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Sum of first $n$ triangle numbers, without induction Background
I wish to calculate $$ S= \sum_{i = 1}^{n}\frac{k(k+1)}{2}$$
I know what the answer is going to be, since this is essentially the sum of the first $n$ triangle numbers.
I.e. $S = (1) + (1+2) + (1+2+3) + \cdots + (1+2+3+\cdots+n)$
All solutions I've seen seem to know in advance what the answer is going to be, and their problem is proving it, which can be done using induction.
Question
However, I was wondering if this can be calculated using formulae for summations instead?
Alternatively, for instances where we do wish to prove it instead of calculate it, are there any other ways besides induction?
For reference
The answer should be $$\frac{n(n+1)(n+2)}{6}$$
|
Another way to deal with your sum, and in general
with the sum of the binomial coefficient in the upper index,
is to consider that
$$
\Delta _{\,n} \left( \begin{gathered}
n \\
m \\
\end{gathered} \right) = \left( \begin{gathered}
n + 1 \\
m \\
\end{gathered} \right) - \left( \begin{gathered}
n \\
m \\
\end{gathered} \right) = \left( \begin{gathered}
n \\
m - 1 \\
\end{gathered} \right)
$$
So
$$
\sum\limits_{a\,\, \leqslant \,k\, \leqslant \,b} {\left( \begin{gathered}
k \\
m \\
\end{gathered} \right)} = \sum\limits_{a\,\, \leqslant \,k\, \leqslant \,b} {\Delta _{\,n} \left( \begin{gathered}
k \\
m + 1 \\
\end{gathered} \right)} = \left( \begin{gathered}
b + 1 \\
m + 1 \\
\end{gathered} \right) - \left( \begin{gathered}
a \\
m + 1 \\
\end{gathered} \right)
$$
and in your case
$$
\sum\limits_{1\,\, \leqslant \,k\, \leqslant \,n} {\left( \begin{gathered}
k + 1 \\
2 \\
\end{gathered} \right)} = \sum\limits_{2\,\, \leqslant \,j\, \leqslant \,n + 1} {\left( \begin{gathered}
j \\
2 \\
\end{gathered} \right)} = \left( \begin{gathered}
n + 2 \\
3 \\
\end{gathered} \right) - \left( \begin{gathered}
2 \\
3 \\
\end{gathered} \right) = \left( \begin{gathered}
n + 2 \\
3 \\
\end{gathered} \right)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prime numbers of the form $(1\times11\times111\times1111\times...)-(1+11+111+1111+...)$ Let
$$R(1) = 1-1,$$
$$R(2) = (1\times11) - (1+11),$$
$$R(3) = (1\times11\times111) - (1+11+111),$$ and so on...
$$R(4)=1355297\quad\text{(a prime number!)}$$
$R(4)$ is the only prime I found of such form up to $R(200)$. Are there anymore primes of such form?
|
*
*Just a very basic observation, not a complete answer. Let's note by
$$a_n=1111...1$$
with $n$ of $1's$. So we have
$$a_n \equiv 1 \pmod{10}$$
And
$$\prod_{k=1}^{n} a_k \equiv 1 \pmod{10}$$
And
$$\sum_{k=1}^{n} a_k \equiv n \pmod{10}$$
And
$$R(n)=\prod_{k=1}^{n} a_k - \sum_{k=1}^{n} a_k \equiv 1 - n \pmod{10}$$
So, whenever $n-1$ is divisible by $2,5$ or $10$, $R(n)$ is not a prime.
*And another one is $$a_n \equiv \sum_{k=1}^{n}1=n \pmod{9}$$
So
$$\prod_{k=1}^{n} a_k \equiv n! \pmod{9}$$
And
$$\sum_{k=1}^{n} a_k \equiv \frac{n(n+1)}{2} \pmod{9}$$
And
$$R(n)=\prod_{k=1}^{n} a_k - \sum_{k=1}^{n} a_k \equiv n! - \frac{n(n+1)}{2} \pmod{9}$$
For $n \geq 6$ with $\frac{n(n+1)}{2}$ divisible by 3, $R(n)$ is not a prime. Out of $n=6t$, $n=6t+1$, $n=6t+2$, $n=6t+3$, $n=6t+4$ and $n=6t+5$ only $n=6t+1$ and $6t+4$ could yield primes. Because $n=6t+1$ is clarified by the case 1 ($n-1 = 6t$ divisible by $2$) we are left with $6t+4$.
*Another basic observation is $$\gcd(a_p,a_q)=1$$
where $p,q$- primes. A short proof is here.
|
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|
Find the value of constants $c_1, c_2, c_3, c_4$ for which function $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable The Problem:
Find the value of constants $c_1, c_2, c_3, c_4$ for which function $f: \mathbb{R} \rightarrow \mathbb{R} $ (written below) is differentiable:
$$
f(x) = \begin{cases}
c_1 \arctan x + c_2 \cos x + \sin x & \quad x < 0\\
\ln(1+x^2) - x \ln 2 + 2^x & \quad 0 \leq x \leq 1\\
c_3 (x-1)(x-2)\cdots (x-2016) + c_4 (x+1) & \quad 1 < x \\
\end{cases}
$$
Potential solution:
First I need to find values for which the function is continuous.
$\lim_{x \rightarrow 0^-}$ of 1st case should be equal to the 2nd case evaluated at $x=0$.
$$\lim_{x \rightarrow 0^-} c_1 \arctan x + c_2 \cos x + \sin x = \ln 1 + 1$$
$$\Rightarrow c_2 = 1 ,\ c_1 \in \mathbb{R}$$
2nd case evaluated at $x=1$ should be equal to $\lim_{x \rightarrow 1^+}$ of the 3rd case.
$$\ln 2 - \ln2 + 2 = \lim_{x \rightarrow 1^+} c_3 (x-1)(x-2)\cdots (x-2016) + c_4 (x+1)$$
$$\Rightarrow c_4 = 1 ,\ c_3 \in \mathbb{R}$$
Solving finds the values of $c_2$ and $c_4$ for which the function is continuous.
To find values for which it is differentiable I first need to find derivatives of each of the parts and then go through the previous process again. But, I don't know how to find the derivative of the 3rd case. Here's what I tried.
The derivative:
$$
f'(x) = \begin{cases}
c_1 \frac{1}{1+x^2} - c_2 \cdot \sin x + \cos x & \quad x < 0\\
2x \cdot \frac{1}{1+x^2} - \ln 2 + \ln 2 \cdot 2^x & \quad 0 \leq x \leq 1\\
???\ \ c_3 (2016 x^{2015} + \ldots) + c_4 \ \ ??? & \quad 1 < x \\
\end{cases}
$$
Repeating the same process with the derivative instead of original function:
$$\lim_{x \rightarrow 0^-} c_1 \frac{1}{1+x^2} - 1 \cdot \sin x + \cos x = 2\cdot0\frac{1}{1+0} - \ln2 + 2^0\ln 2$$
$$\Rightarrow c_1 = -1$$
Here's the problematic part:
$$ 2 \cdot \frac{1}{1+1^2} - \ln 2 + \ln 2 \cdot 2^1 = \lim_{x \rightarrow 1^+} c_3 (2016 x^{2015} + \ldots) + 1$$
$$c_3 = ???$$
Any ideas on how to find this derivative and then its limit? Is there a different approach for solving this problem?
|
$\lim_\limits{x\to 1^+} f'(x)$
We don't need to find the exact derivative of $c_3(x−1)(x−2)⋯(x−2016)+c_4(x+1)$
When we differentiate we get $ c_3 (x-2)\cdots(x-2016) + c_3(x-1) [(x-3)\cdots(x-2016) + (x-2)(x-4)\cdots]+c_4$
And as $x$ approaches $1, c_3(x-1)[\text{everything inside the brackets}] = 0$
leaving $c_3 (2015!) + c_4$
regarding your work on $c_1$
you calculated the derivatives correctly, but you made an error when you evaluated them in a neighborhood of 0.
$\lim_\limits{x=0^+} f'(x) = 2\cdot0\frac{1}{1+0} - \ln2 + 2^0\ln 2 = 0$
|
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|
Does not algebraic multiplicity= geometric multiplicity $\Rightarrow$ the matrix is diagonalizable?
let $A=\left(\begin{array}{ccc} -5 & -1 & 6 \\ -2 & -5 & 8 \\ -1 & -1 & 1 \end{array}\right)$ and $B=\left(\begin{array}{ccc} -9 & 3 & -3 \\ -14 & 4 & -7 \\ -2 & 1 & -4 \end{array}\right)$ matrices above $\mathbb{C}$ are they similar?
So I started with finding the eigenvalues and geometric multiplicity.
For $A$ I got $f_{\lambda}(x)=(x+3)^3$ and $m_{\lambda}(x)=(x+3)^3$
For $B$ I got $f_{\lambda}(x)=(x+3)^3$ and $m_{\lambda}(x)=(x+3)^2$
So the jordan normal form of $A$ is $A=J_{1}(-3),J_{1}(-3),J_{1}(-3)$ or $A=\left(\begin{array}{ccc} -3 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{array}\right)$
And the jordan normal form of $B$ is $B=J_{2}(-3),J_{1}(-3)$ or $B=\left(\begin{array}{ccc} -3 & 1 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{array}\right)$
So $A$ is not similar to $B$ is it correct?
|
Indeed they are not similar. But $A$ is in fact not diagolizable.
It has a size 3 Jordan block:
$(A+3E)^2 \neq 0$ (whereas $(B+3E)^2=0$). This is also what you got using the $m_\lambda$ formulation.
|
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|
Aptitude simplification question Given $$p+q+r=1\;\;\;\;\&\;\;\;p^2+q^2+r^2=2\;\;\;\;\&\;\;\;p^3+q^3+r^3=3$$
find the value of $$p^4+q^4+r^4$$
|
More generally, we have that for all $a,b,c\in\mathbb R$, $n\in\mathbb Z$, $n\ge 0$:
$$S_{n+3}=S_{n+2}S_1-S_{n+1}(ab+bc+ca)+S_{n}(abc),$$
where $S_n=a^n+b^n+c^n$. It can be proved by simply expanding.
Also notice $ab+bc+ca=\frac{(a+b+c)^2-\left(a^2+b^2+c^2\right)}{2}$.
Now apply this for your problem. Let $(a,b,c)=(p,q,r)$ and also let firstly $n=0$ (to find $pqr$) and then $n=1$ (to find $p^4+q^4+r^4$).
|
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|
Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$ Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$
I find out the answer $104$ but my friend find the answer $96$ using calculator.which one is correct?
|
Hint:
let $x=\sqrt(5)+\sqrt(6)$ and $y=\sqrt(5)-\sqrt(6)$
so
$$(x+\sqrt{7})(x-\sqrt{7})(y+\sqrt{7})(-y+\sqrt{7})$$
then you can use the fact of
$$(a-b)(a+b)=a^2-b^2$$
|
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|
How to show $\frac{1}{xy}$ = $\frac{1}{x}\frac{1}{y}$? I'm trying to prove that $\frac{1}{xy}$ = $\frac{1}{x}*\frac{1}{y},\forall x,y\neq 0$ using the field axioms of addition, multiplication, and the distributive law: $(x+y)z = xz+yz$ but am having a hard time doing so.
Things I've tried:
I've tried letting $\frac{1}{xy}$ = $\frac{1}{xy}$1 and $\frac{1}{xy}$=$\frac{1}{xy}$+0 and trying some fancy stuff, but so far I can't get anywhere with those.
Any help is greatly appreciated.
|
Remember the fields axioms (explained below). First you should think about which one to use in the statement of the desired identity. Certainly the demonstration through the axiom (M3). Second, you must realize that the secret of proof is to use the uniqueness of the axiom (M3).
Fixed real numbers $a$, $b$ there are real numbers $\alpha$, $\beta$ and $\gamma$ such that $a\cdot \alpha =1$, $b\cdot \beta =1$ and $(b\cdot b)\cdot \gamma=1$. Let's $u,v$ and $w$ real numbers such that $a\cdot u =1$, $b\cdot v =1$ and $(a\cdot b)\cdot w=1$. The uniqueness assures us that $u=\alpha$, $v=\beta$ and $w=\gamma$. So if $\frac{1}{a}\cdot \frac{1}{b}=\alpha\cdot \beta$ satisfies identity ${a\cdot b}\left(\frac{1}{a}\cdot \frac{1}{b} \right)=1$, then we will have equality $\frac{1}{a\cdot b}=\left(\frac{1}{a}\cdot \frac{1}{b} \right)$. In fact,
\begin{align}
(x\cdot y)\cdot \left( \frac{1}{x}\cdot \frac{1}{y}\right)
&=
(x\cdot y)\cdot \left( \frac{1}{y}\cdot \frac{1}{x}\right)& \textrm{ by (M4)}
\\
&=
\left((x\cdot y)\cdot \frac{1}{y}\right)\cdot \frac{1}{x}& \textrm{ by (M1)}
\\
&=
\left(x\cdot \left(y\cdot \frac{1}{y}\right)\right)\cdot \frac{1}{x}&\textrm{ by (M1) }
\\
&=
\left(x\cdot 1\right)\cdot \frac{1}{x}& \textrm{ by (M3)}
\\
&=
x\cdot \frac{1}{x}& \textrm{ by (M2)}
\\
&=
1&\textrm{ by (M3)}
\end{align}
*
*(M1) $(a\cdot b)\cdot c=a\cdot (b\cdot c), \quad \forall a,b,c\in \mathbb{R}-\{0\}$
*(M2) $\exists ! 1\in \mathbb{R}-\{0\} \textrm{ such that } 1\cdot a= a, \quad \forall a\in\mathbb{R}-\{0\}$
*(M3) $\forall a\in \mathbb{R}-\{0\},\exists !\; \dfrac{1}{a}\in\mathbb{R} \textrm{ such that } a\cdot\dfrac{1}{a}=1$
*(M4) $a\cdot b= b\cdot a, \quad \forall a,b\in \mathbb{R}-\{0\}$
|
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|
Suppose that $q=\frac{2^n+1}{3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$ I am curious for a starting point to prove the following claim
Suppose that $q={2^n+1 \above 1.5pt 3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$
For example take $n=61$. Then $q ={2^{61}+1 \above 1.5pt 3} = 768614336404564651$ which is prime. Then $\binom{2^{61}}{2}-1 = 2658455991569831744654692615953842175$ and it's largest prime factor is $q$. Note ${2^{29}+1 \above 1.5pt 3} =178956971$ is not prime and the largest prime factor of $\binom{2^{29}}{2}-1$ is $3033169$. But even then we could note that $178956971 = 59*3033169$. A counterexample to the claim would suffice to to show it's wrong.
|
First of all, we have to show that $\frac{2^n+1}{3}$ is a factor of $$\binom{2^n}{2}-1=\frac{2^n(2^n-1)}{2}-1=\frac{2^{2n}-2^n-2}{2}$$
To show this, we consider
$$(2^n+1)^2-3(2^n+1)=2^{2n}+2\cdot 2^n+1-3\cdot2^n-3=2^{2n}-2^n-2$$
So, $\binom{2^n}{2}-1$ is divisble by $2^n+1$ and therefore by $\frac{2^n+1}{3}$.
But $(\frac{2^n+1}{3})^2>\frac{2^{2n}+2^n+1}{9}$
Since $9$ is a factor of $\binom{2^n}{2}-1$ for odd $n$ (and $n$ must be odd, otherwise $\frac{2^n+1}{3}$ is not an integer) , there cannot be a larger prime factor $p$ because we would have $$3^2\cdot \frac{2^n+1}{3}\cdot p>2^{2n}+2^n+1>\frac{2^{2n}-2^n-2}{2}$$
To show that $9$ is a factor of $2^{2n}-2^n-2$ (for odd n), which implies that $9$ is a factor of $\binom{2^n}{2}-1$ , we consider that $2^n\equiv 2,5 \ or\ 8\mod 9$. In any case we get $2^{2n}-2^n\equiv 2\mod 9$.
|
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|
Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square
Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square.
We can observe that $49=7^2, 4489=67^2, 444889=667^2, \ldots$
I have tried expanding terms of the sequence, and to express it as a whole square. But it was too tough for me.
Any help will be appreciated.
|
The $n$th term is $T_{n}=9+8.10+8.10^2+...8.10^{n-1}+4.10^n+...+4.10^{2n-1}$:
$$
\begin{align}
T_n &= 9+8.10(1+10+...10^{n-2})+4.10^n(1+...+10^{n-1}) \\
&= 9+8.10(\frac{10^{n-1}-1}{9})+4.10^n (\frac{10^{n}-1}{9}) \\
&=9+8.\frac{10^{n}-10}{9}+4.\frac{10^{2n}-10^n}{9} \\
&=\frac{81+8.10^n-80+4.10^{2n}-4.10^n}{9} \\
&=\frac{1}{9}.(1+4.10^n+4.10^{2n}) \\
&={(\frac{{1+2.10^n}}{3}})^2
\end{align}
$$.
Now $10\equiv 1\pmod{3}\Rightarrow 10^n\equiv 1\pmod{3}\Rightarrow 2.10^n+1\equiv 0\pmod{3}$, thus $3$ divides the numerator and we are done.
|
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|
Solve quadratic fraction I would like to simplify the fraction
$$\frac{x^2-2x}{x^2+x-6}$$
I know from Mathematica that it should equal $\frac3{3x}$ but how do I get there?
|
$x^2+x-6 = (x-2)(x+3)\rightarrow \frac{x(x-2)}{(x+3)(x-2)}=\frac{x}{x+3}$, so your result is incorrect, as the next counterexample shows:
$x=1; \frac{x^2-2x}{x^2+x-6}=\frac{1}{4}\neq 1=\frac{3}{3x}$
|
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|
prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$ If $a,b,c$
are the sides of a triangle and $p,q,r$ are positive real numbers then prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$
After modification. I have to prove $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$
|
We'll prove that your inequality is true for all reals $p$, $q$ and $r$
and $a$, $b$ and $c$ are lengths-sides of triangle.
Indeed,
$$a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)=$$
$$=\frac{1}{2}\left((p-q)^2(a^2+b^2-c^2)+(p-r)^2(a^2+c^2-b^2)+(q-r)^2(b^2+c^2-a^2)\right)\geq0$$
because $\sum\limits_{cyc}(a^2+b^2-c^2)=a^2+b^2+c^2>0$ and
$$\sum\limits_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2)=\sum\limits_{cyc}(2a^2b^2-a^4)=16S^2>0$$
Done!
|
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|
Defining the foci of "slanted" ellipse equation How to define the foci ($F_1,F_2$) coordinates of the slanted ellipse $x^2+4xy+9y^2=9$?
|
Your ellipse is centered at the origin, hence in order to find its foci it is enough to find its vertices, or the direction of its axis. To do that, we may compute the stationary points of the quadratic form $q(x,y)=x^2+4xy+9y^2$ under the constraint $x^2+y^2=1$ through the method of Lagrange's multipliers. $\nabla q = 2\lambda(x,y) $ leads to the system
$$\left\{\begin{eqnarray*}x+2y &=& \lambda x \\ 2x+9y&=&\lambda y \end{eqnarray*} \right.\tag{1}$$
hence the values of $\lambda$ associated with the stationary points are the eigenvalues of the matrix
$$\begin{pmatrix}1 & 2 \\ 2 & 9 \end{pmatrix} \tag{2}$$
i.e. $5\pm 2\sqrt{5}$, and $(1)$ gives that the directions of the axis are $(1,2+\sqrt{5})$ for the major axis and $(1,2-\sqrt{5})$ for the minor axis. It follows that a vertex on the minor axis is given by
$$ V_m = \left(3\cdot\frac{\sqrt{5}-2}{\sqrt{10}},\frac{3}{\sqrt{10}}\right) \tag{3}$$
and a vertex on the major axis is given by:
$$ V_M = \left(3\cdot\frac{\sqrt{5}+2}{\sqrt{10}},-\frac{3}{\sqrt{10}}\right) \tag{4}$$
so:
$$ a^2=\|V_M\|^2 = 3+\frac{6}{\sqrt{5}},\qquad b^2 = \|V_m\|^2 = 3-\frac{6}{\sqrt{5}} \tag{5} $$
and $c^2 = a^2-b^2 = \frac{12}{\sqrt{5}}$. At last,
$$ F_1, F_2 = \pm \frac{c}{a}V_M\tag{6} $$
hence:
$$\boxed{ F_1, F_2 = \color{red}{\left(\pm 3\sqrt{\frac{2}{5}(\sqrt{5}+2)},\mp 3\sqrt{\frac{2}{5}(\sqrt{5}-2)}\right)}} \tag{7} $$
|
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|
Expanding $(2^b-1)\cdot(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})$ Can someone please explain to me how the following works (primarily interested in an explanation of the second step when $2^b$ is expanded?
I understand how each series cancels out to equal $2^n-1$ at the end.
$$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=2^b \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b}) - (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=(2^b + 2^{2b} + 2^{3b} + \cdots + 2^{ab})-(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})\\
&=2^{ab}-1\\
&=2^n-1
\end{align*}$$
|
$$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=2^b \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b}) - (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=(2^b + 2^{2b} + 2^{3b} + \cdots + 2^{ab})-(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})\\
&= \left[2^{ab} - 1\right] + \left[(2^b + 2^{2b} + \cdots + 2^{
(a-1)b})-(2^b+2^{2b}+ \cdots + 2^{(a-1)b}) \right]\\
&= 2^{ab} - 1 +(2^b - 2^b) + (2^{2b} - 2^{2b}) + \cdots \\
&=2^{ab}-1\\
&=2^n-1
\end{align*}$$
|
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|
Evaluate $\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$ Somebody asked this and I think it's quite interesting as I couldn't figure out how to evaluate this but the Wolfram Alpha says its limit is $\frac e2$.
$$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$
Could someone help here?
|
For $0<x<1$: We have $$\frac {1}{x}\ln (1+x)=1-x/2+(x^2/3-x^3/4)+(x^4/5-x^5/6)+...=$$ $$=1-x/2+x^2/3-(x^3-x^4/5)-(x^5-x^6/7)-...$$ Therefore $$1- x/2<\frac {1}{x}\ln (1+x)<1-x/2+x^2/3.$$ Therefore $$e(1-e^{-x/2+x^2/3})<e-(1+x)^{\frac {1}{x}} <e(1-e^{-x/2}).$$
Both $(-x/2+x^2/3)$ and $(-x/2)$ belong to the interval $(-1,0).$ When $y\in (-1,0)$ we have $$1+y<1+y+(y^2/2!+y^3/3!)+(y^4/4!+y^5/5!)+...=e^y$$ $$=1+y+y^2/2!+(y^3/3!+y^4/4)+..<1+y+y^2/2!.$$ So $$-y_1-y_1^2/2=1-(1+y_1+y_1^2/2!))<1-e^{y_1}$$ where $y_1=-x/2+x^2/3.$..... And also $$1-e^{-x/2}<1-(1-x/2!)=x/2.$$ Now $y_1+y_1^2/2=-x/2 +x^2F(x)$ where $F(x)$ is a polynomial, so for some $K>0$ we have $x\in (0,1)\implies |F(x)|<K.$ Therefore $$ e/2-xK<\frac {e-(1+x)^{\frac {1}{x}}}{x}<e/2.$$
|
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|
Angle between edges in a square pyramid
How do you calculate the missing angle? I couldn't figure it out. This is not homework, just a math practice question that I made up to test myself.
|
First proof by analytical geometry.
Let
*
*$A(1,1,0),B(1,-1,0),C(-1,-1,0),D(-1,1,0)$ be the basis' vertices,
*$S(0,0,z)$ be the summit, with unknown $z$ and
*$a=SA=SB=SC=SD$.
*$\alpha$ be the unknown angle between $\vec{SA}$ and $\vec{SC}$.
Then $\tag{1} \cases{\vec{SA}=(1,1,0)-(0,0,z)=(1,1,-z)\\ \vec{SB}=(1,-1,0)-(0,0,z)=(1,-1,-z)\\ \vec{SC}=(-1,-1,0)-(0,0,z)=(-1,-1,-z)} $
A first consequence is that $\tag{2}a^2=\vec{SA}^2=2+z^2.$
Now, using the two forms of the dot product:
$$\tag{3}\vec{SA}.\vec{SB}=\cases{\|SA\|\|SB\|\cos(ASB)=a^2 \cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}a^2\\(1) \times (1) + (1) \times (-1)+(-z) \times (-z)=z^2}$$
giving : $\tag{4}a^2=\sqrt{2}z^2$
With (2} and {4}, we deduce the values $\tag{5}\cases{z^2=2+2\sqrt{2}\\a^2=4+2\sqrt{2}}$
Expressing as in (3), $\vec{SA}.\vec{SC}$ in two ways, that the unknown angle $\alpha$ is such that, using (1),
$$\tag{4}\vec{SA}.\vec{SC}=\cases{\cos(\alpha)a^2\\(1) \times (-1) + (1) \times (-1)+(-z) \times (-z)=z^2-2}$$
Thus $\cos(\alpha)=\dfrac{z^2-2}{a^2}$. Using (4), one obtains:
$\cos(\alpha)=\frac{2\sqrt{2}}{4+2\sqrt{2}}=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1.$
The final result is $\alpha=acos(\sqrt{2}-1)\approx 1.14372 rad.$
i.e. close to $55^{\circ}1/2$.
Second proof by law of cosines (applied twice) (see (https://en.wikipedia.org/wiki/Law_of_cosines))
We take the same conventions as in the previous proof.
In triangle SAB: $\ AB^2=SA^2+SB^2-2 SA.SB \cos(\pi/4).$
giving $2^2=a^2+a^2-2a^2 \cos(\frac{\pi}{4})$ out of which we obtain
$$\tag{5}a^2=2(2+\sqrt{2}).$$
In triangle SAC: $\ AC^2=SA^2+SC^2-2 SA.SC \cos(\alpha)$
giving $(2 \sqrt{2})^2=a^2+a^2-2a^2 \cos(\alpha).$
or $\tag{6}4=a^2(1- \cos(\alpha)).$
Using (5) in (6), we are able to deduce
$$\cos(\alpha)=1-\frac{4}{a^2}=1-\frac{2}{2+\sqrt{2}}=1-\frac{\sqrt{2}}{1+\sqrt{2}}=1-\sqrt{2}(\sqrt{2}-1)=\sqrt{2}-1$$ as before.
|
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|
Find the vlues of the trigonometric functions from the given information
$\tan t = \dfrac{1}{4}$, terminal point of $t$ is in the third quadrant.
So $\tan t = \dfrac{-\sqrt{1-\cos^2 t}}{\cos t} = \dfrac{1}{4} $, because it's in the third quadrant. After solving for $\cos^2 t$ get as an answer that $\cos ^2 t = \dfrac{16}{17}$, but apparently this is wrong. Where lies my mistake?
|
If
$$\tan t = \frac{1}{4}$$
then
$$t = \arctan \left(\frac{1}{4}\right) = 14.03°$$
or $0.244$ radians.
Beware
$$\tan t = \frac{\sin t}{\cos t} = \frac{\sqrt{1 - \cos^2t}}{\cos t}$$
In your formula, there is a minus sign which is wrong.
Hence
$$\frac{\sqrt{1 - \cos^2t}}{\cos t} = \frac{1}{4}$$
$$4\sqrt{1 - \cos^2 t} = \cos t$$
Squaring
$$16 - 16\cos^2 t = \cos^2 t$$
$$16 = 17\cos^2 t$$
$$\cos t = \pm \sqrt{\frac{16}{17}} = \pm 4\sqrt{\frac{1}{17}}$$
Then choose the right sign for what you know.
|
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|
How to calculate $\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$? I need some help with computing this limit:
$$\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$$
I'm guessing that I should do a Taylor expansion of $\cos(\sin x)$ in base, but what should I do with the stuff in the exponent?
|
This answer is a codicil to that of Paramanand Singh. I give here a different calculation of $$\lim\limits_{x\to 0} \frac{\cos(\sin x)-1+\frac{x^2}{2}}{x^4}$$
Frequent use of the replacement of $\sin y$ by $\frac{\sin y}{y}y$ is made.
We have
$$ \frac{\cos(\sin x)-1+\frac{x^2}{2}}{x^4}= \frac{\cos(\sin x)-\cos x}{x^4}
+ \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}.$$ And we calculate each separately.
For the first we use the difference of cosines formula $\cos a-\cos b=2\sin \frac{a+b}{2}\sin \frac{b-a}{2}$
So we have $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin \frac{\sin x+x}{2}}{x}\frac{\sin \frac{x-\sin x}{2}}{x^3}$$
the standard replacement above reduces to the limit of
$$\frac{1}{2}\frac{\sin x+x}{x}\frac{x-\sin x}{x^3}\rightarrow \frac{1}{6}.$$
As for the limit $\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$, many people would be willing to take this as granted but I give here a nice elementary derivation.
First $$\left(\frac{1-\cos x}{x^2}\right)^2=\frac{2-2\cos x-\sin^2 x}{x^4}\rightarrow \frac{1}{4}$$
Second $$\frac{x-\sin x}{x^3}\frac{\sin x}{x}=\frac{x\sin x-\sin^2 x}{x^4}\rightarrow \frac{1}{6}$$
Subtracting the first from the second gives
$$\frac{2\cos x-2 +x\sin x }{x^4}\rightarrow -\frac{1}{12} $$
we can rewrite this as
$$2\frac{\cos x-1 +\frac{x^2}{2} }{x^4}+\frac{\sin x -x}{x^3}\rightarrow -\frac{1}{12} $$ and this gives
$$\frac{\cos x-1 +\frac{x^2}{2} }{x^4}\rightarrow \frac{1}{24} $$
Finally adding them together we get
$$\frac{1}{6}+\frac{1}{24}=\frac{5}{24}.$$
|
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|
Finding the smallest value of the sum
The multiplication of three natural numbers $a$, $b$ and $c$ equals $2016$. What is the smallest value of $a + b + c$?
I started by a prime factorization to find that $2016 = 2^5 \cdot 3^2 \cdot 7$.
What should I do next?
|
If we consider the fact that $7=2\cdot3\cdot\frac76$, then $2016=2^6\cdot3^3\cdot\frac76$, and it becomes easy to represent this number as a product of $\color\red3$ natural numbers which are as close to each other as possible:
*
*$a=2^{6/\color\red3}\cdot3^{3/\color\red3}=12$
*$b=2^{6/\color\red3}\cdot3^{3/\color\red3}=12$
*$c=2^{6/\color\red3}\cdot3^{3/\color\red3}\cdot\frac76=14$
|
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|
Closed form of $k$-th term in the recurrence $D_j = 1 + z^2\left(1-\frac{1}{D_{j-1}}\right)$, where $D_0 = 1+z^2$ I have $D_0=1+z^2$ and $D_{j}=1+z^2(1-1/D_{j-1})$ for $j>0$. I want to write an expression for arbitrary $D_k$, and what I have is below.
$D_{k}=1+z^2\Bigg(1-\bigg(1+z^2\big(1-(1+z^2(1-(\cdots(1+z^2)^{-1}\cdots)^{-1})^{-1})^{-1}\big)^{-1}\bigg)^{-1}\Bigg)$
Is there better notation I can use to make this more precise/clear?
|
If you work out the first few $D_j$'s explicitly, you should see a pattern:
$$D_1=1+z^2\left(1-{1\over1+z^2}\right)=1+z^2\left(z^2\over1+z^2\right)={1+z^2+z^4\over1+z^2}$$
$$D_2=1+z^2\left(1-{1+z^2\over1+z^2+z^4}\right)=1+z^2\left(z^4\over1+z^2+z^4\right)={1+z^2+z^4+z^6\over1+z^2+z^4}$$
|
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|
The integral $\int \frac{x^2(1-\ln(x))}{(\ln(x))^4-x^4} dx$? $$\int \frac{x^2(1-\ln(x))}{\ln^4(x)-x^4} dx$$
I tried to factorize the denominator so that I could apply integration by parts but that didn't help me at all.
|
HINT:
$$\mathcal{I}(x)=\int\frac{x^2\left(1-\ln(x)\right)}{\ln^4(x)-x^4}\space\text{d}x=$$
$$\frac{1}{2}\int\frac{\ln(x)-1}{x^2+\ln^2(x)}\space\text{d}x-\frac{1}{4}\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x+\frac{1}{4}\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x$$
*
*Now, for $\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x$,
substitute $u=\ln(x)-x$ and
$\text{d}u=\left(\frac{1}{x}-1\right)\space\text{d}x$:
$$\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x=\int\frac{1}{u}\space\text{d}u=\ln\left|u\right|+\text{C}=\ln\left|\ln(x)-x\right|+\text{C}$$
*Now, for $\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x$,
substitute $s=\ln(x)+x$ and
$\text{d}s=\left(\frac{1}{x}-1\right)\space\text{d}x$:
$$\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x=\int\frac{1}{s}\space\text{d}s=\ln\left|s\right|+\text{C}=\ln\left|\ln(x)+x\right|+\text{C}$$
|
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|
Prove or disprove $\sum\limits_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$
Let $x,y,z\ge 0$. Prove or disprove
$$\dfrac{x^4+y^4}{x+y}+\dfrac{z^4+y^4}{z+y}+\dfrac{z^4+x^4}{x+z}\le 3\dfrac{x^4+y^4+z^4}{x+y+z}$$
This is what I tried. Without loss of generality, let $x+y+z=1$, then
$$\Longleftrightarrow \sum_{cyc}\dfrac{x^4+y^4}{x+y}\le 3(x^4+y^4+z^4)$$
and
$$(x^4+y^4)=(x+y)(x^3+y^3)-xy(x^2+y^2)=(x+y)(x^3+y^3)-xy(x+y)^2+2x^2y^2$$
which is equivalent to
$$\sum_{cyc}\left((x^3+y^3)-xy(x+y)+\dfrac{2x^2y^2}{x+y}\right)\le 3(x^4+y^4+z^4)$$
or
$$2\sum_{cyc}x^3+2\sum_{cyc}\dfrac{x^2y^2}{x+y}\le 3\sum_{cyc}(x^4+xy(x+y))$$
and now I'm stuck.
|
we have to show that $$\frac{3(x^4+y^4+z^4)}{x+y+z}-\frac{x^4+y^4}{x+y}-\frac{z^4+y^4}{z+y}-\frac{z^4+x^4}{x+z}\geq 0$$
assuming we have $$x=\min(x,y,z)$$ then we set
$$y=x+u,z=x+u+v$$ and we have after some algebra
$$\left( 12\,{u}^{2}+12\,uv+12\,{v}^{2} \right) {x}^{5}+ \left( 30\,{u}
^{3}+45\,{u}^{2}v+75\,u{v}^{2}+30\,{v}^{3} \right) {x}^{4}+ \left( 28
\,{u}^{4}+56\,{u}^{3}v+144\,{u}^{2}{v}^{2}+116\,u{v}^{3}+28\,{v}^{4}
\right) {x}^{3}+ \left( 12\,{u}^{5}+30\,{u}^{4}v+120\,{u}^{3}{v}^{2}+
150\,{u}^{2}{v}^{3}+72\,u{v}^{4}+12\,{v}^{5} \right) {x}^{2}+ \left( 2
\,{u}^{6}+6\,{u}^{5}v+45\,{u}^{4}{v}^{2}+80\,{u}^{3}{v}^{3}+57\,{u}^{2
}{v}^{4}+18\,u{v}^{5}+2\,{v}^{6} \right) x+6\,{u}^{5}{v}^{2}+15\,{u}^{
4}{v}^{3}+14\,{u}^{3}{v}^{4}+6\,{u}^{2}{v}^{5}+u{v}^{6}
\geq 0$$ which is true.
the other cases are analogously
|
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|
Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$ I can't make progress with proving this inequality.
I have tried opening the brackets and using $abc=1$ in order to obtain the following:
$$a^2+b^2+ \frac 1{a^2b^2}+18+9a^2b^2+\frac 9{a^2}+\frac9{b^2}\ge 16 \left (ab+\frac 1a+\frac 1b \right)$$
|
Since the right side of your inequality does not depend on substitution $a\rightarrow-a$,
it's enough to prove it for positive variables.
We'll prove that your inequality is true for all positives $a$, $b$ and $c$ such that $abc=1$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality is equivalent to
$$8\cdot9u^2w^2\leq18w^4+9(9u^2-6v^2)w^2+9(9v^4-6uw^3)$$ or
$f(v^2)\geq0$, where
$$f(v^2)=9v^4-6v^2w^2-6uw^3+u^2w^2+2w^4$$
Since $f'(v^2)=18v^2-6w^2>0$, it remains to prove our inequality for a minimal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$ and $c=\frac{1}{a^2}$.
We obtain $(a-1)^2(9a^6+18a^5+13a^4+8a^3+21a^2+2a+1)\geq0$.
Done!
|
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|
How to find the square roots of $z = 5-12i$ I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rightarrow \quad z_1 = -3+2i \quad z_2 = 3-2i
$$
My question is: can the following method (below) be used to solve the problem above? Motivation for this question: if I were to find the cubic roots of the number given, I couldn't use the first method.
I will use this "other method" it in a different problem.
Find the square roots of $ z = 2i $
The method:
Since $ \rho = 2 $ and $ \theta = \frac{\pi}{2} $ we have to find a complex number such that
\begin{align*}
z_k^2 &= 2i\\
z_k^2 &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\rho_k^2 \left( \cos 2\theta_k + i \sin 2 \theta_k \right) &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\end{align*}
\begin{cases}
\rho^2 &= 2\\
2\theta_k &= \frac{\pi}{2} + 2k\pi
\end{cases}
\begin{cases}
\rho &= \sqrt{2}\\
\theta_k &= \frac{\pi}{4} + 2k\pi
\end{cases}
\begin{align*}
z_0 &= \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 1+i\\
z_1 &= \sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = -1-i
\end{align*}
Using this method for the first example seems like a dead end, specially because of the fact that the angle of $z$ is not as straightforward as the angle of the second example.
Thank you.
|
If you let $\theta$ be the angle in the first case, then using the fact that $\tan \theta = -12/5$, you can find $\cos\theta$ and $\cos\frac{\theta}{2}$ indirectly. One of the roots would be:
$\sqrt{13}(-\frac{3}{\sqrt{13}} + i \frac{2}{\sqrt{13}}) = z_1.$
|
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|
Show that $1^n+2^n+3^n+4^n$ is divisible by 5 if and only if n is not divisible by 4 Show that $$1^n+2^n+3^n+4^n$$ is
divisible by 5 if and only if n is not divisible by 4.
I don't find the relation why must divisible by 5 anyone can give a hint or give some part of proof?
|
If $4 \mid n \implies n = 4k $
$1^{4k} + 2^{4k} + 3^{4k} + 4^{4k} \equiv 1^k + 1^k + 1^k + 1^k \equiv 1 + 1+ 1 + 1 \equiv 4 \space \text{(mod 5)}$
If $4 \nmid n \implies n = 4k + 1 \vee n = 4k + 2 \vee 4k + 3$
For, $n = 4k + 1$ you get:
$1^{4k+1} + 2^{4k+1} + 3^{4k+1} + 4^{4k+1} \equiv 1^k \times 1 + 1^k \times 2 + 1^k \times 3 + 1^k \times 5 \equiv 1 + 2+ 3 + 4 \equiv 10 \equiv 0 \space \text{(mod 5)}$
|
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|
This sequence $\lfloor \sqrt{2003}\cdot n\rfloor $ contains an infinite number of square numbers
Show that: the sequence $\lfloor \sqrt{2003}\cdot n\rfloor $ contains an infinite number of square numbers.
Maybe consider the Pell equation to solve this problem. But how to do this? Thanks
|
CW again. Comment on the pretty idea in the other answers: Suppose we have
$$ x^2 - d y^2 = -k, $$
with small integer $k > 0.$
$$ ( x - y \sqrt d) (x + y \sqrt d) = -k, $$
$$ x - y \sqrt d = \frac{-k}{x + y \sqrt d}, $$
$$ x = y \sqrt d - \frac{k}{x + y \sqrt d}, $$
$$ x^2 = xy \sqrt d - \frac{kx}{x + y \sqrt d}, $$
$$ xy \sqrt d = x^2 + \frac{kx}{x + y \sqrt d}. $$
If $kx < x + y \sqrt d,$ then $\lfloor xy \sqrt d \rfloor = x^2.$
When is $kx < x + y \sqrt d ?$ We know $ x - y \sqrt d < 0,$ so that
$y \sqrt d > x.$ That is,
$$ x + y \sqrt d > 2x. $$
Therefore, if $k = 1$ or $k = 2,$ we do get $\lfloor xy \sqrt d \rfloor = x^2.$
However, as I found when doing $x^2 - 7 y^2 = -3,$ if $k \geq 3,$ this does not work. Let me prove that, need some more time...
This will do. If $x^2 - d y^2 = -k$ and the very mild $x > \sqrt \frac{k}{3},$ but $y \sqrt d \geq 2x,$ then $d y^2 \geq 4 x^2.$ Then $x^2 - d y^2 \leq - 3 x^2.$ However, we made the mild bound $x > \sqrt \frac{k}{3},$ so that $3 x^2 > k.$ It follows that $x^2 - d y^2 < k.$ This contradicts the assumption that $y \sqrt d \geq 2x.$ So, except for a small finite number of cases with $x \leq \sqrt \frac{k}{3},$ we have
$$ y \sqrt d < 2x, $$
$$ x + y \sqrt d < 3x, $$
$$ \frac{1}{x + y \sqrt d} > \frac{1}{3x}. $$ From
$$ xy \sqrt d = x^2 + \frac{kx}{x + y \sqrt d} $$ this says
$$ xy \sqrt d > x^2 + \frac{kx}{3x}, $$
$$ xy \sqrt d > x^2 + \frac{k}{3}. $$
That is, if $k \geq 3,$ unless $x \leq \sqrt \frac{k}{3},$
$$ \lfloor xy \sqrt d \rfloor \neq x^2 . $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Differentiation involving implicit and parametric. I was asked to differentiate the term,
$$X^3 + XY^2 - Y^3$$
For such I reached,
$$3X^2 + 2XY\dfrac{dy}{dx} - 3Y^2\dfrac{dy}{dx}$$
The apparent answer is,
$$3X^2 + Y^2 + 2XY\dfrac{dy}{dx} - 3Y^2\dfrac{dy}{dx}$$
How exactly this is reached as I am new to the whole implicit and parametric function.
|
You forgot about the product rule. Assuming $y=y(x),$
\begin{align*}
\frac{d}{dx} \left( x^3 +xy^2 - y^3 \right) &= 3x^2 + \frac{d}{dx} \left(xy^2 \right) -3y^2 \frac{dy}{dx} \\
&= 3x^2 + 2xy \frac{dy}{dx} + y^2 - 3y^2 \frac{dy}{dx}
\end{align*}
|
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|
Compute $\int^{\pi/2}_0 \frac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$ I have tried solving this for about an hour and will probably resort to head banging in some time:
$$\int ^{\frac{\pi}{2}}_{0} \dfrac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$$
I first divided by $\cos^4 x$ and then subsequently put $\tan x = t$, to get:
$$\int ^{\infty}_{0} \dfrac{1+t^2}{(a^2 + b^2 t^2)^2}dt$$
This has become unmanageable. Neither splitting the numerator, nor Partial fraction (taking t^2 = z and applying partial fraction) seems to work.
|
For a solution that doesn't involve complex analysis: Try the substitution $t = \frac{a}{b} \tan u$. Then $du = \frac{ab}{a^2 + b^2 t^2} \,dt$ and
your integral becomes
$$\frac{1}{a^3 b^3} \int_0^{\pi/2} (b^2 \cos^2 u + a^2 \sin^2 u) \,du$$
which is more doable.
|
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|
Expressing $\sqrt[3]{7+5\sqrt{2}}$ in the form $x+y\sqrt{2}$ Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.
I.e. Show that it is $1+\sqrt{2}$.
|
You can assume that the nested radical can be expressed in $a+b\sqrt{2}$ form. More specifically, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}\tag{1}$$
With your question, we have $$\sqrt[3]{7+5\sqrt{2}}=a+b\sqrt{2}\tag{2}$$
Cubing both sides, we get $$7+5\sqrt{2}=(a^3+6ab^2)+(3a^2b+2b^3)\sqrt{2}\tag{3}$$
And equating corresponding coefficients, we get the following system of equations: $$\begin{cases}a^3+6ab^2=7\\3a^2b+2b^3=5\tag{4}\end{cases}$$
Cross multiplying, we get a multi-variate polynomial. Namely, $$5a^3-21a^2b+30ab^2-14b^3=0\tag{5}$$
Dividing both sides by $b^3$, we get: $$5\frac {a^3}{b^3}-21\frac {a^2}{b^2}+30\frac {a}{b}-14=0\tag{6}$$
Which is also equal to $5\left(\frac ab\right)^3-21\left(\frac {a}{b}\right)^2+30\left(\frac {a}{b}\right)-14=0$. Substituting $a/b$ with $x$, we get the cubic polynomial$$5x^3-21x^2+30x-14=0\tag{7}$$ with $x=1$ as an integer root.
Since $a/b=x$, we have $$\frac ab=1\implies a=b\tag{8}$$
So from $(3)$, we have $a^3+6a(a)^2=7\implies a^3+6a^3=7\implies 7a^3=7\implies a=b=1$
$$\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$$
For practice, you can try to denest $\sqrt[3]{2+\sqrt{5}}$
|
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|
Closed form solution for infinite summation Thomas, Bruckner & Bruckner, Elementary Real Analysis.
Prove that for all r > 1,
$$\frac{1}{r - 1} = \frac{1}{r+1} + \frac{2}{r^2 + 1} + \frac{4}{r^4 + 1} + \frac{8}{r^8 + 1} + \cdots$$
So far I have $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$
$$\sum_{n=1}^\infty \frac{2^n}{r^{2^n} + 1} = \sum_{n=1}^\infty \left(\frac{2^n}{r^{2^n}} - \frac{2^n}{r^{4^n} + r^{2^n}}\right)$$
|
For any $x\in(0,1)$ we have
$$\frac{1}{1-x}= (1+x)(1+x^2)(1+x^4)(1+x^8)\cdots \tag{1} $$
that can be read as every positive integer has a unique representation in base-$2$.
By taking $\frac{d}{dx}\log(\cdot)$ of both sides, we get:
$$ \frac{1}{1-x}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+\cdots \tag{2} $$
and it is enough to evaluate the last identity at $x=\frac{1}{r}$ to prove the given claim, after trivial simplifications.
|
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|
Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
|
$$(x^2-3x+1)^2=4x^2-12x+9$$
$$(x^2-3x+1)^2=4(x^2-3x+1)+5$$
$$(x^2-3x+1)^2-4(x^2-3x+1)-5=0$$
now let $u=(x^2-3x+1)^2$
so
$$u^2-4u-5=0$$
$$(u+1)(u-5)=0$$
then complete the solution
|
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|
Total Expectation problem to solve in 2 ways - regular and by the law of total expectancy Im trying to solve the question in the way of total expectancy (meaning $E[X]=E[E[X|Y]]$).
the question: "A fair cube with 4 sides is thrown twice. each side has a number from 1 to 4. Let's mark Y - maximum of two results. Evaluate E(Y) both by regular way and by the law of total expectation."
So I managed to do it by the regular path:
However, by the law of total expectation I can't figure out how to do that. I found out that if we mark X - result of first throw then:
*
*$P(Y=1|X=1)=P(Y=2|X=1)=P(Y=3|X=1)=P(Y=4|X=1)=\frac{1}{4}$
*$P(Y=1|X=2)=0$, $P(Y=2|X=2)=\frac{1}{2}$, $P(Y=3|X=2)=P(Y=4|X=2)=\frac{1}{4}$
*$P(Y=1|X=3)=P(Y=2|X=3)=0$, $P(Y=3|X=3)=\frac{3}{4}$, $P(Y=4|X=3)=\frac{1}{4}$
*$P(Y=1|X=4)=P(Y=2|X=4)=P(Y=3|X=4)=0$, $P(Y=4|X=4)=1$
so I see that only $Y|X=1$ ~$U[0,4]$ but I can't say anything about distributions $Y|X=2$, $Y|X=3$, $Y|X=4$. thus I can't use the law of total expectation.
Any advice would be appreciated! thanks
|
Who knows if anyone is looking anymore, never mind OP, but this is a concept I'm trying to grasp as well. I think I understand Ami's dissatisfaction with her own solution -- I too started by computing each conditional probability and wondered what I was supposed to have learned -- and so I've come up with my own attempt to sidestep the tedious, point-by-point computation. So --
We begin with $P(Y = y | X)$.
$\circ$ If $y < X$, then clearly $P(Y = y | X) = 0$.
$\circ$ If $y > X$, then $Y = y$ if and only if the second die yields $y$ -- i.e., precisely when $X_2 = y$. Hence, $P(Y = y | X) = 1/4$.
$\circ$ Finally, if $y = X$, then this is because $X_2 \le X$. There are $X$ ways for this to happen, and so $P(Y = y | X) = X/4$.
We can condense this as
\begin{align}
P(Y = y | X) = \frac{1_{\left[ y > X\right]} + X \cdot 1_{\left[ y = X \right]}}{4},
\end{align}
where $1_A$ is the usual indicator function.
This gives us
\begin{align}
E[Y] &= E[E[Y|X]] \\
&= E\left[\sum_{y = 1}^{4} y \cdot \frac{1_{\left[ y > X\right]} + X \cdot 1_{\left[ y = X \right]}}{4} \right] \\
&= \frac{1}{4} E\left[\sum_{y = 1}^{4} y \cdot \left( 1_{\left[ y > X\right]} + X \cdot 1_{\left[ y = X \right]} \right) \right] \\
&= \frac{1}{4} \sum_{x = 1}^4 \frac{1}{4} \left( \sum_{y = 1}^{4} y \cdot \left( 1_{\left[ y > x\right]} + x \cdot 1_{\left[ y = x \right]} \right) \right) \\
&= \frac{1}{16} \left( \sum_{x = 1}^4 \sum_{y = 1}^{4} y \cdot 1_{\left[ y > x\right]} + \sum_{x = 1}^4 \sum_{y = 1}^{4} yx \cdot 1_{\left[ y = x \right]} \right). \\
\end{align}
Now, the first double summation is the sum of all values of $y$ multiplied by the number of $x$-values that are smaller. The second is the sum of squares from $1$ to $4$. Therefore,
\begin{align}
E[Y] &= \frac{1}{16} \left[ \left( 1\cdot 0 + 2 \cdot 1 + 3 \cdot 2 + 4 \cdot 3 \right) + \left( 1^2 + 2^2 + 3^2 + 4^2 \right) \right] \\
&= \frac{20 + 30}{16} \; = \; \frac{25}{8} \\
\end{align}
Is this simpler than the brute-force method? I don't know. But at least it
generalizes -- with appropriate adjustments to the denominator in the conditional probability -- to dice with any number of sides.
|
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|
The limit of $\frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} $ I want to evaluate the following quotient limit:
$$\lim_{n \to \infty}\frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} $$
I know an exact same answer is already in here.
I want to avoid using the fact that the integral is a limit of Riemann Sum, instead please refer to the following exercises that precede the evaluation of this limit.
*
*(a) Let $f$ be an increasing continuous function, defined for all $x \ge 1$, such that $f(x) \ge 0$. Show that
$$f(1) + f(2) + ... + f(n-1) \le \int_{1}^n f(x) dx \le f(2) + ... + f(n)$$
*
*(b) Let $F(x) = \int_{1}^x f(t) dt.$ Assume that $F(n) \to \infty$ as $n \to \infty$, and that
$$\lim_{n \to \infty} \frac{f(n)}{F(n)} = 0$$.
$\qquad$ Show that
$$\lim_{n \to \infty} \frac{f(1) + f(2) + ... + f(n)}{F(n)} = 1.$$
I have actually proved 1(a) and 1(b) as stated above. As with most exercises from Lang, usually the exercises are interconnected, so I would like to use 1(b) to show that the quotient limit tends to something.
I let $f(x) = x^{\frac{1}{3}}$. Of course this function satisfies all the assumptions in 1(a) and 1(b). Similarly we let $F(x) = \int_{1}^x t^{\frac{1}{3}}dt$. Now let's consider the limit again:
$$\lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} $$
We see that the numerator of the quotient is $f(1) + f(2) + ... + f(n)$. Now consider $F(n) = \int_{1}^n t^{\frac{1}{3}}dt$ = $\frac{3}{4} (n^{\frac{4}{3}} - 1).$ But this is not the same as the denominator of the quotient, $n^{\frac{4}{3}}$. In this case how do I modify my solution so that I can actually use 1(b) as above?
|
$$\lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} = \lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{\frac{3}{4} (n^{\frac{4}{3}} - 1)}\cdot\frac{\frac{3}{4} (n^{\frac{4}{3}} - 1)}{n^{4/3}}$$
First term goes to $1$ by your $\frac{f(1) + f(2) + ... + f(n)}{F(n)}\to1$ result, second term goes to $3/4$ by standard limits.
So, your final answer is $3/4$.
|
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|
What is $\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$ when it exists?
For what values of $a$ and $b$ does the following limit exist and what is the limit in those cases?
$$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$
Actually this is an assignment which I need to do, but I have no idea where to start, please help!
|
A more brute-force approach:
$$|x+3| = \begin{cases}
x + 3, & x \geq -3 \\
-(x+3) = -x - 3, & x < -3
\end{cases}$$
so
$$\dfrac{|x+3|(\sqrt {ax+b}-2)}{x} = \begin{cases}
\dfrac{(x+3)(\sqrt {ax+b}-2)}{x}, & x \geq -3 \\
\dfrac{(-x-3)(\sqrt {ax+b}-2)}{x}, & x < -3 \text{.}
\end{cases}$$
For $x$ within a "small" neighborhood of $0$, we need only worry about the $x \geq -3$ case:
$$\begin{align}
\dfrac{(x+3)(\sqrt {ax+b}-2)}{x} &= (x+3)\dfrac{\sqrt {ax+b}-2}{x}\text{.}
\end{align}$$
We're not splitting the fraction since obviously $$\lim_{x \to 0}\dfrac{2}{x}$$
does not exist.
As $x \to 0$, $x + 3 \to 3$.
Being clever, we could notice that if the limit of $\dfrac{\sqrt {ax+b}-2}{x}$ exists as $x \to 0$, this can be written as
$$\lim_{x \to 0}\dfrac{\sqrt {ax+b}-2}{x} = \lim_{x \to 0}\dfrac{f(x)-f(0)}{x-0} = f^{\prime}(0)$$
where $f(x) = \sqrt{ax+b}$, and $f(0) = \sqrt{b} = 2$, so $b = 4$. Furthermore, $$f^{\prime}(x) = \dfrac{a}{2\sqrt{ax+b}} = \dfrac{a}{2\sqrt{ax+4}}\text{.}$$
We then have
$$f^{\prime}(0) = \dfrac{a}{2\sqrt{4}} = \dfrac{a}{4}\text{.}$$
Thus,
$$\lim_{x \to 0}\dfrac{|x+3|(\sqrt {ax+b}-2)}{x} = \dfrac{3a}{4}$$
as long as $b = 4$. Any $a$ will do.
Furthermore, this relies on the existence of $$\lim_{x \to 0}\dfrac{\sqrt {ax+b}-2}{x}$$
perhaps there is a way to relax this assumption.
|
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|
Prove this 3x3 determinant using properties of determinant.
Prove this $3\times 3$ determinant using properties of determinant.
$$\begin{vmatrix}y+z & x & x \\ y & z+x & y \\ z & z & x+y\end{vmatrix} = 4xyz$$
I have been trying to solve this one for over an hour now. I really can't even get started with it. The question specifically says I can't expand, and have to prove this using other properties of 3x3 determinants.
|
The determinant
$$
\det(A) = \det(a_1, a_2, a_3)
$$
is an alternating multi linear form, this means
$$
\det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3) + \alpha \det(a_2, a_2, a_3)
$$
because $\det$ is linear in the first argument as multi linear form.
Because it is an alternating form, if we exchange the first two arguments, the sign changes and we have
$$
\det(a_2, a_2, a_3) = - \det(a_2, a_2, a_3)
$$
which means $\det(a_2,a_2,a_3)$ vanishes.
This leads to
$$
\det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3)
$$
and similar for the other arguments.
We can use this property to simplify your determinant:
\begin{align}
d &=
\begin{vmatrix}
y + z & x & x \\
y & z + x & y \\
z & z & x + y
\end{vmatrix}
=
\begin{vmatrix}
0 & -2z & -2y \\
y & z + x & y \\
z & z & x + y
\end{vmatrix}
=
2
\begin{vmatrix}
0 & -z & -y \\
y & z + x & y \\
z & z & x + y
\end{vmatrix}
\\
&=
2
\begin{vmatrix}
0 & -z & -y \\
y & x & 0 \\
z & 0 & x
\end{vmatrix}
=
-2
\begin{vmatrix}
0 & z & y \\
y & x & 0 \\
z & 0 & x
\end{vmatrix}
=
2
\begin{vmatrix}
0 & y & z \\
y & 0 & x \\
z & x & 0
\end{vmatrix}
\end{align}
If $x = y = z = 0$ then $d = 0$ and the proposed formula holds.
If two of the variables $x,y,z$ are zero, then one of the columns
is zero and we have e.g.
$$
\det(0, a_2, a_3) = \det(0 + a_2, a_2, a_3) = \det(a_2, a_2, a_3) = 0
$$
If e.g. $x = 0$ we have
$$
d =
2
\begin{vmatrix}
0 & y & z \\
y & 0 & 0 \\
z & 0 & 0
\end{vmatrix}
=
2 y z
\begin{vmatrix}
0 & 1 & 1 \\
y & 0 & 0 \\
z & 0 & 0
\end{vmatrix}
=
2 y z
\begin{vmatrix}
0 & 0 & 1 \\
y & 0 & 0 \\
z & 0 & 0
\end{vmatrix}
=
-2 y z
\begin{vmatrix}
0 & 0 & 1 \\
0 & y & 0 \\
0 & z & 0
\end{vmatrix}
= 0
$$
and similar if just $y$ or $z$ vanishes.
If $x,y,z$ do not vanish each, we have
\begin{align}
d
&=
2
\begin{vmatrix}
0 & y & z \\
y & 0 & x \\
z & x & 0
\end{vmatrix}
=
\frac{2}{xyz}
\begin{vmatrix}
0 & xy & xz \\
yz & 0 & xz \\
yz & xy & 0
\end{vmatrix}
=
\frac{2}{xyz}
\begin{vmatrix}
0 & xy & xz \\
yz & 0 & xz \\
yz & 0 & -xz
\end{vmatrix}
\\
&=
\frac{2}{xyz}
\begin{vmatrix}
0 & xy & xz \\
0 & 0 & 2xz \\
yz & 0 & -xz
\end{vmatrix}
=
\frac{4}{y}
\begin{vmatrix}
0 & xy & xz \\
0 & 0 & 1 \\
yz & 0 & -xz
\end{vmatrix}
=
\frac{4}{y}
\begin{vmatrix}
0 & xy & 0 \\
0 & 0 & 1 \\
yz & 0 & 0
\end{vmatrix}
\\
&=
4xyz
\begin{vmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{vmatrix}
=
-4xyz
\begin{vmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{vmatrix}
=
4xyz
\begin{vmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{vmatrix}
\\
&= 4xyz
\end{align}
|
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|
Simplifying $\frac{\;\frac{x}{1-x}+\frac{1+x}{x}\;}{\frac{1-x}{x}+\frac{x}{1+x}}$
$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$
I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?
|
Let me try.
$$\frac{\frac{x}{1-x}+\frac{1+x}{x}}{\frac{1-x}{x}+\frac{x}{1+x}} = \frac{\frac{1}{1-x}-1 + \frac{1}{x} + 1}{\frac{1}{x}-1 + 1-\frac{1}{1+x}} = \frac{\frac{1}{1-x} + \frac{1}{x}}{\frac{1}{x} - \frac{1}{1+x}} = \frac{\frac{1}{x(1-x)}}{\frac{1}{x(1+x)}} = \frac{1+x}{1-x}$$
|
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|
Calculus Integral $\int \frac{dt}{2^{t} + 4}$
I try resolve this integral, with $u = 2^t + 4 $, but can´t ... plz
$$\int \frac{dt}{2^t + 4} = \frac{1}{\ln 2}\int\frac{du}{u(u-4)} = \text{??} $$
|
Hint:
$$\frac{1}{2^x+4}=\frac{2^x-2^x+4}{4(2^x+4)}=\frac{1}{4} \left(1-\frac{2^x}{2^x+4}\right)$$
|
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|
Is $x^2+x+1$ ever a perfect power? Using completing the square and factoring method I could show that the equation $x^2+x+1=y^n$, where $x,y$ are positive odd and $n$ is positive even integers, does not have solution, but I could not show that for positive odd $x,y$ and odd $n>1$ the equation does (does not) have solution.
Thank you for your contribution.
|
Your equation can be rewritten as
$$\frac{x^{3}-1}{x-1} = y^{N}.$$
The Diophantine equation
$$ \frac{x^{n} − 1}{x-1} = y^{q} \quad x > 1, \quad y>1, \quad n>2 \quad q \geq 2 \quad \tag1$$
was the subject matter of a couple of papers of T. Nagell from the $1920's.$ Some twenty-odd years later, W. Ljunggren clarified some points in Nagell’s arguments and completed the proof of the following result.
Theorem. Apart from the solutions
$$\frac{3^{5}-1}{3-1}=11^{2}, \quad \frac{7^{4}-1}{7-1}=20^{2}, \quad \frac{18^{3}-1}{18-1} = 7^{3},$$
the equation in $(1)$ has no other solution $(x, y, n, q)$ if either one of the following conditions is satisfied:
$(1.$ $q = 2$,
$(2.$ $3$ divides $n$,
$(3.$ $4$ divides $n$,
$(4.$ $q =3$ and $n$ is not congruent with $5$ modulo $6$.
Clearly enough, this theorem implies that there are only two solutions to the equation you are considering: $(x=1, y=3, N=1)$ and $(x=18, y=7, N=3)$.
|
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|
Lower bound for cyclic expression in four variables We are given four positive real numbers $a,b,c,d$. I would like to prove that
$$\frac ab+\frac bc+\frac cd+\frac da\ge4$$
To start solving this I assumed
$$a\ge b\ge c\ge d$$
Therefore
$$\frac ab,\frac bc,\frac cd\ge 1,\ \frac da\le 1$$
and the last one is causing the problem.
|
As lab bhattacharjee commented, You can use the AM-GM inequality. By applying it here, we can obtain:
$$
\begin{align*}
\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}}{4} & \ge \sqrt[4]{\frac{abcd}{bcda}}\\
& = \sqrt[4]{1}\\
& = 1
\end{align*}
$$
Therefore:
$$
\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \ge 4
$$
|
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|
Sum of three numbers and the sum of their squares: $x^2+y^2+z^2 \geq \frac{1}{3}$ for $x+y+z=1$ We have $x, y, z \in \mathbb R$ such that $x+y+z=1$. Prove that $x^2+y^2+z^2 \geq \frac{1}{3}$. I am able to do this using the relationship between the power and arithmetic means. Is there a way to not use this relationship?
|
For any real numbers $x,y,z$, we have
$$\begin{align}
0\le(x-y)^2+(y-z)^2+(z-x)^2&\implies2(xy+yz+zx)\le2(x^2+y^2+z^2)\\
&\implies(x+y+z)^2\le3(x^2+y^2+z^2)
\end{align}$$
So if $x+y+z=1$, then ${1\over3}\le x^2+y^2+z^2$.
|
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|
Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$ Let $a,b,c>0$ prove or disprove
$$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$
since
$$(a^2+b^2+c^2)^2\ge 3(a^2b^2+b^2c^2+a^2c^2)\tag{1}$$
other
$$(a+b+c)^2\le 3(a^2+b^2+c^2)\tag{2}$$
Because of the inequality sign in a different direction so we can't $(1)\times (2)$
|
After a quick look:
$$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge \sqrt[3]{\dfrac{(2s+c)(2s^2+c^2)}{9}} \ge \sqrt[4]{\dfrac{(s^4+2s^2c^2)}{3}} \ge \sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$
where $s=\frac{a+b}{2}$ and we assume $c=\min(a,b,c)$.
The left inequality follows from $2(a^2+b^2)\ge (a+b)^2$.
The right inequality follows from $(s^2-p)(s^2+p-2c^2)\ge 0$ where $p=ab$ (note that $s^2\ge p \ge c^2$).
The middle inequality is, as you may have observed, the original inequality where we set $a=b=s$. Since the inequality is homogeneous, we can set $s=1$, i.e. solving the original inequality when $a=b=1$, which is easy.
This method is called Mixing Variables Method.
You can also try to find an intermediate term like in this solution to get an even nicer solution.
|
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|
cubic equation has $3$ distinct roots $\alpha,\beta,\gamma$ equation $x^3-9x^2+24x+c=0$ has $3$ distinct roots $\alpha,\beta,\gamma$, then $\lfloor \alpha \rfloor+\lfloor \beta \rfloor +\lfloor \gamma \rfloor =$
and $\lfloor \alpha \rfloor = \alpha-\{\alpha\},\lfloor \beta \rfloor = \beta-\{\beta\},\lfloor \gamma \rfloor = \gamma-\{\gamma\},0\leq \{\alpha\},\{\beta\},\{\gamma\}<1$
we assume $f(x) = x^3-9x^2+24x+c$, using $f'(x) = 3x^2-18x+24$
put maximum , minimum $f'(x)=0$
$3x^2-18x+24=0$
$x^2-6x+8=(x-4)(x-2)=0$
$x=2,x=4$
put $x=2,3$ in $f''(x)=6x-18=6(x-3)$
$f''(2) = -6<0$ means $x=2$ is a point of local maximum
and $f''(4)=6>0$ means $x=4$ is a point of local minimum
i can not go further
|
Direct from vieta's relation, we get $\alpha^{2}+\beta^{2}+\gamma^{2}=-b/a=9$
|
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|
The ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime
I tried to show that the ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime in $\mathbb{C}[X,Y]$.
I noticed that $X^{2}-Y^{3}-(Y^{2}-X^{3})=(X-Y)(X^{2}+XY+Y^{2}+X+Y)$. Now, if $X-Y$ is in $(X^{2}-Y^{3},Y^{2}-X^{3})$ there exists $f(X,Y),\ g(X,Y)\in \mathbb{C}[X,Y] $ s.t. $$X-Y=f(X,Y)(X^{2}-Y^{3}) + g(X,Y)(Y^{2}-X^{3}).$$ For $Y=0$ it results $X=f(X,0)X^{2} - g(X,0)X^{3}\Rightarrow 1=f(X,0)X-g(X,0)X^{2}$ (False).
A similar argument goes for$X^{2}+XY+Y^{2}+X+Y$. Is this a right solution?
|
$X(X^2-Y^3)+(Y^2-X^3)=-XY^3+Y^2=Y^2(-XY+1)$. $-XY+1, Y^2$
Suppose that $-XY+1=f(X,Y)(X^2-Y^3)+g(X,Y)(Y^2-X^3)$, we deduce that $1=f(0,Y)(-Y^3)+g(0,Y)Y^2=Y^2(-f(0,Y)Y+g(0,Y))$ impossible.
Suppose that $Y^2=f(X,Y)(X^2-Y^3)+g(X,Y)(Y^2-X^3)$. Set $X=Y=1$, we have $1=f(1,1)(1^2-1^3)+g(1,1)(1^2-1^3)=0$ impossible, so $Y^2$ and $-XY+1$ are not in the ideal $I$ generated by $X^2-Y^3$ and $Y^2-X^3$ but their product is in this ideal so $I$ is not a prime ideal.
|
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|
To find number of distinct terms of binomial expansion I have binomial expansion as
$(x+\frac{1}{x} + x^2 + \frac{1}{x^2})^{15}$.
How do i find number of distinct terms in it. Distinct in sense means terms having different powers of $x$?
I have simplified this as $\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$. How do i proceed
Thanks
|
We have that
$$(x^4+x^3+x+1)^n=(x+1)^n\cdot(x^3+1)^n.$$
Now the powers of $x$ in $(1+x)^n$ are the integers in the interval $[0,n]$.
Moreover for $n\geq 2$, the multiplication of $(x^3+1)(x+1)^n$ gives a sum of powers of $x$ which cover the range $[0,n]\cup [3,n+3]=[0,n+3]$.
Hence, for $(x+1)^n\cdot(x^3+1)^n$ the powers of $x$ cover the range $[0,n+3n]=[0,4n]$, which contains $4n+1$ integers.
Therefore
$$\left(x+\frac{1}{x} + x^2 + \frac{1}{x^2}\right)^{15}=\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$$
contains all the powers of $x$ in the range $[-30,4\cdot 15-30]=[-30,30]$. Their number is $61$.
|
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|
distinguishable balls distinguishable boxes where each box contains at least 2 balls Consider $m$ distinguishable balls and $n$ distinguishable boxes where $m > n$ (the boxes and balls are already distinguishable, say they come with preassigned distinct labels).
How many ways are there to distribute the balls into boxes such that each box contains at least $2$ balls?
|
We solve the case of indistinguishable boxes and distinguishable
boxes. The combinatorial species in the first case is
$$\mathfrak{P}_{=n}(\mathfrak{P}_{\ge 2}(\mathcal{Z}))$$
which gives the EGF
$$G(z) = \frac{(\exp(z)-z-1)^n}{n!}.$$
Extracting coefficients we get
$$m! [z^m] G(z)
= m! [z^m] \frac{(\exp(z)-z-1)^n}{n!}
\\ = \frac{m!}{n!} [z^m]
\sum_{k=0}^n {n\choose k} (\exp(z)-1)^k (-1)^{n-k} z^{n-k}
\\ = \frac{m!}{n!}
\sum_{k=0}^n {n\choose k} [z^{m+k-n}] (\exp(z)-1)^k (-1)^{n-k}
\\ = \frac{m!}{n!} \sum_{k=0}^n {n\choose k} (-1)^{n-k}
\times \frac{k!}{(m+k-n)!} {m+k-n\brace k}
\\ = {m\choose n} \sum_{k=0}^n {n\choose k} (-1)^{n-k}
\times \frac{k! (m-n)!}{(m+k-n)!} {m+k-n\brace k}
\\ = {m\choose n} \sum_{k=0}^n {n\choose k} (-1)^{n-k}
\times {m+k-n\brace k} {m+k-n\choose k}^{-1}.$$
We can verify this for some special values like $m=2n$ where we obtain
$$(2n)! [z^{2n}] \frac{(\exp(z)-z-1)^n}{n!}
\\ = (2n)! \frac{1}{n!} \frac{1}{2^n}
= \frac{1}{n!} {2n\choose 2,2,2,\ldots,2}$$
which is the correct value. We also get
$$(2n+1)! [z^{2n+1}] \frac{(\exp(z)-z-1)^n}{n!}
= (2n+1)! \frac{1}{n!} {n\choose 1} \frac{1}{6} \frac{1}{2^{n-1}}
\\ = \frac{1}{(n-1)!} {2n+1\choose 2,2,2,\ldots,2,3}$$
which is correct as well. One more example is
$$(2n+2)! [z^{2n+2}] \frac{(\exp(z)-z-1)^n}{n!}
\\ = (2n+2)! \frac{1}{n!}
\left({n\choose 1} \frac{1}{24} \frac{1}{2^{n-1}}
+ {n\choose 2} \frac{1}{6^2} \frac{1}{2^{n-2}}\right)
\\= \frac{1}{(n-1)!} {2n+2\choose 2,2,2,\ldots,2,4}
+ \frac{1}{2} \frac{1}{(n-2)!} {2n+2\choose 2,2,2,\ldots 2,3,3}.$$
Finally observe that we get the values for distinguishable boxes by
multiplying by $n!$ because the species now becomes
$$\mathfrak{S}_{=n}(\mathfrak{P}_{\ge 2}(\mathcal{Z})).$$
|
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|
How to prove the limit of a sequence by definition: $\lim((n^2 + 1)^{1/8} - n^{1/4}) = 0$? I have to prove the limit of a sequence using $\epsilon - N$ definition: $$\lim_{n\to\infty}((n^2 + 1)^{1/8} - n^{1/4}) = 0$$.
Attempt:
We want to show: $\forall \epsilon>0$ $\exists N \in \mathbb{N}$, s.t. if $n \ge N$, then $|(n^2 + 1)^{1/8} - n^{1/4} - 0|<\epsilon$. So we need to find N as a function of $\epsilon$, s.t. N > $f(\epsilon)$.
$(n^2 + 1)^{1/8} - n^{1/4}$ = $((n^2 + 1)^{1/8} - n^{1/4}) * \frac{(n^2 + 1)^{1/8} + n^{1/4}}{(n^2 + 1)^{1/8} + n^{1/4}}$ = $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}}$.
$\frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}} < \frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}}$, so any n, solving $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}} < \epsilon$, will suffice.
Then $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}}$ = $(n + \frac{1}{n})^{1/4} - n^{1/4}$.
I do not know how to proceed further. I tried the same trick with multiplying the last equation by $\frac{(n + \frac{1}{n})^{1/4} + n^{1/4}}{(n + \frac{1}{n})^{1/4} + n^{1/4}}$, but I did not get anything meaningful.
|
Just keep (fearlessly) your argument going:
$$(n^2 + 1)^{1/8} - n^{1/4} = \frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}}$$
consequently
\begin{align*}(n^2 + 1)^{1/8} - n^{1/4} &= \frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}}\cdot \frac{(n^2 + 1)^{1/4} + n^{1/2}}{(n^2 + 1)^{1/4} + n^{1/2}} \\ & =\frac{(n^2 + 1)^{1/2} - n}{((n^2 + 1)^{1/8} + n^{1/4})\cdot ((n^2 + 1)^{1/4} + n^{1/2})}\end{align*}
and again, doing the same trick
$$(n^2 + 1)^{1/8} - n^{1/4} = \frac{(n^2 + 1) - n^2}{((n^2 + 1)^{1/8} + n^{1/4})\cdot ((n^2 + 1)^{1/4} + n^{1/2})((n^2 + 1)^{1/2} + n)}.$$
The $n^2$ now cancel and you get
$$(n^2 + 1)^{1/8} - n^{1/4} = \frac{1}{((n^2 + 1)^{1/8} + n^{1/4})\cdot ((n^2 + 1)^{1/4} + n^{1/2})((n^2 + 1)^{1/2} + n)}.$$
Now finding your $N$ is easy.
Remark: this is obviously a half joke/half overkill, but works nonetheless.
|
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|
Permutation or combination for repeated chosen items? I have this problem:
Choosing 5 vertices from the picture at random, what are the odds of at least 4 of them belonging to the pyramid?
My book says:
Because we are asked the probability of at least 4 vertices belonging
to the pyramid, we must consider two possibilities: 4 of the chosen
vertices belong to the pyramid or 5 of the vertices belong to the
pyramid. This probability is given by:
$$\frac{^5C_4*^4C_1+^5C_5}{^9C_5}=\frac{1}{6}$$
*
*But I don't understand how this is. Wouldn't it be more logical to do
$$(\frac{5}{9}*\frac{4}{8}*\frac{3}{7}*\frac{2}{6}*\frac{4}{5})+(\frac{5}{9}*\frac{4}{8}*\frac{3}{7}*\frac{2}{6}*\frac{1}{5})
= \frac{1}{126}$$
?
*Why is it that repetitions are not being counted? Shouldn't they be
part of the probability?
*Does the problem assume that the same vertex can be chosen more than
once?
*Why is my solution wrong?
*Can you explain to me my book's solution?
There is a chance my book is wrong, or the problem has an error, is missing information or is not explained properly. That sort of thing happens all the time . I copied the problem exactly as it is in the book.
|
The book's solution is correct. Your approach is viable, but you made an error in its execution.
The author of your text solved the problem by considering which five-element subsets of the vertices contain at least four vertices of the pyramid. That is, the author is making an unordered selection of the vertices. Since there are nine vertices in the diagram, there are
$$\binom{9}{5}$$
ways to select a five-element subset of the vertices. Five of the nine vertices are vertices of the pyramid. Thus, a selection of five vertices which contains at least four vertices of the pyramid either contains four of the five vertices of the pyramid and one of the other four vertices in the figure or it contains all five vertices of the pyramid. Thus, the number of favorable selections is
$$\binom{5}{4}\binom{4}{1} + \binom{5}{5}\binom{4}{0}$$
Dividing the number of favorable selections by the number of possible selections yields the desired probability of
$$\frac{\dbinom{5}{4}\dbinom{4}{1} + \dbinom{5}{5}\dbinom{4}{0}}{\dbinom{9}{5}} = \frac{1}{6}$$
You attempted to solve the problem by making ordered selections of five of the nine vertices. You handled the case in which all five of the selected vertices are vertices of the pyramid correctly. However, in handling the case in which exactly four of the five selected vertices are vertices of the pyramid, you failed to take the order of selection into account. Your calculation for the probability of choosing four of the five vertices of the pyramid and one of the other four vertices in the figure
$$\frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{4}{5}$$
is actually the probability of choosing four vertices of the pyramid, then choosing one of the other four vertices with your fifth selection. However, the vertex that is not in the pyramid could have been chosen first, second, third, fourth, or fifth. As you should check, if you add those five probabilities, you will find that the probability of choosing four vertices of the pyramid and one other vertex in the figure is five times what you calculated. Observe that
$$5 \cdot \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{4}{5} + \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{1}{5} = \frac{1}{6}$$
|
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas.
|
My Work: I will approach in a very short way.we have been given
two characteristic equations of a function and their range .
$$f(x)=x^2-ax+1$$
$$\text{and, we know}$$
$$\lvert x^2-ax+1 \rvert \le \dfrac{1}{4}$$
$$\lvert (x+2)^2-a(x+2)+1 \rvert \le \dfrac{1}{4}$$
from the absolute value,we can write-
$$-\dfrac{1}{4}\le x^2-ax+1 \le \dfrac{1}{4}$$
$$-\dfrac{1}{4}\le (x+2)^2-a(x+2)+1 \le \dfrac{1}{4}$$
Now, Let's see how do these two inequalities come to our help? We have to find the range
of $a$ for which these two inequalities held.It is certain that we will find a point of a if we work on
upper limit of these two inequalities and will find another point if we work on lower limit of these two inequalities.
to make my life easy ,I am counting them as equations.[Look,I am just using the equations to determine the peak values of "$a$"]
So,if we take upper limit,we get-
$$x^2-ax+1 = \dfrac{1}{4}.......(1)$$
$$(x+2)^2-a(x+2)+1 = \dfrac{1}{4}......(2)$$
now if we solve these two equation we will get ,$a^2=7$
Again, if we take lower limit,we get-
$$ x^2-ax+1=-\dfrac{1}{4}.........(3)$$
$$ (x+2)^2-a(x+2)+1=-\dfrac{1}{4}..........(4)$$
If we solve these two equations we will get $a^2=9$
Hence, $ 7\le a^2 \le 9$
|
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How do I solve differentil equation if the begining values are given? How do I solve differentil equation if the begining values are given?
$\frac{dx}{dt}=x^2+5x$ with x(0)=-3. I need to find x(t).
$\int \frac{dx}{x^2+5x}=\int dt$
So when I put it on Symbollab I get that left side is $-\frac{2}{5}arctanh(\frac{2}{5}(x+\frac{2}{5}))$, I already see that is going to be hard to extract x from here.
What have I done?
$\int \frac{dx}{x^2+5x}=\int \frac{dx}{x^2(1+\frac{x}{5})}$ than I use substitution that $1+\frac{x}{5}$ is u, than $-\frac{x^2}{5}du$.
Than I get $\frac{-1}{5}ln(1+\frac{x}{5})=t+C$.
$ln(1+\frac{x}{5})=-5t+C$
$1+\frac{x}{5}=e^{-5t}C$
$\frac{x}{5}=e^{-5t}C-1$
$x=\frac{5}{e^{-5t}C-1}$
$-3=\frac{5}{e^{0}C-1}$
$-3=\frac{5}{C-1}$
Than I calculate C. This all seems nice to me.
But when I put than integral in WolframAlpha I get that $\int \frac{dx}{x^2+5x}=\frac{1}{5}(log(x)-log(x+5))+constant$
Why I don't get when I integrate those results from WolframAlpha or Symbollab?
|
In your calculation the initial step
$$
\int \frac{dx}{x^2+5x}
\ne \int \frac{dx}{x^2(1+x/5)}
= \int \frac{dx}{x^2 + x^3/5}
$$
is wrong.
You can decompose the fraction using polynomial factorization and then look for a sum of the fractions with the factors as denominator
$$
\frac{1}{x^2+5x}
= \frac{1}{x(x+5)}
= \frac{A}{x} + \frac{B}{x+5} \iff \\
1 = A (x+5) + B x = (A+B)x + 5A
$$
Comparing the coefficients gives the equations
$$
A + B = 0 \\
5A = 1
$$
or $A = 1/5$, $B = -1/5$.
Thus
$$
\frac{1}{x^2+5x}
= \frac{1}{5} \left( \frac{1}{x}- \frac{1}{x+5} \right) \\
$$
This is called partial fraction decomposition.
So this leads to the WA solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometry and integrals related picture
In my book it says $ \frac{x}{R} = \tan\theta$
ok, that is pretty obvious, but then it says that it implies that
$$ \frac{dx}{r^2} = \frac{R\,d\theta}{r^2 \cos^2 \theta} = \frac{d\theta}{R}$$
I really cannot understand how $d\theta$ got involved at all. Can anyone please try to explain the connection?
|
You have $\dfrac x R = \tan\theta,$ so $x = R\tan\theta.$ If you know that $$\frac d {d\theta} \tan\theta = \sec^2\theta = \frac 1 {\cos^2\theta},$$ then, assuming $R$ remains constant as $x$ and $\theta$ change, this yields
$$
\frac{dx}{d\theta} = \frac d {d\theta}(R\tan\theta) = R\sec^2\theta = \frac{R}{\cos^2\theta}.
$$
Hence
$$
dx = \frac{R\,d\theta}{\cos^2\theta},
$$
and so
\begin{align}
\frac{dx}{r^2} & = \frac{R\,d\theta}{r^2\cos^2\theta} \\[12pt]
& = \frac{R\,d\theta}{R^2} & & \text{since } \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac R r, \text{ so } r\cos\theta = R, \\[12pt]
& = \frac{d\theta} R.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
To find the range of $\sqrt{x-1} + \sqrt{5-x}$ To find the range of $\sqrt{x-1} + \sqrt{5-x}$.
I do not know how to start?
Thanks
|
Clearly $1\le x\le5$
If $y=\sqrt{x-1}+\sqrt{5-x}>0, y^2=4+2\sqrt{(x-1)(5-x)}$
Now $\sqrt{(x-1)(5-x)}\ge0$
and using AM-GM inequality, $\sqrt{(x-1)(5-x)}\le\dfrac{x-1+5-x}2=?$
|
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|
Find the equation solution $\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$ solve this following equation
$$\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$$
the equation obvious have a root $x=3$. But how to solve other roots?
|
*
*for $x\le\lfloor{\frac{9}{6}}\rfloor$ there is no answer since $\lfloor{\frac{9}{x}}\rfloor$ is at least $6$ and when $\lfloor{x}\rfloor$ becomes zero, the other term is $7$.
*for $\frac{9}{6}<x \le \frac{9}{5}$, we have $\lfloor{\frac{9}{x}}\rfloor=5$ and $\lfloor{x}\rfloor=1$, hence all this range is a valid answer.
*for $\frac{9}{5}<x<2$, we have $\lfloor{\frac{9}{x}}\rfloor=4$ and $\lfloor{x}\rfloor=1$, hence no answer in this range.
*for $2\le x \le \frac{9}{4}$, we have $\lfloor{\frac{9}{x}}\rfloor=4$ and $\lfloor{x}\rfloor=2$, hence all this range is a valid answer.
*for $\frac{9}{4}<x<3$, we have $\lfloor{\frac{9}{x}}\rfloor=3$ and $\lfloor{x}\rfloor=2$, hence no answer in this range.
*for $x=3$, we have $\lfloor{\frac{9}{3}}\rfloor=3$ and it is an answer.
*for $3<x<4$, we have $\lfloor{\frac{9}{x}}\rfloor=2$ and $\lfloor{x}\rfloor=3$, hence no answer in this range.
*for $4\le x \le \frac{9}{2}$, we have $\lfloor{\frac{9}{x}}\rfloor=2$ and $\lfloor{x}\rfloor=4$, hence all this range is a valid answer.
*for $\frac{9}{2}<x<5$, we have $\lfloor{\frac{9}{x}}\rfloor=1$ and $\lfloor{x}\rfloor=4$, hence no answer in this range.
*for $5\le x <6$, we have $\lfloor{\frac{9}{x}}\rfloor=1$ and $\lfloor{x}\rfloor=5$, hence all this range is a valid answer.
*for $x\ge 6$ also there is no answer.
|
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|
If two consecutive numbers are removed from the series $1+2+3+\ldots+n$ the average becomes $99/4$. Find the two numbers. The initial average will be $\frac{n+1}{2}$. If the two numbers are $k$ and $k+1$ then the new average will be $\frac{n(n+1)/2-(2k+1)}{n-2}$. I couldn't figure further even though I got the relation between $n$ and $k$ in many different ways.
If the question is not clear, here is an example to explain it.
If $n=10$, the initial average will be $5\cdot 5$ {$(1+2+\cdots + 10)/10$}
Now if two consecutive numbers like $2,3$ or $8,9$ are removed from this series, the new average changes, and this new average has been given to be $99/4$, however we also don't know the value of $n$, so the question seems to be pretty difficult.
|
This approach finds that the new average lies within $\pm 1$ of the original average. This significantly narrows down possibilities, and the solution can then be found easily by elimination.
After removing the two numbers, the new average, $a$, is given by
$$\begin{align}
\frac {99}4=a&=\frac{\frac{n(n+1)}2-(2k+1)}{n-2}\\
&=\underbrace{\frac {n+1}2}_{\text{original average, $a_0$}}+
\underbrace{\frac {n-2k}{n-2}}_{\in [-1,1] \text{ for } 1\le k\le n-1}
\end{align}$$
Hence $a$ lies within $\pm 1$ of the original average $a_0$ before removal, i.e.
$$a_0-1\;\le\; a=\frac {99}4=24.75\;\le\; a_0+1$$
As $a_0=\frac {n+1}2$, it can only be either an integer or an integer and a half, hence
$24\le a_0\le 25.5$.
$$\begin{array} {lrrrr}
\hline{a_0(n)=\frac{n+1}2} &24&24.5&25&25.5\\
n &47 &48 &49 &\color{red}{50}\\
n-2 &45 &46 &47 &\boxed{48} \\
a-a_0(n)=\frac {n-2k}{n-2}&\frac 34&\frac 14&-\frac14&-\frac34\\
\hline
\end{array}$$
Also, the sum of the remaining numbers $\frac {99}4 (n-2)$ must be integer, so $(n-2)$ must a multiple of $4$, the only candidate for which is $48$.
Hence we conclude that $\color{red}{n=50, k=43}\qquad \blacksquare$
|
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|
How do I find an irrational root of a quartic function? For example, the equation $2x^4 - 3x^3 - 1 = 0$ has no rational roots, but it does have an irrational root between x = 1 and x = 2. If it had a rational root, I would be able to find it, but I am a bit confused on how to find an irrational root.
|
In this case consider:
\begin{align}
2 \, x^4 - 3 \, x^3 -1 &= (a \, x^2 + b \, x + 1)(x^2 + c \, x -1) \\
&= a \, x^4 + (ac + b) \, x^3 + (-a + bc + 1) \, x^2 + (-b + c) \, x - 1
\end{align}
This leads to $b = c$, $a=2$, $1 - a + bc = -1 + b^2 = 0$ or $b = \pm 1$, and
$ac + b = 3b = -3$ which narrows $b$ to $b = -1$. Now,
$$2 \, x^4 - 3 \, x^3 -1 = (a \, x^2 + b \, x + 1)(x^2 + c \, x -1) = (2 \, x^2 - x +1)(x^2 - x - 1).$$
The polynomial $2 \, x^2 - x + 1$ can be factored as $2(x - a)(x - b)$, where $4a = 1 + i \, \sqrt{7}$ and $4b = 1 - i \sqrt{7}$, and $x^2 - x - 1$ can be factored as $(x - \alpha)(x - \beta)$ where $2 \, \alpha = 1 + \sqrt{5}$ and $2 \, \beta = 1 - \sqrt{5}$. This brings the factorization into the form:
$$2 \, x^4 - 3 \, x^3 -1 = 2(x - a)(x - b)(x - \alpha)(x - \beta).$$
The four roots would then be
$$x \in \{ a, b, \alpha, \beta \} = \left\{ \frac{1 + i \, \sqrt{7}}{4}, \frac{1 - i \, \sqrt{7}}{4}, \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2} \right\}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rolling three dice - sum of two results equals third one Is my answer correct?
Three six-sided dice are thrown, what is the probability that the sum of two of them equals the third one?
If order does not matter, there are 9 such possibilities:
*
*(1, 1, 2)
*(1, 2, 3)
*(1, 3, 4)
*(1, 4, 5)
*(1, 5, 6)
*(2, 2, 4)
*(2, 3, 5)
*(2, 4, 6)
*(3, 3, 6)
The total number of possibilities is $\frac{6^3}{3!}=6^2=36$. Therefore, the probability is $\frac{9}{36}$.
|
If you imagine there are three distinguishable dices (either different colors, or you throw them one after the other), then there are $6^3=216$ possible cases.
Now your enumeration counts down to :
$(1,1,2)$ : $3$ cases
$(1, 2, 3)$ : $6$ cases
$(1, 3, 4)$ : $6$ cases
$(1, 4, 5)$ : $6$ cases
$(1, 5, 6)$ : $6$ cases
$(2, 2, 4)$ : $3$ cases
$(2, 3, 5)$ : $6$ cases
$(2, 4, 6)$ : $6$ cases
$(3, 3, 6)$ : $3$ cases
for a total of $45$ cases, and a probability of $\frac{45}{216}=\frac{5}{24}$.
Hope I did it the right way :-(
|
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|
Find best-fit parabola to the given data
Find the parabola $At^2+Bt+C$ that best approximates the data set $t= -1,0,1,2,3$ and $b(t) = 5,2,1,2,5$.
Would I be using least squares such that $x = (A^TA)^{-1}A^Tb$? Thank you in advance.
|
Problem statement
Given a sequence of $m=5$ data points $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$, and the trial function
$$
y(x) = a_{0} + a_{1} x + a_{2} x^{2}.
$$
The linear system is
$$
\begin{align}
\mathbf{A} a &= y\\
%
\left[
\begin{array}{rrr}
1 & -1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 3 & 9 \\
\end{array}
\right]
%
\left[
\begin{array}{rrr}
a_{0} \\
a_{1} \\
a_{2}
\end{array}
\right]
%
& =
%
\left[
\begin{array}{rrr}
5 \\
2 \\
1 \\
2 \\
5
\end{array}
\right]
%
\end{align}
$$
Solution via normal equations
The normal equations are
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} a & = \mathbf{A}^{*} y \\
%
\left[
\begin{array}{rrr}
5 & 5 & 15 \\
5 & 15 & 35 \\
15 & 35 & 99 \\
\end{array}
\right]
%
\left[
\begin{array}{rrr}
a_{0} \\
a_{1} \\
a_{2}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{rrr}
15 \\
15 \\
59
\end{array}
\right]
%
\end{align}
$$
The solution is
$$
\begin{align}
a_{LS} & = \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1}
\mathbf{A}^{*} y \\
%
&=
\frac{1}{70}
\left[
\begin{array}{rrr}
26 & 3 & -5 \\
3 & 27 & -10 \\
-5 & -10 & 5 \\
\end{array}
\right]
%
\left[
\begin{array}{rrr}
15 \\
15 \\
59
\end{array}
\right] \\
%
&=
%
\left[
\begin{array}{rrr}
2 \\
-2 \\
1
\end{array}
\right],
%
\end{align}
$$
$$
y(x) = 2 -2x + x^{2}.
$$
This is an exact fit because
$$
r^{2} =
\left( \mathbf{A} a_{LS} - b \right)^{*}
\left( \mathbf{A} a_{LS} - b \right) = 0.
$$
The plot shows the solution against the data points.
|
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|
Find the following limit with a sum I am trying to solve the following limit, which includes a sum:
$\lim_{n\to\infty} [\frac{1}{n^2}(2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}})]$
So far, the only thing I have been able to do is to bound the limit between 0 and e:
*
*If $(2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}}) = 0 \to \lim_{n\to\infty} \frac{1}{n^2}=0$
*If $(\frac{(n+1)^n}{n^{n-1}}+\cdots+\frac{(n+1)^n}{n^{n-1}}) = n \frac{(n+1)^n}{n^{n-1}} \to \lim_{n\to\infty} \frac{n}{n^2}\frac{(n+1)^n}{n^{n-1}}= \lim_{n\to\infty} \frac{(n+1)^n}{n^n} = \lim_{n\to\infty} (1+\frac{1}{n})^n = e $
$0 < (2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}}) < (\frac{(n+1)^n}{n^{n-1}}+\cdots+\frac{(n+1)^n}{n^{n-1}}) \to\\
0 < \lim_{n\to\infty} [\frac{1}{n^2}(2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}})] < e$
This is fine, but I would like to know if a closer bound or an exact limit could be found.
Thanks in advance!
|
The expression equals
$$\frac{\sum_{k=1}^{n} (k+1)^k/k^{k-1}}{n^2}.$$
Stolz-Cesaro says we should look at
$$\frac{(n+2)^{n+1}/(n+1)^n}{(n+1)^2 - n^2}=\frac{(1+1/(n+1))^n(n+2)}{2n+2}.$$
The last expresion $\to \frac{e}{2}.$ So by S-C, the limit is $e/2.$
|
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|
Find $24 \cot^2 x$
Suppose $x$ is in the interval $[0,\pi/2]$ and $\log_{24 \sin x} (24 \cos x) = \frac{3}{2}$. Find $24 \cot^2 x$.
From the given equation we have $24 \cos{x} = (24 \sin{x})^{\frac{3}{2}}$ and so $24\cot^2{x} = 24^2\sin{x}$. How do we continue?
|
A Solution
Since
$$
(24\cos(x))^2=(24\sin(x))^3\tag{1}
$$
we have
$$
24\sin^3(x)+\sin^2(x)-1=0\tag{2}
$$
We can try the rational root theorem to get
$$
\sin(x)=\frac13\tag{2}
$$
Now use
$$
24\cot^2(x)=24^2\sin(x)=192\tag{4}
$$
Uniqueness of the Solution
Since $\frac{\mathrm{d}}{\mathrm{d}u}\left(24u^3+u^2-1\right)=72u^2+2u$, the function $24u^3+u^2-1$ is increasing for $u\gt0$. This means there can only be one positive solution. Equation $(1)$ says that we need $\sin(x)\ge0$. Furthermore, the maximum of $24u^3+u^2-1$, for $u\le0$, is $-\frac{3887}{3888}$. Thus, there can be no negative solution.
|
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|
Choose an infinite GP from given terms for a particular sum Is it possible to choose an infinite GP from amongst the terms 1, 1/2, 1/4, 1/8, 1/16 ... with a sum a) 1/5? b) 1/7 ?
My approach was simply choosing terms and probable ratios to match the sum...I would be grateful if you could reveal any trick for this question?
|
Since all terms are of the form $\frac{1}{2^n}$, any geometric progression will be with ratio $\frac{1}{2^k}$
The summation of a geometric series is:
$$\sum\limits_{k=0}^\infty ar^k = \frac{a}{1-r}$$
The question then is what choice of $a$ (initial term) and what choice of $r$ (ratio) will give the desired results.
Since $a$ and $r$ will both be of the form $\frac{1}{2^n}$ we have the sum must be of the form
$$\frac{\frac{1}{2^m}}{1-\frac{1}{2^n}} =\frac{2^{n-m}}{2^n-1}$$
for some given $n$ and $m$
The question is, can you find non-negative integers $n$ and $m$ such that $\frac{2^{n-m}}{2^n-1}$ is equal to $\frac{1}{5}$? is equal to $\frac{1}{7}$? If not, then why? If yes, which?
For $\frac{1}{5}$, the question becomes whether we can find $a,b$ such that $5\cdot 2^a = 2^b-1$. Notice that for $a\geq 1$ and $b\geq 1$ the left side is even while the right side is odd. If $b=0$ the right side is zero and the left side is positive. If $a=0$, the left side is equal to $5$, but no integer $b$ satisfies $2^b-1=5$
Thus $\frac{1}{5}$ is impossible.
For $\frac{1}{7}$, we ask if we can find $a,b$ such that $7\cdot 2^a=2^b-1$. From the same logic, we see that they cannot both be positive and $b$ cannot be zero. This leaves $a=0$ and we find that $b=3$ works.
This corresponds back to our $n$ and $m$ from before implying the geometric progression $\frac{1}{8},\frac{1}{64},\frac{1}{512},\dots$ has sum equal to $\frac{1}{7}$
|
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|
Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
|
\begin{align}
x^{2016}-1=&(x^2)^{1008}-1=(x^2-1)((x^2)^{1007}+\dots+x^2+1)\\
&=(x+1)(x-1)(x^{2014}+\dots+x^2+1).
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Is it inequality true Please help me prove the inequality:
$$\sqrt[3]{3+\sqrt[3] 3}+\sqrt[3]{3-\sqrt[3] 3}<2\sqrt[3] 3$$
Thanky for your help and your attention.
|
Let be
$$a=\sqrt[3]{3+\sqrt[3] 3}, b=\sqrt[3]{3-\sqrt[3] 3}$$
now we have
$$a^3+b^3=\left(\sqrt[3]{3+\sqrt[3] 3}\right)^3+\left(\sqrt[3]{3-\sqrt[3] 3}\right)^3=3+\sqrt[3]3+3-\sqrt[3]3=6.$$
Becouese,
$$a^3+b^3>a^2b+b^2a$$
we have
$$ab(a+b)<a^3+b^3=6$$
Now
$$3ab(a+b)<18$$
or
$$3a^2b+3ab^2<18$$
Add in the both sides eqution:
$$a^3+b^3=6$$
and inequality
$$3a^2b+3ab^2<18$$
we have:
$$(a+b)^3<24=8\cdot 3$$
or
$$a+b<2\sqrt[3]3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Pythagorean triples $(a,b,c)$ and divisibility of $a$ or $b$ by $3$. Call a triple of integers $(a, b, c)$ a Pythagorean triple if $a^2 + b^2 = c^2$
, i.e., if $a, b, c \in \mathbb{N^*}$ are the (measures of) sides of a right triangle. Examples of Pythagorean triples are (3, 4, 5), (5, 12, 13), (8, 15, 17) and (3312, 16766, 17090).
Show that if $(a, b, c)$ is a Pythagorean triple
if at least one of a and b is divisible by 3.
I think to begin with $a^2 + b^2 = c^2 \rightarrow a^2 - c^2 = b^2$ to show this.
Then I get stuck to prove this?
|
A number that is not divisible by $3$ may be written either as $3m+1$ or as $3m+2$ for some integer $m$. The square of such a number is then
$$
(3m+1)^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1\\
(3m + 2)^2 = 9m^2 + 12m + 4 = 3(3m^2 + 4m + 1) + 1
$$
and we see that the square of such a number is always of the form $3n+1$ for some integer $n$.
Now, assume that neither $a$ nor $b$ is divisible by $3$. Then by the above, we have
$$
\color{red}{a^2} + \color{blue}{b^2} = \color{red}{3i + 1} + \color{blue}{3j + 1} = 3(i+j) + 2
$$
for some integers $i, j$. In other words, $a^2 + b^2$ will be $2$ more than some multiple of $3$. But on the other side of the equality sign we have $c^2$, which is either a multiple of $3$ or $1$ more than such a multiple. This is a contradiction, and therefore the assumption that neither $a$ nor $b$ is divisible by $3$ must be false.
|
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|
Show series $1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2$ ... does not converge I was wondering if my proof is correct, and if there are any better alternative proofs. Or maybe proof that use nice tricks i might need in the future.
$$1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} \ldots = \sum_{n =1}^\infty \left(\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2}\right) + 1 - \frac{1}{2^2}$$
Now we know that $$\sum_{n = 1}^\infty \frac{1}{2n + 1}$$ diverges and $$\sum_{n = 1}^\infty -\frac{1}{(2n + 2)^2}$$ converges. Hence their sum diverges (I proved this fact). Hence, the series diverges. Any obvious mistake or better way of tackling it? Maybe using partial sums since i am clueless how to use them.
|
First, there seems to be quite a bit of confusion in the comments concerning "grouping". Let's take a look at it, essentially:
\begin{align}
1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} +\cdots &\stackrel{?}{=}
\left(1 - \frac{1}{2^2}\right) + \left(\frac{1}{3} - \frac{1}{4^2}\right) + \left(\frac{1}{5} - \frac{1}{6^2}\right) + \cdots \\ &= \sum_{n=0}^\infty \left(\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2}\right).
\end{align}
Now, when you say that you're "not allowed to group", I guess you're right in that you've generated a somewhat different series but the two series are concretely related. In particular, if the left side converges, then the right side must also converge. This is a special case of the fact that, if a sequence converges, then any subsequence of that sequence also converges. Specifically, if we let $s_n$ denote the $n^{\text{th}}$ partial sum of the series on the left and we let
$$S_n = \sum_{n=0}^n \left(\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2}\right)$$
denote the $n^{\text{th}}$ partial sum of the series on the right, then $S_n = s_{2n}$. Thus, $S_n$ is a subsequence of $s_n$ and, if $s_n$ converges then $S_n$ must converge to the same limit. Taking the contrapositive, if $S_n$ diverges, then $s_n$ must also diverge.
I think you make a mistake, though, at the next step by breaking the series up into two series. You are essentially rearranging the series which is only valid when the series is absolutely convergent.
The approach at this point is to simply combine the fractions to get
$$\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2} = \frac{4 n^2+6 n+3}{4 (n+1)^2 (2 n+1)},$$
to which the limit comparison test is easily applicable.
This is exactly the approach I took in my answer to this question.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $x_{n+1}=x_n+0.8x_n(1-x_n)-0.072$ converges Assume that $x_1$ is a real number and for $n \gt 1$ :
$x_{n+1}=x_n+0.8x_n(1-x_n)-0.072$
a) Assume that $x_1=0.6 $
Prove that $\{x_n\}$ converges and find its limit.
b) Find a set of real numbers like $A$ such that if $x_1 \in A$ then $\{x_n\}$ is convergent.
Note 1 : I know a method to answer the part "a" but its not a precise way. ( Put $L$ instead of every $x_i$ )
Note 2 : I have no idea about part "b".
|
Let
\begin{align}
f(x) &= x+0.8x(1-x)-0.072 \\
&= x+\frac{4}{5}x(1-x) - \frac{9}{125} \\
&= \frac{9}{10}+\frac{9}{25}\left(x-\frac{9}{10}\right)-\frac{4}{5} \left(x-\frac{9}{10}\right)^2
\end{align}
I obtained that last expression by expanding $f$ in a series about $9/10$. You can expand out the last two expressions to see that they are equal.
Subtracting $9/10$ off from both sides, factoring out an $(x-9/10)$ from the right and taking the absolute values, we get
\begin{align}
\left|f(x)-\frac{9}{10}\right| &= \left|x-\frac{9}{10}\right|
\left|\frac{9}{25}-\frac{4}{5}\left(x-\frac{9}{10}\right)\right|.
\end{align}
Thus, the distance between $f(x)$ and $9/10$ will be less than the distance between $x$ and $9/10$ precisely when
$$\left|\frac{9}{25}-\frac{4}{5}\left(x-\frac{9}{10}\right)\right| < 1.$$
This last inequality is equivalent to
$$\frac{1}{10}<x<\frac{13}{5}.$$
Thus, the sequence will converge to $9/10$ whenever $x$ is in that interval.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1997689",
"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $b-d$ If it's given that ->
$a^{3} = b^{2}$ ;
$c^{5} = d^{4}$ ;
$a - c = 19$
Find $b-d$
PS- I have given 4 hours of my life to this question and even then I am NOT able to get even the SLIGHTEST hint . So , i request you all to explain the solution of this question . Thanks in Advance !!
|
Put $a=x^2, b=x^3, c=y^4, d=y^5$:
$$\begin{align}
a^3=b^2\;&\Rightarrow (x^2)^3=(x^3)^2\\
c^5=d^4\; &\Rightarrow (y^4)^5=(y^5)^4\\
\text{Given} \qquad\qquad a-c&=19\\
x^2-y^4&=19\\
(x-y^2)(x+y^2)&=19\end{align}
$$
As $19$ is prime and $y^2>0$, and also given that $x,y$ are integers, possible combinations are
$$
x-y^2=1; x+y^2=19 \;\Rightarrow x=10, y=\pm3\\
\text{or}\\
x-y^2=-19; x+y^2=-1\;\Rightarrow x=-10, y=\pm 3\\$$
Solutions are:
$$
(x,y)=(\pm 10, \pm 3) \\
\Rightarrow b-d=\pm 1000 \pm243\\=\pm757, \pm 1243\\
$$
__
$$\color{lightgrey}{\begin{align}(x,y)=(10,3)\;\Rightarrow b-d&=757\\
(x,y)=(10,-3)\;\Rightarrow b-d&=1243\\
(x,y)=(-10,3)\;\Rightarrow b-d&=-1243\\
(x,y)=(-10,-3)\;\Rightarrow b-d&=-757\end{align}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $(a \cdot b) \text{ mod } n = ((a \text{ mod } n) \cdot (b \text{ mod } n)) \text{ mod } n$
Prove for all $n \in \mathbb{N}, n >1$ and $a,b \in \mathbb{Z}$ that
$$(a \cdot b) \text{ mod } n = ((a \text{ mod } n) \cdot (b \text{ mod
} n)) \text{ mod } n$$
I'm not sure if this is a proof but I tried another way this time:
Let $x=a \text{ mod } n$, let $y= b \text{ mod } n$ where $x,y \in \mathbb{Z}$
$\Rightarrow$
$(x \cdot y) \text{ mod } n= (a \cdot b) \text{ mod } n$
Create a "matching" $p,q \in \mathbb{Z}$ such that:
$a=x+pn$ and $b=y+qn$
$\Rightarrow ((x+pn)(y+qn)) \text{ mod } n= (xy+xqn+ypn+pqn^{2}) \text{ mod } n=(xy+n(xq+yp+pqn)) \text{ mod } n = (x \cdot y) \text{ mod } n$
I hope this is correct and counts as proof?
|
Suppose that:
$$\begin{cases}
a \mod n = r_a\\
b \mod n = r_b\\
(a \cdot b) \mod n = r\\
\end{cases}$$
or equivalently
$$\begin{cases}
a = q_a \cdot n + r_a\\
b = q_b \cdot n + r_b\\
a \cdot b = q \cdot n + r\\
\end{cases}.$$
Then:
$$r_a \cdot r_b = (a-q_a \cdot n)\cdot (b-q_b \cdot n) = \\
= a \cdot b - a \cdot q_b \cdot n - b \cdot q_a \cdot n + q_a \cdot q_b \cdot n^2 = \\
= q \cdot n + r - a \cdot q_b \cdot n - b \cdot q_a \cdot n + q_a \cdot q_b \cdot n^2 = \\
= (q + q_a \cdot q_b \cdot n - a \cdot q_b - b \cdot q_a) \cdot n + r.$$
In other words:
$$r_a \cdot r_b = k \cdot n + r \Rightarrow \\
(r_a \cdot r_b) \mod n = r \Rightarrow \\
r = (a \cdot b) \text{ mod } n = ((a \text{ mod } n) \cdot (b \text{ mod
} n)) \text{ mod } n .$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1998233",
"timestamp": "2023-03-29T00:00:00",
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|
How to show $\sum^{n}_{k=0}{a_kb_k=A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})}}$ by induction?
Let $n\in\mathbb{N}$ and $a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_{n+1}\in\mathbb{R}$ and $A_k=\sum^{n}_{k+1}{a_j}$ with $1\leq k\leq n$.
Show that $\sum^{n}_{k=1}{a_kb_k=A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})}}$ by induction.
I tried several times, however, I have a hard time to resolve this in the induction step. This is what I tried so far and looks most promising to me:
$$\sum^{n+1}_{k=1}{a_kb_k}=a_{n+1}b_{n+1}+\sum^{n}_{k=1}{a_kb_k}$$
$$=a_{n+1}b_{n+1}+\left(A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})}\right)$$
$$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)(b_k-b_{k+1})}\right)$$
$$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)(b_k-b_{k+1})}\right)$$
$$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)b_k}-\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)b_{k+1}}\right)$$
$$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+b_1a_1+\sum^{n}_{k=2}{\left(\sum^{k}_{i=1}{a_i}\right)b_k}-b_{n+1}\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)}-\sum^{n-1}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)b_{k+1}}\right)$$
$$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+b_1a_1-b_{n+1}\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)}\right)$$
$$=a_{n+1}b_{n+1}+\left(b_1a_1+b_{n+1}\left(A_n-\sum^{n}_{k=1}A_k\right)\right)$$
$$=???$$
I'd appreciate any hints or spotted errors!
|
Posting my answer from chat:
$$
\begin{align}
\sum^{n+1}_{k=1}{a_kb_k} &= a_{n+1}b_{n+1}+\sum^{n}_{k=1}{a_kb_k} \\
&= a_{n+1}b_{n+1}+A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})} \\
&= a_{n+1}b_{n+1}+A_nb_{n+1} - A_{n+1} (b_{n+1}-b_{n+2})+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\
&= A_{n+1}b_{n+2} + a_{n+1}b_{n+1} - (A_{n+1} - A_n)b_{n+1}+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\
&= A_{n+1}b_{n+2} + a_{n+1}b_{n+1} - a_{n+1}b_{n+1}+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\
&= A_{n+1}b_{n+2} + \sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})}
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determinant with variables What is the following determinant?
$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}$$
I calculated it as $0$ but I do not think it is right. Thanks in advance.
|
If you are not familiar with the concepts used in Thomas Andrews' (elegant) answer; you can try to compute the determinant by using properties to create zeroes.
For example; subtract the second row from the other rows and then add the first column to the second to get:
$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix} =
\begin{vmatrix}
1 & -1 & 0 & 0 \\
a & 1+b & c & d \\
0 & -1 & 1 & 0 \\
0 & -1 & 0 & 1 \end{vmatrix}
=
\begin{vmatrix}
1 & 0 & 0 & 0 \\
a & 1+a+b & c & d \\
0 & -1 & 1 & 0 \\
0 & -1 & 0 & 1 \end{vmatrix}$$
Now expand the determinant to the first row, reducing it to a $3 \times 3$-determinant. Subtracting $c$ times the second row from the first and $d$ times the third row from the first, you get:
$$\begin{vmatrix}
1+a+b & c & d \\
-1 & 1 & 0 \\
-1 & 0 & 1 \end{vmatrix}=
\begin{vmatrix}
1+a+b+c+d & 0 & 0 \\
-1 & 1 & 0 \\
-1 & 0 & 1 \end{vmatrix}$$
Expanding to the first row again agrees with Thomas Andrews' answer: $1+a+b+c+d$.
|
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|
If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \cos^{2} a)$$ But now, I'm stuck. Solutions are greatly appreciated.
|
$$\sin^3 \alpha+\cos^3 \alpha=(\sin \alpha+\cos \alpha)(\sin^2 \alpha-\sin \alpha\cos \alpha+\cos^2 \alpha)$$
$$1.2^3=\sin^3\alpha+\cos^3 \alpha+3\sin \alpha\cos \alpha(\sin \alpha+\cos \alpha)$$ It follows $$\sin \alpha\cos \alpha=0.22$$ and because $$\sin \alpha+\cos \alpha=1.2$$ one has sum and product of two unknowns so its are solution of the quadratic equation $$X^2-1.2X+0.22=0$$ with positive discriminant $0.14$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2001078",
"timestamp": "2023-03-29T00:00:00",
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|
Find the solutions of the differential equation $y\frac{dy}{dx}+\sqrt{x-xy^4}=0$ Find the solutions of the differential equation $y\frac{dy}{dx}+\sqrt{x-xy^4}=0$
Attempt: If $u=x-xy^4$ then $\frac{du}{dx}=1-(y^4+4xy^3\frac{dy}{dx})$ But I couldn't solve it from here. Do you have any suggestion? Thanks
|
$$ y\frac{dy}{dx} + \sqrt x \sqrt{1-y^4} = 0 \\ y \frac{dy}{dx}=-\sqrt x \sqrt{1-y^4}\\ \int\frac{y}{\sqrt{1-y^4}}dy = - \int \sqrt x dx = -\frac23 x \sqrt x + C \\ \int \frac12\frac{d}{dy} \arcsin(y^2)dy = -\frac{2}{3}x\sqrt x + C \\ \arcsin(y^2) = -\frac43 x\sqrt x + C \\ y = \pm \sqrt{\sin\left(-\frac43 x\sqrt x + C \right)}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solving higher-degree trigonometric equation Is there any method to find value of $\sin (A)$ from $\sin (9A)$ having known value of $\sin (9A) = \sin (30 degree) = 0.5$?
In $\sin (9A)$, being a nine-degree equation, there will be nine-roots. Can we find all nine roots from $\sin (9A) = 0.5$?
|
Using trigoniometric idendities (and not the way that @EmilioNovati is using):
*
*$$\sin\left(9\text{A}\right)=\sin\left(\text{A}\right)\left(1+2\cos\left(2\text{A}\right)\right)\left(1+2\cos\left(6\text{A}\right)\right)$$
*$$\cos\left(2\text{A}\right)=\cos^2\left(\text{A}\right)-\sin^2\left(\text{A}\right)$$
*$$\cos^2\left(\text{A}\right)=1-\sin^2\left(\text{A}\right)$$
*$$\cos\left(6\text{A}\right)=15\sin^4\left(\text{A}\right)\cos^2\left(\text{A}\right)-15\sin^2\left(\text{A}\right)\cos^4\left(\text{A}\right)+\cos^6\left(\text{A}\right)-\sin^6\left(\text{A}\right)$$
*$$\cos^4\left(\text{A}\right)=\left(1-\sin^2\left(\text{A}\right)\right)^2$$
*$$\cos^6\left(\text{A}\right)=\left(1-\sin^2\left(\text{A}\right)\right)^3$$
So, we get:
*
*$$\sin\left(9\text{A}\right)=\frac{1}{2}\cdot\left(1+2\cos\left(2\text{A}\right)\right)\cdot\left(1+2\cos\left(6\text{A}\right)\right)$$
*$$\cos\left(2\text{A}\right)=1-2\cdot\left(\frac{1}{2}\right)^2=\frac{1}{2}$$
*$$\cos\left(6\text{A}\right)=15\cdot\left(\frac{1}{2}\right)^4\cdot\cos^2\left(\text{A}\right)-15\cdot\left(\frac{1}{2}\right)^2\cdot\cos^4\left(\text{A}\right)+\cos^6\left(\text{A}\right)-\left(\frac{1}{2}\right)^6$$
*$$\cos^2\left(\text{A}\right)=1-\left(\frac{1}{2}\right)^2=\frac{3}{4}$$
*$$\cos^4\left(\text{A}\right)=\left(1-\left(\frac{1}{2}\right)^2\right)^2=\frac{9}{16}$$
*$$\cos^6\left(\text{A}\right)=\left(1-\left(\frac{1}{2}\right)^2\right)^3=\frac{27}{64}$$
So:
$$\sin\left(9\text{A}\right)=\frac{1}{2}\cdot\left(1+2\cdot\frac{1}{2}\right)\cdot\left(1+2\cdot-1\right)=-1$$
|
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|
$\int_{0}^{2\pi} \frac{\cos{3\theta}}{5-4\cos{\theta}}\,d{\theta}$ by using complex/real integration How to evaluate $\int_{0}^{2\pi} \frac{\cos{3\theta}}{5-4\cos{\theta}}\,d{\theta}$
by using complex integration?
I assume $z=e^{i{\theta}}$, $\frac{1}{iz}dz=d{\theta}$,
$$\cos{\theta}=\frac{z+z^{-1}}{2}
\quad\mbox{and}\quad \cos3{\theta}=\frac{(z+z^{-1})(z^2+z^{-2}-1)}{2}.$$
Hence
$$\frac{1}{2i}\oint_{|z|=1} \frac{{(z+z^{-1})(z^2+z^{-2}-1)}}{(5-4\frac{z+z^{-1}}{2})z}\,dz$$
and I'm stuck in this. Could you give me a hints or solution?
|
Complex Method
We better use $\cos 3x=Re(e^{3xi})=Re(z^3)$ to make the life easier.
Let $z=e^{xi}$, then the integral is transformed to a contour integration with a unit circle in anti-clockwise direction.
\begin{aligned}
I&=\operatorname{Re} \oint_{|z |=1}\frac{z^3}{5-\frac{4}{2}\left(z+\frac{1}{z}\right)} \cdot \frac{d z}{i z}\\
& = \operatorname{Re}\left(\frac{-1}{i} \oint_{|z |=1} \frac{z^3}{(2 z-1)(z-2)} d z\right) \\
& =\operatorname{Re}\left(-2 \pi\left(\frac{1}{2} \cdot \frac{\frac{1}{8}}{-\frac{3}{2}}\right)\right) \\
& =\frac{\pi}{12}
\end{aligned}
Real Method
$\begin{aligned} \displaystyle \int_{0}^{2 \pi} {\frac{\cos(3 x)}{5 - 4 \cos(x)}} dx & = \int_{0}^{2 \pi} {\frac{4 \cos^{3}(x) - 3 \cos(x)}{5 - 4 \cos(x)}} dx \\ &= \int _ { 0 } ^ { 2 \pi } \left[ - \cos ^ { 2 } x - \frac { 5 } { 4 } \cos x - \frac { 13 } { 16 } + \frac { 65 } { 16 ( 5 - 4 \cos x ) } \right] d x\\ & = \left[ - \frac { x } { 2 } - \frac { \sin 2 x } { 4 } - \frac { 5 } { 4 } \sin x - \frac { 13 } { 16 } x \right] _ { 0 } ^ { 2 \pi } +\frac { 65 } { 16 } \int _ { 0 } ^ { 2 \pi } \frac { d x } { 5 - 4 \cos x }\\&= - \frac { 21 \pi } { 8 } + \frac { 65 } { 16 } \int _ { 0 } ^ { 2 \pi } \frac { d x } { 5 - 4 \cos x } \end{aligned} \tag*{} $
$\displaystyle \textrm{Let }t=\tan \frac{x}{2}, \textrm{ then }d t = \frac { 1 } { 2 } ( 1 + t ^ { 2 } ) d x$
$\displaystyle \begin{aligned} \int _ { 0 } ^ { 2 \pi } \frac { d x } { 5 - 4 \cos x } & =2 \int _ { 0 } ^ { \pi } \frac { d x } { 5 - 4 \cos x } \\ &=2 \int_0^{\infty} \frac{1}{5-4 \cdot \frac{1-t^2}{1+t^2}} \cdot \frac{2 d t}{1+t^2}\\& = 4 \int _ { 0 } ^ { \infty } \frac { d t } { ( 3 t ) ^ { 2 } + 1 } \\&= \frac { 4 } { 3 }\left [ \tan ^ { - 1 } ( 3 t )\right ] _ { 0 } ^ { \infty } \\ &= \frac { 4 } { 3 } \cdot \frac { \pi } { 2 } \\ &= \frac { \pi } { 6 } \end{aligned} \tag*{} $
$\displaystyle \therefore I = - \frac { 21 \pi } { 8 } + \frac { 65 } { 16 } \cdot \frac { 2 \pi } { 3 }=\boxed{\frac{\pi }{12}} \tag*{} $
|
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|
Prove by induction, that $ \sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!}$ If I'm not wrong,
$$\sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!},$$
but I am having trouble proving it by induction...
If $n=1$ the formulas coincide.
Sup $n=k$ is valid: $$\sum_{i=1}^k \frac{i}{(i+1)!}= \frac{1}{(k+1)(k-1)!}.$$
Then if $n=k+1$,
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!}= \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+1)!} $$
$$ = \frac{1}{(k+1)(k-1)!} + \frac{k+1}{(k+1)!} $$
$$ = \frac{k + (k +1)}{(k+1)!} = \frac{2k + 1}{(k+1)!} $$ I really don't see from this how I should arrive to
$$ \sum_{i=1}^{k+1} \frac{i}{(i+1)!} = \frac{1}{(k+2)(k)!} $$
|
For $n=2$ we have
$$
\frac{1}{2!}+\frac{2}{3!}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}
$$
On the other hand,
$$
\frac{1}{(2+1)(2-1)!}=\frac{1}{3}
$$
so your conjecture is wrong.
For $n=3$ we have
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}=
\frac{12}{4!}+\frac{8}{4!}+\frac{3}{4!}=
\frac{23}{24}
$$
and you can make a better conjecture.
|
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|
Calculate the multiplicative inverse of $7$ in $\mathbb{Z}_{12}$ I'm not sure how to do this but I gave it a try:
I know that $7$ must have an inverse because $\text{ gcd }(12,7)=1$
Now, on this, use euclidean algorithm. We have $$7x \equiv 1 \text{ mod } 12$$
$$x \equiv 7^{-1}(\text{ mod } 12)$$
So
$$12 = 7 \cdot 1+5$$
$$7=5 \cdot 1+2$$
$$5=2\cdot 2+1$$
Now we can stop this here and write:
$$1=5-4$$
And now I cannot replace $4$ because there isn't any $4$ in the equations above : /
What to do? There is another, easier way?
|
You have
\begin{align}
\color{red}{12}&=\color{red}{7}\cdot1+\color{red}{5}\\
\color{red}{7} &=\color{red}{5}\cdot1+\color{red}{2}\\
\color{red}{5} &=\color{red}{2}\cdot2+\color{green}{1}
\end{align}
Then
$$
1=\color{red}{5}-\color{red}{2}\cdot2=
\color{red}{5}-2(\color{red}{7}-\color{red}{5})=
-2\cdot\color{red}{7}+3\cdot\color{red}{5}=
-2\cdot\color{red}{7}+3(\color{red}{12}-\color{red}{7})=
-5\cdot\color{red}{7}+3\cdot\color{red}{12}
$$
|
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|
Unique pair of positive integers $(p,n)$ satisfying $p^3-p=n^7-n^3$ where $p$ is prime
Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.
Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.
Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2<n^7-n^3=\text{r.h.s}.$ This means that either $p>n^2$ or $p<n^2$ must hold.
Assuming $p>n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p>n^2)$.
Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.
Finally, we get $(5,2)$ as the solution to the given equation.
But how do I conclude this is the only solution possible? Also, why does'nt $p<n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem.
|
To eliminate the case $p\lt n^2$, note that $p\lt n^2$ implies $p+1\lt n^2+1$ and $p-1\lt n^2-1$, and this gives $p^3-p=p(p+1)(p-1)\lt n^2(n^2+1)(n^2-1)\le n^3(n^2+1)(n^2-1)=n^7-n^3$.
Remark: The paragraph that argues that either $p\gt n^2$ or $p\lt n^2$ isn't really necessary. It's obvious that $p\not=n^2$, since primes cannot be squares.
|
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|
Solve for x if $1 + x + x^2 + x^3 .... = 2$ I've gone as far as rewriting it as
$$\sum_{n=1}^\infty x^{n-1} = 2$$
Not sure where to go from here.
|
$$|x|<1 \to \\1+x+x^2+x^3+x^4+...=2\\ 1+x(1+x+x^2+x^3+x^4+...)=2\\1+x(2)=2\\x=\frac{1}{2}$$
or
$$\\1+x+x^2+x^3+x^4+...=2\\ 1+x+x^2(1+x+x^2+x^3+x^4+...)=2\\1+x+x^2(2)=2\\2x^2+x-1=0\\x=\frac{1}{2} \color{red}{\checkmark}\\x=-1 ,\color{red}{|x|<1}$$
so $x=-1 $ is not acceptable
|
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|
convergence of $\sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{ \sinh \frac{1}{n}}{\sin\frac{1}{n}} \right )$ $ \sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}} \right ) $
Find the value of $\alpha$ for which the series converges
please, I have no idea to approach the solution
thanks.
|
Hint. By using Taylor series expansions, as $n \to \infty$, we have
$$
\frac{\sinh \frac1n}{\sin \frac1n} = 1+\frac{1}{3 n^2}+O\left(\frac{1}{n^4}\right)
$$ and
$$
(n^2+1)\log^\alpha \left(\frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}}\right)=\frac{1}{3^a n^{2\alpha-2}}+O\left(\frac{1}{n^{2\alpha}}\right)
$$ the given series is then convergent iff we have $2\alpha-2>1$.
|
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|
Inconclusive second derivative test rigorous proof I need to find and identify stationary points of the following function:
$f(x,y) = x^4 + 2x^2y^2 - y^4 - 2x^2 + 3$
Second derivative test appears to be inconclusive (Hessian equal to zero) at the point $(0,0)$. By plotting this appears to be a maximum. How does one prove this rigorously?
|
Suppose $(x,y)$ lies in the unit disc centered at the origin (i.e. $x^2+y^2 \le 1$).
\begin{align}f(x,y) - f(0,0) &= x^4+2x^2y^2 - y^4-2x^2\\
&\le x^4 + 2x^2y^2 - y^4 - 2x^4 & x^4 \le x^2\\
&=-(x^2-y^2)^2\\
&\le 0.
\end{align}
|
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|
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Problem Statement:-
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Attempt at solution:-
Let $\alpha=\sqrt{3x^2-7x-30}\;\;\; \text{&} \;\;\;\beta=\sqrt{2x^2-7x-5}$.
Then, we have $$\alpha-\beta=x-5\tag{1}$$
And, we have $$\alpha^2-\beta^2=x^2-25\tag{2}$$
From $(1)$ and $(2)$, we have
$$\alpha+\beta=\dfrac{(x-5)(x+5)}{x-5}=x+5\qquad(\therefore x\neq5)\tag{3}$$
From $(1)$ and $(3)$, we get
$$\alpha=\sqrt{3x^2-7x-30}=x\qquad\qquad\beta=\sqrt{2x^2-7x-5}=5$$
On solving any one of $\alpha=x$ or $\beta=5$, we get $x=6,-\dfrac{5}{2}$.
On putting $x=-\dfrac{5}{2}$, $\alpha=x$ is not satisfied. Hence, $x=6$ is the only solution.
But as in $(3)$, we have ruled out $x=5$, as a solution then we can't put $x=5$ in the original equation. But, if we do put $x=5$ in the original equation we see that it is indeed satisfied. So, I tried a solution which also gives $x=5$ as a solution. So I picked the last solution from the $(2)$ without cancelling $(x-5)$.
$$(x-5)(\alpha+\beta)=x^2-25\implies (x-5)(\alpha+\beta-(x+5))=0$$.
Now, I am pretty much stuck here.
Edit-1:- I just saw now and feel pretty stupid about it.
If we proceed from the last step i.e. $(x-5)(\alpha+\beta-(x+5))=0$, we get
$$(x-5)(\alpha+\beta-(x+5))=0\implies (x-5)=0\;\;\;\text{ or }\;\;\;\alpha+\beta=x+5$$
So we have the following equations to be solved:-
$$\alpha-\beta=x-5\\
\alpha+\beta=x+5\\
x-5=0$$
Which results in $x=5,6$.
Now, since I have put so much effort in posting this question, I might as well ask for better solutions if you can come up with one. And, please don't post a solution which includes a lot of squaring to result into a quartic equation.
|
rewriting as $$\sqrt{3x^2-7x-30}=x-5+\sqrt{2x^2-7x-5}$$
after squaring we get
$$3x^2-7x-30-(x-5)^2-2x^2+7x+5=2(x-5)\sqrt{2x^2-7x-5}$$
simplifying
$$10x-50=2(x-5)\sqrt{2x^2-7x-5}$$
$$5x-25=(x-5)\sqrt{2x^2-7x-5}$$
squaring again
$$- \left( 2\,x+5 \right) \left( x-6 \right) \left( x-5 \right) ^{2}=0$$
from here you will get the possible solutions.
|
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|
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now take $x^2+4x+3$ and factorise that:
$$ x^2+4x+3 $$
Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.
\begin{array}{|c|c|}
\hline
x^2 & \\
\hline
& 3 \\
\hline
\end{array}
Then find HCF of 3:
$$3\\
1 | 3
$$
Enter the values $1x$ and $3x$ into the other two boxes:
\begin{array}{|c|c|}
\hline
x^2 & 1x \\
\hline
3x& 3 \\
\hline
\end{array}
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=3(x+3)
$$
Therefore:
$$x^2+4x+3=(x+1)(x+3)$$
It follows that:
$$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$
Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...
|
Avoid the 'box method' which will only work for quadratics with nice integer roots and doesn't really tell you anything about what's going on.
Suppose we have a quadratic $p(x)$ that factorizes as:
$$
p(x) = (x + s)(x + t)
$$
where we don't know what $s$ and $t$ are yet. If we multiply out the brackets, then we get:
$$
p(x) = x^2 + sx + tx + st = x^2 + (s+t)x + st
$$
Now suppose that we are given $p(x)$ in the form
$$
p(x) = x^2 + bx + c
$$
In order to factorize $p(x)$, all we need to do is find two numbers $s$ and $t$ such that
$$s+t=b$$
and
$$st=c$$
In your case, you want to find $s$ and $t$ such that $s+t=4$ and $st=3$. It shouldn't take you very long to realize that $s=3$ and $t=1$ will do. Then you can immediately factorize the polynomial as $(x+3)(x+1)$.
Slogan: To factorize a quadratic of the form $x^2+bx+c$, just find two numbers that add to give $b$ and multiply to give $c$.
Just for interest
This method should work for all the quadratics you see in Math 1, and I hope that it's a bit clearer what's going on compared to the 'Box Method'. You might be interested in how we solve quadratic equations that don't have nice solutions. For example, suppose we were trying to factorize
$$
x^2+x-1
$$
We want to find two numbers that add together to give $1$ and multiply together to give $-1$. It turns out that the right two numbers are
$$
\frac{1+\sqrt{5}}{2}
$$
and
$$
\frac{1-\sqrt{5}}{2}
$$
How did I work those out? Well, I'll show you a method that was invented by the ancient Babylonians. Suppose we have two numbers $b$ and $c$ and we're trying to find two numbers $s$ and $t$ such that $s+t=b$ and $st=c$.
If we square the first equation, we get
$$
b^2 = (s+t)^2 = s^2 + 2st + t^2
$$
Now since $st=c$, we can subtract $4c$ from the left hand side and $4st$ from the right hand side to get:
$$
b^2-4c = s^2 - 2st + t^2
$$
But $s^2 - 2st + t^2=(s-t)^2$, so we may write
$$
s-t = \sqrt{b^2-4c}
$$
(Here, we are free to assume that $s\ge t$, so this is the positive square root.)
Now we can recover $s$ and $t$:
$$
s = \frac{(s+t) + (s-t)}{2} = \frac{b+\sqrt{b^2-4c}}{2}
$$
$$
t = \frac{(s+t) - (s-t)}{2} = \frac{b-\sqrt{b^2-4c}}{2}
$$
|
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How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to prove such formula's (By principal of mathematical induction). I don't know how to find result of such series.
Please help. I shall be thankful if you guys can provide me general solution (Since I have been told that there exist a general solution by my friend who gave me this question).
|
In general, we have
$$\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)=\frac{n(n+1)(n+2)\dots(n+p+1)}{p+2}$$
Prove by induction,
$$\sum_{k=1}^{n+1}k(k+1)(k+2)\dots(k+p)\\=(n+1)(n+2)\dots(n+p+1)+\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)\\=(n+1)(n+2)\dots(n+p+1)+\frac{n(n+1)(n+2)\dots(n+p+1)}{p+2}\\=\frac{\color{#034da3}{(p+2)}\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}+\color{#034da3}n\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}}{p+2}\\=\frac{\color{#ee8844}{(n+1)(n+2)\dots(n+p+1)}\color{#034da3}{(n+p+2)}}{p+2}$$
And easy enough to check for $n=1$ to see it is true.
$$1\times2\times3\times\ldots\times(1+p)=\frac{1\times2\times3\times\ldots\times(1+p)\times\require{cancel}\cancel{(2+p)}}{\cancel{p+2}}$$
This is slightly off, since my sum ends at $n(n+1)(n+2)(n+3)$, while you end off at $(n-3)(n-2)(n-1)n$. To readjust, have $n=p-3$ in my sum and it will become yours.
|
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|
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the results.
Are there other solutions, simpler approaches?
I found some other questions about the same expression, but they simplify this to another form: Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?, Alternative proof of $\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$ and Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$. Perhaps also Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ can be considered similar. The expression $\cos^6x+\sin^6x$ also appears in this integral Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $ but again it is transformed to a different from then required here.
Note: The main reason for posting this is that this question was deleted, but I still think that the answers there might be useful for people learning trigonometry. Hopefully this new question would not be closed for lack of context. And the answers from the deleted question could be moved here.
|
Factoring from $u^3+v^3=(u+v)(u^2-uv+v^2)$:
\begin{align}
\cos^6x+\sin^6x&=(\cos^2x+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)\\
&=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x-\cos^2x\sin^2x\\
&=1-3\cos^2x\sin^2x
\end{align}
|
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|
Prove that: $\sum\limits_{cyc}\frac{1}{a}\sum\limits_{cyc}\frac{1}{1+a^2}\geq\frac{16}{1+abcd}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+d^2}\right)\geq\frac{16}{1+abcd}$$
I tried Rearrangement, C-S and more, but without success.
|
My second solution
By Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
&\frac{1}{1+a^2} + \frac{1}{1+b^2} + \frac{1}{1+c^2} + \frac{1}{1+d^2}\\
=\ & \frac{(1/a)^2}{1+(1/a)^2} + \frac{(1/b)^2}{1+(1/b)^2} + \frac{(1/c)^2}{1+(1/c)^2} + \frac{(1/d)^2}{1+(1/d)^2}\\
\ge\ & \frac{(1/a + 1/b + 1/c + 1/d)^2}{4 + (1/a)^2 + (1/b)^2 + (1/c)^2 + (1/d)^2}.
\end{align}
It suffices to prove that
$$\frac{(1/a + 1/b + 1/c + 1/d)^3}{4 + (1/a)^2 + (1/b)^2 + (1/c)^2 + (1/d)^2}\ge \frac{16}{1 + abcd}$$
or equivalently
$$\frac{(a+b+c+d)^3}{4 + a^2 + b^2 + c^2 + d^2} \ge \frac{16abcd}{1 + abcd}.\tag{1}$$
By Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prove the case when
$a=b=c \le d$. Let $d = a + s$ for $s \ge 0$. It suffices to prove that
\begin{align}
&a^3 s^4+(13 a^4-16 a^3+1) s^3+(60 a^5-48 a^4+12 a) s^2+(112 a^6-96 a^5-64 a^3+48 a^2) s\\
&\quad +64 a^7-64 a^6-64 a^4+64 a^3 \ge 0.
\end{align}
It is not difficult.
Remarks: I hope to see nice proofs for (1).
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
% https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
|
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|
If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}$, then calculate $14S$.
If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.
The question can be simplified to:
Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^4}$.
As $1+n^2+n^4$ forms a GP,
$$t_n=\frac{n(n^2-1)}{n^6-1}\,.$$
But I can't figure out how to solve further. It would be great if someone could help.
|
HINT:
As $n^4+n^2+1=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$
$$\dfrac{2n}{n^4+n^2+1}=\dfrac{(n^2+n+1)-(n^2-n+1)}{(n^2-n+1)(n^2+n+1)}=?$$
If $f(m)=\dfrac1{m^2-m+1}, f(m+1)=?$
See Telescoping series
|
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|
How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule) How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule)
$$ \lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} $$
$$ \lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]}$$
I suppose in the second I may not take into account arctan and sin as sinx approximately equals x
|
Let us consider $$A= \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3}$$ and use Taylor series around $x=0$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$1+\sin(x)=1+x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\tan(x)=x+\frac{x^3}{3}+O\left(x^5\right)$$ $$1+\tan(x)=1+x+\frac{x^3}{3}+O\left(x^5\right)$$ Now, using the generalized binomial theorem $$\sqrt{1 + \sin(x)}=1+\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{48}+\frac{x^4}{384}+O\left(
x^5\right)$$ $$\sqrt{1 + \tan(x)}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{11 x^3}{48}-\frac{47 x^4}{384}+O\left(x^5\right)$$ $$\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}=\frac{x^3}{4}-\frac{x^4}{8}+O\left(x^5\right)$$ $$A=\frac{1}{4}-\frac{x}{8}+O\left(x^2\right)$$ which not only shows the limit but also how it is approached.
|
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|
Sum first $n$ squares I am trying to find the formula for the sum of square number but I am struggle
with it. I now that it isn't complex but I am young in maths.
What I've done:
Using odd numbers to find the pattern
Using triangle number and the relative
formula:
$$\frac{n(n+1)}{2}$$
1 $\rightarrow$
1+3 $\rightarrow$
1+3+5 $\rightarrow$
.
.
.
1+3+5+...=$n^2$
1+1+1+...=$n-1$
Thanks
|
What you can do is follow Euler's method :
\begin{equation*}
\sum_{k=1}^n ((k+1)^3 - k^3) = (n+1)^3 - 1
\end{equation*}
But also :
\begin{equation*}
\sum_{k=1}^n ((k+1)^3 - k^3) = \sum_{k=1}^n (3k^2 + 3k +1)
\end{equation*}
Thus, knowing $\sum_{k=1}^n k = \frac{n(n+1)}{2}$, we get :
\begin{equation*}
3\sum_{k=1}^n k^2 = (n+1)^3 - 1 - 3 \frac{n(n+1)}{2} - n = \frac{(2n+1)n(n+1)}{2}
\end{equation*}
thus:
\begin{equation*}
\sum_{k=1}^n k^2 = \frac{(2n+1)n(n+1)}{6}
\end{equation*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Quadratic equation 4 need some guidance with a quadratic equation.
Suppose $x^2+20x-4000=0$
Here is what I have done so far;
Using the Quadratic Equation $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where from the above equation, $a=1$, $b=20$, $c=-4000$, we find
$$\begin{align*}x&=\frac{-20+\sqrt{(20)^2-(4)(1)(-4000)}}{2\times 1}\\
&=\frac{-20+\sqrt{16400}}{1}\\
&=108.0624847 \end{align*}$$
and
$$\begin{align*} x&=\frac{-20-\sqrt{20^2-(4)(1)(-4000)}}{2\times1}\\
&=\frac{-20-\sqrt{16400}}{1} \\
&=-148.0624847 \end{align*}$$
Neither of these values for $x$ prove correct when applied to $x^2+20x-4000=0$.
I know something is wrong but I can’t figure out what. Any guidance would be appreciated.
|
Did you figure it out?? I think everyone pointed out where you went wrong, but just to be sure:
$x=\frac{-20+\sqrt{20^2-(4)(1) (-4000)}}{2*\ 1}$
$\frac{1}{2} (-20-20 \sqrt{41})$
= $54.03124237$
And similarly
$x=\frac{-20-\sqrt{20^2-(4)(1) (-4000)}}{2*\ 1}$
=$\frac{1}{2} (-20+20 \sqrt{41})$
$=-74.03124237$
I used Mathematica to check my work:
Letting $answer\text{=}-74.03124237$
$=answer^2+20 answer-4000$
= 0
Similarly,
$answer\text {=} 54.03124237$
$=answer^2+20 answer-4000$
=0
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
|
Let $$P(x)=x^8+x^6-4x^4+x^2+1.$$
then
$$P'(x)=8x^7+6x^5-16x^3+2x.$$
but
$$P(\pm 1)=P'(\pm 1)=0$$
$$\implies P(x)=(x^2-1)^2(x^4+3x^2+1)$$
$$\implies \forall x\in\mathbb R \;\;P(x)\geq 0.$$
|
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"url": "https://math.stackexchange.com/questions/2035806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $\sum\limits_{cyc}\sqrt{a^2+10bc}\geq\frac{1}{2}\sum\limits_{cyc}\sqrt{22a(b+c)}$ Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$\sum\limits_{cyc}\sqrt{a^2+10bc}\geq\frac{1}{2}\sum\limits_{cyc}\sqrt{22a(b+c)}$$
I tried SOS, C-S, Holder, Mixing Variables and more, but without success.
|
The Buffalo Way works although it is an ugly solution.
WLOG, assume that $a+b+c=3$. Write the inequality as
$$\sum_{\mathrm{cyc}} \sqrt{\frac{a^2+10bc}{11}} \ge \sum_{\mathrm{cyc}} \sqrt{\frac{a(b+c)}{2}}.$$
Let $X = \frac{a^2+10bc}{11}, \ Y = \frac{b^2+10ca}{11}, \ Z = \frac{c^2+10ab}{11}$
and $U = \frac{a(b+c)}{2}, \ V = \frac{b(c+a)}{2}, \ W = \frac{c(a+b)}{2}$.
We will use the following bounds:
$$\sqrt{x} \le \frac{(x+1)(x^2+14x+1)}{2(x+3)(3x+1)}, \quad \forall x > 0$$
and
$$\sqrt{x} \ge \frac{2(x+3)(3x+1)x}{(x+1)(x^2+14x+1)}, \quad \forall x > 0$$
which follows from
$$\Big(\frac{(x+1)(x^2+14x+1)}{2(x+3)(3x+1)}\Big)^2 - x = \frac{(x-1)^6}{4(x+3)^2(3x+1)^2}.$$
With the bounds above, it suffices to prove that
$$\sum_{\mathrm{cyc}(X,Y,Z)} \frac{2(X+3)(3X+1)X}{(X+1)(X^2+14X+1)} \ge
\sum_{\mathrm{cyc}(U,V,W)}\frac{(U+1)(U^2+14U+1)}{2(U+3)(3U+1)}.$$
After homogenization, it suffices to prove that $f(a,b,c)\ge 0$
where $f(a,b,c)$ is a homogeneous polynomial with degree $32$.
We use the Buffalo Way. Due to symmetry, assume that $c\le b\le a$.
If $c = 0$, $f(a,b,0)$ is a polynomial with non-negative coefficients. True.
If $c > 0$, let $c=1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$.
$f(1+s+t, 1+s, 1)$ is a polynomial with non-negative coefficients. True.
We are done.
|
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|
Prove a geometry question about angles and radii in five collinear circles? Our teacher give us this question and I worked on it but I couldn't find a way to prove that. Is it possible to help me to prove that?
Thanks.
|
Assumptions
I'm assuming that the following are true:
*
*$A$, $B$, $G$, $H$, $I$, $K$ are collinear
*$B$, $G$, $H$, $I$, $K$ are the centres of their respective circles
*$A$, $C$, $D$, $E$, $F$, $J$ are collinear and the line on which they lie is tangent to all the circles.
Definitions
We know these angles:
*
*$\angle CAB$ = $\angle DAG$ = $\angle EAH$ = $\angle FAI$ = $\angle JAK$. Call this angle $\theta$.
*$\angle ACB$ = $\angle ADG$ = $\angle AEH$ = $\angle AFI$ = $\angle AJK$ = $90^{\circ}$.
And we can define some lengths. Let:
*
*$R_1$, $R_2$, etc. be the lengths of $CB$, $DG$, etc.
*$H_1$, $H_2$, etc. be the lengths of $AB$, $AG$, etc.
Known Information
*
*For right angled trianges, we know that $\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypoteneuse}}$, so:
$$\forall n \in \{1, \dots, 5\}, \sin\theta = \frac{R_n}{H_n} \Rightarrow H_n = \frac{R_n}{\sin\theta}$$
*
*Looking at the diagram, we can also see that:
$$\forall n \in \{1, \dots, 4 \}, H_{n+1} = H_n + (R_n + R_{n+1})$$
Forming a Rule
We can substitute for $H_n$ from the first equation into the second and rearrange:
\begin{align}
&& \frac{R_{n+1}}{\sin\theta} &= \frac{R_n}{\sin\theta} + R_n + R_{n+1} \\
\Rightarrow && R_{n+1} \left(\frac{1}{\sin\theta} - 1\right) &= R_n \left(\frac{1}{\sin\theta} + 1\right) \\
\Rightarrow && R_{n+1} \left(1 - \sin\theta\right) &= R_n \left(1 + \sin\theta \right) \\
\Rightarrow && R_{n+1} &= R_n \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right) \\
\Rightarrow &&R_{n+k} &= R_n {\left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)}^k \\
\end{align}
The result
This means that:
\begin{align}
R_1 \cdot R_5 &= R_1 \cdot \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^4\right) \\
&= \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^2\right) \cdot \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^2\right) \\
&= R_3 \cdot R_3
\end{align}
Extra results
This rule, coupled with the fact that there's nothing special about $R_1$ (it could just as easily have been called $R_0$ or $R_{1000}$) means that if we take any two collections of integers $A = \{A_1, A_2, \dots\}$ and $B = \{B_1, B_2, \dots\}$ and define $S_A = \sum_{a \in A}a$ and $S_B = \sum_{b \in B}b$ we get:
\begin{align}
\prod_{a \in A} R_{a} &= R_0^{|A|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A} \\
&= R_0^{|A| - |B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A - S_B} R_0^{|B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_B} \\
&= R_0^{|A| - |B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A - S_B} \prod_{b \in B} R_{b}
\end{align}
If $|A| \neq |B|$ and $z = \frac{S_A - S_B}{|A|-|B|}$ is an integer, this simplifies to:
$$
\prod_{a \in A} R_{a} = R_z^{|A| - |B|} \prod_{b \in B} R_{b}
$$
Or if $|A| = |B|$ and $S_A = S_B$ (i.e. two equal sized collections with equal sums), we end up with:
$$
\prod_{a \in A} R_{a} = \prod_{b \in B} R_{b}
$$
and your question is a special case of this where $A = \{1,5\}$ and $B = \{3,3\}$.
|
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|
Evaluate exactly the following series that includes fibonacci How can i find the exact value of this?
$$\sum_{n=1}^∞ \frac{f_n}{100^n}$$
Assuming that $ f_1=1$ and $f_2=1$ and $f_n=f_{n-1}+f_{n-2}$
I can approximate it to 0.01010203050813.........
But what is the exact value? Should i use $\lim_{n\to ∞}$ ? And how to solve it?
Thank you.
|
Following the hint of @lulu above, we let
$$
F(x) = \sum_{n = 1}^\infty f_nx^n
$$
We are after $F(\frac1{100})$, and to do that, we need a good expression for $F(x)$. The Fibonacci recurrence relation gives
$$
F(x) = \sum_{n = 1}^\infty f_nx^n = x + x^2 + \sum_{n = 3}^\infty f_nx^n\\
= x + x^2 + \sum_{n = 3}^\infty(f_{n-1} + f_{n-2})x^n\\
= x + x^2 + \left(\sum_{n = 3}^\infty f_{n-1}x^n\right) + \left(\sum_{n = 3}^\infty f_{n-2}x^n\right)
$$
To get a better grip on those two sums, note what happens when we divide them by $x$ and $x^2$ respectively:
$$
\frac1x\sum_{n = 3}^\infty f_{n-1}x^n = \sum_{n = 3}^\infty f_{n-1}x^{n-1} = \sum_{n=2}^\infty f_nx^n = F(x) - x\\
\frac1{x^2}\sum_{n = 3}^\infty f_{n-2}x^n = \sum_{n = 3}^\infty f_{n-2}x^{n-2} = \sum_{n = 1}^\infty f_{n}x^n = F(x)
$$
Each of these lines give us
$$
\sum_{n = 3}^\infty f_{n-1}x^n = x(F(x) - x)\\
\sum_{n = 3}^\infty f_{n-2}x^n = x^2F(x)
$$
Inserting this, we have
$$
F(x) = x + x^2 + x(F(x) - x) + x^2F(x)\\
F(x) = x + x^2 + xF(x) - x^2 + x^2F(x)\\
F(x) -xF(x) - x^2F(x) = x\\
F(x)(1-x-x^2) = x\\
F(x) = \frac{x}{1-x-x^2}
$$
|
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|
Inverse Nasty Integral Intrigued by the original question on a nasty integral, one wonders what functions $f(x)$ exist such that
$$\int_0^\infty f(x)\; dx=\frac {\pi}2$$
Something to do with the area of a half-circle with unit radius perhaps?
Edited To Add
It should be specified that $f(x)$ should not contain $\pi$, and is preferably a rational function.
The intention of posting the question was to take a different approach to the the original nasty integral by first considering the answer and then working backwards, sort of a heuristic approach.
This has yielded an excellent answer posted by @achillehui below, which also addresses the original nasty integral question in a very simple way.
|
Using Glasser's Master theorem, it is very easy to cook up very nasty looking
integral which evaluates to $\frac{\pi}{2}$.
For example, start from the integral
$$\int_{-\infty}^\infty \frac{1}{u^2+1} du = \pi\tag{*1}$$
If one replace the $u$ in integrand by $\displaystyle\;x - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2}$, one obtain
$$\int_{-\infty}^\infty \frac{1}{\left(x - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2}\right)^2 + 1} dx = \pi$$
Expand the integrand and extract the even part, you get something to challenge your friends:
$$\int_0^\infty
\frac{
x^{14}-15x^{12}+82x^{10}-190x^8+184x^6-60x^4+16x^2
}{x^{16}-20x^{14}+156x^{12}-616x^{10}+1388x^8-1792x^6+1152x^4-224x^2+16
}\; dx = \frac{\pi}{2}$$
Notes
The statement about Glasser's Master theorem in above link is slightly off. The coefficient $|\alpha|$ in front of $x$ there need to be $1$. Otherwise, there will be an extra scaling factor on RHS of the identity. When in doubt, please consult the original paper by Glasser,
Glasser, M. L. "A Remarkable Property of Definite Integrals." Math. Comput. 40, 561-563, 1983.
and an online copy of that paper can be found here.
Update
About the nasty integral that trigger all this.
If we replace the $u$ in integrand of $(*1)$ by $t - \frac{1}{t} - \frac{1}{t+1}$, change variable to $x = \frac{1}{t}$ on both intervals $[0,\infty)$ and $(-\infty, 0]$ and combine the result. We get
$$ \int_{-\infty}^\infty
\frac{1}{\left(\frac{1}{x}-x-\frac{x}{x+1}\right)^2+1} \frac{dx}{x^2}
= \int_{-\infty}^\infty \frac{1}{\left(t-\frac{1}{t}-\frac{1}{t+1}\right)^2+1} dt = \pi
$$
With help of a CAS, the integrand in leftmost integral can be expanded to
$$\frac{1}{\left(\left(\frac{1}{x}-x-\frac{x}{x+1}\right)^2+1\right)x^2} =
f(x) \stackrel{def}{=}
\frac{(x+1)^2}{x^6+4x^5+3x^4 - 4x^3 - 2x^2 + 2x + 1}$$
In terms of $f$, we have
$$\int_{-\infty}^\infty f(x) dx = \pi \quad\implies\quad
\int_0^{\infty} \frac{f(x) + f(-x)}{2} dx = \frac{\pi}{2}\tag{*2}$$
With help of a CAS again, one find
$$\frac{f(x) + f(-x)}{2} = \frac{x^8-4x^6+9x^4-5x^2+1}{x^{12}-10x^{10}+37x^8-42x^6+26x^4-8x^2+1}$$
The rightmost integral of $(*2)$ is nothing but the nasty integral in the link.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve for all real values of $x$ and $y$ Solve the system of equations for all real values of $x$ and $y$
$$5x(1 + {\frac {1}{x^2 +y^2}})=12$$
$$5y(1 - {\frac {1}{x^2 +y^2}})=4$$
I know that $0<x<{\frac {12}{5}}$ which is quite obvious from the first equation.
I also know that $y \in \mathbb R$ $\sim${$y:{\frac {-4}{5}}\le y \le {\frac 45}$}
I don't know what to do next.
|
for $$x,y\ne 0$$ we obtain
$$1+\frac{1}{x^2+y^2}=\frac{12}{5x}$$
$$1-\frac{1}{x^2+y^2}=\frac{4}{5y}$$
adding both we get
$$5=\frac{6}{x}+\frac{2}{y}$$
from here we obtain
$$y=\frac{2x}{5x-6}$$
can you proceed?
after substitution and factorizing we get this equation:
$$- \left( x-2 \right) \left( 5\,x-2 \right) \left( 5\,{x}^{2}-12\,x+9
\right)=0
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan \theta = \frac{5}{12}$$
Thus $$\sin\theta=\frac{5}{13}$$
But If I do like this , $$\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$$
$$ 2\sin\theta +2=3\cos \theta$$
$$ (2\sin\theta +2)^2=9\cos^2 \theta$$
$$ 4\sin^2\theta+8\sin\theta +4=9-9\sin^2\theta$$
$$13\sin^2\theta+8\sin\theta-5=0$$
Therefore I get two answers $$\sin\theta=\frac{5}{13} , \sin\theta =-1$$
What is the reason behind this ? Why am I getting two answers in one method and one in another ?
|
In the second solution, when $\sin \theta = -1$, what does $\cos \theta$ equal?
That's right: $0$. And you multiplied through by that. Doing so introduced a new (but wrong) solution.
To be more correct, the SQUARING introduced a second solution, as it often does; that solution happened to be invalid because it corresponded to a denominator that was zero.
A simpler example. Look at
$$
x = 2
$$
and square it to get
$$
x^2 = 4
$$
which has two solutions, $x = 2$ and $x = -2$.
|
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|
Prove the following trignometric inequality for all real $x$ Prove the following trignometric inequality for all $x \in \Bbb R$
$$x^2 \sin(x) + x \cos(x) + x^2 + {\frac 12} >0$$
take $x$ in the form of radians.
This particular question is the seventh question of the 1995 Indian RMO.
|
Put $t=\tan(\frac{x}{2})$. Now substitute it in the original expression. We have $$ x^{2}\sin x + x\cos x + x^{2} + \frac{1}{2}$$ $$= (1+\sin x)x^{2} + x\cos x+\frac{1}{2}$$ $$=(1 + \frac{2t}{1+t^{2}})x^{2} + x(\frac{1-t^{2}}{1+t^{2}}) + \frac{1}{2}$$ $$= \frac{(1+t)^{2}}{1+t^{2}}[x^{2} - x\frac{1-t^{2}}{(1+t)^{2}} + \frac{(1-t^{2})^{2}}{4(1+t)^{4}} + \frac{1+t^{2}}{4}]$$ $$= \frac{(1+t)^{2}}{1+t^{2}}[x-\frac{1-t^{2}}{2(1+t)^{2}}]^{2} + \frac{(1+t)^{2}}{4} > 0. $$ Hope it helps.
|
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|
Finding volume of a solid over region E, bounded below by cone $z = \sqrt{x^2 + y^2}$ and above by sphere $ z^2 = 1- x^2 - y^2$ I set up the triple integral as follows:
$v=\int^1_0\int^{\sqrt{x-1}}_{-\sqrt{x}}\int^{\sqrt{1-x^2-y^2}}_{\sqrt{x^2+y^2}}(1)dzdydx$ Then went to polar coords getting $$\int^{2\pi}_0\int_0^1(\sqrt{1-r^2}r=r^2)drd\theta$$$$2\pi(\frac{-1}{2})[\frac{2}{3}(1-r^2)^{3/2}-1]|^1_0=-\pi[(0-1)-(\frac{2}{3}-1)]=\pi+\frac{1}{3}$$
The answer should be $\frac{\pi}{3}(2-\sqrt{2})$
|
Note that the solid is of revolution, formed by rotating the circular sector between $z=\sqrt {1-x^2}$ and $z=x$ (about the $z$-axis).
From cylinder method, we have that
\begin{align}V&=2\pi \int_{0}^{\sqrt2 \over 2}x|\sqrt {1-x^2}-x|dx\\
&=2\pi({1\over3} - {\sqrt2\over6})
\end{align}
|
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|
Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$ There is supposed to be a clean solution to the integral below, maybe involving some symmetry
$$ \int^1_0 \frac{\ln(1+x)}{x}dx$$
I have tried integration by parts as followed:
$\ln(x+1)=u$ ,$\frac{1}{x+1} dx = du$ and also $\frac{1}{x}dx=dv$, $\ln(x)=v$. Then, the integral becomes
$$\int^1_0 \frac{\ln(1+x)}{x}dx=- \int^1_0 \frac{\ln x}{1+x}dx$$
which does not make this easier.
I have also tried using the identity $\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$. So let $I = \int^1_0 \frac{\ln(1+x)}{x}dx$ and also $I = \int^1_0 \frac{\ln(2-x)}{1-x} $. Then
$$2I= \int^1_0 \ln(1+x)+\ln(2-x)$$
which is not any easier, either.
Any ideas? :)
|
HINT:
use the expansion $$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$$
$$\frac{ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots$$
$$\int \frac{ln(1+x)}{x}\ dx=x-\frac{x^2}{2^2}+\frac{x^3}{3^2}-\frac{x^4}{4^2}+\ldots$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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|
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
|
Three answers:
i) $\frac n0$ is undefined. If you are ever given an expression $\frac 5{x+3}$ you know from the very beginning that it is impossible for these to be $\frac 50$ so it is impossible for $x +3 =0$. $x +3 \ne 0$ and $x \ne -3$.
So whatever you do, you know $x = -3$ will not be an acceptable solution.
So to get rid of the fractions you went from
$A: \frac 5{x+3} + \frac 2{2x +6} = 4$ to
$B: 5(2x+6) + 2(x+3) = 4(x+6)(x+3)$
These two equations are not the same. If $A$ is true then $B$ is true and you can get $B$ from $A$ but if $B$ is true you don't know that $A$ is true. All the solutions to $A$ are solutions to $B$, but $B$ has (maybe) more solutions than $A$ does. In $A$ it is impossible for $x + 3 = 0$ and for $2x + 6 = 0$. But in $B$ it is not.
This is could an "extraneous solution". You did this the correct way to solve but you need to note that when you multiply by an "unknown" to get rid of a denominator, you must check whether the "unknown" adds some "extraneous" solutions.
So you solve $B$ and get $x = -1.5$ or $x =-3$. But you already know that $x=-3$ is not an acceptable answer. It is an acceptable answer to $B$ but not to $A$. So the answer in $x = -1.5$.
ii) When you have $\frac ab + \frac cd = e$ so $ad +cb = ebd$ notice you only need to multiply by the greatest common multiple of $b$ and $d$. You don't have to multiply by all of $b$ and $b$.
To get rid of the denominator in
$\frac 5{x+3} + \frac 2{2x+6} = 4$
we don't have to multiply by $(x+3)(2x+6)$. $2x + 6 = 2(x+3)$ so we only have to multiply by $2x+6$.
$\frac 5{x+3} + \frac 2{2x+6} = 4$
$5*2 + 2 = 4(2x + 3)$.
Indeed we could have only multiplied by $x+3$ to get $5 + \frac 22 = 4(x+3)$.
iii) You could have reduced $\frac 2{2x+6} = \frac 1{x+3}$ from the start.
$\frac 5{x+3} + \frac 2{2x + 6} = 4$
$\frac 5{x+3} + \frac 1{x+3} = 4$
$\frac 6{x+3} =4 $
$6 = 4(x+3)$.
====
Note ii) and iii) show that this is not actually a quadratic problem.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove equilateral triangle in another triangle We got triangle $ABC$ ($AC = BC$). ∠$ACB = 120^\circ$. $AM = MN = NB = \frac 13AB$.
Prove that CNM is equilateral triangle.
Any ideas how to do that? I'd be grateful if someone helps me!
P.S.I am sorry that the drawing is not very accurate!
|
Let us use complex numbers. There is no loss of generality in assuming $C$ as the origin, $X$ axis along $CA$ and $A$ is the point $(1,0)$. Then $B$ is represented by the complex number $e^{\frac{2\pi i }{3}} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$. $M$ and $N$ are respectively
$$m = \frac{1 \cdot e^{\frac{2\pi i }{3}} +2 \cdot 1}{3} = \frac{\frac{3}{2} + \frac{i\sqrt{3}}{2}}{3}$$
and
$$n = \frac{2 \cdot e^{\frac{2\pi i }{3}} +1 \cdot 1}{3} = \frac{i\sqrt{3}}{3}$$
Since
$$m = \left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)n$$
It follows that $M$ can be obtained by rotating $CN$ clockwise by $60^\circ$ and consequently, the triangle $CMN$ is equilateral.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2046651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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|
Calculating $\int_0^{\pi/3}\cos^2x+\dfrac{1}{\cos^2x}\mathrm{d}\,x$ I've been given the following exercise:
Show that the exact value of $$\int_0^{\pi/3}\cos^2x+\frac{1}{\cos^2x}\,\mathrm{d}x = \frac{\pi}{6}+\frac{9}{8}\sqrt{3}$$
Can someone help me with this?
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We can separate the integral into $$\int_{0}^{\frac{\pi}{3}} \cos^{2}(x) + \sec^{2}(x) dx = \int_{0}^{\frac{\pi}{3}}\cos^{2}(x)dx +\int_{0}^{\frac{\pi}{3}} \sec^{2}(x) dx =I_1 + I_2$$. For the first integral we have $$ I_1 = \int_{0}^{\frac{\pi}{3}} \cos^{2}(x) dx =\int_{0}^{\frac{\pi}{3}} \frac{1+\cos 2x}{2} dx = \frac{2x+\sin 2x}{4}$$ For the second integral, we have $$I_2 =\int_{0}^{\frac{\pi}{3}} \sec^{2}(x) dx = \tan x $$. Substitute the limits $0$ to $\frac{\pi}{3}$ We have $\frac{\pi}{6}+\frac{\sqrt3}{8} + \sqrt3 = \frac{\pi}{3} + \frac{9\sqrt3}{8}$. We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2047646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving $(1+x^2)y' - 2xy = (1+x^2)\arctan(x)$ I'm asked to solve the differential equation: $$(1+x^2)y' - 2xy = (1+x^2)\arctan(x).$$
I rewrite it:
$$y' - \frac{2x}{1+x^2}y = \arctan(x).$$
The integrating factor is:
$$e^{-\int{\frac{2x}{1+x^2}}dx} = e^{-\ln{1+x^2}} = (e^{ln{1+x^2}})^{-1} = \frac{1}{1+x^2}.$$
$$\frac{1}{1+x^2}y' - \frac{1}{1+x^2}\frac{2x}{1+x^2}y = \arctan(x)\frac{1}{1+x^2}$$
$$\frac{d}{dx}(\frac{1}{1+x^2}y) = \arctan(x)\frac{1}{1+x^2}$$
So to solve this equation I need to find the primitive of
$$\arctan(x)\frac{1}{1+x^2}$$
But I don't think I can do that yet, is there any other way to solve this differential equation?
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Take into account that
$$
\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}
$$
so that
$$
\int dx\; \frac{\arctan(x)}{1+ x^2} = \int dx\; \arctan(x) \frac{d}{dx}\arctan(x) = \int dx\; \frac{d}{dx}\left(\frac{1}{2}\arctan^2(x)\right) = \frac{1}{2}\arctan^2(x) + c
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2048151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.