Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\int_0^{2\pi}\frac{\sin^2x}{5+4\cos x}\ dx$ Using complex integration Here is the integral I am trying to evaluate:
$$\int_0^{2\pi}\dfrac{\sin^2x}{5+4\cos x}\ dx$$
I am getting the final answer as $-{\pi}/8$ but the correct answer is ${\pi}/4$
Here are the steps I did:
$$\begin{align}
\dfrac{\sin^2x}{5+4\cos x} dx &= ... | So far the steps you showed are correct except for a little typo I corrected.
But you didn't show the calculation of the residues.
With $f(z) =\frac{-1}{4i}\frac{{z^4+1-2z^2}}{{z^2}(2z+1)(z+2)}$ I get
$$\operatorname{Res}_{z=0}f(z)= -\frac{5}{16}i \mbox{ and }\operatorname{Res}_{z=-\frac 12}f(z)= \frac{3}{16}i$$
Hence,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $\prod_{m=1}^\infty \frac{1}{m^2}$ have a closed form? We know that $$\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6},$$ but what about the product of the reciprocal of the squares:
$$\prod_{m=1}^\infty \frac{1}{m^2}?$$
Do we use a different product representation to compute this? Maybe the cosine product
$$\co... | For a sum to converge, it is necessary that the terms converge to $0$ (the neutral element for the sum). $\frac{1}{n^2}$ satisfies this.
For a product to converge to a nonzero value, it is necessary that the factors converge to $1$ (the neutral element for the product). $\frac{1}{n^2}$ does not satisfies this, so the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Why do we use this complicated algorithm for finding values of trigonometric functions? My Mathematics Textbook covers the topic of Values of Trigonometric Functions at Allied Angles using some general formulae first and then goes on to the topic of finding the values of trigonometric functions at allied angles using a... | What, precisely, are these "simple formulae" that you are referring to?
The algorithm here essentially breaks down all the periodicity, symmetry, and reflection-type changes you can do to the angle (some change the value in simple ways).
The point is to normalize the angle you're working with, ultimately, to the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding $\lim_{x\to a} \frac{1}{(a^2-x^2)^2}(\frac{a^2+x^2}{ax}-2\sin( \frac{a\pi}{2})\sin(\frac{x\pi}{2}))$
Without using the L'Hopital Rule, evaluate the following limit.
$$\lim_{x\to a} \frac{1}{(a^2-x^2)^2}\left(\dfrac{a^2+x^2}{ax}-2\sin\left( \frac{a\pi}{2}\right)\sin\left(\frac{x\pi}{2}\right) \right)$$
Wher... | Let, $x-a=z$.
So, the limit changes to
$\lim \limits_{z \to 0} \frac{\frac{a^2+x^2}{ax}-2\text{sin}(\frac{aπ}{2})\text{sin}(\frac{xπ}{2})}{(a^2-x^2)^2}$
$=\lim \limits_{z \to 0} \frac{\frac{(x-a)^2}{ax}+2-2\text{sin}(\frac{aπ}{2})\text{sin}(\frac{xπ}{2})}{(z(x+a))^2} $
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^{\infty} \frac{ dx }{(x^4+c)(x^2+1) } $ Consider $$\int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } $$
where $c \in \mathbb{R}$. Now, it is possible to evaluate this using complex analysis, but I was wondering if there is way to compute this integral without complex analysis.
I was thinking on partial fra... | Assume $c>0$ for convergence.
$$I = \int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) }
= \frac1{c+1}\int_0^{\infty}\left( \frac{ 1}{x^2+1}- \frac{ x^2-1}{x^4+c }\right) dx \\
= \frac{\pi}{2(c+1)} - \frac{1}{c+1}\int_0^{\infty}\frac{ x^2-1}{x^4+c } dx$$
where
\begin{align}
\int_0^{\infty}\frac{ x^2-1}{x^4+c } dx
& =\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that $AD\cdot BD \cdot CD \leq \dfrac{32}{27}$ where $ABC$ is a triangle of circumradius 1 and $D\in (BC)$.
Let triangle $ABC$ of circumradius $1$ and $D$ a point on side $(BC)$.
Prove that $$AD\cdot BD\cdot CD\leq \dfrac{32}{27}.$$
My idea. By letting $\alpha = \dfrac{BD}{BC}$ (of course $0<\alpha <1$) we ge... |
It is a really bad practice to ask a geometry problem without a picture.
Let $OD = x.$ Then $BD\cdot CD = 1 - x^2.$ If $\angle AOD = \alpha,$ then:
$$AD^2 = 1+x^2-2x\cos\alpha.$$
So you need to prove:
$$(1-x^2)\sqrt{1+x^2-2x\cos\alpha}\leq\dfrac{32}{27}.$$
But this is just AM-GM:
$$(1-x^2)(1+x^2-2x\cos\alpha)^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ Let $x \geq 0.$ I need to prove that $x-\sin(x)\geq\dfrac{x^3}{(\pi+x)^2}.$
I tried the derivative, of $f(x)=x-\sin(x)-\dfrac{x^3}{(\pi+x)^2}$ which is $1-\cos(x)-\dfrac{x^2(x+3\pi)}{(\pi+x)^3},$ but it has a complicated formula.
Any ideas, hints?
Edit: sorry, there was a mistak... | Maybe an easier way to do it, though conceptually similar with @Integrand
By multiplying out and simplifying the terms, while noting that $x(x+\pi)^2-x^3=\pi x(x+2\pi)=\pi(x+\pi)^2-\pi^3$, the inequality is equivalent to:
$(\pi-\sin x)(x+\pi)^2 \ge \pi^3, x \ge 0$
But now $\pi-\sin x \ge \pi -1$ and we see that $(\pi+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $a+b=2 \text{ and } ab = -1.$ Determine $a^{10} + b^{10}$
Let $a,b \in \mathbb{R}$ for which $$a+b=2 \text{ and } ab = -1.$$ Determine $a^{10}+b^{10}.$
I tried to approach this the following way:
From $ab=-1$ we get that $a = -\frac{1}{b}$ and substituting this to $a+b=2$ yields in $b^2-2b-1=0$. This quadratic h... | Another way: Following your tack, if you choose $b = 1-\sqrt{2}$ then $a=-1/(1-\sqrt{2}) = 1+\sqrt{2}$ after you rationalize the denominator. So $a$ and $b$ are conjugates.
If you expand $(1+\sqrt{2})^{10} + (1-\sqrt{2})^{10}$ with binomial theorem, all the odd terms will cancel out. The 10th row of Pascal's trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$ Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
So, using AM-GM, or just pop out squares under square roots we can show:
$$\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc +... | The problem has appeared here various times and maybe elsewhere too for sure. The key to this is to observe the relationship between: $(a-b)^2 \ge 0$ and $a^2+ab+b^2 \ge \dfrac{3(a+b)^2}{4}$. They are equivalent. Using the latter but taking the square root first $3$ times for $(a,b), (b,c),(c,a)$ and add up, yielding t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $ \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$, find $|a_1|$
If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$. Find $|a_1|$
I could conclude that $\displaystyle \lim_{n \to \infty}a_n$ must be $\infty$. Bu... | Note that
$$
\frac{a_3}{a_1} = \frac{a_2a_3}{a_1a_2} = \frac21,\quad
\frac{a_5}{a_3} = \frac{a_4a_5}{a_3a_4} = \frac43,\quad
\frac{a_7}{a_5} = \frac{a_6a_7}{a_5a_6} = \frac65,\quad\dots
$$
and therefore
$$
\frac{a_{2k+1}}{a_1} = \frac{a_3}{a_1} \frac{a_5}{a_3}\cdots \frac{a_{2k+1}}{a_{2k-1}} = \frac21 \frac43 \cdots \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int \cos^2(x)\tan^3(x) dx$ using trigonometric substitution How would I integrate to evaluate $\int \cos^2(x)\tan^3(x) dx$ using trigonometric substitution?
I made an attempt by making substitutions such as $$\cos^2(x)=1-\sin^2(x)$$
$$\tan(x) = \frac{\sin(x)}{\cos(x)}$$ and $$\tan^2(x)=1+\sec^2(x)$$
But I co... | Just for fun:
$$\int \cos^2 x \tan^3x \ \mathrm{d} x$$
$$=\int \frac{\tan^3 x}{\sec^2 x} \ \mathrm{d} x$$
Now let $u = \tan^2 x, \mathrm du = 2 \tan x \sec^2 x \ \mathrm dx$:
$$=\int \frac{u \tan x }{\sec^2 x} \cdot \frac{\mathrm{d} u}{2 \tan x \sec^2 x}$$
$$=\frac{1}{2} \int \frac{u}{\sec^4 x} \mathrm d u$$
$$=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Surface integral of piecewise volume boundary? How should I go about solving the surface integral
$$\iint_{\partial V} 2(x^2+y^2) \, dS,$$ where $V$ is the region bounded by the paraboloid $$z=\frac12-x^2-y^2$$ and the cone $$z^2=x^2+y^2?$$
I have done surface integrals before, but I am unsure how to proceed with findi... | I assume the surface in question lies above the plane $z = 0;$ however, if not, you can easily adapt this solution to the case that the surface contains both the upper and lower nappe of the cone.
Graphing the two surfaces shows that the surface for which we would like to compute the surface area resembles an ice cream... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in t... | $$
16x^2 - 9 = 9\left( \left( \tfrac{4x}{3} \right)^2 - 1 \right) = 9(\sec^2\theta - 1) = 9\tan^2\theta.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 2
} |
Sum of complex roots' fractions According to this:
If $\omega^7 =1$ and $\omega \neq 1$ then find value of
$\displaystyle\frac{1}{(\omega+1)^2} +
\frac{1}{(\omega^2+1)^2} +
\frac{1}{(\omega^3+1)^2} +
... + \frac{1}{(\omega^6+1)^2}=?$
First I try like
$\displaystyle\frac{1}{\omega+1} +
\frac{1}{\omega^2+1} +
\frac{1}... | Let $\dfrac1{w+1}=x\implies w=\dfrac{1-x}x$
$$\implies\left(\dfrac{1-x}x\right)^7=1$$
As $x\ne0,$ $$x^6-3x^5+5x^4-\cdots=0$$
We need $$\sum_{r=1}^6x_r^2=\left(\sum_{r=1}^6x_r\right)^2-2\sum_{1\le i< j\le6}x_ix_j=\left(\dfrac31\right)^2-2\cdot\dfrac51$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Proving $\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $ For $a,b,c \in (0,\infty).$ Prove$:$
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $$
My proof by SOS$:$
$$ \left( {a}^{2}+bc \right) \left( ac+{b}... | $$LHS-RHS=\frac{4abc(b-c)^{2}(c-a)^{2}(a-b)^{2}}{(b+c)(c+a)(a+b)(a^{2}+bc)(b^{2}+ca)(c^{2}+ab)}+\frac{(a^{2}+b^{2}+c^{2}-bc-ca-ab)(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}-a^{2}bc-b^{2}ca-c^{2}ab)}{(a^{2}+bc)(b^{2}+ca)(c^{2}+ab)}\geq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3708222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find value of $\cot(\theta-\alpha)$ Let $\theta$ and $\alpha$ be the solutions to $$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=\dfrac{3\cos2x+6}{10}$$
find the value of $\cot(\theta-\alpha)$, where $\dfrac{\pi}{2}<\theta<\alpha<\dfrac{3\pi}{2}$
My attempt :
Let $A=\dfr... | Hint:
For $\sin \dfrac x2=s,\cos\dfrac x2=c$ and $c\ne s$
$$\dfrac{s^2}{1-\dfrac cs}+\dfrac{c^2}{1-\dfrac sc}=\dfrac{s^3-c^3}{s-c}=\cdots=1+\dfrac{\sin x}2$$
Now use $\cos2x=1-2\sin^2x$ to form a Quadratic Equation in $\sin x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate $\iint\frac{dxdy}{(1+x^2+y^2)^2}$ over a triangle
Calculate
$$\iint\frac{dxdy}{(1+x^2+y^2)^2}$$ over the triangle $(0,0)$, $(2,0)$, $(1,\sqrt{3})$.
So I tried changing to polar coordinates and I know that the angle is between $0$ and $\frac{\pi}{3}$ but I couldn't figure how to set the radius because it... | Here is an alternative to Robert's nice way to find how $r(\theta)$ depends on $\theta$.
Let $A(0,0)$, $B(2,0)$ and $C(1,\sqrt{3})$ be the three vertices of the triangle. Imagin, or see the picture below, a ray starting from $A$ and intersects with the side $CB$ at $D$. Suppose the angle $\angle DAB=\theta$. You want t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ Question: Solve the following equation in integers $x,y:$ $$x^2+6xy+8y^2+3x+6y=2.$$
Solution: For some $x,y\in\mathbb{Z}$ $$x^2+6xy+8y^2+3x+6y=2\\\iff x^2+2xy+4xy+8y^2+3x+6y=2\\\iff x(x+2y)+4y(x+2y)+3(x+2y)=2\\\iff(x+4y+3)(x+2y)=2.$$
Now if $(x+4y... | Another solution is to let $k=2y$ and get
$$x^2+3xk+2k^2+3x+3k=2$$
$$\iff (x^2+2xk+k^2)+(k^2+3xk)+3(x+k)=2\iff (x+k)^2+(x+k)(k+3)=2$$
So now we can let $m=x+k$ and get
$$m(m+k+3)=2 \implies (m,k) \in \{(1,-2),(-1,-4),(2,-4),(-2,-2)\}$$
Noting that $k=\frac{2}{m}-m-3$ in our calculations. This gives (after dividing the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}... | You can use the series
$$\ln(3/2) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^nn} = \frac{1}{2} - \frac{1}{8} + \frac{1}{24} - \frac{1}{64} + \dotsc$$
Clearly the partial sums fluctuate closer and closer to $\ln(3/2)$, and each new term has a strictly smaller magnitude than the previous one. Thus, since
$$\sum_{n=1}^{6}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
How to compute the integral $\int_0^1 \frac{x\ln x}{\ln (1-x)}dx$?
How to compute the integral $\int_0^1 \frac{x\ln x}{\ln (1-x)}dx$?
We can write
\begin{align*}
\int_0^1 \frac{x\ln x}{\ln (1-x)}dx=\int_0^1\frac{(1-u)\ln (1-u)}{\ln u}du
=-\int_0^1\frac{1-u}{\ln u}\sum_{n=1}^\infty \frac{u^n}{n}du
\end{align*}
Seeing... | One thig we could do is to expand as series
$$\frac{x}{\log (1-x)}=-1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{24}+\frac{19 x^4}{720}+\frac{3
x^5}{160}+\frac{863 x^6}{60480}+\frac{275 x^7}{24192}+\frac{33953
x^8}{3628800}+O\left(x^9\right)$$ where the coefficients correspond to the absolute value of Gregory coeffic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Does the polynomial $r^3+s^3+t^3-3rst-1$ have rational roots? I am looking at natural numbers $n$ of the form
$$n=a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc) = \det \left(\begin{array}{rrr}a & b & c \\c & a & b \\b & c & a\end{array}\right) $$
with $a,b,c \in \mathbb{Q}$.
Suppose we can find rational numbers $r,s,... | I hope, that this it what you are looking for:
$\gamma:=e^{i\frac{2\pi}{3}}$
$A^3+B^3+C^3-3ABC=D^3 \enspace$ with $\enspace D:=(a+b+c)(a+\gamma b+\gamma^2 c)(a+\gamma^2 b+\gamma c)$
and therefore $\enspace D= a^3+b^3+c^3-3abc\,$ .
So, you'll get $\,A,B,C\,$ expressed by $\,a,b,c\,$ which means, that with rational nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum $\sum_{n=1}^{\infty} \frac{3^n}{5^n-2^{2n}}$. Can someone help me with this sum $\sum_{n=1}^{\infty} \frac{3^n}{5^n-2^{2n}}$. I can't find $S_n$
| You can write $$\frac{3^n}{5^n-2^n} = \frac{(3/5)^n}{1 - (2/5)^n} = \sum_{k=0}^\infty (3/5)^n (2/5)^{kn}$$
so your sum is
$$ \eqalign{ \sum_{n=1}^\infty \sum_{k=0}^\infty (3/5)^n (2/5)^{kn}
&= \sum_{k=0}^\infty \sum_{n=1}^\infty (3/5)^n (2/5)^{kn}\cr
&= \sum_{k=0}^\infty \frac{3 \cdot 2^k}{5^{1+k}-3 \cdot 2^k}}$$
But t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy$ equal to $3\zeta(2)+\zeta(3)-6$ or to infinity? I am comparing two commercial Computer Algebra Sytems on the integral $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy$. One says it equals $3\zeta(2)+\zeta(3)-6=0.1368...$, the other that it does not converge.
which one i... | Thanks to metamorphy and Angina Seng, I can answer completely the question:
*
*the integral diverges, since $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy = \int_0^1\int_0^1 \frac{Li_3(xy)-xy}{(xy)^2}dxdy +\int_0^1\int_0^1 \frac{xy}{(xy)^2}dxdy$, where:
a) the second integral on the rhs diverges
b) the first one on ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating $\cos\frac\pi4$ from the half-angle formula gives $\sqrt{\frac12}$ instead of $\frac{\sqrt{2}}2$. What went wrong? I am using the formula $$\cos\left(\frac x2\right)=\sqrt{\frac{1+\cos(x)}2}$$ to find $\cos\left(\frac{pi}{4}\right)$ but it does not give me the correct result.
$$
\cos\left(\frac{\pi}{4}\righ... | $\sqrt {\frac 12} = \frac{\sqrt 1} {\sqrt 2} = \frac 1 {\sqrt 2} = \frac {\sqrt 2} {\sqrt 2 \cdot \sqrt 2} = \frac {\sqrt 2} {2} $
This process is called 'rationalising the denominator' and you should try to do this when the denominator is an algebraic irrational number.
| {
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"url": "https://math.stackexchange.com/questions/3723653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration of $\frac{1}{u^4 + (4\zeta^2-2)u^2 + 1}$ I am trying to compute
$$I(\zeta) = \int_{-\infty}^{\infty} \frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1}\, du$$
for positive real $\zeta$. Can anyone help?
I'm way out of practice for integrals except for simple stuff like $\int 1/(1+u^2)\, du = \tan^{-1}... | Partial fraction expansion gives
$$
\frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$
and since
$$\int \frac{du}{u^2+a^2}\, = \frac{1}{a}\int \frac{a\, du}{u^2+a^2}\,= \frac{1}{a}\int \frac{a^2\, dx}{a^2x^2+a^2}\, = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all pairs of positive integers $(m, n)$ for which $X^m + X + 1$ divides $X^n + 1$ and pairs $(m,n)$ for which $X^m +X −1$ divides $X^n +1$?
The special case I can think of is when $n=3$, and $m=2$ for the first part. But I don't know if other cases exist. Any help would be appreciated. Thanks!
| Let $r$ be a root of $X^m+X+1.$ The roots of $X^n+1$ lie on the unit circle, so $r$ does as well. Since $\Im(r+r^m) = \Im(-1) = 0$ and $r^m, r$ lie on the unit circle, $r, r^m$ are conjugates, so $r = r^{-m} \Rightarrow r^{m+1} = 1.$ But we also have $r^{m+1} = -r^2 - r,$ so $r^2+r+1 = 0 \Rightarrow r = w, w^2$ where $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=\frac{-3}{2}$,show that $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$ I think that I've done a major part of the problem but I'm stuck at a point.
Here's what I've done :
It's given to us that
$$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(... | If the sum of squares of two real numbers is zero, it implies that both numbers are zero. If you want you can simply prove this using Reductio-Ad-Absurdum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of the following approximation for $|x| \leqslant 1$ we have $1+x \leqslant e^x \leqslant1+x+x^2$ I was going through the text Introduction to Algorithms by Cormen et. Al. where I came across the following approximation...
for $|x| \leqslant 1$ we have $1+x \leqslant e^x \leqslant 1+x+x^2$
Now I could prove the... | For $0\leq x\leq 1$, we have
\begin{align*}
e^x - 1 - x - x^2 & = - \frac{{x^2 }}{2} + \sum\limits_{n = 3}^\infty {\frac{{x^n }}{{n!}}} \le - \frac{{x^2 }}{2} + \sum\limits_{n = 3}^\infty {\frac{{x^2 }}{{n!}}} \\ & = - \frac{{x^2 }}{2} + x^2 \left( {e - 1 - 1 - \frac{1}{2}} \right) = x^2 (e - 3) \leq 0.
\end{al... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\sqrt{\frac{bc}{a(3b+a)}} + \sqrt{\frac{ac}{b(3c+b)}} + \sqrt{\frac{ab}{c(3a+c)}} \ge \frac{3}{2}$. Prove: $\sqrt{\dfrac{bc}{a(3b+a)}} + \sqrt{\dfrac{ac}{b(3c+b)}} + \sqrt{\dfrac{ab}{c(3a+c)}} \ge \dfrac{3}{2}$ with $a, b, c$ are positive real numbers.
Let $a \le b \le c$:
\begin{align*}
\sqrt{\dfrac{bc}{a(3b+a... | From an old Vasile Cirtoaje inequality (see solution here, or here)
If $x,\,y,\,z$ are positive numbers, such that $xyz=1,$ then $$\sqrt{\frac{x}{y+3}}+\sqrt{\frac{y}{z+3}}+\sqrt{\frac{z}{x+3}}\geqslant\frac{3}{2}. \quad (1)$$
Beause $abc=1,$ we can put $x=\frac ab,\,y = \frac bc, \, z=\frac ca.$ Now, the inequality ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x) = x^4 - x^2 + 1$, find the values of $x$ such that $f(f(f(x))) \le x^8$
If $$f(x) = x^4 - x^2 + 1$$ find the values of $x$ such that $f(f(f(x))) \le x^8$
I noticed that $f(\pm 1) = 1 \implies \underbrace{f(f(f(... f(\pm 1)..)))}_{\text{n times}} = 1$. thus, $f(f(f(x))) = x^8$ at $x = \pm 1$. Fortunately, the... | $f(x)$=$x^2(x^2+1/x^2-1)$ ,
Now notice that
$(x^2+1/x^2-1)\ge1$ for all real x . (Apply $AM \ge GM$ for $x^2$ and $1\over x^2$)
That is $f(x)\ge x^2$ , which implies that $f(f(f(x))) \ge x^8$
But in the question it is given that $f(f(f(x))) \le x^8$
Therefore , $f(f(f(x))) = x^8$ , which will happen when $(x^2+1/x^2-1)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing that $\sum\limits_{j=0}^\infty a_j=\sum\limits_{k=0}^\infty b_k$ Show when $\sum\limits_{j=0}^\infty a_j$ is absolute convergent, than $\sum\limits_{k=0}^\infty b_k$ with $b_k:=(a_0+2a_1+...+2^ka_k)/2^{k+1}$ is also absolute convergent,
and even $\sum\limits_{j=0}^\infty a_j=\sum\limits_{k=0}^\infty b_k$
My at... | When you want to prove the absolute convergence of the series $\sum_k b_k$ you should prove that $\sum_k |b_k|$ is finite. When you only have non-negative terms, you can rearrange them in any order. Here
$$
\begin{aligned}
\sum\limits_{k=0}^\infty |b_k| &= \sum\limits_{k=0}^\infty \left |\sum\limits_{i=0}^k \frac{1}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show the $\arcsin$ identity: $ \arcsin(1 - 2x) + 2\arcsin(\sqrt{x}) = \pi / 2$ Can somebody find an elementary proof of the following identity:
$$
\arcsin ( 1 - 2x) + 2 \arcsin(\sqrt{x}) = \frac\pi2
$$
I noticed it while solving the following integral:
$$ I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} = -2 \sqrt{1 - x}\sqrt{x}... | Let $x=u^2$ with $0\le u\le1$ and rewrite the equation $\arcsin(1-2x)+2\arcsin(\sqrt x)=\pi/2$ as
$$\arcsin(1-2u^2)={\pi\over2}-2\arcsin u$$
Think of each side as an angle, $\theta_L$ and $\theta_R$. Clearly $-\pi/2\le\theta_L\lt\pi/2$, that being the range of the arcsine function. Because $u$ is nonnegative, we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the following limit: $\lim\limits_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})$ How can I evaluate following limit $$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=?$$
My first try:
$$\lim_{x\to\infty}(12x^2-2)\to \infty$$
$$\lim_{x\to\infty}(6x\sqrt{3x^2-2})\to \infty$$so
$$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=\in... | Your method is not correct. It's not true that $\infty - \infty =0$. You may proceed by rationalizing the given function as follows:
$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=\lim_{x\to\infty} \frac{(12x^2-2-6x\sqrt{3x^2-2})( (12x^2-2+6x\sqrt{3x^2-2}) ) } { 12x^2-2+6x\sqrt{3x^2-2} } $
$=\lim_{x\to \infty} \frac{(12x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$ with Fubini's theorem I have to solve the following double integral
$$\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$$
with $A= \left[0,+\infty\right[ \times [0,1].$ So far I've tried to solve it integrating w.r.t. $y$ first.
$$\iint_0^1\frac{dy\,dx}{(1+x^2)(1+x^2 y^2)} ... | \begin{align} & \iint\limits_{[0,+\infty) \times [0,1]} \frac{d(x,y)}{(1+x^2)(1+x^2 y^2)} \\[8pt] = {} & \int_0^1 \left( \int_0^\infty \frac{dx}{(1+x^2)(1+x^2y^2)} \right) \, dy \end{align}
So we need partial fractions:
\begin{align}
& \frac 1 {(1+x^2)(1+x^2 y^2)} = \frac {Ax+B} {1+x^2} + \frac{Cx+D}{1+x^2y^2} \\[10pt]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\sum\limits_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$? I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series wh... | Another approach is to use the Anti-difference concept.
Since the digamma function is defined as
$$
\psi \left( z \right) = {d \over {dz}}\ln \Gamma \left( z \right)
$$
its functional equation is
$$
\Delta \psi \left( z \right) = \psi \left( {z + 1} \right) - \psi \left( z \right) = {d \over {dz}}\ln \left( {z\,\Gamma ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
| If you want to go through L'Hospital you may separate $\frac{1}{\cos x}$ the limit of which is $1$ for $x \to 0$:
$$\frac{\tan x - \sin x}{x^3} = \frac 1{\cos x}\cdot \frac{\sin x - \sin x \cos x}{x^3}$$
So, you only need to calculate the limit for $x \to 0$ of
$$\frac{\sin x - \sin x \cos x}{x^3}= \frac{\sin x - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Factoring $ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$
Factor $$ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$$
I used the factor theorem to get factors
$$f(a, b, c)=(a-b)(b-c)(a-c)\;g (a, b, c)$$
for some polynomial $g (a, b, c)$.
How can I continue using this method?
(sorry for the previously messed up question I'm ne... | $$ab^3-a^3b+bc^3-b^3c+ac^3-a^3c$$ $$=ab(a+b)(b-a)+bc (b+c)(c-b)+ac (a+c)(a-c)$$ $$=ab( (a+b+c)-c)(b-a)+bc ((a+b+c)-a)(c-b)+ac ((a+b+c)-b)(a-c)$$ $$=(a+b+c) [ab (b-a)+bc (c-b)+ac (a-c)]-abc (b-a+c-b+a-c)$$ $$=(a+b+c) [ab (b-a)+bc (c-b)+ac (a-c)]+0$$$$=(a+b+c)[(a-b)(b-c)(c-a)]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Remainder when divided by $7$ What would be the remainder when
$12^1 + 12^2 + 12^3 +\cdots + 12^{100}$ is divided by $7$ ?
I tried cyclic approach (pattern method), but I couldn't solve this particular question.
| We have
$$1+\sum_{i=1}^{100}12^i=\frac{12^{101}-1}{12-1}=\frac{12^{101}-1}{11}$$
Since $\gcd(7,11)=1$, we need only find the remainer of $12^{101}-1$ with $7$.
We have $12^{101}\equiv 5^{101}\bmod 7$ and $5^{101}=25^{50}\cdot5\equiv 5\mod7$
Thus $12^{101}-1\equiv 4\mod7$
and as we explained this gives $\frac{12^{101}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that the Diophantine equation $m(m-1)(m-2)(m-3) = 24(n^2 + 9)$ has no solutions
Consider the Diophantine equation $$m(m-1)(m-2)(m-3) = 24(n^2 + 9)\,.$$ Prove that there are no integer solutions.
One way to show this has no integer solutions is by considering modulo $7$ (easy to verify with it).
I am curious ... | Here is one approach that is more motivated. Let $k$ be the integer in between $m-1$ and $m-2.$ So $$k=\frac{(m-1)+(m-2)}{2}=m-\frac{3}{2}.$$ Then the left side of the equation is
\begin{align*}
m(m-1)(m-2)(m-3) &= \left(k+\frac{3}{2}\right)\left(k+\frac{1}{2}\right)\left(k-\frac{1}{2}\right)\left(k-\frac{3}{2}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$?
How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$
This is taken from a definite integral where $x$ varies from $0$ to $1$.
My attempt:
multiplied by conjugate
$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{(\sqrt{1+x}+\sqrt{... | $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{dx}{\sqrt{(\sqrt{1+x}-\sqrt{1-x})^2}}$$
$$=\int \frac{dx}{\sqrt{2-2\sqrt{1-x^2}}}$$
Let $x=\sin\theta\implies dx=\cos\theta d\theta$
$$=\int \frac{\cos\theta d\theta}{\sqrt{2-2\cos\theta}}$$
$$=\int \frac{\cos\theta d\theta}{\sqrt{4\sin^2\frac{\theta}{2}}}\quad \quad \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to prove that these two functions do not intersect? I have this function $\;g(x)-x\;$ on the domain $\;0<x<\frac{1}{3}\;$, where
$g(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 ... | Partial answer
Denote
$$g_1(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)+1}}{2 \sqrt{2} \pi \cosh (\pi x)}$$ and
$$g_2(x) = \frac{18 \sqrt{2} x^2 \left(9 \pi x^3+4 \tanh ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Olympiad inequality proof issue Prove that $(a^2+b^2)^2\geq(a+b+c)(a+b-c)(b+c-a)(c+a-b)\ \forall \ a,b,c\in\mathbb{R^+} $.
I, forgetting to consider whether $a_1$ and $a_2$ are strictly non-negative (don't think they are), found a proof (almost) using the AM-GM inequality with $a_{1}=(a+b+c)(a+b-c)=(a+b)^2-c^2$ and $a_... | We are to prove (continuing from your work)
$$ \left(c^2-(a+b)^2\right)\left(c^2-(a-b)^2\right)+4a^2b^2\geqslant 0\iff\left(c^2-(a^2+b^2)\right)^2\geq 0$$
which is trivially true.
Your proof works if $b+c\geqslant a$, $c+a\geqslant b$ and $a+b\geqslant c$. Otherwise, if exactly one of the three terms containing a minus... | {
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"url": "https://math.stackexchange.com/questions/3742448",
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Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic? Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Show that $a_{n}$ converges as $n\to\infty$. What is the limit? Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic?
MY ATTEMPT
The answer to the first quest... | Let $f(x)=\sqrt{x^2+x}-x$.
By AM-GM we have $\sqrt{x(x+1)}< \dfrac{x+(x+1)}{2}\ $ thus $\ f(x)< \dfrac 12\ $ and $\ a_n< \dfrac 12$
This also means $A=2x+1-2\sqrt{x(x+1)}>0$
Since $f'(x)=\dfrac{A}{2\sqrt{x(x+1)}}$ then $f$ is $\nearrow$ and so is $a_n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Closed-form solution for the determinant of a Vandermonde-like matrix I'm trying to find a closed-form solution $\forall$ odd integer $n\ge 3$ for the determinant of a matrix with some structure on it. After some manipulation, I've reduced it to the following matrix:
$\small\begin{bmatrix}\boldsymbol{t_{1}^{n}-t_{a}^{n... | These seem to be a direct consequence of Schendel's 1891 theorem about "confluent Vandermonde matrices", as explained in (say) "On a Recursion Formula Related to Confluent Vandermonde",
Shui-Hung Hou and Edwin Hou, The American Mathematical Monthly,
Vol. 122, No. 8 (October 2015), pp. 766-772, or here or here. An exam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$
I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all... | Here we use $$ab(a-b)+bc(b-c)+ca(c-a)=a^2(b-c)+b^2(c-a)+c^2(a-b)=-(a-b)(b-c)(c-a)$$
In short we cab write $$\sum ab(a-b)=\sum a^2(b-c)=-(a-b)(b-c)(c-a)~~~~(1)$$
Then $$F=\frac{a^2+b^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-a)(b-c)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$$
$$\implies F=\frac{(a^2+b^2-a^2)(c-b)+(c^2+a^2-b^2)(a-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How can I evaluate $\int _0^1\frac{\tan ^{-1}\left(3\sqrt{\frac{a}{4-a}}\right)}{\sqrt{a}\sqrt{4-a}}\:\mathrm{d}a$ I have the integral:
$$\int _0^1\frac{\tan ^{-1}\left(3\sqrt{\frac{a}{4-a}}\right)}{\sqrt{a}\sqrt{4-a}}\:\mathrm{d}a.$$
If I use $u=3\sqrt{\frac{a}{4-a}}$, I get
$$6\int _0^{\sqrt{3}}\:\frac{\tan ^{-1}\lef... | Using $$ \tan^{-1}(x) = \frac{i}{2} \, \ln\left( \frac{1 - i \, x}{1 + i \, x} \right) $$ then
\begin{align}
\int \frac{\tan^{-1}(x)}{x^2 + a^2} \, dx &= \frac{i}{2} \, \left( \int \frac{\ln(1 - i \, x)}{x^2 + a^2} \, dx - \int \frac{\ln(1 + i \, x)}{x^2 + a^2} \, dx \right).
\end{align}
Using
\begin{align}
\int \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $\sum_{r=1}^{\infty}\left(\frac{2r+5}{r^2+r}\right)\left(\frac{3}{5}\right)^{r+1}$
$\text{Find the value of}$ $$\lim_{n\to \infty}\left(\sum_{r=1}^{n}\left(\frac{2r+5}{r^2+r}\right)\left(\frac{3}{5}\right)^{r+1}\right)$$
$\text{Answer}: \frac{9}{5}$
Firstly I split "linear-upon-quadratic" term:
$$\frac{2r+5}{... | Your idea was right:$$\sum_{r=1}^{\infty}\left(\frac{5}{r}-\frac{3}{r+1}\right)\left(\frac{3}{5}\right)^{r+1}$$
$$=\sum_{r=1}^{\infty}\left(\frac{5}{r}\right)\left(\frac{3}{5}\right)^{r+1}-\sum_{r=1}^{\infty}\left(\frac{3}{r+1}\right)\left(\frac{3}{5}\right)^{r+1}$$
$$=3\sum_{r=1}^{\infty}\left(\frac{1}{r}\right)\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $f(x)-x f'(x)\neq 0$? I have this function:
$f(x)=\cosh ^{-1}\left(\frac{4 a^2 x^2+\left(a^2 x^2-1\right)^2 \cosh (2 \pi x)}{\left(a^2 x^2+1\right)^2}\right),$
where $0<x<\frac{1}{a}$ and $a$ is a positive real number. I want to prove that $\;f(x)-x f'(x)>0$, or at least to prove that $f(x)-x f'(x)\n... | I hope and wish that you will receive simpler answers.
If you compose Taylor series around $x=0$ (I skip the intermediate steps)
$$A=\frac{4 a^2 x^2+\left(a^2 x^2-1\right)^2 \cosh (2 \pi x)}{\left(a^2
x^2+1\right)^2}=1+2 \pi ^2 x^2+\left(\frac{2 \pi ^4}{3}-8 \pi ^2 a^2\right) x^4+O\left(x^6\right)$$
$$f(x)=\cosh ^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How can i solve $\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$ How can i solve following $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$$
My work:
I substituted $x=\tan\theta$, $dx=\sec^2\theta d\theta $
integral becomes
$\int \dfrac{\tan^3\theta+2\tan \theta-7}{\sqrt{\tan^2\theta+1}}\ \sec^2\theta d\theta$
$\int \dfrac{\tan\the... | $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx$$
$$=\int \frac{x(x^2+1)+x-7}{\sqrt{x^2+1}}\ dx$$
$$=\int x\sqrt{x^2+1}\ dx+\int \frac{x}{\sqrt{x^2+1}}\ dx-\int \frac{7}{\sqrt{x^2+1}}\ dx$$
$$=\frac12\int \sqrt{x^2+1}\ d(x^2+1)+\frac12\int \frac{d(x^2+1)}{\sqrt{x^2+1}}-7\int \frac{dx}{\sqrt{x^2+1}}$$
$$=\frac13(x^2+1)^{3/2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Find the approximate sum of the series $\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}$ Prove that the series converges and find the approximate sum. $$\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}$$
To prove that converges what I did is to use the Cauchy's convergence test.
$$\lim_{x\to \infty} \sqrt[n]{a_n} =\lim_{x\t... | Another way to check convergence is the comparison test
$$\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}<\sum_{n=0}^{\infty} \frac{1}{3^n}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$$
This also provides a way to approximate the sum to as many places as required. Lets say you want to approximate the sum to $\epsilon$ accuracy. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sums of powers of cosines and sines shifted by $2\pi/3$ I have stumbled across these two identities
$$
\begin{split}
\cos^2(x)+\cos^2(x+2\pi/3)+\cos^2(x+4\pi/3) &= 3/2,\\
\cos^4(x)+\cos^4(x+2\pi/3)+\cos^4(x+4\pi/3) &= 9/8.
\end{split}
$$
There is also the more intricate
$$
\begin{split}
\cos^2(x)\sin^2(x)+\cos^2(x+2\pi... | This is an answer to question 3.
Let
$$f_n(x):=\cos^n(x)+\cos^n\bigg(x+\frac{2\pi}{3}\bigg)+\cos^n\bigg(x+\frac{4\pi}3\bigg)$$
$$\small g_n(x):=\cos^n(x)\sin^n(x)+\cos^n\bigg(x+\frac{2\pi}3\bigg)\sin^n\bigg(x+\frac{2\pi}3\bigg)+\cos^n\bigg(x+\frac{4\pi}3\bigg)\sin^n\bigg(x+\frac{4\pi}3\bigg) $$
where $n$ is a positive ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
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Evaluating limit of the function at $\frac{\pi}{2}$ I'm trying to solve this
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}$$
I have tried using the L'Hôpital's rule
But I'm stuck at
$$\lim_{x \to \frac{\pi}{2}} \frac{-\sin{x}}{3(x-\frac{\pi}{2})^2}$$
Since the above equation is not in the $\frac{0}{0... | welcome to MSE
AS a hint
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}=\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi}{2})^3}$$now take $x-\frac{\pi}{2}=a $ when $x$ tends to $\frac{\pi}{2}$ ,a tends to zero
$$\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the error terms in the asymptotic expansion of a binomial coefficient are uniformly bounded I'm interested in the asymptotic behavior of
$$ a_{n,k} :=\frac{1}{2^n} \binom{n}{\frac{n}{2} +k},$$
where $k$ is within a constant times $\sqrt{n}$, say
$$ |k| < C \sqrt{n }.$$
Using Stirling's approximation, I found
... | The desired uniform bound follows from essentially the same computation that you might have used to obtain the formula for $a_{n,k}$.
Indeed, the Stirling's approximation tells that
\begin{align}
n! = \sqrt{2\pi} \, n^{n+\frac{1}{2}} e^{-n} \left( 1 + \mathcal{O}\left(n^{-1}\right) \right),
\end{align}
where the implic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\sqrt{8-\sqrt{63}}$ I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
| Note that $63=9 \times 7$ and $8=\frac{1}{2}(9+7)$. Therefore,
$$ 9+7-2\sqrt{9 \times 7} = (\sqrt{9}-\sqrt{7})^2, $$
so that
$$ 8-\sqrt{63} = \frac{1}{2}(16-2\sqrt{63}) = \frac{1}{2}(3-\sqrt{7})^2 $$
and
$$ \sqrt{8-\sqrt{63}} = \frac{3-\sqrt{7}}{\sqrt{2}} = \frac{3\sqrt{2}-\sqrt{14}}{2}. \quad \blacksquare $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Integrate: $\int \frac{x}{\left(x^2-4x-13\right)^2}dx$.
Integrate:
$$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$
Here's my attempt:
I first completed the squares for the denominator:
$$\left(x^2-4x-13\right)^2=(x-2)^2-17 \implies \int \frac{x}{\left(\left(x-2\right)^2-17\right)^2}dx$$
I then used $u$-subsituition:
$$... | Express the second integral as
\begin{align}
\int \frac{2}{\left(u^2-17\right)^2}du
&= -\frac2{17}\int \frac{du}{u^2-17} +\frac2{17}\int \frac{u^2du}{(u^2-17)^2}
\end{align}
with
\begin{align}
\int \frac{u^2du}{(u^2-17)^2}
= -\frac12\int u d\left(\frac1{u^2-17} \right)
= -\frac u{2(u^2-17)}+\frac12\int \frac {ud u}{u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Sum of Continued Fractions Let $x$ be a positive integer.
Consider the following sum (maybe there is a better notation with continued fractions, but I am not aware of it):
$\frac{1}{x} + \frac{1}{x- \frac{1}{x}} + \frac{1}{x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}}} +\frac{1}{x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}} - \frac{1}... | If $\{a_i\}$ is the sequence of terms, and $s_n$ is the sum of first $n$ terms, then we get $$a_{n+1} = \frac{1}{x-s_{n}}$$ for $n\geq1$ and $a_1= \frac{1}{x}$. Rearranging this, $s_n = x-\frac{1}{a_{n+1}}$. Each term is $\geq 1/x$, so their reciprocals are $\leq x$.
| {
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"url": "https://math.stackexchange.com/questions/3758171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $2a^3+27c=9ab$
Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $$2a^3+27c=9ab$$
So far, I let the roots of $x^3+ax^2+bx+c=0$ be $r_1, r_2,$ and $r_3$. $r_1=r_2-d$ and $r_3=r_2+d$ because they form an arithmet... | Let's do this thing
$(r+d)^3 +a(r+d)^2 +b(r+d) + c =(r-d)^2 + a(r-d)^2 + b(r-d) + c = r^3 + ar^2 +br + c = 0$ (read the hidden for the insight as to how I thought to do the next step... or just go on to the next step)
Insight: Now the standard trick is to note that the coefficients of the even powers of $d$ in $(r+d)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find coordinates of a point Q on the graph $\sin (x) + \cos (y) = 0.5$ given that the gradient of its tangent is perpendicular to point P. Note:
Point $P$ is on the $y$-axis and above the $x$-axis
$\frac{-\pi}{6}\le x \le\frac{7\pi}{6}$
$\frac{-2\pi}{3}\le y\le\frac{2\pi}{3}$
What I have done so far:
Solving for $P$:
... | Note that the curve (one of an infinite number of such "loops" satisfying the given equation) is symmetrical about the point $ \ \left(\frac{\pi}{2} \ , \ 0 \right) \ ; $ this is a consequence of the trigonometric function properties $ \ \cos (-y) \ = \ \cos y \ $ and $ \ \sin \left(\frac{\pi}{2} + \xi \right) \ = \ \... | {
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"url": "https://math.stackexchange.com/questions/3761928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Definite integral of $\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx$ using the substitution of $x=2\tanθ$. Can someone help with the integral $\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx$?
I'm supposed to find the definite integral for this using the substitution $x=2\tanθ$.
This is what I've done so far:
$$\longrightarrow \frac{dx}... | Alternate easier method: use reduction formula:
$\color{blue}{\int \frac{dt}{(t^2+a^2)^n}=\frac{t}{2(n-1)a^2(t^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)a^2}\int \frac{dt}{(t^2+a^2)^{n-1}}}$ as follows
$$\int_{-2}^2 \dfrac{5}{(x^2+4)^2}dx$$$$=10\int_{0}^2 \dfrac{dx}{(x^2+2^2)^2}$$
$$=10\left[\frac{x}{2(2-1) 2^2(x^2+2^2)}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Proof that $\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0$ How should I prove
$$\forall n\in\mathbb{N}:\, \frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0?$$
My attempt:
$$\begin{align}\frac{e^{\frac{\pi i}{4}... | To proceed:
$$\begin{align}
\frac{1+i}{2}+\frac{1}{i^{2n}+i^{2n+1}}-i^{n^2}&=\frac{1+i}{2}+\frac{1}{i^{2n}(1+i)}-i^{n^2}\\
&=\frac{1+i}{2}+\frac{(-1)^n}{1+i}\frac{1-i}{1-i}-i^{n^2}\\
&=\frac{1+i}{2}+\frac{(-1)^n(1-i)}{2}-i^{n^2}\\
&=\frac{1+(-1)^n}{2}+i\frac{1-(-1)^n}{2}-i^{n^2}
\end{align}$$
For $n$ even, $4\mid n^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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If $a, b, c\in\mathbb R^+,$ then prove that: $\sum_{cyc} \frac1{c-b}\left(\frac1{\sqrt{a+2b}}-\frac1{\sqrt{a+2c}}\right)\ge\frac3{\sqrt{(a+b+c)^3}}.$ We have:$$\sum_{cyc}\frac1{c-b}\left(\frac1{\sqrt{a+2b}}-\frac1{\sqrt{a+2c}}\right)\\\\=\sum_{cyc}\frac{\sqrt{a+2c}-\sqrt{a+2b}}{(c-b)\sqrt{a+2b}\sqrt{a+2c}}\\\\=\sum_{cy... | Now, by AM-GM and Jensen we obtain:
$$\sum_{cyc}\frac{2}{\sqrt{(a+2b)(a+2c)}(\sqrt{a+2b}+\sqrt{a+2c})}\geq$$
$$\geq\sum_{cyc}\frac{2}{\frac{a+2b+a+2c}{2}\cdot\sqrt{2(a+2b+a+2c)}}=\frac{3}{\sqrt{(a+b+c)^3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
plot of $\sin(x) + \sin(y)= \cos(x) + \cos(y)$ I was playing arround with implicit plots of the form $f(x,y) = g(x,y)$, and I noticed that if you plot in the plane the following equation: $\sin(x) + \sin(y)= \cos(x) + \cos(y)$ you get the following graph:
My question is why does this trigonometric functions give us th... | Using Prosthaphaeresis Formulas
$$2\sin\dfrac{x+y}2\cos\dfrac{x-y}2=\cos\dfrac{x+y}2\cos\dfrac{x-y}2$$
If $\cos\dfrac{x-y}2=0\implies\dfrac{x-y}2=(2n+1)\dfrac\pi2, x-y=(2n+1)\pi$
else $\sin\dfrac{x+y}2=\cos\dfrac{x+y}2\iff\tan\dfrac{x+y}2=1\implies\dfrac{x+y}2=m\pi+\dfrac\pi4\iff x+y=\dfrac{(4m+1)\pi}2$
So we are getti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Roze... | The first problem.
By Holder $$\left(\sum_{cyc}\frac{a^2}{b}\right)^2\sum_{cyc}a^2b^2\geq\left(\sum_{cyc}a^2\right)^3.$$
Thus, it's enough to prove that $$\sum_{cyc}a^2b^2\leq4,$$ which is true by AM-GM:
$$\sum_{cyc}a^2b^2=(a^2+c^2)(b^2+d^2)\leq\left(\frac{a^2+c^2+b^2+d^2}{2}\right)^2=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
prove that $\binom{k+1}{1} S_k(n)+\binom{k+1}{2} S_{k-1}(n)+.........+\binom{k+1}{k+1} S_0(n)=(n+1)^{k+1}-1$ For $n,k \in N$,we define
$S_k(n)=1^k +2^k+... +n^k$
if $(1+x)^p=1+\binom{p}{1}x+\binom{p}{2}x^2+.... +\binom{p}{p}x^p,p\in N$,then we have to prove that
$\binom{k+1}{1} S_k(n)+\binom{k+1}{2} S_{k-1}(n)+...+\b... | By binomial theorem:
$$(1+x)^{m+1}-x^m=1+{n \choose 1}(x)+{n \choose 2} (x^2)+....+{m+1 \choose m} (x^m)$$
putting $x=1,2,3,,..n$ and adding them, we get
$$(n+1)^{n+1}-1=(1+1+1...+1)+{m+1 \choose 1}(1+2+3+...+n)+{m+1 \choose 2}(1^2+2^2+3^2++...+n^2)+{m+1 \choose 3} (1^3+2^3+3^3+4^3+...+n^3)+.....+{m+1 \choose m}(1^m+2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Challenging Problem: $\int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =\pi^a e^{\frac{-b}{\sqrt{c}}}\sin \big({\frac {d}{\sqrt{e}}}\big)$ $$ \int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =\pi^a e^{\frac{-b}{\sqrt{c}}}\sin \big({\frac {d}{\sqrt{e}}}\big) $$ where $a,b,c,d,e$ are positive integers and $c$ ... | Without contour integration (the simplest, for sure)
This is just for your curiosity : you ended with a series of antiderivatives
$$I_k=\int \frac{\sin(nx) }{x+k}dx$$ where $k$ is a complex number. Let $y=x+k$ to make
$$I_k=\int \frac{\sin (n (y-k))}{y}\,dy=\cos (k n)\int\frac{ \sin (n y)}{y}dy-\sin (k n)\int\frac{ \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of a diagonal matrix and a symmetric positive semidefinite matrix Let the following symmetric matrix
$A = \left[ \matrix{a_{12}+a_{13} & -a_{12} & -a_{13} \\
-a_{12} & a_{12}+a_{23} & -a_{23} \\
-a_{13} & -a_{23} & a_{13}+a_{23} } \right]$,... | By Sylvester's criterion, $A+D$ is positive semidefinite if and only if all principal minors of $A+D$ are nonnegative. In other words, $A+D$ is PSD iff
\begin{aligned}
&d_1+a_{12}+a_{13}\ge0,\\
&d_2+a_{12}+a_{23}\ge0,\\
&d_3+a_{13}+a_{23}\ge0,\\
&(d_1+a_{12}+a_{13})(d_2+a_{12}+a_{23})\ge a_{12}^2,\\
&(d_1+a_{12}+a_{13}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding a polynomial $f(x)$ of degree 5 such that $f(x)$ is divisible by $x^3$ and $f(x)+2$ is divisible by $(x+1)^3.$ There is some polynomial $f(x)$ of degree $5$ such that both of these properties hold:
$f(x)$ is divisible by $x^3$.
$f(x)+2$ is divisible by $(x+1)^3.$
Find that polynomial.
I know that because $f(x)$... | If $f(x)+2=ax^5+bx^4+cx^3+2=(x^3+3x^2+3x+1)(dx^2+ex+f)$,
then multiplying out and equating coefficients yields
$a=d, b=e+3d, c=f+3e+3d, 0=d+3e+3f, 0=e+3f,$ and $ 2=f$.
This system is easy to solve for $a, b, c, d, e, $ and $f, $
and then the answer is $f(x)=ax^5+bx^4+cx^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\cos^2(\theta)+\sin^2(\theta) = 1$ $$\cos^2(\theta) + \sin^2(\theta) = 1$$
I solved this by using right triangle,
$$\sin(\theta) = \frac{a}{c}, \quad \cos(\theta) = \frac{b}{c}$$
$$\cos^2(\theta) + \sin^2(\theta) = 1$$
$$\Bigl(\frac{b}{c}\Bigr)^2 + \Bigl(\frac{a}{c}\Bigr)^2 = 1 $$
$$\frac {a^2 + b^2} {c^2} = 1 $... | My start, for my students is the real circunference $\mathcal C$ of center $O(0,0)$ and radius $r$, $P\equiv(x,y)\in \mathcal C$ $$\mathcal C\colon \quad x^2+y^2=r^2$$ $r>0$ and $x=r\cos\theta, y=r\sin\theta$, $\theta\in[0,2\pi[$, that it gives: $$r^2\cos^2\theta+r^2\sin^2\theta=r^2 \iff r^2(\cos^2\theta+\sin^2\theta)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to find the position vector for the point of intersection of a line and the perpendicular line through a point C
How could I find the exact coordinates of the point N for example which is the point of intersection of the line $L=\begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} +t\begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatr... | Your derivation seems correct indeed, by a direct check, for $t=-\frac 6{11}$ we obtain
$$N=\begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} -\frac 6{11}\begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}=\begin{pmatrix} \frac {18}{11} \\ \frac {1}{11} \\ \frac {40}{11} \end{pmatrix}$$
then
$$\overrightarrow{CN}=\begin{pmatrix} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$ I need help with the following question:
Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$
My solution: First I know that $143=11\cdot 13$ then because $\gcd (11,13) = 1$ then $3x^2+18x+95\equiv 0\pmod {143}$ if, and only if $$3x^... | You are allowed to use the quadraditic formula.
$3x^2 + 18x + 95 \equiv 0 \pmod {143}$ means
$x \equiv \frac {18\pm \sqrt{18^2 -4*95*3}}{6} \pmod {143}$.
Now use Chinese remainder theorem on $143 = 13*11$
$\frac {18\pm \sqrt {18^2 -4*95*3}}{6} \pmod {11}\equiv$
$\frac {7\pm {7^2 - 4*7*3}}{6}*12\pmod {11}\equiv$
$2(7\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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understand the proof of $\frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}<\frac{4 n+3}{6} \sqrt{n}$
If $n \in \mathbb{N}^*$, prove that \begin{align*}\frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}<\frac{4 n+3}{6} \sqrt{n}.\end{align*}
I am having trouble understanding the following proof of this problem:
Let
$a_{n}=(... | If $\lambda = \frac 12,$ by using the inequality they provided :
$$\dfrac{\sqrt{n}}{a_n-a_{n-1}}>\frac 23\iff \sqrt{n} >\frac 23\left(a_n-a_{n-1}\right ).$$
Sum this up for and then you will get one side by telescoping. You can do a similar thing for $\lambda=\frac 34$ as well.
| {
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Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not s... | Using the Lagrange multiplier method, we make the expression $F(x,y,\lambda)=x^2+y^2+\sqrt{32}+\lambda(x^2+y^2-2x+2y-2)$ and then calculate partial derivatives on $x$, $y$ and $\lambda$, require them to be $0$, then solve for $x$ and $y$ (and $\lambda$):
$$0=\frac{\partial F}{\partial x}=2x+2\lambda x-2\lambda$$
$$0=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the local and absolute values of a polynomial on a closed finite interval Problem:
Find the local and absolute extreme values of the $y$ on the interval $[0,5]$.
$$ y = x^3 - 9x^2 + 24x - 2 $$
Answer:
To see a plot of this function, click on the link below.
enter link description here
Based upon the plot, I cla... | For $x\in(0,5)$ we have
$$y' = 3x^2 - 18x + 24=0 \implies x_1=2\:,x_2=4 \implies f(x_1)=18\:, f(x_2)=14$$
and since
$$y'' = 6x - 18\implies f''(x_1)<0\:, f''(x_2)>0$$
then on the boundary
$$x_0=0\:,x_3=5 \implies f(x_0)=-2\:, f(x_3)=18$$
Therefore
*
*$x_1$ is a point of local maximum
*$x_2$ is a point of local min... | {
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"question_score": "2",
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Inequality with a High Degree Constraint This question-
Suppose that $x, y, z$ are positive real numbers and $x^5 + y^5 + z^5 = 3$. Prove that $$ {x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \ge 3 $$
The inequality has a high degree constraint which can convert a $5$-degree polynomial to a $0$-degree term and makes i... | We'll prove that $$x^{6.5}+y^{6.5}+y^{6.5}\geq x^6+y^6+z^6,$$ for which it's enough to prove that:
$$\sum_{cyc}\left(10x^{6.5}-10x^6-x^5+1\right)\geq0$$ or
$$5\sum_{cyc}\left(2x^{1.5}-3x+1\right)x^{5}+\sum_{cyc}\left(5x^6-6x^5+1\right)\geq0,$$ which is true by AM-GM.
Now, by C-S and Vasc we obtain:
$$\sum_{cyc}\frac{x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show $\binom{n}{[n/2]} = \binom{n-1}{[(n-1)/2]} + \sum_{i=0}^{[n/2] - 1} \frac{1}{i+1} \binom{2i}{i} \binom{n-2i-2}{[n/2]- i - 1}$ The identity
$$\binom{n}{\lfloor n/2\rfloor} = \binom{n-1}{\lfloor (n-1)/2\rfloor} + \sum_{i=0}^{\lfloor n/2\rfloor - 1} \frac{1}{i+1} \binom{2i}{i} \binom{n-2i-2}{\lfloor n/2\rfloor - i - ... | In trying to evaluate
$$\sum_{q=0}^{\lfloor n/2 \rfloor -1}
\frac{1}{q+1} {2q\choose q}
{n-2q-2\choose \lfloor n/2 \rfloor - q - 1}$$
we get for $n=2m$
$$\sum_{q=0}^{m -1}
\frac{1}{q+1} {2q\choose q}
{2m-2q-2\choose m - q - 1}
\\ = [z^{m-1}] (1+z)^{2m-2} \sum_{q=0}^{m -1}
\frac{1}{q+1} {2q\choose q}
z^q (1+z)^{-2q}.$$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Is there a simpler form for $\sqrt{(-1 +\sqrt{1+8x^2} + 4x^2)/8} \cdot (3 + \sqrt{1+8x^2})/4|x|$ I'm trying to find a simpler form for $y=\sqrt{(-1 +\sqrt{1+8x^2} + 4x^2)/8} \cdot (3 + \sqrt{1+8x^2})/4|x|$
This is a "V"-shaped curve that:
*
*goes through $x=0, y=1$
*asymptotically approaches $y=\frac{|x|+\sqrt{2}}{2... | Let $x$ be a positive quantity, so we have;
$$f(x)=\left(\frac{3+\sqrt{8x^{2}+1}}{8x}\right)\sqrt{\frac{4x^{2}-1+\sqrt{8x^{2}+1}}{2}}$$
Another thing to consider is that for $T_{n-1}=n(n-1)/2$ or the $n$-th triangular number, we have
$$\sqrt{8T_{n-1}+1}=2n-1$$
Now keeping this in mind, If we let $x^2=T_{n-1}=n(n-1)/2$,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc}$ for non-negative $a$, $b$, $c$ I was trying this question with factorization and other similar methods,
Let $a, b, c \geq 0$. Prove that
$$\begin{array}{c}
\sqrt{a^4+a^2b^2+b^4}+\sqrt{b^4+b^2c^2+c^4}+\sqrt{c^4+c^2a^2+a^4} \\[4pt]
\geq a\sqrt{2a^2... | By the Minkowski's inequality, we have
\begin{align}
\mathrm{LHS} &= \sum_{\mathrm{cyc}}\sqrt{(a^2 + b^2/2)^2 + (b^2\sqrt{3}/2)^2}\\
&\ge \sqrt{\left(\sum_{\mathrm{cyc}} (a^2 + b^2/2)\right)^2 +
\left(\sum_{\mathrm{cyc}} b^2\sqrt{3}/2\right)^2 }\\
&= \sqrt{3(a^2+b^2+c^2)^2}.
\end{align}
Since $x\mapsto \sqrt{x}$ is co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Convergence of $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ The problem is:
prove $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ converges absolutely for all $z\neq 0$, but does not converge uniformly near $z=0$.
Proof:
for all $z\neq 0$
$$\left|2^n\sin\frac{1}{3^nz}-2^n\frac{1}{3^nz}\right|
= \left|
2^n\left(
-\frac{(\fra... | Further to my comment, we can use that $\sin(x)\to x$ for $x\to0$ and look at the ratio of terms:
$$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\lim_{n\to\infty}\frac{2^{n+1}\times z\times3^n}{2^n\times z\times 3^{n+1}}=\frac 23<1$$
As you mentioned the problem is as $z\to0$, $\sin\left(\frac 1{3^nz}\right)\not\to0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An uncertainty for $a^3+b^3+c^3-3abc$ I have a doubt regarding $a^3+b^3+c^3-3abc$.
For factoring, it is easy that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$ as $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
But on using AM-GM inequality, we see-
$$ {a^3+b^3+c^3\over 3}\geq abc \Rightarrow a^3+b^3+c^3\ge3abc $$
AM-GM inequ... | AM-GM only work for non-negative nunber. Where the equality cases hold when all of variables are equal.
On the other hand, if $a,b,c\geqslant 0 $ then $a+b+c \geqslant 0$ if and only if $a=b=c=0$ (if one of them are greater $0$ then $a+b+c>0$).
So the equality also follow by AM-GM.
I hope my answer is useful for you. ... | {
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Evaluate $\int\frac{\mathrm{d}x}{{(x^4+2x+10)}^4}$ I am having trouble with this integral $\displaystyle \int \frac{\mathrm{d}x}{{(x^4+2x+10)}^4}$
I know this can be solved using the "ostrogradsky" method. But that is too lengthy.
i tried using integration by parts by multiplying and dividing by $4x^3+2$ but nothing p... | In the same spirit as @lab bhattacharjee, because of the fourth power in denominatour, assume that $$\int\frac{dx}{{(x^4+2x+10)}^4}=\frac{P_n(x)}{{(x^4+2x+10)}^3}$$ Differentiate both sides and remove the denominator to get
$$0=-1+\left(x^4+2 x+10\right) P_n'(x)-6 \left(2 x^3+1\right) P_n(x)\tag 1$$ For me, this differ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why If $x^2=-\frac{1}{3}$ then $x=\pm\frac{i\sqrt{3}}{3}$ If $x^2=-\frac{1}{3}$ then $x=\pm\frac{i\sqrt{3}}{3}$ according to my text book.
I understand that $x=\pm\sqrt{-\frac{1}{3}}$ but do not fully grasp how to express this as $\pm\frac{i\sqrt{3}}{3}$
I understand that $i$ is $\sqrt{-1}$
As far as I can get with my ... | After your edit; compare your simplification with
\begin{align}
x&=\pm\sqrt{-\frac{1}{3}}\\
x&=\pm\frac{\sqrt{-1}}{\sqrt{3}}\\
x&=\pm\frac{i}{\sqrt{3}}\\
x&=\pm i \frac{\sqrt{3}}{{3}}
\end{align}
The last step is achieved since
\begin{align}
\sqrt{3} \sqrt{3} = 3
\end{align}
Thus
$$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does $x^2-8x+17=0$ have nonreal solutions?
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.
Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please... | Also note:
If the solutions are $a, b$ then $(x-a)(x-b) = 0$ and $x^2 -(a+b)x + ab = 0$ and as $x^2 - 8x +17 =0$ we would have $a+b = 8$ and $ab = 17$.
Option 1: $4 + i$ and $4- i$. Then $(4+i) + (4-i) = (4+4) + (i-i) = 8$. So for so good. $(4-i)(4+i) = 16 + 4i - 4i -i^2 = 16 -i^2 = 16 -(-1) =16+1 = 17$. That work... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791855",
"timestamp": "2023-03-29T00:00:00",
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Where is the error in this "proof" that 3=0? I saw this video (link at bottom), with a supposed "proof" that $3=0$. It goes as follows:
Let $x$ be a solution of $$x^2+x+1=0 \tag1$$
Since $x\neq0$, we can divide both sides by $x$:
$$\frac{x^2+x+1}{x}=\frac0x\implies x+1+\frac1x=0 \tag2$$
From $(1)$,
$$x^2+x+1=0\impli... | The problem is $x^3=1$ does not imply that $x=1$. The equation $x^3-1=0$ has three possible roots and the root $x=1$ is an additionally generated root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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Is there an efficient way of showing $\int_{-1}^{1} \ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx = 2$? Hi I recently came across the following integral:
$$
\int_{-1}^{1} \ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx
$$
When an integral calculator finds the antiderivative of the equation (if it even can) it com... | Integrate by parts
\begin{align}
\int_{-1}^{1} \ln \frac{2(1+\sqrt{1-x^2})}{1+x^2}dx=&
- \int_{-1}^1 x \>d\left( \ln \frac{2(1+\sqrt{1-x^2})}{1+x^2} \right)dx\\
=& \int_{-1}^1 \left(\frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}}
+\frac{2x^2}{1+x^2} \right)dx\\
=&\int_{-1}^1 \left( 1+ \frac1{\sqrt{1-x^2}}-\frac2{1+x^2} \right)dx\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Use factor theorem to find all zeros of polynomial $2x^3+3x^2+x+6$ with one known factor $x+2$ I am to find the factors of $2x^3+3x^2+x+6$ where I am told that$x+2$ one of the factors.
Using synthetic division to divide $2x^3+3x^2+x+6$ by $x+2$ I confirm there is no remainder so it is a zero and the new quotient is $2... | $2x^2-x+3=2(x^2-\frac{1}{2}x+\frac{3}{2})$ where the discriminant is $\Delta=(\frac{-1}{2})^2-4\times 1\times \frac{3}{2}=\frac{1}{4}-6<0$. Therefore $x^2-\frac{1}{2}x+\frac{3}{2}$ has no real roots and so it is an irreducible polynomial over $\mathbb{R}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Is $f(x,y)=\frac{xy^3}{x^2+y^6}$ differentiable at $(0,0)$? Is the following function differentiable at $(0,0)$?
$$ \
f(x,y) =
\begin{cases}
\frac{xy^3}{x^2+y^6} & \text{if } (x,y) \ne (0,0), \\
0 & \text{if } (x,y) = (0,0).
\end{cases}
$$
I found that both of the partial derivatives are $0$, and then ... | Recall that continuity is a necessary condition for differentiability since differentiability implies continuity and by $y^3=v \to 0$ using polar coordinates we have
$$\frac{xy^3}{x^2+y^6}=\frac{xv}{x^2+v^2}=\cos\theta\sin \theta$$
| {
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"source": "stackexchange",
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How's my proof, are there any mistakes I have made? Compute $\int \sin(x) \left( \frac{1}{\cos(x) + \sin(x)} + \frac{1}{\cos(x) - \sin(x)} \right)\,dx. $
First, simplify the two fractions into one using a common denominator
\begin{align*}
\int\sin(x)& \left(\frac1{\cos(x)+\sin(x)}+\frac1{\cos(x)-\sin(x)}\right)dx\\
... | Your solution is correct.
Alternatively you could've used the substitution $u=\cos(2x)$.
| {
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Evaluating $\int \ln(2x+3) \mathrm{d}x$ Evaluate $$\int \ln(2x+3)\mathrm{d}x$$
Set $r = 2x+3 \rightarrow \mathrm{d}r = 2\mathrm{d}x$
So integral becomes $\displaystyle \frac{1}{2}\int \ln(r)\mathrm{d}r$
Set $u=\ln(r)$, $\mathrm{d}v=\mathrm{d}r$, so $\mathrm{d}u = \frac{1}{r}\mathrm{d}r$ and $v=r$
$\implies \displaystyl... | Your solution is correct as a constant plus another constant can be represented by a different constant, so $-\frac{3}{2}+C=C_1$.
As an alternative, you could integrate by parts and let $u=\ln(2x+3)$ and $dv=dx$. Then $du=\frac{2}{2x+3}$ and we can take $v=x+\frac{3}{2}$. It follows that
\begin{align}\int \ln(2x+3)\,dx... | {
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"timestamp": "2023-03-29T00:00:00",
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How would you calculate a derivative of $ f(x)= \frac{\sqrt{x+1}}{2-x}$ by the limit definition? I have a function defined as follows:
$$
f(x)= \frac{\sqrt{x+1}}{2-x}
$$
I tried to calculate the derivative using the limit definition using four methods, but I was unsuccessful in any. Could someone help me calculate it a... | Taking the first definition :
$$\lim_{h\to 0} \frac{\frac{\sqrt{(x+h)+1}}{2-(x+h)}-\frac{\sqrt{x+1}}{2-x}}h = \lim_{h\to 0} \frac{\sqrt{(x+h)+1}(2-x)-\sqrt{x+1}(2-(x+h))}{(2-(x+h))(2-x)h}$$$$=
\lim_{h\to 0} \frac{(x+h+1)(2-x)^2-(x+1)(2-x-h)^2}{(2-x-h)(2-x)h( \sqrt{(x+h)+1}(2-x)+\sqrt{x+1}(2-(x+h)) )} $$
$$= \lim_{h\to... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
} |
Solving a recurrence relation when $n$ is a power of $7$ I'm working through some recurrence relation examples and am struggling with this question that assumes inputs of $n$ are powers of $7$.
Essentially we have
$T(n) = T({\lfloor}n/7{\rfloor}) + \log_{3}(n)$
$T(1) = 0$
Using the bottom up method I have found the fol... | It seems more helpful to consider the function
$f(0) = 1$
$f(n + 1) = f(n) + \log_3(7^{n + 1}) = f(n) + (n + 1) \log_3(7)$
Here, we have $f(n) = T(7^n)$.
Then it is immediate that $f(n) = 1 + \sum\limits_{i = 1}^n n \log_3(7) = 1 + \log_3(7) \sum\limits_{i= 1}^n n$. Using the identity $\sum\limits_{i = 1}^n n = \frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Polynomial Question sum of powers Let $f(x)$ be a quadratic polynomial such that $f(-4) = -22,$ $f(-1)=2$, and $f(2)=-1.$ Let $g(x) = f(x)^{16}.$ Find the sum of the coefficients of the terms in $g(x)$ with even exponents. (For example, the sum of the coefficients of the terms in $-7x^3 + 4x^2 + 10x - 5$ with even expo... | First of all, we need to determine what quadratic $f(x)$ is. We can write some equations in the form of a general quadratic($ax^2+bx+c$). We plug in the values stated in the question into our expression to get the following system of linear equations:
*
*$16a-4b+c=-22.$
*$a-b+c=2.$
*$4a+2b+c=-1.$
Now, we can solve ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving $3(1−a+a^2)(1−b+b^2)(1−c+c^2)≥1+abc+a^2b^2c^2$ My task was to prove the question above over real variables.
I thought that this minor inequality should help-
$$ 3(1 − a + a^2)(1 − b + b^2) ≥ 2(1 − ab + a^2 b^2). $$
which is true.
By this inequality, the original inequality is converted to-
$$ (1 - ab)^2 (1 - c)... | Two SOS solutions with the help of computer
*
*According to Vasc's solution in @Michael Rozenberg's answer, we have a simple SOS expression:
\begin{align}
&3(a^2-a+1)(b^2-b+1)(c^2-c+1) - (1 + abc + a^2b^2c^2)\\
=\ & \frac{1}{8}(abc-3c+2)^2 + \frac{3}{8}(abc-2ab+c)^2 + \frac{3}{8}(a-1)^2(b-1)^2(2c-1)^2\\
&\quad + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find positive integers $a$ which there exists an integer $b, 0 \leq b \leq 2002$, such both quadratics $x^2+ax+b$ and $x^2+ax+b+1$ have integer roots. Compute the number of positive integers $a$ for which there exists an integer $b, 0 \leq b \leq 2002$, such that both of the quadratics $x^2+ax+b$ and $x^2+ax+b+1$ have ... | For them to be integers both $a^2-4b$ and $a^2-4(b+1)$ have to be perfect squares. Let them be $k^2$ and $l^2$ respectively. Then $k^2-4=l^2 \implies k^2-l^2=4 \implies (k-l)(k+l)=4$
Since $a$ and $b$ both are integers, then $k$ and $l$ are also trivially integers. These two integers multiply to give $4$. hence they mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving $(a^2 + 1)(b ^2 + 1)(c ^2 + 1) ≥ 2(ab + bc + ca)$ where $a,b,c$ are real numbers. The inequality above seems very compelling for the pqr-method.
So this was my attempt-
$$ LHS = (a^2 + 1)(b^2 + 1)(c^2 + 1) = 1 + a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 + a^2b^2c^2 $$
Now substituting $p = a+b+c$ , $q = ab+bc+... | If you write LHS - RHS as a quadratic in $a$, then the discriminant is:
$$D = 4(b+c)^2 - 4((b^2+1)(c^2+1)-2bc)(b^2+1)(c^2+1).$$
But $(b^2+1)(1+c^2)\geq (b+c)^2$ by C-S and $(b^2+1)(c^2+1)-2bc = b^2c^2+(b-c)^2+1\geq 1,$
so the discriminant is non-positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
When its possible to write a number $a+b\sqrt{c}$ as the square of a sum? I was teaching my high school students how to find the roots of 2nd degree equations like the following:
$\frac{x^2}{2}+\sqrt{3}x-\sqrt{2}=0$
In this case, using the formula we get:
$x=-\sqrt{3}\pm\sqrt{3-2\sqrt{2}}$
To simplify I writted $3-2\sq... | Suppose $a+b\sqrt{c} = (m+n\sqrt{c})^2$, where $a$, $b$, $c$, $m$, and $n$ are integers and further that $c > 0$ (otherwise we are not talking about the real-valued square root or we are talking about $a+0 = a$). Then \begin{align*}
a+b\sqrt{c} &= (m+n\sqrt{c})^2 \\
&= m^2 + cn^2 + 2 m n \sqrt{c} \\
\end{align*}
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find last two digits of $302^{46}$ I need to find the last two digits of $302^{46}$ without resorting to Euler's theorem or Chinese remainder theorem (they have not been introduce so far in the course; I can user Fermat's little theorem though). This is what I tried:
We have to work $\pmod{100}$ and it is easy to see t... | $2^{46} \equiv 4^{23} \equiv 4 \cdot 4^{22} \equiv 4 \cdot 16^{11} \equiv 4 \cdot 16 \cdot 16^{10} \equiv 4 \cdot 16 \cdot 56^{5} \equiv$
$\quad 4 \cdot 16 \cdot 56 \cdot 56^{4} \equiv 4 \cdot 16 \cdot 56 \cdot 36^{2} \equiv$
$\quad 4 \cdot 16 \cdot 56 \cdot 96 \equiv^\text{algorithm complete / now applying disc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
What is the fastest way of calculating whether vectors are linearly independent?
Is the following family $(u_1, u_2, u_3)$ linearly independent?
$$ u_1 := \begin{pmatrix} 1 \\ 3 \\ 5 \\ -1 \end{pmatrix} ,
u_2 := \begin{pmatrix} 1 \\ -1 \\ -3 \\ 3 \end{pmatrix},
u_3 := \begin{pmatrix} 3 \\ 2 \\ 1 \\ 4 \end{pmatrix}$$... | With Gaussian elimination, it is not very long to prove the rank is $2$:
\begin{align}
\begin{bmatrix}
1&1&3\\ 3&-1&2\\ 5&-3&1\\-1&3&4
\end{bmatrix} \rightsquigarrow
\begin{bmatrix}
1&3&1\\-1&2&3\\-3&1&5\\3&4&-1
\end{bmatrix} \rightsquigarrow
\begin{bmatrix}
1&3&1\\ 0& 5 & 4\\-3&1&5\\ 0 & 5& 4
\end{bmatrix} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sq... | For $y=x-2$, you have $$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}=\frac{\sqrt{(y+2)^2-4}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}} = \frac{\sqrt{y^2+4y}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}}$$
When $x$ tends to $2$, $y$ tends to $0$, so you have
$$\frac{\sqrt{y^2+4y}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}} = \frac{2\sqrt{y}\sqrt{1 + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Prove that $\sqrt{8}$ is irrational in different method I tried to prove that $\sqrt{8}$ is irrational. I said let $\sqrt{8}$ be rational then $\sqrt{8}$ = $a/b$ where $a$ and $b$ are relatively prime.
Then $2\sqrt{2}=a/b$ , and $\sqrt{2} =a/(2b)$. it is obvious that $RHS$ is rational and $LHS$ is irrational (assumed t... | let $\sqrt{8}$ is equal to $\frac {a}{b}$ where $(a,b)=1$
Then $8 = \frac {(a^2)}{(b^2)}$ , so $8(b^2)= a^2$
Because of the fact that $(a,b)$ are relatively prime $b$ cannot divide $a$.Hence $8|a^2$.
Then ,let's say $a=2k$ where $k$ is a positive integer.
Then $8(b^2)=(2k)^2$. Then,by simplification $2(b^2)=k^2$ .It is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Limit of $(x+1+\sqrt{(x+1)^2 +1})$ as $x\to-\infty$ This seemed easy at first, but then I confronted my result with the one I got from WolframAlpha and they are different:
$$
\lim_{x\to-\infty}(x+1+\sqrt{(x+1)^2 +1})
$$
WolframAlpha says the limit is equal to 0, though for me it seems to be negative infinity. I tried s... | Your mistake was in writing $\lim_{x\to -\infty}\frac{(x+1)^2+1}{\sqrt{(x+1)^2+1}}=0$. The expression $\frac{(x+1)^2+1}{\sqrt{(x+1)^2+1}}$ equals $\sqrt{(x+1)^2+1}$, which tends to $\infty$ as $x\rightarrow-\infty$.
You can attack this problem using other techniques. One approach is to multiply and divide by the conjug... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
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.