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Make upper bounds for $\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}$. I want to make upper bounds for the following fraction respected to $\beta$ only (not $k$ or $n$).
\begin{align*}\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}
\end{align*}
where $\beta\simeq 0.7$, $0\leq k\leq n$, and $k, n\in\mathbb{N}$.
Since $0\leq k\leq n$ then we get $2^{n-k-1}\geq\frac{1}{2}$. So we have
\begin{align*}
\frac{1}{2^{n-k-1}}\leq2
\end{align*}
Now we should find upper bounds for $\frac{2}{1-2^{\beta-(n-k)}}$. here I tried as follows
\begin{align*}
n-k\geq0 &\Rightarrow \beta-(n-k)\leq \beta \\&\Rightarrow 1-2^{\beta-(n-k)}\geq1-2^{\beta}
\end{align*}
but I could not make as I mentioned upper bounds in this way, because $1-2^{\beta}\leq0$. Any hint can help me.
|
For each integers $0\le k\le n$ put
$$f(n,k)=\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}=\frac 1{2^{n-k-1}-2^{\beta-1}}.$$
If $n=k$ then $f(n, k)$ is negative. If $n-k\ge 1$ then $f(n,k)\le \frac 1{1-2^{\beta-1}}$ and the equality is attained when $n-k=1$. Thus the required tight upper bound is $\frac 1{1-2^{\beta-1}}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove hard inequality Let $a,b,c>0$, prove that: $$\frac{1}{(2a+b)^2}+\frac{1}{(2b+c)^2}+\frac{1}{(2c+a)^2}\geq\frac{1}{ab+bc+ca}$$
I tried to use the inequality $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq\frac{(a+b+c)^2}{x+y+z} \forall x,y,z>0$ but that's all I can do. I think this inequality is too tight to use AM-GM or Cauchy Schwarz alone.
|
Let $x=2a+b,y=2b+c,z=2c+a$, then the inequality becomes:$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}$$ By using $a^2+b^2+c^2\ge ab+bc+ca$,
$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\dfrac{\left(xy\right)^2+\left(yz\right)^2+\left(zx\right)^2}{x^2y^2z^2}\ge\dfrac{x^2yz+xy^2z+xyz^2}{x^2y^2z^2}=\dfrac{xyz\left(x+y+z\right)}{x^2y^2z^2}=\dfrac{x+y+z}{xyz}$$ Then, by using $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}$, $$\dfrac{x+y+z}{xyz}=\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\ge\dfrac{9}{xy+yz+zx}$$ Lastly, use $a^2+b^2+c^2\ge ab+bc+ca$ again, $$\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}\le\dfrac{27}{3xy+3yz+3zx}=\dfrac{9}{xy+yz+zx}$$ Therefore, $\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}$ is correct. Can you get it?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Greatest integer function problem If $\alpha$ is real root of equation $$x^5 -x^3 +x-2$$ , then [$\alpha$^6 ] is equal to ? Here [.] represents greatest integer function . Any hint please!
|
$x^5 - x^3 + x = \frac {x(x^6 + 1)}{x^2 + 1}\\
x(x^4 - x^2 +1) = \frac {x}{x^2+1}(x^6 + 1)\\
\frac {x}{x^2+1}(x^6 + 1) = 2$
We can conclude that $x> 0$
as $(x^6 + 1)> 0$ we need $\frac {x}{x^2+1} > 0$
$\frac {x}{x^2+1} < \frac 12$ for all x
$(x^6 + 1) > 4\\
\alpha^6 > 3$
But is it possible for $\alpha^6 > 4$?
$f(x) = x^5 - x^3 + x - 2$ is an increasing function,
If $f(\alpha) > 0$ then it is not possible for there to be any root to the right of $x^6 = 4$
$\lfloor \alpha^6\rfloor = 3$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate limit with or without the use of l'Hospital's rule or series expansions $$\lim_{x\to0^{+}}\left(\frac{\cos^2x}{x}-\frac{e^x}{\sin x}\right)$$
One can easily calculate this limit by using series expansions for the functions appearing inside the round brackets, yielding $-1$.
My question is: can anyone give another proof of this result without the use of either l'Hospital's rule or series expansions?
|
$\begin{array}\\
\dfrac{\cos^2x}{x}-\dfrac{e^x}{\sin x}
&=\dfrac{1-\sin^2x}{x}-\dfrac{e^x}{x+O(x^3)}\\
&=\dfrac{1-(x+O(x^3))^2}{x}-\dfrac{1+x+O(x^2)}{x+O(x^3)}\\
&=\dfrac{1-x^2+O(x^3)}{x}-\dfrac{1+x+O(x^2)}{x}(1+O(x^2))\\
&=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)(1+O(x^2))}{x}\\
&=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)}{x}\\
&=\dfrac{-x-x^2+O(x^3)}{x}\\
&=-1-x+O(x^2)\\
&\to -1\\
\end{array}
$
|
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|
Show whether $ x^3$ is $O(g(x))$ where $g(x)=x^2 $ I know that $x^3$ is $O(x^3)$ or any exponent higher than 3, but how do I show mathematically that $x^3$ is not $O(x^2)$?
|
If you want to show that it's NOT the case that $x^3=O(x^2)$, then you are showing that $x^2=o(x^3)$, or in other words, that $x^2$ is "little-oh" of $x^3$, which communicates that $x^2$ is of a smaller order of growth than $x^3$.
As gimusi has noted, you could show $x^2=o(x^3)$ via limits as follows
$$\lim_{x\rightarrow \infty} \frac{x^3}{x^2}=\lim_{x\rightarrow \infty} \frac{3x^2}{2x}=\lim_{x\rightarrow \infty} \frac{6x}{2}=\lim_{x\rightarrow \infty}\frac{6}{2}x=\lim_{x\rightarrow \infty}3x=3\lim_{x\rightarrow \infty}x=3\cdot \infty =\infty$$
*Note the use of of L'Hopital's Rule since $\lim_{x\rightarrow \infty} \frac{x^3}{x^2}=\frac{\infty}{\infty}$ in the beginning
We could also show this using the definition of "big-oh." Consider that $f(x)=O(g(x))$ iff there exists constants $c,k$ such that $|f(x)| \le c \cdot |g(x)|$ for all $x>k$.
Let $f(x)=x^3$ and $g(x)=x^2$ and assume $x^3=O(x^2)$. Hence, by definition of big-oh, there exists constants $c,k$ such that
$$|x^3|\le c \cdot |x^2|$$
for all $x>k$. Note that $x^2>0$ and $x^3>0$ as $x\rightarrow \infty$, so we can remove the absolute value signs, having
$$x^3 \le c \cdot x^2$$
Now,
$$\frac{1}{x^2}x^3\le c \cdot x^2\frac{1}{x^2}$$
$$\Rightarrow x \le c$$
By the trichotomy law, for constants $c,k$ we know either $c=k$, $c<k$, or $k<c$.
Case i. $c=k$
By substitution we have $x \le k$. But we already stated $x>k$. Hence, we have contradiction.
Case ii. $c<k$
If $c<k$ and $x \le c$, we have $x < k$. But we already stated $x>k$. Hence, we have contradiction.
Case iii. $k<c$
Recall that $x \le c$ for ALL $x>k$. So whenever $x>c>k$, we have a contradiction.
Thus, if we assume $x^3=O(x^2)$, then in all cases we have a contradiction. Therefore, it is NOT the case that $x^3=O(x^2)$, and it is true that $x^2=o(x^3)$.
|
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|
prove that ${3^{3n}} + 3^{2n} + 3^{n } + 1$ is divided by $4$. by induction I tried to take the 3 out but it is not helping me much.
|
By induction?
Well if we assume $3^{3n} + 3^{2n} + 3^n + 1= 4K$ then
$3^{3(n+1)} +3^{2(n+1)} + 3^{n+1} + 1=$
$3^{3n+3} + 3^{2n + 2} + 3^{n+1} + 1=$
$3^{3n}*27 + 3^{2n}*9 + 3^n*3 + 1 =$
$[3^{3n} + 3^{2n} + 3^n+1] + 26*3^{3n} + 8*3^{2n} + 2*3^n =$
$4K + 4(6*3^{2n} + 2*3^{2n}) + 2(3^{3n} + 3^n) =$
$4[K+(6*3^{2n} + 2*3^{2n})] + 2*3^{n}(3^{2n} + 1)=$
$4[K+(6*3^{2n} + 2*3^{2n})] + 4*3^{n}*\frac {3^{2n} + 1}2$.
The only thing left to prove is that $3^{2n} + 1$ is even and.... c'mon.....!
|
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Markov Chain: Calculating Expectation Reach a Certain Set of States Suppose I have a Markov chain $Z_k$ with $6$ states, as depicted below:
The probability of moving from one node to a neighboring node is $1/2$. For example, the probability of moving from node $1$ to node $2$ is $1/2$ and the probability of moving from node $1$ to node $6$ is $1/2$ etc.
Suppose $P(Z_0=1)=1$. That is we start state $1$. We need to compute two things:
*
*Compute the expected time we first reach the bottom of our pyramid (states $3$, $4$, or $5$). That is compute $E[T_B]$ where $T_B=min\{j: Z_j \in \{3,4,5\}\}$
My attempt: I try listing out all the possibilities
I can go from:
*
*I can go from State $1-2-3$. Time is $2$ when base is reached and probability $\frac{1}{2} \cdot \frac{1}{2}$
*$1-2-1-2-3$. Time is $4$ when base is reached. Probability of occurring is $\frac{1}{16}$.
*$1-2-1-2-1-2-3$. Time is $6$ when base is reached. Probability of occurring is $\frac{1}{64}$
*$1-2-1-2-1-2-1-2-3$. Time is $8$ when base is reached. Probability of occurring is $\frac{1}{64}$.
etc...
Thus my conclusion for these types of sequences expected value is:
$2 \cdot \frac{1}{4}+ 4*\frac{1}{16}+6*\frac{1}{64}+8*\frac{1}{256} ...$
$\sum_{k=1}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{8}{9}$
*Now, I can also go from State $1-6-5$. Time is $2$ when base is reached and probability $\frac{1}{2} \cdot \frac{1}{2}$
*$1-6-1-6-5$. Time is $4$ when base is reached. Probability of occurring is $\frac{1}{16}$.
*$1-6-1-6-1-6-5$. Time is $6$ when base is reached. Probability of occurring is $\frac{1}{64}$
*$1-6-1-6-1-6-1-6-5$. Time is $8$ when base is reached. Probability of occurring is $\frac{1}{64}$.
Thus my conclusion for these types of sequences expected value is:
$2 \cdot \frac{1}{4}+ 4*\frac{1}{16}+6*\frac{1}{64}+8*\frac{1}{256} ...$
$\sum_{k=1}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{8}{9}$
But there are more possibilities:
*$1-2-1-6-5.$ The time is $4$ probability $1/16$
*$1-2-1-6-1-6-5$. The time is $6$ probability $1/64$
10.$1-2-1-6-1-6-1-6-5 etc..$. The time is $8$ with probability $1/256$.
$\sum_{k=2}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{7}{18}$
Still more possibilities....:
Too many possibilities (unfortunately gave up) as its like the Markov Chain restarts when we go back to $1$. Couldn't figure it out. Please let me know what I should do and thank you for the help
|
As an alternative approach in the spirit of your initial attempt, you can condition on the number $2k$ of steps to reach $\{3,4,5\}$ from $1$:
\begin{align}
E(T_B)
&= \sum_{k=1}^\infty 2k\ P(\text{$2k$ steps}) \\
&= \sum_{k=1}^\infty 2k\ 2^k \left(\frac{1}{2}\right)^{2k} \\
&= \sum_{k=1}^\infty k \left(\frac{1}{2}\right)^{k-1} \\
&= \frac{1}{(1-1/2)^2} \\
&= 4
\end{align}
|
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|
Question: particular formula for sums of squares and square of sums What is a general / particular formula (likely combinatorics or number theory) that describes a relationship between the sum of squares and square of sums? I remember once seeing a formula where “the ration between the sum of three numbers squared and the square of the sum of three numbers was ~1/3 or might have been ~2/3?
What is the name of this formula or a more general form of this?
|
For the series of natural number :
$$1^2 + 2^2 + 3^2 +4^2 + ....+n^2 = \frac{n(n+1)(2n+1)}{6}$$
And
$$(1+2+3+4+...n)^2 = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$$
So,
$$ \frac{1^2 + 2^2 + 3^2 +4^2 + ....+n^2}{(1+2+3+4+...n)^2} = \frac{n(n+1)(2n+1)}{6} \times \frac{4}{n^2(n+1)^2}$$
$$ = \frac23 \left(\frac{2n+1}{n(n+1)}\right)$$
In general the ratio would depend upon the number of terms included.
|
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|
Does the fractional equation $\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$ have any solutions? We have a partial fraction equation:
$$\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$$
I multiplied the equation by the common denominator $(x+5)(x-5)$ and got $0=1$. Is this correct?
|
Yes it is equivalent to
$$\frac{2x}{x^2-25}=\frac{2x+1}{x^2-25}\iff 2x=2x+1$$
which indeed has not solutions for $x\in \mathbb R$.
|
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|
Proving $\lim_{x\to 1}\sqrt{2x} = \sqrt 2 $ using $\varepsilon$-$\delta$
Prove that $$\lim_{x\to 1}\sqrt{2x} = \sqrt 2 $$ using $\varepsilon$-$\delta$.
What I've tried is
$$\left|\frac{(\sqrt{2x}-{\sqrt 2})(\sqrt {2x}+{\sqrt 2})}{\sqrt {2x}-{\sqrt 2}}\right|
=\left|\frac{2x-2}{\sqrt {2x}-{\sqrt 2}}\right|$$ and because $|x-1|< \delta$ implies that $x> 0$ so $\sqrt {2x}$ is defined.
$$\frac{|2||x-1|}{\sqrt 2}<\varepsilon$$ then we can choose
$$\delta= \frac{\varepsilon \sqrt 2 }{2}.$$
|
You're aiming to make $|\sqrt{2x} - \sqrt{2}|$ very small. Rationalising the numerator is a good idea, but you seemed to rationalise $|\sqrt{2x} + \sqrt{2}|$ instead. Rather, we have
$$|\sqrt{2x} - \sqrt{2}| = \left|\frac{(\sqrt{2x} - \sqrt{2})(\sqrt{2x} + \sqrt{2})}{\sqrt{2x} + \sqrt{2}}\right| = \frac{|2x - 2|}{\sqrt{2x} + \sqrt{2}}.$$
Now, note that, for all $x \ge 0$,
$$\sqrt{2x} \ge 0 \implies \sqrt{2x} + \sqrt{2} \ge \sqrt{2} \implies \frac{1}{\sqrt{2x} + \sqrt{2}} \le \frac{1}{\sqrt{2}}.$$
Therefore,
$$|\sqrt{2x} - \sqrt{2}| = \frac{1}{\sqrt{2x} + \sqrt{2}}|2x - 2| \le \frac{1}{\sqrt{2}}|2x - 2| = \sqrt{2}|x - 1|.$$
So, if we force $|x - 1| < \frac{\varepsilon}{\sqrt{2}}$, then we have
$$|\sqrt{2x} - \sqrt{2}| \le \sqrt{2}|x - 1| < \varepsilon,$$
as necessary. This suggests choosing $\delta = \frac{\varepsilon}{\sqrt{2}}$, as you did in your working.
However, it is worth noting that such a choice of $\delta$ may not ensure $x$ is positive. In particular, if we look at $\varepsilon = 4\sqrt{2}$, then the corresponding value of $\delta$ would be $4$. This is too big as the number $-1$ is within distance $4$ of $1$, but is not in our domain. So, we should probably limit $\delta \le 1$, in order for $(1 - \delta, 1 + \delta)$ to consist only of positive numbers. That is, choose
$$\delta = \min\left\{1, \frac{\varepsilon}{\sqrt{2}}\right\}.$$
|
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|
Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that
$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$
This problem was proposed by @Ahmad Bow but unfortunately it was closed as off-topic and you can find it here.
Any way, I tried hard on this one but no success yet. here is what I did:
Using the identity
$$H_{n/2}=H_n-n\int_0^1 x^{n-1}\ln(1+x)\ dx, \quad x\mapsto x^2$$
$$H_{n/2}=H_n-2n\int_0^1 x^{2n-1}\ln(1+x^2)\ dx$$
We can write
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\int_0^1\frac{\ln(1+x^2)}{x}\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}\ dx$$
where
\begin{align}
\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}&=\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2}-\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^3}\\
&=\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^2}(1+(-1)^n-\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^3}(1+(-1)^n\\
&=\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^2}(1-(-1)^n-\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^3}(1-(-1)^n\\
&=\frac1{2x}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)
\end{align}
Therefore
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac12\int_0^1\frac{\ln(1+x^2)}{x^2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)\ dx$$
The sum can be done using the following identity
$$ \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$
Differentiate both sides with respect to $a$ then set $a=1/2$ we get
$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$
and the question here is how to calculate the the remaining integral or a different way to tackle the sum $S$ ? Thanks
|
Another proof besides Cornel's one:
Let
\begin{align}
S&=\sum_{n=0}^\infty(-1)^n(H_{n/2}-H_n+\ln2)e^{ix(2n+1)}\\
&=\sum_{n=0}^\infty(-1)^n\int_0^1\frac{y^n \ dy}{1+y}e^{ix(2n+1)}\\
&=\int_0^1\frac{e^{ix}\ dy}{1+y}\sum_{n=0}^\infty\left(-ye^{2ix}\right)^n\\
&=\int_0^1\frac{e^{ix}\ dy}{(1+y)(1+e^{2ix}y)}\\
&=\frac{e^{ix}}{1-e^{2ix}}\left(\ln2-\ln(1+e^{2ix})\right)\\
&=\frac{i}{2\sin x}\left(-\ln(\cos x)-ix\right)\\
&=\frac{x}{2\sin x}-i\frac{\ln(\cos x)}{2\sin x}
\end{align}
substituting $e^{ix(2n+1)}=\cos(x(2n+1))+i\sin(x(2n+1))$ and comparing the real and imaginary parts we get:
$$\sum_{n=0}^\infty(-1)^n(H_{n/2}-H_n+\ln2)\cos(x(2n+1))=\frac{x}{2\sin x}\tag1$$
$$\sum_{n=0}^\infty(-1)^n(H_{n/2}-H_n+\ln2)\sin(x(2n+1))=-\frac{\ln(\cos x)}{2\sin x}\tag2$$
If we follow the same approach, we get
$$\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\cos(x(2n+1))=-\frac{\ln(\sin x)}{2\cos x}\tag3$$
$$\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\sin(x(2n+1))=\frac{\pi/2-x}{2\cos x}\tag4$$
The identity in (1) is discovered by @Ahmad Bow.
|
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|
$ a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$ and $ T = 3.73205080..$? Consider the following sequence :
Let $a_1 = a_2 = 1.$
For integer $ n > 2 : $
$$a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$$
$$ T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$
$$T = ??$$
What is the value of $T$ ?
Is there a closed form or integral for $T$?
I get
$$ T = 3.73205080..$$
The convergeance is fast.
Does anyone recognize this ?
——
Edit
So apparantly $ T = 2 + \sqrt 3 $
Let us generalize.
Take $a_1= 1, a_2 > a_1$
And now the whole sequence depends on $y = a_2$.
We thus define
$$ T(y) = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$
We know $T(1) = T(2) = 2 + \sqrt 3 $.
$$T(3) = 4.4415184401122.. $$
Apparantly $T(3) = \frac{7 + 2 \sqrt 10}{3} $ as found ( no proof ) by lhf.
How about a closed form for $T(y)$ ?
Can all of these rational recursions be transformed into a linear recursion ?
———
Update
See also
About $a_n = \frac{a_{n-1}(a_{n-1} + C)}{a_{n-2}} , t(12,13) = \frac{3}{2}$
|
EDIT: By already knowing the solution to your recurrence a priori, I am able to solve your problem. However, a more interesting question might be why your recurrence is equivalent to
$$a_n=5a_{n-1}-5a_{n-2}+a_{n-3}$$
and how to see this a priori.
Your recurrence is solved by the sequence (as can be checked by direct calculations)
$$a_{n+1} = \frac{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^n + (2-\sqrt 3)^n\cdot(3+\sqrt 3)}{12}.$$
So \begin{split}\frac{a_{n+1}}{a_n}&=\frac{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^n + (2-\sqrt 3)^{n}\cdot(3+\sqrt 3)}{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^{n-1} + (2-\sqrt 3)^{n-1}\cdot(3+\sqrt 3)}
\\
&= (2+\sqrt 3)\cdot\frac{\frac{6}{(2+\sqrt3)^{n}}+3-\sqrt 3+3+\sqrt 3}{\frac{6}{(2+\sqrt3)^{n-1}}+3-\sqrt 3+3+\sqrt 3}\\
&=(2+\sqrt 3)\cdot\frac{\frac{6}{(2+\sqrt3)^{n}}+6}{\frac{6}{(2+\sqrt3)^{n-1}}+6}
\\&\xrightarrow{n\to\infty}(2+\sqrt 3).
\end{split}
I got the closed form for $a_{n+1}$ from A101879 of the OEIS.
|
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|
Limit of an inverse function Let $f:\mathbb R\to \mathbb R$ be an invertible function such that $$\lim_{x\to a} f(x)=b$$
for some $a,b\in \mathbb R$.
Does it follow that $$\lim_{x\to b}f^{-1}(x)= a,$$
where $f^{-1}$ denotes the inverse function of $f$?
Edit: When I consider the $\epsilon,\delta$-definition of the limit, I feel that there should be an example that $\lim_{x\to b}f^{-1}(x)\neq a$ due to the fact that $\epsilon,\delta$-definition is not symmetric (for a given $\epsilon>0$, we find $\delta>0$ such that ....).
However, if we further assume that $f$ is cont., $$b=\lim_{x\to b}x=\lim_{x\to b}f\circ f^{-1}(x)=f(\lim_{x\to b} f^{-1}(x)).$$
It follows that $\lim_{x\to b} f^{-1}(x)=f^{-1}(b)=a$. Thus, one needs a discontinuous function to have a counter example. I wonder whether there is any simple function with this property.
@Floris Claassens'a answer shows that there are some "ugly functions" with this property.
|
If $x=a$, consider the sequence $(a_{n})=(a+\frac{b-a}{2}\frac{1}{n})_{n\geq 1}$ and the sequence $(b_{n})(b+\frac{b-a}{2}\frac{1}{n})_{n\geq1}$. We define $f$ as follows
$$f(x)=\begin{cases}b+\frac{b-a}{2}\frac{1}{2n-1}&\text{ if }x=a+\frac{b-a}{2}\frac{1}{n},\ n>0.\\ b+\frac{b-a}{2}\frac{1}{2n}&\text{ if }x=b+\frac{b-a}{2}\frac{1}{2n},\ n>0.\\2b-a+\frac{b-a}{2}\frac{1}{n}&\text{ if }x=b+\frac{b-a}{2}\frac{1}{2n-1},\ n>0.\\x+b-a&\text{ else}.\end{cases}$$
Note that $f$ is well-defined as for all $m,n\in\mathbb{N}$ we have
\begin{align*}|a+\frac{b-a}{2}\frac{1}{m}-b-\frac{b-a}{2}\frac{1}{n}|&=|a-b+\frac{b-a}{2}\frac{n-m}{mn}|\geq|a-b|-|\frac{a-b}{2}||\frac{1}{m}-\frac{1}{n}|\\&\geq|a-b|-\frac{1}{2}|a-b|>0\end{align*}
Using similar arguments we find that
$$f^{-1}(x)=\begin{cases}a+\frac{b-a}{2}\frac{1}{n}&\text{ if }x=b+\frac{b-a}{2}\frac{1}{2n-1},\ n>0.\\b+\frac{b-a}{2}\frac{1}{2n} &\text{ if }x=b+\frac{b-a}{2}\frac{1}{2n},\ n>0.\\b+\frac{b-a}{2}\frac{1}{2n-1}&\text{ if }x=2b-a+\frac{b-a}{2}\frac{1}{n},\ n>0.\\x+a-b&\text{ else}.\end{cases}$$
is well-defined, and evidently $f^{-1}$ is the inverse of $f$. So $f$ is bijective.
Now let $\varepsilon>0$ and take $\delta=\min(\varepsilon,\frac{b-a}{2})$. For all $x\in\mathbb{R}$ with $|x-a|<\delta$ we have $|f(x)-f(a)|\leq|x+b-a-b|=|x-a|<\varepsilon$. So $\lim_{x\rightarrow a}f(x)=b$.
Furthermore note that $\lim_{n\rightarrow\infty}f^{-1}(b+\frac{b-a}{2}\frac{1}{2n})=b\neq a$, so $\lim_{x\rightarrow b}f^{-1}(x)\neq a$.
For $a=b$ one can actually (contrary to my previous claim) define a similar function. We define
$$f(x)=\begin{cases}a+\frac{1}{2n}&\text{ if }x=a+\frac{1}{n},\ n>0.\\a+\frac{1}{2n-1}&\text{ if }x=a+2+\frac{1}{2n-1},\ n>0.\\a+2+\frac{1}{n}&\text{ if }x=a+2+\frac{1}{2n},\ n>0.\\x&\text{ else}.\end{cases}$$
Using similar arguments we find that $f$ is bijective $\lim_{x\rightarrow a}f(x)=a$, but $\lim_{n\rightarrow \infty}f^{-1}(a+\frac{1}{2n-1})=a+2\neq a$.
|
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|
What are the integer solutions of the equation $(x^2 -1)(y^2 -1)=2(7xy-24)$ Can you help with this one? I've been trying fruitlessly for hours
$$(x^2 -1)(y^2 -1)=2(7xy-24)$$
|
Let me elaborate on my hint given above in the comment section. I will focus on non-negative integer solutions because it is easy to see that if $(x,y)$ is a solution, then so is $(-x,-y)$.
Observe that the given equation can be written as :
\begin{align*}
(xy-7)^2 & =x^2+y^2\\
(xy-6)^2+13 & =(x+y)^2\\
(xy-6)^2-(x+y)^2 & =-13\\
(xy-6+x+y)(xy-6-x-y)&=-13.
\end{align*}
Since $13$ is a prime, so we get the following two systems
\begin{align*}
xy-6+x+y&=13 &&& xy-6+x+y=1\\
xy-6-x-y&=-1 &&& xy-6-x-y=-13
\end{align*}
These systems can be written as:
\begin{align*}
x+y&=7 &&& x+y=7\\
xy&=12 &&& xy=0
\end{align*}
So the only non-negative integer solutions are $(x,y)=(3,4), (4,3),(0,7),(7,0)$.
|
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|
Group with simple relations I'm working on the following algebra problem:
Prove that if elements $a$ and $b$ of some group satisfy the relations
1) $a^5 = b^3 = 1$ and
2) $b^{-1}ab = a^2$
then $a = 1$.
Here is an example of what I've tried:
If $a^5 = 1$, then using relation 2) , multiplying on the right by $a^3$ on both sides gives *$b^{-1}aba^3 = 1$.
If $b^3 = 1$, then using relation 2), multiplying on the left by $b^4$ on both sides gives **$ab = b^4a^2$.
Then, I tried substituting ** into *. Unfortunately, this just gets me relation 1) back again. It seems that my methods to exploit the relations always end up in simply recovering relation 1), rather than showing $a = 1$.
Is there anything more clever I can try here?
Thanks!
|
If $b^{-1}ab=a^2$, $b^3 =1$ and $a^5=1$, then
$$\begin{align*}
b^{-2}ab^2 &= b^{-1}(b^{-1}ab)b\\
&= b^{-1}a^2b\\
&= (b^{-1}ab)^2\\
&= (a^2)^2 = a^4 = a^{-1}.
\end{align*}$$
So therefore,
$$\begin{align*}
a&= b^{-3}ab^3\\
&= b^{-1}(b^{-2}ab^2)b\\
&= b^{-1}(a^{-1})b\\
&= (b^{-1}ab)^{-1}\\
&= (a^2)^{-1}\\
&= a^3.
\end{align*}$$
Therefore, $a=a^3$.
|
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|
How can I solve the following inequality? I have the inequality:
$lg((x^3-x-1)^2) < 2 lg(x^3+x-1)$
And I'm not sure how should I go about solving it. I wrote it like this:
$2lg(x^3-x-1) < 2lg(x^3+x-1)$
$lg(x^3-x-1) < lg(x^3 + x - 1)$ (*)
Here I have the conditions:
$x^3-x-1 > 0$
$x^3+x-1 > 0$
At first this stumped me, but then I realised I can just add the inequalities to get:
$2x^3 -2> 0$
$x^3-1>0$
$x^3>1 \Rightarrow x \in (1, + \infty)$
So that is our condition.
Going back to (*) and raising the inequality to the power of $10$, I got:
$x^3-x-1 < x^3+ x - 1$
$2x>0 \Rightarrow x \in(0, +\infty)$
So, if we also consider the condition, we have
$x \in (1, + \infty) \cap (0, + \infty)$
So $x \in (1, + \infty)$.
The problem with this answer is that it is wrong. My textbook lists the following possible answers:
A. $\mathbb{R}$
B. $(0, + \infty)$
C. $(1, + \infty)$
D. $(0, 1)$
E. Other answer
So I got answer C, but after I checked the back of the book I found that the correct answer should be E. So, what did I do wrong and what is that other answer?
|
The domain it's $$x^3-x-1\neq0$$ and $$x^3+x-1>0.$$
Now, since $$\ln(x^3-x-1)^2=2\ln|x^3-x-1|,$$ we need to solve
$$|x^3-x-1|<x^3+x-1$$ or
$$-x^3-x+1<x^3-x-1<x^3+x-1.$$
The left inequality gives $x>1$ and the right inequality gives $x>0,$ which gives the answer:
$$(1,+\infty)\setminus\{x|x^3-x-1=0\}.$$
|
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|
In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2 {x})(1+\sin^4 {x}) , \cos^2{x}=0$
$(1-\sin^2{x})^2=(1+\sin^2 {x})(1+\sin^4 {x})$
$-3\sin^2{x}=\sin^6{x}$
Which is not possilbe
Is there a trick or something to solve this equation or to know how many solutions are there ?
|
One trick is letting $t=\sin^2(x)$, so that the equation reduces to a polynomial $t^4+(1-t)^3 - 1 = t^4 - t^3 + 3t^2 - 3t = t(t^3-t^2+3t-3) = t(t-1)(t^2+3)=0$
Thus either $t=0,1$ so that $\sin(x) = 0,1,-1$. In $[0, 2\pi)$, we clearly have solutions $x=0, \pi, \frac{\pi}{2}, \frac{3\pi}{2}$.
|
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|
Determining $n$ from its divisors Let $d_i$ be the $i^{th}$ smallest divisor of $n$.
Suppose that $n$ has $3$ prime factors and that:
$$n = d_{13}+d_{14}+d_{15}$$
$$d_{15}+1 = (d_5+1)^3$$
Is it possible to uniquely determine $n$?
I tried doing some stuff with writing $n=p^\alpha q^\beta r^\gamma$ and working modulo different things, but couldn't figure it out.
|
Well, $n<3d_{15}$ and $d_{15}\mid n$, so either $n=d_{15}$ or $n=2d_{15}$. However, in that first case, $d_{14}+d_{13}$ would be zero and that's impossible. Hence,
$$d_{15}=\frac n2.$$
It follows that $n<4d_{14}$ and because $d_{14}\mid n$ and $d_{14}<n/2$, we must have $d_{14}\in\{n/3,n/4\}$. But if $d_{14}=n/4$, then $d_{13}$ must be as well, which is not the case. Therefore,
$$d_{14}=n/3$$
and
$$d_{13}=n-\frac n2-\frac n3=\frac n6.$$
We now have that $d_1=1$, $d_2=2$ and $d_3=3$. So the three prime factors are $2$, $3$ and some other prime, $p$.
Note that
$$1\equiv \frac n2 + 1\equiv (d_5+1)^3\equiv d_5+1\pmod 3,$$
so $3\mid d_5$.
Assume $d_5=6$, then $n=2((6+1)^3-1)=684=2^2\cdot 3^2\cdot 19$. But $684$ has $3\cdot 3\cdot 2=18$ proper divisors, while $n$ has $16$, so $684$ is not a solution and we conclude that $d_5\neq 6$.
Therefore, $d_5\ge 9$, so $d_4=6$ and $4,5,8\nmid n$.
Since $n$ cannot have divisors larger than $n/2$, we conclude that $n$ has exactly $16$ proper divisors. Write $n=2\cdot3^ap^b$ with $a,b\ge 1$, then it follows that
$$(a+1)(b+1)=8.$$
So $\{a,b\}=\{1,3\}$.
If $a=3$, then $9\mid n$ and we must have $d_5=9$. Then $n=2((9+1)^3-1)=1998=2\cdot 3^3\cdot 37$, so $n=1998$ has three prime factors and also satisfies the conditions.
Finally, say $a=1$, so $n=6p^3$ for some prime $p>3$. Then $d_5=p$, but $3\mid d_5$, which is a contradicion.
Conclusion: $n=1998$ is the only integer which satisfies both conditions.
|
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|
Prove convergence of $\sum_{n=1}^{\infty}\frac{\sin^3\left(\frac{\pi n}{n+5}\right)}{\sqrt{n^2+n}-n}$
I need to prove convergence of the series.
$$
\sum_{n=1}^{\infty}a_n,\ \
a_n=\frac{\sin^3\left(\frac{\pi n}{n+5}\right)}{\sqrt{n^2+n}-n}
$$
Well, at first I noticed that $\forall n\in\mathbb{N}\ a_n>0$:
$$
\left.
\begin{aligned}
&0<\frac{\pi n}{n+5}<\pi\Rightarrow\sin^3\left(\frac{\pi n}{n+5}\right)>0\\
&\sqrt{n^2+n}>n\Rightarrow\sqrt{n^2+n}-n>0
\end{aligned}
\right\}\Rightarrow a_n>0
$$
So, then I tried to apply comparison test to this series. However, it was not helpful since no matter how I bounded $\sin^3\left(\frac{\pi n}{n+5}\right)$, I could not find $b_n\geqslant a_n:\sum_{n=1}^{\infty}b_n$ is convergent.
Thus, I would be glad if someone could give me some clue to this problem.
|
We have that
$$\sin^3\left(\frac{\pi n}{n+5}\right)=\sin^3\left(\frac{\pi n+\pi 5-\pi 5}{n+5}\right)=\sin^3\left(\pi-\frac{\pi 5}{n+5}\right)=$$
$$=\sin^3\left(\frac{\pi 5}{n+5}\right) \sim \left(\frac{\pi 5}{n+5}\right)^3$$
and
$$\frac{1}{\sqrt{n^2+n}-n}=\frac{1}{\sqrt{n^2+n}-n}\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}=\frac{\sqrt{n^2+n}+n}{n}\sim 2$$
therefore $a_n \sim \frac1{n^3}$ and the given series converges by comparison test.
|
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|
$|{\sqrt{a^2+b^2}-\sqrt{a^2+c^2}}|\le|b-c|$ where $a,b,c\in\mathbb{R}$ Show that $|{\sqrt{a^2+b^2}-\sqrt{a^2+c^2}}|\le|b-c|$ where $a,b,c\in\mathbb{R}$
I'd like to get an hint on how to get started. What I thought to do so far is dividing to cases to get rid of the absolute value. $(++, +-, -+, --)$
but it looks messy. I'm wondering if there is any nicer way to solve it.
Would love to hear some ideas.
Thanks in advance!
|
For all $a,b,c \in \mathbb{R}$,
\begin{align}
0 &\leq (b-c)^2\\
2bc &\leq b^2 + c^2\\
2a^2bc &\leq a^2b^2 + a^2c^2\\
a^4 + 2a^2bc + b^2c^2 &\leq a^4 + a^2b^2 + a^2c^2 + b^2c^2\\
a^2 + bc&\leq \sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2} \label{1}\tag{1}\\
2a^2 + 2bc&\leq 2\sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2}\\
2a^2 -2\sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2}&\leq -2bc\\
2a^2 +b^2 + c^2 -2\sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2}&\leq b^2 -2bc + c^2\\
(a^2 +b^2) -2\sqrt{(a^2+b^2)(a^2+c^2)} + (a^2 + c^2)&\leq b^2 -2bc + c^2\\
\end{align}
and then take the square root of both sides.
Note that it was not necessary to use absolute values the first time we took a square root $(\ref{1})$ because the right hand side is understood to be positive.
|
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|
Find all positive integers m,n and primes $p\geq5$ such that $m(4m^2+m+12)=3(p^n-1)$ I did it like this. We can manipulate the equation to come to
$\frac{(m^2+3)(4m+1)}{3}=p^n$
Now 3 divides either $(m^2+3) or (4m+1) $
If we assume 3 divides $(m^2+3)$ and it is equal to $(4m+1)$ (my intuition) .we get
$\frac{m^2+3}{3}=4m+1$.solving this we get $m=12$ we can substitute back in equation to get $n=4:p=7$ which is correct. But i am not able to prove that
$\frac{m^2+3}{3}=4m+1$.
Please help me to complete my solution
|
This is a variation on John Omielan's answer, starting with the equation in the form $(m^2+3)(4m+1)=3p^n$. As John notes, each factor $m^2+3$ and $4m+1$ must contain a power of the prime $p$, since $m^2+3$ and $4m+1$ are both greater than $3$ if $m\ge1$.
Now using properties of the $\gcd$ relation, we see that
$$\begin{align}
\gcd(m^2+3,4m+1)
&=\gcd(m^2-12m,4m+1)\\
&=\gcd(m(m-12),4m+1)\\
&=\gcd(m-12,4m+1)\quad\text{(since }\gcd(m,4m+1)=1)\\
&=\gcd(m-12,49)
\end{align}$$
It follows that $p$ can only be $7$, $m$ must be congruent to $5$ mod $7$, and the smaller of the two powers of $7$ is either $7$ or $49$. Now
$$m^2+3\ge{4m+1\over3}\quad\text{for all }m\ge1$$
so if $3\mid4m+1$ we must have $4m+1=3\cdot7$ or $3\cdot49$. The former gives $m=5$, for which $m^2+3=28$ is not a power of $7$, while the latter gives $m=146/4$, which is not an integer. Therefore we must have $3\mid m^2+3$. Now
$${m^2+3\over3}\ge4m+1\quad\text{for }m\ge12$$
Recalling that our gcd analysis showed that $m$ must be congruent to $5$ mod $7$, and that we've already seen that $m=5$ gives $m^2+3=28$, we see that $m=12$ is the only possible solution: $4\cdot12+1=49$ and $12^2+3=147=3\cdot49$. In sum, $m=12$, $p=7$, and $n=4$ is the only solution (with positive integers $m$ and $n$ and prime $p$) to $m(4m^2+m+12)=3(p^n-1)$.
|
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|
Eigenvalues and Eigenvectors of a 3 by 3 matrix I want to find eigenvalues and eigenvectors of this matrix:
$$
\begin{bmatrix}
2 & 3 & 0 \\
3 & 6 & 1 \\
0 & 1 & 6 \\
\end{bmatrix}
$$
So eigenvalues:
$$
\begin{vmatrix}
2-\lambda_1 & 3 & 0 \\
3 & 6-\lambda_2 & 1 \\
0 & 1 & 6-\lambda_3 \\
\end{vmatrix}
$$
then
$$-\lambda^3+14\lambda^2-50\lambda+16=-(8-\lambda)(\lambda^2-6\lambda+2)$$
and the eigen values are: $\lambda_1=3-\sqrt{7}, \lambda_2=3+\sqrt{7}, \lambda_3=8$
So then I want to find eigenvectors for eigenvalue $\lambda_1=3-\sqrt{7}$
I know that:
$$$$
\begin{bmatrix}
2-(3-\sqrt{7}) & 3 & 0 \\
3 & 6-(3-\sqrt{7}) & 1 \\
0 & 1 & 6-(3-\sqrt{7}) \\
\end{bmatrix}
$$$$
and that matrix I multiply with matrix:
$$$$
\begin{bmatrix}
x \\
y \\
z\\
\end{bmatrix}
$$$$
and this gives me:
$$$$
\begin{bmatrix}
(\sqrt{7}-1)x+3y\\
3x+(\sqrt{7}+3)y+z \\
y+(\sqrt{7}+3)z \\
\end{bmatrix}
$$$$
and then I get three equation:
$$\begin{cases}
(\sqrt{7}-1)x+3y=0\\[2ex]
3x+(\sqrt{7}+3)y+z =0\\[2ex]
y+(\sqrt{7}+3)z=0
\end{cases}$$
Now, every such system will have infinitely many solutions, because if e is an eigenvector, so is any multiple of e. So our strategy will be to try to find the eigenvector with $x=?$
How I decide the $x$ value when the correct answer is:
$x=0.87, y=-0.479, z=0.085$.
Yes I know that when I set the $x=0.87$ then the equalitons gives me the correct $y$ and $z$ values.
|
The strategy is that you define one of the variables $x$, $y$, or $z$ as independent variable. Then two other variables will be dependent on it. Let's choose $z$ to be the independent. In this case your eigenvector corresponding to $\lambda_1=3-\sqrt{7}$ is $(\dfrac{(2+\sqrt{7})(4+\sqrt{7})}{3}z,-(3+\sqrt{7})z,z)$.
There is no one correct answer. Once you fix $z$ you obtain another vector.
|
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|
$\frac{a^2+b^2+1}{ab+1}=k$. Find $k$ $a$ and $b$ are positive integers such that
$$\frac{a^2+b^2+1}{ab+1}=k$$
where $k$ is positive integer. Find all values $k$
My work:
$a$ is the root of the equation
$$x^2-kbx+b^2+1-k=0$$
Let $c$ is another root of this equation. Then
$c=kb-a$ and $ac=b^2+1-k$
|
Note that, mod $ab+1$, $0=b^2(a^2+b^2+1)=b^4+b^2+1$. Therefore, $b^4+b^2+1 > 0$ is divisible by $ab+1$, thus $b(b^3+b) \geq ab$. So if $b \neq 0$, then this can be written as $k \leq b^2+1$ which entails (if $a > 0$), $c \geq 0$.
Take now, for a given $k$, $(a,b)$ a solution in non-negative integers with $a \geq b$ and $a+b$ minimal. Assume $b \neq 0$ (hence $ab>0$).
Then by minimality $c \geq a$ (since $(c,b)$ is a solution) thus $a^2 \leq ac = b^2+1-k \leq b^2$, so $a=b$. But this implies $a=b=0$, a contradiction. So $b=0$ and $k=a^2+1$.
The possible values of $k$ are thus exactly all the $x^2+1$ for integers $x$.
|
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|
Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \equiv 2^6 \equiv 4\pmod{10} \\
(...)$$
I can see that this goes 2,4,8,6 and then repeats. I remember that the gist of the exercise is to find the remainder based on this repetition. How do I do that? I know that $572^{42} \equiv 2^{42}\equiv ? \pmod {10}$. How do I simplify that 42 and answer this using that repetition?
|
Applying $\ xy\bmod xz\, =\, x(y\bmod z)\ = $ mod Distributive Law to factor out $\,x = 2\,$ below
$\ \ \ \ \begin{align}(2a)^{\large 2+4N}\!\bmod 10 &\,=\, 2(2\, \color{#c00}2^{\large\color{#c00} 4N} a^{\large 2} \color{#c00}a^{\large\color{#c00} 4N}\bmod 5)\\[.2em]
&\,=\, 2(2a^{\large 2}\bmod 5)\ \ {\rm by}\ \ 5\nmid a\,\Rightarrow\ \color{#c00}{a^{\large 4}\equiv 1}\!\!\!\pmod{5}\\[.3em]
&\,=\, 4\ \ \ {\rm if}\ a\equiv \pm1\!\!\!\pmod{\!5}\end{align}$
|
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|
Linear Algebra: Determine if the set of polynomials span P2 The polynomial given is { ${1 - 2x + x^2, 2 - 3x, x^2 + 1, 2x^2 + x}$}
What I do not understand is, how am I going to solve the augmented matrix if it only spans to P2 (meaning the highest power will be 2?) but I am given 4 equations.
|
It's clear that the system $\{x^2,x,1\}$ generates $P2$, since
$$P2 = \{ax^2+bx+c\ |\ a,b,c\in\Bbb{R}\} = span(x^2,x,1).$$
So if we can show that the system
$$\begin{align}&\text{I. } &1 - &2x &+ x^2 \\ &\text{II. } &2 - &3x & \\ &\text{III. } &1\ \ \ &\ \ + &x^2 \\ &\text{IV. } & &x+ &2x^2\end{align}$$
generates $\{x^2,x,1\}$, then we are done.
First of all, it's easy to see that $\text{II.} = \text{I.} + \text{III.} - \text{IV.}$, so we only need $\text{I., III. }$ and $\text{IV.}$ to generate $\{x^2,x,1\}$, and we can use Gauss-elimination to show this:
$$\begin{matrix}\text{I.} \\ \text{III.} \\ \text{IV.}\end{matrix} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &1 &0 &1 \\ &0 &1 &2 \end{bmatrix}} \stackrel{r_2-r_1}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &0 &2 &0 \\ &0 &1 &2 \end{bmatrix}} \stackrel{r_2/2}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &0 &1 &0 \\ &0 &1 &2 \end{bmatrix}} \longrightarrow \\ \stackrel{r_3-r_2}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &0 &1 &0 \\ &0 &0 &2 \end{bmatrix}} \stackrel{r_3/2}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &0 &1 &0 \\ &0 &0 &1 \end{bmatrix}} \stackrel{r_1+2r_2}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &0 &1 \\ &0 &1 &0 \\ &0 &0 &1 \end{bmatrix}} \stackrel{r_1-r3}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &0 &0 \\ &0 &1 &0 \\ &0 &0 &1 \end{bmatrix}}$$
Following the steps that we took (backwards):
$$\begin{align}x^2 &= \frac{2x^2}{2} = \frac{(2x^2+x)-(x)}{2} = \frac{(2x^2+x)-(2x/2)}{2} = \\ &= \frac{(2x^2+x)-((1-2x+x^2)-(1+x^2))/2}{2} = \\ &= \frac{\text{IV.}-(\text{I.}-\text{III.})/2}{2} = \frac{\text{IV.}}{2} - \frac{\text{I.}}{4} + \frac{\text{III.}}{4}\end{align}$$
$$\begin{align}x &= \frac{2x}{2} = \frac{(1-2x+x^2)-(1+x^2)}{2} = \frac{\text{I.}}{2} - \frac{\text{III.}}{2}\end{align}$$
$$\begin{align}1 &= (1-2x+x^2)+2(x)-(x^2) = \\ &= \text{I.}+2\left(\frac{\text{I.}}{2} - \frac{\text{III.}}{2}\right)-\left(\frac{\text{IV.}}{2} - \frac{\text{I.}}{4} + \frac{\text{III.}}{4}\right) = \\ &= \frac{9}{4}\text{I.}-\frac{5}{4}\text{III.}-\frac{1}{2}\text{IV.}\end{align}$$
So we can express $\{x^2,x,1\}$ out of these, meaning that we could exress any $p \in P2$ with linear combinations of the given polynomials.
|
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|
Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g(x)+5$ where $g(x)=(x+1)(x+2)(x+3)(x+4)$
Let's find the maximum value of g(x)
It can be clearly seen that from $x=-6$ to $x=6$, maximum value of $g(x)$ is at $x=6$.
$$g(6)=5040$$
Let's find the minimum value of $g(x)$
From the sign scheme one can see that negative value of $g(x)$ occurs in the interval $(-4,-3)$ and $(-2,-1)$
Hence intuitively it feels that the minimum value of g(x) would be at $x=-\dfrac{3}{2}$ or at $x=-\dfrac{7}{2}$ as in case of parabola also, minimum value is at average of both the roots.
So $g\left(-\dfrac{3}{2}\right)=-\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{5}{2}=-\dfrac{15}{8}$
At $x=-\dfrac{7}{2}$, $g\left(-\dfrac{7}{2}\right)=-\dfrac{5}{2}\cdot\dfrac{3}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{15}{8}$
So range of $g(x)$ would be $[\dfrac{-15}{8},5040]$
Hence range of $f(x)$=$\left[\dfrac{25}{8},5045\right]$
But it is given that the range is $[a,b]$ where $a,b\in N$
I am stuck here. I am also not able to prove mathematically that at $x=-\dfrac{3}{2}$ or $x=-\dfrac{7}{2}$, minimum value of $g(x)$ will occur.
Please help me in this.
|
The range will be f(X)$\in$[4,5045]
It is simple approach will be solution to quadratic equation as in article. https://www.mathsdiscussion.com/solution-of-quadratic-equation/
|
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|
Please help how to work the inequation (x-1)/(x-5)<0 I am studying for college finals and I cannot solve the inequation (x-1)/(x-5)<0 using the method the professor taught us, which I have to use in the exam.
This is another inequation using said method:
I noticed that the inequation I posted has a different symbol than the pic, but the first one is the only example of this type the professor gave us with the less than symbol.
This is how I tried to solve it:
I appreciate any help you can give. Thanks in advance.
|
Remember if $\frac ab < 0$ then $a$ and $b$ are "different signs".
So if $\frac {x-1}{x-5} < 0$ then either
1) $x - 1 > 0$ and $x -5 < 0$
OR
2) $x-1 < 0$ and $x-5 > 0$.
In case 1) we have $x - 1> 0$ so $x > 1$ and $x-5 < 0$ so $x < 5$. So $x$ is between $1$ and $5$ or $1 < x < 5$
In case 2) we have $x -1 < 0$ and $x< 1$ and $x-5> 0$ so $x > 5$. So $x$ is both less than $1$ and greater than $5$. That's impossible.
So Case 1: is the true case and $1 < x < 5$.
.....
Alternatively. $1 < 5$ always and so $-5 < -1$ always, and $x-5 <x - 1$ always no matter what $x$ is.
So when we know that $x-1$ and $x-5$ are "different signs" we know that $x-5$ must be the negative one (because it is the smaller one) and $x -1$ must be the positive one because it is bigger.
So $x -5 < 0 < x-1$
So $x - 5 < 0$ and $x < 5$. And $0 < x-1$ so $1 < x$. So $1 < x < 5$.
|
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|
Calculate cumulative probability for X~U(0,1) I'm trying to calculate
$$ P\left( |X - \mu_X| \geq k \sigma_X \right) $$
for $X\sim$uniform(0,1). I've calculated that $E[X]=1/2$, Var$[X]=1/12$, know that
$$ f_X(x) = 1, 0 \leq x \leq 1 $$ and calculated that
\begin{equation}
P(X \leq x) = F_X(x) = \begin{cases} 0 & x<0,\\ x & 0\leq x\leq 1\\ 1 & x>1 \end{cases}.
\end{equation}
For my attempt at the problem, I have done:
\begin{equation}
\begin{aligned}
P\left(\left| X-\frac{1}{2}\right|\geq\frac{k}{\sqrt{12}}\right) & = 1 - P\bigg(\frac{1}{2}-\frac{k}{\sqrt{12}}\leq X\leq \frac{1}{2}+\frac{k}{\sqrt{12}} \bigg)\\
& = 1-\int_{\frac{1}{2}-\frac{k}{\sqrt{12}}}^{\frac{1}{2}+\frac{k}{\sqrt{12}}} f_X(x)dx\\
& = 1-\frac{k}{\sqrt{3}}
\end{aligned}
\end{equation}
However, the solutions in my book is:
\begin{equation}
P\left(\left| X-\frac{1}{2}\right|\geq\frac{k}{\sqrt{12}}\right) = \begin{cases}1-\frac{2k}{\sqrt{12}} & k<\sqrt{3}\\ 0& k\geq \sqrt{3} \end{cases}.
\end{equation}
I don't understand how they went from the first step to the last step, and the work they did was omitted. If anyone could elucidate, I would greatly appreciate it.
|
you have done well but there are some points you are missing, first of all:
\begin{equation}
P(X \leq x) = F_X(x) = \begin{cases} 0 & x<0,\\ x & 0\leq x < 1\\ 1 & x \geq1 \end{cases}.
\end{equation}
and another one is in the last step of calculations, where:
\begin{equation}
\begin{aligned}
P\left(\left| X-\frac{1}{2}\right|\geq\frac{k}{\sqrt{12}}\right) & = 1 - P\bigg(\frac{1}{2}-\frac{k}{\sqrt{12}}\leq X\leq \frac{1}{2}+\frac{k}{\sqrt{12}} \bigg)\\
& = 1-\int_{\frac{1}{2}-\frac{k}{\sqrt{12}}}^{\frac{1}{2}+\frac{k}{\sqrt{12}}} f_X(x)dx\\
\end{aligned}
\end{equation}
So if $\frac{1}{2}+\frac{k}{\sqrt{12}} < 1$ and $\frac{1}{2}-\frac{k}{\sqrt{12}} \geq 0$ then $F_X(x) = x$ and :
$1-\int_{\frac{1}{2}-\frac{k}{\sqrt{12}}}^{\frac{1}{2}+\frac{k}{\sqrt{12}}} f_X(x)dx = 1-(\frac{1}{2}+\frac{k}{\sqrt{12}}-(\frac{1}{2}-\frac{k}{\sqrt{12}})) = 1-\frac{2k}{\sqrt{12}} = 1-\frac{k}{\sqrt{3}}$
In this case, the conditions leads to:
$\frac{1}{2}+\frac{k}{\sqrt{12}} < 1 \Rightarrow k < \sqrt{3}$
and if $\frac{1}{2}+\frac{k}{\sqrt{12}} \geq 1$ then $F_X(x) = 1$, So:
$1-\int_{\frac{1}{2}-\frac{k}{\sqrt{12}}}^{\frac{1}{2}+\frac{k}{\sqrt{12}}} f_X(x)dx = 1 - 1 = 0$
In this case, the conditions leads to:
$\frac{1}{2}+\frac{k}{\sqrt{12}} \geq 1 \Rightarrow k \geq \sqrt{3}$
Also, in case of $\frac{1}{2}+\frac{k}{\sqrt{12}} < 0$ we will have $F_X(x) = 0$, So:
$1-\int_{\frac{1}{2}-\frac{k}{\sqrt{12}}}^{\frac{1}{2}+\frac{k}{\sqrt{12}}} f_X(x)dx = 1 - 0 = 1$
But in this case we will see that:
$\frac{1}{2}+\frac{k}{\sqrt{12}} < 0 \Rightarrow k < -\sqrt{3} \Rightarrow \frac{1}{2}-\frac{k}{\sqrt{12}} > 1$
So, it will be unacceptable.
|
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What is the interpretation of the difference? We have the linear maps \begin{equation*}f:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+x_2 \\ x_3 \\ x_1+x_2\end{pmatrix} \ \text{ and } \ h:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+ x_2 \\ x_3 \\ x_1-x_2\end{pmatrix}\end{equation*}
We also have the vectors \begin{equation*}\vec{v}:=\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix} \ \text{ und } \ \vec{w}:=\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}\end{equation*}
I want to check if $f(\vec{v}$ and $f(\vec{w})$ are linearly (in)dependent and then the same for $h(\vec{v})$ and $h(\vec{w})$ and then I want to interpret the difference.
With $f$ we get \begin{equation*}f(\vec{v})=f\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} \ \ \text{ and } \ \ f(\vec{w})=f\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}\end{equation*}
The vectors $f(\vec{v})$ and $f(\vec{w})$ are linearly dependent, since $f(\vec{v})=0 \cdot f(\vec{w})$.
With $h$ we get \begin{equation*}h(\vec{v})=h\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix} \ \ \text{ and } \ \ h(\vec{w})=h\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}\end{equation*}
The vectors $h(\vec{v})$ and $h(\vec{w})$ are linearly independent, since the one vector has zeroes and the other one not, correct?
What is the interpretation of the difference?
|
Yes it is correct, the interpretation is that $v$ belongs to the nullspace of $f$ while by $h$ vectors $v$ and $w$ are mapped into two linearly independent vectors.
|
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|
How do I prove that $\sum_{i=3}^{n} \frac{i-2}{\binom{i}{2}} < \frac n 4$ for all natural $n > 3$? How do I prove:
$$\frac{3-2}{\binom 3 2} + \frac {4 - 2}{\binom 4 2} + \dots + \frac{n-2}{\binom n 2} < \frac n 4$$
I tested this sum on a variety of $n$ from $2$ to $100$ and they all seem to be less than $\frac n 4$, however I can't find a way to prove it.
|
$\displaystyle \sum\limits_{i=3}^n (i-2){\binom i 2}^{-1} = 2\sum\limits_{i=3}^n \frac{i-2}{i(i-1)}=2H_{n-1}-4+\frac{4}{n}$
From the assumption $\displaystyle~2H_{n-1}-4+\frac{4}{n}<\frac{n}{4}~$ we conclude for $~n\to n+1~$ :
$\displaystyle 2H_n-4+\frac{4}{n+1}= \left(2H_{n-1}-4+\frac{4}{n}\right) + \left(\frac{4}{n+1}-\frac{2}{n}\right)< $
$\displaystyle < \frac{n}{4} + \frac{4}{n+1}-\frac{2}{n}<\frac{n+1}{4}$
The right side of the inequation chain means $~n^2-7n+8>0~$,
it's the same as $\displaystyle~n>4+\frac{n}{n-2}~$, which is true for $~n\geq 6~$ .
The cases $~n\in\{3,4,5\}~$ can be checked by hand.
|
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|
Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow \frac{7}{5} \rightarrow \frac{7}{12}$.
Repeating this process appears to map every fraction to
$\phi$ and $\frac{1}{\phi}$:
$$
\begin{array}{ccccccccccc}
\frac{2}{5} & \frac{7}{5} & \frac{7}{12} & \frac{19}{12} & \frac{19}{31} & \frac{50}{31} & \frac{50}{81} & \frac{131}{81} & \frac{131}{212} & \frac{343}{212} & \frac{343}{555} \\
0.4 & 1.40 & 0.583 & 1.58 & 0.613 & 1.61 & 0.617 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Another example:
$$
\begin{array}{ccccccccccc}
\frac{11}{7} & \frac{18}{7} & \frac{18}{25} & \frac{43}{25} & \frac{43}{68} & \frac{111}{68} & \frac{111}{179} & \frac{290}{179} & \frac{290}{469} & \frac{759}{469} & \frac{759}{1228} \\
1.57143 & 2.57 & 0.720 & 1.72 & 0.632 & 1.63 & 0.620 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Q. Why?
|
Let $f$ be the map that takes $a/b$ to $(a+b)/(a+2b)$. We can prove inductively that the $n$th iteration of this process gives
$$f^n(a/b) = \frac{F_{n}a + F_{n+1}b}{F_{n+1}a + F_{n+2}b},$$
where $F_n$ is the $n$th Fibonacci number. Since $b$ is always non-zero, asymptotically, this ratio approaches
$$\lim_{n\rightarrow \infty} \frac{F_{n}a + F_{n+1}b}{F_{n+1}a + F_{n+2}b} = \lim_{n\rightarrow \infty}\frac{F_{n+1}}{F_{n+2}} = \varphi^{-1},$$
by Binet's formula. The argument for the odd convergents is basically identical.
Edit: As M. Winter points out in the comments, the last limit is a little tricky. You can follow the steps outlined in the comments, or here is an alternative. Given fractions $a/c < b/d$, the mediant satisfies the inequality
$$\frac{a}{c} < \frac{a+b}{c+d} < \frac{b}{d}.$$
In our case, we have
$$\frac{F_na}{F_{n+1}a} < \frac{F_na + F_{n+1}b}{F_{n+1}a+F_{n+2}b} < \frac{F_{n+1}b}{F_{n+2}b},$$
so the result follows by the squeeze theorem.
|
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|
Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt[3]{(n^3+2n^2+1)^2} + \sqrt[3]{(n^3+2n^2+1)(n^3-1)} + \sqrt[3]{(n^3-1)^2}}$$
$$\lim\limits_{n \to \infty} \dfrac{2n^2+2}{\sqrt[3]{n^6+4n^5+4n^4+2n^3+4n^2+1} + \sqrt[3]{n^6+2n^5-2n^2-1} + \sqrt[3]{n^6-2n^3+1}}$$
And I saw that we can factor $n^2$ in the denominator and if we do the same in the numerator, we'd get that the limit is equal to $2$. The problem is that my textbook claims that this limit is actually $\dfrac{2}{3}$. I don't see why should I have a $3$ in the denominator since the coefficient of $n^2$ would be $1$ if I would carry out the factoring to detail. So, what did I do wrong?
|
It may be easier with binomial expansion:
$n\left(1+\dfrac2n+\dfrac1{n^3}\right)^{1/3}-n\left(1-\dfrac1{n^3}\right)^{1/3}$
$=n\left[1+\dfrac13\left(\dfrac2n+\dfrac1{n^3}\right)+o\left(\dfrac1{n^2}\right)
-\left(1-\dfrac13\dfrac1{n^3}+o\left(\dfrac1{n^6}\right)\right)\right]$
|
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|
Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statement where $n=2k-1$ to the above, to factor out 15 in order to prove divisibility? Would it be easier to assume $n=k$ is odd and prove $n=k+2$ is divisible by 15?
|
You could go mod $3$ and mod $5$ and conclude, an alternate proof is by induction : of course $4^n + 5^n + 6^n$ is divisible by $15$ when $n=1$. However, note that if $n \geq 3$:
$$
4^n + 5^n + 6^n - 4^{n-2} - 5^{n-2} - 6^{n-2} \\=\color{blue}{(4^n - 4^{n-2})} + \color{green}{(5^n - 5^{n-2})} + \color{red}{(6^n - 6^{n-2})} \\= \color{blue}{15(4^{n-2})} +\color{green}{ 15(8 \times 5^{n-3})} + \color{red}{ 15(14 \times 6^{n-3})}
$$
where terms of same colour are equal by factorization. Thus, the claim follows since the sum of two multiples of $15$ is also a multiple of $15$.
|
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|
Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$ $2\cos^2 x+\sin x=1$
$\Rightarrow 2(1-\sin^2 x)+\sin x=1$
$\Rightarrow 2-2 \sin^2 x+\sin x=1$
$\Rightarrow 0=2 \sin^2 x- \sin x-1$
And so:
$0 = (2 \sin x+1)(\sin x-1)$
So we have to find the solutions of each of these factors separately:
$2 \sin x+1=0$
$\Rightarrow \sin x=\frac{-1}{2}$
and so $x=\frac{7\pi}{6},\frac{11\pi}{6}$
Solving for the other factor,
$\sin x-1=0 \Rightarrow \sin x=1$
And so $x=\frac{\pi}{2}$
Now we have found all our base solutions, and so ALL the solutions can be written as so:
$x= \frac{7\pi}{6} + 2\pi k,\frac{11\pi}{6} + 2\pi k, \frac{\pi}{2} + 2\pi k$
|
$2\cos^2(x)+\sin(x)=1$
Using $2\cos^2(x)-1=\cos(2x)$ we have $\cos(2x)=-\sin(x)=\cos(\pi/2+x).$
Hence $2x=\pm(\pi/2+x)+2n\pi$, $n\in\mathbb{Z}$ and $x=\pi/2+2n\pi$ or $x=-\pi/6+2n\pi/3.$ The second expression can be re-written as $-\pi/6\pm2\pi/3+2k\pi$ or $-\pi/6+2k\pi$ giving the three solutions
\begin{align}
x&=\pi/2+2n\pi\\x&=-\pi/6+2k\pi=11\pi/6+2k\pi\\x&=-5\pi/6+2k\pi=7\pi/6+2k\pi\\
\end{align}
Note that one of the solutions from the second expression, $-\pi/6+2\pi/3+2k\pi=\pi/2+2k\pi$ is absorbed into the first of the three solutions. This happens because this is a double root, tangential to the x-axis - see plot from Wolfrom Alpha:
|
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|
Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$. Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$.
My Attempt:
$$y=\dfrac {x^n}{x-1}$$
$$y=x^n\cdot(x-1)^{-1}$$
Differentiating both sides,
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n\cdot x^{(n-1)}$$
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n \cdot\dfrac {x^n}{x}$$
|
$f(x) = \frac{x^n}{x-1} = \frac{x^n-1+1}{x-1} = x^{n-1} + x^{n-2} + \ldots + x + 1 + \frac{1}{x-1} \Rightarrow$
$f^{(n)}(x) = \left(x^{n-1} + x^{n-1} + \ldots + x + 1\right)^{(n)} + \left(\frac{1}{x-1}\right)^{(n)} = 0 + (-1)^n\frac{n!}{(x-1)^{n+1}} = (-1)^n\frac{n!}{(x-1)^{n+1}}$.
|
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|
why $\lim_{x\to+∞}\dfrac{(1+x)^{3/2}}{x^{1/2}} -x\\$ \begin{align}
& \lim_{x\to+∞}\frac{(1+x)^{3/2}}{x^{1/2}} -x\\[8pt]
= {} &\lim_{x\to+∞} \dfrac{(1+(1/x))^{3/2}-1}{1/x} \\[8pt]
= {} & \lim_{x\to+∞} \dfrac{\frac{3}{2}\cdot\frac{1}{x}}{\frac{1}{x}}
\end{align}
Detailed picture with clearer specifications
|
The manipulations may be:
$${(1+x)^\frac32 \over \sqrt x }-x = {(1+x)^\frac32\over x^\frac12 } - {x^\frac32\over x^\frac12}$$
$$= {x^\frac32\over x^\frac12}\left(\left(1+\frac 1x\right)^\frac32 - 1\right)$$
Using Newton’s general binomial theorem we have $\left(1+\frac 1x\right)^\frac32 = 1 + \binom{3/2}1 \frac 1x + \binom{3/2}2 \frac 1{x^2}\dots$, and since terms including higher powers of $\frac 1x$ are negligibly small ($\to 0$) we may write it as simply $1+\frac 3{2x}$.
Now, we have $$x^{\frac 32 - \frac 12}\cdot\left(1+\frac 3{2x} -1\right)$$
$$=x\cdot \frac 32\cdot \frac 1x=\boxed{\frac 32\cdot \frac 1x\over \frac 1x}$$
|
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|
Show that $\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{MC^2}+\frac{1}{MD^2}\geq2$ for $M$ inside square $ABCD$ with side length $2$
Show that $$\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{MC^2}+\frac{1}{MD^2}\geq2$$ for any point $M$ inside of the square $ABCD$ whose side length is $2$.
I could manage to prove this using analytic geometry. Looking for a geometrical proof and possibly generalization of that.
For an analytic geometry proof, see @MichaelRosenberg's answer to the question Prove:$\frac{1}{(x-1)^2+(y-1)^2}+\frac{1}{(x+1)^2+(y-1)^2}+\frac{1}{(x-1)^2+(y+1)^2}+\frac{1}{(x+1)^2+(y+1)^2}\geq2 $ ,if $-1<x,y<1$ .
|
COMMENT:
An experimental approach;
In a Pythagorean triple (a, b,c) it can be seen that $c^2<2.5 ab$.Now considering the sketch we have:
$MA^2<2.5(S_{AHME} =AE \times AH)$
$MB^2<2.5(S_{HBGM} =AE \times BH)$
$MC^2<2.5(S_{EMFC} =AH \times EC)$
$MD^2<2.5 (S_{MGDF}=BH \times EC)$
$\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{MC^2}+\frac{1}{MD^2}>\frac{EC\times BH+AH\times AH\times EC+AE\times BH+AE\times AH}{2.5(AE\times AH\times EC\times BH)}=\frac{AF(AH+BH)+EC(AH+BH)}{2.5\times S_{AHME}\times S_{MGDF}}=\frac{4}{2.5 S_{AHME}\times S_{MGDF}}=\frac{1.6}{S_{AHME}\times S_{MGDF}}$
The denominator $S_{AHME}\times S_{MGDF}$ is fractional, that is it's value is less than unity; therefore the sum of fractions can be greater than 2. If M is at center of square, then all fractions are equal to $1$ and inequality changes to equality.
|
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|
Show function is continuous at $(0,0)$ Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function with
$$\space f(x, y) = \begin{equation}
\begin{cases}
\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}, & (x, y) \neq (0, 0)\\
0, & (x, y) = 0
\end{cases}\end{equation}$$
Show that f is continious.
I've already shown that f is continous for $(x, y) \neq (0,0)$ but I am having trouble with finding a proof for $(0, 0)$ using epsilon-delta or limits of sequences. Can anyone help?
This is how far I got with simplifying for epsilon delta
$$
\Bigg{|}\frac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{y^2\log(1+x^2y^2)}{\sqrt{y^4*(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{y^2\log(1+x^2y^2)}{y^2\sqrt{(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{\log(1+x^2y^2)}{\sqrt{(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\leq \big{|}\log(1+x^2y^2)\big{|}
$$
But now I am clueless on how to connect
$$
\big{|}log(1+x^2y^2)\big{|} < \epsilon \\
$$
and
$$
\sqrt{x^2+y^2} < \delta
$$
|
Let $|x|,|y| <1$.
$\log (1+x^2y^2) \le x^2y^2 \le x^4 +y^4 \le (x^2+y^2);$
$\epsilon >0$ be given.
Choose $\delta =\epsilon^{1/2}$.
Used
AM-GM: $x^4+y^4 \ge 2x^2y^2$; and
$\log (1+z) \le z$ for $z \ge -1$.
|
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|
How to simplify $(a-b)^2(y-x)+(b-a)(x-y)^2-(b-a)(x-y)$? How to simplify the following?:
$(a-b)^2(y-x)+(b-a)(x-y)^2-(b-a)(x-y)$
I have performed the following, but I think what I have arrived at is not the final step:
$(b-a)((b-a)(y-x)+(x-y)^2-(x-y))$
$ (b-a)((b-a)(y-x)+(y-x)^2+(y-x))$
$ (b-a)((y-x)((b-a)+(y-x)+1)))$
$ (b-a)(y-x)(y+b-a-x+1)$
|
Adjust the signs and binomials right away:
$$(a-b)^2(y-x)+(b-a)(x-y)^2-(b-a)(x-y)=\\
-(b-a)^2(x-y)+(b-a)(x-y)^2-(b-a)(x-y)=\\
(b-a)(x-y)[-(b-a)+(x-y)-1]=\\
(b-a)(x-y)[x-y+a-b-1]=\\
(a-b)(x-y)(y-x-a+b+1).$$
|
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|
Proving that $8^x+4^x\geq 5^x+6^x$ for $x\geq 0$. I want to prove that $$8^x+4^x\geq 6^x+5^x$$ for all $x\geq 0$. How can I do this?
My attempt:
I try by AM-GM: $$8^x+4^x\geq 2\sqrt{8^x4^x}=2(\sqrt{32})^x.$$
However, $\sqrt{32}\approx 5.5$ so I am not sure if $$2(\sqrt{32})^x\geq 5^x+6^x$$ is true.
Also, I try to compute derivatives but this doesn't simplify the problem. What can I do?
|
Note that $8^x=(6^x)^{\log_6(8)}$ so that \begin{equation}\tag 1\label 14^x+8^x-6^x-5^x\geq 8^x-6^x-5^x\to\infty\end{equation} as $x\to\infty$.
I will show that the only non-negative solution to \begin{equation}\tag 2\label 28^x+4^x=6^x+5^x\end{equation} is $x=0$. Note that $$8^x+4^x=6^x+5^x\iff 8^x-6^x=5^x-4^x.$$
By the Mean value Theorem, we have $8^x-6^x=2x\cdot c^{x-1}$ and $5^x-4^x=x\cdot d^{x-1}$ for some $c\in[6,8]$ and $d\in[4,5]$.
So we have that \eqref{1} is equivalent (for suitable $c$ and $d$) to $$2x\cdot c^{x-1}=x\cdot d^{x-1}.$$ For $x>0$ this is equivalent to $$2=\left(\frac{d}{c}\right)^{x-1}$$ which is impossible since $d\le5<6\le c$. Hence, $x=0$ is the only non-negative solution to \eqref{1}.
From the intermediate value Theorem (and from \eqref{2}), it follows that $8^x+4^x-6^x-5^x>0$ for all $x>0$.
|
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|
Probability of k heads being the longest run in a row in n flips? EDIT: I miswrote the question I was trying to originally convey. Below is the correct question.
How can I use the Bernoulli distribution calculate the probability of the longest run of heads:
$0$ heads in a row in $5$ flips being the longest run
$1$ head in a row in $5$ flips being the longest run
$2$ heads in a row in $5$ flips being the longest run
$3$ heads in a row in $5$ flips being the longest run
$4$ heads in a row in $5$ flips being the longest run
$5$ heads in a row in $5$ flips being the longest run
Is there a specific formula that can be use when one is interested in the number of successes in a row?
Not sure if I'm approaching this correctly, but for example if I want the probability that 4 heads appears in a row in 5 flips: HHHHT and THHHH are the only possibilities? So would the probability just be 1/16? I'm looking for a formula I can use.
|
Maximum run of zero heads: The only way this can occur is if all five flips are tails. For a fair coin, this has probability
$$\left(\frac{1}{2}\right)^5 = \frac{1}{32}$$
of occurring.
Maximum run of one head: For this to occur, there must either be one, two, or three heads in the sequence, no two of which are consecutive.
One head in the sequence: There are five positions in which the only head could occur. For a fair coin, this occurs with probability
$$\binom{5}{1}\left(\frac{1}{2}\right)^5 = \frac{5}{32}$$
Two heads in the sequence, with the two heads not occurring consecutively: This can occur in six ways. Let's see why. Line up three tails in a row, which creates four spaces, two between consecutive tails and two at the ends of the row.
$$\square T \square T \square T \square$$
To ensure that no two heads are consecutive, choose two of these four spaces in which to place a head. For instance, choosing the first and third spaces yields the sequence
$$HTTHT$$
The number of ways we can choose two of the four spaces is
$$\binom{4}{2} = 6$$
The actual sequences are HTHTT, HTTHT, HTTTH, THTHT, THTTH, TTHTH. For a fair coin, the probability of having a maximum run of one head if there are two heads in the sequence is
$$\binom{4}{2}\left(\frac{1}{2}\right)^5 = \frac{6}{32} = \frac{3}{16}$$
Three heads in the sequence, with no two of the heads occurring consecutively: This can occur in one way: HTHTH. For a fair coin, the probability of having a maximum run of one head if there are three heads in the sequence is
$$\left(\frac{1}{2}\right)^5 = \frac{1}{32}$$
Since these three cases are mutually exclusive and exhaustive, the probability that the maximum run has one head is
$$\frac{5}{32} + \frac{3}{16} + \frac{1}{32} = \frac{12}{32} = \frac{3}{8}$$
Maximum run of two heads: For this to occur, either two, three, or four heads must occur in the sequence, with no more than two being consecutive.
Two heads occur in the sequence and they are consecutive: The run of two heads must begin in one of the first four positions in the sequence, resulting in one of the sequences HHTTT, THHTT, TTHHT, TTTHH. For a fair coin, this occurs with probability
$$\binom{4}{1}\left(\frac{1}{2}\right)^5 = \frac{4}{32} = \frac{1}{8}$$
Three heads occur in the sequence, with exactly two being consecutive: Place two tails in a row. This creates three spaces, one between the two tails and two at the ends of the row.
$$\square T \square T \square$$
Choose one of these three spaces for the pair of consecutive heads. Choose one of the remaining two spaces for the remaining head. For instance, if we choose the third space for the pair of consecutive heads and the first space for the single head, we get the sequence
$$HTTHH$$
The number of such sequences is $3 \cdot 2 = 6$. For a fair coin, this occurs with probability
$$\binom{3}{1}\binom{2}{1}\left(\frac{1}{2}\right)^5 = \frac{6}{32} = \frac{3}{16}$$
The actual sequences are HHTHT, HHTTH, HTHHT, HTTHH, THHTH, THTHH.
Four heads occur in the sequence, with no more than two being consecutive: This can occur in one way: HHTHH. For a fair coin, this has probability
$$\left(\frac{1}{2}\right)^5 = \frac{1}{32}$$
of occurring.
Since these three cases are mutually exclusive and exhaustive, the probability of a maximum run of two heads occurring is
$$\frac{1}{8} + \frac{3}{16} + \frac{1}{32} = \frac{11}{32}$$
Maximum run of three heads: For this to occur, there must either be three heads or four heads, with exactly three of the heads being consecutive.
Exactly three heads occur, with all of them being consecutive: The first head must occur in one of the first three tosses, giving HHHTT, THHHT, or TTHHH. For a fair coin, this occurs with probability
$$3\left(\frac{1}{2}\right)^5 = \frac{3}{32}$$
Four heads occur, with exactly three of them being consecutive: This can occur in two ways: HHHTH or HTHHH. For a fair coin, this occurs with probability
$$2\left(\frac{1}{2}\right)^5 = \frac{1}{16}$$
Since the two possible cases are mutually exclusive and exhaustive, the probability that the maximum run of heads has length $3$ is
$$\frac{3}{32} + \frac{1}{16} = \frac{5}{32}$$
Maximum run of four heads: This can occur in two ways: $HHHHT$ or $THHHH$. For a fair coin, this has probability
$$2\left(\frac{1}{2}\right)^5 = \frac{2}{32} = \frac{1}{16}$$
of occurring, as you found.
Maximum run of five heads: The only way this can occur is if all five flips are heads. For a fair coin, this has probability
$$\left(\frac{1}{2}\right)^5 = \frac{1}{32}$$
of occurring.
Check: Since the maximum run of heads must have length $0$, $1$, $2$, $3$, $4$, or $5$, the probabilities should add to $1$.
$$\frac{1}{32} + \frac{3}{8} + \frac{11}{32} + \frac{5}{32} + \frac{1}{16} + \frac{1}{32} = 1$$
|
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Let $f(x,y)=\frac{x^2+y^2} x$ In each point of the circle $x^2+y^2-2y=0$ calculate the directional derivative
Let $f(x,y)=\frac{x^2+y^2} x$
a) In each point of the circle $x^2+y^2-2y=0$, what is the value of the directional derivative with respect to a vector $(a,b) \in R^2$
My approach:
the circle is $x^2+(y-1)^2=1$ and can be parametrized as $(cos \theta,sin \theta +1)$
On the other hand $\nabla f(x,y)=(\frac {x^2-y^2}{x^2},\frac {2y} x)$ , if $(x,y)$ is on the circle then $x=cos \theta$ and $y=sin \theta +1$
And $\frac{∂f}{∂v}(x,y)=\nabla f(x,y) \cdot v $, $v=(a,b)$.
I think I should substitute $x=cos \theta$ and $y=sin \theta +1$ in $\nabla f(x,y)$ and do the product is it right?
b) When you are on a point of that circle, in which direction should you move to make $f$ not to change (note it depends on the point).
My approach:
I compute the level curves and for $f(x,y)=k$ I get $(x-\frac k 2 )^2 +y^2=\frac {k^2}4$ which is a circle with center $(\frac k 2,0)$ and radius $\frac k 2$.
Is it correct if for a point in the circle a) I found a level curve that pass through that point f$? In that case, how can I find that level curve and how can i get the direction?
c) On the same circle and for each point , in which direction should you move to get the maximum variation of f?
It is almost the same problem of b) and calculate $\nabla f$
|
We have that
$$x^2+y^2-2y=0 \iff x^2+(y-1)^2=1$$
which can be parametrized by
$$(x,y)=(\cos \theta, \sin \theta +1) \implies \hat v=(-\sin \theta, \cos \theta), \,\theta\in[0,2\pi)$$
and therefore
$$\nabla f(x,y) =\left(\frac{x^2-y^2}{x^2}, \frac{2y}{x}\right)=\left(\frac{-2\sin^2\theta-2\sin \theta}{\cos^2 \theta}, \frac{2\sin \theta +2}{\cos \theta}\right)$$
from which we can evaluate
$$\frac{∂f}{∂\hat v}(x,y)=\nabla f(x,y) \cdot \hat v=\frac{2\sin^3\theta+2\sin^2 \theta}{\cos^2 \theta}+ 2\sin \theta +2=$$
$$=(2\sin \theta+2)\left(\tan^2 \theta+1\right)=2y\left(\frac{(y-1)^2}{x^2}+1\right)$$
which is the answer for point “a”.
For point “b” and “c” recall the geometrical meaning of the gradient and refer to
*
*Find a unit vector that minimizes the directional derivative at a point
|
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|
Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:
$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$
My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\displaystyle\lim_{x\to0}{\frac{\cos{(2x)}\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{(2x)}}}$
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{2x}}}=\displaystyle\lim_{x\to0}{\frac{\cosh(3x^2)\cdot e^{8x^3}-1}{x^2\sin{(2x)}}}\cdot\displaystyle\lim_{x\to0}{\frac{1}{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1}}$
All of my attempts failed.
$$\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1\;\;\text{is continuous &decreasing }$$
Source: Matematička analiza 1, 2. kolokvij
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Using Taylor series from the strating point.
Use the standard series of $\cosh(t)$ and make $t=3x^2$ gives
$$\cosh(x^3)=1+\frac{9 x^4}{2}+O\left(x^{8}\right)$$ Using the binomial expansion or Taylor series again
$$\sqrt{\cosh(x^3) }=1+\frac{9 x^4}{4}+O\left(x^{8}\right)$$
Use the standard series of $e^t$ and make $t=4x^3$ gives
$$e^{4x^3}=1+4 x^3+O\left(x^6\right)$$
Use the standard series of $\tan(t)$ and make $t=2x$ gives
$$\tan(2x)=2 x+\frac{8 x^3}{3}+O\left(x^5\right)$$
Combining everything
$$\frac{\sqrt{\cosh{(3x^2)}}\, e^{4x^3}-1}{x^2\tan(2x)}=\frac {4 x^3+\frac{9 x^4}{4}+O\left(x^6\right) } {2 x^3+\frac{8 x^5}{3}+O\left(x^7\right) }$$ Now, long division to get
$$\frac{\sqrt{\cosh{(3x^2)}}\, e^{4x^3}-1}{x^2\tan(2x)}=2+\frac{9 x}{8}-\frac{8 x^2}{3}+O\left(x^3\right)$$
Use your pocket calculator with $x=0.1$; you should get $2.08865$ while the above truncated expansion gives $2.08583$.
|
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|
Suppose two cubic polynomials f (x) and g(x) satisfy the following: f (2) = g(4); f (4) = g(8); f (8) = g(16); f (16) = g(32) + 64. Suppose two cubic polynomials $f (x)$ and $g(x)$ satisfy the following: If $f(2) = g(4)$; $f (4) = g(8)$;
$f (8) = g(16)$; $f (16) = g(32) + 64$. What is the value of $g(128) − f (64)$?
The first thing that comes to mind is that the polynomials must be in the form $ax^3 + b$ because 4 restrictions are given, but this question was given in an oral competition and I don't think my method is appropriate.
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Let $F(x)=g(2x)-f(x) $. We have $F(2)=0,F(4)=0,F(8)=0$. So
\begin{eqnarray*}
F(x) = \lambda (x-2)(x-4)(x-8).
\end{eqnarray*}
Using $F(16)=-64$ gives
\begin{eqnarray*}
F(x) = -\frac{(x-2)(x-4)(x-8)}{3 \times 7} .
\end{eqnarray*}
So
\begin{eqnarray*}
F(64) = -\frac{62 \times 60 \times 56}{3 \times 7} =- 8 \times 20 \times 62 = \color{red}{-9920}.
\end{eqnarray*}
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\dfrac{dy}{dx}$ if $y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]$
Derivative of $y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]$
Put $\sqrt{x}=\sin\alpha,\;\sqrt{a}=\sin\beta$
$$
y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]=\sin^{-1}\bigg[\sqrt{x(1-a)}-\sqrt{a(1-x)}\bigg]\\
=\sin^{-1}\bigg[\sin\alpha|\cos\beta|-|\cos\alpha|\sin\beta\bigg]
$$
My reference gives the solution $\dfrac{1}{2\sqrt{x}\sqrt{1-x}}$, but is it a complete solution ?
How do I proceed further with my attempt ?
|
Using
$$ (\arcsin u)'=\frac1{\sqrt{1-u^2}}u' $$
one has
\begin{eqnarray}
\frac{dy}{dx}&=&\frac{1}{\sqrt{1-(\sqrt{x-ax}-\sqrt{a-ax})^2}}(\sqrt{x-ax}-\sqrt{a-ax})'\\
&=&\frac1{\sqrt{1-(x-ax)-(a-ax)+2\sqrt{a(1-a)x(1-x)}}}\bigg[\frac{\sqrt{1-a}}{2\sqrt x}+\frac{\sqrt a}{2\sqrt{1-x}}\bigg]\\
&=&\frac{1}{\sqrt{(1-a)(1-x)+ax+2\sqrt{a(1-a)x(1-x)}}}\cdot\frac{\sqrt{(1-a)(1-x)}+\sqrt{ax}}{2\sqrt{x(1-x)}}\\
&=&\frac{1}{2\sqrt{x(1-x)}},
\end{eqnarray}
where
$$ 1-(x-ax)-(a-ax)=(1-a)(1-x)+ax $$
is used.
|
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|
Proving that a list of perfect square numbers is complete Well, I have a number $n$ that is given by:
$$n=1+12x^2\left(1+x\right)\tag1$$
I want to find $x\in\mathbb{Z}$ such that $n$ is a perfect square.
I found the following solutions:
$$\left(x,n\right)=\left\{\left(-1,1^2\right),\left(0,1^2\right),\left(1,5^2\right),\left(4,31^2\right),\left(6,55^2\right)\right\}\tag2$$
Is there a way to prove that this a complete set of solutions? So I mean that the solutions given in formula $(2)$ are the only ones?
My work:
*
*We know that:
$$
1 + 12x^2 \left(1+x \right) \ge 0
\space \Longleftrightarrow \space
x \ge -\frac{1+2^{-2/3}+2^{2/3}}{3}
\approx -1.07245
\tag3
$$
So we know that for $x<-1$ there are definitely no solutions.
|
COMMENT:
n is odd, let $n=(2m+1)^2$ ; m∈Z, then:
$4m(m+1)=12x^2(1+x)$ ⇒ $m(m+1)=3 x^2 (1+x)$
Suppose $m=3t$, then:
$t(3t+1)=x^2(1+x)$
⇒ $n=(2m+1)^2=(6t+1)^2$
For example $n=31^2= (6\times 5 +1)^2$ or $n=55^2=(6\times 9+1)^2$
Therefore there may be more solutions. Now let $m+1=3t$ then:
$(3t-1)3t=3x^2(1+x)$ ⇒ $(3t-1) t=x^2(1+x)$ ⇒ $n=(2m+1)^2= (6t-1)^2$
For example $n=5^2=( 6\times 1 -1)^2$
In this case there may also be some solutions.
Moreover n can be the square of negative numbers, so it can have general forms $n=[±(6t+1)]^2$ or $n=[±(6t-1)]^2$.We have to show whether equation $x^3+x^2=3t^2±t$ can have infinite solutions or limited number of solutions in Z.
|
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|
Error in finding the derivative of the general quadratic function $f(x)=ax^2+bx+c$ Where is the error in this 'proof' that the derivative of the quadratic function is equal to $x(2a+1)$? Note: I use $h$ to denote a small number.
$$f(x)=ax^2+bx+c\\f'(x)=\lim\limits_{h \to 0} \frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h}\\=\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+hx+c-ax^2-bx-c}{h}\\=\lim\limits_{h \to 0}\frac{2ahx+ah^2+hx}{h}\\=\lim\limits_{h \to 0}2ax+ah+x\\=2ax+x\\=x(2a+1)$$
|
You wrote $$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+\color{red}{hx}+c-ax^2-bx-c}{h}$$ instead of $$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+\color{blue}{bh}+c-ax^2-bx-c}{h}.$$
Simplifying, it becomes
$$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+bh+c-ax^2-bx-c}{h}=\lim_{h\to0}\frac{h(2ax+b)}{h}=2ax+b$$
as desired.
|
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|
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Tried dividing by $y$ to get $$y^\prime+\frac{x}{y}=\frac{\sqrt{x^2+y^2}}{y}$$
$$(y^\prime)^2+2\frac{x}{y}+\frac{x^2}{y^2}=\frac{x^2+y^2}{y^2}$$
$$(y^\prime)^2+2\frac{x}{y}=1$$
$$y^\prime=\sqrt{1-2\frac{x}{y}}$$
Tried using $v=\frac{y}{x}$
$$y^\prime=v^\prime x+v$$
$$v^\prime x + v=\sqrt{1-2v^{-1}}$$
Not sure what to do now or if I've even done something right until now.
|
Another way:
$$yy^\prime+x=\sqrt{x^2+y^2}$$
$$ydy+xdx=\sqrt{x^2+y^2}dx$$
$$\frac 12 (dy^2+dx^2)=\sqrt{x^2+y^2}dx$$
$$\frac 12 \frac {d(x^2+y^2)}{\sqrt{x^2+y^2}}=dx$$
Integrate.
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|
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$
Let's see what happens:-
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$
In the same way for further terms, we will get $0$
Let's also confirm for general term
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$
So the whole expression $E$ will be zero
But actual answer is $1$
Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$
Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?
I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.
But I got the following counter thought:-
$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$
As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$
$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$
$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$
Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.
But surprisingly $\dfrac{1}{3}$ is the correct answer.
I am feeling very confused in these two things. Please help me.
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That is not legitimate, because the number of terms inside the limit are growing as $n$ tends to infinity.
Rather, one sees that
\begin{align*}
\sum_{k=1}^{n}\dfrac{n}{n^{2}+n}\leq\sum_{k=1}^{n}\dfrac{n}{n^{2}+k}\leq\sum_{k=1}^{n}\dfrac{n}{n^{2}+1},
\end{align*}
the left and right-sided both tend to $1$.
|
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|
evaluate $\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}$ $$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}
$$
I have tried the following:
$$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}=\\
\lim_{n\to\infty}\left(\frac{n^3}{n^3}\right)\cdot\left(\frac{1-\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}}{1+\frac{2}{n}-\frac{1}{n^3}}\right)^{\left(\frac{n^2}{n^2}\right)\left(\frac{8n-3+\frac{2}{n}-\frac{2}{n^2}}{10+\frac{3}{n}-\frac{1}{n^2}}\right)}$$
So I have a limit of $1^{\infty}$
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Write$$f(n):=\frac{n^3-n^2+2n+3}{n^3+2n_2-1}=\frac{1-\frac1n+o\left(\frac1n\right)}{1+\frac2n+o\left(\frac1n\right)}=1-\frac3n+o\left(\frac1n\right)$$and$$g(n):=\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}=\frac45n+o(1)$$so$$\lim_{n\to\infty}f(n)^{g(n)}=\exp\left(-3\cdot\frac45\right)=\exp\left(-\frac{12}{5}\right).$$
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|
Finding the eigenvalues of a $3 \times 3$ matrix without the determinant I am trying to find the eigenvalues of
$$A = \begin{pmatrix} 0 & -2 & -3 \\ -1 & 1 & -1 \\ 2 & 2 & 5\end{pmatrix}$$
Without using the determinant. I first tried doing something like I did here Finding the eigenvalues of $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ matrix without the determinant
And from the system
$\begin{cases}
-\lambda x_1 & -2x_2 & -3x_3 &= 0 \\
-x_1 &+ \ (1-\lambda)x_2 &- x_3 &= 0 \\
2x_1 &+ \ 2x_2 &+ \ (5 - \lambda)x_3 &= 0
\end{cases}$
Tried to solve for
$\cfrac{-\lambda x_1 - 2x_2}{3} = -x_1 + (1 - \lambda)x_2 = \cfrac{2x_1 + 2x_2}{\lambda - 5}$
But nothings seems to come of that. I then proceeded to try to find $\lambda$ such that
$$\begin{pmatrix} -\lambda & -2 & -3 \\ -1 & 1-\lambda & -1 \\ 2 & 2 & 5-\lambda\end{pmatrix}$$ is non-invertible, by looking at the reduced row echelon form and see for what $\lambda$ the rank is less than 3.
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & \lambda^2 - \lambda - 2 & \lambda - 3 \\ 0 & 2(2-\lambda) & 3 - \lambda \end{pmatrix}
\rightarrow
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 2(2-\lambda) & 3 - \lambda \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 2$. Another operation yields:
$\begin{equation}
\begin{split}
\rightarrow
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 0 & 3 - \lambda \Big (1 + \cfrac{2(2-\lambda}{\lambda ^2 - \lambda - 2} \Big) \end{pmatrix} \rightarrow
\begin{pmatrix} 1 & \lambda - 1 & 1 \\ 0 & 1 & \cfrac{\lambda - 3}{\lambda^2 - \lambda - 2} \\ 0 & 0 & 1 \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 1, 3$ (because these are roots of the polynomial).
Now it happens that the eigenvalues are indeed $\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 3$. But I'm not sure how what I've done makes sense as a mathematical process for finding eigenvalues. Moreover it doesn't even seem to fully work in this other example
$B = \begin{pmatrix} 1 & 5 \\ 3 & 3 \end{pmatrix}$
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 - \lambda & 5 \\ 3 & 3 - \lambda \end{pmatrix} \rightarrow \begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 3 & 3 \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq 5$.
$\begin{equation}
\begin{split}
\begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 0 & \cfrac{-15}{1 - \lambda} + 3 - \lambda \end{pmatrix} \rightarrow \begin{pmatrix} 1 & \cfrac{5}{1 - \lambda} \\ 0 & 1 \end{pmatrix}
\end{split}
\end{equation}$
Where $\lambda \neq -2, 6$.
Here it happens that $B$ has eigenvalues $\lambda_1 = -2, \lambda_2 = 6$, but there is no eigenvalue for $1$. So what's going on here?
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You have to look at the factors you're dividing the rows by to make diagonal elements $1$: their product turns out to be $(+/-)$ the characteristic polynomial, so its roots will be the eigenvalues. You say you want to do this "without using the determinant", but your row reduction procedure (keeping track of those factors) is exactly what you use to compute the determinant.
Thus in your first example, you
interchanged two rows (which multiplies the determinant by $-1$), then divided by $-1$ to get the $1$ in position $(1,1)$, then added multiples of the first row to get $0$'s in the rest of the first column. Then you divided the second row by
$\lambda^2 - \lambda - 2$, and added a multiple of the second row to the third to get a $0$ in position $(3,2)$. Finally you divided the third row by
$$3-\lambda+{\frac { \left( -4+2\,\lambda \right) \left( -3+\lambda
\right) }{{\lambda}^{2}-\lambda-2}}=-{\frac { \left( -1+\lambda
\right) \left( -3+\lambda \right) }{\lambda+1}}
$$
So the product of the factors is
$$ - (\lambda^2 - \lambda - 2) {\frac { \left( -1+\lambda
\right) \left( -3+\lambda \right) }{\lambda+1}} =- (\lambda - 1)(\lambda - 2)(\lambda - 3) $$
and the characteristic polynomial is $(\lambda - 1)(\lambda - 2)(\lambda-3)$
so the eigenvalues are $1, 2, 3$.
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|
Show $x^2+y^2=9z+3$ has no integer solutions
Show $x^2+y^2=9z+3$ has no integer solutions
So I know that $x^2+y^2=3(3z+1)$
And since $3\mid (9z+3)$ and $3\not\mid (3z+1)$ then $9z+3$ cannot be a square since it has a prime divisor which has a power less then $2$.
So does know this isn't a pythagorean triple $(x,y,\sqrt{9z+3})$ give me that there are no solutions or do I have to show more?
|
You've shown that $3|9x + 3$. You've also shown that $9\not \mid 9x+3$.
SO that means $3|x^2 + y^2$ so but $9\not \mid x^2 + y^2$.
If $x$ and $y$ are divisible by $3$ then $x^2 + y^2$ is divisible by $9$. So that's out. If one of $x$ or $y$ is divisible by $3$ but the other isn't then $x^2 + y^2$ is not divisible by $3$ and that's out.
So the only options is neither $x$ nor $y$ is divisible by $3$.
So $x \equiv \pm 1 \pmod 3$ and $y \equiv \pm 1 \pmod 3$. Which means $x^2 \equiv (\pm 1)^2 \equiv 1 \pmod 3$ and $y^2 \equiv (\pm 1)^2 \equiv 1 \pmod 3$.
So $x^2 + y^2 \equiv 2\pmod 3$ and $3\not \mid x^2 + y^2$ so that's out.
So all three options are out.
|
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|
Would this be a legal move in a Matrix? Here I have a matrix:
\begin{pmatrix}
b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{pmatrix}
I was wondering can I do the following operation on this:
$C2 \to C2 \times C2$
So it would give me the following:
\begin{pmatrix}
b^2c^2 & b^2c^2 & b+c \\
c^2a^2 & c^2a^2 & c+a \\
a^2b^2 & a^2b^2 & a+b \\
\end{pmatrix}
This is needed in order to prove the determinant is equal to zero.
|
Since this will (for most $a$ and $b$) change the value of the determinant, lets use co-factor multiplication by your matrix $M$ to prove this, which shouldn't be too bad given it is a $3x3$ matrix. I will denote each $2x2$ co factor as $A_i$ for simplicity.
$$det(M) = (b+c)det(A_1) - (c+a)det(A_2) + (a+b)det(A_3)$$
$$det(M) = (b+c)(a^3c^2b-a^3b^2c)-(c+a)(c^3b^2a-c^3a^2b)+(a+b)(b^3c^2a -b^3a^2c)$$
Hmm... maybe this wasn't the right approach but whatever too late now
$$det(M) = (a^3c^2b^2-a^3b^3c+a^3c^3b-a^3b^2c^2)-(b^3c^3a-b^3a^2c^2+b^3c^2a^2-b^3a^3c)+(c^3b^2a^2-c^3a^3b+c^3b^3a-c^3a^2b^2)$$
Jesus christ... notice that we can take out a $a^3, b^3$ or $c^3$
$$det(M) = a^3(c^2b^2-b^3c+c^3b-b^2c^2)-b^3(c^3a-a^2c^2+c^2a^2-a^3c)+c^3(b^2a^2-a^3b+b^3a-a^2b^2)$$
And then do even better by noticing we can take out a $ab, bc$ or $ac$
$$det(M) = a^3bc(cb-b^2+c^2-bc)-b^3ca(c^2-ac+ca-a^2)+c^3ba(ba-a^2+b^2-ab)$$
Now some terms cancel... thank god....
$$det(M) = a^3bc(-b^2+c^2)-b^3ca(c^2-a^2)+c^3ba(-a^2+b^2)$$
Rearrange some more..
$$det(M) = -a^3bc(b^2-c^2)-b^3ca(c^2-a^2)-c^3ba(a^2-b^2)$$
Factor out a $-abc$
$$det(M) = -abc(a^2(b^2-c^2)+b^2(c^2-a^2)+c^2(a^2-b^2))$$
Distribute
$$det(M) = -abc((a^2b^2-a^2c^2)+(b^2c^2-b^2a^2)+(c^2a^2-c^2b^2))$$
Reorder terms
$$det(M) = -abc(a^2b^2-b^2a^2-a^2c^2+c^2a^2+b^2c^2-c^2b^2)$$
Thus
$$det(M) = -abc(0)$$
$$det(M) = 0$$
I retract my former statement that this wouldn't be too bad, but luckily I denoted the co factors as $A_i$, saving me loads of time.
|
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|
$\int_0^\infty \frac{\sin^n x}{x^m}dx$ could be expressed via $\pi$ or $\log$ I want to show some results first (they were computed by MMA)
$$
\int_0^\infty \frac{\sin^5 x}{x^3} dx =\frac{5}{32}{\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^5 x}{x^5} dx =\frac{115}{384} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^5 x}{x^2} dx =\frac{5}{16}\,{\color{Red}\log}\, \frac{27}{5} \quad
\int_0^\infty \frac{\sin^5 x}{x^4} dx =-\frac{5}{96}(27\,{\color{Red }\log } \,3-25\,{\color{Red}\log }\,5) \\
$$
and
$$
\int_0^\infty \frac{\sin^6 x}{x^4} dx =\frac{1}{8} {\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^6 x}{x^6} dx =\frac{11}{40} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^6 x}{x^3} dx =\frac{3}{16}\,{\color{Red}\log}\, \frac{256}{27} \quad
\int_0^\infty \frac{\sin^6 x}{x^5} dx ={\color{Red}\log}\, \frac{3^\frac{27}{16}}{4} \\
$$
As we can see, in the integral $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$
, if $n-m$ is even, the integral is expressed via $\pi$, and if $n-m$ is odd, the integral is expressed via $\log$. Amazing for me, it seems always true such as
$$\int_0^\infty \frac{\sin^8 x}{x^2} dx =\frac{5\pi}{32}$$ and
$$\int_0^\infty \frac{\sin^8 x}{x^3} dx =\frac{9}{8}\log\frac{4}{3}$$
Do we have a general method to compute $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$ which implies these laws?
My attempt
This post tells us how to compute $\displaystyle\int_0^\infty \frac{\sin^n x}{x^n}dx \tag{*}$
We can compute some other $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$ via $(*)$, such as via the formulas
$$\displaystyle\int_{0}^{\infty}\dfrac{\sin^3 x}{x}\,dx = \dfrac{3}{4}\int_{0}^{\infty}\dfrac{\sin x}{x}\,dx - \dfrac{1}{4}\int_{0}^{\infty}\dfrac{\sin 3x}{x}\,dx$$
and
$$\int_0^\infty \frac{\sin^2 (2x)}{x^2}dx=\int_0^\infty \frac{4\sin^2 x-4\sin^4 x}{x^2}$$
But it's complicated to compute the general cases. Could you please share some ideas of a possible method to show the law mentioned above?
|
The aim of this answer is to give the explicit expression for the value of the integrals. Essentially it is a development of the previous answer. Notice that for convenience I changed $m$ to $m+1$.
We are going to prove:
For all $0\le m<n$:
$$
\int_0^\infty\frac{\sin^n x}{x^{m+1}}dx
=\frac{(-1)^{\left\lfloor\frac{n-m-1}2\right\rfloor}}{2^{n-1}m!}
\begin{cases}
\displaystyle\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}
(-1)^k\binom nk(n-2k)^m\;\log(n-2k),&n-m=0\operatorname{mod}2;\\
\displaystyle\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}
(-1)^k\binom nk(n-2k)^m\;\frac\pi2,&n-m=1\operatorname{mod}2.\\
\end{cases}\tag1
$$
Observe that for $m=0$ and even $n$ both sides of (1) diverge.
A sketch of the proof:
We start with the expression:
$$\begin{align}
\int_0^\infty\frac{\sin^n x}{x^{m+1}}dx
&=\frac1{(2i)^n}\int_0^\infty\frac{(e^{ix}-e^{-ix})^n}{x^{m+1}}dx\\
&=\frac1{(2i)^n}\int_0^\infty\frac{dx}{x^{m+1}}\sum_{k=0}^n(-1)^k\binom nk e^{i(n-2k)x}.\tag2
\end{align}
$$
Integrating (2) by parts $m$ times one arrives at:
$$
\int_0^\infty\frac{\sin^n x}{x^{m+1}}dx=
\frac1{(2i)^nm!}\int_0^\infty\frac{dx}{x}\sum_{k=0}^n(-1)^k\binom nk i^m(n-2k)^m e^{i(n-2k)x}.\tag3
$$
Now observe that $\dfrac{(-1)^{n-k}(2k-n)^m}{(-1)^k(n-2k)^m}=(-1)^{n-m}$. Therefore the expression (3) can be rewritten as:
$$
\int_0^\infty\frac{\sin^n x}{x^{m+1}}dx
=
\begin{cases}
\displaystyle \frac{i^{m-n}}{2^{n-1}m!}\int_0^\infty\frac{dx}x
\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}
(-1)^k\binom nk(n-2k)^m \cos(n-2k)x,&n-m=0\operatorname{mod}2;\\
\displaystyle\frac{i^{m-n+1}}{2^{n-1}m!}\int_0^\infty\frac{dx}x
\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}
(-1)^k\binom nk(n-2k)^m \sin(n-2k)x &n-m=1\operatorname{mod}2.\\
\end{cases}\tag4
$$
The lower line together with the well-known identity $$\int_0^\infty\frac{\sin ax}x dx=\frac\pi2\operatorname{sgn}a$$ immediately gives the corresponding line of (1).
To obtain the upper line of (1) one observes that:
$$
\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}(-1)^k\binom nk(n-2k)^m=0\tag5
$$
for even values of $n-m$. A proof can be found elsewhere. Essentially it is based on the fact that
$$
\int_{-\infty}^\infty\frac{\sin^nx}{x^{m+1}}dx=0,
$$
since the integrated function is odd.
In view of (5) we can rewrite the upper line in r.h.s. of (4) as:
$$
\frac{i^{m-n}}{2^{n-1}m!}
\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}
(-1)^k\binom nk(n-2k)^m \int_0^\infty\frac{\cos(n-2k)x-\cos x}xdx.
$$
The resulting integral is of (generalized) Frullani type, so:
$$
\int_0^\infty\frac{\cos(n-2k)x-\cos x}xdx=-\log(n-2k),
$$
proving the upper line of (1).
|
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"url": "https://math.stackexchange.com/questions/3476409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$
I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.
|
Another possible approach consists in considering the change of variables $x = r\cos(\theta)$ and $y = r\sin(\theta)$, from whence we get
\begin{align*}
\begin{cases}
r^{2}\cos(2\theta) = 7\\\\
r^{2}(2 + \sin(2\theta)) = 74
\end{cases} \Longleftrightarrow r^{2} = \frac{7}{\cos(2\theta)} = \frac{74}{2+\sin(2\theta)}
\end{align*}
Can you take it from here?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the next year of the year $2016$ written as $n^{m-1}(n^{m}-1)$ See that : $2016=2^{6-1}(2^{6}-1)$
Find the next year of the year $2016$ written as $n^{m-1}(n^{m}-1)$ where $n,m>1$ natural numbers
I don't have any ideas to start in this problem
My be we need the factorization of $2016=2^{5}3^{2}.7$
But I need know the relation between factorization primary and second
|
Since $n^{m-1}(n^{m}-1)$ increases quite fast , we can try inputting some values of $m$
For $m=2$ , we get $n(n^2-1) \gt 2016$ , which is true for $n \ge 13. $ So the smallest value is at $n = 13$ , which gives $13 \times 168 = \color{#2dd}{2184 }.$
For $m=3$ , we get $n^2(n^3-1) \gt 2016$ , which is true for $n \ge 5 .$ So the smallest value is at $n = 5$ , which gives $25 \times 124 = \color{#5d0}{3100}.$
For $m=4$ , we get $n^3(n^4-1) \gt 2016$ , which is true for $n \ge 3 .$ So the smallest value is at $n = 3$ , which gives $27\times 80 = \color{#d05}{2160} . $
For $m=5$ , we get $n^4(n^5-1) \gt 2016$ , which is true for $n \ge 3 .$ So the smallest value is at $n = 3$ , which gives $81\times 242 = \color{#20f}{19602 }.$
For $m=6$ , we already have the value $2^5(2^6-1)$ and for $m \gt 6$ , the equation is not true for $n,m \gt 1$
So the next smallest value is $\boxed{2160 }.$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3476997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Normal to the curve $y=x^2$ forming the shortest chord
Find the normal to the curve $y=x^2$ forming the shortest chord
Normal at point $P(x,y)=(t,t^2)$ is $x+2ty=t+2t^3$, which meets the point $Q(m,m^2)$
$$
m+2tm^2=t+2t^3\implies m-t=-2t(m^2-t^2)=-2t(m-t)(m+t)\\
t+m=\frac{-1}{2t}\implies m=-t-\frac{1}{2t}\\
T=\Delta^2=(x_1-x_2)^2+(y_1-y_2)^2\\
=(t-m)^2+(t^2-m^2)^2=(t-m)^2[1+(t+m)^2]=(2t+\frac{1}{2t})^2\big[1+\frac{1}{4t^2}\big]\\
=(\frac{4t^2+1}{2t})^2(\frac{4t^2+1}{4t^2})=\bigg[\frac{4t^2+1}{2t}\bigg]^3\\
\frac{dT}{dt}=3\bigg[\frac{4t^2+1}{2t}\bigg]^2.\frac{2t(8t)-(4t^2+1).2}{4t^2}=3\bigg[\frac{4t^2+1}{2t}\bigg]^2.\frac{16t^2-8t^2-2}{4t^2}\\
=3\bigg[\frac{4t^2+1}{2t}\bigg]^2.2.\frac{4t^2-1}{(2t)^2}=0\\
\implies t^2=\frac{1}{4}\implies t=\pm\frac{1}{2}\\
N:x+y=3/4
$$
But my reference gives the solution $y=\pm\dfrac{x}{\sqrt{2}}+1$, which is asked Which normal to $=^2$ forms the shortest chord?, but I really can't figure out what is going wrong with my attempt ?
|
$\left(\frac{4t^2+1}{2t}\right)^2\frac{4t^2+1}{4t^2}=\frac{(4t^2+1)^3}{(2t)^4}$ and not $\left(\frac{4t^2+1}{2t}\right)^3$. With this correction it should continue like user207119's answer to the linked question.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$. Problem: Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$.
My efforts: $(y-a)^2(x^2-a^2)=x^4+a^4\implies (y-a)^2=\dfrac{x^4+a^4}{x^2-a^2}\implies x=\pm a$ are vertical asymptotes. What are the remaining asymptotes?
|
This doesn't satisfy
the usual definition
of asymptote but ...
If
$(y-a)^2(x^2-a^2)=x^4+a^4
$
then
$\begin{array}\\
(y-a)^2
&=\dfrac{x^4+a^4}{x^2-a^2}\\
&=\dfrac{x^4-a^2x^2+a^2x^2+a^4}{x^2-a^2}\\
&=\dfrac{x^2(x^2-a^2)+a^2(x^2+a^2)}{x^2-a^2}\\
&=x^2+\dfrac{a^2(x^2-a^2+2a^2)}{x^2-a^2}\\
&=x^2+a^2+\dfrac{2a^4}{x^2-a^2}\\
\text{so}\\
y-a
&=\sqrt{x^2+a^2+\dfrac{2a^4}{x^2-a^2}}\\
&=x\sqrt{1+a^2/x^2+\dfrac{2a^4}{x^2(x^2-a^2)}}\\
&\approx x(1+a^2/(2x^2)+\dfrac{2a^4}{2x^2(x^2-a^2)})
\qquad\text{for large }x\\
&= x+a^2/(2x)+O(1/x^3)\\
\end{array}
$
so $y = x+a$
is a "asymptote".
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of $\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$ as $x\to\infty$ I wish to find the limit of $$\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$$ as $x\to\infty$. I think that this limit does not exist.
|
Since
$$
\sin p - \sin q = 2\cos \left( {\frac{{p + q}}
{2}} \right)\sin \left( {\frac{{p - q}}
{2}} \right)
$$
you have that
$$
\begin{gathered}
\left| {\sin p - \sin q} \right| = 2\left| {\cos \left( {\frac{{p + q}}
{2}} \right)\sin \left( {\frac{{p - q}}
{2}} \right)} \right| = \hfill \\
\hfill \\
= 2\left| {\sin \left( {\frac{{p - q}}
{2}} \right)\cos \left( {\frac{{p + q}}
{2}} \right)} \right| \leqslant 2\left| {\frac{{p - q}}
{2}} \right| \hfill \\
\end{gathered}
$$
thus
$$
\left| {\sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right)} \right| \leqslant \left| {\sqrt {x + 2} - \sqrt {x + 4} } \right|
$$
But
$$
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {x + 2} - \sqrt {x + 4} } \right) = 0
$$
so
$$
\mathop {\lim }\limits_{x \to + \infty } \left| {\sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right)} \right| = 0
$$
and
$$
\mathop {\lim }\limits_{x \to + \infty } \sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right) = 0
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding mixed probability density function. Please check my answer. Given random variable $X$ with cumulative distributive function
$$
F_X(x)=
\begin{cases}
0&x<0
\\
\dfrac{1}{4}x^2&0\leq x<1
\\
\dfrac{1}{2}&1\leq x<2
\\
\dfrac{1}{3}x&2\leq x<3
\\
1&x\geq 3
\end{cases}.
$$
Find the probability density function of $X$.
To find the p.d.f, I plot the graph of $F_X(x)$,as below.
Based on picture, I conclude $X$ is mixed random variable.
For continuous random variable, differentiating $F_X(x)$ we have
$$f_X(x)=
\begin{cases}
\dfrac{1}{2}x&0\leq x< 1\\
\dfrac{1}{3}&2\leq x<3
\end{cases}$$
For discrete random variable, we have
$$f(1)=F_X(1^+)-F_X(1^-)=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4},$$
$$f(2)=F_X(2^+)-F_X(2^-)=\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{1}{6}.$$
Now I want check $f(x)$ is p.d.f or not p.d.f.
\begin{eqnarray}
\int\limits_{0}^{1} \dfrac{1}{2}x dx + \int\limits_{2}^{3} \dfrac{1}{3} dx &=&
\left[\dfrac{1}{4}x^2\right]_0^1 + \left[\dfrac{1}{3}x\right]_2^3\\
&=& \dfrac{1}{4}+1-\dfrac{2}{3}\\
&=& \dfrac{7}{12}
\end{eqnarray}
total probability of discrete and continuous:
$$\dfrac{7}{12}+\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{14+6+4}{24}= 1.$$
So, we have p.d.f. for continous variable
$$f_X(x)=
\begin{cases}
\dfrac{1}{2}x&0\leq x< 1\\
\dfrac{1}{3}&2\leq x<3
\end{cases}$$
and for discrete random variable
$f(1)=\dfrac{1}{4}$ and
$f(2)=\dfrac{1}{6}.$
Is my answer correct?
|
As you have correctly observed, $F_X$ is not continuous and thus $X$ cannot have an (absolutely) continuous distribtion. So there is no (measurable) positive function $f$ such that
$$F_X(x) = \int_{-\infty}^x f(t) dt, x \in \mathbb{R}$$
and thus no 'density' exists.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Square and rectangle The following figure shows an infinite checkered mesh, made up of $1$ cm squares, with the vertices of the highlighted squares.
Note that a square and rectangle with vertices in the mesh are highlighted. The square has exactly $5 $ mesh points inside and the rectangle has exactly $2$ mesh points inside.
a) Present a rectangle with vertices in the mesh that has exactly $2019$ mesh points in its
inside.
b) Show that there are two squares with vertices in the mesh and with non-parallel sides, so that there are
exactly $100$ mesh points inside each one.
c) Show that there is a square with vertices in the mesh and that it has exactly $84$ mesh points in its
inside.
Attemp: a. The rectangle with vertices $(0,0), (2020,0), (0, 2), (2020, 2)$ will work.
b. Let $2$ vertices be $(0,0), (ad, bd)$ where $\gcd(a,b) = 1$ WLOG. There are $4(d+1)-4 = 4d$ boundary points and the area is $d^2(a^2+b^2).$ By Pick's formula, $d^2(a^2+b^2) = 100 + 2d - 1 = 99+2d,$ so $d|99$ by taking $\mod d.$ Letting $99 = dk,$ we get $d(a^2+b^2) = k+2,$ so $d| (99/d + 2) \Rightarrow d = 1, 11.$ If $d=1,$ then $a^2+b^2=101$ which has solutions $(\pm 10, \pm 1)$ and permutations. If $d = 11,$ then $a^2+b^2 = 1,$ which has solutions $(\pm 1, 0)$ and permutations. This gives $2$ unique squares.
c. Same principle, but now $d^2(a^2+b^2) = 2d+83,$ so $d|(83/d + 2),$ so $d=1,$ which gives $a^2+b^2=85,$ with solutions $(\pm 9, \pm 2), (\pm 7, \pm 6)$ for a total of $2$ squares up to reflection and rotation.
b) Using Pick's formula should be easy but I don't know how to use it
How can I do this?
|
For part b, the square with vertices at $(1,0)$, $(0,10)$, $(10,11)$, and $(11,1)$ and its mirror image (across the line $y=x$) does the trick: each contains the $100$ points $(i,j)$ with $1\le i,j\le10$.
|
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|
Finding the Fourier series of $\frac{4-2\cos x}{5-4\cos x}$ I am required to find the Fourier Series of the following function, on $[-\pi,\pi]$:
$$f(x)=\frac{4-2\cos(x)}{5-4\cos(x)}$$
This question seemed simple, at first. I managed to compute $a_0$ (which wasn't too easy) and noticed that $b_n=0$ for all $n\in\mathbb{N}$ (since the given $f$ is even). However, computing the coefficients $a_n$ seemed like an impossible mission. So I cheated a little, and looked at the final answer, which is:
$$f(x)\simeq1+\sum_{n=1}^{\infty}\frac{1}{2^n}\cos(nx)\triangleq\mathfrak{F}(x)$$
The problem is $a_n$ - I don't know how to compute them, and therefore I couldn't find the Fourier Series by myself. However, after finding out what the final answer is, I thought of a creative way to show the final answer is actually the Fourier Series of $f(x)$. This is kind of cheating, but I'll show my solution anyway:
Let $z=\frac{\cos(x)+i\sin(x)}{2}, z\in\mathbb{C}.$ Using the fact that $z^n=\frac{\cos(nx)+i\sin(nx)}{2^n}$, and that $|z|=\frac 12<1$, we receive:
$$\mathfrak{F}(x)=\sum_{n=0}^{\infty}\Re{(z^n)}=\Re{\left(\sum_{n=0}^{\infty}z^n\right)}=\Re{\left(\frac{1}{1-z}\right)}$$$$=\Re{\left(\frac{1-\bar{z}}{|1-z|^2}\right)}=\Re{\left(\frac{1-0.5\cos(x)+0.5i\sin(x)}{(1-0.5\cos(x))^2+(0.5\sin(x))^2}\right)}$$$$=\frac{1-0.5\cos(x)}{1-\cos(x)+0.25\cos^2(x)+0.25\sin^2(x)}=\frac{4-2\cos(x)}{5-4\cos(x)}\equiv f(x)$$
We know that $f(x)$ is differentiable everywhere. Therefore, it has only one Fourier Series, and it has to be $\mathfrak{F}(x)$, since it is a Fourier Series that converges to $f(x)$. This might not be a true statement (so please correct me if so), but either way, I would like to hear your thoughts about a more legit solution to the problem (without knowing the Fourier Series in advance).
|
Here is a noncomplex way to do the integral. Multiply top and bottom by $5+4\cos x$:
$$\frac{20+6\cos x +8\cos^2x}{25-16\cos^2 x} = -\frac{1}{2} +\frac{6\cos x}{9+16\sin^2 x} +\frac{65}{2}\frac{1}{9+16\sin^2 x}$$
$$I = \int_{-\pi}^\pi -\frac{1}{2} + \frac{6\cos x}{9+16\sin^2 x} +\frac{65}{2}\frac{1}{9+16\sin^2 x} \:dx = \pi +\frac{1}{2}\arctan\left(\frac{4}{3}\sin x \right)\Biggr|_{-\pi}^\pi + \frac{65}{2}\int_{-\pi}^\pi \frac{\sec^2 x}{9 + 25\tan^2 x} \:dx$$
$$ = \pi + 0 + \frac{13}{6}\arctan\left(\frac{5}{3}\tan x \right)\Biggr|_{-\pi}^\pi $$
That last limit is not simple to evaluate because of the singularities, but just take the limits separately to each one. Each interval of $\pi$ picks up a value of $\pi$ so we have
$$\pi +\frac{13\pi}{3} = \frac{16\pi}{3}$$
|
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|
How to evaluate $\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$ I am trying to calculate this integral, but I find it is very challenging
$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$$
but somehow I have managed to local its closed form to be $$(-1)^n\left(\frac{1}{2}-n\ln 2+\sum_{j=0}^{n-1}H^{*}_j\right)$$
where $n\ge 0$, $H^{*}_0=0$ and $H^{*}_k=\sum_{j=1}^{k}\frac{(-1)^{j+1}}{j}$
I have try
$$\frac{1-x}{1+x}\cdot \frac{x^n}{(x^2-1)^{1/2}}$$
$$-x^n\sqrt{\frac{x-1}{(x+1)^2}}$$
$$-\int_{0}^{1}x^n\sqrt{\frac{x-1}{(x+1)^3}}\mathrm dx$$
from this point I tried to use the binomial to expand $$\sqrt{\frac{x-1}{(x+1)^3}}$$ but it seem not possible
|
My solution is similar to that of @omegadot but I think it may be a little cleaner. For $n\in\mathbb{Z}^+$ let
$$C_n=(-1)^n\int_0^1\frac{x^{n-1}}{x+1}dx$$
Then
$$C_1=-\int_0^1\frac{dx}{x+1}=\left.-\ln(x+1)\right|_0^1=-\ln2$$
$$C_{n+1}-C_n=(-1)^{n+1}\int_0^1\frac{x^n+x^{n-1}}{x+1}dx=(-1)^{n+1}\int_0^1x^{n-1}dx=\frac{(-1)^{n+1}}n$$
So we can sum a telescoping series to get
$$C_n-C_1=\sum_{k=1}^{n-1}\left(C_{k+1}-C_k\right)=\sum_{k=1}^{n-1}\frac{(-1)^{k+1}}k$$
Then since
$$\sqrt{x^4-2x^2+1}=\sqrt{\left(1-x^2\right)^2}=\left|1-x^2\right|=1-x^2=(1+x)(1-x)$$
for $0\le x\le 1$, we have for $n\in\mathbb{Z}^+$
$$\begin{align}\int_0^1\frac{1-x}{1+x}\cdot\frac{x^n}{\sqrt{x^4-2x^2+1}}dx&=\int_0^1\frac{x^n}{(x+1)^2}dx=\left.-\frac1{(x+1)}x^n\right|_0^1+n\int_0^1\frac{x^{n-1}}{x+1}dx\\
&=-\frac12+(-1)^nnC_n\\
&=-\frac12+(-1)^nn\left[-\ln 2-\sum_{k=1}^{n-1}\frac{(-1)^k}k\right]\end{align}$$
This is using a recurrence relation rather than an infinite series. Of course for $n=0$ the integral works out to
$$\int_0^1\frac{dx}{(x+1)^2}=\left.-\frac1{(x+1)}\right|_0^1=-\frac12+1=\frac12$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding the inverse function of: $y=3x^2-6x+1$ I am trying to find the inverse function, $f^{-1}(x)$ of this function, $y=3x^2-6x+1$
Normally the teacher told us to interchange the $x$ and $y$
$x=3y^2-6y+1$ and... now transpose the formula to make $y$ the subject. But this look not possible. Is there another alternative method to find the $f^{-1}(x)?$
|
$y=3x^2-6x+1=3(x^2-2x)+1$;
$y=3(x-1)^2 -3+1=$
$3(x-1)^2-2$;
A parabola opening upwards, vertex $(1,-2)$;
Local inverse for:
1) $x\gt 1$:
$(1/3)(y+2)=(x-1)^2$;
$x= 1+\sqrt{(1/3)(y+2)}$;
Switch:
$y=1+\sqrt{(1/3)(x+2)}$, $x \gt -2$.
Local inverse for
2) $x \lt 1$;
$-\sqrt{(1/3)(y+2)}=x-1$;
Switch:
$y=1-\sqrt{(1/3)(x+2)}$, $x \gt -2$.
Does a global inverse exist?
Is there an inverse function for $(-\epsilon+1, \epsilon +1)$, $\epsilon >0$ ?
|
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"url": "https://math.stackexchange.com/questions/3487759",
"timestamp": "2023-03-29T00:00:00",
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|
Find polynomials $P(x)P(x-3) = P(x^2)$
Find all polynomials $P \in \Bbb R[x]$ such that $P(x)P(x-3) = P(x^2) \quad \forall x \in \Bbb R$
I have a solution but I'm not sure about that. Please check it for me.
It is easy to see that $P(x) = 0, P(x) = 1$ satisfied.
Consider $P(x) \neq c$ :
We have $P(x+3)P(x) = P((x+3)^2)$
If $a$ is a root of $P(x) $ then $a^2$ and $(a+3)^2$ are roots of $P(x)$.
If $|a| > 1$, then $ |a^2| = |a||a| > |a| \Rightarrow P(x)$ has infinitely many roots.
If $0 < |a| < 1$, then $|a^2| = |a||a| < |a| \Rightarrow P(x)$ has infinitely many roots.
If $a = 0 \Rightarrow (a+3)^2 = 9$ is a root of $P(x)$, so $P(x)$ has infinitely many roots.
Hence, every root $a$ of $P(x)$ has $|a| = 1$. So $1 = |(a+3)^2| = (|a+3|^2) \Rightarrow |a+3|=1$.
We have $a + 3 = \cos\alpha + i\sin\alpha + 3 \Rightarrow (\cos\alpha+3)^2 + \sin^2\alpha = 1 \Rightarrow \cos\alpha = -\frac{3}{2}\ (\text{impossible}).$
$P(x) = 0, P(x) = 1$ are all the results.
|
You point out that if $a$ is a root then $a^{2}$ is a root and $a+3$ is a root.
You don't need to divide into three cases to determine that means either infinitely or zero roots.
By induction $a + 3k; k \in \mathbb N$ are roots. If $j\ne k$ then $a+3j \ne a + 3k$ so if there is one root $a$ there are at least countably infinite distinct roots. So there are either zero roots, of infinitely many roots.
Magnitude is irrelevant[1]
Only trivial constant polynomials have infinite or zero (complex) roots so that means $P(x) = c$ for some $c$ where $c^2 = P(x)P(x-3) =P(x^2) = c$.
[1]
Although if $a$ is a root then $a +3$ is a root. And if $|a+3| < 1$ then $|a+3| + |a|=|a+3| + |-a| \ge |(a+3)+(-a)| =|3|=3$ so $|a|\ge 3-|a+3| > 2$. So at least one root is $\ge 1$.
|
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|
How to do partial fractions with a denominator to the power of a variable? How can I take the following sum and simplify it with partial fractions? $$\sum^{\infty}_{k=1}\frac{k-1}{2^{k+1}}$$
I know the denominator can be rewritten as $(2^k)(2)$, but how do I deal with the $2^k$ when doing partial fractions?
Usually, when there is an exponent in the denominator, you put terms all the way through - if you have $x^3$ in the denominator you end up with $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}$. So, what do I do when I have $2^k$ - and I either do not know what $k$ or $k$ goes to infinity?
|
Here's a recurrence for
these power sums.
$\begin{array}\\
s_m(x)
&=\sum_{n=0}^{\infty} n^mx^n\\
&=\sum_{n=1}^{\infty} n^mx^n\\
&=x\sum_{n=1}^{\infty} n^mx^{n-1}\\
&=x\sum_{n=0}^{\infty} (n+1)^mx^{n}\\
&=x\sum_{n=0}^{\infty} x^{n}\sum_{k=0}^m \binom{m}{k}n^k\\
&=x\sum_{k=0}^m \binom{m}{k}\sum_{n=0}^{\infty} x^{n}n^k\\
&=x\sum_{k=0}^m \binom{m}{k}\sum_{n=1}^{\infty} x^{n}n^k\\
&=x\sum_{k=0}^m \binom{m}{k}s_k(x)\\
&=x\sum_{k=0}^{m-1} \binom{m}{k}s_k(x)+xs_m(x)\\
\end{array}
$
so
$\begin{array}\\
(1-x)s_m(x)
&=x\sum_{k=0}^{m-1} \binom{m}{k}s_k(x)\\
\end{array}
$
or
$\begin{array}\\
s_m(x)
&=\dfrac{x\sum_{k=0}^{m-1} \binom{m}{k}s_k(x)}{1-x}\\
s_0(x)
&=\dfrac{1}{1-x}\\
s_1(x)
&=\dfrac{x\sum_{k=0}^{m-1} \binom{m}{k}s_k(x)}{1-x}\\
&=\dfrac{x\dfrac{1}{1-x}}{1-x}\\
&=\dfrac{x}{(1-x)^2}\\
s_2(x)
&=\dfrac{x\sum_{k=0}^{1} \binom{2}{k}s_k(x)}{1-x}\\
&=\dfrac{x(s_0(x)+2s_1(x))}{1-x}\\
&=\dfrac{x(\dfrac{1}{1-x}+2\dfrac{x}{(1-x)^2})}{1-x}\\
&=\dfrac{x((1-x)+2x)}{(1-x)^3}\\
&=\dfrac{x(1+x)}{(1-x)^3}\\
s_3(x)
&=\dfrac{x\sum_{k=0}^{2} \binom{3}{k}s_k(x)}{1-x}\\
&=\dfrac{x(s_0(x)+3s_1(x)+3s_2(x))}{1-x}\\
&=\dfrac{x(\dfrac{1}{1-x}+3\dfrac{x}{(1-x)^2}+3\dfrac{x(1+x)}{(1-x)^3})}{1-x}\\
&=\dfrac{x((1-x)^2+3x(1-x)+3x(1+x))}{(1-x)^4}\\
&=\dfrac{x(1-2x+x^2+3x-3x^2+3x+3x^2)}{(1-x)^4}\\
&=\dfrac{x(1+4x+x^2)}{(1-x)^4}\\
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the remainder when $5^{55}+3^{55}$ is divided by $16$
Find the remainder when $5^{55}+3^{55}$ is divided by $16$.
What I try
$a^{n}+b^{n}$ is divided by $a+b$ when $n\in $ set of odd natural number.
So $5^{55}+3^{55}$ is divided by $5+3=8$
But did not know how to solve original problem
Help me please
|
Continuing your method:
$$5^{55}+3^{55}=(5+3)(5^{54}-5^{53}\cdot 3+5^{52}\cdot 3^2-\cdots +5^2\cdot 3^{52}-5\cdot 3^{53}+3^{54}) \Rightarrow \\
\small 5^{54}-5^{53}\cdot 3+5^{52}\cdot 3^2-\cdots +5^2\cdot 3^{52}-5\cdot 3^{53}+3^{54}\equiv 1-1+1-\cdots+1-1+1\equiv 1\pmod{2}$$
Hence:
$$5^{55}+3^{55}\equiv 8 \pmod{16}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit $\lim\limits_{x \to +\infty} \sqrt[n]{(x+a_1)(x+a_2)...(x+a_n)} - x$ =? What is the limit of $ f(x) = \sqrt[n]{(x+a_1)(x+a_2)...(x+a_n)} - x$ when $x$ goes to plus infinity? The number $n$ is fixed and $a_i$ are some constants.
All I know is the answer:
$(a_1 + a_2 + ... + a_n) / n$
I figured out that I can limit the root from above with the arithmetic mean. And this gives me some nice expression as an upper limit. But OK... this only proves the limit is $\leq (a_1 + a_2 + ... + a_n) / n$.
But how do I limit it from below with the same value. I tried using the harmonic mean but it seems it does not lead me to anything nice. So... any ideas?
|
Notice that $$(x+a_1)(x+a_2)\cdot (x+a_n)=x^n+ b x^{n-1}+\mathcal{O}(x^{n-2})$$
Where $b=a_1+a_2+\dots +a_n$, and
$$\sqrt[n]{x^n+bx^{n-1}+\mathcal{O}(x^{n-2})}=\left(x^n+bx^{n-1}+\mathcal{O}(x^{n-2})\right)^{1/n}=x \left(1+bx^{-1} +\mathcal{O}(x^{-2})\right)^{1/n}$$
Now, we can use the first order Taylor approximation of $(1+x)^q$ around $x=0$ (which is $1+q x+\mathcal{O}(x^2)$) to get
$$x \left(1+b +\mathcal{O}(x^{2-n})\right)^{1/n}= x\left(1+\frac{b}{n x} +\mathcal{O}(x^{-2})\right)=x-\frac{b}{n } +\mathcal{O}(x^{-1}). $$
From this, we can see that the limit is $\frac{b}{n}$.
A more elementary solution: Let $f(x)=\sqrt[n]{(x+a_1)(x+a_2)\cdot (x+a_n)}$, note that
$$\lim_{x\to \infty} \frac{f(x)}{x}=\lim_{x\to \infty} \sqrt[n]{(1+a_1x^{-1})(1+a_2x^{-1})\cdot (1+a_nx^{-1})}=1\tag{1}$$
Note that
$$f(x)-x=\frac{f(x)^n-x^n}{\sum_{k=0}^{n-1} f(x)^{n-1-k} x^{k}}= \frac{f(x)^n-x^n}{f(x)^{n-1}+f(x)^{n-2}x+\dots+f(x)x^{n-2}+x^{n-1}}$$
The numerator is a polynomial of degree $n-1$ with a leading coefficient of $a_1+a_2+\dots+a_n$, the denominator is composed of the sum of $n$ terms $f(x)^{n-1-k}x^k$. Divide both by $x^{n-1}$ to get
\begin{align*}\lim_{x\to \infty}f(x)-x&=\lim_{x\to \infty} \frac{f(x)^n-x^n}{x^{n-1}} \cdot \frac{1}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1}\\
&=(a_1+a_2+\dots+a_n)\cdot \frac{1}{1+1+\dots+1} \\ &
=\frac{a_1+a_2+\dots+a_n}{n}\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int_0^e \left(\operatorname{W}(x)^{2}x-\frac{6x}{8}-\frac{3\operatorname{W}(x)}{8}+\frac{3}{8}\right)\,dx=0$ Hi I was playing with the Lambert function when I wondering myself about that :
Prove that :
$$\int_0^e \left(\operatorname{W}(x)^2 x-\frac{6x}{8}-\frac{3\operatorname{W}(x)}{8}+\frac{3}{8} \right) \, dx=0$$
My try
It's straightforward if we have :
\begin{align}
& \int \left(\operatorname{W}(x)^2 x-\frac{6x}{8}-\frac{3\operatorname{W}(x)}{8}+\frac{3}{8} \right) \, dx \\[8pt]
= {} & \frac{(x (\operatorname{W}(x) - 1) (4 x \operatorname{W}(x)^3 - 3 \operatorname{W}(x)^2 + 3 (x + 1) \operatorname{W}(x) - 3 x))}{(8 \operatorname{W}(x)^2)} \\
& {} + \text{constant}
\end{align}
And after using the fundamental theorem of calculus.
My question
How to prove it using others method?
Thanks in advance for your time.
|
By the definition of the Lambert $W$ function we have, taking $W(x)=u\Rightarrow x=ue^u$, that $$\int_0^e W(x)^2 x \, dx = \int_0^1 u^3 e^{2u} (u+1) \, du = \frac{3}{8} (e^2-1)$$ and $$-\frac{3}{8} \int_0^e W(x) \, dx = \int_0^1 ue^u (u+1) \, du = -\frac{3}{8} (e-1)$$ and trivially $$\int_0^e \left(\frac{3}{8}-\frac{6}{8}x\right) \, dx = -\frac{3}{8} (e-1)e.$$
|
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|
Limit of sum $x^3+x^5+x^7+x^9+...)$ I am asked to give the limit of:
$$ x^3+x^5+x^7+x^9+... \quad x\in(-1,1)$$
So I do the following:
The sum of the first $n$ terms will be equal to:
$$x^3+x^5+x^7+...+x^{3+2(n-1)}$$
I factor out $x^3$, I get:
$$x^3(1+x^2+x^4+..+x^{2(n-1)})$$
I also factor out $x^2$, I get:
$$x^3 x^2(1/x^2+1+x^2+x^3+...+x^{n-1}=x^5(1/x^2+1+x^2+x^3+...+x^{n-1})$$
So I have:
$$x^5/x^2+x^5(1+x^2+x^3+...+x^{n-1})=x^3+\frac{x^5 (1-x^n)}{1-x}$$
So I take the limit when $n$ goes to infinity and I get:
$$x^3+\frac{x^5}{1-x}=\frac{x^3(1-x^2)}{1-x^2}+\frac{(1+x)x^5}{1-x^2}=\frac{x^3-x^5+x^5+x^6}{1-x^2}=\frac{x^3+x^6}{1-x^2}$$
But the right answer is $\frac{x^3}{1-x^2}$
Where did I go wrong?
|
The problem is that when you factor out $x^2$, the remaining powers should still differ by two, because multiplying powers is additive. What you really wanted to do is write $y=x^2$, and it becomes
$$\frac{x^3}{1-y}$$
and we can simply substitute $x^2$ for $y$.
|
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|
Can I apply elementary row operation then find eigen values of a matrix? Suppose if a matrix is given as
$$ \begin{bmatrix}
4 & 6\\
2 & 9
\end{bmatrix}$$
We have to find its eigenvalues and eigenvectors.
Can we first apply elementary row operation . Then find eigenvalues.
Is their any relation on the matrix if it is diagonalized or not.
|
First compute the eigenvalues as follows:
$$
\begin{align}
\det(\lambda I-A)&=\det\left(
\begin{bmatrix}
\lambda & 0 \\
0 & \lambda
\end{bmatrix}
-
\begin{bmatrix}
4 & 6 \\
2 & 9
\end{bmatrix}
\right)
= \det \left(
\begin{bmatrix}
\lambda-4 & -6 \\
-2 & \lambda-9
\end{bmatrix}
\right) \\
&=
(\lambda-4)(\lambda-9)-(-2)(-6) \\
&=
\lambda^2-13\lambda + 24
\end{align}
$$
The eigenvalues are then
$$
\begin{align}
\lambda_{1,2} &= \frac{13 \pm \sqrt{13^2-4\times 24} }{2} \\
&= \frac{13\pm \sqrt{73}}{2}
\end{align}
$$
Now to compute the eigenvector for $\lambda_1$, we need to solve the following
$$
\begin{align}
\begin{bmatrix}
\lambda_1 - 4 & -6 \\
-2 & \lambda_1 -9
\end{bmatrix}
&=
\begin{bmatrix}
\frac{13 + \sqrt{73} }{2} - 4 & -6 \\
-2 & \frac{13 + \sqrt{73} }{2} -9
\end{bmatrix} \\
&=
\begin{bmatrix}
\frac{5 + \sqrt{73} }{2} & -6 \\
-2 & \frac{-5 +\sqrt{73} }{2}
\end{bmatrix}
\end{align}
$$
Apply Gaussian elimination (i.e. $R_2=R_2+2\frac{2}{5+\sqrt{73}}R_1)$, we get
$$
\begin{bmatrix}
\frac{5 + \sqrt{73} }{2} & -6 \\
0 & 0
\end{bmatrix}
$$
So, $x_2$ is a free variable and let $x_2=s$, we
$$
\begin{align}
\frac{5 + \sqrt{73} }{2} x_1 = 6x_2 \implies x_1 = \frac{12}{5+\sqrt{73}}s
\end{align}
$$
Finally,
$$
x=
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
=
\begin{bmatrix}
\frac{12}{5+\sqrt{73}} s \\ s
\end{bmatrix}
=
s
\begin{bmatrix}
\frac{12}{5+\sqrt{73}} \\ 1
\end{bmatrix}
$$
The eigenvector for $\lambda_1$ is
$$
v_1 =
\begin{bmatrix}
\frac{12}{5+\sqrt{73}} \\ 1
\end{bmatrix}
$$
Now you can compute the second eigenvector in the same scenario.
|
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|
$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
x^2+xy+y^2=3 \end{cases}$$
by reducing the system to a system of second degree.
What can I look for in such situations? What is the way to solve this kind of systems? The only thing I see here is that we can factor:
$$\begin{cases} x^2(x^2+y^2)+y^4=21 \\
x(x+y)+y^2=3 \end{cases}$$
|
Multiply the first by $x^2-y^2$ and the second by $x-y$.$$x^6-y^6=21(x^2-y^2),\\x^3-y^3=3(x-y)$$
Then take the ratio
$$x^3+y^3=7(x+y).$$
Adding the two above,
$$2x^3=10x+4y$$
and using $2y=x^3-5x$,
$$4x^2+2x(x^3-5x)+(x^3-5x)^2-12=x^6-8x^4+19x^2-12=0.$$
By trial and error, $x^2=1$ are solutions and we factor as
$$(x^2-1)(x^4-7x^2+12)=(x^2-1)(x^2-3)(x^2-4).$$
|
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|
Solving for the closed form of recurrence relations using characteristic polynomial I know how to find the closed form of some recurrence relations such as
those that are similar to the Fibonacci Sequence. I am not sure how to solve a recurrence relation using the characteristic polynomial when there is a constant involved like
$a_n = 3a_{n-1} -1$ (I know how to solve this using substitution, but I want to know-how using the characteristic polynomial)
or
$a_n = 6a_{n-1} + 7a_{n-2} +3$
In using the characteristic polynomial, how do I treat the constant when factoring?
|
There is always the matrix approach:
$$
\begin{pmatrix}
a_{n} \\ 1
\end{pmatrix}
=
\begin{pmatrix}
3 & -1 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
a_{n-1} \\ 1
\end{pmatrix}
$$
The characteristic polynomial of that matrix is $x^2-4x+3=(x - 3)(x - 1)$ and so $a_n=\alpha 3^n + \beta 1^n$. The coefficients are determined by the initial conditions.
The same approach works for the other recurrence:
$$
\begin{pmatrix}
a_{n} \\ a_{n-1} \\ 1
\end{pmatrix}
=
\begin{pmatrix}
6 & 7 & 3 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
a_{n-1} \\ a_{n-2} \\ 1
\end{pmatrix}
$$
The characteristic polynomial of that matrix is $x^3-7x^2-x+7=(x - 7) (x + 1) (x - 1)$ and so $a_n=\alpha 7^n + \beta (-1)^n + \gamma 1^n$. The coefficients are determined by the initial conditions.
The characteristic polynomial of $a_n = 6a_{n-1} + 7a_{n-2} +3$ is $(x - 7) (x + 1)$ and appears as a factor in the characteristic polynomial of the matrix, as in the first example.
|
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|
Solving logarithm leaving in terms of $p$ and $q$ I would like to check the steps if Part a) is done correctly.
For Part b), how do I continue from below? I seem to stuck for $\log_{10}(5)$…
Here is the problem:
Given that $p = \log_{10} 2$ and $q = \log_{10} 7$, express the following in terms of $p$ and $q$.
a) $\log_{7} 4 = \frac{\log_{10} 4}{\log_{10} 7} = \frac{2 \log_{10} 2}{\log_{10} 7} = \frac{2p}{q}$
b) $\log_{10} \sqrt[3]{\frac{25}{49}} = \log_{10}5^\frac{2}{3} - \log_{10}7^\frac{2}{3} = \frac{2}{3}\log_{10}5 - \frac{2q}{3}$
[Source]
|
Part a is correct and $\log_{10} 5+\log_{10} 2=1$
|
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|
Given that $\piThe left side of the expression reduces to $2\sin x$ and the right side reduces $2+2\sin x$
Their addition gives $2+4\sin x$ . The correct answer is 2
I realize that this has got to do with the fact that $\pi<x<3\pi/2$ , in which case $\sin x$ will be -ve.
But then the expression should be $2-4\sin x$. How does it end up being 2?
|
$$\sqrt{4\sin^4x+\sin^2(2x)}+4\cos^2\left(\frac\pi4-\frac x2\right)$$
$$=\sqrt{4\sin^4x+4\sin^2x\cdot\cos^2x}+2\left(\cos\left(\frac\pi2-x\right)+1\right)$$
$$=2\vert \sin x \vert \sqrt{\sin^2x+\cos^2x}+2\sin x+2$$
$$=-2\sin x+2\sin x+2=2$$
|
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|
We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled?
Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?
Answer is $\frac{137}{60}.$
There is a similar question asked in MSE but I do not understand the method used by Henry.
In particular, if we let $X$ be the minimum value rolled up to and including $5$, then
$$E(X) = \sum_{x=1}^5 xP(X=x) = 1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15 = \frac{137}{60}.$$
It seems that we are using the fact that
$$P(X=x) = \frac{1}{x(x+1)}.$$
I do not understand how to obtain the equation above.
|
First, $X$ is not the minimum value rolled before obtaining a $5$, it is the minimum value rolled up to and including the first roll that comes up $5$, so that $X=5$ is possible.
The event $X=5$ means that $5$ comes up before any of $1$, $2$, $3$, or $4$ (we don't care about $6$). Since each of the five numbers is equally likely to come up first,
$$P(X=5)=\frac15.$$
Now suppose $1\le x\le4$. Now the event $X=x$ means that, among the $x+1$ numbers $1,\dots,x,5$, the number $x$ comes up first, and $5$ second. Thus we have
$$P(X=x)=\frac{(x-1)!}{(x+1)!}=\frac1{(x+1)x}\text{ for }1\le x\le4.$$
|
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|
Proving the inequality $\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$ I am struggling to prove
$$\prod_{n=1}^k \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$
For all $k \geq 1$. Clearly, this is true for $k=1$, so it suffices to show
$$\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$
The obvious way to show this would be to evaluate the product directly, though I don't think that is feasible. I am not sure how else to approach this product. This would be a (much) tighter upper bound than
$$\exp\left ({\sum_{n=1}^k} \frac1{n^2+\ln n} \right )$$
Which comes from the monotone convergence theorem.
|
I have not full solution to this problem.
I want to prove by induction that:
$$ \prod_{n=1}^{k} \left(1 + \frac{1}{n^2 + \ln{n}}\right) < \frac{7}{2} \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)} < \frac{7}{2}$$
It's obvious for $k=1$.
Suppose we proved for $k$:
$$ \prod_{n=1}^{k + 1} \left(1 + \frac{1}{n^2 + \ln{n}}\right) < \frac{7}{2} \left(1 + \frac{1}{(k+1)^2 + \ln{(k+1)}}\right) \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)}$$
So we need to prove that:
$$ \left(1 + \frac{1}{(k+1)^2 + \ln{(k+1)}}\right) \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)} < \frac{(k+2)^2 + \ln(k+2)}{(k+3)^2 + \ln(k+3)} \\
\frac{(k+1)^2 + \ln(k+1) + 1}{(k+2)^2 + \ln(k+2)} < \frac{(k+2)^2 + \ln(k+2)}{(k+3)^2 + \ln(k+3)}
$$
But I don't know how to prove the last inequality, which according to Wolfram is true for $k > -1$.
Probably I will improve my answer after a while.
|
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|
Find the radius of the inscribed circle of a right triangle. Find the radius of the inscribed circle of a right triangle. The triangle's height is $\sqrt{6} + \sqrt{2}$ while the bisector of the right angle is 4.
This seemed like a generic similar triangles problem but it isn't that simple when I tried to solve it. Any help?
EDIT: Here is a picture of the problem
|
Find the radius $r$ of the inscribed circle of a right triangle
$ABC$
given its height $|CD|=h_c=\sqrt6+\sqrt2$
and bisector $|CE|=4$.
Let $I$ be the center of the inscribed circle
and $A_t,B_t,C_t$ its touching points
\begin{align}
\triangle CED,\ \triangle IEC_t,\triangle CB_tI:\quad
\frac{h_c}{\beta_c}
&=
\frac{r}{|IE|}
=
\frac{r}{\beta_c-|CI|}
=
\frac{r}{\beta_c-r\,\sqrt2}
,\\
r&=\frac{\beta_c\,h_c}{\beta_c+h_c\,\sqrt2}
=\frac{2\,\sqrt6}3
.
\end{align}
Edit
Additionally, the side lengths of the triangle can be found as follows.
We know that
\begin{align}
2r&=a+b-c
,\\
r\,(a+b+c)&=c\,h_c
,
\end{align}
so the hypotenuse is found as
\begin{align}
c&=
\frac{2\,h_c\,\beta_c^2}{2\,h_c^2-\beta_c^2}
=
\tfrac{4\sqrt6}3\,(1+\sqrt3)
\approx 8.92284
.
\end{align}
The sizes of the legs can be found
from system
\begin{align}
ab&=c\,h_c
,\\
a+b&=2\,r+c
.
\end{align}
Since $a<b$, we have
\begin{align}
a&=
r+\tfrac12\,\Big(c-\sqrt{(2r+c)^2-4\,c\,h_c}\Big)
=
\tfrac{2\sqrt6}3\,(1+\sqrt3)
,\\
b&=
r+\tfrac12\,\Big(c+\sqrt{(2r+c)^2-4\,c\,h_c}\Big)
=
2\,\sqrt2\,(1+\sqrt3)
.
\end{align}
It follows that
$\angle CAB=30^\circ$,
$\angle ABC=60^\circ$.
|
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|
$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead.
$$\int\sqrt{x\cdot \sqrt[3]{x\cdot \sqrt[4]{x\cdot\sqrt[5]{x\ldots } }}}\,dx\tag{1}$$
$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx\tag{2}$$
I was able to solve (1), as the integrand simplifies to $x^{e-2}$, however, I'm struggling with solving (2).
If we rewrite the roots as powers, we get:
$$\int x^\frac{2}{2}\cdot x^\frac{3}{4}\cdot x^\frac{4}{8}\cdot x^\frac{5}{16}\ldots\,dx$$
combining the powers we get:
$$\int x^{\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\ldots}$$
the exponent is the infinite sum
$$\sum^{\infty}_{n=1}\frac{n+1}{2^n}\tag{3} $$
we can split this into:
$$\sum^{\infty}_{n=1}\frac{n}{2^n}+\sum^{\infty}_{n=1}\frac{1}{2^n} $$
The right sum is well known except here the sum begins at $n=1$, meaning that the right sum evaluates to 1. Messing around with desmos, the integrand appears to be $x^3,x>0$ implying that (3) converges to 3 and the $\sum^{\infty}_{n=1}\frac{n}{2^n}$ converges to 2.
Which is part I'm struggling with. Any ideas?
$$\sum^{\infty}_{n=1}\frac{n}{2^n}$$
|
The binomial theoem gives$$(1-x)^{-2}=\sum_{n\ge0}(n+1)(-1)^nx^n,$$so the exponent is$$-1+(1-1/2)^{-2}=3.$$So the integral is $\int x^3dx=\frac14x^4+C$.
|
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|
An urn of 4 balls with 2 colors. Pick 2 balls and place them back 4 times. What's the probability of picking 2 balls of the same color twice in a row? An urn of 4 balls with 2 colors. Pick 2 balls and place them back 4 times. What's the probability of picking 2 balls of the same color twice in a row?
So the probability of picking 2 balls of the same color is $2\choose1$$2\choose2$/$4\choose2$, but I don't know the probability of getting that twice in a row out of 4 draws
|
Assumptions:
*
*There are two balls of one color and two balls of a second color
*Suppose the colors are Green and Red. A successful outcome occurs in any of these cases when you choose two green followed by two green, two green followed by two red, two red followed by two green, or two red followed by two red.
The probability of picking two balls of the same color in a single draw is as you found $\dfrac{1}{3}$. A single draw will be labeled $S$ for both balls are the same color and $D$ for the balls are different colors. Here are the possible outcomes:
$$\begin{array}{c|c|c}\text{Draws} & \text{Probability} & \text{2-in-a-row?} \\ \hline DDDD & \left(\dfrac{2}{3}\right)^4\left(\dfrac{1}{3}\right)^0 & \text{No} \\ DDDS & \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^1 & \text{No} \\ DDSD & \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^1 & \text{No} \\ DDSS & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{Yes} \\ DSDD & \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^1 & \text{No} \\ DSDS & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{No} \\ DSSD & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{Yes} \\ DSSS & \left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3 & \text{Yes} \\ SDDD & \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^1 & \text{No} \\ SDDS & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{No} \\ SDSD & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{No} \\ SDSS & \left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3 & \text{Yes} \\ SSDD & \left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2 & \text{Yes} \\ SSDS & \left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3 & \text{Yes} \\ SSSD & \left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3 & \text{Yes} \\ SSSS & \left(\dfrac{2}{3}\right)^0\left(\dfrac{1}{3}\right)^4 & \text{Yes}\end{array}$$
Adding this up, I get:
$$3\left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^2+4\left(\dfrac{2}{3}\right)^1\left(\dfrac{1}{3}\right)^3+\left(\dfrac{2}{3}\right)^0\left(\dfrac{1}{3}\right)^4 = \dfrac{7}{27}$$
|
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|
if $mx+3|x+4|-2=0$ has no solutions, solve for $m$
If $mx+3|x+4|-2=0$ has no solutions, which of the following value could be $m$?
(A)5
(B)$-\frac{1}{2}$
(C)-3
(D)-6
(E)$\frac{10}{3}$
my attempt:
$$mx-2=-3|x+4| \\ m^2x^2-4mx+4=9x^2+72x+144 \\ (9-m^2)x^2+(4m+72)x+140=0$$
because the equation has no solutions, therefore
$$(4m+72)^2-4(9-m^2)140<0 \\ 576m^2+576m+144<0 \\ 4m^2+4m+1<0 \\ (2m+1)^2<0$$
maybe I made a mistake but I couldn't find it
|
1) $y_1=3|x-(-4)| \ge 0$ ;
2) $y_2= Mx +2$, where $M=-m$
$M=3=-m$ no intersection (Why?)
No intersection for $3 \ge M >1/2$ (Why?);
$3 \ge -m >1/2$, or
$-3 \le m < -1/2$.
|
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|
Linear differential equations, integrating factor Solve the following differential equation:
$$ dr+(2r \cot\theta+\sin 2\theta)d\theta=0$$
I have tried like this:
$$ \frac{dr}{d\theta}+2r\cot\theta=-\sin{2\theta}$$
\begin{align}
I.F. &=e^{\int{2\cot\theta d\theta}}\\
& =e^{-2\log\sin\theta}\\
& =\frac 1{\sin^2\theta}\\
\end{align}
$$\therefore\frac r{\sin^2\theta}=-\int\frac{\sin 2\theta d\theta}{\sin^2\theta}\\
\implies \frac r{\sin^2\theta}=-2\int{\cot\theta d\theta}\\
\implies \frac r{\sin^2\theta}=2\log\sin\theta+c $$
But in my book the answer is:
$$2r{\sin^2\theta}+{\sin^4\theta}=c$$
Please check out which is correct..
|
You messed up with your multiplication. Multiplying the integral factor to both sides should yield:
$$\dfrac{d}{d\theta}\left(\dfrac{r}{\sin^2\theta}\right)=\left(\dfrac{1}{\sin\theta}\right)^2\dfrac{dr}{d\theta}+2\cot\theta\left(\dfrac{1}{\sin\theta}\right)^2 r=-\sin2\theta\left(\dfrac{1}{\sin\theta}\right)^2=-2\cot\theta$$
and so
$$\dfrac{r}{\sin^2\theta}=-2\int\cot\theta d\theta=-2\log(\sin\theta)+C$$
|
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|
Inverse of $I + A$ I am trying to solve the following exercise in Artin, without breaking into cases for even and odd $k$.
A square matrix $A$ is called nilpotent if $A^k = 0$ for some $k > 0$. Prove that if $A$ is nilpotent, then $I + A$ is invertible.
Here's my attempt.
I claim that the inverse of $I + A$ is
$$(I + A)^{-1} = \sum\limits_{n=0}^{k-1} (-1)^n A^n.$$
We prove this is a right inverse.
\begin{align*}
(I + A)\sum\limits_{n=0}^{k-1} (-1)^n A^n & = \sum\limits_{n=0}^{k-1} (-1)^n (I + A)A^n \\
& = \sum\limits_{n=0}^{k-1} (-1)^n (A^n + A^{n+1})
\end{align*}
If $n$ is even, then $n + 1$ is odd and vice-versa. Hence, the series telescopes:
\begin{align*}
& = (A^0 + A^{1}) - (A^1 + A^2) + (A^2 + A^3) + \ldots \pm (A^{k-1} + A^k) \\
& = I \pm A^k \\
& = I
\end{align*}
It seems that the sign of the final term is dependent on whether $k$ is even or odd. However, it shouldn't matter, because if $A^k = 0$, then $-A^k = 0$.
Is there a better way to formalize this fact?
|
We know that, if $|x|<1$, then
$$\dfrac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 -x^5 + \cdots$$
For just any square matrix, $\mathbf A$, it is not true that
$$\dfrac{1}{1+\mathbf A} =
1 - \mathbf A + \mathbf A^2 - \mathbf A^3 + \mathbf A^4 -\mathbf A^5 + \cdots$$
But, since $\mathbf A$ is nilpotent, this becomes the finite sum
$$(1+\mathbf A)^{-1} =
1 - \mathbf A + \mathbf A^2 - \mathbf A^3 + \mathbf A^4 + \cdots
+ (-1)^{k-1}\mathbf A^{k-1}$$
This can be easily verified by multiplying the right side by $1+\mathbf A$.
|
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|
Why is $(1+\frac{1}{n})^n < (1+\frac{1}{m})^{m+1}$? Why is it that
$$
\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1},
$$
for any natural numbers $m, n$?
I have tried expanding using the binomial series and splitting into cases. I understand why it is trivially true when $m=n$ but I am not sure if there is a rigorous proof for other cases?
|
Both sequences
$$
a_n=\left(1+\frac{1}{n}\right)^n, \quad
b_n=\left(1+\frac{1}{n}\right)^{n+1}
$$
are monotonic. In particular, $a_n$ is increasing and $b_n$ is decreasing.
Hence, if $m,n\in \mathbb N$ and $k=$max$\{m,n\}$, then
$$
a_n\le a_k\le b_k\le b_m.
$$
Why are these sequences monotonic?
We have
$$
\frac{a_{n+1}}{a_n}=\frac{(n+2)^{n+1}n^n}{(n+1)^{2n+1}}=\frac{n+2}{n+1}\cdot\left(\frac{n^2+2n}{(n+1)^2}\right)^n \\ =\frac{n+2}{n+1}\cdot\left(1-\frac{1}{(n+1)^2}\right)^n \ge \frac{n+2}{n+1}\cdot\left(1-\frac{n}{(n+1)^2}\right)=\frac{(n+2)(n^2+n+1)}{(n+1)^3}=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}>1.
$$
In a similar fashion we get that $b_n$ is decreasing:
$$
\frac{b_n}{b_{n+1}}=\frac{\left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}}=\frac{(n+1)^{2n+3}}{n^{n+1}(n+2)^{n+2}}=\frac{n+1}{n+2}\cdot \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1} \\ = \frac{n+1}{n+2}\cdot \left(1+\frac{1}{n^2+2n}\right)^{n+1}\ge \frac{n+1}{n+2}\cdot
\left(1+\frac{n+1}{n^2+2n}\right) =\frac{(n+1)(n^2+3n+1)}{(n+2)(n^2+2n)}=\frac{n^3+4n^2+4n+1}{n^3+4n^2+4n}>1.
$$
|
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|
Find locus of $\Delta ABC$ centroid with orthocentre at origin and side slopes 2, 3 and 5 Let $ABC$ be a triangle with slopes of the sides $AB$, $BC$, $CA$ are $2,3,5$ respectively. Given origin is the orthocentre of the triangle $ABC$. Then find the locus of the centroid of the triangle $ABC$.
Since sides have slopes $2,3,5$ then altitudes must have slopes $\frac{-1}{2}, \frac{-1}{3}$ and $\frac{-1}{5}$ respectively. Then equations of the altitudes are respectively \begin{align*} & 2y+x=0 \\ & 3y+x=0 \\ & 5y+x=0. \end{align*} If $(\alpha, \beta)$ be the coordinates of the centroid, how can I find the locus of the point $(\alpha, \beta)$ from here?
|
Respectively, the vertexes A, B and C are on the three altitude lines you obtained $3y+x=0$, $5y+x=0$ and $2y+x=0$. So, let their coordinates be $A(a,-\frac a3)$, $B(b,-\frac b5)$ and $C(c,-\frac c2)$. Then, use them to match the three side slopes
$$\frac{-\frac b5 + \frac a3 }{b-a}=2,\>\>\>\>\>\>\>
\frac{-\frac b5 + \frac c2 }{b-c}=3,\>\>\>\>\>\>\>
\frac{-\frac a3 + \frac c2 }{a-c}=5$$
which lead to the ratio $a:b:c = 33:35:32$ and the corresponding vertexes in terms of a single parameter $t$
$$A(33t,-\frac {33t}3),\>\>\>\>\>B(35t,-\frac {35t}5),\>\>\>\>\>C(32t,-\frac {32t}2)$$
Then, the coordinates of the centroid are
$$x = \frac{A_x+B_x+C_x}3=\frac{100t}3,\>\>\>\>\>y = \frac{A_y+B_y+C_y}3=-\frac{34t}3$$
Eliminate $t$ to obtain its locus
$$y=-\frac{17}{50}x$$
|
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|
Birthday problem-Probability exactly $2$ triples and $4$ pairs if $20$ people in room Say there are 20 people in a room. What is the probability there are exactly 2 triples and 4 pairs. Is my answer shown below correct?
Assume 365 days in the year.
$P= \dfrac{\binom{365}{2}\binom{363}{4}\binom{20}{3}\binom{17}{3}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2} \cdot 359 \cdot 358 \cdot 357 \cdot 356 \cdot 355 \cdot 354}{365^{20}}$?
Term $365C2$ chooses the $2$ birthdays for the $2$ triples. Each triple has a different birthday. Term $363C4$ chooses the $4$ birthdays for the $4$ pairs. Each pair has different birthdays. Term $20C3$ selects the $3$ people for the first triple and $17C3$ the $3$ people for the second triple. Term $14C2$ picks the $2$ people for first pair, $12C2$ for second pair, $10C2$ the third pair, and finally $8C2$ for the fourth pair. The term $(359 \cdot 358 \cdot 357 \cdot 356 \cdot 355 \cdot 354)$ is the birthdays for the remaining $6$ people, which do not match. I start with $359$ because $6$ birthdays have taken by the $2$ triples and the $4$ pairs.
All this is then divided by the total number of possible birthday selections $365^{20}$.
I am wondering if the selection of the people for the $2$ triples should be $20C6$ instead of $20C3 \cdot 17C3$ as I show. I believe my method is the correct one. Please let me know.
|
You have solved the problem correctly.
There are $\binom{365}{2}$ ways to choose the two days for the triples. There are $\binom{20}{3}$ ways to choose which three people share the earlier of these two birthdays and $\binom{17}{3}$ ways to choose which three of the remaining people share the later of these two birthdays.
If you chose which six people were part of the triples, you would have to multiply the $\binom{20}{6}$ ways of selecting the six people by the $\binom{6}{3}$ ways of selecting which three of those six people had the earlier birthday. Notice that
$$\binom{20}{3}\binom{17}{3} = \frac{20!}{3!17!} \cdot \frac{17!}{3!14!} = \frac{20!}{3!3!14!} = \frac{20!}{6!14!} \cdot \frac{6!}{3!3!} = \binom{20}{6}\binom{6}{3}$$
|
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|
How to multiply $\sqrt{5x^2+2x+1}$ with $\frac{1}{x}$? I believe the answer says, $\sqrt{5+\frac{2}{x}+\frac{1}{x^2}}$, but I don't see how they obtained it. I'm quite rusty with radicals.
|
Just a slight variation to the approaches already posted.
Suppose
$A = \sqrt{5x^2 + 2x + 1}$
$B = \frac{1}{x}$
If $AB = C$, then ${(AB)}^2 = C^2$ or $A^2B^2 = C^2$
$A^2$ = $5x^2 + 2x + 1$
$B^2$ = $\frac{1}{x^2}$
So $C^2 = 5 + \frac{2}{x} + \frac{1}{x^2}$
And to return to C: $$\sqrt{C^2} = \pm\sqrt{5 + \frac{2}{x} + \frac{1}{x^2}}$$
|
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|
Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1]$$
$$x_1 = -1/2,~~~x_2 = -1$$
therefore:
$$2x^2 + 3x + 1 = (x + 1/2)(x+1)$$
Now I check it:
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x \cdot x + x \cdot 1 + \frac{1}{2}\cdot x + \frac{1}{2}\cdot 1$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + \frac{3}{2}x + \frac{1}{2}$$
but:
$$2x^2 + 3x + 1 \ne x^2 + \frac{3}{2}x + \frac{1}{2}$$
Why does't quadratic formula work when $a \ne 1$?
however, I can pull out the 1/2.
$$2x^2 + 3x + 1 \ne \frac{1}{2}[2x^2 + 3x + 1]$$
I feel that this is something I must have missed in grade school...
Does this mean you need to multiple by "a" if $a \ne 1$?**
|
On the quadratic formula:
$$ax^2 +bx+c=0$$
$$r_i =\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$a\cdot(x-r_1)(x-r_2)$$
First, the "$a\cdot$" term in the factored result "$a\cdot(x-r_1)(x-r_2)$" is an often overlooked term when using the quadratic formula to factor a second order polynomial, because usually a=1, or the "$a\cdot$" term gets divided out by 0 on the RHS. So people forget that they need to add this "$a\cdot$" to the factorized result, but its really there, even if it doesn't matter because it will get canceled by the zero on the RHS.
Second, quadratic formula really applies to only the LHS of the expression $ax^2 +bx+c=0$. Its just that algebra textbooks usually only care about factoring the LHS when the RHS is zero. However, its perfectly ok to use the quadratic formula to factor the LHS when the right hand side is not zero. Example: $ax^2+bx+c = g(x) => a\cdot(x-r_1)(x-r_2) = g(x)$
|
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|
If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy} {ab}\cos(\theta)$ If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, prove
that $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos(\theta)=\sin^2(\theta)$
My trial:Let:
$\cos^{-1}(\frac{x}{a})=\alpha$, and $\cos^{-1}(\frac{x}{a})=\beta$
$\sin(\theta)= \sin(\alpha) \cos(\beta)+\cos(\alpha)\sin(\beta)$
$\sin (\theta)=\sqrt{1-\frac{y^2}{b^2}}\bigl(\frac{x}{a}\big)+\sqrt{1-\frac{x^2}{a^2}}\big(\frac{y}{b}\bigr)$
$\sin^2(\theta)=\bigl(1-\frac{y^2}{b^2}\bigr)\frac{x^2}{a^2}+\bigl(1-\frac{x^2}{a^2}\bigr)\frac{y^2}{b^2}+\frac{2xy}{ab}\sqrt{(1-\frac{y^2}{b^2})((1-\frac{x^2}{a^2})}$
$\sin^2(\theta)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{2xy}{ab}\sqrt{(1-\frac{y^2}{b^2})((1-\frac{x^2}{a^2})}-\frac{2x^2y^2}{a^2b^2}$
I could not get the final solution, any idea, maybe there is a better algorithm?
|
Using
\begin{eqnarray*}
\cos(a+b) = \cos(a) \cos(b)- \sin(a) \sin(b).
\end{eqnarray*}
We have
\begin{eqnarray*}
\frac{xy}{ab} - \sqrt{ \left( 1- \frac{x^2}{a^2} \right) \left( 1- \frac{y^2}{b^2} \right) } = \cos( \theta).
\end{eqnarray*}
Now move the $\frac{xy}{ab}$ to RHS, Square and rearrange.
|
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|
Find the $n$-th power of a $3{\times}3$ matrix using the Cayley-Hamilton theorem. I need to find $A^n$ of the matrix $A=\begin{pmatrix}
2&0 & 2\\
0& 2 & 1\\
0& 0 & 3
\end{pmatrix}$ using Cayley-Hamilton theorem.
I found the characteristic polynomial $P(A)=(2-A)^2(3-A)$ from which I got $A^3=7A^2-16A+12$. How to continue?
|
We can compute $A^2$ directly:
$$
A^2 = \left(
\begin{array}{ccc}
2 & 0 & 2 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{array}
\right)\left(
\begin{array}{ccc}
2 & 0 & 2 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{array}
\right)
=
\left(
\begin{array}{ccc}
4 & 0 & 10 \\
0 & 4 & 5 \\
0 & 0 & 9 \\
\end{array}
\right).
$$
From the Cayley-Hamilton theorem, it follows that
\begin{align}
A^3 &= 7A^2 -16A + 12I\\
&= 7\left(
\begin{array}{ccc}
4 & 0 & 10 \\
0 & 4 & 5 \\
0 & 0 & 9 \\
\end{array}
\right) - 16\left(
\begin{array}{ccc}
2 & 0 & 2 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{array}
\right) + 12\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)\\
&= \left(
\begin{array}{ccc}
8 & 0 & 38 \\
0 & 8 & 19 \\
0 & 0 & 27 \\
\end{array}
\right).
\end{align}
Observe the pattern
$$
A^n = \left(
\begin{array}{ccc}
2^n & 0 & -2 \left(2^n-3^n\right) \\
0 & 2^n & -2^n+3^n \\
0 & 0 & 3^n \\
\end{array}
\right).
$$
Clearly this holds for $n=1$. Assume that it holds for some $n\geqslant 1$, then
\begin{align}
A^{n+1} &= AA^{n}\\
&= \left(
\begin{array}{ccc}
2 & 0 & 2 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{array}
\right)\left(
\begin{array}{ccc}
2^n & 0 & -2 \left(2^n-3^n\right) \\
0 & 2^n & -2^n+3^n \\
0 & 0 & 3^n \\
\end{array}
\right)\\
&= \left(
\begin{array}{ccc}
2^{n+1} & 0 & -2 \left(2^{n+1}-3^{n+1}\right) \\
0 & 2^{n+1} & -2^{n+1}+3^{n+1} \\
0 & 0 & 3^{n+1}. \\
\end{array}
\right)
\end{align}
So by induction, this formula holds for all positive integers $n$.
|
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|
Find the remainder when $p(x)$ is divided by $x^2 -1$ Polynomial $p(x)$ leaves a remainder of $4$ when divided by $x-1$ and a remainder of $-2$ when divided by $x+1$.
Find the remainder when $p(x)$ is divided by $x^2 -1$ .
According to Remainder Theorem, when a polynomial $p(x)$ is divided by $(ax+b)$, the remainder is $p\left( -\frac {b}{a}\right)$ .
So, I did the following:
\begin{align}p(1)&=4\\
p(-1)&= -2\end{align}
\begin{align}p(x)&= (x^2-1)q(x) + Ax+ B\\
p(x)&= (x-1)(x+1)q(x) + Ax+B\end{align}
When \begin{align}p(1) &= A(1) + B\\
A+B&=4\tag{1}\end{align}
When \begin{align}p(-1)&= -A+ B\\
-A+B&= -2\tag{2}\end{align}
Doing $(1)+(2)$ gives:
\begin{align}2B&=2\\
B&=1\end{align}
Substitute $B=1$ into $(1)$ gives $A=3$
So, I got the remainder as $3x+ 1$
But, the answer in the book is $x+3$ , which means my values of $A$ and $B$ have been mixed up.
Please tell me where I went wrong
|
$3x+1 = 3(x-1) + 4\\3x+1 = 3(x+1)-2$
So your answer $3x+1$ is correct.
|
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|
Does $\iint_D \frac{x^2}{x^2+y^2} dx dy $ converge on $D= \left\{ (x, y) : x^2+y^2\leq ax \right\} $ ? If yes, what value does it converge to? Powers equal to $2$ entice the polar substitution.
$D= \left\{ (x, y) : x^2+y^2\leq ax \right\}$, so $0 \leq r \leq a \cos \phi.$ For the domain to make sense, we need either $\phi \in [0, \frac{\pi}{2}] \cup [\frac{3 \pi}{2}, 2\pi)$ and $a \geq 0$, or $\phi \in [\frac{\pi}{2}, \frac{3 \pi}{2}]$ and $a \leq 0$.
Everything below was achieved with help from the comments.
$$\iint_D \frac{x^2}{x^2+y^2} dx dy = \iint_{D'} r \cos^2 \phi \ d\phi dr.$$
Case $1$:
Since $(0, 0) \in D $ (equivalently, $r=0$ for any $\phi$) makes the integrand indefinite,
$$
\lim_{\epsilon \rightarrow 0}
\int_{0}^{\frac{\pi}{2}} \cos^2 \phi \ d \phi
\int_{\epsilon}^{a \cos \phi} r \ dr
+
\lim_{\epsilon \rightarrow 0}
\int_{\frac{3 \pi}{2}}^{2\pi} \cos^2 \phi \ d \phi
\int_{\epsilon}^{a \cos \phi} r \ dr
$$
$$
2\lim_{\epsilon \rightarrow 0}
\int_{0}^{\frac{\pi}{2}} \cos^2 \phi \ d \phi
\int_{\epsilon}^{a \cos \phi} r \ dr
$$
$$
2\lim_{\epsilon \rightarrow 0}
\int_{0}^{\frac{\pi}{2}} \left( \frac{a^2 \cos^2 \phi}{2}-\frac{\epsilon^2}{2} \right) \cos^2 \phi \ d \phi
$$
$$
\lim_{\epsilon \rightarrow 0}
\int_{0}^{\frac{\pi}{2}} \left( {a^2 \cos^2 \phi}-{\epsilon^2} \right) \cos^2 \phi \ d \phi
$$
Put $0$ instead of $\epsilon$, because it won't let me integrate:
$$
a^2
\int_{0}^{\frac{\pi}{2}} { \cos^4 \phi} \ d \phi = \frac{a^2}{2} Β \left(\frac{1}{2}, \frac{5}{2} \right) = \frac{a^2}{2} \cdot \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{5}{2})}{\Gamma(3)} = \frac{a^2}{2} \cdot \frac{\sqrt{\pi} \cdot 0.75 \sqrt{\pi}}{2} = \frac{3 a^2 \pi}{16}
$$
For case $2$, the result should be same.
Was this correct?
Formula $Β(m, n)=2\int_0^{\pi/2}\sin^{2m-1} \theta \cos^{2n-1} \theta \ d \theta$ was used, but I'm not sure I can extract it myself, and would be grateful for reference.
The answer would be, that for fixed $a$, it's convergent. (I am not sure about the definition of convergence here, and can't look it up, since the classes were missed, and google is blocked).
For Oliver Jones:
let $r_i=a\frac{i}{i+1}$, $\phi_i=\frac{\delta}{i}.$
$$\iint_{D_\delta} r \cos^2 \phi \ d\phi dr= \lim_{i \rightarrow \infty} \sum_i \left(a\frac{i}{i+1}\right) \left(\cos^2 \frac{\delta}{i} \right)
\left(
a\frac{i+1}{i+2} - a\frac{i}{i+2}
\right)
\left(
\frac{\delta}{i+1}-\frac{\delta}{i}
\right) =
\lim_{i \rightarrow \infty} \sum_i \left(a\frac{i}{i+1}\right) \left(1 \right)
\left(
\frac{a}{(i+1)(i+2)}
\right)
\left(
\frac{-\delta}{i(i+1)}
\right)
=
\sum_{i=0}^{\infty} \left(a^2\frac{1}{i+1}\right) \left(1 \right)
\left(
\frac{1}{(i+1)(i+2)}
\right)
\left(
\frac{-\delta}{(i+1)}
\right).
$$
It converges because of the $4$th power in the denominator.
Or maybe like this:
$$
\lim_{\epsilon \rightarrow 0} \int_{\epsilon}^a x \ dx \int_{-\sqrt{x(a-x)}}^{\sqrt{x(a-x)}} \frac{1}{1+ (y/x)^2} d (y/x)$$
$$
\lim_{\epsilon \rightarrow 0} \int_{\epsilon}^a 2 x \arctan \sqrt{x(a-x)} \ dx $$
How to proceed here, I am not sure.
$$
\lim_{\delta \rightarrow 0}\iint_{\delta^2 \leq r^2+ar \cos \theta + a^2/4 \leq a^2/4} r \cos ^2 \theta \ dr \ d\theta
$$
Let's choose $\theta \in [-\pi/2, \pi/2], a\geq 0$.
Then
$
\delta^2 \leq r^2 - ar + a^2/4 \leq r^2+ar \cos \theta + a^2/4 \leq r^2 + a^2/4 \leq a^2/4.
$
$
0 \leq r^2 - ar + a^2/4 -\delta^2 \Longrightarrow r\in (0, \frac{a-2\delta}{2}).
$
$$
\lim_{\delta \to 0} \int_{-\pi/2}^{\pi/2} \cos ^2 \theta \ d \theta \int_0^{(a-2\delta)/2} r dr = \frac{a^2}{8} В(1/2, 3/2) = \frac{a^2 \pi}{16}
$$
|
I get the same result as your first one. But I don't see the need for special functions, or introducing the $\epsilon$ you use.
Consider the region $x^2+y^2\leq ax$. This is the same as $$\left(x-\frac{a}{2}\right)^2+y^2\leq\frac{a^2}{4}$$
So the region is a disc centered at $\left(\frac{a}{2},0\right)$ with its boundary touching the origin. I'll assume $a>0$. A disc like this is described in polar coordinates by $r\leq a\cos(\theta)$ for $\theta\in[-\pi/2,\pi/2]$.
So the integral is
$$
\begin{align}
&\int_{\theta=-\pi/2}^{\theta=\pi/2}\int_{r=0}^{r=a\cos(\theta)}r\cos^2(\theta)\,dr\,d\theta\\
&=\int_{\theta=-\pi/2}^{\theta=\pi/2}\cos^2(\theta)\int_{r=0}^{r=a\cos(\theta)}r\,dr\,d\theta\\
&=\int_{\theta=-\pi/2}^{\theta=\pi/2}\cos^2(\theta)\frac{1}{2}a^2\cos^2(\theta)\,d\theta\\
&=\frac{a^2}{2}\int_{\theta=-\pi/2}^{\theta=\pi/2}\cos^4(\theta)\,d\theta\\
&=\frac{a^2}{2}\int_{\theta=-\pi/2}^{\theta=\pi/2}\left(\frac{1+\cos(2\theta)}{2}\right)^2\,d\theta\\
&=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac{1+\cos(\varphi)}{2}\right)^2\,d\varphi\\
&=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac14+\cos(\varphi)+\frac{1}{4}\cos^2(\varphi)\right)\,d\varphi\\
&=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac14+\frac{1}{4}\cos^2(\varphi)\right)\,d\varphi\\
&=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac14+\frac{1}{8}\left(1+\cos(2\varphi)\right)\right)\,d\varphi\\
&=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac38+\frac{1}{8}\cos(2\varphi)\right)\,d\varphi\\
&=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\frac38\,d\varphi\\
&=\frac{3a^2}{32}(2\pi)=\frac{3a^2\pi}{16}
\end{align}$$
|
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|
Prove $f(x)=x^3+x$ is injective How can algebraically prove $f(x)=x^{3}+x$ is injective.
I can get to $a^2+b^2+ab=-1$ but I can't go any further.
|
It suffices to note that $f(x)$ is a strictly increasing function on the reals.
That said, if you do want to go that route, we have
$$\begin{align}
f(a) = f(b)
&\iff
a^3+a = b^3 +b
\\&\iff
(a^3-b^3)+(a-b) = 0
\\&\iff
(a-b)(a^2+ab+b^2)+(a-b) = 0
\\&\iff
(a-b)(a^2+ab+b^2+1) = 0
\\&\iff
a = b \,\,\,\text{ or }\,\,\, a^2+ab+b^2+1 = 0
\end{align}$$
Thinking of $a^2+ab+b^2+1 = 0$ as a quadratic equation in $a$, we see that the discriminant is $\Delta = b^2 - 4(b^2+1) = -3b^2 -4 < 0$, so the equation has no solution on the reals.
It follows that the only option is $a=b$, and hence $f$ is injective.
|
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|
How to make a generating function to solve this problem? How many ways are there to put 8 identical balls into 3 boxes so that no box has more than 4 balls in it? But Box 3 can only have up to 2 balls inside it.
I started the problem by trying to write out the polynomials. For the first two boxes I thought it would be (1 + x + x ² + $x^3$ + $x^4$ )² and then for Box 3 it would be (1 + x + x ²)
So altogether, (1 + x + x ²)(1 + x + x ² + $x^3$ + $x^4$ )²
But from here, I am unsure how to write it as a series to find the number of ways for 8 balls.
A computer multiplied it and got $1 + 3 x + 6 x^2 + 9 x^3 + 12 x^4 + 13 x^5 + 12 x^6 + 9 x^7 + 6 x^8 + 3 x^9 + x^{10} + 0+0+ \ldots$
|
Your generating function approach to giving the answer $6$ as the coefficient of $x^8$ in the expansion of $(1+x+x^2)(1+x+x^2+x^3+x^4)^2$ is correct and the possibilities are:
4 + 4 + 0
3 + 4 + 1
4 + 3 + 1
2 + 4 + 2
3 + 3 + 2
4 + 2 + 2
Your method assumes the balls are indistinguishable but the boxes are distinguishable. Other assumptions would produce different results.
|
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|
What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$ Is there closed form for
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
My approach,
In this paper page $95$ Eq $(5)$ we have
$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$
Divide both sides by $x$ then integrate we get
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n=\int\frac{\operatorname{Li}_2\left(\frac{1-x}{2}\right)}{x}\ dx-\operatorname{Li}_3(-x)+\ln2\operatorname{Li}_2(x)-\operatorname{Li}_2\left(\frac12\right)\ln x$$
and my question is how to find the remaining integral? Thanks
Maybe you wonder why I have it as an indefinite integral, I meant so as I am planning to plug $x=0$ to find the constant after we find the closed form of the integral if possible.
I tried Mathematica, it gave
Edit
With help of $Mathematica$ I was able to find
\begin{align}
\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n&=-\frac13\ln^3(2)+\frac12\ln^2(2)\ln(1-x)-\frac12\zeta(2)\ln(x)+\frac32\ln^2(2)\ln(x)\\
&\quad-\ln(2)\ln(x)\ln(1-x)-\frac12\ln(2)\ln^2(x)-\frac12\ln^2(2)\ln(1-x)\\
&\quad-\ln^2(2)\left(\frac{x}{1+x}\right)+\ln(2)\ln\left(\frac{x}{1+x}\right)[\ln(1-x)+\ln(x)]\\
&\quad+\ln(x)\ln(1-x)\ln(1+x)+\ln(x)\operatorname{Li}_2\left(\frac{1-x}{2}\right)+\ln\left(\frac{x}{1+x}\right)\operatorname{Li}_2(x)\\
&\quad+\ln(1+x)\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x}{1+x}\right)\ln\left(\frac{2x}{1+x}\right)-\operatorname{Li}_2\left(\frac{2x}{1+x}\right)\ln\left(\frac{2x}{1+x}\right)\\
&\quad+\operatorname{Li}_2\left(\frac{1+x}{2}\right)\ln\left(\frac{x}{2}\right)-\ln\left(\frac{x}{1+x}\right)\operatorname{Li}_2\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(x)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)\\
&\quad+\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(-x)+\ln(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(3)
\end{align}
|
This is a long comment to https://math.stackexchange.com/a/3523732/198592 which just provides my result for comparison.
Let $\overline{H}_n=\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}$ be the alternating harmonic sum and define the generating function of order $q=0,1,2,...$ as
$$g_{q}(x) = \sum_{n=1}^\infty\frac{\overline{H}_n}{n^q}x^n\tag{1}$$
For $q=2$ I have obtained
$$\begin{align}
{g}_2(x)& =-\operatorname{Li}_3\left(\frac{x+1}{2}\right)-\operatorname{Li}_3(-x)-\operatorname{Li}_3(x)-\operatorname{Li}_3\left(\frac{x}{x+1}\right)+\operatorname{Li}_3\left(\frac{2 x}{x+1}\right)\\
& +\log (2) \operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x+1}{2}\right) \left(\log (x)-\log \left(\frac{2 x}{x+1}\right)\right)\\
& + \operatorname{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right) \log (x)+\operatorname{Li}_2(x) \left(\log \left(\frac{x}{x+1}\right)+\log (x+1)\right)\\
& +\left(\operatorname{Li}_2\left(\frac{x}{x+1}\right)
-\operatorname{Li}_2\left(\frac{2 x}{x+1}\right)\right) \log \left(\frac{2 x}{x+1}\right)+\frac{1}{2} \log ^2(2) \log (x)\\
& +\frac{1}{2} \log ^2\left(\frac{2 x}{x+1}\right) \left(\log \left(\frac{1-x}{2}\right)+\log \left(\frac{1}{x+1}\right)-\log \left(-\frac{x-1}{x+1}\right)\right)\\
& +\log (2) \log (x) \log \left(\frac{2 x}{x+1}\right)-\frac{1}{2} \log (2) \log (x) (\log (x)-2 \log (x+1)+\log (4))\\
& -\frac{1}{12} \pi ^2 \log (x)+\log \left(\frac{1-x}{2}\right) \log \left(\frac{x+1}{2}\right) \log (x)+\frac{7 \zeta (3)}{8}+\frac{\log ^3(2)}{6}
\end {align}\tag{2}$$
Notice the appearance of $\zeta(3)$ which I don't see in your expression.
Here is the graph of the g.f.
I find the following boundary values
$$g_2(+1) = \frac{1}{4} \pi ^2 \log (2)-\frac{\zeta (3)}{4}\simeq 1.40976$$
$$g_2(-1) = -\frac{1}{4} \pi ^2 \log (2)+\frac{5 \zeta (3)}{8}\simeq -0.958987$$
The values at $\pm \frac{1}{2}$ are somewhat too long to be provided here at the moment.
To facilitate comparison here is the Mathematica statement
g2[x_]:=Log[2]^3/6 - 1/12 \[Pi]^2 Log[x] + 1/2 Log[2]^2 Log[x] +
Log[2] Log[x] Log[(2 x)/(1 + x)] +
1/2 (Log[(1 - x)/2] + Log[1/(1 + x)] -
Log[-((-1 + x)/(1 + x))]) Log[(2 x)/(1 + x)]^2 +
Log[(1 - x)/2] Log[x] Log[(1 + x)/2] -
1/2 Log[2] Log[x] (Log[4] + Log[x] - 2 Log[1 + x]) +
Log[x] PolyLog[2, 1/2 - x/2] +
Log[2] PolyLog[2, x] + (Log[x/(1 + x)] + Log[1 + x]) PolyLog[2, x] +
Log[(2 x)/(
1 + x)] (PolyLog[2, x/(1 + x)] -
PolyLog[2, (2 x)/(1 + x)]) + (Log[x] -
Log[(2 x)/(1 + x)]) PolyLog[2, (1 + x)/2] - PolyLog[3, -x] -
PolyLog[3, x] - PolyLog[3, x/(1 + x)] + PolyLog[3, (2 x)/(1 + x)] -
PolyLog[3, (1 + x)/2] + (7 Zeta[3])/8
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3522822",
"timestamp": "2023-03-29T00:00:00",
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|
The real numbers $w, x, y, z$ satisfy the equalities $w+x+y+z=0$ and $w^2+x^2+y^2+z^2=1$. Prove that $-1\le wx+xy+yz+zw\le0$ The real numbers $w, x, y, z$ satisfy the equalities $w+x+y+z=0$ and $w^2+x^2+y^2+z^2=1$. Prove that $-1\le wx+xy+yz+zw\le0$
Supposing $wx+xy+yz+zw>0$. I've proven via contradiction that $wx+xy+yz+zw\le0$.
But suppose $wx+xy+yz+zw<-1$. Then $wx+xy+yz+zw+1<0$. I'm not sure how to factorise the LHS.
|
We have $a^2+b^2 \geq 2ab$ for any two real numbers $a$ and $b$. So
$$
\begin{aligned}
0=(x+y+z+w)^2 &= x^2+y^2+z^2+w^2+2(xy+yz+zw+wx)+2(xz+wy) \\
&=1+2(xy+yz+zw+wx)+2(xz+wy) \\
&\leq 1+2(xy+yz+zw+wx)+(x^2+z^2+y^2+w^2) \\
&=2+2(xy+yz+zw+wx)
\end{aligned}
$$
It follows that
$$-1 \leq xy+yz+zx+zw$$
Equality is achieved when $y=w$ and $x=z$.
This gives two equality cases $\left(\dfrac{1}{2}, -\dfrac{1}{2}, \dfrac{1}{2}, -\dfrac{1}{2}\right)$ and $\left(-\dfrac{1}{2}, \dfrac{1}{2}, -\dfrac{1}{2}, \dfrac{1}{2}\right)$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3524453",
"timestamp": "2023-03-29T00:00:00",
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|
For △ABC, prove $\frac a{h_a} + \frac b{h_b} + \frac c{h_c} \ge 2 (\tan\frac{\alpha}2+ \tan\frac{\beta}2 + \tan\frac{\gamma}2)$
Given $\triangle ABC$, (using the main parameters and notation), prove that $$ \frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2 \cdot \left(\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2}\right)$$
Great, another problem that already has a solution. We have many ways of expressing the area of a triangle, for example $$2 \cdot [ABC] = \frac{abc}{2R} = \sqrt{2(a + b + c) \cdot \sum_{cyc}\frac{c + a - b}{2}} = ah_a = bh_b = ch_c$$
The above equations are used in the solution I have provided below.
I would be greatly appreciated if you could come up with any other solutions.
|
Use the sine rule and the area formuli of the triangle $\frac12 h_a a = \frac12 bc\sin\alpha$, $\frac12 h_b b = \frac12 ca\sin\beta$, $\frac12 h_c c = \frac12 ab\sin\gamma$ to express
$$\frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} = \frac{\sin^2\alpha +\sin^2\beta+\sin^2\gamma}{\sin\alpha \sin\beta\sin\gamma}\tag 1$$
Then, evaluate
$$\sin^2\alpha- 2\sin\alpha \sin\beta\sin\gamma\cdot\tan\frac{\alpha}{2}$$
$$=4\sin^2\frac{\alpha}2\cos^2\frac{\alpha}2-4\sin^2\frac{\alpha}2\sin\beta\sin\gamma
=2\sin^2\frac{\alpha}2\left(2\cos^2\frac{\alpha}2-2\sin\beta\sin\gamma\right)$$
$$=2\sin^2\frac{\alpha}2\left(1+\cos\alpha-\cos(\beta-\gamma)+\cos(\beta+\gamma)\right)
= 4\sin^2\frac{\alpha}2\sin^2\frac{\beta-\gamma}2\tag 2\ge 0$$
and, similarly,
$$\sin^2\beta-2\sin\alpha\sin\beta\sin\gamma\cdot\tan\frac{\beta}{2} =4\sin^2\frac{\beta}2\sin^2\frac{\alpha-\gamma}2\ge 0 \tag 3$$
$$\sin^2\gamma-2\sin\alpha \sin\beta\sin\gamma\cdot\tan\frac{\gamma}{2}=4\sin^2\frac{\gamma}2\sin^2\frac{\alpha-\beta}2 \ge 0\tag 4$$
As a result, $(2)+(3)+(4)$ leads to
$$\sin^2\alpha +\sin^2\beta+\sin^2\gamma\ge
2\sin\alpha \sin\beta\sin\gamma\left(\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}+\tan\frac{\gamma}{2}\right)$$
Substitute the inequality into $(1)$ to obtain,
$$\frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2\left(\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}+\tan\frac{\gamma}{2}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Second order IVP differential equation So the given material and equations I have are
$$(1):x=C_1\cos(t)+C_2\sin(t)$$
$$x \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$$
$$x'\left(\frac{\pi}{3}\right)=0$$
First I sub in the $\frac{\pi}{3}$ and the $\frac{\sqrt{3}}{2}$ into my first equation and get
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$
Then finding $x'$and subbing in the same I get
$$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$
Now I am supposed to add the two equations together to find either $C_1$ or $C_2$ like so
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$
$$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$
But neither of the terms cancel out and I'm not sure if i need to find $x''$ to proceed
|
$$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$
$$ \implies \sqrt{3}C_1=C_2$$
Plug that in first equation.
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{{3}}{2}C_1$$
$$C_1=\frac{\sqrt{3}}{4}$$
$$ \implies C_2=\sqrt{3}C_1=\frac {3} 4$$
|
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"url": "https://math.stackexchange.com/questions/3525969",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ when $a+b+c=0$
Let $a,b,c$ such that $$a + b + c =0$$ and
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$ is defined.
Find the value of $P$.
This is a very queer problem.
|
Elegant approach
Think of $p(a, b, c)$ as a function $p : X\to\mathbb{R}$, where $X = \{(a, b, c)\in\mathbb{R}^3 : a+b+c=0\}$.
If $p$ is a constant function, i.e. if $p(a, b, c)\equiv P$ for some constant $P$, then in particular, $p(a, b, c) = P$ for any values of $a, b, c$ we might care to choose, so long as they satisfy $a+b+c=0$.
So it's as easy as just picking some set of values that will be easy to evaluate, such as $a=1, b=-1, c=0$, and substituting:
\begin{align}
p(a, b, c) &= \frac{a^2}{2a^2+bc} + \frac{b^2}{2b^2+ac} + \frac{c^2}{2c^2+ab} \\ \Rightarrow
p(1, 0, 0) &= \frac{1}{2+0} + \frac{(-1)^2}{2(-1)^2+0} + 0 = \frac{1}{2} + \frac{1}{2} = 1
\end{align}
Therefore $P = 1$.
(In case you don't want to accept from the statement of the problem that $p$ is indeed a constant function, it's straightforward enough to see: because $c = -a-b$, any increment in $a$ will change $p$ by $dp = (\partial{p}/\partial{a} - \partial{p}/\partial{c})\,da$; on the other hand, the symmetry of $p(a, b, c)$ in all three arguments implies $p(a, b, c) = p(c, b, a) \Rightarrow$ $\partial{p}/\partial{a} - \partial{p}/\partial{c} = 0 \Rightarrow$ $dp/da=0$. A similar argument can be made for $b$, to establish $dp/db=0$.)
Brute force approach
If you prefer to ignore the easy route shown above, and brute force this problem, then that is also possible. From $a + b + c = 0$, substitute $c = -(a + b)$, and simplify:
\begin{align}
P &= \frac{a^2}{2a^2+bc} + \frac{b^2}{2b^2+ac} + \frac{c^2}{2c^2+ab} \\
&= \frac{a^2}{2a^2-ab-b^2} + \frac{b^2}{2b^2-ab-a^2} + \frac{(a+b)^2}{(2a^2+4ab+2b^2)+ab} \\
&= \frac{a^2}{(a-b)(2a+b)} + \frac{-b^2}{(a-b)(a+2b)} + \frac{(a+b)^2}{(2a+b)(a+2b)} \\
&= \frac{a^2(a+2b) - b^2(2a+b) + (a+b)(a^2-b^2)}{(a-b)(2a+b)(a+2b)} \\
&= \frac{(a^3+2a^2b) + (-2ab^2-b^3) + (a^3-ab^2 + a^2b-b^3)}{(a-b)(2a+b)(a+2b)} \\
&= \frac{2a^3+3a^2b -3ab^2-2b^3}{(a-b)(2a+b)(a+2b)}
= \frac{(a-b)(2a+b)(a+2b)}{(a-b)(2a+b)(a+2b)} = 1
\end{align}
|
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|
In how many different ways can apples be placed in the box if at least there is one apple of each color?
Suppose there are $4$ red apples, $5$ green apples, and $6$ yellow apples, $9$ of them will be put into a box. In how many different ways can apples be placed in the box if at least there is one apple of each color?
I've tried to solve this problem and got the result of $673596$ different possible compositions. Here's how I try to solve it.
One apple of each color must be in the box, therefore the new sample space is that containing $3$ red apples, $4$ green apples and $5$ yellow apples $(3R, 4G, 5Y)$, and because there are already $3$ apples in the box, I just need to pick the remaining $6$ apples.
The problem now reduced to how much partition of $12$ objects into $4$ part namely $R$ (for red apples), $G$ (for green apples), $Y$ (for yellow apples) and $N$ (for none of the three) are possible, which is.
$$\sum \binom{12}{R,G,Y,N}$$
for $R+G+Y = 6$, and $N = 6$.
My question is whether there is any some kind of generalization of this problem so that I could solve it easily without deliberately looking for every possible arrangement of $R$, $G$ and $Y$ (which is how I try to solve the problem).
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The generating function for the number of ways to arrange $n$ apples is
$$
\begin{align}
&\left(x+x^2+x^3+x^4\right)\left(x+x^2+x^3+x^4+x^5\right)\left(x+x^2+x^3+x^4+x^5+x^6\right)\\
&=\frac{x-x^5}{1-x}\frac{x-x^6}{1-x}\frac{x-x^7}{1-x}\\
&=\left(x^3-x^7-x^8-x^9+x^{12}+x^{13}+x^{14}-x^{18}\right)\sum_{k=0}^\infty\binom{k+2}{k}x^k
\end{align}
$$
The coefficient of $x^n$ is
$$
\scriptsize\binom{n-1}{n-3}-\binom{n-5}{n-7}-\binom{n-6}{n-8}-\binom{n-7}{n-9}+\binom{n-10}{n-12}+\binom{n-11}{n-13}+\binom{n-12}{n-14}-\binom{n-16}{n-18}
$$
which is equal to
$$
\scriptsize\binom{n-1}{2}-\binom{n-5}{2}-\binom{n-6}{2}-\binom{n-7}{2}+\binom{n-10}{2}+\binom{n-11}{2}+\binom{n-12}{2}-\binom{n-16}{2}
$$
where the sum is only taken over the terms where the upper term is greater than or equal to the lower term.
For $n=9$, we get
$$
\binom{8}{2}-\binom{4}{2}-\binom{3}{2}-\binom{2}{2}=18
$$
|
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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.