Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Another way to compute $\lim\limits_{x\to+\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)$ I need to compute as a title a limit with $x\to+\infty$. The only way I found to obtain a result is to use the L'Hôpital's rule:
$$\lim\limits_{x\to+\infty}x^2\log\bigg(\frac{x^2+1}{x^2+3}\bigg)=\lim\limits_{t\to 0}\frac{1}{t^2}\l... | If it is OK for you to calculate directly a derivative then you may proceed as follows:
*
*Set $x^2 = \frac{1}{t}$ and consider $t\to 0^+$
So, you get
\begin{eqnarray*}x^2\log\bigg(\dfrac{x^2+1}{x^2+3}\bigg)
& \stackrel{x^2 = \frac{1}{t}}{=} & \frac{\log \frac{1+t}{1+3t}}{t}\\
& = &\frac{\log (1+t) - 0}{t} - \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
} |
Find minimum of function and minimum point
We have a function $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$$ the task to find minimum of this function and point where is minimum.
I tried to do by mark $t=x+\frac{1}{x}$ and then simplify this function and after that find deriv... | Why is that substitution bad? We get $${t^6-(t^6-6t^4+9t^2-2)\over t^3+(t^3-3t)} = {6t^4-9t^2+2\over 2t^3-3t}$$
$$ = 3t+{2\over 2t^3-3t}$$
Derivative of that is $$3-{12t^2-6\over (2t^3-3t)^2}$$
So we want to solve $$4t^6-12t^4+9t^2= 4t^2-2$$i.e.
$$4t^6-12t^4+5t^2+ 2=0$$
Write $s=t^2$, so we have $$4s^3-12s^2+5s+2=0$$
W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x\to0^+}x(\lfloor \frac{1}{x}\rfloor+\lfloor \frac{2}{x}\rfloor+\cdots+\lfloor \frac{k}{x}\rfloor), \, k \in \mathbb N$. $$M:=x\left(\left\lfloor \frac{1}{x} \right\rfloor+\left\lfloor \frac{2}{x}\right\rfloor+\cdots+\left\lfloor \frac{k}{x}\right\rfloor\right),\, k \in \mathbb N.$$
Using $\lfloor y ... | Another way is, because $x$ is approaching $0^+$, i.e. it becomes very small and is positive, then $\exists n\in \mathbb{N}$ s.t. $n\leq \frac{1}{x}<n+1$ or $n=\left\lfloor \frac{1}{x} \right\rfloor$ then $\forall i\in\mathbb{N}$:
$$i\cdot n \leq \frac{i}{x}<i\cdot(n+1) \Rightarrow i\cdot n \leq \left\lfloor \frac{i}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Finding the modulus and argument of complex numbers My problem:
Find the magnitude and argument of the following.
$$\frac{3+4i}{1-i} + \frac{2-i}{2+3i}$$
My solution so far:
\begin{align}
\frac{3+4i}{1-i} + \frac{2-i}{2+3i} &= \frac{3+4i}{1-i} \times \frac{1+i}{1+i} + \frac{2-i}{2+3i} \times \frac{2-3i}{2-3i}\\
&=... | in your solution, you find, correctly, $a$ and $b$ such that $\frac{3+4i}{1-i} + \frac{2-i}{2+3i} = a+bi$
but you need to find the modulus and the argument of the number. That is, you need to find $r>0$ and $\theta\in[0,2\pi)$ such that
$$\frac{3+4i}{1-i} + \frac{2-i}{2+3i} = r(\cos \theta + i\sin \theta).$$
There is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A mistake on computing $\int \frac{dx}{\sqrt{x+1}+1}$ I have to find $\int \frac{dx}{\sqrt{x+1}+1}$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.
First I write $\frac{1}{\sqrt{x+1}+1}=\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}}{x}-\frac{1}{x}$, therefore $\int \frac{dx}{\sqrt{x+1}+1}... | Note that
$$
\frac{x+2-2\sqrt{x+1}}{x^2}
=\frac
{(x+2)^2-4(x+1)}
{x+2+2\sqrt{x+1}}
\cdot\frac 1{x^2}\ ,
$$
and this has nothing to do with
$$
\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \cdot \frac 1x
=
\frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{(\sqrt{x+1}+1)(\sqrt{x+1}+1)} \cdot \frac 1x
=
\frac 1{(\sqrt{x+1}+1)^2}\ ,
$$
which leads... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
... | Here's an analytical answer with greater emphasis on reasoning than anything specific to probability which might lend greater insight into the problem.
Consider if there was only one coin. The probability to have an even number of heads is $1\over2$, since there are two possible outcomes and we are only interested in o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 7
} |
Hypergeometric representation of Fresnel $S(x)$ I am trying to find a representation for the Fresnel integral
$$S(x)=\int_0^x\sin\frac{\pi t^2}{2}\,\mathrm dt$$
Then with $$\sin x=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
We have
$$S(x)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}\int_0^x\left(\frac{\pi t^2}2\right)^{2n+1}... | Note that the recurrence relation for $(S_n)_{n \in \mathbb{N}_0}$ implies
$$ S_n = \frac{\Gamma\left(n + \frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}\right) }\frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(n + \frac{3}{2}\right)} \frac{\Gamma\left(\frac{7}{4}\right)}{\Gamma\left(n + \frac{7}{4}\right)} \frac{\left(-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
At how many points is this piecewise function discontinuous?
$$f(x) =
\begin{cases}
\frac{3x}{x^2-16} & \text{when $x\leq 2$} \\[2ex]
\frac{x-2}{x^2-5x+6} & \text{when $x\gt 2$}
\end{cases}
$$
At what points is this piecewise function discontinuous?
My solution:
$x=-4, 4, 2, 3$ are the points at which the denominat... | You are correct about $x=-4$ and $x=3$, but you also need to check $x=2$ since that's the point where you jump from one function to the other.
For the piecewise function to be continuous at $x=2$ you need $$lim_{x \to 2} f(x) = f(2).$$ Since $f(x)$ is defined differently from the left and the right of $x=2$ you will n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solve the inequality $x^2 - 3 > 0$ For the inequality $x^2 - 3 > 0$, we have
\begin{align} x^2 - 3 & = (x+\sqrt 3)(x- \sqrt 3) > 0 \end{align}
Therefore,
\begin{align} x > -\sqrt 3 \end{align}
and
\begin{align} x > \sqrt 3 \end{align}
But, this is clearly wrong as we should get $x < -\sqrt 3$ and $x > \sqrt 3$ as the ... | First treat the inequality as if it were to be an equation: $ x^{2} - 3 = 0 $. Thus, $$ x^{2} = 3 $$ and solving for $x$, $$x =\pm \sqrt{3}$$
This makes $x = -\sqrt{3}$ and $x = \sqrt{3}$ the "roots" of that equation.
Now to solve the $inequality$ $ x^{2} - 3 > 0 $, keep in mind the aforementioned "roots" into the foll... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Detemine the unit digit of a number Find the unit digit of the number:
$$3^{7005} \times 6^{8000}$$
My turn:
$$3^{7005}\times 6^{8000}=3^{7005}\times 3^{8000} \times2^{8000}$$ $$3^{13005} \times 2^{8000}$$
But i could not go on ?
| We want the unit digit of the number $3^{7005} \cdot6^{8000}$. This is equivalent to asking for the residue of the number $3^{7005} \cdot6^{8000}$ modulo $10$. Hence, we want some number $n$ with $0 \leq n \leq 9$ such that
$$
n \equiv 3^{7005} \cdot6^{8000} \pmod{10}
$$
Note that $3^4 \equiv 1 \pmod{10}$ and $6^n \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $k=constant$ such that $f(a,\,b,\,c\,)=\frac{3a+2b}{\sqrt{5a^2-ab+b^2}}+\frac{3b+2c}{\sqrt{5b^2-bc+c^2}}\leqq f(a,k\,a+\sqrt[3]{abc}-k\,c,c\,)$ Give $3$ positve numbers $a,\,b,\,c$ such that $abc= 1$ , prove:
$$f\left ( a,\,b,\,c \right )= \frac{3\,a+ 2\,b}{\sqrt{5\,a^{\,2}- ab+ b^{\,2}}}+ \frac{3\,b+ 2\,c}{\sqrt{... | The inequality in your title is incorrect. There is no $k$ for which this is true.
Let us set $a = c$ in your inequality. Then we have
$$ f(a,b,a) \leq f(a, (a^2b)^{1/3}, a).$$
This implies
$$f(a,b,a) \leq f(a, (a^2b)^{1/3}, a) \leq f(a, (a^8 b)^{1/9}, a) \leq f(a, (a^{26} b)^{1/27}, a) \leq ... \leq f(a, a, a).$$
This... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$ via partial fractions $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$
= $ \int \frac{Ax+B}{(x+1)} + \frac{Cx+B}{(x+1)^2} + \frac{Dx+E}{x+2}$
= $\int (Ax+B)(x+1)(x+2) + (Cx+B)(x+2) + (Dx+E)(x+1)^2$
= $ \int Ax^3 + 3Ax^2 + 2Ax + Bx^2 + 3Bx + 2B + Cx^2 + 2Cx + Bx + 2B + Dx^3 + ... | For a partial fraction decomposition, one only requires a polynomial on the resulting fraction numerators to be one less than those of the corresponding denominators. In this case the resulting fractions should be
$$\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}$$
which simplifies the equations significantly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Can the value of the Riemann zeta function at $n=2$ be derived from the Wallis formula for $\pi$? It is well known that the Riemann zeta function, defined for all positive integers $n>1$ by
$$
\zeta(n) = \sum_{m=1}^{\infty} m^{-n}
$$
takes the value $\displaystyle \frac{\pi^2}{6}$ at $n=2$. On the other hand, a product... | There are several proofs for the sine product, e.g. look for “sine product formula proof”.
You can also take a look here, there is a link to the old proof of Euler.
For a proof of the Wallis product it’s enough to look e.g. here.
It’s $~\displaystyle \frac{\pi}{2} = \frac{\pi x}{\sin(\pi x)}|_{x=\frac{1}{2}} = \Gamma(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{x+4}{x^2 + 2x + 5}$ I am having issues with this integral. I am not sure if it is irreducible or not. I can't use the quadratic formula, but I can rearrange the integral to be $\int \frac{x+4} {(x^2 + 2x + 1) + 4}$, but I don't know how to deal with the $+4$.
Here is my work treating the quadratic... | We will integrate first of all change quadratic $(x+1)^2+4$ and after fraction of $x+1+3$ and divide by quadratic and we can integrate it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How can this integral be solved? I am trying to solve (by hand) the integral:$$\int_0^∞\frac1x\left(\frac{(x+a)^2}{((x+a)^2+b)^{3/2}}-\frac{(x-a)^2}{((x-a)^2+b)^{3/2}}\right)\,\mathrm dx,$$
where $a$ and $b$ are real and positive. Mathematica actually generates the answer, but I am stumped on how to solve it by hand an... | Let $I$ denote the integral. By a simple substitution, we find that
$$ I
= \mathrm{PV}\!\int_{-\infty}^{\infty} \frac{1}{x-a} \frac{x^2}{(x^2+b)^{3/2}} \, \mathrm{d}x
= \operatorname{Re}\left( \int_{-\infty + i\epsilon}^{\infty+i\epsilon} \frac{1}{z-a} \frac{z^2}{(z^2+b)^{3/2}} \, \mathrm{d}z \right), $$
where the path... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Question about partial fractions with irreducible quadratic factors Given this rational function: $$\frac{-4x^4-2x^3-26x^2-8x-44}{(x+1)(x^2 +3)^2}$$
The decomposition would look like this: $$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$
And the final answer would be: $$\frac{-4}{x+1} - \frac{2}{(x^2+3... | The five functions
$$f_1(x)=\frac{1}{x+1},\; f_2(x)=\frac{x}{x^2+3}, \; f_3(x)=\frac{1}{x^2+3},\; f_4(x)=\frac{x}{(x^2+3)^2}, \; f_5(x)=\frac{1}{(x^2+3)^2}.$$
is a basis which linearly generates all rational functions of the form
$$\frac{P(x)}{(x+1)(x^2 +3)^2}$$
where $P$ is ANY polynomial of degree $\leq 4$.
In your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Integrate $\int \frac{dx}{\sqrt{x^2-9}}$ by trig substitution $$
\begin{align}
x = 3\sec\theta, dx &= 3\sec\theta\tan\theta d\theta\\\\
\int \frac{dx}{\sqrt{x^2-9}} &= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{(3\sec\theta)^2 - 3^2}} \\\\
& = \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2(\sec^2\theta -1)}} ... | Another approach:
$$I=\int\frac{dx}{\sqrt{x^2-9}}=\frac{1}{3}\int\frac{dx}{\sqrt{\left(\frac{x}{3}\right)^2-1}}$$
Let $u=\frac{x}{3}\implies3du=dx\implies$
$$I=\int\frac{du}{\sqrt{u^2-1}}=\text{arccosh}(u)+C=\text{arccosh}\left(\frac{x}{3}\right)+C$$
But recall the identity for $x\in[1,\infty)$ $$\text{arccosh}(x)=\ln\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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If $\sin(x) - \cos(x) = 1/3$ then determine $\sin(x)\cos(x)$
If
$$\sin(x) - \cos(x) = \frac{1}{3}$$
then determine
$$\sin(x)\cos(x)$$
I know that the expected solution is squaring both sides of equation and solving it this way:
\begin{gather}
\sin^2(x)+\cos^2(x)= 1 \\[4px]
(\sin(x) - \cos(x))^2 = \left(\frac{1}... | You can use $$\sin x = \frac{2t}{1+t^2}; \cos x = \frac {1-t^2}{1+t^2}$$
where $t=\tan (x/2)$, but you just have to know that some value of $t$ works.
Then you get a quadratic to solve for $t$. And once you know $t$ you can solve the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Differentiate $11x^5 + x^4y + xy^5=18$ I am not sure how to differentiate $11x^5 + x^4y + xy^5=18$. I have a little bit of experience with implicit differentiation, but I'm not sure how to handle terms where both variables are multiplied together.
I have tried
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\f... | This is how you could do it:
$11x^5+x^4y+xy^5=18$
Use the $\frac{d}{dx}$ operator (not $\frac{dy}{dx}$)
$55x^4+\frac{d}{dx}(x^4y+xy^5)=0$
Now, use the product rule on each part in the brackets:
$55x^4+y\frac{d}{dx}x^4+x^4\frac{dy}{dx}+y^5\frac{d}{dx}x+x\cdot4y^4\frac{dy}{dx}=0$
$55x^4+y\cdot3x^3+x^4\frac{dy}{dx}+y^5+4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
double integration with a domain Given that $D = \{ 1 \leq x+y \leq 2 , y+1 \leq x \leq 3y+1 \}$ and we want to evaluate $\int \int \limits_D \frac{x+y-1}{y^2 } e^{\frac{x-1}{y}}$ and used the substation $u=x+y$ and $v = \frac{x-1}{y}$ which the answer in the book use it, and get to the integral $\int \limits_{1}^{2} d... | The notations:
$$\begin{cases}u=x+y\\ v = \frac{x-1}{y} \end{cases} \Rightarrow \begin{cases} x=\frac{1+uv}{1+v}\\ y=\frac{u-1}{1+v}\end{cases}$$
The Jacobian:
$$\begin{vmatrix}x_u& x_v\\ y_u&y_v\end{vmatrix}=\begin{vmatrix}\frac{v}{1+v}& \frac{u-1}{(1+v)^2}\\ \frac1{1+v}&\frac{1-u}{(1+v)^2}\end{vmatrix}=\frac{1-u}{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove the inequality $\sum x+6\ge 2(\sum\sqrt{xy}) $ Let $x;y;z\in R^+$ such that $x+y+z+2=xyz$. Prove that $$x+y+z+6\ge 2(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}) $$
This inequality is not homogeneous and look at the condition i thought that
i would substitute the variables $x;y;z$ such as:
+)If i need to solve $x^2+y^2+z^2+... | The condition gives
$$\sum_{cyc}\frac{1}{x+1}=1.$$
Now, let $x=\frac{b+c}{a}$ and $y=\frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{a+b}{c}$ and we need to prove that:
$$\sum_{cyc}\frac{b+c}{a}+6\geq2\sum_{cyc}\sqrt{\frac{(b+c)(a+c)}{ab}}$$ or
$$\sum_{cyc}(a^2b+a^2c+2abc)\geq2\sum_{cyc}a\sqrt{bc(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{... | $$e^{\frac{1}{x}}<\frac{x}{x-1}$$
$$e^{\frac{1}{x}}<\frac{1}{1-\frac{1}{x}}$$
$$e^{-\frac{1}{x}}<\frac{1}{1+\frac{1}{x}}$$
$$\frac{1}{e^{\frac{1}{x}}}<\frac{1}{1+\frac{1}{x}}$$
$$\frac{1}{1+\frac{1}{x}+\frac{1}{2x^2}+...}<\frac{1}{1+\frac{1}{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Prove that $\frac{a + b + c}{b - a} > 3$ with $ab < b^2 < 4ca$.
$a$, $b$, $c$ are three positives such that $ab < b^2 < 4ca$. Prove that $$\large \dfrac{a + b + c}{b - a} > 3.$$
I can't think of a way to get around this problem. Although I can see that based on the given condition, $ax^2 + bx + c = 0$ has no roots, w... | Assuming $f(x) = ax^2+bx+c>0$ for all real $x$ and
$a>0$ and $(b-a)>0$
put $x=-2,$ We get
$f(-2)
=4a-2b+c> 0\Rightarrow 2a+c>2(b-a)$
$\Rightarrow \displaystyle \frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}>1+2=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Continued fraction of $\frac{1+\sqrt{5}}{2}$ I am trying to get a better understanding of continued fractions (CF) and was watching a view tutorial clips e.g. this here and looking through some stackexchange articles. Than found this article, where someone was able to get the CF of $\frac{1+\sqrt{5}}{2}$. Unfortunately... | $$
\rho = \frac{1+\sqrt{5}}2
$$
$$2 < \sqrt{5} < 3 \implies \left\lfloor \rho \right\rfloor = 1$$
$$\rho = 1 - 1 + \frac{1+\sqrt{5}}2$$
$$= 1 + \frac{\sqrt{5}-1}2$$
$$= 1 + \frac{( \sqrt{5}-1)*(1 + \sqrt{5})}{2(1 + \sqrt{5})}$$
$$= 1 + \frac{(5 + \sqrt{5} - \sqrt{5} -1 )}{2(1 + \sqrt{5})}$$
$$= 1 + \frac{4}{2(1 + \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Fewest number of marbles in a bag such that drawing the probability of drawing 2 blue marbles is $\frac{1}{6}$.
Two marbles are randomly selected without replacement from a bag
containing blue and green marbles. Probability of drawing both blue is
$\frac{1}{6}$. If three marbles are drawn, then the probability all... | Divide the second equation by the first:
$$\frac{x-2}{y-2} = \frac{6}{21} = \frac{2}{7}$$
Which when simplified gives:
$$7x = 2y + 10$$
What are the smallest integer values of x and y that solve this equation? Well, observe that a possible solution is $x = 2, y = 2$, but this is clearly wrong since $x = y$ in this case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Series sum with factorial notation The sum of series
$\displaystyle 1+\frac{1}{1!}\cdot \frac{1}{4}+\frac{1\cdot 3}{2!}\cdot \frac{1}{4^2}+\frac{1\cdot 3 \cdot 5}{3!}\cdot \frac{1}{4^3}+\cdots $
what i try:
$$1+\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1\cdot 3\cdot \cdots (2r-1)}{r!\cdot 4^r}$$
$$1+\lim_{n\righta... |
The series under consideration is
\begin{align*}
\color{blue}{1+\sum_{r=1}^\infty\frac{(2r-1)!!}{r!4^r}}&=1+\sum_{r=1}^\infty\frac{(2r)!}{r!4^r(2r)!!}\\
&=1+\sum_{r=1}^\infty\frac{(2r)!}{r!r!}\left(\frac{1}{8}\right)^r\\
&=\sum_{r=0}^\infty\binom{2r}{r}\left(\frac{1}{8}\right)^r\\
&=\left.\frac{1}{\sqrt{1-4z}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Any quicker method to find matrix exponent for a scaled permutation matrix? I am trying to find the eigen value of matrix $e^A$ such that the matrix A is given as
$$
\begin{bmatrix}
a & 0 & 0\\
0 & 0 & a\\
0 & a & 0
\end{bmatrix}
$$
I found $A^2 = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2 \end{bmatrix... | By diagonalization,
$$\exp\begin{pmatrix} 0 & a \\ a & 0 \end{pmatrix} = \begin{pmatrix} \cosh(a) & \sinh(a) \\ \sinh(a) & \cosh(a)\end{pmatrix} $$
is just De Moivre/Euler's formula $e^{i\alpha}=\cos(\alpha)+i\sin(\alpha)$ written in a equivalent way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim\limits_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$ Find the limit, $$\lim_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$$
Where $[.]$ denotes the integral part of $.$?
Efforts: If $x$ is integer,
$$\lim_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$$ $$\lim_{n\to\infty}{x(1+2^2+\dots +n^2)\over n^3}$$... | Applying Stolz–Cesàro theorem with
$a_n=\sum\limits_{k=1}^n [x\cdot k^2]$ and
$b_n=n^3$ (which is strictly monotone and divergent)
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=
\frac{[x(n+1)^2]}{(n+1)^3-n^3}=
\frac{[x(n+1)^2]}{(n+1)^2+(n+1)n+n^2}$$
and we have (using inequality @Winther suggested in the comments $x-1\leq[x]<x$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Showing that the locus of point $N$ is $x^2+y^2=a^2$
Question: A point $P(a\cos\theta,b\sin\theta)$ lies on an ellipse with equation $$\varepsilon:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$The tangent to the ellipse $\varepsilon$ at point $P$ is perpendicular to a straight line $l$ which has passed through its focus and int... | Denote $P$ by $(x_P, y_P)$ instead, then $\dfrac{x_P^2}{a^2} + \dfrac{y_P^2}{b^2} = 1$ and the tangent line at $P$ is$$
t_P: \frac{x_P x}{a^2} + \frac{y_P y}{b^2} = 1. \tag{1}
$$
Because $l$ is perpendicular to $t_P$, then $l$ is given by$$
l: -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} = C_P,
$$
where $C_P$ is a constant d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Shortcut for value of $f(1)$ where $f(x) = \int e^x \left(\arctan x + \frac {2x}{(1+x^2)^2}\right)\,dx$ If $$f(x) = \int e^x \biggr(\arctan x + \frac {2x}{(1+x^2)^2}\biggr)\,dx$$ and $f(0)=0$ then value of $f(1)$ is?
This is actually a Joint Entrance Examination question so I have to do it in two minutes. Is there a s... | Actually there is a formula $$\int e^x (g (x)+g'(x))\,dx = e^x\cdot g (x)+c.$$
Now for $$\int e^x \biggr(\arctan x + \frac {2x}{(1+x^2)^2}\biggr)\,dx $$, do the following manipulation:
$$\int e^x \biggr(\arctan x + \frac {2x}{(1+x^2)^2}\biggr)\,dx =\int e^x \biggr(\arctan x - \frac {1}{1+x^2}+\frac {1}{1+x^2}+\frac {2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6... | Alternatively:
$$\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x=\frac{\sin^6x+\cos^6x+1}{\sin^6 x\cos^6x}=\\
\frac{2^6[(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)+1]}{\sin^62x}=\\
\frac{64(2-\frac34\sin^22x)}{(\sin^22x)^3}=\frac{128-48(1-\cos^22x)}{(\sin^22x)^3}=\\
\frac{80+48\cos^22x}{(\sin^22x)^3}\ge 80.$$
equality o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Upper Bound for Polynomial Using Evenly Spaced Points Suppose that $x \in [0,1]$ and the points $x_1, x_2,\ldots x_n$ are evenly spaced in the interval $[0,1]$. I am trying to find a tight bound for the maximum of:
$$(x - x_1)(x-x_2)\cdots(x-x_n) $$
I realize that the above expression will be smaller in absolute value ... | Let $n\geq3$ and let $x_i:=\frac{i}{n-1}$ so that we want to determine the maximum of the polynomial
$$f_n(x):=\prod_{i=1}^n(x-x_i)=\prod_{i=1}^n\left(x-\frac{1-i}{n-1}\right),$$
on the interval $[0,1]$. If $n$ is odd then for all $x\in(x_1,x_2)$ and $i>2$ we have
$$\frac{1-i}{n-1}<x-x_i<\frac{2-i}{n-1}<0,$$
from whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$28 + p \sqrt{3} = (q + 2 \sqrt{3})^2$, where p and q are integers, find the values of p and of q. This is how much I have solved so far,
$$28 + p \sqrt{3} = (q + 2 \sqrt{3})^2$$
$$ \implies28 + p \sqrt{3} = q^2 + (4 \sqrt{3})q + 12$$
$$\implies p \sqrt{3} = q^2 + (4 \sqrt{3})q - 16$$
$$\implies p = (q^2 + (4 \sqrt{3}... | Just do it.
$28 + p\sqrt{3}=(q+2\sqrt 3)^2 = q^2 +4q\sqrt 3 + 12$
$16 - q^2 = \sqrt 3(4q - p)$.
$16-q^2$ is an integer. So is $4q -p$. So $\sqrt 3\times (4q-p)$ is an integer.
Let $\sqrt 3(4q-p) = M$ then our first thought is $\sqrt 3 = \frac M{4q-p}$. But thats impossible as $\sqrt 3$ is irrational.
The only way... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $. then find the minimum value of $x^3 y + xy^3 +4$.
I assume $ x = r \sin (w)$ and $ y = r\cos(w) $. $ x^3 y + xy^3 +4 = L $ which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my ... | Lagrangian multiplier for constrained function
$$L(x,y)= F(x,y)-\lambda G(x,y) $$
set in the order given, partially differentiating with respect to both variables
$$ \dfrac{F_x}{F_y}=\dfrac{G_x}{G_y}= \lambda $$
$$\dfrac{2x+y}{2y+x}=\dfrac{3x^2y+x^3}{x^3+3y^2} $$
Cross multiplying to simplify we get two conditions
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Proving that $g(x)$ defined as $x^2$ on the rationals and $x^4$ on irrationals, is discontinuous at $2$ Define $g: \mathbb{R} \to \mathbb{R}$ by $$g(x) =\begin{cases} x^2 & \text{if } x \text{ is rational}, \\\\ x^4 & \text{if } x \text{ is irrational}. \end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.
Solution... | Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
$$-\sqrt[4]{5} < x < \sqrt[4]{5}.$$
If you had chosen instead an irrational $x$ such that
$$2 - \min\{\delta, 2 - \sqrt[4]{5}\} < x < 2,$$
then you'd have $x > 2 - (2 - \sqrt[4]{5}) = \sqrt[4]{5}$, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Computation of the Laplacian form of $g_{\sigma, \eta}$ in Huisken and Sinestrari's paper I'm trying understand how to compute evolution equation for $g_{\sigma, \eta}$ defined on page $3$ of this article and I'm reading the Lectures on Mean Curvature Flows by Xi-Ping Zhu, which gives some details about the computation... | Computing in normal coordinates,
\begin{align*}
\Delta g_{\sigma, \eta} &= \nabla_i \nabla_i g_{\sigma,\eta}\\
&= \nabla_i \left[ \frac{(\nabla_i |A|^2 - (1 + \eta) 2 H \nabla_i H) H^{2 - \sigma} - (|A|^2 - (1 + \eta) H^2) (2 - \sigma) H^{1 - \sigma} \nabla_i H}{(H^{2 - \sigma})^2} \right]\\
&= \nabla_i \left[ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
McLaurin expansion of $(a^2+z^2)^{-\frac{1}{2}}$
Show that $$(a^2+z^2)^{-\frac{1}{2}}=\frac{1}{a}+\sum_\limits{n=1}^{\infty}\frac{(-1)^n(2n-1)!!)}{a^{2n+1}(2n)!!}z^{2n},\:|z|<|a|,a>0$$
$$(a^2+z^2)^{-\frac{1}{2}}=\frac{1}{a}(1+\frac{z^2}{a^2})^{-\frac{1}{2}}=\frac{1}{a}(1+\frac{z^2}{a^2})^{-\frac{1}{2}}=\frac{1}{a}+\f... | Hint. This might be useful
\begin{align}
\frac{(2n-1)!!}{(2n)!!} &=\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n}\\\\
&=\frac{1\cdot \color{red}{2}\cdot 3\cdot \color{red}{4} \cdot 5\cdot\color{red}{6}\cdots(2n-1)\cdot \color{red}{2n}}{(2\cdot 4\cdot 6\cdots (2n))^\color{red}{2}}\\\\
&=\frac{(2n)!}{(2^{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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maximum value of $\sum (a-b)^2$
If $a^2+b^2+c^2=5$ and $a,b,c \in \mathbb{R},$ find the maximum
value of $(a-b)^2+(b-c)^2+(c-a)^2$.
My Try: $(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ac)$
$$=10-2(ab+bc+ac)$$
Now this implies $ab+bc+ac\le 5$. So
$$(a-b)^2+(b-c)^2+(c-a)^2 = 10-2(ab+bc+ac) \geq 20.$$
Could some hel... | Here's a more visual approach; for a point $(a,b,c)\in\Bbb{R}^3$ the condition that
$$a^2+b^2+c^2=5,$$
is equivalent to being on the sphere of radius $\sqrt{5}$ centered at the origin, and
$$(a-b)^2+(b-c)^2+(c-a)^2=||(a,b,c)-(b,c,a)||^2,$$
where the point $(b,c,a)\in\Bbb{R}^3$ is obtained from $(a,b,c)$ by a rotation o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Earth's surface area
Here we are trying to calculate the earth's surface area via geodetic coordinates:
$x=(Rp(\lambda)+h)\sin (\lambda)\cos(\phi)$
$y=(Rp(\lambda)+h)\sin (\lambda)\sin(\phi)$
$x=((1-e^2)Rp(\lambda)+h)\cos (\lambda)$
where $p(\lambda)=\frac 1 {\sqrt{1-e^2\cos ^2 \lambda}}$
We first compute the flat... | Consider $$Z=1+\left(\frac{1}{e}-e\right) \tanh ^{-1}(e)$$ Now, by Taylor around $e=0$
$$\tanh ^{-1}(e)=e+\frac{e^3}{3}+\frac{e^5}{5}+O\left(e^7\right)$$
$$Z=1+\left(\frac{1}{e}-e\right)\left(e+\frac{e^3}{3}+\frac{e^5}{5}+O\left(e^7\right)\right)=2-\frac{2 e^2}{3}-\frac{2 e^4}{15}+O\left(e^6\right)=2-\frac{2 e^2}{3}+O\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $f \in L^p(0,\infty)$ and $g(x) = \int_0^\infty \frac{1}{x+y} f(y) \, dy$, show $\lim\limits_{x \to \infty} g(x) = 0$
For $x \in (0, \infty)$, let:
\begin{align*}
g(x) &= \int_0^\infty \frac{1}{x+y} f(y) \, dy \\
\end{align*}
Show that $g(x) \mathop{\longrightarrow}\limits_{x \to \infty} 0$ for $f \in L^p(0,\... | The result is not true for $p =\infty$. In fact if $f \equiv 1$ then $g(x)=\infty$ for all $x$ even though $f \in L^{\infty}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can't solve system of linear equations (that need simplification first) I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$\frac{3x+2}{4} - \frac{x+2y}{2} = \frac{x-3}{12}$$
$$\frac{2... | The first mistake I see is here:
$$9x + 6 - 6x - 12 y = x - 3$$ $$9x -6x -x - 12y = -3 +6$$
The second line should read
$$9x-6x-x-12y=-3-6$$
you forgot to invert the sign when moving $6$ to the other side of the $=$ sign.
You were equally sloppy with the second equation, when you did this:
$$5x - 6x + 8y - 15y = 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Given three non-negative numbers $a,b,c$ so that $a+b+c=3$. Prove that $\prod\limits_{cyc}(\!2+ a^{2}\!)+abc\geqq 28$ .
Given three non-negative numbers $a,\!b,\!c$ so that $a+\!b+\!c\!=\!3$. Prove that $\prod\limits_{cyc}(\!2+ a^{2}\!)+ abc\geqq 28$.
Let $a+ b+ c= 3u= 3, ab+ bc+ ca= \frac{3u}{X}, abc= \frac{u^{3}}{w... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$\prod_{cyc}(2u^2+a^2)+u^3w^3\geq28u^6$$ or $f(w^3)\geq0,$ where
$$f(w^3)=w^6-11u^3w^3+16u^6-24u^4v^2+18u^2v^4.$$
But $$f'(w^3)=2w^3-11u^3\leq0,$$ which says that it's enough to prove our inequality for a maximal value of $w^3$,
which happens ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Maclaurin Series Expansion of $\ln(1+\sin x)$ Hello can anyone help with this question
Show that the Maclaurin series of the function
$$\ln(1+\sin x)$$
up to the term in $x^4$ is
$$x-x^2/2 + x^3/6 - x^4/12 + \ldots$$
So I know the expansion for $\ln(1+x)= x - x^2 + x^3/3 +\dots$ and that of $\sin x= x - x^3/3!+x^5/5... | What you tried is, however, an interesting attempt to solve the given task but not quite right. It is in fact (and please don't ask me why) correct up a few terms if you finish what you tried. Anyway, this is not the standard way of finding a MacLaurin Series of a given function.
Recall, a MacLaurin Series Expansion i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Have I found the correct Laurent series expansion? I am supposed to find the Laurent series expansion for
$$f(z) = \frac{z+1}{z^2-4}$$ in the region $1<|z+1|<3$.
My solution:
$$w=z+1 \Leftrightarrow z=w-1 \Rightarrow f(z) = \frac{w-1+1}{(w-1)^2-4} = \frac{w}{(w+1)(w-3)}$$
Partial fractions: $$\frac{w}{(w+1)(w-3)} = \f... | Seems like the second sum is good on $\mid z+1\mid\gt3$. You should factor out $3$ instead of $w$ (in the denominator).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Taylor series of functions with matrix input Today my professor said something interesting, replace $x$ in $f(x)$ with matrix \begin{bmatrix}x&1\\0&x\end{bmatrix}
Then $$f(\begin{bmatrix}x&1\\0&x\end{bmatrix}) = f(x)\begin{bmatrix}1&0\\0&1\end{bmatrix} + f^{'}(x)\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$
and asked us to ... | I believe that, as mentioned by mathreadler in his comment on the question itself, that there is indeed a typo and that the correct formula may be written
$f \left (\begin{bmatrix} x & 1 \\ 0 & x \end{bmatrix} \right ) = f(x)I + f'(x)N, \tag 0$
with $I$ and $N$ explained in what follows.
I assume $f(x)$ is represented ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$ The function
$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$
Has the following poles of order 2:
$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$
$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{... | If you want an approach that doesn't require as much differentiation, first use partial fractions to write $\dfrac{z^3}{z^4+a^4}=\sum_{k=0}^3\dfrac{c_k}{z-z(k)}$ so $$\frac{z^6}{(z^4+a^4)^2}=\sum_k\frac{c_k^2}{(z-z(k))^2}+2\sum_{k<l}\frac{c_kc_l}{(z-z(k))(z-z(l))}\\=\sum_k\frac{c_k^2}{(z-z(k))^2}+2\sum_{k<l}\frac{c_kc_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Laplace Transforms with Non-repeated Irreducible quadratic factors I'm solving the following differential equation with a Laplace Transform:
$$x″+ 9x = \cos(t) + \delta(t-\pi)$$
The initial conditions are that x(0) and x'(0) are equal to 0.
After after the transform, we obtain:
$$F(S) = \frac{S}{(S^2+9)[(S^2+1)+9]} + \... | Alternative Method
$$
F(s)=\underbrace{\frac{s}{(s^2+1)(s^2+9)}}_{\mathbf{(I)}}+\underbrace{\frac{e^{-\pi s}}{s^2+9}}_{\mathbf{(II)}}
$$
Alternative (and fast) method for $\mathbf{(I)}$:
We can also decompose $\mathbf{(I)}$ using partial fractions by the following way:
$$
\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{Es+F}{s^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convert $\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$ into an algebraic function
Convert trigonometric function into algebraic function.
$$\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$$
My approach is as follows:
$sin2\theta$ is to be calculated
$$\tan^{-1}x=\gamma... | I firstly apologise about how this is written, I really ought to learn how to type this stuff properly. Secondly, if there's any mistakes please do say!
$$\tan(2x) = 2\tan(x)/1-\tan^2(x)$$
so $$\cot(2x) = (1-\tan^2(x))/2\tan(x)$$
swapping $x = \arctan(x)$
$$\cot(2\arctan(x)) = (1-x^2)/2x$$
Okay, that bit is done,
Let ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Equation solution set of $f^{-1}(x)=f(x)$?
let $f(x)=-(x+2)^3-2 $ . Now what is the equation solution set of $f^{-1}(x)=f(x)$ ?
My try : I know that if a function $f$ is increasing in the domain $D$ on
which its inverse function $f^{-1}$ exists, and a lies in
$D$, then $f (a) = f^{-1}(a)$ if and only if $f(a) = a$
N... | No, if you have $f(x) = f^{-1}(x)$, then $f(f(x)) = x$, but $f(f(-3.7)) = 2.913 \ne -3.7$, so it's wrong. Since $f$ is monotone, $f(x)=f^{-1}(x)$ iff $f(f(x)) = x$. Let $$g(x) := f(f(x))-x = -[(-(x+2)^3-2)+2]^3-2 - x = (x+2)^9 - 2 - x = (x+2)((x+2)^8-1)$$
Therefore, $f(f(x)) = x$ iff $g(x) = 0$ iff $x = -2$ or $(x+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac {\sin (2x)}{(\sin x+\cos x)^2}\,dx$ Integrate
$$\int \frac {\sin (2x)}{(\sin x+\cos x)^2} \,dx$$
My Attempt:
$$=\int \frac {\sin (2x)}{(\sin x + \cos x)^2} \,dx$$
$$=\int \frac {2\sin x \cos x}{(\sin x+ \cos x)^2} \,dx$$
Dividing the numerator and denominator by $\cos^2 x$
$$=\int \frac {2\tan x}... | $$I=\int\frac{\sin(2x)}{(\sin(x)+\cos(x))^2}dx=\int\frac{2\sin(x)\cos(x)}{\sin^2(x)+2\sin(x)\cos(x)+\cos^2(x)}dx\tag{1}$$ $$=\int\frac{2\sin(x)\cos(x)+1-1}{2\sin(x)\cos(x)+1}dx=\int dx-\int\frac{dx}{\sin(2x)+1}=x-\int\frac{1-\sin(2x)}{\cos^2(2x)}dx\tag{2}$$ $$=x-\int\sec^2(2x)dx+\int\frac{\sin(2x)}{\cos^2(2x)}dx\tag{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Question:
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Following from... | Suppose $(1+\sqrt3)(1+\sqrt7)=p/q$ for some $p,q\in\Bbb Z^+$. Then we have that $$q(1+\sqrt3)=\frac p{1+\sqrt7}=\frac{p(1-\sqrt7)}{-6}\implies p\sqrt7-6q\sqrt3=p+6q\ne0\tag1$$ This implies that $$p\sqrt7+6q\sqrt3=\frac{(p\sqrt7+6q\sqrt3)(p\sqrt7-6q\sqrt3)}{p\sqrt7-6q\sqrt3}=\frac{7p^2-108q^2}{p+6q}\tag2$$ Adding $(1)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
} |
Finding $\lim_{n\to\infty}\sqrt n\int_1^n \frac{\arctan (x/\sqrt n)}{x^2+n}\,dx$ Let $I_n = \int_{1}^{n} \frac{\arctan \frac{x}{\sqrt n}}{x^{2}+n}\,dx$, for $n \geq 1$. How can I show that $\lim_{n \to \infty} \sqrt nI_n=\frac{\pi}{8}$ ?
I've tried to first solve the integral but I'm really stuck, I couldn't find an i... | Let $\tan^{-1}\frac{x}{\sqrt n} = t$
$\frac{dx}{ 1+ \frac{x}{\sqrt n^2}} \frac{1}{\sqrt n} = dt$
$\frac{dx}{x^2 + n} = \frac{1}{\sqrt n}dt$
So, $I_n = \int^n_{x=1} \ \frac{1}{\sqrt n} tdt = \frac{1}{\sqrt n}\big[\frac{t^2}{2}\big]^n_{x=1}$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[\tan^{-1}(\frac{x}{\sqrt n}))^2\big]^n_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How do you solve $\frac{x-1}{\sqrt{x}+2}=\frac{5}{2}$? I solved it using the quadratic formula and it went like:
\begin{gather}
\frac{x-1}{\sqrt{x}+2}=\frac{5}{2} \\
2(x-1)=5(\sqrt{x}+2) \\
2x-2=5\sqrt{x}+10 \\
2x-12=5\sqrt{x} \\
4x^2+144-48x=25x \tag*{(squaring both sides)}\\
4x^2-73x+144=0
\end{gather}
Then do the us... | If you rewrite the equation as
$$
2(x-1)=5(\sqrt{x}+2)
$$
and then as
$$
2x-12=5\sqrt{x}
$$
you can immedately notice that a necessary condition for a solution is $x\ge6$, because the right-hand side is positive.
With this side condition, you can square: $4x^2-48x+144=25x$, which indeed has the roots $16$ and $9/4$, bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Change of variables for double integral Problem: Using the change of variables $$x=\sqrt2u-\sqrt\frac{2}{3}v,y=\sqrt2u+\sqrt\frac{2}{3}v$$
Calculate the double integral $$\iint_Rx^4-2x^3y+3x^2y^2-2xy^3+y^4dA$$ where $R$ is the region bound by $x^2-xy+y^2=2$.
My work so far: The Jacobian is fairly trivial: $\frac{4}{\sq... | We can save a lot of pain and get the correct answer by completing the square:
\begin{align}
x^4-2x^3 y+3x^2y^2-2xy^3+y^4
&= (x-y)^4 + 2x^3 y -3x^2 y^2 +2x y^3
\end{align}
Pull a common factor $xy$ out of the leftover terms and complete the square again:
\begin{align}
&= (x-y)^4 + xy(2x^2 -3x y +2 y^2) \\
&= (x-y)^4 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding A solution to a matrix equation I have the following problem in a past exam that involves matrix equations:
Suppose that $X$is a $2 \times2$ matrix satisfying:
$X^{2} = 6X +I$
a) Find the values of $\alpha$ and $\beta$ such that $X^{3} = \alpha X + \beta I$
b) Hence find a solution of the matrix equation
$X^{3... | Suppose $X=\begin{bmatrix} a & b \\ c &d \end{bmatrix}$.
It is given that $X^{3} = \begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} X + \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix}$.
Therefore,$\begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} \times \begin{bmatrix} a & b \\ c &d \end{bmatrix}+ \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_0^{\frac{\pi}{2}}(\cos t)^{\frac{5}{2}}{\rm d}t$? Let $\sqrt{\cos t}=:u$，then $t=\arccos u^2$，${\rm d}t=-\dfrac{2u}{\sqrt{1-u^4}}{\rm d}u$. Therefore
\begin{align*}
\int_0^{\frac{\pi}{2}}(\cos t)^{\frac{5}{2}}{\rm d}t&=2\int_0^{1}\frac{u^6}{\sqrt{1-u^4}}{\rm d}u,
\end{align*}
which seems to be the... | Solution
Consider integrating by part. One can obtain
\begin{align*}
\int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t&=\int_0^{\frac{\pi}{2}}\cos^{\frac{3}{2}}t{\rm d}(\sin t)\\
&=\left[\cos^{\frac{3}{2}}t\sin t\right]_0^{\frac{\pi}{2}}+\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos^{\frac{1}{2}}t\sin^2 t{\rm d}t\\
&=0+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral Bounds of Part of an Ellipse
I'm looking through my notes, and I've completed this questions integral, getting $ -144 \int cos^4(\theta)$ and due to the fact that $x=4cos(\theta)$ I calculated that the lower bound is $cos^-1(\frac{2\sqrt{3}}{4})=\frac{\pi}{6}$. The lecture notes confirm that this is correct h... | You need to solve both equations at once.
$$
-2 = 4 \cos\theta\qquad\qquad
-3 \sqrt 3 = 6 \sin\theta
$$
If you try to use inverse functions on the first, you get an angle in the second quadrant: $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$. If you try to use inverse functions on the second, you get a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^4}$
Evaluate $\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$
The answer stated is $\displaystyle {1 \over 6}$.
What I've tried:
$$\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{\cos (\sin x... | For small $x$, $\frac{\cos(\sin x)-\cos x}{\sin x-x}\sim\cos^\prime x\sim -x$, while $\frac{\sin x-x}{x^4}\sim-\frac{1}{6x}$, so the limit is $\frac16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4+x^5)^{17}$? I figured that $(1+x+x^2+x^3+x^4)^{17} = (1-x^6)^{17}*(1-x)^{-17}$ but don't know what else to do.
I would really appreciate any help
| You can pick up an $x^5$ from one term and a $1$ from every other term 17 different ways.
You can pick up an $x^4$ from one term, an $x$ from a second term in $\dfrac{17!}{15!}$ ways.
You can get $x^3$ and $x^2$ in $\dfrac{17!}{15!}$ ways.
You can get $x^3$, $x$, $x$ in $\dfrac{17!}{1!2!14!}$ ways.
You can get $x^2, x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Finding $\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$
$$\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$$
I have to resolve this limit, I tried factoring out x, I tried rewriting $x^3+x$ as
$x^3(1+ \frac{x}{x^3})$ and it doesn't seem to cancel. What should I do?
| \begin{align}
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3+x}}{x}
=&
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3\cdot(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3}\cdot\sqrt[\,3]{(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} \frac{x\cdot\sqrt[\,3]{(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} {\sqrt[\,3]{(1+1/x^2)}}
\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve ODE $y'' = y^3 -y y'$ I want to solve $y'' = y^3 -y y'$ with boundary conditions $y(1) = 1/2$ and $y(2) = 1/3$ but not sure how to start. If someone could give me a complete step by step explanation, it would be greatly appreciated because I want to fully understand it.
Edit:
By the hint $(y'+\frac{y^2}{2})'=y^3... | The integration is not messy. After the hint, you have
$$dx=\frac {dy}{\frac {y^4}4 -\frac {y^2} 2+c_1}=4\frac {dy}{y^4 -2y^2+c_1}=4 \frac {dy}{(y^2-a)(y^2-b)}$$ Now, use partial fraction decomposition
$$\frac {1}{(y^2-a)(y^2-b)}=\frac 1{a-b}\left(\frac 1{y^2-a}-\frac 1{y^2-b} \right)$$ Just continue and, when done, ap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $2\tan^{-1}\frac{\sqrt{a^{2}- 2}}{a}= \tan^{-\,1}a\sqrt{a^{2}- 2} $ Prove
$$2\tan^{-1}\dfrac{\sqrt{a^{2}- 2}}{a}= \tan^{-\,1}a\sqrt{a^{2}- 2} \tag{a p u}$$
I find
$$a\sqrt{a^{2}- 2}= \dfrac{\dfrac{\sqrt{a^{2}- 2}}{a}+ \dfrac{\sqrt{a^{2}- 2}}{a}}{1- \dfrac{\sqrt{a^{2}- 2}}{a}\dfrac{\sqrt{a^{2}- 2}}{a}}$$
This... | Using
$$\tan^{-1}(u) - \tan^{-1}(v) = \tan^{-1} \frac{u - v}{1 + uv}$$
where $u=\frac{\sqrt{a^2-2}}{a}$ and $v = a\sqrt{a^2- 2}$
you get
$$\tan^{-1}\frac{\sqrt{a^2-2}}{a} - \tan^{-1} a\sqrt{a^2- 2} = \tan^{-1} \frac{\frac{\sqrt{a^2-2}}{a} - a\sqrt{a^2-2}}{1+a^2-2} = -\tan^{-1}\frac{\sqrt{a^2-2}}{a} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sub-Sum of Roots of Unity Let $\alpha$ be an algebraic integer of a cyclotomic field, and let $\theta_1, \theta_2, ..., \theta_n$ be roots of unity such that
$$\sum_{i=1}^n \theta_i = 2\alpha.$$
Does there necessarily exists a sub-sum of the $\theta_i$ that equals $\alpha$?
| No. Here we use the cyclotomic polynomial $\Phi_{105}$.
$$
\begin{align}
\Phi_{105}(x)&= x^{48} + x^{47} + x^{46} - x^{43} - x^{42} - 2 x^{41} - x^{40} - x^{39} \\ &\qquad+ x^{36} + x^{35} + x^{34} + x^{33} + x^{32} + x^{31} - x^{28} - x^{26} \\
&\qquad- x^{24} - x^{22} - x^{20} + x^{17} + x^{16} + x^{15} + x^{14} + x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help calculate the limits Help calculate the limits:
1) $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt {k(n-k)}}$$
2) $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{(n-k)\ln{n}} $$
3) $$\lim_{n\to \infty}{n}^p \sin(\pi(\sqrt 2 +1)^n) $$
In the directions to this number it is written: compare $$(\sqrt 2 +1)^n$$ with the wh... | This suppose to be a comment but it getting kinda long so I placed it here in term of an answer instead, although it's not directly answering your question but it may helps... In regard to Rick's comment, I think this is what he meant:
Suppose I have something like:
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \dfrac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Checking the Monotonic function
Let $g''(x)>0$, such that $x \in \mathbb{R} $ and $f(x)=2g(2x^3–3x^2)+g(6x^2–4x^3–3)$, $x \in \mathbb{R} $,then
(A)in interval $x \in (1,\infty) $; $f(x)$ is increasing function
(B)in interval $x \in (-\infty , \frac{-1}{2}) $; $f(x)$ is decreasing function
(C) in interval $x \in (0,\in... | I will just give a hint.
$g''(x)>0$ implies that $g'(x)$ is increasing, meaning $g'(a)>g'(b)$ whenever $b<a$.
Now
$$
\begin{split}
f'(x)&= 2g'(2x^3-3x^2)\cdot (6x^2-6x) + g'(6x^2-4x^3-3) \cdot (12x - 12x^2) \\
&= 12x(x^2-1)g'(2x^3-3x^2) + 12x(1-x^2)g'(6x^2-4x^3-3) \\
&= 12x(x^2-1)g'(2x^3-3x^2) - 12x(x^2-1)g'(6x^2-4x^3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me pleas... | Set $y=x^2+6x+7,$ and note that the resulting system $$y^2+6y+7-x=0,\,\,\,x^2+6x+7-y=0$$ is symmetric. That is, performing the transformation $(x,y)\mapsto(y,x)$ leaves the system unchanged except for a permutation. Thus, for every solution $(x,y),$ it is always the case that $(y,x)$ is also a solution, where $x,y\in\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Contour integration problem with sin and cos so I'm revising contour integration for an upcoming complex analysis exam. I have been asked to integrate $$\int_0^{2\pi}\frac{\sin^2x}{a+b \cos x}dx$$
I thought the sensible thing to do here would be to substitute in $z=e^{ix}$ and take a contour integral around the unit c... | $$
\begin{align}
\int_0^{2\pi}\frac{\color{#C00}{\sin^2(x)}}{\color{#090}{a+b\cos(x)}}\,\color{#00F}{\mathrm{d}x}
&=\oint\frac{\color{#C00}{-\frac{z^4-2z^2+1}{4z^2}}}{\color{#090}{\frac{bz^2+2az+b}{2z}}}\color{#00F}{\frac{\mathrm{d}z}{iz}}\\
&=-\frac1{2i}\oint\frac{z^4-2z^2+1}{bz^2+2az+b}\,\frac{\mathrm{d}z}{z^2}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove $n^{3/2} \le 2^n/n$ for all $n\geq 8$ via Induction? How to prove $n^{3/2} \le 2^n/n$ for all $n \geq 8$
My attempt:
$\textbf{Base case}$ is $n_0=8$: $\quad \left(8^{3/2}=4\right)\leq \left(32=\frac{2^8}{8}\right) \checkmark$
$\textbf{Induction hypothesis}$: $\exists n\in \mathbb{N}_{\geq 8}:(n+1)^{3/2}\... | Consider function $f(x)=4^x-x^5$. Now we have to prove that $f(x)>0$ for $x\geq8$ using induction.
Let start with $f(8)=4^8-8^5=2^{16}-2^{15}\gt0$
Now, for any $x=k\geq8$ let us assume $f(k)=4^k-k^5\geq0$
Now put $x=k+1$, then $f(k+1)=4^{k+1}-(k+1)^5=4\times4^k-(k+1)^5$
$f(k+1)=[4^k-k^5]+[4^k-5k^4]+[4^k-10k^3]+[4^k-10k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to find this kind of relationship: $x-\frac{1}{x}=A$ and $x+\frac{1}{x}=\sqrt{A^2+4}$? Could someone explain how to get from: $x-\frac{1}{x}=A$ to $x+\frac{1}{x}=\sqrt{A^2+4}$ ? It is one of the Algebra II tricks.
Thanks.
| Start by squaring both sides:
$$\begin{align}x-\frac{1}{x}&=A\\\left(x-\frac{1}{x}\right)^2&=A^2\\x^2-2+\frac{1}{x^2}&=A^2.\end{align}$$
Then try adding $4$ to both sides and "reversing" the processes above.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
$$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$
To begin, I proved a lemma:
Lemma. An acute triangle has at most one a... | Hint: Use the formulas $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=\frac{r}{R}+1$$
$$2A=ab\sin(\gamma)=ac\sin(\beta)=ab\sin(\gamma)$$
$$A=sr=\frac{abc}{4R}$$
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
Doing this you will get
$$4A(a+b+c)>a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\,
Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$
Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\i... | Anyway, this integral doesn't require complex analysis: it is well-known that the antiderivatives of $\:\dfrac 1{(1+x^2)^n}\:$ can be calculated recursively. In this case, it is particularly simple:
Use integration parts to calculate the integral of $\frac1{1+x^2}$, setting
\begin{gather}
u=\frac1{1+x^2},\quad \mathrm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Approximate solution: factorial and exponentials If z= $\dbinom{200}{100}/(4^{100})$, what is the value of z?
The options are:
a. $z<1/3$
b. $1/3<z<1/2$
c. $1/2<z<2/3$
d. $2/3<z<1$
How should I go about solving these type of problems?
| Here is an elementary approach without estimating the value of the expression:
\begin{eqnarray*} \frac{1}{4^m}\binom{2m}{m}
& = & \frac{1}{4^m}\cdot \frac{\prod_{i=1}^m 2i \cdot \prod_{i=1}^m (2i-1) }{(m!)^2} \\
& = & \frac{1}{4^m}\cdot 4^m\frac{\prod_{i=1}^m i \cdot \prod_{i=1}^m \left(i-\frac{1}{2}\right) }{\prod_{i=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Compute without calculator, $ \frac{1}{\cos^{2}(10)} + \frac{1}{\sin^{2}(20)} + \frac{1}{\sin^{2}(40)} - \frac{1}{\cos^{2}(45)} $ Compute without calculator,
$$ \frac{1}{\cos^{2}(10^{\circ})} + \frac{1}{\sin^{2}(20^{\circ})} + \frac{1}{\sin^{2}(40^{\circ})} - \frac{1}{\cos^{2}(45^{\circ})} $$
Attempt:
Let $A = \... | \begin{align*}
&\;\frac{1}{\cos^{2}10^{\circ}} + \frac{1}{\sin^{2}20^{\circ}} + \frac{1}{\sin^{2}40^{\circ}} - \frac{1}{\cos^{2}45^{\circ}} \\
=&\;\frac {16\sin^210^\circ\cos^220^\circ}{16\sin^210^\circ\cos^210^\circ\cos^220^\circ}+\frac{4\cos^220^\circ}{4\sin^220^\circ\cos^220^\circ}+ \frac{1}{\sin^{2}40^{\circ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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A problem Dealing with Sampling with replacement Below is a problem from the Schaum book: "Probability and Statistics". I started
the problem but I am confident that I am on the wrong approach. I am hoping somebody will tell me where I went wrong.
Problem:
An urn holds $60$ red marbles and $40$ white marbles. Two sets ... | The sample variance is $n$ times the variance of one marble. This makes $\sigma_s^2=\frac {36}{5}$. The variance of the difference of two samples is twice the variance of one, so $\frac{72}5$. The square root of this is about $3.79$, so a difference of $8$ is about $2$ standard deviations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find minimum value of $P=17x^2+17y^2+16xy$ Let $x,y$ be positive real numbers such that $4x^2+4y^2+17xy+5x+5y\ge 1$. Find minimum value of $$P=17x^2+17y^2+16xy$$
My idea:
I see that $x=y=\frac{\sqrt 2 -1}{5}\rightarrow P=2(3-2\sqrt 2)$
So I proved $P\ge 2(3-2 \sqrt 2)(4x^2+4y^2+17xy+5x+5y)\ge2(3-2\sqrt 2)$
Is my appro... | We have
$$17\,x^{\,2}+ 17\,y^{\,2}+ 16\,xy\geqq 2(\,3- 2^{\,\frac{3}{2}}\,)$$
$$\because\,17\,x^{\,2}+ 17\,y^{\,2}+ 16\,xy- 2(\,3- 2^{\,\frac{3}{2}}\,)- (\,2- 2^{\,\frac{1}{2}}\,)(\,\underbrace{4\,x^{\,2}+ 4\,y^{\,2}+ 17\,xy+ 5\,x+ 5\,y- 1}_{\,\geqq 0}\,)\geqq 0$$
$$\because\,{\rm discriminant}[\,17\,x^{\,2}+ 17\,y^{\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Gravity of a Circular Ring at a co-planar external point. I have a circular ring of unit mass and fixed radius R which lies in the XY plane at point $O$ with coordinates $O:(0,0)$.
I wish to find a formula for the gravitational force at a point $P: (D,0)$ which lies in the same plane as the ring and is at some variable... | In order to express the result in terms of complete elliptic integrals, it is easier to compute the gravitational potential $\phi(D)$ first. Then the (radial) field is given by $F(D) = - \phi'(D)$. Following your approach we find
$$ \phi(D) = - \frac{G M}{2 \pi D} \int \limits_0^{2\pi} \frac{\mathrm{d} \theta}{\sqrt{1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ \int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx$ Evaluate $$ \int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx$$
I have tried Weierstrass substitution and tried to split into two integrations, but it gets really messy.
Is there a better way to approach this problem? I feel that comple... | \begin{align}\int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx
=& \int \frac{a^2\csc^2x+b^2\sec^2x}{a^4\csc^2x+b^4\sec^2x}\,dx\\
=& \int \frac{ d(b^2\tan x -a^2\cot x)}{(a^2+b^2)^2 +(b^2\tan x -a^2\cot x)^2} \\
=& \ \frac1{a^2+b^2}\tan^{-1}\frac{b^2\tan x -a^2\cot x}{a^2+b^2}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$. I need to show that $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$.
This happens if $(x+1)\ln\Big(1 + \frac 2 {2x+1}\Big) > 1$ so let's study the function $f(x) = (x+1)\ln\Big(1 + \frac 2 {2x+1}\Big)$
Edit
We have abandoned the previous metho... | From $f(x) = (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)$ we compute the derivative.
$$
\begin{eqnarray}
f'(x) = & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)' \\
= & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(\frac {2x+3} {2x+1}\Big)' \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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A non-trivial solution for $\lambda_1 + 2 \lambda_2 + \cdots + (p-2)\lambda_{p-2} = 0$ in $\mathbb{Z}_p$ Let $K = \mathbb{Z}_p$, where $p$ is a prime number and $p \neq 2,3$ . Find a non-trivial solution for the system:
$$\left\{ \begin{align}
& \lambda_1 + \lambda_2 + \cdots + \lambda_{p-2} = 0 \ (\mod p) \\
& \la... | If we have $$\lambda_2 + 2\lambda_3 + \cdots + (p-3) \lambda_{p-2} \equiv 0 \pmod{p} \tag{$\ast$}$$ then setting $\lambda_1 = -(\lambda_2 + \cdots+\lambda_{p-2})$ gives a solution $(\lambda_1, \cdots, \lambda_{p-2})$ to the described system of two equations. So, we just need to find a nontrivial solution to the single... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating factor for $(x^2-y^2-y)dx-(x^2-y^2-x)dy=0$ I am having trouble finding the integrating factor for turning the below differential equation into an exact one (Tenenbaum and Pollard, exercise 10, problem 6). Any hints and suggestions would extremely helpful and lead me to the solution.
Solve the differential ... | The part of $\dfrac{\partial P}{\partial y} - \dfrac{\partial Q}{\partial x} = 2(x-y-1)=2(z-1)$ shows that you may assume a new variable $z=x-y$ which which shows the denominator of integrating factor might be of the form
$$\dfrac{\partial I}{\partial x}Q - \dfrac{\partial I}{\partial y}P = P+Q=z$$
so
$$p(z)=\dfrac{2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Test for the convergence of the series $\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}$ Test for the convergence of the series $\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}$
My attempt
$\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}=\frac{-\frac{1}{2}}{1}+\frac{\frac... | Concatenate terms in pairs, with the odd n first. $(\frac{-1}{2n}+\frac{3}{2(n+1)})=\frac{2n-1}{2n^2}\gt \frac{1}{2n}$ The sum then becomes $\gt \sum_{k=1}^\infty \frac{1}{2(1+2k)}$ which diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove if $3\nmid n$, then $n^2$ has a remainder $1$ when divided by $3$? Lets say:
$n^2=3k+1$ ; $n\neq3$.
I'm trying to prove this by induction, therefore:
$(n+1)^2=n^2+2n+1=3k+2n+2$
Any suggestion on how to move foward?
| If 3 does not divide $n$, that means that $n$ has remainder 1 or 2 by division with 3.
So either $n=3k+1$ or $n=3k+2$.
In both cases we have $n^2=(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\mod 3$
or $n^2=(3k+2)^2=3(3k^2+4k)+4=3(3k^2+4k)+3\cdot 1+1\equiv 1\mod 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$ If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$
After we solve for $x$ (its a quadratic), and find that $x=1\pm i$, it's trivial to see the powers of $x$ in the complex plane, but the problem must... | It's enough to prove that
$$x^9-(x^4+x^2+1)(x^6+x^3+1)=1$$ or
$$(x^3-1-(x^4+x^2+1))(x^6+x^3+1)=0,$$ for which it's enough to prove that
$$x^4-x^3+x^2+2=0$$ or
$$x^4-2x^3+2x^2+x^3-2x^2+2x+x^2-2x+2=0$$ or
$$(x^2-2x+2)(x^2+x+1)=0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Calculate $\int_3^4 \sqrt {x^2-3x+2} \, dx$ using Euler's substitution
Calculate $\int_3^4 \sqrt {x^2-3x+2}\, dx$ using Euler's substitution
My try:
$$\sqrt {x^2-3x+2}=x+t$$
$$x=\frac{2-t^2}{2t+3}$$
$$\sqrt {x^2-3x+2}=\frac{2-t^2}{2t+3}+t=\frac{t^2+3t+2}{2t+3}$$
$$dx=\frac{-2(t^2+3t+2)}{(2t+3)^2} dt$$
$$\int_3^4 \sqr... | Using the third substitution of Euler
$$\sqrt{{{x}^{2}}-3x+2}=\sqrt{\left( x-1 \right)\left( x-2 \right)}=\left( x-1 \right)t$$
$$x=\frac{2-{{t}^{2}}}{1-{{t}^{2}}}\Rightarrow dx=\frac{2t}{{{\left( {{t}^{2}}-1 \right)}^{2}}}dt$$
$$\begin{align}
& =-\int_{1/\sqrt{2}}^{\sqrt{2/3}}{\frac{2{{t}^{2}}}{{{\left( {{t}^{2}}-1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral of $\frac{1}{\sqrt{2-x^2}}$ I know that the integral of $\frac{1}{\sqrt{1-x^2}} = arcsin(x)$, but what is the the integral of $\frac{1}{\sqrt{2-x^2}}$? Is this how you do it?
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2}(1-\frac{x^2}{\sqrt{2}})} = \frac{1}{\sqrt{2}} \arcsin\left(\frac x{\sqrt{2}}\right)$$
| This is a simple u-substitution problem. Use $u=\frac{x}{\sqrt{2}}$:
$$
\int\frac{1}{\sqrt{2-x^2}}\,dx=
\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{1-\left(\frac{x}{\sqrt{2}}\right)^2}}\,dx=\\
\frac{\sqrt{2}}{\sqrt{2}}\int\frac{1}{\sqrt{1-\left(\frac{x}{\sqrt{2}}\right)^2}}\frac{d}{dx}\left(\frac{x}{\sqrt{2}}\right)\,dx=
\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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New bound for Am-Gm of 2 variables Today I'm interested by the following problem :
Let $x,y>0$ then we have :
$$x+y-\sqrt{xy}\leq\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$
The equality case comes when $x=y$
My proof uses derivative because for $x\geq y $ the function :
$$f(x)=x+y-\sqrt{xy}-\exp\Big(\frac{x\ln(x)+y... | Idea for a proof :
As the inequality is homogenous we obtain an equivalent inequality like :
$$\left(\frac{1}{x}\right)^{\frac{2}{x^2+1}}+\frac{1}{x}-\frac{1}{x^2}\geq1$$
Using the Bernoulli's first approximation and some others terms in the Newton's expansion of $(1+x)^a$ we have ($0< x\leq 1$).:
$$\left(\frac{1}{x}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
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Find $x_1^3+x_2^3+x_3^3$ for a degree 3 polynomial. I have the polynomial $P(x) = x^3+mx^2-3x+1, m\in \mathbb{R}$. I need to find $x_1^3+x_2^3+x_3^3$ as a $m$ function.
I tried to use Viette equations: $x_1+x_2+x_3 = m, x_1x_2+x_1x_3+x_2x_3 = 3, x_1x_2x_3 = 1$ Then I expanded $(x_1+x_2+x_3)^3 = (x_1^3+x_2^3+x_3^3) + 3... | Since $x_1,x_2,x_3$ are the roots we find that $x_1^3=3x_1-mx_1^2-1$ similarly for $x_2,x_3$ adding the three equations we find that $x_1^3+x_2^3+x_3^3=3(x_1+x_2+x_3)-m(x_1^2+x_2^2+x_3^2)-3$ we see that $(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_2x_3+x_1x_3)$ thus the required value is $x_1^3+x_2^3+x_3^3=3(-m)-m(((-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not.
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, ... | Hint: Use that $$\sum_{i=1}^na^{n-i}b^{i-1}=\frac{a^n-b^n}{a-b}$$ if $$a\ne b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Divide the numbers 1,2,…,n into two sets of equal sum. My approach:
If sum(1 to n) is divisible by 2, then it is possible to divide the set.
If n is even take 1st and last element and add to 1st set, then add 2nd and 2nd last to 2nd set...and so on, In then end to balance out both set I remove 1 from 1st set and add it... | A triangular number is even if and only if its argument is $\in\{0,3\}\bmod 4$.
If the argument is $\equiv 0\bmod 4$, you can specify blocks from the "outside in". For $n=8$ you put $1$ and $8$ into one block, then $2$ and $7$ into the second block, and alternate blocks until you hit the middle:
$1+2+3+4+5+6+7+8=\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I need to define the cubic equation $f(x)= ax^3 + bx^2 + cx + d$ It is given that if cubic function $f(x)$ is divided by $(x^2+4)$ and $(x+3)$, the remainders will be $7x-20$ and $-2$ respectively. Also it is said show that $f(x)$ is $x^3 + 6x^2 + 11x +4$. I tried but can i find the $a,b,c$, and $d$ without knowing the... | Solving in pari/gp:
? chinese(Mod(7*x-20,x^2+4),Mod(-2,x+3))
%1 = Mod(3*x^2 + 7*x - 8, x^3 + 3*x^2 + 4*x + 12)
?
? Mod(a*x^3+b*x^2+c*x+d,(x^2+4)*(x+3))
%2 = Mod((-3*a + b)*x^2 + (-4*a + c)*x + (-12*a + d), x^3 + 3*x^2 + 4*x + 12)
I.e. $3=-3a+b$, $7=-4a+c$, $-8=-12a+d$.
And $\quad a=1\quad\Longrightarrow\quad b=6,c=11,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove that inequality $27+7\left(ab+bc+ca\right)\le 8\left(\sum \sqrt{a+3b}\right)$ For the positive reals $a,b$ and $c$ so that $a+b+c=3$. Show that $$27+7\left(ab+bc+ca\right)\le 8\left(\sqrt{a+3b}+\sqrt{b+3c}+\sqrt{c+3a}\right)$$
That inequality has been created by Imad Zak, i think it is interesting problem.
We ca... | You are actually very close. We can go on from your last inequality, with $x=2u, y=2v, z=2w$, we need to show
$$u+v+w \ge u^2v^2+v^2w^2+ w^2u^2\tag{1}$$
for $u^2+v^2+w^2=3$.
Now, the RHS of (1) is
$$\frac{(u^2+v^2+w^2)^2 - (u^4+v^4+w^4)}2 = \frac{9- (u^4+v^4+w^4)}2$$
So we need to prove
$$u^4+ v^4 + w^4 + 2(u+v+w) \ge9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation The problem is stated as:
Find the polynomial $P(x)$ of minimal degree such that
$$
\sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0.
$$
and the answer in the book is
$$
P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8}... | Note that one has $o(x^6)=o(x^5)$ (as $x\to 0$), but not $o(x^5)=o(x^6)$.
Also, $o((x-x^2)^2)=o(x^5)$ is incorrect: one has $\lim_{x\to 0}\frac{x^5}{(x-x^2)^2}=0$ but $\lim_{x\to 0}\frac{x^5}{x^5}=1$. In general, $o(f(x))=o(g(x))$ does NOT imply that $o(g(x))=o(f(x))$ as discussed in an answer to another question of y... | {
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"url": "https://math.stackexchange.com/questions/3269168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Modular arithmetic: 5th powers (mod 11) Doing a question about modular arithmetic and number theory and I have a mark scheme that doesn't really explain very well.
It say " The 5th powers (mod 11) are 0,1 and 10 ". I can find examples to support this, such as $11^5$ is equal to 0(mod 11), $2^5$ is equal to 10(mod 11).... | Well you can write any number as $11n,11n+1,11n+2,11n+3,...11n+10$
This basically means that a number can leave a remainder from 0 to 10 upon division by 11
So if you know the remainders for the fifth powers of all the numbers upto 11 , you can know which remainders all the numbers will leave upon division by 11.
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factoring out the expression I'm sorry, it must pretty basic, but...
How to factor out this expression $x^2 + 5x - 5 = 0$ ?
(The method I normally use to factor expressions doesn't work there because when we factor expression above so that format will be $(ax + b)(qx + c)$ $b$ and $c$ definitely won't be integers, an... | You can use completing square method to factorize it.
Look at:
$$x^2+5x-5=x^2+ 2x\times\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-5=(x +\frac{5}{2})^2-(\sqrt\frac{45}{2})^2$$
Now use the fact that $a^2-b^2 = (a+b)(a-b)$, so:
$$x^2+ 5x -5 =(x +\frac{5}{2})^2-(\sqrt\frac{45}{2})^2 = (x+\frac{5}{2}+\sqrt\frac{45}{2})(x+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $u=(1+\cos t)(1+\cos 2t)-\sin t\sin 2t$ and $v=\sin t(1+\cos 2t)+\sin 2t(1+\cos t)$, then $u^2+v^2=4(1+\cos t)(1+\cos 2t)$
If
$$u = (1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta \qquad v = \sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)$$
then show that
$$u^2 + v^2 = 4(1+\cos\theta)(1+\co... | You simplified the values incorrectly. See below.
Alternatively:
$$\begin{align}\color{blue}u&=(1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta=\\
&=1+\cos \theta +\cos 2\theta +\cos \theta \cos 2\theta-\sin\theta \sin2\theta=\\
&=1+\cos\theta+\cos2\theta+\cos3\theta=\\
&=1+\cos2\theta+2\cos\theta\cos2\theta= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dividing polynomial $f(x)$ by $(x-a)(x-b)(x-c)$
$f(x)$ is a polynomial with a degree greater than 3. When $f(x)$ is divided by $(x-a)(x-b)(x-c)$, prove remainder is
$$\frac{f(a)(x-b)(x-c)}{(a-b)(a-c)}+ \frac{f(b)(x-a)(x-c)}{(b-c)(b-a)} +\frac{f(c)(x-a)(x-b)}{(c-b)(c-a)}$$
My Try
I tried this using the conventional ... | Because $$f(a)=Aa^2+Ba+C,$$
$$f(b)=Ab^2+Bb+C$$ and $$f(c)=Ac^2+Bc+C.$$
Now, work with the given remainder (it's also a polynomial of the second degree).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f(x)=\frac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$ let $f(x)=\dfrac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$
my try : $x:=\tan t$ then $f(\tan t)=\dfrac{\tan t}{1+\tan^2 t}=\sin t \ \ \ \text{or} \ \ -\sin t$
Now what ?
| I would try to find the inverse of $f$ first - this can be done by switching the $x$ and $y$ variables, as follows:
Square to get rid of the square root:
\begin{align}
y = \frac{x}{\sqrt{x^2+1}} \Rightarrow y^2&=\frac{x^2}{x^2+1} \\
&= 1-\frac{1}{x^2+1}
\end{align}
then rearrange to get $x$ to be the subject variable:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of infinite series with variable range between -1 to 1 Let $S$ denote the infinite sum $2 + 5x + 9x^2 + 14x^3 + 20x^4 + ....$, where $| x | < 1$, then what is the value of $S$?
I am not able to find the generalized form. What is the trick to solve it?
| HINT:
$$S=2+5x+9x^2+14x^3+20x^4+\cdots$$
$$(1-x)S=2+(5-2)x+(9-5)x^2+(14-9)x^3+(20-14)x^4+\cdots$$
Let $T=(1-x)S$
$$T(1-x)=2+(3-2)x+(4-3)x^2+(5-4)x^4+\cdots=1+\dfrac1{1-x}$$ for $|x|<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+... | Since\begin{multline}A(x^2+2)^2+(Bx+C)(x-1)(x^2+2)+(Dx+E)(x-1)=\\=Ax^4+Bx^4-Bx^3+Cx^3+4Ax^2+2Bx^2-Cx^2+Dx^2+\\-2Bx+2Cx-Dx+Ex+4A-2C-E,\end{multline}solve the system$$\left\{\begin{array}{l}A+B=1\\-B+C=-1\\4A+2B-C+D=2\\-2B+2C-D+E=-1\\4A-2C-E=2.\end{array}\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $x^2-5y=3$ has no integer solutions. Here is my attempt:
$x^2-5y=3$ is the same as $2x^2-10y=6$, which is equivalent to $2x^2=10y+6$.
This means that 2 divides $x^2$. Since 2 is prime, then 2 divides x, or $x=2k$ for some integer $k$.
Then we have that $(2k)^2-5y=3$, or $4k^2=5y+3$. This implies that 2 divi... | If $x^2-5y=3$ then $x^2\equiv3\pmod 5,$
but that would contradict that,
for all integers $x$, $x^2\equiv0, 1, $ or $4\pmod5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.