Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Express $x$ in terms of $a$ and $b$: $\sin^{-1} {\frac{2a}{1+a^2}} + \sin^{-1}{\frac{2b}{1+b^2}} = 2\tan^{-1}x$
Find the value of $x$ from the following equation in terms of $a$ and $b$
$$\sin^{-1} {2a\over{1+a^2}} + \sin^{-1}{2b\over{1+b^2}} = 2\tan^{-1}x$$
I tried to expand the LHS using the formula $$\sin^{-1}c+\sin^{-1}d = \sin^{-1}\left(c\sqrt{1+d^2} + d\sqrt{1+c^2}\right)$$
But it didn't work out. Could anyone help me out?
|
Let $\arctan a=u,a=\tan u$
If $-\dfrac\pi2\le2u\le\dfrac\pi2\iff -1\le a\le1$ $$P=\arcsin\dfrac{2a}{1+a^2}=2\arctan a$$
If $2u>\dfrac\pi2,P=\pi-2\arctan a$
If $2u<-\dfrac\pi2,P=-\pi-2\arctan a$
Now use my answer in showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2909590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Range of $y = \frac{x^2-2x+5}{x^2+2x+5}$? How do I approach this problem? My book gives answer as $[\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}]$. I tried forming an equation in $y$ and putting discriminant greater than or equal to zero but it didn't work. Would someone please help me?
I get $x^2 (y-1) + 2x (y+1) + (5y-5) =0$ and discriminant gives $2y^2 - y + 2 \leq 0$, which has complex roots.
|
Hint:
The derivative of $\dfrac{x^2-2x+5}{x^2+2x+5}$ is
$\dfrac{4 (x^2 - 5)}{x^2+2x+5}$ and so the critical points are $\pm \sqrt 5$.
Consider also $\displaystyle\lim_{x\to\pm\infty}\dfrac{x^2-2x+5}{x^2+2x+5}=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2912043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Evaluate $\lim_{x\rightarrow0}{\frac{(x-\arctan(x))\ln(1+2\sin(x))}{(1+\cos{x})(e^x-1-x)^2}}$ using Taylor I want to evaluate the following limit:
$$\lim_{x\rightarrow0}{\frac{(x-\arctan(x))\ln(1+2\sin(x))}{(1+\cos{x})(e^x-1-x)^2}}$$
For example, we have $x-\arctan{x}$. They are both $0$. This seems to be the so called "cancellation of terms". Therefore I apply the Taylor series:
$$\arctan{x}=x-\frac{x^3}{3}+o(x^3)$$
Therefore
$$x-\arctan{x}\sim x-x +\frac{x^3}{3}+o(x^3)=\frac{x^3}{3}+o(x^3)$$
Somebody solved this problem textbook and developed this to the fifth term (not third). Now a few questions arise:
*
*Why do I need to develop this Taylor series to the fifth term (assuming I have to).
*Do I need to develop every function to the same term in my limit?
Any hints?
|
For the direct questions: One must, needs to, determine powers of the expansion beyond constant factors to determine the behavior of the expansions; Not every function has the same powers in an expansion. The result then becomes "at least get a few powers of $x$ of each function".
This becomes more familiar by the example in question. Consider each of the four functions first:
\begin{align}
x - \tan^{-1}(x) &= \frac{x^3}{3} - \frac{x^5}{5} + \frac{x^7}{7} + \mathcal{O}(x^9) \\
\ln(1 + 2 \, \sin(x)) &= 2 x - 2 x^2 + \frac{7 x^3}{3} - \frac{10 x^4}{3} + \mathcal{O}(x^5) \\
1 + \cos(x) &= 2 \cos^{2}\left(\frac{x}{2}\right) = 2 - \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^6) \\
(e^{x} - 1 - x)^{2} &= \frac{x^4}{4} + \frac{x^5}{6} + \frac{5 x^6}{72} + \mathcal{O}(x^7).
\end{align}
With these expansions it is seen that the growth of powers of $x$ are not the same, but at least each expansion has powers of $x^4$ or greater.
Now, for the limit.
\begin{align}
F(x) &= \frac{(x - \tan^{-1}(x) ) \, \ln(1 + 2 \, \sin(x))}{2 \, \cos^{2}\left(\frac{x}{2}\right) \, (e^{x} - 1 - x)^{2}} \\
&= \frac{\left(\frac{x^3}{3} - \frac{x^5}{5} + \frac{x^7}{7} + \cdots\right) \left( 2 x - 2 x^2 + \frac{7 x^3}{3} - \frac{10 x^4}{3} + \cdots\right)}{\left(2 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \right) \left(\frac{x^4}{4} + \frac{x^5}{6} + \frac{5 x^6}{72} + \cdots \right)} \\
&= \frac{\frac{2 x^4}{3} - \frac{2 x^5}{3} + \frac{17 x^6}{45} + \cdots}{\frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{72} + \cdots} \\
&= \frac{\frac{2}{3} - \frac{2 x}{3} + \frac{17 x^2}{45} + \cdots}{\frac{1}{2} + \frac{x}{6} + \frac{x^2}{72} + \cdots} \\
&= \frac{4}{3} \, \left( 1 - \frac{4 x}{3} + \frac{49 x^2}{36} + \cdots \right)
\end{align}
and
\begin{align}
\lim_{x \to 0} F(x) &= \lim_{x \to 0} \frac{4}{3} \, \left( 1 - \frac{4 x}{3} + \frac{49 x^2}{36} + \mathcal{O}(x^3) \right) \\
&= \frac{4}{3}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2912191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Find this integral $I=\int\frac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$ Find the integral
$$I=\int\dfrac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$$
My try: $$(1-3x)(x+1)=-3x^2-2x+1=-3(x+1/3)^2+\dfrac{4}{3}$$
Thus
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$
|
Substitute $u=\sqrt{1-3x}$ thus $\mathrm{d}x=-\dfrac{2\sqrt{1-3x}}{3}\,\mathrm{d}u$
$$I={\displaystyle\int}\dfrac{2\left(u^2-1\right)}{\sqrt{3}u^2\left(4-u^2\right)^\frac{3}{2}}\,\mathrm{d}u$$
$$I=\dfrac{2}{\sqrt{3}}{\displaystyle\int}\left(\dfrac{1}{\left(4-u^2\right)^\frac{3}{2}}-\dfrac{1}{u^2\left(4-u^2\right)^\frac{3}{2}}\right)\mathrm{d}u$$
$$I=\dfrac{2}{\sqrt{3}}{\displaystyle\int}\dfrac{1}{\left(4-u^2\right)^\frac{3}{2}}\,\mathrm{d}u-\dfrac{2}{\sqrt{3}}{\displaystyle\int}\dfrac{1}{u^2\left(4-u^2\right)^\frac{3}{2}}\,\mathrm{d}u$$
Now perform trigonometric substitution $u=2\sin v$ to solve other two integrals
$$I=\dfrac{2}{\sqrt{3}}{\displaystyle\int}\dfrac{2\cos\left(v\right)}{\left(4-4\sin^2\left(v\right)\right)^\frac{3}{2}}\,\mathrm{d}v-\dfrac{2}{\sqrt{3}}{\displaystyle\int}\dfrac{\cos\left(v\right)}{2\sin^2\left(v\right)\left(4-4\sin^2\left(v\right)\right)^\frac{3}{2}}\,\mathrm{d}v$$
$$I=\dfrac{1}{2\sqrt{3}}{\displaystyle\int}\dfrac{1}{\cos^2\left(v\right)}\,\mathrm{d}v-\dfrac{1}{8\sqrt{3}}{\displaystyle\int}\dfrac{1}{\cos^2\left(v\right)\sin^2\left(v\right)}\,\mathrm{d}v$$
can you solve it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2915564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Loney: If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + p = 0$, then $\tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = n\pi$
If $\alpha, \beta, \gamma$ are the roots of the equation $$x^3 + px^2 + qx + p = 0,$$
prove that $$\tan^{-1}\left(\alpha\right) + \tan^{-1}\left(\beta\right) + \tan^{-1}\left(\gamma\right) = n\pi$$
except in one particular case.
This question is from S. L. Loney's 'Plane Trigonometry' page 327 q13.
It may be useful to note that this section utilises $$\tan\left(\alpha + \beta + \gamma + ...\right) = \frac{s_1 - s_3 + s_5}{1 - s_2 + s_4 - \cdots}$$
where
$s_1 =$ the sum of the tangents of the separate angles,
$s_2 =$ the sum of the tangents taken two at a time,
$s_3 =$ the sum of the tangents taken three at a time, and so on.
I do not know where I should start with this question.
|
By Vieta’s formula we have
$$-abc=p$$
$$-(a+b+c)=p$$
$$ab+ac+bc=q$$
By using the formula
$$\tan^{-1}u+\tan^{-1}v=\tan^{-1}\frac{u+v}{1-uv}\mod{\pi}\qquad{uv\ne 1}$$, we have
$$
\begin{align}
\tan^{-1}a+ \tan^{-1} b+ \tan^{-1} c
&= \tan^{-1} \frac{a+b}{1-ab}+ \tan^{-1} c \\
&=\tan^{-1}\frac
{\frac{a+b}{1-ab}+c}
{1-\frac{(a+b)c}{1-ab}} \\
&=\tan^{-1}\frac
{\frac{a+b+c-abc}{1-ab}}
{\frac{1-ab-ac-bc}{1-ab}} \\
&=\tan^{-1}\frac{(a+b+c)-abc}{1-(ab+ac+bc)} \\
&=\tan^{-1}\frac{-p+p}{1-q}\\
&=0 \mod{\pi}
\end{align}
$$
We have assumed that $ab\ne 1$ and $q\ne 1$.
It can be shown that
$$ab\ne 1\implies q\ne 1$$
Substitute $ab=1$ into the first three Vieta’s equations,
$$-c=p$$
$$a+b+c=-p\implies a+b=-p-c=0$$
$$1+(a+b)c=q\implies q=1$$
Therefore, $q\ne1$ is the stronger condition, and $q=1$ should be the only exception that the question expects you to give.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2917298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Linear transformations defined by $T(v) = Av$. Find all of possible $v$ I'm stuck on a problem. The problem is this:
The linear transformation $T : \Bbb{R}^4 \to \Bbb{R}^2$ is defined by $T(v) = Av$, where
$$A = \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix}$$
Find all vectors $v$ such that: $$T(v) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$
So I have forgotten how to do this. Do I:
- reduce $A$ to reduced row-echelon form (why do I do this? Is it because it's easy to solve once you have pivot columns and free variables)?
- rewrite the system of equations
Is this right:
\begin{align}
A &= \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & -5 & -2 & 7 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix} \\
&= \begin{bmatrix} 1 & 0 & \frac{1}{5} & \frac{-29}{5} \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix}
\end{align}
so: $v_4 = t, v_3 = s, v_2 = \frac{-2}{5}s - \frac{7}{5}t, v_1 = \frac{-1}{5}s + \frac{29}{5}t$
$$\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} \frac{-1}{5}s + \frac{29}{5}t \\ \frac{-2}{5}s - \frac{7}{5} t \\ s + 0t \\ 0 + t \end{bmatrix} = s \begin{bmatrix} \frac{-1}{5} \\ \frac{-2}{5} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} \frac{29}{5} \\ \frac{-7}{5} \\ 0 \\ 1 \end{bmatrix} $$
Is this the set of all $v$?
EDIT
I messed up, the first $\frac{7}{5}$ should be a $\frac{-7}{5}$
|
Your way to solve systems of equations is absolutely correct. But you solved the wrong system. What you did is solving the system of equations $Ax=0$. But you need to solve $Ax=\begin{bmatrix} 1 \\ 2 \end{bmatrix}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2918172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
maximum value of expression $(\sqrt{-3+4x-x^2}+4)^2+(x-5)^2$ maximum value of $\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
what i try
$\displaystyle -3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$
$\displaystyle -6x+38+8\sqrt{-3+4x-x^2}$
using derivative it is very lengthy
help me how to solve, thanks in advance
|
Given:
$\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
Denote $x=t+2$, then:
$$f(t)=\bigg(\sqrt{1-t^2}+4\bigg)^2+\bigg(t-3\bigg)^2\;\forall\;t\in[-1\;,1]\\
f'(t)=2\bigg(\sqrt{1-t^2}+4\bigg)\cdot \frac{-t}{\sqrt{1-t^2}}+2(t-3)=0 \Rightarrow\\
-2t-\frac{8t}{\sqrt{1-t^2}}+2t-6=0 \Rightarrow \\
4t=-3\sqrt{1-t^2} \stackrel{t<0}{\Rightarrow} \\
16t^2=9-9t^2 \Rightarrow \\
t=-\frac35.$$
Hence, for $f(t)$:
$$f(-1)=32\\
f(1)=20\\
f(-\frac35)=46.08 \ \text{(max)}.$$
It means $x=t+2=-\frac35+2=\frac75$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2918617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Which pairs of positive integers (,) satisfy $^2−2^=153$? My attempt: Rearrange to $x^2=2^n + 153$ and with $2^n\geq 2\ $ it follows $x^2 \geq 155\ $.
The next square number is 169, so $x = 13$ and $n = 4$. A first solution. Since $2^n$ is even and 153 is odd, $x^2$ will be odd. So any candidate solution will have an even distance of $2m$ from a previous solution and the difference between these solutions is $(x+2m)^2 - x^2 = 4mx + 4m^2$. This difference can be expressed as a difference between two powers of 2, $4mx + 4m^2 = 2^p - 2^n$. My idea was to show that this doesn´t work so that the first solution is the only one.
|
$n$ must be even.
$$ 153 = 3^2 \cdot 17 $$
If $$ x^2 - 2 y^2 $$
is divisible by $3,$ then both $x,y$ are divisible by $3.$ Since this $y$ would be a power of $2$ this is impossible.
So $n$ is even, $n=2k,$ and we actually have $$ x^2 - (2^k)^2 = 153 \; , $$
$$ (x+ 2^k) (x-2^k) = 153 $$
Umm. $$ (x+ 2^k) - (x-2^k) = 2^{k+1} $$
This leads to a finite set of possible $x,$ we can factor $153$ as (ordered pairs)
$$ 153 \cdot 1 $$
$$ 51 \cdot 3 $$
$$ 17 \cdot 9 $$
$$ 153 - 1 = 152 = 8 \cdot 19 $$
$$ 51 - 3 = 48 = 16 \cdot 3 $$
$$ 17 - 9 = 8 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2919551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove that $(x−2y+z)^2 \geq 4xz−8y$ Let $x,y,z$ be nonnegative real numbers such that $x+z\leq2$
Prove that, and determine when equality holds.
$(x−2y+z)^2 \geq 4xz−8y$
Please correct me if my methods are incorrect or would lead nowhere.
I tried expanding the LHS of the inequality getting
$x^2+4y^2+z^2-4xy-4yz+2xz \geq 4xz-8y$
And got lost as to how I should manipulate the inequality to find something true through rough work.
After I tried manipulating
$x+z\leq2$ subtract 2
$x+z-2\leq0$ since $y\ge 0$
$x+z-2\le y$
subtract 2y and add 2 to both sides
$x-2y+z\le 2-y$
And again lost sight of how I could manipulate the inequalities.
|
consider,
$(x-2y+z)^2-4xz+8y=x^2+4y^2+z^2-4xy-4yz+2xz-4xz+8y$
$=x^2+4y^2+z^2-4xy-4yz-2xz+8y=x^2+4y^2+z^2-4y(x+z)-2xz+8y$
$=(x-z)^2+4y^2+4y(2-(x+z))\geq 0$
because $y$ is non negative and $x+z\leq 2$.
Hence that inequality holds.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2921691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
How would I find the common ratio to determine the sum of the given geometric serie: How would I find the common ratio to compute the sum of the given geometric serie: $$\sum_{n=1}^\infty= \frac{(8^n+2^n)}{9^n}$$
|
First note the evaluation of the relevant finite summation:
$$\sum _{n=1}^{N}{c}^{n}={\frac {{c}^{N+1}}{c-1}}-{\frac {c}{c-1}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(0)$$
And then observe that your sum is the sum of two such summations,the key difference being, these are infinite:
$$\sum _{n=1}^{\infty }{\frac {{8}^{n}+{2}^{n}}{{9}^{n}}}=\sum _{n=1}^{
\infty } \left( {\frac {8}{9}} \right) ^{n}+\sum _{n=1}^{\infty }
\left(\frac {2}{9} \right) ^{n}
$$
We then simply set $c_1=\frac {8}{9}$ and $c_2=\frac {2}{9}$, and substitute the infinite upper bound for our sum with a variable $N$, like we see in the evaluation of the finite sum $(0)$:
$$\sum _{n=1}^{
N } {c_1} ^{n}+\sum _{n=1}^{N }
c_2 ^{\,n}={\frac {{c_1}^{N+1}}{c_1-1}}-{\frac {c_1}{c_1-1}}+{\frac {{c_2}^{N+1}}{c_2-1}}-{\frac {c_2}{c_2-1}}$$
Our final step is the part for which one of the fundamental principles of calculus is required, we evaluate the limit for our variable $N$ as it becomes infinite:
$$\sum _{n=1}^{
\infty } \left( {\frac {8}{9}} \right) ^{n}+\sum _{n=1}^{\infty }
\left(\frac {2}{9} \right) ^{n}
=\lim _{N\rightarrow \infty }\Biggl(\sum _{n=1}^{
N } \left( {\frac {8}{9}} \right) ^{n}+\sum _{n=1}^{N }
\left(\frac {2}{9} \right) ^{n}
\Biggr)=\lim _{N\rightarrow \infty }\Biggl({\frac {{(\frac {8}{9})}^{N+1}}{\frac {8}{9}-1}}-{\frac {(\frac {8}{9})}{(\frac {8}{9})-1}}+{\frac {{(\frac {2}{9})}^{N+1}}{\frac {2}{9}-1}}-{\frac {\frac {2}{9}}{\frac {2}{9}-1}}\Biggr)=\frac{58}{7}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2922251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding Multivariable limits using polar coordinates How do I find the limit of this multivariable function as it goes to zero using polar coordinates?
$$ \frac{\sin (x^2 + y^2)}{(x^2 + y^2)^2}
$$
|
Use
\begin{align}
x &= r \cos \theta \\
y &= r \sin \theta
\end{align}
So $x^2 + y^2 = r^2$ hence
\begin{equation}
\frac{\sin (x^2 + y^2)}{(x^2 + y^2)^2}
=
\frac{\sin r^2}{r^4}
\end{equation}
Using L'Hopital twice, we get
\begin{equation}
\frac{\sin r^2}{r^4}
\sim
\frac{2\cos\left(r^2\right)-4r^2\sin\left(r^2\right)}{12r^2}
\rightarrow
+\infty
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2923143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Basis of image of operator I am trying to solve the following question:
I have found the Matrix $M(T)$ as:
\begin{bmatrix}
0&-2&2&0\\
-2&0&0&2\\
2&0&0&-2\\
0&2&-2&0\\
\end{bmatrix}
For $b$, I solve $MX=0$ and get the answer.
Can anyone please help me with the $c$ part?
|
The "image of T" is the set of all vectors v such that Tu= v for some vector u. In particular, $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ is in the image of T if and only if there exist $\begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix}$ such that $\begin{bmatrix}0 & -2 & 2 & 0 \\ -2 & 0 & 0 & 2 \\ 2 & 0 & 0 & -2 \\ 0 & 2 & -2 & 0 \end{bmatrix}\begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix}= \begin{bmatrix}-2x+ 2y \\ -2w+ 2z \\ 2w- 2z \\ 2x- 2y\end{bmatrix}= \begin{bmatrix}a \\ b \\ c \\ d \end{bmatrix}$.
That is, we must have -2x+ 2y= a, -2w+ 2z= b, 2w- 2z= c, and 2x- 2y= d. The critical observation now is that a= -2x+ 2y= -(2x- 2y)= -d and b= -2w+ 2z= -(2w- 2z)= -c.
Any vector in the image of T is of the form $\begin{bmatrix}a \\ b \\ -b \\ -a \end{bmatrix}= \begin{bmatrix}a \\ 0 \\ 0 \\ -a\end{bmatrix}+ \begin{bmatrix}0 \\ b \\ -b \\ 0 \end{bmatrix}= a\begin{bmatrix}1 \\ 0 \\ 0 \\ -1\end{bmatrix}+ b \begin{bmatrix}0 \\ 1 \\ -1 \\ 0 \end{bmatrix}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2926459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Residue of $z_0=1$ for $f(z)=\frac{z^3+5}{z(z-1)^3}$
Consider $$f(z)=\frac{z^3+5}{z(z-1)^3}.$$ I am trying to find the residue of the pole of order $3$, $\ z_0=1$.
I know from calculations that $$\text{Res}(f,1)=\frac{1}{2}\lim_{z\to 1}\frac{\partial^2}{\partial z^2}\left(\frac{z^3+5}{z}\right)=6.$$
I wish to express $f$ as a Laurent series, to find the coefficient of $(z-1)^{-1}$ and hence confirm that Res$(f,1)=6$. I believe the radius of convergence is $0<|z-1|<1$.
I start by taking out a mulitplicative factor of $\frac{1}{(z-1)^2}$, so
\begin{align}
f(z)&=\frac{1}{(z-1)^2}\left(\frac{z^3+5}{z(z-1)}\right) \\
&=\frac{1}{(z-1)^2}\left(-\frac{5}{z}+\frac{6}{z-1}\right) \\
&=\frac{6}{(z-1)^3}-\frac{5}{(z-1)^2}\left(\frac{1}{z}\right)
\end{align}
But I am having difficulty completing the Laurent series as I don't know how to manipulate $\frac{1}{z}$ into a series in terms of $z-1$ that is convergent in $0<|z-1|<1$. Thank you in advanced.
|
First of all, note that$$\frac{z^3+5}{z(z-1)^3}=-\frac5z+\frac6{z-1}-\frac3{(z-1)^2}+\frac6{(z-1)^3}.$$So, what can we do with $-\frac5z$? We have\begin{align}-\frac5z&=-\frac5{1+(z-1)}\\&=-5\left(1-(z-1)+(z-1)^2-(z-1)^3+\cdots\right)\end{align}As you can see, this doesn't matter for compution of the residue, which is $6$, as you wrote.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2931544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$ I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)
Question: If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$
My solution: Let $a \geq b$
From $a^2+b^2 \gt a+b$ we get $a^2-a \gt b-b^2$
Since $a \geq b$ we can get $a^3-a^2 \gt b^2-b^3$ $\Rightarrow$ $a^3+b^3 \gt a^2+b^2$
If this solution is incorrect, please explain why and attach the correct solution. Thank you.
|
Also, we can make the following.
Since by the condition $1>\frac{a+b}{a^2+b^2},$ by C-S we obtain:
$$a^3+b^3>\frac{(a^3+b^3)(a+b)}{a^2+b^2}\geq\frac{(a^2+b^2)^2}{a^2+b^2}=a^2+b^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2934061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
}
|
Finding the result of the polynomial The question is that: Given that $x^2 -5x -1991 = 0$, what is the solution of $\frac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}$
I've tried to factorize the second polynomial like this:
$\frac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}
=\frac{(x-2)^4+x(x-2)}{(x-1)(x-2)}
=\frac{(x-2)((x-2)^3+x)}{(x-1)(x-2)}=\frac{(x-2)^3+x}{(x-1)}$
However I could not solve it with the given equation $x^2 -5x -1991 = 0$. I know that direct substitution may work, but I think that there is a neat solution to this polynomial. Thanks in advance!
Edited:
I've found the way to solve it with the help of lhf
$\frac{(x-2)^4+(x-1)^2-1}{(x-1)(x-2)}=\frac{((x-2)^2+1)((x-2)^2-1)+(x-1)^2}{(x-1)(x-2)}=\frac{(x^2-4x+5)(x-1)(x-3)+(x-1)^2}{(x-1)(x-2)}=\frac{(x^2-4x+5)(x-3)+x-1}{(x-2)}=\frac{(x^2-4x+5)(x-2)-x^2+4x-5+x-1}{(x-2)}=\frac{(x^2-4x+5)(x-2)-(x-2)(x-3)}{(x-2)}=x^2-5x+8$
|
Hint:
$\dfrac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}$ simplifies to $x^2 - 5 x + 8$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2935022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Proving $a\sqrt{7}>\frac{1}{c}$, where $a$ is an integer and $c = \lceil\frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$
Let $a \in \Bbb{N}$, and $c=\lceil \frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$. Prove that $$a\sqrt{7}>\frac{1}{c}$$
So the very original problem sounds like this:
It is given that $a,b \in \Bbb{N}$, $\sqrt{7}-\frac{a}{b}>0$. Prove that $\sqrt{7}-\frac{a}{b}>\frac{1}{ab}$.
Out of the first inequality, I expressed $b>\frac{a}{\sqrt{7}}$. So I thought that the least possible value of $b$ is $\lceil\frac{a}{\sqrt{7}}\rceil$. Also, I changed the inequality that I have to prove to $\frac{a^2+1}{ab}<\sqrt{7}$. I decided to change $b$ with $\lceil\frac{a}{\sqrt{7}}\rceil$, so the value of $\frac{a^2+1}{ab}$ would be as big as possible. I got $\frac{a^2+1}{a\lceil\frac{a}{\sqrt{7}}\rceil}<\sqrt{7}$. Then I made a $c=\lceil \frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$. Therefore, what I had to prove was $$\frac{a^2+1}{a\left(\frac{a}{\sqrt{7}}+c\right)}<\sqrt{7}$$ which is equivalent to $$a^2+1<a^2+ac\sqrt{7}$$ which is equivalent to $$a\sqrt{7}>\frac{1}{c}$$ Somehow, this inequality is not correct. If you could point me a mistake I made, I'd be insanely grateful.
|
If $$0 <\sqrt{7} - \frac{a}{b} \leq \frac{1}{ab}$$ then (rearranging and squaring) $$\frac{a^2}{b^2} < 7 \leq \frac{(a^2 + 1)^2}{(ab)^2}$$ or, rearranging again, $$a^2 < 7b^2 \leq a^2 + 2 + \frac{1}{a^2}.$$
If $a \neq 1$, then $a^2 < 7b^2 < a^2 + 3$, which is impossible by Ross Millikan's remark on quadratic residues modulo $7$. If $a = 1$, then the inequality becomes $1 < 7b^2 < 4$, which is also clearly impossible.
Note that this approach can be generalized to the square root of any number $n$ such that $-2$ is not a quadratic residue modulo $n$ ($-1$ is never a quadratic residue).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2939663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Q. Reducible polynomials in $\mathbb{Z}[x]$ Show that $x^3+ax^2+bx+1$ $\in \mathbb Z[x]$ is reducible on $\mathbb{Z}$ if and only if $a=b$ or $a+b=-2$.
If it is reducible, then it has root in $\mathbb Z$. Be $u$ the root, so i can write it as:
$x^3+ax^2+bx+1=(x-u)(x^2+cx+d)$
That developing the right expression, I get my conditions.
To make the reverse path, can I simply assume a factorization and suppose conditions to reach my original polynomial?
|
If $x^3 + ax^2 + bx + 1$ is reducible in $\Bbb Z[x]$, then we have
$x^3 + ax^2 + bx + 1 = (x^2 + cx + d)(x + e)$
$= x^3 + ex^2 + cx^2 + ecx + dx + ed = x^3 + (c + e)x^2 + (ec + d)x + ed; \tag 1$
we know that $x^3 + ax^2 + bx + 1$ must factor in this form since (i.) a cubic, if reducible, may always be written as the product of a linear and a quadratic polynomial; (ii.) the product of the leading coefficients of the factors is $1$. Choosing each leading coefficient of the factors to be $1$ is possible since this case can always be obtained by reversing the signs of the factors.
Then, comparing coefficients of like terms,
$c + e = a, \tag 2$
$ec + d = b, \tag 3$
$ed = 1; \tag 4$
therefore,
$e = d = 1 \vee e = d = -1; \tag 5$
$e = d = 1 \Longrightarrow c + 1 = b, \; c + 1 = a \Longrightarrow a = b; \tag 6$
$e = d = -1 \Longrightarrow -c - 1 = b, \; c - 1 = a \Longrightarrow a + b = -2; \tag 7$
we can run the logic engine in reverse; if
$a = b, \tag 8$
we simply choose
$e = d = 1, \; c = a - 1; \tag 9$
then it is easy to check that
$(x^2 + (a - 1)x + 1)(x + 1) = x^3 + x^2 + (a - 1)x^2 + (a - 1)x + x + 1$
$= x^3 + x^2 + ax^2 - x^2 + ax - x + x + 1 = x^3 + ax^2 + ax + 1$
$= x^3 + ax^2 + bx + 1; \tag{10}$
if $a + b = -2$, then we pick
$e = d = -1, \; c = a + 1 = -b - 1; \tag{11}$
thus,
$(x^2 + (a + 1)x - 1)(x - 1) = x^3 - x^2 + (a + 1)x^2 - (a + 1)x - x + 1$
$= x^3 - x^2 + ax^2 + x^2 - ax - x - x + 1 = x^3 + ax^2 - (a + 2)x + 1$
$= x^3 + ax^2 + bx + 1. \tag{12}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2942864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Geometric series and polynomials
i) Find a generating function expression of a sequence with terms
$$d_n=\sum_{p=0}^n p^3$$
using operations on the geometric series $\sum_{n\geq 0} x^n$
ii) Derive a polynomial (in $n$) expression for $d_n$.
for i) I got $x(1+4x+x^2)/(1-x)^5$
but I'm confused what to do for ii), how does one derive that?
|
Yes, the generating function is correct: just apply the operator $x\frac{d}{dx}$ to $\sum_{n\geq 0} x^n=1/(1-x)$ three times ($3$ is the exponent of $k$) and then divide by $(1-x)$ (for $\sum_{k=0}^n$):
$$f(x)=\frac{x(1+4x+x^2)}{(1-x)^5}$$
Now, in order to find a polynomial formula for $\sum_{k=0}^n k^3$ we have to extract the coefficient of $x^n$,
$$\begin{align}\sum_{k=0}^n k^3&=[x^n]f(x)=[x^n](x+4x^2+x^3)\cdot(1-x)^{-5}\\&=
[x^{n-1}](1-x)^{-5}+4[x^{n-2}](1-x)^{-5}+
[x^{n-3}](1-x)^{-5}\\
&=(-1)^{n-1}\binom{-5}{n-1}+4(-1)^{n-2}\binom{-5}{n-2}+(-1)^{n-3}\binom{-5}{n-3}\\
&=\binom{n+3}{4}+4\binom{n+2}{4}+\binom{n+1}{4}
=\frac{n^2(n+1)^2}{4}
\end{align}$$
where for the expansion of $(1-x)^{-5}$ we used the Newton's generalized binomial theorem and
$$\binom{-r}k =(-1)^{k}\binom{r+k-1}{r-1}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2942997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Antiderivative of $x\sqrt{1+x^2}$ I am attempting this problem given to me, but the answer key does not explain the answer. The question asks me to find the antiderivative of $x\sqrt{1+x^2}$.
My attempt:
$\frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$
$g'(x)$ must be $x$, therefore $g(x)=\frac{1}{2}x^2$
$f'(x)$ must equal $\sqrt{1+x^2}$, therefore $f(x)=\frac{2}{3}(1+x^2)^{\frac{3}{2}}$
But then I saw that if $g(x)=\frac{1}{3}x^2$, then $g'(x)=\frac{2}{3}x$, and $f(x)=(1+x^2)^{\frac{3}{2}}$, then $\frac{d}{dx}f(g(x))'=g'(x)f'(g(x))=(2/3)x(3/2)\sqrt{1+x^2}=x\sqrt{1+x^2}$.
But I saw this by chance; what would be a more general and effective way to do this? Should I check $this$, and then write $that$, and put together $those$, etc.? Or is the way I just 'saw' the way the numbers should fall together how one would normally approach this?
|
Alternatively:
$$\int{x\sqrt{1+x^2} \,\mathrm{d}x} = \frac12\int{\sqrt{1+x^2} \,\mathrm{d}x^2}=\\
\frac12\int \sqrt{1+t} \, \ \mathrm dt=\frac12\cdot \frac23\cdot (1+t)^{3/2}+C=\frac13(1+x^2)^{3/2}+C.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2943961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Integrate $\int \frac {dx}{\sqrt {(x-a)(x-b)}}$ where $b>a$ Integrate: $\displaystyle\int \dfrac {dx}{\sqrt { (x-a)(x-b)}}$ where $b>a$
My Attempt:
$$\int \dfrac {dx}{\sqrt {(x-a)(x-b)}}$$
Put $x-a=t^2$
$$dx=2t\,dt$$
Now,
\begin{align}
&=\int \dfrac {2t\,dt}{\sqrt {t^2(a+t^2-b)}}\\
&=\int \dfrac {2\,dt}{\sqrt {a-b+t^2}}
\end{align}
|
Proceeding further from where you left
$$
\int \dfrac {2\,}{\sqrt {a-b+t^2}}\ dt
$$
$$
\int \dfrac {2}{\sqrt {(t)^2 + (\sqrt{a-b}) ^2}}\ dt
$$
Using
$$
\int \frac{1}{\sqrt{x^2 + a ^ 2}} = \log({x+\sqrt{x^2+a^2}})
$$
It becomes
$$
2 \cdot {\log (t+\sqrt{t^2 + (\sqrt{a-b}) ^2})}
$$
Then substitute back the value of t .
And you get the required result .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2944173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
Can someone help me finish this: evaluate $S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$ I am asked to find the closed form solution for the below.
$$S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$$
Just writing out the $S_1, S_2, S_3$, I have managed to find a pattern, which is:
$$S_n = \frac{S_{n-1}}{1-x^{2^n}} + \frac{x^{2^n}}{1-x^{2^{n+1}}}$$
I am not sure how to proceed onwards to solve this recurrence relation. Is there a clever trick I can do to solve it?
|
$$\begin{align} \frac{x}{1-x}-S_n &= \\
&= \frac{x}{1-x}-\frac{x}{1-x^2}-\frac{x^2}{1-x^4}- \ldots - \frac{x^{2^{n-1}}}{1-x^{2^{n}}} \\
&=\frac{x^2}{1-x^2}-\frac{x^2}{1-x^4}- \ldots - \frac{x^{2^{n-1}}}{1-x^{2^{n}}} \\
&=\ldots \\
&=\frac{x^{2^{n-1}}}{1-x^{2^{n-1}}}- \frac{x^{2^{n-1}}}{1-x^{2^{n}}} \\
&=\frac{x^{2^{n}}}{1-x^{2^{n}}}. \end{align}$$
So,
$$S_n= \frac{x}{1-x}-\frac{x^{2^{n}}}{1-x^{2^{n}}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2944491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?
|
$-4x^2 -2x -4 = -(4x^2 + 2x + 4)$
The $4x^2 +2x + 4$ must come from some $(2x+b)^2$, to get the right square, and this has linear term $4b$, which should equal $2x$ so $b=\frac{1}{2}$.
Now $(2x + \frac12)^2 = 4x^2 + 2x + \frac14$, so we need an extra $3\frac34$ to get $4$, like we need. So in all
$$-4x^2 -2x -4 = -\left((2x+\frac12)^2 + 3\frac34\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2944950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
}
|
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6} \in \mathbb{Z}$.
Induction step: We want to show that it holds for $n=k+1$.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
|
If you want to use induction you want to show that
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}$$ is an integer. You can expand all the terms and use the known fact about the expression for $k$ to get there.
Another approach is to note that $6$ is a common denominator and say you want to prove that the numerator $2k^3-3k^2+k$ is divisible by $6$. But $2k^3-3k^2+k=(2k-1)(k-1)k$ and one of $k$ or $k-1$ is even and one of the terms is a multiple of $3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
}
|
Let $a
Let $a<b$ and $a,b\in\Bbb R$. Then there is $c\in\Bbb R\setminus\Bbb Q$ such that $a<c<b$.
My attempt:
*
*$a+b$ is irrational
Let $c:=\dfrac{a+b}{2}$
*$a+b$ is rational
Let $x:=\dfrac{a+b}{\sqrt 2}$. Then $x$ is irrational.
*
*$a<x<b$
Let $c:=x$
*
*$b\le x$
Then $x-b<x-a$. Take $x'\in (x-b,x-a)$ such that $x'$ is rational.
Then $a<x-x'<b$ where $x-x'$ is irrational.
Let $c:=x-x'$.
*
*$x\le a$
Then $a-x<b-x$. Take $x'\in (a-x,b-x)$ such that $x'$ is rational.
Then $a<x+x'<b$ where $x+x'$ is irrational.
Let $c:=x+x'$.
My proof is quite short. I'm worried if it's sloppy and contains mistakes. Please help me verify it!
|
If $a, b$ both are rationals then, $\frac{b+a}{2}$ is rational. Then, lets pick $c= \frac{b+a}{2}+\frac{b-a}{2\sqrt{2}}$
If $a$ is rational, and $b$ is irrational, we take $c=\frac{b+a}{2}$
If $a, b$ are irrational:
$\frac{a+b}{2}$ is irrational then pick it as $c$.
Else we pick, $c=\frac{b+a}{2}+ (b-\frac{b+a}{2})/2$. $(b-\frac{b+a}{2})$ is irrational.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2952803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Simple inequality with products of finite geometric series Given numbers $p$ and $q$ with $1\le p < q$, and integers $n$ and $m$ with $1\le n < m$, does this simple inequality hold:
$$\left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\dots+\frac{1}{p^n}\right)\left(1+\frac{1}{q}+\frac{1}{q^2}+\frac{1}{q^3}+\dots+\frac{1}{q^m}\right) < \left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\dots+\frac{1}{p^m}\right)\left(1+\frac{1}{q}+\frac{1}{q^2}+\frac{1}{q^3}+\dots+\frac{1}{q^n}\right)$$
Must be elementary. (Edit: I am happy with any proof or disproof, I just meant to say I thought the inequality in itself should be "elementary".)
When $p$ and $q$ are prime numbers, this says $\sigma_{-1}(p^nq^m)<\sigma_{-1}(p^mq^n)$ where $\sigma_{-1}$ is the abundancy index.
|
Yes, it holds.
*
*If $1=p\lt q$, then the inequality is equivalent to$$(n+1)\cdot\frac{1-\frac{1}{q^{m+1}}}{1-\frac 1q}\lt (m+1)\cdot\frac{1-\frac{1}{q^{n+1}}}{1-\frac 1q},$$i.e.$$\frac{1-\frac{1}{q^{m+1}}}{1+m}< \frac{1-\frac{1}{q^{n+1}}}{1+n}\tag1$$Let $f(x):=\frac{1-\frac{1}{q^{x+1}}}{1+x}$ for $x\ge 1$. Then, $$f'(x)=\frac{-q^{x + 1} + x \ln q + \ln q + 1}{q^{x+1}(x + 1)^2}=\frac{g(x)}{q^{x+1}(x+1)^2}$$where $g(x):=-q^{x + 1} + x \ln q + \ln q + 1$. Then,$$g'(x)=-(q^{x+1}-1)\ln q\lt 0$$So, $g(x)$ is decreasing with $g(1)=0$ implying $g(x)\lt 0$ for $1\lt x$. Since $f'(x)\lt 0$, we have that $f(x)$ is decreasing, from which $(1)$ follows.
*If $1\lt p\lt q$, then the inequality is equivalent to$$\frac{1-\frac{1}{p^{n+1}}}{1-\frac 1p}\cdot\frac{1-\frac{1}{q^{m+1}}}{1-\frac 1q}\lt \frac{1-\frac{1}{p^{m+1}}}{1-\frac 1p}\cdot \frac{1-\frac{1}{q^{n+1}}}{1-\frac 1q},$$i.e.$$\left(1-\frac{1}{p^{n+1}}\right)\left(1-\frac{1}{q^{m+1}}\right)\lt \left(1-\frac{1}{p^{m+1}}\right)\left(1-\frac{1}{q^{n+1}}\right),$$i.e.$$\frac{1-\frac{1}{q^{m+1}}}{1-\frac{1}{p^{m+1}}}\lt \frac{1-\frac{1}{q^{n+1}}}{1-\frac{1}{p^{n+1}}},$$i.e. $$\frac{p^{m+1}(q^{m+1}-1)}{q^{m+1}(p^{m+1}-1)}\lt \frac{p^{n+1}(q^{n+1}-1)}{q^{n+1}(p^{n+1}-1)}\tag2$$Let $h(x):=\frac{p^{x+1}(q^{x+1}-1)}{q^{x+1}(p^{x+1}-1)}$ for $x\ge 1$. Then, we get$$h'(x)=\frac{(\frac pq)^{x+1}(p^{x+1}-1)(q^{x+1}-1)\left(\frac{\ln q}{q^{x+1}-1}-\frac{\ln p}{p^{x+1}-1}\right)}{(p^{x+1}-1)^2}$$Let $i(x):=\frac{\ln x}{x^{c+1}-1}$ where $c\gt 1$ is a real number. Then,
$$i'(x)=-\frac{x^{c+1}((c+1)\ln x-1)+1}{x(c^{x+1}-1)^2}$$For $x\ge 2$, we get $(c+1)\ln x-1\gt (1+1)\ln 2-1\gt 0$ from which $i'(x)\lt 0$ follows. So, for $x\ge 2$, we see that $i(x)$ is decreasing from which $\frac{\ln q}{q^{x+1}-1}-\frac{\ln p}{p^{x+1}-1}\lt 0$. It follows that $h'(x)\lt 0$, so $h(x)$ is decreasing for $x\ge 2$. Now, we can see that $h(1)\gt h(2)$ since this is equivalent to$$\frac{p^{2}(q^{2}-1)}{q^{2}(p^{2}-1)}\gt \frac{p^{3}(q^{3}-1)}{q^{3}(p^{3}-1)},$$i.e.$$q(q+1)(p^2+p+1)\gt p(p+1)(q^2+q+1),$$i.e.$$(q-p) (p + q + 1)\gt 0,$$i.e.$$q\gt p$$Therefore, we see that $h(1)\gt h(2)\gt h(3)\gt\cdots$. It follows that $(2)$ is true for integers $n,m$ such that $1\le n\lt m$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2956935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the domain of the following function The given function is:
$f(x)=\sqrt{\log_{|x|-1}(x^2 + 4x +4)}$
My approach:
The argument $x^2+4x+4>0$ for all $x\neq-2$
Also, the base $|x|-1$ should be greater than 0 and not equal to 1.
$\therefore |x|-1>0$
$\implies |x|>1$
$\implies x>1$ or $x<-1$
And $|x|-1\neq1$
$\implies x\neq2,-2$
Taking all this in consideration, my answer is $D_f= (-\infty,-2)\cup(-2,-1)\cup(1,\infty)$
However, the given answer is $(-\infty,-3]\cup(-2,-1)\cup(2,\infty)$
Where did I go wrong? Thanks in advance.
EDIT:
My new approach:
Case I:
$0<|x|-1<1\implies 1<|x|<2$
Then, $x^2 + 4x+4\leq1\implies x^2+4x+3\leq0$
or $-3\leq x\leq-1$
$\therefore x \in (-2,-1)$---(1)
Case II:
$|x|-1>1 \implies |x|>2$
Then, $x^2+4x+4\geq1 \implies x^2=4x+3\geq0$
or $x\geq-1 $ or $x\leq-3$
$\therefore x \in (-\infty,-3]\cup(2,\infty)$---(2)
From (1) and (2), $x\in(-\infty,-3]\cup(-2,-1)\cup(2,\infty)$
P.S.: Thanks @gimusi
|
HINT
Recall that
*
*for $a>1 \quad \log_a x \ge0 \iff x\ge 1$
*for $0<a<1 \quad \log_a x \ge 0 \iff 0<x\le 1$
therefore in order to have $\log_{|x|-1}(x^2 + 4x +4) \ge 0$ we need to consider two cases
*
*$|x|-1>1 \implies x^2 + 4x +4\ge 1$
*$0<|x|-1<1 \implies 0<x^2 + 4x +4\le 1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2959308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Probability that second ball is magenta
attempt
Notice that in any urn, we have $r-1+n-r = n-1$ balls. We have to pick 2 balls, so the sample space size is ${n-1 \choose 2 }$. Now if we want the second ball to be mangenta, then we must have either $MM$ or $RM$. So the probability of this is
$$ P = \frac{ MM + RM }{{n-1 \choose 2} } = \frac{ \frac{(n-r)^2}{(n-1)^2} + \frac{r-1}{n-1} \cdot \frac{n-r}{n-1} }{{n-1 \choose 2}} $$
but the answer the lecturer gives is $1/2$, does the above answer seems reasonable?
Now, for second part, we want $P(MM)$ which is just
$$ \frac{ \frac{(n-r)^2}{(n-1)^2} }{{n-1 \choose 2}} $$
is this correct?
|
In the first urn there are $(r - 1)$ red balls i.e. $0$ red balls and $(n-r)$ i.e. $(n-1)$ magenta balls. There are two ways the second ball drawn will be magenta $RM$ and $MM$. So, the number of ways to select that (the first term in the square brackets shows for the number of favorable ways of $RM$ and the second one reflects for $MM$) is $[0 + \binom{n-1}{1}* \binom{n-2}{1}]$.
Similarly for the second urn $[\binom{1}{1}*\binom{n-2}{1} + \binom{n-2}{1}*\binom{n-3}{1}]$ ways. So, we end up with this sequence:
$[0 + \binom{n-1}{1}* \binom{n-2}{1}]$ - first urn
$+$ $[\binom{1}{1}*\binom{n-2}{1} + \binom{n-2}{1}*\binom{n-3}{1}]$ - second urn
$+$ $[\binom{2}{1}*\binom{n-3}{1} + \binom{n-3}{1}*\binom{n-4}{1}]$ - third urn
$+...+[\binom{n-3}{1}*\binom{2}{1} + \binom{2}{1}*\binom{1}{1}]$ - third from last urn
$+ [\binom{n-2}{1}*\binom{1}{1} + 0 ] $ - second from last
$+ 0$. - last urn
Note that $0$ is for the last urn which does not have any magenta balls.
Adding all of the above terms will give:
$[(n-2)*\frac{n*(n-1)}{2}]$. And the total number of ways the one urn can be selected is $n$. And selecting two balls without replacement will be $(n-1)*(n-2)$ ways. So, the probability will be:
$[(n-2)*\frac{n*(n-1)}{2}]$/$n*(n-1)*(n-2)$ = $0.5$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2959555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Prove $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$ by induction Here is my attempted proof:
$\forall n \in \mathbb{N}$, let $S_n$ be the statement: $\frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
Base case: $S_1$: $\frac{2^{4(1)}-(-1)^1}{17} = \frac{16+1}{17} = 1 \in \mathbb{N}$
Inductive step: $\forall n \geq 1$, $S_n$ holds
$S_{n+1}$: $\frac{2^{4(n+1)}-(-1)^{n+1}}{17} = \frac{2^{4n} \cdot 2^4 +(-1)^{n}}{17} = \frac{2^{4n} \cdot (2^4+1) - 2^{4n} +(-1)^{n}}{17}$.
Now, we are left with $2^{4n} - \frac{2^{4n}-(-1)^n}{17}$.
We know $\frac{2^{4n}-(-1)^n}{17}$, $2^{4n} \in \mathbb{N}$ and $2^{4n} > \frac{2^{4n}-(-1)^n}{17} \implies 2^{4n} - \frac{2^{4n}-(-1)^n}{17} \in \mathbb{N}$
$$\tag*{$\blacksquare$}$$
Is my proof correct? I also feel there is a more elegant/smoother proof (assuming mine is correct).
|
Your proof is fine.
But note:
$(a - b)(\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k}) =$
$a(\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k}) - (\sum\limits_{k=0}^{n-1} a^kb^{n-1 -k})=$
$(\sum\limits_{k=0}^{n-1}a^{k+1}b^{n-1 - k}) -(\sum\limits_{k=0}^{n-1}a^kb^{n-k}) = $
$(\sum\limits_{j=1}^n a^jb^{n-j})-(\sum\limits_{j=0}^{n-1}a^jb^{n-j})=$
$([\sum\limits_{j=1}^{n-1} a^jb^{n-j}]+a^nb^{n-n}) - (a^0b^{n-0}+[\sum\limits_{j=1}^{n-1} a^jb^{n-j}]) = $
$a^n - b^n$.
So for all integers $a,b$ and $n\ge 1$ we will always have $a-b$ divides $a^n - b^n$
So $2^4 -(-1) = 17$ will always divide $(2^4)^n - (-1)^n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2964749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
If $\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2} = \frac{a} {(x + 2)^2}$, then what is $a$? $$\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2}$$
The expression above is equivalent to $$\frac{a} {(x + 2)^2}$$
where $a$ is a positive constant and $x \neq -2$.
What's the value of $a$?
|
Through basic substitution one finds:
$$a =\frac{2 \left(x^4+5 x^3+7 x^2+3 x-4\right)}{x}$$
not a positive constant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2966822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Mean of two numbers by infinite sequences Consider two numbers $a$ and $b$, and the following sequence alternating between even and odd positions:
$$
a+2b+3a+4b+5a+6b\ldots,
$$
If we ''normalize''
$$
\frac{a+2b+3a+4b+\ldots}{1+2+3+4+\ldots},
$$
it turns out this ratio approaches the mean value of $a$ and $b$: $(a+b)/2$. In general
$$
\frac{a+2^n b+3^n a+4^n b+\ldots}{1^n+2^n+3^n+4^n+\ldots}=\frac{a+b}{2}
$$
for $n\geq1$. However if we use exponential functions instead powers:
$$
\frac{m^1 a+m^2 b+m^3 a+m^4 b+\ldots}{m^1 +m^2 +m^3 +m^4 +\ldots}
$$
for some $m>1$, this ratio oscillates and does not approach any number.
Could someone explain why convergence to the mean is obtained by using sequences of powers, and the ratio diverges for sequences of exponentials?
|
To give an evaluation, to make more rigourous, in the first case we have
$$1^n+2^n+3^n+4^n+\ldots+k^n\sim \frac{k^{n+1}}{n+1}$$
$$a+2^n b+3^n a+4^n b+\ldots+k^nb\sim a\left(\frac{k^{n+1}}{n+1}-2^n\frac{k^{n+1}}{(n+1)2^{n+1}}\right)+2^nb\frac{k^{n+1}}{(n+1)2^{n+1}}=$$$$=\frac12(a+b)\frac{k^{n+1}}{n+1}$$
and therefore
$$\frac{a+2^n b+3^n a+4^n b+\ldots+k^nb}{1^n+2^n+3^n+4^n+\ldots+k^n}\sim \frac{\frac12(a+b)\frac{k^{n+1}}{n+1}}{\frac{k^{n+1}}{n+1}}\to \frac{a+b}{2}$$
On the other hand for $m\neq 1$
$$m^1 +m^2 +m^3 +m^4 +\ldots+m^k =\frac{m^{k+1}-m}{m-1}$$
$$m^1 a+m^2 b+m^3 a+m^4 b+\ldots+m^kb\sim a\left(\frac{m^{k}-m}{m-1}-\frac{m^{k}-m^2}{m^2-1}\right)+b\frac{m^{k+2}-m^2}{m^2-1}=$$
$$=a\frac{m^{k+1}-m}{m^2-1}+b\frac{m^{k+2}-m^2}{m^2-1}$$
and therefore
$$\frac{m^1 a+m^2 b+m^3 a+m^4 b+\ldots+m^kb}{m^1 +m^2 +m^3 +m^4 +\ldots+m^k}\sim \frac {a+bm} {m+1}$$
but for
$$m^1 a+m^2 b+m^3 a+m^4 b+\ldots+m^ka$$
we would obtain
$$\frac{m^1 a+m^2 b+m^3 a+m^4 b+\ldots+m^ka}{m^1 +m^2 +m^3 +m^4 +\ldots+m^k}\sim \frac {am+b} {m+1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2968055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
}
|
Surface area of sphere and cone Given the sphere $x^2 + y^2 + z^2 = 1$ and the cone $z = \alpha \sqrt{x^2 + y^2}$, $\alpha > 0$, $S_1$ is the portion of the sphere inside the cone while $S_2$ is the portion of the cone inside the sphere. I need to find the value of $\alpha$ such that both surface areas are the same (it should be $\alpha = 3/4$).
This confuses me not because of the parametrizations by themselves, but because of the limits of integration and the method of simplyfying $\alpha$ for a numerical value.
For $S_{1}$, we can parametrize the sphere and define the domain as we see fit in order to restrict for the cone. Then, $r_1(u,v) = \left(cos(u)sin(v),sin(u)sin(v),cos(v) \right)$. $S_2$ is similar: $r_2(u,v) =\left(u cos(v), u sin(v), \alpha u \right) $, where z is obtained by plugging in the radius in the given cone's equation.
This is where things get confusing. There are two ways to obtain the element of area: $\mid{r_u * r_v}\mid$ and the one I've had more sucess with in this particular problem: $z = f(x,y)$ -> $\sqrt{1 + z_x^2 + z_y}$. This makes us go back to the initial equations ignoring the parametrizations for now. Also, since the cone is in the positive direction, our other z is simply $z = \sqrt{1 - x ^2 - y^2}$. Thus,
$S_{1}$: $\sqrt{1 + \dfrac{x^2}{1-x^2 -y^2 } + \dfrac{y^2}{1-x^2 -y^2} } = \sqrt{\dfrac{1}{1-x^2 -y^2 }} $
$S_{2}$: $\sqrt{1 + \dfrac{a^2 x^2}{x^2 + y^2} + \dfrac{a^2 y^2}{x^2 + y^2} } = \sqrt{1 + a^2} $
I don't know what to do next. According to the brief solution I have at hand, the limit of integration for both $S_1$ and $S_2$ is $x^2 + y^2 \leq a^2$, but I don't know where that comes from (although a quick sketch of the zone where the sphere and the cone touch shows that the intersection would be indeed a circle) and, even more confusing, is a step in the second integral that I can't understand:
$\int \int_{x^2 + y^2 \leq a^2 } \sqrt{1 +a^2} dxdy = \pi a^2 \sqrt{1+ a^2} = \dfrac{\pi}{\sqrt{1 + a^2}} (?) $ (Where did the last one come from?)
This is more algebra than anything else, but that simplification and one that comes after that (for finding a) seems to assume that $a^2 = \dfrac{1}{1+a^2}$ which, to me, doesn't make any sense. I've spent hours trying to a) understand why the $a^2$ simplification is true and b) the limits of integration without any success whatsoever, which is why I'm posting this exercise that may look somewhat trivial at first glance, but isn't to me.
|
convert to spherical
$x = \rho\cos\theta\sin \phi\\
y = \rho\sin\theta\sin \phi\\
z = \rho\cos \phi$
$z= \alpha \sqrt{x^2 + y^2}$ becomes
$\cos \phi = \alpha \sin \phi\\
\tan\phi = \frac {1}{\alpha}\\
\phi = \arctan \frac 1{\alpha}$
$\int_0^{2\pi}\int_0^{\arctan{\frac 1\alpha}} \sin\phi \ d\phi\ d\theta$
Is the area of the spherical cap.
$\int_0^{2\pi}\int_0^{1} \rho\sin(\arctan{\frac 1\alpha}) \ d\rho\ d\theta$
Is the area of the cone inside the sphere.
$\pi \sin(\arctan{\frac 1\alpha}) = 2\pi(1-\cos\arctan{\frac 1\alpha})$
$\frac {1}{\sqrt{1+\alpha^2}} = 2(1-\frac {a}{\sqrt {1+\alpha^2}})\\
1 + 2\alpha = 2\sqrt {1+\alpha^2}\\
1 + 4\alpha + 4\alpha^2 = 4 +4\alpha^2\\
4\alpha - 3= 0\\
\alpha = \frac 34$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2968365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Proving $\lim_{n\rightarrow-\infty}\frac{3x^2+x}{2x^2+1}=\frac{3}{2}$ Using Definition Definition of Limit of Function As n $\rightarrow\infty$: Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a function and let $B\in\mathbb{R}$. If for all $\epsilon>0$, there exists $N>0$ such that $x<-N\Rightarrow |g(x)-B|<\epsilon$, we write $\lim_{n\rightarrow\infty}g(x)=B$.
I want to prove $\lim_{n\rightarrow-\infty}\frac{3x^2+x}{2x^2+1}=\frac{3}{2}$.
Here is my attempt.
We must find the value $-N<0$ such that: $x<-N\Rightarrow |g(x)-\frac{3}{2}|=|\frac{3x^2+x}{2x^2+1}-\frac{3}{2}|=|\frac{3x^2+x}{2x^2+1}-\frac{3(x^2+\frac{1}{2})}{2(x^2+\frac{1}{2})}|=|\frac{x-\frac{3}{2}}{2x^2+1}|<\epsilon$.
When $N>\frac{3}{2}$, we have: $-x>N\Rightarrow |\frac{-x-\frac{3}{2}}{2x^2+1}|=\frac{-x-\frac{3}{2}}{2x^2+1}<\frac{-x-\frac{3}{2}}{2x^2}<\frac{-x}{2x^2}=\frac{-1}{2x}<\epsilon\Rightarrow -x>\frac{1}{2\epsilon}$. So, choose $N>\frac{1}{2\epsilon}$.
Now, this is the bit where I'm having some issues.
Then, for all $x<-N<\min\{-\frac{1}{2\epsilon},-\frac{3}{2}\}$, we have $|\frac{x-\frac{3}{2}}{2x^2+1}|=\frac{-x+\frac{3}{2}}{2x^2+1}<\frac{-x+\frac{3}{2}}{2x^2}=\frac{\frac{1}{2\epsilon}+\frac{3\epsilon}{2\epsilon}}{\frac{9}{2}}$... I can't quite get this inequality to work the way I need it to.
Any help appreciated.
|
Let $o<\epsilon <1/2$. Then $x <-\frac 1 {\epsilon} $ implies $|g(x)-2|=\frac {|x-3/2|} {2x^{2}+1} \leq \frac {|x|+3/2} {2x^{2}}=\frac {|x|} {2x^{2}}+\frac 3 {4x^{2}}=\frac {1} {2|x|}+\frac 3 {4x^{2}}< {\epsilon /2}+\frac {3\epsilon ^{2}} 4<\epsilon$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2969810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$
This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)
|
With $AM-GM$
\begin{align}
1.(1+\frac{1}{2^3})(1+\frac{1}{3^3})\cdots(1+\frac{1}{n^3})
&\leq\left(\dfrac1n(n+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots+\frac{1}{n^3})\right)^n \\
&\leq\left(\dfrac1n(n+\sum_{n=1}^\infty\frac{1}{n^3}-1)\right)^n \\
&\leq\left(\dfrac1n(n+\zeta(3)-1)\right)^n \\
&\leq\left(1+\dfrac{\zeta(3)-1}{n}\right)^n \\
&< e^{\zeta(3)-1}\\
&<\frac32
\end{align}
Thanks to Winther.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
}
|
Solve differential equation $f''''(x)=f'''(x)f''(x)f'(x)f(x)$ I met this DE recently, and I am utterly befuddled at how to solve it
$$f''''(x)=f'''(x)f''(x)f'(x)f(x)$$
I tried this:
$$\frac{f''''(x)}{f'''(x)}=f''(x)f'(x)f(x)$$
$$\ln|f'''(x)|=c_1+\int f(x)f'(x)f''(x)dx$$
I do not know how to solve the right side, though. Integration by parts?
Plaese help.
|
Hint:
Let $u=\dfrac{df}{dx}$ ,
Then $\dfrac{d^2f}{dx^2}=\dfrac{du}{dx}=\dfrac{du}{df}\dfrac{df}{dx}=u\dfrac{du}{df}$
$\dfrac{d^3f}{dx^3}=\dfrac{d}{dx}\left(u\dfrac{du}{df}\right)=\dfrac{d}{df}\left(u\dfrac{du}{df}\right)\dfrac{df}{dx}=\left(u\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^2\right)u=u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2$
$\dfrac{d^4f}{dx^4}=\dfrac{d}{dx}\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)=\dfrac{d}{df}\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)\dfrac{df}{dx}=\left(u^2\dfrac{d^3u}{df^3}+4u\dfrac{du}{df}\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^3\right)u=u^3\dfrac{d^3u}{df^3}+4u^2\dfrac{du}{df}\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^3$
$\therefore u^3\dfrac{d^3u}{df^3}+4u^2\dfrac{du}{df}\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^3=\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)u\dfrac{du}{df}uf$
$u^2\dfrac{d^3u}{df^3}+4u\dfrac{du}{df}\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^3=fu^3\dfrac{du}{df}\dfrac{d^2u}{df^2}+fu^2\left(\dfrac{du}{df}\right)^3$
$u^2\dfrac{d^3u}{df^3}+(4-fu^2)u\dfrac{du}{df}\dfrac{d^2u}{df^2}+(1-fu^2)\left(\dfrac{du}{df}\right)^3=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2972440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
sum of multiplicative inverses for $p=3k+1$ I want to prove that for any prime $p=3k+1$
$$\sum_{i=1}^ki^{-1}\equiv\sum_{i=1}^k\left(i+\frac{p-1}{2}\right)^{-1}\;(\text{mod}\;p)$$
but I can't seem to get anywhere.
Can anyone tell me how I can approach this kind of problem?
|
Let $p=2h+1$. By Fermat little theorem, we have
\begin{align*} S:=\sum_{i=1}^k i^{-1}- \sum_{i=1}^k (i+h)^{-1}\equiv \sum_{i=1}^k i^{p-2}- \sum_{i=1}^k (i+h)^{p-2} \pmod p
\end{align*}
since $1\le i, i+h <p$. Then,
\begin{align*} S\equiv& -\sum_{i=1}^k\sum_{j= 0} ^{p-3}\binom{p-2}{j} i^j h^{p-2-j} \pmod p\\
\equiv& -\sum_{j= 0} ^{p-3}\binom{p-2}{j}h^{p-2-j} \sum_{i=1}^ki^j \pmod p.\\
\end{align*}
But $\binom{p-2}{j} \equiv (j+1)(-1)^j \bmod p$, then
\begin{align*} S\equiv& -\sum_{j= 0} ^{p-3}(j+1)(-1)^j h^{p-2-j} \sum_{i=1}^k i^j \pmod p\\
\equiv& \sum_{j= 0} ^{p-3}(-h)^{p-2-j} (j+1)\sum_{i=1}^ki^j \pmod p\\
\equiv& 2\sum_{j= 0} ^{p-3}(j+1)\sum_{i=1}^k(2i)^j \pmod p\\
\equiv& 2 \sum_{i=1}^k\sum_{j= 0} ^{p-3}(j+1)(2i)^j \pmod p\\
\end{align*}
Now
\begin{align*} \sum_{j= 0} ^{p-3}(j+1)(2i)^j =\frac{(p-2)(2i)^{p-1}-(p-1)(2i)^{p-2}+1}{(2i-1)^2} &\equiv -\frac{1}{2i(2i-1)} \pmod p\\
& \equiv \frac{1}{2i}-\frac{1}{2i-1} \pmod p.
\end{align*}
Then
\begin{align*} S\equiv& 2 \sum_{i=1}^k \left( \frac{1}{2i}-\frac{1}{2i-1}\right) \pmod p\\
\equiv&2\left(\sum_{i=1}^k\frac{1}{i}-\sum_{i=1}^{2k}\frac{1}{i}\right) \pmod p\\
\equiv&-2\sum_{i=k+1}^{2k}\frac{1}{i}\pmod p\\
\equiv&-2\sum_{i=1}^k\frac{1}{k+i}\pmod p\\
\equiv&-6\sum_{i=1}^k\frac{1}{3i-1}\pmod p\\
\equiv&-6\sum_{i=1}^{p-1}\frac{1}{i}+6\sum_{i=1}^k\frac{1}{3i-2}+6\sum_{i=1}^k\frac{1}{3i}\pmod p
\end{align*}
But
\begin{align*}\sum_{i=1}^k\frac{1}{3i-2} =\sum_{j=1}^k\frac{1}{3(k-j+1)-2}=\sum_{j=1}^k\frac{1}{p-1-3j+1} \equiv -\sum_{j=1}^k\frac{1}{3j}
\pmod p\end{align*}
Then
\begin{align*} S\equiv& -6\sum_{i=1}^{p-1}\frac{1}{i} \pmod p.
\end{align*}
and this is zero by a well-known result. $\square$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2973677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number
I want to show that
*
*If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
*If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_1=F_2=1$, we suppose that $n \geq 3$.
$$\begin{align}
F_{n+1}&=F_n+F_{n-1} \\
F_{n+1}^2&=(F_n+F_{n-1})^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2
\end{align}$$
$$\begin{align}
F_{n-1} &=F_{n-2}+F_{n-3} \\
F_{n-1}^2&=(F_{n-2}+F_{n-3})^2=F_{n-2}^2+2F_{n-2}F_{n-3}+F_{n-3}^2
\end{align}$$
so that
$$
F_{n+1}^2-F_{n-1}^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2-F_{n-2}^2-2 F_{n-2} F_{n-3}-F_{n-3}^2
$$
How can we deduce that the latter is divisible by $4$?
Or do we show it somehow else, for example by induction?
|
Modulo two the Fibonacci sequence cyclically repeats the pattern of length three $0,1,1,0,1,1,0,1,1,\ldots$. So if $F_n$ is even then both $F_{n-1}$ and $F_{n+1}$ are odd. Therefore both $F_{n-1}^2$ and $F_{n+1}^2$ are congruent to $1\pmod 4$ (mod eight actually!). Therefore their difference is a multiple of eight.
Modulo three the cyclically repeating pattern has length eight: $$0,1,1,2,0,2,2,1,0,1,1,2,0,2,2,1,\ldots.$$
Again, we see that if $F_n$ is divisible by three, then $F_{n-1}$ and $F_{n+1}$ are either both $\equiv1\pmod 3$ or both $\equiv-1\pmod3$.
In the former case their cubes are both $\equiv 1\pmod 9$ (check that $1^3\equiv4^3\equiv7^3\pmod9$. In the latter case both cubes are $\equiv -1\pmod9$ ($2^3\equiv5^3\equiv8^3$). Both these facts actually also follow from the fact that the group $\Bbb{Z}_9^*$ is cyclic of order six.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2974491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Prove that $\sqrt[3]{5} + \sqrt{2}$ is irrational I tried with both squaring and cubing the statement, it got messy, here's my latest attempt:
Assume for the sake of contradiction: $\sqrt[3]{5} + \sqrt{2}$ is rational
$\sqrt[3]{5} + \sqrt{2}$ = $\frac{a}{b}$ $a,b$ are odd integers $> 0$ and $ b\neq 0$
${(\sqrt[3]{5} + \sqrt{2})}^3$ = $\frac{a^3}{b^3}$
by multiplying by $b^3$:
${(\sqrt[3]{5} + \sqrt{2})}^3 \times b^3 $ = ${a^3}$
so: $a^3$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}^3$ which means $a$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}$
doing the same thing with $b$ i found :
$\frac{a^3}{{(\sqrt[3]{5} + \sqrt{2})}^3} $ = ${b^3}$
so: $b^3$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}^3$ which means $b$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}$ (wrong)
${(\sqrt[3]{5} + \sqrt{2})}$ is a common divisor for both $a$ & $b$ which is a contradiction, thus $\sqrt[3]{5} + \sqrt{2}$ is irrational. (wrong)
|
You can't do divisibility in irational and rational numbers. When you are operating with divisibility you have to have an integers. It is a relation defined on integer numbers.
Suppose it is rational, then exist rational number $q$ such that $$\sqrt[3]{5} + \sqrt{2}= q$$ so $$ 5 = (q-\sqrt{2})^3 = q^3-3q^2\sqrt{2}+6q-2\sqrt{2}$$
So we have $$\sqrt{2}(\underbrace{3q^2+2}_{\in\mathbb{Q}}) = \underbrace{q^3+6q-5}_{\in\mathbb{Q}}$$
so $$\sqrt{2}= \underbrace{q^3+6q-5\over 3q^2+2}_{\in\mathbb{Q}}$$
A contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2975313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
}
|
Calculating coefficient I have a generating function,
$$ \frac{(1-x^7)^6}{(1-x)^6} $$
and I want to calculate the coefficient of $x^{26}$
Solution for this is,
$$ {26+5 \choose 5} - 6{19+5 \choose 5} + 15{12+5 \choose 5} - 20{5+5 \choose 5} $$
Is there formula for this? If there is, what is called?
If there is no formula, how can I calculate it?
Thanks!
|
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{26}]}&\color{blue}{\frac{\left(1-x^7\right)^6}{\left(1-x\right)^6}}\\
&=[x^{26}]\left(1-x^7\right)^6\sum_{j=0}^{\infty}\binom{-6}{j}(-x)^j\tag{1}\\
&=[x^{26}]\left(1-\binom{6}{1}x^7+\binom{6}{2}x^{14}-\binom{6}{3}x^{21}\right)\sum_{j=0}^\infty\binom{j+5}{5}x^j\tag{2}\\
&=\left([x^{26}]-6[x^{19}]+15[x^{12}]-20[x^5]\right)\sum_{j=0}^\infty\binom{j+5}{5}x^j\tag{3}\\
&\,\,\color{blue}{=\binom{31}{5}-6\binom{24}{5}+15\binom{17}{5}-20\binom{10}{5}}\tag{4}
\end{align*}
in accordance with the claim.
Comment:
*
*In (1) we expand the denominator using the binomial series expansion.
*In (2) we expand the polynomial up to powers of $x^{21}$ since higher powers do not contribute to $[x^{26}]$ and we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.
*In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (4) we select the coefficients accordingly.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2979109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Integral $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}$, may involve beta function Evaluate the integral: $$\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2} dx$$
I have used substitutions like $x^4 = u$, or $x^{-4} = u$. After many hours, I came up with $\frac{2}{1+x^4} = u$ which reduces the integral to :
$$\int_1^2 \frac{2^{3/4}}{8} \left(\frac{u-1}{2-u}\right)^{3/4} du$$
and then taking $\frac{u-1}{2-u} = z$ gives:
$$\frac{2^{3/4}}{8} \int_0^\infty \frac{z^{3/4} }{(1+z)^2} dz = \color{red}{ \frac{2^{3/4}}{8}\beta(\tfrac{7}{4},\tfrac{1}{4})}$$
Last answer comes from using:
$$\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}} dx = \beta(m,n)$$
This answer seems to be correct as online integral calculator gives $0.7005...$ as answer.
What can be alternate approaches
|
Alternative approaches? The integral is a hypergeometric functions, and I would immediately try to get it because then I could always use some identities to simplify if needed.
Let's use the substitution $x^4=v$, which the OP rejected.
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{4} \int_0^1 v^{-3/4} (1-v)^{3/4}(1+v)^{-2} dv$$
We know the general integral for the hypergeometric function:
$$\mathrm {B} (b,c-b)\,_{2}F_{1}(a,b;c;z)=\int _{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\,dx\qquad \Re (c)>\Re (b)>0$$
In this case $a=2$, $b=1/4$, $c=2$, so we get:
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{4} \mathrm {B} \left(\frac14 , \frac74 \right) {_2 F_1} \left(2,\frac14; 2;-1 \right)$$
Using some simplifications:
$$\frac{1}{4}\mathrm {B} \left(\frac14 , \frac74 \right)=\frac{1}{4}\Gamma \left( \frac14 \right) \Gamma \left(\frac74 \right)=\Gamma \left(\frac54 \right)\Gamma \left(\frac74 \right)$$
$${_2 F_1} \left(2,\frac14; 2;-1 \right)={_1 F_0} \left(\frac14; ;-1 \right)=(1-(-1))^{-1/4}=\frac{1}{2^{1/4}}$$
(It's known that ${_1 F_0} \left(a; ;x \right)=(1-x)^{-a}$).
So we get:
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{2^{1/4}}\Gamma \left(\frac54 \right)\Gamma \left(\frac74 \right)$$
Which is the same as the OP's result, but obtained in a more simple way, using the known properties of hypergeometric functions.
This can be further simplified by using the reflection formula for the Gamma function (thanks to @Szeto for reminding me):
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3}{16 \cdot 2^{1/4}}\Gamma \left(\frac14 \right)\Gamma \left(\frac34 \right)$$
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3\pi}{16 \cdot 2^{1/4} \sin \frac{\pi}{4}}$$
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3\pi \cdot 2^{1/4}}{16}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2983736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Calculate $\sum_{n=1}^{\infty} \arctan\bigl(\frac{2\sqrt2}{n^2+1}\bigr) $ $$ \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{2\sqrt2}{k^2+1}= \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{(\sqrt{k^2+2}+\sqrt2)-\sqrt{k^2+2}-\sqrt2)}{(\sqrt{k^2+2}+\sqrt2)(\sqrt{k^2+2}-\sqrt2)+1}= $$
$$\lim_{n \to\infty} \sum_{k=1}^{n} \arctan(\sqrt{k^2+2}+\sqrt2)-\arctan(\sqrt{k^2+2}-\sqrt2) $$
But the elements do not reduce :(
|
Result
$$\pi -\frac{1}{2} \arctan \left(2 \sqrt{2}\right)\simeq 2.52611... $$
Derivation
More complicated than the elegant telescoping approach, but that was the way I found the result.
Observing that
$$\text{arctan}(t) = t \int_{0}^1 \frac{1}{1+t^2 x^2}\,dx$$
and interchanging sum and integral the sum can be written as
$$s =2 \sqrt{2} \int_{0}^1 \sum_{n=1}^\infty \frac{ \left(n^2+1\right)}{\left(n^2+1\right)^2+8 x^2}\,dx$$
The sum can be done:
$$\sum _{n=1}^{\infty } \frac{n^2+1}{\left(n^2+1\right)^2+8 x^2}=\frac{1}{2 \left(x^2+1\right)}\text{Re}\left(\pi \sqrt{-1+i x} (1+i x) \cot \left(\pi \sqrt{-1+i x}\right)-2\right)$$
and, suprisingly, also the final $x$-integral can be done with the result provided.
Corollary 1
With the same method we can easily find
$$s_{c}=\sum _{n=1}^{\infty } \arctan\left(\frac{1}{n^2+1}\right)\simeq 1.03729...\\= \frac{1}{8} \left(3 \pi -4 i \log \left(\sin \left(\sqrt{-1-i} \pi \right)\right)+4 i \log \left(\sin \left(\sqrt{-1+i} \pi \right)\right)\right)\\=\frac{3 \pi }{8}-\arg \left(\sin\pi \left(\sqrt{-1+i} \right)\right)\tag{c1}$$
This can be simplified to the real expression
$$s_{c}=\frac{3 \pi }{8}-\arctan \left(\frac{\cos \left(\sqrt[4]{2} \pi \sin \left(\frac{\pi }{8}\right)\right) \sinh \left(\sqrt[4]{2} \pi \cos \left(\frac{\pi }{8}\right)\right)}{\sin \left(\sqrt[4]{2} \pi \sin \left(\frac{\pi }{8}\right)\right) \cosh \left(\sqrt[4]{2} \pi \cos \left(\frac{\pi }{8}\right)\right)}\right)\tag{c2}$$
or
$$s_{c}=\frac{3 \pi }{8}-\arctan\left(\frac{\tanh \left(\pi \sqrt{\frac{1}{2} \left(\sqrt{2}+1\right)}\right)}{\tan \left(\pi \sqrt{\frac{1}{2} \left(\sqrt{2}-1\right)}\right)}\right)\tag{c3}$$
This is a rather trig-loaded expression.
I could not find a telescoping sum in this case. The factor $2^{\frac{3}{2}}$ in the OP was accurately tuned.
Corollary 2
Also
$$s_{c2} = \sum _{n=1}^{\infty } \arctan \left(\frac{1}{n^2}\right)\\=
\arctan \left(\frac{1-\cot \left(\frac{\pi }{\sqrt{2}}\right) \tanh \left(\frac{\pi }{\sqrt{2}}\right)}{1+\cot \left(\frac{\pi }{\sqrt{2}}\right) \tanh \left(\frac{\pi }{\sqrt{2}}\right)}\right)\\\simeq 1.42474...$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2987659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Proof verification of $\{x_n\} = \left(1 + {1\over 2n}\right)^n$ is an increasing sequence.
Let $n\in \mathbb N$ and:
$$
x_n = \left(1 + {1\over 2n}\right)^n
$$
Show that $\{x_n\}$ is an increasing sequence.
$\Box$ Consider ratio test of two consequent terms $x_n$ and $x_{n+1}$:
$$
\frac{x_{n+1}}{x_n} = \frac{\left(1 + {1\over 2n + 2}\right)^{n+1}}{\left(1 + {1\over 2n}\right)^n}
= \frac{\left(1 + {1\over 2n + 2}\right)^{n}}{\left(1 + {1\over 2n}\right)^n} \cdot\left(1 + {1\over 2n + 2}\right) = \\
= \left(\frac{2n(2n+3)}{(2n+1)(2n+2)}\right)^n \cdot\left(1 + {1\over 2n + 2}\right)
$$
We are done in case this product is greater than $1$.
Denote:
$$
P^n = \left(\frac{2n(2n+3)}{(2n+1)(2n+2)}\right)^n = \left(\frac{4n^2 + 6n}{4n^2 + 6n+2}\right)^n
$$
Split $P$ into partial fractions:
$$
P^n = \left(1 + \frac{1}{n+1} - \frac{2}{2n+1}\right)^n = \left(1 - \frac{1}{(n+1)(2n+1)}\right)^n
$$
Since $\frac{-1}{(n+1)(2n+1)} > -1$ we may apply Bernoulli's:
$$
P^n \ge 1 - \frac{n}{(n+1)(2n+1)}
$$
Thus:
$$
\frac{x_{n+1}}{x_n} \ge \left(1 - \frac{n}{(n+1)(2n+1)}\right)\cdot\left(1+ \frac{1}{2n+2}\right) = \frac{2n^2 + 2n +1}{2n^2+3n+1} \cdot \frac{2n+3}{2n+2} = \\
= \frac{4 n^3 + 10 n^2 + 8 n + 3}{4 n^3 + 10 n^2 + 8 n + 2}
$$
From here it's clear that:
$$
{x_{n+1}\over x_n} > 1
$$
This completes the proof that $x_n$ is monotonically increasing. ${\blacksquare}$
Is this a valid proof? Also I would appreciate any simpler methods to show that.
|
$b_n=(1+1/n)^n$ is increasing. Your sequence is $x_n=\sqrt{b_{2n}}$, so it is increasing.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2990359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Inverse of tridiagonal Toeplitz matrix Consider the following tridiagonal Toeplitz matrix. Let $n$ be even.
$${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}}
{0}&{1}&{}&{}&{}\\
{1}&{0}&{1}&{}&{}\\
{}&{1}&{\ddots}&{\ddots}&{}\\
{}&{}&{\ddots}&{\ddots}&{1}\\
{}&{}&{}&{1}&{0}
\end{array}} \right]$$
What is the inverse $A^{-1}$?
Clearly, $A^{-1}$ is symmetric.
I look for a proof of the following conjecture that $A^{-1}$ is given as follows:
If $A_{i, j}^{-1}$ such that $j$ is odd and $i =1+j + 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$. From which follows by symmetry:
If $A_{i, j}^{-1}$ such that $j$ is even and $i =-1+j - 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$.
All other $A_{i, j}^{-1} = 0$.
Here is an example, computed with Matlab, for $n=10$ which shows the structure:
$${A_{10 \times 10}^{-1}} = \left[ {\begin{array}{*{20}{r}}
0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 & 1 \\
1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 & 0
\end{array}} \right]$$
|
The most direct way to establish the conjecture is to perform the multiplication $E = A^{-1} A$ with the conjectured $A^{-1}$ and show that the result is the unit matrix. Since $A$ is bi-diagonal, each entry of $E$ is the sum of at most two terms, so there is little confusion.
$$E_{ik} = \sum_{j=1}^n A^{-1}_{ij} A_{j k} = A^{-1}_{ij} A_{j, j-1} \delta_{j-1,k} + A^{-1}_{ij} A_{j, j+1} \delta_{j+1,k}\\
= A^{-1}_{i, k+1} A_{k+1, k} + A^{-1}_{i, k-1} A_{k-1, k} \\
= A^{-1}_{i, k+1} + A^{-1}_{i, k-1} $$
Now the diagonal is
$$E_{ii}
= A^{-1}_{i, i+1} + A^{-1}_{i, i-1} $$
From the structure of $A^{-1}$, if $i$ is odd, then $A^{-1}_{i, i+1} = 1 $ and $ A^{-1}_{i, i-1} = 0 $. If $i$ is even, then $A^{-1}_{i, i+1} = 0 $ and $ A^{-1}_{i, i-1} = 1 $. So in all cases, $E_{ii} = 1$.
For $k \ne i $, i.e the off-diagonal terms, we have $E_{ik} = A^{-1}_{i, k+1} + A^{-1}_{i, k-1} $. From the structure of $A^{-1}$, either both of these summands are zero, or one is $+1$ and the other $-1$, as we have values of $(-1)^m$ for an index difference of $2m$. Hence, always $E_{ik} =0$.
This completes the proof. $\qquad \Box$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2991218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
How to show $\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$ using induction I´d like to show that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}<2$$ using the fact that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$$
I guess the answer use transitivity of natural numbers and the inequality $$n+1<2^n$$ In this case, i can use it because i suppose it for natural n such that n>1. I'd like to have something like $$....+(n+1)<2(2^{n+1})$$ because you can divide both sides by $2^{n+1}$, and so, you get $$...+\frac{n+1}{2^{n+1}}<2$$
I´ve noticed that $$2^n+n+1<2^n+2^n=2^{n+1}<2(2^{n+1})$$ and it leads me to think $$...+n+1<2^n+n+1$$ i´ve tried dividing both sides ,of the inequality i´m assuming as a fact, by $2^n/2$ in order to get $2^n$
in the right hand of the inequality and then, add $n+1$ to both sides: $$(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n})\frac{2^n}{2}+n+1<2(\frac{2^n}{2})+n+1=2^n+n+1$$ but at this part i get stuck because if i continue it doesn't lead me to what i´d like to show.
Please, help me. I hope someone here have already proved this before, because i don't find anything similar to this at any part of the web.
|
In order to use induction in the way you tried, you may show a stronger inequality which implies the original one, such as
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}\leq 2-\frac{n+a}{2^n}$$
where $a$ is a non-negative constant to be determined.
The basic step: for $n=1$ is $\frac{1}{2}\leq 2-\frac{1+a}{2}$, that is $a\leq 2$.
Then, at the inductive step, for $n\geq 1$,
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}\leq \left(2-\frac{n+a}{2^n}\right)+\frac{n+1}{2^{n+1}}\stackrel{?}{\leq}2-\frac{n+1+a}{2^{n+1}}$$
and the last inequality simplifies to $a\geq 2$.
So $a=2$ works!!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2992278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Evaluating $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ for a triangle with sides $2$, $3$, $4$
What is
$$\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$$
for a triangle with sides $2$, $3$, and $4$?
One can use Heron's formula to get $\sin A$, etc, and use $\cos A = (b^2+c^2-a^2)/(2bc)$ to get the cosines. But that's lots of calculate.
Is there a better way to get the answer? Thanks!
|
Using an application of the Inscribed Angle Theorem, we get
$$
\begin{align}
2R\sin(A)=a\tag{1a}\\
2R\sin(B)=b\tag{1b}\\
2R\sin(C)=c\tag{1c}
\end{align}
$$
where $R$ is the radius of the circumcircle.
Furthermore, with $s=\frac{a+b+c}2$,
$$
\begin{align}
\text{Area}
&=\sqrt{s(s-a)(s-b)(s-c)}\tag2\\[3pt]
&=\frac12bc\sin(A)\tag3\\
&=\frac{abc}{4R}\tag4
\end{align}
$$
Explanation:
$(2)$: Heron's Formula
$(3)$: triangular area given by Cross Product
$(4)$: apply $\text{(1a)}$ to $(3)$
Therefore,
$$
\begin{align}
\sin(A)+\sin(B)+\sin(C)
&=\frac{a+b+c}{2R}\tag5\\
&=\frac{4s\sqrt{s(s-a)(s-b)(s-c)}}{abc}\tag6
\end{align}
$$
Explanation:
$(5)$: apply $\text{(1a)}$, $\text{(1b)}$, and $\text{(1c)}$
$(6)$: get $R=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$ from $(2)$ and $(4)$
The Law of Cosines says
$$
\begin{align}
\cos(A)&=\frac{b^2a+c^2a-a^3}{2abc}\tag{7a}\\
\cos(C)&=\frac{c^2b+a^2b-b^3}{2abc}\tag{7b}\\
\cos(C)&=\frac{a^2c+b^2c-c^3}{2abc}\tag{7c}
\end{align}
$$
Adding these and factoring yields
$$
\begin{align}
\cos(A)+\cos(B)+\cos(C)
&=\frac{(a+b-c)(a-b+c)(-a+b+c)}{2abc}+1\tag8\\
&=\frac{4(s-a)(s-b)(s-c)}{abc}+1\tag9
\end{align}
$$
Combining $(6)$ and $(9)$ gives
$$
\frac{\sin(A)+\sin(B)+\sin(C)}{\cos(A)+\cos(B)+\cos(C)}=\frac{4s\sqrt{s(s-a)(s-b)(s-c)}}{4(s-a)(s-b)(s-c)+abc}\tag{10}
$$
Plugging $(a,b,c)=(2,3,4)$ into $(10)$ gives
$$
\begin{align}
\frac{\sin(A)+\sin(B)+\sin(C)}{\cos(A)+\cos(B)+\cos(C)}
&=\frac{4\cdot\frac92\sqrt{\frac92\cdot\frac52\cdot\frac32\cdot\frac12}}{4\cdot\frac52\cdot\frac32\cdot\frac12+2\cdot3\cdot4}\\
&=\frac{3\sqrt{15}}{7}\tag{11}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2992867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
System of equations $a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2)$ Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
$$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$
I have a solution
Let $f(x)=\frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for the number of fixed points of $f(f(f(f(x))))$
1) fixed points of $f(x)$
Any fixed point $u$ of $f(x)$ gives a solution $(u,u,u,u)$.
These fixed points are solution of $x^3-x+t=0$ and so :
If $t\in(0,\frac 2{3\sqrt 3})$ : three such solutions
If $t=\frac 2{3\sqrt 3})$ : two such solutions
If $t\in(\frac 2{3\sqrt 3},+\infty)$ : one such solution
2) fixed points of $f(f(x))$ which are not fixed points of $f(x)$
Any fixed point $u$ of $f(f(x))$ which is not fixed point of $f(x)$ gives a solution $(u,f(u),u,f(u))$
These fixed points are solution of $x^2-tx-1=0$ and so :
Always two solutions to this quadratic giving the same solutions to the original equation (just a circular permutation)
3) fixed points of $f(f(f(f(x))))$ which are neither fixed points of $f(x)$, neither fixed points of $f(f(x))$
These are solutions of a degree $12$ or degree $11$ ugly polynomial :
$(t^4-3t^2+1)x^{12}-t^3x^{11}+(-7t^4+18t^2-6)x^{10}$
$+(-t^5+6t^3)x^{9}+(26t^4-48t^2+15)x^{8}+(6t^5-14t^3)x^{7}$
$+(t^6-54t^4+72t^2-20)x^{6}+(-14t^5+16t^3)x^{5}+(-7t^6+64t^4-63t^2+15)x^{4}$
$+(-t^7+14t^5-9t^3)x^{3}+(11t^6-41t^4+30t^2-6)x^{2}+(2t^7-5t^5+2t^3)x^{1}$
$+(t^8-6t^6+11t^4-6t^2+1)$
And I dont know how to determine the number of real roots of this polynomial :
For certain values (for example $t=\frac 38$), this polynomial has $12$ real roots and so $3$ new solutions of original equation (leading to $3+1+3=7$ solutions to the equation (counting as one all circular permutations))
For certain values (for example $t=10$), this polynomial has no real roots and so only $2$ solutions for the equation.
There is certainly a clever transformation for the equation $f(f(f(f(x))))=x$ allowing to count the roots depending on $t$ easily.
But I did not see it up to now.
|
I set your polynomial to $0$ and graphed it in Desmos with $t$ along the x-axis (and $x$ along the y-axis... oops!) For each value of $t$, I drew a vertical line and counted the number of intersections:
\begin{align*}
12 \quad &\text{when}\; 0<t<\frac{\sqrt{5}-1}{2}\\
11 \quad &\text{when}\; t=\frac{\sqrt{5}-1}{2}\\
12 \quad &\text{when}\; \frac{\sqrt{5}-1}{2}<t<0.6432\\
8 \quad &\text{when}\; t \approx 0.6432\\
4 \quad &\text{when}\; 0.6432<t<\frac{\sqrt{5}+1}{2}\\
3 \quad &\text{when}\; t=\frac{\sqrt{5}+1}{2}\\
4 \quad &\text{when}\; \frac{\sqrt{5}+1}{2}<t<2\\
2 \quad &\text{when}\; t=2\\
0 \quad &\text{when}\; t>2\\
\end{align*}
When $t\approx0.6432$, it appears to be tangent at 4 points simultaneously, but I'm not sure.
When $t=2$, Wolfram Alpha can factor your polynomial into the form shown below, so it's clear that this transition occurs at exactly $t=2$. These two roots are already accounted for by $f(f(x))$.
$$ (x^2 - 2 x - 1)^2 (5 x^8 + 12 x^7 - 8 x^6 - 60 x^5 - 38 x^4 + 68 x^3 + 64 x^2 + 12 x + 25) $$
There are two vertical asymptotes at $t=\frac{\sqrt{5}\pm 1}{2}$. I'm guessing 11 intersections means 2 circular permutations, and 3 means 0? Anyway, the way I got the asymptotes was using this equation (sorry about the confusing variables):
$$
\frac{y-x}{y}=\left(\frac{x\left(1-\left(\frac{x}{1-y^2}\right)^2\right)^2}{\left(1-\left(\frac{x}{1-y^2}\right)^2\right)^2-x^2}\right)^2
$$
As $y \to\infty$, the left-hand side goes to 1 and $\frac{x}{1-y^2} \to 0$. Therefore:
$$
\pm x\left(1-0^2\right)^2=\left(1-0^2\right)^2-x^2
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2996282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Prove eigenvalues of a symmetric matrix are in a certain interval I am given a matrix $A=\begin{bmatrix}1&2&0\\0&1&2\\0&0&1\end{bmatrix}$. I am asked to compute $A^tA=\begin{bmatrix}1&2&0\\2&5&2\\0&2&5\end{bmatrix}$
and then to prove that the eigenvalues of $A^tA$ are all in $]0,8[$.
I have no idea how to prove this. I think it has something to do with matrix norms ?
I don't know whether the previous questions are relevant here, but I was asked to compute
$e^{tA}=\begin{bmatrix}e^t&2te^t&2t^2e^t\\0&e^t&2te^t\\0&0&e^t\end{bmatrix}$ and $A^{-1}=\begin{bmatrix}1&-2&4\\0&1&-2\\0&0&1\end{bmatrix}$.
|
If $A^TAx = \lambda x$, then $x^TA^TAx = \lambda x^Tx$, so $\lambda ||x||^2 = ||Ax||^2$, where $||y|| = \sqrt{\sum y_i^2}$ is the norm of the vector $y$. It follows that $\lambda > 0$, since $A$ is invertible, so $||Ax|| \neq 0$ for $x \neq 0$.
To see that $\lambda < 8$, we need to show that for all $x$, we have $||Ax||^2 < 8 ||x||^2$. For this, note that $x = (a,b,c) \implies Ax = (a+2b,b+2c,c)$, in which case $$||Ax||^2 = a^2 + 4b^2 + 4ab + b^2 + 4c^2 + 4bc + c^2 = a^2 + 5b^2 + 5c^2 + 4ab+4bc$$
Take the difference $8||x||^2 - ||Ax||^2$, it is equal to $7a^2+3b^2+3c^2 -4ab-4bc$. Can we prove this is greater than zero for all $(a,b,c)$ non-zero?
Well, we can, by combining the $ab$ and $bc$ nicely into squares.Like this:
$$
(2b^2 -4bc+2c^2) + (4a^2 - 4ab+b^2) + c^2 + 3a^2 = 2(b-c)^2 + (2a-b)^2 + c^2 + 3a^2
$$
which is a positive linear combination of squares. Note that if the RHS equals zero, this forces $c=a=0$ and $b-c = 0$ so $b = 0$. In other words, if $(a,b,c) \neq (0,0,0)$ then the difference is positive, giving $\lambda < 8$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2998090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Solve this Semi-Linear PDE (Partial Differential Equation) with the Characteristic Method I need to solve this linear PDE:
$3u_x - 4u_y = y^2$
The initial condition provided is:
$ u (0,y)= sin(y)$
I need to use the Characteristic Method. I learned the method from this video.
I have reached an answer. However, I am not sure if it is wright.
My intermediate steps are:
First constant: $c_1= y + \frac{4}{3}x $
Second constant: $c_2= \frac{y^3}{3} + 4u $
Using an arbitrary function G to make the relation between both constants,
$c_2 =G(c_1) $, we have that:
$\frac{y^3}{3} + 4u = G(y + \frac{4}{3}x) $
With the initial condition we have:
$G(y) = \frac{y^3}{3} +4sin(y)$
After the definition of $G(y)$ above , I inputed the value of $c_1$ , having:
$G(y + \frac{4}{3}x) = \frac{(y+\frac{4}{3}x)^3}{3}+ 4sin(y+\frac{4}{3}x) $.
Finally, solving for $u$:
$u(x,y) = \frac{(y+\frac{4}{3}x)^3}{12}+sin(y+\frac{4}{3}x) - \frac{y^3}{12}$
A friend of mine solved this problem with a different approach. She reached a different result. There are some comments along her solution that were written in portuguese.
Is this right?
If I did something wrong, what was it?
Thanks in advance!
|
$$u(x,y) = \frac{(y+\frac{4}{3}x)^3}{12}+\sin(y+\frac{4}{3}x) - \frac{y^3}{12}\quad\text{is correct}$$
Expanding leads to :
$$u(x,y)=\sin(y+\frac{4}{3}x)+\frac{y^2x}{3}+\frac{4yx^2}{9}+\frac{16x^3}{81}$$
So, there is no mistake in your calculus. There is a sign mistake in the handwritten page, which at end gives $\sin(y+\frac{4}{3}x)-\frac{y^2x}{3}+\frac{4yx^2}{9}-\frac{16x^3}{81}$.
Unfortunately the handwritten page is not enough readable to see where exactly the mistake occurred.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2998680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Solving $8x^3 - 6x + 1$ using Cardano's method Solve for the first root of $8x^3 - 6x + 1 = 0$
After solving I get $\sqrt[3]{\frac{-1 + \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$, which is not a solution to the cubic equation:
Here's how I come up with:
Using Cardano's method:
let $x = y - \frac{b}{3a}$
Then compress:
Since $b$ is $0$, then it turns out to be $8y^3 - 6y + 1 = 0$.
Divide $8$ to both sides.
$$ y^3 - \frac{3}{4}y + \frac{1}{8} = 0 $$
Let $3st = \frac{-3}{4}$ and $s^3 - t^3 = \frac{-1}{8}$
Now:
\begin{align}
\left(\frac{-1}{4t}\right)^{3} - t^{3} &= \frac{1}{8}\\
...\\
8t^6 - t^3 + \frac{1}{8} &= 0\\
\end{align}
I'll uncompress the equation above so it becomes quadratic:
\begin{align}
8t^2 - t + \frac{1}{8} &= 0\\
...\\
\left(\frac{1 + \sqrt{3}i}{16}\right)\left(\frac{1 - \sqrt{3}i}{16}\right)\\
\end{align}
Then take the cuberoot to get $t$ (only taking the positive root):
$$
\sqrt[3]{\frac{1 + \sqrt{3}i}{16}}
$$
Since $x = y - \frac{0}{24}$, which is similar to $x = y$
and $y = s - t$, then:
$t = \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$
$s = \sqrt[3]{\frac{1 - \sqrt{3}i}{16}}$
\begin{align}
y &= s - t\\
y &= \sqrt[3]{\frac{1 - \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}\\
x &= y + 0\\
x &= \sqrt[3]{\frac{1 - \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}\\
\end{align}
Checking if it is a solution, it turns out that it is not. Note: I'm new to this method so its unclear to me why it dont work and PLEASE dont mark this as a duplicate. Thanks.
|
With free computer algebra system Maxima 5.42.1
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3000859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to solve the limit $\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}$. How to solve this limit??
$$\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}$$
It's a limit, not a series
|
\begin{align}
\lim_{k\to\infty} \frac{(2k)!}{2^{2k}\cdot(k!)^2}
&=\lim_{k\to\infty} \frac{(2k)!}{2^k \cdot 2^k \cdot k! \cdot k!} \\
&=\lim_{k\to\infty} \frac{(2k)!}{(2^k \cdot k!)^2} \\
&=\lim_{k\to\infty} \frac{(2k)(2k-1)\cdots(2)(1)}{(2k)^2 (2k-2)^2 \cdots (4)^2 (2)^2} \\
&=\lim_{k\to\infty} \frac{(2k-1)(2k-3)\cdots(1)}{(2k)(2k-2)\cdots(2)} \\
&=0
\end{align}
In the last step, you can think of the fraction as the infinite product of fractions less than $1$ ($\frac{1}{2} \times\frac{3}{4} \times \frac{5}{6} \times\frac{7}{8}...$), which will decrease to $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3008771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
What is $\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$? What's the result of:
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$$
Is it
$$\frac{1}{|{x}|}-\frac{x^2}{|x|x^2}=\frac{1}{|{x}|}-\frac{1}{|x|}=0$$
or
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^2x}=\frac{1}{|{x}|}-\frac{1}{x}\frac{x^2}{|x|^2}=\frac{1}{|{x}|}-\frac{1}{x}=\left\{\begin{matrix}\frac{2}{x},x<0\\ 0,x>0 \end{matrix}\right.$$
|
Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^3}=\frac{1}{|{x}|}-\frac{x^2}{|x|x^2}=\frac{1}{|{x}|}-\frac{1}{|{x}|}=0$$
For the same reason the second one is wrong.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3010837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
}
|
Find the range of $xy$ under the conditions: $x^2-xy+y^2=9$ and $|x^2-y^2|<9$ Assume $x,y \in \mathbb{R}^+$, and satisfied the following expression:
$$x^2-xy+y^2=9$$
$$\left|x^2-y^2\right| < 9$$
find the range of $xy$
My approach: $x^2-xy+y^2=9$ $\Rightarrow$ $xy+9=x^2+y^2 \geq 2xy$ $\Rightarrow$ $xy \leq 9$
But I don't know how to find the lower bound. please help me..thanks very much.
|
For the lower bound you can use the follwing facts:
*
*$x^2-xy+y^2 = (x-y)^2 +xy \Rightarrow 9-xy = (x-y)^2$
*$x^2-xy+y^2 = (x+y)^2 -3xy \Rightarrow 9+3xy = (x+y)^2$
*$\left|x^2-y^2\right| < 9 \Leftrightarrow (x+y)^2(x-y)^2 < 81$
Plugging 1. and 2. into 3. you get:
$$(x+y)^2(x-y)^2 < 81 \Leftrightarrow (9-xy)(9+3xy) < 81 \Leftrightarrow xy(6-xy) < 0 \stackrel{x,y >0}{\Leftrightarrow} \boxed{xy>6}$$
Together with your upper bound $\boxed{xy \leq 9}$ you get
$$\boxed{6 < xy \leq 9}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3013749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding the Maclaurin series of $e^{\sin x}$ by comparing coefficients I believe I have found a nice way to find the Maclaurin series of $e^{\sin x}$. Please check if there are any mistakes with my working. Is this method well known?
|
(Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $\cos x\approx1-\frac{x^2}{2}+\frac{x^4}{24}$.)
Let $f(x)= e^{\sin x}$, therefore $f'(x)=\cos x*e^{\sin x}=\cos x*f(x)$ by chain rule.
Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f^{'}(x)=b+2cx+3dx^2+4ex^3+5fx^4$.
Sub in known expressions into $f^{'}(x)=\cos x*f(x)$ to get: $b+2cx+3dx^2+4ex^3+5fx^4=(1-\frac{x^2}{2}+\frac{x^4}{24})(a+bx+cx^2+dx^3+ex^4+fx^5)$.
Expand the right hand side to get: $RHS=a+bx+(c-\frac{1}{2})x^2+(d-\frac{b}{2})x^3+(e-\frac{c}{2}+\frac{a}{24})x^4+(f-\frac{d}{2}+\frac{b}{24})x^5$.
Comparing coefficients of $LHS$ and $RHS$ yields:
$b=a$
$2c=b$
$3d=c-\frac{1}{2}$
$4e=d-\frac{b}{2}$
$5f=e-\frac{c}{2}+\frac{a}{24}$.
When $x=0$, $f(x)=1$ and $f(x)=a$, therefore $a=1$. Which can be used to find:
$b=1$
$c=\frac{1}{2}$
$d=0$
$e=-\frac{1}{8}$
$f=-\frac{1}{15}$.
This means the Maclaurin series of $e^{\sin x}=1+x+\frac{1}{2}x^2-\frac{1}{8}x^4-\frac{1}{15}x^5$ which is indeed to correct series.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3015698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Conditional Probability of a Uniform Random Subset.
Question:
Let X = $({1,2,3,..,10})$ Let $Y$ be a uniformly random subset of $X$. Define the events:
A = “$Y$ contains at least $4$ elements",
B = “all elements of $Y$ are even".
What is $Pr(A|B)$?
Answer: 0.1875
Attempt:
I know that P(A $\bigcap$ B) / P(B) is what I have to ultimately find.
For $P(B)$ = $\frac{5}{10}$ = $\frac{1}{2}$
For P(A) => Must have exactly 4, 5, 6, 7, 8, 9, 10 elements
So, I used the binomial for this by doing:
P(A) =
$10\choose4$$.$ $\frac{1}{10}$$^4$$.$$(1-\frac{1}{10}$)$^6$ +
$10\choose5$$.$ $\frac{1}{10}$$^5$$.$$(1-\frac{1}{10}$)$^5$ +
$10\choose6$$.$ $\frac{1}{10}$$^6$$.$$(1-\frac{1}{10}$)$^4$ +
$10\choose7$$.$ $\frac{1}{10}$$^7$$.$$(1-\frac{1}{10}$)$^3$ +
$10\choose8$$.$ $\frac{1}{10}$$^8$$.$$(1-\frac{1}{10}$)$^2$ +
$10\choose9$$.$ $\frac{1}{10}$$^9$$.$$(1-\frac{1}{10}$)$^1$ +
$10\choose10$$.$ $\frac{1}{10}$$^{10}$$.$$(1-\frac{1}{10}$)$^0$ +
= $0.012795$
P(A $\bigcap$B) = $\frac{1}{2}*0.012795$
Pr(A|B) = $\frac{\frac{1}{2}*0.012795}{2}$
Pr(A|B) = $0.012795$
Where did I go wrong with this approach? I find finding P(A) very time consuming, I feel like there has to be a much more simpler approach.
|
$5$ elements of $X$ are even.
To have at least $4$ elements and all elements are even, you have either $4$ elements ($5$ ways, choose one to leave out) or $5$ elements ($1$ way).
$$\frac{1+5}{2^{5}}=\frac{6}{32}$$
Remark:
*
*We are choosing uniformly over all subsets. $P(B)= \frac{2^5}{2^{10}} \ne \frac{5}{10}$.
*We don't have to find $P(A)$, what is of interest is $P(A \cap B)$.
*We do not have $P(A \cap B)=P(A)P(B)$ in general, we need independence.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3019627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.
Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.
To be monotone it must be either increasing or decreasing, so:
$a_n \ge a_{n-1}$ or $a_n \le a_{n-1}$
$\sqrt{n+1}-\sqrt{n} \ge? \sqrt{n}-\sqrt{n-1} $
$\sqrt{n+1}+\sqrt{n-1} \ge? \sqrt{n} + \sqrt{n}$
I know that $\sqrt{n+1} \ge \sqrt{n}$ and $\sqrt{n} \ge \sqrt{n-1}$
But I can't make a statement about their sum since they have opposite signs that is one is greater or equal and the other is less or equal.
Can you help me to figure out the solution?
|
Note that, for $n\geq 1$,
$$ (\sqrt{n+1}+\sqrt{n-1})^2 = n+1+2\sqrt{n^2-1}+n-1 = 2n+2\sqrt{n^2-1}\leq 2n +2n=4n, $$
so that
$$ \sqrt{n+1}+\sqrt{n-1} \leq 2\sqrt{n} \quad \iff\quad \sqrt{n+1}- \sqrt n \leq \sqrt n - \sqrt{n-1}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3021788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Minimum value of the given function
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .
Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$
Hence the point lies on the line y=x+2 it is the minimum sum of distance from the above given two points.
But from here I am not able to get the value of x and hence $\alpha$. Any suggestions?
|
If we write this as $$\sqrt{x^2+(x+1)^2}+\sqrt{(x-2)^2+(x-3)^2}$$ then we are searching for a point $T$ on line $y=x$ for which $TA+TB$ takes minumim where $A(0,-1)$ and $B(3,2)$.
Now this is well know problem froma ancient greek. Reflect $A$ acros this line and get $A'(-1,0)$. By triangle inequality we can see that $TA+TB\geq A'B$ and that minimum is achieved at intersection of lines $y=x$ and line $A'B$ which is $y = {x +1 \over 2}$.
So $\alpha = A'B^2 = 20$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3022822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
}
|
Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not possible. I see why this wouldn't be possible, I'm not sure if it really is the cause which makes that equation illegal though.
But now, apparently, I could do this:
$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x=\lim_{x\to -\infty}\frac{[\sqrt{x^2+5x+3}+x][\sqrt{x^2+5x+3}-x]}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{x^2+5x+3-x^2}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5x+3}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5+3/x}{\sqrt{1+5/x+3/x^2}-1}=-5/2$
which gives me the correct result.
But in the 3th step I used $x^2-x^2=0$, how is that legal?
Also, in the 2nd step I implicitly used:
$-x\sqrt{x^2+5x+3}+x\sqrt{x^2+5x+3}=0$
Which also seems to be fine, but why?
|
I will flesh out much of what would be considered a series of valid steps you may take in order to get something similar to the first step performed in the initial limit calculation. I will note that the second step that was taken is valid because we perform all of the algebra in the limit expression first, then we take a limit of what results.
The following line of reasoning shows that whenever you have $\sqrt{x^2+\beta x+\alpha}$ as a term (added or substracted, not multiplied or divided) in a limit, it may be replaced freely by $|x+\frac{\beta}{2}|$ inside the limit (of course slight adjustments must be made for it to be shown in general).
Note that for every $b>\frac{5}{2}$,
$$\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$
To see why this is the case, here is a series of inequalities. First pick a $b>\frac{5}{2}$. Then pick $x$ so that $x\le\frac{3-b^2}{2b-5}$.
Then
$$(2b-5)x\le 3-b^2,$$ $$2bx+b^2\le 5x+3,$$
$$x^2+2bx+b^2\le x^2+5x+3,$$
$$\sqrt{x^2+2bx+b^2}\le\sqrt{x^2+5x+3},$$
$$\sqrt{x^2+2bx+b^2}+x\le\sqrt{x^2+5x+3}+x,$$
$$\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$
The intuition for this comes largely from working backward. So thus the claim is justified.
So now we can say that
$$\lim_{b\to \frac{5}{2}^-}\lim_{x\to-\infty}\sqrt{x^2+2bx+b^2}+x \le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x,$$
$$\lim_{b\to \frac{5}{2}^-}|x+b|+x=\lim_{b\to \frac{5}{2}^-}-x-b+x=\lim_{b\to \frac{5}{2}^-}-b=-\frac{5}{2}\le\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x.$$
Now since $3<\frac{25}{4}$, we have that.
$$\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x\le \lim_{x\to-\infty}\sqrt{x^2+5x+\frac{25}{4}}+x=\lim_{x\to-\infty}|x+\frac{5}{2}|+x=\lim_{x\to-\infty}-x-\frac{5}{2}+x=\lim_{x\to-\infty}-\frac{5}{2}=-\frac{5}{2}.$$
So $$\lim_{x\to-\infty}\sqrt{x^2+5x+3}+x=-\frac{5}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3024120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
}
|
Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $
Prove the following identities:
$$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\tag {ii}$$
For $(\mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1.
I then tried just multiplying out the brackets and got as far as $$1+ \sec^2x + \frac{2}{\cos^2x \sin^2x}$$
|
(i) Let $C=\cos(2x), S=\sin(2x)=2\sin(x)\cos(x)$
$S^2*RHS= S^2(1+\frac{2}{(S/2)^2})= S^2+8=9-C^2$
$\begin{align}S^2*LHS
&=(2+2\sin^2(x))(2+2\cos^2(x)) \cr
&=(2+(1-C))(2+(1+C))\cr
&=9-C^2 =S^2*RHS
\end{align}$
QED
(ii) Let $t=\tan(x/2),\text{then }\sin(x)=\frac{2t}{1+t^2}\text{ , }\cos(x)=\frac{1-t^2}{1+t^2}$
$1-\tan(x)=1-\frac{2t}{1-t^2}=\frac{1-2t-t^2}{1-t^2}$
$1-\cot(x)=1-\frac{1-t^2}{2t}=\frac{1-2t-t^2}{-2t}$
$\begin{align}
\frac{\cos(x)}{1-\tan(x)} + \frac{\sin(x)}{1-\cot(x)}
&=\frac{(1-t^2)^2 - (2t)^2}{(1-2t-t^2)(1+t^2)} \cr
&=\frac{(1-2t-t^2)(1+2t-t^2)}{(1-2t-t^2)(1+t^2)} \cr
&= \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} = \sin(x) + \cos(x)
\end{align}$
QED
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3027602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Volume between cone and sphere of radius $\sqrt2$ with surface integral Consider the cone $z^2=x^2+y^2$ between $z=0$ and $z=1$. Find the volume of the region above this cone and inside the sphere of radius $\sqrt2$ centered at the origin that encloses the cone.
The straightforward approach to this problem would have been a triple integral in spherical coordinates, but for practice I tried using the cone as a surface and using a surface integral to find the volume between the cone and the sphere. I first parameterized the surface using cylindrical coordinates $x=r\cos\theta$, $y=r\sin\theta$, $z=r$ (since $z^2=x^2+y^2$), and then found the intersection points using the equations $z=r$ and $r^2+z^2=2$:
$$2z^2=2\Rightarrow z=1\Rightarrow r=1$$
Now that the bounds of integration have been found I used the function $z=\sqrt{2-r^2}$ as the function to integrate and set up my surface integral as:
$$\iint_S \sqrt{2-r^2}\Vert\vec{r_r}\times\vec{r_\theta}\Vert\,\text{d}S$$
I then computed the cross product, for $\vec{r}(r,\theta)=\left<r\cos\theta,r\sin\theta,r\right>$:
$$\begin{vmatrix}
\hat{\imath}&\hat{\jmath}&\hat{k}\\
\cos\theta&\sin\theta&1\\
-r\sin\theta&r\cos\theta&0
\end{vmatrix}=\hat{\imath}(-r\cos\theta)-\hat{\jmath}(-r\sin\theta)+\hat{k}(r\cos^2\theta+r\sin^2\theta)$$
So the surface integral becomes
$$\int_0^{2\pi}\int_0^1\sqrt{\left(2-r^2\right)\left(r^2\cos^2\theta+r^2\sin^2\theta+r^2\right)}\,\text dr\,\text d\theta$$
$$=\int_0^{2\pi}\int_0^1\sqrt{\left(2-r^2\right)r^2\left(\cos^2\theta+\sin^2\theta+1\right)}\,\text dr\,\text d\theta$$
$$=\int_0^{2\pi}\int_0^1\sqrt{2r^2(2-r^2)}\,\text dr\text d\theta$$
$$=2\sqrt2\pi\int_0^1r\sqrt{2-r^2}\,\text dr$$
Using a $u$-substitution of $u=2-r^2$:
$$\sqrt2\pi\int_1^2
\sqrt{u}\,\text du=\sqrt2\pi\left.\left(\frac{2u\sqrt u}{3}\right)\right|_1^2=\sqrt2\pi\left(\frac{4\sqrt2-2}{3}\right)$$
$$=\frac{2\pi}{3}\left(4-\sqrt2\right)$$
However, the answer given in the answer key says that the volume is $\frac{4\pi}{3}(\sqrt2-1)$, so I would like to know why my answer is incorrect, whether that be a computational mistake or faulty reasoning as to why a surface integral would work here to find the volume.
EDIT: Here is the posted solution:
|
We can also calculate by cap area and its solid angle.
Denoting radius by $ R={\sqrt 2}$
Cap Area $A= 2 \pi R. R (1-1/\sqrt 2 ) $
Solid Angle of Cap = $\dfrac{A}{ R^2} = 2 \pi (1-1/\sqrt 2 )$
Remaining Solid Angle $V_1 = 4 \pi -\dfrac{A}{ R^2} = 2 \pi (1+1/\sqrt 2 )$
Remaining Volume $V= \dfrac{R^3}{3}. V_1 = \dfrac{4 \pi }{3}(\sqrt 2 +1)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3028125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Factor $10^n -1$ There is an identity or quick way to factor problems of the form $c^x -1$ e.g. $10^n-1$, what is it?
I mean that it can be represented in some way as multiple factors,e.g. $(x-a)(y-b)$, there is a way to expand the difference easily.
|
Since it’s subtraction, you could use the difference of two squares I suppose.
$$a^2-b^2 = (a+b)(a-b) \implies a^c-b^c = \big(a^{\frac{c}{2}}+b^{\frac{c}{2}}\big)\big(a^{\frac{c}{2}}-b^{\frac{c}{2}}\big)$$
In this case, you have $b = 1$, so the factorization is
$$\big(c^{\frac{x}{2}}+1\big)\big(c^{\frac{x}{2}}-1\big) \implies (10^5+1)(10^5-1)$$
A more commonly used idea is factoring the difference of $n^{th}$ powers for all $n$ and the sum of $n^{th}$ powers for all odd $n$ (which isn’t relevant but is very similar to the former).
$$a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+…+a^2b^{n-3}+ab^{n-2}+b^{n-1})$$
This works because through expansion because multiplying $a$ to each term is canceled by multiplying $b$ to the term before it, which leaves only the first term for $a$ and the last term for $b$.
$$ = \underbrace{a^n\color{blue}{+a^{n-1}b+a^{n-2}b^2+…+a^3b^{n-3}+a^2b^{n-2}+ab^{n-1}}}_{a(a^{n-1}+a^{n-2}b+…+ab^{n-2}+b^{n-1})}\underbrace{\color{red}{-a^{n-1}b-a^{n-2}b^2-a^{n-3}b^3-…-a^2b^{n-2}-ab^{n-1}}-b^n}_{-b(a^{n-1}+a^{n-2}b+…+ab^{n-2}+b^{n-1})}$$
$$= a^n-b^n$$
So, for instance, you can factor your example as follows:
$$10^n-1 = (10-1)(10^{n-1}+10^{n-2}+10^{n-3}+…+1)$$
which also bears striking resemblance to the geometric series if written as $\frac{10^n-1}{10-1} = 1+10+10^2+…+10^{n-1}$.
As a side note, you can use
$$(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-a^{n-4}b^3+…+b^{n-1})$$
to factor the sum of $n^{th}$ powers, but this only works for an odd $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3029574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Simplify third degree polynomial equations. Given an equation:
$6x^3 - 13x^2 + 8x + 3 = 0$
Broken down to get one form
$(x + {3\over2})$
How can you divide the prior equation to know it will simplify to
$(x + {3\over2})(6x^2 + 4x + 2) = 0$
|
Following @MartinArgerami's discovery that there was probably a typo.
if we have a degree of three polynomial we can use the Rational Zero Theorem as such:
Ration Zero Theorem: For $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_2x^2 + a_1x + a_0$, every rational zero of $f(x)$, $\frac{p}{q}$ has $p$ as a factor of $a_0$ and $q$ as a factor of $a_n$
We have the polynomial $f(x) = 6x^3 + 13x + 8x + 3$
p) Factors of the constant $a_0 = 3$ : $\pm 1$, $\pm3$
q) Factors of the coefficient $a_n = 6$ : $\pm1$, $\pm2$, $\pm3$, $\pm6$
$\frac{p}{q_1}$ : $\pm1$, $\pm3$
$\frac{p}{q_2}$ : $\pm\frac{1}{2}$, $\pm\frac{3}{2}$
$\frac{p}{q_3}$ : $\pm\frac{1}{3}$, $\pm\frac{3}{3}$
$\frac{p}{q_4}$ : $\pm\frac{1}{6}$, $\pm\frac{3}{6}$
We can see that some of these possible zeros are repeated so we can cross some of them out. Our final possible zero list is the following:
$x = \pm\frac{1}{6}$, $\pm\frac{1}{3}$, $\pm\frac{1}{2}$, $\pm1$, $\pm\frac{3}{2}$, $\pm3$
Using synthetic division we try these possible factors until we find one. Let's try $-3$.
When we try $-3$, we are saying, let us see if $(x+3)$ is a factor of $f(x)$
$\begin{array}{c|rrr}&6&13&8&3\\-3&&-18&15&-69\\\hline\\&6&-5&23&-66\\\end{array}$
We have a remainder of $-66$ so $-3$ is not a factor. We are looking for a remainder of $0$. For the sake of time, let us try $\frac{-3}{2}$ since you have found already that it is a factor.
$\begin{array}{c|rrr}&6&13&8&3\\-\frac{3}{2}&&-9&-6&-3\\\hline\\&6&4&2&0\\\end{array}$
We now have a $0$ remainder and $-\frac{3}{2}$ is a zero.
Thus, $f(x)$ can be written like this:
$f(x) = divisor(x) \times quotient(x) + remainder(x) = d(x)q(x) + r(x) \\
= (x+\frac{3}{2})(6x^2+4x+2)$
you can further factor the quotient to find the remaining zeros of $f(x)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3030976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How many non-negative solutions for $x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$? My solution:
We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$
$\Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - x_{1} \quad (*)$
Consider:
$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} \geq 0, x_{3} \geq 4, x_{4} \geq 0 \quad (**)$
$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} \geq 5, x_{3} \geq 4, x_{4} \geq 6 \quad (***)$
Let $f$ is the function that compute the number of non-negative solutions of an equation.
$\implies f(*) = f(**) - f(***)$
Thus, the number of non-negative solutions of (*) is $\sum_{x_{1} = 2}^{8}( {40 - x_{1} + 3 - 1 \choose 3 - 1} - {25-x_{1}+3 - 1 \choose 3 - 1}) = 3045$
I found that the right answer is 210 by trying some programming script. But I don't know what was wrong with my solution. Please help me. Thank you!
|
Math answer
Note that the given constraints for $x_1, x_2$ and $x_4$ and $\sum\limits_{i = 1}^4 x_i = 40$ allows us to define
$$
\begin{aligned}
x_3 &= 40 - x_1 - x_2 - x_4 \\
&\ge 40 - 8 - 4 - 5 \\
&= 23.
\end{aligned}$$
This renders the constraint $x_3 \ge 4$ redundant. As a result, the required answer is $(8-2+1) \times (4+1) \times (5+1) = 210$.
Julia Programming Script
x1 = 2:8
x2 = 0:4
x4 = 0:5
x3 = [40 - i - j - k for i in x1 for j in x2 for k in x4]
println(minimum(x3)) # returns 23
println(length(x3)) # returns 210
Test this script on Tutorial's Point's online compiler.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3032553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Question about constructing fields
Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(\sqrt 2)^\times$
Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,\,\, x^2+x+2,\,\, x^2+2x+3,\,\, x^2+2x+4,\,\, x^2+3x+3,\,\, x^2+3x+4,\,\, x^2+4x+1,\,\, x^2+4x+2$
For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$
Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(\sqrt 2)^\times$?
|
For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[\sqrt{2}]$.
If $\alpha^2 = 2$ and $\beta = x \alpha + y$ we have
$$\beta^2 + \beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)\alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[\sqrt{2}]$. Since the cardinalities are equal it's also onto.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3034349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
In how many ways 11 items can be distributed among 3 peoples such that, sum of items received by any two is more than the third
In how many ways $11$ items can be distributed among $3$ peoples such that, number of items received by any two person is more than the number of items received by the third person.
I have tried with case by case. But it will be cumbersome when the number of people and no. of items are large. Is there any general strategy to solve this problem, which can be applied for large numbers?
|
We have $x_1+x_2>x_3$ or, $x_3<\frac{x_1+x_2+x_3}{2}$. And it holds for $ x_1, x_2$ also. Hence, $x_i$ can be $5$ maximum(as they are integers) for $i\in \{1,2,3\}$. So, we need to find the number of solution of the equation $$x_1+x_2+x_3=11,~~~ 1\le x_1,x_2,x_3\le 5$$
Which is equivalent to finding coeff. of $x^{11}$ in the expression $(x+x^2+x^3+x^4+x^5)^3$.
$$\begin{align}[x^{11}](x+x^2+x^3+x^4+x^5)^3 &= [x^{11}]~\big(x^3(1+x+x^2+x^3+x^4)^3\big)\\&=[x^{8}]~(1+x+x^2+x^3+x^4)^3\\ &=[x^{8}]~\left( \frac{1-x^5}{1-x}\right)^3\\&=[x^8]~(1-x^5)^3(1-x)^{-3}\\&= [x^8]~(1-3x^5+3x^{10}-x^{15})\left(\sum_{r=0}^{\infty}\binom{-3}{r}x^r \right)\\ &= [x^8]~ (1-3x^5+3x^{10}-x^{15})\left(\sum_{r=0}^{\infty}\binom{3+r-1}{r}x^r\right)\\&=\binom{3+8-1}{8}-3\cdot\binom{3+3-1}{3}=15 \end{align}$$
Generalization
We can generalize for any perimeter $n\in\mathbb{N}$. We need to find the coeff. of $x^n$ in the expression $(x+x^2+\cdots +x^s)^3, s=\lfloor\frac{n}{2}\rfloor$.
Note: It is equivalent to finding all triangles with integer sides, which has perimeter $11$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3042299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$
Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$
Note that $(x^4+x^3+x^2+x+1)(x-1) = x^5 - 1$.
I tried reducing the equation $\mod{5}$, which gave me some information on $x$, but I was not able to utilize this information.
|
If $r= ord _p(x)$ then by Fermat $r\mid p-1$ and by proposition $r\mid 5$.
If $r=5$ then $p-1 =5k$ so $p\equiv_5 1$
If $r=1$ then $x=1$ so $p\mid 5$ and thus $p=5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3042635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Divisor of $x^2+x+1$ can be square number? $$1^2+1+1=3$$
$$2^2+2+1=7$$
$$8^2+8+1=73$$
$$10^2+10+1=111=3\cdot37$$
There is no divisor which is square number.
Is it just coincidence? Or can be proved?
*I'm not english user, so my grammer might be wrong
|
No, for $x=18$ we get $x^2+x+1=343=7^3$.
Here are the first few counterexamples:
$$
\begin{array}{rrl}
x & x^2+x+1 & \text{factorization}\\
18 & 343 & 7^3 \\
22 & 507 & 3 \cdot 13^2 \\
30 & 931 & 7^2 \cdot 19 \\
67 & 4557 & 3 \cdot 7^2 \cdot 31 \\
68 & 4693 & 13 \cdot 19^2 \\
79 & 6321 & 3 \cdot 7^2 \cdot 43 \\
116 & 13573 & 7^2 \cdot 277 \\
128 & 16513 & 7^2 \cdot 337 \\
146 & 21463 & 13^2 \cdot 127 \\
165 & 27391 & 7^2 \cdot 13 \cdot 43 \\
177 & 31507 & 7^2 \cdot 643 \\
191 & 36673 & 7 \cdot 13^2 \cdot 31 \\
214 & 46011 & 3 \cdot 7^2 \cdot 313 \\
\end{array}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3047388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Fallacious moving of powers resulting with a correct trigonometric series identity.
Prove that
$$
\\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1}
( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1}
$$
for all values of $x$.
I came across this joke which said that you could just bring the powers out of the brackets and everything works out right.
$$
\require{cancel}
\sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1}
( \sin^{2(\textbf{n+r+1})}x + \cos^{2(\textbf{n+r+1})}x )\right)
\\ = \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1}
(\cancelto1{\sin^2x + \cos^2x })^{\textbf{n+r+1}}\right)
\\ = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1}
$$
But how would you actually go about proving this? I noticed that this value is also equal to $B(n+1,n+1)$, which may or may not be relevant.
|
Let's prove $$
\\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1}
( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1} \tag1
$$
for all values of $x$.
By differentiating both sides with respect to $x$, on the right hand side one gets $0$, on the left hand side one gets
$$
\sum_{r=0}^n (-1)^r {n \choose r}
\left( 2\cos x\cdot\sin^{2(n+r+1)-1}x - 2\sin x\cdot\cos^{2(n+r+1)-1}x \right)
$$
$$
2\cos x\cdot\sin^{2n+1}x \sum_{r=0}^n (-1)^r {n \choose r}
\sin^{2r}x - 2\sin x\cdot\cos^{2n+1}x \sum_{r=0}^n (-1)^r {n \choose r}
\cos^{2r}x
$$ $$
2\cos x\cdot\sin^{2n+1}x\cdot \left(1-\sin^2x \right)^n-2\sin x\cdot\cos^{2n+1}x\cdot \left(1-\cos^2x \right)^n=0.
$$
Since both sides of $(1)$ clearly agree at $x=0$, then $(1)$ is true for all values of $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3049314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Integral of $(1-x^2)^{1/4}$ I came across this integral while originally solving the integral of $((1-\sqrt{x})/(1+\sqrt{x}))^{1/2}$ which led me to two integrals, one, $\sqrt{x}/(1-x^2)^{1/2}$ which on substituting $(1-x^2)^{1/2}$ as t gives $2(1-t^2)^{1/4}$, which is in the title as well and the other a simple $1/(1-x^2)^{1/2}$ which yeilds $\arcsin(x)$. I can't seem to find a method to solve that integral, would appreciate any help you can offer.
|
Note that:
$$
\int \frac{\sqrt x}{\sqrt{1-x^2}}\,dx = \int (1-t^2)^{1/4}\frac{-2t}{2t\sqrt{1-t^2}}\,dt
$$
But in fact you can work directly with what you originally have:
$$ I= \int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx =\int \sqrt{\frac{1-u}{1+u}}2u\,du = \int 4(z^2-1)\sqrt{2-z^2} \,dz$$
using the substitutions $x=u^2, u=z^2 -1$. Then, let $z=\sqrt{2}\sin{y}$ which gives you:
$$
I= \int 4(2\sin^2{y}-1)\sqrt{2}\cos{y},\ dy = 4\sqrt{2} \int \cos{y}\sin^2{y} -\cos^3{y} \,dz
$$
$$= \frac{4\sqrt{2}}{3} \sin^3{y} - 4\sqrt{2}(\sin{y}-\frac{\sin^3{y}}{3}) + Constant
$$
Finally you can then rewrite the result in terms of the variable $x$ using:
$$
y=\arcsin{\frac{z}{\sqrt{2}}}=\arcsin{\frac{\sqrt{\sqrt{x}+1}}{\sqrt{2}}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3056550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Does the constant $C$ in this solution to a differential equation equal infinity? The problem is $y' = -\frac{1}{t^2} - \frac{1}{t}y + y^2;\ y_p = \frac{1}{t}$. My solution is
$$\begin{align}
y = \frac{1}{t} + B &\implies y' = -\frac{1}{t^2} + B' \\
&\implies -\frac{1}{t^2} - \frac{1}{t}y + y^2 = -\frac{1}{t^2} + B' \\
&\implies -\frac{1}{t^2} - \frac{1}{t} \left(\frac{1}{t} + B \right) + \left(\frac{1}{t} + B\right)^2 = -\frac{1}{t^2} + B' \\
&\implies B' - \frac{1}{t}B = B^2 \\
&\implies L = B^{-1} \\
&\implies L' = -B^{-2} \left(B^2 + \frac{1}{t}B \right) \\
&\implies L' + \frac{1}{tB} = -1 \\
&\implies L' + \frac{1}{t}L = -1 \\
&\implies L_h = \frac{1}{t} \\
&\implies L = \frac{1}{t}\int\frac{-1}{\frac{1}{t}}dt \\
&\implies L = \frac{1}{t} \left(-\frac{1}{2}t^2 + C_{tentative} \right) \\
&\implies L = \frac{C - t^2}{2t} \\
&\implies B = \frac{2t}{C - t^2} \\
&\implies y = \frac{1}{t} + \frac{2t}{C - t^2}
\end{align}$$
Although this solution does not appear equivalent to the answer given in my book, one or two people in chat reviewed this work and could not see anything incorrect. The issue is that the given particular solution, $y_p = \frac{1}{t}$, cannot be obtained by plugging any finite value into the $C$ in my general solution. We could say that the general solution yields the given particular solution when $C =$ infinity, more specifically, when $C =$ some $\aleph$ expression which makes the term $\frac{2t}{C - t^2}$ disappear regardless of $t$, but this feels a bit outside the box for a textbook problem. Is my general solution correct, and if so, how if at all can we derive the given particular solution from it?
|
This is a very common situation. A general solution for a nonlinear $n$-th order differential equation is a solution depending on $n$ independent parameters, but it does not necessarily give you all the solutions. As in this case, some may be obtained as limits where some parameters go to $\infty$. There are also cases where some particular solutions (singular solutions) are obtained as envelopes of the general solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3057450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Proof of this formula for $\sqrt{e\pi/2}$ and similar formulas. \begin{align}
\sqrt{\frac{e\pi}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}
\end{align}
as seen here.
Is there other series that relate $\pi$ and $e$?
Also, it's possible to rewrite the continued fraction above in terms of known functions/numbers?
|
About 2 years ago I discovered a lot of pretty nice series that relate $\pi$ and $e$, for instance :
$$\sum_{n=1}^{\infty}\frac{n^2}{16n^4-1}=\frac{\pi}{32}\cdot\frac{e^{\pi}+1}{e^{\pi}-1}$$
$$\sum_{n=1}^{\infty}\frac{n^2}{4n^4+1}=\frac{\pi}{8}\cdot\frac{e^{\pi}-1}{e^{\pi}+1}$$
$$\sum_{n=1}^{\infty}\frac{n^2}{(4n^4+1)(16n^4-1)}=\frac{\pi}{10}\cdot\frac{e^{\pi}}{e^{2\pi}-1}$$
$$\sum_{n=1}^{\infty}\frac{n^2(32n^4+3)}{(4n^4+1)(16n^4-1)}=\frac{\pi}{4}\cdot\frac{e^{2\pi}+1}{e^{2\pi}-1}$$
$$\sum_{n=1}^{\infty}\frac{n^2(64n^4+11)}{(4n^4+1)(16n^4-1)}=\frac{\pi}{2}\cdot\frac{e^{3\pi}-1}{(e^{2\pi}-1)(e^{\pi}-1)}$$
If you're looking for any mathematical identity that relates $\pi$ and $e$, I can also suggest :
$\cdot$ The Stirling limit : $$\lim_{n\to\infty}\frac{n!e^n}{n^n\sqrt{n}}=\sqrt{2\pi}$$
$\cdot$ The well known integral : $$\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}\text{d}x=\frac{\pi}{e}$$
$\cdot$ Victor Adamchik's integrals : $$\int_{-\infty}^{\infty}\frac{\text{d}x}{(e^x-x+1)^2+\pi^2}=\frac{1}{2}$$ $$\int_{-\infty}^{\infty}\frac{\text{d}x}{(e^x-x)^2+\pi^2}=\frac{1}{1+\Omega}$$
Where $\Omega$ is the mathematical constant defined by $\text{ }\Omega e^{\Omega}=1$.
Hope this helps.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3063482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
\begin{align*}
360 ÷ 2 &= 180 \text{, and } 1 + 8 + 0 = 9\\
180 ÷ 2 &= 90 \text{, and } 9 + 0 = 9\\
90 ÷ 2 &= 45 \text{, and } 4 + 5 = 9\\
45 ÷ 2 &= 22.5 \text{, and } 2 + 2 + 5 = 9\\
22.5 ÷ 2 &= 11.25 \text{, and } 1 + 1 + 2 + 5 = 9\\
11.25 ÷ 2 &= 5.625 \text{, and } 5 + 6 + 2 + 5 = 18 \text{, and } 1 + 8 = 9\\
5.625 ÷ 2 &= 2.8125 \text{, and } 2 + 8 + 1 + 2 + 5 = 18 \text{, and } 1 + 8 = 9
\end{align*}
As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!
|
In general if an integer is divisible by 9 then its digital root is $9$. Any multiple of an integer divisible by $9$ will also be divisible by $9$ and have digital root $9$.
What's slightly surprising here is that this property is also preserved by division by $2$. Note that it doesn't work for, say, division by $3$, since $360/3 = 120$ has digital root $3$.
A relevant property that $2$ has, but $3$ doesn't, is that it's a factor of $10$. Dividing by $2$ is equivalent to multiplying by $5$, which preserves the property, and then dividing by $10$, which also preserves the property because it just removes a final zero, or adds or shifts the decimal point.
More generally, division by an integer $n$ will preserve the property of having digital root $9$ if all the prime factors of $n$ are factors of $10$, that is, it equals $2^a5^b$ for some $a,b \geq1$. This works because division by $2^a5^b$ is equivalent to multiplying by $2^b5^a$ and then dividing by $10^{a+b}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3064917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 3
}
|
Algebraic Closed Form for $\sum_{n=1}^{k}\left( n- 3 \lfloor \frac{n-1}{3} \rfloor\right)$ Let's look at the following sequence:
$a_n=\left\{1,2,3,1,2,3,1,2,3,1,2,3,...\right\}$
I'm trying to calculate:
$$\sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor$$
The problem is, I'm looking for a closed form for this summation:
$$\sum_{n=1}^{k}\left( n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor\right)$$
Is it possible?
|
If $n \equiv 0 \pmod{3}$, i.e. say $n=3s$ (where $s \geq 1$), then $a_n=3s-3\lfloor s-\frac{1}{3}\rfloor=3s-3(s-1)=3$.
If $n \equiv 1 \pmod{3}$, i.e. say $n=3s+1$ (where $s \geq 0$), then $a_n=3s+1-3\lfloor s\rfloor=1$.
If $n \equiv 2 \pmod{3}$, i.e. say $n=3s+2$ (where $s \geq 0$), then $a_n=3s+2-3\lfloor s+\frac{1}{3}\rfloor=2$.
So depending on what $k$ is we can count the number of terms which are $1's$, $2's$ and $3's$.
For $k=1,2$, the sum will be $\color{red}{1}$ and $\color{red}{3}$, respectively. So for $k \geq 3$, we do the following:
If $k=3t$ for $t \geq 1$, then
$$\sum_{n=1}^k\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=\sum_{n=1}^{3t}\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=t(1+2+3)=6t=\color{blue}{2k}.$$
If $k=3t+1$ for $t \geq 1$, then
$$\sum_{n=1}^k\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=\sum_{n=1}^{3t+1}\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=t(1+2+3)+1=6t+1=\color{blue}{2k-1}.$$
Likewise we can get the expressions for $k=3t+2$ as
$$\sum_{n=1}^k\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=\sum_{n=1}^{3t+2}\left(n-3\left\lfloor n-\frac{1}{3}\right\rfloor\right)=t(1+2+3)+(1+2)=6t+3=\color{blue}{2k-1}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3065950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
}
|
Find the directrix of the parabola with equation $y=-0.5x^2+2x+2$
Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$
I did this:
$$a=-0.5, b = 2, c = 2$$
Formula for the directrix is:
$$y=-1/(4a)$$
$$y=-1/(4\cdot(-0.5))=3.5$$
This is not right:
What went wrong? What is the proper way to do it?
|
Complete the square $$\begin{aligned} y&=-0.5x^2+2x+2\\ &=-0.5(x^2-4x)+2\\ &=-0.5(x^2-4x+4-4)+2\\ &=-0.5[(x-2)^2-4]+2\\&=-0.5(x-2)^2+4\end{aligned}$$
or equivalently $$y-4=-0.5(x-2)^2.$$
The directrix is given by $$y-4=-\frac{1}{4a}\quad \text{with}\; a=-0.5$$
or $$y=4+\frac{1}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3066042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
$\frac{7x+1}2, \frac{7x+2}3, \frac{7x+3}4, \ldots ,\frac{7x+2016}{2017}$ are reduced fractions for integers $x\in(0,301)$.
BdMO 2017 junior catagory Question 7. $$\dfrac{7x+1}2, \dfrac{7x+2}3, \dfrac{7x+3}4, \ldots ,\dfrac{7x+2016}{2017}$$
Here $x$ is a positive integer and $x < 301$. For some values of $x$ it is possible to express these given fraction in such fraction where denominator and numerator are co-prime. How many such $x$ is possible?
For an example if $x=4$, then $\dfrac{7x+1}2 = \dfrac{28+1}2 = \dfrac{29}2$. Here $29$ and $2$ are co-primes. But in the third term of this pattern I've noticed that $\dfrac{28+2}3 = \dfrac{30}3$ where $30$ and $3$ are not co-primes. So, $x$ is not $4$.
|
As noted by lab bhattacharjee, we need to see if $7x-1$ is co-prime to all $k=2,3,\ldots,2017$. If $0<x<301$, then $6\le 7x-1\le 7\times 300-1=2099$. Clearly, an integer in the interval $[6,2017]$ is not co-prime to all of the integers in $[2,2017]$ (in other words, $t$ is not co-prime to $t$ for $t\ge 2$). So we have to worry about $x$s such that $2017<7x-1\le 2099$, that is, $289\le x\le 300$. Plus if $x$ is odd, $7x-1$ is even so it is not co-prime to $2$. There are only $6$ integers remaining, and the list of $7x-1$ is as shown below.
*
*$x=290$ so $7x-1=2029$ is prime so it is a good candidate.
*$x=292$ so $7x-1=2043$ is not co-prime to $3$.
*$x=294$ so $7x-1=2057$ is not co-prime to $11$.
*$x=296$ so $7x-1=2071$ is not co-prime to $19$.
*$x=298$ so $7x-1=2085$ is not co-prime to $5$.
*$x=300$ so $7x-1=2099$ is prime so it is another good candidate.
Therefore there are only two good $x$s: $x=290$ and $x=300$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3066236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $3^{2n} +7$ is divisible by 8
Prove by induction that $3^{2n} +7$ is divisible by 8 for $n \in \Bbb N$
So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?
If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$
Assume true for $n=k$ so $$ 8\vert 3^{2k} +7$$
If $n=k+1$
$$3^{2(k+1)} +7$$
$$3^{2k+2} +7$$
If $3^{2n} +7$ is divisble by 8 then $3^{2n} +7=8A$ where $A \in \Bbb Z$, and thus $3^{2k+2} +7=8B$ where $B \in \Bbb Z$
$$3^2 \times 3^{2k}+7$$
$$3^2 \times (8A)=72A$$
$$72A =8(9A)=9B$$
So by induction $3^{2n} +7$ is divisibe by 8 $\forall n \in \Bbb N$
|
We can prove it like this :
$$ 9^n + 7 = 9^n + \sqrt[n]{7}^n $$
And by using factorization :
$$ 9^n + \sqrt[n]{7}^n = (9+7)(9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1})$$
$$ 9^n + \sqrt[n]{7}^n = (16)(9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1})$$
$$ \frac{9^n + 7}{16} = 9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1}$$
We have $16$ divides $9^{n}+7$ and $8$ divides $16$, so $8$ divides $9^{n}+7$ which is the same as' $8$ divides $3^{2n}+7$'
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3067826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
Power of matrix using diagonalization First one
$$\begin{pmatrix}
2& 3\\5
& 1
\end{pmatrix}^{20}$$
Second one
$$A=\begin{pmatrix}
4&0& 0\\0
& 3&0\\2 &0&2
\end{pmatrix}^{20}$$
$$P=\begin{pmatrix}
1&0& 0\\0
& 1&0\\1 &0&1
\end{pmatrix}$$
$$P^{-1}=\begin{pmatrix}
1&0& 0\\0
& 1&0\\-1 &0&1
\end{pmatrix}$$
I have some difficult to solve this problem.
For the first one, the characteristic equation is $(\lambda-1)(\lambda-2)-15=\lambda^2-3\lambda-13=0$ that $\lambda$ is $\frac{3\pm\sqrt{61}}2$ is it right?
For the second one, I find $P$ and $P^{-1}$, but multiplication of $P^{-1}AP$ is not diagonal.
Is there any case that $A$ cannot be represented as a diagonal but power of $A$ can be calculated?
|
$A$ has distinct eigenvalues, it must be diagonalizable.
\begin{align}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\begin{bmatrix}4 & 0 & 0 \\ 0 & 3 & 0 \\ 2 & 0 & 2 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} &= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\begin{bmatrix}4 & 0 & 0 \\ 0 & 3 & 0 \\ 4 & 0 & 2 \end{bmatrix} \\
&=\begin{bmatrix}4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{bmatrix} \end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3068767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find shortest distance from the parabola $y=x^2-9$ to the origin.
Find shortest distance from the parabola $y=x^2-9$ to the origin.
First, I find minima of $\sqrt{x^2+(x^2-9)^2}$, so use derivative and ...
Is have an easier way?
|
Here is an easier way: the shortest distance $r$ is taken at the minimum of
$r^2 = y^2 + x^2 = y^2 + y + 9$ This gets minimal at $y = -0.5$ (take the derivative or write $r^2 = (y+ 0.5)^2 + 8.75 \ge 8.75 $), so $r = \sqrt{y^2 + y + 9} \ge \sqrt{8.75} \simeq 2.9580$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3070891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
}
|
Differentiate $f ( x ) = \frac { \ln \left( x ^ { 2 } \cos ( x ) \right) } { \sqrt { 1 - x ^ { 2 } } }$ could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for me to show my own working of the problem.
|
We have $${d\over dx}\ln {x^2\cos x}={1\over x^2\cos x}\cdot {(2x\cos x-x^2\sin x)}\\{d\over dx}\sqrt{1-x^2}=-{x\over \sqrt{1-x^2}}$$therefore by defining $g(x)=\ln x^2\cos x$ and $h(x)=\sqrt{1-x^2}$ and using $\left({g\over h}\right)'={g'h-gh'\over h^2}$ we finally obtain$$f'(x){=\left({g\over h}\right)'=\left({g'h-gh'\over h^2}\right)(x)\\={\left({1\over x^2\cos x}\cdot {(2x\cos x-x^2\sin x)}\right)\sqrt{1-x^2}+\left({x\over \sqrt{1-x^2}}\right)\ln x^2\cos x\over 1-x^2}\\={\left({2\over x} {-\tan x}\right)\sqrt{1-x^2}+\left({x\over \sqrt{1-x^2}}\right)\ln x^2\cos x\over 1-x^2}\\={\left({2} {-x\tan x}\right)(1-x^2)+x^2\ln x^2\cos x\over x(1-x^2)\sqrt{1-x^2}}}$$ for $|x|<1$ and $x\ne 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3072985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
How to apply CRT to a congruence system with moduli not coprime? $x=1 \pmod 8$
$x=5 \pmod{12}$
8 and 12 are not coprime, I could break it to:
$x=1 \pmod 2$
$x=1 \pmod 4$
and
$x=5 \pmod 3$
$x=5 \pmod 4$
But what are the next steps to solve it? By the way, $x$ should be $17$ not sure how to get that number ...
Thanks in advance.
|
Here is a way:
$$\begin{cases}
x\equiv 1\pmod 8\\x\equiv 5\pmod{12}
\end{cases}\iff \begin{cases}
x -1\equiv 0\pmod 8\\x -1\equiv 4\pmod{12}\end{cases}\iff
\begin{cases}
\frac{x -1}4\equiv 0\pmod 2\\\frac{x -1}4\equiv 1\pmod{3}
\end{cases}$$
Now set $y=\frac{x-1}4$. As $3-2=1$, the solutions of the last system of congruences is
$$ y\equiv 0\cdot 3- 1\cdot 2 =-2\pmod{6},$$
so that, multiplying by $4$,
$$x-1\equiv -8 \iff x\equiv -7\iff x\equiv 17\pmod{24}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3075823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
evaluate this elliptic hyperbloid volume? How can I calculate the volume of this region in cylindrical coordinates?
$D=\{2x^2+y^2=z^2+4,|z| \le 2\}$
I think I got this wrong :
$$\operatorname{Volume} = 2\int_{0}^{2\pi}\int_{0}^{2} \int_{0}^{\sqrt{2r^2\cos^2\theta+r^2\sin^2\theta-4}}rdzdrd\theta$$
The problem is that the region is an ellipse and not a circle. I can integrate for $x^2+y^2=z^2+4$. Could you explain to me how to do it?
I think, what's wrong is also the Jacobian :
$x=2r\cos\theta$
$y=\frac{1}{\sqrt{2}}r\cos\theta$
$|J|=2\sqrt{2}$ ?
so maybe z range like this : $z=\{0,\sqrt{8r^2\cos^2\theta+\frac{1}{2}r^2\sin^2\theta-4}\}$
|
It's a single sheet hyperboloid and its central axis of symmetry is the $z$-axis.
Easiest way to find its volume is to consider that at every $z$ we have an ellipse, and integrate over the areas of those ellipses.
For a constant $z$ we have the ellipse:
$$\frac{x^2}{\frac{z^2+4}{2}}+\frac{y^2}{z^2+4}=1$$
It has semi axes $a=\sqrt{\frac{z^2+4}{2}}$ and $b=\sqrt{z^2+4}$.
Since the area of an ellipse is $\pi a b$, the volume is:
$$\text{Volume}=\int_{-2}^2 \pi a b\,dz = \int_{-2}^2 \pi\cdot\frac{z^2+4}{\sqrt 2}\,dz$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3076175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Determining the area of a right triangle, perimeter given, hypotenuse value given in terms of one of the legs. The problem states:
Right Triangle- perimeter of $84$, and the hypotenuse is $2$ greater than the other leg. Find the area of this triangle.
I have tried different methods of solving this problem using Pythagorean Theorem and systems of equations, but cannot find any of the side lengths or the area of the right triangle. I looked for similar problems on StackExchange and around the internet, but could not find anything.
Does anyone know anything that could help find the side lengths of the triangle and the area as well?
Method that I tried:
*
*Made a system with the values given.
\begin{align}
a+b+c&=84 \\
c&=b+2
\end{align}
*Substituted $c$ with $b+2$.
\begin{align}
a+b+b+2&=84 \\
a + 2b &= 82 & \text{subtracted $2$ from both sides}\\
a + a^2 - 4 &= 82
\end{align}
*$c^2$ is $(b+2)(b+2)$, so I used Pythagorean Theorem to isolate one of the variables.
\begin{align}
a^2+b^2 &=c^2\\
a^2 + b^2 &=(b+2)(b+2)\\
a^2+b^2 &=b^2+2b+4\\
a^2&=2b+4 & \text{ (Subtracted $b^2$ from both sides) }
\end{align}
OR
\begin{align}
a^2-4&=2b
\end{align}
I do not know what to do after this point.
|
$\\ \textbf{Finding triples, given perimeter using Euclid's formula}$ where $P=perimeter$
$$P=(m^2-n^2 )+2mn+(m^2+n^2 )=2m^2+2mn\implies n=\frac{P-2m^2}{2m}\quad where \quad \biggl\lceil\frac{\sqrt{P}}{2}\biggr\rceil\le m \le \biggl\lfloor\sqrt{\frac{P}{2}}\biggr\rfloor$$
Here, the lower limit ensures that $m>n$ and the upper limit insures that $n>0$.
Example:
$$P=84\Rightarrow \biggl\lceil\frac{\sqrt{84}}{2} \biggr\rceil =5 \le m\le\biggl\lfloor\sqrt{\frac{84}{2}}\biggr\rfloor =6:\quad f(84,5)\notin\mathbb{N}\quad f(84,6)=1\Rightarrow
F(6,1)=(35,12,37)$$
There are no other solutions for $P=84$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3076504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$ To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$
I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I proceed to induct on $m$ (and on $m$ only).
I am not perfectly confident in my self, however, for I see two placeholders, $n$ and $m.$ Is this a case where double induction is needed (first on $m$ and then on $n$)?
Whether or not this proof requires double induction, may someone explain when double induction is needed?
Consider any fixed $n, r \geq 0$ and the following two cases (I know that only one case is needed to complete this inductive proof).
CASE 1
\begin{align}
\binom{n + 0}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{0}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r}\binom{0}{0} \\ &= 0 + 0 + \cdots + \binom{n}{r} \\ &= \binom{n}{r}
\end{align}
CASE 2
\begin{align}
\binom{n + 1}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{1}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r-1}\binom{1}{r - (r-1)} + \binom{n}{r}\binom{1}{r - r} \\ &= 0 + 0 + \cdots + \binom{n}{r-1} + \binom{n}{r} \\ &= \binom{n}{r-1} + \binom{n}{r}
\end{align}
INDUCTION
Suppose it is true for $m \leq k.$ Now, consider $$\binom{n + (k + 1)}{r}.$$ It follows from Pascal's Identity that
$$\binom{n + (k+1)}{r} = \binom{n + k}{r} + \binom{n + k}{r-1}$$
And,
\begin{align}
\binom{n + k}{r} + \binom{n + k}{r-1} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\bigg[\binom{k}{r - i} + \binom{k}{r - 1 - i}\bigg] \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k+1}{r-i} \\ &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k+1}{r-i}
\end{align}
Hence, the equality holds for $m = k + 1.$ Given that the equality holds for $m = 0, 1,$ and that if equality holds for $m = k,$ it then holds for $m = k + 1,$ it follows that the equality holds $\forall m \in \mathbb{N}.$
|
For a short reply, your induction proof has tiny problem, in that $r$ can take value $n+k$ for the $m = k+1$ induction step. So when you use induction hypothesis to get ${{n+k}\choose{r}} = \sum_{i=1}^r {n\choose i}{k\choose {r-i}}$, you can actually use it only for $r<n+k$. This is not big problem though as the $r = n+k$ case is trivial.
For the double-induction, I don't think it's necessary here. The reason is that you are actually fixing an arbitrary $n$ first, and then do induction proof. So the induction proof is within the context of the fixed $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3081297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Find range of $x$ if $\log_5\left(6+\frac{2}{x}\right)+\log_{1/5}\left(1+\frac{x}{10}\right)\leq1$
If $\log_5\left(6+\dfrac{2}{x}\right)+\log_{1/5}\left(1+\dfrac{x}{10}\right)\leq1$, then $x$ lies in _______
My Attempt
$$
\log_5\bigg(6+\dfrac{2}{x}\bigg)+\log_{1/5}\bigg(1+\dfrac{x}{10}\bigg)=\log_5\bigg(6+\dfrac{2}{x}\bigg)-\log_{5}\bigg(1+\dfrac{x}{10}\bigg)\leq1\\
\log_5\frac{(6x+2)10}{x(10+x)}\leq1\implies\frac{(6x+2)10}{x(10+x)}\leq5\\
\frac{4(3x+1)}{x^2+10x}\leq1\\
\implies 12x+4\leq x^2+10x\quad\text{or}\quad12x+4>x^2+10x\\
x^2-2x-4\geq0\quad\text{or}\quad x^2-2x-4<0\implies x\in\mathcal{R}
$$
My reference gives the solution $(-\infty,1-\sqrt{5})\cup(1+\sqrt{5},\infty)$, what is going wrong here ?
|
By your work we need to solve
$$\frac{x^2-2x-4}{x(x+10)}\geq0$$ and the domain gives $x>0$ or $-10<x<-\frac{1}{3}.$
The first by the interval's method gives
$$1-\sqrt5\leq x<0$$ or $$x\geq1+\sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
$$\left[1-\sqrt5,-\frac{1}{3}\right)\cup[1+\sqrt5,+\infty).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3084524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
The number of prime pairs of $x^2-2y^2=1$ How to find the number of pairs of positive integers $(x,y)$ where $x$ and $y$ are prime numbers and $x^2-2y^2=1$?
I am not getting any clue here.
|
Since $2y^2$ is even, $x$ must be odd so that $x^2$ can be odd as well. This much should be obvious.
From there it's not too much of a leap to find that $x^2 \equiv 1 \pmod 4$. If $y$ is odd as well, then $y^2 \equiv 1 \pmod 4$, too, but $2y^2 \equiv 2 \pmod 4$, which means that $x^2 - 2y^2 \equiv 3 \pmod 4$, but clearly $1 \equiv 1$, not $3 \pmod 4$.
Therefore $y$ must be even, so that $2y^2 \equiv 0 \pmod 4$, and consequently $x^2 - 2y^2$ can be congruent to $1 \pmod 4$ as is necessary to solve the equation.
Of course 2 is prime, and so is $-2$. If $y = \pm 2$, then $x = \pm 3$. And if $y = 0$, then $x = 1$. The only remaining possibilities are composite even numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3084983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
For $G$ the centroid in $\triangle ABC$, if $AB+GC=AC+GB$, then $\triangle ABC$ is isosceles. (Likewise, for the incenter.)
Let $G$ be the centroid of $\triangle ABC$. Prove that if $$AB+GC=AC+GB$$ then the triangle is isosceles!
Of course, the equality is true, when we have isosceles triangle, but the other way is not trivial for me. I have tried using vectors, even the length of the medians.
|
In the standard notation we obtain:
$$c+\frac{1}{3}\sqrt{2a^2+2b^2-c^2}=b+\frac{1}{3}\sqrt{2a^2+2c^2-b^2}$$ or
$$3(b-c)=\frac{3(b^2-c^2)}{\sqrt{2a^2+2b^2-c^2}+\sqrt{2a^2+2c^2-b^2}},$$ which gives $b=c$ or
$$\sqrt{2a^2+2b^2-c^2}+\sqrt{2a^2+2c^2-b^2}=b+c$$ or
$$\sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}=bc-2a^2,$$ which is impossible for $bc-2a^2\leq0.$
But for $bc>2a^2$ we obtain
$$(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)=(bc-2a^2)^2$$ or
$$(b-c)^2(a+b+c)(b+c-a)=0,$$ which gives $b=c$ again.
The second problem we can solve by the similar way.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3086013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Determine the sequence generated by a generating function $A(z)=2z-1+\frac{1}{2z-2z^2}$
I have no clue how to solve this, I tried looking at other examples but I am just stuck, could anyone be so kind and explain how to solve this step by step?
|
If you want to see the generating function:
$\begin{array}\\
A(z)
&=2z-1+\frac{1}{2z-2z^2}\\
&=2z-1+\frac1{2z}\frac{1}{1-z}\\
&=2z-1+\frac1{2z}\sum_{n=0}^{\infty} z^n\\
&=2z-1+\frac1{2z}+\sum_{n=1}^{\infty} \frac{z^{n-1}}{2}\\
&=2z-1+\frac1{2z}+\sum_{n=0}^{\infty} \frac{z^{n}}{2}\\
&=2z-1+\frac1{2z}+\frac12+\frac{z}{2}+\sum_{n=2}^{\infty} \frac{z^{n}}{2}\\
&=\frac53 z-\frac12+\frac1{2z}+\sum_{n=2}^{\infty} \frac{z^{n}}{2}\\
\end{array}
$
If you want to invert the function,
that is,
find $z$ in terms of $A$:
$\begin{array}\\
A(z)
&=2z-1+\frac{1}{2z-2z^2}\\
&=\frac{(2z-1)(2z-2z^2)+1}{2z-2z^2}\\
&=-\frac{4 z^3 - 6 z^2 + 2 z - 1
}{2z-2z^2}\\
\end{array}
$
so
$-A(2z-2z^2)
=4 z^3 - 6 z^2 + 2 z - 1
$
or
$4 z^3 - (2A+6) z^2 + (2a+2) z - 1
=0
$.
Wolfy gives some very complicated expressions
for $z$ in terms of $A$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3088889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ Problem:
solve equation
$$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$$
I don't look for easy solution (square booth side and things like that...) I look for some tricks for "easy" solution because:
I would like to use substitution, but we have $3x$ and $-3x$, but I can't see it.
Solution:
|
Let $2x^2-3x+5=t^2$. Then:
$$\sqrt{t^2+6x}+t=3x \Rightarrow t^2+6x=t^2-6xt+9x^2 \Rightarrow \\
2t=3x-2 \Rightarrow 4t^2=9x^2-12x+4 \Rightarrow \\
4(2x^2-3x+5)=9x^2-12x+4 \Rightarrow \\
x^2=16 \Rightarrow x=4.$$
Note: The square roots of the LHS of the original equation exist for all $x\in \mathbb R$, however, the LHS is nonnegative, hence $3x\ge 0 \Rightarrow x\ge 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3090819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Is $1111111111111111111111111111111111111111111111111111111$ ($55$ $1$'s) a composite number? This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = \frac{1}9(999...) = \frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
|
More explicitly,
$$\begin{align*}
\frac{10^{55} - 1}{9}
&= \frac{(10^5)^{11} - 1}{9} \\
&= \frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + \cdots + 10 + 1)}{9} \\
&= \frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + \cdots + 1)}{9} \\
&= (10^4 + 10^3 + \cdots + 1)(10^{50} + 10^{45} + \cdots + 1).
\end{align*}$$
The first factor is $11111$, which in turn is $41 \cdot 271$, and the second factor has as its smallest prime factor $1321$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3091832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
}
|
Calculate conditional probability; throwing the cube The symmetrical cube was threw $30$ times.
Calculate probability situation, when in first $20$ throws came out $4$ times number $3$, if in $30$ throws number $3$ came out $7$ times.
|
So in this case we are dealing with conditional probability, I think.
Let's calculate $\mathbf{B}$
$7=7+0+0 \to $ $3$ times, because 7 could be anywhere
$7=6+1+0\to$ $6$ times
$7=5+1+1\to$ $3$ times
$7=5+2+0\to$ $6$ times
$7=4+3+0\to$ $6$ times
$7=4+2+1\to$ $6$ times
$7=3+2+2\to$ $3$ times
$7=3+1+3\to$ $3$ times
So
$\mathbf{B}=36$
Let's calculate $\mathbf{A} \cap\ \mathbf{B}$
$4=3+1\to$ $2$ times
$4=4+0\to$ $2$ times
$4=2+2\to$ $1$ time
$\mathbf{A} \cap\ \mathbf{B}=5$
So
$ P(\mathbf{A} \setminus \mathbf{B})=\left(\frac{5}{36}\right) $
Is it correct answer? Thanks in advance
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3092539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Tough Irrational Equation highschool Have been trying to solve this irrational equation for a day but as it seems, i'm not going anywere with it. Can somebody offer me a tip ? Thanks!
*Tried a "t" substitution for x squared but it still yields a 4th degree polynomial equation instead of an 8th, which I think can be solved by factoring, but i'm sure there's an easier way to do it.
$$\sqrt{\frac{1+2x\sqrt{1-x^2}}{2}} + 2x^2 =1$$
|
We have
$$\sqrt{\frac{x^2+2x\sqrt{1-x^2}+1-x^2}{2}}+2x^2-1=0$$ or
$$\frac{|x+\sqrt{1-x^2}|}{\sqrt2}+2x^2-1=0.$$
Now, $$x=\sqrt{1-x^2}$$ gives $x=\frac{1}{\sqrt2}$ which is not a root of our equation.
Thus, our equation is equivalent to
$$\frac{|2x^2-1|}{\sqrt2|x-\sqrt{1-x^2}|}+2x^2-1=0$$ or since $2x^2-1\leq0,$
$$(2x^2-1)\left(1-\frac{1}{\sqrt2|x-\sqrt{1-x^2}|}\right)=0,$$ which gives $$x=-\frac{1}{\sqrt2}$$ or
$$\sqrt2|x-\sqrt{1-x^2}|=1.$$
Now, if $x-\sqrt{1-x^2}\geq0$ we obtain $x\geq0$ and $x^2\geq1-x^2\geq0$ or $2x^2-1\geq0,$
which with $2x^2-1\leq0$ gives $x=\pm\frac{1}{\sqrt2},$ which is impossible.
Id est, $x-\sqrt{1-x^2}<0$ and it remains to solve
$$\sqrt2(x-\sqrt{1-x^2})=-1.$$
Can you end it now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3094528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Integral with respect to $x +$ constant Is this a valid expression:
$$\int xd(x+5)$$
I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:
$$\int (u - 5)du= \frac{u^2}{2} - 5u + C = \frac{(x+5)^2}{2} - 5(x+5) +C = \frac{x^2}{2} -12.5 + C $$
Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$\int xd(x+5) = \int xdx$$
|
Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)
$$\begin{align}
\int xe^{-x^2}dx &= \int \frac{xe^{-x^2}dx}{1}\\
&= \int \frac{xe^{-x^2}dx}{1}\cdot\frac{\frac{d(-x^2)}{dx}}{\frac{d(-x^2)}{dx}}\\
&= \int \frac{xe^{-x^2}dx\cdot\frac{d(-x^2)}{dx}}{-2x}\\
&= \int \frac{xe^{-x^2}d(-x^2)}{-2x}\\
&= \color{red}{-\frac{1}{2}\int e^{-x^2}d(-x^2)} &[1]\\
&= \color{green}{-\frac{1}{2}\int e^udu} &[2]\\
&= -\frac{1}{2}e^u + C\\
&= -\frac{1}{2}e^{-x^2} + C\\
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3095860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Solving $2\sin\theta\cos\theta + \sin\theta = 0$
The question is to solve the following question in the range $-\pi \le \theta \le \pi$
$$2\sin\theta\cos\theta + \sin\theta = 0$$
I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $\pm2/3\pi$ and the values when $\sin\theta = 0$. Although I missed the early factorisation I don't know what I'm doing to actually arrive at an incorrect answer:
$$\begin{align}
2\sin\theta\cos\theta + \sin\theta &= 0 \qquad\text{(square)} \tag{1} \\
4\sin^2\theta\cos^2\theta + \sin^2\theta &= 0 \tag{2}\\
4\sin^2\theta(1-\sin^2\theta) + \sin^2\theta &= 0 \tag{3} \\
4\sin^2\theta - 4\sin^4\theta + \sin^2\theta &= 0 \tag{4} \\
5\sin^2\theta - 4\sin^4\theta &= 0 \tag{5}
\end{align}$$
and then solving by substitution/the quadratic equation I get $\sin\theta = \pm\sqrt(5)/2$ and $0$ but as this out of bounds for sin so cannot be the answer.
I know using Symbolab that this solution is correct for the quadratic I've generated, so I must be going wrong somewhere above after missing that factoristion. I feel like it is in the squaring step but not sure what would be wrong here...
Thanks a lot for your help.
|
I'll start by graphing this function x-axis is $\theta / \pi $ which shows the function is zero at 5 points.
\begin{align}
2 \cdot \sin(\theta)\cos(\theta) + \sin(\theta) & = 0 \\
sin(\theta) \cdot (2\cdot\cos(\theta)+1) & = 0
\end{align}
So either $\sin(\theta) = 0$ or $2\cdot\cos(\theta)+1 = 0 \Rightarrow \cos(\theta) =
-0.5$
Considering each of these cases then $\theta$ is $-\pi$, $-\dfrac{2}{3}\pi$, $0$, $\dfrac{2}{3}\pi$ or $\pi$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3096324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
If $S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$, then what is $\lfloor S \rfloor$?
If
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$$
then $$\lfloor S \rfloor = \text{?}$$
What I tried:
I know that
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\cdots=\zeta(4)=\frac{\pi^4}{90}\approx 1.1$$
then $\lfloor S \rfloor =1$.
But how do I find with inequality? Please have a look.
|
Hint: Note that
$$
\begin{align}
\frac1{(k-1)^3}-\frac1{k^3}
&=\frac{3k^2-3k+1}{k^3(k-1)^3}\\
&\gt\frac{3k^2-3k}{k^3(k-1)^3}\\
&=\frac3{k^2(k-1)^2}\\
&\gt\frac3{k^4}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=n}^\infty\frac1{k^4}
&\lt\frac13\sum_{k=n}^\infty\left(\frac1{(k-1)^3}-\frac1{k^3}\right)\\
&=\frac1{3(n-1)^3}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3097084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Differentiation uner the integral sign - help me find my mistake This is my integral:
$$I(a)=\int_0^\infty\frac {\ln(a^2+x^2)}{(b^2+x^2)}dx.$$
Taking the first derivative with respect to a:
$$I'(a)=\int_0^\infty \frac {2adx} {(a^2+x^2)(b^2+x^2)}.$$
This is how I did the partial fraction decomposition:
$\frac {2a} {(a^2+x^2)(b^2+x^2)}=\frac {Ax+B} {(a^2+x^2)}+\frac {Cx+D} {(b^2+x^2)}$.
From here I get that $A=C=0$, $B=\frac {2a} {(b^2-a^2)}$ and $D=\frac {-2a} {(b^2-a^2)}$
Is this correct? Because when I try to solve $I'(a)$ using these values for $B$ and $D$ I get a different solution from the textbook?
|
Let’s redo the work$$\frac 1{(a^2+x^2)(b^2+x^2)}=\frac {Ax+B}{a^2+x^2}+\frac {Cx+D}{b^2+x^2}$$Multiplying both sides by the common denominator$$1=(Ax+B)(b^2+x^2)+(Cx+D)(a^2+x^2)$$To find values for $A$, $B$, $C$, and $D$, first set $x^2=-a^2$. Thus$$\begin{align*}1 & =(Ax+B)(b^2-a^2)\\ & =Ax(b^2-a^2)+B(b^2-a^2)\end{align*}$$Therefore, it’s easy to see that $A=0$ and $B=1/(b^2-a^2)$. Now do a similar procedure for the other term by setting $x^2=-b^2$. Thus$$\begin{align*}1 & =(Cx+D)(a^2-b^2)\\ & =Cx(a^2-b^2)+D(a^2-b^2)\end{align*}$$Thus, $C=0$ and $D=1/(a^2-b^2)$. So to sum everything up$$\frac 1{(a^2+x^2)(b^2+x^2)}\color{blue}{=\frac 1{(b^2-a^2)(a^2+x^2)}+\frac 1{(a^2-b^2)(b^2+x^2)}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3101376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Evaluate the integral $\int_1^\infty \left(\frac{1+x}{1+x^2}\right)^2\,dx$ I have to prove that this integral is finite so that the series $\sum_1^\infty \left(\frac{1+n}{1+n^2}\right)^2$ converges , but I am not able to integrate the function . Please help
|
$$
\begin{align}
\int\left(\frac{1+x}{1+x^2}\right)^2\,dx
&=\int\frac{1+2x+x^2}{\left(1+x^2\right)^2}\,dx\\
&=\int\left(\frac{1+x^2}{\left(1+x^2\right)^2}+\frac{2x}{\left(1+x^2\right)^2}\right)\,dx\\
&=\int\frac{1}{1+x^2}\,dx+\int\frac{1}{\left(1+x^2\right)^2}\frac{d}{dx}\left(1+x^2\right)\,dx\\
&=\arctan{x}+\int\left(1+x^2\right)^{-2}\,d\left(1+x^2\right)\\
&=\arctan{x}+\frac{1}{-2+1}\left(1+x^2\right)^{-2+1}+C\\
&=\arctan{x}-\frac{1}{1+x^2}+C.
\end{align}
$$
Wolfram Alpha check. $\int\frac{1}{1+x^2}\,dx=\arctan{x}+C$, by the way, is a well-known table integral.
$$
\begin{align}
\int_1^\infty \left(\frac{1+x}{1+x^2}\right)^2\,dx
&=\lim\limits_{b\rightarrow\infty}\int_1^b \left(\frac{1+x}{1+x^2}\right)^2\,dx\\
&=\lim\limits_{b\rightarrow\infty}\bigg[\arctan{x}-\frac{1}{1+x^2}\bigg]_{1}^{b}\\
&=\lim\limits_{b\rightarrow\infty}\bigg[\arctan{b}-\frac{1}{1+b^2}-\bigg(\arctan{1}-\frac{1}{1+1^2}\bigg)\bigg]\\
&=\frac{\pi}{2}-0-\bigg(\frac{\pi}{4}-\frac{1}{2}\bigg)\\
&=\frac{\pi}{2}-\frac{\pi}{4}+\frac{1}{2}\\
&=\frac{\pi}{4}+\frac{1}{2}\\
&=\frac{\pi+2}{4}.
\end{align}
$$
Wolfram Alpha gives the same answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3102084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Use the definition of limit to show that $\lim_{x \to -1} \frac{x+5}{2x+3}=4$
Use the definition of limit to show that
$\lim_{x \to -1} \frac{x+5}{2x+3}=4$
We have $|\frac{x+ 5}{2x+3} -4|= |\frac{x+5-8x-12}{2x+3}|=|\frac{-7x-7}{2x+3}|=\frac{7}{|2x+3|}|x+1|$
To get a bound on the coefficient of $|x+1|$, we restrict $x$ by the condition $-2<x<0$ [neighborhood of $-1$] . For $x$ in this interval, we have $-1<2x+3<3$, so that
$|\frac{x+ 5}{2x+3} -4|= \frac{7}{|2x+3|}|x+1|<...$
Is that true, please? And I don’t know how can I complete.
|
You way of thinking is on the right track. First, we need to "guess" $\delta$ so we will make a draft of the calculation first.
This is just the "draft" :
\begin{align*}
\big|\frac{x+5}{2x+4}-4\big| &< \varepsilon \\
7\big|\frac{x+1}{2x+3}|&< \varepsilon \\
|x+1| &< \frac{\varepsilon}{7}|2x+3|
\end{align*}
So now, for convenience, I bound $|x+1|<\frac{1}{4}$. You can bound with any convenient number. Observe that
\begin{align*}
-\frac{1}{4} &< x + 1 < \frac{1}{4} \\
-\frac{1}{2} &< 2x + 2 < \frac{1}{2} \\
\frac{1}{2} &< 2x + 3 < \frac{3}{2}
\end{align*}
Now, we are all set. From the calculation above, we can fix $\varepsilon>0$ and set $\delta = \min\{\frac{\varepsilon}{14},\frac{1}{4}\}$
Thus, we will obtain :
\begin{align*}
|x+1| &< \frac{\varepsilon}{14}\\
|x+1| &< \frac{\varepsilon}{7} . \frac{1}{2}\\
|x+1| &< \frac{\varepsilon}{7}|2x+3|\\
|7x+7| &< \varepsilon|2x+3|\\
\big|\frac{x+5}{2x+3}-4\big| &< \varepsilon
\end{align*}
Thus, we have completed the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3103516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$? I need to find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$. Right now I am getting $\frac{0}{0}$ if I don't factor first, or $\frac{2}{0}$ if I do.
Here are my factoring steps:
$\frac {x^2+x} {x^2-x-2}$
$=\frac{x(x+1)}{(x-2)(x+1)}$
replace $x$ with $-1$
$=\frac{-1(-1+1)}{(-1-2)(-1+1)}$
$=\frac{2}{-3 (0)}$
$=\frac{2}{0}$
How can I solve this problem?
|
$$
\begin{align}
x^2-x-2
&=x^2-2x\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-2\\
&=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}-\frac{2\cdot4}{4}\\
&=\left(x-\frac{1}{2}\right)^2-\frac{9}{4}\\
&=\left(x-\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2\\
&=\left(x-\frac{1}{2}-\frac{3}{2}\right)\left(x-\frac{1}{2}+\frac{3}{2}\right)\\
&=(x-2)(x+1)
\end{align}
$$
Now, things are going to cancel out nicely:
$$
\lim\limits_{x \rightarrow -1}\frac {x^2+x} {x^2-x-2}=
\lim\limits_{x \rightarrow -1}\frac {x(x+1)} {(x-2)(x+1)}=
\lim\limits_{x \rightarrow -1}\frac {x} {x-2}=\frac{-1}{-1-2}=\frac{-1}{-3}=\frac{1}{3}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3104145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
}
|
show this inequality $a_{1}+a_{2}+\cdots+a_{n}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$
let $a_{1},a_{2},\cdots,a_{n}\ge 0,n\ge 3$,and such $$a^2_{1}+a^2_{2}+\cdots+a^2_{n}=1$$
show that
$$a_{1}+a_{2}+\cdots+a_{n}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$$
I can prove when $n=3$, it need to prove
$$a_{1}+a_{2}+a_{3}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
where $a^2_{1}+a^2_{2}+a^2_{3}=1$.
since $$ (a_{1}+a_{2}+a_{3})^2\ge 3(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})\tag{1}$$
and
$$1=a^2_{1}+a^2_{2}+a^2_{3}\ge a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}\tag{2}$$
$(1)\times (2)$ we have
$$(a_{1}+a_{2}+a_{3})^2\ge 3(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})^2$$
so we have$$a_{1}+a_{2}+a_{3}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
But for $n\ge 4$ ,I want to show
$$f(a_{1},a_{2},\cdots,a_{n})-f(a_{1},a_{2},a_{3},\cdots,a_{n-2},\sqrt{a^2_{n-1}+a^2_{n}},0)\ge 0$$,where
$$a_{1}=\max(a_{1},a_{2},\cdots,a_{n}).f(a_{1},a_{2},\cdots,a_{n})=\dfrac{1}{\sqrt{3}}(a_{1}+\cdots+a_{n})-(a_{1}a_{2}+\cdots+a_{n}a_{1})$$
and
$$f(a_{1},a_{2},\cdots,a_{n})-f(a_{1},a_{2},a_{3},\cdots,a_{n-2},\sqrt{a^2_{n-1}+a^2_{n}},0)$$$$=\dfrac{1}{\sqrt{3}}(a_{n-1}+a_{n}-\sqrt{a^2_{n-1}+a^2_{n}})+a_{n-2}\sqrt{a^2_{n}+a^2_{n-1}}-a_{n-2}a_{n-1}-a_{n-1}a_{n}-a_{n}a_{1}$$I can't prove it
|
For $n=3$ you have a proof.
We'll prove that for all $n\geq4$ the following stronger inequality is true:
$$\sum_{k=1}^na_k\geq2\sum_{k=1}^na_ka_{k+1},$$ where $a_{n+1}=a_1$.
Indeed, we need to prove that
$$\sum_{k=1}^na_k^2\left(\sum_{k=1}^na_k\right)^2\geq4\left(\sum_{k=1}^na_ka_{k+1}\right)^2,$$ which is true because
$$\sum_{k=1}^na_k^2-\sum_{k=1}^na_ka_{k+1}=\frac{1}{2}\sum_{k=1}^n\left(a_k-a_{k+1}\right)^2\geq0$$ and
$$\left(\sum_{k=1}^na_k\right)^2\geq4\sum_{k=1}^na_ka_{k+1}$$ is true by AM-GM:
$$\sum_{k=1}^na_ka_{k+1}\leq(a_1+a_3+...)(a_2+a_4+...)\leq\left(\frac{\sum\limits_{k=1}^na_k}{2}\right)^2$$ because for odd $n$ we can assume that $a_1=\min\limits_{i}\{a_i\}$.
For example, for $n=5$ it works so:
$$a_1a_2+a_2a_3+a_3a_4+a_4a_5+a_5a_1\leq $$
$$\leq a_1a_2+a_2a_3+a_3a_4+a_4a_5+a_5a_2+a_4a_1=$$
$$=(a_1+a_3+a_5)(a_2+a_4)\leq\left(\frac{a_1+a_3+a_5+a_2+a_4}{2}\right)^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3108726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
&(x+y)(x^2 - xy + y^2)\\
&= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
&= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
&= x^3 + x^2y - x^2y - xy^2 + xy^2 + y^3\\
&= x^3 + y^3\\
\end{split}$$
Q.E.D.
Question:
Spivak says there is an easy proof that, if I use this other theorem:
$$
x^3 - y^3 = (x-y)(x^2 + xy + y^2)
$$
then, I will also allow me to find out $x^n+y^n$ whenever $n$ is odd.
How to do this? I fail to see how.
|
Maybe not easier, but given $$\frac{1-z^n}{1-z}=\sum_{i=0}^{n-1}z^i\\1-z^n=(1-z)\sum_{i=0}^{n-1}z^i\\$$ let $z=\frac{y}{x}$, then $$1-\left(\frac{y}{x}\right)^n=\left(1-\frac{y}{x}\right)\sum_{i=0}^{n-1}y^ix^{-i}$$ multiplying by $x^n$ yields $$x^n-y^n=(x-y)\sum_{i=0}^{n-1}y^ix^{n-1-i}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3110478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
What is the smallest number of $45^\circ$–$60^\circ$–$75^\circ$ triangles in non-trivial substitution tiling? Let base = $45^\circ$–$60^\circ$–$75^\circ$ triangle.
Over at What is the smallest number of bases that a square can be divided into? it was determined that 23 base were needed to make a $45^\circ$–$45^\circ$–$90^\circ$ triangle.
How about non-trivial dissections of base into similar triangles? Start with base and divide it into smaller copies of base. Ideally the method should be specific to base and wouldn't work with other triangles. Also, at least one of the internal triangles should have no edges parallel to the original triangle.
What are the simplest non-trivial dissections of base into similar triangles?
|
Hmm... This doesn't exactly fit your criterion in that it's not unique to the $45^∘–60^∘–75^∘$ triangle, but the central triangle here has no edges parallel to the edges of the original triangle.
The vertices are:
$$
\begin{array}{ccc}
\{0,0\} \\
\{1,0\} \\
\left\{\frac{1}{2} \left(3-\sqrt{3}\right),\frac{1}{2} \left(3-\sqrt{3}\right)\right\} \\
\left\{\frac{1}{22} \left(21-2 \sqrt{3}\right),\frac{1}{22} \left(6+\sqrt{3}\right)\right\} \\
\left\{-\frac{3}{22}
\left(-5+\sqrt{3}\right),0\right\} \\
\left\{\frac{1}{22} \left(6+\sqrt{3}\right),\frac{1}{22}
\left(6+\sqrt{3}\right)\right\} \\
\left\{\frac{1}{286} \left(237-43 \sqrt{3}\right),\frac{1}{143} \left(21-2 \sqrt{3}\right)\right\} \\
\left\{\frac{1}{286} \left(216-41 \sqrt{3}\right),\frac{1}{22} \left(6+\sqrt{3}\right)\right\} \\
\left\{\frac{1}{286} \left(174-37 \sqrt{3}\right),\frac{1}{286} \left(36+17 \sqrt{3}\right)\right\} \\
\left\{\frac{996-197 \sqrt{3}}{1430},\frac{192+43 \sqrt{3}}{1430}\right\} \\
\left\{\frac{996-197
\sqrt{3}}{1430},\frac{306+73 \sqrt{3}}{1430}\right\} \\ \left\{\frac{1}{130} \left(102-19
\sqrt{3}\right),\frac{318+31 \sqrt{3}}{1430}\right\} \\
\end{array}
$$
This is based on the three other four-self-similar-triangle dissections beyond the usual midpoints-of-the-sides one:
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3111813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Show that $f_A$ is an inner product Let $A$ be a $2 \times 2$ matrix with real entries. For $X, Y$ in $R^{2 \times 1}$ let
$f_A(X, Y) = Y^tAX$.
Show that $f_A$ is an inner product on $R^{2 \times 1}$ if and only if $A = A^t$, $A_{11} > 0$, $A_{22} > 0$,
and $det A > 0$.
I was able to solve the first part, assuming the internal product. How to do reciprocal? I could only verify that $f_A(cX+Z, Y) = cf_A(X, Y)+f_A(Z, Y)$.
But I could not complete the other two conditions, for example
$f_A(X, Y) = Y^tAX= Y^tA^tX= \{X^tAY\}^t$,
how to conclude that it is equal to $ { f_A(Y, X) }$ (
in this case there is no conjugation bar because everything is real).
Any tips?
|
You're basically there with (conjugate) symmetry: note that $X^\top A Y = f(Y, X)$. It's also a scalar (well, a $1 \times 1$ matrix), so the transpose is superfluous.
As for the positive-definiteness, this is where we need to start making use of the individual entries of $A$. Suppose $X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$. Then,
$$f_A(X, X) = \begin{bmatrix} x_1 & x_2 \end{bmatrix} \begin{bmatrix} A_{11} & A_{12} \\ A_{12} & A_{22} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = A_{11}x_1^2 + A_{22} x_2^2 + 2A_{12}x_1x_2$$
We want this to be greater than or equal to $0$. Start by completing the square (remember $A_{11}, A_{22} > 0$):
\begin{align*}
f_A(X, X) &= A_{11}\left(x_1^2 + 2\frac{A_{12}}{A_{11}} x_1x_2 \right) + A_{22}x_2^2 \\
&= A_{11}\left(x_1^2 + 2\frac{A_{12}}{A_{11}} x_1x_2 + \left(\frac{A_{12}}{A_{11}} x_2\right)^2\right) + A_{22}x_2^2 - \frac{A_{12}^2}{A_{11}}x_2^2 \\
&= A_{11}\left(x_1 +\frac{A_{12}}{A_{11}}x_2\right)^2 + \frac{A_{22}A_{11} - A_{12}^2}{A_{11}}x_2^2 \\
&= A_{11}\left(x_1 +\frac{A_{12}}{A_{11}}x_2\right)^2 + \frac{\det A}{A_{11}}x_2^2.
\end{align*}
Since $A_{11} > 0$ and $\det A > 0$, it follows that $f_A(X, X) \ge 0$. Further, if $f_A(X, X) = 0$, then both the squares must be $0$, in other words, $x_2 = 0$ and $x_1 + \frac{A_{12}}{A_{11}} x_2 = x_1 = 0$. That is, $X = 0$, completing the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3112211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.