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Seeking methods to solve: $\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt$ Seeking Methods to solve the following two definite integrals:
\begin{equation}
I_(n) = \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt \qquad J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt
\end{equation}
For $n \in \mathbb{R},\:n \gt 1$
The method I took was to take the following integral:
\begin{equation}
\int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)
\end{equation}
Where: $0 \leq k \lt n$. Here let $m = 1$.
Differentiate under the curve with respect to $k$ and taking the limit as $k \rightarrow 0^+$ (via the Dominated Convergence Theorem), i.e.
\begin{align}
\lim_{k \rightarrow 0+} \frac{d}{dk} \left[ \int_0^\infty \frac{t^k}{t^n + 1}\:dt \right] &= \lim_{k \rightarrow 0+} \frac{d}{dk} \left[\frac{1}{n}B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \right] \\
\lim_{k \rightarrow 0+} \int_0^\infty \frac{t^k \ln(t)}{t^n + 1}\:dt &= \lim_{k \rightarrow 0+} \left[\frac{1}{n^2} B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\left[\psi^{(0)}\left(\frac{k + 1}{n} \right) - \psi^{(0)}\left(1 - \frac{k + 1}{n} \right)\right] \right] \\
\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt&= \frac{1}{n^2} B\left(1 - \frac{1}{n}, \frac{1}{n}\right)\left[\psi^{(0)}\left(\frac{1}{n} \right) - \psi^{(0)}\left(1 - \frac{1}{n} \right)\right] \\
&=- \frac{\pi^2}{n^2}\operatorname{cosec}\left(\frac{\pi}{n}\right)\cot\left(\frac{\pi}{n} \right)
\end{align}
Which is our expression for $I_n$. Taking the same approach but differentiating twice with respect to $k$ we arrive at our expression for $J_n$:
\begin{equation}
J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt = \frac{\pi^3}{n^3}\operatorname{cosec}\left(\frac{\pi}{n} \right)\left[\operatorname{cosec}^2\left(\frac{\pi}{n}\right) + \cot^2\left(\frac{\pi}{n}\right) \right]
\end{equation}
And in fact we may generalise:
\begin{equation}
\int_0^\infty \frac{\ln^p(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k\rightarrow 0}\frac{d^p}{dk^p}\left[\frac{1}{n} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\right]
\end{equation}
Where $p \in \mathbb{N}$
This method however was just an extension of another integral. I'm curious, if I had just started with $I_n, J_n$ what alternative methods could be used?
|
Here is an alternative strategy that can be used to find $I^{(1)}_n$ (your $I_n$), $I^{(2)}_n$ (your $J_n$), and $I^{(p)}_n$ (the general case where $p \in \mathbb{N}$).
Writing
$$I^{(1)}_n = \int_0^\infty \frac{\ln t}{1 + t^n} \, dt = \int_0^1 \frac{\ln t}{1 + t^n} \, dt + \int_1^\infty \frac{\ln t}{1 + t^n} \, dt.$$
Enforcing a substitution of $t \mapsto 1/t$ in the second of the integrals appearing to the right yields
$$I^{(1)}_n = \int_0^1 \frac{\ln t}{1 + t^n} \, dt - \int_0^1 \frac{\ln t}{1 + t^n} \, t^{n - 2} dt.$$
Exploiting the geometric sum for the term $1/(1 + t^n)$ leads to
$$I^{(1)}_n = \sum_{k = 0}^\infty (-1)^k \int_0^1 t^{kn} \ln t \, dt - \sum_{k = 0}^\infty (-1)^k \int_0^1 t^{kn + n - 2} \ln t \, dt.$$
After integrating by parts we have
\begin{align}
I^{(1)}_n &= -\frac{1}{n^2} \sum_{k = 0}^\infty \frac{(-1)^k}{(k + 1/n)^2} + \frac{1}{n^2} \sum_{k = 0}^\infty \frac{(-1)^k}{(k + 1 - 1/n)^2}\\
&= \frac{1}{n^2} \sum_{\substack{k = 0\\k \in \text{odd}}} \frac{1}{(k + 1/n)^2} - \frac{1}{n^2} \sum_{\substack{k = 0\\k \in \text{even}}} \frac{1}{(k + 1/n)^2}\\
& \qquad -\frac{1}{n^2} \sum_{\substack{k = 0\\k \in \text{odd}}} \frac{1}{(k + 1 - 1/n)^2} + \frac{1}{n^2} \sum_{\substack{k = 0\\k \in \text{even}}} \frac{1}{(k + 1 - 1/n)^2}.
\end{align}
Shifting the odd indices by: $k \mapsto 2k + 1$, and the even indices by: $k \mapsto 2k$, gives
\begin{align}
I^{(1)}_n &= \frac{1}{4n^2} \sum_{k = 0}^\infty \frac{1}{(k + 1/2 + 1/2n)^2} - \frac{1}{4n^2} \sum_{k = 0}^\infty \frac{1}{(k + 1/2n)^2}\\
& \qquad - \frac{1}{4n^2} \sum_{k = 0}^\infty \frac{1}{(k + 1 - 1/2n)^2} + \frac{1}{4n^2} \sum_{k = 0}^\infty \frac{1}{(k + 1/2 - 1/2n)^2}\\
&= -\frac{1}{4n^2} \left [ \psi^{(1)} \left (1 - \frac{1}{2n} \right ) + \psi^{(1)} \left (\frac{1}{2n} \right ) \right ]\\
& \qquad + \frac{1}{4n^2} \left [\psi^{(1)} \left (1 - \left (\frac{1}{2} - \frac{1}{2n} \right ) \right ) + \psi^{(1)} \left (\frac{1}{2} - \frac{1}{2n} \right ) \right ]\\
&= -\frac{\pi^2}{4n^2} \left [\operatorname{cosec}^2 \left (\frac{\pi}{2n} \right ) - \operatorname{cosec}^2 \left (\frac{\pi}{2} + \frac{\pi}{2n} \right ) \right ]\\
&= -\frac{\pi^2}{4n^2} \left [\operatorname{cosec}^2 \left (\frac{\pi}{2n} \right ) - \sec^2 \left (\frac{\pi}{2n} \right ) \right ]\\
&= -\frac{\pi^2}{n^2} \left [\frac{\cos^2(\pi/2n) - \sin^2 (\pi/2n)}{\{2 \sin (\pi/2n) \cos (\pi/2n)\}^2} \right ]\\
&= -\frac{\pi^2}{n^2} \operatorname{cosec} \left (\frac{\pi}{n} \right ) \cot \left (\frac{\pi}{n} \right ),
\end{align}
as expected. Note we have made use of the reflexion formula for the trigamma function $\psi^{(1)}(z)$.
In a similar way to how $I^{(1)}_n$ was found above, $I^{(2)}_n$ and $I^{(p)}_n$ can also be found.
|
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"url": "https://math.stackexchange.com/questions/3073077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Determining the area of a right triangle, perimeter given, hypotenuse value given in terms of one of the legs. The problem states:
Right Triangle- perimeter of $84$, and the hypotenuse is $2$ greater than the other leg. Find the area of this triangle.
I have tried different methods of solving this problem using Pythagorean Theorem and systems of equations, but cannot find any of the side lengths or the area of the right triangle. I looked for similar problems on StackExchange and around the internet, but could not find anything.
Does anyone know anything that could help find the side lengths of the triangle and the area as well?
Method that I tried:
*
*Made a system with the values given.
\begin{align}
a+b+c&=84 \\
c&=b+2
\end{align}
*Substituted $c$ with $b+2$.
\begin{align}
a+b+b+2&=84 \\
a + 2b &= 82 & \text{subtracted $2$ from both sides}\\
a + a^2 - 4 &= 82
\end{align}
*$c^2$ is $(b+2)(b+2)$, so I used Pythagorean Theorem to isolate one of the variables.
\begin{align}
a^2+b^2 &=c^2\\
a^2 + b^2 &=(b+2)(b+2)\\
a^2+b^2 &=b^2+2b+4\\
a^2&=2b+4 & \text{ (Subtracted $b^2$ from both sides) }
\end{align}
OR
\begin{align}
a^2-4&=2b
\end{align}
I do not know what to do after this point.
|
Let the sides of the right triangle be $x,y,x+2$.
Given,
$2x+y=82 \tag{1}$
$x^2 + y^2 = (x+2)^2 \tag{2}$
$$\implies x^2 + y^2 = x^2 +4x+4 $$
$$\implies y^2 = 4x+4 $$
Now, substitute the value of $x$ from equation (1) in terms of $y,$ you will get a quadratic equation in $y$ whose roots can be easily found and hence, the sides and area.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3076504",
"timestamp": "2023-03-29T00:00:00",
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|
Candy machines and optimal strategy in terms of expected value Problem
We have three candy machines: call them G (good), B (bad) and M (mixed) . G always gives you a candy when you put 1\$. B never gives you a candy when you put 1\$. M gives you a candy with probability $1/2$ when you put 1\$. You want a candy and you approach the three machines but you don't know which one is G, B or M. You use the following strategy. You approach a random machine. If you don't get a candy in $n$ trials, you change a machine. If you don't get a candy from the second machine in $k$ trials, you change to the remaining machine. At most, we can pay for a candy $n+k+1$ \$. We want to calculate the expected cost of obtaining a candy using this strategy. Is it the case that $k=n=1$ is the optimal strategy, i.e. yielding the least expected cost of obtaining a candy?
Attempt of solution
Currently, my thinking is the following. Consider the indicator random variable $X_i = 1$ meaning that we pay 1\$ at stage $i$. Observe that:
*
*$P(X_1 = 1) = 1$ (we always pay 1\$ at the beginning)
*$P(X_2 = 1) = \frac{1}{3} + \frac{1}{3}\frac{1}{2}$ (you pay at stage $2$ iff you don't receive a candy at stage $1$ which happens either when you choose $B$ with probability one third or you choose $M$ with probability one third and then $M$ doesn't give you a candy with probability one half)
*$P(X_3 = 1) = \frac{1}{3} + \frac{1}{3}(\frac{1}{2})^2$ (similar justification as previously but keeping in mind that you consider an event of not obtaining a candy at stages $1$ and $2$)
*$\dots$
*$P(X_{n+1} = 1) = \frac{1}{3} + \frac{1}{3}(\frac{1}{2})^n$
*$P(X_{n+2} = 1) = \frac{1}{6} \frac{1}{2} + \frac{1}{6}(\frac{1}{2})^n$ (= the probability that we don't receive a candy at stages up to $n+1$. This could happen either when we first happen to choose B and then M, in which case the probability of not obtaining a candy in $n+1$ trials is $\frac{1}{6} \frac{1}{2}$, or when we first choose M and than B, in which case the probability of not obtaining a candy in $n+1$ trials is $\frac{1}{6}(\frac{1}{2})^n$)
*$\dots$
*$P(X_{n+k+1} = 1) = \frac{1}{6} (\frac{1}{2})^k + \frac{1}{6}(\frac{1}{2})^n$
We are interested in $E(\Sigma_{i=1}^{n+k+1} X_i)$. Using linearity of expectation, we may write:
$$E(\Sigma_{i=1}^{n+k+1} X_i) = 1 + E(\Sigma_{i=2}^{n+1}X_i) + E(\Sigma_{j=n+2}^{n+k+1}X_j) = $$
$$=1 + \Sigma_{i=1}^n(\frac{1}{3} + \frac{1}{3}(\frac{1}{2})^i) + \Sigma_{j=1}^k(\frac{1}{6}(\frac{1}{2})^j + \frac{1}{6}(\frac{1}{2})^n)=$$
$$= 1 + \frac{1}{3}n + \frac{1}{3}\Sigma_{i=1}^n\frac{1}{2^i} + k \frac{1}{6} \frac{1}{2^n} + \frac{1}{6}\Sigma_{j=1}^k \frac{1}{2^j}$$
Now, we can easily observe that when $n$ is fixed and we make $k$ bigger, $E$ also grows. This excludes strategies $n < k$ from being optimal. We can go somewhat further by observing that:
$$ \Sigma_{i=1}^n\frac{1}{2^i} = \frac{1 - \frac{1}{2}^{n+1}}{1 - \frac{1}{2}}- 1 = 1 - (\frac{1}{2})^n$$
This gives you:
$$E(X) = 1 + \frac{1}{3}n + \frac{1}{3}(1 - (\frac{1}{2})^n) + k \frac{1}{6} \frac{1}{2^n} + \frac{1}{6}(1 - (\frac{1}{2})^k)=$$
$$ 1 + \frac{1}{3}n + \frac{1}{3} - \frac{1}{3}(\frac{1}{2})^n + k \frac{1}{6} \frac{1}{2^n} + \frac{1}{6} - \frac{1}{6}(\frac{1}{2})^k$$
$$6 E(X) = 9 + 2n - 2 (\frac{1}{2})^n + k \frac{1}{2^n} - (\frac{1}{2})^k$$
$$6 E(X) = 9 + 2n + (k-2)\frac{1}{2^n} - (\frac{1}{2})^k$$
This makes it clear that when you keep $k$ fixed and make $n$ bigger, $E$ grows as well. This excludes strategies $k < n$ from being optimal. So the optimal strategy should satisfy $n=k$. So let's assume that in the equation for $E$:
$$6 E(X) = 9 + 2n + (n-2)\frac{1}{2^n} - (\frac{1}{2})^n=$$
$$ 9 + 2n + (n-3)\frac{1}{2^n}$$
I will not go further now, but, at least know, it's clear for me that the right hand side of the above equation assumes minimal value for $n=1$ which yields the strategy $n=k=1$ as optimal.
|
Clearly you always want $k = 1$. If you have tried two machines and gotten candy from neither, the last one must be the Good one, so there's no reason not to try that next. This leaves our choice of $n$. Intuitively it's clear that this one should be 1 as well: after failing to get a candy from the machine once, it's either Bad or Mixed with probability 1/2, while both other machines are Good with probability 1/2, so those machines definitely appear more appealing. But let's make this formal.
The strategy $n = k = 1$ nets you an expected value of $5/3$ tries until you obtain a candy. Now suppose that $n \geq 2$. Write $X$ for the number of tries, and $B, M, G$ for the event that the first machine you try is the Bad, Mixed, Good one. Then:
$$
\begin{align*}
\mathbb{E}(X) &= \underbrace{\mathbb{E}(X \mid B)}_{\geq n + 1}\mathbb P(B) + \underbrace{\mathbb{E}(X \mid M)}_{> 1}\mathbb P(M) + \underbrace{\mathbb{E}(X \mid G)}_{=1}\mathbb P(G)\\
&>(n+1)\frac13+ 1\times \frac13 + 1 \times \frac13\\
& \geq \frac53.
\end{align*}
$$
Here the "strictly greater than" on line 2 stems from the fact that the expected number of tries when the first machine is Mixed is actually strictly greater than 1. Thus $n = k = 1$ is optimal.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
How to compute $u_x(\sqrt {(x^2+y^2)})=$? This is in reference to this question
$g(z) = f(|z|)$ is not holomorphic for a non constant function $f$
If $u$ is a function of single variable $x$ and if I want to differentiate $u(\sqrt {(x^2+y^2)}$ with respect to $x$
then what will be $u_x(\sqrt {(x^2+y^2)})=$?
It is given in the question that linked above
that $u_x(\sqrt {(x^2+y^2)})=u_x\times \frac{x}{\sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false
What is the correct way to solve it?
Any help.
|
What happens when I do this with $u(x)=x^2+x$. Write
$$f(x,y)=u\left(\sqrt{x^2+y^2}\right)=x^2+y^2+\sqrt{x^2+y^2}.$$
Then
$$\frac\partial{\partial x}f(x,y)=2x+\frac{x}{\sqrt{x^2+y^2}}.$$
But $u'(x)=2x+1$, and
$$u'\left(\sqrt{x^2+y^2}\right)\frac\partial{\partial x}\sqrt{x^2+y^2}
=\left(2\sqrt{x^2+y^2}+1\right)\frac{x}{\sqrt{x^2+y^2}}
=2x+\frac{x}{\sqrt{x^2+y^2}}.$$
I don't see any problem....
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show using the formal, limited based definition of integral I am currently trying to figure out the following: for $f(x) = 2x - 5$, I want to show that, using the formal, limited based definition of integral, $\int_{3}^{7}f(x) = 20$ (both the domain and codomain of $f(x)$ are $\mathbb{R}$). However, I already seem to get stuck on the formal definition of the integral, which is (I believe) $\lim_{n \rightarrow \infty}\sum_{i=0}^{n-1}f(t_i)\Delta t$. Can anyone explain to me how to proceed? Or show me the steps I need to follow to get to the right answer?
|
Divide the interval from x= 3 to x= 7 into n equal length subintervals. Each has length (7- 3)/n= 4/n. The x values at the endpoints of the subintervals are $x_0= 3$, $x_1= 3+ 4/n= (3n+ 4)/n$, $x_2= 3+ 2(4/n)= (3n+ 8)/n$, …, $x_i= 3+ i(4/n)= (3n+ 4i)/n$, …, $x_n= 3+ n(4/n)= 7$. At each i, $f(x_i)= 2x_i- 5= 2(3n+ 4i)/n- 5= (6n+ 8i- 5n)/n= (n+ 8i)/n$.
On the ith interval, $f(x_i)\Delta x= [(n+ 8i)/n](4/n)= (4n+ 32i)/n^2= \frac{4}{n}+ \frac{32i}{n^2}$. Add those: $\sum_{i= 0}^{n-1} \frac{4}{n}+ \frac{32i}{n^2}= \frac{1}{n}\sum_{i=0}^{n-1} 4+ \frac{32}{n^2}\sum_{i=0}^{n-1} i$.
Now, adding "4" to itself n times is just 4n while adding "i", with i running from 0 to n-1 means 0+ 1+ 2+ 3+ 4+ …+ n-1.
So what is such sum? To start with lets call it S(n). That is, S(n)= 0+ 1+ 2+ 3+ 4+ …+ n-1. Of course, it is the same if we do the sum in the opposite order: S(n)= n-1+ n-2+ n-3+ n-4+...+ 1+ 0. Add those two "vertically"! That is, 2S(n)= [0+ n-1]+ [1+ n-2]+ [2+ n-3]+ …+ [n-2+ 1]+ [n- 1]. Every term is "n-1" and there are n terms. 2S= n(n-1) so S= n(n-1)/2.
Check that for a few values of n: with n= 5, 0+ 1+ 2+ 3+ 4= 10 and 5(5-1)/2= 20/2= 10. with n= 9, 0+ 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8= 36 and 9(8)/2= 36.
So we have that the sum is n(n-1)/2 and we multiply that by $\frac{32}{n^2}$: $\frac{32}{n^2}\frac{n(n-1)}{2}= \frac{32}{n^2}\frac{n^2- n}{2}= 16(1- \frac{1}{n})$.
Putting those together the "Riemann sum" with n intervals is $4+ 16(1- \frac{1}{n})$. As n goes to infinity, 1/n goes to 0 so that sum goes to $4+ 16(1+ 0)= 20$.
Of course, geometrically, the area under that straight line from x= 3 to x= 7 is a trapezoid with bases of length 2(3)- 5= 1 and 2(7)- 5= 9 and height 7- 3= 4. The area of that trapezoid is (1/2)(1+ 9)(4)= 20 as before.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$ To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$
I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I proceed to induct on $m$ (and on $m$ only).
I am not perfectly confident in my self, however, for I see two placeholders, $n$ and $m.$ Is this a case where double induction is needed (first on $m$ and then on $n$)?
Whether or not this proof requires double induction, may someone explain when double induction is needed?
Consider any fixed $n, r \geq 0$ and the following two cases (I know that only one case is needed to complete this inductive proof).
CASE 1
\begin{align}
\binom{n + 0}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{0}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r}\binom{0}{0} \\ &= 0 + 0 + \cdots + \binom{n}{r} \\ &= \binom{n}{r}
\end{align}
CASE 2
\begin{align}
\binom{n + 1}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{1}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r-1}\binom{1}{r - (r-1)} + \binom{n}{r}\binom{1}{r - r} \\ &= 0 + 0 + \cdots + \binom{n}{r-1} + \binom{n}{r} \\ &= \binom{n}{r-1} + \binom{n}{r}
\end{align}
INDUCTION
Suppose it is true for $m \leq k.$ Now, consider $$\binom{n + (k + 1)}{r}.$$ It follows from Pascal's Identity that
$$\binom{n + (k+1)}{r} = \binom{n + k}{r} + \binom{n + k}{r-1}$$
And,
\begin{align}
\binom{n + k}{r} + \binom{n + k}{r-1} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\bigg[\binom{k}{r - i} + \binom{k}{r - 1 - i}\bigg] \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k+1}{r-i} \\ &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k+1}{r-i}
\end{align}
Hence, the equality holds for $m = k + 1.$ Given that the equality holds for $m = 0, 1,$ and that if equality holds for $m = k,$ it then holds for $m = k + 1,$ it follows that the equality holds $\forall m \in \mathbb{N}.$
|
I think the formula is wrong. If you have $n$ white balls and $m$ red balls then, to choose $r$ balls from then you have to choose $i$ white balls and $r-i$ red balls for some $i$ between $0$ and $\min \{r,n\}$. Hence the sum should be over $0\leq r \leq \min \{r,n\}$. However, if you define $\binom {n} {i}$ to be $0$ when $i >n$ then the formula is correct.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute $ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$
Compute $$ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$$
I did the following:
$$ \lim\limits_{n \to \infty}\frac{\sqrt{\frac{3n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}}{\frac{n}{n^2}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}} = \frac{\sqrt{3}}{\sqrt{1}}=\sqrt3$$
However, the correct answer is different. Why am I wrong?
Thank you for your help.
|
$$
\lim_{n\to\infty}\frac{\sqrt{3n^2 + n-1}}{n+\sqrt{n^2-1}}
$$
Factor out $n$, thus:
$$
\frac{\sqrt{3n^2 + n-1}}{n+\sqrt{n^2-1}} = \frac{n\sqrt{3 + {1\over n} - {1\over n^2}}}{n\left(1 + \sqrt{1- {1\over n^2}}\right)} = \frac{\sqrt{3 + {1\over n} - {1\over n^2}}}{1 + \sqrt{1- {1\over n^2}}}
$$
So your limit becomes:
$$
\lim_{n\to\infty}\frac{\sqrt{3 + {1\over n} - {1\over n^2}}}{1 + \sqrt{1- {1\over n^2}}}
$$
Now what happens to ${1\over n^k}$ when $n\to\infty, k\in\Bbb N$?
|
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|
Find range of $x$ if $\log_5\left(6+\frac{2}{x}\right)+\log_{1/5}\left(1+\frac{x}{10}\right)\leq1$
If $\log_5\left(6+\dfrac{2}{x}\right)+\log_{1/5}\left(1+\dfrac{x}{10}\right)\leq1$, then $x$ lies in _______
My Attempt
$$
\log_5\bigg(6+\dfrac{2}{x}\bigg)+\log_{1/5}\bigg(1+\dfrac{x}{10}\bigg)=\log_5\bigg(6+\dfrac{2}{x}\bigg)-\log_{5}\bigg(1+\dfrac{x}{10}\bigg)\leq1\\
\log_5\frac{(6x+2)10}{x(10+x)}\leq1\implies\frac{(6x+2)10}{x(10+x)}\leq5\\
\frac{4(3x+1)}{x^2+10x}\leq1\\
\implies 12x+4\leq x^2+10x\quad\text{or}\quad12x+4>x^2+10x\\
x^2-2x-4\geq0\quad\text{or}\quad x^2-2x-4<0\implies x\in\mathcal{R}
$$
My reference gives the solution $(-\infty,1-\sqrt{5})\cup(1+\sqrt{5},\infty)$, what is going wrong here ?
|
You forgot to check when $$6+\dfrac{2}{x}>0$$ and $$1+\dfrac{x}{10}>0$$
is true!
|
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|
Proving $3n^2 - 4n \in \Omega(n^2)$ Proving $3n^2 - 4n \in \Omega(n^2)$
Attempt:
$3n^2 - 4n \geq cn^2$
$n(3n-4) \geq cn^2$
$(3n-4) \geq cn$
$3n - 4 - cn \geq 0$
$n(3-c) \geq 4$
$n \geq \frac{4}{3-c}$
would do I choose for $n_0$ and $c$ to satisfy this proof?
|
$3n^2 - 4n \in \Omega(n^2)$ means:
there are $c>0$ and $n_0 \in \mathbb N$ such that $3n^2-4n \ge cn^2$ for $n \ge n_0$.
Since $\frac{3n^2-4n}{n^2}= 3-\frac{4}{n} \to 3$ as $n \to \infty$, there is $n_0$ such that $\frac{3n^2-4n}{n^2} \ge 2$ for $n \ge n_0$.
Hence $3n^2-4n \ge 2n^2$ for $n \ge n_0$.
This shows that $3n^2 - 4n \in \Omega(n^2)$.
|
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|
How to prove binomial coefficient $ {2^n \choose k} $ is even number? Prove:
${2^n \choose k}$ (for integers $k$ & $n$ : $0<k<2^n$) is even number.
I have tried induction but was unable to get any useful results.
|
Provided that you already know $\binom{2^n}{k}$ is an integer, it suffices to show the numerator has more factors of $2$ than the denominator. We have:
$$\binom{2^n}{k}=\frac{(2^n)(2^n-1)\cdots(2^n-k+1)}{k!}.$$
There are at least $n$ factors of $2$ in the numerator because $2^n$ is a factor. So now we need to count the number of factors of $2$ in the denominator, $k!$.
We have $k!=k(k-1)(k-2)\cdots 3\cdot 2 \cdot 1$. At most $\frac{k}{2}$ of these numbers are divisible by $2$. At most $\frac{k}{4}$ of them are divisible by $4$. At most $\frac{k}{8}$ of them are divisible by $8$. And so on. Each number that is divisible by $2$ contributes a factor of $2$, each number that is divisible by $4$ contributes an additional factor of $2$, each number that is divisible by $8$ contributes a third factor of $2$, and so on. So the number of factors of $2$ in $k!$ is no more than
$$\frac{k}{2}+\frac{k}{4}+\frac{k}{8}+\cdots = k.$$
Since $k<n$, the denominator has strictly fewer factors of $2$ than the numerator.
|
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|
$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ Problem:
solve equation
$$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$$
I don't look for easy solution (square booth side and things like that...) I look for some tricks for "easy" solution because:
I would like to use substitution, but we have $3x$ and $-3x$, but I can't see it.
Solution:
|
Render
$(\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5})(\sqrt{2x^2 + 3x +5} - \sqrt{2x^2-3x+5})=6x$
from the difference of squares factorization. So then
$\color{blue}{\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x}$
$\color{blue}{\sqrt{2x^2 + 3x +5} - \sqrt{2x^2-3x+5}=2}$
Add them up:
$2\sqrt{2x^2 + 3x +5} = 3x+2$
Then square and continue.
Be sure to check the roots of your eventual quadratic equation for the proper signs of the square roots. Can you tell from the blue equations that the proper signs will require $x>(2/3)$?
|
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|
Show that if $a,b,c\in \mathbb{N}$ and ${a^2+b^2+c^2}\over{abc+1}$ is an integer it is the sum of two nonzero squares I was reading about the fascinating problem in the IMO ($1988 $ #$6$) that asks:
Let $a$ and $b$ be positive integers such that $(ab+1) | (a^2+b^2)$. Show that ${a^2+b^2}\over{ab+1}$ is a perfect square.
I wondered what would happen with the natural extension of this problem to three constants. Upon exploration with Mathematica, I found that the integral values of ${a^2+b^2+c^2}\over{abc+1}$ all seemed to be expressible as the sum of two nonzero perfect squares. The converse also seems to be true (given that a number is the sum of two nonzero squares, it seems to be expressible in the form of this fraction).
So I pose two questions:
If $a,b,c\in \mathbb{N}$ and $k={{a^2+b^2+c^2}\over{abc+1}}$ is an integer, is $k$ the sum of two nonzero squares?
and
Given that $k$ is expressible as the sum of two nonzero perfect squares, do there exist $a,b,c\in\mathbb{N}$ for which ${{a^2+b^2+c^2}\over{abc+1}}=k$?
|
I PROVED THE CONJECTURE
$$ x^2 + y^2 + z^2 = k + kxyz $$
Notice that we cannot have a solution with $x<0$
Got the other part. Let $x \geq y \geq z \geq 1$ be a solution giving $k$ that minimizes $x+y+z$ among positive solutions giving $k.$ Vieta jumping is just the new solution
$$ (x,y,z) \mapsto (kyz-x,y,z) $$
The equation is
$$ x^2 - (kyz)x + y^2 + z^2 - k = 0. $$
The replacement root $x'$ gives $x + x' = kyz. $ The product is $xx' = y^2 + z^2 - k.$
We have minimized $x+y+z$ among positive triples giving $k.$ We get a new triple with $x' = kyz - x.$ As it is impossible to have $x' <0,$ we have an Alternative:
(A) $kyz - x = 0$
(B) $kyz - x \geq x$
We will show that alternative (B) does not occur for, say, $k > 2.$
ASSUME (B), namely $kyz \geq 2x, $ with $x \geq y \geq z \geq 1.$
$$ x^2 + y^2 + z^2 = k + (kyz)x \geq k + (2x)x= k + 2x^2 $$
$$ \color{red}{ y^2 + z^2 \geq k + x^2}. $$
But
$$ y^2 - (kx)yz + z^2 = k - x^2. $$
Add,
$$ 2y^2 - (kx)yz + 2z^2 \geq 2k . $$
$$ y^2 - \left(\frac{kx}{2} \right)yz + z^2 \geq k . $$
So, $y \geq z \geq 1$ and
$$ \color{red}{y^2 - \left(\frac{kx}{2} \right)yz + z^2 > 0 }. $$
From the quadratic formula,
$$ y > \frac{1}{4} \left( kx + \sqrt {k^2 x^2 - 16} \right) z . $$
When $k \geq 4,$ we see $y > x,$ a contradiction. When $k=3$ and $z \geq 2,$ we still get $y > x.$ Finally, with $k=3$ and $z=1,$ the condition $y^2 + z^2 \geq k + x^2$ reads $y^2 + 1 \geq 3 + x^2$ or $y^2 \geq 2 + x^2,$ so again $y > x.$
These contradictions tell us that alternative (A) actually holds for $k \geq 3.$ That is, $kyz = x,$ so $kyz - x = 0,$ and we have a new solution $(0,y,z)$ with $y \geq z \geq 1$ and this $x=0$ to
$$ x^2 + y^2 + z^2 = k + kxyz. $$ As this $x=0,$ we have reached
$$ \color{red}{ y^2 + z^2 = k } $$
|
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|
What is the value of $1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+... $? I need to find the value of the following series:
$$ 1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+\cdots $$
That is
$$
1+ \frac{1}{2^2}+ \frac{1}{(2^2)^2} +\frac {1}{(2^3)^2}+ \cdots$$
It is the summation of a geometric progression where all the terms are squared. I am unable to go further with it.
|
$$
\begin{align}
1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+\cdots
&=\frac{1}{(2^0)^2}+\frac{1}{(2^1)^2}+ \frac{1}{(2^2)^2}+ \frac{1}{(2^3)^2}+\cdots\\
&=\frac{1}{(2^2)^0}+\frac{1}{(2^2)^1}+ \frac{1}{(2^2)^2}+ \frac{1}{(2^2)^3}+\cdots\\
&=\sum\limits_{n=0}^{\infty}\frac{1}{\left(2^2\right)^n}\\
&=\sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n.
\end{align}
$$
$\sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n$ is a geometric series because it's an expression of the form $\sum\limits_{n=0}^{\infty}ar^n$ where $a=1$ and $r=\frac{1}{4}$. We also know that if $|r|<1$, a geometric series converges (otherwise it diverges which means that the sum does not approach a finite number, in other words, it blows up to either positive or negative infinity) and we can even find what it converges to using the formula $\sum\limits_{n=0}^{\infty}ar^n=\frac{a}{1-r}$. In our case, $\left|\frac{1}{4}\right|<1$. So, we know that our series converges and we can find what it converges to:
$$
\sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}=1\frac{1}{3}.
$$
|
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|
Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$
Calculate $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$
I have tried to put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$$ \int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$
$$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)^2}dt$$
$$ \frac{1}{2}\bigg[-t\ln^2(t)\frac{1}{t^2-1}+\int^{\infty}_{0}\frac{\ln^2(t)}{t^2-1}+2\int^{\infty}_{0}\frac{\ln(t)}{t^2-1}dt\bigg]$$
How can I solve it?
|
Here is yet another slight variation on a theme.
Let
$$I = \int_0^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx$$
then
\begin{align}
I &= \int_0^1 \frac{\ln^2 x}{(1 - x^2)^2} \, dx + \int_1^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx = \int_0^1 \frac{(1 + x^2) \ln^2 x}{(1 - x^2)^2} \, dx \tag1,
\end{align}
after a substitution of $x \mapsto 1/x$ has been enforced in the second of the integrals.
As
$$\frac{1}{1 - x^2} = \sum_{n = 0}^\infty x^{2n}, \qquad |x| < 1,$$
differentiating with respect to $x$ gives
$$\frac{1}{(1 - x^2)^2} = \sum_{n = 1}^\infty n x^{2n - 2}.$$
On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have
$$I = \sum_{n = 1}^\infty n \int_0^1 (x^{2n - 2} + x^{2n}) \ln^2 x \, dx.$$
Integrating by parts twice, we are left with
$$I = \sum_{n = 1}^\infty \left (\frac{2n}{(2n - 1)^3} + \frac{2n}{(2n + 1)^3} \right ).$$
As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have
\begin{align}
I &= \sum_{n = 1}^\infty \left [\frac{1}{(2n - 1)^2} + \frac{1}{(2n - 1)^3} + \frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\
&= \sum_{n = 1}^\infty \left [\frac{1}{(2n - 1)^2} + \frac{1}{(2n - 1)^3} \right ] + \sum_{n = 1}^\infty \left [\frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\
&= \sum_{n = 0}^\infty \left [\frac{1}{(2n + 1)^2} + \frac{1}{(2n + 1)^3} \right ] + \sum_{n = 1}^\infty \left [\frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\
&= 2 + 2 \sum_{n = 1}^\infty \frac{1}{(2n + 1)^2}\\
&= 2 \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}\\
&= 2 \left [\sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} \right ]\\
&= 2 \left (1 - \frac{1}{4} \right ) \sum_{n = 1}^\infty \frac{1}{n^2}\\
&= \frac{3}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\
&= \frac{3}{2} \cdot \frac{\pi^2}{6}\\
&= \frac{\pi^2}{4},
\end{align}
as expected.
|
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|
Exponentiation in Non-Commutative Rings I'm fairly new to abstract algebra and in an exercise I was asked under which conditions it is true that $(ab)^n = a^nb^n$, for $a,b \in R$ and $n$ a positive integer, where $R$ is a ring. It can be easily shown that the statement holds if $R$ is commutative but I am stuck on the reverse implication. I have shown that if $R$ contains no zero divisors then the statement implies commutativity. Indeed, taking $n = 2$
$$(ab)^2 = abab = a^2b^2 \Leftrightarrow a(ba-ab)b = 0 \Leftrightarrow ba = ab$$
Does there exist non-commutative rings with zero divisors such that $(ab)^n =a^nb^n$ or is this property equivalent to commutativity?
|
Hagen von Eitzen gave a noncommutative nonunital ring with this property in the comments. However if your ring has a unit, this is impossible.
The proof below is a mess of algebra, so let me briefly explain the idea.
The problem is that the equalities we are given are homogeneous equations of degree at least $4$, like $abab=a^2b^2$, and we want a homogeneous equation of degree 2, $ab=ba$. Thus we need to somehow dehomogenify these equations, which we can do by substituting $a=1+a$ or $b=1+b$ and seeing what we get. This is the idea of the proof below.
Edit:
I will leave my original proof below, but darij grinberg suggested an excellent way to simplify the calculations by phrasing things in terms of commutators in the comments below.
Observe that the condition $abab=a^2b^2$ for all $a$ and $b$ is the same as saying $$a[a,b]b=0$$ for all $a$ and $b$. Now note that substituting $1+a$ in for $a$ gives $$(a+1)[a+1,b]b=(a+1)[a,b]b=0,$$ since $[1,b]=0$. Similarly, we also get $$a[a,b](b+1)=0\text{, and }(a+1)[a,b](b+1)=0.$$ Now add the first and fourth equations and subtract the middle two to get:
$$a[a,b]b - (a[a,b]b + [a,b]b) - (a[a,b]b+a[a,b]) + (a[a,b]b + a[a,b] + [a,b]b + [a,b]) =0,$$
or, on simplifying, $$[a,b]=0,$$
as desired.
Orignal proof
Consider
$$a^2+2a^2b+a^2b^2=a^2(1+b)^2$$
$$=(a(1+b))^2=(a+ab)^2 = a^2+a^2b + aba + abab.$$
Now use that $a^2b^2 = abab$, and cancel $a^2+a^2b+abab$ from both sides to get
$$ a^2b = aba.$$
By symmetry, we must also have $ba^2=aba$, by considering $(1+b)^2a^2$.
Now consider $(1+a)^2(1+b)^2$.
Expanding this directly, we get
$$(1+2a+a^2)(1+2b+b^2)$$
$$ = 1+2b+b^2 + 2a+4ab + 2ab^2 + a^2 + 2a^2b + a^2 b^2$$
On the other hand, we also have
$$(1+a)^2(1+b)^2=((1+a)(1+b))^2 = (1+a+b+ab)^2,$$
which expands to
$$ 1 + a + b + ab + a + a^2 + ab + a^2b + b + ba + b^2 + bab + ab + aba + ab^2 + abab
$$
$$ = 1 + 2a + 2b + 3ab + a^2 + a^2b + ba + b^2 + bab + aba + ab^2 + abab.
$$
Now use that $bab=ab^2$, $aba=a^2b$, and $abab=a^2b^2$ to rewrite this as
$$ 1 + 2a + 2b + 3ab + a^2 + 2a^2b + ba + b^2 + 2ab^2 + a^2b^2.
$$
Thus we end up with the equality,
$$ 1 + 2a + 2b +4ab + a^2 + b^2 + 2ab^2 + 2a^2b + a^2 b^2$$
$$=1 + 2a + 2b + 3ab + ba + a^2 + b^2 + 2a^2b + 2ab^2 + a^2b^2.
$$
Cancelling $1+2a+2b+3ab+a^2+b^2+2a^2b+2ab^2+a^2b^2$, we end up with
$$ab=ba,$$
as desired.
|
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Maths exam question on consecutive Pythagoras triplets Our 13 year old daughter brought home this maths question that she was asked in a "maths challenge" last week:
Q. The values of the adjacent and opposite sides in a right angle triangle of lengths 3, 4, 5 are consecutive. Obtain another triangle with consecutive adjacent and opposite sides having values between 500 and 1000.
My wife and I spent two hours trying to figure it out before we gave up. The teacher provided the solution (696, 697, 985) but could not explain how to obtain it. Searching on internet we can see solutions to this but none of them that we would expect even a very clever 13 year old to know. Are we missing something obvious?
Thanks,
Adrian
|
$x^2+(x+1)^2=y^2 \implies (x+1)^2 = (y+x)(y-x)$
Note that $a=y+x$ and $b=y-x$ gives $x={a-b\over 2}$ and the equation becomes $(a-b+2)^2=4ab$
$2x^2 > 500^2$, $2(x+1)^2 < 1000^2\implies x \geq 354$ and $x \leq 706$ so $1706\geq a\geq 854$ and $646 \geq b\geq 0$
Let $gcd(a,b)=k$ then $(a_1k-b_1k+2)^2=4a_1b_1k^2$ meaning $k^2 \mid (a_1k-b_1k+2)^2$ meaning $k \mid a_1k-b_1k+2$ meaning $k \mid 2$.
($1$) $k=1$ then $(a-b+2)^2=4ab$ where $a,b$ are coprime and are both square numbers.
$42^2>1706\geq a\geq 854>29^2$ and $26^2>646 \geq b\geq 0$ means there are only $12$ values of $a$ to be checked. Furthermore $a$ cannot be even because if so then $b$ is odd to be coprime with $a$ and the left side becomes odd. So we only needs to check $6$ values of $a$.
Expand equation $b^2-2(3a+2)b+(a+2)^2=0$, $\Delta = 32a(a+1)$ is a square number or simply ${a+1\over 2}$ is a square number.
Plug in values $a=31^2,33^2,...,41^2$ and check ${a+1\over 2}$ we get only one pair ($a=41^2, b=17^2$), this correspond to the poster's $x=696$
($2$) $k=2$ then $(a_1-b_1+1)^2=16a_1b_1$, thus $a_1, b_1$ are both square numbers. $20^2<427\leq a_1\leq853<30^2, $$b_1 \leq 323<18^2$
Expand equation ${b_1}^2-2(9a_1+1)b_1+(a_1+1)^2=0$, $\Delta = 64a_1(5a_1+1)$ is a square number or simply $5a+1$ is a square number.
This time we have $9$ values of $a_1$ to check from $21$ to $29$. However none of the values produce an integral $b_1$. As a result $x=696$, given by the poster, is the only solution.
|
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How do I rearrange this? Hi please could some one explain the steps taken to rearange the below. I've had a dabble but don't seem to be getting it.
$$\frac{C}{\sqrt{C^2 - V^2}} $$
To
$$\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}$$
Any help would be greatly appreciated.
Thanks
|
$$
\begin{align}
\frac{C}{\sqrt{C^2 - V^2}}
&=\frac{C}{\sqrt{C^2 - V^2}}\cdot\frac{1/C}{1/C}\\
&=\frac{C/C}{\frac{1}{C}\sqrt{C^2\bigg(1 - \frac{V^2}{C^2}\bigg)}}\\
&=\frac{1}{\frac{1}{C}\sqrt{C^2}\sqrt{1 - \frac{V^2}{C^2}}}\\
&=\frac{1}{\frac{C}{C}\sqrt{1 - \frac{V^2}{C^2}}}\\
&=\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}.
\end{align}
$$
This result is only true for $C\gt0$. For $C\lt0$, we would get a slightly different answer:
$$\frac{1}{\frac{1}{C}\sqrt{C^2}\sqrt{1 - \frac{V^2}{C^2}}}=\frac{1}{\frac{|C|}{C}\sqrt{1 - \frac{V^2}{C^2}}}=\frac{1}{-\sqrt{1 - \frac{V^2}{C^2}}}=-\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}.$$
|
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how to find the radius of convergence ? $\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$ How can i find the radius of convergence ? i dont know where to start i cant use $\frac{a_n}{a_{n+1}}$ test here.
$\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$
the question looks simple but i dont know how to solve it
i got that $R = 0$ but its wrong because when $x=1$ it is convergent
|
This is harder, but you can use the ratio test by combining the two fractions under the same denominator.
The $n$th term is equal to: $$\frac{x^n(n+1) - x^{n+1}(n)}{n(n+1)}.$$
Now, for:
$$\lim_{n \to \infty} \left(\frac{x^{n+1}(n+2) - x^{n+2}(n+1)}{(n+1)(n+2)} \cdot \frac{n(n+1)}{x^n(n+1) - x^{n+1}(n)} \right) < 1$$
$$\Rightarrow \lim_{n \to \infty} \left(\frac{x^{n+1}(n+2) - x^{n+2}(n+1)}{x^n(n+1) - x^{n+1}(n)} \cdot \frac{n}{n+1} \cdot \frac{n+1}{n+2} \right)<1$$
$$\Rightarrow \frac{x^{n+1} - x^{n+2}}{x^n - x^{n+1}} \cdot 1 \cdot 1 <1$$
$$\Rightarrow |x| <1 \ (*)$$
Therefore, the radius of convergence is $1$.
$(*)$: Going back to the original expression $\frac{x^n}{n} - \frac{x^{n+1}}{n+1}$.
|
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|
Evaluate $\frac{2a}{a^2-4} - \frac{1}{a-2}-\frac{1}{a^2+2a}$
Evaluate
$$\dfrac{2a}{a^2-4} - \dfrac{1}{a-2}-\dfrac{1}{a^2+2a}$$
We have to see what their common term is. Therefrom, we can evaluate the simplified expression by canceling out.
$$a^2 - 4 = (a)^2 - (2)^2 = (a-2)(a+2) \tag {1}$$
$$a^2 +2a = a(a+2)\tag{2}$$
Rewriting the expression
$$\dfrac{2a}{(a-2)(a+2)} - \dfrac{1}{a-2}-\dfrac{1}{a(a+2)}$$
$$\dfrac{2a}{(a-2)(a+2)} - \dfrac{(a+2)}{(a-2)(a+2)}-\dfrac{1}{a(a+2)}$$
Factoring $\dfrac{1}{a+2}$
$$\dfrac{1}{a+2}\biggr (\dfrac{2a}{a-2} - \dfrac{a+2}{a-2}-\dfrac{1}{a}\biggr )$$
This is where I'm stuck. Could you assist me?
Regards
|
Let’s simplify the first fraction.
$$\frac{2a}{a^2-4}=\frac{2a}{(a+2)(a-2)}=\frac{a+2+a-2}{(a+2)(a-2)}$$ which becomes $$\frac{a+2}{(a+2)(a-2)}+ \frac{a-2}{(a+2)(a-2)}=\frac{1}{a+2}+\frac{1}{a-2}$$ so that the entire expression is reduced to $$\frac{1}{a+2}-\frac{1}{a(a+2)}=\frac{1}{a+2}(1-\frac1a)=\frac{a-1}{a(a+2)}$$
|
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|
$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$ How can I go about computing
$$
\sum_{k\ =\ 1}^{n - 1}
\left(1 - \mathrm{e}^{\large 2\pi k\mathrm{i}/n}\right)^{-1}\
{\Large ?}
$$
I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $1/n$, but now I realized that it is the sum of the reciprocals. I've tried using
$$e^{ix}=\cos x+i\sin x,$$
but I didn't get anywhere with that.
|
The answer is $\frac{n-1}{2}$. There are multiple ways to approach it, here is one way;
Let $\displaystyle \alpha_k = e^{i \frac{2\pi k}{n}}$ for $k = 1, \dots, n-1$. We seek to evaluate,
$$\sum_{k=1}^{n-1} \frac{1}{1-\alpha_k}, \ \ (*) $$
We know that $\alpha_1, \dots , \alpha_{n-1}$ are roots of the polynomial,
$$P(z) = 1 + z+ \dots + z^{n-1}$$
This is because $1, \alpha, \dots, \alpha_{n-1}$ are the roots of the polynomial $1 - z^n$ (roots of unity), so when we take away $1$ as a root we obtain $P$. But $\alpha_k$ is a root of $P$ simply means that $P(\alpha_k) = 0$ for each $k$. Our goal now is to find a polynomial that has roots,
$$\frac{1}{1-\alpha_k} = f(\alpha_k) $$
as roots instead of $\alpha_k$, with this polynomial, we find the sum of the roots of that polynomial to obtain $(*)$. Since $f$ is one-to-one we can find the inverse function,
$$f^{-1}(x) = 1 - \frac{1}{x} \Rightarrow f^{-1} \left(\frac{1}{1-\alpha_k} \right) = \alpha_k$$
Therefore the function,
$$Q(x) = P(f^{-1}(x)) = P \left(1 - \frac{1}{x}\right) $$
has roots $\displaystyle \frac{1}{1-\alpha_k}$ for each $k$ (this is the crux of the argument, please convince yourself of this!). Note that $Q$ as defined is not a polynomial yet, but we can use $Q$ as a function to find a polynomial that has the same roots.
Now, we have that,
$$Q(x) = P \left(1 - \frac{1}{x} \right) = 1 + \left(1 - \frac{1}{x} \right) + \dots + \left( 1- \frac{1}{x} \right)^{n-1} $$
Summing this geometric series we obtain,
$$Q(x) = x\left(1 - \left(1-\frac{1}{x} \right)^n\right) $$
With some manipulation we obtain that,
$$R(x) = x^{n-1}Q(x) = x^n - (x-1)^n = nx^{n-1} - \frac{n(n-1)}{2} x^{n-2} + \dots $$
Now we have a polynomial $R$ and I claim that $R$ has roots $\displaystyle \frac{1}{1-\alpha_k}$ for each $k$. This is because,
$$R \left(\frac{1}{1-\alpha_k} \right) = \frac{1}{(1-\alpha_k)^{n-1}} Q \left(\frac{1}{1-\alpha_k} \right) =\frac{1}{(1-\alpha_k)^{n-1}} P\left(f^{-1}\left(\frac{1}{1-\alpha_k} \right) \right) = \frac{1}{(1-\alpha_k)^{n-1}} P(\alpha_k) = 0 $$
Therefore to find $(*)$ we note that it is simply the sum of the roots of $R$, therefore,
$$ \sum_{k=1}^{n-1} \frac{1}{1-\alpha_k} = \frac{n(n-1)/2}{n} = \frac{n-1}{2}$$
This completes the proof.
|
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|
Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
&(x+y)(x^2 - xy + y^2)\\
&= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
&= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
&= x^3 + x^2y - x^2y - xy^2 + xy^2 + y^3\\
&= x^3 + y^3\\
\end{split}$$
Q.E.D.
Question:
Spivak says there is an easy proof that, if I use this other theorem:
$$
x^3 - y^3 = (x-y)(x^2 + xy + y^2)
$$
then, I will also allow me to find out $x^n+y^n$ whenever $n$ is odd.
How to do this? I fail to see how.
|
In this case of $n=3$ your result follows from this other theorem by considering $x^3+y^3=x^3-(-y)^3$. The pattern with odd $n$ is that $-y^n=(-y)^n$.
|
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|
prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ .
I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please help me .!
|
I naturally want to generalize this.
Here's a first cut.
$\begin{array}\\
\dfrac{a^u}{u}+\dfrac{b^v}{v}
+\dfrac{c^{uv}}{uv}
&=\dfrac{va^u+ub^v+c^{uv}}{uv}\\
&=\dfrac{va^u+ub^v+c^{uv}}{u+v+1}\dfrac{u+v+1}{uv}\\
&\ge (a^{uv}b^{uv}c^{uv})^{1/(u+v+1)}\dfrac{u+v+1}{uv}\\
&= (abc)^{uv/(u+v+1)}\dfrac{u+v+1}{uv}\\
&=\dfrac{(abc)^k}{k}
\qquad k = \dfrac{uv}{u+v+1}\\
\end{array}
$
with equality only if
$a^u = b^v = c^{uv}$
or
$a=c^v$ and $b=c^u$.
From now on
I assume that
$u \le v$.
If $k=1$,
so that
$uv = u+v+1$,
or
$(u-1)(v-1) = 2$,
then
$\dfrac{a^u}{u}+\dfrac{b^v}{v}
+\dfrac{c^{uv}}{uv}
\ge abc
$.
If
$uv = k(u+v+1)$,
or
$(u-k)(v-k) = k^2+k$,
then
$\dfrac{a^u}{u}+\dfrac{b^v}{v}
+\dfrac{c^{uv}}{k}
\ge \dfrac{(abc)^k}{k}
$.
Since
$k^2+k = k(k+1)$,
two solutions to
$(u-k)(v-k) = k^2+k$
are
$u=2k, v=2k+1$
and
$u=k+1, v=k^2+2k$.
To find all the integer solutions,
write
$k^2+k = rs$.
Then
$u=r+k, v=s+k$
are solutions.
The ones above are for
$r=1, s=k^2+k$
and
$r=k, s=k+1$.
These are the same for $k=1$.
For $k=2$,
$rs = 6$
so
$(r, s)
=(1, 6),(2, 3)$
and
$(u, v)
=(3, 8), (4, 5)$.
For
$k=3$,
$rs = 12$
so
$(r, s)
=(1, 12), (2, 6).
(3, 4)
$
and
$(u, v)
=(4, 15), (5, 9),
(6, 7)
$.
|
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|
how to solve this two limit tasks Hello i stumbled across this two limits task and i cant find an answer to them:
*
*Find the limit depending on the parameter $A$
$$\lim \limits_{x\to\infty}\left(\left(\sqrt{x+1} - \sqrt[4]{x^2 + x + 1} \right) \cdot x^A\right)$$
I tried by multipliying with
$$\frac{\sqrt{x+1} + \sqrt[4]{x^2 + x + 1}}{\sqrt{x+1} + \sqrt[4]{x^2 + x + 1}}$$
which equals 1 and then
$$\frac{x+1 + \sqrt{x^2 + x + 1}}{x+1 + \sqrt{x^2 + x + 1}}$$ so i can get rid off roots in denominators but i got tangled up,
*$$\lim\limits_{x\to\infty}\left((x+1) - \sqrt[3]{x^3 + x^2} \right)$$
lim in both tasks goes to +infinity
|
You should be aware of Taylor expansion. Substitute $x=1/t$, so $t\to0^+$ and the limit can be rewritten as
$$
\lim_{t\to0^+}\frac{\sqrt{1+t}-\sqrt[4]{1+t+t^2}}{t^{A+1/2}}
$$
The numerator can be rewritten as
$$
1+\frac{1}{2}t-1-\frac{1}{4}{t}+o(t)
$$
So the limit is $1/2$ when $A+1/2=1$. What if $A>-1/2$ or $A<-1/2$?
The second limit can be computed similarly.
|
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|
If $a_n=n\left(1-\frac{1}{n}\right)^{n[\log n]}$, prove $1\leqslant \liminf a_n$ and $\limsup a_n\leqslant e.$
Prove that:
$$\limsup n\left(1-\frac{1}{n}\right)^{n[\log n]} \leqslant e$$
and
$$\liminf n\left(1-\frac{1}{n}\right)^{n[\log n]} \geqslant 1.$$
Attempt. Since for $n>2$:
$$1<[\log n]<\log n+1\leqslant n,$$ we would get:
$$n\left(1-\frac{1}{n}\right)^{n[\log n]}\leqslant n\left(1-\frac{1}{n}\right)^n \to +\infty,$$
$$n\left(1-\frac{1}{n}\right)^{n[\log n]}\geqslant n\left(1-\frac{1}{n}\right)^{n^2} \to 0,$$
so the above estimates are not convenient.
Thanks for the help.
|
First, note that $\log n-1 < [\log n] \leq \log n$. Now, $0<1- \frac{1}{n}<1$, so we have
$$ n \left( 1- \frac{1}{n} \right)^{n \log n} \leq n \left( 1- \frac{1}{n} \right)^{n [\log n]} < n \left( 1- \frac{1}{n} \right)^{n (\log n - 1)}. $$
We now want to use the fact that $\left(1- \frac{1}{n}\right)^{n} \to e^{-1}$ as $n \to \infty$ to simplify the above inequalities. To make the computation easier, write this as $n\log(1- \frac{1}{n}) \to -1$, which can be sharpened to $n\log(1- \frac{1}{n}) = -1 - \frac{1}{2n} + O\left( \frac{1}{n^2} \right)$ using the Taylor series for $\log(1-x).$
Now,
$$\log \left(n \left( 1- \frac{1}{n} \right)^{n \log n} \right) = \log n \left( 1 + n \log \left(1 - \frac{1}{n} \right) \right) = \log n\left(- \frac{1}{2n} + O\left( \frac{1}{n^2}\right) \right) \to 0$$
as $n \to \infty$, and similarly
$$ \log \left( n \left( 1- \frac{1}{n} \right)^{n (\log n - 1)} \right) = \log n \left( 1 + n \log \left(1 - \frac{1}{n} \right) \right) - n \log\left( 1- \frac{1}{n} \right) \to 1$$
as $n\to\infty.$ Taking the exponentials of both of these inequalities gives the desired result.
|
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|
Strategy for solving $\sqrt{3}\cos(2x) - \sin(x)\cos(x) = 1$ I believe the way to solve the following equation is to use the "R-formula":
$$ \sqrt{3}\cos(2x) - \sin(x)\cos(x) = 1 $$
If so, it should be rewritten as:
$ R\cos(x-\mathcal{L})$ or $R(\cos x\cos\mathcal{L}+\sin x\sin\mathcal{L})$.
With coefficients equated as:
$ \sqrt{3} = R\cos\mathcal{L} $
$ ? = R\sin x $
In the top equation, it looks like $ \cos x $ is the coefficient of $ \sin x $. But that doesn't work with R-formula (does it?).
I have spent some time trying to use double-angle identities for $\cos(2x)$ to remove something from the $\sin(x)\cos(x)$ expression, but am uncertain whether that's possible due to the leading $\sqrt{3}$.
I have also taken several suggestions from threads like this one, but repeatedly end up uncertain due to the same double-angle issue and/or coefficient issue.
Is R-formula the right approach to solving this equation? Should a double-angle identity be used first? Or is an entirely different approach correct?
|
One more answer won't do any harm, hopefully. Here's another way to look at this "thing".
Once you replaced $\sin x \cos x$ with $\frac{1}{2}\sin 2x$, your equation becomes
$$\sqrt 3 \cos 2x - \frac{1}{2}\sin 2x = 1.$$
Recalling that cosine and sine are just abscissa and ordinate of points on the circumference of radius $1$, you can find the solutions by intersecting the line
$$r: \sqrt 3 X - \frac{1}{2}Y = 1$$
with the circumference
$$X^2 + Y^2 = 1.$$
This leads to the system of equations
$$
\begin{cases}
\sqrt 3 X - \frac{1}{2}Y = 1\\
X^2 + Y^2 = 1.
\end{cases}
$$
Replacing $Y = 2\sqrt 3 X - 2$ in the second equation yields the quadratic equation
$$13X^2-8\sqrt 3 X +3 =0$$
with solutions
$$ \cos 2x = X = \frac{4\sqrt 3 \pm 3}{13}.$$
Since solutions in terms of $2x$ are in first and fourth quadrant we can write the first set of solutions as
$$2x = \arccos\frac{4\sqrt 3 + 3}{13} + 2k\pi, $$
that is
$$\boxed{x = \frac{1}{2}\arccos\frac{4\sqrt 3 + 3}{13} + k\pi, \ \ k \in \Bbb Z}$$
and the second set of solutions as
$$2x = -\arccos\frac{4\sqrt 3 - 3}{13} + 2k\pi,$$
that is
$$\boxed{x = -\frac{1}{2}\arccos\frac{4\sqrt 3 - 3}{13} + k\pi,\ \ k \in \Bbb Z} $$
|
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|
Am I properly simplifying this geometric progression? I'm studying recurrence relations and am given:
$T(n) = 2 \cdot T(n-1) - 1$
with an initial condition that $T(1) = 3$.
I worked through the first few recurrences:
$T(n-1) = 2^2 \cdot T(n-2) - 2 - 1$
$T(n-2) = 2^3 \cdot T(n-3) - 2^2 - 2 - 1$
and so forth, and came the conclusion that the pattern
$2^k \cdot T(n-k) - 2^{k-1} - 2^{k-2} - ... 2 - 1$
represents this relation. Since we have T(1) = 3, there must be some $n$ and $k$ such that $n-k = 1$, so $k= n - 1$. Substituting:
$2^{n-1} \cdot T(1) - 2^{n-2} - 2^{n-3} - ... 2 - 1$
but T(1) = 3, so...
$2^{n-1} \cdot 3 - 2^{n-2} - 2^{n-3} - ... 2 - 1$
Factoring out a $-1$ I have something that looks for all the world like a geometric progression:
$2^{n-1} \cdot 3 - (1 + 2 + ... + 2^{n-3} + 2^{n-2})$
However, I'm convinced my progression is missing a term, namely $2^{n-1}$. It's outside the progression, being multiplied by three. In my notes, my instructor simplified the progression to:
$\frac{2^{n-1} - 1}{2 - 1}$
I'm confused about where $n-1$ in the exponent of the geometric progression simplification is coming from. I eventually solve the recurrence relation as :
$3 \cdot 2^{n-1} - 2^{n-1} + 1 = 2^n + 1$
Is my understanding of the simplification of the geometric progression correct?
|
You've gotten an answer to your specific question, but I want to explain how to correctly solve the question in a mathematically rigorous manner, and not appealing to handwaving. (Every occurrence of "···" is an instance of lack of rigour.)
Working 1
$T(n) - 2·T(n-1) = -1$.
$2 · ( T(n-1) - 2·T(n-2) ) = 2·-1$. // We do this to match and cancel the $2·T(n-1)$.
$4 · ( T(n-2) - 2·T(n-3) ) = 4·-1$. // We do this to match and cancel the $4·T(n-2)$.
... // Not rigorous! We need to express the pattern rigorously.
Solution 1
Take any integer $n > 1$.
$T(n) - 2·T(n-1) = -1$.
$2^k · ( T(n-k) - 2·T(n-k-1) ) = -2^k$ for every integer $k < n-1$.
$\sum_{k=0}^{n-2} ( 2^k · ( T(n-k) - 2·T(n-k-1) ) ) = \sum_{k=0}^{n-2} -2^k$.
$\sum_{k=0}^{n-2} ( 2^k·T(n-k) - 2^{k+1}·T(n-k-1) ) = \sum_{k=0}^{n-2} -2^k$.
$\sum_{k=0}^{n-2} (2^k·T(n-k)) - \sum_{k=0}^{n-2} (2^{k+1}·T(n-k-1)) = -2^{n-1} + 1$.
$\sum_{k=0}^{n-2} (2^k·T(n-k)) - \sum_{k=1}^{n-1} (2^k·T(n-k)) = -2^{n-1} + 1$.
$2^0·T(n-0) - 2^{n-1}·T(n-(n-1)) = -2^{n-1} + 1$.
$T(n) = 2^{n-1}·T(1) - 2^{n-1} + 1 = 2^n + 1$.
Check that $T(1) = 2^1 + 1$ as well, because the above proof assumed $n > 1$.
Solution 2
Take any integer $n > 1$.
$T(n) - 2·T(n-1) = -1$.
$( T(n) - 2·T(n-1) ) / 2^n = -1/2^n$.
$T(n)/2^n - T(n-1)/2^{n-1} = -1/2^n$.
Take any integer $m > 1$.
$\sum_{n=2}^m ( T(n)/2^n - T(n-1)/2^{n-1} ) = \sum_{n=2}^m -1/2^n$.
$\sum_{n=2}^m (T(n)/2^n) - \sum_{n=2}^m (T(n-1)/2^{n-1}) = -1/2 + 1/2^m$.
$\sum_{n=2}^m (T(n)/2^n) - \sum_{n=1}^{m-1} (T(n)/2^n) = -1/2 + 1/2^m$.
$T(m)/2^m - T(1)/2^1 = -1/2 + 1/2^m$.
$T(m) = 2^m·(T(1)/2-1/2) + 1 = 2^m + 1$.
Check that $T(1) = 2^1+1$ as well.
|
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|
Find maximum value by using AM-GM inequality I have a problem: Find the maximum value of $P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$ with $x,y,z>0$.
Is there anyway to solve this problem by using the AM-GM inequality ? Thank for your answer.
The first way, I tried to use AM-GM twice at the two sum in the denominator, but I get the sum of the fractions with square roots of $xy, yz, zx$ at its denominators.
The other way, I tried to factor the denominator and use both AM-GM and the Schwarz inequality but still get the same with the first way, with higher order
|
For $x=y=z=1$ we get $P=\frac{3}{16}.$
We'll prove that it's a maximal value.
Indeed, by AM-GM $$\sum_{cyc}\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}\leq\sum_{cyc}\frac{x^3y^4z^3}{(x^4+y^4)(2\sqrt{xyz^2})^3}=\sum_{cyc}\frac{\sqrt{x^3y^5}}{8(x^4+y^4)}.$$
Let $\sqrt{\frac{x}{y}}=a$, $\sqrt{\frac{y}{z}}=b$ and $\sqrt{\frac{z}{x}}=c$.
Thus, $abc=1$ and we need to prove that
$$\sum_{cyc}\frac{a^3}{a^8+1}\leq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{1}{2}-\frac{a^3}{a^8+1}\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{2}-\frac{a^3}{a^8+1}-\frac{1}{2}\ln{a}\right)\geq0.$$
Now, let $f(a)=\frac{1}{2}-\frac{a^3}{a^8+1}-\frac{1}{2}\ln{a}.$
Thus, $$f'(a)=-\tfrac{(a-1)(a^{11}(a^4+a^3+a^2+a+1)-9a^8(a^2+a+1)-7a^3(a^4+a^3+a^2+a+1)-a^2-a-1)}{2a(a^8+1)^2}.$$
Since by the Descartes' rule of signs the polynomial
$$
a^{11}(a^4+a^3+a^2+a+1)-9a^8(a^2+a+1)-7a^3(a^4+a^3+a^2+a+1)-a^2-a-1$$
has an unique positive root (this root is $a_1=1.56...$) and $f(a_1)>0$,
we see that in $a_1$ the function $f$ has a local maximum,
which gives that$f$ decreases on $[a_1,+\infty)$ and there is an unique $a_0>a_1$, for which $f(a_0)=0$.
Easy to see that $a_0=2.679...$ and since $a_{min}=1$ and $f(1)=0$,
our inequality is proven for $\max\{a,b,c\}\leq2.5$
Let $a\geq2.5$.
Also, by AM-GM
$$\frac{x^3}{x^8+1}=\frac{x^3}{3\cdot\frac{x^8}{3}+5\cdot\frac{1}{5}}\leq\frac{x^3}{8\sqrt[8]{\left(\frac{x^8}{3}\right)^3\left(\frac{1}{5}\right)^5}}=\frac{1}{8}\sqrt[8]{3^35^5}.$$
Id est,
$$\sum_{cyc}\frac{a^3}{a^8+1}\leq\frac{1}{4}\sqrt[8]{3^35^5}+\frac{2.5^3}{2.5^8+1}=1.0423...<\frac{3}{2}$$ and we are done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $m>1/2$ that minimizes the area of the triangle formed by lines $y=10-2x$, $y=mx$, $y=-x/m$ I am tackling this problem below:
A triangle is formed by the three lines
$$\begin{align}
y &=10-2x \\
y &= mx \\
y &=-\frac{x}{m}
\end{align}$$ where $m>\frac{1}{2}$. Find the value of $m$ for which the area of the triangle is a minimum.
My thoughts are these:
I tried to draw the three graphs nothe rectangular coordinate, and since $m> \frac{1}{2}$, $-\frac{1}{m}>2$, but after that how can I decide the area of triangle formed?
|
Let $\ell_1$ be the line with equation $y = 10-2x$, let $\ell_2$ be the line with equation $y = mx$, and let $\ell_3$ be the line with equation $y = \frac{-x}{m}$. Then the lines $\ell_1$ and $\ell_2$ intersect at
$$P = \left( \frac{10}{m+2}, \frac{10m}{m+2} \right),$$
and the lines the lines $\ell_1$ and $\ell_3$ intersect at
$$
Q = \left( \frac{-10m}{1-2m}, \frac{10}{1-2m} \right).
$$
If $O = (0,0)$, then the line segments $\overline{OP}$ and $\overline{OQ}$ are orthogonal, which means that the area of the triangle is given by
$$
A = \frac{1}{2} \left|OP\right| \left| OQ\right|
= \frac{1}{2} \left( \frac{10 \sqrt{1+m^2}}{m+2} \right) \left( \frac{10 \sqrt{1+m^2}}{1-2m} \right) = \frac{50(1+m^2)}{(m+2)(1-2m)}.
$$
|
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|
In an equilateral $\triangle ABC$ : $ DB^2 + DC^2 + BC^2 = 100 $ I have a question that:
There is a point $D$ inside the equilateral triangle ABC. If
$$ DB^2 + DC^2 + BC^2 = 100 $$
and the area of $DBC$ is $ 5 \sqrt{3} $, find $AD^2$.
This is what I tried: $DB = x ,\: DC = y,\: BC = a$. Then $$ x^2 + y^2 + a^2 = 100 \tag{1}$$
$$ \sqrt{ 2(x^2 y^2 + a^2 x^2 + a^2 y^2 )- (a^4 + x^4 + y^4 ) } = 20 \sqrt{3} \tag{2} $$
I want to find $a$, $x$, and $y$, but I have only two equations. What can I do?
|
Denote $x= AD$, $y= BD$, $z=CD$, $a=BC(=AB=AC)$ and $\alpha=\angle BDC$. So we have: $$(*)\ y^2+z^2+a^2=100,$$
by the area of $\triangle DBC$ condition we have:
$$(**)\ yz\sin\alpha=10\sqrt 3,$$
and by cosine rule on $\triangle DBC$ we have:
$$(***)\ a^2=y^2+z^2-2yz\cos\alpha.$$
Consider the rotation $\mathcal R_{C,60^\circ}$ mapping $B\mapsto A$, $A\mapsto M$ and $D\mapsto N$. Since $CD=CN$ and $\angle DCN=60^\circ$, $\triangle DCN$ is equilateral, and thus $DN=DC=z$ and $\angle DNC=60^\circ$. Since we mapped $BD$ do $AN$, we have $AN=BD=y$. Also we mapped $\triangle BDC$ to $\triangle ANC$, so $\angle ANC=\angle BDC=\alpha$, wherefrom $\angle AND=\angle ANC-\angle DNC=\alpha-60^\circ$. So, in $\triangle AND$ we have $AD=x$, $AN=y$, $DN=z$ and $\angle AND=\alpha-60^\circ$, so by cosine rule:
$$(****)\ x^2=y^2+z^2-2yz\cos(\alpha-60^\circ)= y^2+z^2-yz(\cos\alpha+\sqrt 3\sin\alpha).$$
Multiply $(****)$ by $2$ and substract $(***)$. We get:
$$2x^2-a^2= y^2+z^2-2\sqrt 3yz\sin\alpha.$$
By using $(*)$ and $(**)$ we get:
$$2x^2= a^2+y^2+z^2-2\sqrt 3\cdot 10\sqrt 3=100-60=40,$$
so $AD^2=x^2=20$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $ Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $.
I tried multiplying by the conjugate and I get to an ugly expression. What should I do?
|
So, you arrived at
$$\begin{align}
\sqrt{(x+1/\sqrt x)^3}-\sqrt{x^3}&=\frac{3x^{3/2}+3+x^{-3/2}}{\sqrt{(x+1/\sqrt x)^3}+\sqrt{x^3}}\\\\
&=\frac{3x^{3/2}+3+x^{-3/2}}{x^{3/2}\left(1+\sqrt{(1+1/x^{3/2})^3}\right)}
\end{align}$$
Now let $x\to \infty$. Can you finish now?
|
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|
Variance of a Brownian motion Let $\{X(t), t \geq 0\}$ be a Brownian motion with drift parameter $\mu = 3$ and variance parameter $\sigma^2 = 9$. If $X(0) = 10$, find $P(X(0.5) > 10)$.
First, I calculated the expectation and variance of $X(0.5)).$ Since $X(0.5) - X(0)$ is normal with mean $1.5$ and variance $4.5$, it follows that $E[X(0.5)]
= 1.5 + 10 = 11.5$.
Likewise, we have $\text{Var}(X(0.5) - X(0)) = 4.5$. So I thought that $\text{Var}(X(0.5) - X(0)) = \text{Var}(X(0.5) - 10) = \text{Var}(X(0.5)) = 4.5$. But the answer key says the answer should be $14.5$.
I also tried writing
$$X(t) = 10 + 3t + 3B(t),$$
so
$$\text{Var}(X(0.5)) = \text{Var}(3 B(0.5)) = 9 \cdot \text{Var}(B(0.5))$$
|
It should be $4.5$. There may be an error in the answer key. The variance of the deterministic part is $0$ and doesn't add to the variance of the increment like you wrote.
At $X(0.5)$, we have $X(0.5) = 10 + 3 * 0.5 + 3\sqrt{0.5}Z$, where $Z$ is a standard normal random variable ($m = 0$, $s^2 = 1$). And so to calculate $P(X(0.5) > 10)$, we are calculating $P(10 + 3 * 0.5 + 3\sqrt{0.5}Z > 10)$.
Simplifying we have
\begin{align}
P(11.5 + 3\sqrt{0.5}Z > 10) & = P(3\sqrt{0.5}Z > -1.5) \\
& = P(Z > -\frac{\sqrt{2}}{2})\\
& = 1 - P(Z < -\frac{\sqrt{2}}{2}) \\
& = 1 - \Phi(-\frac{\sqrt{2}}{2})\\
& \approx 0.7602
\end{align}
That should be the answer but be sure to check! Hope this helps.
|
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|
Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$ where a, b, c and d are positive real numbers
I have to prove the following inequality using the Cauchy-Schwarz inequality:
$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$$
where a, b, c and d are positive real numbers.
But I am not able to do it, I am hitting dead-ends with every method I try. Please help!
|
Use $$xy\leq {(x+y)^2\over 4}$$
if $x,y\geq 0$, so
$$\frac{a}{b+c}+\frac{c}{d+a} = {a(a+d)+c(b+c)\over (a+d)(b+c)} \geq 4{a^2+c^2+ad+bc\over (a+b+c+d)^2}$$
and similary $$\frac{b}{c+d}+\frac{d}{a+b}\ge 4{b^2+d^2+ab+dc\over (a+b+c+d)^2}$$
So $$...\geq 4{a^2+c^2+ad+bc+b^2+d^2+ab+dc\over (a+b+c+d)^2}\geq 2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives, such that $F(x-1)+f(1-x)=x^3$ Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives $F:\mathbb{R}\to\mathbb{R}$, such that $F(x-1)+f(1-x)=x^3$ for any $x\in\mathbb{R}$. For $1-x$ instead of $x$, $F(-x)+f(x)=(1-x)^3$, $F(x)+f(-x)=(1+x)^3$ (1).
\begin{align*}
\left(\frac{F(x)-F(-x)}{e^x}\right)' &=\frac{(f(x)+f(-x))-(F(x)+F(-x))}{e^x}\\ &=\left(\frac{F(x)}{e^x}\right)'-\left(\frac{F(-x)}{e^x}\right)'.
\end{align*}
From here on however I don't know how to solve it. I feel that the two relations from (1) or similar ones can be used.
|
So, we're solving for differentiable $g$ such that $g(-x) + g'(x) = (1 - x)^3$. Note that $g'(x) = (1 - x)^3 - g(-x)$, and hence $g$ is twice differentiable with
$$g''(x) = -3(1 - x)^2 + g'(-x) = -3(1 - x)^2 + (1 + x)^3 - g(x)$$
That is, $g$ must satisfy the differential equation
$$y'' + y = -3(1 - x)^2 + (1 + x)^3.$$
This differential equation has the general solution
$$y = A \cos(x) + B \sin(x) + x^3 + 3x - 2.$$
But, of course, you're looking for $f = g'$, so the possible solutions we have are
$$f(x) = -A \sin(x) + B \cos(x) + 3x^2 + 3,$$
with corresponding primitive as above.
Do all such functions work? We have
\begin{align*}f(x) + F(-x) &= -A\sin(x) + B\cos(x) + 3x^2 + 3 + A\cos(-x) + B\sin(-x) + (-x)^3 + 3(-x) - 2 \\
&= (1 - x)^3 + (A + B)\cos(x) - (A + B)\sin(x).
\end{align*}
Therefore, we only get a solution with $A = -B$. Thus, the general solution is
$$f(x) = A(\sin(x) + \cos(x)) + 3x^2 + 3,$$
with corresponding primitive
$$F(x) = A(\sin(x) - \cos(x)) + x^3 + 3x - 2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve these trigonometry equations? I have to work with the following 5 equations:
*
*$(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)=8$
*$(2\cos 2x+1)(2\cos 6x+1)(2\cos 18x+1)=1$
*$\dfrac{\cos 2x}{\sin 3x}+\dfrac{\cos 6x}{\sin 9x}+\dfrac{\cos 18x}{\sin 27x}=0$
*$\dfrac{\cos x}{\sin 3x}+\dfrac{\cos 3x}{\sin 9x}+\dfrac{\cos 9x}{\sin 27x}=0$
*$\dfrac{1}{\cos x\cos 2x}+\dfrac{1}{\cos 2x\cos 3x}+\dfrac{1}{\cos 3x\cos 4x}=0$
These equations have patterns, and I know if we can use the pattern we will solve the equations very easily. I managed to use the pattern on the first equation to find a telescoping series and get this (it is not a full solution but it is the way to solve the first equation):
We have $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ therefore
\begin{align*}
(1)&\Leftrightarrow\dfrac{1}{(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)}=\dfrac18\\
&\Leftrightarrow\left(2\tan x\cdot\dfrac{1}{1-\tan^2x}\right)\cdot\dfrac{1}{1-\tan^22x}\cdot\dfrac{1}{1-\tan^24x}=\dfrac14\tan x\\
&\Leftrightarrow\left(2\tan 2x\cdot\dfrac{1}{1-\tan^22x}\right)\cdot\dfrac{1}{1-\tan^24x}=\dfrac12\tan x\\
\end{align*}
and so on.
However, I can't manage to solve the last four. I can't find the key equalities like $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ in the first equation. So here are my questions:
*
*How to solve equations 2, 3, 4, and 5?
*What is the strategy to find the key equalities like $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ to get a telescoping series for each equation?
Thank you in advance.
|
In the first you can get it by the following way:
$$LS=\frac{2\tan{x}}{\tan{2x}}\cdot\frac{2\tan{2x}}{\tan{4x}}\frac{2\tan{4x}}{\tan{8x}}=\frac{8\tan{x}}{\tan{8x}}$$ and the rest is smooth.
|
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|
Find sum $ \sum\limits_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}} $ Calculate sum
$$ \sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}} $$
I hope to solve this in use of Iverson notation:
my try
$$ \sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}} = \sum_{k,l,m}2^{-l}4^{-m} [2^l \le k < 2^{l+1}][2^{2^m} \le k < 2^{2^m+1}] $$
and now:
$$ [2^l \le k < 2^{l+1}][2^{2^m} \le k < 2^{2^m+1}] \neq 0 $$ if and only if $$2^l \le k < 2^{l+1} \wedge 2^{2^m} \le k < 2^{2^m+1} $$
I can assume that $l$ is const (we know value of $l$) and treat $m$ as variable depence from $l$. Ok so:
$$2^l \le 2^{2^m} \wedge 2^{2^m+1} \le 2^{l+1} $$
but it gives me that $l=2^m$
I think that it is not true (but also I don't see mistake). Even if it is true, how can be it finished?
|
Let's write
$$\sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{\lfloor \log_2(\log_2(k))\rfloor}} = \sum_{i=0}^{n-1} \sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{\lfloor \log_2(\log_2(k))\rfloor}} + \frac{1}{2^{2^n}4^{n}}$$
$$= \sum_{i=0}^{n-1} \sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{i}} + \frac{1}{2^{2^n}4^{n}}$$
Moreover for all $i=0, ..., n-1$,
$$\sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}} = \sum_{j=2^i}^{2^{i+1}-1} \sum_{k=2^j}^{2^{j+1}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}} = \sum_{j=2^i}^{2^{i+1}-1} \sum_{k=2^j}^{2^{j+1}-1} \frac{1}{2^j} = \sum_{j=2^i}^{2^{i+1}-1} \frac{2^j}{2^j} = 2^i $$
You deduce that
$$S = \sum_{i=0}^{n-1} \frac{1}{2^i} + \frac{1}{2^{2^n}4^{n}} = 2 - \frac{1}{2^{n-1}} + \frac{1}{2^{2^n}4^{n}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ $a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?
I need a hint
|
Hint: Use Cauchy Schwarz in Engelform:
$$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\geq \frac{(x+y+z)^2}{a+b+c}$$
It is equivalent to $${z}^{2}{a}^{2}b+{y}^{2}{a}^{2}c+{z}^{2}a{b}^{2}-2\,abcxy-2\,abcxz-2\,a
bcyz+{y}^{2}a{c}^{2}+{x}^{2}{b}^{2}c+{x}^{2}b{c}^{2}
\geq 0$$
and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2\geq 0$$ if $$a,b,c$$ are positive.
|
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|
Confusion in this limits problem
Evaluate $$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3}$$
This is the original method to solve this is:
Taking summation of the square numbers $1^2 + 2^2 + 3^2 +...+n^2 = \frac{1}{6}n(n+1)(2n+1)$
$$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3} = \frac{\frac{1}{6}n(n+1)(2n+1)}{n^3}$$
$$=\lim\limits_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim\limits_{n\to \infty} \frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n})$$
$$=\frac{2}{6} = \frac{1}{3}$$
But when looking at the limit in a different angle I get a different answer,
$$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3} = \lim\limits_{n\to \infty} \frac{1^2}{n^3}+\frac{2^2}{n^3}+...+\frac{1}{n}$$
$$=0+0+...+0 = 0$$
Both the method seem right to me, but why I am getting different answers? What have I done wrong? Please Explain. Thank you!
|
Converting the limit into an integral is one right way to evaluate it.
$$\lim_{n\to \infty}\dfrac{1^2+2^2+\cdots+n^2}{n^3}=\lim_{n\to \infty}\dfrac{1}{n}\sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^2=\int_{0}^{1}x^2\mathrm dx$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show that $\cos(x+y)=\cos(x)\cos(y)−\sin(x)\sin(y)$ by using power series? Original question:Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$
I would like to know an answer to the question linked above by using power series.
I tried to expand the $(x+y)^{2n}$ by using the binomial theorem but it didn't lead me anywhere.
|
Let
$$C_k = \begin{cases}
1 & k \equiv 0 \pmod 4 \\
0 & k \equiv 1 \pmod 4 \\
-1 & k \equiv 2 \pmod 4 \\
0 & k \equiv 3 \pmod 4 \\
\end{cases}$$
$$S_k = \begin{cases}
0 & k \equiv 0 \pmod 4 \\
1 & k \equiv 1 \pmod 4 \\
0 & k \equiv 2 \pmod 4 \\
-1 & k \equiv 3 \pmod 4 \\
\end{cases}$$
Then the series is
$$\cos(t) = \sum_{0 \le k} \frac{C_k}{k!} t^k$$
$$\sin(t) = \sum_{0 \le k} \frac{S_k}{k!} t^k$$
So...
$$\begin{align}
%
& \cos(x + y)
\\
=& \sum_{0 \le k} \frac{C_k}{k!} (x + y)^k
\\
=& \sum_{0 \le k} \sum_{u = 0}^k \frac{C_k}{k!} {k \choose u} x^u y^{k - u}
\\
=& \sum_{0 \le u \le k \le \infty} \frac{C_k}{k!} {k \choose u} x^u y^{k - u}
\\
\\
& \{j = k - u,~ k = u + j\}
\\
\\
=& \sum_{0 \le u,~ 0 \le j} \frac{C_{u + j}}{(u + j)!} {u + j \choose u} x^u y^j
\\
\\
& \left\{ {u + j \choose u} = \dfrac{(u + j)!}{u!j!} \right\}
\\
\\
=& \sum_{0 \le u,~ 0 \le j} \frac{C_{u + j}}{u!j!} x^u y^j
\end{align}$$
And
$$\begin{align}
%
& \cos(x)\cos(y) - \sin(x)\sin(y)
\\
= &
\left(\sum_{0 \le k} \frac{C_k}{k!} x^k \right) \left(\sum_{0 \le k} \frac{C_k}{k!} y^k \right)
- \left(\sum_{0 \le k} \frac{S_k}{k!} x^k \right) \left(\sum_{0 \le k} \frac{S_k}{k!} y^k \right)
\\
= &
\left(\sum_{0 \le k,~ 0 \le j} \frac{C_kC_j}{k!j!} x^ky^j \right)
-\left(\sum_{0 \le k, 0 \le j} \frac{S_kS_j}{k!j!} x^ky^j \right)
\\
= &
\sum_{0 \le k, 0 \le j} \left(\frac{C_kC_j - S_kS_j}{k!j!}\right)x^ky^j
\\
\end{align}$$
Equating coefficients, it just amounts to show
$$\frac{C_{k + j}}{k!j!} = \frac{C_kC_j - S_kS_j}{k!j!}$$
which is
$$C_{k + j} = C_kC_j - S_kS_j$$
(It isn't a coincidence this looks exactly like the original formula.) Theres only 16 possible values for $(j, k) \pmod 4$, so just check all of them with brute force if you can't find a shortcut.
My TA asked me to, so I suppose there should be a good way.
LUL. I think you should be careful with that TA.
|
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|
Define the sequences $x_n$& $y_n$ for $n \geq 1$ using the relations $x_n = x_{n+1} +2y_{n+1}$ & $y_n = y_{n-1} +x_{n-1}$ $?$ Define the sequences $x_n$& $y_n$ for $n \geq 1$ using the relations $x_n = x_{n+1} +2y_{n+1}$ & $y_n = y_{n-1} +x_{n-1}$ $?$
$:$
$n \geq 1$ , $x_0,y_0 \epsilon \mathbb{Z}_{+} $
$x_n = x_{n+1} +2y_{n+1}$
$y_n = y_{n-1} +x_{n-1}$
I came across this problem while solving some problems on limits of sequences,where in this question it asks to define the sequence. So, I tried out by finding the difference $d_x$ and $d_y$ as follows,
$d_x$=$x_{n+1}-x_n$= $-2y_{n+1}$
$d_y$ =$y_n- y_{n-1}$=$x_{n-1}$
then defining the sequence using the difference I got,
$x_n = x_0 -(n-1)2y_{n+1}$
$y_n = y_0 +(n-1)x_{n-1}$
But I know this is not the way because, after it asks to prove the limit.After quite researching I came to know about the recurrence relations, but I don't know anything about that. I'm quite confused!!
|
Hint.
We can arrange it as
$$
\left[
\begin{array}{c}
x_n\\
y_n
\end{array}
\right] =
\left[
\begin{array}{cc}
-1 & -2\\
1 & 1
\end{array}
\right]\left[
\begin{array}{c}
x_{n-1}\\
y_{n-1}
\end{array}
\right]
$$
NOTE
For $A = \left[
\begin{array}{cc}
-1 & -2\\
1 & 1
\end{array}
\right]$
we have
$$
\left[
\begin{array}{c}
x_n\\
y_n
\end{array}
\right] = A^n \left[
\begin{array}{c}
x_0\\
y_0
\end{array}
\right]
$$
The recurrence behavior can be easily depicted by making
$$
A = \left[
\begin{array}{cc}
-1 & -2 \\
1 & 1 \\
\end{array}
\right]
$$
$$
A^2 = \left[
\begin{array}{cc}
-1 & 0 \\
0 & -1 \\
\end{array}
\right]
$$
$$
A^3 = \left[
\begin{array}{cc}
1 & 2 \\
-1 & -1 \\
\end{array}
\right]
$$
$$
A^4 = \left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right]
$$
$$
A^5 = \left[
\begin{array}{cc}
-1 & -2 \\
1 & 1 \\
\end{array}
\right]
$$
etc.
Note the periodicity.
|
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|
Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i solve it Help me please
|
The best way I can seem to think of is differentiating both sides of the equality that gives you the following: $$\dfrac{\mathrm d}{\mathrm dx}\int\dfrac{2x^7+3x^2}{x^{10}-2x^5+1}\mathrm dx=\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{px^3+qx^8}{x^{10}-2x^5+1}\right)$$
$$ \begin{equation}\begin{aligned} \dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{px^3+qx^8}{x^{10}-2x^5+1}\right) &= \dfrac{(3px^{2}+8qx^7)(x^{10}-2x^5+1)-(10x^9-10x^4)(px^3+qx^8)}{(x^{10}-2x^5+1)^2} \\ &= \dfrac{-2qx^{17}-7px^{12}-6qx^{12}+4px^7+8qx^7+3px^2}{(x^{10}-2x^5+1)^2}\end{aligned}\tag1\end{equation}$$
$$\begin{equation}\begin{aligned}\dfrac{\mathrm d}{\mathrm dx}\int\dfrac{2x^7+3x^2}{x^{10}-2x^5+1}\mathrm dx&=\dfrac{2x^7+3x^2}{x^{10}-2x^5+1}\\&=\dfrac{(2x^7+3x^2)(x^{10}-2x^5+1)}{(x^{10}-2x^5+1)^2}\\&=\dfrac{2x^{17}-x^{12}-4x^7+3x^2}{(x^{10}-2x^5+1)^2}\end{aligned}\end{equation}\tag2$$
Equating $(1)$ and $(2)$ and comparing gives us the following system: $$\begin{cases}q=-1\\ 7p+6q=1 \\ p+2q=-1\\ 3p=3\end{cases}\implies \bbox [5px,border:2px solid #C0A000]{\begin{array}pp=+1 \\ q=-1\end{array}}$$
|
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|
Generating function with tough restrictions
In how many ways can a coin be flipped $25$ times in a row so that exactly $5$ heads occur and no more than $7$ tails occur consecutively?
For the heads, I think that it is $\binom{25}{5}$, but I do not know what to do with the tails restriction.
Not sure how to approach this problem. It is from the advanced section in the generating functions chapter in my book.
|
More generally, we might ask how many sequences of flips of length $n$ have $m$ heads and no more than seven tails in a row. We will find a bivariate generating function for this problem.
First suppose we drop the constraint that the sequence must contain exactly $m$ heads, retaining the requirement that there be no more than seven tails in a row; let's say such a sequence is "acceptable". Let $f(z)$ be the generating function for the number of acceptable sequences of length $n$. An acceptable sequence consists of a sequence of zero to seven tails, or a sequence of zero to seven tails followed by a head, followed by an acceptable sequence. In terms of the generating function,this translates to
$$f(z) = (1+z+z^2+ \dots +z^7) (1 + z f(z))$$
We can insert a "marker" for the head in this equation and convert the generating function into a bivariate generating function where the number of acceptable strings of length $n$ with $m$ heads is the coefficient of $u^m z^n$:
$$f(u,z) = (1+z+z^2+ \dots +z^7) (1 + uz f(u,z))$$
Noting that $1+z+z^2+ \dots +z^7 = (1-z^8)/(1-z)$, we can solve the preceding equation for $f(u,z)$:
$$f(u,z) = \frac{(1-z^8)/(1-z)}{1-uz(1-z^8)/(1-z)}$$
In a sense this bivariate generating function is the general solution, but let's see if we can find the answer to the original problem, which is the coefficient of $u^5 z^{25}$ in $f(u,z)$. We start by expanding $f(u,z)$ as a series in $u$.
$$f(u,z) = \frac{1-z^8}{1-z} \sum_{i=0}^{\infty} z^i \left( \frac{1-z^8}{1-z} \right)^i u^i$$
The coefficient of $u^5$ is
$$\begin{align}
[u^5] f(u,z) &= \frac{1-z^8}{1-z} z^5 \left( \frac{1-z^8}{1-z} \right)^5 \\
&= z^5 (1-z^8)^6 (1-z)^{-6} \\
&= z^5 (1 - 6z^8 + 15 z^{16} - O(z^{24})) \sum_{i=0}^{\infty} \binom{6+i-1}{i} z^i
\end{align}$$
From this last equation we see that the coefficient of $u^5 z^{25}$ is
$$[z^{25}][u^5]f(u,z) = \binom{6+20-1}{20} -6 \binom{6+12-1}{12} + 15 \binom{6+4-1}{4} = \boxed{17,892}$$
|
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|
Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a - 2)(c^2 + a^2 - ca - c - a + 1)\\
&\ge (1 - b)\left[\dfrac{3}{4}(c + a)^2 - b + 4\right]\\
&=(1 - b)\left[\dfrac{3}{4}(3 - b)^2 - b + 4\right]\\
&= \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43)
\end{align*}
That means
\begin{align*}
(a - 1)^3 + (b - 1)^3 + (c - 1)^3 &\ge \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43) + (b - 1)^3\\
&= \dfrac{1}{4}(b - 1)(b^2 + 14b - 39)
\end{align*}
Since $c \ge a \ge b \implies b \le \dfrac{a + b + c}{3} = 1$.
And this is where I am stuck right now.
|
My sketchy proof:
First you need to prove that if $x+y+z=0$, then $x^3 + y^3 + z^3 = 3xyz$
Since $a+b+c=3$, we have $(a - 1)+(b-1)+(c-1)=0$, therefore, $S = (a - 1)^3+(b-1)^3+(c-1)^3=3(a - 1)(b-1)(c-1)$
Replace $c = 3 - a -b$, we have $S = 3(a-1)(b-1)(2-a-b)$
Expand this, we have $S/3 = a^2 + b^2 + 4ab - 3(a+b) - ab(a+b) + 2$
We can safely assume that $a \geq b$, since the order $a$, $b$ doesn't matter. So we have: $(a-b)^2 \geq 0$ or $a^2 + b^2 \geq 2ab$. Equality happens when $a = b$.
So $S/3 \geq 6ab - 3(a+b) - ab(a+b) + 2$
Since $c \geq 0$, we have $3 - a - b \geq 0$, or $-(a+b) \geq -3$. Equality happens when $a + b = 3$.
So $S/3 \geq 6ab + 3(-3) + ab(-3) + 2 = 3ab - 7$, or $S \geq 9ab - 21$. Equality happens when $a = b$ and $a + b = 3$, or $a = b = 3/2$. Replace $a$, $b$ in $9ab-21$ with $a = b = 3/2$ we have $S \geq -3/4$
I'm not satisfy with the last step, though.
|
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|
Prove that $ \ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x}$? How can one prove that
$$\ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x} \, ?$$
|
Considering $f(t) = \log t$ and applying Lagrange's theorem in the interval $[x,x+1], x > 0$ you get
$$
f(x+1) - f(x) = f'(\xi_x) (x+1-x) = \frac{1}{\xi_x}, \quad \xi_x \in ]x,x+1[
$$
But, since $f(x+1)-f(x)= \log (x+1)-\log x = \log \left(\frac{x+1}{x}\right)$ and $\frac{1}{\xi_x} > \frac{1}{x+1}$ you can conclude that
$$
\log \left(\frac{x+1}{x}\right) > \frac{1}{x+1}.
$$
|
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|
Proving the identity $(\tan^2(x)+1)(\cos^2(-x)-1)=-\tan^2(x)$ Proving the trigonometric identity $(\tan{^2x}+1)(\cos{^2(-x)}-1)=-\tan{^2x}$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are unsuitable for the situation or if I am not looking at the right angle to tackle this problem.
|
Recall:
$$\begin{align}
\tan^2(x) - \sec^2(x) &= -1 \\
\sin^2(x) + \cos^2(x) &= 1
\end{align}$$
Thus,
$$\begin{align}
\tan^2(x) + 1 &= \sec^2(x)\\
\cos^2(x) - 1 &= -\sin^2(x)
\end{align}$$
We also note that $\cos(x)$ is an even function, and thus $\cos(-x) = \cos(x)$. Thus, the formula becomes:
$$(\tan^2(x) + 1)(\cos^2(-x) - 1) = -\sec^2(x)\sin^2(x) = - \frac{\sin^2(x)}{\cos^2(x)} = -\tan^2(x)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow \frac{\pi}{2}-x$
$\displaystyle I=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\ln(1-k\sin x)dx$
$\displaystyle I =\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\bigg[k\sin x-\frac{k^2\sin^2 x}{2}+\frac{k^3\sin^3 x}{3}-\cdots \bigg]dx$
$\displaystyle I =-2\int^{\frac{\pi}{2}}_{0}\bigg[\frac{k^2\sin^2 x}{2}+\frac{k^4\sin^4 x}{4}+\cdots \bigg]dx$
$\displaystyle I =-\pi\bigg[\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots \bigg]$
How can I find sum of that series. Help me please.
|
For any $R\geq 1$,
$$(R+e^{i\theta})(R+e^{-i\theta}) = (R^2+1)+2R\cos\theta \tag{1}$$
$$ 1+\frac{2R}{R^2+1}\cos\theta = \frac{R^2}{R^2+1}\left(1+\frac{e^{i\theta}}{R}\right)\left(1+\frac{e^{-i\theta}}{R}\right)\tag{2}$$
$$ \log\left(1+\tfrac{2R}{R^2+1}\cos\theta\right) = \log\left(\tfrac{R^2}{R^2+1}\right)+2\sum_{n\geq 1}\frac{(-1)^{n+1}\cos(n\theta)}{n R^n}\tag{3} $$
and for any $n\in\mathbb{N}^+$ we have $\int_{0}^{\pi}\cos(n\theta)\,d\theta=0$, therefore
$$ \int_{0}^{\pi}\log\left(1+\tfrac{2R}{R^2+1}\cos\theta\right)\,d\theta = \pi\log\left(\tfrac{R^2}{R^2+1}\right).\tag{4}$$
Now it is enough to enforce the substitution $\frac{2R}{R^2+1}=\kappa$.
|
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|
Proving by induction of $n$ that $\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} $
$$
\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac3{8}$$
RHS-
$$\frac{1}{2} \ - \frac1{(n+1)2^{n+1}} \ = \frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$\sum_{k=1}^{n+1} \frac {k+2}{k(k+1)2^{k+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
\frac{1}{2} \ - \frac1{(n+1)2^{n+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $\frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$\frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$\frac{2}{(n+1)2^{n+2}} - \frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
then
$$\frac{1}2 -\frac{2}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
and
$$\frac{1}2 -\frac{1}{(n+1)2^{n}} - \frac{1}{(n+2)2^{n+2}}$$
$$\frac{1}2 -\frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$\frac{1}2 -\frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$\frac{1}2 -\frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
|
Your error is just after the sixth step from the bottom:
$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}=\frac{1}2 -\frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
|
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|
How to perform polynomial long division on 1/(1 - x)? How do I perform polynomial long division on $\frac{1}{1 - x}$ to obtain the sequence $1 + x + x^2 + x^3 + \cdots$?
In this video, the teacher went about it in the following way...
$$
\require{enclose}
\begin{array}{r}
1 + x + x^2 + x^3 + \cdots \\
1 - x \enclose{longdiv}{\hspace{10pt}1\hspace{85.5pt}} \\
\underline{-\left(1 - x\right)}\hspace{64.5pt} \\
x\hspace{68.5pt} \\
\underline{-\left(x - x^2\right)}\hspace{42pt} \\
x^2\hspace{46.5pt} \\
\vdots\hspace{52.5pt}
\end{array}
$$
I have always performed polynomial long division wrt the term of highest degree, e.g., to find $\frac{-7 + 5x + x^2}{-1 + 2x}$, I would do the following...
$$
\require{enclose}
\begin{array}{r}
x^2 + 5x \hspace{4pt}- 7\hspace{33pt} \\
2x - 1 \enclose{longdiv}{\hspace{10pt}2x^3 + 9x^2 - 19x + 7\hspace{4pt}} \\
\underline{-\left(2x^3 - x^2\right)}\hspace{50.5pt} \\
10x^2 - 19x + 7\hspace{4pt} \\
\underline{-\left(10x^2 - 5x\right)}\hspace{21.5pt} \\
-14x + 7\hspace{4pt} \\
\underline{-\left(-14x + 7\right)}\hspace{0pt} \\
0\hspace{4pt} \\
\end{array}
$$
When should I use the teachers variation of the conventional method?
|
You do it when $x$ is small compared to $1$. That is not as silly a remark as it sounds like it is. If you stop the video's approach part way through you get a remainder term just like you might with the top down approach. You could write $$\frac 1{1-x}=1+x+x^2+\frac {x^3}{1-x}$$
This is an algebraic fact, valid for all values of $x$ except $x=1$. When you keep going you get an infinite sum on the right, which is useful as long as it converges. It converges when $|x| \lt 1$. If $|x| \ll 1$ the sum converges quickly and you can decide when the error committed by truncating it is acceptable.
In some problems you know that $x$ is small, in which case this is quite useful.
If the division is going to come out even, you can do it either way around. Taking your example
$$\require{enclose}
\begin{array}{r}
-7 + 5x \hspace{4pt}+x^2\hspace{33pt} \\
-1+2x \enclose{longdiv}{\hspace{10pt}7-19x+9x^2+2x^3\hspace{4pt}} \\
\underline{7-14x}\hspace{30.5pt} \\
\hspace {30 pt}- 5x+9x^2\hspace{4pt} \\
\underline{- 5x+10x^2}\hspace{21.5pt} \\
-x^2 + 2x^3\hspace{4pt} \\
\underline{-x^2+2x^3}\hspace{0pt} \\
0\hspace{4pt} \\
\end{array}$$
where I can't get the spacing as nice as you did, but it gives the same result.
|
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|
can not find root of the equation $(0,0)$ with semi-major axis a and semi-minor axis $b$">
I want find out for each point $(x,y)$ in the white region, the nearest point $(a \cos \theta, b \sin \theta )$ on the ellipse curve.I tried with the following approach.
$$
\text {distance} = \sqrt{(x-a \cos \theta)^2+(y-b \sin \theta)^2}
$$
I find out the first derivative of the $$F(\theta)=(x-a \cos \theta)^2+(y-b \sin \theta)^2.$$ The equation after first derivative is:
$$
ax \sec \theta- by \csc \theta=a^2-b^2.
$$
How to calculate theta for this equation?
|
You have $axsec(\theta)- bycsc(\theta)= a^2- b^2$. Let $z= sec(\theta)$. Since $csc(\theta)= \frac{1}{sec(\theta)}$, the equation becomes $axz- \frac{by}{z}= a^2- b^2$. Muliply on both sides to get $axz^2- by= (a^2- b^2)z$.
We can write that as $(ax)z^3+ (b^2- a^2)z- by= 0$ and solve it using the quadratic formula: $z= sec(\theta)= \frac{a^2- b^2\pm\sqrt{(b^2- a^2)^2+ 4abxy}}{2ax}$.
|
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|
Sum of Infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ Prove that the sum of the infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ is 23.
My approach
I got the following term
$S_n=\sum_1^\infty\frac{4n^2}{2^n}-\sum_1^\infty\frac{1}{2^n}$.
For $\sum_1^\infty\frac{1}{2^n}$ the answer is 1 as it forms a geometric series but I am bot able to find the solution to $\sum_1^\infty\frac{4n^2}{2^n}$.
|
Hint. One may used this
$$\sum_{n\ge1}n^{2}x^n=\frac{x}{(1-x)^{2}}+\frac{2x^{2}}{(1-x)^{3}},\qquad |x|<1,$$ proved for example here.
Hope you can take from here.
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_a^b\frac{1}{ x^2}dx$ using limit of a sum definition
From the definition of a definite integral as the limit of a sum,
evaluate $$\int_a^b\frac{1}{ x^2}dx$$
Step 1
To simplify working with $x^2$, divide the interval $[a,b]$ using variable length intervals: $$[a,b]=\bigcup_{j=1}^n\bigg[a+\frac{\sqrt {j-1}(b-a)}{\sqrt n}, a+\frac{\sqrt j(b-a)}{\sqrt n}\bigg]$$
Step 2
Now rewriting the integral to remove the lower limit ($0\leq a\leq b$) : $$\int_a^b\frac1{x^2}dx=\int_0^b\frac{1}{x^2}dx-\int_0^a\frac{1}{x^2}dx$$
Step 3
Dividing $[0,b]$ as described in Step 1:$$[0,b]=\bigcup_{j=1}^n\bigg[\frac{\sqrt {j-1}(b)}{\sqrt n}, \frac{\sqrt j(b)}{\sqrt n}\bigg]$$
Step 4
Using limit of a sum definition on $\int_0^b\frac{1}{x^2}dx$ :
(result can then be used for $\int_0^a\frac{1}{x^2}dx$)
$$\begin{align}
\int_0^b\frac1{x^2} &= \lim_{n\to\infty}\sum_{j=1}^n\bigg(\frac{\sqrt j(b)}{\sqrt n}\bigg)^{-2}\times \bigg[\frac{\sqrt j(b)}{\sqrt n}-\frac{\sqrt {j-1}(b)}{\sqrt n} \bigg] \\
&=\frac1{b}\lim_{n\to\infty}\sqrt n \times \sum_{j=1}^n\frac1{j}\big[\sqrt j - \sqrt {j-1}\big]
\end{align}$$
After this point, I can not seem to find a nice form of $\sum_{j=1}^n\frac1{j}\big[\sqrt j - \sqrt {j-1}\big]$ to work with. Any help would be appreciated.
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We obtain for $0<a\leq b$:
\begin{align*}
\color{blue}{\int_a^b\frac{1}{x^2}\,dx}&=\lim_{n\to\infty}\sum_{j=1}^nf\left(a+j\frac{b-a}{n}\right)\frac{b-a}{n}\\
&=\lim_{n\to\infty}\sum_{j=1}^n\frac{1}{\left(a+j\frac{b-a}{n}\right)^2}\cdot\frac{b-a}{n}\\
&\,\,\color{blue}{=\lim_{n\to\infty}\frac{n}{b-a}\sum_{j=1}^n\frac{1}{\left(\frac{an}{b-a}+j\right)^2}}\tag{1}
\end{align*}
We calcluate the limit (1) by squeezing it with lower and upper bounds which can be easily calculated using telescoping.
We consider the inequality chain
\begin{align*}
\frac{1}{\left(\frac{an}{b-a}+j\right)\left(\frac{an}{b-a}+j+1\right)}
&\leq \frac{1}{\left(\frac{an}{b-a}+j\right)^2}\leq \frac{1}{\left(\frac{an}{b-a}+j-1\right)\left(\frac{an}{b-a}+j\right)}\\
\frac{1}{\frac{an}{b-a}+j}-\frac{1}{\frac{an}{b-a}+j+1}
&\leq \frac{1}{\left(\frac{an}{b-a}+j\right)^2}\leq \frac{1}{\frac{an}{b-a}+j-1}-\frac{1}{\frac{an}{b-a}+j}\tag{2}
\end{align*}
The left-most and right-most part admit telescoping which makes summation and taking the limit an easy job.
We start with the left-most part of (2) and obtain
\begin{align*}
\color{blue}{\lim_{n\to\infty}}&\color{blue}{\frac{n}{b-a}\sum_{j=1}^n\left(\frac{1}{\frac{an}{b-a}+j}-\frac{1}{\frac{an}{b-a}+j+1}\right)}\\
&=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{1}{\frac{an}{b-a}+1}-\frac{1}{\frac{an}{b-a}+n+1}\right)\\
&=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{b-a}{an+b-a}-\frac{b-a}{bn+b-a}\right)\\
&=\lim_{n\to\infty}\left(\frac{n}{an+b-a}-\frac{n}{bn+b-a}\right)\\
&\,\,\color{blue}{=\frac{1}{a}-\frac{1}{b}}\tag{3}
\end{align*}
We continue with the right-most part of (2)
\begin{align*}
\color{blue}{\lim_{n\to\infty}}&\color{blue}{\frac{n}{b-a}\left(\frac{1}{\frac{an}{b-a}}-\frac{1}{\frac{an}{b-a}+n}\right)}\\
&=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{b-a}{an}-\frac{b-a}{bn}\right)\\
&\,\,\color{blue}{=\frac{1}{a}-\frac{1}{b}}\tag{4}
\end{align*}
We finally conclude since (1) is squeezed by (3) and (4)
\begin{align*}
\color{blue}{\int_a^b\frac{1}{x^2}\,dx=\frac{1}{a}-\frac{1}{b}}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why do these laurent series approaches conflict? I was working on a problem of finding the Laurent series of $\frac{1}{z-3}$ that converges where $|z-4| > 1$
So I had one approach, let $u=z-4$ then:
$$\frac{1}{z-3} = \frac{1}{1+u} $$
$$ = \frac{1}{u} - \frac{1}{u^2} + \frac{1}{u^3}...$$
$$ = \frac{1}{z-4} - \frac{1}{(z-4)^2} + ...$$
But this apparently incorrect.
The correct answer is found by noting:
$$ \frac{1}{z-3} = \frac{1}{z-4 + 1} = \frac{1}{z-4} \frac{1}{1 - \frac{-1}{z-4}} = -\frac{1}{(z-4)^2} + ... $$
Where did I go wrong?
|
$$
\begin{align}
\frac1{z-4}\frac1{1+\frac1{z-4}}
&=\frac1{z-4}\left(1-\frac1{z-4}+\frac1{(z-4)^2}-\dots\right)\\
&=\frac1{z-4}-\frac1{(z-4)^2}+\frac1{(z-4)^3}-\dots
\end{align}
$$
The series starts with $\frac1{z-4}$, not $\frac1{(z-4)^2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Numbers of roots in Quadratic Equations If $a,b,c,d$ are real numbers, then show that the equation $$(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$$ has at least two real roots.
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Just line them up and shoot them.
The roots of $x^2 +ax -3b=0$ will also be roots of $(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$. So if $x^2 +a -3b = 0$ has two real roots then $(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$ will have at least two real root.
The roots of $x^2 +ax -3b=0$ will, by the quadratic formula be, $\frac {-a\pm \sqrt{a^2 + 12b}}{2}$. These will be two distinct roots if $a^2 +12b > 0$. This will be one double root if $a^2 =12b$. And this will be two complex if $a^2 + 12b < 0$.
The same holds holds true for the roots of $x^2 -cx+b = 0$. This will have two distinct real root is $c^2 -4b >0$; will have one real root if $c^2 = 4b$; will have two complex roots if $c^2 -4b < 0$.
And $x^2 -dx + 2b=0$ will have two distinct real roots if $d^2 + 8b > 0$; one if $d^2 +8b = 0$; and none if $d^2 +8b < 0$.
So that's lining them up. Now shoot them.
Suppose $a^2 + 12b \le 0$ and $c^2 - 4b \le 0$ and $d^2 + 8b \le 0$.
Then $b \le -\frac {a^2}{12}\le 0$ and $b \ge \frac {c^2}4 \ge 0$ and $b \le \frac {-d^2}8$. So $b =0$ and $a=0$ and $c=0 and $d=0$.
So either $a=b=c=d=0$ and there is one real root $x = 0$.
Or at least one of $a^2 + 12b$ or $c^2 - 4b$ or $c^2 + 8b$ is positive.
And if any $v^2 + kb > 0$ then $\frac {- v\pm \sqrt{v^2 + kb}}2$ will be two distinct real roots.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Basic combinations logic doubt in probability
"If $3$ students are chosen at random from a class with $6$ girls and $4$ boys, what is the probability that all $3$ students chosen will be girls?"
$\left(\dfrac{6}{10}\right)\left(\dfrac{5}{9}\right)\left(\dfrac{4}{8}\right)$
So why can't we use that logic to answer this question?
"A bag holds $4$ red marbles, $5$ blue marbles, and $2$ green marbles. If $5$ marbles are selected one after another without replacement, what is the probability of drawing $2$ red marbles, $2$ blue marbles, and $1$ green marble?"
My answer: $\left(\dfrac{4}{11}\right)\left(\dfrac{3}{10}\right)\left(\dfrac{5}{9}\right)\left(\dfrac{4}{8}\right)\left(\dfrac{2}{7}\right)$
But the correct answer is $\dfrac{(_4C_2) \cdot (_5C_2) \cdot (_2C_1)}{_{11}C_5}$ (where $C$ is a combination).
Why doesn't the logic from the first problem work here?
The draws are without replacement in all cases.
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What you calculated is the probability of selecting two red marbles, two blue marbles, and one green marble in that order. However, if we select blue, green, red, red, blue, we still get two red marbles, two blue marbles, and one green marble.
To correct your attempt, we must multiply by the number of orders in which we could get two red marbles, two blue marbles, and one green marble. Choose two of the five positions for the blue marbles and two of the remaining three positions for the green marbles. The only green marble must go in the remaining position. Then we obtain
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}\left(\frac{4}{11}\right)\left(\frac{3}{10}\right)\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)\left(\frac{2}{7}\right) = \frac{20}{77}$$
Since we do not care about the order in which the marbles are selected, it is simpler to calculate the probability of selecting two of the four red marbles, two of the five blue marbles, and one of the two green marbles when we select five of the eleven marbles, which yields
$$\frac{\dbinom{4}{2}\dbinom{5}{2}\dbinom{2}{1}}{\dbinom{11}{5}} = \frac{20}{77}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that the wave equation takes the form $\frac{\partial^2 u}{\partial r \partial s}$ Show that the wave equation $\frac{\partial^2 u}{\partial t^2} - a^2 \frac{\partial^2 u}{\partial x^2} = 0$ takes the form $\frac{\partial^2 u}{\partial r \partial s} = 0$ under the change of variable $r = x + at$, $s = x - at$.
I don't know how to proceed it.
|
First for $r =x + at$ and $s=x-at$, we have $x=\frac{1}{2}(r+s)$ and $t=\frac{1}{2a}(r-s)$. Using chain rule for partial derivatives.
For $u:=u(r,s)$, we have
\begin{align*}
u_s &= u_x \frac{dx}{ds} + u_t \frac{dt}{ds} = \frac{1}{2}u_x -\frac{1}{2a} u_t \\
&\Rightarrow u_{rs} = \frac{1}{2}\bigg(u_{xx} \frac{dx}{dr} +u_{xt} \frac{dt}{dr}\bigg) -\frac{1}{2a}\bigg( u_{tt}\frac{dt}{dr} + u_{tx} \frac{dx}{dr}\bigg) \\
&= \frac{1}{2}\bigg(\frac{1}{2}u_{xx} +\frac{1}{2a}u_{xt}\bigg) - \frac{1}{2a}\bigg(\frac{1}{2a}u_{tt} + \frac{1}{2}u_{tx}\bigg) \\
&= \frac{1}{4}u_{xx} -\frac{1}{4a^2}u_{tt}
\end{align*}
So $\displaystyle \frac{\partial^2 u}{\partial r\partial s}=0 \iff a^2 \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial t^2}=0 $.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Is it necessarily the case that $m=1$ and that these two divisors are $2^k-1$ and $2^k+1$?
I've tested this up to $y\leq10^{10}$ but I haven't been able to make much progress with standard number theoretic techniques.
If $k=1$ then there are infinitely many solutions of the form $x=y-1$.
Let $(1)$ be the initial version of the problem and let $(2)$ be the supposedly equivalent version of the problem.
$(2)\implies(1)$: Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. We can write $y=m(2^k-1)+1$ for some $m\geq1$. Then
$$2^ky-1=2^k(m(2^k-1)+1)-1=(2^k-1)(2^km+1)$$
so $y-x$ and $y+x$ are two positive divisors of $(2^k-1)(2^km+1)$ which average to $y=m(2^k-1)+1$. By $(2)$, $y-x=2^k-1$ and $y+x=2^k+1$. Then $x=1$ and $y=2^k$.
$(1)\implies(2)$: Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Let $y=m(2^k-1)+1$. We can write the two divisors as $y-x$ and $y+x$ for some $0<x<y$. Thus,
\begin{align*}
y-x&\bigm|2^ky-1,\\
y+x&\bigm|2^ky-1,
\end{align*}
since $2^ky-1=(2^k-1)(2^km+1)$. Manipulating these divisibility relations shows that
\begin{align*}
y-x&\bigm|2^kx-1,\\
y+x&\bigm|2^kx+1,
\end{align*}
where $\gcd(2^kx-1,2^kx+1)=1$. Then $\gcd(y-x,y+x)=1$ so $y^2-x^2\bigm|2^ky-1$. We clearly have $2^k-1\bigm|y-1$. By $(1)$, $x=1$ and $y=2^k$. Then $m=1$ and the two positive divisors were $2^k-1$ and $2^k+1$.
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Too long to comment:
It is necessary that
$$y=\left\lfloor 1+\frac{x-1}{2^k-1}\right\rfloor(2^k-1)+1$$
Proof :
We can write
$$y-1=m(2^k-1)\tag1$$
where $m$ is a positive integer.
Also, $$y^2-x^2\mid 2^ky-1$$
implies $$2^ky-1-(y^2-x^2)\ge 0\tag2$$
From $(1)(2)$, we get
$$2^k(m2^k-m+1)-1-(m2^k-m+1)^2+x^2\ge 0,$$
i.e.
$$(2^k-1)^2m^2-2(2^k-1)(2^{k-1}-1)m-(2^k-2+x^2)\color{red}{\le} 0,$$
i.e.
$$\small\frac{2^{k-1}-1-\sqrt{(2^{k-1}-1)^2+2^k-2+x^2}}{2^k-1}\le m\le \frac{2^{k-1}-1+\sqrt{(2^{k-1}-1)^2+2^k-2+x^2}}{2^k-1}\tag3$$
Since we have
$$\frac{2^{k-1}-1+\sqrt{(2^{k-1}-1)^2+2^k-2+x^2}}{2^k-1}\le \frac{2^{k-1}-1+(2^{k-1}-1+x)}{2^k-1}\tag4$$
and
$$x\lt y=m2^k-m+1\implies \frac{x-1}{2^k-1}\lt m\tag5$$
it follows from $(3)(4)(5)$ that
$$\frac{x-1}{2^k-1}\lt m\le 1+\frac{x-1}{2^k-1}$$
from which
$$m=\left\lfloor 1+\frac{x-1}{2^k-1}\right\rfloor$$
follows.$\quad\blacksquare$
|
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Prove with induction $\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$ Prove with induction the identity
$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$
How can I solve this problem?
Should i set k= (p+1) and n = (p+1), then try to get the left side equal to the right side?
|
When $n=1$,
$$\begin {align}
\sum_{k=1}^n\frac 1{(2k-1)(2k+1)(2k+3)}&=\frac 1{(2.1-1)(2.1+1)(2.1+3)}\\
&=\frac 1{15}\\
&=\frac{1(1+2)}{3(2.1+1)(2.1+3)}\\
&=\frac{n(n+2)}{3(2n+1)(2n+3)}
\end {align}$$
Assume the result to be true for $n=m$
We show it is also true for $n=m+1$,
$$\begin {align}
\sum_{k=1}^{m+1}\frac 1{(2k-1)(2k+1)(2k+3)}&=\sum_{k=1}^{m}\frac 1{(2k-1)(2k+1)(2k+3)}\\
&+\frac 1{(2(m+1)-1)(2(m+1)+1)(2(m+1)+3)}\\
&=\frac {m(m+2)}{3(2m+1)(2m+3)}+\frac{1}{(2m+1)(2m+3)(2m+5)}\\
&=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{m(m+2)}{3}+\frac 1{2m+5}\bigg)\\
&=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{(m^2+2m)(2m+5)+3}{3(2m+5)}\bigg)\\
&=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{2m^3+9m^2+10m+3}{3(2m+5)}\bigg)\\
&=\frac {1}{(2m+1)(2m+3)}\times\frac{(2m+1)(m+1)(m+3)}{3(2m+5)}\\
&=\frac{(m+1)(m+3)}{3(2m+5)(2m+3)}\\
&=\frac{\big(m+1\big)\big((m+1)+2\big)}{3\big(2(m+1)+1\big)\big(2(m+1)+3\big)}\\
\end {align}$$
Hence,
by Principle of Mathematical Induction, the result holds for all $n\geq1$.
|
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|
If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align*}$$
By using this, we get value of $(3\csc3\theta - 4\sec3\theta) =9.95$ by using calculator.
I want know if there's any way to solve this problem without calculator.
|
Consider the right triangle with sides $3,4,5$, whose angle opposite the side $3$ is $9\theta$. Then:
$$\sin 9\theta =\frac35 \Rightarrow 5\sin 9\theta=3; \\
\cos 9\theta =\frac45 \Rightarrow 5\cos 9\theta =4;\\
3\csc 3\theta - 4\sec 3\theta=\frac{3}{\sin 3\theta}-\frac4{\cos 3\theta}=\frac{3\cot 3\theta-4\sin 3\theta}{\sin 3\theta \cos 3\theta}=\\
\frac{5\sin 9\theta \cos3\theta-5\cos 9\theta \sin 3\theta}{\sin 3\theta \cos 3\theta}=\\
\frac{5\sin (9\theta -3\theta)}{0.5\sin 6\theta}=10.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $x^{5} \equiv 2$ mod $221\ $ [Taking modular $k$'th roots if unique] We know that $221 = 17*13$. So we can check if the system has roots to both of those equations separately, which it does:
$x^{5} \equiv 2$ mod $13$ has the solution $6 + 13n$ and $x^{5} \equiv 2$ mod $17$ has the solution $15 + 17n$.
I got these numbers from wolfram, I have no idea how to solve this problem WITHOUT a calculator. And even after finding these numbers. How would one obtain a solution modulo $221$? I was thinking the Chinese Remainder Theorem but I am under the assumption that CRT only applies to problems with powers of $x$ which are $1$.
Thanks.
|
We work all the time modulo $221$ from now on tacitly. Since $2$ is relatively prime to $221$, a/the solution $x$ of $x^5=2$ (modulo $221$ - i write equivalences modulo $221$ as equalities from now on) is also relatively prime to $221$. The Euler indicator function of $221$ is $\varphi(221)=\varphi(17\cdot13)=\varphi(17)\cdot\varphi(13)= 16\cdot 12=192$.
The inverse of $5$ modulo $192$ is $77$, explicitly
$5\cdot 77=385=2\cdot192+1$. From $x^5=2$ we get then (implications in one direction)
$$
\begin{aligned}
x &= (x^{192})^2\cdot x
&&\text{ Euler, since $x^{192}=x^{\varphi(221)}=1$ modulo $221$}
\\
&=x^{385}=(x^5)^{77}=2^{77}=(2^8)^9\cdot 2^5
&&\text{ ($2^8$ is "close" to $221$)}
\\
&=256^9\cdot 32 = 35^9\cdot 32=(35^3)^3\cdot 32
\\
&=42875^3\cdot 32
=1^3\cdot 32=\color{blue}{\boxed{32}}\ .
\end{aligned}
$$
Check (for the other direction):
$$
32^5
=
(2^5)^5=2^{25}
=2^{24}\cdot 2
=(2^8)^3\cdot 2=42875^3\cdot 2=1^3\cdot2=2\ .
$$
|
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|
Find the sum: $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$ Let $0<a<b$, I would like to compute the sum
$$\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right).$$
But first I am worrying that a test convergence might lead to the divergence of this series
What do I miss here?
$$\begin{split}\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)&= \sum_{n=1}^{\infty} \int^{\frac1{a+n}}_{\frac1{b+n}}\frac{dx}{x+1}\\
&= \sum_{n=1}^{\infty} \int_{a+n}^{b+n}\frac{dt}{t(t+1)}~~~~(t= 1/x)\\
&= \sum_{n=1}^{\infty} \int_{a}^{b}\frac{dt}{(t+n)(t+n+1)}\\
&=\int_{a}^{b}dt \sum_{n=1}^{\infty} \frac{1}{t+n}-\frac{1}{t+n+1}~~~(\text{Monotone convergence})\\
&= \int_{a}^{b}\frac{dt}{t+1} ~~~~(\text{by Telescoping sum})\\
&= \ln\left(\frac{b+1}{a+1}\right) \end{split}$$
However the series seems $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$ not to be convergent.
Have I missed something ?
|
Btw., the series does not only seem divergent but it is divergent indeed as can be seen as follows:
$$\ln\left(\frac{b+n+1}{a+n+1}\right) = \ln (b+n+1) - \ln(a+n+1) \stackrel{a < \xi_n < b}{=}(b-a)\frac{1}{\xi_n+n+1} \geq (b-a)\frac{1}{\lceil b\rceil+n+1}$$
Hence, the given sum has a divergent tail of the harmonic series as a minorant.
|
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|
Find the binomial coefificient of $x^8$ in $(1+x^2-x^3)^9$ I was trying to solve it using the multinomial theorem.
I was trying to find which combinations could give me such $x^8$
and I came to the conclusion that it only occours when i take
$(x^2)^4$ or $(-x^3)^2*(x^2)^1$
therefore: $\binom{9!}{0!*4!*0!} + \binom{9!}{0!*1!*2!}$
It just seems just too big of a number.. what am I missing?
|
A comparison. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$.
Multinomial theorem:
\begin{align*}
\color{blue}{[x^8]}&\color{blue}{\left(1+x^2-x^3\right)^9}\\
&=[x^8]\sum_{{k_1+k_2+k_3}\atop{k_1,k_2,k_3\geq 0}}\binom{9}{k_1,k_2,k_3}1^{k_1}\left(x^2\right)^{k_2}\left(-x^3\right)^{k_3}\\
&=[x^8]\sum_{{k_1+k_2+k_3}\atop{k_1,k_2,k_3\geq 0}}\binom{9}{k_1,k_2,k_3}(-1)^{k_3}x^{2k_2+3k_3}\\
&=\binom{9}{5,4,0}(-1)^0+\binom{9}{6,1,2}(-1)^2\tag{1}\\
&=\frac{9!}{5!4!0!}+\frac{9!}{6!1!2!}\\
&=126+252\\
&\,\,\color{blue}{=378}
\end{align*}
Comment:
*
*In (1) we select the coefficient of $x^8$. We also note the sum of the entries of the lower indices of the multinomial coefficient is $9$ (i.e. $5+4+0=6+1+2=9$).
Binomial theorem (twice):
\begin{align*}
\color{blue}{[x^8]}&\color{blue}{\left(1+x^2-x^3\right)^9}\\
&=[x^8]\sum_{k=0}^9\binom{9}{k}\left(x^2+x^3\right)^k\\
&=[x^8]\sum_{k=0}^9\binom{9}{k}x^{2k}\left(1+x\right)^k\\
&=\sum_{k=0}^4\binom{9}{k}x^{2k}[x^{8-2k}]\left(1+x\right)^k\tag{2}\\
&=\sum_{k=0}^4\binom{9}{k}\binom{k}{8-2k}\tag{3}\\
&=\binom{9}{3}\binom{3}{2}+\binom{9}{4}\binom{4}{0}\tag{4}\\
&=252+126\\
&\,\,\color{blue}{=378}
\end{align*}
Comment:
*
*In (2) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and set the upper limit of the sum to $4$ since the exponent of $x^{8-2k}$ is non-negative.
*In (3) we select the coefficient of $x^{8-2k}$.
*In (4) we select the non-zero terms of (3) noting that $\binom{p}{q}=\frac{p(p-1)\cdots(p-q+1)}{q!}=0$ if $q>p$.
|
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|
Generating function of binomial coefficients We want to evaluate the sum $$\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L$$
From this set of notes (page 2, equation 8) we find the formula $$\sum_{n=0}^{\infty}\binom{n}{k}y^n = \frac{y^n}{(1-y)^{n+1}}$$ which suggests that I can do $$\frac{1}{x}\sum_{L=0}^{\infty}\binom{L+1}{2}x^{L+1} = \frac{x^L}{(1-x)^{L+2}}\tag{1}\label{eqn1}$$ since $$\frac{1}{2}L(L+1) = \frac{(L+1)!}{2!((L+1)-2)!}$$
But if I do
\begin{aligned}
\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L & = \sum_{L=0}^{\infty}\frac{L}{2}\frac{d}{dx}x^{L+1} \\
& = \sum_{L=0}^{\infty} \frac{1}{2} \frac{d}{dx} \left[(L+2)x^{L+1} - 2x^{L+1}\right] \\
& = \frac{1}{2} \frac{d^2}{dx^2} \sum_{L=0}^{\infty} x^{L+2} - \frac{d}{dx} \sum_{L=0}^{\infty} x^{L+1} \\
& = \frac{1}{2} \frac{d^2}{dx^2} \frac{x^2}{1-x} - \frac{d}{dx} \frac{x}{1-x} \\
& = \frac{1}{2} \frac{d}{dx} \left[2x(1-x)^{-1} + x^2(1-x)^{-2}\right] - \left[(1-x)^{-1} + x(1-x)^{-2}\right] \\
& = \frac{1}{2} \frac{d}{dx} \frac{2x-x^2}{(1-x)^2} - \frac{1}{(1-x)^2} \\
& = \frac{1}{2} \frac{d}{dx} \left[\frac{x}{(1-x)^2} + \frac{x}{1-x} \right] - \frac{1}{(1-x)^2} \\
& = \frac{1}{2} \left[(1-x)^{-2} + 2x(1-x)^{-3} \right] \\
& = \frac{1+x}{2(1-x)^3}
\end{aligned}
which is nowhere close to (1). Where did I go wrong?
All help is welcome.
|
In my humble opinion, a simpler way would to consider
$$S=\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L\implies T=2S=\sum_{L=0}^{\infty}L(L+1)x^L$$
$$T=\sum_{L=0}^{\infty}L(L-1+2)x^L=\sum_{L=0}^{\infty}L(L-1)x^L+2\sum_{L=0}^{\infty}Lx^L$$
$$T=x^2\sum_{L=0}^{\infty}L(L-1)x^{L-2}+2x\sum_{L=0}^{\infty}Lx^{L-1}$$
$$T=x^2 \left(\sum_{L=0}^{\infty}x^{L} \right)''+2x \left( \sum_{L=0}^{\infty}x^{L}\right)'$$
|
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How to find the correct orthogonal eigenvectors for a basis of an eigenvalue? I have an exercise where I found the two correct eigenvalues $\lambda_{1} = 0$ and
$\lambda_{2} = 6$.
The algebraic multiplicity of $\lambda_{2}$ is 2. Now I try to find $E_{\lambda_{2}}$ with the following matrix. I've already done the elementary operations needed to solve the system.
$(A-6I | 0) = \begin{pmatrix}
1 & 1 & -2 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$
Now we get the following system of linear equations.
$x+y-2z = 0$
$y = y$ (first free variable)
$z = z$ (second free variable)
This basically implies that $x = -y +2z$
So, to get my eigenvectors of my basis I do:
$\begin{pmatrix}
x \\y\\z
\end{pmatrix} = \begin{pmatrix}-1\\1\\0 \end{pmatrix}y +
\begin{pmatrix} 2\\0\\1 \end{pmatrix}z$
therefore $E_{\lambda_{2}} = \bigg\{ \begin{pmatrix}-1\\1\\0 \end{pmatrix} ,
\begin{pmatrix} 2\\0\\1 \end{pmatrix} \bigg\}$.
However, this not what the solution in my textbook is even though we had the same exact matrix.
They have as solution that $E_{\lambda_{2}} = \bigg\{ \begin{pmatrix}1\\-1\\0 \end{pmatrix} ,
\begin{pmatrix} 1\\1\\1 \end{pmatrix} \bigg\}$.
I know that they are different vectors which you can take by multiplying the vectors found in the linear equation $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = ()y + ()z$ by some scalar.
For example I could do
$-1\begin{pmatrix} -1\\1\\0 \end{pmatrix}$ = $\begin{pmatrix} 1\\-1\\0 \end{pmatrix}$ which gives me the first vector they had in their solution.
However, I don't get how they got $\begin{pmatrix} 1\\1\\1 \end{pmatrix}$ as their second vector of the basis. Where did I went wrong?
While writing the question I realized that in the end both my solution and theirs are correct however their solution has the advantage of proposing two orthogonal vectors that are orthogonal with the basis of $E_{\lambda_{1}} = \bigg\{\begin{pmatrix} 1\\1\\-2 \end{pmatrix}\bigg\} $ without resorting to doing Gram-Schmidt afterwards.
So now, I want to know what was the method used to get those orthogonal vectors easily is there a process or was it just trial and error?
|
If you just want a vector that satisfies $x=-y+2z$ and is orthogonal to $\begin{pmatrix}-1\\1\\0\end{pmatrix}$, then solve $x=-y+2z$ and (from the dot product being $0$) $-x+y=0$. You'll find that $x=y=z$ and then might as well pick $\begin{pmatrix}1\\1\\1\end{pmatrix}$. As you mentioned, if you wanted an orthonormal basis, Gram-Schmidt is a process you could follow.
|
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|
$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 = 11 - 2*4$
$4 = 37-3*11$
$1 = 4 -1*3 = 4-1(11-2*4) = 3*4-1*11 = 3(37-3*11)-1*11 = 3*37 -10*11$
$1 = 3*37 -10*11$
$11(10) - 37(3) = -1$
$11(90) - 37(27) = -9$
x = 90
That is my answer. But the correct answer should be:
x = 16
|
We are looking for a solution to $11x\equiv-9\pmod{37}$.
First, solve $11x+37y=1$. As shown in this answer, we use the Extended Euclidean Algorithm:
$$
\begin{array}{r}
&&3&2&1&3\\\hline
1&0&1&-2&3&-11\\
0&1&-3&7&-10&37\\
37&11&4&3&1&0
\end{array}
$$
Thus,
$$
\begin{array}{c}
3\cdot37-10\cdot11=1\\
\Downarrow&\text{multiply by $-9$}\\
90\cdot11\equiv-9\pmod{37}\\
\Downarrow&\text{add $13$ to both sides}\\
16\cdot11+13\equiv4\pmod{37}&90\equiv16\pmod{37}
\end{array}
$$
Therefore, we have
$$
x\equiv16\pmod{37}
$$
|
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|
what is the Prime factorisation of 28! Can someone check my solution for this question?
$28! = 2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times 17^g \times 19^h \times 23^i$
To find a:
$28/2 = 14$
$28/2^2 = 7$
$28/2^3 = 3$
$28/2^4 = 1$
Where $a= 14+7+3+1= 25$ giving $2^{25}$
I repeated this method to find b to i to give a final answer of
$28! = 2^{25} \times 3^{13} \times 5^6 \times 7^4 \times 11^2 \times 13^2 \times 17 \times 19 \times 23$
Is this method/answer correct?
|
Your answer is correct and your method is fine, but not your choice of notation. It is not true that $\dfrac{28}{2^3}=3$. You are interested in integer division here, and therefore what you should write is $\left\lfloor\dfrac{28}{2^3}\right\rfloor=3$.
|
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|
Evaluate $\int _0^{2\pi }\frac{\cos (n\theta) }{a+\cos\theta}\,d\theta$ with $a>1$, $n\in \mathbb{N}-\left\{0\right\}$
Evaluate $$\int _0^{2\pi }\frac{\cos (n\theta) }{a+\cos\theta}\,d\theta,\quad\,a>1$$
I wrote $$f\left(z\right)=\frac{\frac{1}{2}\left(z^n+z^{-n}\right)}{\frac{iz^2}{2}+aiz+\frac{i}{2}}$$
The discriminant is $\Delta =a^2-1>0$
Poles are $z=-a\pm \sqrt{a^2-1}$ and $z=0$
$z=-a+\sqrt{a^2-1}$ and $z=0$ are within the unit circle.
Applying the residue theorem, we compute: $2i\pi \lim _{z\to -a+\sqrt{a^2-1}}\left(\frac{-\frac{i}{2}\left(z^n+z^{-n}\right)}{\frac{1}{2}\left(z+a+\sqrt{a^2-1}\right)}\right)$
The first residue is $\pi \frac{\left(-a+\sqrt{a^2-1}\right)^n+\left(-a+\sqrt{a^2-1}\right)^{-n}}{\sqrt{a^2-1}}$
Now $2i\pi \lim _{z\to 0}\left(\frac{-\frac{i}{2}\left(z^{n+1}+z^{-n+1}\right)}{\frac{1}{2}\left(z+a-\sqrt{a^2-1}\right)\left(z+a+\sqrt{a^2-1}\right)}\right)$ $=\lim _{z\to 0}\left(2\pi \left(z^{n+1}+z^{-n+1}\right)\right)\:$
|
As cosine is an even function, integral from o to 2$\pi$ can be written as 2 times the the integral from 0 to $\pi$, So you would be left with the upper semicircle and hence take $\frac{-ai + i \sqrt{a^2 -1}}{2}$. This is because in the upper plane imaginary part $Im(z) >= 0$.
|
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|
Help with proof that if q is a prime divisor of $\frac{n^p-1}{n-1}$, then either q=p or $q\equiv1 \pmod p$ Here is the proof:
Why does $(\frac{n^p-1}{n-1},n-1)=(p,n-1)$? Unless n=1 I think it should be the geometric sum $\sum_{i=0}^{p-1} n^i$.
|
Assuming $p$ is and odd prime, this is a result of the more general fact
$$\gcd\left(\frac{a^p + b^p}{a + b}, a + b\right) = \gcd(p, a + b)$$ when $\gcd(a,b) = 1$, which can be shown using the binomial theorem, for instance:
$$\begin{align*}\frac{a^p + b^p}{a + b} & = \frac{((a + b) - b)^p + b^p}{a + b} \\ & = \frac{1}{a + b}\left(b^p + \sum_{i = 0}^p \binom{p}{i}(a + b)^i(-b)^{p - i}\right) \\ & = \frac{1}{a + b}\left(b^p + (-b)^p + \sum_{i = 1}^p \binom{p}{i}(a + b)^i(-b)^{p - i}\right) \\ & = \sum_{i = 1}^p \binom{p}{i} (a + b)^{i - 1}(-b)^{p - i}.\end{align*}$$ All terms in the sum but the first are divisible by $a + b$, so we have $$\frac{a^p + b^p}{a + b} \equiv p(-b)^{p - 1} \equiv pb^{p -1} \equiv p \pmod{a + b},$$ where the last congruence uses the fact $\gcd(b,a + b) = \gcd(b^{p - 1}, a + b) = 1$.
Your special case is $a = n$ and $b = -1$.
|
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|
What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ in terms of $x$?I tried factoring $x$ from both polynomials but I don't know what to do next since there'd be a $1$ in the second polynomial.
Any help would be appreciated!
|
Doing long division:
$$P(x)=\frac{x^{7} + x^{27} + x^{47} +x^{67} + x^{87}}{x^3-x}=\frac{x^{86}+x^{66}+x^{46}+x^{26}+x^6}{x^2-1}=\\
\frac{\sum_{i=0}^{9}(x^{86-2i}-x^{84-2i})+2\sum_{i=0}^{9}(x^{66-2i}-x^{64-2i})+3\sum_{i=0}^{9}(x^{46-2i}-x^{44-2i})+4\sum_{i=0}^{9}(x^{26-2i}-x^{24-2i})+5\sum_{i=0}^{2}(x^{6-2i}-x^{4-2i})+5}{x^2-1}=\\
Q(x)+\frac{5}{x^2-1}=Q(x)+\frac{5x}{x^3-x}.$$
|
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|
Smallest square to cover a right triangle I believe smallest possible triangle to cover a square is well studied... But how about square covering a triangle? I read the post here, but I think that I don't want such a general question (and complicated solution), just right triangle.
I first encounter the problem, which is, find the square with minimum length that covers a triangle with sides $5,12,13$ completely. The answer is a square with length$\frac{144}{\sqrt{193}}$ and I believe so, using the method here.
But the problem is, why is it the smallest possible square covering the triangle?
My attempts:
If a square covers the triangle and at the same time they have some vertex(s) coinciding:
If it is $\angle BAC$ touching vertex of square, see here, the square covering it should have side length of at least 12.
If it is $\angle BCA$ touching the vertex of square, see here, the square covering it should also have side length of at least 12.
This leaves us to the problem of $\angle ABC$, see here or below:
It is the smallest because whatever it is rotated clockwise/anti-clockwise through the origin, the side length of square will be larger.
For every square covering a triangle, there is a smaller square covering the same triangle while touching at least a vertex using its vertex(s). But how does it implies that the side length of square is at least $\frac{144}{\sqrt{193}}$? I mean, a square which touches the triangle with its vertex may have a shorter side length, but why is there no square with side length even shorter (doesn't touch the triangle using its vertex)? Why do we need only to consider these three cases?
Any help will be appreciate! Thank you.
|
This answer shows that $\frac{144}{\sqrt{193}}$ is the smallest possible side length.
(This is an answer to the question "why is it the smallest possible square covering the triangle?", but I have to say that this is not an answer to the question "Why do we need only to consider these three cases?". Anyway, I hope this helps.)
Let $A(0,0), B(5,12), C(5,0)$.
First, let us consider the condition that three points $A,B,C$ exist between two parallel lines $L_1$ and $L_2$ whose slope is $\tan\theta\ (0\lt \theta\lt\frac{\pi}{2})$ where the distance between $L_1$ and $L_2$ is $d$.
We may suppose that $L_1$ passes through $C$, so we may write
$$L_1 : y=\tan\theta\ (x-5),\qquad L_2 : y=\tan\theta\ x+\frac{d}{\cos\theta}-5\tan\theta$$
The condition that three points $A,B,C$ exist between $L_1$ and $L_2$ is
$$0\le \tan\theta\times 0+\frac{d}{\cos\theta}-5\tan\theta\quad \text{and}\quad 12\le\tan\theta\times 5+\frac{d}{\cos\theta}-5\tan\theta,$$
i.e.
$$\sin\theta\le\frac d5\quad\text{and}\quad \cos\theta\le\frac{d}{12}\tag1$$
Next, let us consider the condition that three points $A,B,C$ exist between two parallel lines $L_3$ and $L_4$ whose slope is $\tan(\theta+\frac{\pi}{2})\ (0\lt \theta\lt\frac{\pi}{2})$ where the distance between $L_3$ and $L_4$ is $d$.
We may suppose that $L_3$ passes through $A$, so we may write
$$L_1 : y=\tan\left(\theta+\frac{\pi}{2}\right)x,\qquad L_2 : y=\tan\left(\theta+\frac{\pi}{2}\right)x+\frac{d}{\sin\theta}$$
The condition that three points $A,B,C$ exist between $L_3$ and $L_4$ is
$$0\le \tan\left(\theta+\frac{\pi}{2}\right)\times 5+\frac{d}{\sin\theta}\quad \text{and}\quad 12\le\tan\left(\theta+\frac{\pi}{2}\right)\times 5+\frac{d}{\sin\theta},$$
i.e.
$$\cos\theta\le\frac d5\quad\text{and}\quad 12\sin\theta+5\cos\theta\le d\tag2$$
Finally, let us prove that if there exists a pair of $(d,\theta)$ satisfying both $(1)$ and $(2)$, then $d\ge \frac{144}{\sqrt{193}}$.
We have
$$\small\begin{align}&(1)(2)\\\\&\implies\cos\theta\le\frac{d}{12}\quad\text{and}\quad \sqrt{1-\cos^2\theta}\le\frac{d-5\cos\theta}{12}
\\\\&\implies \cos\theta\le\frac{d}{12}\quad\text{and}\quad 1-\cos^2\theta\le \frac{d^2-10d\cos\theta+25\cos^2\theta}{144}
\\\\&\implies \cos\theta\le\frac{d}{12}\quad\text{and}\quad 169\cos^2\theta-10d\cos\theta+\color{blue}{d^2-144}\ge 0
\\\\&\implies \frac{5d+12\sqrt{169-d^2}}{169}\le \cos\theta\quad\text{and}\quad \cos\theta\le\frac{d}{12}
\\\\&\left(\because\ \text{we already know that $d=\frac{144}{\sqrt{193}}\approx 10.4$ is possible, so we may suppose that $\color{blue}{d^2-144}\lt 0$}\right)
\end{align}$$
In order for such $\theta$ to exist, we have to have
$$\begin{align}&\frac{5d+12\sqrt{169-d^2}}{169}\le \frac{d}{12}
\\\\&\implies\sqrt{169-d^2}\le\frac{109}{144}d
\\\\&\implies 169-d^2\ge 0\quad\text{and}\quad 169-d^2\le \left(\frac{109}{144}d\right)^2
\\\\&\implies 0\lt d\le 13\quad\text{and}\quad d\ge\frac{144}{\sqrt{193}}
\\\\&\implies \frac{144}{\sqrt{193}}\le d\le 13\qquad\blacksquare\end{align}$$
|
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|
Proving $\tan^{-1}\frac{1}{2\cdot1^{2}}+\tan^{-1}\frac{1}{2\cdot2^{2}}+\cdots+\tan^{-1}\frac{1}{2\cdot n^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$
By using Mathematical Induction, prove the following equation for all positive integers $n$:
$$\tan^{-1}\frac{1}{2 \cdot 1^{2}} + \tan^{-1}\frac{1}{2 \cdot 2^{2}} +\cdots+\tan^{-1}\frac{1}{2 \cdot n^{2}} = \frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$$
I am able to prove the statement for n=1 and I got the assumption statement when $n = k$ as follows:
$$\tan^{-1}\frac{1}{2 \times 1^{2}}+\tan^{-1}\frac{1}{2 \times 2^{2}}+\cdots+\tan^{-1}\frac{1}{2 \times k^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2k+1}$$
The problem is when I have to prove the case for n=k+1. I'm supposed to prove the following:
$$\tan^{-1}\frac{1}{2 \times 1^{2}}+\tan^{-1}\frac{1}{2 \times 2^{2}}+\cdots+\tan^{-1}\frac{1}{2 \times (k+1)^{2}} = \frac{\pi}{4}-\tan^{-1}\frac{1}{2k+3}$$
I simplified the left hand side - with the assumption - into
$$\frac{\pi}{4}-\left(\tan^{-1}\frac{1}{2k+1}-\tan^{-1}\frac{1}{2(k+1)^{2}}\right)$$
I understand that I am supposed to use the identity
$$\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$$
However, I still am unable to obtain the right-hand side from the expression I have right now.
Is there something I am missing?
|
I don't know what you did wrong but you should get the required result using the given identity:
$$\begin{align}
\arctan{\left(\frac{1}{2k+1}\right)}-\arctan{\left(\frac{1}{2(k+1)^2}\right)}
&=\arctan{\left(\frac{\frac{1}{2k+1}-\frac{1}{2(k+1)^2}}{1+\left(\frac{1}{2k+1}\right)\left(\frac{1}{2(k+1)^2}\right)}\right)}\\
&=\arctan{\left(\frac{2(k+1)^2-(2k+1)}{(2k+1)(2(k+1)^2)+1}\right)}\\
&=\arctan{\left(\frac{2(k^2+2k+1)-(2k+1)}{2(2k+1)(k^2+2k+1)+1}\right)}\\
&=\arctan{\left(\frac{2k^2+4k+2-2k-1}{2(2k^3+4k^2+2k+k^2+2k+1)+1}\right)}\\
&=\arctan{\left(\frac{2k^2+2k+1}{2(2k^3+5k^2+4k+1)+1}\right)}\\
&=\arctan{\left(\frac{2k^2+2k+1}{4k^3+10k^2+8k+3}\right)}\\
&=\arctan{\left(\frac{2k^2+2k+1}{(2k+3)(2k^2+2k+1)}\right)}\\
&=\arctan{\left(\frac{1}{2k+3}\right)}\\
\end{align}$$
|
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|
What is $\cos\left(\frac{\arctan(x)}{3}\right)$ and $\sin\left(\frac{\arctan(x)}{3}\right)$? I know that $$\cos(\dfrac{\pi}{3} - \arctan(x))= \dfrac{1}{2\sqrt{(1+x^2)}} + \dfrac{\sqrt{3}x}{2\sqrt{(1+x^2)}}$$
$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\right)$ = ?
$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\right) = \cos\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\arctan(x)}{3}\right) + \sin\left(\dfrac{\pi}{3}\right)\sin\left(\dfrac{\arctan(x)}{3}\right) = \dfrac{1}{2}\cos\left(\dfrac{\arctan(x)}{3}\right) + \dfrac{\sqrt{3}}{2}\sin\left(\dfrac{\arctan(x)}{3}\right)$
but I can't go further since I don't know how to solve $\sin\left(\dfrac{\arctan(x)}{3}\right)$ and $\cos\left(\dfrac{\arctan(x)}{3}\right)$.
Any suggestion?
|
Let us care about
$$t:=\tan\left(\frac{\arctan(x)}3\right), $$
using the fact that
$$\tan\left(3\frac{\arctan(x)}3\right)=x.$$
By the triple angle formula, this equation writes
$$\frac{3t-t^3}{1-3t^2}=x$$
or
$$t^3-3xt^2-3t+x=0.$$
We depress it with $u:=t-x$, giving
$$u^3-3(x^2+1)u-2x(x^2+1)=0.$$
Now the discriminant is given by
$$(x(x^2+1))^2-(x^2+1)^3=-(x^2+1)^2.$$
As it is negative, the final expression will involve cubic roots of complex numbers, which cannot be expressed without… trigonometry, and you are circling in rounds.
|
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|
Describe the region $\{|z^2 - 1|<1\}$ of the complex plane Since it's complex analysis, I assume there is an easy solution without much algebra. But even in my algebra heavy solution there should be a more streamlined approach... Either approach is appreciated as an answer, or verification that my approach is the standard one, thanks!
My Attempt:
We can equivalently look at the condition $|z^2 - 1|^2 - 1< 0$. Note that $z^2 - 1 = (x^2 - y^2 - 1) +i(2xy)$, so
\begin{align*}
|z^2 - 1|^2 - 1 & = (x^2 - y^2 - 1)^2 + 4x^2 y^2 - 1 \\
& = x^4 + y^4 -2x^2 - 2y^2 + 2x^2 y^2 \\
& = (x^2 + y^2)(x^2 + y^2 - 2) \\
& <0
\end{align*}
Case 1: $x^2 + y^2 < 0$ and $x^2 + y^2 - 2 >0$
This case is not possible
Case 2: $x^2 + y^2 > 0$ and $x^2 + y^2 - 2 <0$
Here, at most one of $x$ and $y$ can be zero, so $z\neq 0$. Further, $x^2 + y^2 <2$ is the open disk radius $2$ centered at $0$
Therefore, The region is the punctured open disk $D(0,2)\setminus \{0\}$
|
I can't say if this is "easier" or not, but here is my approach.
Let $w = z^2$. Then $\{w: |w-1|<1\}$ is a circle of radius $1$ centered at $(1,0)$. The mapping $w: z \mapsto z^2$ transforms the original set into this circle.
In polar form, $z = r_ze^{i\theta_z}$, and $w = r_we^{i\theta_w} = r_z^2e^{i2\theta_z}$.
You can find the polar form of the circle
$$ (x_w-1)^2 + y_w^2 = 1 \implies r_w^2 - 2r_w\cos(\theta_w) = 0 \implies r_w = 2\cos(\theta_w) $$
Converting back to $z$-space gives the polar form
$$ r_z^2 = 2\cos(2\theta_z) \implies r_z = \sqrt{2\cos(2\theta_z)} $$
Note that $w^{1/2}$ has two roots, so the left and right halves of your $z$-plot both map to the same $w$-circle. The right half maps to $-\pi < \arg(w) < \pi$ and the left half maps to $\pi < \arg(w) < 3\pi$
|
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|
Proving $\cos^2x+\sin^2y=1$ is reflexive, symmetric, transitive. I want to make sure that I got the hang of the following relations.
For reflexivity, if $x=y,\cos^2x+\sin^2y=\cos^2x+\sin^2x=1 \implies xRx$, then it is reflexive.
For symmetry, $xRy\implies\cos^2x+\sin^2y=1$ and $yRx\implies\cos^2 y+\sin^2x=1 \implies$ symmetry.
For transitivity, $\cos^2x+\sin^2y=1$ and $\cos^2y+\sin^2z=1$ then $\cos^2x+\sin^2y+\cos^2y+\sin^2z=2 \implies \cos^2x+\sin^2z=2-1=1$ so transitivity holds.
Is this enough to prove the symmetric property? Anti-symmetric is easy because I only need to prove $x=y$ but symmetry needs to be for all $x,y$ but I can't list all possibilities in all questions.
|
Reflexive
$cos^2x+sin^2x=1$
symmetric
Suppose that $cos^2x+sin^2y=1$, $cos^2y+sin^2x=1-sin^2y+1-cos^2x=2-(cos^2x+sin^2y)=2-1=1$
Transitive
$cos^2x+sin^2y=1, cos^2y+sin^2z=1$
$cos^2x+sin^2z=cos^2x+1-cos^2y=cos^2x+sin^2y=1$
|
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|
If $a \equiv 9 \pmod {12}$, find all possible values for $\gcd(a^2+21a+72,252)$ We know that
$$a^2+21a+72 \equiv 9^2 + 21 \cdot 9 + 0 \equiv 6 \pmod {12}$$
So we know that that expression, let's say $\alpha$, is such that $12 \mid \alpha - 6$. But then $12 \mid 2(\alpha - 6)=2a+12$, which means that $\alpha$'s prime factorization contains $2$ as a factor (with multiplicity $1$) and 3 as a factor.
We know that $3$ has multiplicity $2$ because $72=2^3 \cdot 3^2$, $21a = 7 \cdot 3^2 \cdot k, a^2= 3^2k^2$. So $18$ is the lowest possible candidate to gcd. If $a^2 \equiv 5 \pmod 7$, we know that $126$ is the other candidate solution. Is there any way to check whether the system
$$a^2 \equiv 9 \pmod {12} \\ a^2 \equiv 5 \pmod 7$$
has no solution? If it does, does this mean that the gcd is $126$ unless $a^2 \not\equiv 5 \pmod 7$?
|
Note $\ (f,\,\overbrace{ 9\cdot 7\cdot 4}^{\large 252}) = (f,9)(f,7)(f,4) =\, 9\cdot \color{#c00}1\cdot\color{#0a0}2\ $ for $\ f = a^2+21a+72\ $ by
$3\mid a\ \Rightarrow\ 9\mid a^2+21a+72\ \Rightarrow\ (f,9) = \color{}9$
$\!(f,7)\, =\, (f\bmod 7,\,7)\, =\, (a^2\!+2,\,7) =\,\color{#c00} 1,\ $ by $\ x^2\not\equiv -2\pmod{\! 7}$
$\!(f,4) =\, \underbrace{(f(a),4)\, =\, (f(1)\bmod 4,\,4)}_{\large a\ \equiv\ 1\pmod{\!4}} = \color{#0a0}2,\ $ by $\ f\bmod 4\, =\, a^2+a$
|
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|
How to factorized this 4th degree polynomial? I need your help to this polynomial's factorization.
Factorize this polynomials which doesn't have roots in Q.
$ \ f(x) = x^4 +2x^3-8x^2-6x-1 $
P.S.) Are there any generalized method finding 4th degree polynomials factor?
|
Just working through the hint in the accepted answer, and verified with the Wolfram result in the second answer. Just for me, and maybe a check/guide for someone else:
!$x^4+2x^3+x^2−9x^2−6x−1 \\
(x^2+x)^2 - (3x+1)^2 \\
\\
((x^2+x) + (3x+1)) ((x^2+x) - (3x+1)) \\
(x^2+4x+1) (x^2-2x-1) \\
\\
\text{quadratic formula for } (x^2+4x+1)\\
-4/2\pm\sqrt{16-4}/2 \\
-2 \pm\sqrt3 \\
\\
\text{quadratic formula for } (x^2-2x-1) \\
--2/2\pm\sqrt{4-4(-1)}/2 \\
1\pm\sqrt{4+4}/2 \\
1\pm\sqrt2 \\
\\
\text{factored form }\\
(x+2+\sqrt3)(x+2-\sqrt3)(x-1+\sqrt2)(x-1-\sqrt2)\\
$
|
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|
"Dividing" matrix by a vector If I have $\mathbf{Ax}=\mathbf{c}$ where $\mathbf{x} =
\left[\begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{array}\right]$ and $\mathbf{c} = \left[\begin{array}{r}
2x_1-4x_2-x_3-3x_4+2x_5\\
-x_1+2x_2+x_3+x_5\\
x_1-2x_2-x_3-3x_4-x_5\\
-x_1+4x_2-x_3+5x_5
\end{array}\right]$.
To solve the equation for $\mathbf{A}$, I would therefore like to isolate $\mathbf{A}$ - is there a way to do this? Like "dividing" by $\mathbf{a}$. I know that $\mathbf{A}$ is obviously the coefficient matrix but I would like some justification to actually show this like by mathematically solving the equation for $\mathbf{A}$.
|
Notice that $$Ax = A \cdot x =\left[\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\
a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
a_{41} & \cdots & \cdots & \cdots & a_{45}
\end{array}\right]\cdot\left[\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}
\end{array}\right]=\left[\begin{array}{c}
a_{11} x_{1}+a_{12}x_{2}+a_{13}x_{3} +a_{14}x_{4} +a_{15}x_{5}\\
a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}+a_{24}x_{4}+a_{25}x_{5}\\
a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}+a_{34}x_{4}+a_{35}x_{5}\\
a_{41}x_{1}+a_{42}x_{2}+a_{43}x_{3}+a_{44}x_{4}+a_{45}x_{5}
\end{array}\right]$$
$$\left[\begin{array}{c}
a_{11} x_{1}+a_{12}x_{2}+a_{13}x_{3} +a_{14}x_{4} +a_{15}x_{5}\\
a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}+a_{24}x_{4}+a_{25}x_{5}\\
a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}+a_{34}x_{4}+a_{35}x_{5}\\
a_{41}x_{1}+a_{42}x_{2}+a_{43}x_{3}+a_{44}x_{4}+a_{45}x_{5}
\end{array}\right] = \left[\begin{array}{r}
2x_1-4x_2-x_3-3x_4+2x_5\\
-x_1+2x_2+x_3+x_5\\
x_1-2x_2-x_3-3x_4-x_5\\
-x_1+4x_2-x_3+5x_5
\end{array}\right]$ $$
And you can continue from here.
|
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|
Splitting / rearranging sigma sums I am struggling to understand the concept of sigma sum rearrangement. In fact, I don't even know what to call it. That being said, if anybody can recommend sources for me to study this from or let me know what you call this type of problem so that I can study it more, that would be greatly appreciated.
Let $H_{k}$ represent the series of harmonic numbers. Then
\begin{align}
(1&.) \quad \sum\limits_{1 \leq j \leq k} \frac{1}{j} = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{k} \\
(2&.) \quad \sum\limits_{1 \leq j \leq k} \frac{1}{j} = H_{k} \\
\end{align}
Now let:
\begin{align}
(3&.) \quad \sum\limits_{1 \leq k \leq n} kH_{k} \\
(4&.) \quad \sum\limits_{1 \leq k \leq n} k \sum\limits_{1 \leq j \leq k} \frac{1}{j} \\
(5&.) \quad \sum\limits_{1 \leq j \leq k \leq n} k \frac{1}{j} \\
(6&.) \quad \sum\limits_{1 \leq j \leq n} \frac{1}{j} \sum\limits_{j \leq k \leq n} k
\end{align}
Equation $(4.)$ is just a simple substitution but then I begin to get confused. I am used to the following notation:
\begin{align}
\sum_{j=1}^{k} \frac{1}{j}
\end{align}
Thus, I would represent $(4.)$ as:
\begin{align}
\sum_{k=1}^{n} k \sum_{j=1}^{k} \frac{1}{j}
\end{align}
I am thus confused by equation $(5.)$ as I don't know how to represent it in my usual way.
Now to my real question. Equation $(6.)$ contains two visual changes:
1. $\frac{1}{j}$ and $k$ swap places
2. The indices change from $1 \leq k \leq n$ and $1 \leq j \leq k$ change to $1 \leq j \leq n$ and $j \leq k \leq n$
I'm not sure how these indices are changed? Can anyone explain this to me, please? Is there a general method for this?
Thanks!
|
If you look closely, the equation change from 5 to 6 is the reverse of what's happening in 4 to 5. Let's take a look at what's happening to the indices with an example.
Let $n=3$, and we'll figure out what pairs of $(j,k)$ are valid for the index set in each equation.
In equation $4$, going through in order, we have $(1,1),(1,2),(2,2),(1,3),(2,3),(3,3)$. As long as we have those six pairs, all the sums will be the same.
Going from equation $4$ to equation $5$ is an application of the distributive law. If you wrote it out, it would look like: $$1(\frac{1}{1})+2(\frac{1}{1}+\frac{1}{2})+3(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}) = 1\cdot\frac{1}{1} +2\cdot\frac{1}{1}+2\cdot\frac{1}{2} +3\cdot\frac{1}{1}+3\cdot\frac{1}{2}+3\cdot\frac{1}{3}$$
In equation $6$, the indices are the same again, just in a different order: $(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)$
And everything is distributed by fractions instead of coefficients:
$$1\cdot\frac{1}{1} +2\cdot\frac{1}{1}+2\cdot\frac{1}{2} +3\cdot\frac{1}{1}+3\cdot\frac{1}{2}+3\cdot\frac{1}{3}= \frac{1}{1}(1 +2+3) +\frac{1}{2}(2+3) +\frac{1}{3}(3)$$
You can generalize what's happening for any $n$, but it's the same operations. As a general strategy, when notation gets difficult, expand it by trying small numbers or simple examples.
|
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|
using trigonometric identities For proving $$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$
I tried to use that:
\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&=1-\frac{1}{2}\sin^2 (2x)\\
&=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\
&=\frac{3}{4}+\frac{1}{4}\cos 4x
\end{align}
but i can't try more
|
$\displaystyle \sin^4x+\cos^2x=\frac{(1-\cos2x)^2}{4}+\frac{1+\cos 2x}{2}=\frac{3+\cos^22x}{4}=\frac{3+\frac{1+\cos 4x}{2}}{4}=\frac{7+\cos4x}{8}$
$\displaystyle \sin^2x+\cos^4x=\sin^4\left(\frac \pi2-x\right)+\cos^2\left(\frac \pi2-x\right)=\frac{7+\cos4\left(\frac \pi2-x\right)}{8}=\frac{7+\cos4x}{8}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can we show a sum of symmetrical cosine values is zero by using roots of unity? Can we show that
$$\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$\cos\frac{\pi}{5}+\cos\frac{2\pi}{5}+\cos\frac{3\pi}{5}+\cos\frac{4\pi}{5}=0$$
as well, so just out of curiosity, is it true that $$\sum_{k=1}^{n-1} \cos\frac{k\pi}{n} = 0$$
for all $n$ odd?
|
Note that $$\cos(\pi - \alpha)= - \cos(\alpha)$$ Therefore $$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{5\pi}{7})+\cos(\frac{6\pi}{7})=$$
$$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})-\cos(\frac{3\pi}{7})-\cos(\frac{2\pi}{7})-\cos(\frac{\pi}{7})=0$$
The same goes for other natural numbers $n$ instead of $7$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}$ Problem
$$\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}.$$
Attempt
\begin{align*}
&\lim_{x \to 0}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}\\
=&\lim_{x \to 0}\frac{\exp [(\sin x)^x\ln x]-\exp [{x^{\sin x}\ln(\sin x)]}}{x^3}
\\=&\lim_{x \to 0}\frac{\exp [{x^{\sin x}\ln(\sin x)]}(\exp [(\sin x)^x\ln x-{x^{\sin x}\ln(\sin x)}]-1)}{x^3}
\\=&\lim_{x \to 0}\frac{\exp [{x^{\sin x}\ln(\sin x)]} [(\sin x)^x\ln x-{x^{\sin x}\ln(\sin x)}]}{x^3}
\end{align*}
This will help?
|
My Solution
Recall the following formulas
$$\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x-\cdots\tag 1$$
$$e^x=1+x+\frac{1}{2}x^2+\cdots\tag 2$$
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots \tag 3$$
which are the starting point we need to rely on.
From$(1)$,
\begin{align}
\frac{\sin x-x}{x}=\dfrac{x-\dfrac{1}{3!}x^3+o(x^4)-x}{x}=-\frac{1}{6}x^2+o(x^3).\tag{4}
\end{align}
From $(3)(4)$,
\begin{align}
\ln\sin x&=\ln\left(x\cdot \frac{\sin x}{x}\right)=\ln x+\ln\left(1+\frac{\sin x-x}{x}\right)\notag\\
&=\ln x+\frac{\sin x-x}{x}-\frac{1}{2}\left(\frac{\sin x-x}{x}\right)^2+o(x^4)\notag\\
&=\ln x-\frac{1}{6}x^2+o(x^3).\tag{5}
\end{align}
From $(5)$,
$$x\ln\sin x=x(\ln x-\frac{1}{6}x^2+o(x^3))=x\ln x-\frac{1}{6}x^3+o(x^3).\tag{6}$$
From $(2)(6)$,
\begin{align*}
(\sin x)^x&=\exp(x\ln\sin x)=1+x\ln\sin x+\frac{1}{2}(x\ln\sin x)^2+\frac{1}{6}(x\ln\sin x)^3+o(x^3\ln^3 x)\notag\\
&=1+x\ln x+\frac{1}{2}x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x).\tag{7}
\end{align*}
From $(7)$, we obtain a vital result that
\begin{align*}
{\color{red}{f(x):=(\sin x)^x\ln x}}=\ln x+x\ln^2 x+\frac{1}{2}x^2\ln^3 x+\frac{1}{6}x^3\ln^4 x+o(x^3\ln^4 x).\tag{8}
\end{align*}
Morover, by $(1)(2)$, we have
\begin{align*}
x^{\sin x}&=\exp(\sin x\ln x)=1+\sin x\ln x+\frac{1}{2}\sin^2 x\ln^2 x+\frac{1}{6}\sin^3 x\ln^3 x+o(x^3\ln^3 x)\notag\\
&=1+x\ln x+\frac{1}{2} x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x).\tag{9}
\end{align*}
From $(5)(9)$, we oabtain another vital result that
\begin{align*}
&{\color{red}{g(x):=x^{\sin x}\ln\sin x}}\\
=&(1+x\ln x+\frac{1}{2} x^2\ln^2 x+\frac{1}{6}x^3\ln^3 x+o(x^3\ln^3 x))(\ln x-\frac{1}{6}x^2+o(x^3))\\
=&\ln x+x\ln^2 x+\frac{1}{2}x^2\ln^3 x+\frac{1}{6}x^3\ln^4 x-\frac{1}{6}x^2+o(x^3\ln^4 x).\tag{10}
\end{align*}
Now,notice that, by$(8)(10)$,
$$f(x)-g(x)=\frac{1}{6}x^2+o(x^3\ln^4 x)\to 0(x \to 0^+),$$
and, by $(10)$,
$$\frac{e^g(x)}{x}=e^{g(x)-\ln x}=e^{x\ln^2 x+o(x\ln^2 x)}=e^0=1.$$
It follows that
\begin{align*}
\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}&=\lim_{x \to 0+}\frac{e^{f(x)}-e^{g(x)}}{x^3}\\
&=\lim_{x \to 0+}\frac{e^{g(x)}(e^{f(x)-g(x)}-1)}{x^3}\\
&=\lim_{x \to 0+}\frac{e^{g(x)}}{x}\cdot\lim_{x \to 0+}\frac{e^{f(x)-g(x)}-1}{x^2}\\
&=1 \cdot\lim_{x \to 0+}\frac{f(x)-g(x)}{x^2}\\
&=\lim_{x \to 0+}\left(\frac{1}{6}+o(x\ln^4 x)\right)\\
&=\frac{1}{6}.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(1+AB)(A^{3} + B^{3})^{2} = (AB)^{2} + 4AB + 4 $$
$$ A^{6} + B^{6} + BA^{7} + A B^{7} = -2 (AB)^{4} - 2(AB)^{3} + (AB)^{2} + 4AB + 4 $$
I also have tried using$A = (1+x), B = (1-x)$ , and some others, but none solves the problem.
I am now trying $A = (1+x)$ and $(1-x) = -(1+x) + 2 = 2 - A$, so:
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
|
As $-1\le x\le1$
WLOG $x=\cos2t,0\le2t\le\pi,\sin2t=\sqrt{1-x^2}$
$$\implies\sqrt{1+\sin2t}[(2\cos^2t)^{3/2}+(2\sin^2t)^{3/2}]=2+\sin2t$$
As $\sin t,\cos t\ge0$ and $(\sin t+\cos t)^2=1+\sin2t$
$$2\sqrt2(\cos t+\sin t)(\cos^3t+\sin^3t)=2+\sin2t$$
$$\sqrt2(1+\sin2t)(2-\sin2t)=2+\sin2t$$ which is on rearrangement, a Quadratic Equation in $\sin2t(\ge0)$
as $\cos^3t+\sin^3t=(\cos t+\sin t)(\cos^2t+\sin^2t-\sin t\cos t)=\dfrac{(\cos t+\sin t)(2-\sin2t)}2$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Proof that $ \frac{3\pi}{8}< \int_{0}^{\pi/2} \cos{\sin{x}} dx < \frac{49\pi}{128}$ Proof that
$$
\frac{3\pi}{8} <
\int_{0}^{\pi/2} \cos\left(\sin\left(x\right)\right)\,\mathrm{d}x <
\frac{49\pi}{128}
$$
Can somebody give me some instruction how to deal with inequality like that? My current idea is:
I see $\frac{3\pi}{8}$ on the left. So I think that I can prove that
$$ \frac{3}{4}<\cos{\sin{x}} $$
And after take integral:
$$ \frac{3x}{4} \rightarrow \frac{3}{4} \cdot \frac{\pi}{2} = \frac{3\pi}{8} $$
But it is not true because
$$ \cos{\sin{x}} \geqslant \cos{1} \approx 0.5403 < 3/4$$
What have I do in such situation?
|
Follow-up to @rtybase:
$$
\cos x\le 1-\frac{x^2}2+\frac{x^4}{24}\implies
\int_0^{\pi/2}\cos\sin xdx<\int_0^{\pi/2}\left(1-\frac{\sin^2 x}2+\frac{\sin^4 x}{24}\right)dx=\frac{49\pi}{128}.
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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|
Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$. I then differentiate to get $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$
Evaluating this integral by integration by parts and geometric series I get
$$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$
Re-indexing I get
$$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$
$$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ Then using the integral equation for the Dilogarithm I arrive at
$$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
$$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
At this point I could not continue further as I did not know how to simplify the Digamma terms in the sums. I think that by using the Digamma function's relation to the Harmonic Numbers it could be possible to exploit known values of Harmonic sums to arrive at the answer but I could not get the sums in a form where this would work. If anyone could help me continue further or let me know if I am on the right track I would greatly appreciate it. Thank you in advance.
|
Proceed as follows
\begin{align}
&\int_0^1 \ln(1-x)\ln(1+x)\,dx \\
=&\int_0^1\ln2 \ln(1-x)dx +\int_0^1 {\ln(1-x)\ln\frac{1+x}2dx}, \>\>\>{IBP} \\
=& -\ln2 +\int_0^1 x\left( \frac{\ln\frac{1+x}2}{1-x} -\frac{\ln(1-x)}{1+x}\right) dx\\=&-\ln2 +1 -\underbrace{\int_0^1 \ln\frac{1+x}2dx}_{=\ln2-1}+\int_0^1\frac{\ln(1-x)}{1+x}dx +\int_0^1 \underbrace{\frac{\ln\frac{1+x}2}{1-x}dx}_{IBP}\\
=& 2-2\ln2 +2 \int_0^1\frac{\ln(1-x)}{1+x}dx, \>\>\>\>\>
{t=\frac{1-x}{1+x}}\\
=&2-2\ln2 +2\int_0^1 \frac{\ln2-\ln(1+t)+\ln t}{1+t}dt\\
=& 2-2\ln2 +\ln^22-\frac{\pi^2}6
\end{align}
|
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|
Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$ Let $a,b,c,d$ be whole numbers that satisfy
$$2^a + 2^b + 2^c = 4^d$$
What values of $(a,b,c,d)$ would make this equation true?
Here is my work so far.
Without loss of generality, assume $a\ge b\ge c$. Then one trivial solution by inspection is $(1,0,0,1)$. Playing around, I also found a solution at $(3,2,2,2)$. Then I checked $a=5,b=4,c=4$ and found that it also worked.
It seems that there is a family of solutions at $(2n-1,2n-2,2n-2,n)$. I can prove this easily:
\begin{align}
LHS&=2^{2n-1}+2^{2n-2}+2^{2n-2}\\
&=2^{2n-1}+2^{2n-1}\\
&=2^{2n}\\
&=4^n\\
&=RHS
\end{align}
Is this the only solution? If it is, how do I go about proving it?
|
Obviously, $d>0$. WLOG, assume that $a\ge b\ge c$
$2^c(2^{a-c}+2^{b-c}+1)=2^{2d}$
$2^{a-c}+2^{b-c}+1\ge 3$. Therefore, $b-c$ must be $0$ as otherwise $2^{a-c}+2^{b-c}+1$ will be an odd number larger than $1$.
$2^{a-c}+2^{b-c}+1=2^{a-c}+2=2(2^{a-c-1}+1)$ and hence $2^{a-c-1}+1$ must be even. $a-c=1$.
We have $2^c\times 4=2^{2d}$ and hence $b=c=2d-2$, $a=2d-1$.
|
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|
Find $\lim_{x\to\infty}1+2x^2+2x\sqrt{1+x^2}$ Consider the function
$$f(x)=1+2x^2+2x\sqrt{1+x^2}$$
I want to find the limit $f(x\rightarrow-\infty)$
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$, and so we have that
$$\lim_{x\rightarrow-\infty}{(1+2x^2+2x|x|)}=\lim_{x\rightarrow-\infty}(1+2x^2-2x^2)=1$$
However, if you plot the function in Desmos or you do it with a calculator, you will find that $f(x\rightarrow-\infty)=0$
What am I missing?
|
$$\lim_{x\to-\infty}(1+2x^2+2x\sqrt{1+x^2})$$$$=\lim_{x\to-\infty}(\sqrt{1+x^2}+x)^2$$$$=\lim_{x\to-\infty}\frac{1}{(\sqrt{1+x^2}-x)^2}$$$$=\lim_{x\to\infty}\frac{1}{(\sqrt{1+x^2}+x)^2}$$$$=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rationalizing $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$ Problem
Rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$
So, I'm training for the Mexican Math Olimpiad. This is one of the algebra problems from my weekly training.
Before this problem, there was other very similar, after proving it, there's an useful property:
If $a+b+c=0$ then $a^3+b^3+c^3=3abc$
It can be easily proved using the following factorization
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
I tried using this one for the rationalization. I got
$$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}*\frac{(a^2+b^2+c^2-ab-bc-ac)}{(a^2+b^2+c^2-ab-bc-ac)}=\frac{\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}-\sqrt[3]{ab}-\sqrt[3]{bc}-\sqrt[3]{ac}}{a+b+c-3\sqrt[3]{abc}}$$
But i didn't know how to proceed. I tried looking into it with wolfram alpha and i got there is no racionalization, so i assume $a+b+c=0 $ (not $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$, because the first expression wouldn't have sense) if we want a solution (In fact, if this happens, i've already rationalized it).
So my truly question is not the answer, but how to prove that $a+b+c$ needs to be 0
I would appreciate any help, ideas or suggestions, thanks.
|
You can proceed by using $x^3-y^3=(x-y)(x^2+xy+y^2).$
If $a+b+c=0$ and $a^2+b^2+c^2-ab-ac-bc=0$ so the last gives
$$(a-b)^2+(a-c)^2+(b-c)^2=0$$ or $a=b=c,$ which gives $a=b=c=0,$ which is impossible.
Thus, your trick works for all reals $a$, $b$ and $c$ such that $$a+b+c\neq0$$ and
$$(a-b)^2+(a-c)^2+(b-c)^2\neq0,$$ which says that at least two numbers from $\{a,b,c\}$ must be different.
I changed $\sqrt[3]a$ by $a$ and similar, but the essence does not change.
If $a=b=c$ so we work with
$$\frac{1}{3\sqrt[3]a},$$ which is
$$\frac{\sqrt[3]{a^2}}{3a}.$$
|
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|
When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$ When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$, gives $\sum\limits_{n=1}^{\infty} \left(\frac{nx^{n-1}}{2^n}\right)$. so, $n =0$ , becomes $n = 1$. Then, if we were to differentiate $\sum\limits_{n=1}^{\infty} \left(\frac{x}{2}\right)^n$, does it become $\sum\limits_{n=2}^{\infty} \left(\frac{nx^{n-1}}{2^n}\right)$? (from $n= 1$, to $n= 2$?)
If this is how it works,
is the above equation false? and how could we solve $\sum\limits_{n=1}^{\infty} \left(\frac{n^2x^n}{2^n}\right)$?
|
When in doubt, write it out. You don't need to memorize formulas for how the limits change in differentiation of series. There are too many cases to consider anyway. Just look at the first few terms to match the lowest index.
For instance:
$$
\sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n
= \sum_{n=0}^{\infty} \frac{x^n}{2^n}
= 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \cdots
$$
Differentiating the series is going to “kill” the $n=0$ term. So if we differentiate term-by-term, the first nonzero term is $n=1$:
$$
\frac{d}{dx}\sum_{n=0}^{\infty} \frac{x^n}{2^n}
= \sum_{n=1}^{\infty} \frac{nx^{n-1}}{2^n}
= \frac{1}{2} + \frac{2x}{4} + \frac{3x^2}{8} + \cdots
$$
With the first few terms written out, we can see that this series can also be written as
$$
\frac{1}{2} + \frac{2x}{4} + \frac{3x^2}{8} + \cdots
= \sum_{n=0}^\infty \frac{(n+1)x^n}{2^{n+1}}
$$
This re-indexing trick comes in handy when combining power series: If you can make the exponents match, you can combine like terms.
But in contrast:
$$
\sum_{n=1}^{\infty} \frac{x^n}{2^n}
= \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \cdots
$$
Differentiating this series does not kill off the first term since it's not constant.
Instead,
$$
\frac{d}{dx}\sum_{n=1}^{\infty} \frac{x^n}{2^n}
= \sum_{n=1}^{\infty} \frac{nx^{n-1}}{2^n}
= \frac{1}{2} + \frac{2x}{4} + \frac{3x^2}{8} + \cdots
$$
So there's no change to the lower limit.
|
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|
For which $n$ is $\frac{n!}{4}$ equal to $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$? For how many values of $n$, is $\frac{n!}{4}$ equal to $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$?
Further more, is there a way to approximate (or maybe even find the precise answer to) $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$? I tried approximating $n!$ but had no luck.
Thanks in advance!
|
It's not a full solution, but maybe it will help:
Let $x=\sqrt{\frac{n!}{4}}$.
$$ x^2 = \lfloor x\rfloor \cdot (\lfloor x\rfloor + 1)$$
$$ x^2 + \frac14 = (\lfloor x\rfloor + \frac12)^2$$
$$ -\frac12 + \sqrt{x^2+ \frac14} = \lfloor x\rfloor$$
$$ -1 + \sqrt{n!+ 1} = 2\lfloor x\rfloor \in 2\mathbb N$$
So we get a necessary condition
$$ \exists m\in\mathbb N : n! + 1 = (2m+1)^2$$
It's easy to see that it is also a sufficient condition, because
$$ \sqrt{n!} < \sqrt{n! + 1} < \sqrt{n!} + 1$$
$$ \sqrt{n!} < 2m+1 < \sqrt{n!} + 1$$
$$ m < \sqrt{\frac{n!}{4}} < m + \frac12$$
so $\lfloor \sqrt{\frac{n!}{4}}\rfloor = m$.
For low $n$, $n=4$ and $n=5$ satisfy this condition, but I have no proof there is no more solutions.
|
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|
How can i get the least $n $ such that $17^n \equiv 1 \mod(100$)? When I solve the problem:
$17^{2018}\equiv r \pmod{100} $
used Euler theorem since $\gcd (17,100)=1$ and so
$\phi(100)=40$
and so
$17^{40}\equiv 1 \pmod{100}$
But i also found that: $17^{20}\equiv 1 \pmod{100}$
How can i get the least n such that $17^{n}=1\pmod{100} $?
Is there any theorm or generalization to this problem?
Thanks for your help
|
$$17^n\equiv (10+7)^n\equiv 7^n+n\cdot 7^{n-1}\cdot 10 \mod 100 $$
Coincidentally,
$$7^4 \equiv 1 \mod 100$$
$$\rightarrow 7^{-1}\equiv 7^3 \equiv 43 \mod 100$$
So we need to find the smallest $n$ such that,
$$7^n+n\cdot 7^{n-1}\cdot 10 = 7^n+n\cdot 7^{n}\cdot 43\cdot10\equiv 7^n(1+30\cdot n)\equiv 1\mod 100$$
$$7^{-n}\equiv 30n+1 \mod 100$$
However $7^n$ has only 4 possibilities -> ($7,49,43,1$). Only one of these can be of the form $30n+1$.
$$30n+1 \equiv 1 \mod 100$$
Implying that $10|n$
Also, $7^4\equiv 1 \mod 100$ implying that $4|n$
The smallest $n$ satisfying these conditions $n=20$
|
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|
How to solve this congruence of triangles geometrically?
Find x.
Solution with trigonometry:
First there are the missing angles ...
Superior = $180 - 13 - 32 = 135$
Bottom = $180 - 13 - 24 = 143$
By the breast theorem we have to
$$\frac{9 \sqrt{2}}{sen(13)}=\frac{Diagonal}{sen(135)} \to Diagonal=\frac{9 \sqrt{2} \cdot sen(135)}{sen(13)}$$
Also in other triangle...
$$\frac{x}{sen(13)}=\frac{Diagonal}{sen(143)} \to x = \frac{Diagonal \cdot sen(13)}{sen(143)}$$
Using what we know diagonally
$$x = \frac{9 \sqrt{2} \cdot sen(135) \cdot sen(13)}{sen(13) \cdot sen(143)}=\frac{9 \sqrt{2} \cdot sen(135)}{sen(143)}$$
|
In the figure $ x = 3 $. However, the lenght that you seek is $ 5x = 15 $
In your case $$x = \frac{9 \sqrt{2} \cdot sen(135)}{sen(143)} = \frac{9 \sqrt{2} \cdot \frac{1}{\sqrt 2}}{\frac{3}{5}} = 15 $$
|
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|
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2\pi}{7}=\sqrt{7}$$
$$\tan\frac{\pi}{11}+4\sin\frac{3\pi}{11}=\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}=\sqrt{11}$$
$$\tan\frac{6\pi}{13}-4\sin\frac{5\pi}{13}=\tan\frac{2\pi}{13}+4\sin\frac{6\pi}{13}=\sqrt{13+2\sqrt{13}}$$
Does anyone know of any analogous identities for larger primes? I have not been able to find anything similar for $p=17$ or $p=19$.
(I am not asking for proofs of the above equations.)
|
The case of multiples of $\pi/11$ actually involves five "symmetrically equivalent" forms:
$4\sin(5\pi/11)-\tan(2\pi/11)=\sqrt{11}$
$4\sin(\pi/11)+\tan(4\pi/11)=\sqrt{11}$
$4\sin(4\pi/11)-\tan(5\pi/11)=-\sqrt{11}$
$4\sin(2\pi/11)+\tan(3\pi/11)=\sqrt{11}$
$4\sin(3\pi/11)+\tan(\pi/11)=\sqrt{11}$
All are derivable from the quadratic Gauss sum corresponding to the prime number $11$.
These are manifestations of the more compact, "symmetric" relation
$\color{blue}{4\sin(3\theta)-\tan(\theta)=(k|11)\sqrt{11}}$
where $\theta=(2k\pi/11)$ and $(k|11)$ is the Legendre symbol of residue $k$ modulo $11$, with the specific equations quoted above representing $k=1,2,3,4,5$ respectively.
There is a hidden feature in the equation rendered in blue above. In addition to multiples of $\pi/11$ we get one more value of $\theta$ between $0$ and $\pi$ where the function on the left evaluates to $+\sqrt{11}$. Correspondingly there is an additional value of $\theta$ between $\pi$ and $2\pi$, having the same cosine, for which the function value is $-\sqrt{11}$. Now, suppose we plug in $x=2\cos\theta$. Squaring the blue equation, expressing the quantities in terms of $x$ and clearing fractions yields an eighth-degree polynomial equation, which factors as follows:
$(x^5+x^4-4x^3-3x^2+3x+1)(x^3-x^2-x-1)=0$
The quintic factor is just the minimal polynomial for $2\cos(2k\pi/11)$ for $k\in\{1,2,3,4,5\}$. The cubic factor, containing the "extra" roots for $\theta$, has been coupled into the quintic one through the Gauss sum and the "sine-tangent relations" derived from it.
In 2014 the regular hendecagon was discovered to be neusis constructible. The authors "miraculously" found that the neusis construction, which requires finding the roots of the quintic factor given above, can be rendered in terms of cubic roots for which a neusis construction is guaranteed. It turns out that in the construction found by the authors, the distance from the pole of the neusis (a fixed point through which the marked ruler passes) to the straight line that includes one of the marks satisfies the equation
$a^3+a^2+a-1=0.$
This corresponds exactly to the reciprocal of the root of $x^3-x^2-x-1=0$, the latter equation being the coupled cubic factor emerging above from the Gauss sum. The construction is still a bit of a miracle, but we see that its cubic roots do not just appear out of nowhere. They are derived from the Gauss sum!
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"timestamp": "2023-03-29T00:00:00",
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|
How is Wolfram Alpha and the reduction formula arriving at a different result for the integral of $\int \sec^4 x\,dx$ than naive $u$-substitution? I calculated the following on paper for the value of $\int \sec^4 x\,dx$.
$$\int \sec^4 x\,dx=\int \sec^2 x \sec^2 x\,dx=\int (\tan^2 x + 1)(\sec^2 x)\,dx.$$
Let $u = \tan x$, $du = \sec^2 x\,dx$ so
\begin{align}\int \sec^4 x\,dx&=\int u^2 + 1\,du\\&=\frac{1}{3} u^3 + u + C\\&=\frac{1}{3} \tan^3 x + \tan x + C\\&=\frac{1}{3} (\tan x)(\tan^2 x + 1) + C\\&=\frac{1}{3} \tan x \sec^2 x + C\end{align}
Wolfram Alpha, however, gives $\int \sec^4(x)\,dx = \frac13(\cos(2 x) + 2) \tan(x) \sec^2(x) + C$. This is notably not equal to my solution.
According to the "step-by-step solution" from the Wolfram Alpha app, the reduction formula was used to produce $$\frac{1}{3}\tan x \sec^2 x + \frac{2}{3} \int \sec^2 x \,dx$$ then $$\frac{2}{3} \tan x + \frac{1}{3} \tan x \sec^2 x + C$$
Why does the reduction formula produce this added term compared to naive $u$-substitution?
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This is because you made a slight error in $$\frac13\tan^3x+\tan x+C=\frac13\tan x(\tan^2x+3)+C\ne\frac13\tan x(\tan^2x+1)+C.$$ Then you get \begin{align}\frac13\tan x(\tan^2x+3)+C&=\frac13\tan x(\tan^2 x+1)+\frac23\tan x+C\\&=\frac13\tan x\sec^2x+\frac23\tan x\cos^2x\sec^2x+C\\&=\frac13\tan x\sec^2x+\frac23\tan x\cdot\frac{1+\cos2x}2\sec^2x+C\\&=\frac13\tan x\sec^2x(2+\cos2x)\end{align} as given in W|A.
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|
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
We have that $xy^2 + y + 1 \mid x^2y + x + y \implies xy^2 + y + 1 \mid y(x^2y + x + y) - x(xy^2 + y + 1)$
$\iff xy^2 + y + 1 \mid y^2 - x \implies xy^2 + y + 1 \mid xy^2 + y + 1 - x(y^2 - x)$
$\iff xy^2 + y + 1 \mid x^2 + y + 1 \implies xy^2 + y + 1 \le x^2 + y + 1$
$\iff xy^2 \le x^2 \iff y^2 \le x$. And I don't know what to do next.
This problem is adapted from a recent competition.
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You have obtained that $xy^2+y+1 \mid x^2+y+1$, so $x \geq y^2$.
If $x=y^2$, then obviously all pairs of the form $(k^2,k) $ are solutions.
If $x >y^2$, then we let $x=My^2+N$ with $N<y^2$. If $N=0$ then $x=My^2$ so $My^4+y+1 \mid M^2y^4+y+1$ and so $My^4+y+1 \mid My+M-y-1$, which is impossible due to bounding reasons. So $N \geq 1$
Then substituting, we obtain $My^4+Ny^2+y+1 \mid M^2y^4+2MNy^2+y+1+N^2$.
Obviously, $My^4+Ny^2+y+1 \mid M^2y^4+MNy^2+My+M$, so we obtain $My^4+Ny^2+y+1 \mid MNy^2+y+1+N^2-My-M$.
We claim that $MNy^2+y+1+N^2-My-M \geq 0$.
It suffices to show that $MNy^2+y+1+N^2 \geq My+M$.
That's almost obvious, since $MNy^2+y+1+N^2 \geq My^2+y+2 \geq M(y+1)=My+M$ for $y >1$. If $y=1$, then from the initial divisibility we obtain $x=1$, which is a solution that we have obtained before.
So, the claim is proved and :
Since $My^4+Ny^2+y+1 \mid MNy^2+y+1+N^2-My-M$, we have that $My^4+Ny^2 \leq MNy^2-N^2-My-M$, so $MNy^2 \geq My^4 \Rightarrow N \geq y^2$ a contradiction
Finally, all solutions are of the form $(k^2,k)$ with $k$ an integer.
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|
Wish to evaluate $\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)} dx$ We want to evaluate this integral,$\newcommand{\cosec}{\operatorname{cosec}}$
$$\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)}\mathrm dx\tag1$$
$$\int_{0}^{1}[x^4\cosec(\pi x)-3x^2\cosec(\pi x)+2x\cosec(\pi x)]\mathrm dx$$
Apply integration by parts, first part:
$$\int x^4\cosec(\pi x)\mathrm dx=-\frac{x^4}{\pi}\ln[\cot(\pi x/2)]-\frac{4}{\pi}\int x^3\ln[\cot(\pi x/2)]\mathrm dx$$ again we apply integration by parts to this remaining integral,
$$\int x^3\ln[\cot(\pi x/2)]\mathrm dx$$
Integrate this part it is very difficult $\int \ln[\cot(\pi x/2)]\mathrm dx$
I am stuck at this point. I guess there must be another method to deal with this integral $(1)$.
I am looking forward to see how you would go about to evaluates $(1)$
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Too long for comments.
@user10354138 provided a very nice solution for the problem.
Since similar questions would probably happen, I tabulated the expressions of
$$I_n=\int_0^1 x^{n-2}\,\frac{x^2-x}{\sin(\pi x)}\,dx$$
$$\left(
\begin{array}{cc}
n & I_n \\
2 & -\frac{7 \zeta (3)}{\pi ^3} \\
3 & -\frac{7 \zeta (3)}{2 \pi ^3} \\
4 & -\frac{21 \zeta (3)}{2 \pi ^3}+\frac{93 \zeta (5)}{\pi ^5} \\
5 & -\frac{14 \zeta (3)}{\pi ^3}+\frac{279 \zeta (5)}{2 \pi ^5} \\
6 & -\frac{35 \zeta (3)}{2 \pi ^3}+\frac{465 \zeta (5)}{\pi ^5}-\frac{5715 \zeta
(7)}{2 \pi ^7} \\
7 & -\frac{21 \zeta (3)}{\pi ^3}+\frac{930 \zeta (5)}{\pi ^5}-\frac{28575 \zeta
(7)}{4 \pi ^7} \\
8 & -\frac{49 \zeta (3)}{2 \pi ^3}+\frac{3255 \zeta (5)}{2 \pi ^5}-\frac{120015
\zeta (7)}{4 \pi ^7}+\frac{160965 \zeta (9)}{\pi ^9} \\
9 & -\frac{28 \zeta (3)}{\pi ^3}+\frac{2604 \zeta (5)}{\pi ^5}-\frac{80010 \zeta
(7)}{\pi ^7}+\frac{1126755 \zeta (9)}{2 \pi ^9} \\
10 & -\frac{63 \zeta (3)}{2 \pi ^3}+\frac{3906 \zeta (5)}{\pi ^5}-\frac{360045
\zeta (7)}{2 \pi ^7}+\frac{2897370 \zeta (9)}{\pi ^9}-\frac{29016225 \zeta
(11)}{2 \pi ^{11}}
\end{array}
\right)$$
and we see how linear combinations of the $I_n$'s can lead to nice simplifications by cancellations.
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Compute $\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}$ Compute $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}.$$
I started by making the substitution $\arccos x = t$. Hence, $-\frac{dx}{\sqrt{1-x^2}}=dt$.
Now I get that $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}=-\int\limits_{\pi/3}^{2 \pi/3}\frac{dt}{t^2}=\frac{3}{2\pi}-\frac{3}{\pi}=-\frac{3}{2\pi}.$$
However, the result should be $\dfrac{3}{2\pi}$. What did I do wrong?
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Setting $\arccos x = t$ gives us $\frac{-dx}{\sqrt{1-x^2}} = dt$
So our integral is then:
$$\int_{2\pi/3}^{\pi/3} \frac{-dt}{t^2}$$
$$=\int_{\pi/3}^{2\pi/3} \frac{dt}{t^2}$$
$$= \frac{3}{\pi}-\frac{3}{2\pi}$$
$$=\frac{3}{2\pi}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3263310",
"timestamp": "2023-03-29T00:00:00",
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|
The integral $\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi,~ a, b \in \Re ?$ This integral
$$\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi, ~ a, b \in \Re$$
looks suspiciously interesting as it is independent of the parameter $b$. The question is: What is the best way of proving or disproving this?
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How about this, please check it critically
Denote the integral as $$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$
$$(x^2-b^2)^2+a^2x^2=(x^2-r^2)(x^2-s^2) \Rightarrow r=(d+ia)/2,s=(d-ia)/2, d=\sqrt{4b^2-a^2}.$$
$$I=\frac{a^2}{r^2-s^2} \int_{-\infty}^{\infty} \left( \frac{r^2}{x^2-r^2}-\frac{s^2}{x^2-s^2} \right)dx.$$
By considering a semi-circle contour in the upper half plane and applying the residue theorem, we get
$$I=\frac{a^2}{r^2-s^2} 2i\pi \left( r^2 Res \left (\frac{1}{x^2-r^2} \right)_{x=r}-s^2 Res \left ( \frac{1}{x^2-s^2} \right)_{x=-s}\right)=\frac{i\pi a^2}{r-s}=a\pi.$$
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"timestamp": "2023-03-29T00:00:00",
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|
If $a_n=\left[\frac{n^2+8n+10}{n+9}\right]$, find $ \sum_{n=1}^{30} a_n$ I am working on a problem, assuming high school math knowledge.
Let ${a_n}$ be the sequence defined by $$a_n=\left[\frac{n^2+8n+10}{n+9}\right]\,,$$ where $[x]$ denotes the largest integer which does not exceed $x$. Find the value of $ \sum\limits_{n=1}^{30} a_n$.
I honestly do not understand the text in bold. The answer provided is $445$. Could you please explain what I should be looking at here?
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Hint
\begin{align*}
\frac{n^2+8n+10}{n+9}&=\frac{n^2+9n-n-9+19}{n+9}\\
&=n-1+\frac{19}{n+9}\\
\therefore \left\lfloor\frac{n^2+8n+10}{n+9}\right\rfloor&=n-1+\left\lfloor \frac{19}{n+9}\right\rfloor
\end{align*}
As $n$ varies from $1$ to $30$, we will have
\begin{align*}
\left\lfloor\frac{19}{\color{red}{1}+9}\right\rfloor&=\lfloor 1.9\rfloor=1\\
\left\lfloor\frac{19}{\color{red}{2}+9}\right\rfloor&=\lfloor 1.7\rfloor=1\\
\vdots & = \vdots\\
\left\lfloor\frac{19}{\color{red}{11}+9}\right\rfloor&=\lfloor 0.95\rfloor=0\\
\vdots & = \vdots\\
\end{align*}
Hopefully you can take it from here.
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{
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|
unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the number) and when this polynomial is divided by $x^2 $, it leaves $2x + 4$ (again, not sure about the number). From the given conditions, if this polynomial is divided by $(x-1)x^2$, what would be the remainder?
The solution as far as I figured out is this:
first, from the division of $(x-1)^2$, I got that $f(1) = 3$
in the same way from division of $x^2$, I got $f(0) = 4.$
I can write the polynomial as follows:
$f(x) = (x-1)(x)(x) g(x) + ax^2 +bx +c$
$ax^2 + bx + c$ is the remainder. And to find $a,b,c$, I can use the conditions above, so I got $c = 4$ by substituting $x = 0,$ and I got $a+b+4 = 3$ by substituting $x = 1.$
This leaves $a + b = -1,$ and I can't figure out how to continue; please help.
Edit : I made a mistake $f(1)$ should be equal to $4$ and $a+b+c = 4$
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I am going to follow your approach. We have
$$f(x)=(x-1)^2p(x)+x+3$$
So, $f(1)=4$.
We also have that
$$f(x)=x^2q(x)+2x+4$$
And we want to find $a,b,c$ in
$$f(x)=(x-1)x^2g(x)+ax^2+bx+c$$
Plugin $x=1$ you get $4=a+b+c$.
Plugin $x=0$ doesn't give us enough information since this only gives us the remainder after dividing by $x$(instead of $x^2$).
So, instead of plugin, we look $f$ mod $x^2$. We have
$$f(x)=x^2[(x-1)g(x)+a]+(bx+c)$$
Since we are given that the remainder of dividing $f(x)$ by $x^2$ is $2x+4$, we conclude $bx+c=2x+4$, i.e. $b=2$, $c=4$. And therefore, $a=-2$.
We conclude that the reminder of $f$ dividing by $ (x-1)x^2$ is $-2x^2+2x+4$.
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|
Show that $\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}$ The integral $$\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}?$$ can be checked at Mathematica. The question is how to do it by hand?
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Using the substitution $x \mapsto \frac{\pi}2 - x$ we get
$$I = \int_0^{\frac{\pi}2} \frac{\sin x\,dx}{1+\sin2x} = \int_0^{\frac{\pi}2} \frac{\cos x\,dx}{1+\sin2x} = \frac12\int_0^{\frac{\pi}2} \frac{\sin x+\cos x}{1+\sin2x}\,dx = \frac12\int_0^{\frac{\pi}2} \frac{\sin x+\cos x}{(\sin x+\cos x)^2}\,dx$$
so we have
$$I = \frac12\int_0^{\frac{\pi}2} \frac{dx}{\sin x+\cos x} = \begin{bmatrix} t = \tan\frac{x}2 & dx = \frac{2dt}{1+t^2}\\ \sin x = \frac{2t}{1+t^2} & \cos x = \frac{1-t^2}{1+t^2}\end{bmatrix} = -\int_0^1\frac{dt}{t^2-2t-1} $$
We can calculate this as:
$$I = -\int_0^1 \frac{dt}{(t-1)^2+2} =- \int_{-1}^0 \frac{dt}{t^2+2} = \frac1{\sqrt{2}}\operatorname{Artanh}\left(\frac1{\sqrt{2}}\right)$$
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{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given two positive numbers $x,\,y$ so that $32\,x^{6}+ 4\,y^{3}= 1$. Prove that $\frac{(2\,x^{2}+ y+ 3)^{5}}{3(x^{2}+ y^{2})- 3(x+ y)+ 2}\leqq 2048$ . Given two positive numbers $x,\,y$ so that $32\,x^{6}+ 4\,y^{3}= 1$. Prove that
$$p(x)\equiv p= \frac{(2\,x^{2}+ y+ 3)^{5}}{3(x^{2}+ y^{2})- 3(x+ y)+ 2}\leqq 2048$$
My solution in VMF : (and I'm looking forward to seeing a nicer one(s), thanks for your interests !)
$$32\,x^{6}+ 4\,y^{3}= 1\,\therefore\,(y- 2\,x^{2})\left ( x- \frac{1}{2} \right )\leqq 0,\,\left ( x- \frac{1}{2} \right )\left ( y- \frac{1}{2} \right )\leqq 0,\,{y}'= -\,\frac{16\,x^{5}}{y^{2}}$$
Thus, we have
$${p}'(x)=$$
$$= \frac{{\left (\!(\!2 x^{2}+ y+ 3\!)^{5}\!\right )}'\{\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\}- {\left (\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\right )}'(\!2 x^{2}+ y+ 3\!)^{5}}{\left (\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\right )^{2}}=$$
$$\frac{5(\!2 x^{2}+ y+ 3\!)^{4}\left (\!4 x- \frac{16 y^{5}}{x^{2}}\!\right )\{\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\}- \left (\!6 x- 3- \frac{96 x^{5}}{y}+ \frac{48 x^{5}}{y^{2}}\!\right )(\!2 x^{2}+ y+ 3\!)^{5}}{\left (\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\right )^{2}}$$
$$= \frac{\left ( 20\,x(y- 2\,x^{2})(y+ 2\,x^{2})\{\!3(x^{2}+ y^{2})- 3(x+ y)+ 2\!\}- 3[x^{5}(16- 32\,y)+ y^{2}(2\,x- 1)] \right )}{y^{2}\left ( 3(x^{2}+ y^{2})- 3(x+ y)+ 2 \right )^{2}}$$
$$\times (2\,x^{2}+ y+ 3)^{4}$$
Using derivatives, the equation of the tangent line can be stated as follows:
$$p(x)- p\left ( \frac{1}{2} \right )= {p}'(x)\left ( x- \frac{1}{2} \right )\leqq 0\,(\!easy\,to\,see\,immediately\,!\!)\,\therefore\,p(x)\leqq p\left ( \frac{1}{2} \right )\leqq 2048$$
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By AM-GM
$$1=32x^6+4y^3=32x^6+2\cdot\frac{1}{2}+4y^3+2\cdot\frac{1}{2}-2\geq$$
$$\geq3\sqrt[3]{32x^6\cdot\left(\frac{1}{2}\right)^2}+3\sqrt[3]{4y^3\cdot\left(\frac{1}{2}\right)^2}-2=6x^2+3y-2,$$
which gives $$2x^2+y\leq1.$$
Id est, $$\frac{(2x^2+y+3)^5}{3(x^2+y^2)-3(x+y)+2}\leq\frac{4^5}{3(x^2+y^2)-3(x+y)+\frac{3}{2}+\frac{1}{2}}=$$
$$=\frac{1024}{3\left(x^2+y^2-x-y+\frac{1}{2}\right)+\frac{1}{2}}=\frac{1024}{3\left(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\right)+\frac{1}{2}}\leq\frac{1024}{\frac{1}{2}}=2048$$
and we are done!
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|
What is the least common factor in equation $\frac{5}{x+4}=4+\frac{3}{x-2}$ I am attempting to solve for x $\frac{5}{x+4}=4+\frac{3}{x-2}$
I know that I need to find the least common denominator. In this case, since I cannot see a clear relationship among them all I think it's just the product of all 3 denominators:
$\frac{5}{x+4}=4+\frac{3}{x-2}$ = $\frac{5}{x+4}=\frac{4}{1}+\frac{3}{x-2}$
LCD: $(x+4)(1)(x-2)$ = $(x+4)(x-2)$
Is this the LCD?
Because I tried to use this in solving my equation but I arrived at a quadratic. I don't think that my textbook wants me to use quadratics in this section but I'm not sure. Here's how I arrived at that:
$\frac{5}{x+4}=\frac{4}{1}+\frac{3}{x-2}$
$(x+4)(x-2)\frac{5}{x+4}=(x+4)(x-2)\frac{4}{1}+(x+4)(x-2)\frac{3}{x-2}$
Then cancel out common factors:
$(x-2)5=(x+4)(x-2)(4)+(x+4)(3)$
$5x-10=(x+4)(x-2)(4)+3x+12$
And if I multiple out the middle term I'll get a polynomial, which is unexpected so I'm not sure I'm on the right path here... am I?
$5x-10=4x^2-4x-8+3x+12$
$5x-10=4x^2-x+4$
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Yes, you are on the right path. Yes, you do get a quadratic. The quadratic that you should get is $10-6x-4x^2$ and its roots are $1$ and $-\frac52$.
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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.