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Proving Trigonometric Equality I have this trigonometric equality to prove:
$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$
I started with the left hand side, reducing the fractions to common denominator and got this:
$$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos x+\cos x(\frac{\sin x}{\cos x})-\sin x+\sin x(\frac{\sin x}{\cos x})}{1-\left(\frac{\sin^2x}{\cos^2x}\right)}\\=\frac{\cos x+\left(\frac{\sin^2x}{\cos x}\right)}{1-\frac{\sin^2}{\cos^2x}}$$and by finding common denominator top and bottom and then multiplying the fractions i got: $$\frac{\cos^2x}{\cos^3x-\cos x\sin^2x}$$ which is far from the right hand side and I don't know what am I doing wrong. What is the correct way to prove this equality?
|
Here's a hint: $$\frac{\cos x + \frac{\sin x}{\cos x}^2}{1 - \left(\frac{\sin x}{\cos x}\right)^2} = \frac{\frac{\sin x^2 + \cos x^2}{\cos x}}{\frac{\cos x^2 - \sin x^2}{\cos x^2}}.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2299015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
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|
Number equal to the sum of digits + product of digits) Are every number equal to (sum of digits + product of digits) in a given base only two digits long ?
Thought about limiting like this :
$$b^{(n - 1)} \leq N = \text{Product digits} + \text{Sum digits} \leq (b - 1)*n + (b - 1)^{(n)}$$
Obviously the set is finite (when $n \rightarrow +\infty$) but can't figure out how to get more precision about the set
Any idea ?
ie in base 10 : {0,19,29,39,49....,99}
Edit : 'base' abreviation : b. Number of digits for a number N : n
|
Bear with me:
For $k \ge 1; b\ge 2$ $(b - 1)^k \le b^k + (-1)^k$. Pf: induction: For $k =1$ then $(b-1)^1 = b^1 + (-1)^1$. For $k = 2$ then $(b-1)^2 = b^2 -2b +1 < b^2 + 1$. If $(b -1)^n \le b^n + (-1)^n$ then $$(b-1)^{n+1} = (b-1)^n(b-1) \le (b^n + (-1)^n)(b-1) = b^{n+1} + (-1)^nb - b^n +(-1)^{n+1} = b^{n+1} + (-1)^{n+1} - b(b^{n-1} + (-1)^n) \ge b^{n+1} + (-1)^{n+1}$$. With equality holding only if $k = 1$. But for $k = 0 ; (b-1)^0 = 1 > 0 = b^0 - 1^0$.
====
Let $n = \sum_{i=0}^k a_i b^k$ be a $k+1$ digit number and $S = \sum_{i=0}^k a_i$
$n - S = \sum_{i=0}^k a_i (b^k - 1) \ge \sum_{i=1}^k a_i (b - 1)^k$.
(we must can/must index from $1$ on the RHS because $a_0*(b^k -1) = 0$)
Now each $a_i \le (b-1)$ so $P = \prod_{i=0}^k a_i = a_k*(b-1)^k$
So $n - S \ge \sum_{i=1}^{k}a_i(b-1)^k$
$= a_k*(b-1)^k+ \sum_{j=1}^{k-1}a_j(b-1)^k$
$\ge P$
And equality only holds if
i) $\sum_{j=1}^{k-1}a_j(b-1)^k= 0$
either a) all $a_j; 1\le j \le k$ are equal to $0$ or b) $k -1 < 1$ i.e. $k < 2$ and $n $ is a one or two digit number.
If a) then $P =0$ and $n - S=0$ which can only happen if $n$ is a one digit number. ($a_i*b^i > a_i$ if $a_i > 0; i > 0$). So b) must hold.
b) If $n$ is a one- digit number then $n - S = 0 = P$ so $P = 0$ and $n= 0$.
So this only true if $n$ is a two digit number or $n = 0$.
ii) Furthermore, If $n$ is a two-digit number this is only true if:
$n = a_1*b + a_0 = a_1 + a_0 + a_1*a_0; a_1 \ne 0$
$a_1(b - (a_0-1)) = 0$
$a_1 \ne 0$ so $a_0 = b- 1$
So the complete set of solutions for $n$ are $\{0, a_1*b + (b-1)|1 \le a_1 < b\}$
|
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"url": "https://math.stackexchange.com/questions/2304977",
"timestamp": "2023-03-29T00:00:00",
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|
Galois correspondence for the splitting field of $x^3-3$ over $\mathbb{Q}$ I am struggling to find the fixed fields, am I missing a trick here or is it really this nitty gritty?
By the tower law $[\mathbb{Q}(\sqrt[3]{3}, \zeta_3):\mathbb{Q}] = [(\mathbb{Q}(\sqrt[3]{3}, \zeta_3): \mathbb{Q}(\sqrt[3]{3})][\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}] = 2\times3$ = 6, since the minimal polynomial for each respective extension is $\Phi_3$, $x^3-3$.
We also know that $E/F:=\mathbb{Q}(\sqrt[3]{3},\zeta_3)/\mathbb{Q}$ is Galois since each element $\alpha$ has a separable polynomial in $\mathbb{Q}[x]$ which splits completely in $\mathbb{Q}(\sqrt[3]{3},\zeta_3)[x]$, namely $x^3-3$, $\Phi_3$, respectively.
So invoking the fundamental theorem we can write $\operatorname{Gal}(E/F) =[\mathbb{Q}(\sqrt[3]{3}, \zeta_3):\mathbb{Q}] = 6$.
Furthermore, we have the injection $\operatorname{Gal}(E/F) \to S_3$ since $E$ is the splitting field for $x^3-3$.
Hence, $\operatorname{Gal}(E/F) \cong S_3$. Let $\{\alpha_1,\alpha_2,\alpha_3\}:= \{\sqrt[3]{3},\sqrt[3]{3}\zeta_3,\sqrt[3]{3}\zeta_3^2\}$ be an enumeration of the roots
Using the group structure of $S_3$ we know that the following maps are automorphisms of $E$ fixing the base field
\begin{align*}
\sigma_1 = ID,\hspace{0.2cm}
&\sigma_2: \begin{cases}
\alpha_1 \mapsto \alpha_2\\
\alpha_2 \mapsto \alpha_3\\
\alpha_3 \mapsto \alpha_1
\end{cases}
\sigma_3: \begin{cases}
\alpha_1 \mapsto \alpha_3\\
\alpha_2 \mapsto \alpha_1\\
\alpha_3 \mapsto \alpha_2
\end{cases}
\sigma_4: \begin{cases}
\alpha_1 \mapsto \alpha_2\\
\alpha_2 \mapsto \alpha_1\\
\alpha_3 \mapsto \alpha_3
\end{cases}\\
&\sigma_5: \begin{cases}
\alpha_1 \mapsto \alpha_3\\
\alpha_2 \mapsto \alpha_2\\
\alpha_3 \mapsto \alpha_1
\end{cases}
\sigma_6: \begin{cases}
\alpha_1 \mapsto \alpha_1\\
\alpha_2 \mapsto \alpha_3\\
\alpha_3 \mapsto \alpha_2
\end{cases}
\end{align*}
The subgroups of $S_3$ are 3 copies of $S_2$ and $A_3$. These are given by
\begin{align*}
H_1 = \langle \sigma_4 \rangle, \hspace{0.2cm}
H_2 &= \langle \sigma_5 \rangle, \hspace{0.2cm}
H_3 = \langle \sigma_6 \rangle\\
H_4 &= \langle \sigma_2, \sigma_3 \rangle
\end{align*}
The fixed fields are given by
\begin{align*}
E^{\langle \sigma_4 \rangle} &= \mathbb{Q}(\alpha_3,\alpha_1+\alpha_2)\\
E^{\langle \sigma_5 \rangle} &=\mathbb{Q}(\alpha_2,\alpha_1+\alpha_3)\\
E^{\langle \sigma_6 \rangle} &=\mathbb{Q}(\alpha_1,\alpha_2+\alpha_3) \\\
E^{\langle \sigma_2, \sigma_3 \rangle} &= \mathbb{Q}(\alpha_1 + \alpha_2 + \alpha_3) = \mathbb{Q}
\end{align*}
$\Phi_3 = x^2 + x + 1$, so by the Quadratic formula $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, it has roots $\frac{-1 \pm i\sqrt{3}}{2}$. The roots are exactly the roots of unity, so
\begin{align*}
\frac{-1 + i\sqrt{3}}{2} &= \zeta_3\\
\frac{-1 - i\sqrt{3}}{2} &= \zeta_3^2
\end{align*}
The roots of $x^3-3$ are therefore
\begin{align*}
\alpha_1 &=\sqrt{3}\\
\alpha_2 &=\frac{-\sqrt{3} + i3}{2} = \sqrt{3}\zeta_3\\
\alpha_3 &=\frac{-\sqrt{3} - i3}{2} = \sqrt{3}\zeta_3^2
\end{align*}
From this we can easily find the sums
\begin{align*}
\alpha_1 + \alpha_2 &= \sqrt{3} + \frac{-\sqrt{3} + i3}{2}\\
&= \frac{\sqrt{3} + i3}{2}\\
\alpha_1+\alpha_3 &= \sqrt{3} + \frac{-\sqrt{3} - i3}{2}\\
&=\frac{\sqrt{3} - i3}{2}\\
\alpha_2+\alpha_3 &= \frac{-\sqrt{3} + i3}{2} +\frac{-\sqrt{3} - i3}{2}\\
&= -\sqrt{3}
\end{align*}
So the fixed fields are
\begin{align*}
E^{\langle \sigma_4 \rangle} &= \mathbb{Q}(i,\sqrt{3})\\
E^{\langle \sigma_5 \rangle} &=\mathbb{Q}(i,\sqrt{3})\\
E^{\langle \sigma_6 \rangle} &=\mathbb{Q}(\sqrt{3}) \\\
E^{\langle \sigma_2, \sigma_3 \rangle} &= \mathbb{Q}(\alpha_1 + \alpha_2 + \alpha_3) = \mathbb{Q}
\end{align*}
Is my process for finding the fixed fields correct? Thanks for your time!
|
How can $\sigma_4$ and $\sigma_5$ have the same fixed field?
The fixed field of $\sigma_4$ is $\Bbb Q(\alpha_3)=\Bbb Q(\zeta_3^2
\sqrt[3]3)$.
The fixed field of $\sigma_5$ is $\Bbb Q(\alpha_2)=\Bbb Q(\zeta_3
\sqrt[3]3)$.
The fixed field of $\sigma_6$ is $\Bbb Q(\alpha_1)=\Bbb Q(\sqrt[3]3)$.
The fixed field of $\sigma_2$ is $\Bbb Q(\sqrt{-3})=\Bbb Q(i\sqrt3)$.
$E$ contains neither $i$ nor $\sqrt3$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Radius of converge of Laurent series for $\frac{1}{\sin z}$ I want to show that the radius of convergence of Laurent series for $\frac{1}{\sin z}$ is $\pi$
I showed that:
\begin{align}
\frac 1 {\sin z} & = \frac 1 {z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots } = \frac 1 {z\big(1 - (\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots )\big)} \\[10pt]
& = \frac{1}{z} \bigg( 1 + \bigg(\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg) + \bigg(\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg)^2 + \cdots \bigg) \\[10pt]
& = \frac{1}{z} +\frac{z}{3!} + \bigg( \frac{1}{3!}- \frac{1}{5!}\bigg)z^3 + \cdots
\end{align}
This equality holds if and only if:
$$\bigg|\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg| < 1
\iff \bigg|\frac{\sin z}{z} - 1\bigg| < 1$$
How can I show that the last inequality implies $|z| < \pi$ ?
|
Since $\text{Res}\left(\frac{1}{\sin z},z=k\pi\right) = (-1)^m $ for any $k\in\mathbb{Z}$, we have
$$ \frac{1}{\sin z} = \frac{1}{z}+\sum_{m\geq 1}(-1)^m\left(\frac{1}{z-m\pi}+\frac{1}{z+m\pi}\right) \tag{1}$$
over any compact subset of $\mathbb{C}\setminus\pi\mathbb{Z}$. In particular:
$$ \frac{1}{\sin z}-\frac{1}{z} = 2z\sum_{m\geq 1}\frac{(-1)^{m+1}\frac{1}{m^2 \pi^2}}{1-\frac{z^2}{m^2\pi^2}}=2z\sum_{m\geq 1}z^{2m-2}\frac{\eta(2m)}{\pi^{2m}} \tag{2} $$
the radius of convergence of the Laurent series at the origin is $\pi$ (due to $\eta(2m)\approx 1$ for large values of $m$), as expected since $\pi$ is the distance from the origin of the closest singularities (simple poles at $z=\pm \pi$).
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2309549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
AM-GM Inequality concept challenged! If $a,b,c$ are positive numbers and $abc=9$, then find the minimum value of $(1+a)(1+b)(1+c)=?$
Someone's approach is
but for this to be possible, $a=b=c=1$. which doesn't satisfies $abc=9$.
Where does the above method go wrong and what should be the correct answer?
|
By AM-GM
\begin{align}
1+a &\geq 2 \sqrt{a} \\
1+b &\geq 2 \sqrt{b} \\
1+c &\geq 2 \sqrt{c}
\end{align}
Thus
\begin{align}
(1+a)(1+b)(1+c) &\geq 2^3 \sqrt{a} \sqrt{b}\sqrt{c} \\
&= 8.\sqrt{9} \ldots
\end{align}
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2311772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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|
Arclength, finding definite integral Find the length of the segment between $x=5$ and $x=2$ for the following curve.
$$f(x)= \frac{1}{2}x^2-\frac{1}{4}\log\left(x\right)$$
My working so far.
$$f'\left(x\right)=x-\frac{1}{4x}.$$
Then
$$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)\right)^2}\,dx
=\int _2^5\:\sqrt{x^2+\frac{1}{2}+\frac{1}{16x^2}}\,dx.$$
However I have no idea how to find the solution to the definite integral.
|
Hint. We have that
$$1+\left(x-\frac{1}{4x}\right)^2=\frac{(4x)^2+16x^4-8x^2+1}{(4x)^2}=\frac{16x^4+8x^2+1}{(4x)^2}=\left(\frac{4x^2+1}{4x}\right)^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2311852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that the product is never a perfect square
Prove that for nonnegative integers $x_1,\ldots,x_{2011}$ and $y_1,\ldots,y_{2011}$ the product $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_{2011}^2+3y_{2011}^2)$$ is never a positive perfect square.
I thought about generalizing this question to any odd subscript $n$ instead of $2011$. Thus, $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_n^2+3y_n^2)$$ is never a perfect square. For $n = 1$ we have $2x^2+3y^2 = z^2$ and I want to show the only solution is $x = y = z = 0$. If $x$ is even, then $y$ must be even by taking modulo $4$. If $x$ is odd, then $y$ must be odd. I didn't see how to continue from here.
|
Case n=1
Let $z^2=2x^2+3y^2$
$z^2=2x^2 [3]$.
If x is not divisible by 3, then $z^2=2[3]$ which is impossible. Then $x$ is divisible by 3, and consequently $z$ is also a multiple of 3.
Let us look at the prime decomposition of $z$ such that $z^2=2x^2+3y^2$. 3 is a prime factor of $z$ and let us not $p$ its power: $z=3^p z'$
Let us note $x=3^a x'$ and $y=3^b y'$, then $$3^{2p} z'^2 = 2x'^2 3^{2a}+y'^2 3^{2b+1}$$ If we look at this equation modulo $3^{\min(2p, 2a, 2b+1)}$ we get a contradiction.
Indeed, if $2a=2p$ we obtain $z'^2 = 2x'^2+y'^2 3^{(2b+1)-2a}$ which gives $1=2[3]$. If $2a \neq 2p$, dividing by $3^{\min(2p, 2a, 2b+1)}$ leads to an equation $u=v+w$ with exactly 2 numbers among $u$, $v$ and $w$ which are divisible by 3, which is impossible.
This proves the case $n=1$.
General case
For the general case, let us note that the whole product is equal to 0 mod 3 iff one $x_i$ is a multiple of 3. ($2x^2+3y^2$ is 0 modulo 3 iif $2x^2=0[3]$). The product cannot be equal to 1 modulo 3 since we have an odd number of factors (and $2x^2+3y^2$ = 0 or 2). Then the same argument as before shows that $z$ is divisible by 3.
We have $z=3^p z'$ and $3^{2p}z'^2 = \Pi_{i=1}^n (2x_i^2+3y_i^2)$
The factors $2x_i^2+3y_i^2$ are of 2 different types:
*
*the ones with $x_i \neq 0 [3]$ and in that case $(2x_i^2+3y_i^2)= 2[3]$
*the ones with $x_i = 0 [3]$, then the power $p_i$ of 3 in the prime decomposition of $2x_i^2+3y_i^2$ is even or odd. We have two subcases:
*
*if $p_i$ is odd then $(2x_i^2+3y_i^2) / 3^{p_i} = 1[3]$
*if $p_i$ is even then $(2x_i^2+3y_i^2) / 3^{p_i} = 2[3]$
We must have an even number of odd $p_i$, because $2p=\sum_{i} p_i$. Then when we divide the equation by $3^{2p}=\Pi_i 3^{p_i}$ and we take it modulo 3 we have a product off an even number of factors that are 1 modulo 3 and an odd number of factors that are 2 modulo 3. Consequently the right hand side is 2 modulo 3, while the left hand side (a perfect square) is equal to 1 modulo 3. Hence we have a contradiction.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ..........$ $1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ......$
Can anyone help me out how to solve this.
My try : I was thinking about the expansion of $tan^{-1}x$. But in that series positive and negative will come alternatively.
|
A different approach would be to use binomial theorem . Consider the series expansion of $\ln (\frac {1}{1-x}),\ln (\frac {1}{1+x})$ using binomial expression for negative index we have $\ln (\frac {1}{1-x})-\ln (\frac {1}{1+x})=2 (x+\frac {x^3}{3}+\frac {x^5}{5}+.. ) $ now pull out $x $ common from $RHS $ and put $x=\frac {1}{2} $ thus we have $\frac {1}{2x}\ln(\frac {1}{1-x})-\ln(\frac {1}{1+x})=S $ thus $S=\ln (3) $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2313339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).
|
Here's yet another way, for those who aren't comfortable with the $\gtrless$ or $\sim$ notation.
We can use crude rational approximations to $\sqrt 2$ and $\sqrt 3$.
$$\begin{align}
\left(\frac{3}{2}\right)^2 = \frac{9}{4} & \gt 2\\
\frac{3}{2} & \gt \sqrt 2\\
\frac{9}{2} & \gt 3\sqrt 2
\end{align}$$
And
$$\begin{align}
\left(\frac{7}{4}\right)^2 = \frac{49}{16} & \gt 3\\
\frac{7}{4} & \gt \sqrt 3\\
\frac{7}{2} & \gt 2\sqrt 3
\end{align}$$
Adding those two approximations, we get
$$\begin{align}
\frac{9}{2} + \frac{7}{2} = 8 & \gt 3\sqrt 2 + 2\sqrt 3\\
6 + 2 & \gt 3\sqrt 2 + 2\sqrt 3\\
6 - 2\sqrt 3 & \gt 3\sqrt 2 - 2
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2317625",
"timestamp": "2023-03-29T00:00:00",
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|
If $A$ is a non-square matrix with orthonormal columns, what is $A^+$? If a matrix has orthonormal columns, they must be linearly independent, so $A^+ = (A^T A)^{−1} A^T$ . Also, the fact that its columns are orthonormal gives $A^T A = I$. Therefore,
$$A^+ = (A^T A)^{−1} A^T = (I)^{-1}A^T = A^T$$
Thus, $A^+ = A^T$. Am I correct? Thank you.
|
Problem statement
Start with a matrix $$A\in\mathbb{C}^{m\times n}$$ where $m>n$, and
a valid statement for the pseudoinverse matrix
$$
\mathbf{A}^{+} = \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*}
$$
We know (see links) that this matrix is a left inverse:
$$
\mathbf{A}^{+} \mathbf{A} = \mathbf{I}_{n}
\tag{1}
$$
If we add the constrain that the column vectors of $\mathbf{A}$ are orthonormal, we also have
$$
\mathbf{A}^{*} \mathbf{A} = \mathbf{I}_{n}
\tag{2}
$$
Conclusion
The results $(1)$ and $(2)$ suggest the identity
$$
\mathbf{A}^{+} \mathbf{A} = \mathbf{I}_{n} = \mathbf{A}^{*} \mathbf{A}
$$
from which we conclude that
$$
\mathbf{A}^{+} = \mathbf{A}^{*}
$$
Example
$$
\mathbf{A} =
\frac{1}{\sqrt{2}}
\left[
\begin{array}{cr}
i & -1 \\
i & 1 \\
0 & 0 \\
\end{array}
\right],
\qquad
\mathbf{A}^{*} =
\frac{1}{\sqrt{2}}
\left[
\begin{array}{rrc}
-i & -i & 0 \\
-1 & 1 & 0 \\
\end{array}
\right]
$$
$$
\mathbf{A}^{+} \mathbf{A} =
\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right] = \mathbf{I}_{2},
\qquad
\mathbf{A} \mathbf{A}^{+} =
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
\ne \mathbf{I}_{3}
$$
$$
\mathbf{A}^{*} \mathbf{A} =
\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right] = \mathbf{I}_{2},
\qquad
\mathbf{A} \mathbf{A}^{*} =
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
\ne \mathbf{I}_{3}
$$
Background reading
Categorize the pseudoinverse matrix in terms of left and right inverses:
generalized inverse of a matrix and convergence for singular matrix, What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank?
General properties of the pseudoinverse matrix: Moore–Penrose pseudo-inverse Reference.
|
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|
Jordan form dependency on choosing a specific kernel. Let $A=\begin{pmatrix}2 && 2 && 3 \\ 1 && 3 && 3\\ -1 && -2 && -2\end{pmatrix}$
Finding the eigenvalues produce that $\lambda_1=\lambda_2=\lambda_3=1$
The geometry of $A$ is $2$, therefore exists only two blocks, and therefore the only possibility is $J_1(1),J_2(1)$.
This produce the following Jordan matrix: $J=\begin{pmatrix}1 && 1 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{pmatrix}$
Now choosing one basis of kernel of $A-I$, Let $B_1=\{v_1=\begin{pmatrix}3 \\ 3 \\ -3\end{pmatrix},v_2=\begin{pmatrix}0\\1\\-2/3\end{pmatrix}\}$ be such base (you can check it's basis, it's linearly independence and applying both vectors to $A-I$ gives $0$.
We can find the Jordan form, since solutions to $(A-I)w=v_1$ exist
However choosing $B_2=\{u_1=\begin{pmatrix}2 \\ -1 \\ 0\end{pmatrix},u_2=\begin{pmatrix}3 \\0\\-1\end{pmatrix}\}$, solution to $(A-I)w=u_1$ or $(A-I)w=u_2$ don't exist.
How can I find the Jordan form in the 2nd base? how come choosing one basis allows Jordan form to exist (or at-least allows for chaining of generalized eigenvectors) while the other doesn't exist?
EDIT:
$\begin{pmatrix}1 && 2 && 3 \\ 1 && 2 && 3\\ -1 && -2 && -3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}2 \\ -1 \\ 0\end{pmatrix}$ gives no solution since it requires $x+2y+3z$ to have two distinct solutions.
$\begin{pmatrix}1 && 2 && 3 \\ 1 && 2 && 3\\ -1 && -2 && -3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}3 \\ 0 \\ -1\end{pmatrix}$ gives no solution since it requires $x+2y+3z$ to have two distinct solutions.
Therefore if I wanted to chain using $B_2$, no solutions exist, however it's possible to chain using $B_1$. The question is, why does this happen? how can I ensure that I find a generalized eigenvector where this does not happen.
|
The procedure for finding the basis is to start from a generalized eigenvector, that is,
a vector $v_2$ such that $(A-I)^2v_2=0$, but $(A-I)v_2\ne 0$. Since $(A-I)^2=0$, you can take any vector in $\Bbb{R}^3\setminus Ker(A-I)$.
Then $v_1=(A-I)v_2$ will be an eigenvector of $A$.
Note that $Im(A-I)$ is one-dimensional generated by $\begin{pmatrix} 1\\ 1\\ -1\end{pmatrix}$, so $v_1$ will always be
a multiple of $\begin{pmatrix} 1\\ 1\\ -1\end{pmatrix}$.
Note also that $v_2$ cannot be in $Ker(A-I)$, since in that case $(A-I)v_2=0$.
In order to complete the basis, take another vector $v_3$ in $Ker(A-I)$ which is not in the subspace generated by $v_1$.
If you want to search directly for $v_1,v_2$ with $v_1=(A-I)v_2$, you have to note that $v_1$ is in the image of
$A-I$, which is one dimensional, so it is a multiple of $\begin{pmatrix} 1\\ 1\\ -1\end{pmatrix}$.
So in your example you can take $v_2=\begin{pmatrix} 1\\ 0\\ 0\end{pmatrix}$, then $v_1=(A-I)v_2=\begin{pmatrix} 1\\ 1\\ -1\end{pmatrix}$, and take $v_3$ another vector of $Ker(A-I)$, for example $v_3=\begin{pmatrix} 2\\ -1\\ 0\end{pmatrix}$.
Then the matrix of $A$ with respect to the basis $\{v_1,v_2,v_3\}$ will be $\begin{pmatrix} 1&1&0\\ 0&1&0\\0&0& 1\end{pmatrix}$, since $A(v_1)=v_1$, $A(v_2)=v_1+v_2$ and $Av_3=v_3$.
|
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|
Find all real coefficient polynomial such $f(x)|f(x^2-2)$
Find all real coefficient polynomial with the degree is $3$,and such
$$f(x)|f(x^2-2),\forall x\in Z $$
I try: let $f(x)=ax^3+bx^2+cx+d$,then $$f(x^2-2)=ax^6-6ax^4+12ax^2-8a+bx^4-4bx^2+4b+cx^2-2c+d$$,then such
$$(ax^3+bx^2+cx+d)|ax^6-6ax^4+12ax^2-8a+bx^4-4bx^2+4b+cx^2-2c+d,\forall x\in Z $$following I can't it
|
Suppose $f(x) \mid f(x^2 - 2)$. Ghartal's argument has already shown that if $x$ is a root of $f$, then $x^2 - 2$ is also a root of $f$. In fact, by the same argument, we have that for all $x \in \mathbb{C}$, the multiplicity of $x^2 - 2$ as a root of $f$ is at least the multiplicity of $x$ as a root of $f$.
I claim that the above condition is also sufficient to have $f(x)\mid f(x^2 - 2)$. Indeed, we may write $f$ as
$$
f(x) = k\prod_{c \in \mathbb{C}} (x - c)^{m_c}
$$
for some $k \in \mathbb{C}$ (w.l.o.g. $k \neq 0$) and where $m_c \in \mathbb{N}$ is non-zero for finitely many $c \in \mathbb{C}$. Then
$$
f(x^2 - 2) = k\prod_{c \in \mathbb{C}} (x^2 - 2 - c)^{m_c} .
$$
Now by hypothesis we have $m_{c^2 - 2} \geq m_{c}$ for all $c \in \mathbb{C}$. Hence $(x -c)^{m_c} \mid (x^2 - 2 - (c^2 - 2))^{m_{c^2 - 2}}$ for all $c \in \mathbb{C}$. In particular, $f(x) \mid f(x^2 - 2)$. This proves the claim.
To find all degree three polynomials (in $\mathbb{R}[X]$, or $\mathbb{C}[X]$ if you wish) with the given property, it suffices to "trace back" the function $x \mapsto x^2 - 2$ three steps and then take products of the induced linear polynomials; respecting the claim above. The function $x \mapsto x^2 - 2$ does:
$$
+/- \sqrt{3} \mapsto 1 \mapsto - 1 \qquad \text{and} \qquad 0 \mapsto -2 \mapsto 2.
$$
so the possible polynomials of degree at most three are (up to multiplying by a constant):
*
*$(x + 1)$
*$(x - 2)$
*$(x + 1)^2$
*$(x - 2)^2$
*$(x+1)(x-2)$
*$(x-1)(x+1)$
*$(x+2)(x-2)$
*$(x+1)^3$
*$(x - 2)^3$
*$(x - 2)(x+1)^2$
*$(x - 2)^2 (x + 1)$
*$(x-1)(x+1)^2$
*$(x + 2)(x - 2)^2$
*$(x - \sqrt{3})(x-1)(x+1)$
*$(x + \sqrt{3})(x-1)(x+1)$
*$x(x + 2)(x - 2)$
*$(x + 2)(x - 2)(x + 1)$
*$(x - 1)(x + 1)(x - 2)$
The polynomials of degree exact three are the last 11.
|
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|
Find Angle Between Two Curves at Point of Intersection Find angle between these two curves at point of intersection :
$$K_1: x^2y^2 + y^4 = 1$$ and
$$K_2 : x^2 + y^2 = 4 $$
Thanks!
|
To find the points of intersection, we see that the equation of the curve $K_1$ can be written as:
\begin{align*}
x^2 y^2 + y^4 &= 1 \\
y^2 (x^2 + y^2) &= 1
\end{align*}
Since the points of intersection lies on both curve $K_1$ and $K_2$, it must satisfy both the above equation and the equation of $K_2$, meaning:
\begin{align*}
y^2 (x^2 + y^2) &= 1 \\
y^2 \cdot 4 &= 1 \\
y^2 &= \frac{1}{4}
\end{align*}
Plugging this back into the equation for $K_1$, we get:
\begin{align*}
x^2 + \frac{1}{4} &= 4 \\
x^2 &= \frac{15}{4}
\end{align*}
So we have four points of intersection, corresponding to the four combinations of:
$$ x = \pm \frac{\sqrt{15}}{2} \qquad y = \pm \frac{1}{2} $$
The angle of intersection of the curves is the same as the angle between the tangent line of $K_1$ and the tangent line of $K_2$ at the point of intersection.
Looking at the slope of the curve $K_1$, we take the equation of the curve and perform implicit differentiation:
\begin{align*}
x^2 y^2 + y^4 &= 1 \\
\frac{d}{dx} \left[ x^2 y^2 + y^4 \right] &= \frac{d}{dx} 1 \\
2x y^2 + 2 x^2 y \frac{dy}{dx} + 4y^3 \frac{dy}{dx} &= 0 \\
\frac{dy}{dx} &= -\frac{2xy^2}{2x^2 y + 4y^3} \\
&= -\frac{xy}{x^2 + 2y^2}
\end{align*}
Looking at the slope of the curve $K_2$, seeing as it is a circle, we can verify that implicit differentiation gives us:
$$ \frac{dy}{dx} = - \frac{x}{y} $$
Regardless of which of the four intersection points we choose, the slopes have the same sign, as seen by the formulas, so we have two possible pairs of values for the slopes:
$$ (m_1,m_2) = \pm \left(\frac{\sqrt{15}}{17} , \sqrt{15} \right)$$
Turning these slopes into angles of inclinations by taking the arctangent, we get:
$$ (\theta_1, \theta_2) = \pm \left( \arctan \frac{\sqrt{15}}{17} , \arctan \sqrt{15} \right)$$
The angle between the tangent lines is the difference between the inclination angles of the tangent lines, which results in:
$$ \theta = \arctan \sqrt{15} - \arctan \frac{\sqrt{15}}{17} $$
Another way to get an equivalent answer is to deal with the normal vectors of the tangent lines and find the angle between the vectors. Since we have seen that the intersection angle at all four points is the same, the angle we care about can be seen as the angle between the normal vectors:
$$ \vec{n}_1 = \langle \sqrt{15}, 1 \rangle \qquad \vec{n}_2 = \left\langle \frac{\sqrt{15}}{17}, 1 \right\rangle $$
which ends up resulting in:
\begin{align*}
\cos \theta &= \frac{\vec{n}_1 \cdot \vec{n}_2}{\lVert \vec{n}_1 \rVert \lVert \vec{n}_2 \rVert} \\
&= \frac{\frac{15}{17} + 1}{4 \cdot \frac{4 \sqrt{19}}{17}} \\
&= \frac{15 + 17}{16 \sqrt{19}} \\
&= \frac{2}{\sqrt{19}}
\end{align*}
$$ \theta = \arccos \frac{2}{\sqrt{19}} $$
|
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|
To find the last digit of the number $4^{7^5}$? To find the last digit of the number $4^{7^5}$, I use the following method.
first, we know $4\equiv_{10}4,4^2\equiv_{10}6,4^3\equiv_{10}4,...,4^{2n}\equiv_{10}6,4^{2n+1}\equiv_{10}4$
Then$7^1\equiv_21,7^2\equiv_21,7^3\equiv_21,...,7^5\equiv_21$
Therefore, $4^{7^5}\equiv4^{2k+1}\equiv_{10}4$.(because of the remainder 1)
and if to find the last digit of $4^{8^3}$ instead, the answer will be 6. Because $8^n\equiv_20,n\in \mathbb{N^+}$ so $8^n=2k,k\in\mathbb{N^+}$. From above, $4^{2k}\equiv_{10}6$
Is my method correct?
|
Your method is correct. Another way is to consider the easy fact that
$$2^{4n+k}= \begin{cases}2\pmod {10}\text{ for}\space k=1\\4\pmod {10}\text{ for}\space k=2\\8\pmod {10}\text{ for}\space k=3\\6\pmod {10}\text{ for}\space k=0\end{cases}$$ and the calculation $$7^5=16807$$ Then $$4^{7^{5}}=2^{33614}=2^{33612+2}=4\pmod{10}$$
|
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|
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&-8&6\\ 0&-1/2&-3/2&1 \end{array}\right)$$
What am I doing wrong? Steps would be much appreciated.
|
I think Jose comes closest to solving this problem in the way that I would, but his answer doesn't expand too much on the why of it all, so I'm writing up a separate answer.
If there is such a $D$ and $P$ such that $A = P^T D P$, then we say that $A$ is diagonalizable. One can prove that a valid diagonalization of $A$ sets $P:= \left\{\text{normalized eigenvectors of A}\right\}$, an orthogonal matrix (such that $P^T = P^{-1}$), and $D:=diag(\text{eigenvalues of A})$. So indeed, solving this problem reduces to finding the eigenvalues and eigenvectors of $A$. There are many ways to derive the eigensystem of a matrix, but below is my solution.
For an eigenvalue $\lambda$, we must have $\det(\lambda I - A) = 0 \implies \lambda^2 - 6 \lambda - 1 = 0$. The two solutions are $\lambda = 3 \pm \sqrt{10}$.
For $\lambda_1 = 3 + \sqrt{10}$, we have:
$$\begin{aligned}
\mathbf{0} = (\lambda I-A)v &= \begin{pmatrix} 1+\sqrt{10} & -3\\ -3 & -1+\sqrt{10}\end{pmatrix}\begin{pmatrix}v_1\\ v_2 \end{pmatrix}\\
\implies v &= \begin{pmatrix} c\\ (1+\sqrt{10}) c/3\end{pmatrix}
\end{aligned}$$
For $\lambda_2 = 3 - \sqrt{10}$, we have:
$$\begin{aligned}
\mathbf{0} = (\lambda I-A)v &= \begin{pmatrix} 1-\sqrt{10} & -3\\ -3 & -1-\sqrt{10}\end{pmatrix}\begin{pmatrix}v_1\\ v_2 \end{pmatrix}\\
\implies v &= \begin{pmatrix} c \\ (1-\sqrt{10}) c/3\end{pmatrix}
\end{aligned}$$
And while this is incredibly painful to normalize by hand, our matrices ends up being:
$$P = \begin{aligned}
\cfrac{1}{\sqrt{20-2\sqrt{10}}}\begin{pmatrix}\sqrt{10} - 1 & 3\\ 3 & 1-\sqrt{10} \end{pmatrix},\qquad D = \begin{pmatrix}3+\sqrt{10} & 0\\ 0 & 3-\sqrt{10} \end{pmatrix}
\end{aligned}$$
|
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|
Infinitely many positive integers of the form $1998k+1$ such that all digits in their decimal representation are equal
Prove that there exist infinitely many positive integers of the form $1998k+1, k \in \mathbb{N},$ such that all digits in their decimal representation are equal.
We need to find integer solutions to $a \cdot \dfrac{10^n-1}{9} = 1998k+1$. Thus $$a(10^n-1) = 9(1998k+1).$$ How can we find infinitely many integer solutions to this?
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Credit of the full answer to Mlazhinka Shung Gronzalez LeWy.
Let $N = 10^{3n} - 1 = 999\ldots 9$. By factorisation, $N$ is a multiple of $999$, and the other quotient is $$\frac{10^{3n}-1}{10^3-1} = 10^{3(n-1)}+ \cdots + 10^6 + 10^3+1.$$
I further would like the quotient be a multiple of $9$, so that $N$ is a multiple of $999\times 9$. Since $10\equiv 1\pmod 9$, the right hand side
$$10^{3(n-1)}+ \cdots + 10^6 + 10^3+1 \equiv n \pmod 9$$
So $n$ can be any multiple of $9$, i.e. $N = 10^{27m}-1$. Let
$$\begin{align*}
M = \frac N9 &= \frac{10^{27m}-1}{9}\\
10M &=\frac{10^{27m+1}-1}{9}-1
\end{align*}$$
$M$ is a multiple of $999$, and so $10M$ is a multiple of $999$, and so
$$\frac{10^{27m+1}-1}{9}\equiv 1 \pmod{999}$$
But the left hand side is just $10^{27m} + 10^{27m-1} + \cdots + 10^2+10^1+10^0$, which is odd,
$$\frac{10^{27m+1}-1}{9}\equiv 1 \pmod{2}$$
By Chinese remainder theorem,
$$\frac{10^{27m+1}-1}{9}\equiv 1 \pmod{1998}$$
|
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|
Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$ Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$
My proof :D
We have the inequality
$\sum_{cyc}^{ }a^2.\sum_{cyc}^{ }a\geq \sum_{cyc}^{ }a^2b $
which is equivalent to $\sum_{cyc}^{ }b(a-b)^2$ (true)(1)
In another side, by AM-GM, we have: $\sum_{cyc}^{ }a\leq \sum_{cyc}^{ }\dfrac{a^4+1+1+1}{4}=3$ (2)
Thus, from (1) and (2), we've completed our proof.
|
Let $0\le z \le 3$. Then, we have the inequality $z^2 - 2z - 3 \le 0$. We need this because we have $\frac{a^2+b^2+c^2}{3} \le \left(\frac{a^4 + b^4 + c^4}{3}\right)^{1/2} = 1$.
Now, $a^2b+b^2c+c^2a \le (a^2+b^2+c^2)^{1/2}(a^2b^2+b^2c^2+c^2a^2)^{1/2}$, thus we need to prove $a^2b^2+b^2c^2+c^2a^2 \le a^2+b^2+c^2$. We have
$$2(a^2b^2+b^2c^2+c^2a^2) = (a^2+b^2+c^2)^2 - (a^4+b^4+c^4) = (a^2+b^2+c^2)^2 -3.$$
Thus, if we define $a^2+b^2+c^2 = z$, we need to prove $z^2-2z-3 \le 0$, which is what we started with.
|
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|
Forming a quadratic equation with the given roots Question: If $\frac{p^2}{q}$ and $\frac{q^2}{p}$ are the roots of the equation $2x^2+7x-4=0$, find the equation whose roots are $p$ and $q$($p+q$ is real).
My attempt: The required equation is $x^2-(p+q)x+pq$
$(\frac{p^2}{q})(\frac{q^2}{p})=\frac{-4}{2}$
$pq=-2$
$\frac{p^2}{q}+\frac{q^2}{p}=\frac{-7}{2}$
$\frac{p^3+q^3}{pq}=\frac{-7}{2}$
$p^3+q^3=7$
$(p+q)^3-3pq(p+q)=7$
$(p+q)^3+6(p+q)=7$
My problem:I am unable to solve this equation. I think this one is a cubic equation and i do not know how to solve them. Moreover it is not in our syllabus so i there there must be another approach to find $p+q$
|
$$\frac{p^2}{q}+\frac{q^2}{p}=-\frac{7}{2}$$ and $pq=-2$.
Thus, $$p^3+q^3=7$$ or
$$(p+q)((p+q)^2+6)=7$$ or
$$(p+q)^3+6(p+q)-7=0,$$
which gives $p+q=1$ and we get the answer:
$$x^2-x-2=0.$$
|
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|
Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this.
|
Generalization:
For $$\lim_{x\to0}\dfrac{1-\cos ax\cos bx\cos cx}{\sin^2x}=\lim_{x\to0}\dfrac1{1+\cos ax\cos bx\cos cx}\cdot\lim_{x\to0}\dfrac{1-\cos^2ax\cos^2bx\cos^2cx}{\sin^2x}$$
$$=\dfrac12\cdot\lim_{x\to0}\dfrac{1-(1-\sin^2ax)(1-\sin^2bx)(1-\sin^2cx)}{\sin^2x}$$
$$=\dfrac12\cdot\lim_{x\to0}\left(\left(\dfrac{\sin ax}{\sin x}\right)^2+\left(\dfrac{\sin bx}{\sin x}\right)^2+\left(\dfrac{\sin cx}{\sin x}\right)^2+\text{terms containing multiple of }\sin^2\right)$$
$$=\dfrac{a^2+b^2+c^2}2$$
|
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|
Shortest distance between an ellipse and a hyperbola.
Suppose we have a hyperbola $xy=c$ $(c\neq0)$ and an ellipse $x^2/a^2+y^2/b^2=1$ $(a>b>0)$ that do not intersect. What is a quick way to calculate the shortest distance between these two curves?
I thought of setting a point $(a\cos\theta,b\sin\theta)$ on the ellipse and a point $(t,c/t)$ on the hyperbola, then the distance is given by $$\sqrt{(a\cos\theta-t)^2+(b\sin\theta-\frac{c}{t})^2},$$ whose minimum may be found by calculus methods. But the computations look tedious and difficult to carry out. What are some easier ways to do this?
|
There doesn't seem to be a nice closed-form solution to this, but the following might be helpful..
For an ellipse $\frac {x^2}{a^2}+\frac {y^2}{a^2}=1$, the equation of the normal at point $P(a \cos\theta, b\sin\theta)$ on the ellipse is
$$\frac {a\sin\theta}{b\cos\theta}x-y=a\left(\frac{a^2-b^2}{ab}\right)\sin\theta\tag{1}$$
A hyperbola which just touches or does not intersect at all with the ellipse above has the equation $xy=\frac {mab}2$ where $m\ge 1$. The equation of the normal at point $Q (v,\frac {mab}{2v})$ on the hyperbola is
$$\frac {2v^2}{mab}x-y=v\left(\frac {2v^2}{mab}-\frac {mab}{2v^2}\right)\tag{2}$$
The minimum distance between the ellipse and hyperbola is the distance $PQ$ when $(1)=(2)$, i.e. both normals are the same line.
As the coefficients of $y$ are the same in both $(1),(2)$, equating coefficients of $x$ in $(1),(2)$ gives
$$v=a\sqrt \frac{m\tan\theta}{2}\tag{3}$$
This relationship ensures that the tangents and normals at $P,Q$ are parallel to each other respectively (but the normals are not necessarily the same line).
Putting $(3)$ in $(2)$ gives
$$\left(\frac ab \tan\theta\right)x-y=a\sqrt{\frac{m\tan\theta} 2}\left(\frac ab\tan\theta-\frac ba\cot\theta\right)\tag{4}$$
To ensure that both normals are the same line, we need to equate RHS of $(1),(4)$. This gives
$$\left(\frac{a^2-b^2}{ab}\right)\sin\theta=\sqrt{\frac{m\tan\theta}2}\left(\frac ab \tan\theta-\frac ba\cot\theta\right)$$
which is equivalent to
$$(a^2-b^2)\sin\theta=\sqrt{\frac{m\tan\theta}2}\left(\frac{a^2\sin^2\theta-b^2\cos^2\theta}{\sin\theta\cos\theta}\right)\tag{4}$$
Solve numerically $\theta$ in $(4)$, find corresponding value $v$ in $(3)$, then calculate $PQ$. This should give the minimum distance between the ellipse and hyperbola.
See desmos implementation here.
In the trivial case where $a=b$ (i.e. ellipse is a circle), then $\theta=\frac \pi 4$ and $v=a\sqrt{\frac m2}$ . This gives $P=\left(\frac a{\sqrt{2}}, \frac a{\sqrt{2}}\right)$ and $Q=\left(a\sqrt{\frac m2}, a\sqrt{\frac m2}\right)$ and the distance $PQ=a(\sqrt m-1)$.
|
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Proof that $x^3+y^4=z^{31}$ has infinitely many solutions This is a question from RMO 2015.
Show that there are infinitely many triples (x,y,z) of integers such that $x^3+y^4=z^{31}.$
This is how I did my proof:
Suppose $z=0,$ which is possible because $0$ is an integer. Then $x^3+y^4=0 \Rightarrow y^4=-x^3.$ Now, suppose $y$ is a perfect cube such that it is of the form $a^3$ where $a$ is any integer. Then $(a^3)^4=-x^3 \Rightarrow a^{12}=-x^3,$ whereby $x=-(-a)^4.$ Hence there exists infinitely many triples {x,y,z}={$-(-a)^4, a^3, 0$}, which satisfy $x^3+y^4=z^{31}$ for every integer $a$.
However the solution that they have given is quite different from mine. What I want to know is that, is my solution valid, and is this a convincing method to do proofs of such kind?
Thanks in advance!
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A small variation on the 'official' solution: instead of proving that the equation $12r+1=31k$ has infinitely many positive integer solutions we find a particular one, let's say $k=7,r=18$; it follows that $x=2^{4\cdot 18},y=2^{3\cdot 18},z=2^7$ is a particular solution for our equation and from here it is easy to notice that
$$x=2^{4\cdot 18}\cdot k^{4\cdot 31},y=2^{3\cdot 18}\cdot k^{3\cdot 31},z=2^{7}\cdot k^{3\cdot 4}$$
is also a solution for the given equation, for any positive integer $k$.
|
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|
Calculate the sum of the series $\sum_{1\leq aCalculate the sum of the series:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c}$$
My attempt:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c} = \sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}$$
Is it equal?
What's next?
From Wolfram Mathematica I know that $\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}= \frac{1}{1624}$.
|
$$
S=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}
=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \left(1-\frac{1}{2^{b-1}}\right) \frac{1}{3^b5^c}
=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1}\frac{1}{3^b5^c}-2\sum _{c=3}^{\infty } \sum _{b=2}^{c-1}\frac{1}{6^b5^c},
$$
let us say $S_1-2S_2$. Then
$$
S_1=\frac{1}{9}\sum _{c=3}^{\infty } \sum _{b=0}^{c-3}\frac{1}{3^b5^c}=\frac{1}{6}\sum _{c=3}^{\infty }\left(1-\frac{1}{3^{c-2}}\right)\frac{1}{5^c}=\frac{2}{3\cdot 5^2}-\frac{1}{6}\sum _{c=3}^{\infty }\frac{1}{3^{c-2}5^c}
$$
hence
$$
S_1=\frac{2}{3\cdot 5^2}-\frac{1}{6\cdot 3\cdot 5^3}\sum _{c=3}^{\infty }\frac{1}{15^{c-3}}=\frac{2}{3\cdot 5^2}-\frac{1}{14\cdot 6\cdot 5^2}.
$$
You can calculate $S_2$ similarly.
|
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|
Help with polynomial problem, please I am stuck on the following:
If the equation
$$x^4-ax^3+2x^2-bx+1=0$$
has a real root, then show that
$$a^2 + b^2 \geq 8\,.$$
Can I have a hint to go ?
EDIT
My solution after considerable effort:
Since $\,x=0\,$ is not a root of the polynomial we can divide it by $\,x^2$. This yields
$$x^2-ax+2-\frac bx +\frac 1{x^2}\:=\:0 \\[3ex]
\implies\;\left(x-\frac{a}{2}\right)^2 + \left(\frac{1}{x} - \frac{b}{2}\right)^2\:=\:\frac{a^2}{4}+\frac{b^2}{4}-2\:=\:\frac {a^2+b^2-8}{4} \;\geqslant\;0$$
So
$$a^2+b^2-8\:\geqslant\:0\;\;\iff\;\; a^2+b^2\:\geqslant\:8$$
|
since $x=0$ isn't a root of the polynomial ; we can divide it by $x^2 $ yields
$$x^2-ax+2-b/x +1/x^2=0$$
$$\left(x-\frac{a}{2}\right)^2 + \left(\frac{1}{x} - \frac{b}{2}\right)^2= \frac{a^2}{4}+\frac{b^2}{4}-2 = \frac {a^2+b^2-8}{4} \;\;\gt\;\;0$$
So
$$\frac {a^2+b^2-8}{4} \ge 0 \;\;\implies\;\; a^2+b^2-8 \ge 0 \;\;\iff\;\; a^2+b^2 \ge 8$$
|
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|
How to prove this inequality $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge 1.5$? Given $a$, $b$, and $c$ are three positive real numbers. How to prove that sum of $\frac{a}{b+c}$, $\frac{b}{c+a}$ and $\frac{c}{a+b}$ is greater than or equal to $1.5$?
|
by Cauchy Schwarz we have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}\geq \frac{(a+b+c)^2}{2(ab+bc+ca)}\geq \frac{3}{2}$$ if and only if $$a^2+b^2+c^2\geq ab+bc+ca$$ which is true.
|
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|
Which functions have a compositional inverse which is the antiderivative? Which functions $f$ of one variable have a compositional inverse which is the antiderivative of $f$?
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Following up Arthur's suggestion and assuming a form $a x^{b}$ for $f$:
Let $g(x) = f^{-1}(x)$ and let $g'(x) = f(x)$
$$
f = a x^b \Rightarrow g = \frac{a}{b+1} x^{b+1}
$$
I accept I've ignored a possible integration constant for $g$ above.
$$
f (g (x)) = x
\Rightarrow
f (\frac{a}{b+1} x^{b+1}) = x
$$
$$
a \left(
\frac{a}{b+1} x^{b+1}
\right)^{b}
=
x
$$
$$
\frac{a^{b+1}}{(b+1)^{b}}
x^{b(b+1)}
=
x
$$
Which can be solved with two assumptions:
$$
b(b+1) = 1
\Rightarrow
b^{2} + b - 1 = 0
$$
and
$$
\frac{a^{b+1}}{(b+1)^{b}}
=1
\Rightarrow
a = (b+1)^{\frac{b}{b+1}}
$$
The first gives
$$
b =
\frac{-1 \pm \sqrt{5}}{2}
$$
which can be used to obtain $a$.
So this form for $f$ seems solvable.
I wonder what other function forms are possible.
|
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|
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
|
$$\sqrt{x}+\frac 1{\sqrt{x}} = \frac{(x+1)}{\sqrt{x}}$$
$$\left(\frac{x+1}{\sqrt{x}}\right)^2 = \frac{(x^2+2x+1)}x = \frac{(x^2-5x+1)+7x}x = 7x/x = 7$$($x$ is not equal to $0$)
Therefore, $$\sqrt{x}+\frac 1{\sqrt{x}} = \sqrt{7}$$
|
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|
Solve $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$ I came across an interesting equation with two variables $x,y\in\mathbb{R}$, $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$
This, once expanded, can be simplified to $3x^2 - 3xy + 3y^2 - 3y + 1=0.$
How can one proceed to solve it algebraically? The solution according to Wolfram is $(x,y)=(\frac{1}{3},\frac{2}{3}).$
|
Solving the quadratic equation $x^2+(y-1)^2+(x-y)^2=\frac13$ (where the unknown is $x$) gives you$$x=\frac{\left(3 y-\sqrt{3} \sqrt{-9 y^2+12y-4}\right)}6.$$But $-9y^2+12y-4=-(3y-2)^2$. Therefore it is a real number greater than or equal to $0$ if and only if $y=\frac23$ and, when that happens, $x=\frac13$.
|
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|
Prove that $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$ where $\mathcal{C}$ is the unit circle On the generalization of a recent question, I have shown, by analytic and numerical means, that
$$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$$
where $\mathcal{C}$ is the unit circle. Thus, $z=e^{i\theta}$ and $dz=iz~d\theta$. There remains to prove it, however.
What I have done: consider the absolute value part of the integrand,
$$
\begin{align}
|1+z+z^2+\cdots+z^{2n}|^2
&=(1+z+z^2+\cdots+z^{2n})(1+z+z^2+\cdots+z^{2n})^*\\
&=(1+z+z^2+\cdots+z^{2n})(1+z^{-1}+z^{-2}+\cdots+z^{-2n})\\
&=(1+z+z^2+\cdots+z^{2n})(1+z^{-1}+z^{-2}+\cdots+z^{-2n})\frac{z^{2n}}{z^{2n}}\\
&=\left(\frac{1+z+z^2+\cdots+z^{2n}}{z^n} \right)^2\\
&=\left(\frac{1}{z^n}\cdots+\frac{1}{z}+1+z+\cdots z^n \right)^2\\
&=(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2\\
\end{align}
$$
We now return to the integral,
$$
\begin{align}\frac{1}{2\pi i}\int_C |1+z+z^2+\cdots z^n|^2dz
&=\frac{1}{2\pi}\int_0^{2\pi}(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2 (\cos\theta+i\sin\theta)~d\theta\\
&=\frac{1}{2\pi}\int_0^{2\pi}(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2 \cos\theta~d\theta
\end{align}$$
where we note that the sine terms integrate to zero by virtue of symmetry. This is where my trouble begins. Clearly, expanding the square becomes horrendous as $n$ increases, and even though most of the terms will integrate to zero, I haven't been able to selectively find the ones that won't.
The other thing I tried was to simplify the integrand by expressing it in terms of $\cos\theta$ only using the identity
$$\cos n\theta=2\cos (n-1)\theta\cos\theta-\cos(n-2)\theta$$
but this too unfolds as an algebraic jungle very quickly. There are various other expressions for $\cos n\theta$, but they seem equally unsuited to the task. I'll present them here insofar as you may find them more helpful than I did.
$$
\cos(nx)=\cos^n(x)\sum_{j=0,2,4}^{n\text{ or }n-1} (-1)^{n/2}\begin{pmatrix}n\\j\end{pmatrix}\cot^j(x)=\text{T}_n\{\cos(x)\}\\
\cos(nx)=2^{n-1}\prod_{j=0}^{n-1}\cos\left(x+\frac{(1-n+2j)\pi}{2n} \right)\quad n=1,2,3,\dots
$$
where $\text{T}_n$ are the Chebyshev polynomials. Any suggestions will be appreciated.
|
$$
\frac{1}{2\pi i}\int_0^{2\pi}\sum_{l=0}^{2n}e^{il\theta}\sum_{m=0}^{2n}e^{-im\theta}ie^{i\theta}d\theta=\frac{1}{2\pi}\sum_{l,m=0}^{2n}\int_0^{2\pi}\exp(i\theta(l-m+1))d\theta
$$
I think the last integral is zero unless $l-m+1=0$, which happens $2n$ times when $l$ and $m$ go between 0 and 2n.
|
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|
If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$ If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$
I considered the function $f(x)=ax^2+\frac{b}{x}$ and I put $f'(x)=0$ to find $x^3=\frac{b}{2a}$
I am stuck here.
|
We have $x=(\frac {b}{2a})^{\frac {1}{3}} $. From second derivative we find that at this point we have a minima as $f''(x)>0$ . Thus rearranging original equation we have $ax^3+b\geq cx $ plugging in value we have $\frac {3b}{2}\geq c.(\frac {b}{2a})^{\frac {1}{3}} $ cubing and rearranging $27ab^2\geq 4c^3$. Hence the proof.
|
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|
$\sin(2\arcsin(\frac{1}{3}))=? $ A broad one I have written it as $$\sin(2\arcsin(\frac{1}{3}))=2\sin(\arcsin(\frac{1}{3}))\cos(\arcsin(\frac{1}{3})).$$Then, solved for $$\cos(\arcsin(\frac{1}{3}))=\frac{\sqrt{10}}{3}$$ $$\arcsin(\frac{1}{3})=\theta$$$$\sin\theta=\frac{1}{3}\Rightarrow\cos\theta=\frac{\sqrt{10}}{3}.$$But noticed something weird (1) it didn't satisfy the Pythagorean theorem as $\sqrt{10}\approx3.1622,$ which means that the adjacent side to the angle $\theta$ is greater than the hypotenuse, (which is just 3) why this method (using right triangle to determine $\cos(\theta)$) didn't work here although the answer ($\frac{\sqrt{10}}{3}$) satisfied the range of the cosine function, which is $R(\cos\theta)=\Bbb{R}$? After it (2) used this formula$$\arcsin(x)=\arccos(\sqrt{1-x^2});$$$$\arcsin(\frac{1}{3})=\arccos(\sqrt{1-\frac{1}{9}})=\arccos(\frac{\sqrt{8}}{3});$$then, checked if it satisfies or not $D(\arccos(\theta))=[-1;1],$ satisfied, since $\frac{\sqrt{8}}{3}\approx0.942.$ Then proceeded on my problem $$2\sin(\arcsin(\frac{1}{3}))\cos(\arcsin(\frac{1}{3}))=\frac{2}{3}\cdot\frac{\sqrt{8}}{3}=\frac{2\sqrt{8}}{9}.$$ Anyway, it didn't turn out to be the right answer. Where is my mistake?
|
$$\sin2\arcsin\frac{1}{3}=2\cdot\frac{1}{3}\cdot\sqrt{1-\frac{1}{9}}=\frac{4}{9}\sqrt2$$
|
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|
Reduction of order of differential equation! Given that y=x is a solution to the following differential equation
$$(x^2-x+1)\frac{d^2y}{dx^2}-(x^2+x)\frac{dy}{dx}+(x+1)y=0$$
Find a linearly independent solution to the new differential equation by reducing the order. Write a general solution.
I am stuck somewhere here.
My work is below!
Suppose that $f(x)=x$ is a non trivial solution to the differential equation then by the transformation $y=vx$
$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$
$$\frac{d^2y}{dx^2}=2\frac{dv}{dx}+x\frac{d^2v}{dx^2}$$
$$ (x^2-x+1)(2\frac{dv}{dx}+x\frac{d^2v}{dx^2})-(x^2+x)(v+x\frac{dv}{dx})+(x+1)(vx)=0$$
$$ 2((x^2-x+1))\frac{dv}{dx}+x((x^2-x+1))\frac{d^2v}{dx^2}-(x^2+x)(x)\frac{dv}{dx})=0$$
$$x((x^2-x+1))\frac{d^2v}{dx^2}+(-x^3+x^2-2x+2)\frac{dv}{dx}=0 $$
By letting $$w=\frac{dv}{dx}$$ ,$$\frac{dw}{dx}=\frac{d^2v}{dx^2}$$
$$x(x^2-x+1))\frac{dw}{dx}+(-x^3+x^2-2x+2)w=0 $$
$$\frac{dw}{dx}+\frac{(-x^3+x^2-2x+2)}{x(x^2-x+1))}w=0 $$
$$\int \frac{dw}{w}+ \int \frac{2}{x} -1- \frac{2x-1}{x^2-x+1} dx=0 $$
$$ ln|{w}|+ ln|{x}^2| -x- ln{x^2-x+1} =c $$
$$ ln|{w}|+ ln|{x}^2|- ln{x^2-x+1} =c+x $$
$$\frac{wx^2}{x^2-x+1}=ce^x$$
Choosing c=1 as the solution to the differential equation.
$$\frac{dv}{dx}=\frac{e^x(x^2-x+1)}{x^2}$$
Is this right?
|
We are given
$$\tag 1 (x^2-x+1) y'' - (x^2 + x) y' + (x+1) y = 0$$
You made a good Reduction of Order substitution using $y = v x \implies y' = v + v' x$ and we reduce $(1)$ to
$$\tag 2 x(x^2-x+1)\dfrac{d^2v}{dx^2}+(-x^3+x^2-2x+2)\dfrac{dv}{dx}=0$$
Next, you correctly let $\dfrac{dv}{dx} = w$ and reducing $(2)$ we get a Separable Equation that we can integrate as
$$\tag 3 \displaystyle \int \dfrac{1}{w} ~dw = \int -\dfrac{(-x^3+x^2-2x+2)}{x(x^2-x+1))}~dx = \int \left(\dfrac{2x-1}{x^2-x+1}-\dfrac{2}{x}+1\right)dx=0$$
Integrating $(3)$ results in
$$\ln w = \ln(x^2-x+1) -2 \ln x + x + c$$
Solving for $w$
$$\tag 4 w = \dfrac{c~e^x(x^2-x+1)}{x^2}$$
From the original substitution $\dfrac{dv}{dx} = w$, we now have a Separable Equation and can integrate
$$\displaystyle \int dv = \int \dfrac{c~e^x(x^2-x+1)}{x^2}~dx$$
We get
$$v(x) = c ~ e^x \left(1 - \dfrac{1}{x}\right) + k$$
We now substitute back our original $y = v x$ and have
$$y(x) = c ~ e^x(x - 1) + k x$$
|
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|
Significance of the Triangle Inequality After working through a few problems regarding the Cauchy Integral Formula, I'm still a little confused on the significance of the triangle inequality. Why do we use it and what information does it tell us?
See the following example below.
Determine $\int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx$
First, we'll create the contour $$ \begin{cases} C_1: Rt & -1 \le t \le 1 \\ C_2:R e^{i\theta} & \ \ \, 0 \le \theta \le \pi \\ C=C_1+C_2 \end{cases} $$
$$ x^2 +1 =0 \Rightarrow x=\pm i \ \ \ ; \ \ \ \ \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx$$
$$ \Rightarrow \int_C \frac{z^2}{(z+1)^2}dz = \int_C \frac{z^2}{[(z+i)(z-i)]^2}
= \int_C \frac{z^2}{(z+i)^2 (z-i)^2} $$
$z=-i$ is outside the contour and therefore irrelevant, so choosing $z=i$ for the integration.
$$ \Rightarrow z_0=i \ \ \text{and} \ \ f(z)=\frac{z^2}{(z+i)^2} $$
$$ f'(z)=\frac{d}{dz}\left[ \frac{z^2}{(z+i)^2} \right] = \frac{2z}{(z+i)^2}-\frac{2z^2}{(z+i)^3}$$
Therefore, by the Cauchy Integration Formula,
$$ \int_C \frac{f(z)}{(z-i)^2}dz = 2\pi i f'(i) = 2\pi i \left[ \frac{2i}{(2i)^2}-\frac{2i^2}{(2i)^3} \right] = \pi \left[1 + \frac{2}{4i^2} \right] = \pi - \frac{\pi}{2} = \frac{\pi}{2} $$
The following part is where I'm a little confused. I know the work, but am not quite sure why I'm doing it or what the end result is telling us. So, by the triangle inequality,
$$ \left| z^2 +1 \right|^2 \le \left( \left|z^2 \right| +1 \right) ^2 = z^4 +2\left| z^2 \right| +1 \Rightarrow \left| \int_C \frac{z^2}{(z+1)^2}dz \right| \le \frac{R^2}{R^4 +2R +1} $$
$$ \Rightarrow \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx \le \lim_{R \rightarrow \infty} \int_C \frac{z^2}{(z+1)^2}dz \\ \le \lim_{R \rightarrow \infty} \frac{R^2}{R^4 +2R +1} = 0 $$
$$ \therefore \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2} \left[ \frac{\pi}{2} \right] = \frac{\pi}{4} $$
Any help in deciphering the part in the work I pointed out would be very helpful, thank you for your time.
|
A fix on this argument, motivated by Fightclub's observation, is this: We know that for every $R>1$:
$$ \underbrace{\int_{C_1} \frac{z^2}{(z^2+1)^2}dz}_{\int_{-R}^R \frac{x^2}{(x^2+1)^2}dx} + \int_{C_2} \frac{z^2}{(z^2+1)^2}dz = \pi/2 $$
Hence:
$$\lim_{R\rightarrow\infty} \int_{-R}^R \frac{x^2}{(x^2+1)^2}dx + \lim_{R\rightarrow\infty} \int_{C_2} \frac{z^2}{(z^2+1)^2} dz = \pi/2 $$
Thus:
$$ \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)^2}dx + \lim_{R\rightarrow\infty} \int_{C_2} \frac{z^2}{(z^2+1)^2} dz = \pi/2$$
So it suffices to show the second integral has a limit of 0. We get:
$$ \left|\int_{C_2} \frac{z^2}{(z^2+1)^2}dz\right| \leq (\pi R)\max_{z \in C_2} \left[\frac{|z|^2}{|z^2+1|^2}\right] $$
which follows because $C_2$ is half the perimeter of a circle of radius $R$.
But on $C_2$ we get $|z|=R$. Also, by the reverse triangle inequality we get:
$$ |z^2+1| \geq |z^2| - 1 = R^2-1$$
So for all $z \in C_2$ we get (for all $R>1$):
$$ \frac{|z|^2}{|z^2+1|^2} \leq \frac{R^2}{(R^2-1)^2} $$
and indeed
$$ (\pi R)\max_{z \in C_2} \frac{|z|^2}{|z^2+1|^2} \leq \frac{\pi R^3}{(R^2-1)^2} \rightarrow 0 $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to properly simplify and trig substitute this integral? $\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then substitute this $x$ value in and
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, d\theta $
Take the root, simplified to $\int\frac { {5-(5 \sin\theta)}}{5 \sin\theta}\,d\theta $
Factor out 5 and cancel, simplified to $\int\frac { {5(1-( \sin\theta))}}{5 (\sin\theta)}\, d\theta $
Divide by $\sin\theta$ to get $\int\frac { (1-( \sin\theta))}{ (\sin\theta)}\, d\theta $
Simplified to $\int\frac {1}{ \sin\theta}d\,\theta - 1 $
I am stuck here and I can't figure it out. Supposedly the answer simplifies to $\int-5 \sin x \tan x \,dx$
|
If $x = 5\sin\theta$ then $dx = 5\cos\theta\, d\theta.$ That was neglected.
You also replaced $\sqrt{25 - (5\sin\theta)^2}$ with $5 - 5\sin\theta.$ That is also not correct. Notice that, for example,
$$
\sqrt{5^2 - 3^3} = \sqrt{25-9} = \sqrt{16} = 4 \ne 5-3, \text{ so } \sqrt{5^2-3^2} \text{ differs from } 5-3.
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Stability of a fixed point in a nonlinear system with no linear part The following nonlinear system has a fixed point in the origin. I want to know if this fixed point is stable:
\begin{align*}
&\dot{\alpha}=\alpha^2-2\beta(\alpha+\beta)\\
&\dot{\beta}=\beta^2-2\alpha(\alpha+\beta)
\end{align*}
Note that the linearization of the system leads to $\dot{\alpha}=\dot{\beta}=0$ to first order, so this method is useless. To find out the stability, I wrote this equations as:
\begin{align*}
\dot{\alpha}=\left(\alpha,\beta\right)
\cdot A\cdot
\left(\begin{array}{c}
\alpha\\
\beta
\end{array}\right),\quad
A=\left(\begin{array}{cc}
1 & -1\\
-1 & -2
\end{array}\right)
\end{align*}
\begin{align*}
\dot{\beta}=\left(\alpha,\beta\right)
\cdot B\cdot
\left(\begin{array}{c}
\alpha\\
\beta
\end{array}\right),\quad
B=\left(\begin{array}{cc}
-2 & -1\\
-1 & 1
\end{array}\right)
\end{align*}
Since $A$ and $B$ are real and symmetric, they are diagonalizable, with real eigenvalues. It turns out that $A$ and $B$ have one eigenvalue positive and the other negative (in fact, $A$ and $B$ have the same eigenvalues), explicitly:
\begin{align*}
&\text{Eigenvalues of $A$:}\quad \frac{1}{2} \left(-1-\sqrt{13}\right),\frac{1}{2} \left(\sqrt{13}-1\right)\\
&\text{Eigenvalues of $B$:}\quad \text{same of $A$}
\end{align*}
If, for example, the eigenvalues of $A$ were both positive, then $\dot{\alpha}$ would be positive for all points except origin, and then the origin would be unstable (because trajectories starting out of the origin would increase their $\alpha$'s to infinity). But when both $\dot{\alpha}$ and $\dot{\beta}$ are positive or negative depending on the point $(\alpha,\beta)$, as in this case, how can you determine the stability of the fixed point?
EDIT: The phase portrait is:
|
This is mostly a recap of the observations made in the comments, plus some more analysis, because I think it's a nice problem to analyse.
First, both the functional form of the system and the reflection symmetry (see also the phase plane) suggest it's a good idea to introduce $x = \alpha+\beta$, $y = \alpha-\beta$, to obtain
\begin{align}
\dot{x} &= \frac{1}{2}(y^2 - 3 x^2), \tag{1a}\\
\dot{y} &= 3 x y.\tag{1b}
\end{align}
We can simplify system (1) somewhat by rescaling $y \to \sqrt{3} y$ and $t \to \frac{1}{3} t$, yielding
\begin{align}
\dot{x} &= \frac{1}{2}(y^2 - x^2), \tag{2a}\\
\dot{y} &= x y.\tag{2b}
\end{align}
The phase plane of system (2) looks like this:
What's immediately obvious, is that this phase plane is highly symmetric. It seems to be invariant under rotation over an angle of $\frac{2 \pi}{3}$ (= 120 deg), and rotation over an angle of $\frac{2 \pi}{6}$ seems to keep the shape of the orbits invariant, but changes their flow direction. You can check that both of these observations are indeed correct by considering
\begin{equation}
\begin{pmatrix} \xi_1 \\ \eta_1 \end{pmatrix} := R(\frac{2\pi}{3}) \begin{pmatrix} x \\ y \end{pmatrix}\quad \text{and} \quad \begin{pmatrix} \xi_2 \\ \eta_2 \end{pmatrix} := R(\frac{2\pi}{6}) \begin{pmatrix} x \\ y \end{pmatrix},
\end{equation}
with
\begin{equation}
R(\theta) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}
\end{equation}
is the matrix that rotates a vector about the origin over angle $\theta$. Substituting $\xi{1,2}$ and $\eta_{1,2}$ in system (2), you obtain
\begin{align}
\dot{\xi_1} &= \frac{1}{2}(\eta_1^2 - \xi_1^2), \\
\dot{\eta_1} &= \xi_1 \eta_1,
\end{align}
and
\begin{align}
\dot{\xi_2} &= -\frac{1}{2}(\eta_2^2 - \xi_2^2), \\
\dot{\eta_2} &= -\xi_2 \eta_2,
\end{align}
which implies the above observations.
It seems a good idea to 'factor out' this rotational symmetry present in system (2). Introducing polar coordinates $x = r \cos \theta$, $y = r \sin \theta$, we obtain the dynamical system
\begin{align}
\dot{r} &= -\frac{1}{2} r^2 \cos(3\theta),\tag{3a}\\
\dot{\theta} &= \frac{1}{2} r \sin(3\theta),\tag{3b}
\end{align}
which is indeed invariant under $\theta \to \theta + \frac{2 \pi}{3}$. Now, we rewrite the above in terms of the new angle $\phi := 3 \theta$ to obtain
\begin{align}
\dot{r} &= -\frac{1}{2} r^2 \cos(\phi), \tag{4a}\\
\dot{\phi} &= \frac{3}{2} r \sin(\phi). \tag{4b}
\end{align}
Now, the nice thing is that we can reinterpret system (4) as being the `polar representation' of some Cartesian system. That is, if we introduce $X$ and $Y$ by $X := r \cos \phi$ and $Y := r \sin \phi$, we can rewrite system (4) in terms of $X$ and $Y$ to obtain
\begin{align}
\dot{X} &= -\frac{1}{2}(X^2 + 3 Y^2), \tag{5a}\\
\dot{Y} &= X Y. \tag{5b}
\end{align}
The phase plane of system (5) looks as we would have expected:
Moreover, system (5) turns out to be Hamiltonian, i.e. of the form
\begin{align}
\dot{X} &= \frac{\partial H}{\partial Y},\\
\dot{Y} &= -\frac{\partial H}{\partial X},
\end{align}
with
\begin{equation}
H(X,Y) = -\frac{1}{2} Y(X^2 + Y^2).
\end{equation}
Therefore, orbits lie on level sets of $H$, i.e. those curves where $H = E$ (= constant), which can be used to express $X$ in terms of $Y$ as
\begin{equation}
X = \pm \sqrt{-\frac{2 E + Y^3}{Y}}.
\end{equation}
As a final remark, the instability of the origin can be derived from system (5) as follows. Consider the horizontal line $\ell = \left\{ (X,Y) \vert Y=0 \right\}$, i.e. the $X$-axis. Substituting $Y=0$ in system (5), we obtain $\dot{Y} = 0$; therefore, the line $\ell$ is invariant under the flow. In other words, every point on $\ell$ stays on $\ell$ for all time.
It so happens that the origin $(0,0)$ lies on the line $\ell$. Taking a point $(-\epsilon,0) \in \ell$ arbitrarily close to (and directly to the left of) the origin, we see that this point will flow away from the origin, because the flow on $\ell$ is given by $\dot{X} = -\frac{1}{2} X^2$. This holds for all $\epsilon > 0$; hence, we have found an unstable flow direction at the origin, spanned by $(-1,0)$. Therefore, the origin is a (nonlinearly) unstable fixed point of system (5), and in extension of system (1).
|
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|
Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$ Show that
$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$
Starting from the left hand side (LHS)
\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}
Can anyone help me to continue from here
|
(a+b)^3= a^3 + b^3 + 3ab(a + b)
|
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|
Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?
|
Without Vieta's formulas, the polynomial can be derived the hard way by elimination. From the first and last equations:
$$
b+c=6-a \\
bc = 6 / a
$$
Substituting the above into the middle equation:
$$
11=a(b+c)+bc=a(6-a)+6/a \;\;\iff\;\; a^3-6 a^2+11a - 6 = 0
$$
By inspection, the latter equation has $a=1$ as a root, then factoring out $a-1$ gives:
$$a^3-6 a^2+11a - 6=(a-1)(a^2-5a+6)=(a-1)(a-2)(a-3)$$
Since the original system is symmetric in $a,b,c$ it follows that the solutions are $\{1,2,3\}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$ I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have tried expanding the RHS as a sum. The main issue I'm having is working out how to get a $\pi$ on the LHS or removing it from the right to equate the two sides.
$$\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$
|
Let $\dfrac\theta2+\dfrac\pi4=y\implies\theta=2y-\dfrac\pi2$
$$\sin\theta=\cdots=-\cos2y,\cos\theta=\cdots=\sin2y$$
$$\dfrac{1+\sin\theta}{\cos\theta}=\dfrac{1-\cos2y}{\sin2y}=\dfrac{2\sin^2y}{2\sin y\cos y}=?$$
$$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin2y}{1+\cos2y}=?$$
See also : Solve $\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = 2$,
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to find this simplification integral involving products of roots? Consider the following integral
$$
f = \int_0^1 \frac{1}{\sqrt{-\frac{1}{2} \, t^{2} + 1} \sqrt{-t^{2} + 1}} \mathrm \,dt.
$$
If we change variable by letting $x^2=t^2/(2-t^2)$, then we have
$$
f = \int_0^1 \sqrt{2} \cdot\sqrt{\frac{1}{1-x^{4}}} \mathrm \,dx,
$$
which is a simpler form.
I read this in a book and wonder how can we come up with this sort of simplification? Is it just experience or is there systematic way to do it?
Note: The integral is the complete elliptic integral of the first kind $K(1/\sqrt{2})$.
|
$$x^2=\frac{t^2}{2-t^2}$$
$$t=x \sqrt{\frac{2}{x^2+1}}$$
$$dt = \sqrt{2} \sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right) dx$$
So the integrand becomes is
$$\frac{\sqrt{2} \left(\sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right)\right)}{\sqrt{1-\frac{2 x^2}{x^2+1}} \sqrt{1-\frac{x^2}{x^2+1}}}$$
and finally
$$\frac{\sqrt{2}}{\sqrt{1-x^4}}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
What is a simple (or not) way of finding the lengths of the diagonals of this rhombus? I recently taught a group of geometry students about properties of rhombuses, and gave them a set of homework exercises created by a previous instructor which included the following problem.
If a rhombus has sides of length $25$ mm and a $45^{\circ}$ angle, determine the lengths of the diagonals.
Here is an image for reference.
I ultimately removed this problem because I haven't found a simple solution at a high school geometry level (i.e. some algebra, no trig). I've solved the problem two ways, which I will post as an answer, one way involved simple trig and the other way was geometrical but messy (involved solving a system of nonlinear equations and a fourth degree polynomial).
I'm curious if anyone can solve this problem a different way. I'd like to know if there is nice solution involving only geometry, but I would also be interested in seeing any novel approaches anyone comes up with.
|
As I mentioned in the question, here are the two ways that I have found to solve this problem.
Method 1: Trigonometry
If we draw in the diagonals, we get the following image, and we know that each diagonal bisects the other as well as the angles of the rhombus:
That means that we can say
\begin{align*}
a &= 25\sin(22.5^{\circ})\text{ mm}\\
b &= 25\cos(22.5^{\circ})\text{ mm},
\end{align*}
meaning that the diagonals are
\begin{align*}
2a &= 50\sin(22.5^{\circ})\text{ mm}\approx 19.13 \text{ mm}\\
2b &= 50\cos(22.5^{\circ})\text{ mm} \approx 46.19 \text{ mm}.
\end{align*}
Method 2: (Messy) Geometry
Here I started by drawing a line perpendicular to the top connecting to the lower right angle. Since the upper right angle is $45^{\circ}$ we know that this line makes a $45^{\circ}$ angle as well, meaning we have an isosceles triangle:
By the Pythagorean Theorem, we know that
$$2x^{2} = 25^{2} \implies x = \frac{25}{\sqrt{2}} \text{ mm}.$$ Therefore the area of the rhombus is given by
$$\text{Area} = 25 \text{ mm } \times \frac{25}{\sqrt{2}} \text{ mm} = \frac{25^{2}}{\sqrt{2}} \text{ mm}^{2}.$$
Furthermore, going back the the variables introduced in the first image, we can say that
$$a^{2} + b^{2} = 25^{2} \quad \text{and} \quad 4\times \frac{1}{2}ab = \frac{25^{2}}{\sqrt{2}}.$$
Solving for $a$ in the second equation yields
$$ a = \frac{25^{2}}{2b\sqrt{2}},$$
which when substituted into the first equation gives us
$$b^{2} + \frac{25^{4}}{8b^{2}} = 25^{2},$$
which upon multiplying by $b^{2}$ can be written as
$$b^{4} - 25^{2}b^{2} + \frac{25^{4}}{8} = 0.$$
Now, this equation is quadratic in $b^{2}$, so we can just apply the quadratic formula to obtain
$$b^{2} = \frac{25^{2}\pm\sqrt{25^{4} - \frac{25^{4}}{2}}}{2} = \frac{25^{2}\pm\sqrt{\frac{25^{4}}{2}}}{2} = \frac{25^{2}\pm\frac{25^{2}}{\sqrt{2}}}{2} = \frac{25^{2}}{2}\left(1 \pm \frac{\sqrt{2}}{2}\right),$$
and so $$b = \frac{25\sqrt{2\pm\sqrt{2}}}{2} \approx 23.096, 9.567.$$
Back substituting then we obtain
$$a = \frac{25^{2}}{2\left(\frac{25\sqrt{2\pm\sqrt{2}}}{2}\right)\sqrt{2}} = \frac{25}{\sqrt{4\pm 2\sqrt{2}}}\approx 9.567, 23.096.$$
Since we chose $b$ to be the longer diagonal then we can say
\begin{align*}
2a &= \frac{25}{\sqrt{4+ 2\sqrt{2}}}\text{ mm} \approx 19.13 \text{ mm}\\
2b &= \frac{50\sqrt{2+\sqrt{2}}}{2}\text{ mm} \approx 46.19 \text{ mm}.
\end{align*}
|
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|
Proving right angle triangle A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.
|
My idea :
$r = \Delta/s = 1 \implies \Delta = s \implies \Delta^2 = s^2 \implies (s-a)(s-b)(s-c) = s \implies (b+c-a)(c+a-b)(a+b-c) = 4(a+b+c) $.
If all $a,b,c$ are even or odd , then the parity doesn't satisfy. Thus that means $a+b+c$ is even. So $s$ is an integer. Thus by the above argument, we have $ s-a|s,s-b|s,s-c|s $. Now let $r_a,r_b,r_c$ be the exradii of the triangle $ABC$. We have our formula $r_a= \Delta/s-a = s/s-a $. Thus $r_a,r_b,r_c$ are integers. Now also we have the equation :
$\frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c} = \frac{1}{r} \implies\frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c} = 1 $
and this has only one solution namely $ 2,3,6$ and hence you get the values of sides as $3,4,5$.
|
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|
Find coefficient of $x^2$ in a complicated expansion
Find the coefficient of $x^2$ in the expansion of $(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$
I noticed that since we only care about the coefficient of $x^2$, only the coefficients of $x^2$ inside the square brackets, as well as the constant terms, will matter. From there I am stuck when it comes to expanding $(1+2x+3x^2)^6$ as I can't see an easy way to use the binomial theorem (for example by factorising $1+2x+3x^2$, which I can't do as there are no real roots). Is there a way to find the coefficient of $x^2$ in this expansion (specifically $(3x^2+2x+6)^6$ as I can get the answer from there)?
|
Let $A=\big[(1+2x+3x^2)^6+(1+4x^3)^5\big]$.
$$\begin{align}
\color{green}{[x^2]}A
&= \color{green}{[x^2]}(1+2x+3x^2)^6+\color{green}{[x^2]}(\underbrace{1+4x^3}_{\color{blue}{\text{no $x,x^2$ terms}}})^5\\
&=\color{green}{[x^2]}\bigg[\binom 6{5,0,1}1^5(2x)^0(3x^2)^1+\binom 6{4,2,0}1^4(2x)^2(3x^2)^0\bigg]+\color{blue}0\\
&=\frac{6!}{5!\;0!\;1!}\cdot 1^52^03^1+\frac {6!}{4!\;2!\;0!}\cdot 1^42^23^0\\
&=78\\\\
\color{green}{[x^0]}A
&=\color{green}{[x^0]}(1+2x+3x^2)^6+\color{green}{[x^0]}(1+4x^3)^5\\
&=1+1\\
&=2\\\\
\color{green}{[x^2]}(4-x^2)A
&=4\cdot \big(\color{green}{[x^2]}A\big)-\big(\color{green}{[x^0]}A\big)\\
&=4\cdot 78-2\\
&=\color{red}{310}
\end{align}$$
See wiki entry on the Multinomial Theorem for more details.
|
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|
For which positive integer $k$ does the series $\sum_{n=1}^\infty\frac{\sin(n\pi/k)}{n}$ converge?
For which positive integer $k$ does the series $$\sum_{n=1}^\infty\frac{\sin(n\pi/k)}{n}$$ converge?
The cases when $k=1$ or $k=2$ are trivial. For $k>2$, I don't see how to approach it. If one looks at the partial sums:
$$
s_k=\frac{\sin(\frac{\pi}{k})}{1}+\frac{\sin(2\cdot\frac{\pi}{k})}{2}+\cdots+\frac{\sin((k-1)\cdot\frac{\pi}{k})}{k-1}
$$
$$
s_{2k}-s_k=
(-1)\frac{\sin(\frac{\pi}{k})}{k+1}+(-1)\frac{\sin(2\cdot\frac{\pi}{k})}{k+2}+\cdots+(-1)\frac{\sin((k-1)\cdot\frac{\pi}{k})}{k+k-1}
$$
$$
s_{3k}-s_{2k}=\frac{\sin(\frac{\pi}{k})}{2k+1}+\frac{\sin(2\cdot\frac{\pi}{k})}{2k+2}+\cdots+\frac{\sin((k-1)\cdot\frac{\pi}{k})}{2k+k-1}\\
\vdots
$$
one might tend to sum the terms vertically so that the alternating test can be used. But unfortunately this series does not converge absolutely.
|
$$\sum_{n=1}^\infty\frac{\sin (n\pi/k)}{n} = \sum_{n=1}^\infty\frac{\pi}{n\pi}\sin2\pi \frac{n}{2}\frac1{k}$$
The right hand side is a sawtooth wave of period $2$, at $t = 1/k$:
$$f(t) = \frac{\pi}{2}-\frac{\pi}2t, \quad 0<t<2\\
f(t+2) = f(t)\\
f(2n) = 0, \quad n\in\mathbb Z$$
So specifically,
$$\sum_{n=1}^\infty\frac{\sin (n\pi/k)}{n} = f\left(\frac1k\right) = \frac{\pi}{2} - \frac{\pi}{2k}$$
The Fourier series coefficients of $f(t)$ is given by
$$\begin{align*}
b_n &= \frac22\int_{0}^2\frac{\pi}{2}(1-t)\sin\frac{2\pi n t}{2}dt\\
&= \frac\pi2\int_0^2\sin\frac{2\pi n t}{2}dt - \frac\pi2\int_0^2t\sin\frac{2\pi n t}2dt\\
&= 0 + \frac1{2n}\int_0^2 t\ d\cos\frac{2\pi n t}{2}\\
&= \frac1{2n}\left[t\cos\frac{2\pi n t}2\right]_0^2 + \int_0^2\cos\frac{2\pi nt}2 dt\\
&= \left(\frac1{2n}\cdot2\cos2\pi n-0 \right) + 0\\
&= \frac1n
\end{align*}$$
|
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|
the integral of $1/(x\sqrt{4x+1})$ I integrated $\frac{1}{x\sqrt{4x + 1}}$ and got I $\ln\left|\frac{\sqrt{4x+1}-1}{2\sqrt{x}}\right| + K$, where K is a constant term. But my book gives the answer as $\ln\left|\frac{\sqrt{4x+1}-1}{\sqrt{4x+1} + 1}\right| + K$. Where did I make a mistake?
Here is how I solved it.
|
Let $\sqrt{4x+1}=t$.
Hence, $\frac{4}{2t}dx=dt$ and $x=\frac{t^2-1}{4}$.
Thus,
$$\int\frac{1}{x\sqrt{4x+1}}dx=2\int\frac{1}{t^2-1}dt=\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt=\ln\left|\frac{t-1}{t+1}\right|+C$$
|
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"timestamp": "2023-03-29T00:00:00",
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Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$ Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$
I need to specify the value of $f(x,y)$ which makes the given function continuous.
I had tried to give a arbitral relationship between $y$ and $x$, but as $(x, y)$ gets closer to $(0, 0)$ the arbitral movement cannot always be represented as a functional representation between $y$ and $x$.
How could I solve this problem rigorously?
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For $\lim_{(x,y)\rightarrow (0,0)} \frac{sin(x^2 +y^2}{x^2+y^2}$, using $r^2=x^2+y^2$ we get $\lim_{r\rightarrow 0} \frac{sin(r^2)}{r^2}$=$\lim_{r \rightarrow 0} \frac{2 \cdot r \cdot \cos(r^2)}{2r}$=$\lim_{r \rightarrow 0} \cos(r^2)=1$, therefore $\lim_{(x,y)\rightarrow (0,0)} \frac{sin(x^2 +y^2}{x^2+y^2}=1$.
|
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Find the minimum value of $P=\sum _{cyc}\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}$
For $x>0$, $y>0$, $z>0$ and $x+y+z=3$ find the minimize value of $$P=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}$$
We have: $P=\left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{\left(z+1\right)^2\left(z^2+1\right)}+\frac{1}{\left(y+1\right)^2\left(y^2+1\right)}+\frac{1}{\left(x+1\right)^2\left(x^2+1\right)}\right)$
$\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{2\left(z^2+1\right)^2}+\frac{1}{2\left(x^2+1\right)^2}+\frac{1}{2\left(y^2+1\right)^2}\right)$
$\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{9}{2\left(\left(z^2+1\right)^2+\left(y^2+1\right)^2+\left(x^2+1\right)^2\right)}\right)$
I can't continue. Help
|
Let $x=y=z=1$.
Hence, $P=24$.
We'll prove that it's a minimal value.
Indeed, by C-S
$$\sum_{cyc}\frac{(x+1)^2(y+1)^2}{z^2+1}=\sum_{cyc}\frac{(x+1)^2(y+1)^2(x+y)^2}{(z^2+1)(x+y)^2}\geq\frac{\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2}{\sum\limits_{cyc}(z^2+1)(x+y)^2}.$$
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2\geq24\sum\limits_{cyc}(z^2+1)(x+y)^2.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $u=1$ and$$\sum\limits_{cyc}(x+1)(y+1)(x+y)=\sum_{cyc}(x^2y+x^2z+2x^2+2xy+2x)=$$
$$=9uv^2-3w^3+2u(9u^2-6v^2)+6uv^2+6u^3=3(8u^3+uv^2-w^3);$$
$$\sum\limits_{cyc}(z^2+1)(x+y)^2=2\sum_{cyc}(x^2y^2+x^2yz+x^2u^2+xyu^2)=$$
$$=2(9v^4-6uw^3+3uw^3+9u^4-6u^2v^2+3u^2v^2)=6(3u^4-u^2v^2+3v^4-uw^3).$$
Id est, it's enough to prove that $f(w^3)\geq0$, where
$$f(w^3)=(8u^3+uv^2-w^3)^2-16(3u^6-u^4v^2+3u^2v^4-u^3w^3).$$
Now, $$f'(w^3)=-2(8u^3+uv^2-w^3)+16u^3=2w^3-2uv^2\leq0,$$
which says that $f$ is decreasing function.
Thus, it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Let $y=x$ and $z=3-2x$.
Hence, $$\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2\geq24\sum\limits_{cyc}(z^2+1)(x+y)^2$$
gives
$$(x-1)^2(x^4-2x^3-11x^2+24x+4)\geq0.$$
which is obvious.
Done!
|
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|
Common complex roots
If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots then show that $a=b=c$.
My attempts:
Observing $-2$ is a root of $x^3+3x^2+3x+2=0\implies x^3+3x^2+3x+2=(x+2)(x^2+x+1)=0$
Hence $ax^2+bx+c=0$ can have complex roots in common, comming from $(x^2+x+1)=0$
Both the roots of $(x^2+x+1)=0$ and $ax^2+bx+c=0$ are common should imply $a=b=c$ not only this but $a=b=c=1$.
Is this solution correct?
|
If $a$, $b$, and $c$ are not real, then the question as stated is false. The roots of $x^3 + 3x^2 + 3x + 2$ are $-2$ and $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$; and the polynomial $x^2 + (\frac{5}{2} + \frac{\sqrt{3}}{2} i)x + (1 + \sqrt{3} i) = (x + 2)(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i)$ shares two roots with $x^3 + 3x^2 + 3x + 2$ but does not have $a = b = c$.
I think that the strongest statement that can be made is that if $a, b, c \in \mathbb{C}$ have the same argument, then the statement must be true. If this is the case, then we have $a = Ae^{i\theta}, b = B e^{i\theta}, c = e^{i \theta}$ for the same $\theta$ in each case and $A, B, C \in \mathbb{R}$. In such a case, the polynomial $a x^2 + bx + c = e^{i \theta}(A x^2 + Bx + C)$ will have the same roots as $Ax^2 + Bx + C$; and since this latter polynomial is real, its roots must be complex conjugates of each other. This implies that $A x^2 + Bx + C$ is a multiple of $(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i)(x + \frac{1}{2} - \frac{\sqrt{3}}{2} i) = x^2 + x + 1$, and therefore that $A = B = C$. (Note, however, that they are not necessarily equal to 1.)
|
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|
If a is a non real root of $x^7 = 1$, find the equation whose roots are $ a + a^6 , a^2 + a^5, a^3 + a^4$ If a is a non real root of $ x^7 = 1$, find the equation whose roots are $a + a^6 , a^2 + a^5, a^3 + a^4$. This is one of the questions I have encountered while preparing for pre rmo. I feel the question requires the concept of the nth roots of unity and de moivre's theorem. But i actually couldnt work it out. Any help will be appreciated. Thanks in advance.
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Like factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$,
we can prove $a^r+a^{7-r}=a^r+a^{-r}=2\cos\dfrac{2r\pi}7$ for $r=1,2,3$
Now if $7x=2r\pi,$
$\cos4x=\cos(2r\pi-3x)=\cos3x\ \ \ \ (1) $
So, the equation whose roots are $\cos\dfrac{2r\pi}7$ for $r=0,1,2,3$
$$8c^4-8c^2+1=4c^3-3c$$
But $(1)\implies4x=2m\pi\pm3x$ where $m$ is any integer
$4x=2m\pi+3x\implies x=2m\pi,\cos x=1$
So, the equation whose roots are $\cos\dfrac{2r\pi}7$ for $r=1,2,3$
$$\dfrac{8c^4-4c^3-8c^2+3c+1}{c-1}=0$$
Can you take it from here?
|
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|
Proving an inequality of trigonometric functions I am trying to prove that $$(\cos x)^{\cos x} > (\sin x)^{\sin x}$$ for $x \in \left(0, \frac{\pi}{4}\right)$.
I have plotted the graph of $y=(\cos x)^{\cos x} - (\sin x)^{\sin x}$ and see that my conjecture is supported.
Therefore, I tried to prove an equivalent statement $$ \cos x \ln \cos x> \sin x \ln \sin x$$.
I tried to use Calculus to prove the statement.
Let $$f(x)= \cos x \ln \cos x-\sin x \ln \sin x$$
then $$f'(x)=-\sin x\ln \cos x -\cos x \ln \sin x -\sin x - \cos x.$$
From here I don't know how to proceed.
Any help to prove the inequality will be much appreciated!
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(sin x + cos x)^cos x = sin x^cos x + cos x^cos x + P1( sin x, cos x)
(sin x + sin x) ^sin x = sin x^sin x + cos x^sin x + P2( sin x, cos x)
Where P1 and P2 are polynomials with factors sin x and cos x.
But cos x > sin x in range 0 < x < π /4 , therefore:
(sin x + cos x)^cos x > (sin x + sin x) ^sin x;
⇒ sin x^cos x + cos x^cos x + P1( sin x, cos x) > sin x^sin x + cos x^sin x
*
*P2( sin x, cos x)
⇒ cos x^cos x > sin x^sin x + cos x^sin x - sin x^cos x + P2(sin x, cos x) - P1( sin x, cos x)
Lim (cos x^sin x - sin x^cos x ) =1 when x → 0
Lim(cos x^sin x - sin x^cos x ) = 0 when x → π /4
⇒ 0 < (cos x^sin x - sin x^cos x ) < 1 in range 0 < x < π /4
Lim [ P2( sin x, cos x) - P1( sin x, cos x) ] = 0 when x → 0
Lim [ P2( sin x, cos x) - P1( sin x, cos x) ] = 0 when x → π /4
⇒ [ P2( sin x, cos x) - P1( sin x, cos x) ] ≈ 0 in range 0 < x < π /4
Therefore :in range 0 < x < π /4, cos x^cos x > sin x^sin x
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If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{x^2-4)^2}=2x^2-4+2x\sqrt{x^2-4}$
Now,
$a^2-ax=2x^2-4+2x\sqrt{x^2-4}-x^2-x\sqrt{x^2-4}=x^2+x\sqrt{x^2-4}-4=xa-4$
Why am I not getting the intended value?
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HINT: write $a$ in the form $$a=\frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}=\frac{(\sqrt{x+2}+\sqrt{x-2})^2}{4}$$
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|
Calculating limits of areas of circle
If $r$ is the radius of circle then I calculated $f(t)$ and $g(t)$ to be
$$f(t)=\frac{1}{2}r^2t-\frac{1}{2}r^2\sin t=\frac{1}{2}r^2(t-\sin t)$$
$$AB^2=4r^2\sin^2 \frac{t}{2}= 2AC^2+2AC^2\cos(\pi-t)$$
$$g(t)=\frac{1}{2}AC^2\sin (\pi-t)-\frac{1}{2}r^2(t-\sin t )= \frac{r^2\sin^2 \frac{t}{2}}{1-\cos t}-\frac{1}{2}r^2(t-\sin t)$$
$$\frac{f(t)}{g(t)} = \frac{1}{\frac{2\sin^2 \frac{t}{2}\sin t}{(1-\cos t)(t-\sin t)}-1}$$
Thus $$\ \lim_{x\rightarrow 0 } \frac{f(t)}{g(t)} = 0$$
Is this working correct? Intuitively we can see that as $t$ approaches $0$, line segment $AB$ becomes smaller and thus $f(t)$ approaches $0$ and $g(t)$ approaches the area $ABC$ giving a limit of $0$ which is my answer.
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Alternatively:
$$\begin{align}
f(t)=&\frac{1}{2}r^2t-\frac{1}{2}r^2\sin{t}=\frac{1}{2}r^2(t-\sin t). \\
AC=&r\tan{\frac{t}{2}}. \\
S_{ABC}=&\frac{1}{2}AC^2\sin{(\pi-t)}=\frac{1}{2}r^2\tan^2{\frac{t}{2}}\sin{t}=\frac{1}{2}r^2\frac{1-\cos t}{1+\cos t}\sin t. \\
g(t)=&S_{ABC}-f(t)=\frac{1}{2}r^2(\frac{1-\cos t}{1+\cos t}\sin t-t+\sin t)= \\
=&\frac{1}{2}r^2\frac{2\sin t-t(1+\cos t)}{1+\cos t}\\
\lim_\limits{t\to 0} \frac{f(t)}{g(t)}=&\lim_\limits{t\to 0} \frac{(t-\sin{t})(1+\cos t)}{2\sin t-t(1+\cos t)}=2.\end{align}$$
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|
inequality involving the product of lengths of edges of quadrilateral Is there a constant $C$ such that for every quadrilateral with sides $a,b,c,d$ and area $S$, the following inequality holds:
$$abcd\ge C\cdot S^2$$
My attempt:
It is known, that $$S^2\le (s-a)(s-b)(s-c)(s-d)=\frac{1}{16}(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)$$ with possible equality, so we have to have
$${abcd}\ge\frac{C}{16}(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)$$
Of course $C\le 1$ (take $a=b=c=d$). Let's try to prove that $C=1$ works fine, since I found no counterexample.
The inequality we want to prove is equivalent to
$$a^4+b^4+c^4+d^4+8abcd\ge 2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)$$
and I stucked.
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There is no positive constant $C$. Take a non degenerate convex quadrilateral $ABCD$ and then move $D$ to $A$ along the initial side $DA$. Then the product $abcd$ goes to zero (because $d=|DA|$ goes to zero and $c=|DC|$ remains bounded). Moreover $S$ is greater or equal to the area of the triangle $ABC$ which is positive and does not change.
$$0\leftarrow abcd\geq C\cdot S^2\geq C|\triangle ABC|^2>0$$
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|
A right angle at the focus of a hyperbola
$P$ is a point on a hyperbola. The tangent at $P$ cuts a directrix at point $Q$. Prove that $PQ$ subtends a right angle to the focus $F$ corresponding to the directrix.
I have tried to use the general equation of the hyperbola and gradient method to show, but too many unknowns and I can't continue. I tried to show $m_1 m_2 = -1$, but I stuck halfway.
Note (From @Blue). This property holds for all conics, except circles, which have no directrix. For ellipses and hyperbolas, the property holds for either focus-directrix pair. A proof incorporating this level of generality would be nice to see.
We can restate the property in a way that includes the circle as a limiting case:
$P$ is a point on a conic with focus $F$. The line perpendicular to $\overline{PF}$ at $F$ meets the tangent at $P$ in a point on the directrix corresponding to $F$; if $P$ is a vertex, then the perpendicular, tangent, and directrix are parallel, meeting at a point "at infinity". In the case of a circle, the perpendicular is parallel to the tangent (so that they "meet" in a point on a "directrix at infinity").
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Here is a solution for parabola. You can use the similar way for hyperbola.
Let $y^2=2px$ be an equation of our parabola, $P(x_1,y_1)$.
Hence, $F\left(\frac{p}{2},0\right)$ and $x=-\frac{p}{2}$ is an equation of the directrix.
If $x_1=\frac{p}{2}$ then since $yy_1=p(x+x_1)$ is an equation of the tangent,
$Q\left(-\frac{p}{2},0\right)$ and $\measuredangle QFP=90^{\circ}$.
Let $x_1\neq\frac{p}{2}$.
Hence, we can calculate slopes: $$m_{PF}=\frac{y_1}{x_1-\frac{p}{2}}.$$
$x_Q=-\frac{p}{2}$.
Thus, $yy_Q=p\left(-\frac{p}{2}+x_1\right)$, which gives $Q\left(-\frac{p}{2},\frac{p\left(x_1-\frac{p}{2}\right)}{y_1}\right)$ and
$$m_{FQ}=\frac{\frac{p\left(x_1-\frac{p}{2}\right)}{y_1}-0}{-\frac{p}{2}-\frac{p}{2}}=\frac{\frac{p}{2}-x_1}{y_1}$$
and since
$$m_{PF}\cdot m_{FQ}=\frac{y_1}{x_1-\frac{p}{2}}\cdot\frac{\frac{p}{2}-x_1}{y_1}=-1,$$
we are done!
For hyperbola we obtain:
Let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be an equation of our hyperbola, $P(x_1,y_1)$.
Hence, $F\left(\sqrt{a^2+b^2},0\right)$ and $x=\frac{a^2}{\sqrt{a^2+b^2}}$ is an equation of the directrix.
If $x_1=\sqrt{a^2+b^2}$ then since $\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1$ is an equation of the tangent,
$Q\left(\frac{a^2}{\sqrt{a^2+b^2}},0\right)$ and $\measuredangle QFP=90^{\circ}$.
Let $x_1\neq\sqrt{a^2+b^2}$.
Hence, we can calculate slopes: $$m_{PF}=\frac{y_1}{x_1-\sqrt{a^2+b^2}}.$$
$x_Q=\frac{a^2}{\sqrt{a^2+b^2}}$.
Thus, $\frac{a^2}{\sqrt{a^2+b^2}}\cdot\frac{x_1}{a^2}-\frac{y_Qy_1}{b^2}=1$, which gives $Q\left(\frac{a^2}{\sqrt{a^2+b^2}},\frac{b^2}{y_1}\left(\frac{x_1}{\sqrt{a^2+b^2}}-1\right)\right)$ and
$$m_{FQ}=\frac{\frac{b^2}{y_1}\left(\frac{x_1}{\sqrt{a^2+b^2}}-1\right)-0}{\frac{a^2}{\sqrt{a^2+b^2}}-\sqrt{a^2+b^2}}=\frac{\frac{b^2}{y_1}\left(x_1-\sqrt{a^2+b^2}\right)}{a^2-(a^2+b^2)}=\frac{x_1-\sqrt{a^2+b^2}}{-y_1}$$
and since
$$m_{PF}\cdot m_{FQ}=-1,$$
we are done!
|
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Divergent series $\sum _{n=1}^{\infty } \frac{1}{\sqrt{k}} \cos (kx)$ Working in the series above and using maclaurin euler formula i get
$$\sum _{n=1}^{\infty } \frac{\sqrt{2 \pi } \sqrt{\frac{2 \pi n}{x \sqrt{\frac{4 \pi ^2 n^2}{x^2}-1}}+1}}{\sqrt[4]{x^2-4 \pi ^2 n^2}}+\frac{\sqrt{\frac{\pi }{2}}}{\sqrt{x}}=\sum _{n=1}^{\infty } \frac{1}{\sqrt{k}} \cos (k x)$$
but series diverge , it maybe something wrong?
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I am not sure that your approach will get you the result.
This is my hint. The given series does not converge if $x$ is a integer multiple of $2\pi$, i.e $x=2\pi m$ with $m\in\mathbb{Z}$ because
$$\sum _{k=1}^{\infty } \frac{\cos (k 2\pi m)}{\sqrt{k}} =\sum _{k=1}^{\infty } \frac{1}{\sqrt{k}}=+\infty$$
Otherwise, note that (see Prove $\frac{1}{2} + \cos(x) + \cos(2x) + \dots+ \cos(nx) = \frac{\sin(n+\frac{1}{2})x}{2\sin(\frac{1}{2}x)}$ for $x \neq 0, \pm 2\pi, \pm 4\pi,\dots$)
$$\sum_{k=1}^n\cos(kx)= \frac{\sin((2n+1)x/2)}{2\sin(x/2)}-\frac{1}{2}$$
and use Dirichlet's Test.
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Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!
|
Hint: by the AM-HM (arithmetic-harmonic mean) inequality:
$$
\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \;\;\iff\;\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c}
$$
Let $x=a+b+c \in (0,\frac{3}{2}]\,$, then the expression to be minimized can be written as:
$$
a+b+c+\frac1a+\frac1b+\frac1c \ge a+b+c+\frac{9}{a+b+c} = x + \frac{9}{x}
$$
The function $f(x)=x + \frac{9}{x}$ is decreasing on $(0,\frac{3}{2}]\,$, so $f(x) \ge f(\frac{3}{2})=\frac{15}{2}\,$ for $x \in (0,\frac{3}{2}]$.
The minimum value of $\frac{15}{2}$ is attained when $x=\frac{3}{2}$ and AM=HM i.e. $a=b=c=\frac{x}{3}=\frac{1}{2}\,$.
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|
Factorization of polynomial with prime coefficients I'm interested in the problem linked with this answer.
Let $ f(x) = a_n + a_1 x + \dots + a_{n-1} x^{n-1} $ be polynomial with distinct $a_i$ which are primes.
(Polynomials like that for $n= 4 \ \ \ \ f(x) = 7 + 11 x + 17 x^2 + 19 x^3 $)
*
*Is it for some $n$ possible that $x^n-1$ and $f(x)$ have some common divisors?
(Negative answer would mean that it is possible to generate circulant non-singular matrices with any prime numbers)
In other words
*
*$x^n-1$ has roots which lie (as complex vectors) symmetrically in complex plane on the
unit circle, can such root be also a root of $f(x) = a_n + a_1 x +
\dots + a_{n-1} x^{n-1}$ in general case where $a_i$ are constrained
as above?
|
This is certainly possible. The easiest way is to use pairs of twin primes and $n=4$. Such as
$$
f(x)=7+5x+11x^2+13x^3
$$
where $f(-1)=0$ and $x+1$ is a common factor of $f(x)$ and $x^4-1.$
Extending the same idea to third roots of unity. Consider
$$
f(x)=7+5x+17x^2+29x^3+31x^4+19x^5.
$$
Because $7+29=5+31=17+19=36$ we easily see that the third roots of unity $\omega=e^{\pm 2\pi i/3}$ are zeros of $f(x)$ as $f(\omega)=36(1+\omega+\omega^2)=0$. Therefore $f(x)$ has a common factor $\Phi_3(x)=x^2+x+1$ with $x^3-1$ and therefore also with $x^6-1$.
For an example of an odd $n$ I found the following. As
$$53=3+13+37=5+17+31=11+19+23$$ is the sum of three disjoint triples of primes, we can, as above, show that
$$
f(x)=3+5x+11x^2+13x^3+17x^4+19x^5+37x^6+31x^7+23x^8
$$
has the common factor $x^2+x+1$ with $x^9-1$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis of symmetry $y=x$ ?
$$\left(
\begin{array}{cc}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
For a clockwise rotation of $\frac{\pi}{4}$, $\sin{-\frac{\pi}{4}}=\frac{-1}{\sqrt{2}}$ and $\cos{-\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$
$$\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
$$X=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$Y=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$y=x^2$$
$$\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)=\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)^2$$
$$\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{x^2}{2}+\frac{2xy}{2}+\frac{y^2}{2}$$
$$-\sqrt{2}x+\sqrt{2}y=x^2+2xy+y^2$$
$$x^2+2xy+y^2+\sqrt{2}x-\sqrt{2}y=0$$
Have I made a mistake somewhere?
|
Let us start with general conic section
$$Ax^2+Bxy+Cy^2+Dx+Ey+F = 0$$
or equivalently, we can write it as
$$\begin{pmatrix} x & y & 1 \end{pmatrix}\begin{pmatrix} A & B/2 & D/2\\ B/2 & C & E/2\\ D/2 & E/2 & F \end{pmatrix}\begin{pmatrix} x\\ y\\ 1 \end{pmatrix}=0$$
(we will denote the above 3x3 matrix with $M$)
So, let's say you are given a conic section $v^\tau M v = 0$ and let's say we want to rotate it by angle $\varphi$. We can represent appropriate rotation matrix with
$$Q_\varphi=\begin{pmatrix} \cos \varphi& -\sin\varphi & 0\\ \sin\varphi & \cos\varphi & 0\\ 0 & 0 & 1\end{pmatrix}$$
Now, $Q_\varphi$ represents anticlockwise rotation, so we might be tempted to write something like $$(Q_\varphi v)^\tau M (Q_\varphi v) = 0$$ to get conic section rotated by angle $\varphi$ anticlockwise. But, this will actually produce clockwise rotation. Think about it - if $v$ should be a point on the rotated conic, then $Q_\varphi v$ is a point on conic before rotation, thus, the last equation actually means that the new conic rotated anticlockwise will produce the old conic.
So, let us now do your exercise. You have conic $y = x^2$, so matrix $M$ is given by $$ M =\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & -1/2\\ 0 & -1/2 & 0\end{pmatrix}$$ and you want to rotate your conic clockwise by $\pi/4$, so choose $$Q_{\pi/4}=\begin{pmatrix} \cos \frac\pi4& -\sin\frac\pi4 & 0\\ \sin\frac\pi4 & \cos\frac\pi4 & 0\\ 0 & 0 & 1\end{pmatrix}.$$
Finally, we get equation $$\begin{pmatrix} x & y & 1 \end{pmatrix}\begin{pmatrix} \cos \frac\pi4& -\sin\frac\pi4 & 0\\ \sin\frac\pi4 & \cos\frac\pi4 & 0\\ 0 & 0 & 1\end{pmatrix}^\tau\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & -1/2\\ 0 & -1/2 & 0\end{pmatrix}\begin{pmatrix} \cos \frac\pi4& -\sin\frac\pi4& 0\\ \sin\frac\pi4 & \cos\frac\pi4 & 0\\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix} x\\ y\\ 1 \end{pmatrix}=0$$
or simplified $$x^2-2xy+y^2-x\sqrt 2-y\sqrt 2 = 0.$$
|
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|
Inverse of a non-trivial exponential function I am asked to determine the inverse function of this function,
$$f(x)=2^{x}+3^{x}$$
The inverse function can not be found explicitly, since there is no way to explicitly clear x, but this does not mean that it has no inverse.
I could show a way to find the inverse of this function
Edit How can we find the inverse of this function with the problems that it presents in the clearance of the x?
|
You could, for example, write the solution to $2^x + 3^x = y$ as a series in powers of $y-2$:
$$\eqalign{x &= \frac{y-2}{\ln(6)} - \left(\ln(3)^2 + \ln(2)^2\right) \frac{(y-2)^2}{2 \ln(6)^3}\cr +& \left(2 \ln(3)^4 - \ln(3)^3\ln(2)+6\ln(3)^2\ln(2)^2-\ln(3)\ln(2)^3 + 2 \ln(2)^4\right)\frac{(y-2)^3}{6 \ln(6)^5}\cr - &\left(3 \ln(3)^6 - 4 \ln(3)^5 \ln(2) + 18 \ln(3)^4 \ln(2)^2 - 10 \ln(3)^3 \ln(2)^3 + 18 \ln(3)^2 \ln(2)^4 - 4 \ln(3)\ln(2)^5 + 3 \ln(2)^6 \right) \frac{(y-2)^4}{12 \ln(6)^7}
\cr + &\ldots }$$
|
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|
Find n so that the following converges $\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} - \frac 1 {13x + 1} \right) dx$ Question
Determine $n$ such that the following improper integral is convergent
$$ \int_1^{+ \infty} \left(
\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
\right) dx
$$
I'm not sure how to go about this.
Working
This is convergent if
$$
\lim_{b \to + \infty} \int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx
$$
exists.
The indefinite integral is
\begin{equation*}
\begin{aligned}
\int \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right)dx
& = \int \left(\frac{nx^2}{x^3 + 1}
\right) - \int \left(\frac{1}{13x + 1} \right)dx \\
&= \frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1)
\end{aligned}
\end{equation*}
Which gives
\begin{equation*}
\begin{aligned}
&\lim_{b \to + \infty}
\int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx \\
&= \lim_{b \to + \infty} \left[\frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1) \right]_1^b \\
&= \lim_{b \to + \infty}
\left(\left[\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) \right] - \left[\frac{n}{3} \cdot \ln(2) - \frac{1}{13} \cdot \ln(14) \right] \right)
\\ &= \lim_{b \to + \infty}
\left(\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) - \frac{n}{3} \cdot \ln(2) + \frac{1}{13} \cdot \ln(14) \right)
\\ &= \lim_{b \to + \infty}
\left(
\frac{n}{3}
\left(
\ln(b^3 + 1) - \ln(2)
\right)
- \frac{1}{13}
\left(
\ln(13b + 1)
- \ln(14)
\right)
\right)
\\ &= \lim_{b \to + \infty}
\left(
\frac{n}{3}
\left(
\ln \left(\frac{b^3 + 1}{2}\right)
\right)
- \frac{1}{13}
\left(
\ln \left(\frac{13b + 1}{14}\right)
\right)
\right)
\end{aligned}
\end{equation*}
I've tried to use L'Hopital's from here as I have the form $(+ \infty ) - ( +
\infty)$. But things went pretty south.
So I'm sure there's a better approach.
|
A simpler approach:
Since we know that:
$$\int_1^\infty\frac1{x^a}~\mathrm dx<\infty\iff a>1$$
It follows that if we know
$$\lim_{x\to\infty}\frac{\frac{nx^2}{x^3+1}-\frac1{13x+1}}{1/x^a}=c\ne0$$
Then the integral converges iff $a>1$.
For $n=\frac1{13}$, we find that using $a=2$ satisfies the limit, so it converges for $n=\frac1{13}$.
For $n\ne\frac1{13}$, we find that using $a=1$ satisfies the limit, so it diverges for $n\ne\frac1{13}$
|
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|
Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
using L'Hopital's rule we get
$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$
$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$
is this possible to do without Lhopita's rule
|
$$
\begin{aligned}
\lim _{x\to \infty }\left(\arctan\left(\frac{1+x}{4+x}\right)-\frac{\pi \:}{4}\right)x
& = \lim _{t\to 0}\left(\frac{\arctan \left(\frac{t+1}{4t+1}\right)-\frac{\pi }{4}}{t}\right) \\
& = \lim _{t\to 0}\left(\frac{\left(\frac{\pi \:}{4}-\frac{3}{2}t+o\left(t\right)\right)-\frac{\pi \:}{4}}{t}\right)\\
& = \color{red}{-\frac{3}{2}}
\end{aligned}
$$
Solved with Taylor expansion:
$$\arctan \left(\frac{t+1}{4t+1}\right) = \frac{\pi }{4}-\frac{3t}{2}+\frac{15t^2}{4}-\frac{33t^3}{4}+15t^4+\ldots \:$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
show that $xyz\in N$ if $x^n+y^n+z^n\in Z$ Let $x,y,z\in R$ and such for any postive integers $n$ have
$$a_{n}=x^n+y^n+z^n\in Z$$
show that $xyz\in Z$
I have use
$$a_{n+3}=(x+y+z)a_{n+2}-(xy+yz+xz)a_{n+1}+xyza_{n}$$
since $2(xy+yz+xz)=(x+y+z)^2-(x^2+y^2+z^2)\in Z$
but $xy+yz+xz$ can't integers,because $2(xy+yz+xz)$ is integers
|
$2(xy+xz+yz)=(x+y+z)^2-(x^2+y^2+z^2)\in\mathbb Z$ and since
$$(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+xz+yz)-3xyz,$$ we see that $6xyz\in\mathbb Z$.
Now, $$x^4+y^4+z^4=$$
$$=(x+y+z)^4-4(x+y+z)^2(xy+xz+yz)+2(xy+xz+yz)^2+4(x+y+z)xyz$$ or
$$3(x^4+y^4+z^4)=$$
$$=3(x+y+z)^4-12(x+y+z)^2(xy+xz+yz)+6(xy+xz+yz)^2+12(x+y+z)xyz,$$
which says $6(xy+xz+yz)^2\in\mathbb Z$ and from here $xy+xz+yz\in\mathbb Z$.
Thus, $3xyz\in\mathbb Z$ and $4(x+y+z)xyz\in\mathbb Z$.
Let $xyz\in\mathbb Z$ is wrong.
Then $x+y+z$ divided by $3$.
But
$$x^5+y^5+z^5=$$
$$=(x+y+z)^5-5(x+y+z)^3(xy+xz+yz)+5(x+y+z)(xy+xz+yz)^2+$$
$$+5((x+y+z)^2-(xy+xz+yz))xyz,$$
which says that $5((x+y+z)^2-(xy+xz+yz))xyz\in\mathbb Z$,
which gives $xy+xz+yz$ divided by $3$.
Now, from $$x^6+y^6+z^6=$$
$$=(x+y+z)^6-6(x+y+z)^4(xy+xz+yz)+9(x+y+z)^2(xy+xz+yz)^2-$$
$$-2(xy+xz+yz)^3+(6(x+y+z)^3-4(x+y+z)(xy+xz+yz))xyz+3x^2y^2z^2$$ we obtain $3x^2y^2z^2\in\mathbb Z$, which is contradiction.
Thus, $xyz\in\mathbb Z$ and we are done!
|
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|
Horizontal line(s) that intersect $f(x)=x-2+\frac{5}{x}$ in two points. Compute exactly the value(s) of $q$ for which the horizontal line $y = q$
intersects the graph of $f(x)$ in two points that are located on a distance $4$ from each other.
While I found the two lines as $y = 4$ and $y = -8$
, I do not know how to find it by any mathematical means.
|
Intersection of $y$ and $f$:
$$
q = x - 2 + 5/x \iff \\
qx = x^2 -2x + 5 \wedge x \ne 0
$$
So we remember to discard a solution $x=0$ and solve the quadratic equation:
$$
0 = x^2 - (2 + q)x + 5 = (x - (1+q/2))^2 - (1+q/2)^2 + 5 \iff \\
(x-(1+q/2))^2 =(1+q/2)^2 - 5 \iff \\
x = (1+q/2) \pm \sqrt{(1+q/2)^2 -5}
$$
As usual we have no, one or two solutions, depending on the argument of the square root. To get two intersections we need
$$
(1 + q/2)^2 - 5 > 0 \iff \\
(1 + q/2)^2 > 5 \iff \\
\lvert 1 + q/2 \rvert > \sqrt{5}
$$
For the positive case this means
$$
q > 2 (\sqrt{5}-1) = 2.47\dotsb
$$
For the negative case
$$
-(1 + q/2) > \sqrt{5} \iff \\
1 + q/2 < - \sqrt{5} \iff \\
q < -2(\sqrt{5} + 1) = -6.47\dotsb
$$
Now we apply the condition on the distance of the intersection points:
$$
\lvert x_1 - x_2 \rvert = 4 \iff \\
4 = x_1 - x_2 = 2 \sqrt{(1+q/2)^2 - 5} \iff \\
4 = (1 + q/2)^2 - 5 \iff \\
9 = (1 + q/2)^2 \iff \\
\pm 3 = 1 + q/2 \iff \\
q/2 = 2 \vee q/2 = -4 \iff \\
q = 4 \vee q = -8
$$
|
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|
Find the sum of all values of $a$ satisfying that there exist positive integers $a,b$ satisfying $(a-b) \sqrt{ab}=2016$
Find the sum of all values of $a$ satisfying that there exist positive integers $a,b$ satisfying $$(a-b) \sqrt{ab}=2016$$
my try
let $a= x^2$ , $b= y^2$ $\to$ $(a-b) \sqrt{ab}$=$(x^2-y^2)(xy)=2016$
is this approach appropriate for this problem ? please elaborate your help
|
A start of a solution.
The prime factorization of $2016$ is $2^5\cdot3^2\cdot7$.
In order for $\sqrt{ab}$ to be a integer, both $a$ and $b$ must be multiples of squares and the same factors, i.e. $a=zx^2,b=zy^2, x>y$. Thus:
$$(a-b)\sqrt{ab}=z^2(x^2-y^2)xy = 2016$$
Use the fact that if neither of two numbers is divisible by 3, the difference of their squares must be. Thus, $z$ can not be a multiple of $3$, and since it cannot be a multiple of $7$, it must be one of $1,2,4$.
So now we must solve:
$$(x^2-y^2)xy = 2016 = 2^5\cdot 3^2\cdot7$$
$$(x^2-y^2)xy = 504 = 2^3\cdot 3^2\cdot7$$
$$(x^2-y^2)xy = 126 = 2^1\cdot 3^2\cdot7$$
|
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|
ratio between the area of square $wxyz$ and the area of square $ abcd$ equal?
$ABCD$ is a square and $H$ is an interior point, which divides it for four triangles. If $W$, $X,$ $Y$ and $Z$ are the centroids of triangles $AHD$, $AHB$, $BHC$ and $CHD$ respectively ,
then what is the ratio between the area of the square $WXYZ$ and the area of the square $ABCD$ ?
Can anyone provide me a hint or a help to go ?
Thank you very much
|
$XW=YZ=\frac{1}{3}BD$, $XY=WZ=\frac{1}{3}AC$ and $XYZW$ is square.
Thus, the ratio is $\frac{2}{9}$ because $S_{ABCD}=\frac{1}{2}AC\cdot BD$ and
$$\frac{S_{XYZW}}{S_{ABCD}}=\frac{\frac{1}{3}AC\cdot\frac{1}{3}BD}{\frac{1}{2}AC\cdot BD}=\frac{2}{9}$$
For example, let $M$ is a midpoint of $BH$.
Hence, $$\frac{XY}{AC}=\frac{MY}{MC}=\frac{1}{3}$$
|
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|
How to simplify an expression involving several square roots without a calculator? $$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
This type of questions are common in the university entrance examinations in our country but the calculators are not allowed can someone help me to find the way to simplify the expression.
|
Hint: apply $\frac{1}{\sqrt{a} - \sqrt{b}}=\frac{\sqrt{a} + \sqrt{b}}{a - b}$.
|
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|
How to find out $ k $ in this problem If $ k $ is an non negative integer, such that $ 24^k $ divides $ 13! $, what is $ k $ then?
Explanation needed, as I am new to factorials.
|
The highest power of a prime $p$ in $n!$ is given by $$\sum_{k=1} \Biggl \lfloor \frac{n}{p^k}\Biggr \rfloor $$
$24^k = (2^3 \times 3)^k = 2^{3k} \times 3^k$
Highest Power of $2$ in $13!$
Let the highest power be $x$.
$x= \lfloor \frac{13}{2} \rfloor + \lfloor \frac{13}{2^2} \rfloor + \lfloor \frac{13}{2^3} \rfloor = 10 $
Similarly
Highest power of $3$ in $13!$ =$ 5$
Since $24^k= 2^{3k} \times 3^k$
$0 \le k \le \frac{10}{3}$ (Since $2^{3k}$ is present in $24^k$ and maximum power of $2$ is $10$)
and $0 \le k \le 5$ (Since maximum power of $3$ in $13!$ is $5$)
Combining the above two inequalities
$0 \le k \le 3$
So $k=0,1,2,3$
You can read more about the formula used here
|
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|
Deteminant of a matrix If $P= \left(\begin{array}{ll}
A & B \\
C & D \\
\end{array}\right)$, where A, B, C, D are tridiagonal matrices. Then how to define its determinant ?
|
See Determinants of Block Matrices.
A block matrix is a matrix that is defined using smaller matrices, called blocks. For example,
$P= \left(\begin{array}{ll}
A & B \\
C & D \\
\end{array}\right)$
where $A$, $B$, $C$, and $D$ are themselves matrices, is a block matrix. In the specific example
$A= \left(\begin{array}{ll}
a_{(1,1)} & a_{(1,2)} \\
a_{(2,1)} & a_{(2,2)} \\
\end{array}\right)$, $B= \left(\begin{array}{ll}
b_{(1,1)} & b_{(1,2)}& b_{(1,3)} \\
b_{(2,1)} & b_{(2,2)}& b_{(2,3)} \\
\end{array}\right)$, $C= \left(\begin{array}{ll}
c_{(1,1)} & c_{(1,2)} \\
c_{(2,1)} & c_{(2,2)} \\
c_{(3,1)} & c_{(3,2)} \\
\end{array}\right)$, $D= \left(\begin{array}{ll}
d_{(1,1)} & d_{(1,2)}& d_{(1,3)} \\
d_{(2,1)} & d_{(2,2)}& d_{(2,3)} \\
d_{(3,1)} & d_{(3,2)}& d_{(3,3)} \\
\end{array}\right)$
Therefore, it is the matrix $P$
$$P= \left(\begin{array}{ll}
p_{(1,1)} & p_{(1,2)} & \dots & p_{(1,n)} \\
p_{(2,1)} & p_{(2,2)} & \dots & p_{(2,n)} \\
\vdots & \vdots & \ddots & \vdots \\
p_{(n,1)} & p_{(n,2)} & \dots & p_{(n,n)}
\end{array}\right)=
\left(\begin{array}{ll}
a_{(1,1)} & a_{(1,2)} & b_{(1,1)} & b_{(1,2)}& b_{(1,3)}\\
a_{(2,1)} & a_{(2,2)} & b_{(2,1)} & b_{(2,2)}& b_{(2,3)} \\
c_{(1,1)} & c_{(1,2)} & d_{(1,1)} & d_{(1,2)}& d_{(1,3)} \\
c_{(2,1)} & c_{(2,2)} & d_{(2,1)} & d_{(2,2)}& d_{(2,3)} \\
c_{(3,1)} & c_{(3,2)} & d_{(3,1)} & d_{(3,2)}& d_{(3,3)} \\
\end{array}\right)$$
See Blok Matrix
Where the determinant of a matrix $P$ of arbitrary size $n \times n$ can be defined by the a Leibniz formul or the Laplace formula.
|
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|
If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then differentiating we get
$$\frac{f'(x)}{f(x)}=\sum_{k=0}^n \frac{A_k}{x+k}$$
hence
$$\frac{f'(x)}{f(x)}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$
any clue here?
|
Multiplying by $x+7$ and for $x = -7$ you have $A_7 = -\frac{n!}{7!(n-7)!}$
|
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|
How to solve $n$-th derivative of logarithm function? How to slove the $n$-th derivative of logarithm function
$$\frac{d^n}{dx^n} \ln \left(\frac{p}{ax+b}+\frac{1-p}{b}\right) $$
where $p \in [0,1]$, $a$ and $b$ are positive numbers
|
These are the first derivatives. Try to find a general formula for the $n-$th
$$
\begin{array}{r|r}
\text{order} & \text{derivative}\\
\hline
1 & -\frac{a}{b+a x}+\frac{a-a p}{b-a (p-1) x}\\
2 & \frac{a^2}{(b+a x)^2}-\frac{(a-a p)^2}{(b-a (p-1) x)^2} \\
3 & 2\left(-\frac{ a^3}{(b+a x)^3} +\frac{ (a-a p)^3}{(b-a (p-1) x)^3}\right)\\
4 & 6\left(\frac{ a^4}{(b+a x)^4}-\frac{ (a-a p)^4}{(b-a (p-1) x)^4} \right)\\
5 & 24\left(-\frac{a^5}{(b+a x)^5} +\frac{(a-a p)^5}{(b-a (p-1) x)^5}\right)\\
6 & 120 \left(\frac{a^6}{(b+a x)^6}-\frac{(a-a p)^6}{(b-a (p-1) x)^6}\right) \\
7 & 720 \left(-\frac{a^7}{(b+a x)^7}+\frac{(a-a p)^7}{(b-a (p-1) x)^7}\right) \\
8 & 5040 \left(\frac{a^8}{(b+a x)^8}-\frac{(a-a p)^8}{(b-a (p-1) x)^8}\right) \\
9 & 40320 \left(-\frac{a^9}{(b+a x)^9}+\frac{(a-a p)^9}{(b-a (p-1) x)^9}\right) \\
10 & 362880 \left(\frac{a^{10}}{(b+a x)^{10}}-\frac{(a-a p)^{10}}{(b-a (p-1) x)^{10}}\right) \\
11 & 3628800 \left(-\frac{a^{11}}{(b+a x)^{11}}+\frac{(a-a p)^{11}}{(b-a (p-1) x)^{11}}\right) \\
12 & 39916800 \left(\frac{a^{12}}{(b+a x)^{12}}-\frac{(a-a p)^{12}}{(b-a (p-1) x)^{12}}\right) \\
\end{array}
$$
I my humble opinion it is
$$f^{(n)}(x)= (n-1)! \left[(-1)^n\left(\frac{a}{a x+b}\right)^n+(-1)^{n+1} \left(\frac{a(1-p)}{b+a (1-p) x}\right)^n\right]$$
|
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|
Find $\lvert m\rvert$ given $\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$ I've already found the solution, but I was wondering if there is a faster or alternative method. My solution is found below:
$$\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$$
$$(m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}=3$$
$$\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$(m+9)-3(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+3(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}-(m-9)=27$$
$$-3(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+3(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}=9$$
$$-(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}=3$$
$$-(m+9)^\frac{1}{3}(m-9)^\frac{1}{3}\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)=3$$
$$-(m+9)(m-9)\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$-(m+9)(m-9)(27)=27$$
$$m^2-81=-1$$
$$m=\pm4\sqrt{5}$$
$$\lvert m\rvert=4\sqrt{5}$$
In the third line, I cubed both sides, and then in the forth line, I expanded the left side of the equation using the binomial theorem. Is there a faster or alternative way to do this type of question? If in the question there was a higher-index root (ie. instead of the cube roots there is a higher index), I don't think my method would work, because it would take too long to apply the binomial theorem. How would solve a question in this form?
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Let $\sqrt[3]{m+9}=a$, $-3=b$ and $-\sqrt[3]{m-9}=c$.
Hence, we have
$$a+b+c=0.$$
But $$a^3+b^3+c^3-3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=$$
$$=(a+b)^3+c^3-3ab(a+b+c)=(a+b+c)((a+b)^2-(a+b)c+c^2-3ab)=$$
$$=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
We see that $$ a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}((a-b)^2+(a-c)^2+(b-c)^2),$$
which says that $a^2+b^2+c^2-ab-ac-bc=0$ for $a=b=c$ only.
In our case it gives
$$\sqrt[3]{m+9}=-3=-\sqrt[3]{m-9},$$ which is impossible.
Thus, our equation it's
$$a^3+b^3+c^3-3abc=0$$ or
$$m+9-27-(m-9)-9\sqrt[3]{m^2-81}=0$$ or
$$m^2-81=-1,$$
which gives the answer:
$$\{4\sqrt{5},-4\sqrt{5}\}$$
|
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|
Area between arc and line I have been working on a problem for a few days now. This was a challenge problem on a lecture for Trigonometry. I managed to find an equation for the radius, but wasn't able to solve it.
Problem: Line $AB$ is drawn such that $\overline{AB} = 20$. Minor arc $AB$ is drawn with endpoints $AB$ such that the length of arc $AB$ is $21$. Find the area of the region bounded by the arc and the line.
Progress: First, draw the full diagram as shown below.
We can compute the area of $\angle{ABC}$. Using the Law of Cosines, we have that
\begin{align*}
\cos C &= \frac{a^2+b^2-c^2}{2ab} \\
&= \frac{2r^2-400}{2r^2}\\
&= 1-\frac{200}{r^2}
\end{align*}Therefore, $\angle ACB = \cos^{-1}\left(1-\frac{200}{r^2}\right)$. The length of the arc in terms of $\angle ACB$ is
\begin{align*}
&\frac{\angle ACB}{360}2\pi r \\
&= \frac{\angle ACB}{180}\pi r \\
&= \frac{\cos^{-1}\left(1-\frac{200}{r^2}\right)}{180}\pi r
\end{align*}We know that
\begin{align*}
&21 = \frac{\cos^{-1}\left(1-\frac{200}{r^2}\right)}{180}\pi r
\implies \\&\frac{3780}{\pi r} = {\cos^{-1}\left(1-\frac{200}{r^2}\right)}
\implies\\ &\cos\left(\frac{3780}{\pi r}\right) = 1-\frac{200}{r^2}
\end{align*}
However, I couldn't solve this equation. Any help would be appreciated. I can also visualize a calculus approach involving finding the area between two curves, but I want to solve it in an elementary way if possible.
|
hint
Let $t $ be the angle $\angle ACB .$
Then
$$21=rt $$
and
$$r\sin(t/2)=10 .$$
the area we want is the difference
$$S=\frac {r^2}{2}t-\frac{20.h}{2} $$
with
$$h=r\cos(t/2) $$
hence
$$S=\frac {21}{2}r-10\sqrt {r^2-100} $$
with $r $ satisfying
$$r\sin (\frac {21}{2}r)=10$$
|
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|
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) = \frac{(1 + \sin\theta + i\cos\theta)^2}{(1 + \sin\theta)^2 + \cos^2\theta}$$
$$=\frac{(1 + \sin\theta)^2 + \cos^2\theta + 2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
thus
$$x = \frac{(1 + \sin\theta)^2 + \cos^2\theta }{(1 + \sin\theta)^2 + \cos^2\theta} $$
$$y= \frac{2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
According to the binomial theorem,
$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k$$
we get
$$z = \frac{1}{(1 + \sin\theta)^2 + \cos^2\theta}\sum_{k=0}^n \binom{n}{k} ((1 + \sin\theta)^2 + \cos^2\theta)^{n-k}\cdot(2i(1 + \sin\theta)\cos\theta)^k$$
...and that is where I'm stuck. What do you think? Thanks for the attention.
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we know that $a+ib=\sqrt{a^{2}+b^{2}}\cdot e^{i\arctan \frac{b}{a}}\Rightarrow \frac{a+ib}{a-ib}=e^{2i\arctan \frac{b}{a}}$ and $\cos \theta +i\sin \theta =e^{i\theta }$ then:
$\left( \frac{1+\sin \theta +i\cos \theta }{1+\sin \theta -i\cos \theta } \right)^{n}=e^{i\cdot 2n\arctan \frac{\cos \theta }{1+\sin \theta }}=\cos \left( 2n\arctan \frac{\cos \theta }{1+\sin \theta } \right)+i\sin \left( 2n\arctan \frac{\cos \theta }{1+\sin \theta } \right)$
|
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|
Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$
Moving the things in RHS to LHS:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$
Writing everything above a common denominator:
$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$
Multiplying both sides with the denominator to cancel the denominator:
$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$
Multiplying the first two factors in every term:
$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$
Simplifying the first factors in every term:
$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$
Multiplying factors again:
$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$
Removing the parenthesis yields:
$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$
Which results in:
$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$
which is not correct. The correct answer is $x = \frac{5}{2}$.
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But as I suggested in my comment, the better approach is to first simplify each side separately . . .
\begin{align*}
&\frac{1}{x-1} - \frac{1}{x-2} = \frac{1}{x-3} - \frac{1}{x-4}\\[4pt]
\implies\;
&\frac{(x-2)-(x-1)}{(x-1)(x-2)}=\frac{(x-4)-(x-3)}{(x-3)(x-4)}\\[4pt]
\implies\;
&\frac{-1}{(x-1)(x-2)}=\frac{-1}{(x-3)(x-4)}\\[4pt]
\implies\;&(-1)(x-3)(x-4)=(-1)(x-1)(x-2)\\[4pt]
\implies\;&(x-3)(x-4)=(x-1)(x-2)\\[4pt]
\implies\;&x^2-7x+12=x^2-3x+2\\[4pt]
\implies\;&4x=10\\[4pt]
\implies\;&x=\frac{5}{2}\\[4pt]
\end{align*}
Also note, as warned about in Mark Bennet's reply, we have to worry about canceling algebraic factors that might actually be equal to zero. In the steps above, the factors
$$(x-1),\;(x-2),\;(x-3),\;(x-4)$$
were canceled in the cross-multiplication step, but that was safe since, based on the original equation, none of those factors had the potential to be zero.
|
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|
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
We know that product is maximum when difference between $x$, $y$ and $z$ is minimum.
So, I assumed $x=3$, $y=4$ and $z=4$.
Now putting this value in $xyz+xy+yz+zx$ I got my answer $88$. But actual answer is $78$. Where am I doing it wrong?
|
the way the question is worded, $x$ and $y$ and $z$ cannot be equal (they must instead be distinct) so $y$ and $z$ cannot be $4$.
so the correct answer is choosing $5$, $4$ and $2$ for $x$, $y$ and $z$.
|
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|
Solution for the inequality $x \lt \frac 1 x$ I am unable to find correct solution for the inequality:
$x \lt \frac 1 x, x \in \Bbb R, x \ne 0$.
My solution first takes $x > 0$, then we have:
$x < \frac 1x$ (multiply both sides by $x$)
$\iff x^2 < 1$ (taking square root)
$\iff x < 1 \lor x < -1$
$\iff 0 < x < 1$ since $x > 0$.
Then similarly, for $x < 0$,
$x < \frac 1x$ (multiply both sides by $x< 0$)
$\iff x^2 > 1$ (taking square root)
$\iff x > 1 \lor x > -1$
$\iff -1 < x < 0 $ since $x < 0$
But this does not agree with the solution $(x < -1) \lor (0 < x < 1)$
Which part of my solution is incorrect above?
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$$x<\frac { 1 }{ x } \\ x-\frac { 1 }{ x } <0\\ \frac { { x }^{ 2 }-1 }{ x } <0\\ \frac { x\left( x-1 \right) \left( x+1 \right) }{ { x }^{ 2 } } <0\\ x\left( x-1 \right) \left( x+1 \right) <0\\ x\in \left( -\infty ;-1 \right) \cup \left( 0;1 \right) $$
|
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|
Is it possible to solve for $\theta$ in this multivariable equation? Question: Given the equation $$c^2=a^2+(\frac{pq}{\sqrt{q^2\sin^2{\theta}+p^2\cos^2{\theta}}})^2-2a\frac{pq}{\sqrt{q^2\sin^2{\theta}+p^2\cos^2{\theta}}}\cos{\theta},$$
is it possible to solve for $\theta$ in terms of $a,c,p,$ and $q$?
Original Problem: I am attempting to develop an algorithm to solve an extension of a spherical codes problem. This image shows a partial cutaway of a spherical codes solution (a center sphere with as many smaller spheres packed onto it as possible, cut away to show the inner sphere).
I'd like to extend this problem to ellipsoids, so essentially to take an ellipsoid with given dimensions and pack as many spheres of another given dimension on its surface as possible. The equation above is a central part of my solution to this problem, and is derived from the law of cosines with $\theta$ as the opposite angle to side $c$ and the side $b$ substituted by the large fraction ($\frac{pq}{...}$).
Attempts: I've plugged the equation into Matlab's solve feature, Mathematica's Reduce function (both to no avail), and Mathematica's Solve function , which returned this:
$x = \arccos{\sqrt{(-((a^2 p^2 pq^2)/(a^4 p^4 - 2 a^2 c^2 p^4 +
c^4 p^4 - 4 a^2 p^2 pq^2 - 2 a^4 p^2 q^2 +
4 a^2 c^2 p^2 q^2 - 2 c^4 p^2 q^2 + 4 a^2 pq^2 q^2 +
a^4 q^4 - 2 a^2 c^2 q^4 +
c^4 q^4)) + (c^2 p^2 pq^2)/(a^4 p^4 - 2 a^2 c^2 p^4 +
c^4 p^4 - 4 a^2 p^2 pq^2 - 2 a^4 p^2 q^2...}}$ (truncated, else it would continue for 21 total lines)
However, after substituting variables into the full expression above, I determined that it too returned false values (e.g. $\arccos(n)$ where $n>1$).
I was wondering if there is a way to solve this by hand or if it is impossible to do so? Thank you in advance.
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For this problem, I should write two equations
$$c^2=a^2+x^2-2ax \cos(\theta) \tag 1$$
$$x=\frac{pq}{\sqrt{q^2 \sin^2(\theta)+p^2\cos^2(\theta)}}\tag 2$$ From $(1)$ we get $$\cos(\theta)=\frac{a^2-c^2+x^2}{2 a x}\tag 3$$ Replacing in $(2)$, squaring, simplifying and so on, we end with $$Ax^4+Bx^2+C=0\tag 4$$ (which is a "simple" quadratic equation in $x^2$) where $$A=q^2-p^2$$
$$B=-2 \left(q^2 \left(a^2+c^2\right)+p^2 (a^2-c^2) \right)$$ $$C=q^2 \left(\left(a^2-c^2\right)^2+4 a^2 p^2\right)-p^2 \left(a^2-c^2\right)^2$$ For the quadratic, we have $$\Delta=B^2-4AC=16a^2q^2\left(a^2p^2+(p^2-c^2)(p^2-q^2)\right)$$ then $x$ and, back to $(3)$, $\cos(\theta)$ and $\theta$.
|
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|
Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove
$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$
I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for $\sin(x)-(x-\frac{x^3}{6})$ i.e. $x^5/5!-x^7/7!+x^9/9!+\dots$.
I don't see why $x^5/5!-x^7/7!+x^9/9!+\dots$ should be positive for all positive real $x$. Any idea?
|
For any $x>0$ we have $\sin(x)<x$, hence by applying $\int_{0}^{t}(\ldots)\,dx$ to both sides we get $1-\cos t < \frac{t^2}{2}$. By applying $\int_{0}^{x}(\ldots)\,dt$ to both sides we get $x-\sin x<\frac{x^3}{6}$, which can be rearranged as $\sin(x)>x-\frac{x^3}{6}$. By performing the same trick again we also get $\sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}$ and the wanted inequality is proved for any $x>0$.
An equivalent approach is noticing that
$$ \iint_{0\leq a \leq b \leq x}(a-\sin a)\,da\,db,\qquad \iiint_{0\leq a\leq b\leq c\leq x}(a-\sin a)\,da\,db\,dc $$
are clearly positive.
We are dealing with odd functions, hence the reversed inequality holds over $\mathbb{R}^-$.
|
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Partial fractions - integration $$\int \frac{4}{(x)(x^2+4)} $$
By comparing coefficients,
$ 4A = 4 $,
$A = 1$
$1 + B = 0 $,
$B= -1 $
$xC= 0 $,
$C= 0 $
where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$.
So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$.
And my final answer is
$\ln|x| - x \ln |x^2 + 4| + C$.
However my answer is wrong , the answer is - $\ln|x| - \frac{1}{2} \ln |x^2 + 4| + C$.
Where have I gone wrong?
|
I'm not necessarily sure how you got a factor of $x$ in the second term, but here is my go at it:
$$\frac{4}{x(x^2+4)}= \frac{A}{x} + \frac{Bx+C}{x^2+4}=\frac{A(x^2+4)+x(Bx+C)}{x(x^2+4)}=\frac{Ax^2+Bx^2+Cx+4A}{x(x^2+4)}$$
Then, by equating the numerator of the first and last expressions we have:
$$4=x^2(A+B) +Cx +4A$$
Equating coefficients:
$$A+B=0,C=0,4A=4\Rightarrow A=1,B=-1 \text{ and } C=0$$
Then we have:
$$\int{\frac{4}{x(x^2+4)}}\,dx=\int{\frac{1}{x}-\frac{x}{x^2+4}\,dx} = \int\frac{dx}{x}\, - \int{\frac{x}{x^2+4}\,dx}$$
In the second integral, we let $u=x^2+4$ then $du=2x dx\Rightarrow xdx=\frac{1}{2}du$. Hence
$$\int{\frac{dx}{x}} - \frac{1}{2}\int{\frac{du}{u}}=\ln \left|x\right|+\frac{1}{2}\ln\left|u\right|+C=\ln|x|-\frac{1}{2}\ln|x^2+4|+C$$
as required.
|
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|
Obtain the value of $\int_0^1f(x) \ dx$, where $f(x) = \begin{cases} \frac {1}{n}, & \frac{1}{n+1}
Is the function Riemann integrable? If yes, obtain the value of $\int_0^1f(x) \ dx$
$f(x) =
\begin{cases}
\frac {1}{n}, & \frac{1}{n+1}<x\le\frac{1}{n}\\
0, & x=0
\end{cases}$
My attempt
$f$ is bounded and monotonically increasing on $[0,1]$. Also, $f$ has infinite discontinuities but only one limit point. Therefore $f$ is Riemann integrable. Now, to calculate the integration
$\int_0^1f(x) \ dx=\int_{1/2}^{1}1 \ dx + \int_{1/3}^{1/2}\frac{1}{2} \ dx + \int_{1/4}^{1/3}\frac{1}{3} \ dx+...$
$=\sum_{n=1}^\infty \frac{1}{n^2}-\frac{1}{n}+\frac{1}{n+1}$
How do I proceed from here? How do I calculate these summations? I know $\sum \frac{1}{n}$ is $\log 2$, but not the other two summations.
|
since $(0,1) = \bigcup_{n=1}^{\infty} ( \frac1{n+1} , \frac1{n })$. (pairwise disjoint union)
$$\int_{0}^1f(x)dx = \sum_{n=1}^{ \infty} \int_{1/n+1}^{1/n}f(x)dx = \sum_{n=1}^{ \infty} \int_{1/n+1}^{1/n}\frac1ndx=\sum_{n=1}^{ \infty} \frac1{n^2 }- \frac1{(n+1)n}= \frac{\pi^2}{6}-1
$$
Given that $ \frac1{(n+1)n}= \frac1{n}- \frac1{n+1}$ then
$$\sum_{n=1}^{ \infty} \frac1{(n+1)n}=\sum_{n=1}^{ \infty} (\frac1{n}- \frac1{n+1})=1$$
|
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|
Integration with logarithmic expression. I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$
I have almost the same solution where there's a $$3\ln|\csc\theta-\cot\theta|+3\cos\theta+C.$$
The problem is when the substitution comes in. I end up having
$$3\ln\frac{|3-\sqrt{9-x^2}|}{2x}+\sqrt{9-x^2}$$ and when I tried to simplify it further, it resulted to $$3\ln\left|3-\sqrt{9-x^2}\right|-3\ln|2x|+\sqrt{9-x^2}.$$ I hope you could help me to tell where i did wrong.
By the way, I set $a=3$ and $x=\frac{3}{2}\sin\theta$.
|
It is easy to mess up with the constants in such integrals. But we can temporarily ignore them because a simple scaling of the variable and of the integrand can bring us to
$$I:=\int\frac{\sqrt{1-t^2}}tdt.$$
This calls for the substitution $1-t^2=u^2$, giving $dt/t=-u\,du/2t^2$ and
$$I\propto\int\frac{u^2}{u^2-1}du=\int\frac{u^2-1+1}{u^2-1}du=u-\text{artanh u}=\sqrt{1-t^2}-\text{artanh}\sqrt{1-t^2}.$$
Now back to the original problem, we now know that the solution will be of the form
$$\int\frac{\sqrt{9-4x^2}}xdx=\lambda\left(\sqrt{1-\frac49x^2}-\text{artanh}\sqrt{1-\frac49x^2}\right).$$
Differentiating, we get
$$\frac{\sqrt{9-4x^2}}x=-\lambda\frac{\dfrac49x}{\sqrt{1-\dfrac49x^2}}\left(1-\frac1{1-\left(1-\dfrac49x^2\right)}\right)=\lambda\frac{\sqrt{1-\dfrac49x^2}}x$$ and $\lambda=3$.
The final solution is
$$\sqrt{9-4x^2}-3\text{ artanh}\frac{\sqrt{9-4x^2}}3.$$
Note:
We can use the following transformation
$$\text{artanh}\sqrt{1-t^2}=\log\sqrt{\frac{1+\sqrt{1-t^2}}{1-\sqrt{1-t^2}}}=\log\left|\frac{1+\sqrt{1-t^2}}t\right|.$$
|
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|
How to find $\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}$ I came across this problem a few days ago and I have not been able to solve it. Wolfram Alpha says the answer is 1/2 but the answer I came up with is 0. Can anyone see what is wrong with my work and/or provide the correct way of solving this problem?
$$\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x[(x)(1+\frac{1}{x})]^x}{[(x)(1+\frac{1}{x})]^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x^{x+1}(1+\frac{1}{x})^x}{x^x(1+\frac{1}{x})^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x^{x+1}e}{x^xe} $$
$$\lim_{x \to \infty} \frac{x-x}{1} $$
$$\lim_{x \to \infty} \frac{0}{1} $$
$$0$$
I understand my mistakes may be simple and trivial, but I'm trying to learn. Thank you for your help!
|
Consider $$y=\frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}=\frac{ex^{x+1}}{(x+1)^x}-x$$ Now
$$z=\frac{ex^{x+1}}{(x+1)^x}\implies \log(z)=1+(x+1)\log(x)-x\log(x+1)=1+\log(x)-x \log\left(1+\frac 1x\right)$$ Now, using Taylor $$\log\left(1+\frac 1x\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
x^3}+O\left(\frac{1}{x^4}\right)$$ makes $$\log(z)=\log \left({x}\right)+\frac{1}{2 x}-\frac{1}{3
x^2}+O\left(\frac{1}{x^3}\right)$$ $$z=e^{\log(z)}=x+\frac{1}{2}-\frac{5}{24 x}+O\left(\frac{1}{x^2}\right)$$ which finally makes $$y=\frac{1}{2}-\frac{5}{24 x}+O\left(\frac{1}{x^2}\right)$$ showing the limit and how it is approached.
Using $x=10$ which is really small, the exact value would be $$\frac{100000000000 e}{25937424601}-10\approx 0.480153$$ while the above approximation would simply give $\frac{23}{48}\approx 0.479167$.
|
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|
Infinite power summation where no common ratio can be seen Problem in book states to evaluate the sum:
$$\sum_{i=0}^\infty \frac{i^2}{4^i}$$
I can come up with the series S = $\frac{1^2}{4^1} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + \frac{4^2}{4^4} + \frac{5^2}{4^5} + ...$
And 4S = $ 1 + \frac{2^2}{4^1} + \frac{3^2}{4^2} + \frac{4^2}{4^3} + \frac{5^2}{4^4} + \frac{6^2}{4^5} + ...$
If I subtract S from 4S I get 3S = $ 1 + \frac{3}{4^1} + \frac{5}{4^2} + \frac{7}{4^3} + \frac{9}{4^4} + \frac{11}{4^5} + ...$
I cannot find a common ratio to plug in to $\frac{1}{1-r}$ in order to figure this out. I asked my professor for some assistance, and he said the answer should be $\frac{20}{27}$, but I cannot figure out what steps he took to get that. He didn't explain it very well, and all other resources I've checked only show how to compute a geometric sum when there is a common ratio like $\frac{1}{2}$.
How do I determine a common ratio to figure out the sum? I tried division by sequential terms, and that gave me a common ratio of $\frac{5}{16}$, but that left me with an answer of S = $\frac{16}{27}$, which isn't correct.
What am I doing wrong? Or what am I missing?
|
Consider $$\sum_{i=0}^\infty {i^2}{x^i}=\sum_{i=0}^\infty {[i(i-1)+i]}{x^i}=\sum_{i=0}^\infty {i(i-1)}{x^i}+\sum_{i=0}^\infty {i}{x^i}=x^2\sum_{i=0}^\infty {i(i-1)}{x^{i-2}}+x\sum_{i=0}^\infty {i}{x^{i-1}}$$ that is to say
$$\sum_{i=0}^\infty {i^2}{x^i}=x^2\left(\sum_{i=0}^\infty {x^{i}} \right)''+x\left(\sum_{i=0}^\infty {x^{i}} \right)'$$ and you know that
$$\sum_{i=0}^\infty {x^{i}} =\frac{1}{1-x}\qquad \text{for}\qquad x <1$$ Now, things look to be simple.
When finished, replace $x$ by $\frac 14$ to get the final result.
|
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|
Let $a = \frac{9+\sqrt{45}}{2}$. Find $\frac{1}{a}$ I've been wrapping my head around this question lately:
Let
$$a = \frac{9+\sqrt{45}}{2}$$
Find the value of
$$\frac{1}{a}$$
I've done it like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}}$$
I rationalize the denominator like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \times (\frac{9-\sqrt{45}}{9-\sqrt{45}})$$
This is what I should get:
$$\frac{1}{a} = \frac{2(9-\sqrt{45})}{81-45} \rightarrow \frac{1}{a}=\frac{18-2\sqrt{45}}{36})$$
Which I can simplify to:
$$\frac{1}{a}=\frac{2(9-\sqrt{45})}{36}\rightarrow\frac{1}{a}=\frac{9-\sqrt{45}}{18}$$
However, this answer can't be found in my multiple choice question here:
Any hints on what I'm doing wrong?
|
$$\frac { 1 }{ a } =\frac { 9-\sqrt { 45 } }{ 18 } =\frac { 3\left( 3-\sqrt { 5 } \right) }{ 18 } =\frac { 3-\sqrt { 5 } }{ 6 } $$
|
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|
In expansion of $(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ . Find the coefficient of $x^{28}$ I am not able to apply binomial theorem here
$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$
Please help me to find the coefficient of$ x^{28}$
Any help will be appreciated.
|
\begin{eqnarray*}
(1+x+x^{2}+x^{3}+\cdots+x^{27})(1+x+x^{2}+x^{3}+\cdots+x^{14})^{2} \\
= (1+x+x^{2}+\cdots+x^{27}) (1+2x+3x^{2}+\cdots +14x^{13}+15x^{14} +14x^{15} \cdots +2x^{27}+ x^{28}) \\
= \cdots+x^{28}( \underbrace{1+2 +\cdots + 15}_{120} + \underbrace{14+ \cdots +2}_{104})+ \cdots
\end{eqnarray*}
EDIT
|
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|
Solve in positive integers, $ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$ Solve in positive integers, $$ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$$
My attempt :
$ x^3y^3+x^6-y^6+3xy(x^2-y^2)^2$
$=x^3y^3+3x^2y^2(x^2-y^2)+3xy(x^2-y^2)^2+(x^2-y^2)^3$
$=(xy+(x^2-y^2))^3$
$=(x^2+xy-y^2)^3 = 1$
so $x^2+xy-y^2 = 1$
Please suggest how to proceed.
Thank you, AmateurMathGuy and lhf.
Please check my work on Induction for Fibonacci sequence.
$$t_{k+2}-3t_{k+1}+t_{k}=0$$
$$y_{k+2}-3y_{k+1}+y_{k}=0$$
$(t_1,y_1)=(3,1)\to(x_1,y_1)=(1,1)$
$(t_2,y_2)=(7,3)\to(x_2,y_2)=(2,3)$
$(t_3,y_3)=(18,8)\to(x_3,y_3)=(5,8)$
$(t_4,y_4)=(47,21)\to(x_4,y_4)=(13,21)$
Since $1, 2, 3, 5, 8, 13, 21$ are in Fibonacci sequence, we predict that $(x_n,y_n)=(F_{2n-1},F_{2n})$
and will prove by Induction.
$(x_n,y_n)=(F_{2n-1},F_{2n})$ is true for $n=1, 2, 3, 4$
Suppose that $(x_k,y_k)=(F_{2k-1},F_{2k})$ is true,
Since $y_{k+1}-3y_k+y_{k-1}=0$, so $y_{k+1}=3y_k-y_{k-1}=3F_{2k}-F_{2k-2}$
$=3F_{2k}-(F_{2k}-F_{2k-1})=2F_{2k}+F_{2k-1}=F_{2k}+F_{2k+1}=F_{2k+2}$
$x_{k+1}=\frac{t_{k+1}-y_{k+1}}{2}=\frac{(3t_k-t_k)-3y_{k-1}-y_{k-1}}{2}$
$=3\left(\frac{t_k-y_k}{2}\right)-\left(\frac{t_k-1-y_{k-1}}{2}\right)=3x_k-x_{k-1}$
Similarly, $x_{k+1}=F_{2k+1}$, so $(x_n,y_n)=(F_{2n-1},F_{2n})$
Answer : $(x_n,y_n)=(F_{2n-1},F_{2n})$
|
Cassini's identity for the Fibonacci numbers is
$$
F_{n-1}F_{n+1}-F_{n}^{2}=(-1)^{n}
$$
Therefore
$$
1
=F_{2n-1}F_{2n+1}-F_{2n}^{2}
=F_{2n-1}(F_{2n-1}+F_{2n})-F_{2n}^{2}
=F_{2n-1}^2+F_{2n-1}F_{2n}-F_{2n}^{2}
$$
and $x=F_{2n-1}$, $y=F_{2n}$ are solutions of $x^2+xy-y^2 = 1$.
Similarly,
$$
F_{2n-1}^2-F_{2n-1}F_{2n-2}-F_{2n-2}^{2} = 1
$$
and $x=-F_{2n-1}$, $y=F_{2n-2}$ are solutions of $x^2+xy-y^2 = 1$. But the OP only wants positive solutions.
It remains to prove that these are the only solutions.
|
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|
Show that $7 |(n^6 + 6)$ if $7 ∤ n$, $∀ n ∈ ℤ$
Show that $7 |(n^6 + 6)$ if $7 ∤ n$, $∀ n ∈ ℤ$
I need to prove this by the little Fermat's theorem.
My attempt
$n^6 \equiv -6 \pmod 7$
To show $7 ∤ n$ I need to show that $N$ is not congruent to $0$ mod $7$.
as $-6 \equiv 1\pmod 7$
$n^6 \equiv 1\pmod7$
But now, How can I show $N$ is not congruent to $0$ mod $7$ ?
|
Because $n^6+6=n^6-1+7$ and $n^6-1$ divisible by $7$.
$$n^6-1=(n-1)(n+1)(n^2-n+1)(n^2+n+1)$$
and now easy to check $n\equiv\pm1,\pm2\pm3(\mod7)$.
|
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|
Constrained Optimisation Problem: Confusion with Algebra/Contradiction I completed the following constrained optimisation problem:
Maximum and minimum values of $f(x,y) = x^2 + 2xy + y^2$ on the ellipse $g(x,y) = 2x^2 + y^2 - xy - 4x = 0$.
$\nabla f = \lambda \nabla g$
$\therefore (2x + 2y, 2x + 2y) = \lambda(4x - y - 4, 2y - x)$
We have a system of 3 equations in 3 unknowns:
$2x + 2y = \lambda(4x - y - 4)$,
$2x + 2y = \lambda(2y - x)$,
$2x^2 + y^2 - xy - 4x = 0$.
$\\$
$2x + 2y = \lambda(4x - y - 4)$
$\implies \lambda = \dfrac{2x + 2y}{4x - y - 4} \forall \ 4x - y - 4 \not= 0$
$\\$
$2x + 2y = \lambda(2y - x)$
$\implies \dfrac{2x + 2y}{2y - x} \forall \ 2y - x \not= 0$
$\\$
$\therefore \dfrac{2x + 2y}{4x - y - 4} = \dfrac{2x + 2y}{2y - x} \forall \ y \not= \dfrac{x}{2}, y \not= 4x - 4$ Remember this part: This is where my confusion lies.
But if we continue with our reasoning, we get the following (correct) points of potential maxima/minima:
$(x,y) = (0,0)$, $(x,y) = (1, -1)$, $(x,y) = (2,2)$, and $(x,y) = \left(\dfrac{2}{7}, \dfrac{-6}{-7}\right)$.
And, finally, using these points we find that $f_{max} = 16$ at $(x,y) = (2,2)$, and $f_{min} = 0$ at $(x,y) = (0,0)$ and $(x,y) = (1,-1)$.
But here's the problem: We had above that $y \not= \dfrac{x}{2}$; therefore, $(x,y) = (0,0)$ would get us $0 \not= \dfrac{0}{2} \implies 0 \not= 0$, which is obviously false. Also, as we just saw, $(0,0)$ is a solution for $f_{min}$!
The algebra here is confusing me, since we cannot have $y = \dfrac{x}{2}$, but $(0,0)$ is apparently the minima. Is this not a contradiction? Did I do something wrong? Or am I supposed to discard the solution $(0,0)$ and just have $f_{min} = 0$ at $(1, -1)$? What am I misunderstanding?
I would greatly appreciate it if people could please take the time to clarify this.
|
You don't have to solve like that and exclude $x=2y$ or other limitations.
The system
$
\left\{ {\begin{array}{*{20}{l}}
{2x + 2y = \lambda \left( {4x - y - 4} \right)} \\
{2x + 2y = \lambda \left( {2y - x} \right)} \\
{2{x^2} + {y^2} - xy - 4x = 0}
\end{array}} \right.
$
can be solved looking at the first two equations
which say
$\lambda \left( {4x - y - 4} \right)= \lambda \left( {2y - x} \right) $
and then
$4x-y-4=2y-x\to y=\frac{1}{3} (5 x-4)$
substitute in the third
$2 x^2-\frac{1}{3} (5 x-4) x-4 x+\frac{1}{9} (5 x-4)^2=0$
expand and simplify
$7 x^2-16 x+4=0\to x_1=\dfrac{2}{7};\;x_2=2$
$y_1=-\dfrac{6}{7};\;y_2=2$
Plug these values in the first equation to get
$4+4=\lambda(8-2-4)\to \lambda=4$
Substitute in the second
$2x+2y=4(2y-x) \to 6(x-y)=0 \to x=y$
which can be plugged in the third and give
$2x^2+x^2-x^2-4x^2=0\to x_3=0;\;y_3=0$
this last $(0,0)$ is the minimum for the given function, while according to my calculations $(2,2)$ gives the maximum
|
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|
Finding divisibility rules Find the divisibility test for 13.
Let's first find a test for small 3-digit numbers first and then find a general test. We'll be finding tests which involves separation of last digit. These tests will be similar to the tests of 19, 23, 29, etc.
Let $\overline { abc }$ be any number such that $\overline { abc } =100a+10b+c$. Now assume that $\overline { abc }$ is divisible by 13. Then
\begin{align}
\overline { abc } &\equiv 0 \pmod{13}\\
100a+10b+c &\equiv 0 \pmod{13}\\
10\left( 10a+b \right) +c &\equiv 0 \pmod{13}\\
10\overline { ab } +c &\equiv 0 \pmod{13}.
\end{align}
Now that we have separated the last digit from the number, we have to find a way to use it:
What does coefficient of $\overline { ab }$ mean? And why does it have to be 1?
make the coefficient of $\overline { ab }$ 1.
In other words, we have to find an integer $n$ such that $10n\equiv 1 \bmod{13}$.
It can be observed that the smallest $n$ which satisfies this property is 4. Now we can multiply the original equation by 4 and simplify it:
\begin{align}
10\overline { ab } +c &\equiv 0 \pmod{13} \\
40\overline { ab } +4c &\equiv 0 \pmod{13}\\
\overline { ab } +4c &\equiv 0 \pmod{13}.
\end{align}
Aha! We have found out that if $\overline { abc } \equiv 0 \bmod{13},$ then $\overline { ab } +4c\equiv 0 \bmod{13}$. In other words, to check if a 3-digit number is divisible by 13, we can just remove the last digit, multiply it by 4, and then add it to the rest of the two digits.
Now that we have found out the test for 3-digit numbers, let's find out a general divisibility test. Let $\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 }{ x }_{ 1 } } $ be any $n$ digit number. Then
\begin{align}
\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 }{ x }_{ 1 } } &\equiv 0 \pmod{13}\\
10\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +{ x }_{ 1 } &\equiv 0 \pmod{13}\\
40\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +4{ x }_{ 1 } &\equiv 0 \pmod{13}\\
\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +4{ x }_{ 1 } &\equiv 0 \pmod{13}.
\end{align}
Hence, a number is a multiple of 13 if we add 4 times the last digit to the rest of the number and the resulting number is still divisible by 13. $_\square$
link of original article: https://brilliant.org/wiki/divisibility-rules/
|
Although you haven't asked a question, here is a quicker way to reach the same result. Using the division algorithm, write your number as $N=10m + d$ where $0\le d \le 9$.
Then $13 \mid N\iff 13 \mid 4N \iff 13\mid 40m+4d \iff 13\mid m+4d$
(The first equivalence in the chain follows because $4$ and $13$ are relatively prime.)
|
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Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\left\lceil{\frac{1}{n^2}}\right \rceil = 5 - 4$$
Therefore, for $n \in \mathbb R$, where $n \neq 0$
$$4 < \left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil \le 5$$
$$4 < \left\lceil{4}+\frac{1}{n^2}\right \rceil \le 5$$
$$0 < \left\lceil\frac{1}{n^2}\right \rceil \le 1$$
I feel kind of without purpose once I hit point. Should I be scrapping the inequality and solving for $n$ instead?
|
Your proof is probably okay if you specify all statements are $\iff$ results. That way one can reverse direction. Otherwise you are assuming what you need to prove.
$[\frac {4n^2 + 1}{n^2}] = 5 \iff$
$[4 + \frac 1{n^2}] = 5 \iff$
$[\frac 1{n^2}] = 5 - 4$ (actually do you know if that is true? I don't feel comfortable with that step. I'd suggest you first prove that if $k\in \mathbb Z$ then $[k + y] = k + [y]$. [$*$].)
$\iff 0< \frac 1{n^2} \le 1$.
And then continue.
$\iff 1 \le n^2$ which as $n$ is a non-zero integer is true.
And we would be done.
But this is a backwards proof and only works because every step was $\iff$. If any one step was not reversible, the proof would fail[$**$]. To do a direct proof, we take a new sheet of paper and work forward.
$n\ne 0; n\in \mathbb Z \implies$
$n^2 > 0; n^2\ne 0; n^2 \in \mathbb Z \implies$
$n^2 \ge 1 \implies$
$0 < \frac 1{n^2} \le 1 \implies$
$[\frac 1{n^2}] = 1 \implies$
$[4 + \frac 1{n^2}] = 4 + 1 \implies$ [$*$]
$[\frac {4n^2 + 1}{n^2}] = 5$.
======
[$*$]
Let $[k + y] = m$ were $k \in \mathbb Z$.
then $m - 1 < k +y \le m$
so $(m-k) -1 < y \le (m-k)$
so $[y] = m-k = [k+y] - k$.
====
P.S.
As GEdgar points out a more elegant proof would be
$[\frac {4n^2 +1}{n^2}] = 5 \iff$
$4 < \frac {4n^2 + 1}{n^2} \le 5 \iff$
$4n^2 < 4n^2 + 1 \le 5n^2 \iff$
$0 < 1 \le n^2$ which is true.
To make this a forward proof:
$n \ne 0 \implies n^2 \ge 1$ so
$0< 1 \le n^2$
$4n^2 < 4n^2 + 1 \le 5n^2$
$4 < \frac{4n^2 + 1}{n^2} \le 5$
$[\frac {4n^2 + 1}{n^2} ] = 5$.
[$**$]
A proof that $1 = 5$:
$1 = 5 \implies$
$1 -3 = 5-3 \implies$
$-2 = 2\implies$
$(-2)^2 = 2^2\implies$
$4 =4 $ which is true.
So $1 = 5$. Notice every step of the proof is valid except the conclusion.
$-2 = 2 \implies (-2)^2 = 2^2$ but
$-2 = 2 \not \Leftarrow (-2)^2 = 2^2$.
A Conclusion $\to $ true statement is only valid if we do: conclusion $\Leftarrow ... \Leftarrow$ true statement.
But conclusion $\implies ... \implies$ true statement, is NEVER valid.
|
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|
Find all of the points of the form $(x, −1)$ which are $4$ units from the point $(3, 2)$
Find all points of the form $(x, −1)$, which are $4$ units from the point
$(3, 2)$.
I understand the distance formula, I think. Also, I don't know how to format for math on here yet so I apologize for that.
\begin{align}
&\text{dist} = 4 = \sqrt{(x-3)^2 + (2-(-1))^2)} \\
&\implies 16 = (x-3)^2 + (2 + 1)^2 \tag{square both sides} \\
&\implies 16 = x^2 -6x + 18 \tag{expand} \\
&\implies 0 = x^2 -6x +2 \tag{zero on the left}
\end{align}
Next should be finding the factors which would give me the answers but I have no idea what would work for this.
$$(x + \text{something})(x - \text{something}).$$
I must be doing something wrong or missing something. Please correct me.
|
Suppose you have two points in a 2-D coordinate system such as $(x_{1}, y_{1}) = (3, 2)$ and $(x_{2}, y_{2}) = (x, -1)$ and distance between them is 4 units. Hence you can write as
\begin{align}
& d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} \\
\implies & 4 = \sqrt{(3-x)^{2} + (2-(-1))^{2}} \\
\implies & 16 = 9 + x^{2} - 6x +9 \tag{squaring both sides}\\
\implies & x^{2} - 6x + 2 = 0
\end{align}
This is a quadratic equation. The solution to a quadratic equation of the form $ax^{2} + bx + c$ can be found as $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$. Hence comparing the standard form and the equation above, we have $a = 1, b = -6$ and $c = 2$. Hence the roots will be
$$x = \dfrac{6 \pm \sqrt{36 - 8}}{2} = \dfrac{6 \pm 2\sqrt{7}}{2} = 3 \pm \sqrt{7}$$
|
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|
What is $\lim\limits_{n\to\infty}\sum\limits^n_{i=0}F_{2i}-\phi F_{2i-1}$? Firstly, the question is:
Evaluate
$$(1-0\phi)+(2-1\phi)+(5-3\phi)+(8-5\phi)+\cdots=\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}$$
where $F_0=1,$
$F_1=1$ and
$F_n=F_{n-1}+F_{n-2}$ for all integer n (even n negative).
I attempted the basic manipulations with Fibonacci numbers but ultimately I reached this:
$$\begin{align}
&\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}\\
=&\lim_{n\to\infty}\sum^n_{i=0}F_{2i-1}+F_{2i-2}-\phi F_{2i-2}-\phi F_{2i-3}\\
=&\lim_{n\to\infty}1+\sum^n_{i=0}(F_{i}-\phi F_{i})+F_{n+1}\\
=&\lim_{n\to\infty}1+\sum^n_{i=0}(\frac{1-\sqrt{5}}{2}F_{i})+F_{n+1}\\
\end{align}
$$
which probably doesn't get me anywhere.
(I'd also like the sum
$$\lim_{n\to\infty}\sum^n_{i=0}F_{2i-1}-\phi F_{2i-2}$$
but if the method works for this then there's no reason to explicitly calculate this for me)
|
Since $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_i \\ F_{i-1} \end{pmatrix} = \begin{pmatrix} F_{i+1} \\ F_i \end{pmatrix},$$ you are looking for the limit $$\lim_{N \rightarrow \infty} \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} F_{2i} \\ F_{2i-1} \end{pmatrix} = \sum_{i=0}^N \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{2i} \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$ Since $$\begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1-\phi & 1 \end{pmatrix} = (1-\phi) \cdot \begin{pmatrix} 1 & -\phi \end{pmatrix},$$ this is the series $$\sum_{i=0}^{\infty} (1-\phi)^{2i} \cdot \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \sum_{i=0}^{\infty} (1 - \phi)^{2i} = \phi.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve this trigonometric equation ${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0$
Solve the following equation. $${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0.$$
I can't figure out the way to solve this equation. This was my attempt
\begin{array}{l}
{\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0\\
\Leftrightarrow \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - 3(\frac{{\sin x}}{{\cos x}} - \frac{{\cos x}}{{\sin x}}) = 0\\
\Leftrightarrow \frac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\cos }^2}x + {{\sin }^2}x}} - 3\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}} = 0\\
\Leftrightarrow \frac{{1 + 2{{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0\\
\Leftrightarrow \frac{4}{{(1 - \cos 2x)(1 + \cos 2x)}} + 2 + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0\\
\Leftrightarrow \frac{4}{{1 - {{\cos }^2}2x}} + 2 + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0
\end{array}
I don't know how to solve after that. Can anyone help me?
|
As an alternative to Michael Rozenberg's answer, let $u=\tan x$. The equation becomes
$$u^2+{1\over u^2}-3\left(u-{1\over u}\right)=0$$
Multiplying through by $u^2$ to clear denominators, we wind up with
$$u^4-3u^3+3u+1=0$$
The quartic factors into two quadratics:
$$u^4-3u^3+3u+1=(u^2-u-1)(u^2-2u-1)$$
The rest follows as in Michael's answer.
|
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|
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n - L \rvert < \epsilon$
So $\lvert \frac{\sqrt {n^2 +2}}{4n+1} - \frac {1}{4} \rvert$ simplifies to $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}$
Now $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4} < \epsilon$ is
simplified to $\frac {4\sqrt {n^2 +2}}{16n} < \epsilon$ Then I would square everything to remove the square root and simplify fractions but I end up with $n> \sqrt{\frac{1}{8(\epsilon^2-\frac{1}{16}}}$
We can't assume $\epsilon > \frac{1}{4}$ so somewhere I went wrong. Any help would be appreciated.
|
Update If you really need to use a complete $\epsilon$-definition type proof, then using this method will only require that you additionally prove $\lim_{n\to\infty} \frac{n}{4n+1} = \frac{1}{4}$, which is considerably easier. (Just note that $\left|\frac{n}{4n+1}-\frac{1}{4}\right| = \frac{1}{4(4n+1)}$)
We will bound the terms eventually as follows:
For any $\epsilon > 0$ there exists some $N$ so that for all $n\geq N$: $$n^2+2 < n^2(1+\epsilon)$$
(i.e., if $2 < N^2\epsilon$), also clearly:
$$(1-\epsilon)n^2 < n^2+2$$
Putting these facts together, for $n\geq N$
$$\sqrt{1-\epsilon}\frac{n}{4n+1} = \frac{\sqrt{n^2(1-\epsilon)}}{4n+1} < \frac{\sqrt{n^2+2}}{4n+1} < \frac{\sqrt{n^2(1+\epsilon)}}{4n+1} = \sqrt{1+\epsilon}\frac{n}{4n+1} $$
Now taking limits we have:
$$
\lim_{n\to\infty} \sqrt{1-\epsilon}\frac{n}{4n+1}
\leq \lim_{n\to\infty} \frac{n^2+2}{4n+1}
\leq \lim_{n\to\infty} \sqrt{1+\epsilon}\frac{n}{4n+1}
$$
so
$$\frac{\sqrt{1-\epsilon}}{4} \leq \lim_{n\to\infty} \frac{n^2+2}{4n+1} \leq \frac{\sqrt{1+\epsilon}}{4}$$
as this holds for all $\epsilon > 0$, let $\epsilon \to 0^+$:
$$\frac{1}{4} = \lim_{\epsilon\to 0^+}\frac{\sqrt{1-\epsilon}}{4} \leq \lim_{n\to\infty} \frac{n^2+2}{4n+1} \leq \lim_{\epsilon\to 0^+}\frac{\sqrt{1+\epsilon}}{4} = \frac{1}{4}$$
so the result follows from the sandwich theorem.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding a substitution that eliminates the squared term from a cubic equation So I am a little stumped here, and it could be simple. I'm just not exactly sure how to approach this.
The question reads:
Find $\omega_0$ in the set of Complex numbers, such that the substitution $z = \omega - \omega_0$ reduces the cubic equation $z^3 + Az^2 + Bz +C = 0$ into $\omega^3 -m\omega -n =0$
... where I'm assuming those constants are real numbers.
My first attempt was to just do a straight substitution of $z= \omega - \omega_0$ into the first equation, then expand it, and try to solve that down for what $\omega_0$ should be, but I started to realize that that's probably not the way.
Am I just missing something here?
|
Letting $z=\omega - \omega_0$ and getting
$$ z^3 + Az^2 + Bz + C = \omega^3 + A'\omega^2 +B'\omega +C'$$
Can be accomplished by doing the following synthetic divisions
\begin{array}{r|ccccc}
& 1
& A
& B
& C
\\
-\omega_0 & 0
& -\omega_0
& \omega_0^2 - A\omega_0
& -\omega_0^3 + A\omega_0^2 - B\omega_0
\\
\hline
& 1
& -\omega_0 + A
& \omega_0^2 - A\omega_0 + B
& \color{red}{C'=-\omega_0^3+A\omega_0^2-B\omega_0+C}
\\
-\omega_0 & 0
& -\omega_0
& 2\omega_0^2 - A\omega_0
\\
\hline
& 1
& -2\omega_0+A
& \color{red}{B'=3\omega_0^2-2A\omega_0+B}
\\
-\omega_0 & 0
& -\omega_0
\\
\hline
& 1
& \color{red}{A' = -3\omega_0 + A}
\end{array}
So, if we want $A'=0$, then we need $\omega_0 = \frac 13A$
Letting $\omega_0 = \frac 13A$, we get this.
\begin{array}{r|ccccc}
& 1
& A
& B
& C
\\
-\frac 13A & 0
& -\frac 13A
& -\frac 29A^2
& \frac{2}{27}A^3 - \frac 13AB
\\
\hline
& 1
& \frac 23A
& -\frac 29A^2 + B
& \color{red}{C'=\frac{2}{27}A^3-\frac 13AB+C}
\\
-\frac 13A & 0
& -\frac 13A
& -\frac 19A^2
\\
\hline
& 1
& \frac 13A
& \color{red}{B'=-\frac 13A^2+B}
\\
-\frac 13A & 0
& -\frac 13A
\\
\hline
& 1
& \color{red}{A' = 0}
\end{array}
|
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|
Finding derivative of $\frac{x}{x^2+1}$ using only the definition of derivative I think the title is quite self-explanatory. I'm only allowed to use the definition of a derivative to differentiate the above function. Sorry for the formatting though.
Let $f(x) = \frac{x}{x^2+1}$
$$
\begin{align}
f'(x)&= \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac {\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{\frac{x+h}{x^2+2hx+h^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{\frac{(x+h)(x^2+1)-x(x^2+2hx+h^2+1)}{(x^2+1)(x^2+2hx+h^2+1)}}{h}\\
&= \lim_{h\to 0} \frac{\frac{hx^2-2hx-h^2+h}{(x^2+1)(x^2+2hx+h^2+1)}}{h}
\end{align}
$$
I'm currently stuck with simplify the fraction so that I can finally find the derivative. I'd really appreciate some advice on how to proceed with the problem.
|
Your last step is wrong, the actual steps will be,
$$=\lim_{h\to 0} \frac{\frac{-\color{red}{(hx^2-h+h^2x)}}{(1+x^2)(x^2+2hx+h^2+1)}}{\color{red}{h}}$$
$$=\lim_{h\to 0} \frac{\frac{-\color{blue}{(x^2-1+hx)}}{(1+x^2)(x^2+2hx+h^2+1)}}{\color{blue}{1}}$$
$$=\frac{1-x^2}{(1+x^2)^2}$$
|
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|
Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...
|
[ EDIT ] The question has been changed to require $\,x,y,z \ge 0\,$ now, which renders the following not applicable any longer.
(Too long for a comment.) Let $x=a+2, y=b+2, z=c+2\,$, then the conditions rewrite as:
$$\require{cancel}
\begin{cases}
\begin{align}
a b + b c + a c + 4(a + b + c) + \cancel{12} &= \cancel{12} \\
a b c + 2(a b + a c + b c) + 4(a + b + c) + \cancel{8} &= a+b+c + \cancel{8}
\end{align}
\end{cases}
$$
$$\require{cancel}
\iff\;\; \begin{cases}
\begin{align}
a b + b c + a c + 4(a + b + c) &= 0 \\
a b c + 2(a b + a c + b c) + 3(a + b + c) &= 0
\end{align}
\end{cases}
$$
Let $u = a+b+c\,$, then:
$$\require{cancel}
\begin{align}
a b + b c + a c &= -4 u \\
a b c &= 5u\end{align}
$$
Therefore $a,b,c$ are the roots of $t^3-ut^2-4 ut-5u\,$, and the problem reduces to finding for what values of $u$ the cubic has three real roots (other than $u=0$ which corresponds to $x=y=z=2\,$).
|
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|
For what values of $a$ does the system have infinite solutions? Find the solutions. The system is
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
ax+y+z & = & 1+a \\
x-y+z & = & 2+a
\end{array}
\right.$$
After row reducing I got
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
-(1+a)y+0 & = & 1+a \\
(1-a)z& = & 2-a + (1-a)(1+a)
\end{array}
\right.$$
In order to have an infinite set of solutions, I want to have $0=0$ in the last row, meaning that $a=1\Rightarrow 0=1,$ which is not what I want. Have I made any arithmetical mistake? Did this computation times now.
|
Something has gone wrong with the row reduction. First, we eliminate $x,$ which gives us $$\left\{\begin{array}{rcl}x+ay+z & = & 1 \\(1-a^2)y+(1-a)z & = & 1 \\-(1+a)y & = & 1+a.\end{array}\right.$$ Next, we exchange the second and third equations, then use the (now) second equation to eliminate $y$ from the (now) third equation, which gives us $$\left\{\begin{array}{rcl}x+ay+z & = & 1 \\-(1+a)y & = & 1+a\\(1-a)z & = & 2-a^2.\end{array}\right.$$
In order to obtain infinite solutions, we need at least one equation to be $0=0,$ with the other equations still being consistent. Taking $a=-1$ readily gives us $$\left\{\begin{array}{rcl}x-y+z & = & 1 \\0 & = & 0\\2z & = & 1,\end{array}\right.$$ which satisfies the desired properties. If $a\neq-1,$ then we can further reduce the second equation, giving us $$\left\{\begin{array}{rcl}x+ay+z & = & 1 \\y & = & -1\\(1-a)z & = & 2-a^2.\end{array}\right.$$ Then, we can eliminate $y$ from the first equation, giving us $$\left\{\begin{array}{rcl}x+z & = & 1+a \\y & = & -1\\(1-a)z & = & 2-a^2.\end{array}\right.$$ Now, the only equation with a left-hand side that can be forced to be $0$ by choosing a value for $a$ is the third one--in particular, when $a=1$--but then the right-hand side isn't $0.$ Finally, when $a\neq\pm1,$ we can further reduce the third equation, then eliminate $z$ from the first equation, giving us $$\left\{\begin{array}{rcl}x & = & 1+a-\frac{2-a^2}{1-a} \\y & = & -1\\z & = & \frac{2-a^2}{1-a}.\end{array}\right.$$ Thus, any $a$ but $\pm1$ will determine a unique solution, $a=1$ gives no solution, and $a=-1$ gives infinitely-many solutions.
|
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|
Distinct integer solutions to $a^2+b^2=c^2+d^2$ I'm trying to find integer solutions to
$$a^2+b^2=c^2+d^2$$
with values $a> c > d > b>0$
Or in other words, two triangles with integer legs and equal hypotenuse lengths, not necessarily integer. Seems like a Diophantine equation to me, but I only learned how to solve Diophantine equations in the form of Pell's equations. I couldn't find anything on this equation when I checked Wikipedia. It is similar to a Pythagorean quadruple, although not quite, so that's not helpful either. How do I find integer solutions to this?
|
Strong Hint:
$$\begin{align} a^2 + b^2 &= c^2 + d^2 \\ \implies a^2 + b^2 - c^2 &= d^2 \\ \implies a^2 + (b + c)(b - c) &= d^2 \\ \implies (b + c)(b - c) &= d^2 - a^2 \\ \implies (b + c)(b - c) &= (d + a)(d - a) \end{align}$$.
$$\therefore a^2 + b^2 \neq \{p : p = \text{prime number}\} \iff a^2 + b^2 = c^2 + d^2$$
Now what separates a prime number from any other (composite) number?
Solution to finding distinct integer solutions:
$$a^2 + b^2 = c^2 + d^2 = \{n \iff d(n) \geq 4 : d(n) = \text{number of divisors of $n$}, \ \forall n\in \mathbb{R}\}$$
|
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Limit of infinite sum. Let $x$ be a real number such as $0 < x < 1$, compute the following series
$$\sum^\infty _{n=2} \frac{x^n}{n-1}$$ $$\sum^\infty _{n=2} \frac{x^n}{n+1}
$$
For instance, I made something like this : $$\frac{x^n}{n-1} = \frac{x^n}{n(1-\frac{1}{n})} \sim _{n \rightarrow \infty} \frac{x^n}{n}(1+\frac{1}{n} + o(\frac{1}{n})) = \frac{x^n}{n-1} + o(\frac{1}{n})$$
and finaly we have something like this :
$$ \sum_{n=2} ^\infty \frac{x^n}{n-1} \sim \sum^\infty _{n=2}\frac{x^n}{n} = -ln(1-x)$$
However something let me think that this result couldn't be the good one.
NB : sorry for my bad english
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Let $n^\prime = n-1$:
$\sum\limits_{n=2}^\infty = \sum\limits_{n^\prime=1}^\infty {x^{n^\prime + 1} \over n^\prime} = x \sum\limits_{n^\prime}^\infty {x^{n^\prime} \over n^\prime} = -x \ln(1-x)$.
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"url": "https://math.stackexchange.com/questions/2429613",
"timestamp": "2023-03-29T00:00:00",
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Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$. Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$.
So I am really stuck on this one.. I immediately defeated when I saw that I need to compute $\binom {22} {a,b,c}$ where $a,b,c$ is the coefficients of $x^2,-x,1$. So what should be the next step I need to take?
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Using that $(a+b+c)^3\equiv a^3+b^3+c^3\pmod 3$ repeatedly, we have, in mod $3$,
$$\begin{align}&(x^2-x+1)^{22}\\\\&=(x^2-x+1)(x^2-x+1)^3((x^2-x+1)^9)^2\\\\&\equiv (x^2-x+1)(x^6-x^3+1)(x^{18}-x^9+1)^2\\\\&\equiv (x^2-x+1)(x^6-x^3+1)(x^{36}+x^{27}+x^9+1)\\\\&\equiv x^{44}-x^{43}+x^{42}-x^{41}+x^{40}-x^{39}+x^{38}-x^{37}+x^{36}+x^{35}-x^{34}+x^{33}-x^{32}+x^{31}\\\\&\qquad -x^{30}+x^{29}-x^{28}+x^{27}+x^{17}-x^{16}+x^{15}-x^{14}+x^{13}-x^{12}+x^{11}-x^{10}+x^9\\\\&\qquad +x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1\end{align}$$
It follows that the zero terms in mod $3$ are $x^{18},x^{19},x^{20},x^{21},x^{22},x^{23},x^{24},x^{25},x^{26}$.
Next, using that $(a+b+c)^2\equiv a^2+b^2+c^2\pmod 2$ repeatedly, we have, in mod $2$,
$$\begin{align}&(x^2-x+1)^{22}\\\\&=(x^2-x+1)^{16}(x^2-x+1)^4(x^2-x+1)^2\\\\&\equiv (x^{32}+x^{16}+1)(x^{8}+x^4+1)(x^4+x^2+1)\\\\&\equiv x^{44}+x^{42}+x^{38}+x^{34}+x^{32}+x^{28}+x^{26}+x^{22}+x^{18}+x^{16}+x^{12}+x^{10}+x^6+x^2+1\end{align}$$
It follows that the zero terms in mod $6$ are $x^{19},x^{20},x^{21},x^{23},x^{24},x^{25}$.
Therefore, the total number of coefficients that are divisible by $6$ is $\color{red}{6}$.
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"url": "https://math.stackexchange.com/questions/2429982",
"timestamp": "2023-03-29T00:00:00",
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Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd].
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??
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Notice $$n^2 = \frac{(n-1)n + n(n+1)}{2}
\quad\implies\quad (-1)^{n-1} n^2 = (-1)^{n-1}\frac{(n-1)n}{2} - (-1)^{n}\frac{n(n+1)}{2}$$
The sum is a telescoping sum and
$$\require{cancel}\sum_{k=1}^n (-1)^{k-1} k^2 = \color{red}{\cancelto{0}{\color{grey}{(-1)^{1-1}\frac{(1-1)1}{2}}}}- (-1)^n \frac{n(n+1)}{2}
= (-1)^{n-1}\frac{n(n+1)}{2}
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2430887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.