Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Urn with balls - Finding the joint distribution
Consider an urn containing $4$ red and $6$ white balls. We draw two times with replacement. Let $X$ be the number of red balls we draw, and $Y$ the number of white balls we draw. Find the joint probability distribution of $X$ and $Y$.
Since we draw exactly two times:
$$... | Hint: $~~0.36+{0.24}+0.16=1-{0.24}$
It's because $\mathsf P(X{=}1,Y{=}1)= 2\cdot\tfrac{4}{10}\cdot\tfrac{6}{10}=0.48$ by reason that we are selecting 1 red and 1 white ball in any order of extraction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\cos x-\sin 3x=\cos 2x$? My attempt:
$$\begin{align} (\cos x- \cos 2x) - \sin 3x &= 0 \\
2\sin \frac{3x}{2}\sin \frac{x}{2} - 2 \sin \frac{3x}{2}\cos \frac{3x}{2}&=0\\
\sin\frac{3x} {2} \left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)&=0 \end{align}$$
Now either $\sin\frac{3x} {2} = 0$ or $\left(\sin\fra... | We have $\sin\frac12x=\cos\frac32x=\sin(\frac12\pi-\frac32x)$. The general solution to this is$$\tfrac12x=n\pi+(-1)^n(\tfrac12\pi-\tfrac32x)\quad (n\in\Bbb Z).$$The cases of even or odd $n$ then give rise to two sets of solutions:$$x=(k+\tfrac14)\pi\quad\text{or}\quad x=(2k-\tfrac12)\pi\quad(k\in\Bbb Z).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the Limit.? What is the limit of
$Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2} $
for $n > 2$ and $Y_{1}<Y_{2}$.
I have tried solving, but the limit tends to exist between $y_{1}$ and $y_{2}$ and hence I am not able to evaluate what would be the ratio between which they would lie?
| $Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2}$ is homogenous linear diference equation which has characteristic equation $x^2=\frac{1}{3}x+\frac{2}{3}$ or $3x^2-x-2=0$, $(3x+2)(x-1)=0$. Solution is $$Y_{n}=C_1(-\frac{2}{3})^n+C_2(1)^n=C_1(-\frac{2}{3})^n+C_2.$$
Constants $C_1$, $C_2$ can be found from $Y_1, Y_2$:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question about convergence of series using comparison test I am wondering how to determine whether the following series converges using the comparison test:
$$\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$$
In class, my professor used the idea that $\frac{\sin x}{x}\to 1$ as $x$ gets large.
So, my prof... | What we are using is limit comparison test and since
$$\frac{\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}}{\frac{1}{n}}=\frac{\sin{\frac{2}{\sqrt{n}}}}{\frac{2}{\sqrt{n}}}\frac{2}{3+2\frac{\sqrt n}n} \to \frac23$$
as you noticed the series diverges.
Note that limit comparison test also works for the two limiting cases
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882523",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Volume with spherical polar coordinates Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.
I convert to spherical polar coordinates.
$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$
$$\iiint_{K} (\sqrt {4-r^2\s... | Using spherical coordinates, you would have to split up $K$ into two regions,
$$K_1=\left\{(r,\theta,\phi)\mid 0\le r\le2,0\le\theta\le\frac{3\pi}4,0\le\phi\le\frac\pi6\right\}$$
$$K_2=\left\{(r,\theta,\phi)\mid0\le r\le\sqrt{\csc\phi},0\le\theta\le\frac{3\pi}4,\frac\pi6\le\phi\le\frac\pi2\right\}$$
(where $x=r\cos\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Proving Vector Identities Let r=$(x,y,z)$ and $r=$||r||.
(A) Prove that $\nabla^2r^3=12r$.
(B) Is there a value of $p$ for which the vector field f(r) = r/$r^p$ is solenoidal?
What I have tried:
For part (a) I think that $r^3=\sqrt{27} $ but I am unsure of what to do next in terms of the del operator.
| $$
\nabla^2r^3=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})(x^2+y^2+z^2)^{3/2}\\
= \frac{\partial }{\partial x }(3x(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial y }(3y(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial z }(3z(x^2+y^2+z^2)^{1/2}) \\
=\frac{3(2x^2+y^2+z^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884835",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How many 4 digit number can be formed by using 1,2,3,4,5,6,7 if at least one digit is repeated My approach for this was selecting 4 numbers from which one is being repeated
$$\boxed{A}\boxed{B}\boxed{C}\boxed{A} $$
$$\boxed{A}\boxed{B}\boxed{A}\boxed{A} $$
$$\boxed{A}\boxed{A}\boxed{A}\boxed{A} $$
and permitting them... | You’ve miscounted all three of your types with repetitions. This is most obvious in the case of the $AAAA$ type: you counted $\binom74\cdot\frac{4!}{4!}=35$ of them, but there are clearly only $7$: $1111$, $2222$, $3333$, $4444$, $5555$, $6666$, and $7777$. For your $ABAA$ type you give a figure of $\binom74\cdot\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $1998x + 5y + 3z$ given $\frac{1}{x} + \frac{1}{y} = \frac{3}{z}$ Suppose $x,y,z$ are positive integers so that $y$ and $z$ are prime and $x = yz.$ If they satisfy $$\frac{1}{x} + \frac{1}{y} = \frac{3}{z},$$ find $1998x + 5y + 3z.$
I first moved $\frac{1}{y}$ over to the other side and expanded out, but from ... | You are on the right track. After that you should have
$$\frac{1}x=\frac{3}z-\frac{1}y\Longrightarrow \frac{1}{yz}=\frac{3y-z}{yz}\Longrightarrow 1=3y-z\Longrightarrow 3y=z+1$$
Now note that if $y$ is odd, then $z$ has to be even. And since $z$ is a prime, $z$ has to be $2$, but in this case we will have $y=1$ which is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$
My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b... | I have an alternative proof.
We need to prove that
$$
\frac{a}{b} +
\frac{b}{c} +
\frac{c}{a} \geqslant
\frac{a + 2}{b + 2} +
\frac{b + 2}{c + 2} +
\frac{c + 2}{a + 2}
$$
Here we can write this inequality in $2$ forms:
$$
\frac{c}{a} -
\frac{c + 2}{a + 2} \geqslant
\frac{a + 2}{b + 2} -
\frac{a}{b} +
\frac{b + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Inequality with integers I am trying to find an analytical way to prove that for integers $n\ge 6$,
$$\sqrt[n]{n+1}\le \frac{1}{n}+\frac{1}{\sqrt[n]{n+1}}+\frac12$$
| Denote $x = \sqrt[n]{n+1}$. The desired inequality is written as
$\frac{-2nx^2 + (n+2)x + 2n}{2nx} \ge 0$.
It suffices to prove that $-2nx^2 + (n+2)x + 2n \ge 0$ which is written as
$$2n(x - x_1)(x - x_2)\le 0$$
where
$$x_1 = \frac{n+2 + \sqrt{17n^2 + 4n + 4}}{4n},
\quad x_2 = \frac{n+2 - \sqrt{17n^2 + 4n + 4}}{4n}.$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Showing that a matrix raised to a power is equal to itself Let $F$ be a field of cardinality $q=2^k$ for some $k \in \mathbb{N}$ and let
$$E= \begin{bmatrix}
1 & 0 & & & & & &\\
1 & 0 & & & & & &\\
& & \ddots & & & & & &\\
& & & 1 & 0 & & &\\
& & & 1 & 0 & & \\
& & & & & & 1 & 0 & a_{n-1}a_{n-2}\\
& & & & & & 1& 0 & a_... | Write $x=a_{n-1}$ and $y=a_{n-2}$ for notational ease. Let $M$ represent the matrix formed by the first $n-1$ columns of $E$ with zeroes in column $n$, and let $N=E-M$ be the $n$th column of $E$ with zeroes in the first $n-1$ columns. It is not hard to check that
$$M^2=M,\ N^2=(x+1)N,\ NM=0.$$
In particular, when we ex... | {
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Functions which satisfy $f(m^2+n^2)=f(m)^2+f(n)^2$ Find all $f$: $f(m^2+n^2)$=$f(m)^2+f(n)^2$ for all $m,n\in \mathbb{N_0}$ where
$f(1)>0$,
$f\in\mathbb{N_0}$ for all inputs
I did $f(0)=2f(0)^2\implies f(0)=0$
$f(1)=f(1)^2\implies f(1)=1$
From this we can get $f(2)=2, f(4)=4$ etc. easily and also use some tricks like... | We see that $f(2n^2) = 2f^2(n)$. So $f(50)=2f^2(5)=2(25)=50$
So $f(50)=50$. But $50 = 1+49$. So $f(50) = 1 + f^2(7)$.
So $f(7)=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3900226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fitting circle in $f(x)=x^3-3x^2$ what is the radius?
The whole problem is the above.
I tried to solve it analytically but I couldnot find enough equation to solve it.
Assume the center is $(a,-r)$ where $r$ is the radius of the circle.
The points on the sphere $(x-a)^2+(y+r)^2=r^2$
and $(x,f(x))$ must satisfy it so i... |
In figure suppose AC is straight line.We first find the angle BAC the curve makes with x axis:
$y=x^3-3x^2$ ⇒ $y'=3x^2-6x$
⇒ $tan (\widehat{BAC})=y'$ at point A(3, 0):
$tan (\widehat{BAC})=3^3-6\times 3=9$
⇒ $ \widehat{BAC} ≈83^o$
⇒$ \widehat{BAC}=\widehat BAD ≈41.5^o$
So the center of circle is on line $y=(m=tan 41.5... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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A bag contains 7 blue balls and 3 green balls. What is the probability of the following events? A bag contains 7 blue balls and 3 green balls.
I am not sure about my answer but please allow me to share.
*
*If 5 balls are drawn in succession at random without replacement, what is the probability of dra... |
If five balls are drawn in succession at random without replacement, what is the probability of drawing one green, one blue, one green, one blue in order?
You have correctly calculated that the probability that the sequence GBGB occurs in the first four draws is
$$\Pr(GBGB) = \frac{3}{10} \cdot \frac{7}{9} \cdot \fra... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the obvious mistake of this double integration? There is a definite double integral in Cartesian coordinates as follows:
$$\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 2\sqrt{4-x^2}\ dydx\Rightarrow \int_{-2}^{2} 2y\sqrt{4-x^2}\rvert_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \ dx \Rightarrow 4\int_{-2}^{2} (4 - x^2)dx ... | The problem is that $(1-\cos^2)^\frac{3}{2}=|\sin^3\theta|$ and not $\sin^3\theta$. Then you have to consider intervals for $\theta$ to discard absolute value.
The second problem is at the second to the last line, the change of variable $u=\cos\theta \Rightarrow d\theta = -\frac{du}{\sin\theta}$ and you forgot a negati... | {
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"timestamp": "2023-03-29T00:00:00",
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A confusion in the $\epsilon-\delta$ proof of $\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0$ I want to prove
$$
\lim_{x\to 2}\frac{x - 2}{1 + x ^ 2} = 0
$$
The proof is: Let $\epsilon > 0$ be given. Then
$$\left|\frac{x - 2}{1 + x ^ 2} - 0\right| = \frac{|x - 2|}{1 + x ^ 2} \leqslant |x - 2| < \epsilon$$
provided $|x - 2| ... | You have made an incorrect statement which I think needs to be corrected:
it is not true that you "have"
$$ \left\lvert \frac{x - 2}{1 + x ^ 2} - 0\right\rvert < \epsilon \tag1 $$
at the point in the proof you are questioning.
Instead, this inequality is the thing you have to prove at that point.
The real point of dif... | {
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"timestamp": "2023-03-29T00:00:00",
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Help solving the limit of the sequence: $\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$ Given the following sequence,
$$\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$$
I am asked to determine to which $l \in \mathbb{R}$ the sequence converges... | Use$$\ln l=\lim_{n\to\infty}2n^2\underbrace{\frac{1-1/(2n^2)}{1+3/n}}_{\sim1}\underbrace{\ln\underbrace{\frac{1-1/(2n^2)}{1+1/(2n^2)}}_{1-1/n^2+o(1/n^2)}}_{\sim-1/n^2}=-2\implies l=e^{-2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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What is the minimal area under the curve of parabola $y=ax-bx^2$ between $x = 0$ and $x=\frac{a}{b}$ such that it passes through (1,1)?
Let $a > b >0$. The area under a parabola $y=ax−bx^2$ between its two roots, $x=0$ and $x=\frac{a}{b}$, is given by $\frac{a^3}{6b^2}$. Determine the $a$ and $b$ such that the parabol... | Hint. Since $b=a-1>0$, you have to minimize the function
$$f(a)=\frac{a^3}{6b^2}=\frac{a^3}{6(a-1)^2}$$
over $(1,+\infty)$.
You may also let $a=b+1$ and minimize the function
$$g(b)=\frac{a^3}{6b^2}=\frac{(b+1)^3}{6b^2}$$
over $(0,+\infty)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the condition for system of linear equations to be solvable It is clear that if the system of linear equations
$$
\left\{
\begin{array}{c}
x_1-x_3=c_1 \\
x_2-x_1=c_2 \\
x_3-x_2=c_3
\end{array}
\right.
$$
is solvable, then we have $c_1+c_2+c_3=0$.
How could we prove for the backward direction? That is, how cou... | The augmented form of our system is
$$
\left[\begin{array}{rrr|r}
1 & 0 & -1 & c_{1} \\
-1 & 1 & 0 & c_{2} \\
0 & -1 & 1 & c_{3}
\end{array}\right]
$$
Row-reducing gives
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & c_{1} \\
-1 & 1 & 0 & c_{2} \\
0 & -1 & 1 & c_{3}
\end{array}\right]
\xrightarrow{R_2+R_1\to R_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$ Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
I couldn't do this question and hence I looked at the solution which goes as follows:
if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$
I state that $y=lx$. Hence the system becomes:... | $\begin{align}
x^2+2xy & = 5 \\
y^2-3xy & = -2
\end{align}$
$\begin{align}
y = & \frac{5-x^2}{2x} \\
x = & \frac{2+y^2}{3y}
\end{align}$
$\begin{align}
y = & \frac{5-\left( \frac{2+y^2}{3y} \right)^2}{2 \left(\frac{2+y^2}{3y}\right)} \\
\end{align}$
$\begin{align}
y = & \frac{5 \cdot(9y^2)-(2+y^2)^2}{2 \cdot 3y (2+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$n$-th derivative of $f(x)=x\cos(x)$ Can someone help me prove that the $n$-th derivative of $f(x)=x\cos(x)$ is:
$$f^n(x)=x\cos\left(x+\frac{n\pi}{2}\right)+n\cos\left(x+(n-1)\frac{\pi}{2}\right)?$$
Thanks, and stay safe.
| Just using induction using the product rule:
$f(x) = x\cos x$ so
$f'(x) = \cos x + x(-\sin x) = \cos x - x\sin x$
$f^2(x) = -\sin x - \sin x - x\cos x= -2\sin x - x\cos x$
$f^3(x) = -2\cos x-\cos x - x(-\sin x) = -3\sin x +x\sin x$.
By pattern recognition we sort of see that $f^n = \pm n \operatorname{trig} x \pm x\ope... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for all $n \in \mathbb{N}$, $\sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$ $$ \sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$$
To be honest, I can't really get started,
I would like to ask everyone to give me an idea of how to solve it, give me a starting push, thank you.
| we rewrite the sum like this ;$$\sum_{k=0}^{k=n-1}\binom{n+k-1}{n-1}\frac{1}{2^{n+k}}$$
this is just the coefficient of $x^{n-1}$ in expansion $$\frac{{(1+x)}^{n-1}}{2^n}+\frac{{(1+x)}^n}{2^{n+1}}...+\frac{{(1+x)}^{2n-2}}{2^{2n-1}}$$
Recognise this as a GP: hence we seek the coefficient of $x^{n-1}$ in $$\frac{1}{2^{n-... | {
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There are $5$ different apples, $4$ different oranges, and $3$ different bananas. Number of ways to select at least one of each kind?
There are $5$ different apples, $4$ different oranges, and $3$ different bananas. Number of ways to select at least one of each kind?
My first approach: I selected one of each kind fi... | There are a total of $5 + 4 + 3 = 12$ fruits. You can either select or not select each fruit. Thus, if we did not have the restriction that at least one fruit of each type must be selected, there would be $2^{12} = 4096$ possible ways to select the fruits (including the choice of not selecting any of them). Your fir... | {
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"timestamp": "2023-03-29T00:00:00",
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Determine the multiplicity of a root in determinant Consider the determinant
$$\det(A(x,z))=\begin{vmatrix}1-x&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{vmatrix}.$$
It is easy to see that if we write the determinant as a bivariate polynomial $p(x,z)$, then $x=0$ and $z=0$ are its roots, i.e. $p(0,z)=p(x,0)=0$. But how... | Adding or subtracting one row to another does not change the value of the determinant. So $$\det\begin{bmatrix}1-x&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{bmatrix}=\det\begin{bmatrix}1&1&1&1+z\\0&x&0&-z\\0&0&z&z\\0&0&x&x+xz\end{bmatrix}=x^2z^2$$
NB It is $x=0$ and $z=0$ that are the roots of the polynomial.
EDIT: L... | {
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Finding $abc$ for distinct reals satisfying $\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$ The problem is
Let $a$, $b$, $c$ be three distinct real numbers satisfying
$$\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$$ Find $abc$.
I can solve this in the case of $a, b, c \ge 0$ by assuming $a =\max\{a, b, c\}$. In that c... | As @fleablood observed, none of $a,b,c$ can be zero. Hence, notice that the condition is equivalent to $$\frac{1+ab}{a}=\frac{1+bc}{b}=\frac{1+ca}{c}\iff \frac1a+b=\frac1b+c=\frac1c+a$$ Yet this implies that $$\frac1a+b=\frac1b+c\iff b-c=\frac{a-b}{ab}$$ Similarly $$\frac1b+c=\frac1c+a\iff c-a=\frac{b-c}{bc}\quad\text{... | {
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$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $ Did I do it right? How to do the last step? I need to find:
$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $$
So I calculate:
$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| = \li... | $$\begin{split}
\sum_{k=1}^{2n} (-1)^k \sqrt{k} &= \sum_{k=1}^{n} \left((-1)^{2k-1} \sqrt{2k-1} + (-1)^{2k} \sqrt{2k}\right)\\
&= \sum_{k=1}^{n} \left( \sqrt{2k}-\sqrt{2k-1}\right)\\
&= \sum_{k=1}^{n} \frac{1}{\sqrt{2k}+\sqrt{2k-1}}
\end{split}$$
Thus $$\sum_{k=1}^n\frac{1}{2\sqrt{2k}}\leq \sum_{k=1}^{2n} (-1)^k \sqrt{... | {
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Prove using mathematical induction Sequence $b_1, b_2,...$ where $b_1=b_2=1$ and for every natural number $k$ bigger than $2$, $b_k = b_{k-1}+b_{k-2}$.
Prove using mathematical induction that, for every $k \in \mathbb N, b_k \le (\frac{7}{4})^{k-1} $
I started with if $k=3$, $k_3=1+1=2$ and $(\frac{7}{4})^{k-1} =(\... |
I don't know what to do next.
$ \left(\dfrac{7}{4}\right)^{k+1}+\left(\dfrac{7}{4}\right)^{k}=\left(\dfrac74\right)^k\left(\dfrac74+1\right)=\left(\dfrac74\right)^k\left(\dfrac{11}4\right)=\left(\dfrac74\right)^k\left(\dfrac{44}{16}\right)$
$<\left(\dfrac74\right)^k\left(\dfrac{49}{16}\right)=\left(\dfrac74\right)^k\... | {
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Range of quadratic function using discriminant
Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.
Let $2x^2-2xy+y^2=a$.
Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.
We divide both side by $y^2$ and let $t=\frac{x}{y}$.
Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.
Since its ... | Method$\#1:$
For real $t,$ $$a^2-7a-4\ge0$$
If $x=ty$
$$4=x^2-2xy-3y^2=y^2(t^2-2t-3)$$
$$t^2-2t-3=\dfrac4{y^2}>0$$
$$\implies(t-3)(t+1)>0$$
Either $t<-1$ or $t>3$
So, the values of $a$ must satisfy this condition as well
Method$\#2:$
$$4=(x-y)^2-(2y)^2$$
WLOG $y=\tan t, x-y=2\sec t\implies x=2\sec t+\tan t$
$$a=2(2\se... | {
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$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan... | $$L=\lim_{x \to 0}\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
You need to use Taylor series:
$$L=\lim_{x \to 0}\frac{\ln (1+x (x-\dfrac {x^3}3+...)-(1+x^2+\dfrac {x^4}{2}+o(x^4))+1}{x^4}$$
$$L=\lim_{x \to 0}\frac{ (x^2-\dfrac {x^4}3-\dfrac {x^4}2+o(x^4))-x^2-\dfrac {x^4}{2}+o(x^4)}{x^4}$$
$$L=\lim_{x \to 0}... | {
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$\lim_{x \to + \frac{1}{2}} [ \tan(\pi x^2)+(2x-1)4^x+6x-4 ] \cot(x-\frac{1}{2})$ I've tried to solve this limit:
$\lim_{x \to + \frac{1}{2}} \left [ \tan(\pi x^2)+(2x-1)4^x+6x-4 \right ] \cot(x-\frac{1}{2})$
the first parentesis should tend to $\frac{\sqrt{2}}{2}-1$ but I don't know how to procede.
The final result s... | Let $~f(x) = \left[\tan(\pi x^2)+(2x-1)4^x+6x-4 \right]~$ and
let $~g(x) = \tan[x - (1/2)].$
Note that $g(x)$ and each term of $f(x)$ are continuous functions.
You want $\lim_{x\to (1/2)} \frac{f(x)}{g(x)}.$
Before applying L'Hopital's rule, first check $f(1/2)$ and $g(1/2)$.
$$f(1/2) = \tan(\pi/4) + (0)4^{(1/2)} + 3 ... | {
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Determinant and inverse matrix calculation of a special matrix Is there any smart way of calculating the determinant of this kind matrix?
\begin{pmatrix}
1 & 2 & 3 & \cdots & n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmat... | If it’s the determinant you are looking for, then what if you subtract column $j+1$ from column $j$ to get the matrix
$$\begin{pmatrix}
-1 & -1 & -1& \cdots & n \\
1& -1 & -1& \cdots & n-1 \\
1& 1& -1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1& 1& 1& \cdots & 1 \end{pmatrix}_{n\times n}$$
Then ... | {
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Solve integral $\int^1_0 \frac{1-x^2}{{(1+x^2)}\sqrt{1+x^4}}dx$ using subsituition $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ Hi this question has been posed in my integration book where it has been asked to solve it using the given substitution or any other substitution, I've found identical question posted here {1} with ... | Let's begin with the substitution suggested in the question. We have $$\cos t=\frac{\sqrt{1+x^4}} {1+x^2}$$ ($\theta$ is replaced by $t$ to reduce burden of typing mathjax). Differentiating we get
\begin{align}
-\sin t \, dt&=\dfrac {(1+x^2)\cdot \dfrac{4x^3} {2\sqrt{1+x^4}} - 2x\sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\... | {
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"source": "stackexchange",
"question_score": "6",
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A question related to $f(n)=n^4+n^2+1$ If $$f(n)=n^4+n^2+1$$ then we have to evaluate $$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}$$ which, when run in Desmos, returns 6... | By using Sophie German Identity
$$f(n)=n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$$
$$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}=\frac{7\times 13\times 21... | {
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Prove by definition that $\frac{1-2n^3}{4n^3-2n^2-1} \to -\frac12$ $a_n = \dfrac{1-2n^3}{4n^3-2n^2-1}$
My attempt (of the scratch):
$\left| \dfrac{1-2n^3}{4n^3-2n^2-1} + \dfrac{1}{2}\right|=\left| \dfrac{2-4n^3+4n^3-2n^2-1}{8n^3-4n^2-2}\right|=\left|\dfrac{-2n^2+1}{8n^3-4n^2-2} \right|=\dfrac{2n^2-1}{8n^3-4n^2-2}<\dfra... | For large $n$, we have $4n^3 - 2n^2 - 1 > n^3$, in which case $\frac{n^2}{4n^3-2n^2-1} < \frac{n^2}{n^3} = \frac{1}{n}$.
| {
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Given rank-$1$ matrix $A$, how to compute $A^{100}$?
$$A = \begin{bmatrix} 6 & 4\\ -6 & -4\end{bmatrix}$$ Find $A^{100}$.
I tried to find it using diagonalization, but as it is a singular matrix so one of eigenvectors came out zero. How $A^{100}$ can be calculated of same matrix?
| $$|\lambda I - A| = \lambda^2 - 2\lambda$$
A matrix satisfies its characteristic equation:
$$A^2 -2A =0$$
$$A^2 = 2A$$
$$A^4 = 4A^2 = 8A$$
$$A^8 = 16A^4= 2^7 A$$
$$A^{16} = (A^4)^4 = 2^{12}A^4=2^{15}A$$
$$A^{32}= (A^16)^2 = 2^{30}A^2 = 2^{31}A$$
$$A^{64} = (A^16)^4 = 2^{60} A^4 = 2^{63}A$$
$$A^{100}= A^{64+32+4} = 2^{6... | {
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Convergence of the recursive sequence $x_{n+1} = \frac{1}{4-x_n}$ I'm doing exercise 2.4.1 in the book Understanding Analysis by Stephen Abbott. I'd like to ask, if my proof is rigorous and technically correct.
(a) Prove that the sequence defined by $x_1 = 3$ and
\begin{align*}
x_{n+1} = \frac{1}{4 - x_n}
\end{ali... | An alternative solution.
It is clear from the form of the recursion that there is a sequence $\{y_n\}_{n=1}^\infty$ such that $x_n=\frac{y_n}{y_{n+1}}$.
$$\frac{y_{n+1}}{y_{n+2}}=\frac{1}{4-\frac{y_n}{y_{n+1}}}=\frac{y_{n+1}}{4y_{n+1}-y_{n+2}}\implies y_{n+2}=4y_{n+1}-y_n$$.
The base case is $y_1=3$ and $y_2=1$.
$$\lam... | {
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How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$ I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1... | Making the problem more general
$$I_n=\int_0^t \frac {dx}{\sum_{k=0}^n x^k}=\int_0^t \frac {1-x}{1-x^{n+1}}\,dx$$
$$x=y^{\frac{1}{n+1}} \quad \implies \quad I_n=\frac 1{n+1}\int_0^{t^{n+1}} \frac{y^{-\frac{n}{n+1}}-y^{-\frac{n-1}{n+1}}} {1-y} \,dy$$ Using
$$\int \frac {y^a}{1-y}\,dy=\frac{y^{a+1} }{a+1}\,\, _2F_1(1,a+1... | {
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Evaluate $\lim\limits_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$ How to evaluate:
$$\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$$
Personally, i always have trouble to evaluate sum of trignometric series, if you know a paper to... | Let $P_n(X)=\frac{(X+i)^{2n+1}-(X-i)^{2n+1}}{2i}$, then
$$ \begin{aligned} P_n(z)=0&\iff \left(\frac{z+i}{z-i}\right)^{2n+1}=1 \\
&\iff \exists k\in[\![1,2n+1]\!],\frac{z+i}{z-i}=e^{\frac{2ik\pi}{2n+1}} \\
&\iff\exists k\in[\![1,2n+1]\!],z=i\frac{e^{\frac{2ik\pi}{2n+1}}+1}{e^{\frac{2ik\pi}{2n+1}}-1} \\
&\iff\exists k\i... | {
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Find the absolute maximum and minimum of $f(x) = \left|\cos^2(x) - \frac{3}{4}\right|$ on $[0,\pi]$
Find the absolute maximum and minimum of $f(x) = \left|\cos^2(x) - \frac{3}{4}\right|$ on $[0,\pi]$
I have learnt that in order to find the absolute maximum, I need to check the value of the function at the:
*
*side... | Well, I will give you some (big) hints:
*
*Notice that when $x\in\left[0,\frac{\pi}{6}\right]$ we have $\left|\cos^2\left(x\right)-\frac{3}{4}\right|=\cos^2\left(x\right)-\frac{3}{4}$;
*Notice that when $x\in\left[\frac{\pi}{6},\frac{5\pi}{6}\right]$ we have $\left|\cos^2\left(x\right)-\frac{3}{4}\right|=\frac{3}{4}... | {
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Trigonometric problem (problem from a Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two angles are $135^\circ$ and the sum of their tangents are $5$. Calculate the angles.
Is there a shorter/simpler solution than the one pres... | Take $\tan$ of both sides of first equation, $$\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$
$$\Rightarrow \tan 135 = -1 = \dfrac{5}{1-\tan \alpha \tan \beta} $$
$$\Rightarrow \tan \alpha \tan \beta=6$$
Setting $\tan \alpha = x$, $\tan \beta = y$ we get two equations : $$x+y=5 \... | {
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"source": "stackexchange",
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Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number? Is there a way to show that
$$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$
is a rational number?
I found $\alpha=3$ from doing simplifications. But, I would like to known a different approach.
| Note $\sqrt[3]{20-14 \sqrt{2}}=\sqrt2 \cdot \sqrt[3]{5 \sqrt{2}-7}$ and assume $\sqrt[3]{5\sqrt{2}\pm7}=p\sqrt2\pm q$. Then
$$\alpha
=(p\sqrt2+q)+\sqrt2(p\sqrt2-q)=(q+2p)+(p-q)\sqrt2
$$
For $a$ to be rational requires $p=q=\frac{\sqrt[3]{5 \sqrt{2}+7}}{\sqrt2+1}
= \sqrt[3]{\frac{ 5 \sqrt{2}+7}{(\sqrt2+1)^3 }}=1$ to be ... | {
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Inequality for a two variables function I have tried to solve this exercise but I am having difficulties.
I have to find a function greater than $$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}}$$ as the parameter $\alpha$ changes.
That's what I tried:
Since $$sin(xy)\leq1$$
$$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}} \l... | $x^2, y^2 \geq 0$ thus $x^2 + y^2 \geq 0$ and $x^2+y^2\geq y^2$ because $x^2 \geq 0$.
Then put both sides to the power $\frac{3}{2}$
$(x^2+y^2)^\frac{3}{2}\geq (y^2)^\frac{3}{2}=|y|^3$ instead of $y^2$
and we can do this because $y^2 \geq 0$ and $x^2 + y^2 \geq 0$ and $\frac{d}{dz}[z^\frac{3}{2}]=\frac{3}{2}z^\frac{1}{... | {
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've fou... | (This has a fairly obvious improvement
of the inequality.)
Here is a simple algebraic proof of the stronger inequalities
$x^2+y^2=1
\implies x^3+y^3
\ge \dfrac{x+y}{2} \ge \sqrt{xy}.$
More generally,
without the restriction on
$x^2+y^2$,
this gives
$x^3+y^3
\ge \dfrac{(x+y)(x^2+y^2)}{2}
$.
Proof that this inequality i... | {
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"source": "stackexchange",
"question_score": "2",
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Plotting graph given parametric form satisfying $x=f(t)$ and $y=-f(-t)$ I had to plot the curve given by the parametric definition as follows. In attempting to simplify doing so, here is what I see holds between $x$ and $y$.
Plot the curve given by $$x=\frac{(t+2)^2}{t+1} ,y=\frac{(t-2)^2}{t-1}$$
$$\boxed{x=f(t)\\ y... | write $x$ as $\dfrac{(1+t+1)^2}{t+1} = t+1+2+\dfrac{1}{t+1}$ and $y = t-1 -2 + \dfrac{1}{t-1} $
$$\begin{align} x+y &= 2t \Big(\dfrac{t^2}{t^2-1} \Big) \\
y - x &= 6 - \dfrac{2}{t^2-1}\\
x+y &= t^3 \cdot (6+x-y) \end{align}$$
and now you can just put value of $t$ in terms of $x,y$
to get $$(8+x-y)^3 = (x+y)^2 (6+x-y) ... | {
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Maclaurin expansion for sech$(x)$ I am a bit unsure where I have gone wrong in working this out.
Sech$(x)=2/(e^x+e^{-x}).$
Maclaurin expansions:
$e^x = 1+ x + x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 - x^4/24;$
so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$ (I think this is where I have gone wrong.)
The actual... | The simplest way t get it is to write $\;\operatorname{sech} x=\frac1{\cosh x}$ and to use the MacLaurin expansion of $\cosh x$ at order $4$:
$$\cosh x=1+\frac{x^2}2+\frac{x^4}{24}+o(x^4)$$
You obtain the expansion of its reciprocal dividing $1$ by the MacLaurin expansion of $\cosh x$ along increasing powers, up to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Calculating $\lim_{(x,y)\rightarrow (0,0)}\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2}$ if it exists I'm trying to calculate the above limit. I ran a few paths and found out that the limit is $\pi$ (which I also confirmed through WolframAlpha), but to prove it I use polar coordinates to get an expressio... | $$\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2} = \frac{1 - \cos(\pi x y ) }{x^2 + y^2}+\pi\frac{ \sin (\pi(x^2 + y^2))}{\pi(x^2 + y^2)}$$
For first
$$\left|\frac{1 - \cos(\pi x y ) }{x^2 + y^2}\right|\leqslant\left|\frac{2\sin^2(\pi \frac{x^2 + y^2}{2} ) }{x^2 + y^2}\right| \to 0$$
While second is $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A clarification in a limit While reading this answer I do not understand that how can we claim that there exist a polynomial with the properties, even if it exist then how is it same for the function and the inverse.
| Start with $ f ( x ) = x + a _ 2 x ^ 2 + a _ 3 x ^ 3 + a _ 4 x ^ 4 \cdots $ and $ f ^ { − 1 } ( x ) = x + A _ 2 x ^ 2 + A _ 3 x ^ 3 + A _ 4 x ^ 4 \cdots $ and substitute into $ f ^ { - 1 } ( f ( x ) ) = x $ (or into $ f ( f ^ { - 1 } ( x ) ) = x $ for the same final result):
$$ \eqalign { x & = ( x + a _ 2 x ^ 2 + a _ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the integer solutions to $4^x - 9^y = 55$ I want to find the integer solutions of:
$$ 4^x - 9^y = 55$$
For now, I see that $x = 3, y = 1$ is an integer solution to the equation. How can I rigorously prove there are no other solutions for $x, y$ integers?
I tried to solve for $y$, but to no avail. WolframAlpha tell... | This factors as $(2^x + 3^y)(2^x-3^y) = 55$, so $2^x + 3^y \in \{1, 5, 11, 55\}$. This leaves us only finitely many cases of $x$ and $y$ to check: $x \le 5$ (because $2^6$ is bigger than $55$) and $y \le 3$ (because $3^4$ is bigger than $55$).
A check of all $(x,y) \in \{0,1,2,3,4,5\} \times \{0,1,2,3\}$ gives only one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Manipulating an ODE into a full differential Find the general solution of:
$$(x-2)\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + \frac{4y}{x^2} = 0$$
If I first multiply through by $x^2$ then I can write the above as
$$x^2(x-2)\frac{d^2y}{dx^2} + 3x^2\frac{dy}{dx} + 4y $$
$$ =\left[ x^3\frac{d^2y}{dx^2} + 3x^2\frac{dy}{dx}\righ... | Hint:
$$\frac{d}{dx}\left( x^3\frac{dy}{dx}\right) -2x^2 \frac{d^2y}{dx^2} +4y = 0$$
$$\frac{d}{dx}\left( x^3\frac{dy}{dx}\right) -2x^2 \frac{d^2y}{dx^2} \color {red}{-4xy'+4xy'} +4y = 0$$
$$\left( x^3y'\right)' -(2x^2 y')' +4(xy)' = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the missing angle in triangle In the below triangle, we are looking for the value of angle $φ$.
We are given $α=30, β=18, γ=24$ and also that $CD=BD$.
I have solved it with trigonometry (sine law) and found the required angle to be 78 but I need to solve it with Geometry only.
What I have tried so far:
First of a... | Consider a regular $30$-gon $X_1X_2X_3X_4X_5X_6X_7X_8X_9X_{10}X_{11}X_{12}X_{13}X_{14}X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}X_{22}X_{23}X_{24}X_{25}X_{26}X_{27}X_{28}X_{29}X_{30}$ and place it on the plane so that $X_1 \equiv A$, $X_6\equiv B$, and that $X_2$ and $C$ lie on different halfplanes determined by the li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 1
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Jacobian Transformations of Two Random Variables ( probability density function) Let ${Y}_{1}$ and ${Y}_{2}$ be independent,each with density $\frac{1}{{y}^{2}}$ , $y>1$. Consider the transformation ${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$. Find probability density function of random vari... | When defining joint support like this, you need to consider to specify the marginal support of which variable first. If you specify $u_1$ first, since
$$ u_1 = \frac {y_2} {y_1 + y_2} $$
then we see that
when $y_1$ fixed, $y_2 \to +\infty$, $u_1 \to 1$;
when $y_1 \to +\infty$, $y_2$ fixed, $u_1 \to 0$.
And obviously $0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? The question:
How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? At which points do these tangent lines touch the curve?
My attempt:
To find the slope of the tangent line to the curve, we need to... | hint
The equation of the tangent line to the curve at the point $ (a,\frac{a}{a+1}) $ is
$$y=\frac{1}{(a+1)^2}(x-a)+\frac{a}{a+1}$$
it passes through the point $ (1,2) $ if
$$2=\frac{1}{(a+1)^2}(1-a)+\frac{a}{a+1}$$
So,
$$2(a+1)^2=(1-a)+a(a+1)$$
or
$$a^2+4a+1=0$$
thus
$$a=-2\pm \sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$ I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:
$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{... | If the numerator is $x^{2r},$
$$x^{2r}=(1+x^2-1)^r=?$$
Had the numerator been $x^{2r+1},$
$x^{2r+1}=x(1+x^2-1)^r$ first substitute $x^2-1=y$
Now for $S=\dfrac{d(x/(x^2-1)^n)}{dx},$
$S=1/(x^2-1)^n-2(n+1)x^2(x^2-1)^{n+1}$
Write $x^2=x^2-1+1$ and integrate both sides to find
$$\dfrac x{(x^2-1)^n}=I(n)-2(n+1)(I (n) - I(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the period of this sinusoid? I'm stuck trying to find the period of this sinusoid and would really like some pointers to different ways to approach this problem.
$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3})$$
I would like to turn the product into a sum and do something about that $2$ in the expon... | You get 7.5 if you simply apply the LCM to $\frac{4\pi t}{5} = 2\pi$ (period = 2.5) and $\frac{8\pi t}{3} = 2\pi$ (period = 0.75).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solution verification of $\frac{1}{(4n^2)!} >\frac{1}{(4(n+1)^2)!}$ inequality \begin{align}
\frac{1}{(4n^2)!} &>\frac{1}{(4(n+1)^2)!}\\
\frac{(4(n+1)^2)!}{(4n^2)!} &>\frac{1}{1}\\
\frac{(4(n^2+2n+1))!}{(4n^2)!} &>1\\
\frac{(4n^2+8n+4)!}{(4n^2)!} &>1\\
\iff \frac{(4n^2+8n+4)}{(4n^2)}&>1\\
4n^2+8n+4&>4n^2\\
8n &> -4\\
n... | That looks correct, but it is much easier to see that\begin{align}4(n+1)^2>4n^2&\implies(4(n+1)^2)!>(4n^2)!\\&\implies\frac1{(4(n+1)^2)!}<\frac1{(4n^2)!}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.
As an instance, I will justify an example inequality just to show you the intended strategy. Consider th... | Here is a simple and direct way using just some rough estimates:
For $x>2$ you have surely
$$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} \stackrel{x>2}{<} \frac{5x^2 + 3x^2 + x^2}{x^3 - (x^2 +1)} = \frac{9x^2}{x^3 - (x^2 +1)}$$
Since you want to have a $10$ in the numerator, note that
$$\frac 1{10}x^3 > x^2+1 \Leftrightarrow x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve Surface Area Formula For Radius Ok. I have a Surface Area Formula for a cylinder, but I am wanting to solve for the radius. The only information I have is the height of the cylinder, which is 8 inches. I know I can use the Quadratic Formula to convert the formula to solve for the radius, but I get stuck when d... | I'd just complete the square:
$$ A = 2\pi r^2 + 2\pi rh $$
$$ \frac{A}{2\pi} = r^2 + rh $$
To complete the square on the right-hand side, we need to add $\frac{h^2}{4} $ to both sides:
$$ \frac{A}{2\pi} + \frac{h^2}{4} = r^2 + rh + \frac{h^2}{4} $$
$$ \frac{A}{2\pi} + \frac{h^2}{4} = \left(r + \frac{h}{2} \right)^2 $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to solve this? I am having difficulty in the very last step of the problem. The general solution of $\vert\sin x\vert = \cos x$ is -
(A) $2n\pi+{\pi\over4}$, $n\in I$
(B) $2n\pi±{\pi\over4}$, $n\in I$
(C) $n\pi+{\pi\over4}$, $n\in I$
(D) None of these
So what I did was - I made a case for when sinx is greater than ... | We will solve $\sin^{2}(x) = \cos^{2}(x)$, which is obtained upon squaring the equation (so we do not have to deal with sign issues).
$$\sin^{2}(x) = \cos^{2}(x)$$
$$\cos^{2}(x) -\sin^{2}(x) = 0$$
$$\cos(2x) = 0$$
$$x = \frac{\pi}{4} + \pi n, \frac{3\pi}{4} + \pi n$$
$$x = \frac{\pi}{4} + 2\pi n, \frac{5\pi}{4} + 2\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to derive the steps of this intergral for the product of two exponential terms? For homework, I was looking for closed form solutions of exponential functions from this wiki site.
https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
But I was not sure how to derive the steps for this particular ... | Start completing the square
$$ax^2+\frac b {x^2}=a\left(x+\sqrt{\frac{b}{a}}\frac{1}{x}\right)^2-2\sqrt{ab}$$
$$x+\sqrt{\frac{b}{a}}\frac{1}{x}=t \implies x=\frac{1}{2} \left(t\pm\sqrt{t^2+4 \sqrt{\frac{b}{a}}}\right)$$
Making the calculation for the $\color{red}{-}$ branch.
This makes now the integrand to be
$$\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$
Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longri... | $$x^{x^{x^\cdots}}= y \implies x^y = y \implies1 = yx^{-y} = ye^{-y\log (x)}$$
which means
$$y = \frac{-W(-\log (x))}{\log (x)}$$
where $W$ is the principal branch of the Lambert-W function. Now, the problem of the successive derivatives is "much" simpler (have a look here for the derivatives of $W(t)$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Substituting $x = \sec(u)$ to evaluate $\int\sqrt{x^2-1}\ dx$: but why is $\tan(u) = \sqrt{x^2 - 1}$ rather than $\tan(u) = -\sqrt{x^2 - 1}\ $? I tried evaluating $\int\sqrt{x^2-1}\ dx$ using the substitution $x = \sec(u).$
I know my method gets to the same answer as wolframalpha, namely:
$\int\sqrt{x^2-1}\ dx = \frac{... | Let's differentiate and see by using which sign we get the integrand.
$\begin{align}\frac{d}{dx}\left[\frac{x}{2}\sqrt{x^2-1}-\frac12\ln|x+\sqrt{x^2-1}| + c\right]& = \frac{1}{2}\sqrt{x^2-1}+\frac x2\frac{x}{\sqrt{x^2-1}}-\frac{1+\frac{x}{\sqrt{x^2-1}}}{2(x+\sqrt{x^2-1})}\\
\\& = \frac{2x^2-1}{2\sqrt{x^2-1}}-\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
If I would start from the right expression I woul... | Apply the identities
$$\cos4\theta = 2\cos^22\theta -1,\>\>\>\>\>\cos2\theta = 1-2\sin^2\theta $$
$$\sin3\theta =\sin\theta (2\cos2\theta+1)$$
in simplifying the RHS
\begin{align}
RHS=&1+\sin\theta -\sin\theta (2\cos2\theta+1)
-\frac12(1-2\sin^2\theta)+\frac12(2\cos^22\theta-1)\\
=&\sin^2\theta -2\sin\theta\cos2\theta+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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2D system of hyperbolic equation (LeVeque) Again I need help. This time it's exercise 18.1 from LeVeque Finite Volume Methods.
Consider the system $q_t+Aq_x+Bq_y=0$ with $A=\begin{pmatrix}3&1\\1&3\end{pmatrix}$ and $B=\begin{pmatrix}0&2\\2&0\end{pmatrix}$.
Show that these matrices are simultaneously diagonalizible and... | Setting $p = R^{-1}q$, we have
$$
p_t + \Lambda_x p_x + \Lambda_y p_y = 0 \, .
$$
Since the $\Lambda^\alpha$ matrices are diagonal, this system is decoupled.
Applying the method of characteristics componentwise yields
$$
p(x,y,t) = R^{-1} q(x,y,t) = \begin{bmatrix}
F_1(\lambda^x_1 x -t, \lambda^y_1 x - \lambda^x_1 y) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A weird contour integral calculation I have function $\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}dx$.
Here to use residue thm, I rewrite the integral as $\int_{-1}^{1}\frac{\sqrt{1-z^2}}{1+z^2}dz$ with the poles $z=i$ and $-i$. However there is a given condition $-1<x<1$, so it means $z^2<1$. This is the problem because w... | By your comments in my previous question, we can also deal in another way (but I am about to tell you that is more tedious, even if apparently at the beginning it looks easier).
With $x = \cos\theta$ we get
$$-\int_{\pi}^0 \frac{\sin^2(\theta)}{1 + \cos^2(\theta)}\ \text{d}\theta$$
We can close the contour by doubling ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculating divergence for parabolic coordinates. I'm trying to calculate divergence for parabolic coordinates with a goal to check if i understand all notions of vector analysis correctly.
Definition of parabolic coordinates:
$$x = uv \\ y = \frac{1}{2}(u^2 - v^2)$$
Firstly, I calculate $du$ and $dv$:
$$
dx = v du + u... | Thanks for clarifying. OK, so the vector field is dual to the $1$-form $\omega = F\omega_1 + H\omega_2$, where $\omega_1 = \sqrt{u^2+v^2}\,du$ and $\omega_2=\sqrt{u^2+v^2}\,dv$ give an orthonormal coframe. [Note that the metric $dx^2 + dy^2 = \omega_1^2 + \omega_2^2$.]
Of course we have $\star\omega_1 = \omega_2$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$. My solution Let $n\in\mathbb{Z}^{+}$, then there exists $k\in\mathbb{Z}_0^+$, such that $n=5k + i, i\in\{0,1,2,3,4\}$. Now analyzing by cases we have:
*
*If $i=0$, then
\begin{align*}
n = 5k \Rightarrow n = 5k + 4(0).
\... | You also can prove it by induction on $n>11$.
Here is how the inductive step goes:
Suppose that for some $n>11$, one has $\;n=5x+4y$, $x,y \ge 0$.
*
*If $y >0$, as $1=5-4$, you can write $n+1=5(x+1)+4(y-1)$, and $y_1\ge 0 $;
*If $y=0$, $n=5x$, and since $n>11$, you have $x\ge 3$. Now you also can write $1=16-15$, so... | {
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"url": "https://math.stackexchange.com/questions/4006307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to evaluate the integral $I=\int_0^\infty \frac{\sqrt{x}\arctan(x)}{1+x^2}dx$ with residue theorem? let's consider the following integral
$$I=\int_{0}^{+\infty}{\frac{\sqrt{x}\arctan{x}}{1+x^{2}}\,\mathrm{d}x}$$
I know exactly how to evaluate it with sneaky methods and obtained $I=\frac{\pi\sqrt{2}}{8}(\... | Notice that : \begin{aligned}I=2\int_{0}^{+\infty}{\frac{x\arctan{x}}{2\sqrt{x}\left(1+x^{2}\right)}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\frac{x^{2}\arctan{\left(x^{2}\right)}}{1+x^{4}}\,\mathrm{d}x}\\ &=\int_{-\infty}^{+\infty}{\frac{x^{2}\arctan{\left(x^{2}\right)}}{1+x^{4}}\,\mathrm{d}x}\\ I&=\int_{0}^{1}{\int_{-\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomials/ remainder theorem What must be added to p(x)=x^4+2x^3-2x^2+x-1 to make it exactly divisible by g(x)=x^2+2x-3?
The book says that the remainder will be a linear polynomial and so the expression that must be added to p(x) will be of the form ax+b; the book essentially says we add the remainder. The divisor h... | The division algorithm says that if $p(x)$ is a polynomial of degree $n$ and $g(x)$ is a polynomial of degree $m\le 0$ then there is a unique quotient polynomial $q(x)$ of degree $n-m$ and a unique remainder polynomial of degree $< m$ so that: $p(x) = q(x)g(x) + r(x)$.
So if $p(x) = x^4+2x^3-2x^2+x-1$ is of degree $4$... | {
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General Inequality Problem. The question goes like this :
Let $a,b,c,d$ be positive reals and given that $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$
My approach goes like this:
I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$
Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$
Although ... | Tangent line detailed:
You can rewrite your inequality as $$f(a)+f(b)+f(c)+f(d)\geq {1\over 8}$$ where $f(x)= 6x^3-x^2$. Now we see that we have equality case if $a=b=c=d={1\over 4}$ so it is reasonable to calculate tangent line for $f$ at $x= {1\over 4}$ which is $g(x)={5x-1\over 8}$.
If you draw a graph for $f$ and $... | {
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Prove That $n(n+1)(n+2)(n+3)$ is never a square of a positive integer. $n \in \mathbb{N}$, bigger than zero.
My proof:(A Proof by Contradiction)
Alert: There is a lot of variables.
Suppose that:
$n(n+1)(n+2)(n+3)=x^{2}$, with $x \in \mathbb{N}$
$$n(n+3)(n+2)(n+1)=x^{2}$$
$$(n^{2}+3n)(n^{2}+3n+2)=x^{2}$$
let $n^{2}+3n =... | Hint: Show that $ n(n+1)(n+2)(n+3) + 1$ is a perfect square.
Hence, the original expression cannot be the square of a positive integer (but could be the perfect square 0).
| {
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Prime Number Sums The prime numbers $a,b$ and $c$ are such that $a+b^2=4c^2.$ Determine the sum of all possible values of $a+b+c.$
My Attempt
$a+b^2=4c^2$.
$a=4c^2-b^2$.
$a=(2c+b)(2c-b)$.
After this, I tried testing cases but I'm not totally sure how to account for everything.
Thanks!
| Note that $c=2,3$ yield solutions: $a=7,b=3,c=2 \text{ and }a=11,b=5,c=3$
As has been noted in the comments, $2c-b=1$ is necessary for $a$ to be prime.
For prime numbers $c\ge 5$, $c$ must have the form $6k\pm 1$. Note that for $c=6k-1$ we have $2c-b=1=12k-2-b \Rightarrow b=12k-3$. Since $3\mid 12k-3$, $b$ cannot be pr... | {
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How do I bypass the limit for the tan function on the calculator? So basically, my partner and I are creating problems for a project. She created one where we aren't sure what the correct answer is.
Problem:
$Find \ cos\frac{\theta}{2} \ if \ tan \ \theta = \frac{3}{4}; \pi < \theta<\frac{3\pi}{2}$
I'll show you what w... | You cannot do the inverse tangent because the range of inverse tangent is $(-\frac{\pi}{2},\frac{\pi}{2})$, and so your answer for step 2 (and also the answer from your calculator) is already incorrect.
I would suggest you do this:
Draw the triangle in the third quadrant (because it is in the range $(\pi, \frac{3\pi}{2... | {
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Prove $a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$
If $a,b,c,d$ are real numbers and $a+b+c+d=0$. prove:
$$a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$$
Using Euler's identity: if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$. in this problem substitute $x=a+b,y=c,z=d:$
$$(a+b)^3+c^3+d^3=3cd(a+b)$$
$$a^3+b^3+c^3+d^3=3(a+b)(cd-ab)=3(acd-a^2b+bcd... | $$3(acd-a^2b+bcd-ab^2) = 3(cd(a + b) - ab(a+b)) = 3(cd(a + b) + ab(c+d)) = 3(abc + abd + acd + bcd)$$
| {
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Nonlinear recurrence relation over $\mathbb{C}$ A sequence in $\mathbb{C}$ is given: $(a_n)_{n=1}^{\infty}$ s.t.
$a_1 = i$ and $a_{n+1} = \frac{3}{2 + a_n}$.
Assume $\lim(a_n)$ exists.
Using the fact that $\lim(a_n) = \lim(a_{n+1})$ and solving the resulting quadratic gives that $\lim(a_n)\in \{-3,1\}$.
I was trying to... | A specific approach to convergence is to examine how $a_{n}-1$ changes in each iteration.
\begin{align}
a_{n+1} - 1
&= \frac{3}{2+a_n} - 1 \\
&= \frac{1-a_n}{3-(1-a_n)}
\end{align}
Then if $|a_n-1|$ is small relative to $3$ it is apparent that $|a_{n+1}-1|$ decreases by a factor of nearly $1/3$ on each iteration and ... | {
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Why does this pattern work: $1 \cdot{1} = 1, 11 \cdot{11} = 121, 111 \cdot{111} = 12321\ldots$ I have recently learned about pattern that goes like this:
$$\begin{align}
1^2 &= 1\\
11^2 &= 121\\
111^2 &= 12,321\\
1,111^2 &= 1,234,321\\
11,111^2 &= 123,454,321.
\end{align}$$
It is a very cool pattern, but after a bit it... | Since OP is asking for an algebraic explanation, I will try a different approach, possibly trickier than the other answer which is elegant but not formal.
We consider the polynomial $(1+x+x^2+\dots +x^n)^2$: by expanding the square we get
$$\begin{array}{|c|c|c|c|c|c|c|}
\hline
1 & x & x^2 &\cdots& x^{n-1}& x^n\\
\hli... | {
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Factoring $a^4+b^4+(a-b)^4$ I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.
I rewrite it as:
$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not nega... | It's a variation of Candido's identities, its proof without words are as follows:
Now
\begin{align}
[x^2+y^2+(x+y)^2]^2 &= 2[x^4+y^4+(x+y)^4] \\
[a^2+(-b)^2+(a-b)^2]^2 &= 2[a^4+(-b)^4+(a-b)^4] \\
[2a^2-2ab+2b^2]^2 &= 2[a^4+b^4+(a-b)^4] \\
4(a^2-2ab+b^2)^2 &= 2[a^4+b^4+(a-b)^4] \\
a^4+b^4+(a-b)^4 &= 2(a^2-ab+... | {
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Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$ I accidentally found out that the two integrals below
$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluat... | Since $\int_0^\infty\frac{f(x)dx}{1+x^2}=\int_0^1\frac{f(x)+f(1/x)}{1+x^2}dx$, $\int_0^\infty\tfrac{\arctan x^kdx}{1+x^2}=\tfrac{\pi}{2}\int_0^1\tfrac{dx}{1+x^2}=\tfrac{\pi^2}{8}$ for all $k\in\Bbb R$.
| {
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"source": "stackexchange",
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Impossible integral?
$$\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2$$
I saw this problem in a calculus exam. $f$ is assumed to be continuous. Using the Fundamental Theorem of Calculus, I calculated $f(t)= 3t^2+8t+8 $. But when I integrate $f$ it gives me $\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2-5$:
$$\frac {d}{dx}\int_{0}^{... | Suppose $f(t)=at^2+bt+c$
we have
$$\int_0^{x^2-1} \left(a t^2+b t+c\right) \, dt=$$
$$=x^6+x^4+3 x^2\equiv \frac{a x^6}{3}+x^4 \left(\frac{b}{2}-a\right)+x^2 (a-b+c)-\frac{a}{3}+\frac{b}{2}-c$$
and
$$
\begin{cases}
\frac{a }{3}=1\\
\frac{b}{2}-a=1\\
a-b+c=3\\
-\frac{a}{3}+\frac{b}{2}-c=0\\
\end{cases}
$$
the system has... | {
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Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6...
Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6, find distance of focus from tangent at P.
Let P and ... | As you rightly obtained, $t^2 = 2 \implies t = \pm \sqrt2$
If we take $t = \sqrt2$ and rewrite,
$P(2a, 2a \sqrt2), Q(\frac{a}{2}, - a \sqrt2)$
Slope of tangent at point $Q = \frac{2a}{y} = - \sqrt2$
So equation of tangent line $y + a \sqrt2 = -\sqrt2(x-\frac{a}{2})$
or $y = - \sqrt2 x - \frac{a}{\sqrt2} $
Point $R$ whe... | {
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Generalization of $\sqrt{n}+\sqrt{n+2005}=m$ I was solving the math Olympiad of Belgium the year $2005$, and the last problem was :
If $n$ is an integer then find all values for $n$ for which $\sqrt{n}+\sqrt{n+2005}$ is an integer as well.
My question is just to generalize this problem, so instead of $2005$ , we can ... | In general, for non-negative integers $a, b$, the sum $\sqrt a + \sqrt b$ is an integer if and only if both $\sqrt a$ and $\sqrt b$ are integers (exercise).
Solution:
Squaring, we see that $ab$ is the square of a rational number, and hence we may write $a = m^2 k$, $b = n^2 k$, with $k$ square-free. Then the sum $\sq... | {
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How can I construct a sequence of smooth functions $g_n$ s.t their integrals converge to zero? Define $g_n(x)$ to be zero on $[0,1-\frac{1}{n}]$ and $g_n(x)=ax^2+bx+c$ on $[1-\frac{1}{n},1]$
I want the $g_n$ functions to be in $C^2[0,1]$ and statifies $g_n(0)=0$ and $g_n(1)=1$, $g_n'(0)=0$ and their integrals to conver... | As it was pointed out in the comments, the functions $g_n$ are not even continuous en $[0,1]$. Regarding the second part, it is true that $\int_0^1 g_n(x) dx \to 0$ and your justification is correct.
| {
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Find the whole number solutions for W+X+Y+Z = 15 where W, X, Y, Z ≤ 6 I worked out a solution but don't know if its the right one. Is this the right way to approach the problem? Any help would be appreciated.
First, the number of non-negative integer solutions for $W+X+Y+Z = 15$ can be calculated using stars and bars:
... | The “$\le6”$ constraints are annoying to work with. To fix it,we can substitute $$A=6-W, \quad B=6-X, \quad C=6-Y, \quad D = 6 - Z.$$
Now, the constraints become $A+B+C+D\le 4\times 6 - 15=9$, and $A,B,C,D\ge 0$. Now, you can solve it by standard stars and bars method.
So the answer is $$\binom{9+4-1}{4-1}=220.$$
Where... | {
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Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.
By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 root... | We may apply Sturm's theorem https://en.wikipedia.org/wiki/Sturm%27s_theorem.
Let $\mathrm{rem}(p(x),q(x))$ denote the remainder of $p(x)$ divided by $q(x)$ for two polynomials $p(x), q(x)$.
The Sturm sequence is given by:
$p_0(x) = x^5-5x^3+4x-1$
$p_1(x) = p_0'(x) = 5x^4 - 15x^2 + 4$
$p_2(x) = -\mathrm{rem}(p_0, p_1) ... | {
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Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$.
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.
I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and ... | Write $a = 2^x \cdot m$ and $b = 2^y \cdot n$ where $m$ and $n$ are odd. WLOG $x \geq y$. Now,
$$2^{3x} \cdot m^3 + 2^{3y} \cdot n^3 = 2^{3y} \cdot (2^{3x-3y} \cdot m^3 + n^3) = 2^c$$ If $x > y$, we have a contradiction since in that case an odd number divides LHS. So, $x = y$ and so you would arrive at your argument.
| {
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Let $f(x)=\frac{9^x}{9^x+3}$. Evaluate $\sum_{i=1}^{1995}f(\frac{i}{1996})$.
Let $f(x)=\frac{9^x}{9^x+3}$. Evaluate $\sum_{i=1}^{1995}f(\frac{i}{1996})$.
This problem seems extremely hard until you find out that $f(x)+f(1-x)=1$ and you can then evalute $f(\frac{1}{1996})+f(\frac{1995}{1996}) =1, \dots, f(\frac{998}{1... | There are two aspects which can lead us from
\begin{align*}
f(x)=\frac{9^x}{9^x+3}\tag{1}
\end{align*}
to the functional equation $f(x)+f(1-x)=1$.
Simplification: We can reduce the number of occurrences of $x$ in (1) by factoring out $9^x=3^{2x}$. We obtain
\begin{align*}
f(x)=\frac{9^x}{9^x+3}=\frac{1}{1+3^{1-2x}}\ta... | {
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A limit which exists in polar coordinates but not in Cartesian coordinates? Let's have look at the function
$$ f(x,y) \begin{cases}
\frac{y(x^2+y^2)}{y^2+(x^2+y^2)^2} & (x,y)\neq(0,0)
\\0 & (x,y)=(0,0)\end{cases}.$$
Switching to polar coordinates gives
$$ f(r,\theta)=\begin{cases}
\frac{r^3 \sin \theta}{r^2\sin^2\theta... | The limit seems to exist in polar coordinates since you only tested the curves with fixed $\theta$. This is equivalent to only testing the lines $y=mx$ (and the line $x=0$) in Cartesian coordinates.
Indeed for these lines we have:
$$\lim_{(x,y) \to (0,0)}\frac {y(x^2+y^2)}{y^2 + (x^2 + y^2)^2} = \lim_{x\to 0}\frac {mx^... | {
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Proving that $f$ verifies $f(1−x) + f(x) = 1$ I have a function which is defined for $x \in (0,1)$ and for $p>1$ with the expression
\begin{align*}
f(x) = \sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
\end{align*}
I would like to show that $f(1-x) + f(x) = 1$.
Wh... | I have actually found a way to solve the question from EDIT1 and EDIT2
So
\begin{align*}
F(1-x) + F(x)
&= G(x) \Big( H(1-x) + H(x) \Big) \\
&= \Bigg(\prod_{s=0}^{2p-1}(x-p+s)\Bigg)\Bigg(\sum_{k=0}^{2p-1}\frac{(-1)^{k}}{x-p+k}\binom{2p-1}{k}\Bigg)\\
&= \sum_{k=0}^{2p-1} (-1)^{k}\binom{2p-1}{k}\Bigg(\prod_{s=0, s \neq k... | {
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Is $\tan(\alpha/2)=(1-\cos\alpha)/\sin \alpha, \,?$ Is it true that
$$\bbox[5px,border:2px solid #138D75]{\tan\left(\frac {\alpha}2\right)=\frac{1-\cos\alpha}{\sin \alpha}, \quad ?} \tag1$$
My solution:
First, we note that the expression $\tan \left ( \frac{\alpha}{2} \right )$ makes sense when $\alpha \neq \pi + 2k\pi... | We could multiply both sides by $\sinα$
Then on the LHS use the double angle for $\sinα = 2\sin(α/2) \cos(α/2)$ and put $\tan(α/2)=\sin(α/2)/\cos(α/2)$ then the LHS becomes $2\sin^2(α/2)$.
The RHS is the same as $1-\cosα = 1-(1-2\sin^2(α/2))$ from a double angle formula for $\cosα$, so it's true.
| {
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There's 8 black balls and 7 white balls. 3 of the balls are drawn at random. Probability of drawing 2 of one color and 1 of the other color? A bin has 8 black balls and 7 white balls. 3 of the balls are drawn at random. What is the probability of drawing 2 of one color and 1 of the other color?
Here's what I tried:
Cas... | Case $1$: $2$ black balls and $1$ white ball. The correct probability will be
$P(2B,1W) = \displaystyle \frac{ 3 \cdot8 \cdot 7 \cdot 7}{15 \cdot 14 \cdot 13} = \frac{28}{65}$.
$\big[\text{It is basically } \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{7... | {
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Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity $a(x+b)^3+c\equiv4x^3-24x^2+48x+d$
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity
$$a(x+b)^3+c\equiv4x^3-24x^2+48x+d$$
I have already found $a=4$ and $b=-2$ but I'm struggling to find $c$ and $d$.
... | $4x^3 - 24x^2+48x + d = 4(x^3 - 6x^2 + 12x - 8) + 32+ d = 4(x-2)^3 + 32+ d\implies a = 4, b = -2, c = 32+d$. Note that there are more than one right answer for this. You can take for example $c = 3, d = -29$ or $c = 0, d = -32$ as they are just two answers among many !
| {
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"answer_id": 1
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Induction for binomial coefficients I would like some help to prove the following equality :
$$\sum_{i=0}^n \binom{n}i^2=\binom{2n}n$$
I wanted to do a proof by induction :
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=1}^{n+1} \binom{n+1}i^2=1 + \sum_{i=0}^{n} \binom{n+1}{i+1}^2=1+\sum_{i=0}^{n} \bigg(\binom{n+1}i+\bino... | One way to proceed is to prove a more general identity by induction, and then deduce the identity in the problem statement as a corollary.
The more general identity is called "Vandermonde's Identity":
$$\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} = \binom{m+n}{k} \tag{1}$$
for non-negative integers $k$, $m$ and $n$. We g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the integral of $\frac{1}{x(x^n+a^n)}$ I was working on the problems in Mathematical Methods for Physics and Engineering by Riley,Hobson & Bence.
In Problem 2.34 (d) I'm supposed to find this integral: $$J=\int\frac{dx}{x(x^n+a^n)}.$$
I used partial fractions and arrived at the form
$$J=\frac{1}{a^n}\left[\log ... | We seek a decomposition of the form
$$\frac{1}{x(x^n + a^n)} = \frac{A}{x} + \frac{Bx^{n-1}}{x^n + a^n} = \frac{(A+B)x^n + Aa^n}{x(x^n+a^n)}.$$ Hence the choice $A = a^{-n}$, $B = -A = -a^{-n}$, yields
$$\frac{1}{x(x^n+a^n)} = \frac{1}{a^n} \left(\frac{1}{x} - \frac{x^{n-1}}{x^n + a^n}\right).$$
The rest is straightfo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4045238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
prove for bell number using induction on n hi guys I have to prove this equality $$B_n=e^{-1}\sum_{k=0}^{\infty}\frac{k^n}{k!},$$ that is called bell equality only using induction on $n$ . How can i do this? I have tried by substituting the recursive formula $\sum\limits_{k=0}^{n} \binom{n}{k} B_{k}$ in the first one... | Using the Bell´s formula we have that $$B(n)=\sum\limits_{k=0}^nS(n,k).$$
This is the total number of ways to put $n$ balls in an arbitrary number of
boxes (no empty boxes remaining). To count them we look at the number of balls (at this parameter
we will call it $a$, and it can be from $0$ to $(n - 1)$) that accompany... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$ I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to so... | $$ \begin{align}f(n+1)-f(n)&=\frac{(n+1)^5-n^5}5+\frac{(n+1)^3-n^3}3+\frac7{15}
\\&=\frac{5n^4+10n^3+10n^2+5n+1}5+\frac{3n^2+3n+1}3+\frac7{15}\\&=\text{integer}+\frac15+\frac13+\frac 7{15}\end{align}$$
is an integer for all integers $n$ and so is $f(0)$. The claim follows by induction.
Here's a generalization for you:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{... | Hint: $\,a+b\mid a^3+b^3\Rightarrow 35\mid 16^3\!+\!19^3,\,17^3\!+\!18^3\Rightarrow 35\mid x.\,$ $x$ is even so $\,2\cdot 35\mid x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Need help simplifying a set of equations (and understanding how to solve it) i have three algebraic expressions, each using the others. in these equations a, b, c and t are known and plugged in later:
$x = a^{-1}(t + y + z)$
$y = b^{-1}(t + x + z)$
$z = c^{-1}(t + x + y)$
i have managed to successfully solve the equati... | You can write your system as
$$\left\{\begin{align}
ax = (y+z) + t\\
by = (x+z) + t\\
cz = (x+y) + t
\end{align}\right.$$
and then in matrix form you'd get
$$\begin{align}
&\pmatrix{a&0&0\\0&b&0\\0&0&c}\pmatrix{x\\y\\z}
=
\pmatrix{0&1&1\\1&0&1\\1&1&0}\pmatrix{x\\y\\z}
+
t\pmatrix{1\\1\\1}\\
\implies
&\underbrace{\pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate the sum of all irrational roots of $4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$
Calculate the sum of all irrational roots of
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$
I'm not even sure how to begin here, I tried raising it to the power of three, tried writing $8x^{3}+ 3$ with $x^{3}+ y^{3}= \left ( x+ y \right )\left ( x^{2}-... |
*
*If we assume "cube root" is the real-valued root, then we have
$$\begin{cases}4\sqrt[3]{8x-3}-8x^3-3=0 \\ x\in\mathbb R \end{cases} \Longleftrightarrow 512 x^9 + 576 x^6 + 216 x^3 - 512 x + 219=0$$
$$\Longleftrightarrow (2 x - 1) (4 x^2 + 2 x - 3) (64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73)=0$$
Then using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Probability that the product of two numbers in $\{0,1,...,10^{n - 1}\}$ contains $2n - k$ digits, $k = 0,1,...$ How to find the probability that the product of two numbers in ${0,1,...,10^n - 1}$ contains $2n-k$ digits, $k = 0,1,...$? As I understand, the sum of all the combinations of two numbers is equal to $10^{2n}$... | This will give an approximate answer, which will be very close. The region of interest is the square $[0,10^n-1] \times [0,10^n-1]$ in the $xy$ plane. The numbers with $2n-k$ digits are the numbers from $10^{2n-k-1}$ to $10^{2n-k}-1$. Instead of counting lattice points, we can view $x$ and $y$ as continuous variable... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the system of equations: $32y+32x^3=6x+17$, $16z+32y^3=6y+9$, $8x+32z^3=6z+5$ where $x,y,z\in \mathbb{R}$ Solve the system of equations: $$\begin{cases} 32y+32x^3=6x+17 \\16z+32y^3=6y+9 \\8x+32z^3=6z+5 \end{cases}$$ where $x,y,z\in \mathbb{R}$ (Bulgaria 1960)
I attempted to solve this question as follows:
$32y+32... | Just for simplicity's sake, make a change of variables $$(a,b,c) = (2x,2y,2z)$$
We have the system of equations
\begin{align}
16b + 4a^3 &= 3a+17 \\
8c + 4b^3 &= 3b+9 \\
4a + 4c^3 &= 3c+5 \\
\end{align}
or
\begin{align}
16(b-1) &= - 4a^3+3a+1 \\
8(c-1) &= - 4b^3+3b+1 \\
4(a-1) &=-4c^3+ 3c+1 \\
\end{align}
or
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
please help me finish the proof of this equation problem: prove that $ 2(\sqrt{2}+\sqrt{6})\sqrt[4]{3}\int_{0}^{1}\frac{1}{\sqrt{x}\sqrt{1-x}\sqrt{(\sqrt{2}+\sqrt{6})^{2}-x}} dx = \beta (\frac{1}{6},\frac{1}{3}) $
My attempt:
use the relationship between hypergeometric series and elliptic integral, we have
$F_{1}(\alph... | $\frac1{\sqrt{2} + \sqrt{6}} = \lambda^*(3),$ where $\lambda^*$ is defined here.
Now, this seems to imply that
$$K(\frac1{\sqrt{2} + \sqrt{6}}) = \frac{3^{\frac14} \Gamma^3(1/3)}{2^{7/3}\pi},$$ as per this page.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.