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Finding first three terms of a GP whose sum and sum of squares is given? We have to find three numbers in G.P. such that their sum is $\frac{13}{3}$ and the sum of their squares is $\frac{91}{9}$.
Here's what I have tried to do:-
$$a+ar+ar^2=\frac{13}{3}$$
$$a^2+a^2r^2+a^2r^4=\frac{91}{9}$$
$$\frac{a^2+a^2r^2+a^2r^4}{a+ar+ar^2}=\frac{91}{9}*\frac{3}{13}$$
$$\frac{a+ar^2+ar^4}{1+r+r^2}=\frac{7}{3}$$
We have not been taught how to solve a biquadratic equation in school, so how should I proceed from here on?
|
Take 3 numbers $a/r$ ,a,ar $a/r$+a+ar=$13/3$ and $a^2/r^2$+$a^2$+$a^2r^2$=$91/9$ from this take $a^2$ common so on sim plifying we get $a^2/r^2$(1+$r^2+r^4$)=$(91/9$ Then square 1 st condition by using idw
Identity $(a+b+c)^2$=$a^2+b^2+c^2+2ab+2bc+2ac=$169/9$ from this we get $91/9$+$2a(a/r+a+ar)$=$169/9$ so $91/9$+2a$(13/3)$=$169/9$ from here a=1 plug it in any of the condition you get r as 1/3 or 3 thus terms are 1,3,1/3.thanks!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $a$ and $b$ be real numbers satisfying $a^3 - 3ab^2 = 47$ and $b^3 - 3a^2 b = 52$. Find $a^2 + b^2$. Let $a$ and $b$ be real numbers satisfying $a^3 - 3ab^2 = 47$ and $b^3 - 3a^2 b = 52$. Find $a^2 + b^2$.
I observed that adding the two has a suspicious similarity to (a+b)^3, but I cannot relate the two. Can someone help me?
Any help is appreciated!
|
Good try with $(a+b)^3$, but you will find it better to expand
$$\eqalign{(a^2+b^2)^3
&=a^6+3a^4b^2+3a^2b^4+b^6\cr
&=(a^3-3ab^2)^2+(b^3-3a^2b)^2\ .\cr}$$
If you have begun studying complex numbers, you can use the hint from Joey Zou to explain the "miracle" factorisation in the last line. However if you don't know about complex numbers yet, just expand the last line to check that it is correct.
|
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|
Solve for $x$: $x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$ Solve for $x$: $$x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$$
My attempt: I have changed this equation into the fractional part function. so that we know $0 \leq \{x\}<1$. I have final equation $x^2 +7x-x\{x+2\}-\{2x-2\}=14$. How to proceed now?
|
I would let $x=y+z$, with integer $y$ and $0\leq z<1,z=\{x\}$. Then either
$$(y+z)(y+2)+2y-2+3y+3z=12$$
or
$$(y+z)(y+2)+2y-1+3y+3z=12$$
So, for example, $z(y+5)$ is an integer, so only some fractions are allowed.
Also, 12 is between $y(y+2)+2y-2+3y$ and $(y+1)(y+2)+2y-2+3y+3$, for the first case, or something similar for the second case; and $y$ is an integer.
|
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|
double factorial series problem In the problem the sum is given as $$\sum\limits_{n=1}^\infty \frac{(2n-1)!!}{(2n)!!}$$
and then when I try to solve it using Gauss's test I get
$$\frac{a_{n}}{a_{n+1}}=\frac{2n+2}{2n+1}$$
but in the solution there is given:
$$\frac{a_{n}}{a_{n+1}}=\frac{2n}{2n-1}$$
my reasoning was:
$$\frac{a_{n}}{a_{n+1}}=\frac{\frac{(2n-1)!!}{(2n)!!}}{\frac{(2n+1)!!}{(2n+2)!!}}=\frac{1\cdot3\cdot...\cdot(2n-1) \times 2\cdot4\cdot...\cdot2n\cdot(2n+2)}{2\cdot4\cdot...\cdot2n \times 1\cdot3\cdot(2n-1)\cdot(2n+1)}=\frac{2n+2}{2n+1}$$
I believe that I made a mistake, but I don't know where?
|
$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2n)!!^2} = \frac{1}{4^n}\binom{2n}{n} $$
but the RHS behaves like $\frac{1}{\sqrt{\pi n}}$, hence the series is diverging. Also without the exact asymptotics:
$$ \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\prod_{k=1}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right) \geq \frac{1}{4n}.$$
|
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|
For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7 For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7.
I'm not sure how to do this proof so any help would be appreciated.
|
If $n=1$, then
$$
4^{n} + 10\cdot 9^{2n-2} = 4+10 = 14,
$$
divisible by $7$;
if $n \geq 1$ is an integer such that $4^{n} + 10\cdot 9^{2n-2} = 7k$ for some integer $k \geq 1$, then
$$
4^{n+1} + 10\cdot 9^{2(n+1) - 2} = 4\cdot 4^{n} + 10\cdot 9^{2n} = 4(4^{n} + 10\cdot 9^{2n-2}) = 28k,
$$
divisible by $7$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Finding Angles Given Two Triangles with Equal Perimeters The two right triangles shown below have equal perimeters
The hypotenuse of
the orange triangle is one leg of the green triangle stacked on top of it. If the smallest angle of the orange triangle is 20 degrees, what are the angles of the green right triangle?
|
I think I have a solution but it will be difficult to explain. I will use the following variables:
$$
\begin{split} a&=\text{base of orange triangle} \\ b&=\text{height of orange triangle} \\ c&= \text{"left" side of green triangle} \end{split}
$$
Then the hypotenuse of the orange triangle is $\sqrt{a^2+b^2}$ and the hypotenuse of the green triangle is $\sqrt{a^2+b^2+c^2}$. Since the perimeters of both triangles are equal we must have:
$$
\begin{split} &&a+b=c+\sqrt{a^2+b^2+c^2} \\ &\implies& a+b-c=\sqrt{a^2+b^2+c^2} \\ &\implies& (a+b-c)^2=a^2+b^2+c^2 \\ &\implies& a^2+b^2+c^2+2ab-2ac-2bc=a^2+b^2+c^2 \\ &\implies& ab=ac+bc \\ &\implies& c=\frac{ab}{a+b} \end{split}
$$
So now we have all sides in terms of $a$ and $b$. But we can relate $a$ and $b$ using the angle we are given. So:
$$
\tan(20)=\frac{b}{a} \implies b=a\tan(20)
$$
Using this we get:
$$
\begin{split} a^2+b^2=a^2+a^2\tan^2(20)=a^2(1+\tan^2(20))=\frac{a^2}{\sin^2(20)} \end{split}
$$
Now we can get an expression for $a^2+b^2+c^2$ purely in terms of $a$:
$$
\begin{split} a^2+b^2+c^2&= \frac{a^2}{\sin^2(20)}+\frac{a^2b^2}{\frac{a^2}{\sin^2(20)}+2ab}\\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)}{\frac{a^2}{\sin^2(20)}+2a^2\tan(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)\sin^2(20)}{a^2+2a^2\tan(20)\sin^2(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^2\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \\ &=a^2\left[ \frac{1}{\sin^2(20)}+\frac{\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \right] \\ &= a^2K^2 \end{split}
$$
where $K^2$ is just the number in the square brackets (the only reason I made it $K^2$ instead of $K$ is because we will need to take square roots). Finally, letting $\theta$ be the smallest angle of the green triangle, we get:
$$
\begin{split} \cos(\theta)&=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}} \\ &=\frac{\frac{a}{\sin(20)}}{aK} \\ &=\frac{1}{K\sin(20)} \end{split}
$$
You can work out that $K\approx 2.9262$ and then plugging this in gives us $\theta\approx 2.34^{\circ}$.
|
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|
Is my method for finding the Basis for the Row(A) correct?
$$A=\begin{pmatrix}-1&0&2\\3&2&0\\0&1&3 \end{pmatrix}$$
Find a basis for $\text{Row} (A)$
$$Row(A)=\text{span} \left[\begin{pmatrix}-1\\0\\2\end{pmatrix},\begin{pmatrix}3\\2\\0\end{pmatrix},\begin{pmatrix}0\\1\\3\end{pmatrix}\right]$$
$$\begin{pmatrix}-1&3&0\\0&2&1\\2&0&3 \end{pmatrix}$$
Upon performing Reduced Row Echleon Form (RREF)
I get:
$$\begin{pmatrix}1&-3&0\\0&2&1\\0&0&0 \end{pmatrix}$$
The first and third column has a pivot so the basis must be the corresponding vectors so therefore:
$$\text{basis: } Row(A)=\text{span} \left[\begin{pmatrix}-1\\0\\2\end{pmatrix},\begin{pmatrix}0\\1\\3\end{pmatrix}\right]$$
Is this flawless, and is there a faster method? Anyway I can improve my answer?
|
Elementary row operations preserve the row space of $A$ (but change the column space). If you want to find a basis for $\mathrm{row}(A)$, perform elementary row operations on $A$ until you reach the reduced row echelon form. The non-zero rows of the reduced row echelon form will form a basis for $\mathrm{row}(A)$. In fact, you don't even have to perform row operations until you reach the reduced row echelon form - you can stop whenever it is clear what is the basis for $\mathrm{row}(A)$.
In your example,
$$ \left( \begin{matrix} -1 & 0 & 2 \\ 3 & 2 & 0 \\ 0 & 1 & 3 \end{matrix} \right) \xrightarrow{R_1 = (-1) \cdot R_1}
\left( \begin{matrix} 1 & 0 & -2 \\ 3 & 2 & 0 \\ 0 & 1 & 3 \end{matrix} \right) \xrightarrow{R_2 = R_2 - 3R_1}
\left( \begin{matrix} 1 & 0 & -2 \\ 0 & 2 & 6 \\ 0 & 1 & 3 \end{matrix} \right) $$
and from the last matrix it is already clear that
$$ \mathrm{row}(A) = \mathrm{span} \{ (1, 0, -2), (0, 1, 3) \}. $$
If you want to find the column space of $A$, you can do the same with column operations (which preserve the column space but change the row space).
|
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|
Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~x^2+y^2=44^2-90xy+x^2y^2$
But from here I didn't know what to do. Could you help me in solving this equation?
|
$$xy+x+y=44$$
$$\implies xy+x+y+1=45=3\times15$$
$$\implies (y+1)(x+1)=3\times15$$
therefore $y=2$ and $x=14$ , also it satisfies the second identity. Hence $x^2+y^2=196+4=200.$
|
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|
Linear Transformation: Orthogonal Projections
Define $\mathbf{u_1} =$ $\begin{align} \begin{bmatrix}
0 \\
0 \\ 1 \\1\\
\end{bmatrix}\end{align}$ and $\mathbf{u_2} =$ $\begin{align} \begin{bmatrix}
1 \\
1 \\ 0 \\0\\
\end{bmatrix}\end{align}$. Define $\mathbf{W} = \textrm{Span} \{\mathbf{u_1},\mathbf{u_2}\}$. Define two other vectors
$\mathbf{u_3}$ and $\mathbf{u_4}$, such that the set $\mathbf{F} =
\{\mathbf{u_1},\mathbf{u_2},\mathbf{u_3},\mathbf{u_4}\}$ forms an
orthogonal basis for $\mathbb{R}^4$.
Let $T: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ be the linear
transformation which is orthogonal projection onto $\mathbf{W}$.
*
*What is $T(\mathbf{u_1})$ and $T(\mathbf{u_3})$?
*Find the matrix $M_T$ which represents the transformation $T$ with respect to the $\mathbf{F}$ coordinates.
Hi there, I am preparing for my linear algebra final examination. I stumbled across this question, however, no solutions were provided for this question. I am worried that I am studying the wrong thing.
My approach to the questions:
First, I computed the vectors $\mathbf{u_3}$ and $\mathbf{u_4}$. After finding the null space for $\begin{align} \begin{bmatrix}
1 & 1 & 0 & 0\\
0 & 0 & 1 & 1\\
\end{bmatrix}\end{align}$, I obtained that $\mathbf{u_3} =$ $\begin{align} \begin{bmatrix}
-1 \\
1 \\ 0 \\0\\
\end{bmatrix}\end{align}$ and $\mathbf{u_4} =$ $\begin{align} \begin{bmatrix}
0 \\
0 \\ -1 \\1\\
\end{bmatrix}\end{align}$.
*
*$T(\mathbf{u_1}) = \frac {\mathbf{u_1} \cdot \mathbf{u_1}}{\mathbf{u_1} \cdot \mathbf{u_1}}\mathbf{u_1} + \frac {\mathbf{u_1} \cdot \mathbf{u_2}}{\mathbf{u_2} \cdot \mathbf{u_2}}\mathbf{u_2} = \mathbf{u_1}$.
*$T(\mathbf{u_3}) = \frac {\mathbf{u_3} \cdot \mathbf{u_1}}{\mathbf{u_1} \cdot \mathbf{u_1}}\mathbf{u_1} + \frac {\mathbf{u_3} \cdot \mathbf{u_2}}{\mathbf{u_2} \cdot \mathbf{u_2}}\mathbf{u_2} = \mathbf{0}$ (since $\mathbf{u_3}$ is orthogonal to both $\mathbf{u_1}$ and $\mathbf{u_2}$)
*We are working on the $\mathbf{F}$ coordinates. Note that $\mathbf{u_1}, \mathbf{u_2} \in \mathbf{F}$ and $\mathbf{u_3}, \mathbf{u_4} \in \mathbf{F^\perp}$. Thus, each basis vectors is an eigenvector for the projection. Hence, the matrix $M_T$ is given by:
$$\begin{align} \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix} \end{align}$$
Are my solutions and reasonings correct? Any insight on these would be highly appreciated. Thank you!
|
I think that your solution is correct, other than the fact that you computed $\mathbf{u}_3$ and $\mathbf{u}_4$. Note that the completion of an orthogonal linearly independent set to an orthogonal base is not unique; however, the rest of the question can be answered even without knowing who the $\mathbf{u}_i$-s are.
|
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|
What are roots of $x^{3}+3x^{2}+4x+1$? There are no divisor of 1 in this polynomial for which would be satisfied $x^{3}+3x^{2}+4x+1=0$.
How to find roots here?
|
Put $x = y - 1$, this yields:
$$y^3 + y - 1 = 0$$
Then compare with the identity:
$$\begin{split}
(a+b)^3 \equiv a^3 + 3 a^2 b + 3 b a^2 + b^3\Longrightarrow \\
(a+b)^3 - 3 a b (a+b) - (a^3 + b^3) \equiv 0
\end{split}
$$
So, if we can find two numbers $a$ and $b$ such that:
$$-3 a b = 1$$
and
$$a^3 + b^3 = 1$$
then $y = a + b$ will be a solution.
If we put $A = a^3$ and $B = b^3$, then we require that:
$$A B = -\frac{1}{27}$$
$$A + B = 1$$
Now solving for $A$ and $B$ requires solving a quadratic equation. When extracting $a$ and $b$, you must make sure that $a b = -\frac{1}{3}$, so the choice of the phase factor multiplying the cube root for $A$ fixes the phase factor for $b$.
|
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|
Is it true that $ \sum_{t = 1}^T \frac{T-t}{ t+ \sqrt{T-t}} \in O(T) $? Is the following true?
$$
\sum_{t = 1}^T \frac{T-t}{ t+ \sqrt{T-t}} \in O(T)
$$
|
The idea is similar to A.S.' answer, I just want to make the calculation more explicit and clearer.
In fact we can show that
$$\limsup_{T \to \infty} \frac{1}{T}\sum_{t = 1}^T \frac{T - t}{t + \sqrt{T - t}} = \infty, \tag{$*$}$$
thus the conjecture is false.
We prove $(*)$ by considering a subsequence of $\{1, 2, \ldots\}$.
Let $T = n^2, n = 1, 2, \ldots$. We then partition the summation in $(*)$ into each sub-block $\{in + 1, \ldots, (i + 1)n\}$ of $\{1, 2, \ldots, n^2\}$ for $i = 0, 1, \ldots, n - 1$. Since the summand $\frac{T - t}{t + \sqrt{T - t}}$ is a nonincreasing function of $t$, it follows that
\begin{align*}
& \frac{1}{T}\sum_{t = 1}^T \frac{T - t}{t + \sqrt{T - t}} \\
= & \frac{1}{n^2}\sum_{i = 0}^{n - 1}\sum_{k = in + 1}^{(i + 1)n} \frac{n^2 - k}{k + \sqrt{n^2 - k}} \\
\geq & \frac{1}{n^2}\sum_{i = 0}^{n - 1}\frac{n^2 - (i + 1)n}{(i + 1)n + \sqrt{n^2 - (i + 1)n}} \times n \\
= & \sum_{i = 0}^{n - 1}\frac{1 - \frac{i + 1}{n}}{\frac{i + 1}{n} + \frac{1}{n}\sqrt{1 - \frac{i + 1}{n}}} \times \frac{1}{n} \\
\geq & \sum_{i = 0}^{n - 1}\frac{1 - \frac{i + 1}{n}}{\frac{i + 1}{n} + \frac{1}{n}} \times \frac{1}{n} \\
\geq & \sum_{i = 0}^{n - 1}\frac{1 - \frac{i + 1}{n}}{\frac{i + 1}{n} + \frac{i + 1}{n}} \times \frac{1}{n} \\
= & \frac{1}{2}\sum_{i = 0}^{n - 1}\frac{1 - \frac{i + 1}{n}}{\frac{i + 1}{n}} \times \frac{1}{n} \\
\geq & \frac{1}{2}\int_{\frac{1}{n}}^1 \left[\frac{1}{x} - 1\right] dx \\
= & \frac{1}{2}\log n - \frac{1}{2} + \frac{1}{2n}.
\end{align*}
The last expression tends to $\infty$ as $n \to \infty$. Therefore, $(*)$ holds.
|
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|
How to evaluate the following integral ? What substitution will be helpful? $$\int\frac{cotx }{(1-sinx)(secx+1)}dx $$
we can write this as
$$\int \frac {cosecx.cotx}{(cosecx-1)(secx+1)}dx $$
Now $cosecx=t$ gives $cosecx.cotx=-dt $ which appears in the numerator, what to do about $secx+1$ ?
|
$$\frac{\cot x}{(1-\sin x)(1+\sec x)}=\frac{1+\sin x}{\sin x(1+\cos x)}=\frac{1}{\sin x(1+\cos x)}+\frac{1}{1+\cos x}$$
Enforcing the Wiereatrass Substitution $\tan(x/2)=u$ and $dx=\frac{2}{u^2+1}\,du$ reveals
$$\begin{align}
\int \frac{\cot x}{(1-\sin x)(1+\sec x)}\,dx&=\int \left(\frac{(u^2+1)^2}{4u}+\frac{u^2+1}{2}\right)\,\frac{2}{u^2+1}du\\\\
&=\int \left(\frac12 u +\frac12 u^{-1}+1\right)\,du\\\\
&=\frac14 \tan^2\left(\frac{x}{2}\right)+\frac12 \log\left(\tan \left(\frac{x}{2}\right)\right)+\tan\left(\frac{x}{2}\right)+C
\end{align}$$
|
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|
Find all distinct $a,b,c$ such that $a +1/b = b + 1/c = c+1/a$ Could anyone advise me how to find all values of $a,b, c$ such that $a + \dfrac{1}{b} = b + \dfrac{1}{c} = c + \dfrac{1}{a} \ , $ where $a,b,c $ are distinct real numbers?
I have found that $a, b, c$ can take on either $1,-2, $ or $-\dfrac{1}{2} $, so there are at least 6 sets of solutions. Thank you.
|
Easy parameterization:
$$
(a,b,c) = \left(a, \frac{1}{1-a}, 1-\frac{1}{a}\right),
$$
where $a$ is any real, $a\ne 0, a\ne \frac{1}{2}, a\ne 1$.
Then $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=1$.
|
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|
How can I find the rank of $A^6$ if $A$ is a matrix of order $3$ and rank of $A^3$ is $2$? I have tried it in the following manner. We know that $$rank(XY)\ge rank(X)+rank(Y)-n$$ where $X$ and $Y$ are two matrices of order $n$ and also $$rank(XY) \le \min{\{rank(X),rank(Y)\}}$$ If we take $X=Y=A^3$ then using the above two inequalities we obtain $1\le rank(A^6)\le 2$ i.e. rank of $A^6=1 \text{ or } 2$. Is it correct at all? I have failed to obtain a definite rank of $A^6$ by the above process. Is it ok?
|
Assume $A \in M_3(\mathbb{C})$. Let $a,b,c$ be the root of characteristic polynomial of $A$ or say eigenvalue of $A$. So $a^3,b^3,c^3$ will be eigenvalue of $A^3$. As kernel of $A^3$ is one dimensional one of the eigen value of $A^3$ will be $0$, say $c^3 = 0$. So $c=0$.
First note $a,b$ simultaneously can't be $0$, in that case characteristic polynomial of $A$ must be $x^3$ and $A^3=0$.
Now we will show neither $a$ nor $b$ is zero. Assume $b=0$ and $a\neq 0$. Using Jordan form, in that case matrix of $A$ w.r.t some basis will be one of the following form
\begin{align}
\begin{pmatrix}
a&0 &0\\
0&0&1\\
0&0&0
\end{pmatrix}, \begin{pmatrix}
a&0 &0\\
0&0&0\\
0&0&0
\end{pmatrix}
\end{align}
In either case we will have $A^3$ has rank $1$, since in those case we will have
\begin{align}
A^3 &= \begin{pmatrix}
a^3&0 &0\\
0&0&0\\
0&0&0
\end{pmatrix}
\end{align}
So we have neither $a$ nor $b$ equal to $0$.
Now consider the case when $a=b$. In that case matrix of $A$ w.r.t some basis will be one of the following form
\begin{align}
\begin{pmatrix}
a&0 &0\\
0&a&0\\
0&0&0
\end{pmatrix}, \begin{pmatrix}
a&1 &0\\
0&a&0\\
0&0&0
\end{pmatrix}
\end{align}
As $a \neq 0$, In either case $A^3$ and $A^6$ both will have rank $2.
Now consider the last case $a,b$ are distinct non zero number. Then there exist a eigenbasis so that matrix of $A$ w.r.t that basis will be of the following form
\begin{align}
\begin{pmatrix}
a&0 &0\\
0&b&0\\
0&0&0
\end{pmatrix}
\end{align}
In this case also $A^3$ and $A^6$ both will have rank $2$.
If $A\in M_3(\mathbb{R})$, use the fact that $\{v_1,v_2,...v_k\} \subset \mathbb{R}^n$ is Linearly independent over $\mathbb{R}$ iff they are Linearly independent over $\mathbb{C}$. So that Real rank of $A$ and complex rank of $A$ remain same.
|
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|
Proof: $f(x)=x^2+x-4$ is continuous at $x=2$ Proof: Let $\epsilon>0$ and $\delta =\min\{1,\epsilon/6\}$.Then if $|x-2|<\delta$, $|x-2|<1$, so $|x+3|<6$.
Thus, $$|f(x)-f(c)|= |x^2+x-4-(2^2+2-4)| = |x^2+x-6|=|x-2||x+3|<6\delta \leq \epsilon$$
Doing this problem as practice. Thoughts? comments?
|
Late answer but let us see if $\delta= \min \lbrace 1,\frac{\epsilon}{6} \rbrace $ works or not. First case is if $\delta=1.$ If $|x-2|<1,$ then $$|f(x)-f(2)|=|x^2+x-6|=|x-2||x+3|<(1)\left(6\right)<\frac{\epsilon}{6}(6)=\epsilon$$ since $|x-2|<1$ implies both $|x-2| < \frac{\epsilon}{6}$ and $|x+3|<6.$
Now let $\delta = \frac{\epsilon}{6}.$ Proceeding like before, we have $$|f(x)-f(2)|=|x^2+x-6|=|x-2||x+3|<\frac{\epsilon}{6}(6)=\epsilon$$ since $|x-2|<\frac{\epsilon}{6}$ implies $|x-2|<1$ and $|x+3|<6.$
|
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|
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{n+1}2$$
I tried to prove that that the difference is a multiple of $5$.
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=2(3^{3n+1}\cdot 13+2^n)$$
Therefore it's enough to prove that $3^{3n+1}\cdot 13+2^n$ is a multiple of $5$. But if I do again this method applied to this "new problem" is get something similar. I think that there exist a different method to do this using induction.
|
Or we could just expand and rearrange$$
\begin{align}
3^{3n+1}+2^{n+1}
&= 3\cdot27^n + 2\cdot2^n \\
&= 3\cdot(25+2)^n + 2\cdot2^n \\
&= 3\left(5k+2^n \right) + 2\cdot 2^n \tag{Using Binomial Theorem}\\
&= 5\cdot k'+3\cdot2^n+2\cdot2^n \\
&= 5\cdot k'+5\cdot 2^n \\
\end{align}
$$
Hence proved
|
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|
Find polynomial $f(x)$ based on divisibility properties of $f(x)+1$ and $f(x) - 1$ $f(x)$ is a fifth degree polynomial. It is given that $f(x)+1$ is divisible by $(x-1)^3$ and $f(x)-1$ is divisible by $(x+1)^3$. Find $f(x)$.
|
If $(x-1)^3$ divides $f(x)+1$, then $(x-1)^2$ divides $f'(x)$.
If $(x+1)^3$ divides $f(x)-1$, then $(x+1)^2$ divides $f'(x)$.
As deg$\,f=5$, then deg$\,f'=4$, and hence $f'(x)=a(x-1)^2(x+1)^2=a(x^4-2x^2+1)$,
for some $a\in\mathbb R$.
Thus
$$
f(x)=\frac{a}{5}x^5-\frac{2a}{3}x^3+ax+b,
$$
for some $b\in\mathbb R$.
Now, as $(x-1)^3$ divides $g(x)=f(x)+1$, in particular $f(1)+1=g(1)=0$. Thus
$$
-1=f(1)=\frac{a}{5}-\frac{2a}{3}+a+b. \tag{1}
$$
Similarly,
as $(x+1)^3$ divides $h(x)=f(x)-1$, in particular $f(-1)-1=g(-1)=0$. Thus
$$
1=f(-1)=-\frac{a}{5}+\frac{2a}{3}-a+b. \tag{2}
$$
Adding $(1)$ and $(2)$ we obtain that $b=0$, and thus $a=-15/8$.
Hence
$$
f(x)=-\frac{3}{8}x^5+\frac{5}{4}x^3-\frac{15}{8}x.
$$
|
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|
Evaluating $\lim_\limits{x\to 1 }\bigl( (2^x x + 1)/(3^x x)\bigr)^{\tan(\pi x/2)}$ I have to calculate limit
$$\lim_{x\to 1 } \left(\frac{2^x x + 1}{3^x x}\right)^{\tan(\frac{\pi x}{2})}.$$
I know $\tan(\frac{\pi x}{2})$ is undefined in $x = 1$, but can I just put $x = 1$ into $\frac{x\cdot 2^x + 1}{x\cdot3^x}$ and get
$$\lim_{x\to 1 } (1)^{\tan(\frac{\pi x}{2})} = 1.$$
Is the answer $1$ correct?
It's forbidden to use L'Hôpital's rule.
|
For $x$ near $0$, $a^x=1+x\log(a)+O\left(x^2\right)$.
Furthermore, if $\lim\limits_{n\to\infty}\left|b_n\right|=\infty$ and $c=\lim\limits_{n\to\infty}a_nb_n$, then $\lim\limits_{n\to\infty}\left(1+a_n\right)^{b_n}=e^c$.
Therefore,
$$
\begin{align}
\lim_{x\to1}\left(\frac{2^xx+1}{3^xx}\right)^{\tan\left(\frac{\pi x}2\right)}
&=\lim_{x\to0}\left(\frac{2^{x+1}(x+1)+1}{3^{x+1}(x+1)}\right)^{\tan\left(\frac\pi2(x+1)\right)}\\
&=\lim_{x\to0}\left(\frac{2(1+x)\left(1+x\log(2)+O\left(x^2\right)\right)+1}{3(1+x)\left(1+x\log(3)+O\left(x^2\right)\right)}\right)^{-1/\tan\left(\frac\pi2x\right)}\\
&=\lim_{x\to0}\left(\frac{3+x(2+2\log(2))+O\left(x^2\right)}{3+x(3+3\log(3))+O\left(x^2\right)}\right)^{-1/\tan\left(\frac\pi2x\right)}\\
&=\lim_{x\to0}\left(1-\frac x3\left(1+\log\left(\frac{27}4\right)\right)+O\left(x^2\right)\right)^{-1/\tan\left(\frac\pi2x\right)}\\[9pt]
&=e^{\frac2{3\pi}\left(1+\log\left(\frac{27}4\right)\right)}
\end{align}
$$
|
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|
Prove $ 5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3) $ if $a^2+b^2+c^2=3$ Let $a,b,c\in\mathbb{R}$ such that $a^2+b^2+c^2=3$. Prove that:
$$
5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3)
$$
I tried to homogenize the inequality to get:
$$
5(a^4+b^4+c^4)+(a^2+b^2+c^2)^2≥\frac{8}{\sqrt3}(a^3+b^3+c^3)\sqrt{(a^2+b^2+c^2)}
$$
I hoped that one don't needs the condition anymore to prove this, but I couldn't get any further.
|
Hint:
It is enough to show $f(x)=5x^4+3-8x^3+2(x^2-1)=(x-1)^2(5x^2+2x+1)\ge0$ which is obvious.
|
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|
A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)^2=4^2\\x^2+\frac{1}{x^2}+2x\times \frac{1}{x}=16\\x^2+\frac{1}{x^2}=16-2 $$
2:solving quadratic equation ,and putting one of roots$$x+\frac{1}{x}=4\\\frac{x^2+1}{x}=4\\x^2+1=4x\\x^2-4x+1=0\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(+1)}}{2}=\\x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\\x=2+\sqrt3 \to x^2=4+4\sqrt3+3=7+4\sqrt3\\x^2+\frac{1}{x^2}=7+4\sqrt3+\frac{1}{7+4\sqrt3}=\\7+4\sqrt3+\frac{1}{7+4\sqrt3}\cdot\frac{7-4\sqrt3}{7-4\sqrt3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{49-16\cdot 3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{1}=14$$
3: visual approach .assume side length of a square is $x+\frac{1}{x}=4$
now I am looking for new idea to proof.Any hint will be appreciated.(more visual proof - geometrical - trigonometrical - using complex numbers ...)
|
Here is a mildly disguised version of the same idea.
Let $x=e^u$
Then we have $$\cosh u=2$$
And we want $$x^2+\frac {1}{x^2}=2\cosh 2u=2(2\cosh^2 u-1)=2(8-1)=14$$
|
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|
Alternate approaches to solve this Integral Evaluate $$I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\:dx$$
I have used parts taking first function as Integrand and second function as $1$ we get
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}-\int \frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right) \times x dx$$
now $$\frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)=\frac{1}{2\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)} \times \left(\frac{\frac{1+\sqrt{x}}{-2\sqrt{x}}-\frac{1-\sqrt{x}}{2\sqrt{x}}}{(1+\sqrt{x})^2}\right)=\frac{-1}{2\sqrt{x}\sqrt{1-x}(1+\sqrt{x})}$$ so
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\int\frac{xdx}{2\sqrt{x}\sqrt{1-x}(1+\sqrt{x})} $$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int \frac{\sqrt{x}dx}{\sqrt{1-x}(1+\sqrt{x})}$$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int\frac{\sqrt{x}(1-\sqrt{x})dx}{(1-x)^{\frac{3}{2}}}$$ $\implies$
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int\frac{\sqrt{x}dx}{(1-x)^{\frac{3}{2}}}-\frac{1}{2}\int\frac{xdx}{(1-x)^{\frac{3}{2}}}$$
Now $$\int\frac{xdx}{(1-x)^{\frac{3}{2}}}=\int\frac{(x-1+1)dx}{(1-x)^{\frac{3}{2}}}$$ and if we split is straight forward to compute.
Also $$\int\frac{\sqrt{x}dx}{(1-x)^{\frac{3}{2}}}$$ can be evaluated using substitution $x=sin^2y$
I need any other better approaches to evaluate this integral.
|
First rationalize:
$$\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$$
$$=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\times\sqrt{\frac{1-\sqrt{x}}{1-\sqrt{x}}}$$
$$=\frac{\sqrt{1-\sqrt{x}}}{\sqrt{1+\sqrt{x}}}\times\frac{\sqrt{1-\sqrt{x}}}{\sqrt{1-\sqrt{x}}}$$
$$=\frac{1-\sqrt{x}}{\sqrt{1-x}}$$
$$=\frac{1}{\sqrt{1-x}}-\frac{\sqrt{x}}{\sqrt{1-x}}$$
Then to do the integral, for second half let $x=t^2$ so $dx=2tdt$:
$$\int\frac{1}{\sqrt{1-x}}-\frac{\sqrt{x}}{\sqrt{1-x}}dx$$
$$=-2\sqrt{1-x}-\int\frac{2t^2}{\sqrt{1-t^2}}dt$$
$$=-2\sqrt{1-x}+\int2\frac{1-t^2}{\sqrt{1-t^2}}-\frac{2}{\sqrt{1-t^2}}dt$$
$$=-2\sqrt{1-x}+\int2\sqrt{1-t^2}dt-2\arcsin t$$
$$=-2\sqrt{1-x}+t\sqrt{1-t^2}+\int\frac{1}{\sqrt{1-t^2}}dt-2\arcsin \sqrt{x}$$
$$=-2\sqrt{1-x}+\sqrt{x}\sqrt{1-x}+\arcsin t-2\arcsin \sqrt{x}+c;c\in\mathbb{R}$$
$$=-2\sqrt{1-x}+\sqrt{x}\sqrt{1-x}-\arcsin \sqrt{x}+c$$
|
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|
Finding the Maximum value.
Maximize $xy^2$ on the ellipse $b^2x^2 +a^2y^2= a^2b^2$
The steps I tried to solve:
$$\nabla f = (y^2,2yx)\lambda\qquad g = (2xb^2,2y^2a^2)\lambda$$
$$y^2= 2xb^2\lambda$$
$$2yx= 2y^2a^2\lambda$$
$$
\left.
\begin{array}{l}
\text{}&y^2= 2xb^2\lambda\\
\text{}&
\end{array}
\right\}
*a^2y
$$
$$
\left.
\begin{array}{l}
\text{}&2yx= 2y^2a^2\lambda\\
\text{}&
\end{array}
\right\}
*b^2x
$$
$$y^3a^2= 2yxa^2b^2 \lambda$$
$$2yx^2b^2 = 2y^2a^2b^2x\lambda$$
$\color{maroon}{\mathbf{Equalize}}$
$$2a^2b^2xy = 2a^2b^2xy^2$$
$$y=1$$
My main problem is which equations does one set each equal to. If anyone knows which one equals the other this will lead me to the right place in finding the solution.
|
Since you started off using Langrange multipliers, let’s continue down that path. Using the correct value for $g_y$, we have $$\begin{align}
y^2 &= 2b^2\lambda x \\
2xy &= 2a^2\lambda y
\end{align}$$ which upon eliminating $\lambda$ gives $$
a^2y^3 = 2b^2x^2y,
$$ so either $y=0$ or $a^2y^2=2b^2x^2$. Substituting this into the constraint: $$b^2x^2+2b^2x^2=a^2b^2 \\
\text{or} \\
x^2=\frac13a^2.$$ Plugging this back into the constraint and solving for $y$: $$\frac13a^2b^2+a^2y^2=a^2b^2 \\
\text{or} \\
y^2=\frac23b^2.$$ $f$ is an even function of $y$, so it doesn’t matter which root we use to test these points against $f$. $$
f\left(+\sqrt{\frac13}a,\pm\sqrt{\frac23}b\right) = {2\sqrt3 \over 9}ab^2 \\
\text{and} \\
f\left(-\sqrt{\frac13}a,\pm\sqrt{\frac23}b\right) = -{2\sqrt3 \over 9}ab^2
$$ We still have $y=0$ as a candidate, but $f(x,0)=0$, so this doesn’t yield an extremum.
|
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|
Number of Polynomials with Integer Coefficients that are bounded by $x^2$ and $x^4+1$ What is the number of polynomials $p(x)$ with integer coefficients, such that
$x^2≤p(x)≤x^4+1$ for all real numbers $x$?
|
Indeed, no polynomial with degree $3$ can satisfy this property because it has to have a negative value at some point.
Also, it is clear that (by taking very large $x$), the degree of this polynomial must be $2,3$ or $4$.
So this polynomial has to have a degree of $2$ or $4$.
Case 1. deg$p$ is $2$.
Let $p(x)=ax^2+bx+c$. Since $0 \le p(0)=c \le 1$, we either have $c=0$ or $c=1$.
If $c=0$, we have $p(x)=ax^2+bx$. For $p(x) \ge x^2$ to hold for all $x$, we must have $a \ge 1$.
Also, the determinant of $p(x)-x^2$ must be $\le 0$ for all $x$.
The determinant is just $b^2$, so $b=0$. Now $p(x)=ax^2$.
Since $x^4+1 \ge ax^2$, we take $x=1$ so that $a \le 2$. AM-GM gives $x^4+1 \ge 2x^2 \ge x^2$, so $a=1,2$ works.
The solution set for this case is $p(x)=x^2$ and $p(x)=2x^2$.
Case 2. deg$p$ is $4$.
Denote the coefficient of $x^4$ as $l$, and the constant term as $e$.
Clearly, $0 \le l \le 1$ by taking a large $x$. Since $l$ is nonzero, $l=1$.
Again, we have $0 \le e \le 1$ by taking $x=0$.
The coefficient of $x^3$ must be $0$. Otherwise, $x^4+1-p(x)$ can take negative values as it is a degree $3$ polynomial.
Let $p(x)=x^4+ax^2+bx+e$.
SubCase 2-1. $e=1$.
We have $-ax^2-bx \ge 0$ for all $x$. We must have $a \le 0$. Since the determinant of $-ax^2-bx$ is less than $0$, we have $b^2 \le 0$, so $b=0$.
We now have $x^4+ax^2+1 \ge x^2$. Take $x=1$ to get $a \ge -1$.
AM-GM gives $x^4+1 \ge x^4-x^2+1 \ge x^2$, so both $a=0$ and $a=-1$ work.
The solution set for this subcase is $p(x)=x^4+1$ and $x^4-x^2+1$.
SubCase 2-2. $e=0$.
We have $-ax^2-bx+1 \ge 0$. We must have $a \le 0$, and the determinant $b^2+4a \le 0$.
Also, setting $x=1$ gives $a+b \ge 0$.
Now we have $b^2 \le -4a \le 4b$, giving $b(b-4) \le 0$, so $b=0,1,2,3,4$.
If $b=0$, we have $a \ge 0$ and $a \le 0$, so $a=b=0$. This gives $p(x)=x^4$, which fails.
If $b=1$, we have $4a \ge -1$, so $-\frac{1}{4} \le a \le 0$. This gives $a=0$.
This gives $p(x)=x^4+x$, which fails at $x=2$.
If $b=2$, we have $4a \ge -4$, so $-1 \le a \le 0$.
If $a=0$, we have $p(x)=x^4+2x$, which fails at $x=1$.
If $a=-1$, we have $p(x)=x^4-x^2+2x$, which fails at $x=-1$.
If $b=3$, we have $4a \ge -9$, so $-\frac{9}{4} \le a \le 0$, giving $a=-2,-1,0$.
If $a=0$, we have $p(x)=x^4+3x$, which fails at $x=1$.
If $-2 \le a \le -1$, we have $p(x)=x^4+ax^2+3x$, which fails at $x=-1$.
If $b=4$, we have $-4 \le a \le 0$.
If $a=0$, we have $p(x)=x^4+4x$, which fails at $x=1$.
If $-4 \le a \le -1$, we have $p(x)=x^4+ax^2+4x$, which fails at $x=-1$.
Therefore, there are no solutions in this subcase.
The final answer is $x^2$, $2x^2$, $x^4+1$, $x^4-x^2+1$.
Hopefully, someone can find a cleaner solution..
|
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|
Partial sum of $\sum \frac {1} {k^2}$ It is pretty well-known that $\sum_{k = 1}^{\infty} \frac {1} {k^2} = \frac {\pi^2} {6}$. I am interested in evaluating the partial sum $\sum_{k = 1}^{N} \frac {1} {k^2}$. Here is what I have done so far. Since we have
$$\sum_{k = 1}^{N} \frac {1} {k^2} = 1 + \sum_{k = 2}^{N} \frac {1} {k^2}
< 1 + \sum_{k = 2}^{N} \frac {1} {k^2 - 1} =\\= 1 + \frac {1} {2} \left ( \sum_{k = 2}^{N} \frac {1} {k - 1} - \sum_{k = 2}^{N} \frac {1} {k + 1} \right )
= 1 + \frac {1} {2} \left ( \frac {3} {2} - \frac {2N + 1} {N^2 + N} \right )
= \frac {7} {4} - \frac {2N + 1} {2 N^2 + 2 N}$$
and
$$\sum_{k = 2}^{N} \frac {1} {k^2 - 1} - \sum_{k = 2}^{N} \frac {1} {k^2} < \int_{2}^{N} \left ( \frac {1} {x^2 - 1} - \frac {1} {x^2} \right ) \textrm {d}x = \int_{2}^{N} \frac {\textrm {d}x} {x^2 - 1} - \int_{2}^{N} \frac {\textrm {d}x} {x^2} < \frac {1} {N^2 - N} - \frac {1} {2} + \log \sqrt {3},$$
we have
$$\frac {7} {4} + \frac {1} {2} - \log \sqrt {3} - \frac {2 N^2 + N + 1} {2N^3 - 2N} < \sum_{k = 1}^{N} \frac {1} {k^2} < \frac {7} {4} - \frac {2N + 1} {2 N^2 + 2 N}$$
But how to obtain a more precise expression without using analytic methods such as Euler summation?
|
The Euler summation formula gives, for all $x\ge 1$,
$$
\sum_{n\le x}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{1}{x}+O(x^{-2}),
$$
by taking $f(t)=t^{-2},\, f'(t)=-2t^{-3}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integral with inverse trig functions and u substitution I've been trying to find the integral of $\dfrac{dx}{x \sqrt{x^2-4}}$. Currently, following the example problem, I have it as $\dfrac{1}{2} S \dfrac{1\,du}{x \sqrt{4 [(\frac{x}{2})^2-1]}}$. From what I've got it looks like it's going to the formula matching $\sec^{-1}(x)$. But at this point I'm not sure what to do now about the 4 in there that will get me to the $1/2$ coefficient. I have $u = x/2$, and $du=\frac{1}{2} \, dx$. The answer given is $-\frac{1}{2}\csc^{-1}(\frac{x}{2})+C$. The opposite trig function from what I thought and with a negative somehow.
|
\begin{align}
& \frac{dx}{x \sqrt{x^2-4}} = \frac{dx}{x\sqrt{4\left(\left(\frac x 2\right)^2 - 1\right)}} = \frac 1 {\sqrt 4} \cdot \frac{dx}{x\sqrt{\left( \frac x 2 \right)^2 - 1}} = \frac 1 2 \cdot \frac{dx/2}{\frac x 2 \sqrt{\left( \frac x 2 \right)^2 - 1}} \\[10pt]
= {} & \frac 1 2 \cdot \frac{du}{u\sqrt{u^2 -1}} = \frac 1 2 \cdot \frac{\sec\theta\tan\theta \, d\theta}{\sec\theta\sqrt{\sec^2\theta - 1}} = \frac{d\theta} 2.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do you solve $\frac{|x^2+5x+6|}{|x|-3} = 1$ How do you solve
$ \frac{|x^2+5x+6|}{|x|-3} = 1 $ ?
I have tried rearranging, polynomial division, multiplying both sides by a a fraction to simplify to no avail for the last hour.
|
$$ \frac{|x^2+5x+6|}{|x|-3} = 1 \implies |x^2+5x+6|=|x|-3$$
Solving Algebraically:
Case$1$: When, $x <-3$,
$$ x^2+5x+6=-x-3 \implies x^2+6x+9=0 \implies x=-3 \text{ [No solution]. }$$
Case$2$: When, $-3 < x \leq -2$,
$$ -x^2-5x-6=-x-3 \implies x^2+4x+3=0 \implies (x+3)(x+1)=0 \implies x=-3,-1 \text{ [No solution from this case]. }$$
Case$3$: When, $-2 < x \leq 0$,
$$ x^2+5x+6=-x-3 \implies x^2+6x+9=0 \implies x=-3 \text{ [No solution from this case]. }$$
Case$4$: When, $x > 0$ but $x \neq 3$,
$$ x^2+5x+6=x+3 \implies x^2+4x+3=0 \implies \implies (x+3)(x+1)=0 \implies x=-3,-1 \text{ [No solution from this case]. }$$
So, the answer is no solution.
Solving Geometrically:
As mentioned by Laars Helenius.
You can plot both $|x^2+5x+6|$ and $(|x|-3)$ on the same plot and check for the intersecting point, and that will be your answer. But you have to check whether those solution points in your domain. In this case, you will get one point $-3$, but your function domain is $\mathbb{R}\setminus \{-3,3\}$. So, no solution.
|
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|
Simplify the expression $(a+1)(a^2+1)(a^4+1)\cdots(a^{32}+1)$ How do I simplify this expression?
$$\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)$$
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Oh this is so cute!
We know that $(x + y)(x - y) = x^2 - y^2$ and therefore for for any $(a^n - 1)(a^n + 1) = (a^{2n} - 1)$.
So if we simply multiply $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)$ by $(a-1)$ we get
$(a -1)\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$(a^2 - 1)\left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
And the whole thing collapses like dominoes:
$\left( a^{4}-1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$\left( a^{8}-1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$ \left( a^{16}-1 \right)\left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$ \left( a^{32}-1 \right) \left( a^{32}+1 \right)=$
$ \left( a^{64}-1 \right)$
So $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right) = \frac {a^{64} - 1}{a - 1}$.
But here's are two other another interesting observations:
$(a - 1)(a^n + a^{n-1} + ...+a + 1) = (a^{n+1} + a^n + ... + a^2 + a) - (a^n + a^{n-1} +... + a + 1) = (a^{n+1} - 1)$
So $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right) = \frac {a^{64} - 1}{a - 1} = (a^{63} + a^{62} + .... + a^2 + a + 1)$
Which we can verify directly by noting
$(a^n + 1)( a^{n-1} + ...+ a + 1) = a^{2n - 1} + a^{2n - 2} + ... + a + 1$
so $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$(a^3 + a^2 + a + 1) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$(a^7 + a^6+.. + a + 1) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
...
$(a^{63} + a^{62}+.. + a + 1) $
Read up on proof by induction and on geometric series. They're fun stuff.
|
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|
Prove $\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+c^3+abc}+\frac{1}{b^3+c^3+abc} \leq \frac{1}{abc}$ I have to apologize in advance that the question is missing some vital information (like, $a,b,c$ are positive? natural? rational?) I took it off some video off youtube.
In the video, an 11 year old was in the process of solving this question: prove $\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+c^3+abc}+\frac{1}{b^3+c^3+abc} \leq \frac{1}{abc}$.
I don't really see how we can solve it in a simple way without using differential multivariable calculus. And even with calculus - it isn't so simple and not very pleasant. If anyone knows the question in its full form and knows how to answer it simply, I will be very happy to see it.
In the video, the child said "wlog $abc=1$" and defined $x=a^3$, $y=b^3$, $z=c^3$ and so he had to prove that
$\frac{1}{x+y+1}+\frac{1}{x+z+1}+\frac{1}{y+z+1} \leq 1$
The $abc=1$ part seems to be not so general, but if that helps you solve the question in any simple way - have at it.
|
We have $$\sum \frac{1}{a^3+b^3+abc} \le \sum \frac{1}{a^2b+ab^2+abc} = \frac{1}{a+b+c} \sum \frac{1}{ab} = \frac{1}{abc}$$
Note that $$a^3+b^3 \ge a^2b+ab^2 \iff (a-b)^2(a+b) \ge 0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The number of integers $n$ such that the quadratic equation $nx^2+(n+1)x+(n+2)=0$ has rational roots is The number of integers $n$ such that the quadratic equation $nx^2+(n+1)x+(n+2)=0$ has rational roots is
$(A)0\hspace{1cm}(B)1\hspace{1cm}(C)2\hspace{1cm}(D)3$
The condition for the rational roots is that the discriminant should be greater than or equal to zero and discriminant should be a perfect square.
Discriminant$=(n+1)^2-4n(n+2)=n^2+2n+1-4n^2-8n=-3n^2-6n+1$
$-3n^2-6n+1\geq 0.......................(1)$
I dont know how to apply the perfect square condition here.With the equation $(1)$,i get only one integer $n$.But the correct answer given in my book is $2$.Two integer $n$ are possible.$n=-1,n=-2$.
Please help me.Thanks
|
Complete the square:
$$-3n^2-6n+1 = -3n^2-6n-3+3+1 = -3(n+1)^2+4.$$
For this to be nonnegative, we must have $(n+1)^2 \le \frac{4}{3}$. So $n=0, -1, -2$ are the only solutions, and $n=0$ gives a linear, not a quadratic, equation (though this special case would probably stop me from using this problem).
|
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|
least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
The least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
$\bf{My\; Try::}$ Let $$K = 2x^2+y^2+2xy+2x-3y+8$$
So $$\displaystyle y^2+(2x-3)y+2x^2+2x+8-K=0$$
Now For real values of $y\;,$ We have $\bf{Discriminant\geq 0}$
So $$(2x-3)^2-4(2x^2+2x+8-K)\geq 0$$
So $$-4x^2-20x-23+4K\geq0\Rightarrow 4x^2+20x+23-4K\leq0$$
Now How can I Solve after that, Help me
Thanks
|
\begin{align}
&2x^2+y^2+2xy+2x-3y+8\\
&=\left(x^2+y^2+\frac94+2xy-3x-3y\right)+x^2+5x+\frac{23}{4}\\
&=(x+y-3/2)^2+x^2+5x+\frac{23}4\\
&=(x+y-3/2)^2+\left(x+\frac52\right)^2-\frac12\\
&\geq-\frac12\\
&(x=-\frac52,\space y=4)
\end{align}
|
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|
Seemingly Simple Integral $\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$. Evaluate $$\int_0^1 f(x) dx$$ where
$$f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}$$
I started off with the substitution $x=\sin y$, which resulted in the integrand reducing to
$$\sin^2y\cdot \ln (\sin y) dy$$
Then I used the property of definite integrals that
$$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$
Then too it wasn't getting simplified.
I tried $e^z=\sin x$, but this gave no headway because after a while I reached a complete full-stop. How should I go about this?
|
$$
\begin{aligned}
I &\stackrel{x=\sin \theta}{=} \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta \ln (\sin \theta)}{\cos \theta} \cdot \cos \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \sin ^{2} \theta \ln (\sin \theta) d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)\\& \stackrel{IBP}{=} \left[\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right) \ln (\sin \theta)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)d(\ln(\sin \theta) )\\& =-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\theta}{2} d(\ln (\sin \theta))}_{J}+\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\sin 2 \theta}{4} d(\ln (\sin \theta))}_{K}
\end{aligned}
$$
Integration by parts yields
$$\begin{aligned} J &=\left[\frac{\theta}{2} \ln (\sin \theta)\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta =\frac{\pi}{4} \ln 2 \end{aligned}$$
$$
\begin{aligned}
K &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 \theta}{\sin \theta} \cos \theta d \theta =\frac{1}{4} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) d \theta =\frac{\pi}{8} \\
\end{aligned}
$$
Now we can conclude that
$$\boxed{I =\frac{\pi}{8}(1-\ln 4)}$$
|
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Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must factorize the denominator:
$$x^3+2x^2 = (x+2)\cdot x^2$$
Great. So now we write three fractions:
$$\frac{A}{x^2} + \frac{B}{x} + \frac{C}{x+2}$$
Eventually we conclude that
$$A(x+2)+B(x+2)(x)+C(x^2) = 5x^2+3x-2$$
So now we look at what happens when $x = -2$:
$$C = 12$$
When $x = 0$:
$$A = -1$$
And now we are missing $B$, but we can just pick an arbitrary number for $x$ like... $1$:
$$B = -1$$
We replace the values here:
$$\int \frac{-1}{x^2}dx + \int \frac{-1}{x}dx + \int \frac{12}{x+2}dx$$
Which results in
$$\frac{1}{x}-\ln(x)+12\ln(x+2)+K$$
But I fear the answer actually is
$$\frac{1}{x}+2\ln(x)+3\ln(x+2)+K$$
Can you tell me what did I do wrong, and what should I have done?
|
The value of $A$ is right. $A = -1$. When you plug in $x = -2$, you've forgotten to divide it by $x^2$, so you get $C = 12$. Actually, $C = 3$.
\begin{align}
-(x+2) + Bx(x+2) + 3x^2 &= 5x^2+3x-2 \\
3x^2-x-2 + Bx(x+2) &= 5x^2+3x-2 \\
Bx(x+2) &= 2x^2 + 4x = 2x(x+2)\\
B &= 2
\end{align}
\begin{align}
& \int \frac{-1}{x^2}dx + \int \frac{2}{x}dx + \int \frac{3}{x+2}dx \\
=& \frac{1}{x}+2\ln(x)+3\ln(x+2)+K
\end{align}
|
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|
Let z = 1 + i. Find the real and imaginary parts of z^19 Let z = 1 + i. Find the real and imaginary parts of $z^{19}$
|
It is quickly determined that $\frac{z}{\sqrt{2}} = \frac{1+i}{\sqrt{2}} = e^{\pi i/4}$ and
\begin{align}
\left(\frac{z}{\sqrt{2}} \right)^{19} &= e^{19 \pi i/4} = e^{4 \pi i + 3 \pi i/4} = \cos(4\pi) \, \left( \cos\left(\frac{3 \pi}{4}\right) + i \, \sin\left(\frac{3\pi }{4} \right) \right) \\
&= \frac{-1 + i}{\sqrt{2}}.
\end{align}
From this: $z^{19} = 2^{9} \, (-1 + i)$.
|
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|
Find BC/AB if I lies in the circumcircle of AEF Let $ABC$ be a right triangle with angle $B=\frac{\pi}{2}$. Let $E$ and $F$ be the midpoint of $AB$ and $AC$ respectively. If $I$ the in centre of $ABC$ lies on the circumcircle of $AEF$, find the ratio $BC/AB$.
My guess would be that $AB=BC$ but I am unable to prove it. I tried angle chasing but that doesn't seem to lead me anywhere. Or is this possible by using coordinates?
|
Without loss of generality you may choose coordinates
$$A=\begin{pmatrix}0\\0\end{pmatrix}\qquad
B=\begin{pmatrix}1\\0\end{pmatrix}\qquad
C=\begin{pmatrix}1\\\lambda\end{pmatrix}\qquad
E=\begin{pmatrix}1/2\\0\end{pmatrix}\qquad
F=\begin{pmatrix}1/2\\\lambda/2\end{pmatrix}$$
Now the barycentric coordinates of the incenter are $a:b:c$ and we have $a=\lambda$, $b=\sqrt{1+\lambda^2}$ and $c=1$. So the incenter has coordinates
$$I = \frac{aA+bB+cC}{a+b+c}=
\frac{1}{\lambda+\sqrt{1+\lambda^2}+1}\begin{pmatrix}
\sqrt{1+\lambda^2}+1\\\lambda
\end{pmatrix}$$
Now you want $A,E,F,I$ cocircular. Looking at this answer of mine, you can check that by computing the determinant
$$\begin{vmatrix}
x_1^2+y_1^2 & x_1 & y_1 & 1 \\
x_2^2+y_2^2 & x_2 & y_2 & 1 \\
x_3^2+y_3^2 & x_3 & y_3 & 1 \\
x_4^2+y_4^2 & x_4 & y_4 & 1
\end{vmatrix}=0$$
So in this case:
$$\begin{vmatrix}
0 & 0 & 0 & 1 \\
1/4 & 1/2 & 0 & 1 \\
1/4+\lambda^2/4 & 1/2 & \lambda/2 & 1 \\
\frac{\left(\sqrt{1+\lambda^2}+1\right)^2+\lambda^2}
{\left(\lambda+\sqrt{1+\lambda^2}+1\right)^2} &
\frac{\sqrt{1+\lambda^2}+1}{\lambda+\sqrt{1+\lambda^2}+1} &
\frac{\lambda}{\lambda+\sqrt{1+\lambda^2}+1} &
1
\end{vmatrix}=0$$
Developing the determinant by the first row simply means restricting things to the lower left $3\times3$ determinant. You can further simplify the equation by multiplying rows by their denominator.
$$\begin{vmatrix}
1 & 2 & 0 \\
1+\lambda^2 & 2 & 2\lambda \\
\left(\sqrt{1+\lambda^2}+1\right)^2+\lambda^2 &
\left(\sqrt{1+\lambda^2}+1\right)
\left(\lambda+\sqrt{1+\lambda^2}+1\right) &
\lambda\left(\lambda+\sqrt{1+\lambda^2}+1\right)
\end{vmatrix}=0$$
Unfortunately the square roots haven't canceled yet, so I'll ask a computer algebra system for help here.
sage: PR1.<x,b> = QQ[]
sage: m = matrix([
... [1,2,0],
... [1+x^2,2,2*x],
... [(b+1)^2+x^2,(b+1)*(x+b+1),x*(x+b+1)]])
sage: d = m.det()
sage: ideal([d, x^2 + 1 - b^2]).variety(QQbar)
[{b: -1, x: 0},
{b: 0, x: -1*I},
{b: 0, x: 1*I},
{b: 1, x: 0},
{b: 1.666666666666667?, x: 1.333333333333334?}]
So there are five algebraic solutions. Two have a complex value for $\lambda$, leading to a zero length $b$ which is not geometrically sensible since it would imply $A=C$. (Speaking about lengths and points at complex coordinates in the same setup would be tricky business, and probably should start with a clear definition of the terms involved.)
Two have $\lambda=0$, so $B=C$, in which case many of the objects involved are not well defined.
The only remaining solution is for $\lambda=\frac43$ as Lucian wrote.
|
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|
What is the sum of the digits of the sum of the digits? Problem
Let $(10^{2016}+5)^2=225N$. If $S$ is the sum of the digits of N, then find the sum of the digits of S
Attempt
Let's look at some smaller cases. We have $\dfrac{(10^{3}+5)^2}{225} = 4489$,$\dfrac{(10^{4}+5)^2}{225} = 444889$, and $\dfrac{(10^{5}+5)^2}{225} = 44448889$. Thus we see the pattern and $S = 4*2015+8*2014+9 = 24181$ and the sum of the digits of $S$ is $\boxed{16}$.
Question
Prove the result in the solution above by induction or some other method. That is, show that for $n > 1$
$$\dfrac{(10^{n}+5)^2}{225} =\underbrace{44\ldots4}_\text{n-1 4's}\underbrace{88\ldots8}_\text{n-2 8's}9.$$
|
hint
*
*$U_{n+1}=\frac{(10^{n+1}+5)^2}{225}=\frac{(10^{n}+5+10^n*9)^2}{225}$
= $\frac{(10^{n}+5)^2}{225}+\frac{2*(10^{n}+5)*(10^n*9)}{225}+\frac{(10^{n+1}+5)^2}{225}=U_{n}+10^{n-1}(44*10^{n-1}+4)$ after simplification
|
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|
How can one show the inequality $\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge a+b+c$ How can one show the inequality
$$\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge a+b+c$$
Where $a,b,c$ are real and $ab+bc+ac$ is no equal to zero
This is what I did:
By Cauchy we have:
$$\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge\frac{(a+b+c)^2}{\sqrt{b^2-bc+c^2}+{\sqrt{a^2-ac+c^2}}+{\sqrt{a^2-ab+b^2}}}$$
So I need to prove that $${\sqrt{b^2-bc+c^2}+{\sqrt{a^2-ac+c^2}}+{\sqrt{a^2-ab+b^2}}}\ge(a+b+c)$$
We see that for example $b^2-bc+c^2=(b-c)^2+bc$
I still stuck here!
|
Hint
Use Holder inequality:
$$\left(\sum_{cyc}\dfrac{a^2}{\sqrt{b^2-bc+c^2}}\right)^2\left(\sum_{cyc}a^2(b^2-bc+c^2)\right)\ge (a^2+b^2+c^2)^3$$
so we need prove to
$$(a^2+b^2+c^2)^3\ge (a+b+c)^2\sum_{cyc}a^2(b^2-bc+c^2)$$
This is not hard to prove it.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $x^{10}y^6z^{-21}$ in $(3x^2 -5y^{\frac{1}{2}} + z^{-3})^{24}$ with Binomial theorem. Find $x^{10}y^6z^{-21}$ and $x^{16}y^4z^{-18}$ in $(3x^2 -5y^{\frac{1}{2}} + z^{-3})^{24}$ with Binomial theorem.
Well, I believe I found $x^{10}y^6z^{-21}$ and $x^{16}y^4z^{-18}$ .
But, I'm a bit hesitated about the solution, because I didn't use the fact: $i+j+k = 24$ and I believe I have to use it. What do you guys think? Here's my solution:
|
It is also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an algebraic expression.
Instead of using the multinomial version we could also successively apply the binomial theorem. We obtain
\begin{align*}
[x^{10}&y^6z^{-21}](3x^2-5y^{\frac{1}{2}}+z^{-3})^{24}\tag{1}\\
&=[x^{10}y^6z^{-21}]\sum_{k=0}^{24}\binom{24}{k}3^kx^{2k}(-5y^{\frac{1}{2}}+z^{-3})^{24-k}\tag{2}\\
&=3^5\binom{24}{5}[y^6z^{-21}](-5y^{\frac{1}{2}}+z^{-3})^{19}\\
&=3^5\binom{24}{5}[y^6z^{-21}]\sum_{k=0}^{19}\binom{19}{k}(-5)^ky^{\frac{1}{2}k}z^{-3(19-k)}\tag{3}\\
&=3^5(-5)^{12}\binom{24}{5}\binom{19}{12}\\
\end{align*}
Comment:
*
*In (2) we select $k=5$ in order to obtain the coefficient of $[x^{10}]$.
*In (3) we select $k=12$ in order to obtain the coefficient of $[y^{6}]$ as well as $[z^{-21}]$.
Note: If we consider a summand $$[x^{10}y^6z^{-21}]a_{ijk}x^{2i}y^{\frac{1}{2}j}z^{-3k}$$ of the trinomial (1) with coefficient $a_{ijk}$, the following equations have to be fulfilled in order to find a possible solution
\begin{align*}
i+j+k&=24\\
2i&=10\\
\frac{1}{2}j&=6\\
-3k&=-21
\end{align*}
The last three equations give $i=5,j=12$ and $k=7$ fulfilling $i+j+k=24$.
We start (as we could always do) the second example with a plausibility check regarding the exponents
\begin{align*}
[x^{16}&y^4z^{-18}](3x^2-5y^{\frac{1}{2}}+z^{-3})^{24}
\end{align*}
We get the following equations
\begin{align*}
i+j+k&=24\\
2i&=16\\
\frac{1}{2}j&=4\\
-3k&=-18
\end{align*}
The last three equations give $i=8,j=8$ and $k=6$ not fulfilling $i+j+k=24$.
We conclude
\begin{align*}
[x^{16}&y^4z^{-18}](3x^2-5y^{\frac{1}{2}}+z^{-3})^{24}=0
\end{align*}
|
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|
Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$
My questions:
$1)$ Is there an easy way to see that $x=-8$ is a root too?
$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$
|
Ніnt:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=- \left( x+8 \right) \left( x+2 \right) \left( x-1 \right) ^{2}.$$
|
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|
Evaluating Indefinite Integral $$∫ \frac{1}{(1-x^3)^{1/3}}\, dx$$
I tried substituting $1-x^3$ as $t^3$ but I am not able to calculate it after that.
Thanks!
|
For solving
$$
I=∫\frac{1}{\sqrt[3]{1-x^3}}\, \mathrm{d}x
$$
Let $t=\dfrac{x}{\sqrt[3]{1-x^3}}$,
solved
$$
\begin{aligned}
x&=\frac{t}{\sqrt[3]{t^3+1}}\\
\mathrm{d}x&=\frac{1}{(t^3+1)\sqrt[3]{t^3+1}}\mathrm{dt}\\
\end{aligned}
$$
and
$$
\frac{1}{\sqrt[3]{1-x^3}}=\sqrt[3]{t^3+1}
$$
then
$$
I = \int\frac{\sqrt[3]{t^3+1}}{(t^3+1)\sqrt[3]{t^3+1}}\mathrm{dt}=\int\frac{1}{t^3+1}\mathrm{d}t
$$
The next part is simple.
$$
\begin{aligned}
I
&= \frac{1}{3}\int\frac{1}{1+(-1)^{2/3} t}+\frac{1}{1-(-1)^{1/3}t}+\frac{1}{t+1}\mathrm{d}t\\
&= \frac{1}{3}\left(\ln (t+1)-(-1)^{1/3} \ln \left((-1)^{1/3}-t\right)+(-1)^{2/3} \ln \left(t+(-1)^{2/3}\right)\right)\\
&=\frac{1}{3} \left(\ln \left(\frac{x}{\sqrt[3]{1-x^3}}+1\right)+(-1)^{2/3} \ln \left(\frac{x}{\sqrt[3]{1-x^3}}+(-1)^{2/3}\right)-(-1)^{1/3} \ln \left((-1)^{1/3}-\frac{x}{\sqrt[3]{1-x^3}}\right)\right)
\end{aligned}
$$
Finally, take the derivation and verify that the result is correct.
D[1 / 3 (-(-1)^(1 / 3) Log[(-1)^(1 / 3) - x / (1 - x^3)^(1 / 3)] + Log[1 + x / (1 - x^3)^(1 / 3)] + (-1)^(2 / 3) Log[(-1)^(2 / 3) + x / (1 - x^3)^(1 / 3)]), x] // FullSimplify
|
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|
Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$ Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
Partial integration can't solve the integral $\int \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$.
What substitution (or other methods) would you suggest?
|
For $x \in [1,\infty)$, you have $$0 \le \frac{1}{x\sqrt[3]{x^2+1}} \le \frac{1}{x^\frac{5}{3}}$$ and $\int_1^\infty \frac{dx}{x^\frac{5}{3}}$ is convergent.
|
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|
A misunderstanding concerning $\pi$ The very well-known expression
$$\frac {\pi} {4} = 1 - \frac {1} {3} + \frac {1} {5} - \frac {1} {7} + \cdots$$
puts me face to face with a contradictory position. Let
$$s_N = \sum_{k = 0}^{N} \frac {1} {4k + 1} - \sum_{k = 0}^{N} \frac {1} {4k + 3}.$$
Then it is obvious that
$$\frac {\pi} {4} = \lim_{N \to \infty} s_N.$$
By Euler-MacLaurin summation formula,
$$\sum_{k = 0}^{N} \frac {1} {4k + 1} = \int_{0}^{N} \frac {dx} {4x + 1} + \frac {1} {2} \left(1 + \frac {1} {4N + 1} \right) + o \left (\frac {1} {N^2} \right)$$
and
$$\sum_{k = 0}^{N} \frac {1} {4k + 3} = \int_{0}^{N} \frac {dx} {4x + 3} + \frac {1} {2} \left(\frac {1} {3} + \frac {1} {4N + 3} \right) + o \left (\frac {1} {N^2} \right).$$
We then have
$$s_N = \frac {1} {4} \log \left(3 - \frac {6} {4N + 3} \right) + \frac {1} {3} + \frac {1} {(4N + 1) (4N + 3)} + o \left (\frac {1} {N^2} \right)$$
and
$$\lim_{N \to \infty} s_N = \frac {\log 3} {4} + \frac {1} {3}.$$
But $\frac {\log 3} {4} + \frac {1} {3} \ne \frac {\pi} {4}$. How come? Where have I done the mistake?
|
such mistakes are sometimes difficult to track down! however, your idea is an interesting one. it leads to the following expansion:
$$
S=\sum_{k=0}^{\infty} \frac1{4k+1}-\frac1{4k+3} \\
=1 -\sum_{k=1}^{\infty} \frac1{4k-1}-\frac1{4k+1}
$$
hence
$$
1-S =\sum_{k=1}^{\infty} \frac1{4k-1}-\frac1{4k+1} \\
=2\sum_{k=1}^{\infty}\frac1{(4k)^2}\left(1-\frac1{(4k)^2}\right)^{-1} \\
=2\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac1{(4k)^{2n}} \\
=2\sum_{n=1}^{\infty}\frac1{4^{2n}}\sum_{k=1}^{\infty}\frac1{k^{2n}} \\
=2\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{4n}}
$$
now, using the expression for the zeta function at even integers in terms of the Bernouilli numbers:
$$
2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}
$$
we obtain
$$
1-S = \sum_{k=1}^{\infty} (-1)^{n+1} \frac{B_{2n}}{(2n)!}\left(\frac{\pi}2\right)^{2n}
$$
and, rearranging, with $S=\frac{\pi}4$, we obtain
$$
\frac{\pi}4 = \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!}\left(\frac{i\pi}2\right)^{2n}
$$
|
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|
Let $x^2+kx=0;k$ is a real number .The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set. Let $f(x)=x^2+kx;k$ is a real number.The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set.
The equation $x^2+kx=0$ has solutions $x=0,-k$.
So the solutions of the equation $f(f(x))=0$ should be $x=0,-k$.
$f(f(x))=(x^2+k x)^2+(x^2+k x)k=0$ gives
$x^4+k^2x^2+2kx^3+x^2k+x k^2=0$
$x^4+2kx^3+(k^2+k)x^2+xk^2=0$
$x(x^3+2kx^2+(k^2+k)x+k^2)=0$ has the solutions $x=0,x=-k$
Therefore $x^3+2kx^2+(k^2+k)x+k^2=0$ has the solution $x=-k$.
But putting $x=-k$ in the equation $x^3+2kx^2+(k^2+k)x+k^2=0$ gives me nothing.It just gives me $0=0$.
I am stuck here.Please help me.Thanks.
|
Let me try this way
$$f(x)=x^2+kx$$
$$f(f(x))=f(x)^2+kf(x)=f(x)(f(x)+k)$$
From $f(f(x))=f(x)(f(x)+k)$ the zeros of $f(x)$ are the zeros of $f(f(x))$. This makes you looking for other two remaining zeros since $f(f(x))$ is of 4-th degree.
Obviously they are contained in $f(x)+k$. Since you want to have the same real zeros as for $f(x)$ even for $f(x)+k$ this is possible only if $k=0$.
Other than that $f(x)+k$ may have two strictly complex solutions, and this is the case when $x^2+kx+k$ has two strictly complex solutions. A simple condition is $k^2-4k < 0$ or $0<k<4$
This finally gives $0\leq k<4$
(The problem you have encountered comes from the fact that under the given conditions $-k$ is the zero of $f(f(x))$, and this is why you have got $0=0$.)
|
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|
Hessian of a function Given this equation: $f(x,y,z) = \sqrt{1+x^2+y^2+z^2}$
I tried to calculate the Hessian --> for example $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{1+x^2+y^2+z^2}}$ The second derivativ respect to x is hard to calculate for me: I tried the product rule: $x*(1+x^2+y^2+z^2)^{-1/2}$.
Then i get: $\frac{1}{1+x^2+y^2+z^2} - \frac{x^2}{(1+x^2+y^2+z^2)^{3/2}}$
But this is wrong.
Could anyone help?
|
What you get it's quite correct, because $$\frac{\partial f}{\partial x^2}=\frac{1}{\sqrt{1+x^2+y^2+z^2}}-\frac{x^2}{(1+x^2+y^2+z^2)^{3/2}}$$
And making some algebra you get $$\frac{\partial f}{\partial x^2}= \frac{1+y^2+z^2}{(1+x^2+y^2+z^2)^{3/2}}$$
I think that the other may be similar to this
|
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|
How to use Mathematical Induction to prove $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}$? $$\frac{1}{1 \cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}$$
What I have so far in the induction is:
$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$
But I'm not sure where to go from there . . .
|
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{k+1}{k+2}$$
|
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|
Identity involving $\arcsin$ and $\sin$. $$\sin z = \sin \left(\arcsin \frac{\pi}{6} - \arccos \frac{\pi}{6} \right)$$
Is there a specific expansion of the expression on the right hand side of the equation? To evaluate the value of $z$?
|
As pointed out by @Bye_World, this looks like the trigonometric identity:
$$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \tag{1}$$
Since you already have:
$$arcsin \frac{\pi}{6} \Rightarrow \sin \alpha = \frac{\pi}{6} \tag{2}$$
$$arccos \frac{\pi}{6} \Rightarrow \cos \beta = \frac{\pi}{6} \tag{3}$$
You can then calculate the opposite sides:
$$\cos \alpha = \sqrt{1 - (\frac{\pi}{6})^2} \tag{4}$$
$$\sin \beta = \sqrt{1 - (\frac{\pi}{6})^2} \tag{5}$$
If you replace $(2, 3, 4, 5)$ in $(1)$, you now have:
$$\sin(\alpha - \beta) = \frac{\pi^2}{18} - 1 \tag{6}$$
The value of $z$ is therefore the $\arcsin$ of $(6)$.
|
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|
How to find a straight line orthogonal to a curve? We want to find the equation of line passing through $(3,6)$ and cutting the curve $y=\sqrt{x}$ orthogonally .
I thought that it means we have to find a normal through that point equation of normal is $y=mx -2am -am^3 $.
Putting $(3,6)$ in, we get $(m+2)(m^2-4m+3)$.
Then the equation of the straight line is coming $y+ 2x +4=0$.
EDIT not by OP: How to find the equation of straight line orthogonal to the curve.
|
You need this result:
If two curves $f(x)$ and $g(x)$ (or a curve and a line), pass each other orthogonally at $x=c$, then $f(c)=g(c)$ and $f'(c)g'(c)=-1$.
Let the equation of the line be $g(x)=ax+b$ and let $f(x)=\sqrt{x}$. Then we get the following system of equations:
$$3a+b=6$$
$$ac+b=\sqrt{c}$$
$$a \cdot \frac{1}{2\sqrt{c}} = -1$$
Subtracting the first from the second, we get $(c-3)a=\sqrt{c}-6$, thus $$a=\frac{\sqrt{c}-6}{c-3}$$
Therefore we get the following equation in $c$:
$$\frac{\sqrt{c}-6}{c-3} \cdot \frac{1}{2\sqrt{c}} = -1$$
$$\sqrt{c}-6 = -2\sqrt{c}(c-3)$$
Let $\sqrt{c}=x$, we get the following equation in $x$: $$x-6 = -2x(x^2-3)$$
$$x-6 = -2x^3+6x$$
$$2x^3-5x-6 =0$$
We notice that $x=2$ is a root, use the rational root theorem.
$$(x-2)(2x^2+4x+3)=0$$
$$x=2 \vee 2x^2+4x+3=0$$
The last equation has $b^2-4ac=16-24=-8$, so there are no real roots.
Thus $c=x^2=4$. Therefore $a=-4$ and $b=18$.Thus $y=-4x+18$.
|
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|
Minimal c satisfying $x+y-(xy)^c \geq 0$ for all $x,y\in [0,1]$ What is the minimal real $c$ satisfying $x+y-(xy)^c \geq 0$ for all $x,y \in [0,1]$?
Experimentally (though my experiments weren't necessarily accurate enough) I reached as low as $c=\tfrac{13}{32}$, where $c=\tfrac{12}{32}$ violates it.
|
Your minimal value also does violate it. Consider $x=y=0.02$, then
$$x+y-(xy)^{13/32}=-0.00164730...$$
Consider the implicit function $$x+y=(xy)^c$$
It is easy to show that the plot is symmetric over $y=x$. Changing coordinates to polar and considering function $r(\theta)$ we can show that $\theta=\tfrac\pi4$ is the maximum (see below). Your inequality then is equivalent to $r\geqslant r(\theta)$. Consider the path $x=y$ and do the substitution $x=y=t$ to get
$$2t=t^{2c}$$
Variable $t$ is then the maximum of $r(\theta)$. Find the function $t(c)$:
$$2=t^{2c-1}$$
$$t=2^{1/(2c-1)}$$
We can use hyperreal numbers to find its zero
$$2^{1/(2c-1)}=0$$
$$\frac{1}{(2c-1)}=-\infty$$
$$2c-1=0^-$$
$$c=\frac12^-$$
so the minimal $c$ is $$c=\frac12$$
Function $r(\theta)$
For $x+y=(xy)^c$ we substitute $x=r\cos(\theta)$, $y=r\sin(\theta)$ to get
$$r(\sin\theta+\cos\theta)=(r^2\sin\theta\cos\theta)^c$$
$$\sin\theta+\cos\theta=(r^{2-1/c}\sin\theta\cos\theta)^c$$
$$(\sin\theta+\cos\theta)^{1/c}=r^{2-1/c}\sin\theta\cos\theta$$
$$r^{2-1/c}=\frac{(\sin\theta+\cos\theta)^{1/c}}{\sin\theta\cos\theta}$$
$$r=\left(
\frac{(\sin\theta+\cos\theta)^{1/c}}{\sin\theta\cos\theta}
\right)^{c/(2c-1)}$$
Its derivative is equal to
$$r'=\frac{c}{2c-1}\cdot
\\
\left(
\frac{(\sin\theta+\cos\theta)^{1/c}}{\sin\theta\cos\theta}
\right)^{(-c-1)/(2c-1)}\cdot
\\
\left(
\frac{
\frac1c(\sin\theta+\cos\theta)^{1/c-1}(\cos\theta-\sin\theta)\sin\theta\cos\theta-(\sin\theta+\cos\theta)^{1/c}(
\cos\theta\cos\theta-\sin\theta\sin\theta
)
}{
(\sin\theta\cos\theta)^2
}
\right)$$
And is equal to $0$ when
$$\left(
\frac{(\sin\theta+\cos\theta)^{1/c}}{\sin\theta\cos\theta}
\right)^{(-c-1)/(2c-1)}=0$$
$$\sin\theta+\cos\theta=0$$
$$\theta=\pi n-\tfrac\pi4$$
None of them belong to $[0;\pi/2]$, second one:
$$(\sin\theta+\cos\theta)^{-1}(\cos\theta-\sin\theta)\sin\theta\cos\theta=c(
\cos\theta\cos\theta-\sin\theta\sin\theta
)$$
$$(\cos\theta-\sin\theta)\sin\theta\cos\theta=c\cos(2\theta)(\sin\theta+\cos\theta)$$
Plug $\theta=\tfrac\pi4$ to get
$$\cos\theta=\sin\theta,\cos(2\theta)=0 \implies 0=0$$
|
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|
Convergence of $\sum \limits _{n=1}^{\infty} (-1)^{n} \frac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$ $\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$
How to check this? I've tried using Leibniz test, it's easy to prove that this one is monotonous, but its limit is rather not $0$.
|
You had the right idea to try to use the alternating series test, however you should double check your calculations for the limit of the terms:
$\dfrac{2^n n!}{5 \cdot 7 \cdots (2n+3)}$ $= \dfrac{3 \cdot 2^nn!}{1 \cdot 3 \cdot 5 \cdots (2n+3)}$ $= \dfrac{3 \cdot 2^nn!}{1 \cdot 3 \cdot 5 \cdots (2n+3)} \cdot \dfrac{2^{n+2}(n+2)!}{2 \cdot 4 \cdots (2n+4)}$
$= \dfrac{3 \cdot 2^{2n+2}n!(n+2)!}{(2n+4)!}$ $= \dfrac{3 \cdot 2^{2n+2}(n+2)!^2}{(n+2)(n+1)(2n+4)!}$ $= \dfrac{3 \cdot 2^{2n+2}}{(n+2)(n+1)} \cdot \dbinom{2n+4}{n+2}^{-1}$
$\sim \dfrac{3 \cdot 2^{2n+2}}{(n+2)(n+1)} \cdot \dfrac{\sqrt{\pi(n+2)}}{2^{2n+4}}$ $= \dfrac{\tfrac{3\sqrt{\pi}}{4}}{(n+1)\sqrt{n+2}} \to 0$ as $n \to \infty$,
Then, by the alternating series test, the series converges.
In fact, the series converges absolutely, as we have shown that the terms behave like $n^{-3/2}$.
|
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Generating functions - deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ I would like some help with deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ using generating functions.
I have managed to do this for $1^2 + 2^2 + 3^2 +\cdots$ by putting
$$f_0(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 +\cdots$$
$$f_1(x) = x \frac{d}{dx}[f_0(x)] = \frac{1}{(1-x)^2} = 0 + x + 2x^2 + 3x^3 +\cdots$$
$$f_2(x) = x \frac{d}{dx}[f_1(x)] = \frac{x^2+x}{(1-x)^3} = 0^2 + 1^2x + 2^2x^3 + 3^2x^3+\cdots,$$
and I assume I'm supposed to be able to do something similar in this case, but things get trickier when it's bounded by n and I keep getting stuck.
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To answer otherwise, you can use the known fact that the sum of the first $n$ positive odd numbers is equal to $n^2$. So you have
$$\begin{cases}1=1^2\\1+3=2^2\\1+3+5=3^2\\1+3+5+7=4^2\\................\\1+3+5+…….+(2n-1)=n^2 \end{cases}$$ Therefore
$$\sum_{i=1}^{i=n}n^2=n+(n-1)3+(n-2)5+(n-3)7+………+2(2n-3)+(2n-1)$$ That is to say
$$\sum_{i=1}^{n}n^2=\sum_{k=0}^{n-1}(n-k)(2k+1)=\frac{n(n+1)(2n+1)}{6}$$
NOTE.- Quite different story is the sequence of the corresponding inverse squares; one has the Euler’s formula $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{(\pi)^2}{6}$$
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|
Proving that an equation has a solution
If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}=0$$
has at least one solution in the interval $(-1,1)$
My solution is as follows, I could not be sure on its correctness. If you have any other solutions, could you please share them?
My attempt:
Let$$f(x)=\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2} = \frac{a}{(x+1)(x-(\frac{-1+\sqrt5}{2}))(x-(\frac{-1-\sqrt5}{2}))}+\frac{b}{(x-1)(x^2+x+2)}$$
Then, the two summands have distinct signs in the interval $(\frac{-1+\sqrt5}{2},1)$. Therefore, $f$ might take on the value zero in this interval.
We know that $f(x)\rightarrow-\infty$ as $x \rightarrow 1^-$ and $f(x)\rightarrow\infty$ as $x \rightarrow (\frac{-1+\sqrt5}{2})^+$.
Then, for a sufficiently small $\delta>0$, $f(1-\delta)<0$ and $f(\frac{-1+\sqrt5}{2}+\delta)>0$. Also, $f$ is continuous on $[\frac{-1+\sqrt5}{2}+\delta,1-\delta]$. Thus, from the Intermediate Value Theorem, there exist at least one $c\in(\frac{-1+\sqrt5}{2}+\delta,1-\delta)$ such that $f(c)=0$.
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Sketch of solution. May be unstable. :-)
Rewrite $f(x)$ as
$$
f(x) = \frac{a(x^3+x-2)+b(x^3+2x^2-1)}{(x^3+2x^2-1)(x^3+x-2)}
$$
and denote the numerator by
$$
g(x) = (a+b)x^3+2bx^2+ax-(2a+b)
$$
and the denominator by
$$
h(x) = (x^3+2x^2-1)(x^3+x-2)
$$
Note that $h(x)$ has a zero (and therefore $f(x)$ has a pole) at $x_0 = \frac{\sqrt{5}-1}{2}$. In this vicinity, $h(x)$ goes from positive to negative.
Now,
$$
g(x_0) = \frac{3\sqrt{5}-9}{2}
$$
which is a negative number, while $g(1) = 2b$, a positive number. Therefore, in the interval $(x_0, 1)$, $f(x)$ goes from a large positive number to a large negative number. Since there are no other poles in this interval, there must be a zero in that interval.
|
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|
What is $\frac{1}{1+\sqrt[3]{2}}$ in $\mathbb{Q}(\sqrt[3]{2})$? Since $\mathbb{Q}(\sqrt[3]{2})$ is a field, any number $\neq 0$ has a reciprocal.
How then to write $\frac{1}{1+\sqrt[3]{2}}$ as a number $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ with fractions $a,b,c \in \mathbb{Q}$?
In general, how to "rationalize denominators" in a cubic field such as $\mathbb{Q}(\sqrt[3]{2})$ ?
$$ \frac{1}{a + b\sqrt[3]{2} + c\sqrt[3]{4}}$$
|
Use the expansion $(1+x^3)=\left( x+1 \right) \left( {x}^{2}-x+1 \right)$.
Then
$$
\frac{1}{1+\sqrt[3]{2}}=\frac{1-\sqrt[3]{2}+(\sqrt[3]{2})^2}{(1+\sqrt[3]{2})(1-\sqrt[3]{2}+(\sqrt[3]{2})^2)}=\frac{1-\sqrt[3]{2}+(\sqrt[3]{2})^2}{3}=\frac{1}{3}-\frac{\sqrt[3]{2}}{3}+\frac{(\sqrt[3]{2})^2}{3}
$$
|
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|
Nice way to solve $\int\int \frac{1}{1-(xy)^2} dydx$? This is something I've been thinking about lately;
$$\int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx$$
Solutions I've read involve making the substitutions: $x= \frac{sin(u)}{cos(v)}$ and $y= \frac{sin(v)}{cos(u)}$. This reduces the integral to the area of a right triangle with both legs of length $\frac{\pi}{2}$. My problem is that coming up with this substitution is not at all obvious to me, and realizing how the substitution distorts the unit square into a right triangle seems to require a lot of reflection. My approach without fancy tricks involves letting $u = xy$ and then the integral "simplifies" accordingly:
$\begin{align*} \int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx &= \int_0^1\frac{1}{x}\int_0^x \frac{1}{1-u^2}dudx\\
&= \int_0^1\frac{1}{2x}\int_0^x \frac{1}{1-u}+\frac{1}{1+u}dudx\\
&= \int_0^1\frac{1}{2x}ln\left(\frac{1+x}{1-x}\right)dx
\end{align*}$
If I've done everything right this should be $\frac{\pi^2}{8}$ but I haven't figured out how to solve it.
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Here is an alternative derivation. The form of the integrand suggest expanding it a geometrical series
$$\frac{1}{1-(xy)^2} = \sum_{n=0}^\infty x^{2n}y^{2n}$$
Now integrating term by term we get
$$\int_0^1\int_0^1\frac{{\rm d}x{\rm d}y}{1-(xy)^2} = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$
and from the well known result $\frac{\pi^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=0}^\infty\frac{1}{(2n+1)^2} + \sum_{n=1}^\infty\frac{1}{(2n)^2}$ we get
$$\int_0^1\int_0^1\frac{{\rm d}x{\rm d}y}{1-(xy)^2} = \frac{\pi^2}{8}$$
|
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|
Sine of argument with large n approximation I have worked an integral and reduced the integral to
$$\frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}$$
I want to show that for $$n\rightarrow \infty$$ the above equation reduces to
$$\frac{1}{2}$$
Evidently, this means the $2$ sine functions must cancel each other. But what is a good way to do this? Large $n$ results in sine toggling between $-1$ and $1$. Notice that the either sine function will have a sign opposite to the other.
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First notice that
$$\begin{align}
\sin(\frac{3n\pi}{2})&=\sin(\frac{n\pi}{2}+n\pi) \\
&=\sin(\frac{n\pi}{2})\cos(n\pi)+\cos(\frac{n\pi}{2})\sin(n\pi) \\
&=\sin(\frac{n\pi}{2})\cos(n\pi)+0 \\
&=(-1)^n\sin(\frac{n\pi}{2})
\end{align}$$
and hence
$$\sin(\frac{n\pi}{2})-\sin(\frac{3n\pi}{2})=(1-(-1)^n)\sin(\frac{n\pi}{2})$$
The above expression is $0$ or $2$ depending on $n$ is even or odd. Now, you can find a bound easily
$$0 \le (1-(-1)^n)\sin(\frac{n\pi}{2}) \le 2$$
And finally
$$\frac{1}{2} \le \frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}\le \ \frac{1}{2} + \frac{1}{ n \pi}$$
Now, you can guess what happens next! :)
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|
How to prove $2\sqrt{2+\sqrt{3}}=\sqrt{2}+\sqrt{6}$? My calculator and I were arguing one day about the cosine of some number.
The calculator said "$\cos(\frac x2)=\sqrt{2}+\sqrt{6}$".
I said "That's absurd because $\cos(\frac x2)=\sqrt{\frac{1+\cos(x)}2}$, which evaluates to $2\sqrt{2+\sqrt{3}}$ for this particular $x$!"
So I finally decided to do this the non-radical way and found the decimal approximations to be equivalent.
Which is weird and probably not a coincidence. So why is this, and does this have other values that work out like this?
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A way of showing this that has not yet been given by anyone is to obtain a "nice equation" (i.e. with integer coefficients) having $\;\sqrt{2}+\sqrt{6}\;$ as a solution and then solving this equation by standard methods.
Start by setting $\;x = \sqrt{2}+\sqrt{6}.\;$ Subtracting $\sqrt 2$ from both sides and then squaring both sides gives $\;x^2 - 2\sqrt{2}x + 2 = 6.\;$ Isolate the radical term and square again, to get $\;x^2 - 4 = 2\sqrt{2}x,\;$ followed by $\;x^4 - 8x^2 + 16 = 8x^2,\;$ or $\;x^4 - 16x^2 + 16 = 0.$
This equation is quadratic in $x^2,$ so using the quadratic formula gives
$$x^2 \; =\; \frac{16 \pm \sqrt{256 - 64}}{2} \;=\; \frac{16 \pm 8\sqrt{4 - 1}}{2} \;=\; 4(2 \pm \sqrt{3})$$
Therefore, we get
$$x \; = \; \pm2\sqrt{2 \pm \sqrt{3}}$$
Since $\;\sqrt{2} + \sqrt{6}\;$ is clearly greater than $3$ (note that $\sqrt{2} > 1$ and $\sqrt{6} > 2$), it follows that the value of $x$ that corresponds to $\;\sqrt{2} + \sqrt{6}\;$ is the expression $\;2\sqrt{2 \pm \sqrt{3}}.$
|
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|
Mathematically, how does one find the value of the Ackermann function in terms of n for a given m? Looking at the Wikipedia page, there's the table of values for small function inputs. I understand how the values are calculated by looking at the table, and how it's easy to see that 5,13,29,61,125 is $2^{n+3}-3$, but how does one go about calculating this "iterative" formula without pattern identification?
I started by looking at 61 (Ackermann 3,3) as being $2*(2*(2*(2*1+3)+3)+3)+3$
, which all I'm doing is expanding the recursive formula, but I have no idea that's simplified to create $2^{n+3}-3$ rather than just looking at patterns. This is not homework, just curiosity.
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$$A(0,n) = n+1 \;\text{(by definition)}$$
$$A(1,n) \rightarrow A(0,A(1,n-1)) \rightarrow A(1,n-1)+1 \rightarrow A(1,n-2)+2\Rightarrow A(1,0)+n$$
$$\rightarrow A(0,1)+n \rightarrow 2+n = \color{red}{2+(n+3)-3}$$
$$A(2,n) \rightarrow A(1,A(2,n-1)) \rightarrow A(2,n-1)+2 \rightarrow A(2,n-2)+4 \Rightarrow A(2,0)+2n$$
$$\rightarrow A(1,1)+2n \rightarrow 2n+3 = \color{red}{2(n+3)-3}$$
$$A(3,n) \rightarrow A(2,A(3,n-1)) \rightarrow 2(A(3,n-1)+3)-3 \rightarrow 4(A(3,n-2)+3)-3 $$
$$\Rightarrow 2^n(A(3,0)+3)-3 \rightarrow 2^n(A(2,1)+3)-3 = 2^n(2^3)-3 = \color{red}{2^{n+3}-3} $$
$$A(4,n) \rightarrow A(3,A(4,n-1)) \rightarrow 2^{A(4,n-1)+3}-3 \rightarrow 2^{2^{A(4,n-2)+3}}-3 \rightarrow 2^{2^{2^{A(4,n-3)+3}}}-3 $$
$$\Rightarrow\,(^{n}2)^{A(4,0)+3}-3 \rightarrow (^{n}2)^{A(3,1)+3}-3 \rightarrow (^{n}2)^{2^3}-3 \,=\, \color{red}{{^{n+3}}2-3}$$
$$\text{Assume}\;A(m,n) = 2[m](n+3)-3,\; \text{and note} \;2[m]2=4 \;\forall m>0$$
$$A(m+1,0) \rightarrow A(m,1) \rightarrow 2[m]4-3 = 2[m](2[m]2)-3 = \color{red}{2[m+1]3-3}$$
$$A(m+1,n+1) \rightarrow A(m,A(m+1,n)) \rightarrow 2[m](2[m+1](n+3)-3+3)-3\\
= 2[m](2[m+1](n+3))-3 = \color{red}{2[m+1](n+4)-3}$$
$$\mathbf{QED}$$
Note: single right arrow represents a single iteration of Ackermann function, and a double arrow represents many (usually $n$ iterations)
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|
How to prove $\sum\left(\frac{a}{b+c}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$ The question is to prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$$
$$a,b,c>0$$
I tried Cauchy, AM-GM, Jensen, etc. but had no luck. Thank you.
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$$\sum_{cyc}\frac{a^2}{(b+c)^2}-\frac{3(a^2+b^2+c^2)}{4(ab+ac+bc)}=$$
$$=\sum_{cyc}\left(\frac{a^2}{(b+c)^2}-\frac{3a^2}{4(ab+ac+bc)}\right)=$$
$$=\sum_{cyc}\frac{a^2(4ab+4ac-2bc-3b^2-3c^2)}{4(ab+ac+bc)(b+c)^2}=$$
$$=\sum_{cyc}\frac{a^2((a-b)(3b+c)-(c-a)(3c+b))}{4(ab+ac+bc)(b+c)^2}=$$
$$=\frac{1}{4(ab+ac+bc)}\sum_{cyc}(a-b)\left(\frac{a^2(3b+c)}{(b+c)^2}-\frac{b^2(3a+c)}{(a+c)^2}\right)\geq0$$
because
$sign(a-b)=sign(a(3b+c)- b(3a+c))=sign(a-b)=sign((a+c)^2-(b+c)^2).$
Done!
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|
Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$.
Should I use AM-GM for the expression in the middle of the inequality? We have $a+b \geq 2\sqrt{ab}$ etc.?
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Use AM-HM Inequality for both.
$$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}} \Rightarrow \frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$$
Similarly, you get $$\frac{1}{a}+\frac{1}{c}\ge \frac{4}{a+c}$$ and $$\frac{1}{c}+\frac{1}{b}\ge \frac{4}{c+b}$$
Now add the three and get the left inequality.
For right inequality, $$\frac{\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}}{3}\ge \frac{3}{\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}}$$ or
$$\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\ge \frac{9}{a+b+c}$$
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Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi$ How to compute
$$\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi ?$$
The answer I am provided with is $\dfrac {(-1)^n\pi r^n} {x^{n+1}}$ for $\ x>r$, but I have no idea whether this is actually correct and how to get this.
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Suppose we seek to evaluate
$$\int_0^{2\pi} \frac{(r\cos\phi+x)\cos(n\phi)}{r^2+2xr\cos\phi+x^2}
\; d\phi.$$
Introduce $z=\exp(i\phi)$ so that $dz=iz \; d\phi$ to get
$$\int_{|z|=1}
\frac{(r(z+1/z)/2+x)(z^n+1/z^n)/2}{r^2+xr(z+1/z)+x^2}
\frac{dz}{iz}
\\ = \frac{1}{4i} \int_{|z|=1}
\frac{(r(z+1/z)+2x)(z^n+1/z^n)}{(r^2+x^2)z + xrz^2 + xr}
\; dz.$$
The denominator may be factored manually and we get
$$(xz+r)(rz+x)$$
so the poles are at $z=0$ and
$$\rho_0 = -r/x
\quad\text{and}\quad
\rho_1 = -x/r$$
and with $r \lt x$ only $\rho_0$ is inside the contour.
We get for the residue at $\rho_0$
$$\frac{1}{4i} \left.
\frac{(r(z+1/z)+2x)(z^n+1/z^n)}{(r^2+x^2) + 2xrz}
\right|_{z=-r/x}
\\ = \frac{(-1)^n}{4i}
\frac{(-r^2/x - x + 2x)((r/x)^n+(x/r)^n)}
{r^2+x^2-2r^2}
\\ = \frac{(-1)^n}{4ix} ((r/x)^n+(x/r)^n).$$
Supposing that $n$ is a non-negative integer the only
remaining contribution is from the pole at zero of
$$\frac{1}{4i} \int_{|z|=\epsilon}
\frac{1}{z^n}
\frac{r(z+1/z)+2x}{(r^2+x^2)z + xrz^2 + xr}
\; dz.$$
Observe that
$$\frac{1}{(xz+r)(rz+x)}
= \frac{1}{x^2-r^2}\frac{1}{z+r/x}
- \frac{1}{x^2-r^2}\frac{1}{z+x/r}.$$
and that
$$[z^n] \frac{1}{z-\beta} =
\frac{1}{\beta} [z^n] \frac{1}{z/\beta-1}
= - \frac{1}{\beta^{n+1}}.$$
so that on performing coefficient extraction we obtain
$$\frac{1}{4i} \frac{1}{x^2-r^2}
\left(-r(-x/r)^{n-1}+r(-r/x)^{n-1}
- r(-x/r)^{n+1}+r(-r/x)^{n+1}
\\ - 2x(-x/r)^{n}+2x(-r/x)^{n}\right).$$
This is
$$\frac{1}{4i} \frac{1}{x^2-r^2} (-x/r)^{n-1}
(-r-r(-x/r)^2-2x(-x/r))
\\+ \frac{1}{4i} \frac{1}{x^2-r^2} (-r/x)^{n-1}
(r+r(-r/x)^2+2x(-r/x))
\\ = \frac{1}{4i} \frac{1}{x^2-r^2} (-x/r)^{n-1}
(-r+x^2/r)
\\+ \frac{1}{4i} \frac{1}{x^2-r^2} (-r/x)^{n-1}
(-r+r^3/x^2)
\\ = \frac{1}{4i} \frac{1}{r} (-x/r)^{n-1}
- \frac{1}{4i} \frac{r}{x^2} (-r/x)^{n-1}
\\ = - \frac{1}{4i} \frac{1}{x} (-x/r)^{n}
+ \frac{1}{4i} \frac{1}{x} (-r/x)^{n}.$$
Finally collect the contributions from both poles to get
$$2\pi i \times \frac{(-1)^n}{4ix}
\left(-(x/r)^n + (r/x)^n + (r/x)^n + (x/r)^n\right)
\\ = 2\pi i \times \frac{(-1)^n}{4ix}
\times 2 (r/x)^n
\\ = \frac{\pi (-1)^n}{x} \left(\frac{r}{x}\right)^n.$$
|
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How can I count solutions to $x_1 + \ldots + x_n = N$? I am interested in how many non-negative integer solutions there are to:
$$x_1 + \ldots + x_N = B$$
where at least $K$ of the variables $x_1, \ldots , x_N \geq C$
For example when:
$B = 5, N = 3, K = 2, C = 2$
I want to count the solutions to:
$$x_1 + x_2 + x_3 = 5$$
where at least $2$ of the variables are $\geq 2$.
I found the total number of candidate solutions using the $\binom{B+N-1}{B} = 21$
However, only $9$ of them have two variables $\geq 2$.
\begin{align*}
2+0+3& =5\\
2+1+2& =5\\
3+0+2& =5\\
1+2+2& =5\\
3+2+0& =5\\
0+2+3& =5\\
0+3+2& =5\\
2+3+0& =5\\
2+2+1& =5
\end{align*}
I feel there is a connection to the Associated Stirling numbers of the second kind. But I can't place it :(
EDIT:
Here is my code for enumerating them all to count the number of ways of select B elements from a set of N (uniformly with replacement), such that you have at least C copies of K elements - also shows the output for this question I'm asking here as it's the core piece. Obviously can't be run for very large values of the parameters - that's why I'm here :) Code is here
Here is another example for B = 6, N = 3, C = 2 and K = 2 there are 16 solutions:
\begin{align*}
0+2+4& = 6\\
0+3+3& = 6\\
0+4+2& = 6\\
1+2+3& = 6\\
1+3+2& = 6\\
2+0+4& = 6\\
2+1+3& = 6\\
2+2+2& = 6\\
2+3+1& = 6\\
2+4+0& = 6\\
3+0+3& = 6\\
3+1+2& = 6\\
3+2+1& = 6\\
3+3+0& = 6\\
4+0+2& = 6\\
4+2+0& = 6\\
\end{align*}
There are a number of different and correct solutions below. I don't know which to accept.
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You could move to new variables $x_i' = x_i - C \ge 0$ for the variables with the minimum value $C$. Then the equation transforms to
$$
x'_1 + \ldots x'_K + x_{K+1} + \ldots x_N = B - K \, C
$$
assuming WLOG GRUMBL MUMBL that those variables are the first $K$ ones.
This is a problem of the form
$$
n_1 + \ldots + n_N = n \quad (n_i \ge 0) \quad (*)
$$
which is related to compositions from combinatorics. A composition of $n$ is a way to write $n$ as sum of integers $n_i \ge 1$. We need weak composition of $n$, which allows terms $n_i \ge 0$, but only those which have $N$ terms. The number of compositions of $n$ into $N$ parts is $\binom{n-1}{N-1}$, the number of weak compositions of $n$ into $N$ parts follows from looking at the number of compositions of $n+N$ into $N$ parts (and subtracting $1$ for each of the $N$ parts on both sides), so equation $(*)$ has
$$
\binom{n+N-1}{N-1} = \binom{B-K\,C+N-1}{N-1}
$$
solutions.
Your example turns into
$$
x'_1 + x'_2 + x_3 = 5-2\cdot 2 = 1
$$
and should have $\binom{1+3-1}{3-1}=\binom{3}{2}=3!/(2!1!)=3$ solutions.
We find
$$
(x'_1, x'_2, x_3) \to (x_1, x_2, x_3) \\
(1,0,0) \to (3,2,0) \\
(0,1,0) \to (2,3,0) \\
(0,0,1) \to (2,2,1)
$$
The found solution in $x'_i$ has then to be transformed back via $x_i = x'_i + C$, see the right side above.
Update:
The above assumed the constraints were at fixed positions and attached the constraints $x_i \ge C$ to the variables $1$ to $K$.
It should be clear that if the constraints were assigned to a different subset of variables indexed by $I = \{ I_1, \ldots, I_K \} \subset \{1, \ldots, N \}$ we end up with the same solutions, albeit permutated, along to the permutation of the indices.
E.g. for the example we have:
$$
(x_1, x'_2, x_3') \to (x_1, x_2, x_3) \\
(1,0,0) \to (1,2,2) \\
(0,1,0) \to (0,3,2) \\
(0,0,1) \to (0,2,3)
$$
and
$$
(x_1', x_2, x_3') \to (x_1, x_2, x_3) \\
(1,0,0) \to (3,0,2) \\
(0,1,0) \to (2,1,2) \\
(0,0,1) \to (2,0,3)
$$
These are all $9$ solutions.
What we need here are combinations, as we are interested only in the chosen subset, not in what order the elements were chosen.
There are $\binom{N}{K}$ $K$-combinations from a set of $N$ elements.
Note: This part needs $C \ge 1$, otherwise the resulting tuples are not distinct. There might be further conditions to watch, to justify this factor, see the comments.
Result:
There should be
$$
\binom{N}{K} \binom{B-K\,C +N-1}{N-1}
$$
solutions to the original problem. Caution: This might degrade to an upper bound, see the comments.
|
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|
Find all non negative integers x,y,z so we get a whole square Find all $x,y,z\in\mathbb{N_0}$ so that there exists a $k\in\mathbb{N}$ so that
$$4^x+4^y+4^z=k^2$$.
We can transform this problem to:
Find all $a,b\in\mathbb{N_0}$ so that there exists a $t\in\mathbb{N}$ so that
$$4^a+4^b+1=t^2$$
where $a=x-z, b=y-z, k=2^z \cdot t$.
We see that $3|t^2 \Rightarrow 3|t$ and so we have that $a+b\equiv2 \mod3$.
Looking at $\mod 5$ we can see that both of $a$ and $b$ can't be both even at the same time. So one of them has to be odd.
|
Consider ,
$4^{x}+4^{y}+4^{z}=k^{2}.$
Assume that $x \geq y \geq z$.
It is immediately obvious that $k$ is even.
Let $k=2m$ where $m$ is an integer.
Thus we have,
$4^{x}+4^{y}+4^{z}=4m^{2}$, assuming that $x,y,z \geq 2$, we have,
$4^{x-1}+4^{y-1}+4^{z-1}=m^{2}$, and so we return to our intial equation.
If we repeat this procedure until at least one of $x,y,z$ are $0$, we shall reach the equation you described.
$1+4^{p}+4^{q}=n^{2}$ where $p,q,n$ are arbitrary integers. (*)
WLOG let $p>q$, so we have,
$4^{q}(4^{p-q}+1)=(n-1)(n+1)$.
(*) revealed that $n$ is odd and so $n-1,n+1$ are consecutive even integers. Thus one of these integers must be divisible by a power of $4$, and the other must be of the form $2b$ where $b$ is odd.
And so we have,
$4^{q-1}(4^{p-q}+1)=\frac{(n-1)}{2}\frac{(n+1)}{2}$. (Credit to Samiron)
Equating, the odd and even parts, we are left with,
$\frac{n-1}{2}=4^{q-1} \Rightarrow n=2(4^{q-1})+1,$ and
$\frac{n+1}{2}=4^{p-q}+1 \Rightarrow n=2(4^{p-q})+1$.
Finally,
$4^{p-q}=4^{q-1} \Rightarrow \frac{p+1}{2}=q$
Hence (*) becomes,
$1+4^{p}+4^{\frac{p+1}{2}}=n^{2} \Rightarrow (1+2^{p})^{2}=n^{2}, \Rightarrow n=2p+1$ , and we are done.
NB: If we equate,
$\frac{n-1}{2}=4^{p-q}+1 \Rightarrow n=2(4^{p-q})+3$ and,
$\frac{n+1}{2}=4^{q-1} \Rightarrow n=2(4^{q-1})-1$
Hence,
$2(4^{q-1})-1=2(4^{p-q})+3 \Rightarrow 2(4^{q-1})=2(4^{p-q})+4 \Rightarrow 4^{q-1}=4^{p-q}+2 \Rightarrow 4^{q-1}-4^{p-q}=2$
And so we have a pair of numbers that are multiples of $4$, that differ by $2$. This is clearly not possible. Hence this case yields no solutions.
|
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|
Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n + n4^{n-1} + 1)\cdot(4+1) + 2\cdot(2^n + n2^{n-1} + 1) + 1
\\
=
&
(4k+1)\cdot(4+1) + 2(2r+1) + 1
\\
= &16k+4k+4 +1+4r+2+1
\\
=
&20k + 4r + 8 = 4(5+r+2)
\end{align}
But i've only proved it is multiple of $4$.
|
You only need modular arithmetic here: both $3$ and $5$ have order $2$ modulo $8$, i.e. $3^r\equiv3^{r\mod 2},\enspace 5^r\equiv5^{r\mod 2}\pmod 8$. Now
*
*If $n$ is odd, $5^n\equiv 5$ and $3^{n-1}\equiv 1\mod8$, so
$$5^n + 2*3^{n-1}+ 1\equiv 5+2+1\equiv 0\mod8.$$
*If $n$ is even, $5^n\equiv 1$ and $3^{n-1}\equiv 3\mod8$, so
$$5^n + 2\cdot3^{n-1}+ 1\equiv 1+2\cdot 3+1\equiv 0\mod8.$$
|
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|
If $S=\{z\in\mathbb C:z\neq0\text{ and }\left|z+\frac{1}{z}\right|=a\}$ find $\min\limits_{z\in S}|z|$
If $a$ is a positive real number and $S=\{z\in \mathbb C : z\neq 0\ \text{ and } \vert z+\frac{1}{z}\vert = a\}$, Find $\min\limits_{z\in S}\vert z\vert$ and $\min\limits_{z\in S}\vert\frac{1}{z}\vert$
I first tried using the triangle inequality to get:
$$\vert z\vert^2-a\vert z\vert+1\ge 0$$
$$\vert z\vert\in\big(-\infty,\frac{a-\sqrt{a^2-4}}{2}\big)\cup\big(\frac{a+\sqrt{a^2-4}}{2},\infty\big)$$
since $\vert z\vert\ge 0$ and $a^2>a^2-4$,
$$\vert z\vert\in\big(0,\frac{a-\sqrt{a^2-4}}{2}\big)\cup\big(\frac{a+\sqrt{a^2-4}}{2},\infty\big)$$
So $\min\vert z \vert=0$. But the answer given is $\min\vert z\vert=\frac{\sqrt{a^2-4}-a}{2}$
|
Let $z=re^{i\phi}$.$\,\,\,$ If $|z+z^{-1}|=a$, then
$$r^2+r^{-2}+2\cos(2\phi)=a^2 \tag 1$$
We can write $(1)$ in a more convenient form as
$$\left(r+\frac1r\right)^2=a^2-4\sin^2(\phi) \tag 2$$
Note that $|a|\ge 2$ since the minimum of $r+\frac1r$ is $2$.
Now, differentiating both sides of $(2)$ with respect to $\phi$ yields
$$\left(2r-\frac{2}{r^3}\right)\frac{dr}{d\phi}=-4\sin(2\phi)$$
whereupon setting $\frac{dr}{d\phi}$ to $0$ reveals that $2\phi = \ell \pi$ for integer $\ell$.
For $r\le 1$, the minimum for $r$ occurs when $2\phi =0$. Substituting $2\phi=0$ into $(2)$ reveals
$$r_{min}+\frac1{r_{min}}=|a| \implies r_{min}=\frac{|a|-\sqrt{a^2-4}}{2}$$
|
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|
How to find $P(-1)$ for $\frac{P(2x)}{P(x+1)}=8-\frac{56}{x+7}$ and $P(1)=1$? $P(x)$ is a polynomial such that $P(1)=1$ and $\frac{P(2x)}{P(x+1)}=8-\frac{56}{x+7}$ and $P(-1)$ is rational. How to find $P(-1)$?
|
$8-\frac{56}{x+7}=\frac{8x+56-56}{x+7}=\frac{8x}{x+7}=\frac{P(2x)}{P(x+1)}$.
Expanding, we get,
$8xP(x+1)=(x+7)P(2x)$
It can be seen that $2x|P(2x)$, so $P(x)=xQ(x)$ for some polynomial $Q$.
Also, $x+7|P(x+1)$, so $P(x)=(x+6)R(x)$ for some polynomial $R$.
Combining the above results, $P(x)=x(x+6)S(x)$. We attempt to substitute it back in.
$\frac{8x}{x+7}=\frac{2x(2x+6)S(2x)}{(x+1)(x+7)S(x+1)}$
Simplifying and multiplying out, we get:
$(2x+2)S(x+1)=(x+3)S(2x)$.
Continuing the argument, we see that $x+2|S(x)$.
As mrprottolo noticed by comparing coefficients it is of degree 3, the polynomial is:
$$P(x)=\alpha x(x+2)(x+6)$$
which can also be verified by substitution. Plugging in $x=1$ to find $\alpha$, we have $\alpha=\frac{1}{21}$. Plugging in $-1$, $P(-1)=\frac{1}{21}\times(-1)\times(2-1)\times(6-1)=-\frac{5}{21}$.
|
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|
Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}>0$
This is not the starting inequality.
Is there something wrong in this method?
|
For the second inequality multiply by $a$ to get
$$\ln \left(1 + \frac 1a\right)^a < 1$$ or
$$\left(1 + \frac 1a\right)^a < e $$
which is correct since function $f(a) = \left(1 + \frac 1a\right)^a$ is strictly increasing and ha a limit at $+\infty$ equal to $e$.
To prove the first part use the function $g(a) = \left(1 + \frac 1a\right)^{a+1}$ which converges to $e$ from above.
|
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|
How to compute $\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used? Thanks
$$\lim\limits_{x \to 1^+} \left(\frac{\tan \sqrt {x-1}}{\sqrt{x^2-1}}\right)$$
|
$$
\lim_{x \to 1^+} \left(\frac{\tan\sqrt{x-1} }{\sqrt{x^2-1}}\right) = \lim_{x \to 1^+} \left(\frac{\tan\sqrt{x-1} }{\sqrt{x-1}\sqrt{x+1}}\right) = \lim_{x \to 1^+} \left(\frac{\sin\sqrt{x-1} }{\sqrt{x-1}} \cdot \frac{1}{\cos\sqrt{x-1}} \cdot \frac{1}{\sqrt{x+1}}\right) =$$
$$ \lim_{x \to 1^+} \left(\frac{\sin\sqrt{x-1} }{\sqrt{x-1}} \right) \cdot \lim_{x \to 1^+} \frac{1}{\cos\sqrt{x-1}} \cdot \lim_{x \to 1^+}\frac{1}{\sqrt{x+1}} = 1 \cdot 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}
$$
|
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|
Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$ Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
I found the general term of the sequence.
It is $\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$
So the sequence becomes $\sum_{k=1}^{1999}\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$
I tried telescoping but i could not split it into two partial fractions.And this raised to $\frac{1}{2}$ is also troubling me.What should i do to find the answer?
|
Given $$1+\frac{1}{k^2}+\frac{1}{(k+1)^2} = 1+\frac{1}{k^2}+\frac{1}{(k+1)^2}-\frac{2}{k(k+1)}+\frac{2}{k(k+1)}$$
So $$1^2+\left[\frac{1}{k}-\frac{1}{(k+1)}\right]^2+2\left[\frac{1}{k}-\frac{1}{(k+1)}\right]=\left[1+\frac{1}{k}-\frac{1}{k+1}\right]^2$$
|
{
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|
Approximation of $\frac{x}{\sqrt{x^2+R^2}}$ How do you prove this statement?
If $x\gg R$ then
$$\frac{x}{\sqrt{x^2+R^2}}\cong 1-\frac{1}{2}\left(\frac{R}{x}\right)^2$$
I have no ideas even how to start.
|
Compare these two functions at $x=\infty$. If their quotient $f/g$ tends to $1$, it means $f\sim g$:
$$\lim_{x\to\infty} \frac{\frac{x}{\sqrt{x^2+R^2}}}{1-\frac{1}{2}\left(\frac{R}{x}\right)^2}=
\lim_{x\to\infty} \frac{\frac{x}{\sqrt{x^2+R^2}}}{\frac{2x^2-R^2}{2x^2}}=
\lim_{x\to\infty} \frac{x\cdot 2x^2}{\sqrt{x^2+R^2}(2x^2-R^2)}=
\lim_{x\to\infty} \frac{2}{\sqrt{1+R^2/x^2}(2-R^2/x^2)}=1$$
|
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|
How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
|
$$x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)=\frac{x\left(\left(\sqrt{x^2-x}\right)^2-\left(\sqrt{x^2-1}\right)^2\right)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
$$=\frac{x(1-x)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
Since we're searching for the limit as $x\to -\infty$, let $x<0$. Then:
$$=\frac{\frac{1}{-x}(x(1-x))}{\sqrt{\frac{x^2}{(-x)^2}-\frac{x}{(-x)^2}}+\sqrt{\frac{x^2}{(-x)^2}-\frac{1}{(-x)^2}}}=\frac{x-1}{\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}}\stackrel{x\to -\infty}\to -\infty$$
Because $\sqrt{1-\frac{1}{x}}\stackrel{x\to -\infty}\to 1$ and $\sqrt{1-\frac{1}{x^2}}\stackrel{x\to -\infty}\to 1$ and $x-1\stackrel{x\to -\infty}\to -\infty$.
|
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|
Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx-0.25\int\frac{1}{2x^2+x+3}\,dx.$$
The left one is pretty straight forward with $\ln|\cdot|$,
Problem: does anyone have some "technique" to solve the right integral? hints would be appreciated too.
Edit: maybe somehow: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$
|
When you have a second degree polynomial in the denominator with $\Delta <0$, you can proceed like this: find the solutions, in this case $\frac{-1}{4} \pm \frac{i\sqrt {23}}{4}$, and then write the polynomial in the usual way
$$2\Big(x+\frac{1}{4} + \frac{i\sqrt {23}}{4}\Big)\Big(x+\frac{1}{4} - \frac{i\sqrt {23}}{4}\Big) $$
then you can use the fact that $(a-b)(a+b)=a^2-b^2$ to deduce
$$2x^2+x+3=2\Big(x+\frac{1}{4} \Big)^2+\frac{23}{8}.$$
Then apply a substitution to get $\frac{1}{1+y^2}$.
|
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|
Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
|
Setting $t=\sqrt[3]{35-x^3}$, we have
$$tx(t+x) = 30$$
and
$$t^3+x^3=35 \implies (t+x)(t^2+x^2-tx)=35$$
This gives us
$$\dfrac{t^2+x^2-tx}{tx} = \dfrac76 \implies 6t^2-13tx+6x^2 = 0 \implies 6t^2 - 9tx - 4tx + 6x^2 = 0$$
This gives us
$$3t(2t-3x)-2x(2t-3x) = 0 \implies (2t-3x)(3t-2x) = 0 \implies t = \dfrac{2x}3, \dfrac{3x}2$$
Trust you can finish it from here easily.
We then obtain $$t^3 = \dfrac8{27}x^3, \dfrac{27}8x^3 \implies 35-x^3 = \dfrac8{27}x^3, \dfrac{27}8x^3 \implies x^3=8,27$$This gives us $$x=2,2\omega,2\omega^2,3,3\omega,3\omega^2$$
|
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|
Solve $ \int {\frac{(x-1)dx}{(x-2)(x+1)^2 x^2 }} $ $$ \int {\frac{(x-1)dx}{(x-2)(x+1)^2 x^2 }} $$
I don't know about decomposition of fractions a lot,but I know that it is method which I have to use in this example. Please, help me.
|
$$ \frac{x-1}{(x-2) x^2 (x+1)^2} = \frac A{x-2}+\frac{Bx+C}{x^2} + \frac{D(x+1)+E}{(x+1)^2} $$
$$ \frac{x-1}{ x^2(x+1)^2} = A + \frac{Bx+C}{x^2}(x-2) + \frac{D(x+1)+E}{(x+1)^2}(x-2). $$
$x\to 2: A = \frac1{36}.$
$$ \frac{x-1}{(x-2)(x+1)^2} = \frac A{x-2}x^2+Bx+C + \frac{D(x+1)+E}{(x+1)^2}x^2 $$
$x\to 0: C = \frac12.$
$$ \frac{x-1}{(x-2)x^2} = \frac A{x-2}(x+1)^2+\frac{Bx+C}{x^2}(x+1)^2 + D(x+1)+E$$
$x\to -1: E = \frac23.$
$$ \frac{x-1}{(x-2)x(x+1)^2} = \frac A{x-2}x+\frac{Bx+C}{x} + \frac{D(x+1)+E}{(x+1)^2}x $$
$x\to\infty: A + B + D = 0.\quad (1)$
$$ \frac{x-1}{(x-2) x^2 (x+1)^2} = \frac A{x-2}+\frac{Bx+C}{x^2} + \frac{D(x+1)+E}{(x+1)^2}$$
$x\to 1: -A+B+C+\frac{2D+E}4 = 0. \quad (2)$
$$(1)-(2): 2A-C+ \frac D2-\frac E4 = 0,$$
$$D=-4A+2C+\dfrac E2 = -\frac19+1+\frac13 = \frac{11}9$$
$B=-(D+A) = -(\frac{11}9+\frac1{36}) = -\frac 54$
|
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|
Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$ Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$
Let $P(x)=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$
By using $AM-GM$ inequality on the first two terms give me
$8(27)^{\log_{6}x}+27(8)^{\log_{6}x}\geq2\sqrt{8\times 27\times(27\times 8)^{\log_6x}}\geq 12\sqrt{6x^3}$
$P(x)=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3\geq12\sqrt{6x^3}-x^3$
I found the maximum value of $12\sqrt{6x^3}-x^3$ by differentiating and equating it to zero.I found that $12\sqrt{6x^3}-x^3$ is maximum at $x=6$.And its maximum value is $216$.
Can the maximum value of $12\sqrt{6x^3}-x^3$ be the maximum value of $P(x)?$I am confused because P(x)is greater than or equal to $12\sqrt{6x^3}-x^3$.$P(x)$ may be greater than the maximum value of $12\sqrt{6x^3}-x^3$.
|
Angelo Mark has provided a good hint, but it seems that the OP needs more.
We can write
$$\begin{align}8\cdot 3^t+27\cdot 2^t-6^t&=8\cdot 3^t+27\cdot 2^t-2^t\cdot 3^t-8\cdot 27+8\cdot 27\\&=8(3^t-27)-2^t(3^t-27)+8\cdot 27\\&=(3^t-27)(8-2^t)+8\cdot 27\end{align}$$
Note here that $(3^t-27)(8-2^t)$ is non-positive.
|
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|
Which integers can be represented as the most pair of difference of two squares? Let $f(x)$ be the number of $a,b,x\in \mathbb N$ where $a^2-b^2=x$.
For example, 1971 is not only $986^2-985^2$ but also $50^2-23^2$, $114^2-105^2$, $330^2-327^2$. So, $f(1971)=4$. Is there some sort of limit to $f(x)/x$ or $f(x)/ln(x)$ or similar? For which kind of $x$?
|
Partial answer. If $x$ is odd, then $a=(x+1)/2$, $b=(x-1)/2$, satisfy $x=a^2-b^2$.
If $x$ is even, and $a$, $b$ exist, so that $x=a^2-b^2$, then $x=(a-b)(a+b)$, and clearly, both $a+b$, $a-b$ are even, and hence $x$ is divisible by $4$. Then, $a=(x+4)/4$ and $y=(x-4)/4$, satisfy $x=a^2-b^2$.
Note that in the first case $x\ge 3$, and in the second $x\ge 8$.
So $x$ is either odd $\ge 3$ or multiple of 4 and $x\ge 8$.
|
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|
How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$
So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
However I'm stucked from here on. Thank you in advance!
|
Let $A=\cos x.$ We have $\cos 2x=2A^2-1.\quad$$ \cos 3 x= 4A^3-3A.\quad$$ \cos 4x=2\cos^22x-1=8A^4-8A^2+1.\quad$$\cos 6x=2(\cos 3x)^2-1=32A^6-48A^4+18A^2-1.$ From the expressions for $\cos 2x,\; \cos 4x,\;\cos 6x$ we have $32A^6=\cos 6x+6\cos 4x+15\cos 2x+10$. So $\cos^6x=\sum_{n=0}^{\infty}(-1)^nx ^{2n}2^{-5}(6^{2n}+6.4^{2n}+15.2^{2n}+10)/(2n)!.$
|
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|
Evaluating the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$
Evaluate $$\int_{0}^{\infty}\frac{dx}{1+x^3}.$$
I tried integration by partial fraction. My work is below:
$$\int_{0}^{\infty}\frac{dx}{1+x^3}=\frac{1}{3} \int_{0}^{\infty}\frac{1}{x+1}+\frac{1}{3} \int_{0}^{\infty}\frac{2-x}{x^2-x+1}.$$
It seems that the result would go to infinity. But the answer is $\dfrac{2\pi}{3\sqrt3}$.
Did I do something wrong?
|
Hint. You may first integrate over the finite set $[0,M]$ then let $M \to \infty$:
$$\int_{0}^M\frac{dx}{1+x^3}=\frac{1}{3} \int_{0}^M\frac{1}{x+1}\:dx+\frac{1}{3} \int_{0}^M\frac{2-x}{x^2-x+1}\:dx,$$ you will see that the $\log$ terms cancel.
Edit. Let $M>0$. We have $$\frac{1}{3} \int_{0}^M\frac{1}{x+1}\:dx=\frac{1}{3}\log (M+1)$$
$$\frac{1}{3} \int_{0}^M\frac{2-x}{x^2-x+1}\:dx=-\frac{1}{6}\log (M^2-M+1)-\frac1{\sqrt3}\arctan\left(\frac{2 M-1}{\sqrt{3}}\right)+\frac{\pi}{6\sqrt3}.$$
You may observe that, as $M \to \infty$,
$$
\frac{1}{3}\log (M+1)-\frac{1}{6}\log (M^2-M+1)=\frac{1}{6}\log \left(\frac{1+2/M+1/M^2}{1-1/M+1/M^2} \right) \to 0.
$$
|
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|
how to find hyperbola equation knowing tangent line and point I have a problem.
A hyperbola passes through point $(3,2)$ and $9x+2y-15=0$ is a tangent line. Find the equation of hyperbola.
|
Let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be the hyperbola.It passes through $(3,2)$.
So $\frac{9}{a^2}-\frac{4}{b^2}=1................(1)$
Also tangent $9x+2y-15=0$ is given.
Put $y=\frac{-9x+15}{2}$ in the equation of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{x^2}{a^2}-\frac{1}{b^2}\times\frac{81x^2+225-270x}{4}=1$
$(\frac{1}{a^2}-\frac{81}{4b^2})x^2+\frac{270}{4b^2}x-(1+\frac{225}{4b^2})=0$
This is a quadratic equation in $x$,whose discriminant should be zero,as tangent cuts hyperbola at one point.Put $b^2-4ac=0$
$\frac{18225}{4b^4}=-4(\frac{1}{a^2}-\frac{81}{4b^2})(1+\frac{225}{4b^2}).....(2)$
Solve $(1)$ and $(2)$ to find $a^2$ and $b^2$,and substitute in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to get the equation.
Hope it helps.
|
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|
$\lim \frac{x^2-\sin^2{x}}{\tan(3x^4)}$ as $x$ goes to $0$
Calculate $\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)}$
How does one calculate this limit?
Is it valid to say, since $\sin^2{x}$ is approximated by $x^2$ as $x \to 0$, we have:
$\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)} =\lim_{x \to 0} \frac{x^2-x^2}{\tan(3x^4)} =0$
|
No, it is not valid to approximate $\sin^2 x$ by $x^2$, since there are higher order terms that are still significant. Using the expansion
$$\sin x = x - \frac{x^3}{6} + O(x^5)$$
we have that
$$\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$$ so that
$$\frac{x^2 - \sin^2 x}{\tan(3x^4)} = \frac{\frac 1 3 x^4 + O(x^6)}{\tan(3x^4)}$$
leading to an overall limit that is non-zero.
|
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Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.
That's what I've tried:
Let a Cauchy-Schwarz Inequality be :
\begin{array}
(((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2 )\left(\left(\cfrac{ab}{\sqrt{a}}\right)^2 +\left(\cfrac{ac} {\sqrt{b}}\right)^2+\left(\cfrac{bc}{\sqrt{c}}\right)^2\right)& \ge (ab+ac+bc)^2 \\
(a+b+c)(ab^2+\cfrac{a^2 c^2}{b}+b^2 c) &\ge (ab+ac+bc)^2 \\
\end{array}
However how should I now factorize the left hand side of the inequality as $(a^2+b^2+c^2) \cdot Something $ ?
I think I've made the problem harder than it needs to be .
|
If you insist on C-S, let
$$
u=(a,b,c),\quad v=(b,c,a).
$$
Then,
$$
\text{RHS}=u\cdot v\underbrace{\leq}_{\text{C-S}}||u||\times||v||=\text{LHS}.
$$
|
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|
Show the following inequality holds I want to show that $( 1 + x_1 ) (1 + x_2 )... ( 1 + x_n ) \ge ( 1 + (x_1 x_2 ... x_n ) ^\frac {1} {n } ) ^ n $ for all $x_i > 0$.
I started by taking logarithm on both sides and trying to use the concavity of logarithm but this reverses the inequality. I don't really know how to start, any help is apprecaited.
|
by induction:
$n=2, (1+x_1)(1+x_2)=1+x_1+x_2+x_1x_2 \ge 1+2\sqrt{x_1x_2}+x_1x_2=(1+(x_1x_2)^{\frac{1}{2}})^2$
when $n=k, (1+x_1)(1+x_2)...(1+x_k) \ge (1+(x_1x_2...x_k)^{\frac{1}{k}})^k$ holds,
$n=k+1, (1+x_1)(1+x_2)...(1+x_k)(1+x_{k+1}) \ge (1+(x_1x_2...x_k)^{\frac{1}{k}})^k(1+x_{k+1}) $
now, we prove
$(1+(x_1x_2...x_k)^{\frac{1}{k}})^k(1+x_{k+1}) \ge (1+(x_1x_2...x_kx_{k+1})^{\frac{1}{k+1}})^{k+1}$
let $a=(x_1x_2...x_k)^{\frac{1}{k}},x=x_{k+1} \implies (1+a)^k(1+x) \ge (1+(a^kx)^{\frac{1}{k+1}})^{k+1} \\ \iff \ln{(1+a)^k}+\ln{(1+x)} \ge (n+1) \ln{((1+(a^kx)^{\frac{1}{k+1}})} $
$f(x)=\ln{(1+a)^k}+\ln{(1+x)}-\ln{((1+(a^kx)^{\frac{1}{k+1}})}$
$f'(x)=\dfrac{1}{x}(\dfrac{x}{1+x}-\dfrac{y}{1+y}),y=(a^kx)^{\frac{1}{k+1}}$
$\dfrac{x}{1+x}$ is mono increasing function,$\implies \\ y=x(x=a), f'(x)=0, \\ y>x (x<a),f'(x)<0,\\ y<x(x>a),f'(x)>0 \implies \\f(x=a)=f_{min}=0 \implies f(x) \ge 0$
QED
|
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|
Why this system have one solution Let $b\in (1,2),x\in (0,\frac{\pi}{2})$,if such
$$\begin{cases}
2b^2+b-4=2\sqrt{4-b^2}\cos{x}\\
2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}
\end{cases}$$
show that:$$x=\dfrac{\pi}{6}$$
Here is what I already got.
First of all, one should notice equation $x=\dfrac{\pi}{6}$,
$$2b^2+b-4=\sqrt{12-3b^2}$$
then $b$ such
$$b^3-3b+1=0$$
But this kind of proof does not fit my appetite as it not only involves some additional theorem but also not very nice as the simple nice form of the question of itself.
|
Solving this is mainly just fiddly algebra.
$$2b^2+b-4=2\sqrt{4-b^2}\cos{x}$$
$$2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}$$
The second can be rewritten as:
$$2b^2-4=2b\left(\cos x\cos\frac{\pi}{18}-\sin x\sin\frac{\pi}{18}\right)-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}$$
Rearranging the first one gives:
$$\cos x=\frac{2b^2+b-4}{2\sqrt{4-b^2}}$$
and hence:
$$\sin x=\sqrt{1-\frac{(2b^2+b-4)^2}{4(4-b^2)}}=\frac{\sqrt{4(4-b^2)-(2b^2+b-4)^2}}{2\sqrt{4-b^2}}$$
Putting this together gives:
$$2b^2-4=2b\left(\frac{2b^2+b-4}{2\sqrt{4-b^2}}\cos\frac{\pi}{18}-\frac{\sqrt{4(4-b^2)-(2b^2+b-4)^2}}{2\sqrt{4-b^2}}\sin\frac{\pi}{18}\right)-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}$$
By repeated rearranging and squaring you can get rid of all the square roots and eventually end up with a high degree polynomial in $b$ to solve. It is unlikely to factorize nicely. Numerically technique would most likely be required to determine the roots of the equation.
|
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|
Solve $\sin 2x=-\cos x$ I'm working on solving the problem $\sin(2x)=-\cos(x)$ but I got stuck.
I got the following:
$\sin 2x=-\cos x \Leftrightarrow \sin 2x=\cos(x+\pi )\Leftrightarrow \sin 2x=\sin\left(\frac{\pi }{2}-(x+\pi)\right)\Leftrightarrow \sin 2x= \sin\left(-\frac{\pi }{2}-x\right)$
then I did
$2x =-\frac{\pi }{2}-x +2\pi k \Leftrightarrow 3x=\frac{\pi }{2} +2\pi k\Leftrightarrow x_1=\frac{\pi }{6}+\frac{2\pi}{3} k\ $
$2x =\pi -(-\frac{\pi }{2}-x) +2\pi k \Leftrightarrow x_2=\frac{3\pi}{2}+2\pi k$
None of these answers are correct.
The correct answer is $x=\frac{\pi}{2}+k\pi$ and $x=-\frac{\pi}{2}\pm\frac{\pi}{3}+2 k\pi$
I don't know what I'm doing wrong in my attempt.
|
$$\sin 2x = -\cos x $$
$$2\sin x\cos x = -\cos x$$
$$2\sin x\cos x + cos x= 0$$
$$\cos x (2\sin x + 1) = 0$$
$$\cos x = 0 \text{ or } \sin x = -\frac{1}{2}$$
$$x = (2n + 1)\frac{\pi}{2} \text{ or } x = m\pi + (-1)^{m+1}\frac{\pi}{6}$$
where $m,n\in\Bbb Z$
|
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|
Interval for area bounded by $r = 1 + 3 \sin \theta$ I'm trying to calculate the area of the region bounded by one loop of the graph for the equation
$$
r = 1 + 3 \sin \theta
$$
I first plot the graph as a limaçon with a maximum outer loop at $(4, \frac{\pi}{2})$ and a minimum inner loop at $(-2, -\frac{3 \pi}{2})$. I then note the graph is symmetric with respect to the $\frac{\pi}{2}$ axis and the zero for the right half is at $\theta = \arcsin(-\frac{1}{3})$.
So, I chose the interval $[\arcsin(-\frac{1}{3}),\frac{\pi}{2}]$ to calculate the area which can then be multiplied by $2$ for the other half. The problem is that the answer in the book seems to use $\arcsin(\frac{1}{3})$ instead, note the change of sign.
Just to make sure I'm not misunderstanding where I went wrong, I get the answer
$$
\frac{11 \pi}{4} - \frac{11}{2} \arcsin(-\frac{1}{3}) + 3 \sqrt 2
$$
Whereas the book gets
$$
\frac{11 \pi}{4} - \frac{11}{2} \arcsin(\frac{1}{3}) - 3 \sqrt 2
$$
It's a subtle change of sign but I'd really like to understand where I went wrong.
|
$$
\begin{align}
\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12r^2\,\mathrm{d}\theta
&=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12(1+3\sin(\theta))^2\,\mathrm{d}\theta\\
&=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12\left(1+6\sin(\theta)+9\sin^2(\theta)\right)\mathrm{d}\theta\\
&=\int_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\frac12\left(1+6\sin(\theta)+9\left(\frac{1-\cos(2\theta)}2\right)\right)\mathrm{d}\theta\\
&=\left[\frac{11}4\theta-3\cos(\theta)-\frac98\sin(2\theta)\right]_{\arcsin(-1/3)}^{\pi-\arcsin(-1/3)}\\
&=\frac{11}4\left(\pi+2\arcsin\left(\frac13\right)\right)+3\sqrt2
\end{align}
$$
This is the area of the outer loop
The sum of both loops is
$$
\begin{align}
&\int_0^{2\pi}\frac12\left(1+6\sin(\theta)+9\left(\frac{1-\cos(2\theta)}2\right)\right)\mathrm{d}\theta\\
&=\left[\frac{11}4\theta-3\cos(\theta)-\frac98\sin(2\theta)\right]_0^{2\pi}\\
&=\frac{11}2\pi
\end{align}
$$
so the area of the inner loop is
$$
\frac{11}4\left(\pi-2\arcsin\left(\frac13\right)\right)-3\sqrt2
$$
|
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|
A $\log \Gamma $ identity: Where does it come from? $$\log \Gamma (n)=n\log n -n +\frac{1}{2} \log \frac{2\pi}{n}+\int_0^\infty \frac{2\arctan (\frac{x}{n})}{e^{2\pi x}-1} \,\mathrm{d}x$$
Is an identity that is derived from using Sterling's approximation. I can't quite figure out how it was used, and was wondering for a proof.
|
Proposition : $$\int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right]$$
Proof : Let $ \displaystyle \text{I} (a) = \int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x$
$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{e^{-2ar \pi x}}{1+x^2} \mathrm{d}x $
$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \int_{0}^{\infty} e^{-x(2ar \pi + y)} \sin y \ \mathrm{d}y \ \mathrm{d}x $
$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin y}{2ar\pi + y} \mathrm{d}y $
$\displaystyle = - \sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin x}{2ar\pi + x} \mathrm{d}x$
Substitute $\displaystyle x \mapsto 2 r \pi x$
$\displaystyle \implies \text{I}(a) = - \sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin 2 r \pi x}{x + a} \mathrm{d}x $
$\displaystyle = -\int_{0}^{\infty} \dfrac{1}{x+a} \sum_{r=1}^{\infty} \dfrac{\sin(2 r \pi x)}{r} \mathrm{d}x $
$\displaystyle = -\pi \int_{0}^{\infty} \dfrac{1}{x+a} \left(\dfrac{1}{2} - \{x\} \right) \mathrm{d}x \quad \left( \because \sum_{r=1}^{\infty} \dfrac{\sin (2 r \pi x)}{r} = \dfrac{\pi}{2} - \pi \{x\} \right) $
$\displaystyle = -\pi \left[ \int_{0}^{\infty} \dfrac{\mathrm{d}x}{2(x+a)} - \int_{0}^{\infty} \dfrac{\{x\}}{x+a} \mathrm{d}x \right] $
$\displaystyle = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - \text{A}(n) \right]$
where $\displaystyle \text{A}(n) = \int_{0}^{n} \dfrac{\{x\} }{x+a} \mathrm{d}x$
Now,
$\displaystyle \text{A}(n) = \int_{0}^{n} \dfrac{\{x\} }{x+a} \mathrm{d}x $
$\displaystyle = \sum_{k=0}^{n} \int_{k}^{k+1} \dfrac{x-k}{x+a} \mathrm{d}x $
$\displaystyle = \sum_{k=0}^{n} \left[1 - (k+a)\log \left(\dfrac{k+a+1}{k+a}\right) \right] $
$\displaystyle = n - \sum_{k=0}^{n} \left[ (k+a+1) \log (k+a+1) - (k+a) \log (k+a) - \log (k+a+1) \right] $
$\displaystyle = n + a\log a - (a+n) \log (a+n) +\log(a \cdot (a+1) \cdot \ldots \cdot (a+n)) - \log a $
$\displaystyle \implies \text{I} (a) = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - n - a\log a + (a+n) \log (a+n) - \log(a \cdot (a+1) \cdot \ldots \cdot (a+n)) + \log a \right] $
Note that $\displaystyle \lim_{n \to \infty} \dfrac{ n! n^t}{t \cdot (t+1) \cdot \ldots \cdot (t+n)} = \Gamma(t) $
$\displaystyle \implies \text{I} (a) = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - n - a\log a + (a+n) \log (a+n) - a\log(n) - \log (n!) + \log(\Gamma(a)) + \log a \right] $
Simplifying using Stirling's Approximation and $\displaystyle \lim_{n \to \infty } n \log \left(1 + \dfrac{a}{n} \right) = a $, we have,
$\displaystyle \text{I} (a) = -\pi \left[\log(\Gamma(a+1)) - \dfrac{1}{2} \log(2a \pi) - a (\log(a) -1) \right] $
$\displaystyle = \pi \left[ \dfrac{1}{2} \log(2a \pi) + a (\log(a) -1) - \log(\Gamma(a+1)) \right] \quad \square $
Now,
Applying Integration By Parts on the proposition and simplifying, we get,
$ \displaystyle \int_{0}^{\infty} \tan^{-1} \left(\dfrac{t}{a}\right)\dfrac{\mathrm{d}t}{e^{2\pi t} - 1} = \dfrac{1}{2} \left[ \log(\Gamma(a)) - \dfrac{\log(2a)}{2} - \left( a - \dfrac{1}{2} \right)\log (a) +a\right] $
$$\therefore \log(\Gamma(a)) = 2\int_{0}^{\infty} \tan^{-1} \left(\dfrac{t}{a}\right)\dfrac{\mathrm{d}t}{e^{2\pi t} - 1} + \dfrac{\log(2a)}{2} + \left( a - \dfrac{1}{2} \right)\log (a) -a \quad \square $$
|
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|
How to show $\phantom{d}_d C_0+\phantom{d}_d C_4 + \cdots = 2^{d-2} + 2^{\frac{d}{2}-1} \cos(\frac{d \pi}{4}) $? I want to show following identities
\begin{align}
&\phantom{d}_d C_0+\phantom{d}_d C_4 + \cdots = 2^{d-2} + 2^{\frac{d}{2}-1} \cos(\frac{d \pi}{4}) \\
&\phantom{d}_d C_1+\phantom{d}_d C_5 + \cdots = 2^{d-2} + 2^{\frac{d}{2}-1} \sin(\frac{d \pi}{4}) \\
& \phantom{d}_d C_2+\phantom{d}_d C_6 + \cdots = 2^{d-2} - 2^{\frac{d}{2}-1} \cos(\frac{d \pi}{4}) \\
& \phantom{d}_d C_3+\phantom{d}_d C_7 + \cdots = 2^{d-2} - 2^{\frac{d}{2}-1} \sin(\frac{d \pi}{4})
\end{align}
Can you give me any hint or computation for this identities?
|
HINT:
As the subscript of each subsequent binomial coefficient differ $4,$
If $x^4=1\implies x=\pm1,\pm i$
Set $x=\pm1,\pm i$ one by one in the following identity $$(1+x)^n=\sum_{r=0}^n\binom nr x^r$$
Find $$\dfrac{(1+1)^n\pm(1-1)^n\pm(1+i)^n\pm(1-i)^n}4$$
$1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)$
$(1+i)^n=2^{n/2}\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)^n=2^{n/2}\left(\cos\dfrac{n\pi}4+i\sin\dfrac{n\pi}4\right)$
Similarly, $(1-i)^n=2^{n/2}\left(\cos\dfrac{n\pi}4-i\sin\dfrac{n\pi}4\right)$
Can you take it from here?
|
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|
If $A,B,A+I,A+B$ are idempotent matrices how to show that $AB=BA$?
If $A,B,A+I,A+B$ are idempotent matrices how to show that $AB=BA$ ?
MY ATTEMPT:
$A\cdot A=A$
$B\cdot B=B$
$(A+I)\cdot (A+I)=A+I$ or, $A\cdot A+A\cdot I+I\cdot A+I\cdot I=A+I$ which implies $A\cdot I+I\cdot A=0$ (using above equations)
$(A+B)\cdot (A+B)=A+B$ or,$A\cdot A+A\cdot B+B\cdot A+B\cdot B=A+B$ which implies $A\cdot B+B\cdot A=0$
What's next?
|
Let $a$ and $b$ be elements of a (possibly noncommutative) ring. If $a$, $b$, $a+1$ and $a+b$ are idempotent, then $ab = ba$.
Proof. From $a^2 = a$ and $(a+1)^2 = a+ 1$ one gets $a^2 + a + a + 1 = a + 1$, whence $a^2 + a = 0$. Since $a^2 = a$, it follows that $a + a = 0$ and finally $a = -a$. Now, from $(a+b)^2 = a + b$ one gets $a^2 + ab + ba + b^2 = a + b$. Again, since $a^2 = a$ and $b^2 = b$, it follows $ab + ba = 0$, and since $a = -a$, we finally get $ab -ab= 0$, that is $ab = ba$.
|
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|
Solve $\log_{1/3}(x^2-3x+3)≥0$ I want to solve $$\log_{1/3}(x^2-3x+3)≥0$$
Now I know the result is: $x ∈ <1;2>$, but i am not sure how to get it.
My thoughts: $\frac{1}{3}$ to the power of positive number $= (x^2-3x+3)$, now I would solve $x^2-3x+3$ with the help of discriminant to get the points where $x$ is zero, and say the answer is for the positive intervals. Is this correct / is there easier way to do it?
|
$$\log_{1/3}(x^2-3x+3) \ge 0$$
This occurs when:
$$x^2-3x+3 \le 1$$
$$x^2-3x+2 \le 0$$
Factorizing, we get:
$$(x-1)(x-2) \le 0$$
Solving the inequality, we get:
$$1 \le x \le 2$$
ALTERNATE SOLUTION
Consider the following equation:
$$\log_{1/3}(x^2-3x+3) = 0$$
If we solve, we get:
$$x^2 -3x + 2 = 0$$
$$(x-1)(x-2) = 0$$
Now, if we have $x < 1$, $x^2 - 3x + 3 \ge 1$
This means that a possible value is $\log_{1/3}5$, which would a value less than $0$.
Similarly if $x > 2$, $x^2 - 3x + 3 \ge 1$.
When $1 < x < 2$, we have that $0 < x^2 - 3x + 3 < 1$, meaning only fractional answers exist.
This means that a possible value is $\log_{1/3}\frac{1}{3}$, which would a value greater than $0$.
|
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|
Limit: $\lim_{n \to \infty} \frac{3^n+5^n+10^n}{-2^{n+1}+5^{n+1}+10^{n+1}}$ Limit: $$\lim_{n \to \infty} \frac{3^n+5^n+10^n}{-2^{n+1}+5^{n+1}+10^{n+1}}$$
I know the limit is $\frac{1}{10}$, but I am not sure how to get to it. Any help would be greatly appreciated.
|
$$\lim _{ n\to \infty } \frac { 3^{ n }+5^{ n }+10^{ n } }{ -2^{ n+1 }+5^{ n+1 }+10^{ n+1 } } =\lim _{ n\to \infty } \frac { { 10 }^{ n }\left( \frac { { 3 }^{ n } }{ { 10 }^{ n } } +\frac { { 5 }^{ n } }{ { 10 }^{ n } } +1 \right) }{ { 10 }^{ n+1 }\left( -\frac { { 2 }^{ n+1 } }{ { 10 }^{ n+1 } } +\frac { { 5 }^{ n+1 } }{ { 10 }^{ n+1 } } +1 \right) } =\\ ==\lim _{ n\to \infty } \frac { { \left( \frac { 2 }{ 10 } \right) }^{ n }+{ \left( \frac { 5 }{ 10 } \right) }^{ n }+1 }{ { 10 }\left( { -\left( \frac { 2 }{ 10 } \right) }^{ n+1 }+{ \left( \frac { 5 }{ 10 } \right) }^{ n+1 }+1 \right) } =\frac { 1 }{ 10 } $$
|
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|
Find invertible matrices without Jordan form Here is the statement of the problem: Let $ A $ be the following matrix with entries from $ \mathbb{R} $ $$ \begin{pmatrix} 1 & 2 & -4 & 4 \\ 2 & -1 & 4 & -8 \\ 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 3\end{pmatrix} $$
Let $ A^t $ be its transpose. Find an invertible matrix $ P \in M_4\left(\mathbb{R}\right) $ such that $ PAP^{-1}=A^t $
By using Jordan form, there are invertible matrices $ B,C $ so that $ BAB^{-1}=J=CA^tC^{-1} $. In other words, $ \left(C^{-1}B\right)A\left(C^{-1}B\right)^{-1}=A^t $, which is what we want. So it suffices to find such matrices. However, it seems one should calculate a lot to get the desired matrices. My question is ...
Is it possible to answer this problem without using Jordan form?
Let $ B = \begin{pmatrix} 0 & 2 \\ 2 & -2 \end{pmatrix} $, then we have $ A-I_4=\begin{pmatrix} B & -B^2 \\ I_2 & -B \end{pmatrix} $. Perhaps it's a useful identity?
|
Normally, for "small" computational problems, it is usually best to work them in a straightforward manner, even if it is a bit tedious, because that is usually faster than finding a shortcut. In this problem however, you seem to have found the key to the shortcut.
By using the useful identity in your question, we have:
\begin{align*}A - I_4 &= \begin{bmatrix}B&-B^2\\I_2&-B\end{bmatrix} \\ &= \begin{bmatrix}B&B\\I_2&I_2\end{bmatrix} \begin{bmatrix}I_2& \\ &-B\end{bmatrix} \\ &= \begin{bmatrix}B&\\ &I_2\end{bmatrix} \begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix} \begin{bmatrix}I_2& \\ &-B\end{bmatrix} \ \ \color{red}{(1)}\end{align*}
Since $B^T = B$, by transposing both sides of equation $\color{red}{(1)}$, we get:
\begin{align*}A^T-I_4 &= \begin{bmatrix}I_2& \\ &-B\end{bmatrix}^T \begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix}^T \begin{bmatrix}B&\\ &I_2\end{bmatrix}^T \\ &= \begin{bmatrix}I_2& \\ &-B\end{bmatrix} \begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix} \begin{bmatrix}B&\\ &I_2\end{bmatrix} \ \ \color{red}{(2)}\end{align*}
Since $B$ is invertible, we can rearrange equation $\color{red}{(1)}$ as follows:
\begin{align*}\begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix} &= \begin{bmatrix}B&\\ &I_2\end{bmatrix}^{-1}(A-I_4)\begin{bmatrix}I_2&\\ &-B\end{bmatrix}^{-1} \\ &= \begin{bmatrix}B^{-1}&\\ &I_2\end{bmatrix}(A-I_4)\begin{bmatrix}I_2&\\ &-B^{-1}\end{bmatrix} \ \ \color{red}{(3)}\end{align*}
Then, substituting equation $\color{red}{(3)}$ into equation $\color{red}{(2)}$ yields:
\begin{align*}A^T-I_4 &= \begin{bmatrix}I_2& \\ &-B\end{bmatrix} \begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix} \begin{bmatrix}B&\\ &I_2\end{bmatrix} \\ &= \begin{bmatrix}I_2& \\ &-B\end{bmatrix}\begin{bmatrix}B^{-1}&\\ &I_2\end{bmatrix}(A-I_4)\begin{bmatrix}I_2&\\ &-B^{-1}\end{bmatrix}\begin{bmatrix}B&\\ &I_2\end{bmatrix} \\ &= \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}(A-I_4)\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix} \\ &= \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}A\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix} - \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}I_4\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix} \\ &= \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}A\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix} - I_4 \ \ \color{red}{(4)}\end{align*}
Adding $I_4$ to both sides of equation $\color{red}{(4)}$ yields $A^T = \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}A\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix}$.
Therefore, we can set $P = \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}$ to get $A^T = PAP^{-1}$.
|
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|
Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.
I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps.
On this topic we learned also about Cauchy inequality, but I have no experience with it.
The answer according to Mathematica is when $A=B=C=60$.
Any ideas?
|
If you don't like to my first solution, here is a more simple solution.Assume without loss of generality that A,B,C belongs to the first quadrant.And so, it's easy to see that:
\begin{align*}
x+y\geq 2\sqrt{xy} \tag{1}
\end{align*}
\begin{align*}
x+z\geq 2\sqrt{xz} \tag{2}
\end{align*}
\begin{align*}
y+z\geq 2\sqrt{yz} \tag{3}
\end{align*}
Multiplying (1),(2),(3), we get:
$$ (x+y)(x+z)(y+z) \geq 8xyz$$
$$ \frac{1}{8} \geq \frac{xyz}{(x+y)(x+z)(y+z)}$$
$$ \displaystyle \frac{1}{8} \geq \sqrt{\frac{xy}{(x+z)(y+z)}}\sqrt{\frac{xz}{(x+y)(y+z)}}\sqrt{\frac{yz}{(x+y)(x+z)}} $$
By Ravi transformation and Law of cosines, we get
$$ \frac{1}{8} \geq sen\frac{\alpha}{2} sen\frac{\beta}{2} sen\frac{\gamma}{2}$$
And this is equivalent to:
$$ \frac{1}{8} \geq cos \left(\frac{\pi}{2}-\frac{\alpha}{2}\right) cos \left(\frac{\pi}{2}-\frac{\beta}{2}\right) cos \left(\frac{\pi}{2}-\frac{\gamma}{2}\right)$$
Take the substitution $\displaystyle \frac{\pi}{2}-\frac{\alpha}{2}=A$,$\displaystyle \frac{\pi}{2}-\frac{\beta}{2}=B$,$\displaystyle \frac{\pi}{2}-\frac{\gamma}{2}=C$, we get $\displaystyle A+B+C=\pi$ and:
$$ \frac{1}{8} \geq cosAcosBcosC$$
|
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|
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
$729=9^3$
For any number to be divisible by $9$, the sum of the digits have to be divisible by $9$. The given number is divisible by $9$.
Then I tried dividing the given number by $9$. The quotient was like $123456789012\cdots$
So the quotients sum is $(45\times 8)+1=361$
Is this the way to proceed? Is there a shorter way?
|
According to Wolfy,
or by repeated use of
$x^3-1
=(x-1)(x^2+x+1)
$,
$\frac{x^{81}-1}{x-1}
= (x^2+x+1) (x^6+x^3+1) (x^{18}+x^9+1) (x^{54}+x^{27}+1)
$.
Since
$x^{3n} \equiv 1 \bmod (x^3-1)$,
$x^{6n}+x^{3n}+1
\equiv 3 \bmod (x^3-1)
$.
Therefore
the right 3 factors
are all
$\equiv 3 \bmod (x^3-1)
$.
Therefore the whole product
$\equiv 27(x^2+x+1)
\bmod (x^3-1)
$.
Since
$\begin{array}\\
x^2+x+1
&=((x-1)+1)^2+(x-1)+2\\
&=(x-1)^2+2(x-1)+1+(x-1)+2\\
&=(x-1)^2+3(x-1)+3\\
\end{array}
$
we have
$\begin{array}\\
27(x^2+x+1)
&=27((x-1)^2+3(x-1)+3)\\
&=27(x-1)^2+81(x-1)+81\\
&\equiv 81 \bmod 729
\qquad\text{setting }x=10
\text{ since }27\cdot 9^2, 81\cdot 9
\equiv 0 \bmod 729\\
\end{array}
$
Therefore
the answer is
$81$.
|
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|
Find all natural numbers $x,y$ such that $3^x=2y^2+1$. Find all natural numbers $x,y$ such that
$$3^x=2y^2+1$$
solutions are $(1,1)$, $(2,2)$, $(5,11)$. I found that parity of both is same and If $x$ Is odd it is of the form $4k+1$.
|
Here's a somewhat simpler proof using Pell equations
(as Eric Towers suggested might be possible).
If $x$ is odd, say $x=2k+1$, then $3z^2 = 2y^2 + 1$ where $z = 3^k$.
Hence $(2y)^2 - 6z^2 = -2$ and we have
$$
2y + 3^k \sqrt{6} = (2+\sqrt{6}) (5 + 2\sqrt{6})^m
$$
for some $m=0,1,2,\ldots$. The first two solutions $m=0$, $m=1$ give
$z=1$, $z=9$ which recovers the known solutions with $x=1$ and $x=5$.
Suppose $k>2$. Then $z = 3^k \equiv 0 \bmod 27$, and by computing powers of
$5 + 2 \sqrt{6}$ mod $27$, we find that $27 \mid z$ if and only if
$m \equiv 4 \bmod 9$. But then $z$ is always a multiple of the $m=4$ solution
(in general if $2m+1 \mid 2m'+1$ then the $m$-th $z$ divides the $m'$-th one),
and the $m=4$ solution has $z = 8721 = 3^3 17 \, 19$. So $z$ can never be
a power of $3$ once $k>2$.
If $x$ is even, say $x=2k$, then $2y^2 = 3^{2k} - 1 = (3^k-1) (3^k+1)$.
Hence $3^k-1$ is either a square or twice a square, and the former is
impossible (no square is congruent to $-1 \bmod 3$). So $3^k-1 = 2{y'}^2$
for some integer $y'$, giving a smaller solution $(k,y')$ to $3^x = 2y^2 + 1$.
Continuing in this fashion eventually yields a solution with $x$ odd.
But that solution cannot be $(x,y) = (5,11)$, because
$(3^{10} - 1) / 2 = 2^2 11^2 61$ is not a square. This leaves
the $(x,y) = (1,1)$ solution, and indeed $(x,y) = (2,2)$ also works
as we know $-$ but $x=4$ does not because $(3^4 - 1)/2 = 40$
is not a square either. This completes the proof that the three solutions
with $x=1,2,5$ are the only ones.
|
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|
Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}{x+4} & \quad -4 < x < 0 \\
\frac{x}{x+4} & \quad x < −4
\end{array}
\right.
$$
Solving,
$$4>\frac{x}{x+4}$$
$$4x+16>x$$
$$3x>-16$$
$$x>-\frac{16}{3}=-5.3$$
and
$$4>-\frac{x}{x+4}$$
$$4x+16>-x$$
$$5x>-16$$
$$x>-3.2$$
The answer is $x<-5.3$ and $x>-3.2$. What am I doing wrong?
|
Your problem is very simple. :)
Why admin delete my ANS?????????????????????
Here's my Hint:
$$\left | \dfrac{x}{x+4} \right |<4 \overset{x \ne -4}{\rightarrow} -4<\dfrac{x}{x+4}<4 \Leftrightarrow \left\{\begin{matrix}
&-4<\dfrac{x}{x+4} \\
& \dfrac{x}{x+4}<4
\end{matrix}\right.
\Leftrightarrow
\left\{\begin{matrix}
&\dfrac{5x+16}{x+4}>0 \\
& \dfrac{3x+16}{x+4}>0
\end{matrix}\right.$$
$$\Leftrightarrow
\left\{\begin{matrix}
&x \in\left ( -\infty;-4 \right )\cup \left ( -\dfrac{16}{5};+\infty \right ) \\
& x \in\left ( -\infty;-\dfrac{16}{3} \right )\cup \left ( -4;+\infty \right )
\end{matrix}\right.
\Rightarrow \boxed{x \in \left ( -\infty;-\dfrac{16}{3} \right )\cup \left ( -\dfrac{16}{5};+\infty \right).}$$
|
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|
Is $7^{101}+8^{101}$ divisible by 25? If not, what is $ 7^{101} + 8^{101} \bmod 25$ What I derived is: $$\begin{align}7^{101}+8^{101} &\equiv (5+2)^{101}+ (5+3)^{101} \\
&\equiv 2^{101}+101\cdot5\cdot2^{100}+3^{101}+101\cdot 5\cdot 3^{100} \\
&\equiv 2^{101}+(100+1)\cdot5\cdot 2^{100}+3^{101}+(100+1)\cdot5\cdot3\cdot2^{100}\\
& \equiv 2^{101}+5\cdot 2^{100}+3^{101}+5\cdot3^{100} \pmod {25}.
\end{align}$$
Then I don't know what's the next step.
|
By Euler's theorem, for any $n$ coprime with $25$, $n^{\varphi(25)}=n^{20}\equiv1\mod25$, hence $n^k\equiv n^{k\bmod20}\mod25$, so that
$$7^{101}+8^{101}\equiv 7+8=15\mod 25.$$
|
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|
Solve the equation $x^{4}-2x^{3}+4x^{2}+6x-21=0$ Solve the equation $$x^{4}-2x^{3}+4x^{2}+6x-21=0$$ given that two of its roots are equal in magnitude but opposite in sign.
I don't know how to solve it. The roots are given as $\pm\sqrt{3},1\pm i\sqrt{6}. $
|
You are given the information that some polynomial $x^2-c$ divides $x^4-2x^3+4x^2+6x-21$. Now polynomial division gives:
$x^4-2x^3+4x^2+6x-21 = (x^2-c) \cdot (x^2-2x+c+4) + 2 (3-c) x+ (c^2+4c-21).$
But the remainder is supposed to be zero, so we get:
$2 (3-c) x+ (c^2+4c-21) = 0$
and in particular $c=3$. Thus,
$x^4-2x^3+4x^2+6x-21 = (x^2-3) \cdot (x^2-2x+7)$
and the roots are easy to find.
|
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|
how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then
$f(x)=f\left( \frac{x+1}{2} \right)+f\left( \frac{x-1}{2} \right)$.
But how to find all functions?
|
We can solve this one in the same manner as here.
So, it is easy to see that $f(0)=0$ and $f$ is odd.
Let $f(1)=0$. If we put $y=1$ we get $$f(x^2-1) = (x-1)f(x)$$ and if we put $-x$ instead of $x$ we get $$f(x^2-1) = -(x+1)f(-x) = (x+1)f(x)$$ So we have $$(x-1)f(x)= (x+1)f(x)\implies f(x)=0\;\;\;\forall x$$ and we are done.
|
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"timestamp": "2023-03-29T00:00:00",
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How to simplify this? What did I do wrong? $$(3-\sqrt3) (2-\sqrt3)-\sqrt3\cdot\sqrt{27}.$$
So I simplified the $\sqrt3\cdot\sqrt{27}$ part into
$$\sqrt 3\cdot\sqrt{9\cdot3} = \sqrt3\cdot3\cdot\sqrt3=4\cdot\sqrt3$$
Then I multiplied the brackets :
$$6+\sqrt3-2\cdot\sqrt3-3\cdot\sqrt3 =6-4\cdot\sqrt3.$$
Then minus the $4\cdot\sqrt3$ we got
$$6-8\cdot\sqrt3$$
However the answer should be $-5\cdot\sqrt3$
How can I get that answer, what did I do wrong?
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Your mistake was in the equality $$\sqrt 3 \cdot 3 \cdot \sqrt 3 = 4\cdot \sqrt3$$ which is false.
$$\sqrt{3}\sqrt{9\cdot 3} = \sqrt{3}\cdot 3\cdot \sqrt{3} = \sqrt{3}\cdot \sqrt{3} \cdot 3 = 3\cdot 3 = 9 \neq 4\cdot \sqrt 3$$
Also, you made a mistake in multiplying the brackets:
$$(3-\sqrt 3)(2-\sqrt 3) = 6-3\sqrt 3 -2\sqrt 3 + 3$$
which is not what you got.
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How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
|
There are only three possibilities for the divisibility of an integer by $3$, which are: no remainder, a remainder of $1$, or a remainder of $2$. But if we multiply $2$ by itself over and over again, the no remainder option is impossible, as that would mean that $2$ is a multiple of $3$, which it is not.
The thing is also that we can multiply reminders. So $2$ leaves a remainder of $2$, and $2 \times 2 = 4$, which leaves a remainder of $1$. And $1 \times 2 = 2$, which leaves a remainder of $2$ again.
Therefore the powers of $2$ alternate remainders on division by $3$ according to the parity of the exponent: $2^n \equiv 1 \pmod 3$ if $n$ is even and $2^n \equiv 2$ or $-1 \pmod 3$ if $n$ is odd.
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.
The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequality. How do I show the other inequality?
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We have to use the word "triangle" at some point.
$$ a^2+b^2+c^2 \le a(b+c)+b(a+c)+c(a+b)=2(ab+ac+bc)=6$$
hence
$$(a+b+c)^2\le 12$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Error when approximating $\int_0^{1}(x^{2}+x)dx$ with midpoint rule. Task is to define the exact error when approximating $\int_0^{1}(x^{2}+x)dx$ with midpoint rule using n subintervals. I know the error term is $E(f)=\frac{1}{24}(b-a)f^{''}(\varepsilon)h^{2}$ but im not sure if this is the exact form and if it is how do you get it?
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I believe i know the answer now if someone needs it. $M_n=h(f(\frac{1}{2n})+f(\frac{3}{2n})+f(\frac{5}{2n})+,..,+f(\frac{1-\frac{1}{n}+1}{2}))$
$M_n=\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{2k+1}{2n})=\frac{1}{n}\sum_{k=0}^{n-1}(\frac{2k+1}{2n})+\frac{1}{n}\sum_{k=0}^{n-1}(\frac{2k+1}{2n})^{2}$
$M_n=\frac{1}{2n^{2}}\sum_{k=0}^{n-1}(2k+1)+\frac{1}{4n^{3}}\sum_{k=0}^{n-1}(2k+1)^{2}$
$=\frac{1}{2n^{2}}(2\sum_{k=0}^{n-1}k+\sum_{k=0}^{n-1}1)+\frac{1}{4n^{3}}(4\sum_{k=0}^{n-1}k^{2}+4\sum_{k=0}^{n-1}k+\sum_{k=0}^{n-1}1)$
Using sum formulas: $\sum_{i=m}^{n}1=n+1-m$, $\sum_{i=0}^{n}i=\frac{n(n+1)}{2}$, $\sum_{i=0}^{n}i^{2}=\frac{n(n+1)(2n+1)}{6}$
$\longrightarrow M_4=\frac{1}{2n^{2}}(2\frac{(n-1)n}{2}+n)+\frac{1}{4n^{3}}(4\frac{(n-1)n(2n-1)}{6}+2(n-1)n+n)$
$=\frac{n^{2}}{2n^{2}}+\frac{1}{4n^{3}}(\frac{4(2n^{3}-3n^{2}+n)}{6}+\frac{12n^{2}-6n}{6})$
$=\frac{1}{2}+\frac{8n^{3}-12n^{2}+4n+12n^{2}-6n}{24n^{3}}=\frac{1}{2}+\frac{8n^{3}-2n}{24n^{3}}=\frac{5}{6}-\frac{1}{12n^{2}}$ So the correct error is $\frac{1}{12n^{2}}$
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{
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve this limit involving cube root and infinity? $\lim_{x\to\infty} \left((x^3+x^2)^{1/3} - (x^3-x^2)^{1/3})\right)$ How can I solve this limit? I know the answer is $2/3$.
I tried factorisation, but solving the complicated denominator using L'Hopital's Rule returns a wrong answer, $0$.
$$ \lim_{x\to\infty} \left((x^3+x^2)^{1/3} - (x^3-x^2)^{1/3})\right) $$
Thanks.
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You can use the identity $$a^3−b^3=(a−b)(a^2+ab+b^2)$$ with $a=(x^3+x^2)^{1/3}$ and $b=(x^3-x^2)^{1/3}$. So, you want to multiply (numerator and denominator) with $a^2+ab+b^2$. This will give you \begin{align}a-b&=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}=\frac{a^3-b^3}{a^2+ab+b^2}\\[0.2cm]&=\frac{x^3+x^2-(x^3-x^2)}{(x^3+x^2)^{2/3}+(x^3+x^2)^{1/3}(x^3-x^2)^{1/3}+(x^3-x^2)^{2/3}}\\[0.2cm]&=\frac{2x^2}{(x^{3})^{2/3}(1+\frac1x)+(x^3)^{1/3}(1+\frac1x)(x^3)^{1/3}(1-\frac1x)+(x^3)^{2/3}(1-\frac1x)}\\[0.2cm]&=\frac{2x^2}{x^2(1+\frac1x)+x(1+\frac1x)x(1-\frac1x)+x^2(1-\frac1x)}\\[0.2cm]&=\frac{2}{1+\frac1x+(1+\frac1x)(1-\frac1x)+1-\frac1x}\\[0.2cm]&=\frac{2}{3-\frac1{x^2}}\end{align} Now let $x\to +\infty$ to conclude that $$\lim_{x\to +\infty}\frac{2}{3-\frac1{x^2}}=\frac2{3-0}=\frac23$$
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Subsets and Splits
Fractions in Questions and Answers
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