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Different ways of computing $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$
Possible Duplicate:
On the sequence $x_{n+1} = \sqrt{c+x_n}$
I am wondering how many different solutions one can get to the following question:
Calculate $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$
Post your favorite solution please.
| Consider the sequence $$x_n = \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots \text{ n times}}}}$$ i.e. $$x_{n+1} = \sqrt{1+x_n}.$$
For instance, $x_1 = \sqrt{1}$, $x_2 = \sqrt{1+\sqrt{1}}$, $x_3 = \sqrt{1+\sqrt{1+\sqrt{1}}}$.
We will prove that the sequence is monotonically increasing and is bounded above by $\phi$ where $\phi = \dfrac{1+\sqrt{5}}{2}$ satisfies $\phi^2 = 1 + \phi$ and $\phi > 0$.
Boundedness
First we will prove that the sequence is bounded above by $\phi$. The proof follows from induction.
For $n=1$, $x_n = 1 = \dfrac{1+1}{2} < \dfrac{1+\sqrt{5}}{2} = \phi$. Hence, the claim is true for $n=1$.
Now assume it is true for some $k \in \mathbb{N}$.
$x_k < \phi \implies 1+x_k < 1 + \phi = \phi^2$. (Since $\phi = \dfrac{1+\sqrt{5}}{2}$ satisfies $\phi^2 = 1 + \phi$).
Hence, $x_{k+1} = \sqrt{1+x_k} < \phi$ and thereby the sequence is bounded above.
By a similar argument, we also have that $x_n$ is bounded below by $1$. Hence, we now have that $$1 \leq x_n < \phi.$$
Monotonically increasing
We will now prove that the sequence is monotonically increasing i.e. $x_n < x_{n+1}$.
Consider $x_{n+1}^2 - x_n^2$. We have $x_{n+1}^2 - x_n^2 = (1+x_n) - x_n^2 = \dfrac54 - \left( x_n - \dfrac12\right)^2$.
The quadratic equation $f(x) = 1 + x - x^2$ is positive whenever $x \in \left( \dfrac{1-\sqrt{5}}{2}, \dfrac{1+\sqrt{5}}{2}\right)$.
Since we proved earlier that $x_n \in \left[1,\dfrac{1+\sqrt{5}}{2}\right)$, we have that $x_{n+1}^2 - x_n^2 = 1 + x_n - x_n^2 > 0$.
Since $x_n$'s are positive, we have that $$x_{n+1} > x_n.$$
Hence, now combining both we get that $$1 \leq x_n < x_{n+1} < \phi$$
Hence, by monotone sequence theorem, (or equivalently by completeness of $\mathbb{R}$), the sequence $\{x_n\}_{n=1}^{\infty}$ converges.
Once we know that the sequence converges, we can then make use of the following limit rules.
*
*If $y_n \rightarrow L$, then $y_{n+1} \rightarrow L$.
*If $y_n \rightarrow L$, then $1+y_n \rightarrow 1+L$.
*If $z_n \rightarrow L$, and $z_n > 0$, then $L \geq 0$.
*If $z_n \rightarrow L$, and $z_n > 0$, then $\sqrt{z_n} \rightarrow \sqrt{L}$.
Let us now assume that $x_n \rightarrow L$, then making use of the above limit rules, we get that $$L = \sqrt{1+L}$$ Since $L \geq 0$ (because $x_n > 0)$, solving the quadratic we get that $$L = \dfrac{1+\sqrt{5}}2$$
| {
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"url": "https://math.stackexchange.com/questions/156131",
"timestamp": "2023-03-29T00:00:00",
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Putting ${n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ in a closed form As the title says, I'm trying to transform $\displaystyle{n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ into a closed form. My work:
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = \displaystyle\sum_{p=0}^{n}\binom{n}{p}\exp\left(\frac{p\cdot2i\pi}{5} \right)$
$\displaystyle=\binom{n}{0} + \binom{n}{1}\exp\left(\frac{1\cdot2i\pi}{5} \right) + \binom{n}{2}\exp\left(\frac{2\cdot2i\pi}{5} \right) + \binom{n}{3}\exp\left(\frac{3\cdot2i\pi}{5} \right) + \binom{n}{4}\exp\left(\frac{4\cdot2i\pi}{5} \right) + \binom{n}{5} + \cdots = \left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] + \exp\left(\frac{2i\pi}{5} \right)\left[\binom{n}{1} + \binom{n}{6} + \binom{n}{11} \right ] + \exp\left(\frac{4i\pi}{5} \right)\left[\binom{n}{2} + \binom{n}{7} + \binom{n}{12} \right ] + \exp\left(\frac{6i\pi}{5} \right)\left[\binom{n}{3} + \binom{n}{8} + \binom{n}{13} \right ] + \exp\left(\frac{8i\pi}{5} \right)\left[\binom{n}{4} + \binom{n}{9} + \binom{n}{14} \right ] + \cdots$
I'll recall $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = k$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = u$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = v$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = w$ and $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = z$. Thus
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = k + u\cdot\exp\frac{2i\pi}{5} + v\cdot\exp\frac{4i\pi}{5} + w\cdot \exp\frac{6i\pi}{5} + z\cdot\exp\frac{8i\pi}{5} = k + u\cdot\left (\cos\frac{2\pi}{5} + i\cdot\sin\frac{2\pi}{5} \right ) + v\cdot\left (\cos\frac{4\pi}{5} + i.\sin\frac{4\pi}{5} \right ) + w\cdot\left (\cos\frac{6\pi}{5} + i.\sin\frac{6\pi}{5} \right ) + z\cdot\left (\cos\frac{8\pi}{5} + i.\sin\frac{8\pi}{5} \right )$
Noting that $\cos\frac{2\pi}{5} = \cos\frac{8\pi}{5}$, $\cos\frac{4\pi}{5} = \cos\frac{6\pi}{5}$, $\sin\frac{2\pi}{5} = -\sin\frac{8\pi}{5}$ and $\sin\frac{4\pi}{5} = -\sin\frac{6\pi}{5}$:
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = k + \left(u + z\right)\cos\frac{2\pi}{5} + i\cdot\left(u - z \right)\sin\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5} + i\cdot\left(v - w \right)\sin\frac{4\pi}{5} = \left(k + \left(u + z\right)\cos\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5}\right) + i\cdot\left(\left(u - z \right)\sin\frac{2\pi}{5} + \cdot\left(v - w \right)\sin\frac{4\pi}{5} \right)$
But $\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = \left(2\cos\left(\frac{\pi}{5} \right)\cdot\exp\left(\frac{i\pi}{5}\right)\right)^n = \left(2^n\cos^n\left(\frac{\pi}{5} \right)\cdot\exp\left(\frac{ni\pi}{5}\right)\right) = \left(2^n\cos^n\frac{\pi}{5} \right)\left(\exp\left(\frac{ni\pi}{5}\right)\right) = \left(2^n\cos^n\frac{\pi}{5} \right)\left(\cos\frac{n\pi}{5} + i.\sin\frac{n\pi}{5} \right ) = \left(2^n\cos^n\frac{\pi}{5}\cos\frac{n\pi}{5} \right) + i\cdot \left(2^n\cos^n\frac{\pi}{5}\sin\frac{n\pi}{5} \right)$
So,
$\displaystyle k + \left(u + z\right)\cos\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5} = 2^n\cos^n\frac{\pi}{5}\cos\frac{n\pi}{5}$
and
$\displaystyle\left(u - z \right)\sin\frac{2\pi}{5} + \left(v - w \right)\sin\frac{4\pi}{5} = 2^n\cos^n\frac{\pi}{5}\sin\frac{n\pi}{5}$
and I'm stuck here. I noted that $k + u + v + w + z = 2^n$ but I couldn't isolate $k$. So, any help finishing this result will be fully appreciated. Thanks.
| As you know, $\sum_{j=0}^n {n \choose j} = 2^n$, and if $\omega$ is a primitive $5$'th root of unity
$\sum_{j=0}^n {n \choose j} \omega^j = (1 + \omega)^n$. Now $\sum_{i=0}^4 \omega^{ij} = 5$ if $j$ is divisible by $5$ and $0$ otherwise. So
$$\sum_{k} {n \choose {5k}} = \frac{1}{5} \sum_{i=0}^4 \sum_{j=0}^n {n \choose j} \omega^{ij} = \frac{1}{5} \sum_{i=0}^4 (1+\omega^i)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/156635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Limit finding of an indeterminate form: $\lim\limits_{x\to0} \frac{x^3}{\tan^3(2x)}$ Here is the limit I'm trying to find out:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)}$$
Since it is an indeterminate form, I simply applied l'Hopital's Rule and I ended up with:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)} = \lim_{x\rightarrow 0}\frac{6\cos^3(2x)}{48\cos^3(2x)} = \frac{6}{48} = 0.125$$
Unfortuntely, as far as I've tried, I haven't been able to solve this limit without using l'Hopital's Rule. Is it possibile to algebrically manipulate the equation so to have a determinate form?
| Another idea using $\,\,\displaystyle{\frac{\sin x}{x}\underset{x\to 0}{\longrightarrow}1\,,\,\cos kx\underset{x\to 0}\longrightarrow 1\,\,(k=\text{a constant})\,\,,\,\sin 2x=2\sin x\cos x}$:
$$\frac{x^3}{\tan^3 2x}=\frac{x^3}{\frac{\sin^32x}{\cos^32x}}=\cos^32x\frac{x^3}{\left(2\sin x\cos x\right)^3}=\frac{1}{8}\frac{\cos^32x}{\cos^3x}\left(\frac{x}{\sin x}\right)^3\underset{x\to 0}\longrightarrow \frac{1}{8}\cdot\frac{1}{1}\cdot 1^3=\frac{1}{8}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What other substitutions could I use to evaluate this integral? Consider the integral
$$ \int x^2\sqrt{2 + x} \, dx$$
I need to find the value of this integral, yet all its (seemingly) possible substitutions don't allow me to cancel appropriate terms. Here are three substitutions and their outcomes, all of which cover both terms in the integrand, as well as the composed function inside the root.
$\boxed{\text{Let }u = (2 + x)}$
$$u = 2 + x$$
$$du = 1 \space dx$$
$$dx = du$$
$$\int x^2 \sqrt{u} \space du$$
$\boxed{\text{Let } u = \sqrt{2+x}}$
$$ u = \sqrt{2 + x}$$
\begin{align*} du &= \frac{1}{2}(2 + x)^{-\frac{1}{2}} \space dx \\
&= \frac{1}{2\sqrt{2 + x}} \space dx
\end{align*}
$$ dx = 2\sqrt{2 + x} \space du$$
$$ 2\int ux^2\sqrt{2 + x} \space du $$
$\boxed{\text{Let }u = x^2}$
$$ u = x^2$$
$$ du = 2x \space dx$$
$$ dx = \frac{1}{2x} \space du$$
$$ \frac{1}{2} \int \frac{u\sqrt{2 + x}}{x} \space du$$
As you can see, none of these allow me to move forward into integrating with respect to $u$. What other substitutions can I use that would help me move on with this integral?
| You could even do this problem by parts actually.
Let $u = x^2$ and $dv = \sqrt{x + 2}$. So $du = x$ and $v = \dfrac{2}{3}(x+2)^{3/2}$. So
$$
\int x^2\sqrt{x + 2} dx = \dfrac{2}{3}x^2(x+2)^{3/2} - \dfrac{2}{3}\int x (x+2)^{3/2}
$$
Now let $s = x$ and $dt = (x+2)^{3/2}$. So $ds = 1$ and $t = \dfrac{2}{5}(x+2)^{5/2}$ Then
$$
\int x (x+2)^{3/2} = \dfrac{2}{5}x(x+2)^{5/2} - \dfrac{2}{5} \int (x+2)^{5/2} = \dfrac{2}{5}x(x+2)^{5/2} - \dfrac{4}{35} (x+2)^{7/2}.
$$
So putting it all together we get
$$
\int x^2\sqrt{x + 2} dx = \dfrac{2}{3}x^2(x+2)^{3/2} - \dfrac{4}{15} x(x+2)^{5/2} + \dfrac{8}{105} (x+2)^{7/2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$ I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
| If you don't mind I give the details of @Vasea Igor's answer.
Define $f(a,b):=\frac{a^2}{b-1}+\frac{b^2}{a-1}$, where $a,b>1$. Then
$$
\frac{\partial}{\partial a}f(a,b)=\frac{2a(a-1)^2-(b-1)b^2}{(b-1)(a-1)^2}=\frac{2a^3-4a^2+2a-b^3+b^2}{(b-1)(a-1)^2},
$$
$$
\frac{\partial}{\partial b}f(a,b)=-\frac{a^2}{(b-1)^2}+\frac{2b}{a-1}=-\frac{a^3-a^2-2b^3+4b^2-2b}{(b-1)^2(a-1)}.
$$
The real solutions of the system
$$
2a^3-4a^2+2a-b^3+b^2=0,
$$
$$
a^3-a^2-2b^3+4b^2-2b=0
$$
are
$$
(a,b)={(2,2),(0,0),(1,0),(0,1),(1,1)}.
$$
Since $a,b>1$ the only candidate is $(a,b)=(2,2)$.
Now we compute $H(a,b)$ the Hesse Matrix of $f(a,b)$ at $a=2, b=2$.
$$
H(a,b)=\left[ \begin {array}{cc} \frac{2}{b-1}+2\,{\frac {{b}
^{2}}{ \left( a-1 \right) ^{3}}}&-2\,{\frac {a}{ \left( b-1 \right) ^{
2}}}-2\,{\frac {b}{ \left( a-1 \right) ^{2}}}\\ -2\,
{\frac {a}{ \left( b-1 \right) ^{2}}}-2\,{\frac {b}{ \left( a-1
\right) ^{2}}}&2\,{\frac {{a}^{2}}{ \left( b-1 \right) ^{3}}}+\frac{2}{a-1}\end {array} \right] .
$$
Now we get
$$
H(2,2)=\left[ \begin {array}{cc} 10&-8\\ -8&10\end {array}
\right] .
$$
The eigenvalues of $H(2,2)$ are $18,2>0$. It means $H(2,2)$ is positive definite, which implies the function $f$ has a local minimum (and at the same time, global minimum) at $a=2, b=2$ which is $f(2,2)=8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Limit of $ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}},m\rightarrow\infty$ I would like to show that $$ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}\rightarrow_{m\rightarrow \infty}\frac{\pi}{2}$$
Using integrals:
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} \leq \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}} \leq m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}}+\frac{m}{(m+1)\sqrt{2m+1}}$$
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} = \frac{\pi}{2}-\arctan \left(1+\frac{1}{m} \right)=\frac{\pi}{4}+o(1)$$
The result I get is:
$$ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}\rightarrow_{m\rightarrow \infty}\frac{\pi}{4}$$
Where did I go wrong?
| The initial inequality appears to be set up correctly so the problem may lie with your evaluation of that integral. The result comes much quicker via Riemann Sums:
$$\sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}= \frac{1}{m} \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{(n/m)^2-1}} \to \int^{\infty}_1 \frac{1}{x\sqrt{x^2-1} } dx = \sec^{-1} x \bigg|^{\infty}_1= \frac{\pi}{2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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calculate generally the determinant of $A = a_{ij} = \begin{cases}a & i \neq j \\ 1 & i=j \end{cases}$ calculate generally the determinant of $A = a_{ij} = \begin{cases}a & i \neq j \\ 1 & i=j \end{cases} = \begin{pmatrix}
1 & a & a & · & a \\
· & · & · & · \\
a & a & a & · & 1 \\
\end{pmatrix}$
Any hints?
| First add all collumns to the first one and so $ \displaystyle{ C_1 \to \left((n-1)a+1, (n-1)a+1 , \cdots , (n-1)a+1 \right)^T }$
now the determinant $\det A=[(n-1)a+1] \cdot$
\begin{pmatrix}
1 & a & a & \cdots & a \\
1& 1& a & \cdots &a \\
· & · & · & · \\
1 & a & a & \cdots & 1 \\
\end{pmatrix}
subtracking now from each row the first row you get an upper triangular matrix with principal diagonal $\displaystyle{\operatorname{diag} (1, 1-a, 1-a, \cdots ,1-a)}$ so we get that
$$ \det A= [(n-1)a +1] (1-a)^{n-1}.$$
P.S Can someone fix the latex and write the coefficinent $[(n-1)a+1]$ in front of the determinant? Thank's!
| {
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How to solve this quadratic congruence equation How to solve $x^2+x+1\equiv 0\pmod {11}$ ?
I know that in some equations like $ax\equiv b\pmod d$ if $(a,d)=1$ then the equation has one and only one solution $x\equiv ba^{\phi(d)-1}\pmod d$. Any help will be appreciated. ;)
| You use the quadratic formula!
No, really. But you need to interpret the terms correctly: rather than "dividing" by $2a$ (here, $2$) you need to multiply by a number $r$ such that $2r\equiv 1\pmod{11}$ (namely, $r=6$). And rather than trying to find a square root, here $\sqrt{b^2-4ac} = \sqrt{1-4} = \sqrt{-3}$, you want to find integers $y$ such that $y^2\equiv -3\pmod{11}$. It may be impossible to do so, but if you can find them, then plugging them into the quadratic formula will give you a solution; and if you cannot find them, then there are no solutions.
Now, as it happens, $-3$ is not a square modulo $11$; so there are no solutions to $x^2+x+1\equiv 0\pmod{11}$. (You can find out if $-3$ is a square by using quadratic reciprocity: we have that $-1$ is not a square modulo $11$, since $11\equiv 3\pmod{4}$. And since both $3$ and $11$ are congruent to $3$ modulo $4$, we have
$$\left(\frac{-3}{11}\right) = \left(\frac{-1}{11}\right)\left(\frac{3}{11}\right) = -\left(-\left(\frac{11}{3}\right)\right) = \left(\frac{11}{3}\right) = \left(\frac{2}{3}\right) = -1,$$
so $-3$ is not a square modulo $11$).
(You can also verify that this is the case by plugging in $x=1,2,\ldots,10$ and seeing that none of them satisfy the equation).
On the other hand, if your polynomial were, say $x^2+x-1$, then the quadratic formula would say that the roots are
$$\frac{-1+\sqrt{1+4}}{2}\qquad\text{and}\qquad \frac{-1-\sqrt{1+4}}{2}.$$
Now, $5$ is a square modulo $11$: $4^2 = 16\equiv 5\pmod{11}$. So we can take $4$ as one of the square roots, and taking "multiplication by $6$" as being the same as "dividing by $2$" (since $2\times 6\equiv 1\pmod{11}$), we would get that the two roots are
$$\begin{align*}
\frac{-1+\sqrt{5}}{2} &= \left(-1+4\right)(6) = 18\equiv 7\pmod{11}\\
\frac{-1-\sqrt{5}}{2} &= \left(-1-4\right)(6) = -30\equiv 3\pmod{11}
\end{align*}$$
and indeed, $(7)^2 + 7 - 1 = 55\equiv 0\pmod{11}$ and $3^2+3-1 = 11\equiv 0\pmod{11}$.
We can definitely use this method when $2a$ is relatively prime to the modulus; if the modulus is not a prime, though, nor an odd prime power, then there may be more than $2$ square roots for any given number (or none). But for odd prime moduli, it works like a charm.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral as a limit of a sum How do I get
$$
\int_a^b \frac{1}{x}dx = \ln\left(\frac{b}{a}\right)$$
as a limit of sum. The constant width partition of the interval $(a,b)$ doesn't seem to work.
| The constant width partition works fine. No need for fancy partitions. To wit:
Let $x_k = a+\frac{k}{N}(b-a)$, with $k = 0,...,N$. We have, of course, $x_0=a$, $x_N = b$.
Using the Taylor series expansion of $x \mapsto \ln x$, and the mean value theorem, we obtain the bounds, for $x \geq y$:
$$ \ln y + \frac{x-y}{y} - \frac{(x-y)^2}{2 y^2} \leq \ln x \leq \ln y + \frac{x-y}{y}.$$
Letting $y=x_k$, $x=x_{k+1}$, we get:
$$\ln x_k + \frac{1}{N x_k} - \frac{1}{2 N^2 x_k^2} \leq \ln x_{k+1} \leq \ln x_k + \frac{1}{N x_k}.$$
Rearranging gives:
$$- \frac{1}{2 N^2 x_k^2} \leq \ln x_{k+1} - \ln x_k - \frac{1}{N x_k} \leq 0.$$
Now sum the inequality over $k = 0,...,N-1$, to get:
$$- \frac{1}{2 N^2} \sum_{k=0}^{N-1}\frac{1}{x_k^2} \leq \ln \frac{b}{a} - \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} \leq 0.$$
Since $x_k\geq a$, we obtain:
$$- \frac{1}{2 N a^2} \leq \ln \frac{b}{a} - \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} \leq 0.$$
Now let $N \to \infty$, to obtain the desired result,
$$ \lim_{N \to \infty} \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} = \int_a^b \frac{1}{t} dt = \ln \frac{b}{a}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Two algebra questions I have two questions that I need help with:
1) How many single digit even natural number solutions are there for the equation $A+B+C+D = 24$ such that $A+B > C+D$
A)20 B)11 C)16 D)24
2) Three positive real numbers $x,y,z$ are such that $x+y+Z = 1$. which of the following inequalities best discribe the relation between $XY,YZ,ZX$.
A) $xy+yz+zx > 1/3$
B) $xy+yz+zx <1/3$
C) $xy+yz+zx \le 1/3$
D) $xy+yz+zx \le 2/3$
| Brian's answer is fine. If we are allowed to use cauchy-schwarz inequality
then $$xy+yz+zx\le \sqrt{x^2+y^2+z^2}. \sqrt{x^2+y^2+z^2}=x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=1-2(xy+yz+zx)$$
Hence $xy+yz+zx\le \frac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Why do the endpoints of the Maclaurin series for arcsin converge? The series $$\sum_{n=0}^\infty {{-\frac {1} 2} \choose n} \frac{(-1)^n}{2n+1}$$ is an endpoint for the Maclaurin series for arcsin(x). (The other endpoint is just the negative of this one.) I played around with this a bit and turned it into three (potentially useful) forms :
$\sum_{n=0}^\infty \frac12\frac34\frac56\cdots\frac{2n-1}{2n} \frac{1}{2n+1}$
$\sum_{n=0}^\infty \left( 1-\frac 1 2 \right) \left( 1-\frac 1 4 \right)\cdots \left( 1-\frac 1 {2n} \right)\frac{1}{2n-1}$
$\sum_{n=0}^\infty \dfrac{(2n)!}{2^{2n}(n!)^2}\dfrac{1}{2n+1}$,
but I'm not sure where one could go from here.
| Here's another approach.
We write the shorthand $m!$ for $\Gamma(m+1)$, where $\Gamma$ is the gamma function.
Let $a_n = \displaystyle\frac{(-1)^n}{2n+1} {-\frac{1}{2} \choose n}$.
Then
$$\begin{eqnarray*}
\frac{a_{n+1}}{a_n} &=& -\frac{2n+1}{2n+3} \,
\frac{{-\frac{1}{2}\choose n+1}}{{-\frac{1}{2}\choose n}} \\
&=& -\frac{2n+1}{2n+3} \,
\frac{(-\frac{1}{2})!}{(n+1)!(-\frac{1}{2}-n-1)!}\,
\frac{n!(-\frac{1}{2}-n)!}{(-\frac{1}{2})!} \\
&=& -\frac{2n+1}{2n+3}\,
\frac{-\frac{1}{2}-n}{n+1} \\
&=& \frac{(2n+1)^2}{2(n+1)(2n+3)} \\
&=& 1-\frac{3}{2n} +O\left(\frac{1}{n^2}\right).
\end{eqnarray*}$$
Since $3/2>1$ the series converges by Raabe's test.
(Note that the $a_n$s are positive, so we don't bother taking the absolute value.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
All integer solutions for $x^4-y^4=15$ I'm trying to find all the integer solutions for $x^4-y^4=15$.
I know that the options are $x^2-y^2=5, x^2+y^2=3$, or $x^2-y^2=1, x^2+y^2=15$, or $x^2-y^2=15, x^2+y^2=1$, and the last one $x^2-y^2=3, x^2+y^2=5$.
Only the last one is valid. $x^2+y^2=15$ is not solvable since the primes which have residue $3$ modulo $4$ is not of an equal power.
One particular solution for $x^2-y^2=3, x^2+y^2=5$, is $x_0=2, y_o=1$. How do I get to an expression of a general solution for this system?
Thanks!
| $x^4-y^4=15\implies (x^2-y^2)(x^2+y^2)=15$. Factorization of $15=1.15$ or $3.5$. Case $1$, $(x^2-y^2)=1\implies (x+y)=1$ and $(x-y)=1$ which have only integer solutions $0,1$ but then $(x^2+y^2)\neq15$, therefore, factorization is $3.5$ which implies $(x^2+y^2)=5$ and $(x^2-y^2)=3$ which gives $x^2=4$ and $y^2=1\implies x=\pm2$ and $y=\pm1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/166070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Recurrence relation $C_n = n + 1 + \dfrac{2}{n}\sum\limits_{k=0}^{n-1}C_k$. A Discrete Mathematics book from which I'm self-studying ("Discrete Mathematics and Its Applications", by Kenneth Rosen) asks me to do the following:
Given the following recurrence relation:
$$C_n = n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k$$
The book asks me to show that the sequence $\{C_n\}$, with base case $C_0 = 0$, also satisfies the recurrence relation $nC_n=(n+1)C_{n-1}+2n$ for $n=1,2,\cdots$.
I tried to solve it by induction. For this, I wrote the second recurrence relation for $n+1$:
$$(n+1)C_{n+1} = (n+2)C_{n} + 2n + 2$$
Then, assuming that the first recurrence relation holds for $n$, I tried to substitute $C_n = n + 1 + \frac{2}{n}\sum\limits_{k=0}^{n-1}C_k$ in the above equation, to see if I obtain $C_{n+1} = n + 2 + \frac{2}{n+1}\sum\limits_{k=0}^{n}C_k$.
$$\begin{align*}
(n+1)C_{n+1} &= (n+2)C_{n} + 2n + 2\\
(n+1)C_{n+1} &= (n+2)\left( n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k \right ) + 2n + 2\\
(n+1)C_{n+1} &= n(n+2) + n + 2 + \frac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k + 2n + 2\\
(n+1)C_{n+1} &= n^2 + 5n + 4 + \frac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k\\
(n+1)C_{n+1} &= (n+1)(n+4) + \dfrac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k\\
C_{n+1} &= n+4 + \dfrac{2(n+2)}{n(n+1)}\sum_{k=0}^{n-1}C_k
\end{align*}$$
From this point, I'm not sure how to proceed.
| You can get rid of the sum on the RHS by subtracting:
For $C_{n+1}$:
\begin{align*}
C_{n+1} &= n + 2 + \dfrac{2}{n+1}\sum_{k=0}^{n}C_k \\
(n + 1) C_{n+1} &= (n + 1)(n + 2) + 2 \sum_{k=0}^{n}C_k \tag{1}
\end{align*}
For $C_n$:
\begin{align*}
C_n &= n + 1 + \dfrac{2}{n}\sum_{k=0}^{n-1}C_k \\
n C_n &= n(n + 1) + 2 \sum_{k=0}^{n-1}C_k \tag{2}
\end{align*}
Subtract $(2)$ from $(1)$ to get:
$$
(n + 1) C_{n+1} - n C_{n} = 2n + 2 + 2 C_{n}
$$
Simplify to get the recurrence you want:
$$
(n + 1) C_{n+1} = (n + 2)C_{n} + 2n + 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/168214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$ Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
| By AM>GM
$$
\frac{a+b+c}{3}=1 \Rightarrow abc\le 1 \\
\Rightarrow \frac{1}{2ab^2+1} = \frac{1}{2abc\frac{b}{c}+1}\ge\frac{1}{2\frac{b}{c}+1}=\frac{c}{2b+c} \\
S =
\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge\frac{c}{2b+c}+\frac{a}{2c+a}+\frac{b}{2a+b}
$$
$$
\begin{align}
3S-3 & \ge \frac{3c-2b-c}{2b+c}+\frac{3a-2c-a}{2c+a}+\frac{3b-2a-b}{2a+b}\\
& = 2\left(\frac{c-b}{2b+c}+\frac{a-c}{2c+a}+\frac{b-a}{2a+b}\right) \\
& = \frac{2}{D}\left(3ab^2+3bc^2+3ca^2-9abc\right)\\
& = \frac{6abc}{D}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}-3\right) \\
& \ge 0
\end{align}
$$
where for clarity we simply write $D$ for the positive denominator, and the last inequality is again by AM>GM
$$
1=\left(\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}\right)^{1/3}\le\frac{1}{3}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right)
$$
Finally $3S-3\ge 0 \Rightarrow S\ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/168704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Turn fractions into $\mathbb Z_7$ elements I had to perform a division between two polinomials $2x^2+3x+4$ and $3x+4$, my book suggests to do this operation without worrying about the modulo. So my result is $(3x+4)(\frac{2}{3}x+\frac{1}{9})+\frac{32}{9}$. Unfortunately my book fails to explain how should I perform the conversion from fractions to $\mathbb Z_7$ elements, like it's kind of obvious, it only reports $\frac{1}{3}=5$, $\frac{2}{3} = 10 = 3$, $\frac{32}{9}=2$. Will you please break it down for me?
| Deal first with $1/3$. In the modulo $7$ world, this is a number $x$ such that $3x\equiv 1\pmod{7}$.
How do we find such an $x$? Well $7$ is small, so try all the possibilities for $x$. These are $1$ (well, not really!), $2$, $3$, $4$, $5$, and $6$. Try $x=2$. Does it work? Let's check. We have $(3)(2)\equiv 6\pmod{7}$, doesn't work.
Try $x=3$. We have $(3)(3)=9\equiv 2\pmod{7}$, doesn't work. Try $x=4$. We have $(3)(4)=12\equiv 5\pmod{7}$, doesn't work. Try $x=5$. We have $(3)(5)=15\equiv 1\pmod{7}$. Bingo!
By the way, our first trial almost worked. We have $(3)(2)\equiv 6\equiv -1\pmod{7}$. So if we replace $2$ by $-2$, the product will be congruent to $1$. Thus $x\equiv -2\equiv 5\pmod{7}$.
Now for the $2/3$, this is easy, it is twice $1/3$. Modulo $7$, we get $10$, which is congruent to $3$.
We can get to $32/9$ in many ways. For example, to get $1/9$, square our $1/3$ found above. We get $25$, which is congruent to $4$ modulo $7$. But $32$ is congruent to $4$, so for $32/9$ we will want $(4)(4)$, which is congruent to $2$ modulo $7$.
There are much easier ways. For example, $32$ is congruent to $4$, and $9$ is congruent to $2$, so the fraction should be congruent to $\dfrac{4}{2}$, which is $1$.
Remark: The "guess and test" approach described above is only practical for relatively small primes. For larger primes, one needs a more efficient approach. You will probably learn one soon, the Extended Euclidean Algorithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Pythagorean Theorem for imaginary numbers If we let one leg be real-valued and the other leg equal $bi$ then the Pythagorean Theorem changes to $a^2-b^2=c^2$ which results in some kooky numbers.
For what reason does this not make sense? Does the Theorem only work on real numbers? Why not imaginary?
| In the vein of @AndréNicolas' comment that "ideas like this can be useful" ...
Heron's formula for the area, $W$, of a (non-obtuse) triangle in terms of the lengths of its sides is equivalent to the Pythagorean Theorem for Right-Corner Tetrahedra:
$$W^2 = X^2 + Y^2 + Z^2$$
where $W$ is the "hypotenuse-face" opposite three mutually-perpendicular edges that form right triangles with areas $X$, $Y$, $Z$.
Specifically, if we position the right corner at the origin in 3-space, choose the other vertices at $(x,0,0)$, $(0,y,0)$, $(0,0,z)$, and write
$$\begin{align}
a^2 = y^2 + z^2 \qquad b^2 &= z^2 + x^2 \qquad c^2 = x^2 + y^2 \\
X = \frac{1}{2} y z \qquad Y &= \frac{1}{2} z x \qquad Z = \frac{1}{2} x y
\end{align}$$
then
$$x^2 = \frac{1}{2}\left(-a^2 + b^2 + c^2\right) \qquad y^2 = \frac{1}{2}\left(a^2-b^2+c^2\right) \qquad z^2 = \frac{1}{2}\left(a^2+b^2-c^2\right)$$
whence
$$\begin{align}
W^2 = X^2 + Y^2 + Z^2 &= \frac{1}{4}\left(y^2 z^2+z^2x^2+x^2y^2\right)\\
&=\frac{1}{16}\left(-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2\right) \\
&=\frac{1}{16}\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)
\end{align}$$
which is Heron's Formula for $W$ in terms of side-lengths $a$, $b$, $c$.
Note that I wrote the equivalence is for "non-obtuse" triangles. This is because "hypotenuse-face" $W$ cannot have an obtuse angle; to see this, write $\theta$ for the angle opposite side $a$, so that
$$\cos \theta =\frac{-a^2+b^2+c^2}{2b c} = \frac{x^2}{bc}$$
This value is non-negative (and, hence, $\theta$ is non-obtuse) for any edge-length $x$ ... well, any real edge-length $x$. However, if we allow $x$ to be an imaginary length, then $\theta$ can be obtuse, and then Heron's Formula can be seen to apply to all triangles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Proof of $\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$ Numerically it seems to be true that
$$
\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}.
$$
Any ideas how to prove this?
| Here is an another approach, which I show only a heuristic calculation:
$$\begin{align*}
\int_{0}^{\infty} \frac{\sin x}{\sqrt{x}} \; dx
&= \int_{0}^{\infty} \left( \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} t^{-1/2} e^{-xt} \; dt \right) \sin x \; dx \\
&\stackrel{\ast}{=} \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} t^{-1/2} \int_{0}^{\infty} e^{-xt} \sin x \; dx \; dt \\
&= \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} \frac{t^{-1/2}}{1+t^2} \; dt \\
&= \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} \frac{2du}{1+u^4}.\qquad(t = u^2)
\end{align*}$$
Now it is not hard to show that
$$ \int_{0}^{\infty} \frac{2du}{1+u^4} = \frac{\pi}{\sqrt{2}}.$$
Indeed, you may use the equality
$$ \begin{align*}
\frac{2 du}{1+u^4}
&= \frac{2u^{-2} \; du}{u^2 + u^{-2}}
= \frac{1 + u^{-2} \; du}{u^2 + u^{-2}} - \frac{1 - u^{-2} \; du}{u^2 + u^{-2}}\\
&= \frac{d\left(u - u^{-1}\right)}{\left(u - u^{-1}\right)^2 + 2} - \frac{d\left(u + u^{-1}\right)}{\left(u + u^{-1}\right)^2 - 2}
\end{align*}$$
and hence deduce that
$$ \begin{align*}
\int_{0}^{\infty} \frac{2 du}{1+u^4}
&= \int_{0}^{\infty} \frac{d\left(u - u^{-1}\right)}{\left(u - u^{-1}\right)^2 + 2} - \int_{0}^{\infty} \frac{d\left(u + u^{-1}\right)}{\left(u + u^{-1}\right)^2 - 2}\\
&= \int_{-\infty}^{\infty} \frac{dv}{v^2 + 2} - \color{blue}{\int_{\infty}^{\infty} \frac{dw}{w^2 - 2}} \qquad \begin{pmatrix}v = u - u^{-1} \\ w = u + u^{-1}\end{pmatrix}\\
&= \frac{\pi}{\sqrt{2}} + 0.
\end{align*}$$
as claimed, where the blue-colored integration is taken along the curve starting from $+\infty$ to $2$, and then turning back to $+\infty$, which makes it cancel out. Therefore we obtain the desired result.
The problem in this calculation is that the starred equality is almost unable to be justified by any simple means.
| {
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"url": "https://math.stackexchange.com/questions/171970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 4
} |
Evaluating $\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$ How do I integrate this expression:
$$\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$$.I got this in a book.I do not know how to evaluate integrals of this type.
| Use the famous equation
$$ a\cos x + b\sin x = \pm \sqrt{a^2 + b^2}\cos\left(x - \arctan(b/a)\right)$$
and let $\alpha = \arctan(b/a)$
to get
$$
\begin{aligned}
\mathcal{J}&=\int\dfrac{\ell \sin x + m\cos x}{(a^2 + b^2)\cos^2\left(x - \alpha\right)}\,\mathrm{d}x\\
&=\frac{1}{a^2 + b^2}\int\dfrac{\ell\sin(x - \alpha+\alpha) + m\cos(x-\alpha + \alpha)}{\cos^2\left(x-\alpha\right)}\,\mathrm{d}x\\
&=\frac{1}{a^2+b^2}\int\dfrac{\ell\left[\sin(x-\alpha)\cos\alpha + \sin\alpha\cos(x-\alpha)\right]+m\left[\cos(x-\alpha)\cos\alpha-\sin(x-\alpha)\sin\alpha\right]}{\cos^2\left(x-\alpha\right)}\,\mathrm{d}x\\
&=\frac{1}{a^2+b^2}\int\frac{\sin(x-\alpha)\left(\ell\cos\alpha-m\sin\alpha\right) + \cos(x-\alpha)\left(\ell\sin\alpha+m\cos\alpha\right)}{\cos^2\left(x-\alpha\right)}\,\mathrm{d}x\\
&=\dfrac{\ell\cos\alpha-m\sin\alpha}{a^2+b^2}\int\sec(x-\alpha)\tan(x-\alpha)\,\mathrm{d}x+\dfrac{\ell\sin\alpha+m\cos\alpha}{a^2+b^2}\int\sec(x-\alpha)\,\mathrm{d}x\\
&=\dfrac{\ell\cos\alpha-m\sin\alpha}{a^2+b^2}\sec(x-\alpha) + \dfrac{\ell\sin\alpha+m\cos\alpha}{a^2+b^2}\ln\left|\sec(x-\alpha)+\tan(x-\alpha)\right|+C.
\end{aligned}
$$
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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About the Legendre differential equation Consider the Legendre differential equation $$ (1-x^2) y'' - 2xy' + n(n+1)y = 0 $$ Then its solution is given by $$ y = c_1 P_n (x) + \text{an infinite series} $$ In fact $y = c_1 P_n (x) + c_2 Q_n (x) $ where $P_n$ is Legendre polynomials and $Q_n $ is Legendre function of the second kind. Here I want to prove that 'an infinite series' above can be written by $c_2 Q_n (x)$ for some constant $c_2$.
| The infinite series you are talking about stems from the recurrence relation that can be derived when solving the Legendre differential equation: plugging in
$$y = \sum_{j=0}^{\infty}a_jx^j $$
into
$$[(1-x^2)y']' + n(n+1)y = 0$$
you get
$$\frac{a_{k+2}}{a_k} = \frac{k(k+1) - n(n+1)}{(k+1)(k+2)}$$
Obviously, this leads to separate solutions with all even or all odd powers of $x$. When $n$ is even, then starting with $a_0 \not= 0$, the series terminates and you get (a multiple of) $P_n$. The same holds for odd $n$ starting with $a_1 \not= 0$.
But when n is even and we start with $a_1 \not= 0$, then the series does not terminate. For instance, for $n=0$, we get
$$\frac{a_{k+2}}{a_k} = \frac{k(k+1)}{(k+1)(k+2)} = \frac{k}{k+2}$$ which clearly leads to (taking $a_1=1$)
$$y= \sum_{k=0}^{\infty}\frac{1}{2k+1}x^{2k+1}$$
Now, as
$$ln(1+x) = \sum_{j=0}^{\infty}\frac{(-1)^{j+1}}{j}x^j$$
it is easy to derive
\begin{equation}
\begin{split}
ln(\frac{1+x}{1-x}) &= ln(1+x) - ln(1-x)
\\ &=\sum_{j=0}^{\infty}\frac{(-1)^{j+1}}{j}x^j + \sum_{j=0}^{\infty}\frac{x^j}{j}
\\ &=\sum_{j=0}^{\infty}\frac{(-1)^{j+1} + 1}{j}x^j
\\ &= 2x + 2\frac{x^3}{3} + 2\frac{x^5}{5} + ...
\end{split}
\end{equation}
which is twice what we are looking for. So
$$Q_0(x) = \frac{1}{2} ln(\frac{1+x}{1-x})$$
Similarly,
$$Q_1(x) = \frac{x}{2} ln(\frac{1+x}{1-x}) - 1$$
And the rest of the $Q_n$ can then be determined by using the recurrence relation
$$(n+1)Q_{n+1}(x) = (2n+1)xQ_n(x) - nQ_{n-1}(x)$$
| {
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"url": "https://math.stackexchange.com/questions/176357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine whether the following series converges or diverges I'm stuck on this problem:
$$1 + {1\cdot2\over1\cdot3} + {1\cdot2\cdot3\over1\cdot3\cdot5}+ {1\cdot2\cdot3\cdot4\over 1\cdot3\cdot5\cdot7} +\cdots$$
I've simplified the numerators $n!$ but can't figure out how to represent the denominators.
How do I go about solving this?
| $$ 1 + {1\cdot2\over1\cdot 2\cdot3}2^1{(1)} + {1\cdot2\cdot3\over1\cdot2\cdot3\cdot4\cdot5}2^2(1\cdot2)+ {1\cdot2\cdot3\cdot4\over 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}2^3(1\cdot2\cdot3) +\cdots
= \sum_{n=1}^{\infty}{n!(n-1)!2^{n-1} \over (2n-1)!}$$
It converges by ratio test
$$ \lim_{n\rightarrow \infty} {(n+1)!n!2^{n}\over (2n+1)!}\times {(2n-1)!\over n!(n-1)! 2^{n-1}} = \lim_{n\rightarrow \infty} {2(n+1)n \over (2n+1)(2n)} = {1\over 2} < 1$$
For the value of convergence
$$1\cdot 3 \cdot 5 \cdots n=(2n-1)!!=\frac{2^n}{\sqrt{\pi}}\Gamma(n+\frac12)$$
The sum can be represented as
$$ \sum_{n=1}^{\infty}{n! \over (2n-1)!!} = \sum_{n=1}^{\infty}{n! \sqrt \pi\over 2^n \Gamma(n + {1 \over 2})} = \sum_{n=1}^{\infty}{n! \sqrt \pi\over 2^n \left ( (2n)! \sqrt\pi \over n! 4^n\right )} = \sum_{n=1}^\infty {2^n (n!)^2 \over (2n)!}$$
We have the generating function
$$ \frac{ 4\,\left(\,{\sqrt{x+4}}-{\sqrt{x}}\,{\rm{arcsinh}}(\frac{{\sqrt{x}}}{2})\right) } { \sqrt{(x+4)^3} } =1-\frac{x}{2}+\frac{x^2}{6}-\frac{x^3}{20}+\frac{x^4}{70}-\frac{x^5}{252}+\frac{x^6}{924}-...$$
Choosing $x= -2$
$$ {4 \left ( \sqrt 2 + {\pi \over 2 \sqrt 2}\right ) \over 2 \sqrt 2} = 1 + \sum_{n=1}^\infty {2^n (n!)^2 \over (2n)!} $$
$$ \sum_{n=1}^\infty {2^n (n!)^2 \over (2n)!} = 1 + {\pi \over 2}$$
Reference #1
Reference #2
| {
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"url": "https://math.stackexchange.com/questions/176743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Quadratic equation with absolute value Prepping for the GMAT, I came across the following question:
What is the product of all solutions of:
$$x^2 - 4x + 6 = 3 - |x - 1|?$$
First, I set up two equations, ie:
$$x^2 - 4x + 6 = 3 - (x - 1),$$
and
$$x^2 - 4x + 6 = 3 - (-1) \times (x - 1).$$
These factor down to $3$ solutions: $1$, $2$ and $4$. And $8$ is correct solution in the back of the prep book.
However, when plugging $4$ back into the original equation, it reduces to $6 = 3$, so $4$ does not seem to be a solution. Also, when graphing both, they only intersect at $1$ and $2$.
What part of my process (and seemingly the practice books process) is wrong?
| The solutions to
$$x^2-4x+6 = 3-(x-1)$$
are only valid when $x-1 \ge 0$, i.e. when $x \ge 1$. Likewise, the solutions to
$$x^2-4x+6 = 3+(x-1)$$
are only valid when $x-1 \le 0$, i.e. when $x \le 1$.
Solving the first equation gives $x=1,2$, both of which are valid. Solving the second gives $x=1,4$. Notice that in this latter case $4$ is not valid since $4 \nleq 1$, and so the only solutions to $x^2-4x+6=3-\left|x-1\right|$ are $x=1,2$. (The textbook is wrong!)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate $\frac{1}{\sqrt{1+x^2}}$ using substitution? How you integrate $\frac{1}{\sqrt{1+x^2}}$ using following substitution? $1+x^2=t$ $\Rightarrow$ $x=\sqrt{t-1} \Rightarrow dx = \frac{dt}{2\sqrt{t-1}}dt$... Now I'm stuck. I don't know how to proceed using substitution rule.
| $$A=\int\frac{1}{\sqrt[]{1+x^2}}$$
Let, $x = \tan\theta$
$dx = \sec^{2}\theta{d\theta}$
substitute, $x$, $dx$
$$A=\int\left(\frac{1}{\sec\theta}\right){\sec^{2}\theta{d\theta}}$$
$$A=\int{\sec\theta{d\theta}}$$
$$A=\int{\sec\theta\left(\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$
$$A=\int{\left(\frac{\sec^2\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$
Let, $(\sec\theta + \tan\theta) = u$
$(\sec^2\theta + \sec\theta\tan\theta)d\theta = du$
$$A=\int\frac{du}{u}$$
$$A=\ln{u}+c$$
$$A=\ln{\vert\sec\theta + \tan\theta\vert}+c$$
$$A=\ln{\vert\sqrt[]{1+\tan^2\theta} + \tan\theta\vert}+c$$
$A=\ln{\vert\sqrt[]{1+x^2} + x\vert}+c$, where $c$ is a constant
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/179129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Evaluating $\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$ using $\frac{dy}{dx}=\frac{2}{(x+3)^2}$ where $y=\frac{x+1}{x+3}$ Find derivative of $$y= \frac{ax+b}{cx+d}$$
I found it to be $$\frac{dy}{dx}=\frac{a}{cx+d}-\frac{c(ax+b)}{(cx+d)^2}$$
Use it to evaluate:
$$\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$$
I figured that here $y=\frac{x+1}{x+3}$ and $$\frac{dy}{dx}=\frac{1}{x+3}-\frac{(x+1)}{(x+3)^2}$$
and using the technique I learned from my last question I did this:
$$\frac{dy}{dx}=\frac{(x+3)}{(x+3)^2}-\frac{(x+1)}{(x+3)^2}=\frac{2}{(x+3)^2}$$
which I could then substitute back, having changed the limits by substituting $1$ into $y$ and then $0$ into $y$:
$$y|_{x=1}=\frac{x+1}{x+3}=\frac{1}{2}$$
$$y|_{x=0}=\frac{1}{3}=\frac{1}{3}$$
$$2\int_0^1{\frac{dy}{dx}}\ln(y)dx=2\int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$
This gives me:
$$2\int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$
$$=2\left[y(\ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2} = 2\left[\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-\frac{1}{2}\right)\right]-\frac{1}{3}\left[\ln\left(\frac{1}{3}\right)-\frac{1}{3}\right]\\$$
$$\ln\left(\frac{1}{2}\right)-1-\frac{2}{3}\ln\left(\frac{1}{3}\right)+\frac{2}{9}$$
The problem is I am supposed to end up with something else. Can anyone spot any issues with this?
EDIT: This is the answer I am supposed to be getting:
$$\frac{1}{6}\ln(3)-\frac{1}{4}\ln(2)-\frac{1}{12}$$
| The solution I have now is:
$$\frac{1}{2}\int_\frac{1}{3}^\frac{1}{2}\ln(y)dy=\frac{1}{2}\left[y(ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2}$$
$$=\frac{1}{2}\left[\frac{1}{2}\ln(\frac{1}{2})-\frac{1}{2}-\frac{1}{3}\ln(\frac{1}{3})+\frac{1}{3}\right]\\$$
$$=\frac{1}{2}\ln\left(\frac{1}{2}\right)-\frac{1}{4}-\frac{1}{3}\ln\left(\frac{1}{3}\right)+\frac{1}{6}$$
$$=-\frac{1}{4}\ln(2)+\frac{1}{6}\ln({3})-\frac{6}{24}+\frac{4}{24}$$
$$=-\frac{1}{4}\ln(2)+\frac{1}{6}\ln({3})-\frac{1}{12}$$
| {
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"url": "https://math.stackexchange.com/questions/179475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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No closed form for the partial sum of ${n\choose k}$ for $k \le K$? In Concrete Mathematics, the authors state that there is no closed form for
$$\sum_{k\le K}{n\choose k}.$$
This is stated shortly after the statement of (5.17) in section 5.1 (2nd edition of the book).
How do they know this is true?
| Proof of the answer by nczksv.
$$
\begin{array}{l}
f = \dfrac{(1+x)^n}{x^K \cdot (1-x)}
= \left(\dfrac{1+x}{x}\right)^n \cdot \dfrac{x^{n-K}}{1-x}\\
= \left(4^n+1\right)^n \cdot 4^n \cdot \dfrac{x^{n-K}}{4^n - 1}\\
= \dfrac{\left(4^n+1\right)^n \cdot 4^n }{(4^n - 1)(4^n)^{(n-K)}}
\end{array}
$$
Let $A = 4^n - 1$, mod = $A+1$. Consider both flooring function and mod function,
Note that
$$\forall r > 1, ~
\left\lfloor
\frac{(A+1)^r}{A} \right\rfloor
\equiv \left\lfloor (A+1)^{r-1} + \frac{(A+1)^{r-1}}{A} \right\rfloor
= \left\lfloor \frac{(A+1)^{r-1}}{A} \right\rfloor
= \frac{A+1}{A}$$
we have
$$
\begin{array}{l}
f = \dfrac{\left(A+2\right)^n }{A \cdot (A+1)^{n-K-1}} \\
= \dfrac{ \sum_{k=0}^{K} {n \choose k} (A+1)^{n-k} + \sum_{k=K+1}^{n} {n \choose k} (A+1)^{n-k} }{A \cdot (A+1)^{n-K-1}}\\
= \dfrac{ \sum_{k=0}^{K} {n \choose k} (A+1)^{K+1-k} }{A} +
\dfrac{ \sum_{k=K+1}^{n} {n \choose k} (A+1)^{K+1-k} }{A}\\
= \sum_{k=0}^{K} {n \choose k} + \dfrac{ \sum_{k=0}^{K} {n \choose k} }{A} +
\dfrac{ \sum_{k=K+1}^{n} {n \choose k} (A+1)^{K+1-k} }{A}
= \sum_{k=0}^{K} {n \choose k} + 0
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Use the identity $\cos(A-B) -\cos(A+B) = 2\sin(A)\sin(B)$ to prove another identity and evaluate a sum Use the identity $$\cos(A-B) -\cos(A+B) = 2\sin(A)\sin(B)$$ to prove that:
$$2\sin(j\theta)\sin(\frac{1}{2}\theta)=\cos((j-\frac{1}{2})(\theta))-\cos((j+\frac{1}{2})\theta).$$
This seemed almost too easy, so I am wondering if I am missing something?
I just let $A = j\theta$ and $B = \frac{1}{2}\theta$, and substituted it directly into the identity. Since the question has asked me to use the identity, I am assuming that we can use it directly.
Have I missed anything?
The next part of the question asks:
Deduce that
$$\sum^n_{j=1}\sin(j\theta) = \frac{\cos(\frac{1}{2}\theta)-\cos((n+\frac{1}{2})\theta)}{2\sin(\frac{1}{2}\theta)}, \text{ if $\theta$ is not a multiple of $2\pi$}$$
Writing this out in general terms I see that the middle terms all cancel out (i have omitted the denominator here):
$$\cos((j-\frac{1}{2})\theta)-\cos((j+\frac{1}{2})\theta) + \cos(((j+1)-\frac{1}{2})\theta)-\cos(((j+1)+\frac{1}{2})\theta) + \cos(((j+2)-\frac{1}{2})\theta)-\cos(((j+2)+\frac{1}{2})\theta)+\dotsb+ \cos((n-\frac{1}{2})\theta)-\cos((n+\frac{1}{2})\theta)$$
I notice that if $\theta$ is a multiple of $2\pi$, then I would lose the 1st term, thus $\theta$ cannot be a multiple of $2\pi$.
Can anyone give me some direction to make this more mathematically rigorous? I feel like I am just a few steps away from answering this correctly.
Thanks in advance!
| Your proof of the second identity is correct. If $\sin \left( \frac{1}{2}\theta \right) \neq 0$ (what does this mean in terms of $\theta$?) this identity is equivalent to
$$\begin{equation*}
\sin \left( j\theta \right) =\frac{\cos \left( \left( j-\frac{1}{2}\right)
\theta \right) -\cos \left( \left( j+\frac{1}{2}\right) \theta \right) }{
2\sin \left( \frac{1}{2}\theta \right) }.
\end{equation*}$$
The first terms of the numerators of the sum, which you didn't wrote in your proof, are $$\cos \left( \frac{1}{2} \theta \right) -\cos \left(
\frac{3}{2} \theta \right) +\cos \left( \frac{3}{2}
\theta \right) -\cos \left( \frac{5}{2} \theta\right) +\cdots
$$
Evaluation of the sum, which is a telescoping sum, as noticed by André Nicolas
\begin{eqnarray*}
\sum_{j=1}^{n}\sin \left( j\theta \right) &=&\sum_{j=1}^{n}\frac{\cos
\left( \left( j-\frac{1}{2}\right) \theta \right) -\cos \left( \left( j+
\frac{1}{2}\right) \theta \right) }{2\sin \left( \frac{1}{2}\theta \right) }
\\
&=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) }\sum_{j=1}^{n}\cos
\left( \left( j-\frac{1}{2}\right) \theta \right) -\cos \left( \left( j+
\frac{1}{2}\right) \theta \right) \\
&=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) }
\sum_{j=1}^{n}a_{j}-a_{j+1},\qquad a_{j}=\cos \left( \left( j-\frac{1}{2}
\right) \theta \right) \\
&=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) }\left(
a_{1}-a_{n+1}\right) ,\qquad \text{See below} \\
&=&\cdots
\end{eqnarray*}
For the telescoping sum $\sum_{j=1}^{n}a_{j}-a_{j+1}$ we have
\begin{eqnarray*}
\sum_{j=1}^{n}a_{j}-a_{j+1} &=&\left( a_{1}-a_{2}\right) +\left(
a_{2}-a_{3}\right) +\cdots +\left( a_{n-1}-a_{n}\right) +\left( a_{n}-a_{n+1}\right) \\
&=&a_{1}-a_{n+1}.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality $a+b+c \geqslant abc +2$ Assuming $a,b,c \in (0, \infty)$, we need to prove that:
$$a+b+c \geqslant a b c+2 \quad \text{if} \quad ab+bc+ca=3$$
Can you give me an idea, please? This inequality seem to be known, but I didn't manage to solve it.
| Solve for $c$ from the constraint:
$$
c = \frac{3- a b}{a+b}
$$
and substitute back into the inequality:
$$
a + b + \frac{3-a b}{a+b} \geqslant 2 + \frac{a b(3-a b)}{a+b} \implies \frac{3 + a^2 b^2 + a^2 + b^2 -2 a b -2a -2b+3}{a+b} \geqslant 0
$$
Multiply both sides by $(a+b)$:
$$
3 + a^2 b^2 + a^2 + b^2 -2 a b -2a -2b+3 \geqslant 0
$$
The left-hand-side can be reduced to the shifted sum of squares:
$$
(a b-2)^2 + (a+b-1)^2 - 2 \geqslant 0
$$
It is easy to see that the left-hand-side attains its minimum at $a=b=1$, where the inequality is saturated.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Congruence inequality Given $n>2$, by calculation or otherwise deduce that $5^{2^{n-3}} \neq -1 \pmod {2^n}$
Note:The problem arose when I tried to deduce $\langle5\rangle \cap \langle2^n-1\rangle=\{1\}$ in the group $\mathbb{Z}^{*}_{2^n}$, I have showed $\operatorname{ord}(5)=2^{n-2}$
| We show by induction that if $n \ge 3$, then $5^{2^{n-3}}\equiv 1+2^{n-1}\pmod{2^n}$. And it is clear that $1+2^{n-1}\not\equiv -1 \pmod{2^n}$ if $n \ge 3$.
The result holds when $n=3$. Now we do the induction step. Suppose that we know that for a certain $k$, we have $5^{2^{k-3}}\equiv 1+2^{k-1}\pmod{2^k}$. We show that
$5^{2^{k-2}}\equiv 1+2^{k}\pmod{2^{k+1}}$.
By assumption, $5^{2^{k-3}}=1+2^{k-1} +t2^k$ for some integer $t$. Square both sides, and simplify modulo $2^{k+1}$. We get
$$5^{2^{k-2}}\equiv (1+2^{k-1})^2=1+2^k+2^{2k-2}\pmod{2^{k+1}}.$$
But $2^{2k-2}$ is divisible by $2^{k+1}$, since $2k-2 \ge k+1$ when $k \ge 3$. The result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Writing 1/3 as a sum of other numbers Is it possible to write $0.3333(3)=\frac{1}{3}$ as sum of $\frac{1}{4} + \cdots + \cdots\frac{1}{512} + \cdots$ so that denominator is a power of $2$ and always different?
As far as I can prove, it should be an infinite series, but I can be wrong.
In case if it can't be written using pluses only, minuses are allowed as well.
For example, $\frac{1}{2}$ is $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$
So, what about $\frac{1}{3}$?
| More generally we can do this for any fraction $\frac{m}{n}$. We know that the sum of the infinite geometric series
$$ S=a+ar+ar^2+ar^3+\dots=\frac{a}{1-r} $$
if $ |r| < 1 $. So we could take $r=-\frac{1}{n-1}$ and $a=\frac{m}{n-1}$ giving
$$ S=\frac{\frac{m}{n-1}}{1+\frac{1}{n-1}}=\frac{\frac{m}{n-1}}{\frac{n}{n-1}}=\frac{m}{n} $$
We can in fact be very general with both $r$ and $a$. Just suppose $r=\frac{s}{t}$ (with $|s|>|t|$ to ensure $|r|<1$), and let $a=1$ for now then
$$ S=\frac{1}{1-\frac{s}{t}}=\frac{t}{t-s} $$
So if instead we take $a=\frac{m(t-s)}{nt}$ we have again that $S=\frac{m}{n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that $ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $ I would like to show that:
$$ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $$
Using AM-GM inequality:
$$ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3\frac{(a+2b)(b+2c)(c+2a)}{(a+2c)(b+2a)(c+2b)} $$
It suffices to show that:
$$ \frac{(a+2b)(b+2c)(c+2a)}{(a+2c)(b+2a)(c+2b)} \geq1 $$
$$ \Longleftrightarrow ab^2+bc^2+ca^2-(a^2b+b^2c+c^2a)\geq0$$
$$ \Longleftrightarrow x+y+z\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
$$ xyz=1$$
($x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$)
How can I prove this last inequality?
Is there any simpler proof?
| Or maybe just do it by Holder's Inequality?
$$\left(\sum_{cyc} \left(\frac{a+2b}{a+2c} \right)^3 \right) \left(\sum_{cyc}(a+2b)(a+2c)\right)\left(\sum_{cyc}(a+2c)\right)^2 \ge \left(\sum_{cyc}(a+2b)\right)^4$$
As $\displaystyle \sum_{cyc}(a+2c) = 3(a+b+c) = \sum_{cyc}(a+2b)$, and $$\displaystyle \sum_{cyc}(a+2b)(a+2c) = a^2 + b^2 + c^2 + 8(ab + bc + ca) = (a+b+c)^2 + 6(ab+bc+ca) \le 3(a+b+c)^2$$ because $3(ab + bc+ca) \le (a+b+c)^2$, hence we have that $$\sum_{cyc} \left(\frac{a+2b}{a+2c} \right)^3 \ge 3$$ which finishes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \frac{1}{2}\cot^{-1}\frac{2\sqrt[3]{4}+1}{\sqrt{3}}+\frac{1}{3}\tan^{-1}\frac{\sqrt[3]{4}+1}{\sqrt{3}}=\dfrac{\pi}{6}. $ Prove that $$ \frac{1}{2}\cot^{-1}\frac{2\sqrt[3]{4}+1}{\sqrt{3}}+\frac{1}{3}\tan^{-1}\frac{\sqrt[3]{4}+1}{\sqrt{3}}=\dfrac{\pi}{6}. $$
| Let $y^3=4$ and $x^2=3$
So, we need to prove $ \frac{1}{2}\cot^{-1}\frac{2y+1}{x}+\frac{1}{3}\tan^{-1}\frac{y+1}{x}=\dfrac{\pi}{6}. $
or $3\cot^{-1}\frac{2y+1}{x}+2\tan^{-1}\frac{y+1}{x}=\pi$
or $3(\frac{\pi}{2}-\tan^{-1}\frac{2y+1}{x})+2\tan^{-1}\frac{y+1}{x}=\pi$
as $\tan^{-1}a+\cot^{-1}a=\frac{\pi}{2}$
or we need to prove, $3\tan^{-1}\frac{2y+1}{x}-2\tan^{-1}\frac{y+1}{x}=\frac{\pi}{2}$
As $2\tan^{-1}a=\tan^{-1}\frac{2a}{1-a^2}$
and $3\tan^{-1}a=\tan^{-1}\frac{3a-a^3}{1-3a^2}$
$2\tan^{-1}\frac{y+1}{x}=\tan^{-1}\frac{2x(y+1)}{x^2-(y+1)^2}$
$=\tan^{-1}\frac{-2x(y+1)}{y^2+2y-2}$ as $x^2=3$
$3\tan^{-1}\frac{2y+1}{x}=\tan^{-1}\frac{3x^2(2y+1)-(2y+1)^3}{x(x^2-3(2y+1)^2)}$
$=\tan^{-1}\frac{y^2-y+2}{x(y^2+y)}$ as $x^2=3$ and $y^3=4$
or $\tan^{-1}\frac{y^2-y+2}{x(y^2+y)}-\tan^{-1}\frac{-2x(y+1)}{y^2+2y-2}$ $=\frac{\pi}{2}$
Now, if $\tan^{-1}a-\tan^{-1}b=\frac{\pi}{2}$,
$\tan^{-1}a=\frac{\pi}{2}+\tan^{-1}b=\cot^{-1}(-b)=\tan^{-1}(-\frac{1}{b})$
$\implies ab=-1$
So, we need to show $\frac{y^2-y+2}{x(y^2+y)}\frac{-2x(y+1)}{y^2+2y-2}=-1$ using $x^2=3$ and $y^3=4$
or $\frac{y^2-y+2}{(y^2+y)}\frac{-2(y+1)}{y^2+2y-2}=-1$
As $y^3=4$, $(y+1)(y^2-y+2)=(y+1)(y^2-y+1)+y+1=y^3+1+y+1=6+y$
and $(y^2+y)(y^2+2y-2)$ $=y^4+y^3(2+1)+y^2(2-2)-2y=y^3(y+3)-2y=4(y+3)-2y=2y+12=2(y+6)$
So, $\frac{y^2-y+2}{(y^2+y)}\frac{-2(y+1)}{y^2+2y-2}=\frac{(6+y)(-2)}{2(6+y)}=-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$
Let $a,b,c$ be positive numbers . Prove the following inequality:
$$\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3.$$
What I tried:
I used Cauchy-Schwarz in the following form
$\sqrt{Ax}+\sqrt{By}+\sqrt{Cz} \leq \sqrt{(a+b+c)(x+y+z)}$ for:
$$A=11a, \quad{} B=11b, \quad{} C=11c$$
and
$$x=\frac{1}{5a+6b}, \quad{} y=\frac{1}{5b+6c}, \quad{} z=\frac{1}{5c+6a}$$ but still nothing. Thanks for your help :)
I tried something else:
$$\large\frac{\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}}}{3} \leq \sqrt{\frac{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}}{3}}$$ and what we have to prove become:
$$\large\sqrt{\frac{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}}{3}} \leq 1 \Leftrightarrow \sqrt{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}} \leq \sqrt{3}$$
Another attempt
$$\large\sqrt{\frac{1}{xy}} \leq \frac{\frac{1}{x}+\frac{1}{y}}{2}=\frac{x+y}{2xy}$$
and $y=1$ and $\displaystyle x=\frac{5a+6b}{11a}$. So:
$$\large\sqrt{\frac{11a}{5a+6b}} \leq \frac{\frac{5a+6b}{11a}+1}{2 \cdot \frac{5a+6b}{11a}}=\frac{8a+3b}{5a+6b}.$$ Now we have to prove:
$$\sum_{cyc}{\frac{8a+3b}{5a+6b}} \leq 3.$$
But still nothing .
| We can prove the more general inequality for certain values of the ratio between b and a:
$$\sqrt{\frac{(a+b)x}{ax+by}}+\sqrt{\frac{(a+b)y}{ay+bz}}+\sqrt{\frac{(a+b)z}{az+bx}} \leq 3$$
if 0.8152 < b/a < 1.2267
Re-working the LHS and applying Cauchy-Schwarz we get:
$$(\sqrt{az + bx}\sqrt{\frac{(a + b) x}{(a z + b x) (a x + b y)}}
+ \sqrt{ax + by}\sqrt{\frac{(a + b) y}{(a x + b y) (a y + b z)}}
+ \sqrt{ay + bz}\sqrt{\frac{(a + b) z}{(ay + bz) (a z + b x)}})^2
\leq
(a y + b z + a z + b x + a x + b y)(\frac{(a + b) x}{(a z + b x) (a x + b y)}+\frac{(a + b) y}{(a x + b y) (a y + b z)}+\frac{(a + b) z}{(ay + bz) (a z + b x)})
=\frac{(a + b)^3 (x + y + z) (y z + x y + xz)}{(a x + b y) (b x +
a z) (a y + b z)}
$$
We are now looking for an upper bound of this last expression. Denote this upper bound by k and consider the expression:
$$G=(a + b)^3 (x + y + z) (y z + x y + zx) - k (a x + b y) (b x + a z) (a y + b z)$$
This needs to be always negative if k is an upper bound.
Simplifying it turns out that we need a value of k such that:
$$G=((a+b)^3-a^2 b k)(y^2 z+x z^2+x^2 y)+((a+b)^3-a b^2 k)(xy^2+yz^2+zx^2)+(3 (a+b)^3-a^3 k-b^3 k)xyz<=0$$
Applying AM/GM we get:
$$(a^2 b k-(a+b)^3)(y^2 z+x z^2+x^2 y)+(a b^2 k-(a+b)^3)(xy^2+yz^2+zx^2)>=3((a^2 b k-(a+b)^3)+(a b^2 k-(a+b)^3))xyz$$
In order for this to hold we need to assume that (A1) $$a^2 b k-(a+b)^3>=0$$ and (A2) $$a b^2 k-(a+b)^3>=0$$
So now we can choose k so that the coefficients before xyz in the last two expressions are equal. In other words we need k such that:
$$3(a + b)^3 - a^3 k - b^3 k =
3 ((a^2 b k - (a + b)^3) + (a b^2 k - (a + b)^3))$$
It is easy to see that k=9 we completes the proof if our assumptions (A1) and (A2) hold.
They hold for a small range of b/a ratios - approximately 0.8152 < b/a < 1.2267.
In order for A1 and A2 to hold for k=9 we can rewrite them in the form $$\frac{(a + b)^3}{a^2 b}<=9 \quad and \quad \frac{(a + b)^3}{a b^2}<=9$$
Letting x=b/a this means that we need the positive values of x for which both:$$
1/x + 3 x + x^2 - 6<=0 \quad and \quad 1/x^2 + 3/x + x - 6<=0$$
Solving the resulting cubics we get:$$
2 - \sqrt{3} Cos(\pi/18) + 3 Sin(\pi/18)\le\frac{b}{a}\le-1 + 3 Cos(\pi/9) - \sqrt{3} Sin(\pi/9)
$$
or numerically 0.8152 < b/a < 1.2267. This is the range of possible valus for b/a for which our inequality always holds.
You can also note that the product of the interval ends is equal to 1.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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FFT processor which can only be used once Given an 8 point FFT processor, which can be used only once, compute the DFTs of the sequence.
$$x_1(n)=[1,8,6,7,4,2,3,1]$$
$$x_2(n)=[1,4,3,2,8,7,6,1]$$
| Let $y(n)=x_1(n)+jx_2(n)$.
$$\therefore y(n)=[(1+1j), (8+4j), (6+3j), (7+2j), (4+8j), (2+7j), (3+6j), (1+1j)]$$
Now we can calculate the DFT of $y(n)$ by using the single 8 point FFT processor.
$$\begin{align}
\therefore Y(k) & = \sum_{n=0}^{N-1}{y(n)e^{-{2\pi{}jnk\over N}}} \\
& =\begin{bmatrix}
(32 + 32j) \\
\left(-6 - \sqrt{2} +j \left(- 8 \sqrt{2} -10\right)\right) \\
(4 - 2j) \\
(- \sqrt{2} +j \left(- 4 \sqrt{2} -4\right)) \\
(-4 + 4j) \\
(-6 + \sqrt{2} +j \left(-10 + 8 \sqrt{2}\right)) \\
(-12 + 2j) \\
(\sqrt{2} +j \left(-4 + 4 \sqrt{2}\right))
\end{bmatrix}
\end{align}$$
Since DFTs are linear,
$$\begin{align}
Y(k) & =DFT\big(x_1(n)+jx_2(n)\big) \\
& =X_1(k)+jX_2(k) \tag{1}\\
\end{align}$$
where $X_1(k)$ and $X_2(k)$ are the DFTs of $x_1(n)$ and $x_2(n)$ respectively.
Taking conjugate on both sides we have,
$$Y^*(k) = X_1^*(k)-jX_2^*(k)$$
where $X^*$ denotes complex conjugation.
Also trivially,
$$Y^*(N-k) = X_1^*(N-k)-jX_2^*(N-k)$$
where $N$ is the length of the DFT, which is $8$ here.
By the property of symmetry, if $x(n)$ consists of real numbers only, then,
$$X(k)=X^*(N-k)$$
where $N$ is the length of the DFT.
$$\therefore Y^*(N-k)=X_1(k)-jX_2(k) \tag{2}$$
From equations (1) & (2), we have,
$$\begin{align}
X_1(k) & ={Y(k)+Y^*(N-k)\over 2}\\
X_2(k) & ={Y(k)-Y^*(N-k)\over 2j}\\
\text{and, } Y^*(N-k) & =\begin{bmatrix}
(32 - 32j) \\
(\sqrt{2} -j \left(-4 + 4 \sqrt{2}\right))\\
(-12 - 2j) \\
(-6 + \sqrt{2} -j \left(-10 + 8 \sqrt{2}\right)) \\
(-4 - 4j) \\
(- \sqrt{2} -j \left(- 4 \sqrt{2} -4\right)) \\
(4 + 2j) \\
\left(-6 - \sqrt{2} -j \left(- 8 \sqrt{2} -10\right)\right)
\end{bmatrix}\\
\therefore X_1(k) & = \begin{bmatrix}
32\\
-3 +j \left(- 6 \sqrt{2} -3\right)\\
-4 - 2j\\
-3 +j \left(- 6 \sqrt{2} + 3\right)\\
-4\\
-3 +j \left(-3 + 6 \sqrt{2}\right)\\
-4 + 2j\\
-3 +j \left(3 + 6 \sqrt{2}\right)
\end{bmatrix}\\
\text{and, }X_2(k) & =\begin{bmatrix}
32\\
-7 - 2 \sqrt{2} +j \left(\sqrt{2} + 3\right)\\
- 8j\\
-7 + 2 \sqrt{2} +j \left(-3 + \sqrt{2}\right)\\
4\\
-7 + 2 \sqrt{2} +j \left(- \sqrt{2} + 3\right)\\
8j\\
-7 - 2 \sqrt{2} +j \left(-3 - \sqrt{2}\right)
\end{bmatrix}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Auxiliary trigonometric identities [Identidades Trigonométricas Auxiliares] I need to show two auxiliary trigonometric identities:
1) $\sec^2x = \tan ^2x + 1 (\cos x \neq 0)$
2) $\csc^2x = \cot^2x +1 (\sin x \neq 0)$
How could I do it?
[Original Portuguese]
Identidades Trigonométricas Auxiliares
Sendo preciso mostrar a demonstração das duas fórmulas das IDENTIDADES TRIGONOMÉTRICAS AUXILIARES:
1) sec²x = tg²x + 1 (cos x ≠ 0)
2) cossec²x = cotg²x = 1 (sen x ≠ 0)
Como seria possível fazê-lo???
| Starting from $\sin^2 x + \cos^2 x = 1$, you get, dividing by $\cos^2 x$ (if this is $\ne 0$)
\[ \frac{\sin^2x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x} = \frac 1{\cos^2 x} \]
which gives
\[ \tan^2 x + 1 = \mathrm{sec}^2 x \]
Dividing the first identity by $\sin^2 x$ we get
\[ \frac{\sin^2x}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} = \frac 1{\sin^2 x} \]
which gives
\[ 1 + \cot^2 x = \mathrm{csc}^2 x \]
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proofs using linear congruences We have just covered solving linear congruences, and I am confused about how to use them in proofs. I understand that the linear congruence $$cx \equiv b \pmod m$$ has a unique solution $\bmod m$ if $\gcd(c,m) = 1$, but a general approach to problems escapes me.
Sample Questions
Prove that if $x^2 \equiv n \pmod {65}$ has a solution, then so does $x^2 \equiv -n \pmod {65}$.
Show that if $n \equiv 7 \pmod 8$, then $n$ is not the sum of three squares.
My Work
For the first one, $x^2 \equiv n \pmod {65}$ implies that $65 | (x^2 - n)$, so (I believe) that means that $n = b^2$ for some $b$, so $65 | (x^2 - b^2)$. We proved a result that said that if $a^2 \equiv b^2 \pmod p$ for some prime $p$, then $a \equiv b$ or $a \equiv - b \pmod p$. However, I don't think that is the right track to go down here because $65$ is not prime, and I'm unsure about assuming $n = b^2$ for some $b$.
For the second one, suppose to the contrary that $n = a^2 + b^2 + c^2$ for some $a, b, c$. Then $n \equiv 7 \pmod 8$ implies that $8 | (n - 7)$ so $n = 8k + 7$ for some $k$. So substituting in gives me $a^2 + b^2 + c^2 = 8k + 7$, and I'm unsure how to proceed from here.
| Hint $\rm(1)\ \ \ x^2 \equiv n,\,\ y^2\equiv -1\:\Rightarrow\: -n\equiv x^2y^2\equiv (xy)^2.\ $ But $\rm\:mod\ 65\!:\ {-}1 \equiv 64\equiv (\_ )^2.$
$\rm(2)\ \ mod\ 4\!:\ x^2\!+\!y^2\!+\!z^2 \equiv\, 3\:\Rightarrow\:x,y,z\:$ odd, by $\rm\:odd^2\equiv 1,\ even^2\equiv 0.\:$ Therefore we deduce $\rm\phantom{(2)\ \ } mod\ 8\!:\ x^2\!+\!y^2\!+\!z^2\equiv\:7\:\Rightarrow\:x,y,z\:$ odd $\rm\:\Rightarrow\:x^2\!+\!y^2\!+\!z^2\equiv 3,\:$ by $\rm\:odd^2\!\equiv\{\pm1,\pm3\}^2\equiv 1.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$? $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \cdots
=
\ln\left(3\right)
$$
| Let's try one line proof using Riemann sums. So, our limit may be written as
$$\lim_{n\to\infty}\sum_{k=1}^{n} \left(\frac1{3k-2}+\frac1{3k-1}+\frac1{3k}-\frac1k\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}\frac1{1+\frac{k}{n}}=\int_0^2\frac{1}{1+x}=\ln3.$$
Q.E.D.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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By using properties of determinants show that determinant is equal to $(1+a^2+b^2)^3$ $$\begin{vmatrix}1+a^2-b^2&2ab&-2b\\
2ab&1-a^2+b^2&2a\\
2b&-2a&1-a^2-b^2\end{vmatrix}=(1+a^2+b^2)^3$$
I have been trying to solve the above determinant. But unfortunately my answer is always coming as:
$$1+3a^2+3a^4+a^6+3a^2b^4+3b^2+4a^2b^2+a^4b^2+b^6+3b^4$$
Please help me to solve this problem.
| Assuming $a,b\in\mathbb{R}$ (and if not, proving it for real $a,b$ should allow you it to extend it to complex $a,b$), each row vector has the norm $(1+a^2+b^2)$ and the row vectors are all orthogonal so the matrix divided by $(1+a^2+b^2)$ is an orthogonal matrix and the absolute value of its determinant is 1 which immediately implies your result.
| {
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How to represent $(43.3)_7$ in base-8? I am trying to represent $(43.3)_7$ in base-8.
But in only one integer digit by truncating the rest and using the numerical unsigned representation.
| $43.3_7 = 4*7 + 3 + \frac 37=$
$4(2*3 + 1) + 3 + \frac 37*\frac 88=$
$8*3 + (4+3) + \frac{\frac {24}7}8=$
$3*8 + 7 + \frac 38 + \frac 37\frac 18=$
$3*8 + 7*8^0 + 3*8^{-1} + \frac {24}7*8^{-2}=$
$3*8 + 7*8^0 + 3*8^{-1} + 3*8^{-2}+ \frac 37*8^{-2} =$
.... via induction....
$3*8 + 7*8^0 + 3*8^{-1} + 3*8^{-2} + 3*8^{-3} + ...... = $
$37.\overline 3_8$
....
It's worth noting that just like calculating in base 10.
$\frac 17 = $
$7 $ divided into $1$
or $7$ divided into $\frac 88 = \frac 18 + $ $7$ divided into $\frac 18$
or $\frac 18 + $ $7$ divided into $\frac 8{8^2} =\frac 18 + \frac 1{8^2} + $ $ 7$ divided into $ \frac 1 {8^3}$ and so on .... $=$
$\frac 18 + \frac 1{8^2} + \frac 1{8^3} + .... = $
$.111111...._7$ so
$\frac 37 = 0.\overline 3_8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$ How do I prove that :
$$2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x).$$
It seems to be easy , but I have no idea:)
thanks :)
| We start the same way as uforoboa: Expanding the left hand side, you are trying to show that
$$2x^4 + 2y^4 + 2z^4 + 4(x^2y^2+y^2z^2+z^2x^2) \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$$
This equivalent to
$$2x^4 + 2y^4 + 2z^4 +x^2y^2+y^2z^2+z^2x^2 \geq 3(x^3y+y^3z+z^3x)$$
By AM-GM one has $x^3y \leq 1/2(x^4 + x^2y^2)$, $y^3z \leq 1/2(y^4 + y^2z^2)$, and $z^3x \leq 1/2(z^4 + z^2x^2)$. Inserting this into the above it suffices to show that
$$2x^4 + 2y^4 + 2z^4 +x^2y^2+y^2z^2+z^2x^2 \geq {3 \over 2}x^4 + {3 \over 2}y^4 + {3 \over 2}z^4 +{3 \over 2}x^2y^2+{3 \over 2}y^2z^2+{3 \over 2}z^2x^2$$
This is the same as showing
$$x^4 + y^4 + z^4 \geq x^2y^2+y^2z^2+z^2x^2$$
This follows from AM-GM on each term on the right, or just expanding $(x^2 - y^2)^2 + (y^2 - z^2)^2 + (x^2 - z^2)^2 \geq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove/disprove $\sum x^2y^2 \ge \sum x^3y$ Prove $x^2y^2+y^2z^2+z^2x^2 \ge$ or $\le x^3y+y^3z+z^3x$ where $x,y,z$ are real numbers.
Actually, I have reached here from this problem:
Inequality. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$
| Define $f(x,y,z)=x^2y^2+y^2z^2+z^2x^2-x^3y-y^3z-z^3x$. Then $f(1,1,2)=-2$ and $f(1,1,-1)=4$. Therefore, neither the inequality nor its reverse is true.
For positive numbers, we also have $f(3,2,1)=-16$ and $f(4,6,1)=24$. So even for all positive numbers the inequality is not true.
| {
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Showing that $\lim_{k\rightarrow 0}\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} = \int_0^1\frac{dx}{\sqrt{(1-x^2)}}$ So I'm trying to show that:
$$\lim_{k\rightarrow 0}\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} = \int_0^1\frac{dx}{\sqrt{(1-x^2)}}$$
I guess this boils down to a solid understanding of uniform convergence.
There's also the second issue that the theorem in Rudin's PMA which discusses the exchange of a definite integral and a limit says that the the sequence of integrands must converge to the limit integrand uniformly on a closed interval, in this case that would be $[0,1]$. But of course our function is only defined on $[0,1)$, thus should I be considering the interval $[0,1-\epsilon]$?
As far as proving uniform convergence, I was looking at the the sequence:
$$M_{a_n} = \sup_{x\in [0,1-\epsilon]}|\frac{1}{\sqrt{(1-x^2)(1-(a_n)^2x^2)}} - \frac{1}{\sqrt{(1-x^2)}}|$$
for any sequence $a_n\rightarrow 0$.
And trying to prove that the sequence $M_{a_n}$ goes to zero, but it doesn't seem to upon pulling out the factor $\frac{1}{\sqrt{1-x^2}}$.
To sum up, this problem with its complexities is just a bit beyond my level of understanding and comfort with analysis. Could someone help me sort through it? Thanks.
| What you want to show is that
$$\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\to \frac{1}{\sqrt{1-x^2}}$$
on uniformly $[0,1-\epsilon]$ for all $\epsilon>0$ and that
$$\lim\limits_{\epsilon\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\to \int_0^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$ uniformly in $k$. The first fact gives us
$$\lim\limits_{k\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}=\int_0^{1-\epsilon}\frac{dx}{\sqrt{1-x^2}}$$
for all $\epsilon>0$, and since integration is continuous
$$\lim_{\epsilon\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{1-x^2}}=\int_0^{1}\frac{dx}{\sqrt{1-x^2}}$$
while adding the second fact gives
$$\lim\limits_{\epsilon\to 0}\lim\limits_{k\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}=\lim\limits_{k\to 0}\int_0^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$
and from the these three equalities we can conclude the desired result.
To prove the first fact, note that
$$\begin{align}
\sup_{x\in [0,1-\epsilon]}\left|\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}-\frac{1}{\sqrt{1-x^2}}\right| &\le\left(\sup_{x\in [0,1-\epsilon]}\frac{1}{\sqrt{1-x^2}}\right)\sup_{x\in [0,1-\epsilon]}\left|\frac{1}{\sqrt{1-k^2x^2}}-1\right|\\
&\le\left(\sup_{x\in [0,1-\epsilon]}\frac{1}{\sqrt{1-x^2}}\right)\left|\frac{1}{\sqrt{1-k^2(1-\epsilon)^2}}-1\right|\to 0\\
\end{align}$$
giving uniform convergence. For the second fact, note that
$$\int_{1-\epsilon}^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\le \frac{1}{\sqrt{1-k^2}}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-k^2}}\int_{0}^1\chi_{[1-\epsilon,1]}(x)\frac{dx}{\sqrt{1-x^2}}$$
where $\chi_{[1-\epsilon,1]}$ is the indicator function for the interval $[1-\epsilon,1]$. The function $\chi_{[1-\epsilon,1]}(x)\frac{1}{\sqrt{1-x^2}}$ converges pointwise to $0$ except at $x=1$ and is dominated by $\frac{1}{\sqrt{1-x^2}}$ which has finite integral, thus by the Dominated Convergence Theorem we see that $\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\to 0$. If we bound $k$ below $1$ (say $k\le 1/2$) we see that
$$\frac{1}{\sqrt{1-k^2}}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\leq \frac{4}{3}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\to 0$$
which is clearly uniform in $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do we solve the equation? How do we solve the following equation in the set of real numbers?
$$(x+1)\cdot \sqrt{x+2} + (x+6)\cdot \sqrt{x+7}=(x+3)\cdot (x+4).$$
I wrote the given equation has the form
\begin{equation*}
(x+1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x+7} - 3) = (x-2)(x+4)
\end{equation*}
This equation is equivalent to
\begin{equation*}
(x-2)\left(\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4\right) = 0.
\end{equation*}
But I can not prove that the equation
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 = 0
\end{equation*}
has no solution. Detail
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 <0, \forall x \geqslant -2.
\end{equation*}
| We have
$$\dfrac{x+1}{\sqrt{x+2}+2}<\dfrac{x+2}{\sqrt{x+2}+2}\leqslant \dfrac{x+2}{2}$$
and
$$\dfrac{x+6}{\sqrt{x+7}+3}<\dfrac{x+6}{3}\leqslant \dfrac{x+6}{2}.$$
Therefore
$$\dfrac{x+1}{\sqrt{x+2}+2}+\dfrac{x+6}{\sqrt{x+7}+3}<\dfrac{x+2}{2}+\dfrac{x+6}{2}=x+4$$
Thus the equation
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2} + 2} + \dfrac{x+6}{\sqrt{x+7} + 3} - x - 4 = 0
\end{equation*}
has no solution.
Thus, the given equation has only root $x = 2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/202670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Calculate a sum involving nth root of unity Calculate
$$1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1}$$
Where $\epsilon$ is nth root of unity.
There is a hint that says: multiply by $(1-\epsilon)$
Doing this multiplication I get:
$$1+\epsilon+\epsilon^{2}+\epsilon^{3}+\cdots+\epsilon^{n-1}-n$$
The answer is: $-\frac{n}{1-\epsilon} $if $ \epsilon\neq1$ and $\frac{n(n+1)}{2}$ if $\epsilon=1$.
I can't see how to get this result...
| Of course, when $\epsilon=1$,
$$
\begin{align}
&1+2\epsilon+3\epsilon^2+4\epsilon^3+\dots+n\epsilon^{n-1}\\
&=1+2+2+4+\dots+n\\
&=\frac{n(n+1)}{2}
\end{align}
$$
When $\epsilon\ne1$
$$
\left.
\begin{align}
1+\epsilon+\epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{1-\epsilon^n}{1-\epsilon}\\
\epsilon+\epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon-\epsilon^n}{1-\epsilon}\\
\epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon^2-\epsilon^n}{1-\epsilon}\\
\epsilon^3+\dots+\epsilon^{n-1}&=\frac{\epsilon^3-\epsilon^n}{1-\epsilon}\\
&\vdots\\
\epsilon^{n-1}&=\frac{\epsilon^{n-1}-\epsilon^n}{1-\epsilon}\\
\end{align}
\right\}\mbox{ $n$ equations}
$$
Summing these $n$ equations yields
$$
\begin{align}
1+2\epsilon+3\epsilon^2+4\epsilon^3+\dots+n\epsilon^{n-1}
&=\frac{\frac{1-\epsilon^n}{1-\epsilon}-n\epsilon^n}{1-\epsilon}\\
&=\frac{n}{\epsilon-1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/206523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
$$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$
by Mathematical induction，n$\ge $1
| I didn't prove it by mathematical induction.
Set $a_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$, then by calculation, $a_{n+1}-a_{n}=\frac{1}{2n+1}-\frac{1}{2n+2}$, $a_{n}$ is strictly monotonically increasing.
So we can obtain that
$$\begin{align*}
a_{n}&=a_{1}+\sum_{k=1}^{n-1}(a_{k+1}-a_{k})\\
&=\frac{1}{2}+\sum_{k=1}^{n-1}(\frac{1}{2k+1}-\frac{1}{2k+2})\\
&=\sum_{k=0}^{n-1}(\frac{1}{2k+1}-\frac{1}{2k+2})\\
&=\sum_{k=0}^{n-1}\int_{0}^{1}(x^{2k}-x^{2k+1})dx\\
&=\int_{0}^{1}\sum_{k=0}^{n-1}(x^{2k}-x^{2k+1})dx\\
&=\int_{0}^{1}\frac{1-x^{2n}}{1+x}dx\\
&=\ln2-\int_{0}^{1}\frac{x^{2n}}{1+x}dx
\end{align*}$$
When $x\in[0,1]$, $\frac{x^{2n}}{2}\leq\frac{x^{2n}}{1+x}\leq x^{2n}$, integraling at both sides, we can get that $(0<)\frac{1}{2(2n+1)}\leq\int_{0}^{1}\frac{x^{2n}}{1+x}dx\leq\frac{1}{2n+1}$, so for arbitrary $n$, $a_{n}<\ln2<\frac{25}{36}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 6
} |
Special Differential Equation I ended up with a differential equation that looks like this:
$$\frac{d^2y}{dx^2} + \frac 1 x \frac{dy}{dx} - \frac{ay}{x^2} + \left(b -\frac c x - e x \right )y = 0.$$
I tried with Mathematica. But could not get the sensible answer. May you help me out how to solve it or give me some references that I can go over please? Thanks.
| $\dfrac{d^2y}{dx^2}+\dfrac{1}{x}\dfrac{dy}{dx}-\dfrac{ay}{x^2}+\left(b-\dfrac{c}{x}-ex\right)y=0$
$\dfrac{d^2y}{dx^2}+\dfrac{1}{x}\dfrac{dy}{dx}-\left(ex-b+\dfrac{c}{x}+\dfrac{a}{x^2}\right)y=0$
Let $y=\dfrac{u}{\sqrt{x}}$ ,
Then $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x}}\dfrac{du}{dx}-\dfrac{u}{2x\sqrt{x}}$
$\dfrac{d^2y}{dx^2}=\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{2x\sqrt{x}}\dfrac{du}{dx}-\dfrac{1}{2x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}=\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}$
$\therefore\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}+\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}-\dfrac{u}{2x^2\sqrt{x}}-\left(ex-b+\dfrac{c}{x}+\dfrac{a}{x^2}\right)\dfrac{u}{\sqrt{x}}=0$
$\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\left(ex-b+\dfrac{c}{x}+\dfrac{4a-1}{4x^2}\right)\dfrac{u}{\sqrt{x}}=0$
$\dfrac{d^2u}{dx^2}-\left(ex-b+\dfrac{c}{x}+\dfrac{4a-1}{4x^2}\right)u=0$
The above ODE is hypergeometric only when the special cases below:
$1$. $e=0$
$2$. $b=0$ and $c=0$
$3$. $c=0$ and $a=\dfrac{1}{4}$
Other than the above special cases the above ODE is not hypergeometric.
Unfortunately it also not belongs to any confluent forms of Heun’s equation.
Therefore to solve the above ODE generally is extremely difficult.
One of the main reason is that the coefficient of $u$ has too many terms or contains too high power terms. The similar situation also appear in Titchmarsh's ODE.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/211803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
expanding $(a+b+c+...)^x$ where $x$ is natural neatly Suppose that $a,b,c,d,...$ are unknown variables.
One wishes to expand $(a+b+c..)^x$ where $x$ is natural number ina neat manner (for e.g. using combination, sigma etc.).
What would be some way?
Also, what would be the number of terms that would allow writing neatly?
| One can expand $(x_1+x_2+x_3+\dots +x_m)^n$ by using the multinomial theorem. It states that
$$(x_1+x_2+x_3+\dots +x_m)^n=\sum_{k_1+k_2+\dots+k_m=n}\binom{n}{k_1,k_2,k_3,\dots,k_m}\prod_{t=1}^mx_t^{k_t}$$
$$=\sum_{k_1+k_2+\dots+k_m=n}\binom{n}{k_1,k_2,k_3,\dots,k_m}x_1^{k_1}x_2^{k_2}x_3^{k_3}\cdots x_m^{k_m}$$
Where $$\binom{n}{k_1,k_2,k_3,\dots,k_m}=\frac{n!}{k_1!k_2!\cdots k_m!}$$
What this basically means is take all the sets $\{k_1,k_2,\dots,k_m\}$ such that it's elements add up to $n$ and each element is a whole number, then find $\binom{n}{k_1,k_2,k_3,\dots,k_m}$ for each of those sets, multiply the resultant by $x_1^{k_1}x_2^{k_2}x_3^{k_3}\cdots x_m^{k_m}$ and add it all together.
For example: $(a+b+c+d)^3$. The coefficient of $a^2b$ will be $$\binom{3}{2,1,0,0}=\frac{3!}{2!1!0!0!}=\frac{6}{2\cdot1\cdot1\cdot1}=3$$
While the coefficient of $abc$ will be
$$\binom{3}{1,1,1,0}=\frac{3!}{1!1!1!0!}=\frac{6}{1\cdot1\cdot1\cdot1}=6$$
Doing this for the rest of the terms we get
$$a^3+3 a^2 b+3 a b^2+b^3+3 a^2 c+6 a b c+3 b^2 c+3 a c^2+3 b c^2+c^3+3 a^2 d+6 a b d+3 b^2 d+6 a c d+6 b c d+3 c^2 d+3 a d^2+3 b d^2+3 c d^2+d^3$$
Note that in the expansion you can see smaller binomial expansions (e.g. the first 4 terms $a^3+3 a^2 b+3 a b^2+b^3$ is just $(a+b)^3$). In fact $(x_1+x_2+x_3+\dots +x_m)^n$ will contain every binomial expansion $(x_i+x_j)^n$ where $i\ne j$ and $1\le i,j\le m$. This follows from the fact that you can choose a set which contains exactly 2 non-zero elements $k_i,k_j$ such that $k_i+k_j=n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/213481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the limit $ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2}$ I cannot find the following limit:
$$ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2} \, . $$
Please, help me.
| Here is a proof using Tools 1-3 below and, I think, nothing else.
Tool 1: $2\cos(a)\cos(b)=\cos(a+b)+\cos(a-b)$.
Tool 2: $1-\cos(a)=2\sin^2(\frac{a}2)$.
Tool 3: Let $u(z)=\frac{\sin(z)}z$, then $u(z)\to1$ when $z\to0$. (You said you knew that.)
Using Tools 1 and 2, one gets
$$
\frac{1-\cos x\cos 2x}{x^2}\stackrel{\text{(Tool 1)}}{=}\frac{1-\cos 3x}{2x^2}+\frac{1-\cos x}{2x^2}\stackrel{\text{(Tool 2)}}{=}\frac{\sin^2(x/2)}{x^2}+\frac{\sin^2(3x/2)}{x^2},
$$
hence
$$
\frac{1-\cos x\cos 2x}{x^2}=\frac{u^2(\frac{x}2)+9u^2(\frac{3x}2)}4.
$$
Now, Tool 3 yields $u(\frac{x}2)\to1$ and $u(\frac{3x}2)\to1$, hence the limit is $\frac{1^2+9\cdot1^2}4=\frac52$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/214444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How I convert Decimal Using negetive Base 2? I dont have any idea how i convert deciaml using negetive base 2?
i googling today long time but couldn't find any solution.
can any tell me process?
i need to learn it .
| One can use the standard algorithm for converting from one base to another. Divide by the target base $b$ to get an integer quotient and a remainder in the chosen range of digits, generally but not always $\{0,\dots,|b|-1\}$; replace the original number by the quotient and repeat; continue until you get a quotient of $0$, and read the remainders in reverse order. For example, to convert $-19$ to base $-2$:
$$\begin{align*}
-19&=-2\cdot 10+\color{red}{1}\\
10&=-2\cdot(-5)+\color{red}{0}\\
-5&=-2\cdot3+\color{red}{1}\\
3&=-2\cdot(-1)+\color{red}{1}\\
-1&=-2\cdot1+\color{red}{1}\\
1&=-2\cdot0+\color{red}{1}\\
0&=-2\cdot0
\end{align*}$$
Reading the remainders (in red) from the bottom up, we find that $-19=111101_{-2}$. As a check,
$$\begin{align*}
1\cdot(-2)^5&+1\cdot(-2)^4+1\cdot(-2)^3+1\cdot(-2)^2+0\cdot(-2)^1+1\cdot(-2)^0\\
&=-32+16-8+4+1\\
&=-19\;.
\end{align*}$$
Added: Here are a couple more examples.
$$\begin{align*}
4&=-2\cdot(\color{blue}{-2})+\color{red}{0}\\
\color{blue}{-2}&=-2\cdot\color{green}{1}+\color{red}{0}\\
\color{green}{1}&=-2\cdot0+\color{red}{1}\;,
\end{align*}$$
so $4=100_{-2}$.
$$\begin{align*}
7&=-2\cdot(\color{blue}{-3})+\color{red}{1}\\
\color{blue}{-3}&=-2\cdot\color{green}{2}+\color{red}{1}\\
\color{green}{2}&=-2\cdot(\color{blue}{-1})+\color{red}{0}\\
\color{blue}{-1}&=-2\cdot\color{green}{1}+\color{red}{1}\\
\color{green}{1}&=-2\cdot0+\color{red}{1}\;,
\end{align*}$$
so $7=11011_{-2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/216800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$ Considering the following formulae:
(i) $1+2+3+..+n = n(n+1)/2$
(ii) $1\cdot2+2\cdot3+3\cdot4+...+n(n+1) = n(n+1)(n+2)/3$
(iii) $1\cdot2\cdot3+2\cdot3\cdot4+...+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$
Find and prove a 'closed formula' for the sum
$1\cdot2\cdot3\cdot...\cdot k + 2\cdot3\cdot4\cdot...\cdot(k+1) + ... + n(n+1)(n+2)\cdot...\cdot (k+n-1)$
generalizing the formulae above.
I have attempted to 'put' the first 3 formulae together but I am getting nowhere and wondered where to even start to finding a closed formula.
| The pattern looks pretty clear: you have
$$\begin{align*}
&\sum_{i=1}^ni=\frac12n(n+1)\\
&\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\
&\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;,
\end{align*}\tag{1}$$
where the righthand sides are closed formulas for the lefthand sides. Now you want
$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\;;$$
what’s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.
Added: The general result, of which the three in $(1)$ are special cases, is $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k)\;.\tag{2}$$ For $n=1$ this is $$k!=\frac1{k+1}(k+1)!\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then
$$\begin{align*}\sum_{i=1}^{n+1}i(i+1)&(i+2)\dots(i+k-1)\\
&\overset{(1)}=(n+1)(n+2)\dots(n+k)+\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\\
&\overset{(2)}=(n+1)(n+2)\dots(n+k)+\frac1{k+1}n(n+1)(n+2)\dots(n+k)\\
&\overset{(3)}=\left(1+\frac{n}{k+1}\right)(n+1)(n+2)\dots(n+k)\\
&=\frac{n+k+1}{k+1}(n+1)(n+2)\dots(n+k)\\
&=\frac1{k+1}(n+1)(n+2)\dots(n+k)(n+k+1)\;,
\end{align*}$$
exactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\dots(n+k)$, and the rest is just algebra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/219693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
What are good ways to find the limits of this function? Compute: $\displaystyle\lim_{x\rightarrow 2}\frac{\sqrt{x+1}-\sqrt{ 1-x}}{x}$
| I assume you are interested in $$\lim_{x \to 0} \dfrac{\sqrt{1+x} - \sqrt{1-x}}x$$
Multiply and divide by $\sqrt{1+x} + \sqrt{1-x}$ to get
$$\dfrac{\sqrt{1+x} - \sqrt{1-x}}x \times \dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \dfrac{(1+x)-(1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \dfrac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}\\ = \dfrac2{\sqrt{1+x} + \sqrt{1-x}}$$
Now can you finish it off?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/220448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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As shown in the figure: Prove that $a^2+b^2=c^2$ Geometry: Buildings in the triangle
Other triangles with the same property:
$1.$ 12 18 6 12 30 102
$2.$ 15 30 15 15 15 90
$3.$ 24 30 54 24 6 42
$4.$ 30 10 40 30 20 50 (proposed problem in sense of clockwise)
$5.$ 36 12 6 12 18 96
$6.$ 36 18 6 36 6 78
$7.$ 42 6 36 42 12 42
$8.$ 60 6 57 30 3 24
$9.$ 60 24 12 12 6 66
Using matlab we can find all triangles (integer solutions) with this property sums of squares:
| Denote the length of the lower side (the hypotenuse of the big triangle) by $h$. Then the side containing $c$ is of length $h\sin 40^\circ=h\cos 50^\circ$ (Do you see why?)
Now let's find the length of segment from the rightmost bottom vertex to the one at the bottom of $b$ (denote it by $l_1$): Using the law of sines, we get $\frac{h\sin 40^\circ}{\sin 110^\circ}=\frac{l_1}{\sin20^\circ}$, so $$l_1=\frac{h\sin 40^\circ\sin20^\circ}{\sin 110^\circ}$$
Denote by $l_2$ the length of the segment between the bottom of $b$ and the bottom of $a$. Then $\frac{l_1+l_2}{\sin50^\circ}=\frac{h\sin 40^\circ}{\sin 80^\circ}$, so $$l_2=\frac{h\sin 40^\circ\sin50^\circ}{\sin 80^\circ}-l_1=h\sin 40^\circ\left(\frac{\sin50^\circ}{\sin 80^\circ}-\frac{\sin20^\circ}{\sin 110^\circ}\right)$$
Now, we can express $a,b,c$ in terms of $h$:
$$\frac{a}{\sin30^\circ}=\frac{h-l_1-l_2}{\sin(180^\circ-30^\circ-(180^\circ-40^\circ-40^\circ))}=\frac{h-l_1-l_2}{\sin50^\circ}
\\ \frac{b}{\sin30^\circ}=\frac{h-l_1}{\sin(180^\circ-30^\circ-(180^\circ-40^\circ-70^\circ))}=\frac{h-l_1}{\sin80^\circ}
\\ \frac{c}{\sin30^\circ}=\frac{h}{\sin(180^\circ-50^\circ-30^\circ)}=\frac{h}{\sin100^\circ}$$
Now you can simplify all those expressions and substitute. (To calculate the angles, I used the fact that the sum of angles in a triangle is $180^\circ$ and each time I looked at a different triangle)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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The quadratic form $x^2 + ny^2$ via prime factors Elementary algebra shows that the product of two numbers in the form $x^2 + ny^2$ again has the same form, since if $p = (a^2 + nb^2)$ and $q = (c^2 + nd^2)$,
$$pq = (a^2 + nb^2)(c^2 + nd^2) = (ac \pm nbd)^2 + n(ad \mp bc)^2$$
My question is: Assuming that a number $z$ can be factored into primes of the form $x^2 + ny^2$, does every representation of $z$ in this form arise from repeated applications of this formula to the prime factors?
| Alright, the answer to the actual question asked is yes, as follows. I am taking a prime $p$ with $\gcd(p,n) = 1.$ Then I am demanding $u^2 + n v^2 = p.$ Next, I am taking some number, composite or prime, call it $Q,$ and demand about $pQ$ rather than $Q$ itself,
$$ pQ = r^2 + n s^2. $$
First, we get
$$ \left( \frac{u}{v} \right)^2 \equiv \left( \frac{r}{s} \right)^2 \equiv -n \pmod p. $$ Choose $\pm s$ so that
$$ \left( \frac{u}{v} \right) \equiv \left( \frac{r}{s} \right) \pmod p. $$
So we have
$ u s \equiv v r \pmod p,$ or
$$ -us + vr \equiv 0 \pmod p. $$
Next,
$$ p^2 Q = (ur + n v s)^2 + n (-us + v r)^2, $$ and so
$$ Q = \left( \frac{ur + n v s}{p} \right)^2 + n \left(\frac{-us + v r}{p} \right)^2 $$ in integers.
Combine this once again with $p = u^2 + n v^2$ and you get back to $pQ = r^2 + n s^2$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove this property for an arbitrary circle Prove that in an arbitrary circle, the point on the circle closest to the origin must lie on the extended line connecting the circle's centre and the origin.
|
Let the equation of the circle be $(x-a)^2+(y-b)^2=r^2$
So, any point on the circle can be $P(r\cos \theta+a,r\sin \theta+b)$
If the distance$(d)$ from the point $(r\cos \theta+a,r\sin \theta+b)$ from the origin,
$d^2=(r\cos \theta+a)^2+(r\sin \theta+b)^2=r^2+a^2+b^2+2r(a\cos \theta+b\sin \theta)=r^2+a^2+b^2+2r\sqrt{a^2+b^2}\cos(\theta-\arctan\frac b a)$
This will be minimum if $\cos(\theta-\arctan\frac b a)=-1$
$\implies \theta-\arctan\frac b a=(2n+1)\pi$ where $n$ is any integer.
So, $\tan \theta=\tan\{(2n+1)\pi+\arctan\frac b a\}=\frac b a\implies b\cos \theta=a\sin \theta$
Now the area of the triangle with vertices $(0,0),(a,b),(r\cos \theta+a,r\sin \theta+b)$ is
$$\frac 1 2\det\begin{pmatrix} 0 & 0 & 1 \\ r\cos \theta+a & r\sin \theta+b & 1 \\ a & b&1\end{pmatrix}=\frac 12 \left(b(r\cos \theta+a)-a(r\sin \theta+b)\right)=\frac{r(b\cos \theta-a\sin \theta)}2=0$$
So, $(0,0),(a,b),(r\cos \theta+a,r\sin \theta+b)$ are co-linear for the minimum distance of $P$ from the origin $O(0,0)$ for any $a,b,r$.
A little generalization:
If the distance of $P(r\cos \theta+a,r\sin \theta+b)$ from $Q(c,d)$ is $D,$
$D^2=(r\cos \theta+a-c)^2+(r\sin \theta+b-d)^2$
$=r^2+(a-c)^2+(b-d)^2+2r\{(a-c)\cos \theta+(b-d)\sin \theta\}$
Applying the same method,the distance will be minimum is $(b-d)\cos \theta=(a-c)\sin \theta$
Now the area of the triangle with vertices $(c,d),(a,b),(r\cos \theta+a,r\sin \theta+b)$ is
$$\frac 1 2\det\begin{pmatrix} c & d & 1 \\ r\cos \theta+a & r\sin \theta+b & 1 \\ a & b&1\end{pmatrix}$$
$$=\frac 12\det\begin{pmatrix} c-a & d-b & 1-1 \\ r\cos \theta+a-a & r\sin \theta+b-b & 1-1 \\ a & b&1\end{pmatrix}=\frac{r\{(a-c)\sin\theta-(d-b)\cos\theta\}}2=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integral of cosec squared ($\operatorname{cosec}^2x$, $\csc^2x$) According to my sheet of standard integrals,
$\int \csc^2x \, dx = -\cot x + C$.
I am interested in a proof for the integral of $\operatorname{cosec}^2x$ that does not require differentiating $\cot x$. (I already know how to prove it using differentiation, but am interested in how one would calculate it without differentiation.)
| Question: $\displaystyle\int{\csc^2(x)dx} = $
Using substitution: $u = \dfrac{1}{\sin^2(x)}$
$\dfrac{du}{dx} = \dfrac{-2\cos(x)}{\sin^3(x)} \quad \longrightarrow du = \left(\dfrac{-2\cos(x)}{\sin(x)}\right)\times \left(\dfrac{1}{\sin^2(x)}\right)dx$
$\therefore -\dfrac{1}{2}\tan(x)du = \dfrac{1}{\sin^2(x)} dx \qquad (1)$
We still must find what is $\tan(x)$ in terms of $u$, hence:
$u = \dfrac{\sin^2(x) + \cos^2(x)}{\sin^2(x)} = 1+ \cot^2(x)$
$u-1 = \cot^2(x)$
$\sqrt{u-1} = \cot(x)$
$\dfrac{1}{\sqrt{u-1}} = \tan(x)$
Therefore substitute in $(1)$ and our integral becomes:
$-\dfrac{1}{2}\displaystyle\int{(u-1)^{-\frac{1}{2}}}du = -\dfrac{1}{2}\dfrac{\sqrt{(u-1)}}{\dfrac{1}{2}} + C = -\sqrt{\left(\dfrac{1}{\sin^2(x)} - 1)\right)} + C $
$= -\sqrt{\left(\dfrac{1-\sin^2(x)}{\sin^2(x)}\right)} + C = -\cot(x) + C$
Hope this makes sense,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/239808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How do you compute the limit with multiple variables without fail? $$\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2}$$
I don't know what to do doesn't it give always $0$?
Whether $x=0$ $x=y$ or $y=0$ or $x=y^2$ it always give $0$ since it goes to $(0,0)$
| $\left | \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2} \right |\leq \dfrac{|x^3-y^3| (2x^2+3y^2)}{x^2+2y^2} \leq \dfrac{|x^3-y^3| (2x^2+4y^2)}{x^2+2y^2}\leq 2|x^3-y^3|$
Where we used the inequality $|\sin (x)| \leq |x|$
ADDED Let $f(x,y)= x^3-y^3$, now knowing that $ \lim _{ (x,y) \rightarrow (0,0)} f(x,y)= 0$
It means that for $\epsilon >0$ there is a $\delta >0$ such that $ \forall (x,y) \in \{ (x,y ) | \sqrt{ x^2+y^2} < \delta \}$ we have $ | x^3-y^3| < \frac{ \epsilon }{2}$.
Therefore for that $\delta >0$ we have $\left |\dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2} \right | < \epsilon$, whenever $ (x,y) \in \{ (x,y ) | \sqrt{ x^2+y^2} < \delta \}$. Since $\epsilon $ was arbitary we conclude be definition of the limit that
$ \lim _{ (x,y) \rightarrow (0,0)} \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2}=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/251884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor Polynomial of $\arctan$ of given Degree and Error
Replace the following function by its taylor polynomial of the given
grade, and approximate the error in the given interval:
$$f(x) = \arctan(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$$
My solution and thoughts
We only need the first three derivatives and the fourth one for the error
$$
f'(x) = \frac{1}{1+x^2} \\
f''(x) = \frac{2x}{(x^2+1)^2} \\
f'''(x) = \frac{6x^2-2}{(x^2+1)^3} \\
f^{(iv)}(x) = \frac{24x(x^2-1)}{(x^2+1)^4}
$$
And by definition we know that
$$
T_3(f,x,0) = \sum\limits_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k
$$
we get
$$
T_3(f,x,0) = x - \frac{x^3}{3}
$$
We know that
$$R_3(x) = \frac{f^{(iv)}(c)}{4!}x^4, |x| \le \frac{1}{10}$$
by plugging in the maximal value of $x = c = \frac{1}{10}$ we get for any $c$:
$$
\left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right| \le \\
\le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4}
$$
$$
\Rightarrow \left|R_3(x)\right| \le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4 \cdot 4!}\cdot\left|x^4\right| \Leftrightarrow \\
\Leftrightarrow \left|R_3(x)\right| \le \frac{10}{101^4}
$$
Is this solution correct? How can I make it more formally right? (I know I lack some mathematical formalism).
| Your work looks good to me, except for a mistake in the last step (probably a typo). Other than that, it's pretty much exactly the way I would hope one of my students would solve this problem.
The mistake is that you lost the $99$ in the final step, so that at the end you should have $$
\left|R_3(x)\right| \le \frac{99 \cdot 10}{101^4}.$$
(Also, the final implication is $\Rightarrow$, not $\Leftrightarrow$. Since you're substituting a value for $x$ here the two inequalities are not equivalent.)
As a side note, there's an easier way to calculate the Taylor series about $0$ for $f(x) = \arctan x$.
Since $$\frac{d}{dx} \arctan x = \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 \pm \ldots,$$ via the geometric series formula when $|x| < 1$,
$$\arctan x = C + x - \frac{1}{3} x^3 + \frac{1}{5}x^5 - \frac{1}{7} x^7 \pm \ldots.$$
Since $\arctan 0 = 0$, we have $C = 0$. This gives you $$T_3(f,x,0) = x - \frac{1}{3}x^3$$ rather painlessly.
You do still need to find the fourth derivative of $\arctan x$ to use the remainder formula for Taylor polynomials, however.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/252360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to multiply polynominals I can't figure out how to multiply these polynominals $$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)$$
I tried multiplying like this
$$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)$$
$$3x^4-7x^3+(5x^2)(2x^2)(-6x^2)+(5x)(7x)(-4x)+(8)(9)$$
$$3x^4-7x^3-60x^2-140x+72$$
It says the answer is $$-12x^6-19x^5-14x^4-68x^3+28x^2+69x+72$$ but how did they get it?
| To multiply two polynomials, distribute each term in the left set of parentheses over the entire collection of terms in the right set of parentheses. Your example would start out like:
$$
\begin{align*}
& (\color{red}{5x^2+3x^4-7x^3+5x+8})(2x^2-4x+9-6x^2+7x)\\
= & \color{red}{5x^2}(2x^2-4x+9-6x^2+7x)\\
& \color{red}{+ 3x^4}(2x^2-4x+9-6x^2+7x)\\
& \color{red}{- 7x^3}(2x^2-4x+9-6x^2+7x)\\
& \color{red}{+ 5x}(2x^2-4x+9-6x^2+7x)\\
& \color{red}{+ 8}(2x^2-4x+9-6x^2+7x).
\end{align*}
$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/253522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}$, is the following true?
Using the previous question*, arithmetics of bordes, and the sandwich theorem, find the limits of the following sequenes:
b. $\left(\dfrac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}$
(*) the previous question is to prove that $\lim_{n\to \infty }\left(1+\dfrac{x}{n}\right) = e^x$
Looking at the question it looked quite easy. Since we proved in class that $\lim (a_n)^k = (\lim a_n)^k$ I can easily say the limit of the inside is $1$, therefore the limit of everything is $1^n=1$ (This is a good time to notice that I don't know how to use anything but inline equations here):
$$\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}=\left(\dfrac{1+\dfrac{8}{n}-\dfrac{1}{n^{2}}}{1-\dfrac{4}{n}-\dfrac{5}{n^{2}}}\right)^{n}$$
due to arithmetics of borders:
$$\begin{align*}
\lim\left(\frac{1+\dfrac{8}{n}-\dfrac{1}{n^{2}}}{1-\dfrac{4}{n}-\dfrac{5}{n^{2}}}\right)^{n}&=\left(\frac{\lim\left(1+\dfrac{8}{n}-\dfrac{1}{n^{2}}\right)}{\lim\left(1-\dfrac{4}{n}-\dfrac{5}{n^{2}}\right)}\right)^{n}\\\\
&=\left(\frac{\lim1+\lim\dfrac{n}{8}-\lim\dfrac{1}{n^{2}}}{\lim1-\lim\dfrac{4}{n}-\lim\dfrac{5}{n^{2}}}\right)^{n}\\\\
&=1^{n}\\\\
&=1
\end{align*}$$
But for some reason this feels wrong to me. It doesn't use the previous question or the sandwich theorem, and the solution feels to trivial to be true.
Is there anything wrong here?
| Note that for any $\epsilon > 0$,
$$1 + \dfrac{8-\epsilon}{n} \le 1 + \dfrac{8}{n} - \dfrac{1}{n^2} \le 1 + \dfrac{8}{n}$$
for sufficiently large $n$. Therefore
$$ \eqalign{e^{8-\epsilon} &= \lim_{n \to \infty} \left(1 + \dfrac{8-\epsilon}{n}\right)^n \le \liminf_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2}\right)^n \cr&\le
\limsup_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2}\right)^n \le
\lim_{n \to \infty} \left(1 + \dfrac{8}{n}\right)^n = e^8}$$
and taking $\epsilon \to 0+$ we get
$$\lim_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2} \right)^n = \lim_{n \to \infty} \left(1 + \frac{8}{n} \right)^n = e^8$$
Similarly,
$$\lim_{n \to \infty} \left(1 - \frac{4}{n} - \frac{5}{n^2} \right)^n = e^{-4}$$
So
$$\lim_{n \to \infty} \left(\frac{n^2 - 8 n - 1}{n^2 - 4 n - 5}\right)^n =
\dfrac{\displaystyle \lim_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2} \right)^n}
{\displaystyle \lim_{n \to \infty} \left(1 - \dfrac{4}{n} - \dfrac{5}{n^2} \right)^n} = e^{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
5 linear equations in 5 unknowns I need an example of 5 linearly independent equations with 5 variables. How can I write such a equation set. As an example:
0 0 0 0 1 | -4
0 0 1 -1 0 | 3
1 0 0 0 1 | 2
0 7 -8 0 0 | -14
2 0 0 0 0 | 2
But this is wrong. I need to build a correct one.
| You can choose any $\hat{x}_{1} , \ldots , \hat{x}_{5}$, and use the system
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{bmatrix}
=
\begin{bmatrix}
\hat{x}_{1} \\
\hat{x}_{2} \\
\hat{x}_{3} \\
\hat{x}_{4} \\
\hat{x}_{5} \\
\end{bmatrix}.
$$
If you want something that is less obvious, you can start multiplying the rows by nonzero constants:
$$
\begin{bmatrix}
2 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{bmatrix}
=
\begin{bmatrix}
2 \hat{x}_{1} \\
\hat{x}_{2} \\
\hat{x}_{3} \\
\hat{x}_{4} \\
\hat{x}_{5} \\
\end{bmatrix}
$$
(in this example, we multiply the first row by $2$), or adding multiples of one row to another row:
$$
\begin{bmatrix}
2 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & -3 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{bmatrix}
=
\begin{bmatrix}
2 \hat{x}_{1} \\
\hat{x}_{2} \\
\hat{x}_{3} - 3 \hat{x}_{5} \\
\hat{x}_{4} \\
\hat{x}_{5} \\
\end{bmatrix}
$$
(in this example, we add $-3$ times the fifth row to the third row).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/255750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving square root of a square is the same as absolute value Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:
$\begin{align}
f(x) &= \sqrt {x^2} \\
f(-1) &= \sqrt {(-1)^2} \\
f(-1) &= \sqrt {1} \\
f(-1) &= 1
\end{align}$
$\begin{align}
g(x) &= x \\
g(-1) &= -1
\end{align}$
thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:
$y = \sqrt {x^2}$:
$y = x$:
$y = \mid x \mid$:
From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$?
| When $x < 0$, $|x| = -x > 0$. $-x \ne x$ (unless $x=0$) and $(-x)^2 = x^2$. There are two "square roots" of any positive number $y$, i.e. numbers whose square is $y$, and the positive one is called $\sqrt{y}$. So $\sqrt{x^2} = -x = |x|$ when $x < 0$, and $\sqrt{x^2} = x = |x|$ when $x \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/258876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 0
} |
Find the upper and lower limits of $xf(x)$, as $x\rightarrow \infty$ Define $$f(x)=\int_{x}^{x+1}\sin(t^2)dt$$
Find the upper and lower limits $xf(x)$, as $x\rightarrow \infty$.
I find the answer as $+1, -1$ since $|\sin(x)| \le 1$. (Of course I calculated that function)
Is that right or did I miss something?
==========================================================
I solved this way.
$2xf(x)=\cos(x^2)-\cos[(x+1)^2]+r(x)$ where $r(x)=\frac{\cos(x+1)^2}{x+1}-2x\int_{x^2}^{(x+1)^2}\frac{cos(u)}{4u^{3/2}}du$
Therefore $xf(x)=\frac{1}{2}{\cos(x^2)-\cos(x+1)^2}+\frac{r(x)}{2}$
Using trigonomeric formula: $2\sin(a)\sin(b)=\cos(a-b)-\cos(a+b)$
Rewrite $xf(x)=±\sin(x^2+x+\frac{1}{2})\sin(x+\frac{1}{2})+\frac{r(x)}{2}$.
As $x\rightarrow \infty, r(x) \rightarrow 0$.
Suppose $x^2=2k\pi$ for integer $k$.
To achieve $±1$, we have to show that $x+\frac{1}{2} \rightarrow 2n\pi+\frac{\pi}{2}$ for some $n$ as $x\rightarrow \infty$.
For each $n$, there exists $k$ such that $\sqrt{2\pi k}+\frac{1}{2} < 2n\pi+\frac{\pi}{2} <\sqrt{2\pi (k+1)} +\frac{1}{2} $
Distance between $\sqrt{2\pi k}+\frac{1}{2}$ and $2n\pi+\frac{\pi}{2}$ is at most $\sqrt{2\pi (k+1)} +\frac{1}{2} -\sqrt{2\pi k}+\frac{1}{2}$.
As $k \rightarrow \infty$ the distance becomes arbitrary small.
Therefore $ x \rightarrow \infty$, $xf(x)=±\sin(2n\pi+\frac{\pi}{2})=±1$.
| $$
\begin{align}
f(x)
&=\int_x^{x+1}\sin(t^2)\,\mathrm{d}t\\
&=-\int_x^{x+1}\frac1{2t}\,\mathrm{d}\cos(t^2)\\
&=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}
-\int_x^{x+1}\cos(t^2)\frac1{2t^2}\,\mathrm{d}t\\
&=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}+O\left(\frac1{x^2}\right)\tag{1}
\end{align}
$$
So
$$
\begin{align}
xf(x)
&=\frac{\cos(x^2)}{2}-\frac{\cos((x+1)^2)}{2(1+1/x)}+O\left(\frac1{x}\right)\\
&=\frac12\left(\cos(x^2)-\cos((x+1)^2)\right)+O\left(\frac1{x}\right)\tag{2}
\end{align}
$$
Therefore, mapping $x\mapsto\sqrt x$ on the right, we get
$$
\limsup_{x\to\infty}\,xf(x)
=\limsup_{x\to\infty}\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\tag{3 sup}
$$
and
$$
\liminf_{x\to\infty}\,xf(x)
=\liminf_{x\to\infty}\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\tag{3 inf}
$$
When $x$ is big, $\sqrt x$ varies much slower than $x$:
$$
\begin{align}
\sqrt{x+2\pi}-\sqrt x
&=\frac{2\pi}{\sqrt{x+2\pi}+\sqrt x}\\
&\le\frac\pi{\sqrt x}\tag{4}
\end{align}
$$
Thus, for any $\epsilon>0$, choose an $x_0>\dfrac{\pi^2}{\epsilon^2}$ so that $2\sqrt{x_0}+1$ is an odd multiple of $\pi$. Then $(4)$ guarantees that for $x\in[x_0,x_0+2\pi]$, $2\sqrt x+1$ is within $2\epsilon$ of an odd multiple of $\pi$.
Choose the $x\in[x_0,x_0+2\pi]$ which is $0\bmod{2\pi}$ and we get that
$$
\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\ge1-\epsilon\tag{5 sup}
$$
Choose the $x\in[x_0,x_0+2\pi]$ which is $\pi\bmod{2\pi}$ and we get that
$$
\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\le-1+\epsilon\tag{5 inf}
$$
These last two inequalities show that
$$
\limsup_{x\to\infty}\,xf(x)=1\tag{6 sup}
$$
and
$$
\liminf_{x\to\infty}\,xf(x)=-1\tag{6 inf}
$$
A Simpler Approach
As shown in $(2)$,
$$
\begin{align}
xf(x)
&=\frac12\left(\cos\left(x^2\right)-\cos\left((x+1)^2\right)\right)+O\!\left(\frac1x\right)\\
&=\underbrace{\sin\left(x^2+x+\tfrac12\right)}_{\text{frequency $\frac{x+1/2}\pi$}}\underbrace{\sin\left(x+\tfrac12\right)}_{\text{period $2\pi$}}+O\!\left(\frac1x\right)\tag7
\end{align}
$$
So, as $x\to\infty$, the curve looks like a rapidly oscillating sine wave which has an oscillating amplitude with period $2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the sum of all solutions $2x + 3y = n$ has exactly $2011$ non-negative integral solutions. Determine the SUM of the possible values of $n$.
| Let us verify for $n=m[2,3]+k$ where $0\le k<[2,3]$
If $k=0\implies 2x+3y=6m\implies 3(2m-y)=2x,$ so $y$ must even $=2z$(say),
So, $x=3(m-z)\implies 0\le z\le m$ so $z$(hence $y$) can assume $m+1$ values in non-negative integers.
If $k=1\implies 2x+3y=6m+1\implies 3(y-2m)+1=2x,$ so $y$ must odd $=2z+1$(say), where $z\ge 0$
So, $x+1=3(m-z)\implies (m-z)\ge 1\implies 0\le z\le m-1$ so $z$(hence $y$) can assume $m$ values in non-negative integers.
$\implies n=6(m+1)+1=6m+7$ will have $m+1$ solutions in non-negative integers.
So, we observe that $6m,6m+2,6m+3,6m+4,6m+5,6m+7$ will have $m+1$ solutions in non-negative integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/262452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Help in proving $ \left( 1 + \frac{1}{n} \right)^{n} \leq \sum\limits_{k=0}^{n} \frac{1}{k!} < 3 $. I am trying to prove this statement for all $ n \geq 1 $ using induction:
$$
\left( 1 + \frac{1}{n} \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3.
$$
I said:
*
*Base case $ n = 1 $:
$$
\left( 1 + \frac{1}{1} \right)^{1} \leq \sum_{k=0}^{1} \frac{1}{k!} < 3,
$$
which is okay.
*Induction step: Suppose that $ \displaystyle \left( 1 + \frac{1}{n}
\right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3 $ for a given $ n \in
\mathbb{N} $.
Transition from $ n \to n + 1 $:
$$
\displaystyle \left( 1 + \frac{1}{n + 1} \right)^{n+1}
= \left( 1 + \frac{1}{n + 1} \right)^{n} \left( 1 + \frac{1}{n + 1} \right)
= \ldots \text{Help} \ldots
< 3.
$$
I need some guidance for proof-writing (-thinking) in orders.
| For all $ n \in \mathbb{N} $, we have
\begin{align}
\left( 1 + \frac{1}{n} \right)^{n}
&= \sum_{k=0}^{n} \binom{n}{k} \left( \frac{1}{n} \right)^{k} \quad (\text{By the Binomial Theorem.}) \\
&= \sum_{k=0}^{n} \frac{n!}{k!(n - k)!} \cdot \frac{1}{n^{k}} \quad (\text{By the definition of the binomial coefficient.}) \\
&= \sum_{k=0}^{n} \frac{1}{k!} \cdot \frac{n!}{(n - k)!} \cdot \frac{1}{n^{k}} \\
&= \sum_{k=0}^{n} \frac{1}{k!} \left( \prod_{i=n-k+1}^{n} i \right) \frac{1}{n^{k}} \quad (\text{By cancellation of terms.}) \\
&\leq \sum_{k=0}^{n} \frac{1}{k!} \left( \prod_{i=n-k+1}^{n} n \right) \frac{1}{n^{k}} \quad (\text{As $ i \leq n $ for all $ i \in \{ n - k + 1,\ldots,n \} $.}) \\
&= \sum_{k=0}^{n} \frac{1}{k!} \cdot n^{k} \cdot \frac{1}{n^{k}} \\
&= \sum_{k=0}^{n} \frac{1}{k!} \\
&\leq 1 + \sum_{k=0}^{n-1} \frac{1}{2^{k}} \quad (\text{By comparison of terms.}) \\
&< 1 + \sum_{k=0}^{\infty} \frac{1}{2^{k}} \\
&= 1 + 2 \quad (\text{Sum of a well-known convergent geometric series.}) \\
&= 3. \quad (\text{Voilà!})
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/264458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Evaluate $\int {2x\over x^2-1}dx$ My friend evaluated this before he went to bed: $$\int {2x\over x^2-1}dx$$
The answer was $\log(x^2-1)$.
I just can't figure out how that works. I know that $\int \frac1x dx = \log|x|$, so what just happened to $2x$?
| $\frac{2x}{x^2-1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac1{x-1}+\frac1{x+1}$
Alternatively, using Partial Fraction Decomposition, let $\frac{2x}{x^2-1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac A{x+1}+\frac B{x-1}$ where $A,B$ are arbitrary constants.
So, $2x=(A+B)x+B-A$
Comparing the constant terms in either of the identity, $B-A=0\implies A=B$
Comparing the coefficients of $x,A+B=2\implies A=B=1$
So, $\frac{2x}{x^2-1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac 1{x+1}+\frac 1{x-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/266684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
} |
Compute the series Compute the series
$$1)\space\sum_{n=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1}\right)\frac{1}{n(n+1)}$$
$$2)\space\sum_{n=1}^{\infty}\left(1-\frac{1}{2}+\frac{1}{3}-\cdots(-1)^{n}\frac{1}{n+1}\right)\frac{1}{n(n+1)}$$
| For the first part, recognize that $\frac {1}{n(n+1)} = \frac {1}{n} - \frac {1}{n+1}$. So the sum can we written as
$$ \sum_{n=1}^{\infty} ( 1 + \frac {1}{2} + \frac {1}{3}+ \ldots + \frac {1}{n+1} ) ( \frac {1}{n} - \frac {1}{n+1}) $$
Break this up into the separate terms of the right parenthesis, and we find that consecutive terms cancel out. Note that we are canceling terms that are directly next to each other, so this step is justified. We'd obtain
$$\sum_{n=1}^{\infty} \big[ (1 + \frac {1}{2} + \ldots + \frac {1}{n+1}) - (1 + \frac {1}{2} + \ldots + \frac {1}{n}) \big] (\frac {1}{n}) = \sum_{n=1}^{\infty} \frac {1}{(n+1)n}$$
Finally, evaluate this as telescoping series $\sum_{n=1}^{\infty} \frac {1}{(n+1)n} = \sum_{i=1}^{\infty} \frac {1}{n} - \frac {1}{n+1} = 1$.
With these ideas, you should try the second part by yourself. At the most, you will need to evaluate $\sum_{n=1}^\infty \frac {(-1)^n}{n+1}$, which you can do by using the MacLaurin expansion $\ln(1+x) = x - \frac {x^2}{2} + \frac {x^3}{3} - \frac {x^4}{4}$ for a suitable value of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/267088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify these products of permutations Can someone please explain how to get these answers? Everytime I think I understand the method, I end up getting a completely different answer to the one provided. How exactly do you go about simplifying these?
[Note: Read from left to right]
$$(i)\;\;\; (1,2,3).(2,4)(1,3,5) = (1,4,2,5)$$
$$(ii)\;\;\; (1,3,4)(2,5).(2,3,4)(1,5) = (1,4,5,3,2) $$
| To find, for example, $(1,2,3)\cdot(2,4)(1,3,5)$ you need to apply it on every $1\leq i\leq 5$:
$$\begin{array}{l}
[1](1,2,3)\cdot(2,4)(1,3,5)=[2](2,4)(1,3,5)=4\\
[2](1,2,3)\cdot(2,4)(1,3,5)=[3](2,4)(1,3,5)=5\\
[3](1,2,3)\cdot(2,4)(1,3,5)=[1](2,4)(1,3,5)=3\\
[4](1,2,3)\cdot(2,4)(1,3,5)=[4](2,4)(1,3,5)=2\\
[5](1,2,3)\cdot(2,4)(1,3,5)=[5](2,4)(1,3,5)=1
\end{array}$$
Hence, by writing it as a permutation, starting with one, you have: $(1,4,2,5)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The area of a triangle determined by the bisectors. How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture:
$AQ$ is the bisector of the angle $\angle BAC$, $BR$ -bisector for $\angle ABC$ and $CP$ -bisector for $\angle ACB$. Now, it must calculated the area for the triangle $PQR$ knowing that $AB=c$, $BC=a$ and $CA=b$.
I tried to use the angle bisector theorem for every bisectors but I didn't obtained anything.
Thanks :)
| We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, \quad (I)$$
Using the angle bisector theorem we get:
$$BP=\frac{ac}{a+b},\quad (1)$$
$$BR=\frac{ac}{b+c}, \quad (2)$$
$$CR=\frac{ab}{b+c},\quad (3)$$
$$CQ=\frac{ab}{a+c},\quad (4)$$
$$AQ=\frac{bc}{a+c},\quad (5)$$
and
$$AP=\frac{bc}{a+b}. \quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=\frac{1}{2}ab\sin\gamma, \quad (7)$$
$$A_{PBR}=\frac{1}{2}BP\cdot BR\sin\beta, \quad (8)$$
$$A_{RCQ}=\frac{1}{2}CR\cdot CQ\sin\gamma, \quad (9)$$
and
$$A_{QAP}=\frac{1}{2}AQ\cdot AP\sin\alpha. \quad (10)$$
Let $R$ be the circumradius, we know that:
$$\sin \alpha = \frac{a}{2R}, \quad (11)$$
$$\sin \beta = \frac{b}{2R}, \quad (12)$$
$$\sin \gamma = \frac{c}{2R}, \quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get:
$$A_{PQR}=\frac{1}{2} \cdot \frac{abc}{2R}-\frac{1}{2} \frac{a^2c^2b}{(a+b)(b+c)2R}-\frac{1}{2} \cdot \frac{a^2b^2c}{(b+c)(a+c)2R}-\frac{1}{2} \cdot \frac{b^2c^2a}{(a+b)(a+c)2R}, \Rightarrow$$
$$A_{PQR}=\frac{abc}{4R}[1-\frac{ac}{(a+b)(b+c)}-\frac{ab}{(b+c)(a+c)}-\frac{bc}{(a+b)(a+c)}], \Rightarrow$$
$$A_{PQR}=\frac{abc}{2R}[\frac{abc}{(a+b)(b+c)(a+c)}], \Rightarrow$$
$$A_{PQR}=A_{ABC}[\frac{2abc}{(a+b)(b+c)(a+c)}]$$
Using Heron's formula we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve this inequality $\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}$ Prove that
$ \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdot \frac{9}{10}\cdot \frac{11}{12}\cdot \frac{13}{14}...\cdot \frac{91}{92}\cdot \frac{93}{94}\cdot \frac{95}{96}\frac{97}{98}\cdot \frac{99}{100} <\frac{1}{10}$
| The product of the first 31 fractions is $0.100923...$ while the product of the first 32 fractions is $0.099346...$, so it looks like you only need the first 32 of the 50 fractions to get below $1/10$. Using more fractions (all less than 1) will only make that smaller.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/274753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
calculate $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos2x}{\cos x-\sin x}$ $$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(2x)}{\cos(x)-\sin(x)}=\lim_{x\rightarrow\frac{\pi}{4}}\frac{2\cos^{2}(x)-1}{\cos(x)-\sqrt{1-\cos^{2}(x)}}$$
$$t=\cos(x)$$
$$\lim_{x\rightarrow\frac{\pi}{4}}\frac{2t^{2}-1}{t-\sqrt{1-t^{2}}}=\lim_{x\rightarrow\frac{\pi}{4}}\frac{(2t^{2}-1)(t+\sqrt{t^{2}-1})}{2t^{2}-1}$$
$$\lim_{x\rightarrow\frac{\pi}{4}}(2t^{2}-1)(t-\sqrt{t^{2}-1})=3(\sqrt{2}-1)$$
Please help me to find an error. Correct answer is $\sqrt{2}$. Thanks
| $$ \lim_{x\to\frac{\pi}{4}}\frac{\cos2x}{\cos x-\sin x}=
\lim_{x\to\frac{\pi}{4}}\frac{\cos^2 x-\sin^2 x}{\cos x-\sin x}=$$
$$ \lim_{x\rightarrow\frac{\pi}{4}}\frac{(\cos x- \sin x)(\cos x+\sin x)}{\cos x-\sin x}=$$
$$= \lim_{x\rightarrow\frac{\pi}{4}}{\cos x+\sin x}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is this possible If $r_1, r_2, t_1,$ and $t_2$ are real numbers and if $\left|r_{1}\right|<\left|r_{2}\right|$, $\left|t_{1}\right|<\left|t_{2}\right|$
and
$$
\left|\left(r_{1}-r_{2}\right)\left(t_{1}-t_{2}\right)\right|>\left|\left(r_{1}+r_{2}\right)\left(t_{1}+t_{2}\right)-2\left(r_{1}r_{2}+t_{1}t_{2}\right)\right|,
$$
does it hold that
$
\left|\left(r_{1}+r_{2}\right)-\left(t_{1}+t_{2}\right)\right|\ge\left|\left|r_{1}\right|-\left|t_{1}\right|\right|+\left|\left|r_{2}\right|-\left|t_{2}\right|\right|?
$ Thanks in advance.
| The second inequality $$|r_{1}+r_{2}-(t_{1}+t_{2})|>||r_{1}|-|t_{1}||+||r_{2}|-|t_{2}||$$holds if $r_{1}> t_{1}>0$, $r_{2}> 0> t_{2}$, then we have $$|r_{1}+r_{2}-(t_{1}+t_{2})|=|r_{1}-t_{1}|+|r_{2}-t_{2}|>||r_{1}|-|t_{1}||+||r_{2}|-|t_{2}||$$
By assumption $r_{1}<r_{2}$, $0<t_{1}<-t_{2}$, then the left hand of the first inequality become $$(r_{2}-r_{1})(t_{1}-t_{2})=r_{2}t_{1}-r_{1}t_{1}-r_{2}t_{2}+r_{1}t_{2}$$
The right hand side now is depended on whether $$-2t_{1}t_{2}<2r_{1}r_{2}-(r_{1}+r_{2})(t_{2}+t_{1})$$
If this is true when we want to have $$r_{2}t_{1}-r_{1}t_{1}-r_{2}t_{2}+r_{1}t_{2}>2r_{1}r_{2}+2t_{1}t_{2}-t_{1}r_{1}-t_{2}r_{2}-t_{1}r_{2}-t_{2}r_{1}$$
Re-arranging the terms we need to have $r_{2}t_{1}+r_{1}t_{2}>r_{1}r_{2}+t_{1}t_{2}$.
Let $r_{2}=50,r_{1}=1,t_{2}=-4,t_{1}=3$, then left hand become $150-4=146$, right hand become $50-12=38$. While $-2t_{1}t_{2}=24$,$2r_{1}r_{2}-(r_{1}+r_{2})(t_{2}+t_{1})=100+50=150$.
Now go back to the original inequality we have:
$$49*7=343>|51*-1-2*(50-12)|=127$$ and $$|51-(-1)|=52>48=2+46$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/276768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Optimization of $2x+3y+z$ under the constraint $x^2+ y^2+ z^2= 1$
Let $x, y$ and $z$ be real numbers such that $x^2+ y^2+ z^2= 1$. Find the
maximum and minimum values of $2x + 3y + z$ .
How can I able to solve the problem? Thanks for your help.
| Using Lagrange Multiplier, $$f(x,y,z,\lambda)=2x+3y+z-\lambda(x^2+y^2+z^2-1)$$
$$\frac{\delta f}{\delta x}=2-2\lambda x$$ etc,
For the extreme values of $f, \frac{\delta f}{\delta x}=0$ etc,
$\implies x=\frac1{\lambda},y=\frac3{2\lambda},z=\frac1{2\lambda}\implies 2x+3y+z=\frac7{\lambda}$
But $x^2+y^2+z^2=1\implies \lambda^2=\frac72$
If $\lambda=\sqrt{\frac72},2x+3y+z=\frac7{\lambda}=\sqrt {14}$
If $\lambda=-\sqrt{\frac72},2x+3y+z=\frac7{\lambda}=-\sqrt {14}$
Alternatively, as $x,y,z$ are real, we can set $x=\cos A\cos B,y=\cos A\sin B,z=\sin A$
So, $$2x+3y+z=2\cos A\cos B+3\cos A\sin B+\sin A=\cos A(2\cos B+3\sin B)+\sin A$$
Let us find the extreme values of $F(C)=a\cos C+b\sin C$ (This can derived at least in two more ways without using calculus, see below)
So, $F'(C)=-a\sin C+b\cos C$
For the extreme values of $F, F'(C)=0\implies a\sin C-b\cos C=0\iff \frac a{\cos C}=\frac b{\sin C}=\pm \sqrt{a^2+b^2}$
$F''(C)=-a\cos C-b\sin C$
If $\frac a{\cos C}=\frac b{\sin C}=\sqrt{a^2+b^2}, F''(C)=-\sqrt{a^2+b^2}<0$
So, $F_{max}=\sqrt{a^2+b^2}$
Similarly, $F_{min}=-\sqrt{a^2+b^2}$
So, $-\sqrt{2^2+3^2}\le 2\cos B+3\sin B\le \sqrt {2^2+3^2}$
$\implies -\sqrt {13}\cos A+\sin A\le \cos A(2\cos B+3\sin B)+\sin A\le \sqrt {13}\cos A+\sin A$
Again, $\sqrt {13}\cos A+\sin A\le \sqrt{13+1}=\sqrt{14}$
and $-\sqrt {13}\cos A+\sin A\ge -\sqrt{13+1}=-\sqrt{14}$
$\implies -\sqrt {14} \le \cos A(2\cos B+3\sin B)+\sin A\le \sqrt {14}$
$\implies -\sqrt {14} \le 2x+3y+z\le \sqrt {14}$
[
(i)Let $y=a\cos C+b\sin C\implies (y-a\cos C)^2=(b\sin C)^2=b^2-b^2\cos^2C$
$\implies (a^2+b^2)\cos^2C-2ya\cos C+y^2-b^2=0$ which is a Quadratic equation in $\cos C$. As $\cos C$ is real, $$(-2ya)^2\ge4(a^2+b^2)(y^2-b^2)\implies y^2\le a^2+b^2\implies -\sqrt{a^2+b^2}\le y\le \sqrt{a^2+b^2}$$
(ii) Putting $a=R\cos D,b=R\sin D$ where $R>0$ so that $R^2=a^2+b^2$
So, $a\cos C+b\sin C=R\cos D\cos C+R\sin D\sin C=\sqrt{a^2+b^2}\cos(C-D)$
As $-1\le \cos(C-D)\le 1,$ $$-\sqrt{a^2+b^2}\le a\cos C+b\sin C\le \sqrt{a^2+b^2}$$
]
| {
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"url": "https://math.stackexchange.com/questions/276861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$ Evaluate
$$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
| New answer (Jun 6, 2022). About 9 and half years since my first solution, it suddenly dawned on me that I can compute this integral in another way:
Note that, for $0 < x < 1$,
$$ \frac{1}{1-x} + \frac{1}{\log x} = \int_{0}^{1} \frac{1-x^s}{1-x} \, \mathrm{d}s. $$
Plugging this to OP's integral and interchanging the order of integration, we get
\begin{align*}
I := \int_{0}^{1} \left( \frac{1}{1-x} + \frac{1}{\log x} \right)^2 \, \mathrm{d}x
&= \int_{0}^{1}\int_{0}^{1} \int_{0}^{1} \frac{1-x^s}{1-x} \cdot \frac{1-x^t}{1-x} \, \mathrm{d}x \, \mathrm{d}s \,\mathrm{d}t
\end{align*}
Let us study the innermost integral. Performing integration by parts,
\begin{align*}
&\int_{0}^{1} \frac{1-x^s}{1-x} \cdot \frac{1-x^t}{1-x} \, \mathrm{d}x \\
&= \underbrace{\left[ \frac{x}{1-x}(1-x^s)(1-x^t) \right]_{x=0}^{x=1}}_{=0} - \int_{0}^{1} \frac{(s+t)x^{s+t} - sx^s - tx^t}{1-x} \, \mathrm{d}x \\
&= (s+t)\psi_0(s+t+1) - s\psi_0(s+1) - t\psi_0(t+1), \tag{1}
\end{align*}
where $\psi_0(\cdot)$ is the digamma function and we utilized its integral representation in the last step. To make use of this formula, note that the change of variables $(x, y) = (s+t, s-t)$ yields
$$ \forall f \in C([0,2]) \ : \quad \int_{0}^{1}\int_{0}^{1} f(s+t) \, \mathrm{d}s\mathrm{d}t
= \int_{0}^{1} x[f(x) + f(2-x)] \, \mathrm{d}x. $$
Using this and by integrating both sides of $\text{(1)}$, we get
\begin{align*}
I
&= \int_{0}^{1} x \bigl[ x \psi(x+1) + (2-x)\psi(3-x) \bigr] \, \mathrm{d}x - 2 \int_{0}^{1} x \psi(x+1) \, \mathrm{d}x \\
&= \int_{0}^{1} x(2-x) \bigl[ \psi(3-x) - \psi(x+1) \bigr] \, \mathrm{d}u \\
&= \int_{0}^{1} x(2-x) \left[ \frac{1}{2-x} + \psi(2-x) - \psi(x+1) \right] \, \mathrm{d}u \\
&= \frac{1}{2} + 2 \int_{0}^{1} (1-x) \log ( \Gamma(2-x)\Gamma(x+1)) \, \mathrm{d}x \tag{int. by parts}
\end{align*}
By noting that $\log ( \Gamma(2-x)\Gamma(x+1))$ is symmetric about $x = \frac{1}{2}$ and using the reflection formula, this further reduces to
\begin{align*}
I
&= \frac{1}{2} + \int_{0}^{1} \log ( \Gamma(2-x)\Gamma(x+1)) \, \mathrm{d}x \tag{by symmetry} \\
&= \frac{1}{2} + \int_{0}^{1} \left( \log(1-x) + \log x + \log \pi - \log\sin(\pi x) \right) \, \mathrm{d}x \\
&= \bbox[color:navy;padding:5px;border:1px dotted navy;]{\log(2\pi) - \frac{3}{2}}
\end{align*}
Old answer (Jan 18, 2013). Here is an another approach using the principle of analytic continuation:
Let $I$ denote the integral. Applying the substitution $x = e^{-t}$, the integral is recast as
\begin{align*} I
&= \int_{0}^{\infty} \left\{ \frac{1}{(1-e^{-t})^{2}} - \frac{2}{t(1-e^{-t})} + \frac{1}{t^2} \right\} e^{-t} \, \mathrm{d}t \\
&= \int_{0}^{\infty} \left\{ \frac{e^{t}}{(e^{t} - 1)^{2}} - \frac{1}{t^2} \right\} \, \mathrm{d}t + \int_{0}^{\infty} \left\{ \frac{1 + e^{-t}}{t^2} - \frac{2}{t(e^{t}-1)} \right\} \, \mathrm{d}t.
\end{align*}
It is easy to observe that the first integral evaluates as
\begin{align*}
\int_{0}^{\infty} \left\{ \frac{e^{t}}{(e^{t} - 1)^{2}} - \frac{1}{t^2} \right\} \, \mathrm{d}t
&= \left[ \frac{1}{t} - \frac{1}{e^{t} - 1} \right]_{0}^{\infty}
= -\frac{1}{2}.
\end{align*}
We thus focus on the second integral. We do so by first introducing the regularized version
$$ F(s) := \int_{0}^{\infty} \left\{ \frac{1 + e^{-t}}{t^2} - \frac{2}{t(e^{t}-1)} \right\} e^{-st} \, \mathrm{d}t. $$
Differentiating $F(s)$ twice, we get
\begin{align*}F''(s)
&= \int_{0}^{\infty} \left\{ 1 + e^{-t} - \frac{2t}{(e^{t}-1)} \right\} e^{-st} \, \mathrm{d}t \\
&= \frac{1}{s} + \frac{1}{s+1} - 2\sum_{n=1}^{\infty} \frac{1}{(n+s)^2} \\
&= \frac{1}{s} + \frac{1}{s+1} - 2\psi'(s+1),
\end{align*}
where $\psi(\cdot)$ refers to the digamma function. Integrating both sides and utilizing the condition $F'(+\infty) = 0$ and the formula $\psi_0(s) = \log s + o(1)$ together, we get
$$ F'(s) = \log s + \log(s+1) - 2\psi_{0}(s+1). $$
Integrating both sides again, we have
$$ F(s) = s \log s + (s+1)\log(s+1) - 2s - 1 - 2\log\Gamma(s+1) + C. $$
To determine the constant $C$, we rearrange the terms as
$$ F(s) = \left\{ (s+1)\log\left(\frac{s+1}{s}\right) - 1 \right\} + 2\left\{ \left(s+\frac{1}{2}\right)\log s - s - \log\Gamma(s+1) \right\} + C. $$
Then by the Stirling's formula, we have
$$ 0 = F(+\infty) = -\log(2\pi) + C $$
and thus $C = \log (2\pi)$. Therefore
$$I
= -\frac{1}{2} + F(0)
= \bbox[color:navy;padding:5px;border:1px dotted navy;]{\log(2\pi) - \frac{3}{2}}$$
as desired.
| {
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"url": "https://math.stackexchange.com/questions/281406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
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Two different solutions to integral Given the very simple integral
\begin{equation}
\int -\frac{1}{2x} dx
\end{equation}
The obvious solution is
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln{|x|} + C
\end{equation}
However, by the following integration rule
\begin{equation}
\int \frac{1}{ax + b} dx = \frac{1}{a} \ln{|ax + b|} + C
\end{equation}
the following solution is obtained
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2}\ln{|-2x|} + C
\end{equation}
Why are these solutions different? Which is correct?
The second solution can be simplified
\begin{equation}
-\frac{1}{2}\ln{|-2x|} + C = -\frac{1}{2}\ln{|-2|} -\frac{1}{2}\ln{|x|} + C= -\ln{\frac{1}{\sqrt{2}}} - \frac{1}{2}\ln{|x|} + C
\end{equation}
but they still differ.
| Both of the results is Ok. Note that there is no need that two constants $C$ in the first result and $C$ in the second one are the same. Here, we have $C_1=C_2\times0.5\ln|2|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
| multiply abc
$a+b+c\geq abc $
$a^2+b^2+c^2\geq abc $
$a^2+b^2+c^2-(a+b+c)$
$⇔(a-\frac12)^2+(b-\frac12)^2+(c-\frac12)^2\geq \frac34$
therefore
$a^2+b^2+c^2\geq \frac34+abc$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Fibonacci Proof
Prove that: $$F_1F_2+F_2F_3+F_3F_4+\cdots+F_{2n-1}F_{2n}=F_{2n}^2$$
I set it up so:
$$F^2(2k) + F(2k+1)F(2k+2) = F^2(2k+2)$$
I've tried:
$$F(2k)^2 + F(2k+1)*F(2k+2) = F(2k+2)*F(2k)+F(2k+2)*F(2k+1)$$
$$F(2k)^2 = F(2k+2)*F(2k)$$
$$F(2k+1)*F(2k) - F(2k)*F(2k-1) = F(2k)*F(2k)+F(2k)*F(2k+1)$$
$$-F(2k)*F(2k-1) = F(2k)^2$$
??? Is there something wrong with my initial induction hypothesis ?
| $$F_{r+2}^2-F_r^2=(F_{r+2}-F_r)(F_{r+2}+F_r)=F_{r+1}(F_{r+2}+F_r)$$
So, $$F_{r+2}^2-F_r^2=F_{r+2}F_{r+1}+F_{r+1}F_r$$
We can utilize this fact to prove the required proposition without using induction.
Put $r=2n-2,2n-4,\cdots,4,2,0$ to get,
$$F_{2n}^2-F_{2n-2}^2=F_{2n}F_{2n-1}+F_{2n-1}F_{2n-2}$$
$$F_{2n-2}^2-F_{2n-4}^2=F_{2n-2}F_{2n-3}+F_{2n-3}F_{2n-4}$$
$$\cdots$$
$$F_{6}^2-F_{4}^2=F_{6}F_{5}+F_{5}F_{4}$$
$$F_{4}^2-F_{2}^2=F_{4}F_{3}+F_{3}F_{2}$$
$$F_{2}^2-F_{0}^2=F_{2}F_{1}+F_{1}F_{0}$$
Adding them we get, $$F_{2n}^2-F_0^2=\sum_{0\le m\le 2n-1}F_mF_{m+1}$$
But $$F_0=0\implies F_{2n}^2-F_0^2=F_{2n}^2$$
and $$ \sum_{0\le m\le 2n-1}F_mF_{m+1}=\sum_{1\le m\le 2n-1}F_mF_{m+1}+F_0F_1=\sum_{1\le m\le 2n-1}F_mF_{m+1}$$
So, $$F_{2n}^2=\sum_{1\le m\le 2n-1}F_mF_{m+1}$$
Similarly putting $r=2n-1,2n-3,\cdots,3,1$ and adding them we can derive $$\sum_{1\le m\le 2n}F_mF_{m+1}=F_{2n+1}^2-F_1^2=F_{2n+1}^2-1\text{ as }F_1=1$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Review on double integral Forgot how to do these :
$$\displaystyle\int_0^1\int_0^1\frac{\text{d}x\text{d}y}{1-x^2y^2}$$
| Consider the integral over $x$:
$$\int_0^1 dx \; \frac{1}{1-x^2 y^2} $$
Make the substitution $x = \sin{\theta}/y$, $dx = \cos{\theta}/y \: d \theta$:
$$\begin{align} &= \frac{1}{y} \int_0^{\arcsin{y}} d \theta \sec{\theta} \\ &= \frac{1}{y} [\log{(\sec{\theta} + \tan{\theta})]_{0}^{\arcsin{y}}} \\ &= \frac{1}{2 y} \log{\left ( \frac{1+y}{1-y} \right )} \end{align} $$
Now you can do the integral over $y$:
$$\begin{align} \int_0^1 dy \: \int_0^1 dx \: \frac{1}{1-x^2 y^2} &= \frac{1}{2} \int_0^1 dy \: \frac{1}{y} \log{\left ( \frac{1+y}{1-y} \right )}\\ &= \sum_{n=0}^{\infty} \int_0^1 dy \: \frac{y^{2 n}}{2 n+1} \\ &= \sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}\\ &= \frac{\pi^2}{8} \\\\ \end{align} $$
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 0
} |
Approaches to integrate $\int_0^1 \frac{x}{\sqrt{a+bx+cx^2}} dx$ I search for a good approach (or a general solution technique) to integrate
$$\int_0^1 \frac{x}{\sqrt{a+bx+cx^2}} dx$$
My knowledge on integration ends with rational functions.
| $\displaystyle \int\frac{x}{\sqrt{a+bx+cx^2}}dx = \frac{1}{2c}\int\frac{\left(2cx+b\right)-b}{\sqrt{a+bx+cx^2}}dx$
$ \displaystyle = \frac{1}{2c}\int\frac{2cx+b}{\sqrt{a+bx+cx^2}}dx-\frac{b}{2c}\int\frac{1}{\sqrt{a+bx+cx^2}}dx$
for First one put $a+bx+cx^2 = t^2$ and $\left(2cx+b \right)dx = 2tdt$
$ \displaystyle = \frac{1}{c}\sqrt{a+bx+cx^2}-\frac{b}{2c}.\frac{1}{\sqrt{c}}\int\frac{1}{\sqrt{x^2+\frac{b}{c}.x+\frac{a}{c}}}dx$
$\displaystyle = \frac{1}{c}\sqrt{a+bx+cx^2}-\frac{b}{2c\sqrt{c}}\int\frac{1}{\sqrt{\left(x+\frac{b}{2c}\right)^2+\left(\frac{\sqrt{b^2-4ac}}{2c}\right)^2}}dx$
$\displaystyle = \frac{1}{c}\sqrt{a+bx+cx^2}-\frac{b}{2c\sqrt{c}}\ln \left|\left(x+\frac{b}{2c}\right)+\sqrt{x^2+\frac{b}{c}.x+\frac{a}{c}}\right|+\mathbb{C}$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
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} |
A definite integral with hyperbolic cosines I want to show that $$ \int_{0}^{\infty} \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} \ dx = \frac{\cos ( \frac{a}{2} ) \cos ( \frac{b}{2})} {\cos (a) + \cos (b)} \ , \ |a|+|b| < \pi.$$
I thought one approach would be to integrate the appropriate function around a rectangle with vertices at $z=R$, $z=R+i$, $z=-R+i$, and $z=-R$.
I tried $\displaystyle f(z) = \frac{e^{(a+b)z}}{\cosh (\pi z)}, f(z) = \frac{e^{az} \cosh (bz)}{\cosh (\pi z)}$, and $\displaystyle f(z) = \frac{\cosh (az) \cosh (bz)}{\cosh (\pi z)}$.
None of these three choices for $f(z)$ worked.
EDIT:
As AD. stated in the comments, the integral can be rewritten as
$$ \frac{1}{2} \int_{0}^{\infty} \frac{\cosh \big((a+b)x\big) + \cosh \big((a-b) x \big)}{\cosh (\pi x)} \ dx. $$
By integrating $ \displaystyle f(z) = \frac{e^{\alpha z}}{\cosh (\pi z)}$ around the rectangle described above, one can show that $$\int_{0}^{\infty} \frac{\cosh (\alpha x)}{\cosh (\pi x)} \ dx = \frac{1}{2} \sec \left(\frac{\alpha}{2} \right) \ , \ |\alpha| < \pi. $$
Therefore,
$$ \begin{align} &\int_{0}^{\infty} \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} \ dx \\ &= \frac{1}{4} \left[\sec \left(\frac{a+b}{2} \right) + \sec \left( \frac{a-b}{2}\right) \right] \\ &= \frac{1}{4} \left(\frac{1}{\cos (\frac{a}{2}) \cos (\frac{b}{2}) - \sin (\frac{a}{2}) \sin (\frac{b}{2})} + \frac{1}{\cos (\frac{a}{2}) \cos (\frac{b}{2}) + \sin (\frac{a}{2}) \sin (\frac{b}{2})}\right) \\ &= \frac{1}{2} \frac{\cos (\frac{a}{2}) \cos(\frac{b}{2})}{\cos^{2}(\frac{a}{2})\cos^{2} (\frac{b}{2}) - \sin^{2} (\frac{a}{2}) \sin^{2} (\frac{b}{2})} \\ &= \frac{2 \cos (\frac{a}{2}) \cos(\frac{b}{2})}{\big(1+\cos(a)\big) \big(1+\cos(b) \big) - \big(1-\cos(a)\big) \big(1-\cos(b)\big)} \\ &= \frac{\cos (\frac{a}{2}) \cos (\frac{b}{2})}{\cos(a) + \cos (b)} . \end{align}$$
| I have a way of showing this without contour integration in the complex plane. There is a bit of a trick involved and, frankly, Mathematica misleads. It should be noted that the condition $|a|+|b| < \pi$ is needed for the integral to converge. Basically, rewrite the $\cosh$'s as exponentials:
$$\begin{align} \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} &= 2 \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{1+e^{-2 \pi x}} e^{-\pi x} \\ &= \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \int_{0}^{\infty} dx \: (e^{a x}+e^{-a x}) (e^{b x}+e^{-b x}) e^{-(2 k+1) \pi x} \\ \end{align} $$
Evaluating the integrals, we get
$$= \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{(2 k+1)\pi -(a+b)} + \frac{1}{(2 k+1)\pi +(a+b)}\right ] $$
$$ + \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{(2 k+1)\pi -(a-b)} + \frac{1}{(2 k+1)\pi +(a-b)} \right ] $$
Here I note that $a+b$ and $a-b$ should not be some multiple of $\pi$, so that the above sums behave properly.
To get the sums into a somewhat familiar form, I rearrange them a bit to get
$$= \frac{1}{4 \pi} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{k +\left (\frac{1}{2} - \frac{a+b}{2 \pi} \right )} + \frac{1}{k +\left (\frac{1}{2} + \frac{a+b}{2 \pi}\right )}\right ] $$
$$ + \frac{1}{4 \pi} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{k +\left (\frac{1}{2} - \frac{a-b}{2 \pi} \right )} + \frac{1}{k +\left (\frac{1}{2} + \frac{a-b}{2 \pi}\right )}\right ] $$
Now, here is the interesting part (at least to me). Let
$$f(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k+z} $$
This looks like it should be a trig function of some sort. It is not; rather, it is something called a Hurwitz-Lerch transcendent, which does not look like it will be much help. That said, it almost looks like a trig function, so I instead considered the following:
$$\begin{align} f(z) + f(1-z) &= \sum_{k=0}^{\infty} \frac{(-1)^k}{k+z} + \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1-z}\\ &= \sum_{k=0}^{\infty} \frac{(-1)^k}{z+k} + \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{z-(k+1)}\\ &= \sum_{k=-\infty}^{\infty} \frac{(-1)^k}{z+k} \\ &= \frac{\pi}{\sin{\pi z}}\\ \end{align}$$
This is very helpful, because we have precisely this functional form above, e.g.,
$$\frac{1}{2} - \frac{a+b}{2 \pi} = 1 - \left ( \frac{1}{2} + \frac{a+b}{2 \pi} \right ) $$
$$\frac{1}{2} - \frac{a-b}{2 \pi} = 1 - \left ( \frac{1}{2} + \frac{a-b}{2 \pi} \right ) $$
So we get for the integral:
$$\begin{align} \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} &= \frac{1}{4} \left [ \frac{1}{\sin{\left ( \frac{\pi}{2} - \frac{a+b}{2} \right )}} + \frac{1}{\sin{\left ( \frac{\pi}{2} - \frac{a-b}{2} \right )}} \right ] \\ &= \frac{1}{4} \left [ \frac{1}{\cos{\left ( \frac{a+b}{2} \right )}} + \frac{1}{\cos{\left ( \frac{a-b}{2} \right )}} \right ] \\ &= \frac{\cos{\frac{a}{2}} \cos{\frac{b}{2}}}{\cos{a} + \cos{b}} \end{align}$$
QED
| {
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"source": "stackexchange",
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"answer_count": 2,
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Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2} $$
Change of variable
$$ t=\frac{x}{n}$$
$$ I_n = \frac{1}{2n}\int_0^n \frac{\left(1-\frac{x}{n}\right)^n \mathrm dx}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}$$
Taylor expansions:
$$ \left(1-\frac{x}{n}\right)^n = e^{-x} \left(1-\frac{x^2}{2n}+\frac{3x^4-8x^3}{24n^2}+\mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
$$ \frac{1}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)} = 1+\frac{x}{n}+\frac{x^2}{2n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) $$
$$ \frac{\left(1-\frac{x}{n}\right)^n }{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}=e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
So
$$ I_n = \frac{1}{2n} \int_0^n e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right) \mathrm dx $$
$$ I_n = \frac{1}{2n} \left(1+\frac{1}{n}+\frac{1}{2n^2}\times 2-\frac{1}{2n}\times 2 - \frac{1}{2n^2} \times 6 + \frac{1}{8n^2} \times 24 - \frac{1}{3n^2} \times 6+ \mathcal{O} \left(\frac{1}{n^3} \right) \right) $$
$$ I_n = \frac{1}{2n}-\frac{1}{2n^3}+\mathcal{O} \left(\frac{1}{n^4} \right)$$
For example Wolfram gives:
$$ 1-1000^2+ 2\times1000^3\int_0^{\pi/4} \tan(x)^{1000} \mathrm dx \approx 4.9\times 10^{-6}$$
I'm quite sure of my work, I would just like to know if everything is correct!
| May be this can give you an asymptotics. Consider even case
$$
I_{2k}=\int\limits_{0}^{1}\frac{t^{2k}}{t^2+1}dt=\int\limits_{0}^{1}(-1)^k\left(\sum\limits_{j=0}^{k-1}(-1)^{j+1}t^{2j}+\frac{1}{t^2+1}\right)dt=\\ (-1)^k\left(-\sum\limits_{j=0}^{k-1}\frac{(-1)^j}{2j+1}+\frac{\pi}{4}\right)=\sum\limits_{j=k}^\infty\frac{(-1)^{j+k}}{2j+1}=\sum\limits_{j=0}^\infty\frac{(-1)^{j}}{2j+2k+1}
$$
and the odd case
$$
I_{2k+1}=\int\limits_{0}^{1}\frac{t^{2k+1}}{t^2+1}dt=\int\limits_{0}^{1}(-1)^k\left(\sum\limits_{j=0}^{k-1}(-1)^{j+1}t^{2j+1}+\frac{t}{t^2+1}\right)dt=(-1)^k\left(-\sum\limits_{j=0}^{k-1}\frac{(-1)^j}{2j+2}+\frac{1}{2}\ln2\right)=\sum\limits_{j=k}^{\infty}\frac{(-1)^{j+k}}{2j+2}=\sum\limits_{j=0}^\infty\frac{(-1)^{j}}{2j+2k+2}
$$
Hence
$$
I_n=\sum\limits_{j=0}^\infty\frac{(-1)^{j}}{2j+n+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
determinant $s=n+1$ Need help
$A=\begin{vmatrix}
s&s&s &\cdots & s&s\\
s&1&s &\cdots & s&s\\
s&s&2 &\cdots & s&s\\\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\
s&s&s &\cdots & n-1&s\\
s&s&s &\cdots & s&n\\
\end{vmatrix}$
a) calculate $\det A$ when $s =n+1$
Prove that in that case is invertible
b) It may happened that $s=4n$ and $\det A=26\times 3^5$
in a) $\det A = -1$ right? how to prove the invertile?
b) plz some hint
| Adding a multiple of one row to another row in a square matrix does not change the determinant (link). Hence, by subtracting the first row from all the others, we find
$$A=\begin{pmatrix}
s&s&s &\cdots & s&s\\
s&1&s &\cdots & s&s\\
s&s&2 &\cdots & s&s\\
\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\
s&s&s &\cdots & n-1&s\\
s&s&s &\cdots & s&n\\
\end{pmatrix}$$
and
$$B=\begin{pmatrix}
s&s&s &\cdots & s&s\\
0 & 1-s & 0 & \cdots & 0 & 0 \\
0 & 0 & 2-s & \cdots & 0 & 0 \\
\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\
0 & 0 & 0 & \cdots & n-1-s & 0 \\
0 & 0 & 0 & \cdots & 0 & n-s \\
\end{pmatrix}$$
have the same determinant. The determinant of $B$ can be "read off" using the Leibniz Formula; the only non-zero contribution to the sum comes from the identity permutation. Hence $$\det(A)=\det(B)=s\prod_{i=1}^n (i-s).$$ In the case when $s=n+1$, we have
\begin{align*}
\det(A) &= (n+1) \times -n \times -(n-1) \times \cdots \times -1 \\
&= (-1)^{n-1} (n+1)!.
\end{align*}
Since $\det(A) \neq 0$ the matrix $A$ is invertible.
In the case when $s=4n$, we have
\begin{align*}
\det(A) &= 4n \times (1-4n) \times (2-4n) \times \cdots \times (n-4n).
\end{align*}
For no value of $n$ does this equal $26 \times 3^5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculus integration problem: $\int \sin^5 (x) \cos^2 (x)\,dx$ What's the integration of $$\int \sin^5 (x) \cos^2 (x)\,dx?$$
| $$ \int \sin^5 (x) \cos^2(x) dx $$
$$= \int(\sin^2(x))^2 \cos^2(x) \sin(x) dx$$
$$=-\int(1 - \cos^2(x))^2 cos^2(x) (-sin(x) dx) $$
Let $u = \cos(x)$
$\implies du = -\sin(x) dx$
$$= -\int(1 - u^2)² u² (du)$$
$$= -\int(1 - 2u^2 + u^4) u^2 du $$
$$= -\int(u^2 - 2u^4+ u^6) du$$
$$= -\left(\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7}\right) + C$$
$$= -u^3\left(\frac{1}{3} - \frac{2u^2}{5} +\frac{ u^4}{7}\right) + C $$
$$= -\cos^3(x) \left(\frac{1}{3} - \frac{2\cos^2(x)}{5} + \frac{\cos^4(x)}{7}\right) + C $$
$$= -\cos^3(x)\frac{15\cos^4(x) - 42\cos^2(x) + 35}{105} + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
General expression for exponentiating a matrix? $A= \begin{pmatrix}
\cos(x) & \sin(x) \\
-\sin(x) & \cos(x) \\
\end{pmatrix}$
I found that
$A^2 = \begin{pmatrix}
\cos^2(x)-\sin^2(x) & 2\cos(x)\sin(x) \\
-2\cos(x)\sin(x) & \cos^2(x)-\sin^2(x) \\
\end{pmatrix}$
$A^2 = \begin{pmatrix}
\cos(2x) & \sin(2x) \\
-\sin(2x) & \cos(2x) \\
\end{pmatrix}$
and
$A^3 = \begin{pmatrix}
\cos^2(x)[\cos^2(x)-3\sin^2(x)] & \sin(x)[3\cos^2(x)\sin^2(x)] \\
-\sin(x)[3\cos^2(x)+\sin^2(x)] & \cos(x)[-3\sin^2(x)+\cos^2(x)] \\
\end{pmatrix}$
$A^3 = \begin{pmatrix}
\cos(2x)cos(x) - \sin(x)\sin(2x) & \sin(x)\cos(2x) + \cos(x)\sin(2x) \\
-\cos(x)sin(2x)-\sin(x)\cos(2x) & \cos(x) + \cos(2x) -\sin(x)\sin(2x) \\
\end{pmatrix}$
but I am unable to find a general solution. I probably need to perform some algebraic manipulations or whatnot.
| In general, following formula holds: (Proof of this formula uses induction.)
$$A^n =\begin{pmatrix}
\cos (nx) & -\sin(nx) \\
\sin(nx) & \cos(nx)
\end{pmatrix}$$
In fact, matrix $\begin{pmatrix}
\cos (x) & -\sin(x) \\
\sin(x) & \cos(x)
\end{pmatrix}$ behave similar to $\cos x + i \sin x$.
| {
"language": "en",
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"answer_count": 2,
"answer_id": 1
} |
complex power series Sorry for taking this from another question, but the second part was never answered, and I'm not sure how to get there. From: Prove the following equation of complex power series.
Show that for $|z| \lt 1$ with $z \in \Bbb C$, we have
$$ \sum_{k=0}^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \frac{z}{1-z} $$
$$ \sum_{k=0}^\infty \frac{2^k{z^2}^k}{1+{z^2}^{k}} = \frac{z}{1-z} $$
My guess is that the second one is obtained by differentiating the first one or something like that, but I can't manage to prove the first one.
| For the first identity, you're trying to prove that
$$
0 = \frac{z}{1-z} - \sum_{k=0}^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \lim_{K\to\infty} \bigg( \frac{z}{1-z} - \sum_{k=0}^K\frac{{z^2}^k}{1-{z^2}^{k+1}} \bigg)
$$
for $|z|<1$. If you work out the exact value of the expression on the right-hand side for small values of $K$, you get
\begin{align*}
\frac{z}{1-z} - \sum_{k=0}^0 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z}{1-z} - \frac{z}{1-z^2} = \frac{z^2}{1-z^2} \\
\frac{z}{1-z} - \sum_{k=0}^1 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z^2}{1-z^2} - \frac{z^2}{1-z^4} = \frac{z^4}{1-z^4} \\
\frac{z}{1-z} - \sum_{k=0}^2 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z^4}{1-z^4} - \frac{z^4}{1-z^8} = \frac{z^8}{1-z^8}
\end{align*}
and so on, suggesting a pattern that successfully be proved by induction. Then you only need to show that
$$
\lim_{K\to\infty} \frac{z^{2^{K+1}}}{1-z^{2^{K+1}}} = 0
$$
for $|z|<1$. The second identity can be proved similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Problem with simple integral I'm trying to solve this simple integral:
$$\frac12 \int \frac{x^2}{\sqrt{x + 1}} dx$$
Here's what I have done so far:
*
*$\displaystyle t = \sqrt{x + 1} \Leftrightarrow x = t^2 - 1 \Rightarrow dx = 2t dt$
*$\displaystyle \frac12 \int \frac{x^2}{\sqrt{x + 1}} dx = \int \frac{t (t^2 - 1)^2}t dt$
*$\displaystyle \int (t^2 - 1)^2 dt = \frac15 t^5 - \frac23 t^3 + t + C$
*$\displaystyle \frac15 t^5 - \frac23 t^3 + t + C = \frac15 \sqrt{(x + 1)^5} - \frac23 \sqrt{(x + 1)^3} + \sqrt{x + 1} + C$
WolframAlpha tells me steps 1 and 3 are right so the mistake must be somewhere in steps 2 and 4, but I really can't see it.
| $$\begin{align}\frac{x^2}{\sqrt{x+1}} &= \frac{x^2-1}{\sqrt{x+1}} + \frac{1}{\sqrt{x+1}}\\&=(x-1)\sqrt{x+1} + \frac{1}{\sqrt{x+1}}\\&=(x+1)^{3/2} - 2\sqrt{x+1} + \frac{1}{\sqrt{x+1}}\end{align}$$
I think you can integrate each of these terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$? Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$ for $k \ge 1$?
Suppose $p,q,r$ are positive integers satisfying $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}<1$, prove that $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\le \dfrac{41}{42} $ ?
| For the second question, assume without loss of generality that $p\le q\le r$. Let $$d=1-\left(\frac1p+\frac1q+\frac1r\right)\;;$$ $d$ is rational, so let $d=\dfrac{m}n$ in lowest terms. If $$1>\frac1p+\frac1q+\frac1r>\frac{41}{42}\;,$$ then $d<\dfrac1{42}$, so $n>42$. Clearly $n\mid pqr$, so $pqr>42$, and since $4^3=64$, we must have $2\le p\le 3$.
Suppose that $p=2$; then $qr>21$, and clearly $q\ge 3$. If $q=3$, then $r>7$, so $$\frac1p+\frac1q+\frac1r\le\frac12+\frac13+\frac18=\frac{23}{24}\le\frac{41}{42}\;.$$ If $q=4$, then $r\ge 6$, and $$\frac1p+\frac1q+\frac1r\le\frac12+\frac14+\frac16=\frac{22}{24}\le\frac{41}{42}\;.$$ And if $q=5$, then $r\ge 5$, and $$\frac1p+\frac1q+\frac1r\le\frac12+\frac15+\frac15=\frac9{10}\le\frac{41}{42}\;.$$ Thus, $p\ne 2$.
If $p=3$, then $qr>14$. If $q=3$, then $r\ge 5$, and $$\frac1p+\frac1q+\frac1r\le\frac13+\frac13+\frac15=\frac{13}{15}\le\frac{41}{42}\;.$$ If $q=4$, then $r\ge 4$, and $$\frac1p+\frac1q+\frac1r\le\frac13+\frac14+\frac14=\frac5{6}\le\frac{41}{42}\;.$$ Thus, $p\ne 3$, and it is impossible to have $d<\dfrac1{42}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Finding an explicit solution of a differential equation
Find an explicit solution of an initial value problem $$\frac{dy}{dx}=\frac{2x}{1+2y}, \; y(2)=0$$
Attempt: I have no problem finding the general solution which is
$y+y^2=x^2+C$
Then, I find the implict solution which I think is correct, but I am not sure
$y+y^2=x^2-4$
Now, how should I go about finding an explicit solution? Thanks for your help.
| Complete the square on the right hand side i.e.
\begin{align}
y+y^2 & = x^2-4\\
y^2 + y + \dfrac14 & = x^2-4 + \dfrac14\\
\left( y + \dfrac12 \right)^2 & = x^2-\dfrac{15}4\\
y + \dfrac12 & = \pm \sqrt{x^2 - \dfrac{15}4}\\
y & = - \dfrac12 \pm \dfrac{\sqrt{4x^2-15}}2
\end{align}
Further, we need $y(2) = 0$ and hence $y = - \dfrac12 + \dfrac{\sqrt{4x^2-15}}2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A log improper integral Evaluate :
$$\int_0^{\frac{\pi}{2}}\ln ^2\left(\cos ^2x\right)\text{d}x$$
I found it can be simplified to
$$\int_0^{\frac{\pi}{2}}4\ln ^2\left(\cos x\right)\text{d}x$$
I found the exact value in the table of integrals:
$$2\pi\left(\ln ^22+\frac{\pi ^2}{12}\right)$$
Anyone knows how to evaluate this?
| First of all, substituting $x\to\frac\pi2-x$ yields
$$\int_0^{\frac\pi2} \ln(\cos^2(x)) \, dx = \int_0^{\frac\pi2} \ln(\sin^2(x)) \, dx = 2 \int_0^{\frac\pi2} \ln(\sin(x)) \, dx = -\pi\ln(2)$$
where the log-sine integral can be computed using the identity
$$\ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$$
Using the same substitution,
$$\int_0^{\frac\pi2} \ln^2(\cos^2(x)) \, dx = \int_0^{\frac\pi2} \ln^2(\sin^2(x)) \, dx = 4 \int_0^{\frac\pi2} \ln^2(\sin(x)) \, dx$$
In the identity above, square both sides and integrate:
$$\begin{align}
\ln^2(\sin(x)) &= \ln^2(2) + 2\ln(2) \sum_{k=1}^\infty \frac{\cos(2kx)}k + \left(\sum_{k=1}^\infty \frac{\cos(2kx)}k\right)^2 \\[2ex]
\int_0^{\frac\pi2} \ln^2(\sin(x)) \, dx &= \frac\pi2 \ln^2(2) - 2\ln(2) \int_0^{\frac\pi2} (\ln(2) + \ln(\sin(x)) \, dx \\[1ex]
& \qquad + \int_0^{\frac\pi2} \sum_{k=1}^\infty \frac{\cos^2(2kx)}{k^2} \, dx + 2 \int_0^{\frac\pi2} \sum_{i<j}\frac{\cos(2ix)\cos(2jx)}{ij} \, dx\\[2ex]
&= \frac\pi2 \ln^2(2) + \frac12 \sum_{k=1}^\infty \frac1{k^2} \int_0^{\frac\pi2} (1+\cos(4kx)) \, dx \\[1ex]
&\qquad + \sum_{i<j} \frac1{ij} \int_0^{\frac\pi2} (\cos(2(i+j)x) + \cos(2(i-j)x)) \, dx \\[2ex]
&= \frac\pi2\ln^2(2) + \frac\pi4 \sum_{k=1}^\infty \frac1{k^2} \\[2ex]
&= \frac\pi2\ln^2(2) + \frac{\pi^3}{24}
\end{align}$$
and so
$$\int_0^{\frac\pi2} \ln^2(\cos^2(x)) \, dx = \boxed{2\pi\ln^2(2) + \frac{\pi^3}6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 4
} |
The Least Natural number $n$ which has $18$ divisors The Least Natural number $n$ which has $18$ divisors
My Try:: $18 = 2\times 3 \times 3 = 3^2 \times 2$
Now How can I solve it
Thanks
| The smallest prime power $n=p^k$ with 18 divisors would need $k=17$ and hence be $2^{17}$.
If $n=p^kq^m$ with two primes $p,q$ and $k\ge m\ge1$, then it has $(k+1)(m+1)$ divisors, which allows $k=8,m=1$ or $k=5,m=2$ and suggests $n=2^83$ or $n=2^53^2$. (Of course, it is best to take the smallest possible primes and use the biggest exponent with the smallest prime).
If $n=p^kq^mr^l$ correspondingly with $k\ge m\ge l\ge 1$, then we have $(k+1)(m+1)(l+1)$ divisors, which allows $k=m=2$, $l=1$ only and suggests $n=2^23^25$.
Four or more distinct prime divisors are not possible as that would produce either $2^4=16$ or at least $3\cdot 2^3=24$ divisors, but never $18$.
Choose the smallest number among $2^{17}$, $2^83$, $2^53^2$ and $2^23^25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Simplification of $\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$ Simplify the expression $$\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$$
I have done this way : $(1-\cos\alpha \cos\beta)^2 = 1-2\cos\alpha \cos\beta +\cos^2\alpha \cos^2\beta$
Please guide further....
| $$\begin{align}
&\phantom{=\;}( 1 - \cos\alpha \cos\beta)^2 - \sin^2\alpha \sin^2\beta \\
&= ( 1 - \cos\alpha \cos\beta)^2 - (\sin\alpha \sin\beta)^2 &(1) \\
&= ( 1 - \cos\alpha \cos\beta - \sin\alpha \sin\beta)( 1 - \cos\alpha \cos\beta + \sin\alpha \sin\beta) &(2) \\
&= \left( 1 - (\cos\alpha \cos\beta + \sin\alpha \sin\beta)\right)\left( 1 - (\cos\alpha \cos\beta - \sin\alpha \sin\beta) \right) &(3)\\
&= \left( 1 - \cos(\alpha-\beta)\right)\left( 1 - \cos(\alpha+\beta) \right) &(4) \\
&= 2 \sin^2\left(\frac{\alpha-\beta}{2}\right) \cdot 2 \sin^2\left(\frac{\alpha+\beta}{2}\right) &(5) \\
&= 4 \sin^2\left(\frac{\alpha-\beta}{2}\right)\sin^2\left(\frac{\alpha+\beta}{2}\right) &(6) \\
&= \left( 2 \sin\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha+\beta}{2}\right) \right)^2 &(7) \\
&= \left( \cos\beta - \cos\alpha \right)^2 &(8)
\end{align}$$
so that
$$
\sqrt{( 1 - \cos\alpha \cos\beta)^2 - \sin^2\alpha \sin^2\beta} = \left| \cos\beta - \cos\alpha \right|$$
Steps:
*
*Regroup
*Difference of squares: $(x-y)^2 = (x-y)(x+y)$
*Regroup
*Angle Addition formulas for cosine
*Half-angle formulas for sine
*Simplification
*Regroup
*Product-to-Sum (well, -Difference here) "Prosthaphaeresis" formula
See Wikipedia's "List of Trigonometric Identities" page for various formulas (especially the one for Step 8, which isn't so well known but comes in handy).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Verifying Trigonometric Identities: $2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$ Verify that $$ 2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$$
| It's generally easier to turn more-complicated stuff into less-complicated stuff; and one common suggestion is always "first convert to sine and cosine":
$$\begin{align}
\frac{1-\tan^2\theta}{1+\tan^2\theta}
&= \frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}} \\
&=\frac{\left(\cos^2\theta - \sin^2\theta\right)/\cos^2\theta}{\left(\cos^2\theta + \sin^2\theta\right)/\cos^2\theta}\\
&=\frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta} &\text{denom = 1}\\[6pt]
&= \cos^2\theta - \sin^2\theta &(1)\\
&= \cos^2\theta - \left( 1 - \cos^2\theta \right) \\
&= 2 \cos^2\theta - 1 &(2)
\end{align}$$
You can jump from (1) to (2) by recognizing both as forms of $\cos 2\theta$.
With a little more fluency in interrelations between more of the trig functions, you might also have taken this route:
$$\begin{align}
\frac{1-\tan^2\theta}{1+\tan^2\theta} &= \frac{1-\tan^2\theta}{\sec^2\theta} \\
&=\left( 1 - \tan^2\theta \right) \cos^2\theta \\
&=\cos^2\theta - \sin^2\theta \\[6pt]
&= 2\cos^2\theta - 1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ I was solving some problems and I came across this question:
Q: The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ is
(a) 0 (b) 2
(c) 4 (d) none of these
Solution: put $y = x + (3+5)/2 = x+4$
the equation becomes
=> $(y-1)^4 + (y+1)^4 = 16$ ---- (i)
=> $2{y^4 + 6(y)^2 + 1 } = 16$ --------(ii)
My question is how was (i) converted to (ii)? I just couldn't get it. Please help?
| What you need is the expansion of $(a+b)^4$. This is a special case of the very important Binomial Theorem. We have
$$(a+b)^4=a^4+\binom{4}{1}a^3b+\binom{4}{2}a^2b^2+\binom{4}{3}ab^3+b^4.$$
This simplifies to $a^4+4a^3b+6a^2b^2+4ab^3+b^4$.
Put $a=y$, $b=1$, then $a=y$, $b=-1$, and add. There is a pleasant amount of cancellation.
Remarks: $1.$ The symmetrizing move $y=x+4$ is a useful idea.
$2$. For reasons of familiarity, we change the name, and study $(x+1)^4+(x-1)^4$. This function is symmetric about $x=0$. Our function is not $16$ at $x=0$, and by symmetry there are just as many solutions of $(x+1)^4+(x-1)^4=16$ with $x\gt 0$ as there are with $x\lt 0$. So let's see how many there are with $x\gt 0$.
$3.$ The solution $x=1$ is obvious. It is reasonably clear that there are no solutions with $0\lt x\lt 1$. And past $x=1$, our function is increasing. So there is exactly one positive solution, and therefore altogether there are two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Trig substitution integral I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$
The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substitution: $x = 2\tan\theta$, I get an answer that is slightly different but not equivalent, and I've looked at this over and over and I couldn't quite figure out what I did wrong.
$$x = 2\tan\theta$$
$$dx = 2\sec^2\theta d\theta$$
$$\int{ \frac {5x + 1}{x^2 + 4}dx} = \frac{1}{4}\int{\frac{10\tan\theta + 1}{\sec^2\theta} 2\sec^2\theta \,d\theta}$$
$$ = \frac{1}{2}\int{10 \tan\theta + 1}\space d\theta$$
$$ = 5 \ln|\sec\theta| + \frac{\theta}{2} + C$$
We know $\theta = \displaystyle\arctan\left(\frac x2\right)$ and since $\tan\theta = \displaystyle\frac{x}{2}$, we can draw a triangle to see that $\sec\theta = \displaystyle\frac{\sqrt{x^2 + 4}}{2}$.
$$5 \ln|\sec\theta| + \frac{\theta}{2} = 5\ln\left({\frac{\sqrt{x^2 + 4}}{2}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) $$
$$= \frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$
But $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) \neq \frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\arctan\left(\frac x2\right)$$
There seems to be a small difference between the answer provided by Alpha and the trig substitution method, but I cannot see where I made the mistake.
|
$$\frac{5}{2}\ln({\frac{x^2 + 4}{4}}) + \frac{1}{2}\arctan({x/2}) + c_1 \neq \frac{5}{2}\ln(x^2 + 4) + \frac{1}{2}\arctan(x/2) + C$$
Note that $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) = \frac 52 \ln\left(x^2 + 4\right) - \frac 52 \left (\ln 4\right) = \frac 52 \ln(x^2 + 4) + c_2$$
So put $c_1 + c_2 = C$. Then the answers are equal. Solutions to an integral consist of a family of solutions $F(x) + C$, which differ only by a constant. That is, if $F(x) + C$ is the solution after computing an intregral, so is $F(x) + C_i$, for any constant $C_i \neq C$.
In short, you're both correct!
| {
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"url": "https://math.stackexchange.com/questions/310808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove by Induction $\frac{a_n-\sqrt{A}}{a_n+\sqrt{A}} = \left[\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right]^{2^{n-1}} $ $a_1=\frac{1}{2}(a_0+\frac{A}{a_0})$;
$a_2=\frac{1}{2}(a_1+\frac{A}{a_1})$; and
$a_{n+1}=\frac{1}{2}(a_n+\frac{A}{a_n})$ for $n \geq 2$; where $a\gt 0$, $A\gt 0$.
Prove:
$$\frac{a_n-\sqrt{A}}{a_n+\sqrt{A}} =
\left[\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right]^{2^{n-1}}
$$
Remarks:(This question i was doing for my exam tomorrow; and i just got stuck)
I have started with applying Second Principle of Mathematical Induction.
I have never done this type of question earlier.
Any help or hint will be appreciated.
| Let us start:
\begin{align}
\frac{a_n -\sqrt{A}}{a_n+\sqrt{A}} &\underbrace{=}_{*}
\frac{\frac{1}{2}a_{n-1}+ \frac{1}{2}\frac{A}{a_{n-1}} -\sqrt{A}}{\frac{1}{2}a_{n-1}+ \frac{1}{2}\frac{A}{a_{n-1}} +\sqrt{A}} \underbrace{=}_{**}
\frac{\frac{1}{2}a_{n-1}^ 2+ \frac{1}{2}{A} -\sqrt{A}a_{n-1}}{\frac{1}{2}a_{n-1}^ 2+ \frac{1}{2}{A} +\sqrt{A}a_{n-1}} \\
&\underbrace{=}_{***} \frac{a_{n-1}^ 2+ {A} -2\sqrt{A}a_{n-1}}{a_{n-1}^ 2+ {A} +2\sqrt{A}a_{n-1}}
=\frac{(a_{n-1}-\sqrt{A})^2}{(a_{n-1}+\sqrt{A})^2} \\
&=\Bigr(\frac{a_{n-1}-\sqrt{A}}{a_{n-1}+\sqrt{A}}\Bigl)^2\\
&\underbrace{=}_{****}\Bigr(\Bigl(\Bigl(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\Bigl)^2\Bigr)^2 \ldots \Bigr)^2\\
&= \Bigl(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\Bigl)^{2^{n-1}}
\end{align}
Here, I used the recurrence relation at $*$, multiplied enumerator and denominator with $a_{n-1}$ at $**$ and with 2 at $***$ and used induction at $****$.
Background:
Why is this of any interest?
Well first of all, note that the sequence $a_n$ converges towards $\sqrt{A}$. One possible derivation of the sequence is the application of Newton's method to a function that has a root at $\sqrt{A}$. This could be $f(x) = x^2-A$. Newtons's method then reads:
\begin{align}
x_{k+1} & = x_k -\frac{f(x_k)}{f'(x_k)} = x_k -\frac{x_k^2-A}{2x_k} =\frac{x_k^2+A}{2x_k}\\
&= \frac{1}{2} \Bigl(x_k+\frac{A}{x_k}\Bigr)
\end{align}
Which is exactly your sequence. So if we want to estimate
$$ \frac{a_n-\sqrt{A}}{a_n+\sqrt{A}}=c$$
where c is the value we derived from the prooved formula, we can use this information to find out, how good we approximate $\sqrt{A}$.
\begin{align}
&\frac{a_n-\sqrt{A}}{a_n+\sqrt{A}}=c \\
\Leftrightarrow&a_n-\sqrt{A} = c(a_n+\sqrt{A}) \\
\Leftrightarrow& a_n(1-c) = (1+c)\sqrt{A} \\
\Leftrightarrow& a_n= \frac{1+c}{1-c}\sqrt{A} \\
\end{align}
So just by having a starting value $a_0$, we already know good solution $a_n$ will be, in sense of approximating $\sqrt{A}$. This means, that we can choose a $n$ a priori, for which $|a_n-\sqrt{A}|<\epsilon$, for a given $\epsilon$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definite integration evaluation of $\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$ OK, so the question says evaluate the integral
$$\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What I do is use the property that $\int_a^bf(x)dx=\int_a^bf(b+a-x)dx$ and this gives me ($I$ is the value of the integral)
$$\frac{2I}{\pi}=\int_{0}^{\pi}\frac{1}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What should I do ahead to get the value I need? Any tips? (Thanks in advance)
| $$
\begin{aligned}
\because I&=\int_0^\pi \frac{x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} \\& \stackrel{x\mapsto\pi-x}{=} \int_0^\pi \frac{\pi-x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} d x \\
&=\pi \int_0^\pi \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2}-I \\
\therefore I&=\frac{\pi}{2} \int_0^\pi \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} \\
&=\pi \int_0^{\frac{\pi}{2}} \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2}
\end{aligned}
$$
By my post, $$
\boxed{I =\pi\left[\frac{\pi\left(a^2+b^2\right)}{4 a^3 b^3}\right]=\frac{\pi^2\left(a^2+b^2\right)}{4 a^3 b^3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/312145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help with the inequality $\sum_{k=1}^{1006} \sqrt{k \cdot (2014-k)}<506^2\pi$ This question's been solved, come and look if you want to check out some hardcore solutions
Here's an inequality that needs to be proven:
Prove that
$\sqrt{1\cdot 2013} + \sqrt{2\cdot 2012} + \sqrt{3\cdot 2011} + \dots + \sqrt{1006\cdot 1008}$ < $506^2$$\pi$
Thanks
| $$\sum_{k=0}^{n-1} \sqrt{k \cdot (2n-k)} = 2n \left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(1-\dfrac{k}{2n} \right)} \right) = (2n)^2 \left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(1-\dfrac{k}{2n} \right)} \cdot \dfrac1{2n}\right)$$
$$\underbrace{\left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(1-\dfrac{k}{2n} \right)} \cdot \dfrac1{2n}\right) < \int_0^{1/2} \sqrt{x(1-x)}dx}_{\text{Since $\sqrt{x(1-x)}$ is a monotone increasing function for $x \in [0,1/2)$}} = \int_0^{\pi/4} 2\sin^2(y) \cos^2(y) dy = \dfrac{\pi}{16}$$
Hence,
$$\sum_{k=1}^{n-1} \sqrt{k \cdot (2n-k)} < (2n)^2 \dfrac{\pi}{16} = \left(\dfrac{n}2 \right)^2 \pi$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the standard matrix for a linear transformation If $T: \Bbb R^3→ \Bbb R^3$ is a linear transformation such that:
$$
T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6 \\ -14 \\ \end{bmatrix}$$ $$ T\Bigg (\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} -6\\ -40 \\ -2 \\ \end{bmatrix}
$$
Then the standard matrix for T is...
I'm not exactly sure how to approach this problem.
Could anyone explain how to solve this problem?
| Note that if $$
\begin{pmatrix}-2 \\ 3 \\ -4 \\ \end{pmatrix} = -2\begin{pmatrix}1 \\ 0\\ 0 \\ \end{pmatrix} +3\begin{pmatrix}0 \\ 1\\ 0 \\ \end{pmatrix} -4\begin{pmatrix}0 \\ 0\\ 1 \\ \end{pmatrix}
$$ so $$
T\begin{pmatrix}-2 \\ 3 \\ -4 \\ \end{pmatrix} = -2\times T\begin{pmatrix}1 \\ 0\\ 0 \\ \end{pmatrix} +3\times T\begin{pmatrix}0 \\ 1\\ 0 \\ \end{pmatrix} -4\times T\begin{pmatrix}0 \\ 0\\ 1 \\ \end{pmatrix}
=-2T(\epsilon_1)+3T(\epsilon_2)-4T(\epsilon_3)$$ Now do the same for two other vectors to find out two relations written by to $T(\epsilon_i)$. Here, you have a system of 3 equations and 3 unknowns $T(\epsilon_i)$ which by solving that you get $T(\epsilon_i)_1^3$. Now use that fact that $$T\begin{pmatrix}x \\y \\ z \\ \end{pmatrix} = xT(\epsilon_1)+yT(\epsilon_2)+zT(\epsilon_3)$$ to find the original relation for $T$. I think by its rule you can find the associated matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/313798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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How to solve following system of non-linear algebraic equations I have a system
$$
5x^2 - 5y^2 - 3x + 9y = 0,
$$
$$
5x^3 + 5y^3 - 13xy - 15x^2 - y^2 = 0.
$$
How to solve it?
| First, you can observe that if $y=0$, then $x=0$, which gives out a solution.
Let us assume $y\neq 0$ now.
Take $\dfrac{x}{y}=\lambda$.
Then the first equation is:
$$(5\lambda^2-5)y^2+(9-3\lambda)y=0$$
which is
$$(5\lambda^2-5)y+(9-3\lambda)=0$$
So
$$y = \dfrac{3\lambda-9}{5\lambda^2-5}$$
The second one is:
$$(5\lambda^3+5)y^3-(15\lambda^2+13\lambda+1)y^2=0$$
which is
$$(5\lambda^3+5)y-(15\lambda^2+13\lambda+1)=0$$
Thus
$$y = \dfrac{15\lambda^2+13\lambda+1}{5\lambda^3+5}$$,
According to these, we have
$$\dfrac{3\lambda-9}{5\lambda^2-5}=\dfrac{15\lambda^2+13\lambda+1}{5\lambda^3+5}$$
which is equal to:
$$6\lambda^4+11\lambda^3-7\lambda^2-8\lambda+4=0$$
Factorize it:
$$(\lambda+1)(\lambda+2)(2\lambda-1)(3\lambda-2)=0$$
plug in $y = \dfrac{3\lambda-9}{5\lambda^2-5}$.
And $\lambda\neq -1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculus Bonus Problem solve algebraically
$$\lim_{x\rightarrow 0}\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x+x^2}$$
without using L'hopital's rule
This is a bonus question for my Math Intensive Major Cal 1 class, I would like to know how to solve it since it stumped me on the test, and after many hours of working at this problem it has stumped me again and again
| We will use the formula
$$
a-b=\frac{a^{12}-b^{12}}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}
$$
with
$$
a=\sqrt[3]{1+x^2}\quad\mbox{and}\quad b=\sqrt[4]{1-2x}.
$$
This gives
$$
\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}=\frac{(1+x^2)^4-(1-2x)^3}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}=\frac{6x-8x^2+\ldots}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}
$$
$$
\frac{2x(3-4x+\ldots)}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}.
$$
So
$$
\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x+x^2}=\frac{2(3-4x+\ldots)}{(1+x)(a^{11}+a^{10}b+\ldots+ab^{10}+b^{11})}.
$$
Now note that both $a$ and $b$ tend to $1$, so your limit is
$$
\frac{2\cdot 3}{12}=\frac{1}{2}.
$$
| {
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"url": "https://math.stackexchange.com/questions/315539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help understanding the solution to #83 Section 9.1 Calculus 9e, by Larson.
Find $a_n$ of $1, \; \frac{-1}{1\times 3}, \; \frac{1}{1\times3\times5}, \frac{-1}{1\times 3\times 5 \times 7}$
I solved it as $\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!}$
The solution guide says: $\dfrac{\left(-1\right)^{n-1}2^n n!}{\left(2n\right)!} $
I am unable to figure out how $2^n$ gets into the numerator.
if you multiply $\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!}$ by $\dfrac{2n}{2n}$ results in $\dfrac{\left(-1\right)^{n-1}2n}{\left(2n\right)!}$
Please, help me understand how $2n$ goes to $2^n n!$
Have learned from the comments that the denomator is $\left(2n-1\right)!!$ Sorry, still not sure how to solve this problem. This is my first time learning calculus and haven't seen any double factorials. Hate this about this Larson Calculus book, great problems, but they are like Sherlock Novels, always some missing knowledge not showen that is needed to solve the later problems.
| No different results here but a somewhat different way of working through it.
\begin{array}{|c|c|c|c|c|}
\hline
\mathbf{n}&1& 2 &3& 4 & 5\\
\hline
a_n&1&\frac{-1}{1\cdot3}&\frac{1}{1\cdot3\cdot5}&\frac{-1}{1\cdot3\cdot5\cdot7}&\frac{1}{1\cdot3\cdot5\cdot7\cdot9}\\
\end{array}
Terms like $1 \cdot 3 \cdot 5 \cdot 7$ are similar to factorials, except they are
missing the even numbers. But notice that $2^2 \cdot2!$ = $2\cdot4$, that $2^3 \cdot3! = 2\cdot4\cdot6$, that $2^4\cdot4! = 2\cdot 4\cdot6\cdot8$, and so on. So for example $1\cdot3\cdot5\cdot7\cdot 2^33! = 7!$ and $1\cdot3\cdot5\cdot7\cdot9\cdot 2^44! = 9!$. Let's multiply each $a_n$ by $$\frac{2^{n-1}(n-1)!}{2^{n-1} (n-1)!}$$ so that each denominator becomes a factorial. Each $a_n$ multiplies out to $$\frac{(-1)^{n-1}2^{n-1}(n-1)!}{(2n-1)!}$$ We know that $$\frac{2^{n-1}(n-1)!}{(2n-1)!}=\frac{2^nn!}{2n!}$$ when $n\ge1$. So$$\frac{(-1)^{n-1}2^{n-1}(n-1)!}{(2n-1)!}=\dfrac{\left(-1\right)^{n-1}2^n n!}{\left(2n\right)!}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x_{n+1} = \sqrt{3x_n}$ converges for $x_1 = 1,x_1 = 27$ Proof Prove $x_{n+1} = \sqrt{3x_n}$ converges for $x_1 = 1,x_1 = 27$. (Separate problems for $x_1 = 1$ and $x_1 = 27$.)
EDIT: Took out bad algebra.
| There's no need to prove this using induction as you can find the value that it converges to!
Begin by letting $x_1=a$ and finding the first few terms, as follows:
\begin{align*}
x_1 &= a \\
x_2 &= \sqrt{3}\cdot\sqrt{a} = \sqrt{3a} \\
x_3 &= \sqrt{3}\cdot\sqrt{\sqrt{3a}} = \sqrt{3\sqrt{3a}}\\
x_4 &= \sqrt{3}\cdot\sqrt{\sqrt{3\sqrt{3a}}} = \sqrt{3\sqrt{3\sqrt{3a}}}\\
\end{align*}
As you can see, there is an evident pattern in the terms. It can be represented by $x=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}$.
At $x_1=1$, $x=\sqrt{3x}$, which gives an obvious solution of $x=3$.
At $x_1=27$, the equation simplifies to the same $x=\sqrt{3x}$ because the infinite square root makes the 27 negligible, which gives an obvious solution of $x=3$.
| {
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"source": "stackexchange",
"question_score": "6",
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Show that the curve $x^2+y^2-3=0$ has no rational points
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.
Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.
Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.
Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.
In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.
Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.
Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.
Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.
We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.
| It is actually sufficient to analyze the curve modulo 3 (or 2, even), where the question becomes whether $-1 \equiv 2$ is a square modulo 3. You can reduce the proof to this step directly by clearing denominators/divisibility arguments, but this curve is related to the quaternion algebra generated by the square roots of -1 and 3 which ramifies at the prime 3. Here, I'm making use of the Albert-Brauer-Noether-Hasse Theorem.
| {
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