dfrac1k +\frac15\,\sum_{\substack{k\in[20]\\5\nmid k}}\frac1k+\frac1{25}\,\sum_{k=1}^4\,\frac1k\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given matrix $A^2$, how to find matrix $A$?
Let $$A^2 = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$$ Knowing that $A$ has positive eigenvalues, what is $A$?
What I did was the following:
$$A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$$
so
$$A^2 = \begin{pmatrix}
a^2 + bc & ab+bd \\
ac+cd & bc+d^2
\end{pmatri... | Computing matrix powers can be done with diagonalization.
The eigenvalues of $A^2$ have sum $5$ (trace) and product $4$ (determinant), so they are $1$ and $4$.
The corresponding eigenvectors of $A^2$ are $\pmatrix{1\\-2}$ and $\pmatrix{1\\1}$, respectively.
Therefore, $A^2$ is diagonalized as follows: $\pmatrix{1&1\\-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the equation of the normal to the parabola $y^2=4x$ that passes through $(9,6)$
Let $L$ be a normal to the parabola $y^2 = 4x$. If $L$ passes through the point $(9, 6)$, then $L$ is given by
(A) $\;y − x + 3 = 0$
(B) $\;y + 3x − 33 = 0$
(C) $\;y + x − 15 = 0$
(D) $\;y − 2x + 12 = 0$
My attempt: Let $(h,k)$ be... | The equation of the normal is $\color{red}{y=\frac{-kx}{2}+\frac{9k}{2}+6}$. The point $(h,k)$ lies on it so we get
$$k=\frac{-kh}{2}+\frac{9k}{2}+6.$$
But $k^2=4h$ (since $(h,k)$ lies on the parabola as well), so we get
$$k^3-28k-48=0 \implies (k-6)(k+2)(k+4)=0 \implies k=6,-2,-4.$$
These value can be used to find th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}$ Evaluate $$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}$$
My attempt: $$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}=\lim_{x\to+\infty} \frac{3x^2+\cos{\sqrt{x}}}{x^3\sin{\frac{1}{x}}+\f... | $$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}=\lim_{x\to+\infty} \frac{3x^3(1+\frac{x\cos{\sqrt{x}}}{3x^3}) }{x^3 \left( \frac{\sin{\frac{1}{x}}}{\frac{1}{x}} +\frac{1}{x^3} \right)} = 3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Functions of several variables If $f(x,y) = x^2 + xy + y^2 - 3x + 4y - 5$. I know the domain is $\mathbb R^2$. How to determine the image of f is my issue.
| Have a look at my answer here: https://math.stackexchange.com/a/3619647/399263
You can always translate conics to cancel terms in $x,y$.
$f(x+a,y+b)=x^2+(-3+b+2a)x+xy+y^2+(2b+4+a)y+[\cdots]$
Solve $\begin{cases}2a+b-3=0\\2b+4+a=0\end{cases}\iff \begin{cases}a=\frac{10}3\\b=-\frac {11}3\end{cases}$
$$f(x+a,y+b)=\overbra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\alpha,\beta,\gamma$ are the roots of $x^3+x+1=0$, then find the equation whose roots are: $(\alpha-\beta)^2,(\beta-\gamma)^2,(\gamma-\alpha)^2$ Question:
If $\alpha,\beta,\gamma$ are the roots of the equation, $x^3+x+1=0$, then find the equation whose roots are: $({\alpha}-{\beta})^2,({\beta}-{\gamma})^2,({\gamma... | Let $a,b,c$ be the roots of $x^3+x+1=0$ so we have $a+b+c=0, ab+bc+ca=1,abc=-1$, so $a^2+b^2+c^2=-2$ and $c^3=-c-1$
We would explore a transformation from $x$ to $y$ to get the required cubic equation of $y$.
Let $$y=(a-b)^2=a^2+b^2-2ab\implies y=-2-c^2+2/c \implies c=\frac{3}{1+y}$$
Replacing $c$ by $x$ we get the req... | {
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"source": "stackexchange",
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Optimizing a quadratic in one variable with parameterized coefficients Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:
$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$
I want to find $0 < \theta < \frac{\pi}2$ for which I can later ta... | This is a similar but simple approach which gives the same result. Seeing that $y=\tan(\theta)$ can take any positive value, we go for maximizing $x$ and get $\max(x) = 1123$.
We have: $105 = x\tan\theta-\frac{g}{2} \frac{x^2\sec^2\theta}{(110)^2}$
Let $\ a=105,\ y=\tan\theta,\ b = \frac{g}{2} \frac{1}{(110)^2}.\quad$ ... | {
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"source": "stackexchange",
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Proof of inequality by Muirhead We have to prove:
$$\frac{\sqrt{pq}}{p+q+2r}+\frac{\sqrt{pr}}{p+r+2q}+\frac{\sqrt{pr}}{p+r+2q}\leq\frac{3}{4}$$
By multiplying it all out we get the following equivalent:
\begin{align*}
4\sum_{cyc}{\sqrt{pq}(p+r+2q)(q+r+2p)}\leq \\
3(p+r+2q)(p+q+2r)(r+q+2p)
\end{align*}
Let us put:... | Two last lines in your proof are wrong.
Since $$(5,1,0)\succ(4,2,0),$$ by Muirhead $$\sum_{sym}x^5y\geq\sum_{sym}x^4y^2,$$
but you wrote a reversed inequality.
By the way, there is a proof of your inequality without expanding.
Indeed, we need to prove that
$$\sum_{cyc}\frac{yz}{2x^2+y^2+z^2}\leq\frac{3}{4}$$ or
$$\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756545",
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Iterates of $\frac{\sqrt{2}x}{\sqrt{x^2 +1}}$ converge to $\text{sign}(x)$. In this post, a comment states that if $f(x):= \dfrac{\sqrt{2}x}{\sqrt{x^2 +1}}$ and $F_n:=\underbrace{f\circ \dots\circ f}_{n\text{ times}}$, then the pointwise limit $\lim\limits_{n \to \infty} F_n$ is equal to the sign function
$$
\text{sign... | You can find the $n^{th}$ iterate of $f$ explicitly.
$$f(x)=\dfrac{\sqrt2x}{\sqrt{x^2+1}},$$
$$f(f(x))=\frac{\sqrt2\dfrac{\sqrt2x}{\sqrt{x^2+1}}}{\sqrt{\dfrac{2x^2}{x^2+1}+1}}=\frac{2x}{\sqrt{3x^2+1}},$$
$$f(f(f(x)))=\frac{\sqrt2\dfrac{2x}{\sqrt{x^2+1}}}{\sqrt{\dfrac{4x^2}{3x^2+1}+1}}=\frac{2\sqrt2x}{\sqrt{7x^2+1}}$$
a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integer exponent equation
Show that $$(2^a-1)(2^b-1)=2^{2^c}+1$$ doesn't have solution in positive integers $a$, $b$, and $c$.
After expansion I got
$$2^{a+b}-2^a-2^b=2^{2^c}\,.$$
Any hint will be appreciated.
| $$2^{a+b}=2^a+2^b+2^{2^c}$$
Case $1$:
If $a=b$, then $$2^{2a}=2^{a+1}+2^{2^c}$$
If $a+1$ and $2^c$ are distinct, then surely, the hamming weight on the RHS is $2$ but the hamming weight on the left is $1$.
Hence we must have $a+1=2^c$
$$2^{2a}=2^{a+2}$$
Hence $2a=a+2$, hence $a=2$, but since $a+1=2^c$, we have $3=2^c$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac 12 \cos^{-1}x^2$ Let the above expression be equal to $\phi$
$$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$
$$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$
$$\frac{1+\t... | A bit late answer but I thought worth mentioning it.
First note that we can substitute $y=x^2$ and consider $0<y\leq 1$. Furthermore, the argument of $\arctan$ can be simplified as follows:
$$\frac{\sqrt{1+y}+\sqrt{1-y}}{\sqrt{1+y}-\sqrt{1-y}}=\frac{1+\sqrt{1-y^2}}{y}$$
Now, setting $y = \cos t$ for $t \in \left[0,\fra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a_2,a_3,\cdots,a_n$ be positive real numbers and $s=a_2+a_3+\cdots+a_n$. Show that $\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}I tried this problem in this way:-
Let the real numbers are $b_2,b_3,\cdots,b_n$. such that $(b_2,b_3,\cdots,b_n)$is a permutation of the numbers given $(a_2,a_3,\cdots,a_n)$. Hence $s=b_2+b_3+... | Inspired by the solution published in the American Mathematical Monthly:
We can suppose that $\ \forall k \ , \ 0<a_k<1 $.
$\displaystyle \sum_{k=2}^n a_k^{1-\frac{1}{k}} - s = \sum_{k=2}^n \left( a_k^{1-\frac{1}{k}}-a_k\right) = \sum_{k=2}^n a_k^{\frac{1}{2}}\left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right... | {
"language": "en",
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How can $\sin(2nx)$ be expressed as a polynomial? Let $P_n (u)$ be a polynomial in $u$ of degree $n$. Then, prove that for every positive integer $n$, $\sin(2nx)$ is expressible as
$(\cos x)(P_{2n-1} (\sin x))$ for some $P_{2n-1}$.
I tried by starting as
$\sin(2nx)=2\sin(nx)\cos(nx),$ and ta-da, it satisfies for $n=1,2... | Hint:
$$ \sin(2x)=(\cos x)(2 \sin x)$$
$$ \sin(4x) = 2 \sin(2x) \cos (2x) = 2(\cos x)(2 \sin x)(1- 2 \sin^2x) = (\cos x)(4\sin x - 8 \sin^3x)$$
$$ \sin(6x) = \sin(4x + 2x)= \sin(4x)\cos(2x) +\sin(2x)\cos(4x)$$
$$= \sin(4x) (1-2\sin^2 x) + \sin(2x) ( 1- \sin^2(2x))$$
$$ = \sin(4x) (1- 2\sin^2 x) + \sin(2x) - \sin^3(2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761405",
"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx$ without complex analysis This particular integral evaluates to,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\frac{\pi }{8}\ln \left(2\right)-\frac{3\pi }{8}+\frac{\pi }{3}\ln \left(2+\sqrt{3}\right)... | As said in comments, consider
$$I(a)=\int _0^{\infty }\frac{\log \left(ax^3+1\right)}{\left(x^2+1\right)^2}\,dx$$
$$I'(a)=\int _0^{\infty }\frac{x^3}{\left(x^2+1\right)^2 \left(a x^3+1\right)}\,dx$$
Using partial fraction decomposition, the integrand is
$$\dfrac{ a(2a^2-1)x^2+3a^2x-a(a^2+2)}{(a^2+1)^2\left(ax^3+1\right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx$? I am trying to evaluate
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx \quad (1)$$
The typical way to confront this kind of integrals are the conjugates i.e:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx = $$... | Let $\sqrt{\frac{1+x}{1-x}}=t.$
Thus, $$x=\frac{t^2-1}{t^2+1},$$ $$dx=\frac{2t(t^2+1)-(t^2-1)2t}{(t^2+1)^2}dt=\frac{4tdt}{(t^2+1)^2}$$ and we need to calculate $$\int\frac{4t(t+1)}{(t-1)(t^2+1)^2}dt.$$
Now, easy to show that:
$$\int\frac{4t(t+1)}{(t-1)(t^2+1)^2}dt=\frac{2t}{t^2+1}+\ln\frac{(t-1)^2}{t^2+1}+C.$$
Can you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)... | $$\cos(2x)=\cos (x+x) =\cos x \cos x-\sin x \sin x=\cos^2x-\sin^2x$$
$$I =\int \frac{1}{\cos 2x+3}dx= \int \frac{\sec^2x}{\sec^2x(\cos^2x-\sin^2x+3)}dx = \int \frac{\sec^2x}{1-\tan^2x+3\sec^2x}dx $$ Substitute $t=\tan x$
so that $dt=\sec^2x dx$
$$I=\int \frac{1}{1-t^2+3(1+t^2)}dt=\int \frac{1}{4+2t^2} dt=\frac{1} {2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How to solve this ODE: $x^3dx+(y+2)^2dy=0$? I am trying to solve $$ x^3dx+(y+2)^2dy=0 \quad( 1)$$
Dividing by $dx$, we can reduce the ODE to seperate variable form, i.e
$$ (1) \to (y+2)^2y'=-x^3 $$
Hence,
$$ \int (y(x)+2)^2y'(x) dy = \int -x^3dx = - \frac{x^4}{4} + c_1$$
This LHE seems to be easy to solve using integ... | The coefficient of ${\rm d}x$ depends only on $x$ and the coefficient of $y$ depends only on ${\rm d}y$. So $$0 = x^3\,{\rm d}x + (y+2)^2\,{\rm d}y = {\rm d}\left(\frac{x^4}{4} + \frac{(y+2)^3}{3}\right),$$which says that your solutions are described by $$\frac{x^4}{4} + \frac{(y+2)^3}{3} = c,$$for some $c \in \Bbb R$.... | {
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"url": "https://math.stackexchange.com/questions/3765155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$ What methods would work best to find $\displaystyle \int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$
As usual with this kind of integral i tried to differentiate with the respect of a parameter
$$\int _0^1\frac{\ln \left(ax^3+1\right)}{x+1}\:dx$$
$$\int _0^1\... | Instead of using Feynman's trick at once use the following substitution first
$$\underbrace{\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx}_{x=\frac{1-t}{1+t}}$$
$$=\ln \left(2\right)\underbrace{\int _0^1\frac{1}{1+x}\:dx}_{\ln \left(2\right)}+\int _0^1\frac{\ln \left(1+3x^2\right)}{1+x}\:dx-3\underbrace{\int _0^1\fra... | {
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"source": "stackexchange",
"question_score": "4",
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How to integrate $\int {2\over (x^2+2)\sqrt{x^2+4}}dx$?
Solve the following indefinite integral:
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx$$
My approach:
I used the substitution: $x=2\tan t$, $dx=2\sec^2t dt$
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx=\int \frac{2}{(4\tan^2t+2)\sqrt{4\tan^2t+4}}\cdot 2\sec^2t\ dt$$
$$=\in... | Alternatively, let $ x = 2\sinh t \Rightarrow \frac{dx}{dt} = 2\cosh t$.
The integral becomes
$$\begin{array} {r c l }
\displaystyle \int \frac2{(4\sinh^2 t + 2)(2\cosh t)} \cdot 2\cosh t \, dt
&=& \displaystyle \frac12 \int \frac1{2\sinh^2 t + 1} \, dt \\
&=& \displaystyle \frac12 \tan^{-1} (\tanh t) + C \\
&=& \disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative? I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hen... | You wrote, $\frac{1}{2^1} - \frac{1}{2^2} + \frac{1}{2^3} - \frac{1}{2^4} + \cdots = \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) $ ,
This is not correct,
$\frac{1}{2^1} - \frac{1}{2^2} + \frac{1}{2^3} - \frac1{2^4} + \cdots = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{2^n} = \sum_{n=1}^{\infty} \f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove complex numbers $a$ and $b$ are antipodal under stereographic projection $\iff a \overline{b} = -1$ I'm trying to prove the following statement:
Given $a, b \in \mathbb{C}$, prove that $a$ and $b$ correspond to antipodal points on the Riemann sphere under stereographic projection if and only if $a \overline{b} =... | For the direct implication, one could also use the inverse function of $f$, $\phi$:
$$ \phi (x,y,u) = \frac{x+iy}{1-u}$$
for $(x,y,u)\not= (0,0,1)$, $x^2+y^2+u^2=1.$
If $ P = (x,y,u)$ and $Q=(-x,-y,-u)$, then
$$ \phi(P)\overline{\phi(Q)} = \frac{x+iy}{1-u} \cdot \frac{-x+iy}{1+u} = -\frac{x^2+y^2}{1-u^2} = -1$$
The i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Find $\lim _{x \to \infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}$
I just started studying and in our first maths-exam we had to find the following limit:
$\lim \limits_{x \to \infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}$
The other tasks where rather easy, so I feel like I'm missing something obvious here, but I don't even really... | I would divide both numerator and denominator by $7^{x+1}$:
\begin{align*}
\frac{5^x+7^{x+1}}{7^x+5^{x+1}} &= \frac{\frac{5^x}{7^{x+1}} + \frac{7^{x+1}}{7^{x+1}}}{\frac{7^x}{7^{x+1}} + \frac{5^{x+1}}{7^{x+1}}}\\
&= \frac{\frac{1}{7}\left(\frac{5}{7}\right)^x + 1}{\frac{1}{7} + \left(\frac{5}{7}\right)^{x+1}}
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factorize $a^2-2ab+a^2b-2b^2$? I have been stuck on factorizing this:
$$a^2-2ab+a^2b-2b^2$$
I thought I could solve it by making $(a+b)$ as one factor but it didn't work then I tried to add and deduct some terms which that didn't lead me to anything either.
I don't really know what to do next.
| $a^2-2ab+a^2b-2b^2=$
$=-2b^2+a(a-2)b+a^2.$
In order to factorize the last polynomial, we have to solve the following quadratic equation:
$-2b^2+a(a-2)b+a^2=0$
that is equivalent to
$2b^2-a(a-2)b-a^2=0.$
$\Delta=a^2(a-2)^2+8a^2=a^2(a^2-4a+12)\ge0,$
$b=\frac{a(a-2)\pm\sqrt{a^2(a^2-4a+12)}}{4}=\frac{a(a-2)\pm a\sqrt{a^2-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + ... | Write $$(x - 1)^{100} + (x - 2)^{200}=k(x)(x-2)(x-1)+ax+b$$
SInce this is valid for all $x$ it is valid also for $$x=1: \;\;\; 1=a+b$$ and $$x=2: \;\;\; 1=a2+b$$
So $a=0$ and $b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $x\sin A+y\sin B+z\sin C=x^2\sin2A+y^2\sin2B+z^2\sin 2C=0$, show $x^3\sin3A+y^3\sin3B+z^3\sin 3C=0$ I'm having trouble with a question that came in one of my exam and is about complex numbers and trigonometry:
If $$x \sin A +y \sin B+z\sin C=0$$ and $$x^2 \sin 2A + y^2 \sin 2B + z^2 \sin 2C=0$$ then prove that $$x^... | If $\sin{C}=0$, so $$x\sin{A}+y\sin(180^{\circ}k-A)=0,$$ where $k$ is an integer number.
Thus, easy to show (just consider two cases: $k$ is odd and $k$ is even) that $$x^3\sin3A+y^3\sin(540^{\circ}k-3A)=0.$$
Now, let $\prod\limits_{cyc}\sin{A}\neq0.$
Thus, the first condition gives $$z=-\frac{x\sin{A}+y\sin{B}}{\sin(A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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How to compute $\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$?
How to compute the following integral?
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$$
So what I did is to change $\sin(x)$ to $\cos(x)$ with cofunction identity, which is $\sin(\frac{\pi}{2} -x) ... | As proved here
For all $n\geq 2$ we shall show that
$$I(n)=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n-1}x+\cos^{2n-1}x}dx=\frac{\pi}{2n-1}\sum_{k=0}^{n-2}{n-2\choose k}\operatorname{csc}\left(\frac{(2\pi(n-k-1)}{2n-1}\right)$$. Just set $n=2$ we have $$I(1)=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin^3x+\cos^3x}dx =... | {
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"url": "https://math.stackexchange.com/questions/3784579",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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CDF in Probability $$f(x)=\left\{\begin{array}{ll}
C & \text { for }-3 \leq x<3 \\
Dx & \text { for } 3 \leq x<5 \\
0 & \text { otherwise }
\end{array}\right.$$
Given that $P(-3 \leq x < 3) = \frac{2}{3}P(3 \leq x < 5)$, What is the cumulative distribution function $CDF(x)$.
So far I have tried to solve the following:
... | $\int_{-3}^{x}f(t)dt \neq \frac{2}{3} \int_{3}^{x}f(t)dt $, this need not be true
but
$\int_{-3}^{3}f(x)dx = \frac{2}{3} \int_{3}^{5}f(x)dx $
other equation will be
$\int_{-\infty }^{\infty }f(x)dx = 1 $
Get $C,D$ from above two equations
Finally CDF = $\int_{-\infty }^{x }f(t)dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784861",
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"source": "stackexchange",
"question_score": "1",
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The value of the following product is?
Evaluate the following product:
$$\newcommand{\T}[1]{\frac{\sin\frac{\theta}{#1}}{\tan^2\frac{\theta}{#1}\tan\frac{2\theta}{#1} + \tan\frac{\theta}{#1}}} \\
P(\theta) = \T{2} \times \T{2^2} \times \T{2^3} \times .... \infty$$
For $\theta = \frac \pi 4$
Simplified, $P(\theta)$ is... | Carrying on from the answer:
$$\begin{gather}
T(\theta, n) = \frac{\sin t}{\tan^2t\tan 2t + \tan t} \\
= \frac{\cos t}{\tan t \tan 2t + 1} \\
= \frac{\cos t(1-\tan^2t)}{1+\tan^2t} \\
= \cos t \cos 2t \\
= \cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r}
\end{gather}$$
Now,
$$P(\theta) = \cos\theta \cos \frac \theta ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Convergence of series with $a_{n + 1} = \sin{a_n}$ Given the sequence $a_n \mspace{10mu},\mspace{10mu}a_{n+1}=\sin(a_n)\mspace{10mu} and \mspace{10mu} a_1=1. $
Find if the series $\sum_{k=1}^{\infty}a_k$ converges or diverges.
First I found that $a_n$ is monotone decreasing sequence. If $a_n\in[0;\pi/2]$ then $\sin... | As you mentioned, $(a_n)$ decreases to $0$. Therefore, you have
$$a_{n+1}^{-2} - a_n^{-2} = \sin(a_n)^{-2} - a_n^{-2} = \left(a_n - \frac{a_n^3}{6} + o(a_n^4)\right)^{-2} - a_n^{-2}$$ $$=a_n^{-2} \left[\left(1 - \frac{a_n^2}{6} + o(a_n^3)\right)^{-2} - 1 \right] = a_n^{-2} \left(1 + \frac{a_n^2}{3} + o(a_n^2) - 1 \rig... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of all possible values of $a$ such that the following equation $(x - a)^2 + (x^2 - 3x + 2)^2 = 0$ has real root in $x$ :-
Find the sum of all possible values of $a$ such that the following equation $(x - a)^2 + (x^2 - 3x + 2)^2 = 0$ has real root in $x$ :-
What I Tried :- I know $(x^2 - 3x + 2) = (x - 1)... | Hint:sum of two non negative terms can be zero if they are both zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
I tried going along the path of computing $(x+y+z)... | Hint: Calculate $${(xy)(yz)\over xz}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}... | $$x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=\frac{\left(x+\frac{2}{3}x^3\right)^2-\frac{4}{9}(x^2+1)^3}{x + \frac{2x^{3}}{3}+ \frac{2(x^2+1)^{\frac{3}{2}}}{3}}=$$
$$=\frac{-\frac{1}{3}x^2-\frac{4}{9}}{x + \frac{2x^{3}}{3}+ \frac{2(x^2+1)^{\frac{3}{2}}}{3}}=\frac{-\frac{1}{3x}-\frac{4}{9x^3}}{\frac{1}{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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condition on complex numbers to form cyclic quadrilateral. Consider complex numbers $z,z^2,z^3,z^4$ in that order which form a cyclic quadrilateral . If $\arg z=\alpha$ and $\alpha$ lies in $[0,2\pi]$.Find the values $\alpha$ can take.
I encountered this question in one competitive exam.i tried using the property of c... | By Ptolemy we obtain:
$$|z-z^2|\cdot|z^3-z^4|+|z-z^4|\cdot|z^2-z^3|=|z-z^3|\cdot|z^2-z^4|$$ or
$$|z|+|z^2+z+1|=|(z+1)^2|.$$
Now, we can use a triangle inequality.
Id est, for $|z|=r$ we obtain:
$$(\cos\alpha,\sin\alpha)||(r^2\cos2\alpha+r\cos\alpha+1,r^2\sin2\alpha+r\sin\alpha),$$ which gives
$$\sin\alpha(r^2\cos2\alph... | {
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"url": "https://math.stackexchange.com/questions/3794703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Five roots of $x^5+x+1=0$ and the value of $\prod_{k=1}^{5} (2+x_k^2)$
*
*Here, $x_{k}$ are five roots of $x^{5} + x + 1 = 0$.
*I know two roots are $\omega, \omega^{2}$ and next I can find a cubic dividing it by $x^{2} + x + 1$ and using the connection of $3$ roots with the coefficients of this cubic( Vieta's form... | Note that $\prod_{k=1}^5 (2+x_k^2) = \prod_{k=1}^5 (\sqrt{2}+ix_k)(\sqrt 2 - ix_k)$.
This is the product of all roots of a polynomial, whose roots are exactly $\sqrt{2} \pm ix_k$ for $k=1,...,5$.
Note that if $x^5+x+1$ has roots $x_1,...,x_5$, then $p(y) = (-iy+\sqrt 2i)^5 + (-iy+\sqrt 2i) + 1$ has roots $\sqrt 2 + ix_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Find the inequality with the best possible $k= constant$ (with the condition $x^{2}+ y^{2}\leq k$).
Find the inequality with the best possible $constant$
*
*Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{7}$. Prove that
$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1}{1+ xy}\leq \frac{3}... | For example.
Let $x$ and $y$ be non-negative numbers such that $x^2+y^2\leq\frac{2}{15}.$ Prove that:
$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1.5}{1+ xy}\leq \frac{3.5}{1+ \left ( \frac{x+ y}{2} \right )^{2}},$$
where $\frac{2}{15}$ is a best such constant.
It's interesting that:
For any non-negatives $x$ an... | {
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"url": "https://math.stackexchange.com/questions/3804775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $\tan\theta =\cos2\alpha\tan\phi$ then prove that $\tan(\phi-\theta)=\frac{\tan^2\alpha\sin2\phi}{1+\tan^2\alpha\cos2\phi}$ If $\tan\theta =\cos2\alpha\tan\phi$ then prove that $\tan(\phi-\theta)=\frac{\tan^2\alpha\sin2\phi}{1+\tan^2\alpha\cos2\phi}$
I have tried in a few different ways applying the formula $\tan(\p... | $\tan(\phi-\theta)=\frac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}$.
If we replace $\tan\theta=\cos 2\alpha \tan \phi$ we get
\begin{align*}
\frac{\tan\phi-\cos 2\alpha\tan\phi}{1+\tan\phi\cos2\alpha\tan\phi}&=\frac{\tan\phi-(\cos^2\alpha-\sin^2 \alpha)\tan\phi}{1+\tan^2\phi(\cos^2\alpha-\sin^2\alpha)}\\
&=\frac{2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
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Sum of Telescopic Series Question :
$$\sum_{r=1}^n\frac{(r^2-r-1)}{(r+1)!}=?$$
I tried but I got $(-1)!$ in the end. This is what I tried:
$$
\require{cancel}
\frac{r^2-1}{(r+1)!}-\frac{r}{(r+1)!}\\
\frac{\cancel{(r+1)}\cancel{(r-1)}}{\cancel{(r+1)}r\cancel{(r-1)}(r-2)!}-\frac{\cancel{r}}{(r+1)\cancel{r}(r-1)!}\\
T_r... | $$\begin{eqnarray*}\frac{r^2-r-1}{(r+1)!}&=&\frac{r^2-1}{(r+1)!}-\frac{r}{(r+1)!}\\&=&\frac{(r-1)(r+1)}{(r+1)!}-\frac{r}{(r+1)!}\\&=&\frac{r-1}{r!}-\frac{r}{(r+1)!}\end{eqnarray*}$$
Therefore:
$$\begin{eqnarray*}\sum_{r=1}^n\frac{(r^2-r-1)}{(r+1)!}&=&\sum_{r=1}^n\left[\frac{r-1}{r!}-\frac{r}{(r+1)!}\right]\\&=&\frac{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
System of Three Equations - Prove that at least two numbers' absolute value is equal. Question :
Prove that at least two of the numbers $a,b,c$ must have the absolute values equal in order that the system of following equations in $x,y$ may be consistent
$$ a^2x+b^2y+c^2=0$$
$$a^4x+b^4y+c^4 =0$$
$$x+y+1=0$$
I found $... | $$a^2x+a^2y+a^2=0$$
$$\implies (a^2-b^2)y+(a^2-c^2)=0$$
$$a^4x+a^4y+a^4=0$$
$$\implies (a^4-b^4)y+(a^4-c^4)=0$$
If $|a|\neq |b|$ and $|a|\neq |c|$, then ${a^4-b^4\over a^2-b^2}={a^4-c^4\over a^2-c^2}$ which imlies $a^2+b^2=a^2+c^2$ which is $|b|=|c|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating two integrals involving $\tan^{-1}\left(\frac{\sqrt{x(1-x)}}{x+\frac12}\right)$
I want to show
$$I:=\int_0^{1}\tan^{-1}\left(\frac{\sqrt{x(1-x)}}{x+\frac12}\right)dx=\frac{\pi}{8}$$
and
$$J:=\int_0^{1}\frac{1}{1-x}\tan^{-1}\left(\frac{\sqrt{x(1-x)}}{x+\frac12}\right)dx=\pi\log \frac{3}{2}$$
My work:
Let us... | We can use an Euler substitution to simplify both integrals, namely: $\frac{\sqrt{x(1-x)}}{x}=t\Rightarrow x=\frac{1}{1+t^2}$.
$$I=\int_0^1\arctan\left(\frac{\sqrt{x(1-x)}}{\frac{1}{2}+x}\right)dx=\int_0^\infty\arctan\left(\frac{2t}{3+t^2}\right)\left(\frac{1}{1+t^2}\right)'dt$$
$$\overset{IBP}=\int_0^\infty \left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF
I have done this problem by inspection as $$\frac{x^2+6}{6}=[x] \implies x>0.$$
Let [x]=0, then $x$ is non real. Let $[x]=1$, then $x=0$ which contradicts. Let $[x]=2$, it gives $x=\sqrt{6}$, in agrrement. Simi... | $$\left\lfloor x \right\rfloor\le x$$
$$-6\left\lfloor x \right\rfloor\ge -6x$$
$$x^2-6\left\lfloor x \right\rfloor+6\ge x^2-6x+6$$
$$0\ge x^2-6x+6$$
$$x\in[3-\sqrt{3},3+\sqrt{3}] \ \ \ \& \ \ \left\lfloor x \right\rfloor\in\{1,2,3,4\}$$
so let's consider the cases:
$1^{\circ}$ $\left\lfloor x \right\rfloor=1\Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Show that $\frac{\left(2n\right)!}{2^{n}n!} \ge \frac{\left(2n\right)!}{\left(n+1\right)^{n}}$. I am attempting to solve this by induction, yet I am stuck on the last part. Below is my attempted solution.
Show that $\frac{\left(2n\right)!}{2^{n}n!} \ge \frac{\left(2n\right)!}{\left(n+1\right)^{n}}, \forall n \in \mathb... | By the AM-GM inequality you have
$$\sqrt{1 \cdot n} \leq \frac{n+1}{2} \\
\sqrt{2 \cdot (n-1)} \leq\frac{n+1}{2} \\
...\\
\sqrt{n \cdot 1} \leq \frac{n+1}{2} $$
Multiplying everything together you get
$$n! \leq \left( \frac{n+1}{2}\right)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve system of equations $\begin{align} \begin{cases} |y|=|x-3| \\(2\sqrt{z}+2-y)y=1+4y \\ x^2+z-4x=0\end{cases} \end{align}$ Solve system of equations$$\begin{align}
\begin{cases}
|y|=|x-3| \\
(2\sqrt{z}+2-y)y=1+4y \\
x^2+z-4x=0
\end{cases}
\end{align}$$
*
*$$y=3-x;z=4x-x^2$$
Then t... | In this case $4x-x^2\geq0,$ which gives $0\leq x\leq4.$
Now, after substitution $z=4x-x^2$ rewrite the second equation in the form:
$$2(3-x)\sqrt{4x-x^2}=(x-4)^2,$$ which gives $x=4$ or
$$2(3-x)\sqrt{x}=\sqrt{(4-x)^3},$$ which gives also $0<x<3$ and after squaring of the both sides we obtain:
$$4x(3-x)^2=(4-x)^3$$ or $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why are the functions $y=x^{\sin^2(x)}+x^{\cos^2(x)}$ and $y=(x-2\sqrt{x}+1)\cos^2(2x)+2\sqrt{x}$ so similar? I was just playing around with functions and plotting them in Desmos and I found that these functions are remarkably similar despite their totally different expressions. The functions are
$$y=x^{\sin^2(x)}+x^{\... | We have that
$$f(x)=x^{\sin^2(x)}+x^{\cos^2(x)}=x^{1-\cos^2(x)}+x^{\cos^2(x)}$$
$$g(x)=(x-2\sqrt{x}+1)\cos^2(2x)+2\sqrt{x}=$$
$$=(x-2\sqrt{x}+1)\cos^2(2x)+2\sqrt{x}(\cos^2(2x)+\sin^2(2x))=$$
$$=(x+1)\cos^2(2x)+2\sqrt{x}\sin^2(2x)$$
with
$$2\sqrt x \le f(x), g(x) \le x+1$$
and since, as noticed in the graph linked to yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Intuitive explanation for $\lim\limits_{n\to\infty}\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}=e^{-x^2}$ In this post I noticed (at first numerically) that:
$$\lim\limits_{n\to\infty}\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}=e^{-x^2}$$
This can b... | Consider first
$$ \begin{align}
\frac{(n+5)!}{n!} &=(n+5)(n+4)\cdots (n+1)\\
&= n^5 (1+5/n)(1+4/n) \cdots (1+1/n) \\
& \approx n^5 \left(1 + \frac{5+4+\cdots+1}{n}\right) \tag1\\
& = n^5 \left(1 + \frac{5\times6}{2n}\right) \\
\end{align}$$
Or, in general
$$ \frac{(n+x)!}{n!} \approx n^x \left(1 + \frac{x(x+1)}{2n}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Find a matrix $A$ such that $X$ generates the subspace $W$ (the solution space of the system $AX=0$.) Consider $W$ a subspace of $\mathbb{R}^{5}$ generated by:
\begin{align*}
X=\left \{(1,-1,0,5,1), (1,0,1,0,-2),(-2,0,-1,0,-1)\right \}
\end{align*}
Find a system of linear equations $AX=0$ such that W be the solution sp... | Hint: The kernel is the orthogonal complement of the row space. So $A$ needs row space equal to $W^\perp$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$ without using LHS and RHS
Prove that
$$\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$$
My Attempt
$$\Rightarrow \cos^2(2x)-\sin^2(2x) =8\cos^4(x)-8\cos^2(x)+1$$
Add 1 to both sides
$$\cos^2(2x)-\sin^2(2x) +1=8\cos^4(x)-8\cos^2(x)+2$$$$\Rightarrow \cos^2(2x)=4\cos^4(x)-4\cos^2(x)+1$$
$$\R... | For example: $$8\cos^4x-8\cos^2x+1=1-8\cos^2x\sin^2x=1-2\sin^22x=\cos4x.$$
Also, we can use your idea:
$$\cos4x=2\cos^22x-1=2(2\cos^2x-1)^2-1=8\cos^4x-8\cos^2x+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3826516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding $x$ when $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$ Find $x$ if $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1.$
I was thinking of trying to substitute some number $y$ written in terms of $x$ than solving for $y$ to solve for $x.$ However, I'm not sure what $y$ to input, so can someone give me a hint?
| $$\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$$
$$ \sqrt[3]{5 - 3x} = 1-\sqrt{3x - 4}$$
$$ 5 - 3x = (1-\sqrt{3x - 4})^3$$
$$ 5 - 3x = (1-3x)\sqrt{3x-4}+9x-11$$
$$ -4(3x-4)= (1-3x)\sqrt{3x-4}$$
$$ -4(3x-4)= (1-3x)\sqrt{3x-4}$$
$$ -4\sqrt{3x-4}\cdot \sqrt{3x-4}= (1-3x)\sqrt{3x-4}$$
so $x=4/3$ or: $$ -4 \sqrt{3x-4}=1-3x$$
which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$
Prove this trigonometric identity:
$$\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$$
I've simplified it until
$$\frac{2\cos^2\theta}{1-\sin\th... | First, let's make a common denominator of $\frac{1+\cos\theta}{\sec\theta-\tan\theta}+\frac{\cos\theta-1}{\sec\theta+\tan\theta}$.
Ready, set, go!
$$\require{cancel}\begin{aligned}\color{blue}{\frac{1+\cos\theta}{\sec\theta-\tan\theta}}+\color{red}{\frac{\cos\theta-1}{\sec\theta+\tan\theta}}&=\frac{\color{blue}{\left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Minimum of a function without calculus. $a=\frac{{(1+t^2)}^3}{t^4}$
Find the the minimum value of $a$. .$$a=\frac{{(1+t^2)}^3}{t^4}$$.
Instead of calculus i tried using the AM-GM inequality.,as follows:we have $$3+\frac{1}{t^4}+\frac{3}{t^2}+3\left( \frac{t^2}{3}\right)\ge 3+{\left(\frac{1}{9}\right)}^{1/5}$$ which ... | Using the AM-GM inequality, we have
$$1+t^2 = 1+\frac{t^2}{2}+\frac{t^2}{2} \geqslant 3\sqrt[3]{1 \cdot \frac{t^2}{2} \cdot \frac{t^2}{2}} = 3\sqrt[3]{\frac{t^4}{4}},$$
therefore
$$(1+t^2)^3 \geqslant \frac{27}{4}t^4,$$
or
$$\frac{(1+t^2)^3}{t^4}\geqslant \frac{27}{4}.$$
Equality occur when $\frac{t^2}{2} = 1,$ or $t =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine a in the system such that the system is consistent. The system is:
$x_1 + 2x_2 - x_3 = 2, $
$2x_1 - x_2 + x_3 = 1 $,
$-x_1 + 4x_2 -2x_3 = a $
To begin solving the system, I did Row 2 - 2(Row 1). Row 3 + Row 1, then Row 3 + Row 2.
This left me with
\begin{bmatrix}1&2&-1&2\\0&-5&3&-3\\0&0&0&a-1\end{bmatrix}
If ... | Generally you don't need to continue reducing to reduced echelon form since you can see that for the system to be consistent you need $a-1=0$ so $a=1$. But there is a a mistake.
Row 3 + Row 1 gives
$$\begin{bmatrix}1 & 2 & -1 & 2\\0 & -5 & 3 & -3\\0 & 6 & -3 & a+2\end{bmatrix}$$
so Row 3 + Row 2 will give
$$\begin{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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There are triads of perfect squares that are consecutive terms of arithmetic progression? Prove that there exist infinitely many triples of positive integers $ x , y , z $ for which the numbers $ x(x+1) , y(y+1) , z(z+1) $ form an increasing arithmetic progression.
$ \bigg( $ It is equivalent to find all triples of $ 4... | We have $$x^2+x+z^2+z=2(y^2+y)$$ or $$(x+z+1)^2+(x-z)^2=(2y+1)^2$$
We got Pythagorean triples.
Let $x+z+1=m^2-n^2$ and $x-z=2mn$, where $m>n$ and $m$ and $n$ have a different parity.
Thus, $$(x,y,z)=\left(\frac{m^2-n^2+2mn-1}{2},\frac{m^2+n^2-1}{2},\frac{m^2-n^2-2mn-1}{2}\right).$$
Also, we can always take $\frac{m^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\fr... | Area of triangle ABC $=\frac{abc}{4R}=0.25$ where $R=1$ and a>0, b>0, c>0
$abc=1$
$\frac{1}{a}+\frac{1}{b}\geq\frac{2}{\sqrt{ab}}=\sqrt c$
$\frac{1}{b}+\frac{1}{c}\geq\frac{2}{\sqrt{bc}}=\sqrt a$
$\frac{1}{a}+\frac{1}{c}\geq\frac{2}{\sqrt{ac}}=\sqrt a$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \sqrt a+ \sqrt b + \sqrt c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Calculate the Taylor polynomial of a definite integral I want to determine the Taylor polynomial of degree 1 to $f$ where $x=0$
$$f(x)=\int_0^{\sin x} \frac{e^t}{1+t^3} \,dt$$
This is my attempt:
$$f'(x)=\frac{e^{\sin x}}{1+\sin^3 x}$$
$$P(x)=f(0)+f'(0)x$$
$$f(0)=\int_0^0 \frac{e^t}{1+t^3}\,dt=0$$
$$f'(0)=\frac{e^{\sin... | You have
$$f(x)=\int_0^{a(x)} g(t) \, dt \implies f'(x)=g(a(x))\, a'(x)$$ by the fundamental theorem of calculus.
So, for your case
$$f'(x)=\frac{e^{\sin (x)}\cos (x)}{1+\sin^3(x)} $$
Now, it is a nice problem of series composition
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^9\right)$$
$$e^{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Count all positive integers less than 100000 with at least two consecutive 1s? Okay, aforementioned was the problem:
We need to count all the positive integers with at least two consecutive ones.
The way I thought about it was:
There are $10^{5}$ ways to have numbers that are less than 100 000. Now from these, I need t... | Exactly 2 1's?
What is the number of 5 digit numbers with 2 1's?
$1^2\cdot9^3\cdot{5\choose 2}$
How many of those have the 2 1's next to each other? How many locations are there for the first 1?
$1^2\cdot9^3\cdot{4\choose 1}$
3 ones
$1^3\cdot9^2\cdot{5\choose 3}$
And one of ${5\choose 3}$ has the 1's all separated by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$
if $x^5=1$ with $x\neq 1$ then find value of $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$$
So my first observation was x is a non real fifth root of unity. Also $$x^5-1=(x-1)(1... | Note that
$$
\frac1{1+x^n}=\frac12\frac{1+x^{5n}}{1+x^n}=\frac{1-x^n+x^{2n}-x^{3n}+x^{4n}}2\tag1
$$
Applying $(1)$ gives
$$
\begin{align}
\frac{x}{1+x^2}
&=\frac{x-x^3+x^5-x^7+x^9}2\\
&=\frac{x-x^3+1-x^2+x^4}2\tag2
\end{align}
$$
$$
\begin{align}
\frac{x^2}{1+x^4}
&=\frac{x^2-x^6+x^{10}-x^{14}+x^{18}}2\\
&=\frac{x^2-x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Where is the mistake when finding $\int x \sqrt{4+5 x} \ dx$? We have to find the integral of $\frac{dy}{dx}=x \sqrt{4+5x}$. This is from Morris Kline's book, Chapter $7$, exercise $5$, question $1$m.
I try to solve like following:
Let $u=4+5x$. Then $x=\frac{u-4}{5}$. Hence $\frac{d y}{d x}=\left(\frac{u-4}{5}\right) ... | Answer:
$I=\int_{}^{} x\sqrt{5x+4}$
$\Rightarrow $ $I=\frac{1} {5}\int_{}^{} 5x\sqrt{5x+4}$
$\Rightarrow $ $I=\frac{1} {5}\int_{}^{} (5x+4-4)\sqrt{5x+4}$
$\Rightarrow $ $I=\frac{1} {5}\int_{}^{} (5x+4)^{\frac{3}{2}} -\frac{4}{25}\int_{}^{} 5\sqrt{5x+4} $
$\Rightarrow $$I= \frac{1}{5} [\frac{2}{25} (5x+4)^{\frac{5}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $a_{n+1} = \frac{1}{2}(a_n+b_n), $ $b_{n+1} = \frac{1}{2}(b_n+c_n) $and $c_{n+1} = \frac{1}{2}(c_n+a_n)$ are convergent,
Question: Let $a_0,b_0,c_0$ be real numbers.
Define the sequences $(a_n)_n,(b_n)_n, (c_n)_n$ recursively by
$$a_{n+1} = \frac{1}{2}(a_n+b_n), \quad b_{n+1} = \frac{1}{2}(b_n+c_n) \quad \te... | Hint: $$a_{n+1}+b_{n+1}+c_{n+1}=a_n+b_n+c_n=\ldots=a_0+b_0+c_0$$
$$|a_{n+1}-b_{n+1}|+|b_{n+1}-c_{n+1}|+|c_{n+1}-a_{n+1}|=\frac12\,(|a_n-b_n|+|b_n-c_n|+|c_n-a_n|)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A Stirling number identity Let $s(n,j)$ denote the signed Stirling numbers of the first kind and $S(n,j)$ the Stirling numbers of the second kind.
I need the following (probably trivial) identity
$\sum\limits_{j = 0}^n {s(n,j)S(m + j,k)} = 0$ for $k < n$ and $\sum\limits_{j = 0}^n {s(n,j)S(m + j,n)} = {n^m},$ but don... | We seek to evaluate (note that this is zero by inspection when $k\gt
n+m$):
$$\sum_{j=0}^n (-1)^{n+j} {n\brack j} {m+j\brace k}$$
where $k\le n.$ Using standard EGFs this becomes
$$n! [z^n] \sum_{j=0}^n (-1)^{n+j}
\frac{1}{j!} \left(\log\frac{1}{1-z}\right)^j
(m+j)! [w^{m+j}] \frac{(\exp(w)-1)^k}{k!}
\\ = (-1)^n n! m! ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
What I tried...
The equation $x^2=y-6$ is of a parabola. To find the slope of the tangent to the parabola at the point $(1,7)$,
$$\... | By rearranging the equation of the circle, we quickly find that the centre of the circle is $(-8,-6)$ (as you have illustrated).
Let $O$ be the centre of the circle, and $P$ be the point where the line $y=2x+5$ and circle touch. Since the line and circle touch at $P,$ the line through $O$ and $P$ must be the line perpe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Prove that $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$
For any reals $a$, $b$, $c$ and $d$ prove that:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$$
C-S in the IMO 2001 stile does not help here:
\... | First, we give some auxiliary results (Facts 1 through 3). The proofs are easy and thus omitted.
Fact 1: Let $a, b$ be reals. Then $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\le \sqrt{\frac{207+33\sqrt{33}}{512}}$.
Fact 2: Let $\gamma$ be real. Then $\frac{\gamma}{1 + \gamma^2} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Differentiation: $y = 9x + \frac{3}{x}$ Differentiate $y = 9x + \frac{3}{x}$
I think the first step is to turn $\frac{3}{x}$ into a more "friendly" format, so $x$ to the power of something maybe? How do I get $x$ with an index from $\frac{3}{x}$?
| You can also apply the definition (whenever $x\neq 0$):
\begin{align*}
f(x) = 9x + \frac{3}{x} \Rightarrow f'(x) & = \lim_{h\to 0}\frac{1}{h}\left(9(x + h) - 9x + \frac{3}{x + h} - \frac{3}{x}\right)\\\\
& = \lim_{h\to 0}\frac{1}{h}\left(9h - \frac{3h}{x(x + h)}\right)\\\\
& = \lim_{h\to 0}\left(9 - \frac{3}{x(x+h)}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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can't solve $a-2b=3$ & $5b=4a+1$ $$a-2b=3\tag1\label1$$
$$5b=4a+1\tag2\label2$$
Coming to \eqref{1}, $a=2b+3$. Substituting that in \eqref{2}, we get
$5b=4(2b+3)+1$.
$5b=8b+12+1$.
$3b=-13$.
$b=-13/3$.
Substituting value of $b$ in \eqref{1}.
$a-2(-13/3)=3$.
$a-(-26/3)=3$.
Adding $26/3$ to both sides.
$a-0=3+26/3$.
$a=35... | No, you messed up where you said "Adding 26/3 to both sides." If you check again, you will see that this should actually be "subtracting 26/3 to both sides" since $$a-\left(-\frac{26}{3}\right)=a+\frac{26}{3}=3\implies a=-\frac{17}{3}$$.
Your answer and algebra for $b$ are correct, giving $a=-\frac{17}{3}, b=-\frac{13}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of first $n$ terms of the series $1+\frac{1}{4}+3+1+\frac{1}{4}+3+1+\frac{1}{4}+3+\ldots \ldots $ Question: What is the sum of the first $n$ terms of the divergent series :
$$
1+\frac{1}{4}+3+1+\frac{1}{4}+3+1+\frac{1}{4}+3+\ldots \ldots
$$.
My trials: $(1+1+1+\dots)+(\frac{1}{4}+\frac{1}{4}+\frac{1}{4... | The series is divergent
$S_n=\frac {n}{3}\times \frac{17}{4}$, if $n=3k,k\in Z^+$
$S_n=\frac {n+1}{3}\times \frac{17}{4}-3$, if $n=3k-1,k\in Z^+$
$S_n=\frac {n+2}{3}\times \frac{17}{4}-\frac{13}{4}$, if $n=3k-2,k\in Z^+$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A box contains 7 white and 5 black balls. A box contains 7 white and 5 black balls. If 3 balls are drawn simultaneously at random, what is the probability that they are not all of the same colour? Calculate the probability of the same event for the case where the balls are drawn in succession with replacement between d... | How about counting the complementary event--i.e., the three balls drawn are of the same color? This seems easier. There are two mutually exclusive cases: either all the balls drawn are white, or they are all black. In the first case, there are clearly $$\binom{7}{3} = \frac{7!}{3! 4!} = 35$$ ways to pick three whit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3856261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Simplify $\frac{x^3+1}{x+\sqrt{x-1}}$
Simplify $$A=\dfrac{x^3+1}{x+\sqrt{x-1}}.$$
Firstly, $x-1\ge0$ and $x+\sqrt{x-1}\ne0:\begin{cases}x-1\ge0\\x+\sqrt{x-1}\ne0\end{cases}.$ The first inequality is equivalent to $x\ge1$. Can we use that in the second inequality? I mean can we say that $x+\sqrt{x-1}>0$ because $x>0$ ... | The answer to your first question is yes, you can assume that $x\geq 1$ because of $\sqrt{x-1}$, as you said.
The second aim is also true: $x+\sqrt{x-1}\geq 0 $ because $x> 0$ and $\sqrt{x-1}\geq 0$.
So, the domain of your equation is $[1,+\infty)$.
Then, one thing you can do to simplify $A$ is to multiply and divide ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.
Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - ... | The systematic way is to compute the inverse of $2-x$ mod $x^4-2$ using the extended Euclidean algorithm:
$$
1= \frac{1}{14}(x^4-2) + \frac{1}{14} (x^3 + 2 x^2 + 4 x + 8)(2-x)
$$
(courtesy of WA).
Therefore,
$$
\frac{1}{2 - \alpha} = \frac{1}{14} (\alpha^3 + 2 \alpha^2 + 4 \alpha + 8)
$$
where $\alpha=\sqrt[4]{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $ Calculate below limit
$$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $$
Using L'Hôpital's rule might be too tedious. I wonder if there is a trick given the ... | Using binomial first order expansion we have
*
*$\sqrt{x^3+1}=\sqrt{2+(x^3-1)}=\sqrt 2\sqrt{1+\frac{x^3-1}2}\sim \sqrt 2+\sqrt 2\frac{x^3-1}4$
*$\sqrt{x^4+1}\sim \sqrt 2+\sqrt 2\frac{x^4-1}4$
*$\sqrt{x+1}\sim \sqrt 2+\sqrt 2\frac{x-1}4$
*$\sqrt{x^2+1}\sim \sqrt 2+\sqrt 2\frac{x^2-1}4$
then
$$\frac{x^2-1 + \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3862143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplify expression involving fractions for mathematical induction proof. I have this expression
$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$
that I need to manipulate into looking like
$1-\frac{1}{k+2}$
as the final step in my mathematical induction proof process.
My idea was to go something like this:
\begin{align*}
1-\fra... | The first part is right, but:
$$1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$$
$$= 1 - \left(\frac{k+2}{(k+1)(k+2)} -\frac{1}{(k+1)(k+2)} \right)$$
$$= 1 - \frac{k+2 \color{red}{-} 1}{(k+1)(k+2)}$$
and you can continue from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3862458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$ We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$
I just did the question above in the following way:
$t=\frac{\sqrt{6}+\sqrt{3}+1}{2}$
From Heron w... | We are to make use of $$ \cos (90+\theta) = -\sin \theta$$
Let $a=\sqrt{6}$, $b=\sqrt{3}$, $c=1$. So we assumed $a>b>c$ or, $A>B>C$.
Now use cosine-law,
$$ \cos A = \dfrac{b^2+c^2-a^2}{2bc}$$
and show by calculation that
$$ \cos A = -\sin B $$
where, $\sin B = \sqrt{1-\cos^2 B}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Present value of deferred annuity with varying amounts An annual annuity pays the amount 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 (in dollars), the first payment occurring at the end of the second year. The present value at $=0$ of this annuity is 25 dollars at an annual effective rate $i$.
Another annual annuity pays the amoun... | Let $$p(v)=v^2+2v^3+3v^4+4v^5+5v^6+6v^7+5v^8+4v^9+3v^{10}+2v^{11}+v^{12}$$
Since $p(0)=0$, $p$ is strictly increasing on $v\geq0$, and $p(v)\to\infty$ as $v\to\infty$, there is a unique $r>0$ such that $p(r)=25$.
Now, $$\begin{align}
p(v)&=v^2(1+2v+3v^2+4v^3+5v^4+6v^5+5v^6+4v^7+3v^8+2v^9+v^{10})\\
&=v^2(1+v+v^2+v^3+v^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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.