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Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is exactly divisible by $x^2+1$ I have been trying this question for a long time but I am not getting it. So please help me and try to make it as fast as possible
|
Long division gives
$$
\frac{x^4+x^3+8x^2+ax+b}{x^2+1}=x^2+x+7+\frac{(a-1)x+b-7}{x^2+1}
$$
Therefore, the remainder is zero iff $(a-1)x+b-7$ is the zero polynomial, that is, iff
$$
.\begin{cases}
a-1=0\\
b-7=0\\
\end{cases}
$$
Hence, for $a=1$ and $b=7$, we have that $x^4+x^3+8x^2+ax+b$ is divisible by $x^2+1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determinant of a matrix and chech whether it is non negative definite or not
Let $V = \{ f : [0,1] \to \mathbb R | f$ is a polynomial of degree less than or equal to n $\}$. Let $f_j(x) = x^j$ for $0\leq j \leq n$ and let $A$ be the $(n+1) \times (n+1)$ given by $a_{ij} = \int_0^1 f_i(x)f_j(x) dx$. Then which of the followings are true.
*
*$\dim V = n$
*$\dim V >n$
*$A$ is non negative definite.
*$\det A >0$.
Clearly the option 1. is false and 2. is correct
I knnow the matrix $A$ is
$$
\begin{bmatrix}
a_{00} & a_{01} & a_{02} & \cdots & a_{0n}\\
a_{10} & a_{11} & a_{12} & \cdots & a_{1n}\\
a_{20} & a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n0} & a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{bmatrix}
$$
$$A =
\begin{bmatrix}
1 & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n+1}\\
\frac{1}{2} & \frac{1}{3} &\frac{1}{4} & \cdots & \frac{1}{n+2}\\
\frac{1}{3}& \frac{1}{4} & \frac{1}{5} & \cdots & \frac{1}{n+3}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
\frac{1}{n+1} & \frac{1}{n+2} & \frac{1}{n+3} & \cdots & \frac{1}{2n+1}\\
\end{bmatrix}
$$
I am unable to find the determinant of the matrix $A$, so please help to find the determinant of the matrix and the matrix $A$ is non negative definnite or not. Thank you
|
Hint: for any real vector $v$, express $v^T A v$ as an integral.
|
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|
How to easily solve this trigonometric equation? Given equation:
$$\frac{\sin(x) + \sin(5x) - \sin(3x)}{\cos(x) + \cos(5x) - \cos(3x)} = \tan(3x),$$
what is the easiest way to solve it? I know it can be solved by expanding each $\sin(nx)$ and $\cos(nx)$ terms, but is there an easier way?
|
$$\frac { 2\sin { \left( \frac { x+5x }{ 2 } \right) \cos { \left( \frac { x-5x }{ 2 } \right) -\sin { \left( 3x \right) } } } }{ 2\cos { \left( \frac { x+5x }{ 2 } \right) \cos { \left( \frac { x-5x }{ 2 } \right) } -\cos { \left( 3x \right) } } } =\frac { \sin { \left( 3x \right) \left( 2\cos { 2x-1 } \right) } }{ \cos { \left( 3x \right) \left( 2\cos { 2x-1 } \right) } } =\tan (3x)$$
|
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|
New series formula for $\arctan(x)$? Ln(x)? I discovered this equation, but have no idea if it has been previously discovered. Please help determine if it has been previously developed. Or please prove that the equation is not correct.
$$\sum_{n=0}^\infty \frac{x^{2n+1}}{(x^2+1)^{n+1}}\cdot\frac{(2n)!!}{(2n+1)!!}=\arctan(x),$$
for $|x|\leq \pi$, or possibly all $x$.
Likewise, using the same method
for $x> .001$, or possibly x > 0.
$$\sum_{n=1}^\infty \frac{x^{n}-1}{(1+x)^{n}}\cdot\frac{(1)}{(n)}=Ln(x),$$
all follows from dx/dx =1.
|
Evidently from marty cohen's answer, the formula for $\arctan$ was previously known, and as shown below, the result for $\log(x)$ is not too hard to derive. That you have discovered them yourself is no less impressive however.
Substituting $x\mapsto4x^2$ in $(2)$ from this answer, we get
$$
\sum_{k=0}^\infty\frac{4^kx^{2k}}{\binom{2k}{k}}=\frac1{1-x^2}\left[1+\frac{x}{\sqrt{1-x^2}}\sin^{-1}(x)\right]\tag{1}
$$
Integrating $(1)$ gives
$$
\begin{align}
\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}}
&=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\\
&=\frac{\tan^{-1}\left(\frac{\large x}{\sqrt{1-x^2}}\right)}{\sqrt{1-x^2}}\tag{2}
\end{align}
$$
Substituting $x\mapsto\frac{x}{\sqrt{1+x^2}}$ in $(2)$ yields
$$
\frac1{\sqrt{1+x^2}}\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}\left(1+x^2\right)^k}
=\sqrt{1+x^2}\tan^{-1}(x)\tag{3}
$$
and therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}\left(1+x^2\right)^{k+1}}
=\tan^{-1}(x)}\tag{4}
$$
$$
\begin{align}
\sum_{n=1}^\infty\frac{x^n-1}{n(1+x)^n}
&=\log\left(1-\frac1{1+x}\right)-\log\left(1-\frac{x}{1+x}\right)\\
&=\log(x)\tag{5}
\end{align}
$$
|
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|
What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
|
$$\dfrac{3\sqrt3-8}{1-2\sqrt3}=\dfrac{8-3\sqrt3}{2\sqrt3-1}$$
Now $$\dfrac{3\sqrt3-4}{7-2\sqrt3}-\dfrac{8-3\sqrt3}{2\sqrt3-1}=\dfrac{(3\sqrt3-4)(2\sqrt3-1)-(7-2\sqrt3)(8-3\sqrt3)}{(7-2\sqrt3)(2\sqrt3-1)}$$
$$=\dfrac{26\sqrt3-52}{(7-2\sqrt3)(2\sqrt3-1)}$$
Now the denominator $>0,$ and $26\sqrt3-52=26(\sqrt3-2)<0$
|
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|
Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \cdot \frac{1}{2\sqrt{x+2}} = \frac{1}{(2x+6)\sqrt{x+2}}$$
$$g'(x) = 1$$
$$g(x) = x$$
So:
$$\bigg[\arctan(\sqrt{x+2}) \cdot x\bigg]_{-1}^{1} - \int_{-1}^{1} \frac{x}{(2x-6)\sqrt{x+2}}\ dx$$
How should I proceed with the new integral?
|
$$\int_{-1}^{1}\tan^{-1}(\sqrt{x+2})dx=\int_{1}^{3}\tan^{-1}\sqrt{x}\,dx$$
if$f$ be differentiable, increasing and one-to-one on $[a,b]$ such that $f(a)=\alpha$ and $f(b)=\beta$ then
$$\int_{a}^{b}f(x)dx+\int_{\alpha}^{\beta}f^{-1}(x)dx=b\beta-a\alpha$$
therfore
$$\int_{1}^{3}\tan^{-1}\sqrt{x}\,dx+\int_{\large\frac{\pi}{4}}^{\large\frac{\pi}{3}}\tan^{2}x\,dx=\pi-\frac{\pi}4$$
we have
$$\int_{1}^{3}\tan^{-1}\sqrt{x}\,dx=\frac{5\pi}6-\sqrt{3}+1$$
|
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|
Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:
*
*for all $x$ and $y$
*for no $x$ and $y$
*only when $x \neq y \neq 0$
*only when $x = y \neq 0$
|
Suppose that $x$ and $y$ are not $0$.
We know that $$x^2 + y^2 \geq 2xy. \quad (1)$$
Two cases:
*
*$x>0$ and $y<0$ (or $x<0$ and $y>0$), then $\frac{x^2+y^2}{2xy} < 0$, and you cannot have $\sin^2(\theta) < 0$
*$x>0$ and $y>0$ (or $x<0$ and $y<0$). Then $(1)$ becomes
$$\frac{x^2+y^2}{2xy} \geq 1.$$
The only solution is to have $\frac{x^2+y^2}{2xy} = 1$.
|
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|
Solving $2x^4+x^3-11x^2+x+2 = 0$ I am having no idea how I can solve this problem.
I need help!
Here's the problem
$2x^4+x^3-11x^2+x+2 = 0$
I am learning Quadratic Expressions and this is what I need to solve, and I can't understand how :C
|
$2x^4+x^3-11x^2+x+2=0$
Note that the coefficients: $2,1,-11,1,2$ are symmetrical.
$2(x^4+1)+(x^3+x)-11x^2=0$
$2(x^4+4x^3+6x^2+4x+1)-7(x^3+x)-23x^2=0$
$2(x^4+4x^3+6x^2+4x+1)-7(x^3+2x^2+x)-9x^2=0$
$2(x+1)^4-7(x+1)^2x-9x^2=0$
$2\left(\frac{(x+1)^2}{x}\right)^2-7\left(\frac{(x+1)^2}{x}\right)-9=0$
|
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|
Inequality involving ArcTan How to prove that for $x\in[0, +\infty]$ the following inequality is true: $$\arctan x\geq\frac{3 x}{1+2\sqrt{1+x^2}}?$$
I don't have idea from where to start, so any hint is welcome. Thanks in advance.
|
Let's say $x=\tan \theta$, so that:
$$\theta \geq \frac{3\tan \theta}{1+2\sqrt{1+\tan^2 \theta}}$$
$\sqrt{1+\tan^2\theta}=\sec\theta$ because the range of $\arctan x$ always has $\cos \theta > 0$:
$$\theta \geq \frac{3\tan \theta}{1+2\sec\theta}$$
Multiply both the numerator and denominator by $\cos \theta$:
$$\theta \geq \frac{3\sin\theta}{\cos\theta+2}$$
Now, these two functions intersect at $\theta=0$. Also, $\theta$ always has a derivative of $1$, so if we can prove that the derivative of $\frac{3\sin\theta}{\cos\theta+2}$ is always $\leq 1$, we can prove this inequality for $\theta \geq 0$.
$$\frac{d}{d\theta}\frac{3\sin\theta}{\cos\theta+2}=\frac{(\cos\theta+2)\frac{d}{d\theta}(3\sin\theta)-3\sin\theta\frac{d}{d\theta}(\cos\theta+2)}{(\cos\theta+2)^2}$$
$$\frac{d}{d\theta}\frac{3\sin\theta}{\cos\theta+2}=\frac{3\cos^2 \theta+6\cos\theta+3\sin^2\theta}{(\cos\theta+2)^2}=\frac{3+6\cos\theta}{(\cos\theta+2)^2}$$
Now, we need to prove that this is less than or equal to $1$, which is the same as proving:
$$3+6\cos\theta \leq (\cos\theta+2)^2$$
Expand the right side:
$$3+6\cos\theta \leq \cos^2\theta+4\cos\theta+4$$
Subtract both sides by the left side:
$$0 \leq \cos^2\theta-2\cos\theta+1$$
Factor the right side:
$$0 \leq (\cos\theta-1)^2$$
Now, this inequality is obviously true because $0 \leq u^2$ for all $u \in \Bbb{R}$, so we have proven that the derivative of $\frac{3\sin\theta}{\cos\theta+2}$ is always $\leq 1$, so since $\theta$ and that funciton intersect at $\theta=0$, we can conclude that:
$$\theta \geq 0 \implies \theta \geq \frac{3\sin\theta}{\cos\theta+2}$$
$$\theta \geq 0 \implies \theta \geq \frac{3\tan \theta}{1+2\sqrt{1+\tan^2 \theta}}$$
$$x \geq 0 \implies \arctan x \geq \frac{3x}{1+2\sqrt{1+x^2}}$$
|
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|
Converting $\cos\phi$ into $\frac{1−t^2}{1+t^2}$, given that $t = \tan\frac{\phi}{2}$
I have to figure out the working to convert $\cos\phi$ into $\dfrac{1−t^2}{1+t^2}$, given that $t = \tan\dfrac{\phi}{2}$.
It would be amazing if someone could help I've been trying to do it for hours.
|
If $t = \tan \frac \phi 2$, then by drawing a right triangle, one can see that $$\sin \frac \phi 2 = \frac{t}{\sqrt{t^2+1}} \quad \text{ and } \quad \cos \frac \phi 2 = \frac{1}{\sqrt{t^2+1}}$$
So, by using $\cos (2x) = 1 - 2 \sin^2x$ and letting $x = \frac \phi 2$, we get
$$\cos \phi = 1 - 2 \cdot \left(\frac{t}{\sqrt{t^2+1}}\right)^2 = 1-\frac{2t^2}{t^2+1} = \frac{1-t^2}{1+t^2}$$
|
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|
How many subsets contain no consecutive elements? How many subsets of $\{1,2,...,n\}$ have no two consecutive numbers ?
Here is the solution :
The subsets are interpreted as $n$-words from the alphabet $\{0,1\}$. Let $a_n$ be the number of words with no consecutive ones. Then, a word can start from $0$ and proceed in $a_{n-1}$ ways or start with $10$ and proceed in $a_{n-2}$ ways. Therefore, $a_{n} = a_{n-1} + a_{n-2}$. $a_1 = 2, a_2 = 3$. So, $a_n = F_{n+2}$.
I have no trouble understanding the part of the argument linking $a_n$ with the Fibonacci numbers. But, I have trouble understanding the bijective argument.
What is a $n$ word ? And, how is the bijective constructed here ? How is $a_n$ linked to the question ?
|
Choose the first value in the set:
$$\sum_{q=1}^n z^q = z \sum_{q=0}^{n-1} z^q = z\frac{1-z^n}{1-z}.$$
Choose some number of gaps that are at least two:
$$\sum_{p=0}^{n-1} \left(\frac{z^2}{1-z}\right)^p
= \frac{1-z^{2n}/(1-z)^n}{1-z^2/(1-z)}.$$
Sum the contributions that end in at most $n$ and extract the
coefficient:
$$[z^n] \frac{1}{1-z} z\frac{1-z^n}{1-z}
\frac{1-z^{2n}/(1-z)^n}{1-z^2/(1-z)}
\\ = [z^{n-1}] \frac{1-z^n}{1-z}
\frac{1-z^{2n}/(1-z)^n}{1-z-z^2}.$$
Eliminate the terms that do not contribute to $[z^{n-1}],$ first
$$[z^{n-1}] \frac{1-z^n}{1-z}
\frac{1}{1-z-z^2}.$$
and second
$$[z^{n-1}] \frac{1}{1-z} \frac{1}{1-z-z^2}
= [z^{n-1}]
\left(\frac{2+z}{1-z-z^2}-\frac{1}{1-z}\right).$$
Extracting coefficients we obtain
$$F_{n-1} + 2 F_n - 1
= F_n + F_{n+1} - 1 = F_{n+2} - 1.$$
Remark. Here we have not counted the empty set as pointed out in the comments. The answer is $$F_{n+2}$$ if the empty set is included, which it should be.
|
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|
construct polynomial from other polynomials If I have a polynomial, P, with root $a$ and a polynomial, Q, with root $b$, is there a way to construct polynomial R such that $a+b$ is a root of R?
Here's a concrete example. a = $\sqrt2$. $P(x) = x^2 -2$ and $P(a) = 0$.
b = $\sqrt3$. $Q(x) = x^2 - 3$ and $Q(b) = 0$.
From another method, I know that $x^4 -10x^2 + 1$ has $\sqrt2 + \sqrt3$ as a root.
Is there a way to construct $R(x)$ from $P(x)$ and $Q(x)$? $P$, $Q$, and $R$ are polynomials with integer coefficient.
Thank you.
|
Gaussian elimination is the key. The ring
$$ \mathbb{Q}[x,y]/(P(x),Q(y)) $$
is a vector space over $\mathbb{Q}$ with dimension $\partial P\cdot \partial Q$, and a base given by $x^i y^j$ for $i\in[0,\partial P-1],j\in[0,\partial Q-1]$. By representing $1,(x+y),(x+y)^2,\ldots,(x+y)^{\partial P\cdot\partial Q}$ in such a base we get $\partial P\cdot\partial Q+1$ elements in a vector space with dimension $\partial P\cdot\partial Q$: it follows that $(x+y)^{\partial P\cdot \partial Q}$ can be represented as a linear combination of $1,(x+y),\ldots,(x+y)^{\partial P\cdot \partial Q-1}$ with rational coefficients: that provides a polynomial that vanishes at $a+b$.
In our concrete example, $P(x)=x^2-2$ and $Q(y)=y^2-3$, so from:
$$\begin{pmatrix} 1 \\ x+y \\ (x+y)^2 \\ (x+y)^3 \\ (x+y)^4\end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 11 & 9 & 0 \\ 49 & 0 & 0 & 20\end{pmatrix}\begin{pmatrix}1 \\ x \\ y \\ xy\end{pmatrix}$$
we get $(x+y)^4 = 10(x+y)^2 - 1$, and $R(t)=\color{red}{t^4-10t^2+1}$ is a polynomial that vanishes at $\sqrt{2}+\sqrt{3}$. By studying the rank of the previous matrix, we also have that $R(t)$ is the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$.
|
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|
Find $\lim\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}$ The question arise in connection with this problem
Prove that
$$\lim_{n\rightarrow \infty}\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}=1$$
Thanks to answer @Vincenzo Oliva. I forgot the Stolz-Cesàro theorem
|
Note that the sequence you're interested in can be written as $a_n=\dfrac1{\sqrt{n}}\left(\sqrt{4n-1}+\sum\limits_{k=1}^{2n-1}\sqrt{2k-1}-\sqrt{2k}\right).$ Applying the Stolz-Cesàro theorem, we have \begin{align}\lim_{n\to\infty} a_n&=\lim_{n\to\infty}\frac{\sqrt{4n+3}-\sqrt{4n-1}+\sqrt{4n-1}-\sqrt{4n}+\sqrt{4n+1}-\sqrt{4n+2}}{\sqrt{n+1}-\sqrt{n}} \\ &= \lim_{n\to\infty} 2\frac{\sqrt{1+3/(4n)}-1+\sqrt{1+1/(4n)}-\sqrt{1+1/(2n)}}{\sqrt{1+1/n}-1} \\ &=\lim_{n\to\infty} \frac2n\frac{3/8+1/8-1/4}{1/(2n)}=1.\end{align}
|
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What's the mistake on my answer for this inequality $ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $ Good evening to everyone! I have the following inequality: $$ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $$. I don't know what's wrong with my answer: $$ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} \rightarrow \left(x+1\right)\sqrt{x^2+4}>\left(x+2\right)\sqrt{x^2+1} \rightarrow \left(x+1\right)^2\left(x^2+4\right)>\left(x+2\right)^2\left(x^2+1\right) \rightarrow \left(x^2+2x+1\right)\left(x^2+4\right)>\left(x^2+4x+4\right)\left(x^2+1\right) \rightarrow 2x^3-4x<0 \rightarrow x\left(2x^2-4\right)<0 \rightarrow $$ $x$ belongs to $(-\infty,0)$ and $2x^2-4<0 \rightarrow x_1 = 1-\sqrt{2}$ and $x_2=1+\sqrt{2}$ therefore the final result is $x$ belongs to $(1-\sqrt{2},0) $. But my answer sheet shows that $x$ belongs to $(0,\sqrt{2})$. Where am I wrong? Thanks for any responses!
|
ALthough in fact $a>b$ does not necessarily mean that $a^2 > b^2$ even if $a$ and $b$ are both positive, the real place you went wrong is in solving $2x^2-4 = 0$. The answer to that is $x = \pm \sqrt{2}$ not $1 \pm \sqrt{2}$.
The $-\sqrt{2}$ zero is spurious; it was introduced by the quaring operation, and for $x=-\sqrt{2}$ you will find that the right hand side is positive and the left hand side negative (although the match in absolute value). So the interesting interval is $(0,2)$ and it is easy to check, by trying $x=1$ that the left side minus the right side is positive in that interval.
|
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|
Find the power series representation of $e^{-x^2}$ I know that the Maclaurin expansion of $e^x$ is $$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
But i'm not sure how to find the Maclaurin series here
I tried this
$$
f'_{(0)}=-2xe^{-x^2}=0
$$
And that follows to every derivative that follows, so how can I get a power series out of it?
|
That is easier than what you think.
General expansion:
$$e^A = 1 + A + \frac{1}{2}A^2 + \frac{1}{6}A^3 + \cdots$$
So if $A = -x^2$ you get
$$e^{-x^2} = 1 - x^2 + \frac{1}{2}(-x^2)^2 + \frac{1}{6}(-x^2)^3 + \cdots$$
So
$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$
To find the general series expansion, you just have to remember the series expansion of $e^A$:
$$e^A = \sum_{k = 0}^{+\infty} \frac{A^k}{k!}$$
So again: if $A = -x^2$ you gain finally
$$\boxed{e^{-x^2} = \sum_{k = 0}^{+\infty} \frac{(-x^2)^k}{k!} = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} x^{2k}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1843724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
$x^3 +y^2 +z =100z+10y+x$ What is the largest and smallest integer that satisfies this equation. $x^3 + y^2 +z=\overline{zyx}$, where $\overline{zyx}$ denotes the sequence of the digits.
$x^3 +y^2 +z =100z+10y+x$,where $x,y,z>0$
The maximum value of $x$, $y$, $z$ individually can only be $9$.
$\text{Maximum value}= 9^3 + 9^2 + 9 = 819$
So, $100z+10y+z<819$
Then I'm lost here a bit. What do I do next?
|
$$x^3+y^2+z=100z+10y+x$$
$$x(x+1)(x-1)=y(10-y)+99z$$
No needs to check $x=1$ and $y$ have symmetrical sense (e.g. if $y=1$ satisfies so as $y=9$).
Also, $9\leq y(10-y) \leq 25$ hence $$9(1+11z)\leq x(x+1)(x-1) \leq 25+99z$$
$$108\leq x(x+1)(x-1) \leq 916$$
$$\because \quad 3\times 4\times 5 < 108 < 4\times 5\times 6 \implies x\geq 5$$
Now,
$$4\times 5\times 6 \leq y(10-y)+99z \leq 8\times 9\times 10$$
$$120-25 \leq 99z \leq 720-9 \implies 1\leq z \leq 7$$
This can further confine to $5\leq x \leq 8$; and in turn, furthermore confine to $1\leq z\leq 5$.
Simple search gives
$$\overline{zyx}=135,\, 175,\, 518,\, 598$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1845102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Summing $3+7+14+24+37...$ up to $10$ terms
What is $3+7+14+24+37...$ up to $10$ terms?
I only see that difference between two consecutive terms is in AP ie $7-3=4,14-7=7,24-14=10$ and so on. Any ideas on how to do it?
|
$x=1 \phantom{n} 2 \phantom{n} 3 \phantom{n} 4$
$y=3 \phantom{n} 7 \phantom{n} 14 \phantom{n} 24$
$y=m_2(x-x_1)(x-x_2)+m_1(x-x_1)+y_1$
$m_1$ is slope
$m_1=(y_2-y_1)/(x_2-x_1)$
$m_1=(7-3)/(2-1)$
$m_1=4$
$m_2$ changes in slope
$m_2=((y_3-y_2)-(y_2-y_1))/(x_3-x_1)$
$m_2=((14-7)-(7-3))/(3-1)$
$m_2=3/2$
$y=3/2*(x-2)(x-1)+2*(x-1)+3$
$y=1.5x^2-0.5x+4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1846049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
}
|
The value of $x^2+y^2+z^2+w^2$ Let$x,y,z,w$ satisfy
$$\frac{x^2}{2^2 - 1^2} +\frac{y^2}{2^2 - 3^2} +\frac{z^2}{2^2 - 5^2} +\frac{w^2}{2^2 - 7^2} =1$$
$$\frac{x^2}{4^2 - 1^2} +\frac{y^2}{4^2 - 3^2} +\frac{z^2}{4^2 - 5^2} +\frac{w^2}{4^2 - 7^2} =1$$
$$\frac{x^2}{6^2 - 1^2} +\frac{y^2}{6^2 - 3^2} +\frac{z^2}{6^2 - 5^2} +\frac{w^2}{6^2 - 7^2} =1$$
$$\frac{x^2}{8^2 - 1^2} +\frac{y^2}{8^2 - 3^2} +\frac{z^2}{8^2 - 5^2} +\frac{w^2}{8^2 - 7^2} =1$$
My work
$$\frac{x^2}{t - 1^2} +\frac{y^2}{t - 3^2} +\frac{z^2}{t - 5^2} +\frac{w^2}{t - 7^2} =1$$
where $t $ satisfy $4,16,36,64$
$$f(t)=0$$
$$f(t) = (t – 1)(t – 9)(t – 25)(t – 49)–x^2(t – 9)(t – 25)(t – 49) –y^2(t – 1)(t – 25)(t – 49) – z^2(t–1)(t–9)(t–49) – w^2(t–1)(t–9)(t–25)$$
then I compared the coefficient with different value of $t$ .
I want to know that is there any easier alternative methods for this .
|
I multiplied the equations out to get:
$$\begin{pmatrix} -4725 & 2835 & 675 & 315 \\
2079 & 4455 & -3465 & -945 \\
-3861 & -5005 & -12285 & 10395 \\
32175 & 36855 & 51975 & 135135 \\
\end{pmatrix}
\begin{pmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{pmatrix} =
\begin{pmatrix} -14175 \\
31185 \\
-135135 \\
2027025 \\
\end{pmatrix}$$
Then inverted the matrix and solved to get:
$$\begin{pmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{pmatrix} =
\begin{pmatrix} 10.76660156 \\
10.15136719 \\
8.797851563 \\
6.284179688
\end{pmatrix}
$$
giving
$$x^2+y^2+z^2+w^2=36$$
- so there probably is an algebraic short cut to that.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1846993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
|
$$
\begin{align}
x^4-11x^2y^2+y^4
&=\left(x^2-y^2\right)^2-(3xy)^2\\
&=\left(x^2-3xy-y^2\right)\left(x^2+3xy-y^2\right)\\
&=\left(x-\tfrac{3-\sqrt{13}}2y\right)\left(x-\tfrac{3+\sqrt{13}}2y\right)\left(x+\tfrac{3-\sqrt{13}}2y\right)\left(x+\tfrac{3+\sqrt{13}}2y\right)
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1847983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
}
|
Solving $\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$ Does anyone have some tips for me how to go about the problem in the image?
$$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$$
I know it's supposed to be simple, but I can't figure out why the solution is:
$90^{\circ}+ 720^{\circ}k$ where $k=1,2,3,...$
Intuitively, I understand the answer, but that's not the point, I would like to understand the correct mathematical solution. Thank you very much!
|
Notice:
$$\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)=\frac{\sqrt{2}\cos\left(\frac{\theta}{2}\right)}{\sqrt{2}}+\frac{\sqrt{2}\sin\left(\frac{\theta}{2}\right)}{\sqrt{2}}=\sqrt{2}\left[\frac{\cos\left(\frac{\theta}{2}\right)}{\sqrt{2}}+\frac{\sin\left(\frac{\theta}{2}\right)}{\sqrt{2}}\right]=$$
$$\sqrt{2}\left[\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\theta}{2}\right)+\cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\theta}{2}\right)\right]=\sqrt{2}\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)$$
So:
$$\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)=\sqrt{2}\Longleftrightarrow$$
$$\sqrt{2}\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\sqrt{2}\Longleftrightarrow$$
$$\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=1\Longleftrightarrow$$
$$\frac{\pi}{4}+\frac{\theta}{2}=\frac{\pi}{2}+2\pi n\Longleftrightarrow$$
$$\frac{\theta}{2}=\frac{\pi}{4}+2\pi n\Longleftrightarrow$$
$$\theta=\frac{\pi}{2}+4\pi n$$
Where $n\in\mathbb{Z}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1849934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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|
Evaluation of $\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
$\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
Put $\displaystyle x = \frac{1}{u},$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and changing limits, We get
$$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{1}{u(1+u^2)}du$$
So $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{t}{1+t^2}dt = \frac{1}{2}\int_{\frac{1}{e}}^{e}\frac{2t}{1+t^2}dt$$
So $$f(x) = \frac{1}{2}\ln (1+t^2)|_{\frac{1}{e}}^{e} = \ln(e) = 1$$
My question is how can we solve it in some shorter way, Help required, Thanks
|
Just differentiate the $f(x)$ wrt $x$, you will se that $f'(x)=0$ which means $f(x)=constant=p$. SO you can find $p$ by just putting $x=\frac{\pi}{4}$. So
$$p=\int_{\frac{1}{e}}^{1} \frac{tdt}{t^2+1}+\int_{\frac{1}{e}}^{1} \frac{dt}{t(t^2+1)}=\int_{\frac{1}{e}}^{1}\frac{dt}{t}=0-(-1)=1$$
SO $f(x)=1$ which is the final answer
Hope this will be helpful !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1850568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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|
Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
Find the number of solutions of the equation
$$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
where $\lfloor\,\cdot\,\rfloor$ represents the floor function.
My work:
I use the fact that
$$\lfloor nx \rfloor =\sum_{k=0}^{n-1} \left\lfloor x +\frac kn \right\rfloor.$$
So the equation becomes
$$\lfloor x \rfloor +\sum_{k=0}^{1} \left\lfloor x +\frac k2 \right\rfloor +\sum_{k=0}^{3} \left\lfloor x +\frac k4 \right\rfloor +\sum_{k=0}^{7} \left\lfloor x +\frac k8 \right\rfloor \\
\qquad {}+\sum_{k=0}^{15} \left\lfloor x +\frac k{16} \right\rfloor +\sum_{k=0}^{31} \left\lfloor x+\frac k{32} \right\rfloor \\
= \lfloor x \rfloor + \left\lfloor x+\frac 12 \right\rfloor + \left\lfloor x+\frac 64 \right\rfloor + \left\lfloor x+\frac{28}{8} \right\rfloor \\
\qquad {}+ \left\lfloor x+\frac{120}{16} \right\rfloor + \left\lfloor x+\frac{496}{32} \right\rfloor$$
What should I do next?
|
Alt. solution: Write the fractional part of $x$ in binary:
$$
x = n + 0.b_1 b_2 b_3 \ldots = n + \sum_{i=1}^\infty \frac{b_i}{2^i},
$$
where $n \in \mathbb{Z}$ and $b_i \in \{0,1\}$.
Also give your function $\mathbb{R} \to \mathbb{R}$ a name:
$$
f(x) = \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor
$$
Then
\begin{align*}
f(x)
&= \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor \\
&= n + (2n + b_1) + (4n + 2b_1 + b_2) + \cdots + (32n + 16b_1 + 8b_2 + 4b_3 + 2b_4 + b_5) \\
&= 63n + 31b_1 + 15b_2 + 7b_3 + 3b_4 + b_5.
\end{align*}
We can choose $b_i \in \{0,1\}$ arbitrarily,
so it follows that the image of $f$ is exactly
the set of integers whose remainder is between $0$ and $31 + 15 + 7 + 3 + 1 = 57$, mod $63$.
In particular, $12345 \equiv 60 \pmod{63}$, so $12345$ is not in the image of $f$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1851042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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|
Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $7$?
I used the binomial formula: $\dfrac {(7+3)^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot 3^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot + 35 \cdot 3^{34} \cdot 7}{7} + \dfrac {3^{35}}{7}$
Therefore, $10^{35}$ will have the same remainder as $3^{35}$ when divided by $7$.
I was stuck here, and I wanted to try to reverse engineer the answer. I know the remainder is $5$. Therefore I should be able to write $3^{35} -5 + 5$ as $7k+5$. However, I have no idea how to factor out a $7$ from $3^{35}-5$, or from $10^{35}-5.$ How could I find this $k$ non-explicitly?
|
Do you know Fermat little theorem?
for prime $p, n^p \equiv n \mod p\\
n^{p-1}\equiv 1 \mod p$
$10^35 = (10^6)^5(10^5)\\
10^35 \equiv (10^5) \mod 7\\
10^35 \equiv 5 \mod 7$
Alternative:
You could say, $10^{35} -1 = (10^7 - 1)(10^{28} + 10^{21} + 10^{14} + 10^7 + 1)$
And the remainder of the product equals the product of the remainders.
$(10^7 - 1)(10^{28} + 10^{21} + 10^{14} + 10^7 + 1) / 7\\
(3-1)(3^4 + 3^3 + 3^2 + 3 + 1) = 242$
and the remainder of 242/7 = 4
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1851382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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|
Find the determinant using colum or row operations I find problem in simplification. When I tried to simplify I ended up doing the regular process of finding the determinant value. The matrix is $\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ ab & bc & ca \end{pmatrix}$.
|
$$\det\left(\begin{array}{ccc}1&1&1\\a&b&c\\ab&bc&ca\end {array}\right)=abc\det\left(\begin{array}{ccc}\frac{1}{a}&\frac{1}{b}&\frac{1}{c}\\1&1&1\\b&c&a\end {array}\right)=abc\det\left(\begin{array}{ccc}\frac{1}{a}-\frac{1}{b}&\frac{1}{b}&\frac{1}{c}-\frac{1}{b}\\0&1&0\\b-c&c&a-c\end {array}\right)=$$
$$abc\det\left(\begin{array}{ccc}\frac{1}{a}-\frac{1}{b}&\frac{1}{c}-\frac{1}{b}\\b-c&a-c\end {array}\right)=c(b-a)(a-c)-a(b-c)^{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1852938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
number of non differentiable points in $g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
If $\displaystyle f(x) = \lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}\;,n\in \mathbb{N}$
and $\displaystyle g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
Then number of points where $g(x)$ is not differentiable.
$\bf{My\; Try::}$ We can write $$f(x) =\lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}= \left\{\begin{matrix}
\frac{2}{x+1}\;\;,&x+1<-1\Rightarrow x<-2 \\
\frac{x^2}{x^2+1}\;\;, & -1<x+1<1\Rightarrow -2<x<0\\
\frac{2}{x+1}\;\;, & x+1>1\rightarrow x>0 \\
1\;\;, & x+1=1\Rightarrow x=0\\
\frac{3}{2}\;\;,& x+1=-1\Rightarrow x=-2
\end{matrix}\right.$$
Now How can i solve after that help Required, Thanks
|
Hint: using the formula $\tan ^{ 2 }{ \frac { x }{ 2 } = } \frac { 1-\cos { x } }{ \sin { x } } $ we can transform $g(x)$ function so :$$g\left( x \right) =\tan \left( \frac { 1 }{ 2 } \arcsin \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } } \right) \right) =\\ =\pm \sqrt { \frac { 1-\cos { \left( \arcsin \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } } \right) \right) } }{ \sin { \left( \arcsin \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } } \right) \right) } } } =\\ =\pm \sqrt { \frac { 1-\sqrt { 1-{ \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } } \right) }^{ 2 } } }{ \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } } \right) } } =\\ =\pm \sqrt { \frac { 1-\sqrt { 1+1+2f^{ 2 }\left( x \right) +f^{ 4 }\left( x \right) -4f^{ 2 }\left( x \right) } }{ 2f\left( x \right) } } =\\ =\pm \sqrt { \frac { 1-\sqrt { 1+{ \left( 1-f^{ 2 }\left( x \right) \right) }^{ 2 } } }{ 2f\left( x \right) } } $$
hope it will help
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1853193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Generalized Harmonic Number Summation $ \sum_{n=1}^{\infty} {2^{-n}}{(H_{n}^{(2)})^2}$
Prove That $$ \sum_{n=1}^{\infty} \dfrac{(H_{n}^{(2)})^2}{2^n} = \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 $$
Notation : $ \displaystyle H_{n}^{(2)} = \sum_{r=1}^{n} \dfrac{1}{r^2}$
We can solve the above problem using the generating function $\displaystyle \sum_{n=1}^{\infty} (H_{n}^{(2)})^2 x^n $, but it gets rather tedious especially taking into account the indefinite polylogarithm integrals involved. Can we solve it using other methods like Euler Series Transform or properties of summation?
|
second approach suggested by Cornel Ioan Valean using summation by parts and lets start with the following sum:
with ${N \in \mathbb{N}_{\ \geq\ 1}}$
\begin{align}
\sum_{n=1}^N\frac{\left(H_{n-1}^{(2)}\right)^2}{2^n}=\sum_{n=1}^N\frac{\left(H_n^{(2)}\right)^2}{2^n}-2\sum_{n=1}^N\frac{H_n^{(2)}}{n^22^n}+\sum_{n=1}^N\frac1{n^42^n}\tag{1}
\end{align}
on the other hand:
\begin{align}
\sum_{n=1}^N\frac{\left(H_{n-1}^{(2)}\right)^2}{2^n}=\sum_{n=1}^{N-1}\frac{\left(H_{n}^{(2)}\right)^2}{2^{n+1}}=\sum_{n=1}^{N}\frac{\left(H_{n}^{(2)}\right)^2}{2^{n+1}}-\frac{\left(H_{N}^{(2)}\right)^2}{2^{N+1}}\tag{2}
\end{align}
from $(1)$ and $(2)$ we reach
$$\sum_{n=1}^N\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=4\sum_{n=1}^N\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^N\frac{1}{n^42^n}-2\frac{\left(H_{N}^{(2)}\right)^2}{2^{N+1}}$$
letting $N$ approach $\infty$ we get
$$\sum_{n=1}^\infty\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=4\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{1}{n^42^n}-0$$
I was able here to prove $$\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42
\end{align*}$$
which follows $$\sum_{n=1}^\infty\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=2\operatorname{Li_4}\left(\frac{1}{2}\right)+\frac14\zeta(4)+\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1855024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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|
Is it possible to identify this sequence? Interested by this question, $j$ being a positive integer,
I tried to work the asymptotics of
$$S^{(j)}_n=\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+j)}=\frac{\, _2F_1\left(j,-n;j+1;-\frac{1}{n}\right)}{j}$$
I quickly noticed (not a proof) that the asymptotics write
$$S^{(j)}_n=(-1)^j\left(\left(\alpha_0-\beta_0e\right)-\frac{\left(\alpha_1-\beta_1e\right)}{2n}+\frac{\left(\alpha_2-\beta_2e\right)}{24n^2}\right)+O\left(\frac{1}{n^3}\right)$$
in which the $\alpha_k$'s and $\beta_k$'s are all positive whole numbers depending on $j$.
What I found is that $$\alpha_0=(j-1)!\qquad \qquad \beta_0=\text{Subfactorial}[j-1]$$ $$\alpha_1=(j+1)!\qquad \qquad \beta_1=\text{Subfactorial}[j+1]$$ $$\alpha_2=(1+3j)(j+2)!$$ but I did not find any formal representation for $\beta_2$.
I give below a list of values ot this last coefficient as a function of $j$.
$$\left(
\begin{array}{cc}
j & \beta_2 \\
1 & 11 \\
2 & 60 \\
3 & 443 \\
4 & 3442 \\
5 & 29667 \\
6 & 281824 \\
7 & 2936915 \\
8 & 33374022 \\
9 & 411167963 \\
10 & 5462660068 \\
11 & 77886959691 \\
12 & 1186630738810 \\
13 & 19242660629363 \\
14 & 330973762825032
\end{array}
\right)$$
It seems that $\frac{\beta_2}{\text{Subfactorial}[j+2]}$ is close to a straight line but it is not.
Is there any way to identify what is this sequence ?
|
Notice
$$S_n^{(j)}
= \int_0^1 \left(1+\frac{x}{n}\right)^n x^{j-1} dx
= \int_0^1 e^{n\log\left(1+\frac{x}{n}\right)} x^{j-1} dx
= \int_0^1 e^{x - \frac{x^2}{2n} + \frac{x^3}{3n^2} + O(n^{-3})} x^{j-1}dx\\
= \int_0^1 e^x \left[1 - \frac{x^2}{2n} + \frac{8x^3+3x^4}{24n^2} + O(n^{-3})\right]x^{j-1} dx
$$
Compare with expansion in question, we get
$$(-1)^j(\alpha_2 - \beta_2 e) = 8A_{j+2}+3A_{j+3}
$$
where
$$\begin{align}A_{j}
&= \int_0^1 e^x x^j dx
= \left(\int_{-\infty}^1 - \int_{-\infty}^0\right) e^x x^j dx
= \underbrace{\int_0^\infty e^{1-y} (1-y)^j dy}_{x=1-y} - \underbrace{\int_0^\infty e^{-y} (-y)^j dy}_{x = -y}\\
&= (-1)^j\left(e \int_0^\infty e^{-y} (y-1)^j dy - j!\right)
\end{align}
$$
Using the fact
$$\int_0^\infty e^{-y} (y-1)^n dy = !n \stackrel{def}{=} n! \left(\sum_{s=0}^n \frac{(-1)^s}{s!}\right) = \left\lfloor\frac{n!}{e}+\frac12\right\rfloor$$
is the $n^{th}$ derangement number, we find
$$(-1)^j\left(\alpha_2 - \beta_2 e\right)
= (-1)^j\left\{8[e!(j+2) - (j+2)!] - 3[e!(j+3) - (j+3)!]\right\}\\
\implies
\begin{cases}
\alpha_2 &= 3\times(j+3)! - 8\times(j+2)! = (3j+1)(j+2)!\\
\beta_2 &= 3\,\times\,!(j+3) - 8\,\times\,!(j+2)
\end{cases}
$$
|
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"url": "https://math.stackexchange.com/questions/1855581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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|
Quadratic Functional equations. Suppose $f$ is a quadratic ploynomial, with leading cofficient $1$, such that $$f(f(x) +x) = f(x)(x^2+786x+439)$$ For all real number $x$. What is the value of $f(3)$?
|
Suppose $f(x) = x^2 + bx + c$
Then
$$f(f(x)+x) = (f(x)+x)^2 + b(f(x)+x) + c = f(x)^2 + 2xf(x) + x^2 + bf(x) + bx + c = f(x)^2 + 2xf(x) + bf(x) + f(x) = f(x)(f(x) + 2x + b+ 1)$$
Equating factors yields
$$f(x) + 2x + b+ 1 = x^2 + 786 x + 439$$
Then letting $x=3$ we have
$f(3) = 2799-b$
Finding $b$ is a trivial matter by expanding out the previous equation.
There may be an indirect method for determining $b$ but this is as far as I can get without solving for $f(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1856052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Factor polynomial over $C, Q, R$, if one comlex root is given The polynomial is: $P(x)=x^6+x^4-x^3+x^2+1$. I need to factor it over $C, Q, R$ if one complex root is $\sqrt[3]{1}$. Also find all fields in which $P$ is reducible.
Now, I know how to find one factor of P using given complex root, and I end up with $(x^2+x+1)(x^4-x^3+x^2-x+1)$, but how should I go about $(x^4-x^3+x^2-x+1)$?
How can I find all fields in which $P$ is reducible?
|
A start: To factor the quartic, it may be useful to find its roots.
Rewrite the quartic as $x^2(x^2+\frac{1}{x^2}-\left(x+\frac{1}{x}\right)+1)$.
Let $t=x+\frac{1}{x}$. Then to solve our quartic we first solve
$$t^2-2-t+1=0.$$
Another way: Our quartic is $\frac{x^5+1}{x+1}$. So its roots are certain $10$-th roots of unity.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Indefinite Integral - How to do questions with square roots?
$$\int \frac{dx}{x^4 \sqrt{a^2 + x^2}}$$
In the above question, my first step would be to try and get out of the square root, so I would take $ t^2 = a^2 + x^2 $. But that gets me nowhere. How would you solve this, and if you are going to take a substitution what is the logic behind that substitution?
|
You can use trigonometric functions. Substitute $x=a*tan(u)$ and $dx=a*\frac{du}{{cos}^{2}u}$ this gives:
$$\int \dfrac{dx}{x^4 \sqrt{a^2 + x^2}} =\int \dfrac{du}{a^4\cos(u){\tan}^{4}(u)}$$
Working further on the solution of Battani and by simplifying:
$${ \sin{ \left( \arctan { \frac { x }{ a } } \right) } }=\frac{x}{\sqrt{x^2+a^2}}$$
This gives:
$$\frac { 1 }{ { a }^{ 4 } } \left[ -\frac { 1 }{ 3\sin ^{ 3 }{ \left( \arctan { \frac { x }{ a } } \right) } } +\frac { 1 }{ \sin { \left( \arctan { \frac { x }{ a } } \right) } } \right] +C\\=\frac{1}{a^4}\left[ -\frac{(x^2+a^2)^{3/2}}{3x^3}+\frac{\sqrt{x^2+a^2}}{x}\right]$$
Simplifying further:
$$\frac{1}{a^4}\left[ -\frac{(3x^2-{x^2+a^2})\sqrt{x^2+a^2})}{3x^3}\right]=$$
$$-\frac{(2x^2+a^2)\sqrt{x^2+a^2})}{3a^4x^3}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1856844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.
Let's look at
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3-y^3 \\
\iff & 4x^3+4y^3\geq (x+y)^3 \\
\iff & x^3+y^3\geq \frac{1}{4}(x+y)^3
\end{split}$$
is the proof valid? is there a shorter way?
|
Is this proof valid?
Your proof is valid and correct. However, it would be better if you prove that an equivalent statement is true, from an inequality, instead of doing the opposite.
$$\begin{split} & x^3+y^3\geq \frac{1}{4}(x+y)^3 \\
\iff & 4x^3+4y^3\geq (x+y)^3 \\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3-y^3 \\
\iff & 3x^3+3y^3\geq 3xy^2+3yx^2\\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff &(x-y)^2(x+y)\geq 0 \\
\end{split}$$
Which is clearly true when $x+y \geq 0$
Is there a shorter method ?
The given inequality is equivalent to
$$4(x^3+y^3)(x+y)=(1+1)(1+1)(x^3+y^3)(x+y) \geq (x+y)^4$$ which is true by Hölder's inequality
|
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"url": "https://math.stackexchange.com/questions/1856948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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|
Find $a + b + c$, given that $(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$
Let $a,b,c$ be real number such that $$(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$$
Find $a + b + c$.
The answer is: $-\frac{5}{2}$. Please give me some clues or solution.
Thanks.
|
Using CS-inequality: let $x = a+b+c \implies \left(x+\frac{19}{2}\right)^2 = \left(1\sqrt{a+1}+2\sqrt{b+2}+3\sqrt{c+3}\right)^2\leq 14(x+6)\implies \left(x+\frac{5}{2}\right)^2 \leq 0\implies x = -\dfrac{5}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION
Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT
*
*Typed
$\newcommand{\dd}{\; \mathrm{d}}\int \frac{\sqrt{64x^2-256}}x \dd x \to
\int \frac{\sqrt{64(x^2-4)}}x \dd x \to
\int \frac{8\sqrt{x^2-4}}x \dd x$
Use $x=a\sec\theta$, $\dd x=a\sec\theta \tan\theta \dd \theta$.
$a=2$ $\to$ $x=2\sec\theta$, $\dd x=2\sec\theta \tan\theta \dd \theta$.
$=\int \frac{8\sqrt{4\sec^2\theta-4}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8\sqrt{4(\sec^2\theta-1)}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta $
$=\int \frac{8\sqrt{4\tan^2\theta}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8(2\tan\theta)}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta$
$=\int 16\tan^2\theta \dd \theta \to
16\int\tan^2\theta \dd \theta \to
\underset{\text{trig. formula}}{\underbrace{16(\theta+\tan\theta)+C}}$
$\Rightarrow 16(\tan\theta-\theta)+C = 16\tan\theta-16\theta$
$x=2\sec\theta$, $\sec\theta= \frac x2$
$\boxed{16\tan\left(\frac{\sqrt{x^2-4}}2\right) -16\sec^{-1}\left(\frac x2\right)+C}$
*
*Handwritten
|
$$\dfrac{\sqrt{64x^2-256}}x=8x\cdot\dfrac{\sqrt{x^2-4}}{x^2}$$
Let $\sqrt{x^2-4}=y\implies x^2-4=y^2\implies x\ dx= y\ dy$
$$\int\dfrac{\sqrt{64x^2-256}}xdx=8\int\dfrac{y^2dy}{y^2+4}=8\int dy-32\int\dfrac{dy}{y^2+4}=?$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1857798",
"timestamp": "2023-03-29T00:00:00",
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|
Convergence of the series $\sum \frac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$
To prove that nature of the following series : $$\sum \dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$$
they use in solution manual :
My questions:
*
*I don't know how to achieve ( * ) could someone complete my attempts for ( * ) and is it correct if i use (**) to prove that the series is convergent :
$$\fbox{$\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}-\dfrac{(-1)^{n}}{n}+O\left(\dfrac{1}{n^{\frac{4}{3}}}\right)$}\quad (*)$$
My thoughts :
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}
\end{align}
note that :
$$(1+x)^{\alpha}=1+\alpha x+O(x^{2})$$
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1-\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)+O\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)^{2} \right)\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} -\dfrac{(-1)^{n}}{n}+\dfrac{(-1)^{n}}{n^{\frac{4}{3}}} +\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}\times O\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)^{2} \\
&=\ldots\ldots \\
&= \mbox{ I'm stuc here i hope someone complete my attempts }
\end{align}
Or i should use :
note that :
$$(1+x)^{\alpha}=1+O(x)$$
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+O\left( \dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}\right) \right)\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{1}{n}+\dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right) \\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right) \\
\end{align}
$$\fbox{$\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} =\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right)$}\quad (**) $$
|
Fix $N>0
$ and $N
$ even (you can also take $N
$ odd, the proof is essentially the same). We have $$\sum_{n=1}^{N}\frac{\left(-1\right)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+\left(-1\right)^{n}}=\sum_{n=1}^{N/2}\frac{1}{\left(2n\right)^{\frac{2}{3}}+\left(2n\right)^{\frac{1}{3}}+1}-\sum_{n=1}^{N/2-1}\frac{1}{\left(2n-1\right)^{\frac{2}{3}}+\left(2n-1\right)^{\frac{1}{3}}-1}
$$ $$=\sum_{n=1}^{N/2}\frac{\left(2n-1\right)^{2/3}-\left(2n\right)^{2/3}+\sqrt[3]{2n-1}-\sqrt[3]{2n}-2}{\left(\left(2n\right)^{\frac{2}{3}}+\left(2n\right)^{\frac{1}{3}}+1\right)\left(\left(2n-1\right)^{\frac{2}{3}}+\left(2n-1\right)^{\frac{1}{3}}-1\right)}-\frac{1}{\left(N-3\right)^{\frac{2}{3}}+\left(N-3\right)^{\frac{1}{3}}-1}
$$ $$=\sum_{n=1}^{N}\frac{\left(-1\right)^{n}}{n^{2/3}\left(1+O\left(1\right)\right)}+\sum_{n=1}^{N}\frac{\left(-1\right)^{n}}{n\left(1+O\left(1\right)\right)}$$ $$-\sum_{n=1}^{N/2}\frac{2}{n^{4/3}\left(1+O\left(1\right)\right)}-\frac{1}{\left(N-3\right)^{\frac{2}{3}}+\left(N-3\right)^{\frac{1}{3}}-1}
$$ and so the series converges.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Linear combination issue I have 4 vectors:
$u_1 = \begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix} $, $\; u_2 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_3 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_4 = \begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix} $
and I wanna express the following vector in terms of them:
$\; v = \begin{pmatrix}
2 \\
3 \\
4 \\
\end{pmatrix} $
I'm working this way:
First put vectors in a matrix and then put in rref:
$ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
1&1&1&1&3\\
1&0&0&0&4
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&0&0&1&1\\
0&-1&-1&0&2
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&-1&-1&0&2\\
0&0&0&1&1
\end{array}
\right] $ =>
$ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&0&0&0&4\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $
but this result (4 -2 1) is meaningless to me (because $4u_1 -2u_2 + u_3 \ne v$)... does it make any sense? Can it represent some sort of coeficients of the linear combination?
If I swap $u_3$ for $u_4$ in the matrix, the (4 -2 1) is the same but it DOES make sense, because $4u_1 -2u_2 + u_4 = v$
How can I write v as combination of u's ? If I did the right way, how does this make sense?
|
You need three linearly independent vectors to form a basis for three dimensions.
As $u_2$ and $u_3$ are in fact identical and the same, the redundant and duplicated one is unnecessary and also not needed.
Reapply your process with only three column vectors: $u_1, u_2, u_4$, and $v$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the orthocentre of a trinagle. Now, I know this has been asked here but my question is something else so please bear with me.
Question:- If the vertices of a triangle are represented by $z_1, z_2, z_3$ respectively then show that the orthocentre of the $\triangle ABC$ is
$$\dfrac{(a \sec A)z_1+(b \sec B)z_2+(c \sec C)z_3}{a \sec A+b \sec B +c \sec C }$$ $$OR$$ $$\dfrac{z_1\tan A+z_2\tan B+z_3\tan C}{\tan A+\tan B+\tan C}$$
Attempt at a solution:-
Now, from simple trigonometry, to be more specific, from projection law, we can see that
$$\frac{BL}{LC}=\dfrac{c\cos B}{b\cos C}=\dfrac{c\sec C}{b \sec B}$$
So we get from the section formula the coordinates of $L$ as follows
$$L \equiv \left(\dfrac{b\sec B\cdot z_2+c\sec C\cdot z_3}{b\sec B+c\sec C}\right)$$
Now, I don't know much about properties of triangles and found out the following method in a book for finding the ratio in which $H$ divides $AL$
$$\begin{equation}
\begin{aligned}
\dfrac{AH}{HL}
&=\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)} \\
&=\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}} \\
&=\dfrac{c \cos A}{c\cos B\cos C}=\dfrac{a \cos A}{a\cos B\cos C} \\
&=\dfrac{b\cos C +c\cos B}{a\sec A}\cdot\dfrac{1}{\cos B\cos C} \\
&=\dfrac{b \sec B+c\sec C}{a\sec A}
\end{aligned}
\end{equation}$$
Now I don't understand what was used in the derivation of the required ratio above, is it similarity of triangles but i don't think the triangles are similar.
P.S.:- Corrected the link to the question mentioned at the top.
|
$$\dfrac{AH}{HL}=\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}$$
Because the two triangles have the common height $BL$.
$$\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}=\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}}$$
$$\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}=\dfrac{\dfrac 12 AB\cdot NH}{\dfrac 12BL\cdot HL}=\dfrac{\dfrac{1}{2}AB\cdot BH\sin{\alpha}}{\dfrac{1}{2}BL\cdot BH \sin{\beta}}$$
$$\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}}=\dfrac{c \cos A}{c\cos B\cos C}$$
See $\triangle{ABM}$ and $\triangle{BMC}$ :
$$\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}}=\frac{AB\sin\alpha}{\color{red}{BL}\sin\beta}=\frac{c\sin(90^\circ-A)}{\color{red}{c\cos B}\sin(90^\circ-C)}=\frac{c\cos A}{c\cos B\cos C}$$
$$\dfrac{c \cos A}{c\cos B\cos C}=\dfrac{a \cos A}{a\cos B\cos C}$$
Because $\frac cc=\frac aa=1$.
$$\dfrac{a \cos A}{a\cos B\cos C}=\dfrac{b\cos C +c\cos B}{a\sec A}\cdot\dfrac{1}{\cos B\cos C}$$
$$\dfrac{a \cos A}{a\cos B\cos C}=\dfrac{\color{red}{a}}{a\sec A}\cdot \dfrac{1}{\cos B\cos C}=\dfrac{\color{red}{CL+BL}}{a\sec A}\cdot \dfrac{1}{\cos B\cos C}$$
Now $CL=b\cos C$ and $BL=c\cos B$.
$$\dfrac{b\cos C +c\cos B}{a\sec A}\cdot\dfrac{1}{\cos B\cos C}=\dfrac{b \sec B+c\sec C}{a\sec A}$$
$$\dfrac{b\cos C +c\cos B}{a\sec A}\cdot\dfrac{1}{\cos B\cos C}=\frac{\frac{b\cos C}{\cos B\cos C}+\frac{c\cos B}{\cos B\cos C}}{a\sec A}=\dfrac{b \sec B+c\sec C}{a\sec A}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
|
One method for differentiating $m(x)$ is treating $m(x)=\frac{f(x)}{g(x)}$ with
$f(x)=x$ and $g(x)=\sqrt{4x-3}$ and using the quotient rule for differentiation.
Then $m'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}=\frac{\sqrt{4x-3}-(x\frac{2}{\sqrt{4x-3}})}{4x-3}$.
Simplyfying we obtain the desired result $m'(x)=\frac{2x-3}{\sqrt{(4x-3)^{3}}}$.
|
{
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"url": "https://math.stackexchange.com/questions/1860814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Circle Problem:Which of the following are true If the circle $x^2+y^2+2gx+2fy+c=0$ cuts the three circles $x^2+y^2−5=0$, $x^2+y^2−8x−6y+10=0$ and $x^2+y^2−4x+2y−2=0$ at the extremities of their diameters, then which of the following are true
?
$c=-5$
$fg=147/25$
$g+2f=c+2$
$4f=3g$
Can't think of a simple method.Help please!
|
The center of $x^2+y^2−5=0, x^2+y^2−8x−6y+10=0$ and $x^2+y^2−4x+2y−2=0$ is $(0,0), (4,3)$ and $(2,-1)$ respectively.
So, $$(x^2+y^2+2gx+2fy+c)-(x^2+y^2-5)=0$$
$$(x^2+y^2+2gx+2fy+c)-(x^2+y^2−8x−6y+10)=0$$
$$(x^2+y^2+2gx+2fy+c)-(x^2+y^2−4x+2y−2)=0$$
passes through $(0,0), (4,3)$ and $(2,-1)$ respectively.
(For example, $(x^2+y^2+2gx+2fy+c)-(x^2+y^2-5)=0$ represents a line passing through the intersection points of the two circles.)
Therefore, we get
$$c=-5$$
$$c+6 f+8 g+40 = 0$$
$$c-2 f+4 g+12 = 0$$
Solving the system gives $f=-21/10, g=-14/5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1860994",
"timestamp": "2023-03-29T00:00:00",
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|
Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$
My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equation with respect to $x$,
$2b^2x+2a^2yy'=0\\
y'=-\frac{b^2}{a^2}{\cdot}\frac{x}{y}$
Differentiating again with respect to x,
$y''=-\frac{b^2}{a^2}{\cdot}\frac{y-xy'}{y^2}$
a) At $(0,b)$
$y'=0\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{b-(0)(0)}{b^2}=-\frac{b}{a^2}$
$\displaystyle{\kappa=\frac{|-\frac{b}{a^2}|}{\left[1+\left(0\right)^2\right]^\frac{3}{2}}=\frac{b}{a^2}}$
b) At $(a,0)$
$y'=\infty\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{\frac{y}{y'}-x}{\frac{y^2}{y'}}=\infty$
I am not sure, if my solution to part (b) is correct. The book says, the answer must be $\frac{a}{b^2}$.
|
I solved it as follows:
$\begin{align}
y'&=-\frac{b^2}{a^2}\cdot\frac{x}{y}\\
y''&=-\frac{b^2}{a^2}\cdot{\frac{y-xy'}{y^2}}\\
&= -\frac{b^2}{a^2}\cdot{\frac{y-x\cdot{\left(-\frac{b^2}{a^2}\cdot\frac{x}{y}\right)}}{y^2}}\\
&=-\frac{b^2}{a^2}\cdot{\frac{a^{2}y^{2}+b^{2}x^{2}}{a^{2}y^{3}}}\\
&=-\frac{b^2}{a^2}\cdot{\frac{a^{2}b^{2}}{a^{2}y^{3}}}\\
&=-\frac{b^{4}}{a^{2}}\cdot{\frac{1}{y^{3}}}
\end{align}$
The radius of curvature $\kappa$ is given by,
$\begin{align}
\kappa&=\frac{|y''|}{[1+y'^{2}]^\frac{3}{2}}\\
&=\frac{|-\frac{b^{4}}{a^{2}}\cdot{\frac{1}{y^{3}}}|}{\left[1+\left(-\frac{b^2}{a^2}\cdot\frac{x}{y}\right)^{2}\right]^\frac{3}{2}}\\
&=\frac{a^{4}b^{4}}{\left[a^{4}y^{2}+b^{4}x^{2}\right]^\frac{3}{2}}
\end{align}$
|
{
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"url": "https://math.stackexchange.com/questions/1863982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Help solving this recurrence relation I wanted to resolve the determinant of the next (nxn) matrix via recurrence relations:
$$
\begin{vmatrix}
a & 1 & 0 & 0 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0\\
1 & a & 1 & 0 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0 \\
0 & 1 & a & 1 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & a & 1 & 0 &.... 0 & 0 & 0 & 0 & 0\\
.. & .. & .. & .. & .. & .. &..... & .. & .. & ..\\
0 & 0 & 0 & 0 & 0 & 0 & .... 0 & 1 & a & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & .... 0 & 0 & 1 & a & 1\\
0 & 0 & 0 & 0 & 0 & 0 & .... 0 & 0 & 0 & 1 & a\\
\notag
\end{vmatrix}
$$
After analyzing the matrix I found the recurrence relation:
$$
D_{n}-a*D_{n-1} +D_{n-2}=0
$$
So the polynomial that describes this recurrence is:
$$
P(\lambda) = \lambda^2 - a * \lambda + 1
$$
The roots will be:
$$
\lambda_1 = \frac{a}{2} + \frac{\sqrt{a^2-4}}{2}\\
\lambda_2 = \frac{a}{2} - \frac{\sqrt{a^2-4}}{2}
$$
To resolve the recurrence I need 2 constants (C1 & C2) that satisfy:
$$
D_n = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2})^n + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})^n
$$
With the initial conditions
$$
D_1 = a\\
D_2 = a^2 - 1
$$
The problem is I don't know how to resolve the equation system generated by substituting the initial conditions on the function.
Any type of help is appreciated.
|
To solve for the constants we just evaluate the expression for $D_n$ at $n=1$ and $n=2$, as well as use the information that $D_1=a$ and $D_2=a^2-1$ to get
$$C_1\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)+C_2\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)=a$$
$$C_1\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2+C_2\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)^2=a^2-1$$
Now, since $a$ is just some constant, we can treat this as a linear system and write it as
$$A\vec{c}=\left[\begin{array}{cc}
\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right) & \left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right) \\
\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2 & \left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)^2
\end{array}\right]
\left[\begin{array}{c}
C_1 \\
C_2
\end{array}\right]=
\left[\begin{array}{c}
a \\
a^2-1
\end{array}\right]$$
Which has solution
$$
\left[\begin{array}{c}
C_1 \\
C_2
\end{array}\right]
=
\frac{\text{adj }A}{\det A}\left[\begin{array}{c}
a \\
a^2-1
\end{array}\right]
$$
Where
$$\det A=\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)^2-\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2$$
$$=\frac{2}{a-\sqrt{a^2-4}-2}$$
and
$$\text{adj }A=\left[\begin{array}{cc}
\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)^2 & -\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right) \\
-\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2 & \left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)
\end{array}\right]$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculation of the squared Euclidean norm Just a simple question. I'm reading an article and come across an equation that I cannot replicate as done in the original.
\begin{align}
&\|\mathbf{x} - \mathbf{\alpha} \|^2 - \|\mathbf{x} - \mathbf{\beta}\|^2\\
&= \|\mathbf{x}\| \|\mathbf{x}\| - 2 \|\mathbf{\mathbf{\alpha}}\| \|\mathbf{x}\| + \|\mathbf{\alpha}\| \|\mathbf{\mathbf{\alpha}}\| - \|\mathbf{x}\| \|\mathbf{x}\| +2 \|\mathbf{\beta}\| \|\mathbf{x}\| - \|\mathbf{\beta}\| \|\mathbf{\beta}\|\\
&= \mathbf{\alpha}^T \mathbf{\alpha} - \beta^T \beta + 2(\sqrt{\mathbf{\beta}\cdot\mathbf{\beta}}-\sqrt{\mathbf{\alpha}\cdot\mathbf{\alpha}})\|\mathbf{x}\|\\
&= \mathbf{\alpha}^T \mathbf{\alpha} - \beta^T \beta + 2(\sqrt{\mathbf{\beta}\cdot\mathbf{\beta}}-\sqrt{\mathbf{\alpha}\cdot\mathbf{\alpha}})(\sqrt{\mathbf{x}\cdot\mathbf{x}}),
\end{align}
where I used the fact that $$\|a\|\|a\| = \sqrt{a\cdot a}\sqrt{a\cdot a} = \sqrt{a^T a}\sqrt{a^T a} = a^Ta.$$
However, the article gives $$2(\beta-\alpha)^T \mathbf{x} + \alpha^T \alpha - \beta^T\beta $$
|
Your transition from the first line to the second is incorrect. We should have
$$
\|x - \alpha\|^2 - \|x - \beta\|^2 = \\
(x - \alpha)^T(x - \alpha) - (x - \beta)^T(x - \beta) = \\
\|x\|^2 + \|\alpha\|^2 - \|x\|^2 - \|\beta\|^2 - x^T\alpha - \alpha^Tx + x^T\beta + \beta^Tx =\\
\|\alpha\|^2 - \|\beta\|^2 - 2\alpha^Tx + 2 \beta^Tx =\\
\alpha^T\alpha - \beta^T\beta + 2(\beta - \alpha)^Tx
$$
which is the desired result.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1865476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to prove $\prod\limits_{i=1}^{n}(x-4i+2)(x-4i+1)>\prod\limits_{i=1}^{n}(x-4i+3)(x-4i)$ for all $x\in\mathbb{R}$? I would like to prove that for $n\in\mathbb{N}$ we have $f_n(x):=\prod\limits_{r=1}^{n}(x-4r+2)(x-4r+1)>\prod\limits_{r=1}^{n}(x-4r+3)(x-4r)=:g_n(x)$ for all $x\in\mathbb{R}$ (actually it would suffice for $n$ even). My attempt was to form pairs on each side in such a way that we always obtain the same function plus a constant, i.e. for $n=2k$ even:
$$
f_n(x)=(x-2)(x-8k+1)\cdot(x-3)(x-4n+2)\cdot...\cdot(x-4k-2)(x-4k+1)=\\((x-4k-0.5)^2-(4k-1.5)^2)\cdot((x-4k-0.5)^2-(4k-2.5)^2)\cdot...\cdot((x-4k-0.5)^2-(1.5)^2)
$$
And similarly for $g_n(x)$. If we then substitute $x=0.5y+4k+0.5$ we see that the inequality is equivalent to:
$$
(y^2-3^2)(y^2-5^2)\cdot...\cdot(y^2-(2k-5)^2)(y^2-(2k-3)^2)>(y^2-1^2)(y^2-7^2)\cdot...\cdot(y^2-(2k-7)^2)(y^2-(2k-1)^2)
$$
for all $y\in\mathbb{R}$. I think now it would be enough to expand both sides and probably we would end up with a polynomial in $y^2$ where all coefficients are positive. However, computing all the coefficients will be quite tedious. How to prove it more elegantly?
Edit:
As several comments and answers use this same idea: simply reducing it to $(x-4r+2)(x-4r+1)=(x-4r)^2+3(x-4r)+2>(x-4r)^2+3(x-4r)=(x-4r+3)(x-4r)$ doesn't work as $a>c,b>d$ doesn't imply $ab>cd$ in general. But it was my first reflex too :)
|
$\begin{array}\\
(x-4i+2)(x-4i+1)-(x-4i+3)(x-4i)
&=((x-4i)^2+3(x-4i)+2)-((x-4i)^2+3(x-4i))\\
&=2\\
\end{array}
$
so each individual term is greater,
so their product is greater.
|
{
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"url": "https://math.stackexchange.com/questions/1866847",
"timestamp": "2023-03-29T00:00:00",
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|
Find the maximum value that the quantity $2m+7n$ can have
Find the maximum value that the quantity $2m+7n$ can have such that there exist distinct positive integers $x_i$ $(1 \leq i \leq m)$, $y_j$ $(1 \leq j \leq n)$ such that the $x_i$'s are even, the $y_j$'s are odd, and $\displaystyle \sum_{i = 1}^m x_i+\sum_{j=1}^n y_j = 1986$.
We see that $\displaystyle \sum_{i = 1}^m x_i \geq 2+4+\cdots+m = \dfrac{(m+2)\frac{m}{2}}{2} = \dfrac{(m+2)m}{4}$ and $\displaystyle \sum_{j=1}^n y_j \geq n^2$. What do we do from here to maximize $2m+7n$?
|
Alternatively, rewrite $m(m+1)+n^2\leq 1986$ as $(2m+1)^2+(2n)^2\leq 4\cdot 1986+1=7945$. By the Cauchy-Schwarz Inequality,
$$\begin{align}
(2m+7n)+1=(2m+1)+\frac{7}{2}(2n)
&\leq \sqrt{1^2+\left(\frac{7}{2}\right)^2}\,\sqrt{(2m+1)^2+(2n)^2}
\\
&\leq\sqrt{\frac{53}{4}}\,\sqrt{7945}<325\,.
\end{align}$$
Thus, $$2m+7n<324\,.$$ The equality condition for equality case of the Cauchy-Schwarz Inequality is $\frac{2m+1}{1}=\frac{2n}{7/2}$ and $(2m+1)^2+(2n)^2=7945$, which gives
$$(m,n)=\frac{1}{106}\left(-53\pm 2\sqrt{421085},\pm7\sqrt{421085}\right)\,.$$
The closest positive-integer pair $(m,n)$ that obeys $(2m+1)^2+(2n)^2\leq 7945$ is $$(m,n)=(11,43)\,,$$ for which $2m+7n=323$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof that $n = 3k + 5l$ for $n > 7$
Show that for every n greater than $7$, there are non-negative integers $k$ and $l$ such that $$n = 3k+ 5l.$$
So induction seems like a possibility.
$n = 3k + 5l$ and so $n + 1 = 3k + 5l + 1$.
What can be done in such a case? Can I get a hint?
|
By strong induction: We can write
\begin{align*}
8 & = 1 \cdot 3 + 1 \cdot 5\\
9 & = 3 \cdot 3 + 0 \cdot 5\\
10 & = 0 \cdot 3 + 2 \cdot 5
\end{align*}
Let $n \geq 10$. Assume that we can write each integer $m$ such that $8 \leq m \leq n$ in the form $m = 3k + 5l$ for some non-negative integers $k, l$. We wish to show that $n + 1$ can also be written in this form. Since $n + 1 \geq 10 + 1 = 11$, $n + 1 - 3 = n - 2 \geq 8$, by the induction hypothesis, there exist non-negative integers $k'$ and $l'$ such that $n - 2 = 3k' + 5l'$. Therefore, $$n + 1 = n - 2 + 3 = 3k' + 5l' + 3 = 3(k' + 1) + 5l'$$ Hence, each integer $n > 7$ can be expressed in the form $n = 3k + 5l$, where $k$ and $l$ are non-negative integers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1868321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
What approach should I use to solve integrals like this? $$\int{\sqrt{1-{x^3}}}dx$$
I tried with $t=x^3$ but then I have the $3x^2$ dt that I can't get rid of.
|
For any real number of $x$ ,
When $|x|\leq1$ ,
$\int\sqrt{1-x^3}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{3n}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{3n+1}}{4^n(n!)^2(1-2n)(3n+1)}+C$
When $|x|\geq1$ ,
$\int\sqrt{1-x^3}~dx$
$=i\int\sqrt{x^3-1}~dx$
$=i\int x^\frac{3}{2}\sqrt{1-\dfrac{1}{x^3}}~dx$
$=i\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(2n)!}{4^n(n!)^2(1-2n)x^{3n}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{i(2n)!x^{\frac{3}{2}-3n}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{i(2n)!x^{\frac{5}{2}-3n}}{4^n(n!)^2(1-2n)\left(\dfrac{5}{2}-3n\right)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{i(2n)!}{2^{2n-1}(n!)^2(2n-1)(6n-5)x^{3n-\frac{5}{2}}}+C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1871702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x + \csc x + \sec x}=\frac{t(1- t)}{1+ t^2}$$
$$\implies \int \frac{\tan {\frac{x}{2}}(1-\tan {\frac{x}{2}})}{1+\tan^2 {\frac{x}{2}}}{dx}$$
I think, I have made the things more difficult. How can I proceed further? Is there any better substitution for it?
|
$\displaystyle\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}=$
$\displaystyle\int\frac{\sin x\cos x}{1+\sin x+\cos x}\,dx=\int\frac{\sin x\cos x}{1+\sin x+\cos x}\cdot\frac{1-(\sin x+\cos x)}{1-(\sin x+\cos x)}\,dx$
$=\displaystyle\int\frac{\sin x\cos x(1-(\sin x+\cos x))}{1-(\sin x+\cos x)^2}\,dx=\int\frac{\sin x\cos x-\sin^2 x\cos x-\cos^2x\sin x}{-2\sin x\cos x}\,dx$
$\displaystyle=-\frac{1}{2}\int(1-\sin x-\cos x)\,dx=\frac{1}{2}(-x-\cos x+\sin x)+C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1871800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Positive semidefinite ordering for covariance matrices Suppose that X and Z are matrices with the same number of rows. Let
$$ D = \left[\begin{array}{cc} X' X & X'Z \\ Z'X & Z'Z \end{array} \right]^{-1} - \left[\begin{array}{cc} (X' X)^{-1} & 0 \\ 0 & 0 \end{array} \right],$$ where all inverses are assumed to exist and the zeros represent zero matrices of suitable dimensions. How can we prove that $D$ is positive semidefinite?
|
$U = \left(\begin{matrix} X^{T}X & X^TZ \\ Z^TX & Z^TZ \end{matrix}\right) = \left( \begin{matrix} X^T \\ Z^T\end{matrix}\right) \left(\begin{matrix} X & Z \end{matrix} \right) $ is clearly positive semidefinite and since we are given it is invertible it must be positive definite.
Consequently for any invertible matrix $P$ the matrix $PUP^T$ is also positive definite and in particular for $P=\left( \begin{matrix} I & 0 \\ -Z^TX(X^TX)^{-1} & I \end{matrix} \right)$ the matrix $$PUP^T = \left( \begin{matrix} X^TX & 0 \\ 0 & Z^TZ - Z^TX(X^TX)^{-1}X^TZ\end{matrix}\right)$$ is positive definite and hence both $X^TX$ and $Z^TZ - Z^TX(X^TX)^{-1}X^TZ = W$(say) must be symmetric positive definite matrices.
Inverting the above we have $(P^{-1})^{T} \left(\begin{matrix} X^{T}X & X^TZ \\ Z^TX & Z^TZ \end{matrix}\right)^{-1} P^{-1} = \left( \begin{matrix} (X^TX)^{-1} & 0 \\ 0 & 0\end{matrix}\right) + \left(\begin{matrix} 0 & 0 \\ 0 & W^{-1}\end{matrix}\right). $
Since $P^{-1} = \left( \begin{matrix} I & 0 \\ Z^TX(X^TX)^{-1} & I \end{matrix} \right) $ we also have $(P^{-1})^T \left( \begin{matrix} (X^TX)^{-1} & 0 \\ 0 & 0\end{matrix}\right) P^{-1} = \left( \begin{matrix} (X^TX)^{-1} & 0 \\ 0 & 0\end{matrix}\right) $.
So we have the identity
$$ (P^{-1})^{T}\left(\left(\begin{matrix} X^{T}X & X^TZ \\ Z^TX & Z^TZ \end{matrix}\right)^{-1} - \left( \begin{matrix} (X^TX)^{-1} & 0 \\ 0 & 0\end{matrix}\right) \right) P^{-1} = \left(\begin{matrix} 0 & 0 \\ 0 & W^{-1}\end{matrix}\right)$$
that is
$$
\left(\begin{matrix} X^{T}X & X^TZ \\ Z^TX & Z^TZ \end{matrix}\right)^{-1} - \left( \begin{matrix} (X^TX)^{-1} & 0 \\ 0 & 0\end{matrix}\right) = P^T \left( \begin{matrix} 0 & 0 \\ 0 & W^{-1} \end{matrix} \right) P. $$
Since $W$ is symmetric positive definite matrix, so is $W^{-1}$ and the right hand side is positive semi-definite.
|
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|
Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$ I am working on the following exercise. Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$.
I have a couple of ideas for going about this exercise.
$1)$ By moving $1$ to the other side of the equation we obtain:
$y^4-1=x^4-x^3+x^2-x \rightarrow (y^2-1)(y^2+1)=x(x-1)(x^2+1)$.
The LHS gives two consecutive integers I am able to see, but other than that I am stuck and not sure where to go from here.
$2)$ We notice that $1-x+x^2-x^3+x^4=\Phi_{10}(x)=y^4$.
We now have that $\Phi_{10}(x) \geq 0$ since $y=\pm \sqrt[4]{1-x+x^2-x^3+x^4}$.
In this case, with some computation, $x={0,1}$ are the only possibilities that will force $y \in \mathbb{Z}$. From there, I use induction to show that $x \geq 2$ and $x \leq -1$ do not yield a perfect fourth.
In either case, I keep running into some issues. If someone could please other a hint to help me continue through this exercise, that would be very helpful. Thank you.
Update:
Hello, my progress on the following problem is as follows:
Suppose that $x=y$, then we get $y^3-y^2+y-1=0 \rightarrow y=1, \pm i$. Thus we get two solutions $(x,y)={(1,1),(1,-1)}$, the $y=-1$ since $y^4$ was in the original equation.
Next, using @Batominovski hint we can write:
$(2x^2-x)^2 < (x^4-x^3+x^2-x+1) < 4(x^4-x^3+x^2-x+1) \leq (2x^2-x+2)^2 \forall x \in \mathbb{R}$ which is true for $x=0$.
Since the multiplication of a constant (in this case $k=4$) does not affect a solution from existing, we have that $x=0$ is a solution, consequently we find two more solutions are $(x,y)=(0,1),(0,-1)$.
|
One has $$1-x+x^{2}-x^{3}+x^{4}=y^{4}\iff(2x^2+1)^2=4y^4+4(x^3+x)-3$$ Notice that necessarily $y$ must be odd and $x^3+x$ is even when $x$ is even or odd. Hence one has $$(2x^2+1)^2\equiv 4y^4-3 \pmod 8$$ The solutions of this congruence are
$$(x,y)=(x,1),(x,3),(x,5),(x,7)\text { with } x=0,1,2,3,4,5,6,7$$ we can verify that $\color{red}{(x,y)=(0,\pm1),(1\pm 1)}$ are the only solutions because for $y=1+8m,3+8m,5+8m,7+8m$ we have for $m=0$ these only solutions (easier to verify for other candidates with $m\gt 1$making $1-x+x^{2}-x^{3}+x^{4}=\frac{x^5+1}{x+1}$)
|
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|
Fundamental group of $S^2 - \{N,S\}$ I would like to calculate the fundamental group $\Pi_1$ of $S^2 - \{N,S\}$, where $N,S$ are the north respectively south pole of $S^2$, but do not see yet how to do this.
|
The map \begin{align}\phi:S^2\setminus \{N,S\}&\to S^1\times (-1,1)\\ \begin{pmatrix}x\\ y\\ z\end{pmatrix}&\mapsto \frac{1}{\sqrt{x^2+y^2}}\begin{pmatrix}x\\ y\\ z\sqrt{x^2+y^2}\end{pmatrix}\end{align} is a diffeomorphism.
Its inverse is $$\begin{pmatrix}a\\ b\\ c\end{pmatrix}\mapsto \begin{pmatrix}a\sqrt{1-c^2}\\ b\sqrt{1-c^2}\\ c\end{pmatrix}$$
$S^1$ is a deformation retract of $S^1\times(-1,1)$.
|
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|
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac{\pi}{12} $
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{3} -\sqrt{2}}{1+ \sqrt{6}}\right)$$
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) -\tan^{-1}{\sqrt3} + \tan^{-1} {\sqrt2} $$
$$\implies\frac{\pi}{2} -\frac{\pi}{3}=\frac{\pi}{6}$$
Another possibility is
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) +\tan^{-1}{\sqrt3} - \tan^{-1} {\sqrt2} $$
How to solve this ?
|
$$\dfrac{\sqrt{5-2\sqrt6}}{1+\sqrt6}=\dfrac{\sqrt3-\sqrt2}{1+\sqrt3\cdot\sqrt2}$$
$$\implies\arctan\dfrac{\sqrt{5-2\sqrt6}}{1+\sqrt6}=\arctan\sqrt3-\arctan\sqrt2$$
$$\arctan\sqrt3=\dfrac\pi3$$ and
$$\arctan\sqrt2=\text{arccot}\dfrac1{\sqrt2}=\dfrac\pi2-\arctan\dfrac1{\sqrt2}$$
|
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|
If 'x' is real then $\frac{x^2-x+c}{x^2+x+2c}$ can take all real values if? the question asks for the interval in which c lies so that $\frac{x^2-x+c}{x^2+x+2c}$ gives all real values for all x belongs to R.
how to proceed in this problem?
|
Let $y = \frac{x^2-x+c}{x^2+x+2c} $ be a value taken by the given fraction. This gives
\begin{equation*}
x^2(y-1) +x(y+1) +c(2y-1) = 0
\end{equation*}
should have a real root. Thus
\begin{equation*}
(y+1)^2 -4c(y-1)(2y-1) \geq 0
\end{equation*}
and hence
\begin{equation*}
y^2(1-8c) + 2y(1+6c)+(1-4c) \geq 0
\end{equation*}
If the given expression takes all values, then the above should be true for all $y$. Thus we have
\begin{equation*}
(1+6c)^2 - (1-8c)(1-4c) \leq 0
\end{equation*}
and $8c \leq 1$. The above expression simplifies to $24c+4c^2 \leq 0$ and hence $-6 \leq c \leq 0$.
when $c=-6$, the graph is: Clearly, $y=1$ is an asymptote and hence the expression misses the value 1. Again, the same phenomenon is observed when $c=0$. Thus the required range is $-6 < c < 0$
For negative values outside the range, the graph is and for positive values of $c$, the graph is and for values within the range, the graph is
|
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|
Find the least squares solution to $A\vec{x} = \vec{b}$ using the transpose of $A$
Find the least squares solution to $A\vec{x} = \vec{b}$ using the transpose of $A$. Let $A = \begin{bmatrix}1&1&0\\1&0&-1\\0&1&1\\-1&1&-1\end{bmatrix}$ and $\vec{b} = \begin{bmatrix}2\\5\\6\\6\end{bmatrix}$
The formula says $x = (A^TA)^{-1} * A^T b$ so if they want me to use the transpose of $A$ should I do instead $x = (A^{T^T}A)^{-1} * (A^{T^T})b$?
|
$$A^T A = \left(
\begin{array}{ccc}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3 \\
\end{array}
\right)$$
$$(A^T A )^{-1} = \left(
\begin{array}{ccc}
\frac{1}{3} & 0 & 0 \\
0 & \frac{1}{3} & 0 \\
0 & 0 & \frac{1}{3} \\
\end{array}
\right)$$
$$A^T b = (1,14,-5)$$
$$(A^T A )^{-1} A^T b = \left(\dfrac{1}{3},\dfrac{14}{3},-\dfrac{5}{3}\right)$$
|
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|
Computing $\lim_{n \to \infty} \left(\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\cdots+\frac{n}{(n+n)^2}\right)$
Calculate $$\lim_{n \to \infty} \left(\dfrac{n}{(n+1)^2}+\dfrac{n}{(n+2)^2}+\cdots+\dfrac{n}{(n+n)^2}\right).$$
I tried turning this into a Riemann sum, but didn't see how since we get $\dfrac{1}{n} \cdot \dfrac{n^2}{(n+k)^2}$, which I don't see how relates.
|
I thought it might be instructive to present a solution that relies on the Euler-Maclaurin Summation Formula (EMSF). To that end, we proceed.
Note that we can write
$$\sum_{k=1}^{n}\frac{n}{(n+k)^2}=\sum_{k=n+1}^{2n}\frac{n}{k^2}$$
From the EMSF, we have
$$\begin{align}
\sum_{k=n+1}^{2n}\frac{n}{k^2}&=n\left(\int_n^{2n} \frac{1}{x^2}\,dx+\frac12\left(\frac{1}{(2n)^2}-\frac{1}{n^2}\right)+\frac16 \frac{\frac{-2}{(2n)^3}-\frac{-2}{n^3}}{2}-\frac{1}{30}\frac{\frac{-24}{(2n)^5}-\frac{-24}{n^5}}{24}+O\left(\frac{1}{n^7}\right)\right)\\\\
&=\frac12 -\frac{3}{8n}+\frac{7}{48n^2}-\frac{31}{960n^4}+O\left(\frac{1}{n^6}\right)
\end{align}$$
Obviously, the limit as $n\to \infty$ is $\frac12$ as expected. And we need not have carried out terms beyond those that were of order $n^{-1}$. However, the EMSF provides a powerful tool for analyzing a broader array of problems in which asymptotic expansions are developed.
|
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|
Can anyone please help me to find the generalized equation for $x$,$y$ and $z$? $$
A= P(9+2\cos(x)+2\cos(y)+2\cos(z))
$$
$$
B= P(9+2\cos(x-2\pi/3)+2\cos(y-2\pi/3)+2\cos(z-2\pi/3))
$$
$$
C= P(9+2\cos(x+2\pi/3)+2\cos(y+2\pi/3)+2\cos(z+2\pi/3))
$$
|
Observation: replacing $A,B,C$ with $A/p$, etc., we have
\begin{align}
A&= (9+2\cos(x)+2\cos(y)+2\cos(z))\\
B&= (9+2\cos(x-2\pi/3)+2\cos(y-2\pi/3)+2\cos(z-2\pi/3))\\
C&= (9+2\cos(x+2\pi/3)+2\cos(y+2\pi/3)+2\cos(z+2\pi/3))
\end{align}
Summing, we get
\begin{align}
A+B+C&= 27 + 2[\cos(x) + \cos(x-2\pi/3) + \cos(x + 2\pi/3] + \ldots \end{align}
where the ellipses denote similar expressions in $y$ and $z$. But this expression in brackets is always zero, so these equations cannot have a solution unless $A + B + C = 27$.
|
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|
If $x^{x^{x^{16}}}=16$ calculate the value of $x^{x^{x^{12}}}$ I have done the following but I'not satisfied.
if $x^a=a$ then by substitution follows that $x^a=x^{x^a}=x^{x^{x^a}}$ etc.
So $x^{x^{x^{16}}}=16$ is equivalent to $x^{16}=16$, and $x=2^{\frac{1}{4}}$
Substituting $x$ to $x^{x^{x^{12}}}$ returns $2$.
|
It's obvious that if $x=\sqrt[16]{16}$, then $x^{16} = 16$, and $x^{x^{16}} =16$ and $x^{x^{x^{16}}} = 16$, so $x=\sqrt[16]{16}$ solves your equation.
So then $x^{12} = {16^{0.75}} = 8$, and $x^{x^{12}} = x^8 = \sqrt{16} = 4$, and $x^{x^{x^{12}}} = x^4 = \sqrt[4]{16} = 2$. Hence $x^{x^{x^{12}}} = 2$
Ok hang on, I've got this beautiful approach!
Since $x^{x^{x^{16}}} = 16$, it follows that $x^{x^{x^{x^{x^{x^{16}}}}}} = 16$, and that $x^{x^{x^{x^{x^{x^{x^{x^{x^{16}}}}}}}}} = 16$, and so on , so forth, leading to the equation: $x^{x^{x^{x^{...}}}} = 16$. Now, if we look at $x$ exponentiated by both these equal exponents (I'm avoiding complex and negative values of $x$ then), then we get: $x^{x^{x^{x^{...}}}} = x^{16}$. From the previous two equations we get $x^{16}=16$, which give us $x=\sqrt[16]{16}$.
|
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|
The integral $\int_0^1 x^n \log(2-x) \, dx$ In this question I had stuck on the integral:
$$\int_0^1 x^n \log(2-x) \, {\rm d}x$$
Claude Leibovici says that $\displaystyle \int_0^1 x^n \log(2-x) \, {\rm d}x=\frac{\, _2F_1\left(1,n+2;n+3;\frac{1}{2}\right)}{2 (n+1) (n+2)}=a_n+b_n \log 2$ where $a_n, b_n$ are sequences of rational numbers. The statement seems to be true since if we run the integral for some values of $n$ we get that:
$$\begin{array}{||c|c|c|c|c||}
\hline
\text{Integral}&n & a_n & b_n & \text{Value} \\
\hline
\int_0^1 x \log(2-x) \, {\rm d}x&1 & -\frac{5}{4} & 2 & -\frac{5}{4}+ 2 \log 2 \\\\
\int_0^1 x^2 \log(2-x) \, {\rm d}x&2 & -\frac{16}{9} & \frac{8}{3} & -\frac{16}{9} + \frac{8}{3} \log 2 \\\\
\int_0^1 x^3 \log(2-x) \, {\rm d}x&3 & -\frac{131}{48} & 4& -\frac{131}{48} + 4 \log 2 \\
\hline
\end{array}$$
It seems that $a_n$ is a sequence of negative rational numbers and $b_n$ a sequence of positive rational numbers. However I don't see a pattern of how to connect those numbers. Does anyone see?
Maybe it is worth it to take a look at the more general case:
$$\int_0^1 x^n \log(\alpha -x) \, {\rm d}x \; \quad \mathbb{N} \ni \alpha \geq 2$$
That would be more tedious. What I had come as a solution but I could not see how to proceed was something like this:
\begin{align*}
\int_{0}^{1}x^n \log(\alpha -x) \, {\rm d}x &= \int_{0}^{1}x^n \log \left [ \alpha \left ( 1 - \frac{x}{\alpha} \right ) \right ] \, {\rm d}x \\
&=\int_{0}^{1}x^n \left [ \log \alpha + \log \left ( 1- \frac{x}{\alpha} \right ) \right ] \, {\rm d}x \\
&=\log \alpha \int_{0}^{1} x^n \, {\rm d}x + \int_{0}^{1} x^n \log \left ( 1- \frac{x}{\alpha} \right ) \, {\rm d}x \\
&= \frac{\log \alpha}{n+1} - \int_{0}^{1} x^n \sum_{n=1}^{\infty} \frac{\left ( \frac{x}{\alpha} \right )^n}{n} \, {\rm d}x \\
&= \frac{\log \alpha}{n+1} - \sum_{n=1}^{\infty} \frac{1}{n \alpha^n} \int_{0}^{1} x^{2n} \, {\rm d}x\\
&= \frac{\log \alpha}{n+1} - \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n \alpha^n (2n+1)} \\
&= ??
\end{align*}
It seems that:
$$\sum_{n=1}^{\infty} \frac{1}{n \alpha^n (2n+1)} = -2\sqrt{\alpha} \; {\rm arctanh} \left ( \frac{1}{\sqrt{\alpha}} \right ) - \log \left ( \frac{\alpha -1}{\alpha} \right ) +2$$
I guess that won't be too difficult to prove taking into account that:
$${\rm arctanh}(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} $$
|
We can easily develop a recursive formula for the integral of interest. To that end we proceed.
Let $I_n(\alpha)=\int_0^1 x^n \log(\alpha -x)\,dx$. Integrating by parts with $u=x^n$ and $v=(x-\alpha)\log(\alpha -x)-x$ yields
$$\begin{align}
I_n(\alpha)&=(1-\alpha)\log(\alpha-1)-1-n\int_0^1 x^{n-1}\left((x-\alpha)\log(\alpha -x)-x\right)\,dx\\\\
&=(1-\alpha)\log(\alpha-1)-nI_n(\alpha)+n\alpha I_{n-1}(\alpha)-\frac{1}{n+1}\\\\
&=\frac{(1-\alpha)\log(\alpha-1)+n\alpha I_{n-1}(\alpha)-\frac{1}{n+1}}{n+1}
\end{align}$$
Therefore, we find that
$$\bbox[5px,border:2px solid #C0A000]{I_n(\alpha)=\frac{(1-\alpha)\log(1-\alpha)}{n+1}-\frac{1}{(n+1)^2}-\frac{2n}{n+1}I_{n-1}(\alpha)}$$
For $\alpha =2$ this yields
$$\bbox[5px,border:2px solid #C0A000]{I_n(2)=\frac{2n}{n+1}I_{n-1}(2)-\frac{1}{(n+1)^2}}$$
|
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|
Proofs involving summations I am working on the following proof.
Prove that $\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} \geq \frac{n(n+1)(2n+1)}{6}$.
My first question is if the above is equivalent to saying "Prove that:$\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} \geq \displaystyle \sum_{n=1}^{t} \frac{n(n+1)(2n+1)}{6}$." If this was in fact correct, I examined the partial sums and found
$1+\frac{33}{4}+\frac{163}{5}+\frac{171}{2}+ \cdots \geq 1+2^2+3^2+4^2+ \cdots$. From here I see that when $n=1$ the inequality turns into an equality. My plan from here was to use some kind of induction to show that $\forall k_{i}, i,j \in \mathbb{N}_, k_{i} \geq n_{j}$.
Other than this method I do not see another way to go, could someone offer a hint of a different route or some help with my current route.
For my current route I am confused on the bounds for the second summation $\displaystyle \sum_{n=1}^{t} \frac{n(n+1)(2n+1)}{6}$.
Thank you.
|
I think what you are asking is $\frac{2k^4 + 1}{k + 2} \ge \frac{k(k+1)(2k+1)}{6}$.
$\frac{k(k+1)(2k+1)}{6}= \frac{k(k+1)(k+2)(2k+1)}{6(k+2)}=\frac{\frac 16(2k^4 + 7k^3 +7k^2 + 2k)}{k+2}$
So your question is true if and only if $2k^4 + 1 \ge \frac 16(2k^4 + 7k^3 +7k^2 + 2k)$
Which is true if and only if $ 10k^4 + 6 \ge 7k^3 + 7k^2 + 2k$
Which is true if and only if $10k + 6/k^4 \ge 7 + 7/k + 2/k^2$.
Which is true for $k = 1$ and as the LHS increases and the RHS decreases, this is always true.
Which is something I think the authors of the problem didn't notice.
So $\sum_{k=1}^{n}\frac{2k^4 + 1}{k + 2} = \sum_{k=1}^{n-1}\frac{2k^4 + 1}{k + 2}+ \frac{2n^4 + 1}{n+2}$
$\ge 0 + \frac{2n^4 + 1}{n+2} = \frac{2n^4 + 1}{n+2}$
$\ge \frac{n(n+1)(2n+1)}{6}$
|
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|
Contour inegration involving a confluent hypergeometric function with a pole at zero I have the following integral
\begin{equation}
\int_0^1 t^{-5/4} {_1}F_1(-1/4;1/2;t) dt
\end{equation}
I know that it only has a pole at zero since the confluent hypergeometric function is analytic in the entire complex plane but I dont know what contour to use.
|
This integration does not have a convergent result.
$$
\int t^{-5/4} {_1}F_1(-1/4;1/2;t) dt
= \int t^{-5/4} \sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k} \frac{t^k}{\Gamma(k+1)} dt \\=\sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k} \frac{\int t^{k-\frac{5}{4}}dt}{\Gamma(k+1)} =\sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k (k-\frac{1}{4})} \frac{t^{k-\frac{1}{4}}}{\Gamma(k+1)} $$
Since $k-\frac{1}{4} = \frac{3}{4}+k-1 = \frac{\Gamma(-\frac{1}{4})(\frac{3}{4})_k}{\Gamma(\frac{3}{4})(-\frac{1}{4})_k}$
$$=\sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k}\frac{\Gamma(-\frac{1}{4})(\frac{3}{4})_k}{\Gamma(\frac{3}{4})(-\frac{1}{4})_k} \frac{t^{k-\frac{1}{4}}}{\Gamma(k+1)} =\frac{\Gamma(-\frac{1}{4})}{\Gamma(\frac{3}{4}) \sqrt[4]{t}} \sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k}\frac{(\frac{3}{4})_k}{(-\frac{1}{4})_k} \frac{t^{k-\frac{1}{4}}}{\Gamma(k+1)} = \frac{\Gamma(-\frac{1}{4})}{\Gamma(\frac{3}{4}) \sqrt[4]{t}} {_2}F_2(-\frac{1}{4},-\frac{1}{4};\frac{1}{2},\frac{3}{4};t)$$
When $t \to 1$, $\frac{\Gamma(-\frac{1}{4})}{\Gamma(\frac{3}{4}) \sqrt[4]{1}} {_2}F_2(-\frac{1}{4},-\frac{1}{4};\frac{1}{2},\frac{3}{4};1) \approx -4.75045$ by wolframalpha.
When $t = 0$, ${_2}F_2(-\frac{1}{4},-\frac{1}{4};\frac{1}{2},\frac{3}{4};0)= 1$ and $\frac{\Gamma(-\frac{1}{4})}{\Gamma(\frac{3}{4})} =-4$. But on the denominator, the $\sqrt[4]{t} =0 $. So it becomes $\frac{-1}{0}$ form, which means the lower limit of integration doesnot exists.
So $\int_0^1 t^{-5/4} {_1}F_1(-1/4;1/2;t) dt$ does not converge.
|
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|
$\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square
Prove that there exist infinitely many positive integers $n$ such that $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square. Obviously, $1$ is the least integer having this property. Find the next two least integers with this property.
The given condition is equivalent to $(2n+2)(2n+1) = 12p^2$ where $p$ is a positive integer. Then since $\gcd(2n+2,2n+1) = 1$, we have that $2n+2 = 4k_1$ and $2n+1 = k_2$. We must have that $k_1$ is divisible by $3$ or that $k_2$ is divisible by $3$. If $k_1$ is divisible by $3$ and $k_2$ is not, then we must have that $k_1$ is divisible by $9$ and so $2n+2 = 36m$. Then we need $3mk_2$ to be a perfect square where $k_2+1 = 36m$. Thus if $3mk_2 = r^2$, we get $m = \dfrac{1}{72}\left(\sqrt{48r^2+1}+1\right)$.
|
From the equation $(2n+1)(2n+2) = 12p^2$, we know that there are two possibilities:
*
*$2n+2$ is of the form $4x^2$, $2n+1$ is of the form $3y^2$.
*$2n+2$ is of the form $12x^2$, $2n+1$ is of the form $y^2$.
This follows from $2n+1, 2n+2$ being relatively prime. Thus, solutions correspond to integer solutions of the equations $4x^2-3y^2 = 1$ and $12x^2-y^2 = 1$. We claim that the former equation has infinitely many solutions. These correspond to the solutions of $a^2-3b^2 = 1$ where $a$ is even.
The theory of Pell equations can be used to show that the solutions of $a^2-3b^2 = 1$ are given by the powers $a+b\sqrt{3} = (2+\sqrt{3})^i$ for $i\ge 0$. Even without this theory in hand, we can check directly that these give solutions:
$$
a^2-3b^2 = (a+b\sqrt{3})(a-b\sqrt{3}) = (2+\sqrt{3})^i(2-\sqrt{3})^i =
\big[(2+\sqrt{3})(2-\sqrt{3})\big]^i = 1
$$
It only remains to check that infinitely many of these solutions have $a$ as even. An easy inductive argument shows that $a$ is even precisely when $i$ is odd, so there are infinitely many solutions.
The first two nontrivial solutions are $(2+\sqrt{3})^3 = 26+15\sqrt{3}$ and $(2+\sqrt{3})^5 = 362+209\sqrt{3}$. These values of $a$ correspond to $n=337$ and $n=65521$. To check that these are indeed the smallest nontrivial solutions to $(2n+1)(2n+2)=12p^2$, there are two approaches. The first is to apply the theory of Pell equations to note that there are no solutions to $a^2-3b^2=1$ other than those given above, and that there are no solutions at all to $12x^2-y^2=1$. The second method is to check by brute force that no other values less than $65521$ yield solutions.
|
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|
How to find $\sqrt[3]{0.5964}$ using logarithms? \begin{align*}
\sqrt[3]{0.5964} &= \left(0.5964\right)^{\frac{1}{3}}\\
\log \sqrt[3]{0.5964} &= \frac{1}{3} \log 0.5964\\
&= \frac{1}{3} \cdot\overline{1}.7755\\
&= \frac{1}{3}\cdot\left(\overline{3} +2.7755\right)\\
&= \overline{1}.9252
\end{align*}
Can anyone explain what happened in line 4
Many Thanks
|
My guess is that $$\overline{1}.7755=-1+0.7755=-3+2.7755$$ Then divide by $3$ to get $$-1+0.9252=\overline{1}.9252$$ This is a way of keeping a positive mantissa.
|
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|
A question about substitute equivalent form into limit: $\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$ When I had the calculus class about the limit, one of my classmate felt confused about this limit:
$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$
What he thought that since $x^2 > x$ and $x^2 > 3x$ when $x \to \infty$ so the first square root must be $x$ and same for the second. Hence, the limit must be $0$.
It is obviously problematic.
And what I thought is that make prefect square under the limit, though I know the right solution is to rationalize the numerator.
After perfect-squaring, $$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}$$ I assert that since there is a perfect square and a square root. As $x \to \infty$, the constant does not matter. So
$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}= \lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2}-\sqrt{\left(x+\frac{3}{2}\right)^2}\\ = \frac{1}{2}-\frac{3}{2} = -1 $$
But when I conquer this limit: this post.
Here is my argument:
$\sqrt{ax^2+bx+c} =O(\sqrt a(x+\frac{b}{2a}))$ when $x \to \infty$
$$\lim_{x\to\infty}\frac{\sqrt{x^2-2x+3}+\sqrt{4x^2+5x-6}}{x+\sqrt{x^2-1}} =\frac{x-1+2(x+5/4)}{x+x} = \frac{3}{2}$$
Very concise get this answer but it gets downvoted.
I do not know what is wrong with my strategy. But in general case: for instance this problem about cubic root my strategy seems to work really efficient:
$$\lim_{x\to \infty} \sqrt[3]{x^3+6x^2+9x+1}-\sqrt[3]{x^3+5x^2+x+1}$$
My solution is:
$$\lim_{x\to \infty} \sqrt[3]{x^3+bx^2+cx+d} = \lim_{x\to \infty} \sqrt[3]{\left(x+\frac{b}{3}\right)^3}$$
So the limit becomes: $$\lim_{x\to \infty} \sqrt[3]{(x+2)^3+O(x)}-\sqrt[3]{\left(x+\frac{5}{3}\right)^3 + O(x)} =\lim_{x\to \infty} (x+2) -\left(x+\frac{5}{3}\right) = \frac{1}{3} $$
This result gets verified by wolframalpha.
To put all into a nutshell, what is wrong with my solution to these three problem. Is there any counterexample to this substitution. Any help, you will be appreciated.
|
Consider the general case $$A=\sqrt{x^2+ax+b}-\sqrt{x^2+cx+d}=x\left(\sqrt{1+\frac a x+\frac b {x^2}}-\sqrt{1+\frac c x+\frac d {x^2}}\right)$$ and use the fact that, for small $y$, using the generalized binomial theorem or Taylor series, $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ For the first radical, replace $y$ by $\frac a x+\frac b {x^2}$ and by $\frac c x+\frac d {x^2}$ for the second radical. You then obtain $$A=x\left(1+\frac{a}{2
x}+\frac{\frac{b}{2}-\frac{a^2}{8}}{x^2}+O\left(\frac{1}{x^3}\right)-\left(1+\frac{c}{2
x}+\frac{\frac{d}{2}-\frac{c^2}{8}}{x^2}+O\left(\frac{1}{x^3}\right)\right)\right)$$ $$A=\frac{a-c}{2}+\frac{-a^2+4 b+c^2-4 d}{8 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.
Doing the same with $$B=\sqrt[3]{x^3+ax^2+bx+c}-\sqrt[3]{x^3+dx^2+ex+f}$$ and using $$\sqrt[3]{1+y}=1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)$$ you should arrive to $$B=\frac{a-d}{3}+\frac{-a^2+3 b+d^2-3 e}{9 x}+O\left(\frac{1}{x^2}\right)$$
Doing the same with $$C=\sqrt[p]{x^p+a_1x^{p-1}+a_2x^{p-2}+\cdots}-\sqrt[p]{x^p+b_1x^{p-1}+b_2x^{p-2}+\cdots}$$ the limit will just be $$\frac{a_1-b_1} p$$
Your strategy works well for the limit because you just ignore the terms of degrees lower than $p-1$.
|
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|
How to prove the solution of these inequalities is empty? Prove: There does not exist 4 unit vectors $\mathbf{v}_1$, $\mathbf{v}_2$, $\mathbf{v}_3$, $\mathbf{v}_4$ in $\mathbb{R}^3$ such that
$$
\left \{ \
\begin{array}{ll}
\dfrac{4}{3} < \left \|\mathbf{v}_i + \mathbf{v}_j \right \| _2 ^2 < \dfrac{8}{3} \, , \\
\dfrac{4}{3} < \left \|\mathbf{v}_i - \mathbf{v}_j \right \| _2 ^2 < \dfrac{8}{3} \, ,
\end{array}
\ \ \forall\, 1\le i<j\le 4
\right.
.
$$
Note: It's preferred to prove it with algebra rather than high-level geometric concept.
|
Without loss of generality, take $$\mathbf{v}_1=(0,0,1)$$
Let $\displaystyle \mathbf{v}_{k}=\left( \sqrt{1-z_k^2}\cos \theta_{k},\sqrt{1-z_k^2}\sin \theta_{k},z_{k} \right)$ where $k=2,3,4$
$$|\mathbf{v}_1-\mathbf{v}_k|^2=2(1-z_k)$$
$$\frac{4}{3}< 2(1-z_k) < \frac{8}{3}$$
$$-\frac{1}{3} < z_{k} < \frac{1}{3}$$
To minimize $|\mathbf{v}_m-\mathbf{v}_n|$ for $2 \le m < n \le 4$, take $z_{k}=z$ and $\displaystyle \theta_{k}=\frac{2\pi k}{3}$
(We don't need to check for the case when $\theta_{m}=\theta_{n}$ since $|\mathbf{v}_{m}-\mathbf{v}_{n}|^2 < (\frac{2}{3})^2$ is smaller than the greatest lower bound.)
Now
$$|\mathbf{v}_m-\mathbf{v}_n|^2=3(1-z^2)$$
But $$\frac{8}{3} < 3(1-z^2) \le 3$$
That is $|\mathbf{v}_j-\mathbf{v}_k|$ is strictly larger than the least upper bound.
Contradiction.
|
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|
Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$.
Given
$$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$
$$a,b \in (0,\frac{\pi}{2})$$
Show that $$a+2b=\frac{\pi}{4}$$
Does exist any faster method of proving that, other than expanding $\sin{(a+2b)}$?
Thank You!
|
Just for fun... Adding a picture for the first time, so bear with me.
$\sin b = 1/\sqrt{10} \implies \cos b = 3/\sqrt{10}$ and $\sin 2b = 3/5$.
Now consider an isosceles right triangle $ABC$ as in the picture, with $AB = AC = 4$. Pick $D$ on $AC$ so that $AD = 3$, and $DE$ perpendicular to $BC$. $BD = 5$ by the Pythagorean theorem.
$\sin \angle DBA = 3/5$, $\angle ABC = \pi/4$, so it suffices to show $a = \angle DBE$. Now the area of $\triangle ABC$ is $8$ and that of $\triangle ABD$ is $6$, so the area of $\triangle DBC$ is $2$, and $BC = 4\sqrt 2$. From the formula for the area, $DE = 1/\sqrt 2$ , and $EB = 7/\sqrt2$ (by the Pythagorean theorem), $\tan \angle DBE = 1/7$, and the conclusion follows.
ADDED: $DE = 1/\sqrt2$ follows, more directly, from the fact that $\triangle ECD$ is isosceles and right-angled - no need for all the nonsense about areas.
|
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|
How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
|
You have $$\sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right)\text{.}$$ Before determining what this converges to, it is worth establishing for which values of $(a,b)$ it converges at all. To that end, consider the Ratio Test:
$$\left\lvert\frac{a^{n+1}\sum_{m=0}^{n+1}b^m}{a^n\sum_{m=0}^nb^m}\right\rvert=\left\lvert a\left(1+\frac{b^{n+1}}{\sum_{m=0}^nb^m}\right)\right\rvert\to\begin{cases}\lvert ab\rvert&\lvert b\rvert>1\\\lvert a\rvert&\lvert b\rvert=1\\\lvert a\rvert&\lvert b\rvert<1\\\end{cases}$$
So it converges when:
*
*$\lvert ab\rvert<1$ and $\lvert b\rvert>1$
*$\lvert a\rvert<1$ and $\lvert b\rvert\leq1$
And diverges when:
*
*$\lvert ab\rvert>1$ and $\lvert b\rvert>1$
*$\lvert a\rvert>1$ and $\lvert b\rvert\leq1$
This test has been inconclusive when:
*
*$\lvert ab\rvert=1$ and $\lvert b\rvert>1$
*$\lvert a\rvert=1$ and $\lvert b\rvert\leq1$
Let's look at the last case first. If $\lvert a\rvert=1$, then you have $\sum_{n=0}^\infty {\pm_n\left(\sum_{m=0}^nb^m\right)}\text{.}$ The terms of such a series, $\pm_n\left(\sum_{m=0}^nb^m\right)$, do not approach $0$ no matter what $b$ is. So your series will diverge.
What if $\lvert b\rvert>1$ and $\lvert a\rvert=\frac{1}{\lvert b\rvert}$? Since $\lvert b\rvert>1$ we may write $\sum_{m=0}^nb^m=\frac{b^{n+1}-1}{b-1}$, and your series is $\sum_{n=0}^\infty {\pm_n\left(\frac{b^{n+1}-1}{b^n(b-1)}\right)}\text{.}$ Again using $\lvert b\rvert>1$, the terms of this series do not converge to $0$, so this series would be divergent.
Now we know the only situations where your series converges are:
*
*$\lvert ab\rvert<1$ and $\lvert b\rvert>1$ (which implies $\lvert a\rvert<1$)
*$\lvert a\rvert<1$ and $\lvert b\rvert\leq1$ (which implies $\lvert ab\rvert<1$)
When $b\neq1$, both situations give:
$$\begin{align}
\sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right)
&=\sum_{n=0}^\infty a^n\frac{1-b^{n+1}}{1-b}\\
&=\frac{1}{1-b}\left(\sum_{n=0}^\infty a^n-b\sum_{n=0}^\infty (ab)^n\right)\\
&=\frac{1}{1-b}\left(\frac{1}{1-a}-\frac{b}{1-ab}\right)\\
&=\frac{1}{(1-a)(1-ab)}\\
\end{align}$$
And when $b=1$ with $\lvert a\rvert<1$,
$$\begin{align}
\sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right)
&=\sum_{n=0}^\infty (n+1)a^n\\
&=\left.\sum_{n=0}^\infty \frac{d}{dx}x^{n+1}\right|_{x=a}\\
&=\left.\frac{d}{dx}\sum_{n=0}^\infty x^{n+1}\right|_{x=a}\\
&=\left.\frac{d}{dx}\left(\frac{1}{1-x}-1\right)\right|_{x=a}\\
&=\left.\frac{1}{(1-x)^2}\right|_{x=a}\\
&=\frac{1}{(1-a)^2}\\
\end{align}$$
Which conveniently agrees with the formula $\frac{1}{(1-a)(1-ab)}$.
|
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|
Problem solving recurrence equation for $a_n = a_{n-1} + 2a_{n-2}, \,\,\, a_0=a_1=1$ I want to try the generating function $A(x)$ and a formula for $a_n$ for the following sequence:
$$a_n = a_{n-1} + 2a_{n-2}, \,\,\, a_0 = a_1 = 1$$
So I multiplied the above equation on both sides by $x^n$ and summed them for each $n \geq 2$. So:
$$\sum_{n=2}^{\infty} a_nx^n = \sum_{n=2}^{\infty} a_{n-1}x^n + 2\sum_{n=2}^{\infty} a_{n-2}x^n$$
$$\Longrightarrow \sum_{n=2}^{\infty} a_nx^n= \frac{1}{x}\sum_{n=1}^{\infty} a_{n}x^n + 2\frac{1}{x^2}\sum_{n=0}^{\infty} a_{n}x^n$$
$$\Longrightarrow A(x) - x - 1 = \frac{A(x)-1}{x} + 2\frac{A(x)}{x^2}$$
$$\Longrightarrow x^2A(x) - x^3 - x^2 = xA(x)-x + 2A(x)$$
$$\Longrightarrow A(x)(x^2-x-2) = x^3 + x^2 + x$$
$$\Longrightarrow A(x) = \frac{x(x^2 + x + 1)}{(x-2)(x+1)} = \frac{1}{3} \cdot \frac{1}{1+x} + \frac{14}{3} \cdot \frac{1}{2-x}$$
Now if I search for the series expansion, I find:
$$A(x) = \sum_{n=0}^{\infty} (\frac{1}{3} \cdot (-1)^n - \frac{14}{3} \cdot \frac{1}{2^{n+1}})x^n$$
$$\Longrightarrow a_n = \frac{1}{3} \cdot (-1)^n - \frac{14}{3} \cdot \frac{1}{2^{n+1}}$$
However, this is obviously wrong. I checked on Wolfram, and the correct answer should be:
$$a_n = \frac{1}{3} ((-1)^n+2^{n+1})$$
I redid the whole thing twice, but I got the same result, so I guess that there is some flaw in my logic, rather than the calculations (I could be wrong though).
|
I corrected the mistake in the second line of the calculations and got the following:
$$\Longrightarrow \sum_{n=2}^{\infty}a_{n}x^{n} = \sum_{n=2}^{\infty}a_{n-1}x^{n} + 2\sum_{n=2}^{\infty}a_{n-2}x^{n}$$
$$\Longrightarrow \sum_{n=2}^{\infty}a_{n}x^{n} = x\sum_{n=1}^{\infty}a_{n}x^{n} + 2x^2\sum_{n=0}^{\infty}a_{n}x^{n}$$
$$\Longrightarrow A(x) - x - 1= x(A(x)-1) + 2x^2A(x)$$
$$ \Longrightarrow A(x) = \frac{-1}{2x^2 + x -1 } = -\frac{2}{3} \cdot \frac{1}{2x - 1} + \frac{1}{3} \cdot \frac{1}{x + 1}$$
Which gives me the series expansion:
$$ A(x) = -\frac{2}{3}\sum_{n = 0}^{\infty} -2^{n}x^n + \frac{1}{3} \sum_{n=0}^{\infty} (-1)^nx^n$$
$$ = \sum_{n = 0}^{\infty} \frac{1}{3} ( 2^{n+1} + (-1)^n)x^n $$
$$\Longrightarrow \forall n \in \mathbb{N}: a_n = \frac{1}{3} ((-1)^n + 2^{n+1})$$
Which works. Thank you, guys.
|
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|
How do I integrate this. How can I find the integral of
$$\int\frac{x+2}{(x^2+2x+2)\sqrt{x+1}} dx $$
So that the answer=$\sqrt{2}\tan^{-1}(\frac{x}{\sqrt{2(x+1)}})+c$
|
First of all, just substitute $u=\sqrt{x+1}$ and $\text{d}u=\frac{1}{2\sqrt{x+1}}\space\text{d}x$, then we get:
$$\text{I}=\int\frac{x+2}{(x^2+2x+2)\sqrt{x+1}}\space\text{d}x=2\int\frac{u^2+1}{u^4+1}\space\text{d}u$$
Now, use partial fractions:
$$\frac{u^2+1}{u^4+1}=\frac{1}{2(u^2+u\sqrt{2}+1)}-\frac{1}{2(u\sqrt{2}-u^2-1)}$$
So, we get:
$$\text{I}=2\int\frac{u^2+1}{u^4+1}\space\text{d}u=\int\frac{1}{u^2+u\sqrt{2}+1}\space\text{d}u-\int\frac{1}{u\sqrt{2}-u^2-1}\space\text{d}u=$$
$$\int\frac{1}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}}\space\text{d}u-\int\frac{1}{-\left(u-\frac{1}{\sqrt{2}}\right)^2-\frac{1}{2}}\space\text{d}u$$
Now, for $\int\frac{1}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}}\space\text{d}u$, substitute $s=u+\frac{1}{\sqrt{2}}$ and $\text{d}s=\text{d}u$ and for $\int\frac{1}{-\left(u-\frac{1}{\sqrt{2}}\right)^2-\frac{1}{2}}\space\text{d}u$, substitute $w=u-\frac{1}{\sqrt{2}}$ and $\text{d}w=\text{d}u$:
$$\text{I}=2\int\frac{1}{2s^2+1}\space\text{d}s+2\int\frac{1}{2w^2+1}\space\text{d}w$$
|
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|
Prove that $ABC$ is right-angled Prove that if $\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$, then $ABC$ is right-angled.
I only found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$, but I have no idea what to do next.
Thank you in advance for your answers!
|
$$\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$$
Use $\cos^2{A} =\frac{1+\cos2A}2$. Then
$$\cos2A+\cos2B+\cos2C=-1$$
Use $\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C$
Then $$-1-4\cos A\cos B\cos C=-1$$
$$4\cos A\cos B\cos C=0$$
Then $\angle A$ or $\angle B$ or $\angle C$ is $\frac{\pi}{2}$
|
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|
Find $\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$
Find
$$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$$
My attempt :-
$$3(x-1)+2\sqrt{2x^2 - 7x - 4}$$
$$\implies 3x - 3 + 2\sqrt{2x^2 - 7x - 4} + 2x^2 - 2x^2 -7x + 7x - 1 +1 $$
$$\implies 2x^2 -7x - 4 + 2\sqrt{2x^2 - 7x - 4} + (- 2x^2 + 10x + 1) $$
let $y = \sqrt{2x^2 - 7x - 4}$
$$\therefore y^2 + 2y + (- 2x^2 + 10x + 1)$$
solving for $y$ :-
$$y = {-2\pm \sqrt{4 - 4\times(- 2x^2 + 10x + 1)} \over 2} $$
$$\implies y = {-1\pm \sqrt{2x^2 - 5x}} $$
$$\therefore y^2 + 2y + (- 2x^2 + 10x + 1) = (y + 1 + \sqrt{2x^2 - 5x})(y + 1 - \sqrt{2x^2 - 5x})$$
$$\therefore \sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}} = \sqrt{(y + 1 + \sqrt{2x^2 - 5x})(y + 1 - \sqrt{2x^2 - 5x})\over 2}$$
Sadly this does not remove the square root.
So the question is how can i factorize $3(x-1)+2\sqrt{2x^2 - 7x - 4}$ to remove that square root ?
just some hints are fine with me, thanks.
|
$$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}=$$
$$=\sqrt{\frac{3(x-1)+2\sqrt{2x^2 - 7x - 4}}2}=$$
$$=\sqrt\frac{\color{red}{3x-3}+2\sqrt{(2x+1)(x-4)}}2=$$
$$=\sqrt{\frac{\color{red}{2x+1}+2\sqrt{(2x+1)(x-4)}+\color{red}{x-4}}2}=$$
$$=\sqrt{\frac{(\sqrt{2x+1}+\sqrt{x-4})^2}2}=$$
$$=\frac{\sqrt{2x+1}+\sqrt{x-4}}{\sqrt2}$$
|
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|
If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.
If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
positive integer $k$.
This is just curiosity no motivation just rummaging through some old question in a notebook. A simple counter example would work.
Note if $\mathfrak a(n) =\binom{2^n}{2}+1$ then $7$ divides $\mathfrak a(n)$ for $n=2,5,8,11,14,17,\ldots$ and from this I am guessing that $n=3k+2$
|
$$\binom{2^n}2+1=\frac{2^n(2^n-1)}2+1=2^{2n-1}-2^{n-1}+1$$
Now, let's apply mod $7$. Note that $2^3\equiv 1\pmod 7$.
If $n=3k$ we have $4-4+1\neq 0$.
If $n=3k+1$ we have $2-1+1\neq 0$.
This proves your statement, but we can check easily if the "iff" statement is true:
If $n=3k+1$ we kave $1-2+1=0$.
So $7$ divides $\binom{2^n}n+1$ if and only if $n\equiv 2\pmod 3$.
|
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|
Is this a new method for finding powers? Playing with a pencil and paper notebook I noticed the following:
$x=1$
$x^3=1$
$x=2$
$x^3=8$
$x=3$
$x^3=27$
$x=4$
$x^3=64$
$64-27 = 37$
$27-8 = 19$
$8-1 = 7$
$19-7=12$
$37-19=18$
$18-12=6$
I noticed a pattern for first 1..10 (in the above example I just compute the first 3 exponents) exponent values, where the difference is always 6 for increasing exponentials. So to compute $x^3$ for $x=5$, instead of $5\times 5\times 5$, use $(18+6)+37+64 = 125$.
I doubt I've discovered something new, but is there a name for calculating exponents in this way? Is there a proof that it works for all numbers?
There is a similar less complicated pattern for computing $x^2$ values.
|
For a little bit more, see the answer "General method for indefinite summation" which explains how exactly this representation using forward differences allows you to easily find the formula for indefinite summation of powers. Applied to your case you get:
0, 1, 8, 27
1, 7, 19
6, 12
6
and hence:
$n^3 = 0 \binom{n}{0} + 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$.
which immediately gives:
$
\newcommand\lfrac[2]{{\large\frac{#1}{#2}}}
$
$\sum_{k=0}^{n-1} k^3 = 0 \binom{n}{1} + 1 \binom{n}{2} + 6 \binom{n}{3} + 6 \binom{n}{4} = n\lfrac{n-1}{2}(1+\lfrac{n-2}{3}(6+\lfrac{n-3}{4}(6)))$
$\ = \lfrac{n^2 (n-1)^2}{4}$.
and then, if you prefer the indices to end at $n$:
$\sum_{k=1}^n k^3 = \lfrac{(n+1)^2 n^2}{4}$.
As you can see, hardly any computation was necessary to get this result!
|
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|
Expressing products as the sum of two squares Express $(1+x^2)(1+y^2)(1+z^2)$ as the sum of two squares.
I have no clue where to start, I know that we should not post just for solution but a hint will definitely help.
|
By Lagrange's identity
$$ (a^2+b^2)(c^2+d^2) = (ad-bc)^2 + (ac+bd)^2 \tag{1}$$
it follows that numbers that are a sum of two squares form a semigroup.
A straightforward proof of $(1)$ is to consider that $a^2+b^2$ is the norm of $a+bi$ in $\mathbb{C}=\mathbb{R}[i]$.
The norm is multiplicative and $(a-bi)(c+di)=(ac+bd)+(ad-bc)i$. An equivalent alternative is to exploit Binet's theorem $\det(AB)=\det(A)\det(B)$ through the matrices
$$ A = \begin{pmatrix}a & -b \\ b & a\end{pmatrix},\qquad B = \begin{pmatrix}c & d \\ -d & c\end{pmatrix},\quad AB=\begin{pmatrix}ac+bd & ad-bc \\ bc-ad &ac+bd\end{pmatrix}.\tag{2} $$
One way or another, since
$$ (i+x)(i+y)(i+z)=(xyz-x-y-z)+i(xy+xz+yz-1)\tag{3} $$
it follows that:
$$ \boxed{\phantom{\sum_{i=0}^{10}}(1+x^2)(1+y^2)(1+z^2)=\color{red}{(xyz-x-y-z)^2+(xy+xz+yz-1)^2}.\phantom{aa}}\tag{4} $$
|
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|
Prove that $a\equiv b\pmod p\Rightarrow a^p \equiv b^p \pmod {p^2}$ Suppose $p \nmid b$ and $p \nmid a$. Let $a^p \equiv b^p (\text{mod}(p))$. Prove that $a^p \equiv b^p (\text{mod}(p^2))$.
Below is my proof. I'm interested in two things.
1) Verifing this proof
2) Finding a simpler proof (preferably one that doesn't involve the Binomial Theorem).
Suppose $a^p \equiv b^p (\text{mod}(p))$.
Since $a^{p - 1} \equiv b^{p - 1} \equiv 1 (\text{mod}(p))$ By Fermat's Little Theorem, we have $a^p \equiv a \equiv b^p \equiv b (\text{mod}(p))$.
Since $a \equiv b (\text{mod}(p))$ we can write $ a = pk + n$ and $b = pm + n$ where $k$ and $m$ are non-negative integers and $ 1 \leq n \leq p - 1$.
We then have $a^p = (pk + n)$ and $b^p = (pm + n)^p$.
Using the Binomial Theorem we can expand these. Looking at these expanded binomials modulo $p^2$ we see that any terms with powers of $p$ larger than 2 are 0 modulo $p^2$. Thus we are only concerned with the last two terms in the expansion. The expansions are displayed below:
$a^p = (pk + n)^p \equiv \frac{p!}{(p-1)!}n^{p-1}(pk) + n^p \equiv n^p\mod{p^2}$
and
$b^p = (pm + n)^p \equiv \frac{p!}{(p-1)!}n^{p-1}(pm) + n^p \equiv n^p\mod{p^2}$
Thus $a^p \equiv b^p \mod{p^2}$
I would really appreciate any critique on the proof including the formatting. Thanks.
|
Simpler proof:
By Fermat $p \ | \ a-b$, then $a \equiv b$ in modulo $p$
\begin{align*}
\frac{a^p-b^p}{a-b} &= (a^{p-1} + a^{p-2}b + \cdots b^{p-2}a + b^{p-1})\\
&\equiv (b^{p-1} + b^{p-2}b + \cdots b^{p-2}b + b^{p-1})\\
&\equiv p \ b^{b-1} \equiv 0
\end{align*}
in modulo $p$
Since $\frac{a^p-b^p}{a-b}$ and $a-b$ are divisible by $p$. The product $a^p-b^p$ is divisible by $p^2$.
|
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|
why exist $a_{i},a_{j},a_{k}(1\le iAssmue that postive real numbers $a_{1},a_{2},a_{3},a_{4}$ such
$$2(a^4_{1}+a^4_{2}+a^4_{3}+a^4_{4})<33a_{1}a_{2}a_{3}a_{4}$$
show that
There exist $a_{i},a_{j},a_{k}(1\le i<j<k\le 4)$ are side lengths of a triangle
if $a,b,c$ is trangle three side ,then we have
$$\sum a^4<2\sum a^2b^2$$
|
Let $a_4\geq a_3\geq a_2\geq a_1$ and $a_4=a$, $a_3=b$, $a_2=c$, $a_1=d$ and $c=xd$.
Hence, $x\geq1$.
Let $a\geq b+c$, $b\geq c+d$ and $f(a)=2(a^4+b^4+c^4+d^4)-33abcd$.
Hence, $f'(a)=8a^3-33bcd\geq8(b+c)^3-33bcd>0$, which says that
$$f(a)\geq f(b+c)=2(b+c)^4+2b^4+2c^4+2d^4-33(b+c)bcd$$
Let $g(b)=2(b+c)^4+2b^4+2c^4+2d^4-33(b+c)bcd$.
Hence, $g'(b)=8(b+c)^3+8b^3-66bcd-33c^2d$ and $g''(b)=24(b+c)^2+24b^3-66cd>0$,
which says that $g'(b)\geq8(2c+d)^3+8(c+d)^3-66(c+d)cd-33c^2d>0$,
which says that $$2(a^4+b^4+c^4+d^4)-33abcd\geq g(b)\geq g(c+d)=$$
$$=2(2c+d)^4+2(c+d)^4+2c^4+2d^4-33(2c+d)(c+d)cd=$$
$$=d^4\left(2(2x+1)^4+2(x+1)^4+2x^4+2-33(2x+1)(x+1)x\right)=$$
$$=3d^4(x-1)(3x+2)(4x^2+2x-1)\geq0$$
which is contradiction!
This says that $a<b+c$ and $a$,$b$ and $c$ are sides-lengths of triangle or
$b<c+d$, which says that $b$, $c$ and $c$ are sides-lengths of triangle.
Done!
|
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|
Prove that $z^n + 1/{z^n}$ is a rational number Knowing that $z + \frac{1}{z} = 3$ ($z \in \mathbb{R}$), prove that $z^n + \frac{1}{z^n} \in \mathbb{Q}$ ($n \in \mathbb{N}$ and $n \geq 1$).
I've tried to find a general formula for $z^n + \frac{1}{z^n}$.
$$\left(z + \frac{1}{z}\right)^n = \binom{n}{0}z^n + \binom{n}{1}z^{n-2} + ... + \binom{n}{n-1}z^{2-n} + \binom{n}{n}z^{-n}$$
$$\left(z + \frac{1}{z}\right)^n = \binom{n}{0}(z^n + z^{-n}) + \binom{n}{1}(z^{n-2} + z^{2-n}) + ...$$
However, this formula seems to be wrong and I currently have no idea how to correct it and what to do next.
Thank you in advance for your help!
|
Here is an uglier solution:
$$z+\frac{1}{z}=3 \Rightarrow z^2-3z+1=0 \Rightarrow z=\frac{3\pm \sqrt{8}}{2}$$
Note that $\frac{3-\sqrt{8}}{2}=\frac{1}{\frac{3+\sqrt{8}}{2}}$, thus $\{ z, \frac{1}{z} \} = \{ \frac{3+ \sqrt{8}}{2}, \frac{3- \sqrt{8}}{2} \}$.
Then
$$z^n+\frac{1}{z^n}=\left( \frac{3 + \sqrt{8}}{2} \right)^n + \left( \frac{3 - \sqrt{8}}{2} \right)^n= \frac{1}{2^n} \left(\left( 3 + \sqrt{8} \right)^n + \left( 3 - \sqrt{8} \right)^n \right)\\
=\frac{1}{2^n} \left(\sum_{k=0}^n \binom{n}{k} 3^{n-k} \sqrt{8}^k+\binom{n}{k} 3^{n-k} (-1)^k\sqrt{8}^k\right)\\
$$
Now, just observe that when $k$ is odd, the term inside the sum is $0$. This shows that
$$z^n+\frac{1}{z^n}=
\frac{1}{2^n} \left(\sum_{k=0}^{\frac{n}{2}} \binom{n}{2k} 2 \cdot 3^{n-2k} 8^k\right)\\
$$
|
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|
How to prove each element of the following sequence is a perfect square? Sequence $\{a_n\}$ satisfies the following formula:
$a_{n+2}=14a_{n+1}-a_n+12$,
and $a_1=1, a_2=1$.
It is easy to check that $a_3=25$ and $a_4=361$.
The question is how to prove each element of the sequence $\{a_n\}$ is a perfect square?
|
As Umberto P. commented, it is sufficient to prove by induction that
$$a_{n+2}=b_n^2\tag1$$
where $b_{n+2}=4b_{n+1}-b_{n}$ with $b_0=1, b_1=5$.
The base case : $a_2=1=b_0^2$ and $a_3=25=b_1^2$
Suppose that $a_{n+2}=b_n^2$ and $a_{n+1}=b_{n-1}^2$.
Then, using that
$$b_n^2-4b_nb_{n-1}+b_{n-1}^2=6\tag2$$
we get
$$\begin{align}a_{n+3}&=14a_{n+2}-a_{n+1}+12\\&=14b_{n}^2-b_{n-1}^2+12\\&=16b_n^2-8b_nb_{n-1}+b_{n-1}^2-2b_n^2+8b_nb_{n-1}-2b_{n-1}^2+12\\&=(4b_n-b_{n-1})^2-2(b_n^2-4b_nb_{n-1}+b_{n-1}^2-6)\\&=b_{n+1}^2\qquad\blacksquare\end{align}$$
Let us prove $(2)$ by induction.
The base case : $b_1^2-4b_1b_0+b_0^2=25-20+1=6$
Suppose that $b_n^2-4b_nb_{n-1}+b_{n-1}^2=6$. Then,
$$\begin{align}b_{n+1}^2-4b_{n+1}b_{n}+b_{n}^2&=(4b_{n}-b_{n-1})^2-4(4b_{n}-b_{n-1})b_n+b_n^2\\&=16b_{n}^2-8b_{n}b_{n-1}+b_{n-1}^2-16b_n^2+4b_nb_{n-1}+b_n^2\\&=b_n^2-4b_nb_{n-1}+b_{n-1}^2\\&=6\qquad\blacksquare\end{align}$$
|
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Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
|
$$(x+y)^2=3^2$$
$$x^2+2xy+y^2=9$$
$$x^2+y^2=9-2xy$$
square both sides
$$x^4+2x^2y^2+y^4=(9-2xy)^2$$
$$x^4+y^4=(9-2xy)^2-2x^2y^2=\color{red}{-9}$$
|
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|
Find a and b such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$. Find $a$ and $b$ such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$.
That is what the question states, I think you should factor
Any Ideas on how to begin?
|
By factorization,
$$x^2+8x+7 \equiv (x+7)(x+1) \equiv (x+a)(x+b)$$
$$\therefore \quad (a,b)=(1,7) \quad \text{or} \quad (7,1)$$
|
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|
Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$? Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which yields:
$$\frac{1}{3} > \sqrt{x}\sqrt{1 -x}.$$
Squaring it again we get inequality:
$$9x^2 - 9x + 1 > 0,$$
solution to which is a domain $[0, \frac{3 - \sqrt{5}}{6}) \cup (\frac{3 + \sqrt{5}}{6}, 1]$. But, solution to original inequality is just $[0, \frac{3 - \sqrt{5}}{6})$. What am I omitting/where I'm doing mistakes?
|
The number $\sqrt{1-x}-\sqrt{x}$ can be negative and, in this case, the inequality doesn't hold. Squaring both sides can be done only if both sides are nonnegative. Thus your inequality becomes
$$
\begin{cases}
0\le x\le 1 \\[8px]
\sqrt{1-x}-\sqrt{x}\ge0 \\[4px]
1-x-2\sqrt{x(1-x)}+x>\dfrac{1}{3}
\end{cases}
$$
The second inequality becomes $1-x\ge x$, so $x\le1/2$. The third inequality is
$$
\sqrt{x(1-x)}<\frac{1}{3}
$$
so $9x^2-9x+1>0$ that has the solutions
$$
x<\frac{3-\sqrt{5}}{6}\qquad\text{or}\qquad x>\frac{3+\sqrt{5}}{6}
$$
that should be combined with $0\le x\le 1/2$. Thus the solution set of the original equation is
$$
\left[0,\frac{3-\sqrt{5}}{6}\right)
$$
because $(3+\sqrt{5})/6>1/2$.
A different strategy is to write the inequality as
$$
\sqrt{3}\sqrt{1-x}>1+\sqrt{3}\sqrt{x}
$$
With the constraint $0\le x\le1$, this becomes
$$
3-3x>1+2\sqrt{3}\sqrt{x}+3x
$$
or
$$
3x+\sqrt{3}\sqrt{x}-1<0
$$
Setting $t=\sqrt{x}$ (with $0\le t\le1$), this is quadratic: $3t^2+\sqrt{3}t-1<0$. So we have the solution set
\begin{cases}
0\le t \le 1 \\[8px]
\dfrac{-\sqrt{3}-\sqrt{15}}{6}<t<\dfrac{-\sqrt{3}+\sqrt{15}}{6}
\end{cases}
that becomes
$$
0\le t<\dfrac{-\sqrt{3}+\sqrt{15}}{6}
$$
and therefore
$$
0\le x<\left(\dfrac{-\sqrt{3}+\sqrt{15}}{6}\right)^2=\frac{3-\sqrt{5}}{6}
$$
|
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|
Doubt about asymptotic analysis $\frac{1}{x^{a}*(4+9x)^{b+1}}\sim \frac{1}{9^{b+1}x^{a+b+1}}$ I don't understand why is correct
$$a,b\in \mathbb{R}$$
$$x \mapsto \infty $$
$$\frac{1}{x^{a}*(4+9x)^{b+1}}\sim \frac{1}{9^{b+1}x^{a+b+1}}$$
I would write
$$x \mapsto \infty $$
$$\frac{1}{(4+9x)^{b+1}}\sim \frac{1}{9x^{b+1}}$$
so
$$\frac{1}{x^{a}*(4+9x)^{b+1}}\sim \frac{1}{9x^{a+b+1}}$$
Where I made a mistake?
Someone can help me?
Thanks
|
The problem lies in
$$\frac{1}{(4+9x)^{b+1}}\sim \frac{1}{9x^{b+1}}.$$
It should be
$$\frac{1}{(4+9x)^{b+1}}\sim \frac{1}{(9x)^{b+1}}$$
because
$$\lim_{x\to\infty}\frac{(9x)^{b+1}}{(4+9x)^{b+1}}=1.$$
|
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can we express $x=f(y)$, if $y=\exp(x/1.45)(1-\exp(-x))$? Given $y=\exp(x/1.45)(1-\exp(-x))$, I need to express $x = f(y)$.
I tried taking log of both side but didnt get any fruitful expression.
Is it possible? If yes, how?
|
Here are the details of the Lagrange Inversion computation. We have
$$y = \exp(bx)(1-\exp(-x)).$$
where $b=20/29.$ Putting $\exp(x) = t$ so that $x = \log t$ we get
$$y = t^b (1-1/t).$$
We seek to compute
$$[y^n] t(y) =
\frac{1}{2\pi i}
\int_{|y|=\epsilon} \frac{1}{y^{n+1}} t(y) dy.$$
Using the definition we put with $w$ being complex
$$y = w^b (1-1/w) = w^{b-1} (w-1)$$
so that
$$dy = (b w^{b-1} - (b-1) w^{b-2}) \; dw.$$
We get for the integral
$$\frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{w^{(b-1)(n+1)}}
\frac{1}{(w-1)^{n+1}}
\\ \times w \times
w^{b-2} (bw - (b-1))
\; dw.$$
This has two pieces, the first is
$$\frac{b}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{w^{(b-1)n-1}}
\frac{1}{(w-1)^{n+1}}
\; dw
\\ = \frac{b}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+(w-1))^{(b-1)n-1}}
\frac{1}{(w-1)^{n+1}}
\; dw
$$
and the second is
$$\frac{b-1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{w^{(b-1)n}}
\frac{1}{(w-1)^{n+1}}
\; dw
\\ = \frac{b-1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+(w-1))^{(b-1)n}}
\frac{1}{(w-1)^{n+1}}
\; dw.$$
Extracting coefficients yields
$$(-1)^n
\left(b {(b-1)n-2+n\choose n}
- (b-1) {(b-1)n-1+n\choose n}\right)
\\ = (-1)^n
\left(b {bn-2\choose n}
- (b-1) {bn-1\choose n}\right).$$
This is one when $n=0.$ Continuing we have
$$\frac{(-1)^n}{n!}
(b (bn-2)^{\underline{n}} - (b-1) (bn-1)^{\underline{n}})
\\ = \frac{(-1)^n}{n!} (bn-1)^{\underline{n}}
\left(b \frac{bn-(n+1)}{bn-1} - (b-1)\right)
\\ = \frac{(-1)^{n+1}}{n!} (bn-1)^{\underline{n}}
\frac{1}{bn-1}
\\ = \frac{(-1)^{n+1}}{n!} (bn-2)^{\underline{n-1}}.$$
We thus obtain
$$t = 1 + \sum_{n\ge 1} \frac{y^n}{n!} (-1)^{n+1}
\prod_{j=0}^{n-2} (nb-2-j)
\\ = 1 + \sum_{n\ge 1} \frac{y^n}{n!} (-1)^{n+1}
\prod_{j=2}^{n} (nb-j).$$
Remark. Observe that the accepted answer and this post solve
slightly different equations which both yield the correct result.
|
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|
Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$ Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$
I find out the answer $104$ but my friend find the answer $96$ using calculator.which one is correct?
|
Let $x=\sqrt7$ and $y=\sqrt5+\sqrt6$ and $z=\sqrt5-\sqrt6$. It becomes:
$$(y+x)(y-x)(x+z)(x-z)$$
$$=(y^2-x^2)(x^2-z^2)$$
$$=-x^4+x^2(y^2+z^2)-y^2z^2$$
$$=-7^2+7((\sqrt5+\sqrt6)^2+(\sqrt5-\sqrt6)^2)-(\sqrt5+\sqrt6)^2(\sqrt5-\sqrt6)^2$$
$$=-49+7(5+2\sqrt5\sqrt6+6+5-2\sqrt5\sqrt6+6)-(5-6)^2$$
$$=-49+7\times22-1$$
$$=104$$
|
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Calculus 2: Partial fractions problem. Finding the value of a constant I encountered the following problem.
Let $f(x)$ be a quadratic function such that $f(0) = -6$ and
$$\int \frac{f(x)}{x^2(x-3)^8} dx $$
is a rational function.
Determine the value of $f'(0)$
Here's what I tried. I decomposed the fraction integrand below
$$\frac{f(x)}{x^2(x-3)^8} = \frac{A_1}{x} + \frac{A_2}{x^2} +\sum_{i=1}^8 \frac{B_i}{(x-3)^i}$$
By finding a common denominator, I determined
$$f(x) = A_1x(x-3)^8 + A_2(x-3)^8 +x^2 \sum_{j=1}^8 [B_j(x-3)^{8-j} ]$$
$$f'(x) = A_1(x-3)^8 + 8A_2(x-3)^7 + D(x) $$
where $D(x)$ is a function such that $D(0) = 0$. (These are all the remaining terms that go away when we plug $0$ into $f'(0)$).
I used the information that $f(0) = -6$ to get the equation
$$f(0) = A_2(-3)^8 = -6 \rightarrow A_2 = \frac{-2}{3^7}$$.
This leaves us with
$$f'(0) = A_1(-3)^8 +8\cdot \frac{-2}{3^7} \cdot (-3)^7 $$
$$ f'(0)= 3^8 \cdot A_1 + 16$$
I was told $f'(0) = 16$, however I cannot convince myself the value of $A_1$. I would think that the fact that $f(x)$ is a quadratic function should come into play here. Please let me know what you think.
|
Consider that your partial fraction decomposition is equal to $\frac {A_1}{x}+\frac {B_1}{x-3}+h(x),$ where $h(x)$ is the sum of the other terms. The anti-derivative of each of those other terms is a rational function. But $\int (\frac {A_1}{ x}+\frac {B_1}{x-3})\;dx=A_1\ln |x|+B_1\ln |x-3|$ is not a rational function unless $A_1=B_1=0.$ So $A_1=0$ and you are finished.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $ My answer: $2^\frac{-1}{\sqrt{2}} < x < 1$
Textbook answer: $2^\frac{-12}{17} < x < 1$
The only difference between my resolution and the Textbook one is that I solved by saying that
$$\text{(I) }\sqrt{1 - 8\cdot (\log_{1/4}{x})^2} > 1 -3\cdot \log_{1/4}x $$
Is true when II OR III are true:
$$\text{(II) } 1 - 3\cdot \log_{1/4}x > 0 \quad \land \quad 1-8\cdot (\log_{1/4}{x})^2 > (1 - 3\cdot \log_{1/4}x)^2 $$
$$\text{(III) } 1 - 3\cdot\log_{1/4}x < 0 \quad \land \quad 1- 8\cdot(\log_{1/4}{x})^2 > 0 $$
And the Textbook said that (I) is true when:
$$ \text{(IV) }1- 8\cdot(\log_{1/4}{x})^2 > (1 - 3\cdot \log_{1/4}x)^2 \ge 0 $$
Sorry if this is a dumb question, any help is appreciated. I can't really understand if my conditions (II or III) and their condition (IV) are both right or if one of them is wrong.
|
Initially we have $$\sqrt { 1-8\cdot (\log _{ 1/4 }{ x } )^{ 2 } } <1-3\cdot \log _{ 1/4 } x\\ \log _{ 1/4 }{ x } =t\\ \\ 1-8t^{ 2 }>0\\ \left( t-\frac { 1 }{ 2\sqrt { 2 } } \right) \left( t+\frac { 1 }{ 2\sqrt { 2 } } \right) <0\\ t\in \left( -\frac { 1 }{ 2\sqrt { 2 } } ,\frac { 1 }{ 2\sqrt { 2 } } \right) \\ \log _{ 1/4 }{ x } \in \left( -\frac { 1 }{ 2\sqrt { 2 } } ,\frac { 1 }{ 2\sqrt { 2 } } \right) \\ -\frac { 1 }{ \sqrt { 2 } } <\log _{ 2 }{ x } <\frac { 1 }{ \sqrt { 2 } } \\ { 2 }^{ -\frac { 1 }{ \sqrt { 2 } } }<x<{ 2 }^{ \frac { 1 }{ \sqrt { 2 } } }\\ \\ $$
On the other hand, by solving an inequality we get $$ 1-8t^{ 2 }<1-6t+9t^{ 2 }\\ 17t^{ 2 }-6t>0\\ t\left( 17t-6 \right) >0\\ t\in \left( -\infty ,0 \right) \cup \left( \frac { 6 }{ 17 } ,+\infty \right) \\ \log _{ 1/4 }{ x } \in \left( -\infty ,0 \right) \cup \left( \frac { 6 }{ 17 } ,+\infty \right) \\ -\infty <\log _{ 1/4 }{ x } <0\quad \Rightarrow \quad 0<\log _{ 2 }{ x } <+\infty \quad \Rightarrow \quad 1<x<+\infty \\ \frac { 6 }{ 17 } <\log _{ 1/4 }{ x } <+\infty \quad \Rightarrow \quad -\infty <\log _{ 2 }{ x } <-\frac { 12 }{ 17 } \Rightarrow \quad { 2 }^{ -\frac { 12 }{ 17 } }<x<1\\ \\ x\in \left( { 2 }^{ -\frac { 1 }{ \sqrt { 2 } } },{ 2 }^{ \frac { 1 }{ \sqrt { 2 } } } \right) \cap \left( 1,+\infty \right) \cap \left( { 2 }^{ -\frac { 12 }{ 17 } },1 \right) $$
And note that $${ 2 }^{ -\frac { 1 }{ \sqrt { 2 } } }\approx \quad 0.612547326536\\ { 2 }^{ -\frac { 12 }{ 17 } }\approx 0.61306742163\\ { 2 }^{ \frac { 1 }{ \sqrt { 2 } } }\approx 1.63252691944\\ \\ $$
so the answer is
$$\left( { 2 }^{ -\frac { 12 }{ 17 } },1 \right) $$
|
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|
Suppose that $q=\frac{2^n+1}{3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$ I am curious for a starting point to prove the following claim
Suppose that $q={2^n+1 \above 1.5pt 3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$
For example take $n=61$. Then $q ={2^{61}+1 \above 1.5pt 3} = 768614336404564651$ which is prime. Then $\binom{2^{61}}{2}-1 = 2658455991569831744654692615953842175$ and it's largest prime factor is $q$. Note ${2^{29}+1 \above 1.5pt 3} =178956971$ is not prime and the largest prime factor of $\binom{2^{29}}{2}-1$ is $3033169$. But even then we could note that $178956971 = 59*3033169$. A counterexample to the claim would suffice to to show it's wrong.
|
Most probably yes, because:
$$\binom{2^n}{2}-1=\frac{2^n(2^n-1)}{2}-1=2^{n-1}(2^n-1)-1=$$
$$=2^{n-1}(2^n+1 - 2)-1 = 2^{n-1}(2^n+1) - 2^n-1=(2^n+1)(2^{n-1}-1)=$$
$$=3q(2^{n-1}-1)$$
Additionally:
$$2 \equiv -1 \pmod{3}$$
$$2^n \equiv (-1)^n \pmod{3}$$
$$2^n + 1 \equiv (-1)^n +1 \pmod{3}$$
So $n$ is odd, further leading to
$$2^{n-1}-1=\left(2^{\frac{n-1}{2}}-1\right)\left(2^{\frac{n-1}{2}}+1\right)$$
|
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|
Find integral $\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$ I have to find definite integration $$\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$$
I tried to subtitute $x = \sin u$ and $x = e^u$
but got no result . Please help me in proceeding.
|
The first step is to turn the one integral into two as follows:
\begin{align}
\int_{0}^{1} \frac{\ln(x)}{1-x^2} \, dx = \frac{1}{2} \, \left[ \int_{0}^{1} \frac{\ln(x)}{1-x} \, dx + \int_{0}^{1} \frac{\ln(x)}{1+x} \, dx \right]
\end{align}
then using the integrals
\begin{align}
\int_{0}^{1} \frac{\ln(x)}{1-x} \, dx &= \left[Li_{2}(1-x) \right]_{0}^{1} = - \zeta(2) = - \frac{\pi^2}{6} \\
\int_{0}^{1} \frac{\ln(x)}{1+x} \, dx &= - \frac{\pi^2}{12}
\end{align}
the desired integral becomes
$$\int_{0}^{1} \frac{\ln(x)}{1-x^2} \, dx = - \frac{\pi^2}{8}.$$
|
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|
What is the value of $ \sum\limits_{k=0}^{n-1}\binom {n-k-1}{j-1} \binom {r+k}{j+k}$?
What is the value of $$ \sum_{k=0}^{n-1}\binom {n-k-1}{j-1} \binom {r+k}{j+k}$$ where $r \ge j \ge 1$?
I know that
$$ \sum_{k=0}^{n}\binom {n-k}{m}\binom{r+k}{s} = \binom {n+r+1}{m+s+1} \text{ where }n,m \ge0,\text{ and }s\ge r\ge 0$$
Please notice that second term of the first summation can be replaced by
$$ \binom {r+k}{r-j} $$
Then the first summation becomes
$$ \sum_{k=0}^{n-1}\binom {n-k-1}{j-1} \binom {r+k}{r-j} $$ which is identical to the second summation but doesn't satisfy the condition in second summation that is $ r \ngeq r - j$ since both $ r,j \ge 1 $
Is it possible to reduce the first summation similar to the second summation?
|
$$
\begin{align}
\sum_{k=0}^{n-1}\binom{n-k-1}{j-1}\binom{r+k}{j+k}
&=\sum_{k=0}^{n-1}\binom{n-k-1}{j-1}\binom{r+k}{r-j}
\end{align}
$$
which looks like $\binom{n+r}{r}$, but we are missing some $-j\le k\lt0$.
For example, let $n=6$, $j=3$, and $r=4$
$$
\begin{align}
\sum_{k=0}^{n-j}\binom{n-k-1}{j-1}\binom{r+k}{r-j}
&=\overbrace{\binom{5}{2}\binom{4}{1}}^{k=0}+\overbrace{\binom{4}{2}\binom{5}{1}}^{k=1}+\overbrace{\binom{3}{2}\binom{6}{1}}^{k=2}+\overbrace{\binom{2}{2}\binom{7}{1}}^{k=3}\\
&=95
\end{align}
$$
But $\binom{10}{4}=210$. Where is the missing $115$?
$$
\begin{align}
\sum_{k=-j}^{-1}\binom{n-k-1}{j-1}\binom{r+k}{r-j}
&=\overbrace{\binom{8}{2}\binom{1}{1}}^{k=-3}+\overbrace{\binom{7}{2}\binom{2}{1}}^{k=-2}+\overbrace{\binom{6}{2}\binom{3}{1}}^{k=-1}\\
&=115
\end{align}
$$
|
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|
Solve quadratic fraction I would like to simplify the fraction
$$\frac{x^2-2x}{x^2+x-6}$$
I know from Mathematica that it should equal $\frac3{3x}$ but how do I get there?
|
The numerator factors as $x^{2}-2x=x(x-2)$.
By the quadratic formula, the denominator has roots $2$ and $-3$. So the denominator factors as $x^{2}+x-6=(x-2)(x+3)$.
Hence
$\frac{x^{2}-2x}{x^{2}+x-6}=\frac{x(x-2)}{(x-2)(x+3)}=\frac{x}{x+3}$.
|
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|
prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$ If $a,b,c$
are the sides of a triangle and $p,q,r$ are positive real numbers then prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$
After modification. I have to prove $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$
|
Note that substitution $p'=p-t, q'=q-t, r'=r-t$, where $t$ is any real number leads to an equivalent form
$$a^2(p'-q')(p'-r')+b^2(q'-p')(q'-r')+c^2(r'-p')(r'-q')\ge 0.$$
This, as you proved already, is equivalent to
$$(a^2 p'^2+b^2 q'^2 + c^2 r'^2) \ge p'r'(a^2+c^2-b^2)+q'r'(b^2+c^2-a^2)+p'q'(a^2+b^2-c^2).$$
However, since the inequality is symmetric with respect to $p,q,r$, we can assume that $r = \min\{p,q,r\}$. Putting $t=r$ we get $r'=0$ and so the inequality reduces to
$$a^2 p'^2+b^2 q'^2 \ge p'q'(a^2+b^2-c^2),$$
where $p', q'$ are nonnegative (because $p,q\ge r$).
Triangle inequality yieds $c>|a-b|$, so $c^2 > a^2-2ab+b^2$ and therefore
$$a^2+b^2-c^2 < 2ab.$$
Since $p',q'$ are nonnegative, we have that
$$p'q'(a^2+b^2-c^2) \le 2p'q'ab.$$
Thus it suffices to prove that $$a^2p'^2+b^2q'^2 \ge 2p'q'ab.$$
But this is obvious: $$a^2p'^2 + b^2q'^2 - 2p'q'ab = (ap'-bq')^2 \ge 0.$$
|
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|
$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Why does this hold?
|
First, none of them can be in $]-\infty,-2[$ or in $]2,+\infty[$, otherwise it would mean $a^4+b^4+c^4+d^4 > 16$
If some number (let us say $a$) is in $]-2,0[\cup]0,2[$. Then, since $a^4+b^4+c^4+d^4 = 16$ , we have $|b|,|c|,|d|<2$ and $m=\max \{|a|,|b|,|c|,|d|\} < 2$. However, since $a^5+b^5+c^5+d^5\leq m(a^4+b^4+c^4+d^4) < 32$, this is impossible.
So all of them are in $\{2\} \cup\{-2\}$, and $m = 2$.
None of these numbers can be $-2$, because it would mean the other ones are null with $a^4+b^4+c^4+d^4 = 16$.
So at least one of these numbers is $2$. The rest follows.
|
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|
Expanding $(2^b-1)\cdot(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})$ Can someone please explain to me how the following works (primarily interested in an explanation of the second step when $2^b$ is expanded?
I understand how each series cancels out to equal $2^n-1$ at the end.
$$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=2^b \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b}) - (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=(2^b + 2^{2b} + 2^{3b} + \cdots + 2^{ab})-(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})\\
&=2^{ab}-1\\
&=2^n-1
\end{align*}$$
|
You can perhaps understand this better if you set $t=2^b$, so your product becomes
$$
(t-1)(1+t+t^2+\dots+t^{a-1})
$$
Distribute:
$$
t(1+t+t^2+\dots+t^{a-1})-(1+t+t^2+\dots+t^{a-1})
$$
Push $t$ in the first term inside the parenthesis:
$$
(t+t^2+t^3+\dots+t^{a})-(1+t+t^2+\dots+t^{a-1})
$$
Regroup:
$$
\underbrace{(t+t^2+t^3+\dots+t^{a-1})+t^{a}}_{t+t^2+t^3+\dots+t^{a}}
\;
\underbrace{{}-1-(t+t^2+\dots+t^{a-1})}_{-(1+t+t^2+\dots+t^{a-1})}
$$
Cancel the opposite terms:
$$
t^a-1
$$
Substitute back $t=2^b$:
$$
(2^b)^a-1=2^{ab}-1
$$
Conclusion:
$$
(2^b-1)(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})=2^{ab}-1
$$
|
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|
Finding singularities and residues I am trying to solve this question.
$$\frac{(\pi- z)(z^4 -3z^2)}{\sin^{2}z}$$
The above is the given function and I am suppose to find the singular points and their corresponding residues.
My approach was as follows.
$\sin^2z = 0$ means $z = n\pi$ for $n \in \mathbb{N}$
But $z = 0$ is a removable singularity as,
$$\frac{(\pi- z)(z^2 -3)}{\frac{\sin^{2}z}{z^2}} $$
the value of $$\lim_{z\to0}\frac{\sin^2z}{z^2} = 1$$
But I am not sure wheather other singularities are essential or non-removable and how to go about finding their residues.
Please give me hint on how to solve the problem.
Thanks
|
Let $z=w+n\pi$ with $n\in\mathbb{Z}$. Then $\sin^2(z)=\sin^2(w+n\pi)=((-1)^n\sin(w))^2=\sin^2(w)$. Therefore
$$f(z):=\frac{(\pi- z)(z^4 -3z^2)}{\sin^{2}z}=\frac{(\pi- (w+n\pi))(w+n\pi)^2((w+n\pi)^2 -3)}{\sin^{2}w}\\=\frac{(\pi- (w+n\pi))(w+n\pi)^2((w+n\pi)^2 -3)}{w^2(1+O(w^2))}\\
=\frac{(\pi- (w+n\pi))(w+n\pi)^2((w+n\pi)^2 -3)}{w^2}\cdot (1-O(w^2)).$$
because
$\sin(w)=w-\frac{w^3}{6}+ O(w^5)$ implies that $$\sin^2(w)=w^2-2w\cdot\frac{w^3}{6}+ O(w^6)=w^2-\frac{w^4}{3}+ O(w^6)=w^2+O(w^4).$$
Now we are you able to classify the singularities $\{n\pi\}_{n\in\mathbb{Z}}$ of $f$.
Note that for $n\not=0,1$, the numerator is different from zero at $w=0$. Therefore in that case the order of $n\pi$ is TWO. The residue is the coefficient of $1/w$ of
$$
\frac{(\pi- (w+n\pi))(w+n\pi)^2((w+n\pi)^2 -3)}{w^2}\cdot (1-O(w^2))$$
that is the coefficient of $w$ of the numerator
$$(\pi- (w+n\pi))(w+n\pi)^2((w+n\pi)^2 -3).$$
Therefore
$$\mbox{Res}(f;n\pi)=3n(3n-2)\pi^2-(5n-4)n^3\pi^4.$$
For $n=0$ then
$$f(z)=\frac{(\pi-w)w^2(w -3)}{w^2(1+O(w^2))}=\frac{(\pi-w)(w -3)}{1+O(w^2)}$$
which implies that $z=0$ is a removable singularity and $\mbox{Res}(f;0)=0$.
For $n=1$ then
$$f(z)=\frac{w(w+\pi)^2((w+\pi)^2 -3)}{w^2(1+O(w^2))}=\frac{(w+\pi)^2((w+\pi)^2 -3)}{w(1+O(w^2))}$$
which implies that $z=\pi$ has order ONE and
$$\mbox{Res}(f;\pi)=[(w+\pi)^2((w+\pi)^2 -3)]_{w=0}=\pi^4-3\pi^2.$$
|
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|
Find all integers $x$ that can't be written as $x=a+b+c$, where $a$ divides $b$ and $b$ divides $c$ with $aFind all integers $x$ that can't be written as $x=a+b+c$, where $a$ divides $b$ and $b$ divides $c$ with $a<b<c$
This implies that $a$ divides $c$.
I can also rewrite $c=zb$ and $b=ya$ so $c=zya$ with $z\neq0$ and $y\neq0$
So $x=a(1+z+zy)$
$z$ cannot be equal to $1$ because $c=zb$ and $b<c$
$y$ cannot be equal to $1$ because $b=ya$ and $a<b$
So I can state that none of the prime number can be write under that form if $a\neq1$.
But what then?
|
There is no solution for $x=1,2,3,4,5,6,8,12,24$ and solutions for all other $x$.
You should probably consider the case when $a=1$ first. There is a solution to $x-1=z(1+y)$ when $x-1$ is non-prime and bigger than $4$ (since $1+y>2$.)
On to the general equation.
If $X$ has an odd factor $d>5$, then $d-1=2\left(1+\frac{d-3}{2}\right)$, and $ X=\frac{X}{d} + \frac{2X}{d} + \frac{2X(d-3)}{d}$.
So we can reduce the question to $X$ which can be written as $2^n$, $3\cdot 2^n$ and $5\cdot 2^n$.
There is no solution for $1,2,3,4,5,6$.
Since $2\cdot 5-1=3\cdot 3$, we get that $5\cdot 2^n = 2^{n-1}+3\cdot 2^{n-1}+3\cdot 2^{n-1}$.
Since $3\cdot 2^5-1=5\cdot 19$, we have that:
$$3\cdot 2^{n+5}=2^{n}+5\cdot 2^n + 18\cdot 5\cdot 2^n$$
$$x=2^{n+4}=2^n + 5\cot 2^{n+1} + 5\cdot 2^{n+2}$$
Left to check: $x=8,12,24,48$. $48=3(2^4)=3+15+30$.
There is no solution for $8$, since if $c>4$ then $c\geq 6$ and $a+b+c>8$.
For $x=12$, we can't have $a=1$ since $x-1$ is prime. If $a>1$ then $12/a \leq 6$ and $1+y+yz=12/a$ has no solution.
$x=24$ is the last case. For all divisors $a\mid x$, you have either $x/a-1$ is prime, or $x/a-1<6$.
So there is no solution for $x=1,2,3,4,5,6,8,12,24$ and solutions for all other $x$.
|
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|
$f(x+2xy)=f(x)+2f(xy)$; find $f(1992)$ given $f(1991)=a$
A function $f$ satisfies
$$f(x+2xy) = f(x) + 2f(xy)$$
for all $x,y\in\mathbb{R}$. If $f(1991)=a$, what is $f(1992)$? Give an answer dependent on $a$.
I have seen that $f(x)=x$ is solution, but then $f(1992)$ isn't dependent on $a$...
|
For any $x \neq 0$, let $y = \frac{1}{2x}$. We then see that
$$ f(x + 1) = f(x) + 2f\left( \frac{1}{2} \right). $$
Thus by induction, we can prove that
$$
f(n) = f(1) + 2(n-1) f\left( \frac{1}{2} \right)
$$
for all natural numbers $n \geq 1$.
In particular,
$$
f(3) = f(1) + 4f\left( \frac{1}{2} \right).
$$
But taking $x = y = 1$ in the functional equation also gives use that $f(3) = 3f(1)$, so that we get that
$$
3f(1) = f(1) + 4f\left( \frac{1}{2} \right)
$$
which implies that
$$
2f\left( \frac{1}{2} \right) = f(1)
$$
and so we have that
$$ f(n) = f(1) + (n - 1) f(1) = nf(1) $$
for all natural numbers $n \geq 1$.
We thus have that
$$
a = f(1991) = 1991 f(1)
$$
and hence that
$$
f(1992) = 1992 f(1) = \frac{1992a}{1991}.
$$
|
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"url": "https://math.stackexchange.com/questions/1925363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Equal slicing of my spherical cake Recently I baked a spherical cake ($3$ cm radius) and invited over a few friends, $6$ of them, for dinner. When done with main course, I thought of serving this spherical cake and to avoid uninvited disagreements over the size of the shares, I took my egg slicer with parallel wedges(and designed to cut $6$ slices at a go; my slicer has $5$ wedges) and placed my spherical cake right in the exact middle of it before pressing uniformly upon the slicer.
What should the relative placement of my wedges of the slicer be like so that I am able to equally distribute the cake among my $6$ friends?The set-up of the egg slicer looks something like this :
|
Let me recast Lovsovs' answer in another perspective.
Consider at first to split an hemi-sphere into $n$ parts of equal volume.
Let designate as $h(k,n)$ the relative position of the $k$-th cut along the radius of the emisphere starting from the base circle
(corresponding to $k=0$).
$$
0 \leqslant h(k,n) \leqslant 1\quad \left| {\;0 \leqslant k \leqslant n} \right.
$$
Thus we shall have
$$
\frac{2}
{3}\pi R^{\,3} \frac{k}
{n} = \pi \int_{y\; = \;0}^{h\,R} {\left( {R^{\,2} - y^{\,2} } \right)dy} = \pi \left( {R^{\,2} R\,h - \frac{1}
{3}R^{\,3} h^{\,3} } \right)
$$
that means
$$
h^{\,3} - 3h + 2\frac{k}{n} = 0
$$
We will proceed and solve such depressed cubic equation
according to the method indicated in this work by Alessandra Cauli, refer also to the answer to this post.
Putting apart the case $k=0$, for $\;1 \leqslant k \leqslant n$ we define
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}
$$
where $p$ and $q$ are respectively the coefficients of $h^1$ and $h^0$.
The 2nd radical is
$$
\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}} = \left( {\frac{k}
{n}} \right)^{\,2} - 1 = - \frac{{n^{\,2} - k^{\,2} }}
{{n^{\,2} }} \leqslant 0
$$
which being non-positive tells us that there are three real solutions.
Completing the calculation for $u$
$$
\begin{gathered}
u = \sqrt[{3\,}]{{ - \frac{k}
{n} + i\frac{1}
{n}\sqrt {n^{\,2} - k^{\,2} } }} = \frac{1}
{{\sqrt[{3\,}]{n}}}\sqrt[{3\,}]{{n\,e^{\,i\,\alpha } }} = \hfill \\
= e^{\,\,i\,\alpha \,/\,3} \quad \left| \begin{gathered}
\;1 \leqslant k \leqslant n \hfill \\
\;\alpha = \arctan _{\text{4Q}} \left( { - k,\sqrt {n^{\,2} - k^{\,2} } } \right) = \pi - \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} - 1} } \right) = \hfill \\
\;\;\;\; = \pi - \beta \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
We take then
$$
v = - \frac{p}
{{3\,u}} = \frac{1}
{u}\quad \quad \omega = e^{\,i\,\frac{{2\pi }}
{3}} = 1/2\left( { - 1 + i\sqrt 3 } \right)
$$
and arrive to obtain the three solutions as
$$
\begin{gathered}
1 \leqslant k \leqslant n\quad 0 \leqslant \beta = \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} - 1} } \right) < \frac{\pi }
{2} \hfill \\
\left\{ \begin{gathered}
h_{\,1} = e^{\,i\,\alpha /3} + e^{\, - \,i\,\alpha /3} = 2\cos \left( {\frac{\pi }
{3} - \frac{\beta }
{3}} \right) \hfill \\
h_{\,2} = e^{\,i\,\alpha /3 + 2\pi /3} + e^{\, - \,i\,\alpha /3 - 2\pi /3} = - 2\cos \left( {\frac{\beta }
{3}} \right) \hfill \\
h_{\,3} = e^{\,i\,\alpha /3 - 2\pi /3} + e^{\, - \,i\,\alpha /3 + 2\pi /3} = 2\cos \left( {\frac{\pi }
{3} + \frac{\beta }
{3}} \right) \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
of which we can easily determine that only $h_3$
respect the physical conditions of our problem.
Passing now to split the entire sphere, imagine it consisting of the two halves, placed base-to-base, one on the positive and one on the negative $h$ axis.
So, when $n$ is even we will place the cuts at $k = 0,\; \pm 2, \cdots ,\; \pm \left( {n - 2} \right)$,
while for $n$ odd at $k = \pm 1,\; \pm 3, \cdots ,\; \pm \left( {n - 2} \right)$.
In conclusion, always as a ratio to the radius, the cutting blades shall be positioned at:
$$
\left\{ \begin{gathered}
2 \leqslant n\quad 0 \leqslant k \leqslant \left\lfloor {\frac{n}
{2}} \right\rfloor - 1 \hfill \\
\beta \left( {k,n} \right) = \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} - 1} } \right) \hfill \\
h_{\,T} (k,n) = \pm 2\cos \left( {\frac{\pi }
{3} + \frac{{\beta (2k + \bmod (n,2),\,n)}}
{3}} \right) \hfill \\
\end{gathered} \right.
$$
(if you can accept that$\pm 0=0$).
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
How to show that the harmonic function $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions? How can I prove that $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions?
|
For any $n\geq 2$ we have that
$$ \frac{1}{n}<\frac{1}{n}+\frac{1}{3n^3}+\frac{1}{5n^5}+\ldots=\text{arctanh}\frac{1}{n} = \frac{1}{2}\log\frac{n+1}{n-1}\tag{1}$$
hence we may exploit a telescopic sum:
$$\begin{eqnarray*} H_n = 1+\sum_{k=2}^{n}\frac{1}{k} &<& 1+\frac{1}{2}\sum_{k=2}^{n}\left(\log(k+1)-\log(k-1)\right)\\&=&1+\frac{1}{2}\left(\log(n+1)+\log(n)-\log(2)\right)\\&<&\left(1-\frac{\log 2}{2}\right)+\log(n+1).\tag{2} \end{eqnarray*}$$
The inequality
$$ \frac{1}{n}<\log\left(n+\frac{1}{2}\right)-\log\left(n-\frac{1}{2}\right)\tag{3}$$
leads to the sharper (and simpler) bound
$$ \boxed{H_n < \color{red}{ \log(2n+1)}} \tag{4} $$
through the same technique.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1927913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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|
Is there a quick way to find the remainder when this determinant is divided by $5$?
Find the remainder when the determinant $\begin{vmatrix} { 2014 }^{ 2014 } & { 2015 }^{ 2015 } & { 2016 }^{ 2016 } \\ { 2017 }^{ 2017 } & { 2018 }^{ 2018 } & { 2019 }^{ 2019 } \\ { 2020 }^{ 2020 } & { 2021 }^{ 2021 } & { 2022 }^{ 2022 } \end{vmatrix}$ is divided by $5$.
I'm aware that this problem has a number-theoretic solution involving congruence relations. But considering that this was asked as a multiple-choice question in a test, what should be the best way to approach problems like this?
The options were $(a)\quad1\quad (b)\quad2\quad (c)\quad3\quad (d)\quad 4$
|
Note that, modulo $5$,
\begin{align}
2014 &\equiv-1, & 2015 & \equiv 0, & 2016 &\equiv 1,\\
2017 & \equiv2, & 2018 &\equiv-2, & 2019 &\equiv-1, \\
2020 &\equiv 0, & 2021 &\equiv 1, & 2022 & \equiv2.
\end{align}
Furthermore $2$ and $-2$ have order $4$ modulo $5$, so the determinant is congruent to
$$\begin{vmatrix}
1 & 0 & 1 \\ 2 & (-2)^2 & -1 \\ 0 & 1 & 2^2
\end{vmatrix}=\begin{vmatrix}
1 & 0 & 1 \\ 2 & -1 & -1 \\ 0 & 1 & -1
\end{vmatrix}=\begin{vmatrix}
1 & 0 & 1 \\ 0 & -1 & 2 \\ 0 & 1 & -1
\end{vmatrix}=(-1)^2-2=-1\equiv 4.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1931365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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|
Prove or disprove $\sum\limits_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$
Let $x,y,z\ge 0$. Prove or disprove
$$\dfrac{x^4+y^4}{x+y}+\dfrac{z^4+y^4}{z+y}+\dfrac{z^4+x^4}{x+z}\le 3\dfrac{x^4+y^4+z^4}{x+y+z}$$
This is what I tried. Without loss of generality, let $x+y+z=1$, then
$$\Longleftrightarrow \sum_{cyc}\dfrac{x^4+y^4}{x+y}\le 3(x^4+y^4+z^4)$$
and
$$(x^4+y^4)=(x+y)(x^3+y^3)-xy(x^2+y^2)=(x+y)(x^3+y^3)-xy(x+y)^2+2x^2y^2$$
which is equivalent to
$$\sum_{cyc}\left((x^3+y^3)-xy(x+y)+\dfrac{2x^2y^2}{x+y}\right)\le 3(x^4+y^4+z^4)$$
or
$$2\sum_{cyc}x^3+2\sum_{cyc}\dfrac{x^2y^2}{x+y}\le 3\sum_{cyc}(x^4+xy(x+y))$$
and now I'm stuck.
|
Something different than brute force would be the usage of Schur's inequality and your transformation of $x^4+y^4=(x+y)(x^3+y^3)-xy(x+y)^2+2x^2y^2$
$$\sum_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$$
$$(x+y+z)\sum_{cyc}\frac{x^4+y^4}{x+y}\le 3(x^4+y^4+z^4)$$
$$\sum_{cyc}(x^4+y^4+\frac{(x^4+y^4)z}{x+y})\le 3(x^4+y^4+z^4)$$
$$\sum_{cyc}((x^3+y^3)z-xyz(x+y)+\frac{2x^2y^2z}{x+y})\le x^4+y^4+z^4$$
$$\sum_{cyc}(x^3z+y^3z)+2xyz\sum_{cyc}\frac{xy}{x+y}\le x^4+y^4+z^4+2xyz(x+y+z)$$
Bcs of the AM-HM inequality we have $x+y+z\ge \frac{2xy}{x+y}+\frac{2yz}{y+z}+\frac{2zx}{z+x}$ and the other part we obtain from Shurs inequality for $t=2$:
$x^2(x-y)(x-z)+y^2(y-z)(y-x)+z^2(z-x)(z-y)\ge0$. Add those two together and we get our desired inequality.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1932583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
How to find the square roots of $z = 5-12i$ I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rightarrow \quad z_1 = -3+2i \quad z_2 = 3-2i
$$
My question is: can the following method (below) be used to solve the problem above? Motivation for this question: if I were to find the cubic roots of the number given, I couldn't use the first method.
I will use this "other method" it in a different problem.
Find the square roots of $ z = 2i $
The method:
Since $ \rho = 2 $ and $ \theta = \frac{\pi}{2} $ we have to find a complex number such that
\begin{align*}
z_k^2 &= 2i\\
z_k^2 &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\rho_k^2 \left( \cos 2\theta_k + i \sin 2 \theta_k \right) &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\end{align*}
\begin{cases}
\rho^2 &= 2\\
2\theta_k &= \frac{\pi}{2} + 2k\pi
\end{cases}
\begin{cases}
\rho &= \sqrt{2}\\
\theta_k &= \frac{\pi}{4} + 2k\pi
\end{cases}
\begin{align*}
z_0 &= \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 1+i\\
z_1 &= \sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = -1-i
\end{align*}
Using this method for the first example seems like a dead end, specially because of the fact that the angle of $z$ is not as straightforward as the angle of the second example.
Thank you.
|
All you need to do is to convert your $z$ to polar form:
$$z=5-12i=\sqrt{12^2+5^2}e^{i(\arctan(-\frac{12}{5})+k\pi)}=13e^{i(-1.176+k\pi)}$$
Hence, one answer of $z^{\frac{1}{2}}$ is
$$z_1=13^{\frac{1}{2}}e^{i\frac{-1.176}{2}}=\sqrt{13}e^{i(-0.588)}$$
Converting this to Cartesian will give:
$$z_1=\sqrt{13}\left(\cos(-0.588)+i\sin(-0.588)\right)=3-2i$$
which is one of your original results.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find a derivable function $f$ for which $f(x) - f(x-1) =\alpha ( f(x-1) - f(x-2) )$ Find a derivable function $f$ for which $f(x) - f(x-1) =\alpha ( f(x-1) - f(x-2) )$.
My initial conditions would be:
$$\begin{align*}
f(0) &= 0\\
f(1) &= \beta
\end{align*}$$
and $\alpha < 1$
and my domain $[0,+\infty[$
Basically, if looping on integers, every increment will be $\alpha$ times the previous increment, but I want a derivable function.
for example, for $\beta = 0.5$ and $\alpha = 1$:
$$\begin{align*}
f(2) &= 1.5\\
f(3) &= 1.75\\
f(4) &= 1.875
\end{align*}$$
I want to be able to evaluate $f(5.7)$ for instance.
|
Starting from your equation
$$
f(x) - f(x - 1) = \alpha \left( {f(x - 1) - f(x - 2)} \right)
$$
put
$$
g(x) = f(x) - f(x - 1)
$$
from which you get the general solution for $g(x)$
$$
g(x) = \alpha g(x - 1)\quad \Rightarrow \quad g(x) = \alpha ^{\,x} + c
$$
with $c$ being a constant (actually, any periodic function of period $1$).
Then you get $f(x)$ as
$$
\begin{array}{l}
f(x + 1) - f(x) = g(x + 1)\quad \Rightarrow \\
\Rightarrow \quad f(x) = \left( {\sum\limits_{k = 0}^{x - 1} {g(k + 1)} } \right) + d = \left( {\sum\limits_{k = 0}^{x - 1} {\alpha ^{\,k + 1} + c} } \right) + d = \\
= \alpha \frac{{1 - \alpha ^{\,x} }}{{1 - \alpha }} + cx + d \\
\end{array}
$$
with $d$ being again a constant or any periodic function of period $1$,
and adjusting for the initial conditions
$$
f(x) = \alpha \frac{{1 - \alpha ^{\,x} }}{{1 - \alpha }} + \left( {f(1) - f(0) - \alpha } \right)x + f(0)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the smallest real value of $x$ I don't know why my answer is different from the answer sheet.
Find the smallest real value of $x$ that satisfies the equation: $(x+5)(x^2-x-11)=x+5$
Here is what I did. This equation can be rewritten as $(x+5)(x^2-x-12)=(x+5)(x-4)(x+3)=0$, which give $$(x+5)=0$$ or $$(x-4)=0$$ or $$(x+3)=0$$ Then the smallest real number I got is -5. But the answer sheet says 3.
|
If $x \neq -5$ then divide by $x+5$ and have
$$x^2 - x - 11 = 1$$
$$x^2 - x - 12 =0$$
Solve it:
$$x = \frac{1\pm\sqrt{1 + 48}}{2}$$
Hence you get two solutions:
$$x_1 = +4 ~~~~~~~ x_2 = -3$$
So.....
In any case
$x = -5$ solves the equation too.
So.....
|
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.