Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Show that, $2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$ Show that,
$$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$
There is a mixed of sin and tan, how can I simplify th... | It would be easier to attack if you would substitute $\tan^{-1}{\frac17}\;$ for $\sin^{-1}{\frac1{5\sqrt2}}\;$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to compute $\int_0^1\frac{\ln(x)}{1+x^5}dx$?
Let $\phi$ denote the golden ratio $\phi=\frac{1+\sqrt5}{2}$.
How can I prove this sum?
$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\phi}{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]=\left(\frac{2\pi}{5}\right)^2$$
My try:... | Consider the function $f(x)=\frac12(3x^2-1)$ for $x\in(-1,1)$ and its periodic extension with period $2$. Since $f(x)=f(-x)$ we can write
$$f(x)=\sum_{k=0}^{\infty}a_n\cos n\pi x$$
$$\int_{-1}^1\frac12(3x^2-1)dx=0=2a_0$$
$$\begin{align}\int_{-1}^1\frac12(3x^2-1)\cos n\pi x\,dx&=\left[\frac1{n\pi}\frac12(3x^2-1)\sin n\p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$
I was thinking of using mathematical induction for this. That is,
We prove by induction on $n$. The cas... | A non-induction version, just for diversification. Let's note $$S_n=\sum_{k=1}^{n} \frac{1}{k}$$
which is $$S_n=\ln(n)+\gamma +\varepsilon_n$$
Now
$$\sum_{k=1}^{n}\left ( \frac{1}{k} + \frac{2}{k+n} \right )= \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n} $$
$$ = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1... | {
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How to solve this equation algebraically Solve the following simultaneous equations on the set of real numbers:
\begin{cases}x^2 + y^3 = x+1 \\ x^3+y^2=y+1\end{cases}
Thanks for helping!
| Subtract the two equations to get
$(x^2 -y^2) -(x^3 -y^3)-(x-y)=0$
$\Rightarrow (x-y)(x+y-x^2-y^2 -xy-1)=0$
So, either $x-y=0$ or $x+y-x^2-y^2 -xy-1=0$
*
*For the first case,
We have $x^3+x^2-x-1=(x-1)(x^2 -1)=(x-1)^2 (x+1)=0 \Rightarrow x=1 $ or $x=-1$
So, $x=y=1$ or $x=y=-1$
*And for the second case,
$x+y-x^2-y... | {
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Number of polynomials which are divisible by $x+1$ Let $a,b,c,d$ be four integers (not necessarily distinct) in the set ${1,2,3,4,5}$ . The number of polynomials $f(x)=x^4+ax^3+bx^2+cx+d$ which are divisible by $x+1$ are:
$(A)$ Between 55 and 65
$(B)$ Between 65 and 85
$(C)$ Between 86 and 105
$(D)$ More than 105
I see... | On approach would be to brute-force the result by counting. Notice that you've reached the necessary and sufficient condition $a+c = 1+b+d$.
Since $1\le a,b,c,d \le 5$, then $a+c \in \{2, \dots, 10\}$, so $b+d \in \{1, \dots, 9\}$. Notice that $b+d=1$ is impossible, because it would force either $b$ or $d$ to be $0$ wh... | {
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Prove that a complex equation has a solution of module 1
Prove that the equation
$$z^n + z + 1=0 \ z \in \mathbb{C}, n \in \mathbb{N} \tag1$$
has a solution $z$ with $|z|=1$ iff $n=3k +2, k \in \mathbb{N} $.
One implication is simple: if there is $z \in \mathbb{C}, |z|=1$ solution for (1) then $z=cos \alpha + ... | If $z^n + z + 1=0 $ then $z^n = -(1+z)$ and $$|z|^n = |1+z|.$$
If $|z| = 1$, then $z = \cos a + i \sin a$, with $a \in [0, 2\pi]$. Moreover:
$$1^n = |1 + \cos a + i \sin a| \Rightarrow \sqrt{(1+\cos a)^2 + \sin^2 a} = 1 \Rightarrow \\
1 + \cos^2 a + 2 \cos a + \sin^2 a = 1\Rightarrow
\cos a = -\frac{1}{2}\\ \Rightarro... | {
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Find $\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms Find $S=\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms
I first multiplied and divided $S$ with $1\cdot3\cdot5$
$$\frac{S}{15}=\frac{1}{1\cdot3\cdot5\cdot7}+\frac{1\cdot3}{... | Alpha finds the sum to be $\frac {32}{129}$, giving an answer to your question of $1$. The way to do it by hand is telescoping the series.
| {
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Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with t... | So we have to deal with
$$ S=\sum_{k\geq 1}(-1)^{k+1} \sum_{n=\binom{k}{2}+1}^{\binom{k+1}{2}}\frac{1}{n} = \sum_{k\geq 2} (-1)^k A_k $$
and to prove convergence it is enough to show that $\{A_k\}_{k\geq 1}$ is decreasing (from some point on) and convergent to zero. The last claim is straightfoward to prove, since $A_k... | {
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Integrate $ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $ $$ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $$
I thought of substituting $ x-\frac{1}{x} $ as $t$ but it gets stuck midway. I am close but I think I need to sustitute something else here.
| Given expression is also equal to:
$$\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})(\sqrt{\frac{1}{x^2}+x^2})}dx$$Which further reduces to
$$\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})\sqrt{(x-\frac{1}{x})^2-2}}dx$$Now let $x-\frac{1}{x}=t$ The rest is evident.
| {
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Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $... | HINT:
Let $\sqrt{1+e^x}=y\implies e^x=y^2-1\implies e^x\ dx=2y\ dy$
$$\int\sqrt{1+e^x}\ dx=\int\dfrac{2y^2}{y^2-1}dy$$
Now $y^2=y^2-1+1$
$$\dfrac{2y^2}{y^2-1}=2+\dfrac{y+1-(y-1)}{y^2-1}=2+\dfrac1{y-1}-\dfrac1{y+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find $(x+y)$.
We know that $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find the expression $(x+y)$.
My work so far:
$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$$
$$\color{red}{\left(x-\sqrt{x^2+1}\right)\cdot}\left(x+\sqrt{x^2+... | $t\mapsto t+\sqrt{t^2+1}$ is a strictly increasing function, hence for each $x$ there is at most one $y$ that makes the equation true. On the other hand, $$(x+\sqrt{x^2+1})(-x+\sqrt{x^2+1})=(x^2+1)-x^2=1$$ suggests that $y=-x$ is a valid such choice. We conclude that $x+y=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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continued fraction $F(x)$ that is a generating function of central binomial coefficients Given the following continued fraction
$$F(x) =\cfrac{1}{x+\cfrac{2^2(2^2-1)}{6x+\cfrac{3^2(3^2-1)}{12x+\cfrac{4^2(4^2-1)}{20x+\cfrac{5^2(5^2-1)}{30x+\ddots}}}}}=\frac{1}{\sqrt{x^2+4}}$$
Then $$\frac{1}{x}F\left(\frac{1}{x}\right)=... | $F(x)$ can be rewritten as $\displaystyle\;\frac{1}{\frac{2}{P(x)} - x}$ where
$\displaystyle\;\def\CF{\mathop{\LARGE\mathrm K}}
P(x) = \cfrac{1\cdot 2}{1\cdot 2 x + \cfrac{ (1 \cdot 2)(2\cdot 3)}{2\cdot 3 x + \cfrac{(2\cdot 3)(3\cdot 4)}{3\cdot 4 x + \ddots} }}
$.
The CF $P(x)$ has the form $
\displaystyle\;
\CF_{\el... | {
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Discriminant formula issue Could you please explain to me, how to get the formula of discriminant ? How can I visualize it, any articles, lectures?
I can memorize it $b^2 - 4ac$ But, want to understand it.
Thanks.
| Consider the polynomial
$$ax^2+bx+c$$
First I would like to make this polynomial a multiple of a polynomial with its first term a perfect square and the second term even. To do this, we multiply by $\frac{4a}{4a}.$
$$\begin{align} \frac{1}{4a}(4a^2x^2 + 4abx+4ac) &= \frac{1}{4a}((2ax)^2+2b(2ax)+4ac) \\
& = \frac{1}{4a... | {
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"timestamp": "2023-03-29T00:00:00",
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Decouple a system of two second order differential equations I have a system of second-order differential equations that I want to decouple. they are,
$\ddot{x} = \frac{\omega_1^2}{2} x + \omega_2 \dot{y}$
and
$\ddot{y} = \frac{\omega_1^2}{2} y - \omega_2 \dot{x}$
I am thinking that I should use some transformation, bu... | Differentiate the first wrt $t$ to gain an expression for $\ddot y$:
$\dddot{x} = \frac{\omega_1^2}{2} \dot x + \omega_2 \ddot{y}$
Substitute $\ddot{y} = \frac{\omega_1^2}{2} y - \omega_2 \dot{x}$ to get:
$\dddot{x} = \frac{\omega_1^2}{2} \dot x + \frac{\omega_1^2 \omega_2}{2} y - \omega_2^2 \dot{x}$
Rearrange: $\frac{... | {
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Find the missing digits in the expansion of $34!$ If $34!=\overline{295232799cd96041408476186096435ab000000}$ then find the value of $a,b,c$ and $d.$
My Attempt: I can find that $b=0$ because it has seven five integers.
$\lfloor{\frac{34}{5}}\rfloor+\lfloor{\frac{34}{25}}\rfloor=6+1=7$
Also I think $a$ can be found usi... | This is a partial solution...
Use all the divisibility rules.
But first, you can notice that $32!=2^31\times5^7\times\cdots$, so there are $7$ zeros at the end of the number. So $b=0$, and $a\ne 0$.
$34!$ is divisible by $9$, so:
$$4+a+c+d=0\pmod 9.$$
It's divisible by $7$, so:
$$000-000+5a0-643+609-618+847-140+604-c... | {
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If $a^3=1$, is $G$ abelian? If $G$ is a group that satisfies $a^3=1$ for every $a\in G$, then is $G$ abelian?
This is an exercise I found in Jacobson's Basic Algebra. It is analogous to the question: If $G$ is a group that satisfies $a^2=1$ for every $a\in G$, then $G$ is abelian. I tried to multiply on both sides of $... | The multiplicative group of matrices
$$G = \left\{ \begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix} \middle\vert\, a,b,c \in \mathbb{F}_3\right\} \subset \operatorname{Mat}(3 \times 3, \mathbb{F}_3)$$
is a counterexample (it is isomorphic to the group in Joanpemo's answer).
For all those who are not willing to check t... | {
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Using induction to prove for $n ≥ 1, $ $1 \times 5+2\times6+3\times7 +\cdots +n(n + 4) = \frac 16n(n+1)(2n+13).$ This is a very interesting problem that I came across in an old textbook of mine. So I know its got something to do with mathematical induction, which yields the shortest, simplest proofs, but other than tha... | Base Case: For $n = 1$, we have:
$$
1 \times 5 = 5 = \frac{1}{6}(1)(1 + 1)(2(1) + 13)
$$
which works.
Inductive Hypothesis: Assume that the claim holds for $n' = n - 1$, where $n \geq 2$.
It remains to show that the claim holds for $n' = n$. Indeed, observe that:
\begin{align*}
&1 \times 5 +\cdots + n(n + 4) & \\
&= [1... | {
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When one of these two equations has integral sol, why p=3 (mod 4)? Let p is not 7 and p be an odd prime.
AND one of two equations
$x^2+7y^2=p, x^2-7y^2=p$
has integral sol (x,y). then show that $p \equiv3 \pmod4$
what i have done is this, in $\pmod7$, two equations are $x^2 \equiv p \pmod7$
and $p$ can be only $1$ or ... | Suppose $x^2+7y^2=p$. Applying modulo 4:
$$x^2+3y^2=p \pmod 4 \tag1$$ then $$x^2-y^2=p \pmod 4 \tag2$$ Because $p,4$ are coprime from (2) we get $x \ne y \pmod 4$.
Because $a^2 \in \{0, 1\} \pmod 4 \ \forall a$, we have $x^2=0,y^2=1$ or $x^2=1,y^2=0$.
From the first case we get $p = 3 \pmod 4$ and from the second cas... | {
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Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.
Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.
I tried to solve the problem above by doing a change of variables to the spherical coordinate system, that is,
$x= \rho \cos \theta \sin \varphi$
$y = \rho \sin \theta ... | We will assume that the problem refers to the region
above the paraboloid $z=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=2$.
Since the volume can be found by integrating the area of the horizontal slices of the region,
and each of the horizontal slices is a circular disc,
$\displaystyle V=\int_0^1\pi(r(z))^2dz+\int_1^... | {
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"timestamp": "2023-03-29T00:00:00",
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Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$ Let $a,b,c$ be side lengths of a triangle. What are the minimum and maximum of $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}?$$
When $a=b=c$, the value is $9$. In addition, we can write $a=x+y,b=y+z,c=z+x$ since they are side lengths of a triangle. The e... | If $a=b=1$ and $c\rightarrow2^-$ so $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}\rightarrow8,$$
but $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}>8$$ it's
$$\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)>0$$ or
$$(a+b-c)(a+c-b)(b+c-a)>0,$$
which says that the minimum does not exist and
$$\inf\frac{(a+b+c)(2ab+... | {
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"timestamp": "2023-03-29T00:00:00",
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Confusion in rotation matrix - rotation about $y$ axis The rotation matrix about y axis should look like
$$\left[
\begin{array}{ccc}
\cos\frac{\pi}{2} & 0 &\sin\frac{\pi}{2}\\
0 & 1 & 0\\
-\sin\frac{\pi}{2} & 0 &\cos\frac{\pi}{2}\\
\end{array}
\right]
=
\left[
\begin{array}{ccc}
0 & 0 &1\\
0 & 1 & 0\\
-1& 0 &0\\
\end{a... | I believe that you are viewing the problem backwards. The matrix
$$
A=\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
-1 & 0 & 0
\end{pmatrix}
$$
rotates a vector into a new, rotated frame, according to what you are describing with your images. However, what you are asking is, "What matrix brings me from the rotated frame ba... | {
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"timestamp": "2023-03-29T00:00:00",
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Decomposition of a rational function in partial fractions I am trying to decompose the following rational function: $\dfrac{1}{(x^2-1)^2}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^2-1)^2 = (x+1)^2(x-1)^2$
Therefore $\exists A, B, C, D$ s.t: $\dfrac{1}{(x^2-1)^2} = \dfrac{1}{(x-1)^... | We have
$$1=(x-1)^2(Ax+B)+(x+1)^2(Cx+D)$$
You already have
$$B-A=C+D=\frac 14$$
Substituting $x=0$ gives
$$1=B+D$$
Substituting $x=2$ gives
$$1=2A+B+9(2C+D)$$
Solving these gives
$$A=\frac 14,\quad B=\frac 12,\quad C=-\frac 14,\quad D=\frac 12$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$ without L'hôpital I have this $\lim_{}$.
$$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$$
Indetermation:
$$\lim_{x\to 0} \frac{\ln(0+4)-\ln(4)}{0}$$
$$\lim_{x\to 0} \frac{0}{0}$$
Then i started solving it:
$$\lim_{x\to 0} \frac{\ln\frac{(x+4)}{4}}{x}$$
$$\lim_{x\to 0} \frac{... | You want
$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}
$.
One of the definitions
of $\ln$ is that
$\ln(x)
=\int_1^x \frac{dt}{t}
$.
Therefore
$\ln(x+4)-\ln(4)
=\int_4^{x+4} \frac{dt}{t}
$.
If
$x > 0$
then,
since
$\frac14
\ge \frac{1}{t}
\ge \frac1{x+4}
$
for
$4 \le t \le 4+x$,
$\frac{x}{4}
\ge \int_4^{x+4}\frac{dt}{t}
\g... | {
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} |
Show that $x$ and $Q x$ are equidistant $$Q= \begin{bmatrix} \cos x & -\sin x\\ \sin x & \cos x\end{bmatrix}$$
Given x belongs to $\mathbb{R^2}$, show $Qx$ and $x$ are equidistant.
I've tried dot producting $Qx$ and seeing whether they are equal. I just can't seem to get it.
| Of course, @thanasissdr answer is the well educated one, but arguing more naïvely:
As written in my comment:
$$d((a,b),(0,0)=\sqrt{(a-0)^2+(b-0)^2}=\sqrt{a^2+b^2}=\|(a,b)\|$$
Thus we compare the norms.
We have:
$$\|(a,b)\|=\sqrt{a^2+b^2}$$
and
\begin{align*}\|Q(x,y)\|& =\|(\cos(x)a+\sin(x)b,-\sin(x)a+\cos(x)b)\|\\ & =\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplifying derivative result I am doing the derivative of
$$f(x) = \frac{x^2 -4x +3}{x^2-1}$$
So my result is the following
$$f'(x) = \frac{4x^2 -8x +4}{(x^2-1)^2}$$
I am sure the answer is correct, but in my solutions book and In Wolfram Alpha they simplify until
$$f'(x) = \frac{4}{(x+1)^2}$$
And I don't know why, wh... | You are correct. Now note that
$$\frac{4x^2-8x+4}{(x^2-1)^2}=\frac{4(x^2-2x+1)}{((x-1)(x+1))^2}=\frac{4(x-1)^2}{(x-1)^2(x+1)^2}=\frac{4}{\color{red}{(x+1)^2}}$$
This is not $$\frac{4}{(x^2-1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving Chinese Remainder Theorem Algebraically I am doing a practice problem for my final which asks:
Solve the following Chinese Remainder Theorem:
$$
x \equiv 2 \pmod{3}, \\
x \equiv 3 \pmod{5}, \\
x \equiv 5 \pmod{7}, \\
x \equiv 7 \pmod{11} \\
x \equiv 11 \pmod{13}
$$
From the first I can conclude that $x = 3k + ... | \begin{align}
x &\equiv 2 \pmod 3 \\
x &= 2 + 3a \\
\hline
x &\equiv 3 \pmod 5 \\
2+3a &\equiv 3 \pmod 5 \\
3a &\equiv 1 \pmod 5 \\
a &\equiv 2 \pmod 5 \\
a &= 2 + 5b \\
x &= 2 + 3(2 + 5b)\\
x &= 8 + 15b \\
\hline
x &\equiv 5 \pmod 7 \\
8 + 15b &\equiv 5 \pmod 7 \\
1 + b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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find all primes $p$ and $q$ such that $p \cdot q | 2^p + 2^q$
I have to find all prime numbers $p,q$ such that $p\cdot q | 2^p + 2^q$.
I don't know from what I have to start.
| If $p=q$, then, $p^2|2^{p+1}$. So, $p=q=2$.
Now,suppose $p\ne q$.
If $p=2$, we have $q|2+2^{q-1}$. By Fermat's Little Theorem(FLT), $q=3$. Similarly, $q=2,p=3$ is another solution.
Now, suppose $p$ and $q$ are odd primes. Let $p-1=2^ms$ and $q-1=2^nr$, where $r$ and $s$ odd numbers.
$2^q\equiv -2^p\equiv -2\pmod p$. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort ... | $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
| If you know your Pascal's triangle, you know that
$$\binom{20}{1} + \binom{20}{3} +..+ \binom{20}{19}= \sum_{k=0}^{19} \binom{19}{k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Find the $k$ such that $2^{(k-1)n+1}$ does not divide $\frac{(kn)!}{n!}$. Find all positive integers $k$ such that for any positive integer $n$, $2^{(k-1)n+1}$ does not divide $\frac{(kn)!}{n!}$.
From olympiad problem
I'm curious So far no one to solve this problem,Maybe it is very difficult May be other reasons? But I... | The number of factors $2$ in a number $a!$ is:
$$
\left\lfloor \frac a 2 \right\rfloor
+
\left\lfloor \frac a 4 \right\rfloor
+
\left\lfloor \frac a 8 \right\rfloor
+ \cdots
$$
So we have to find all $k$ such that for every $n$, we have
$$
kn - n + 1 >
\sum_{x=1} \left\lfloor \frac{kn}{2^x} \right\rfloor
-
\sum_{x=1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove that $a_1+\frac{a_2^2}{a_1+a_2}+\frac{a_3^2}{a_1+a_2+a_3}>b_1+\frac{b_2^2}{b_1+b_2}+\frac{b_3^2}{b_1+b_2+b_3}.$ Suppose $a_1>a_2>a_3>0$ and $b_1>b_2>b_3>0$ and $a_1>b_1,a_2>b_2,a_3>b_3$.
I want to prove that
$$a_1+\frac{a_2^2}{a_1+a_2}+\frac{a_3^2}{a_1+a_2+a_3}>b_1+\frac{b_2^2}{b_1+b_2}+\frac{b_3^2}{b_1+b_2+b_3}... | Let $f(a,b,c)=a+\dfrac{b^2}{a+b}+\dfrac{c^2}{a+b+c}$.
Suppose $a_1 \geq a_2 \geq a_3$ and $b_1 \geq b_2 \geq b_3$ and $a_1 \geq b_1,a_2 \geq b_2,a_3 \geq b_3$.
I prove $f(a_1,a_2,a_3) \geq f(a_1,a_2,b_3) \geq f(a_1,b_2,b_3) \geq f(b_1,b_2,b_3)$.
The first inequality follows from the monotonicity of $\dfrac{x^2}{r+x}$ f... | {
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"url": "https://math.stackexchange.com/questions/1824457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
It is in the exercises of the AM-GM inequality chapter of a book, and that is why I believe it will be solved by that. Can anyone give me a proof ... | We use AM-GM three times
$$\frac{a^2 +a^2+a^2+a^2 +b^2+c^2}{6} \geq a (abc)^{1/3}$$
$$\frac{b^2 +b^2+b^2+b^2 +a^2+c^2}{6} \geq b (abc)^{1/3}$$
$$\frac{c^2 +c^2+c^2+c^2 +b^2+a^2}{6} \geq c (abc)^{1/3}$$
Summing these inequalities and dividing by 6 gives
$$a^2+b^2+c^2 \geq (a+b+c)(abc)^{1/3}$$
Now using that $abc=1$, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Restricted equality involving prime numbers Given three real numbers such that $a + b + c = 0$, it can be proved that
\begin{align*}
\frac{a^{5} + b^{5} + c^{5}}{5} & = \frac{a^{3} + b^{3} + c^{3}}{3}\cdot \frac{a^{2} + b^{2} + c^{2}}{2}\\
\frac{a^{7} + b^{7} + c^{7}}{7} & = \frac{a^{5} + b^{5} + c^{5}}{5}\cdot \frac{a... | $p=3,5$ are the only possible solutions.
To check this, substitute $a=2, \ b,c=-1$.
Then
$\displaystyle \frac{2^{p+2}-2}{p+2}=3\frac{2^p-2}{p}$.
This equation can (after some effort) be rewritten as $2^p(p-6)=-4p-12$.
The left hand side can only be negative if $p<6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Area of the triangle formed by circumcenter, incenter and orthocenter Lets say we have $\triangle$$ABC$ having $O,I,H$ as its circumcenter, incenter and orthocenter. How can I go on finding the area of the $\triangle$$HOI$.
I thought of doing the question using the distance (length) between $HO$,$HI$ and $OI$ and then ... | A área do triângulo HOI do triângulo ABC de lados a=|BC|, b=|AC| e c=|AB| é dada por:
$AreaHOI=|\frac{1}{4}.\frac{\left|\begin{array}{} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \\ \end{array} \right| }{\left|\begin{array}{} 0 & a^2 & b^2 &1\\ a^2 & 0 & c^2&1 \\ b^2 & c^2 & 0&1 \\1&1&1&0 \end{array} \right| }|$
Deduç... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Evaluate $\int \frac{1}{3+4\tan x} \, dx$ I'm am trying to evaluate $$\int \frac{1}{3+4\tan x} \, dx.$$
I tried to use the universal trig substitution, that is
$$t = \tan{x \over 2}, \quad \tan x = {2t \over 1-t^2}, \quad dx = {2 \over t^2 + 1}dt$$
and after manipulation I got the integral to be
$$\int \frac{2(1 - t^... | Note that $$\frac{1}{3 + 4 \tan x} = \frac{\cos x}{3 \cos x + 4 \sin x} = f(x).$$
Next, observe that $$\log (3 \cos x + 4 \sin x) + C = \int \frac{4 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \, dx = \int g(x) \, dx,$$ via the obvious substitution $u = 3 \cos x + 4 \sin x$. Now the goal is to find a linear combination of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove that $\Gamma(-k+\frac{1}{2})=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}$. I was able to prove that
$$
\Gamma\left (k+\frac{1}{2} \right )=\frac{1\cdot 3\cdot 5\cdots(2k-1)}{2^k}\sqrt{\pi}.\tag{$k\geq 1$}$$
using the Legendre's duplication formula. But I can't do the same to
$$\Gamma\left ( -k+\frac... | If you are willing to accept that for $z \notin \Bbb Z$
$$\Gamma (z) \Gamma (1-z) = \frac \pi {\sin \pi z}$$
then letting $z = \frac 1 2 - n$ leads to
$$\Gamma \left( \frac 1 2 - n \right) \Gamma \left( \frac 1 2 + n \right) = \frac \pi {\sin \left( \frac \pi 2 - n \pi \right)} = \frac \pi {\cos n \pi} = (-1)^n \pi$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Solve Sum of Arccos I am working through a situation with trying to fit an equilateral triangle into a square, and I have boiled it down to the following equation:
$$\arccos\left(\frac{\frac{L}{2}+x}{S}\right) + \arccos\left(\frac{\frac{L}{2}-x}{S}\right) = \frac{2}{3}\pi\;\mathrm{rad}$$
I need to solve this equation f... | Notice that: $\cos(A+B)=\cos A \cos B - \sin A \sin B$
In your example $A=\arccos(\frac{\frac{L}{2}+x}{S})$, and $B=\arccos(\frac{\frac{L}{2}-x}{S})$.
The solution follows from the fact that $\cos(\arccos(x))=x$
A specific solution:
$\arccos\left(\frac{\frac{L}{2}+x}{S}\right) + \arccos\left(\frac{\frac{L}{2}-x}{S}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
... | Let's go through the fundamental ideas here from the beginning:
If you multiply a matrix $$A=\pmatrix{a_{11} & a_{12} & \cdots & a_{1m} \\ a_{21} & a_{22} & \cdots & a_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nm}}$$ by the vector $e_1 = \pmatrix{1 \\ 0 \\ \vdots \\ 0}$, what do you get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\f... | Easier method: use the inequalities
$$|a+b|\le|a|+|b|\quad\hbox{and}\quad |a+b|\ge|a|-|b|\ .$$
If
$$|x-1|<\frac1{10}$$
then
$$|x+1|=|(x-1)+2|\le|x-1|+2<\frac{21}{10}$$
and
$$|x+3|=|4+(x-1)|\ge4-|x-1|>\frac{39}{10}\ .$$
Therefore
$$\frac{|x^2-1|}{|x+3|}=|x-1|\frac{|x+1|}{|x+3|}<\frac1{10}\frac{21/10}{39/10}=\frac{21}{39... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Tips for integrating $\int \frac{dx}{1+\cos(x)}$ I tried the following
$$
\int \frac{dx}{1+\cos(x)}=\int \frac{1-\cos(x)}{1-\cos^2(x)}\,dx=
\int \frac{1-\cos(x)}{\sin^2(x)}\,dx\\
=\int \frac{1}{\sin^2(x)}\,dx-\int \frac{\cos(x)}{\sin^2(x)}\,dx=\int \csc^2x\,dx-\int \cot(x)\csc(x)\,dx
$$
Which I looked up and found to ... | Here is a method similar to C. Dubussy's solution except no substitution needs to be made.
Recalling $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$, the integral may be written as
\begin{align*}
\int \frac{dx}{1 + \cos x} &= \int \frac{dx}{\left (\cos^2 \frac{x}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Prove that $\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}$ using given condition If $$\tan(\frac{\pi}{4}+\frac{\theta}{2})=\tan^3(\frac{\pi}{4}+\frac{\alpha}{2})$$, then prove that $$\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}$$
I tried using the fact that
$\frac{\cos A}{1-\sin A... |
Notations and assumptions
$$
a=\tan{\theta\over2},\quad b=\tan{\alpha\over2},\quad p=\sin{\theta},\quad q=\sin\alpha
$$
Consequence of our notation
$$
{1+a\over1-a}=\left({1+b\over1-b}\right)^3\cdots\spadesuit
$$
An identity
$$
\sin z=2\sin{z\over2}\cos{z\over2}=2\tan{z\over2}\cos^2{z\over2}={2\tan{z\over2}\over\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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A solvable quintic with the root $x=(\sqrt[5]{p}+\sqrt[5]{q})^5$ - what are the other roots? I derived a two parameter quintic equation with the root:
$$x=(\sqrt[5]{p}+\sqrt[5]{q})^5,~~~~~p,q \in \mathbb{Q}$$
$$\color{blue}{x^5}-5(p+q)\color{blue}{x^4}+5(2p^2-121pq+2q^2)\color{blue}{x^3}-5(2p^3+381p^2q+381pq^2+2q^3)\... | I see what you are getting at. Since there are actually five solutions $\alpha$ to the equation,
$$\alpha^5 = p$$
then it makes sense that $p^{1/5}+q^{1/5} = z$ should be the root of a $25$th deg eqn, correct? This is essentially given by your equation slightly modified as,
Eq.1:
$$z^{25}-5(p+q)z^{20}+5(2p^2-121pq+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Eliminate the parameter given $x = \tan^{2}\theta$ and $y=\sec\theta$ $x = \tan^{2} (\theta)$ and $y = \sec (\theta)$
knowing that $\tan^{2} (\theta) = (\tan (\theta))^2 = \dfrac{\sin^{2}\theta}{\cos^{2}\theta}$
and that $\sec(\theta) = \dfrac{1}{\cos(\theta)}$ $\to$ $y=\dfrac{1}{\cos(\theta)}$
For $y$ we can get an e... | Recall the circular identity $$\sin^2 \theta + \cos^2 \theta = 1.$$ Dividing both sides by $\cos^2 \theta$ gives $$\tan^2 \theta + 1 = \sec^2 \theta.$$ Consequently, $$x + 1 = y^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int_0^\infty \frac{dx}{x^2+2ax+b}$ For $a^2<b$, is there an identity of evaluating the following integral?
$$\int_0^\infty \frac{dx}{x^2+2ax+b}$$
What about:
$$\int_0^\infty \frac{dx}{(x^2+2ax+b)^2}$$
My attempt is using partial fractions and completing the square, but I still failed to obtain a nice r... | For the first one:
$$(x+a)^2=x^2+2ax+a^2$$
So
$$(x+a)^2+b-a^2=x^2+2ax+b$$
$$=(b-a^2)\left(\frac{(x+a)^2}{b-a^2}+1 \right)$$
$$=(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}} \right)^2+1 \right)$$
For $b-a^2 \neq 0$. In which case, for evaluating the integral, we enforce the substitution $\tan {u}=\frac{x+a}{\sqrt{b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove the equation has infinitely many solutions
Let $a,b$ be coprime integers. Show that the equation $ax^2+by^2=z^3$ has an infinite set of solutions $(x,y,z)$ with $x,y,z \in \mathbb{Z}$ and $x,y$ mutually coprime (in each solution).
I thought of doing like $x = a^4$ and $y = b^4$, but then we get $a^9+b^9 = z^3$,... | $$ (A x^2 + B y^2)^3 = A (A x^3 - 3 B x y^2)^2 + B (3 A x^2 y - B y^3)^2 $$
We have
$$ U = (Ax^2 - 3By^2), \; \; \; V = (3Ax^2 - B^2). $$
To begin with, we certainly need the restrictions
$$ \gcd(A,B) = 1, \gcd(x,y) = 1, \gcd(A,y) = 1, \gcd(B,x) = 1. $$
Getting there: the prime $2.$ If $A$ is odd but $B$ even, we mu... | {
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"url": "https://math.stackexchange.com/questions/1848313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given MGF of X, find MGF of $ Y=X_1\dot\ X_2 \dot\ X_3$ Let $X_1$, $X_2$, $X_3$ be a random sample from a discrete distribution with probability funciton
$p(0)= 1/3$
$p(1) = 2/3$
Calculate moment generating function, $M(t)$, of $Y=$$X_1$$X_2$$X_3$
My Work
$M_x(t) = \frac{1}{3} + \frac{2}{3}e^t$
then $E[e^{t(X_1X_2X_3... | If $X_1\,,\, X_2$ and $X_3$ be independent , then
$$P(Y=0)=\left( \begin{matrix}
3 \\
1 \\
\end{matrix} \right){{\left( \frac{1}{3} \right)}^{1}}{{\left( \frac{2}{3} \right)}^{2}}+\left( \begin{matrix}
3 \\
2 \\
\end{matrix} \right){{\left( \frac{1}{3} \right)}^{2}}{{\left( \frac{2}{3} \right)}^{1}}+\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find this function, and what method to use? The function is $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}$ and they are asking us to find out what $f(-x)$ is?
| A general method for solving equations of the type:
$$f(g(x))=z(x)$$
Let:
$$u=x-\frac{1}{x}$$
Here we let $u$ be equal to $g(x)$.
$$ux=x^2-1$$
$$x^2-ux-1=0$$
Using the quadratic formula we have:
$$x=\frac{u \pm \sqrt{u^2+4}}{2}$$
Note in a way we found an inverse for $x-\frac{1}{x}$ even though it does not have an inve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ tha... | $x^2 + y^2$ is essentially the square of the distance of a point $(x, y)$ from the origin. Now, the question is asking us to find two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ which lie on the circle and are at a minimum and maximum distance from the origin.
It makes sense that these two points will lie on the line ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integral value of $n$ that makes $n^2+n+1$ a perfect square. Find all integers $n$ for which $n^2+n+1$ is a perfect square.
By hit and trial we get $n=-1,0$ but could someone suggest any genuine approach as how to approach this problem?
| We have: $n^2 + n + 1 = m^2 \implies n^2 + n + (1-m^2) = 0 \implies \triangle = 1^2 - 4(1)(1-m^2) = k^2 \implies 1-4(1-m^2) = k^2 \implies 4m^2 - k^2 = 3 \implies (2m+k)(2m-k) = 3\implies 2m+k = 3, 2m-k = 1$ or $2m+k = -1, 2m-k = -3$. Either case gives $n^2 + n = 0 \implies n = -1, 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Prove that $\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$ if $(a+b+c)^2(a^2+b^2+c^2)=27$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b+c)^2(a^2+b^2+c^2)=27$. Prove that:
$$\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$$
A big problem here around $(a,b,c)=(1.6185...,0.71686...,0.4926...)$.
... | There seems to be bugs in the segment after where I marked "***". Needed for check.
When I saw the form $\sqrt{a^2+3b^2}$ I thought of the absolute value of a complex number.
So let
$$u=a+\sqrt{3}bi$$
$$v=b+\sqrt{3}ci$$
$$w=c+\sqrt{3}ai$$
And now what you want to prove becomes
$$|u|+|v|+|w|\geq6$$
$$u+v+w=(1+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 1,
"answer_id": 0
} |
Statement regarding primes $ \le n$ Following is the statement I believe is true, but can't prove.
Let $n$ be a natural. Let the primes less than equal to $\sqrt{n}$ be $p_1,p_2,...,p_k$. Let $\alpha_i$ be the greatest natural number such that $p_i^{\alpha_i} \le n$. Out of these $k$ primes let us take $2 \le r \le k$... | Let $n = 124$, $p_1 = 2$, $p_2 = 5$. Then $\alpha_1 = 6$, since
$$2^6 = 64 \leq 124 < 2^7 = 128$$
and $\alpha_2 = 2$ since
$$5^2 = 25 \leq 124 < 5^3 = 125.$$
Now, $2^6 + 5^2 = 89$. Choose $b_1 = 2$ and $b_2 = 2$. Then
$$89 < 2^2\cdot 5^2 = 100 \leq 124 = n.$$
The answer below was for an earlier version of the question,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove: $\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$ Show that if $m$ and $n$ are integers with $0\leq m<n$ then $$\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$$
Attempts:
*
*$(-1)^{k}\binom{n}{k}$ is the coefficient of $x^{k}$ in the expansion of $(1-x)^{n}$
*And $\b... | For those who enjoy integrals here is another approach using the
Egorychev method as presented in many posts by @FelixMarin and also by
@MarkusScheuer.
Suppose we seek to verify that
$$\sum_{k=m+1}^n (-1)^k {n\choose k} {k-1\choose m} = (-1)^{m+1}.$$
First proof. Introduce
$${k-1\choose m} = {k-1\choose k-1-m} =
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
On the proof $\tan 70°-\tan 20° -2 \tan 40°=4\tan 10°$ I am currently studying in class 10 and I am unable to do this problem. $$\tan 70 ° -\tan 20° -2 \tan 40° =4\tan 10°$$ Can anybody please help me.
Thanks!
| \begin{align*}
\tan 70-\tan 20 & =\tan 70-\cot 70\\
& =\tan 70-\frac{1}{\tan 70}\\
& = \frac{\tan^2 70-1}{\tan 70}=-\frac{1-\tan^2 70}{\tan 70}\\
& = -\frac{2(1-\tan^2 70)}{2\tan 70}=-\frac{2}{\frac{2\tan 70}{1-\tan^2 70}}\\
& = \frac{-2}{\tan 140}= \frac{-2}{-\tan 40}= \frac{2}{\tan 40}=2\cot 40
\end{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
integrate $\int \frac{x^4-16}{x^3+4x^2+8x}dx$
$$\int \frac{x^4-16}{x^3+4x^2+8x}dx$$
So I first started with be dividing $p(x)$ with $q(x)$ and got:
$$\int x-4+\frac{8x^2+32x-16}{x^3+4x^2+8x}dx=\frac{x^2}{2}-4x+\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx$$
Using partial sum I have received:
$$\int \frac{8x^2+32x-16}{x^3+4... |
$\displaystyle \int \dfrac{1}{x^2+4x+8}dx$
How do I continue from here?
Observe that by writing
$$
x^2+4x+8=(x+2)^2+4
$$ and by making the change of variable
$$
t=2(x+2), \quad dx=\frac12dt, \quad x^2+4x+8=4(t^2+1)
$$ you are led to evaluate
$$
\int \dfrac{1}{x^2+4x+8}dx=\frac18\int\frac1{t^2+1}dt
$$ which is classic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int\frac{x+1}{\sqrt{1-x^2}} \; dx$ without using trigonometric substitution I want to solve:
$$\int\frac{x+1}{\sqrt{1-x^2}} \; dx$$
I know how to solve this using trigonometric substitution, but how can I solve the integral in an other way ?
| Given that the answer for $\int\frac1{\sqrt{1-x^2}}dx$ will be $\sin^{-1}x$ you can't hope to solve the integral without any clue about trig functions. However, there is an easy way to deduce the answer just knowing that sine (and cosine) are the solutions to the harmonic differential equation
$$
\frac{d^2 y}{dx^2} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 1
} |
inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ How can I prove the inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ ?
The derivative of $f(x):=\sqrt{\cos x}-\cos(\sin x)$ is very unpleasant, so the standard method is probably not be the right choice...
| The only way I can think about is Taylor expansions (tedious but doable) as Henning Makholm commented. $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ $$\sqrt{\cos(x)}=1-\frac{x^2}{4}-\frac{x^4}{96}-\frac{19 x^6}{5760}+O\left(x^8\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$ $$\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$$
$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\int \frac{(1- \cos2x)^2}{2.(1+\cos^2 2x)}{dx}$$
$$\frac{1}{2} \int \left[1-\frac{2 \cos2x}{1+\cos^22x}\right] dx$$
What should I do nex... | HINT:
$$\cos^22x=1-\sin^22x$$
Set $\sin2x=y$
Otherwise,
$$\dfrac{\sin^4x}{\sin^4x+\cos^4x}=\dfrac1{1+\cot^4x}$$
Let $\cot^2x=u\implies dx=-\dfrac{du}{1+u^2}$
Method$\#1:\dfrac2{(1+u^4)(1+u^2)}=\dfrac{1+u^4+1-u^4}{(1+u^4)(1+u^2)}=?$
Method$\#2:$ Writing $u^2=y$
$$\dfrac1{(1+y^2)(1+y)}=\dfrac{Ay+B}{1+y^2}+\dfrac C{1+y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Proving that $x^{16} > 5$ when given a polynomial of degree $15$. I am unable to prove the following
If $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x = 7$ prove that $x^{16} > 15$.
| Notice that the left hand is equal to
$(x^3-x)(x^{16}-1)/(x^4-1)$; hence the equation will be transfer into:
$(x^{16}-1)=7(x^4-1)/(x^3-x)=7(x^2+1)/x$; notice that $(x^2+1)/x$ is graeter than 2, hence the right hand is greater than 7*2, therefor $x^{16}>15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $n$ is a positive integer, then $(-2^n)^{-2} + (2^{-n})^2 = 2^{-2n+1}$ I'm not sure why $$(-2^n)^{-2} + (2^{-n})^2=2^{-2n+1}$$
I have been going over this equation for a while now, noticing, and have successfully got quite far in the equation, finding that
$$ (-2^n)^{-2} + (2^{-n})^2 \implies \frac{1}{(-2^n)^2} + \... | So, we've got $\dfrac{1}{-2^n \cdot -2^n} + \dfrac{1}{2^n \cdot 2^n} = \dfrac{1}{2^n \cdot 2^n} + \dfrac{1}{2^n \cdot 2^n} = \dfrac{2}{2^{2n}}$ since adding two of the same things up is the same as multiplying by $2$.
Alright, so you have $\frac{2}{2^{2n}}$. That's good work! Now use the property that $\frac{a}{b} = a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equa... | $$\sin x + \sin y = 1\Rightarrow 2\sin\frac{x+y}{2}\cos \frac{x-y}{2}=1\\\cos x+\cos y=0\Rightarrow 2\cos\frac{x+y}{2}\cos \frac{x-y}{2}=0$$
It follows $$ \cos \frac{x+y}{2}=0\text{ and } \cos \frac{x-y}{2}=0\iff $$ The first ($\cos \frac{x+y}{2}=0$) gives, from the first given equation,
$$\begin{cases}x+y=\pi\\x-y=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.... | Set
$$t=\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
$$t^2=6-2\sqrt5+2\sqrt{(6-2\sqrt5)(6+2\sqrt5)}+6+2\sqrt5$$
$$t^2=12+2\sqrt{36-20}=12+2(4)=20$$
$$t^2=20\implies t=2\sqrt5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Proof that $P(x)=x-\frac{1}3 x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has radius of convergence $1$
Proof that
$P(x)=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has
radius of convergence $1$
First of all, I need to convert this to a series:
$$\sum_{k=1}^\infty \frac{x^{2k-1}(-1)^k}{1-2k}$$
(I hope thi... | Suppose $\{a_n\}$ is the sequence of terms of this series. Then it is clear that $\lim \sup (|a_n|)^{1/n} = 1 $ so that the radius of convergence is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Which answer is correct? Finding the limit of a radical as $x$ approaches infinity. When I take $$\lim_{x \to -∞} \sqrt{x^2+7x}+x,$$ I multiply by the conjugate over the conjugate to get
$$\lim_{x \to -∞}\frac{7x}{\sqrt{x^2+7x}-x},$$ and multiply by either $\frac{\frac{1}{x}}{\frac{1}{x}}$ or $\frac{\frac{1}{-x}}{\frac... | $$\lim_{x \to -\infty} \sqrt{x^2+7x}+x=$$
$$=\lim_{x \to -\infty} |x|\sqrt{1+\frac{7}{x}}+x=$$
$$=\lim_{x \to -\infty} -x\sqrt{1+\frac{7}{x}}+x=$$
$$=\lim_{x \to -\infty} -x\Big(\sqrt{1+\frac{7}{x}}-1\Big)=$$
$$=\lim_{x \to -\infty} -x\Big(1+\frac{7}{2x}+O\Big(\frac{1}{x^2}\Big)-1\Big)=$$
$$=\lim_{x \to -\infty} -\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to ... | A posible way: Let $A,B,C$ a triangle and $AD$ the height. The problem is $AD=1-x$, $BD=x$, $CD=1+x$ and you want minimize $BA+AC$, but note that the area is fixed ($(1-x)(2x+1)/2$), then this sum is minime if the triangle is right in A.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
For which $a$, equation $4^x-a2^x-a+3=0$ has at least one solution.
Find all values of $a$ for which the equation $4^x-a2^x-a+3=0$ has at least one solution.
$\bf{My\; Try::}$ We can write it as $$2^{x}-a-\frac{a}{2^x}+\frac{3}{2^x}=0$$
So $\displaystyle \left(2^x+\frac{3}{2^x}\right)=a\left(1+\frac{1}{2^x}\right).$
... | Put $t = 2^x$. The equation is $t^2-at-a+3 = 0$. This quadratic should have a positive root. Hence $a^2 + 4(a-3) \geq 0$ which gives $(a+6)(a-2) \geq 0$. Thus $a \geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational... | Knowing that x=1 is a solution we know that $(x-1) $ is a factor so we force it to factor out.
$x^4-4x^3+6x^2-4x+1=$
$(x-1)x^3-3x^3 +6x^2-4x+1=$
$(x-1)x^3-(x-1)3x^2+3x^2-4x+1=$
$(x-1)x^3-(x-1)3x^2 +(x-1)3x -x +1=$
$(x-1)x^3-(x-1)3x^2+(x-1)3x-(x-1)=$
$(x-1)[x^3-3x^2+3x-1]=$
Now, we may not know that 1 is a double root y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur. The question is two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur.
My atte... | You already have
$$(m^2-4)x^2+(2mc)x^2+c^2+36=0$$
Solving this gives
$$x=\frac{-mc}{m^2-4}=\frac{-4m}{m^2-4}$$
So, for $m=\pm\frac{2\sqrt{13}}{3}$, the coordinates we want are
$$\color{red}{\left(\mp\frac{3}{2}\sqrt{13},-9\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$
How does one show that $$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$$ for each nonnegative integer $n$?
I tried using the Snake oil technique but I guess I am applying it incorrectly. With the snake oil technique we have $$F(x)= \sum_{n=0}^{\infty}\le... | Suppose we seek to verify that
$$\sum_{k=0}^n {n+k\choose k} \frac{1}{2^k} = 2^n.$$
In the following we make an effort to use a different set of integrals
from the answer by @MarkusScheuer, for variety's sake, even if this is
not the simplest answer.
The difficulty here lies in the fact that the binomial coefficients o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 15,
"answer_id": 13
} |
Multiplicate a number per an elevated parenthesis I have a little problem in calculating the second derivative of a function, because I don't know how to compute this operation:
$(2x)*(x+2)^2$
I have to calculate the $(x+2)^2$ in this way: $(x^2+4+4x)$ and then multiply every term per $(2x)$ or there is another and fas... | Use the product rule $(fg)' = f'g + g'f$ with $f = 2x$ and $g = (x+2)^2$ so you get $$(2x(x+2)^2)' = (2x)'(x+2)^2 + ((x+2)^2)'(2x)$$
giving $$2(x+2)^2 + 4x(x+2) = 2(x+2)(x+2+2x) = 2(x+2)(3x+2)$$ and then differentiating again with the product rule with $f = x+2$ and $g = 3x+2$ gives the derivative of $(x+2)(3x+2)$ as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying Ramanujan-type Nested Radicals Ramanujan found many awe-inspiring nested radicals, such as...
$$\sqrt{\frac {1+\sqrt[5]{4}}{\sqrt[5]{5}}}=\frac {\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{2}-1}{\sqrt[5]{125}}\tag{1}$$$$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\tag{2}$$$$\s... | Considering
$$(\sqrt[4]{a} \pm \sqrt[4]{b})^{4}=
a+b+6\sqrt{ab} \pm 4\sqrt[4]{ab}(\sqrt{a}+\sqrt{b})$$
Making a factor in the form of $\sqrt{m}+\sqrt{n}$,
$$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{a}}{\sqrt{b}}
\implies \frac{a}{b}=5$$
Then
$$\left( \frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1} \right)^{4}=
\frac{(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$
I needed to solve the following equation:
$$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first:... | You concluded from
$$\dfrac{(\tan\theta + \tan 2\theta)(2-\tan\theta\tan2\theta)}{1-\tan\theta\tan2\theta}=(\tan\theta\tan2\theta)\dfrac{\tan\theta + \tan 2\theta}{1-\tan\theta\tan2\theta}$$
(by “canceling” $\dfrac{\tan\theta + \tan 2\theta}{1-\tan\theta\tan2\theta}$ from both sides) that
$$2-\tan\theta\tan2\theta=\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the general solution for: $\frac{dy}{dx} + \frac{y}{x}=\frac{1}{x^2}$
Find the general solution for: $$\frac{dy}{dx} + \frac{y}{x}=\frac{1}{x^2} $$
I don't want to solve this using an integrating factor, I wanted to try solve this with a substitution $y=vx$
With the sub of $y=vx$ it implies that $dy = xdv+vdx$... | Try the natural substitution $y(x)=x^n f(x)$.
Note that
$y' = x^n f' + n x^{n-1}f$.
Therefore,
$x^n f' + n x^{n-1}f + x^{n-1}f = 1/x^2$
or
$$x^n f' + (n+1)x^{n-1}f = \frac{1}{x^2}.$$
This differential equation simplifies significantly if we let $n=-1$, which we are free to do.
Then
$f'/x = 1/x^2$
or
$$f' = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$? What is the limit of
$\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$?
I attempted the problem via L^Hopital's Rule so I rewrote it as
$$y=\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$$
then took the natural of both sides
$$\ln(y)=\ln(\frac{\cos... | Using Taylor series $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ So $$\cos(x)-1+\frac{x^2}{2}=\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ $$\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}=\frac{1}{24}-\frac{x^2}{720}+O\left(x^4\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ Calculate $\det(A)$ and $\det(A)^5$:
$$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$
But how can I determine $\det A^5$ easily/fast? I know tha... | If calculations are done mentally but not jotted down, I think the quickest way to calculate $\det A$ is to note that
\begin{align}
A&=a(e_1+e_2+e_3+e_4)(e_1+e_2+e_3+e_4)^T\\
&+(b-a)(e_2+e_3+e_4)(e_2+e_3+e_4)^T\\
&+(c-b)(e_3+e_4)(e_3+e_4)^T\\
&+(d-c)e_4e_4^T.
\end{align}
Therefore $A$ has the $LDL^T$ decomposition
$$
A... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $n$ is an integer then $4$ does not divide $n^2 - 3$ Having some trouble proving this. Was going to attack with using the contrapositive of this statement but can't seem to show that $n$ isn't an integer.
| You just need to look at two cases: $n$ is odd and $n$ is even.
*
*If $n$ is even, then $n \equiv 0 \textrm{ or } 2 \pmod 4$ but $n^2 \equiv 0 \pmod 4$ regardless, which means that $n^2 - 3 \equiv 1 \pmod 4$, and therefore $$\frac{n^2 - 3}{4} = m + \frac{1}{4},$$ where $m$ is some integer. For example, $n = 10$ give... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to prove that $\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$? As stated in the question. Thank you!
$$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$
| Base case: Assume $a=b=0$. Then $\frac{0^00^0}{0!0!}\le\frac{(0+0)^{(0+0)}}{(0+0)!}$ goes to $1 \le 1$, which is true.
Inductive case: Suppose $a=b$. Increase $b$ to equal $a+k$ for positive $k$. Then $\frac{a^a(a+k)^{a+k}}{a!(a+k)!}\le\frac{(a+a+k)^{(a+a+k)}}{(a+a+k)!}$ rearranges to $\frac{(2a+k)!}{a!(a+k)!} \le \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 ... | $x=32+81k\equiv 32+17k \equiv 59 \pmod {64}$
$k\equiv \frac{27}{17}\equiv\frac{27}{81}\equiv \frac{1}{3}\equiv \frac{129}{3}\equiv43 \pmod {64}$
$x=32+81(43+64k')=3515+5184k'$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Composition relation of R1 ∘ R2
Let $R_1$ and $R_2$ be the relations on $\{1, 2, 3, 4, 5\}$ defined by
$$R_1 = \{(1,1),(2,3),(2,4),(3,5),(5,2),(5,5)\}$$
$$R_2 = \{(1,1),(2,2),(2,3),(2,5),(4,3),(5,5)\}$$
The answer for this is below but I'm not sure how they arrived at this answer.
Answer:
$$R_2 \circ R_1 = \{(1,1),(... | An alternative is through matrix representations of relations ($a_{ij}=1$ if $(i,j)$ is present in the relation, $0$ otherwise) with composition of relations replaced by matrix product (in the same order as in the composition, with boolean addition convention: $1+1=1$). This can be very useful on a computer. Here, we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions?
Question Find the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions
My attempt:
If we try to... | The equation $y^2=4(x-2)$ implies that $x\geq2$. That means the curve defined by $y^2=4(x-2)$ (a parabola) has no point for $x<2$. So, if you are looking for its intersection with another curve (here $x^2+y^2=k$) you should look over $x>2$.
For example, if the goal is having no intersection you should find $k$ such tha... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$ Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$
My attempt:I used two ways but I get to a wrong answer.
My first way:We know that $\frac{a}{b}+\frac{b}{a} \ge 2$ wh... | $\sum\limits_{cyc}\frac{a}{b+c}=\sum\limits_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+s)^2}{\sum\limits_{cyc}(ab+ac)}\geq2$
Because the last inequality it's $(a-c)^2+(b-d)^2\geq0$. Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.
| For the curve $ax^2+2hxy+by^2+2gx+2fy+c = 0$, the center is given by
\begin{align*}
ax+hy+g &= 0\\
hx+by+f &= 0
\end{align*}
and the equation of the ellipse with the axes through the center parallel to the axes is
\begin{align*}
ax^2+2hxy+by^2 + c' = 0
\end{align*}
where $c' = gx_c + fy_c + c$, where $(x_c, y_c)$ is ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$ I'm reading a introductory book on mathematical proofs and I am stuck on a question.
Let $a, b, c, d$ be positive real numbers, prove that if $\frac{a}{b} < \frac{c}{d}$, then $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$.
| HINT: Note that
$$\begin{align*}
\frac{a+c}{b+d}-\frac{a}b&=\frac{a+c}{b+d}\cdot\frac{b}b-\frac{a}b\cdot\frac{b+d}{b+d}\\
&=\frac{b(a+c)-a(b+d)}{b(b+d)}\\
&=\frac{ab+bc-ab-ad}{b(b+d)}\\
&=\frac{bc-ad}{b(b+d)}\;.
\end{align*}$$
You’d like to show that this difference is positive, so you’d like to know that $bc>ad$. Use ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Complex number fraction Find the real and imaginary number:
$$\frac{1}{z}=\frac{2}{2+j3}+\frac{1}{3-j2}$$
How do I invert $\frac{1}{z}$ to $z$ so that I can start solving it?
|
Solve a more general way, assume $q\in\mathbb{C}$:
$$q=\Re[z]+\Im[z]i$$
So, we get:
$$\frac{1}{q}=\frac{\overline{q}}{q\overline{q}}=\frac{\overline{q}}{|q|^2}=\frac{\Re[q]-\Im[q]i}{\Re^2[q]+\Im^2[q]}=\frac{\Re[q]}{\Re^2[q]+\Im^2[q]}-\frac{\Im[q]}{\Re^2[q]+\Im^2[q]}\cdot i$$
Now, in your case:
$$\frac{2}{2+3i}+\frac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof verification: $a_1=2$, $a_{n+1}=3+\frac{a_n}{2}$ is increasing and bounded For base case, $a_2=3+\frac{a_1}{2}=4>a_1$
For $n=k$, let $a_{k+1}>a_k$
Adding 3 on both sides after dividing by 2,
$$3+\frac{a_{k+1}}{2}>3+\frac{a_k}{2}$$
$$a_{k+2}>a_{k+1}$$
Hence the sequence is increasing.
How can I show that the seque... | $$\begin{align}
a_{n+1}&=3+\frac {a_n}2\\
a_{n+1}-6&=\frac 12 \left(a_n-6\right)\\
b_n&=\frac 12 b_{n-1}=\cdots =\frac 1{2^{n-1}}b_1&&(b_n=a_n-6)\\
a_n-6&=\frac 1{2^{n-1}}(\underbrace{\;a_1\;}_{=2}-6)\\
\color{red}{a_n}&\color{red}{=6-\frac 8{2^n}}
\end{align}$$
which is increasing and bounded (at $6$) as $n\to \infty$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Neglecting Terms Based On Their Relevant Magnitude If we have an expression like:
$$\frac{1}{l}[ \frac{1}{a} - \frac{1}{a+l}]$$
and we know that $l<<a$ then we can convert the above expression to
$$ \frac{1}{a(a+l)}$$
and then neglect the l from the denominator. So we end up with:
$$ \frac{1}{a^2} $$
But this is a ve... | Hint. One may write, by the Taylor series expansion, as $\dfrac{l}{a} \to 0$,
$$
\frac{1}{l}\left[ \frac{1}{a} - \frac{1}{a+l}\right]=\frac{1}{a(a+l)}=\frac1{a^2}\cdot\frac{1}{1+\dfrac{l}{a}}=\frac1{a^2}\left[1-\frac{l}{a}+O\left(\frac{l^2}{a^2}\right)\right]=\frac1{a^2}+O\left(\frac{l}{a^3}\right). \tag1
$$ Similarly,... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the remainder when the product of the primes between 1 and 100 is divided by 16? The product of all the prime numbers between 1 and 100 is equal to $P$. What is the remainder when $P$ is divided by 16?
I have no idea how to solve this, any answers?
| Skip the first prime $2$ and look for the product modulo $8$.
The twenty-four odd primes $<100$ are
$$3,5,7,11,13,17, 19,23,29,31,37,41, 43,47,53,59,61,67, 71,73,79,83,89,97. $$
Modulo $8$, these are
$$3,5,7,3,5,1,3,7,5,7,5,1,3,7,5,3,5,3,7,1,7,3,1,1$$
so their product is $$1^53^75^67^6\equiv 3\pmod 8 $$
(where we mig... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
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Find all pairs $(a,b)$ of positive integers, such that $\frac{b^2+ab+a+b-1}{a^2+ab+1}$ is integer.
Find all positive integers $a$ and $b$ such that $$\frac{b^2+ab+a+b-1}{a^2+ab+1}$$ is integer.
My work so far:
If $a=1, b\in\mathbb N$ then
$$\frac{b^2+ab+a+b-1}{a^2+ab+1}=\frac{b^2+2b}{b+2}=b \in \mathbb Z$$
| You have $a(a+b)+1=a^2+ab+1$ divides $$\left(b^2+ab+a+b-1\right)+\left(a^2+ab+1\right)=(a+b)(a+b+1)\,.$$ Because $\gcd\big(a(a+b)+1,a+b\big)=1$, we must have $a(a+b)+1\mid a+b+1$. The rest should be easy.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$ If $\frac{x}{y}$ + $\frac{y}{x}$ = 3 Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$
Any Ideas on how to begin ?
| Hint: Look at:
$$\left(\frac{x}{y} + \frac{y}{x}\right)^2=3^2=9$$
Why to look at this? We know if we expand we will have a $(\frac{x}{y})^2=\frac{x^2}{y^2}$ term and a $(\frac{y}{x})^2=\frac{y^2}{x^2}$ term so this might be worth a try.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a,b,c$ are in Arithmetic Progression, find the other two vertices of the triangle. A triangle $ABC$ is given where vertex $A$ is $(1,1)$ and the orthocentre is $(2,4)$. Also sides $AB$ and $BC$ are members of the family of the lines $ax+by + c = 0$, where $a,b,c$ are in Arithmetic Progression..Find the other two ve... | The arithmetic progression condition amounts to $a+c=2b$.
Now, consider $AB$. Because $A$ is on this line, we have $a+b+c=0$, and as you deduced, we get $ax-a=0\implies x=1$. Thus, the $x$-coordinate of point $B$ must be 1 as well i.e. $B=(1,m)$.
Then, consider point $C$. Because $AB$ is on the line $x=1$, and the orth... | {
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"timestamp": "2023-03-29T00:00:00",
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Maxmum distance from a point to a sphere Which point of the sphere $x^{2} + y^{2} + z^{2} = 1$ is at the maximum distance from the point $\left(\, 2,1,3\, \right)$ ?.
I know that the point is outside the sphere. Should we proceed with maxima and minima concept ?.
| WLOG any point on the sphere $(\cos u,\sin u\cos t,\sin u\sin t)$
$$D^2=(\cos u-2)^2+(\sin u\cos t-1)^2+(\sin u\sin t-3)^2$$
$$2^2+1^2+3^2+\cos^2u+\sin^2u(\cos^2t+\sin^2t)-4\cos u-2\sin u(\cos t+3\sin t)$$
$$=2^2+1^2+3^2+\cos^2u+\sin^2u-4\cos u-2\sin u(\cos t+3\sin t)$$
$\cos t+3\sin t=\sqrt{1^2+3^2}\cos\left(t-\arccos... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\left|\sqrt2-(a/b)\right|\geq1/(3b^2)$ This is Problem 4.26 on p.58 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition).
Prove that $\left|\sqrt2-\dfrac ab\right|\geq\dfrac1{3b^2}$ for all positive integers $a,b$.
Here's what I've done so far. Split into cases:
*
*(i)... |
Prove that $$\left|\sqrt2-\dfrac ab\right|>\dfrac1{3b^2}$$ for integers $a$ and $b$ with $b\neq 0$.
Without loss of generality, assume that $b>0$. Note that
$$\left|\sqrt{2}-\frac{a}{b}\right|=\frac{\left|a^2-2b^2\right|}{b(a+\sqrt{2}b)}\,.\tag{*}$$
If $a<0$, then
$$\left|\sqrt{2}-\frac{a}{b}\right|>\sqrt{2}>\frac{1... | {
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"url": "https://math.stackexchange.com/questions/1899973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
By really long division i got :-
$$Q=ax^{15} + (a+b)x^{14} + \dots +(610a + 377b) $$
$$R = x(987a+610b)+1+610a+377b$$
since remainder is $0 $,
$$987a+610b = 0$$
$$1+610a+377b =... | Still another approach (and a quite compact one). Since
$$ \frac{1}{1-x-x^2}=\sum_{n\geq 0}F_{n+1} x^n \tag{1}$$
it is not difficult to write down the Taylor series of $\frac{1+bx^{16}+a x^{17}}{1+x-x^2} $:
$$\frac{1+bx^{16}+a x^{17}}{1+x-x^2}=\sum_{n\geq 0}(-1)^n\left(F_{n+1} x^n + bF_{n+1} x^{n+16} + a F_{n+1} x^{n+1... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
Find value of x, where $\frac{3+\cot 80^{\circ} \cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$ $$\frac{3+\cot 80^{\circ}\cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$$ Then find $x$
My Try:
Using $\cot 80=\tan 10$ and $\cot 20=\frac{1}{\tan 20}$
we have LHS as $$\frac{3+\frac{\tan ... | Let $\alpha = 20^\circ, \,\, u = \tan(\alpha)$, then
$$ LHS = \frac{1 + 3\tan(80^\circ)\tan(20^\circ)}{\tan(80^\circ) + \tan(20^\circ)} = \frac{1 + 3\tan(\alpha + 60^\circ)\tan(\alpha)}{\tan(\alpha + 60^\circ) + \tan(\alpha)} .$$
notice that
$$ \tan(\alpha + 60^\circ) = \frac{\sqrt{3} + \tan{\alpha}}{1 - \sqrt{3} \tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find a law, and prove it by induction. From the following equalities, I'm supposed to find a law and prove it by induction.
$1=1$; $1-4=-(1+2)$; $1-4+9=1+2+3$, and $1-4+9-16=-(1+2+3+4)$.
I supposed the law was $\sum^{n}_{k=1} (-1)^{k+1}k^2=(-1)^{n+1}\sum^n_{k=1} k$.
Let's say it's valid for a $n\in \mathbb{N}$. For $n... | although this is not an inductive proof it is worth noting, in passing, that this rather pretty result follows from the fact that the difference of the squares of two consecutive integers is equal to their sum. thus, for example:
$$
9^2-8^2 = 9+8\\
7^2-6^2 = 7+6\\
5^2-4^2 = 5+4\\
3^2-2^2= 3+2\\
1^2-0^2= 1+0
$$
adding t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\sum\limits_{k=0}^n(-1)^{n+k}{n\choose k}{ {n+k}\choose n} \frac{1}{k+2}$ I am trying to evaluate the following sum :
$S=\displaystyle\sum_{k=0}^n(-1)^{n+k}{n\choose k} {{n+k}\choose n} \frac{1}{k+2}$
which is the same as
$\displaystyle\sum_{k=0}^n(-1)^{n+k}{n\choose k} {{n+k}\choose k} \frac{1}{k+2}$
My app... | $$
\begin{align}
&\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n+k}{n}\frac1{k+2}\\[6pt]
&=\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n+k}{k}\left(\frac1{k+1}-\frac1{(k+1)(k+2)}\right)\tag{1}\\[6pt]
&=\frac1{n+1}\sum_{k=0}^n(-1)^{n+k}\binom{n+1}{k+1}\binom{n+k}{k}\\
&-\frac1{(n+1)(n+2)}\sum_{k=0}^n(-1)^{n+k}\binom{n+2}{k+2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
cardinality of cartesian product of the infinite set of natural numbers One of the problems in my discrete math course states that we need to prove that $\mathcal{N}\times\mathcal{N}$ is countable specifically when there's a function $f:\mathcal{N}\times\mathcal{N}\to\mathcal{N}$ defined as follows:
$$f(a,b)=\frac{1}{2... |
How is the transition achieved?
$$f(a,b+1)=\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1}\color{blue}{+1})(a+b+1)+a=$$
$$=\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1})(a+b+1) + \color{red}{\dfrac{1}{2}}(\color{blue}{+1})(a+b+1)+a$$
$$=\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1})(a+b\color{orange}{+1}) + \colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality using induction Prove $3^n \geq 2n^2 +1$ for $n = 1,2,\ldots$ using induction
This is what I have so far
Base case - $n=1,3^1 \geq 2^1 + 1 = 3$ true
Induction step - Assume true for some n, then,
$ 3*3^n \geq 3*(2n^2 +1)=6n^2 +3 $
I have to somehow manipulate and show its $\geq 2*(n+1)^2 +1$
| First, show that this is true for $n=1$:
$3^1\geq2\cdot1^2+1$
Second, assume that this is true for $n$:
$3^n\geq2n^2+1$
Third, prove that this is true for $n+1$:
$3^{n+1}=$
$3\cdot\color\red{3^n}\geq$
$3\cdot(\color\red{2n^2+1})=$
$6n^2+3=$
$2n^2+4n^2+3\geq$
$2n^2+4n+3=$
$2n^2+4n+2+1=$
$2(n^2+2n+1)+1=$
$2(n+1)^2+1$
Pl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Contradictory result when testing Linear independence using Gaussian elimination Consider a set of vectors - (2,3,1) , (1,-1,2) and (7,3,8).
I want to find if its linearly dependent or independent.
Putting it as:
\begin{equation}
2a + b + 7c = 0 \\
3a - b + 3c = 0 \\
a + 2b + 8c = 0
\end{equation}
If I use Gaussian ... | There's no contradiction: you reduced echelon form is wrong. Let me do it again:
\begin{align*}
&\begin{bmatrix}
2&1&7\\
3&-1&3\\
1&2&8
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&2&8\\
2&1&7\\
3&-1&3
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&2&8\\
0&-3&-9\\
0&-7&-21
\end{bmatrix}
\rightsquigarrow
\begin{bm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $f^{(100)}(x)$ where $f(x)=\frac{1}{4x^2-1}$ Find $f^{(100)}(x)$ where $f(x)=\frac{1}{4x^2-1}$. I found first,second and third derivative:
$$f'(x)=\frac{-8x}{(4x^2-1)^2}$$
$$f''(x)=\frac{96x^2+8}{(4x^2-1)^3} $$
$$f'''(x)=\frac{-384x(4x^2+1)}{(4x^2-1)^4}$$
I can't seem to find any rule between them. Anyone has any... | $$
f(x) = \frac{1}{2}\left(\frac{1}{2x-1} - \frac{1}{2x+1}\right)
$$
Now if $g(x) = \frac{1}{2x-1} = (2x-1)^{-1}$, then check (by induction) that
$$
g^{(n)}(x) = (-2)^nn!(2x-1)^{-(n+1)}
$$
and if $h(x)=(2x+1)^{-1}$, then
$$
h^{(n)}(x) = (-2)^nn!(2x+1)^{-(n+1)}
$$
Hence,
$$
f^{(n)}(x) = \frac{(-2)^nn!}{2} \left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $[2^n \sqrt{2}],[2^{n+1} \sqrt{2}],\ldots, [2^{2n} \sqrt{2}]$ contains at least one even numbers for every integer $n\ge 1$? I tried as follows:
If not, denote $x=\{2^{n-1}\sqrt{2}\}$, then
$$1-\frac{1}{2^{n+1}}<x<1.$$
Denote $y=[2^{n-1}\sqrt{2}+1]$ and assume $|\sqrt{2}-p/q|<1/q^2$, then
$$\fra... | By a basic trick of Diophantine approximation we have, for all positive integers $p,q$, the inequality
$$
\left|\frac pq-\sqrt2\right|\cdot\left|\frac pq+\sqrt2\right|=\left|\frac{p^2}{q^2}-2\right|\ge\frac1{q^2}.
$$
In particular, if $3/2>p/q>\sqrt2$, we get the estimate
$$
\left|\frac pq-\sqrt2\right|\ge\frac1{3q^2}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
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Fractions in Questions and Answers
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