Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Non-homogeneous recurrence relation, how to solve? Solve the following non-homogeneous recurrenece relation:
$a_1 = 0, a_2= 0, a_3=1$, and $a_n = a_{n-1}+a_{n-2} + 1$
This somehow seems familiar with the Fibonacci sequence, since $a_4$ will be $2$, $a_5$ will be $4$, and so on. But how does one "solve" such task?
Thanks!
| You can use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, run the recurrence backwards to get $a_0 = -1$ (starting at 0 is nicer all around). Shift the recurrence, multiply by $z^n$ and sum over $n \ge 0$, recognize resulting sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
&= \sum_{n \ge 0} a_{n + 1} z^n + \sum_{n \ge 0} a_n z^n + \sum_{n \ge 0} z^n \\
\frac{A(z) - a_0 - a_1 z}{z^2}
&= \frac{A(z) - a_0}{z} + A(z) + \frac{1}{1 - z}
\end{align*}$
Solve for $A(z)$, write as partial fractions:
$\begin{align*}
A(z)
&= \frac{1 - 2 z}{1 - 2 z + z^3} \\
&= \frac{1}{1 - z - z^2} - \frac{1}{1 - z}
\end{align*}$
Now, we know that the Fibonacci numbers $F_n$ satisfy:
$\begin{equation*}
F_{n + 1} = F_{n + 1} + F_n \qquad F_0 = 0, F_1 = 1
\end{equation*}$
As above, this gives the generating function:
$\begin{align*}
F(z)
&= \frac{z}{1 - z - z^2}
\end{align*}$
Using the fact:
$\begin{align*}
\sum_{n \ge 0} F_{n + 1} z^n
&= \frac{F(z) - F_0}{z} \\
&= \frac{1}{1 - z - z^2}
\end{align*}$
by extracting coefficients we see:
$\begin{align*}
a_n
&= [z^n] A(z) \\
&= F_{n + 1} - 1
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What Is The Smallest Solution to $7x^5=11y^{13}$? I started teaching myself Number Theory from a pretty basic textbook and I got completely stuck with this problem.
Let $x$ and $y$ be two non-zero natural numbers such that $7x^5=11y^{13}$ .
The lowest possible value for $x$ has a prime factorization of the form $a^cb^d$ . What's the value of the sum $a+b+c+d$ ?
Can anyone show a solution ? I'm much more interested in the way of thinking rather than the result.
| From $7x^5 = 11y^{13}$, we have that $11$ divides $x$ and $7$ divides $y$. Let $x=7^a 11^b m$ and $y=7^c11^dn$, where $$\gcd(m,7)=\gcd(m,11)=\gcd(n,7)=\gcd(n,11)=1$$
This gives us that
$$7^{5a+1}11^{5b}m^5 = 7^{13c}11^{13d+1}n^{13}$$
This means $5a+1=13c$, $5b=13d+1$ and $m^5=n^{13}$. Since we are after the smallest possible $x$, we have $m=n=1$, $a=5,c=2$ and $b=8,d=3$. This gives us that
$$x=7^5\cdot11^8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Simplify the Complex Fraction I am having trouble with the following complex fraction. I have simplified everything for the most part, but I am stuck on the last part and need to know what I have to do next.
$$\frac{\frac{2}{3}+\frac{4}{5}}{\frac{5}{6}-\frac{1}{2}}$$
(I hope this converts over to the way it needs to be displayed.
I have simplified it to
$\frac{\frac{10}{15}+\frac{12}{15}}{\frac{10}{12}-\frac{6}{12}}$
which equals to
$\frac{\frac{22}{15}}{\frac{4}{12}}$
simplified to
$\frac{4}{12}×\frac{15}{22}$
At this point do I cross multiply and then add the two results? I can't figure out what to do next. Please help me.
| The correct answer is $22/5$:
$$
\frac{\frac{2}{3}+\frac{4}{5}}{\frac{5}{6}-\frac{1}{2}}=
\frac{\frac{10+12}{15}}{\frac{5-3}{6}}=
\frac{\frac{22}{15}}{\frac{2}{6}}=
\frac{\frac{22}{15}}{\frac{1}{3}}=
\frac{22}{15}\cdot \frac{3}{1}=\frac{22}{5}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Laurent Series of $(z^2 + 3z + 2)e^{\frac{1}{z+1}}$ I want to find the Laurent series of $(z^2 + 3z + 2)e^{\frac{1}{z+1}}$ around $z_0 = -1$. However, since this is not a fraction in the form $\frac{a}{z-b}$, I am not sure how to calculate it.
| Note that in $\mathbb{C} \backslash \{ -1 \} $
$$
e^{\frac{1}{z+1}} = \sum_{n=0}^{\infty} \frac{1}{n! (z+1)^n}
$$
Then
\begin{align*}
(z^2+3z+2)e^{\frac{1}{z+1}} & =\left[(z+1)^2+(z+1)\right] \sum_{n=0}^{\infty} \frac{1}{n! (z+1)^n} \\
& =\left[(z+1)^2 \sum_{n=0}^{\infty} \frac{1}{n! (z+1)^n} \right] + \left[(z+1)\sum_{n=0}^{\infty} \frac{1}{n! (z+1)^n}\right] \\
& = \left[(z+1)^2 + (z+1) + \sum_{n=0}^{\infty}\frac{1}{(n+2)!(z+1)^n} \right] + \left[(z+1) + \sum_{n=0}^{\infty}\frac{1}{(n+1)!(z+1)^n}\right] \\
& = 2(z+1)+(z+1)^2 + \sum_{n=0}^{\infty}\left(\frac{1}{(n+1)!}+\frac{1}{(n+2)!}\right)\frac{1}{(z+1)^n} \\
& = 2(z+1)+(z+1)^2 + \sum_{n=0}^{\infty}\left(\frac{n+3}{(n+2)!}\right)\frac{1}{(z+1)^n}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$3 \times 3$ Analogue of Zorn's Vector Matrices A Zorn vector matrix is a $2 \times 2$ matrix whose diagonal elements are scalars and whose off-diagonal elements are 3-vectors. I read about them on this Wikipedia page for split-octonions. The product of two Zorn matrices is given by
$\begin{pmatrix} \alpha & \mathbf a \\ \mathbf b & \beta \end{pmatrix} \begin{pmatrix} \alpha' & \mathbf a' \\ \mathbf b' & \beta' \end{pmatrix} = \begin{pmatrix} \alpha \alpha' + \mathbf a \cdot \mathbf b' & \alpha \mathbf a' + \beta' \mathbf a + \mathbf b \times \mathbf b' \\ \alpha' \mathbf b + \beta \mathbf b' + \mathbf a' \times \mathbf a & \beta \beta' + \mathbf a' \cdot \mathbf b \end{pmatrix}$
Zorn matrices were developed as a representation of the octonions; as with the octonions, multiplication of Zorn matrices is non-associative. The determinant of a Zorn matrix is given by
$\det \begin{pmatrix} \alpha & \mathbf a \\ \mathbf b & \beta \end{pmatrix} = \alpha \beta - \mathbf a \cdot \mathbf b$
and satisfies $(\det AB) = (\det A)(\det B)$ for all Zorn matrices $A$ and $B$. If the off-diagonal elements of $A$ and $B$ are all parallel vectors (say, only the first component is nonzero), then the cross products vanish and Zorn multiplication behaves like ordinary matrix multiplication.
Is there a $3 \times 3$ analogue of the Zorn matrix algebra that preserves the form of multiplication for $2 \times 2$ blocks and has a norm that reduces to the $2 \times 2$ determinant defined above?
*
*
$\begin{pmatrix} \alpha & \mathbf a & \mathbf 0 \\ \mathbf b & \beta & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & 1 \end{pmatrix}
\begin{pmatrix} \alpha' & \mathbf a' & \mathbf 0 \\ \mathbf b' & \beta' & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & 1 \end{pmatrix} =
\begin{pmatrix} \alpha \alpha' + \mathbf a \cdot \mathbf b' & \alpha \mathbf a' + \beta' \mathbf a + \mathbf b \times \mathbf b' & \mathbf 0 \\ \alpha' \mathbf b + \beta \mathbf b' + \mathbf a' \times \mathbf a & \beta \beta' + \mathbf a' \cdot \mathbf b & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & 1 \end{pmatrix} $
*
$\det \begin{pmatrix} \alpha & \mathbf a & \mathbf 0 \\ \mathbf b & \beta & \mathbf 0 \\ \mathbf 0 & \mathbf 0 & 1 \end{pmatrix} =
\det \begin{pmatrix} \alpha & \mathbf 0 & \mathbf a \\ \mathbf 0 & 1 & \mathbf 0 \\ \mathbf b & \mathbf 0 & \beta \end{pmatrix} =
\det \begin{pmatrix} 1 & \mathbf 0 & \mathbf 0 \\ \mathbf 0 & \alpha & \mathbf a \\ \mathbf 0 & \mathbf b & \beta \end{pmatrix} = \alpha \beta - \mathbf a \cdot \mathbf b$
*$\det AB = (\det A)(\det B)$ for all $A,B$ of the form $\begin{pmatrix} \alpha & \mathbf a & \mathbf c \\ \mathbf b & \beta & \mathbf e \\ \mathbf d & \mathbf f & \gamma \end{pmatrix}$
Thanks!
| There is no analogue, or at least not a natural one.
Forms permitting composition were classified in Schafer (1970). The classification implies that if a unital algebra over a field of characteristic not dividing $6$ admits a nondegenerate form $N$ of degree $3$ such that $N(xy)=N(x)N(y)$ for all $x,y$, then it must have dimension $1,2,3,5$ or $9$ over that field (note that this generalizes Hurwitz's theorem that composition algebras only exist in dimensions $1,2,4,8$).
Since a $3\times3$ analogue of the Zorn vector-matrix algebra would have dimension $6\cdot 3 + 3 = 21$, and you require that $\det(AB) = \det(A) \det(B)$ for all $A, B$ which implies that $\det$ would be a form (presumably of degree $3$) admitting composition, it follows that any possible examples would be either degenerate or in characteristic $2$ or $3$. In particular there is no such algebra over $\mathbb{R}$ or $\mathbb{C}$ where $\det$ is a degree-$3$ homogeneous polynomial function of the vector-matrix's entries, like in the $2\times 2$ case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving that $5^n-1$ is divisible by $4$ for $n\geq 0$ by induction I hope this is not counted as a duplicate, as I would like to know if my proof is valid:
$P(n): 5^n - 1$ is divisible by $4$ for $n \ge 0$.
Base Step: $P(0): 5^0-1 = 1-1 = 0 = 0\times 4$.
Induction Supposition: $P(k): 5^k-1$ is divisible by $4$.
Prove: $P(k+1): 5^{k+1}-1$ is divisible by $4,$ or equivalently $5^{k+1}-1 = 4r$, for some integer $r$.
$5^{k+1}-1$
$= 5^k\times5-1$ by Exponent Laws
$= 5\times4r$ by I.H.
$=4(5r)$ which was to be shown.
| For $n=1$ we have $5^{n} - 1 = 5 - 1 = 4.$
Suppose there is an $n \geq 1$ such that $5^{n} - 1 = 4k$ for some $k$. Then
$$5^{n+1} - 1 = 5(5^{n} - 1) + 4 = 20k + 4,$$
so
$$4 \mid (5^{n+1} - 1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
First number $\ge n$ that is divisible by $k$? Is there a good way to compute the first value $\ge n$ that is divisible by $k$?
Right now I am computing $\left\lfloor\frac{n}{k}\right\rfloor k$ but it doesn't always work.
| Start with
$u(n, k)
=k\lceil \frac{n}{k} \rceil
$
being the smallest multiple of $k$
that is $\ge n$.
We now use
$\lceil \frac{n}{k} \rceil
=\lfloor \frac{n+k-1}{k} \rfloor
$.
To see this,
write $n = ak+b$
where
$0 \le b \le k-1$.
Then
$\lceil \frac{n}{k} \rceil
=a
$
if $b=0$
and
$a+1$ if $b > 0$.
But
$\lfloor \frac{n+k-1}{k} \rfloor
=\lfloor \frac{ak+b+k-1}{k} \rfloor
=a+\lfloor \frac{b+k-1}{k} \rfloor
=a$
if $b=0$
and $a+1$
if $b > 0$
Therefore
$u(n, k)
=k \lfloor \frac{n+k-1}{k} \rfloor
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find $\int\limits_{0}^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$ I have to find $$\int_0^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$$
I tried to do it by substitution $t = \tan(\frac{x}{2})$
Then we have that $$\cos(x) = \frac{1-t^2}{1+t^2} \quad dx = \frac{2\,dt}{1+t^2}$$ but then also limits of integration are changing so we have
$$\int\limits_{t(0)}^{t(2\pi)} \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2dt}{1+t^2} = \int\limits_0^0 \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2\,dt}{1+t^2} = 0$$
I figured out that it is not correct because $\tan(\frac{\pi}{2})$ is not defined and $t(\pi) = \tan(\frac{\pi}{2})$ and $\pi \in [0, 2\pi]$. How can I "repair" that and do it right?
| Here's a easier way
Please look my link why it's possible.
$I=2 \int ^{\pi}_0 \frac{1}{5-3\cos x}.dx$
After substitution,
$I=2 \int ^{b\to \infty}_0 \frac{1}{4t^2+1}.dt$
Which gives,
$I= \left|\tan^{-1}(4t)\right|^{b\to \infty}_0 $
From here it's pretty easy. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 6
} |
series convergence by decomposing it into smaller series I have to determine if the series $$ \sum_{n=1}^{\infty} \frac{(-1)^{n} +1}{2}. (\frac{1+2i}{5})^{n} +\frac{(-1)^{n+1}+1}{2}.(\frac{2}{3})^{n} $$ is convergent or not. This is what I tried:
I decomposed the above expression in 4 parts and wrote the series as: $$ \sum_{n=1}^{\infty}{a}_{n} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{2}.(\frac{1+2i}{5})^{n} + \sum_{n=1}^{\infty} \frac{1}{2}.(\frac{1+2i}{5})^{n} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2}.(\frac{2}{3})^{n} + \sum_{n=1}^{\infty}\frac{1}{2} (\frac{2}{3})^{n} $$
and then determined the convergence of each of these four series by regular convergence tests- root test here, and all of them were found to be convergent, and then I concluded that their sum series must also be convergent.
So, my question: Is my method right? and if it is, can somebody please suggest some other , better method of solving this problem and in case it is notcan somebody point out what is wrong?
Thanks in advance.
| $$\sum_{n=1}^\infty \frac{(-1)^{n}}{2}.(\frac{1+2i}5)^n=\frac12\sum_{n=1}^\infty (\frac{-1-2i}5)^n=\frac12.\frac{-\frac{1+2i}5}{1+\frac{1+2i}{5}}$$ $$\sum_{n=1}^\infty \frac{1}{2}.(\frac{1+2i}5)^n=\frac12.\frac{\frac{1+2i}5}{1-\frac{1+2i}{5}}$$ $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2}.(\frac23)^n=-\frac12\sum_{n=1}^\infty (\frac{-2}3)^n=-\frac12 .\frac{-\frac23}{1+\frac23}$$
$$\sum_{n=1}^\infty \frac12(\frac{2}3)^n=\frac12.\frac{\frac23}{1-\frac23}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
how to prove by induction the $ (1+x)^{n}>1+nx+nx^2$ Prove by induction the formula
$ (1+x)^{n}>1+nx+nx^2$ for $x>0$ real number and $n\ge 3$
my try : multiply both sides by $(1+x)$
gives $ (1+x)^{n+1}>1+(n+1)x+(2n+nx)x^2$
have I done something wrong or what I do next?
$(2n+nx)$ must turn into $(n+1)$
| You are almost there. Note that $2n+x \geq 2n \geq n+1$.
Hence, the induction step becomes:
We have
$$(1+x)^n > 1 + nx + nx^2$$
Multiplying by $1+x$, we obtain
\begin{align}
(1+x)^{n+1} & = (1+x)^n (1+x)\\
& > (1+nx+nx^2)(1+x)\\
& = 1+ nx + nx^2 + x + nx^2 + nx^3\\
& = 1+(n+1)x + (n+n)x^2 + nx^3\\
& \geq 1+(n+1)x + (n+n)x^2 & (\because nx^3 \geq 0)\\
& \geq 1+(n+1)x + (n+1)x^2 & (\because n \geq 1)
\end{align}
which gives us the induction step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Laurent series for $\frac{1}{z(z-1)}$ in the domain $1 < |z-2| < 2$
Expand $f(z) = \cfrac{1}{z(z-1)}$ in a Laurent series valid for $1 < |z-2| < 2$
First of all I wrote the fraction in partial fraction form:
$\cfrac{1}{z(z-1)} = \cfrac{-1}{z} + \cfrac{1}{z-1}$
Then, I tried manipulating the fraction to get (z-2) part:
$\cfrac{-1}{z} = \cfrac{-1}{(z-2)+2} = \cfrac{1}{2}\cfrac{-1}{1+(\cfrac{z-2}{2})}$
Now, $|\cfrac{z-2}{2}| < 1$
Thus, I can apply the series expansion of:
$\cfrac{1}{1+x} = 1 - x + x^2 - ...$, when $|x| < 1$
Thus:
$\cfrac{-1}{z} = \cfrac{-1}{2}[1 - (\cfrac{z-2}{2})^2 + ...]$
$\implies \cfrac{-1}{z} = \cfrac{-1}{2}\sum_{n = 0}^{\infty}(-1)^n(\cfrac{z-2}{2})^n$
$\cfrac{-1}{z} = \sum_{n=0}^{\infty}(-1)^{n+1}\cfrac{(z-2)^n}{2^{n+1}}\tag{1}$
But, how can I manipulate the second fraction?
Thanks
| $$\frac{-1}{z} = \frac{1}{-2-(z-2)} = -\frac12\;\; \frac{1}{1 - (\frac{2-z}{2})}$$
$$= -\sum_{n=0}^{\infty}(-1)^n\frac{(z-2)^n}{2^{n+1}} $$
$$\frac{1}{z-1} = \frac{1}{1-(2-z)} =\frac{1}{z-2}\;\; \frac{1}{1 - ({2-z})^{-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int_{-\infty}^\infty\frac{1}{(x^2+a^2)^3}dx=\frac{3\pi}{8a^5}$ for $a>0$ I've been trying to show that $\int_{-\infty}^\infty\frac{1}{(x^2+a^2)^3}dx=\frac{3\pi}{8a^5}$ for $a>0$ using complex analysis methods.
But for some reason I can't get it to come out.
Perhaps someone could figure out where I am going wrong.
Since there are no poles on the real axis, I know that
$\int_{-\infty}^\infty\frac{1}{(x^2+a^2)^3}dx=2\pi i\cdot\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right).$
To calculate $\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right)$
I used the fact that on a small enough disk centered at $ai$,
$\frac{1}{(z+ai)^3}=\sum\limits_{k=0}^\infty c_k(z-ai)^k$.
Thus
$\frac{1}{(z^2+a^2)^3}=\frac{\frac{1}{(z+ai)^3}}{(z-ai)^3}=\sum\limits_{k=0}^\infty c_k(z-ai)^{k-3}$.
So $\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right)=c_2.$
Where $c_2=\frac{d^2}{dz^2}\frac{1}{(z+ai)^3}\bigg|_{z=ai}=\frac{12}{(2ia)^5}=\frac{3}{8ia^5}.$
But that gives me
$\int_{-\infty}^\infty\frac{1}{(x^2+a^2)^3}dx=2\pi i\cdot\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right)=2\pi i\cdot\frac{3}{8ia^5}=\frac{3\pi}{4a^5}.$ Which is off by $\frac{1}{2}$. I must be making a silly mistake somewhere, but I can't seem to find it.
Any help would be appreciated.
| There is a mistake in your calculus of $c_2$, it is rather
$$\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right)=\color{red}{\frac{1}{2!}}\frac{d^2}{dz^2}\frac{1}{(z+ai)^3}\bigg|_{z=ai}$$ due to the Taylor expansion around $z=ia$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve $\int \limits_{0}^{\infty} \frac{\cos(x)}{\cosh(x)} dx$ without complex integration. Solve $$\int \limits_{0}^{\infty} \frac{\cos(x)}{\cosh(x)} dx$$ without complex integration.
This integral can be very easily solved with contour integration, but how can you solve it without taking a tour in the complex plane?
| Expand $ \operatorname{sech}{x} $ in an infinite series in $e^{-x}$:
$$ \frac{2}{e^x(1+e^{-2x})} = 2 \sum_{k=0}^{\infty} (-1)^k e^{-(2k+1)x} $$
Interchange the order of integration, and you are left with integrals of the form
$$ \int_0^{\infty} e^{-\alpha x} \cos{x} \, dx, $$
which can be done using integration by parts, to avoid complex numbers totally, with the result $ \alpha/(1+\alpha^2) $. Hence the integral is equal to
$$ \sum_{k=0}^{\infty} (-1)^k \frac{1+2k}{1+2k+2k^2}. $$
Okay, and now it is time once again for "Special Functions and Pray". Using the usual tricks, one can express infinite sums in terms of the digamma function as follows: first, write the summand in partial fractions as
$$ \frac{1+2k}{1+2k+2k^2} = \frac{1}{2} \left( \frac{1}{k+a} + \frac{1}{k+a^*} \right), $$
where $a$ is the complex root of the denominator, $(1+i)/2$. Now we have the identity
$$ \sum_{k=0}^{K} \frac{1}{k+a} = \psi(K+a+1)-\psi(a+1), $$
where $\psi = (\log{\Gamma})'$ as usual. However, this is not good enough: we have to deal with the $(-1)^k$. This we do by splitting into the even and odd cases: some algebra shows the result we want is
$$ \sum_{k=0}^K \frac{(-1)^k}{k+a} = \frac{1}{2} (-1)^K \psi{\left(\frac{a}{2}+\frac{K}{2}+1\right)}
-\frac{1}{2} (-1)^K \psi{\left(\frac{a}{2}+\frac{K}{2}+\frac{1}{2}\right)}
-\frac{1}{2}\psi{\left(\frac{a}{2}\right)}
+\frac{1}{2} \psi{\left(\frac{a}{2}+\frac{1}{2}\right)}. $$
Now, this sum actually converges by comparison with the alternating harmonic series, the details of which I omit. Hence we can take the limit as $k \to \infty$. The problem is what happens to the terms
$$ \frac{1}{2} (-1)^K \left( \psi{\left(\frac{a}{2}+\frac{K}{2}+1\right)}
- \psi{\left(\frac{a}{2}+\frac{K}{2}+\frac{1}{2}\right)} \right), $$
but a slight abuse of Stirling's approximation shows that this term is $O(\log{n}-\log{(n+1/2)})=o(1)$, so in fact we can just ditch it. Hence the answer is
$$ -\frac{1}{4}\psi{\left(\frac{a}{2}\right)}
+\frac{1}{4} \psi{\left(\frac{a}{2}+\frac{1}{2}\right)} + c.c., $$
because we have a half in the partial fractions. Then we need to compute
$$-\frac{1}{4} \psi{\left(\frac{1}{4}+\frac{i}{4}\right)}
+\frac{1}{4} \psi{\left(\frac{3}{4}+\frac{i}{4}\right)}
-\frac{1}{4} \psi{\left(\frac{1}{4}-\frac{i}{4}\right)}
+\frac{1}{4} \psi{\left(\frac{3}{4}-\frac{i}{4}\right)},$$
but staring at this for long enough, you can pair the terms so that you can apply the identity
$$ \psi(z)-\psi(1-z) = -\pi \cot{\pi z}, $$
which gives you
$$ \frac{1}{4} \pi \cot \left(\left(\frac{1}{4}+\frac{i}{4}\right) \pi \right)+\frac{1}{4} i \pi \coth \left(\left(\frac{1}{4}+\frac{i}{4}\right) \pi \right), $$
and applying some trigonometric identities eventually settles this into the form
$$ \tfrac{1}{2}\pi \operatorname{sech}{\tfrac{1}{2}\pi}, $$
which is thankfully the right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
Why is $1/i$ equal to $-i$? When I entered the value $$\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).
Is there a fault in my calculator or $\frac{1}{i}$ really equals $-i$? If it does then how?
| I always like to point out that this fits well into a pattern you see when "rationalising the denominator", if the denominator is a root:
$$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2}\sqrt{2}$$
$$\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{1}{17}\sqrt{17}$$
$$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{1}{a}\sqrt{a}$$
$$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{1}{\sqrt{-1}}\cdot \frac{\sqrt{-1}}{\sqrt{-1}} = \frac{1}{-1}\sqrt{-1} = - i.$$
In this vein, it is almost more suggestive to write
$$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$$
$$\frac{1}{i} = \frac{i}{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 10,
"answer_id": 4
} |
Prove this Complicated Inequality
Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 + (a + b + c)^2 \le 4$. Prove that $$\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.$$
Let $x = a + b, y = b + c, t = a + c$ Then the INEQ becomes,
$$\frac{ab + 1}{(x)^2} + \frac{bc + 1}{(y)^2} + \frac{ca + 1}{(t)^2} \ge 3.$$
Any further hints?
| By AM-GM
$$\sum_{cyc}\frac{ab+1}{(a+b)^2}=\frac{1}{2}\sum_{cyc}\frac{2ab+2}{(a+b)^2}\geq\frac{1}{2}\sum_{cyc}\frac{2ab+a^2+b^2+c^2+ab+ac+bc}{(a+b)^2}=$$
$$=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{c^2+ab+ac+bc}{(a+b)^2}=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}\geq$$
$$\geq\frac{3}{2}+\frac{1}{2}\cdot3\sqrt[3]{\prod_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}}=\frac{3}{2}+\frac{3}{2}=3.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $ appears to disagree with $\int_0^1 \frac{dx}{4-x^2}$ In question 1280454, t was asked how to find
$$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $$
and of course, you can write this as
$$ \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{4 - \frac{k^2}{n^2}}
= \int_0^1 \frac{dx}{4-x^2}
$$ by using the variable $x=\frac{k}{n}$ in a Reimann integral.
However,
$$\int_0^1 \frac{dx}{4-x^2} = \frac18\left(\ln(5)-\ln(3) \right) \approx 0.06385$$
while each of the $n$ terms in
$$
\sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}}$$ exceeds $\frac{1}{4n}$ so the sum must be greater than $\frac{1}{4}$ in all cases. (Experimentatlly, the limit is about $0.275$)
What gives???
| Otherwise from your method you should take $ a$ as 2 not 4 thus the integral evaluates to be
$$
\int_0^1 \frac{dx}{4-x^2} = \frac1{2a}\left(\ln(\frac{a+(x=1)}{a-(x=1)})-\ln(\frac{a+(x=0)}{a-(x=0)}) \right) = \frac1{2*2}\left(\ln(\frac{2+1}{2-1})-\ln(\frac{2}{2}) \right) = \frac1{4}\ln3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Finding the limit of a function with 2 variables Please help to solve the limit.
$$\lim_{(x,y)\rightarrow(0,0)}\dfrac{x^3+y^3}{x^2+y^2}$$
I tried to solve it but...
$$x:=0 \Rightarrow \dfrac{x^3+y^3}{x^2+y^2} = y$$
$$y:=0 \Rightarrow \dfrac{x^3+y^3}{x^2+y^2} = x$$
| Hints: (i) If $x^2+y^2$ is involved, switching to polar can be useful. Let $x=r\cos\theta$ and $y=r\sin\theta$.
(ii) If one does not wish to use the approach (i), note that
$$|x^3+y^3|=|x+y|(x^2-xy+y^2)\le |x+y|(2x^2+2y^2).$$
(iii) Alternately, let $\epsilon \gt 0$. If $(x,y)$ is close enough to $(0,0)$, then $|x|\lt \epsilon$ and $|y|\lt \epsilon$. Thus $|x^3|\lt \epsilon x^2$ and $|y^3|\lt \epsilon y^2$, and therefore
$$\frac{|x^3+y^3|}{x^2+y^2}\le \frac{|x^3|+|y^3|}{x^2+y^2}\lt \frac{\epsilon x^2+\epsilon y^2}{x^2+y^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For what values of $n$ , does $7 \mid 5^n+1$ $7 \mid 5^n+1$ implies $5^n+1=7a$ for some integer $a$
i.e $5^n=7a-1$
Now , $5^n$ is an integer which always ends with $5$ [for any integer $n$].
Thus , $7a-1$ must also end with $5$.But , this is only possible when , $a$ is an integer ending with $8$ as, $7 \times 8=56$ and $6-1=5$.
But , for all digits ending with $8$ say $8$ itself , $5^n$ does not exist.
So , how to solve this sum ? Please help me .
Thank you.
| $5^n\equiv 1,5,4,-1,2,3,1\pmod{\! 7}$ for $n=0,1,2,3,4,5,6$, respectively.
This pattern continues and $\, 5^n\equiv -1\iff n\equiv 3\, $ mod $6$.
More rigorously: $3$ is the least nonnegative $c$ giving $5^c\equiv -1\pmod{\! 7}$.
Let $n=3+k$ with $k\ge 0$. We'll show $5^n\equiv -1\pmod{\! 7}$ iff $k=6m$ for some $m\ge 0$.
$5^{n}\equiv 5^{3+k}\equiv -5^{k}\equiv -1\pmod{\! 7}\iff 5^k\equiv 1\pmod{\! 7}$.
This is true iff $k=(\text{ord}_7 5)m=6m$ for some $m\ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove: if $n\mid 7^n+6^n$ and $n>1$, then $13\mid n$ Prove: if $n\mid 7^n+6^n$ and $n>1$, then $13\mid n$
Let $p$ be the least prime number such that $p\mid n$.
And I want to show that $p=13$
Let $d$ be the least number such that: $14^d\equiv 0 \pmod {p}$
And by Fermat's little theorem I have: $14^{p-1}\equiv 0 \pmod{p}$
Here I'm stuck
| OP poses two questions: If $n\mid 7^n+6^n$, first show $13\mid n$, and second show that $13$ is the smallest prime factor of $n$
Start by noting that any prime factor of $n$ cannot be a prime factor of either $7^n$ or $6^n$, as a prime factor which divides $n$ and either of $7^n$ or $6^n$ must also divide the other. That is impossible because $\gcd(7^n,6^n)=1$. Thus, none of $2,3,7$ can be factors of $n$. In particular, this means that $n$ must be odd, a fact which is also apparent because the sum $7^n+6^n$ is odd.
Next, recall that for any odd exponent $k$, it is the case that $(a+b)\mid (a^k+b^k)$. Since $n$ is restricted to being odd, this means that $(7+6)\mid (7^n+6^n)$. This shows that $13\mid n$, answering OP's first query.
Since we know that $2,3,7$ are not divisors of $n$, we need only show that $5$ and $11$ are not divisors of $n$ to establish that $13$ is the smallest prime divisor of $n$
The case of $5$ is simple: $7^n+6^n \equiv 2^n+1^n \bmod 5$, and $2^n+1^n \equiv 0 \bmod 5 \iff n=4k+2$. That requires $n$ to be even, and $n$ is restricted to being odd, so $5\mid n$ is ruled out.
The case of $11$ takes a bit more work. Working modulo $11$, and recalling that when a modulus $p$ is prime, $a^k \bmod p \equiv a^{k \bmod (p-1)} \bmod p$
$$\begin {array}{c|c|c|c|c|c|c|c|c|c|c|c|} \\
n\equiv &1&2&3&4&5&6&7&8&9&10& \bmod 10 \\
6^n \equiv &6&3&7&9&10&5&8&4&2&1& \bmod 11 \\
7^n \equiv &7&5&2&3&10&4&6&9&8&1& \bmod 11 \\
\sum \equiv &2&8&9&1&9&9&3&2&10&2& \bmod 11 \\
\end {array}$$
Since the sum $7^n+6^n \not \equiv 0 \bmod 11$ for all $n$, it is the case that $11 \not\mid n$.
This shows that no prime smaller than $13$ divides $n$, answering OP's second query.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Cauchy product: how to calculate? do you know what is the method to find out the limit of this:
0 < q < 1
$$\sum_{n=0}^{\infty} (n+1)(n+2)q^n$$
The $q^n$ is ok, that is the geometric series, it is $\frac{1}{1-q}$.
But how do I do the Cauchy product? For me it seems, this is divergent...
Thank you!
| To do this without differentiation, note that
$$
\begin{eqnarray}
\sum_{n=0}^{\infty}(n+1)q^n
&=& 1+2q+3q^2+4q^3+\ldots \\
&=& 1+(1+1)q+(1+1+1)q^2+(1+1+1+1)q^3+\ldots \\
&=& (1+q+q^2+\ldots) + (q + q^2 + q^3+\ldots) + (q^2+q^3+q^4+\ldots)+\ldots \\
&=&(1+q+q^2+\ldots)(1+q+q^2+\ldots) \\
&=&\frac{1}{(1-q)^2}.
\end{eqnarray}
$$
Similarly, it follows that
$$
\begin{eqnarray}
\sum_{n=0}^{\infty}\frac{1}{2}(n+2)(n+1)q^n &=& 1+3q+6q^2+10q^3+15q^4+\ldots \\
&=& 1+(2+1)q+(3+2+1)q^2+(4+3+2+1)q^3+\ldots \\
&=& (1+2q+3q^2+\ldots) + (q+2q^2+3q^3+\ldots) + (q^2+2q^3+3q^4+\ldots) + \ldots \\
&=& (1+2q+3q^2+\ldots)(1+q+q^2+\ldots) \\
&=& \frac{1}{(1-q)^3}.
\end{eqnarray}
$$
Generally speaking, this is a convolution trick:
$$
\sum_{n}a_n x^n \cdot \sum_{n}b_n x^n = \sum_{n}\left(\sum_{j=0}^{n}a_j b_{n-j}\right)x^n\equiv\sum_{n} \left(a *b\right)_n x^n,
$$
and $(1 * 1)_n = n+1$, and $(1 * (1 * 1))_n = \frac{1}{2}(n+2)(n+1)$, and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If the range of a function is $[a,b]$ ,where $a,b \in \mathbb N$ ,find $f(a+b)$. Let $f(x)= (x+1) (x+2) (x+3) (x+4) + 5$ where $x \in [-6,6]$. If the range of this function is $[a,b]$ ,where $a,b \in \mathbb N$ ,find $f(a+b)$.
ATTEMPT:
Sketching the graph the we know that minima of this function is between $-1$ and $-2$, and since the function is strictly increasing in the interval $[-1,6]$, maxima will be at $6$.
But how to get the minima?
| $\bf{My\; Solution::}$ Let $\displaystyle \left(x+\frac{1}{2}\right) = t\;,$ and $\displaystyle -\frac{11}{2}\leq t \leq \frac{13}{2}$.
Then expression convert into
$$\displaystyle f(t) = \left(t-\frac{3}{2}\right)\cdot \left(t-\frac{1}{2}\right)\cdot \left(t+\frac{1}{2}\right)\cdot \left(t+\frac{3}{2}\right)=\left(t^2-\frac{9}{4}\right)\cdot \left(t^2-\frac{1}{4}\right)+5$$
So $$\displaystyle f(t) = t^4-\frac{5}{2}t^2+\frac{9}{16}+5$$ and $$\displaystyle0 \leq t^2\leq \frac{169}{4}$$
So we get $$\displaystyle f(t) = \left(t^2-\frac{5}{4}\right)^2+\frac{89}{16}-\frac{25}{16}\Rightarrow f(t) = \left(t^2-\frac{5}{4}\right)^2+4$$ and $$\displaystyle0 \leq t^2\leq \frac{169}{4}$$
So $\displaystyle \bf{f(t)_{Min.} = 4}$ when $\displaystyle t^2=\frac{5}{4}$
And $\displaystyle \bf{f(t)_{max.} = \left(\frac{169}{4}-\frac{5}{4}\right)^2+4=(41)^2+4 = 1681+4 = 1685,}$ When $\displaystyle t^2= \frac{169}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Basic Logarithm question - I can't get both answers from quadratic Here's the Question :
If $xy$ = $64$ and $\log_x y + \log_y x = \frac{5}{2}$, find $x$ and $y$
I can get this to
$$log_x y + \frac{1}{\log_x y} \frac{5}{2}$$
let $\log_x y = N$
$$N + \frac{1}{N} = \frac{5}{2}$$
Multiply by 2
$$2N + \frac{2}{N} = 5$$
Multiply by N
$$2N^2 + 2 = 5N$$
$$2N^2 - 5N + 2 = 0$$
$$(2N - 1)(N - 2)$$
Giving :
$$N = \frac{1}{2}$$
$$N = 2$$
Therefore :
$$\log_x y = \frac{1}{2} $$
$$\log_x y = 2$$
Giving
$$x^2 = y$$
$$x^{\frac{1}{2}} = y$$
Part of the original question :
$$xy = 64$$
As $x^2 = y$
$$x * x * x = 64$$
$$x^3 = 64$$
Therefore:
$$x = 4$$
$$y = 16$$
I can't seem to solve for $y = x^{\frac{1}{2}}$ though
Solving for $x^{\frac{1}{2}} = y$
$$x^{\frac{1}{2}} * x^{\frac{1}{2}} = 64$$
$$x^{\frac{1}{2} + \frac{1}{2}} = 64$$
$$x= 64$$
$$xy= 64$$
$$64y= 64$$
Therefore
$$x = 64$$
$$y = 1$$
This is wrong though.
Answer :
$$(4,16) or (16,4)$$
I don't see how they got the second part. The first part makes sense but I'm not able to solve for $y = x^{\frac{1}{2}}$
| If $$y=x^{1/2}$$ and $$xy=64$$ then $$x^{3/2}=64 = 2^6$$ so $$x=2^{6\times 2/3} = 2^4 =16$$ and $$y=16^{1/2}=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A line through the point P(8, -7) is a tangent to the circle C at the point T. Find T
Circle C equation $(x+5)^2+(y-9)^2=25$
A line through the point P(8, -7) is a tangent to the circle C at the point T.
Find T.
I tried simultaneous equations:
1. $(x+5)^2+(y-9)^2=25$
2. $y = m(x-8)-7$
3. differentiate circle: $\dfrac {dy} {dx} = \dfrac{-(x+5)} {y-9}$
where x, y represent co-ordinates of T
But it does not seem to be getting anywhere.
By using first 3 equations I get horrible mix of $x, x^2, x^3, y^2, y, xy$
Many thanks in advance
| Let the co-ordinates of the point $T\equiv(a, b)$ then lines PT & CT are normal to each other then $$\left(\frac{b-(-7)}{a-8}\right)\times \left(\frac{b-9}{a+5}\right)=-1$$ $$a^2+b^2-3a-2b-103=0 \tag 1$$
now, satisfy the equation of circle by point $T\equiv(a, b)$, we get
$$(a+5)^2+(b-9)^2=25$$ $$a^2+b^2+10a-18b+81=0 \tag 2$$ subtracting eq(1) from (2), we get $$13a-16b+184=0 $$$$=> b=\frac{13a+184}{16} \tag3$$
Now, in right $\Delta PTC$, we have $$(PT)^2=(CP)^2-(CT)^2$$ $$ (a-8)^2+(b+7)^2=(8+5)^2+(-7-9)^2-(5)^2$$ Substituting the value of $b$ from eq (3) in above expression, we get a quadratic equation in terms of $a$ as follows
$$(a-8)^2+\left(\frac{13a+184}{16}+7 \right)^2=400$$ $$17a^2+144a+64=0$$ By solving above quadratic equation, we get $a=-8 => b=5 $ & $a=\frac{-8}{17} => b=\frac{189}{17}$ thus we get two points of tangency on the circle as follows $$T\equiv\left(-8, 5\right)$$ & $$T\equiv\left(\frac{-8}{17}, \frac{189}{17}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Show that the following expansion is valid for first $n$ terms The question is as follows, I've gone ahead and attempted as much as I could of it twice, however the final answer that I'm receiving does not match.
I'm omitting the partial fractions steps; if they are required I'll gladly put them up.
$$\frac{3x^2+2x}{x^3-2x^2+3x-6} \equiv \frac{5}{7(x-2)} + \frac{5x-4}{7(x^2+3)}$$
Thus for what range of values of $x$ is the expansion valid?
$$-\frac{1}{3}x - \frac{2}{3}x^2 - \frac{2}{9}x^3+...$$
In order to bring the terms as unity (pardon is I'm using the wrong terminology) I did the following $$7(x-2) = 14(\frac{1}{2}x-1) $$ $$\Rightarrow 7(x^2+3) = 3(\frac{1}{3}x+1)^{\frac{1}{2}}$$
And expanded accordingly using $1 + nx + \frac{n(n-1)(x^2)}{2!} + \frac{n(n-1)(n-2)(x^3)}{3!} + ...$
And my expansions boiled down equivalently to $$(14*3)(1-\frac{1}{2}x+\frac{1}{4}x^2-\frac{1}{8}x^3)(1+\frac{1}{6}x-\frac{1}{72}x^2+\frac{1}{432}x^3)$$
However seeing this last bit, I realised that I must've messed up immensely somewhere.
Essentially I wanted to boil down the expansion to the one above and determine the values for which $-1 < x < 1$ in order for the expansion to be valid.
| You seem to have the wrong partial fractions. I get
\begin{eqnarray*}
\frac{3x^2+2x}{x^3-2x^2+3x-6} &=& \frac{16}{7(x-2)} + \frac{5x
+24}{7(x^2+3)} \\
&=& \frac{-8}{7} \frac{1}{1-\frac{x}{2}} + \frac{5x+24}{21} \frac{1}{1-\frac{-x^2}{3}}. \\
&& \\
\text{Using the series for $\;\frac{1}{1-x}$} && \text{for both of the above fractions} : \\
\frac{-8}{7} \frac{1}{1-\frac{x}{2}} &=& \frac{-8}{7} \left(1 + \frac{x}{2} + \left(\frac{x}{2}\right)^2 + \left(\frac{x}{2}\right)^3 + \cdots \right)\qquad\text{for $|x| \lt 2$} . \\
&& \\
\frac{5x+24}{21} \frac{1}{1-\frac{-x^2}{3}} &=& \frac{5x+24}{21} \left(1 - \frac{x^2}{3} + \left(\frac{x^2}{3}\right)^2 - \left(\frac{x^2}{3}\right)^3 + \cdots \right) \\
&=& \frac{1}{21} \left(24 + 5x - 8x^2 - \frac{5x^3}{3} + \cdots \right)\qquad\text{for $|x| \lt \sqrt{3}$} . \\
&& \\
\text{Summing,}\quad \frac{3x^2+2x}{x^3-2x^2+3x-6} &=& -\frac{1}{3}x - \frac{2}{3}x^2 - \frac{2}{9}x^3+\cdots . \\
&& \\
\end{eqnarray*}
From the radii of convergence of the partial fractions, this series is valid for $|x|\lt \sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Complex Numbers (Geometric Representations) What is the geometrical interpretation of this operation:
Multiplication by $\frac{\left(1-i\right)}{\sqrt{2}}$
Attempt:
multiplication by −i = rotate by −π/2
| $$\frac{(1-i)}{\sqrt{2}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i=\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i=$$
$$\left|\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i\right|e^{\arg\left(\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i\right)i}=$$
$$\sqrt{\left(\frac{1}{2}\sqrt{2}\right)^2+\left(\frac{1}{2}\sqrt{2}\right)^2}e^{-\tan^{-1}\left(\frac{\frac{1}{2}\sqrt{2}}{\frac{1}{2}\sqrt{2}}\right)i}=$$
$$\sqrt{\frac{1}{2}+\frac{1}{2}}e^{-\tan^{-1}(1)i}=e^{-\frac{\pi}{4}i}$$
$----$
$$\frac{(1-i)}{\sqrt{2}}=\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i=e^{-\frac{\pi}{4}i}=\cos\left(-\frac{\pi}{4}\right)+\sin\left(-\frac{\pi}{4}\right)i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve for real value of $x$: $|x^2 -2x -3| > |x^2 +7x -13|$ Here I have a question:
Solve for real value of $x$:
$$|x^2 -2x -3| > |x^2 +7x -13|$$
I got the answer as $x = (-\infty, \frac{1}{4}(-5-3\sqrt{17}))$ and $x=(\frac{10}{9},\frac{1}{4}(3\sqrt{17}-5)$
Please verify it if it is correct or not. Thanks
| hint: Use $|A| > |B| \iff A^2 > B^2 \iff (A-B)(A+B) > 0$. Apply this property to $A = x^2-2x-3, B = x^2+7x-13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $\cos(6x)= 32\cos^6x -48\cos^4x +18\cos^2x -1$ After writing down $\cos6x$= $Re (\cos x + i\sin x)^6$, I used the binomial theorem to expand the expression. Very soon it got really tedious and after trying $5$ times, fruitlessly, to arrive at the given expression, I gave up. Is there a shorter way around this?
| It might be tedious, but really there's no reason you shouldn't get to the answer.
Taking (for reduced finger strain) $c=\cos x$ and $s=\sin x$, and noting that $s^2 = 1-c^2$
$$\begin{align} \cos 6x &= \Re((c+is)^6) \\
&= \Re(c^6+6ic^5s-15c^4s^2-20ic^3s^3 +15c^2s^4+6ics^5-s^6)\\
&= c^6-15c^4s^2+15c^2s^4-s^6\\
&= c^6-15c^4(1-c^2)+15c^2(1-2c^2+c^4)-(1-3c^2+3c^4-c^6)\\
&= c^6(1+15+15+1)+c^4(-15-30-3) + c^2(15+3) -1\\
&= 32c^6-48c^4+18c^2-1\\
&= 32\cos^6 x-48\cos^4 x+18\cos^2 x-1\\
\end{align}$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Finding a Basis for this subspace Set $V=\mathbb{R}^{2x3}$ and let $U$ be a subspace of $V$ defined by:
\begin{equation*}
U=\{B=(b_{ij})\in V\mid b_{11} + b_{12} + b_{13} = -4(b_{21} + b_{22} + b_{23})\}.
\end{equation*}
I would greatly appreciate it if someone could help me understand how to find a basis for $U$ and the dimension of $U$.
| We have $b_{11} + b_{12} + b_{13} = -4(b_{21} + b_{22} + b_{23})$ that is equivalent to $$b_{11} = - b_{12} - b_{13} - 4b_{21} - 4b_{22} - 4b_{23}.$$
Here $b_{12},b_{13},b_{21},b_{22},b_{23}$ are declared to be free variables and the general solution is
$$
(- b_{12} - b_{13} - 4b_{21} - 4b_{22} -
4b_{23},b_{12},b_{13},b_{21},b_{22},b_{23}),
$$
where $b_{12},b_{13},b_{21},b_{22},b_{23} \in \mathbb{R}$.
Now we can find $5$ base vectors by substituting $(1,0,0,0,0)$, $(0,1,0,0,0)$, ..., $(0,0,0,0,1)$ for $b_{12},b_{13},b_{21},b_{22},b_{23}$ in the expression above. In matrix notation it should look as follows:
$$
\begin{pmatrix}
-1 & 1 & 0 \\
0 & 0 & 0
\end{pmatrix},
\begin{pmatrix}
-1 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix},
\begin{pmatrix}
-4 & 0 & 0 \\
1 & 0 & 0
\end{pmatrix},
\begin{pmatrix}
-4 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix},
\begin{pmatrix}
-4 & 0 & 0 \\
0 & 0 & 1
\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx}$ $$\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx}$$
Assume that $x^2 - 6x + 12 = (x - 3)(x - 3) + 3 = (x - 3)^2 + 3$,
then $t = x - 3 \rightarrow dt = dx$,
since $$\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx} = \int_{0}^{3} \frac{12}{t^2 + 3}\,dt$$.
However, I am unsure as to how to continue.
| Note that with your substitution the limits of integration should be changed from -3 to 0.
\begin{align*}
\int \frac{12}{t^2 +3}dt
&=\int\frac{12}{3(\frac{t^2}{3}+1)}dt\\
&=\int\frac{12}{3(\frac{t^2}{(\sqrt{3})^2}+1)}dt\\
&=\int\frac{12}{3\left(\left(\frac{t}{\sqrt{3}}\right)^2+1\right)}dt\\
&=\frac{1}{3}\int\frac{12}{\left(\frac{t}{\sqrt{3}}\right)^2+1}dt\\
\end{align*}
Then you can apply another substitution where $u=\frac{t}{\sqrt3}$ and then use the fact that
$\int\frac{1}{1+x^2}dx=\arctan x+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Derivative of a Rational function $f(x)=\sqrt{2x-5\over3x+1}$ I'm trying to find the derivative of,
$$f(x)=\sqrt{2x-5\over3x+1}$$
I think I can change this into
$$f(x)= \left({2x-5 \over 3x+1}\right)^{1\over2}
\\ =[(2x-5)(3x+1)^{-1}]^{1 \over 2}$$
Am I not right??
From now, how do I continue??
| You have right, from here you have to use this formula: $(u^n)'=n\cdot u^{n-1}\cdot u'$, where $u=(2x-5)(3x+1)^{-1}\Rightarrow u'=\frac{17}{(3x+1)^2}$
So you'll obtain: $\frac{1}{2} \cdot u^{-\frac{1}{2}} \cdot u'=\frac{1}{2}\cdot (2x-5)^{-\frac{1}{2}}(3x+1)^{\frac{1}{2}}\cdot\frac{17}{(3x+1)^2}=\frac{17}{2\sqrt{2x-5}(3x+1)^{\frac{3}{2}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\frac{x}{\sqrt{yz}+\sqrt{3}}+\frac{y}{\sqrt{xz}+\sqrt{3}}+\frac{z}{\sqrt{yx}+\sqrt{3}}\leq \frac{1}{4\sqrt{3}xyz}$ Let $x;y;z>0$ such that: $xy+yz+zx=1$. Prove that:
$\frac{x}{\sqrt{yz}+\sqrt{3}}+\frac{y}{\sqrt{xz}+\sqrt{3}}+\frac{z}{\sqrt{yx}+\sqrt{3}}\leq \frac{1}{4\sqrt{3}xyz}$
I think:
$1=xy+yz+zx\geq 3\sqrt[3]{x^2y^2z^2}\Rightarrow xyz\leq \frac{1}{3\sqrt{3}}\Rightarrow \frac{1}{4\sqrt{3}xyz}\geq \frac{3}{4}$
Then, we need to prove that:
$\frac{x}{\sqrt{yz}+\sqrt{3}}+\frac{y}{\sqrt{xz}+\sqrt{3}}+\frac{z}{\sqrt{yx}+\sqrt{3}}\leq \frac{3}{4}$
But i don't know how :(
Thanks :)
| I can slightly simplify the question as follows.
Put $a=\sqrt{yz}$, $b=\sqrt{xz}$, and $c=\sqrt{xy}$. Then $x=\frac{bc}a$, $y=\frac{ac}b$, and $z=\frac{ab}c$, $a^2+b^2+c^2=1$ and we have to prove
$$\frac{bc}{a^2+a\sqrt{3}}+\frac{ac}{b^2+b\sqrt{3}}+\frac{ab}{c^2+c\sqrt{3}}\leq \frac{1}{4\sqrt{3}abc}.$$
$$\frac{1}{a^3+a^2\sqrt{3}}+\frac{1}{b^3+b^2\sqrt{3}}+\frac{1}{c^3+c^2\sqrt{3}}\leq \frac{1}{4\sqrt{3}a^2b^2c^2}.$$
Remark. Let $f(t)=\frac{1}{t^3+\sqrt{3}t^2}$. Unfortunately, $f’’(t)=\frac{2(6t^2+8\sqrt{3}t+9)}{t^4(t+\sqrt{3})^3}>0$ provided $t>0$, so we cannot apply Jensen inequality.
If we wish to use a common denominator then we have to prove .
$$(a^3b^3+a^3c^3+b^3c^3)+\sqrt{3}(a^3b^2+b^3a^2+a^3c^2+c^3a^2+b^3c^2+c^3b^2)+3(a^2+b^2+c^2)\le$$
$$\frac{1}{4\sqrt{3}}(abc+\sqrt{3}(ab+bc+ac)+3(a+b+c)+3\sqrt{3})$$
Since $a^2+b^2+c^2=1$
$$(a^3b^3+a^3c^3+b^3c^3)+\sqrt{3}(a^3b^2+b^3a^2+a^3c^2+c^3a^2+b^3c^2+c^3b^2)+\frac{9}{4}\le$$ $$ \frac{1}{4\sqrt{3}}(abc+\sqrt{3}(ab+bc+ac)+3(a+b+c)).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Tell if a sum is convergent $\sum\limits_{n=1}^\infty \frac{2}{n(n+1)}$ $$\sum\limits_{n=1}^\infty \frac{2}{n(n+1)}$$
I tried to solve this by saying that $$\frac{2}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}$$
I then made two sums like this:
$$\sum\limits_{n=1}^\infty \frac{2}{n} - \sum\limits_{n=1}^\infty \frac{2}{n+1}$$
And since we know that $1/n$ is divergent I said that $\sum\limits_{n=1}^\infty \frac{2}{n(n+1)}$ would also be divergent, which for some reason was wrong. So my question is simply what am I doing wrong?
| Here are the steps,
$$ \sum\limits_{n=1}^\infty \frac{2}{n(n+1)} $$
$$= 2\sum\limits_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right) $$
$$=2\lim\limits_{m\to\infty}\sum\limits_{n=1}^m \left(\frac{1}{n}-\frac{1}{n+1}\right) $$
$$=2\lim\limits_{m\to\infty} \left(1-\frac12+\frac12-\frac13 + \cdots +\frac{1}{m}-\frac{1}{m+1}\right) $$
$$=2\lim\limits_{m\to\infty} \left(1-\frac{1}{m+1}\right) $$
$$= 2(1-0) = 2$$
Therefore, this series converges to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Proving that a polynomial of the form $(x-a_1)\cdots(x-a_n) + 1$ is irreducible over $\mathbb{Q}$ I want to prove that for any set of distinct integers $a_1,\ldots,a_n$, the polynomial
$$h = (x-a_1)\cdots(x-a_n) + 1$$ is irreducible over the field $\mathbb{Q}$, except for the following special cases which are reducible:
$$\left.\begin{cases}
a_1 = a\\
a_2 = a+2
\end{cases}\right\} \implies h = (x-a-1)^2$$
and
$$\left.\begin{cases}
a_1 = a\\
a_2 = a+1\\
a_3 = a+2\\
a_4 = a+3
\end{cases}\right\} \implies h = ((x-a-1)(x-a-2)-1)^2$$
| If $h(x)$ is reducible over $\mathbb{Q}$, then by Gauss's Lemma, $h(x)$ is reducible over $\mathbb{Z}$. We can find two monic $p(x), q(x) \in \mathbb{Z}[x]$, both with $\deg(p), \deg(q) < n$, such that $h(x) = p(x)q(x)$.
Notice for $1 \le k \le n$,
$$
p(a_k)q(a_k) = h(a_k) = 1 \land p(a_k), q(a_k) \in \mathbb{Z}
\quad\implies\quad p(a_k) = q(a_k).
$$
This implies $p(x)$ and $q(x)$ coincide on more points than their degree and hence they are equal to each other, i.e.,
$$\prod_{k=1}^{n} (x - a_k) = h(x) - 1 = p(x)^2 - 1 = (p(x)-1)(p(x)+1).$$
A consequence of this is $n = 2\ell$ is even. Furthermore, relabel $a_k$ is required, we can assume
$$
p(x) - 1 = \prod\limits_{k=1}^\ell (x-a_k)
\quad\text{ and }\quad
p(x) + 1 = \prod\limits_{k=1}^\ell (x-a_{k+\ell}).
$$
If $h(x)$ is reducible over $\mathbb{Q}$, so does $h(x + a)$ for any constant $a$.
Using a suitable choice of $a$, we only need to study the special case where one of the $a_k$, say $a_0 = 0$. Under this assumption, we have
$$p(0) - 1 = (-1)^\ell\prod_{k=1}^\ell a_k = 0
\quad\implies\quad
p(0) + 1 = (-1)^\ell\prod_{k=1}^\ell a_{k+\ell} = 2.
$$
Since $2$ is a prime, there aren't too much choice for $a_{k+\ell}$, they can only be $\pm 1$ or $\pm 2$. The are only $4$ possibilities
and only $3$ of them leads to sensible solution.
$$\require{cancel}
\begin{array}{|r:l|}
\hline
p(x) + 1 & p(x)-1\\
\hline
(x-2)(x-1) & x(x-3)\\
\cancel{(x-2)(x-1)(x+1)} & \cancel{x(x^2-x-1)}\\
(x+2) & x\\
(x+2)(x+1) & x(x+3)\\
\hline
\end{array}
$$
Putting the offset $a$ back, this leads to following $3$ possibilities for $h(x)$:
$$\begin{align}
(x-a)(x-a-1)(x-a-2)(x-a-3) + 1 &= ((x-a)(x-a-3) + 1)^2\\
(x-a)(x-a+2) + 1 &= (x-a+1)^2\\
\cancel{(x-a)(x-a+1)(x-a+2)(x-a+3) + 1} &= \cancel{((x-a)(x-a+3) + 1)^2}
\end{align}
$$
The $3^{rd}$ set of possibility is not a new one. It can derived from the $1^{st}$ possibility by substitution $a \mapsto a - 3$. This leaves us with two possibilities and it is easy to see they are equivalent to the two exceptions mentioned in question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
} |
Find period of $g$ if $f(x) = \sqrt{\frac{8}{1+x} + \frac{8}{{1-x}}}$ and $g(x) = \frac{4}{f(\sin x)}+\frac{4}{f(\cos x)}$ The given question
$$f(x) = \sqrt{\frac{8}{1+x} + \frac{8}{{1-x}}}$$
$$g(x) = \frac{4}{f(\sin x)}+\frac{4}{f(\cos x)}$$
find period of $g(x)$?
What I have done
putting $\sin(x)$ and $\cos(x)$ in $f(x)$,
we get $\large{\frac{4}{\cos(x)}}$ and $\large{\frac{4}{\sin(x)}}$ respectively.
Putting these in $g(x)$ we get $\sin(x) + \cos(x)$ which should have period $2\pi$ but the solution provided says that period of $g(x)$ is $\frac{\pi}{2}$.
Why?
| \begin{align}
f(\sin x) &= \sqrt{\dfrac{8}{1+\sin x} + \dfrac{8}{1-\sin x}} \qquad \qquad \\
&= 2\sqrt{2}\sqrt{\dfrac{2}{1-\sin^2 x}} \\
&= \dfrac{4}{\lvert \cos x \rvert}
\end{align}
Similarly
$$f(\cos x) = \dfrac{4}{\lvert \sin x\rvert}.$$
Thus
$$g(x) = \lvert \cos x \rvert + \lvert \sin x \rvert,$$
and the period of $g(x)$ is $\dfrac{\pi}{2}$.
In effect $\lvert \cos x \rvert$ and $\lvert \sin x \rvert$ are both $\pi-$periodic but one also has :
$$\lvert \cos(x+\pi/2)\rvert = \lvert \sin x\rvert,$$
which implies $$g(x) = g(x+\pi/2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial derivatives and finding the maximum, is there a simple argument that imply this inequality (perhaps using symmetry somehow?).
Thanks !
| $$\frac{(x+y)^2}{x^2+xy+y^2} = 1 + \frac{xy}{x^2+xy+y^2}$$
$$xy \le \frac{(x^2 + y^2)}{2} $$
$$\therefore \frac{3xy}{2} \le \frac{(x^2 + y^2 + xy)}{2}$$
$$\therefore \frac{xy}{x^2+xy+y^2} \le \frac 13$$
$$\therefore \frac{(x+y)^2}{x^2+xy+y^2} \le \frac 43 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
area of il-defined triangle $AA'$, $BB'$, and $CC'$ are straight lines drawn from the angular points of a triangle through any point $O$ within the triangle and cutting the opposite sides at $A'$, $B'$, and $C'$. $AP$, $BQ$, and $CR$ are cut off from $AA'$, $BB'$, and $CC'$ and are equal to $OA'$, $OB'$, and $OC'$.
Prove that the area of triangle $\Delta PQR$ equals area of triangle $\Delta A'B'C'$.
Context: early 1900 school book based on Euclid. At the end of Book II [pythagorus, length of side opposite angle, area of triangle = SQRT(s(s-a)(s-b)(s - c))] the question:
| Let $(x,y,z)$ be the barycentric coordinates of the point $O$ w.r.t. the given triangle $\Delta ABC$. They can be defined as
$$
\begin{aligned}
x &= OA':AA'\ ,\\
y &= OB':BB'\ ,\\
z &= OC':CC'\ ,\\
1 &= x+y+z\ ,
\end{aligned}
$$
as suggested in the following picture:
We can then write $O=xA+yB+zC$.
For a quick guide i often use Barycentric coordinates for the impatient, Max Schindler, Evan Chen.
First solution, we blindly use barycentric coordinates:
Let us compute $P$. We have $A(1,0,0)$ and
$A'(0:y:z)=(0,\frac y{y+z},\frac z{y+z})$, the mid point of $AA'$ is
$M=\frac 12(A+A')
%=\left(\frac 12,\frac y{2(y+z)},\frac z{2(y+z)}\right)
$, and finally $P$ is the reflection of $O$ in $M$, so
$$
P=2M-O=A+A'-O
=
\left(1+0-x,\ 0+\frac y{y+z}-y,\ 0+\frac z{y+z}-z\right)
=
\left(1-x,\ \frac {xy}{y+z},\ \frac {xz}{y+z}\right)
\ .
$$
In this situation, the points $A'$, $B'$, $C'$; $P$, $Q$, $R$ have the coordinates
$$
\begin{aligned}
A' &= (0:y:z)=\frac 1{y+z}(0,y,z)
&
P &= \frac 1{y+z}((1-x)^2,xy,xz)
\\
B' &= (x:0:z)=\frac 1{z+x}(x,0,z)
&
Q &= \frac 1{x+z}(xy,(1-y)^2,yz)
\\
C' &= (x:y:0)=\frac 1{x+y}(x,y,0)
&
R &= \frac 1{x+y}(xz,yz,(1-z)^2)
\end{aligned}
$$
The formula for the area $[S_1S_2S_3]$ of a triangle $\Delta S_1S_2S_3$ with vertices having barycentric coordinates $S_k(x_k,y_k,z_k)$ as a part of the area $[ABC]$ of $\Delta ABC$:
$$
\frac{[S_1S_2S_3]}{[ABC]}=
\begin{vmatrix}
x_1 & y_1 & z_1\\
x_2 & y_2 & z_2\\
x_3 & y_3 & z_3
\end{vmatrix}
\ .
$$
In our case we compute:
$$
\begin{aligned}
\frac{[A'B'C']}{[ABC]}
&=
\frac1{(y+z)(z+x)(x+y)}
\begin{vmatrix}
0&y&z\\
x&0&z\\
x&y&0
\end{vmatrix}
=
\\
&=
\frac
{2xyz}{(y+z)(z+x)(x+y)}\ ,
\\[3mm]
%
\frac{[PQR]}{[ABC]}
&=
\frac1{(y+z)(z+x)(x+y)}
\begin{vmatrix}
(1-x)^2 & xy & xz\\
xy & (1-y)^2 & yz\\
xz & yz & (1-z)^2
\end{vmatrix}
\\
&=
\frac
{2xyz}{(y+z)(z+x)(x+y)}\ .
\end{aligned}
$$
The first determinant is simple. For the second one we may force the factor $x$ in the first row, and in the first column, the same applies for $y,z$ and compute rather
$$
x^2y^2z^2
\begin{vmatrix}
(1-x)^2/x^2 & 1 & 1\\
1 & (1-y)^2/y^2 & 1\\
1 & 1 & (1-z)^2/z^2
\end{vmatrix}
\ .
$$
$\square$
Second solution, direct computations, but we keep the road closer to geometric operations and intuition.
We have first for instance
$$\frac{[AC'B']}{[ABC]}=\frac{AC'}{AB}\cdot\frac{AB'}{AC}
=\frac{yz}{(x+y)(x+z)}
=\frac{yz(y+z)}{(x+y)(y+z)(z+x)}
\ .
$$
This allows a quick computation of the ratio $[A'B'C']:[ABC]$ by expressing the area $[A'B'C']$ as $[ABC]-\sum [AB'C']$, and we get
$$
\begin{aligned}
\frac{[A'B'C']}{[ABC]}
&=
1
-\sum_{\text{cyclic}}
\frac{yz(1-x)}{(1-x)(1-y)(1-z)}
\\
&=
\frac{2xyz-x-y-z+1}{(1-x)(1-y)(1-z)}
=
\frac{2xyz}{(1-x)(1-y)(1-z)}
\end{aligned}
\ ,
$$
as in the first solution.
Let us compute now the contribution of $[OQR]:[ABC]$. We have, using signed areas this time:
$$
\frac{[OQR]}{[ABC]}
=
\frac{[OQR]}{[OBC]}
\frac{[OBC]}{[ABC]}
=
\frac{OQ}{OB}\cdot
\frac{OR}{OC}\cdot
\frac{OA'}{AA'}
=
\frac{1-2y}{1-y}\cdot
\frac{1-2z}{1-z}\cdot
\frac{x}{1}\ .
$$
Adding the three signed areas (since $1-2x$, $1-2y$, $1-2z$ may become negative), we obtain
$$
\begin{aligned}
\frac{[PQR]}{[ABC]}
&=
\sum_{\text{cyclic}}
\frac{[OQR]}{[ABC]}
=
\sum_{\text{cyclic}}
\frac{x(1-x)(1-2y)(1-2z)}{(1-x)(1-y)(1-z)}
\\
&=
\sum_{\text{cyclic}}
\frac{x(1-x)((1-y)-y)((1-z)-z)}{(1-x)(1-y)(1-z)}
\\
&=
\sum x-\sum\frac{xy}{1-y}-\sum\frac{xz}{1-z}+\sum\frac{xyz}{(1-y)(1-z)}
\\
&=
1-\sum\underbrace{\frac{(x+z)y}{1-y}}_{=y}
+ \frac{xyz}{(1-x)(1-y)(1-z)}\sum(1-x)
\\
&=
1-1+\frac{xyz}{(1-x)(1-y)(1-z)}\cdot 2
\ ,
\end{aligned}
$$
and computations lead to the same result.
$\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
| Suppose that $\sqrt{2} = a/b$, with $a,b$ positive integers. Meaning $a = b\sqrt{2}$. Consider $$A = \{ m \in \Bbb Z \mid m > 0 \text{ and }m\sqrt{2} \in \Bbb Z \}.$$
Well, $A \neq \varnothing$, because $b \in A$. By the well-ordering principle, $A$ has a least element, $s$. And $s,s\sqrt{2} \in \Bbb Z_{>0}$. Then consider the integer: $$r= s\sqrt{2}-s.$$
We have $r =s(\sqrt{2}-1) < s$, and $r > 0$. But $r\sqrt{2} = 2s-s\sqrt{2}$ is again an integer. Hence $r \in A$ and $r < s$, contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 0
} |
A problem while deriving the equation of an ellipse While deriving the equation of an ellipse,
let $0 < k < a$, $(x, y) \in R ^ 2$
(1): $\sqrt{(x + k) ^ 2 + y ^ 2} + \sqrt{(x - k) ^ 2 + y ^ 2} = 2a$
(2): $\sqrt{(x + k) ^ 2 + y ^ 2} = 2a - \sqrt{(x - k) ^ 2 + y ^ 2}$
(3): $(\sqrt{(x + k) ^ 2 + y ^ 2}) ^ 2 = (2a - \sqrt{(x - k) ^ 2 + y ^ 2}) ^ 2$
(4): $a \sqrt{(x - k) ^ 2 + y ^ 2} = a ^ 2 - kx$
The 4 equations above are all equivalent because,
*
*(1)(2) and (3)(4) only have simple arithmetic applied, thus (1) $\Leftrightarrow$ (2) and (3) $\Leftrightarrow$ (4)
*Since both sides in (2) are $\ge$ 0, (2) $\Leftrightarrow$ (3)
Having 2 sets $A$ and $B$ defined as follows,
$A = \{(x, y)|f(x, y) = g(x, y)\}$
$B = \{(x, y)|f(x, y)^2 = g(x, y)^2\} = \{(x, y)|(f(x, y) + g(x, y))(f(x, y) - g(x, y)) = 0\}$
$A \subset B$
$A = B$ if $f(x, y) \ge 0$ and $g(x, y) \ge 0$
This explains (2) $\Leftrightarrow$ (3).
My book goes further from (4) that,
(5): $(a \sqrt{(x - k) ^ 2 + y ^ 2}) ^ 2 = (a ^ 2 - kx) ^ 2$
(6): $(a ^ 2 - k ^ 2) x ^ 2 + a ^ 2 y ^ 2 = a ^ 2 (a ^ 2 - k ^ 2)$
(7): $b ^ 2 = (a ^ 2 - k ^ 2), b > 0, b ^ 2 x ^ 2 + a ^ 2 y ^ 2 = a ^ 2 b ^ 2$
(8): $\frac{x ^ 2}{a ^ 2} + \frac{y ^ 2}{b ^ 2} = 1, a > b > 0, k ^ 2 = a ^ 2 - b ^ 2$
It's clear to me that from (5) to (8) are all equivalent statements, that when plotted on $R ^ 2$ with $(x, y)$ pairs that satisfy those statements, they will all be the same curve.
But it's not clear to me why (4) $\Leftrightarrow$ (5)? For them to be equivalent, both sides of (4) should be $\ge$ 0. But apparently the right side of (4) can anytime be negative when $x$ is big enough.
Could you please clarify?
| In order for $x\gt k$ to be max in (1), $y=0$ is needed. Then, solving
$$x+k+x-k=2a$$
gives you $x=a$. This implies that $x\le a$. Hence, since $0\lt k\lt a\Rightarrow 1\lt\frac ak\Rightarrow a\lt\frac{a^2}{k}$, one has
$$x\le a\lt\frac{a^2}{k}\Rightarrow a^2-kx\gt 0.$$
So, (4)$\iff$(5).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
confusion in using Lebiniz integral rule I was trying this question -
Let $$f: (0,\infty )\rightarrow \mathbb{R}$$ and $$F(x) =
\int_{0}^{x}tf(t)dt$$ If $F(x^2)= x^{4} + x^{5} $, then the value of $\sum_{r=1}^{12}f(r^{2})$ is
I applied chain rule for differentiation in $F(x^2)$ to get $$2xF'(x^2)=4x^3+5x^4$$ then used Leibniz rule in $F(x^2)$ to get
$$F'(x^2)=2x(x^2)f(x^2)$$
and substituted it in above equation to get $f(r^2)$ then trying to sum it up but in the solution the correct equation is $ 2x(x^2)f(x^2)= 4x^3 + 5x^4 $ but according to my method the equation for $f(r^2)$ is $2x(2x)(x^2)f(x^2)= 4x^3 + 5x^4$, what am I doing wrong?
| Since $F(x^2)=x^4+x^5=(x^2)^2+(x^2)^{5/2}$, we have $F(x)=x^2+x^{5/2}$, and $f(x)=x^{-1}F'(x)=x^{-1}(2x+\frac{5}{2}x^{3/2})=2+\frac52\sqrt{x}$. It follows that $f(x^2)=2+\frac52\sqrt{x^2}=2+\frac52|x|$. Hence
$$
\sum_{r=1}^{12}f(r^2)=\sum_{r=1}^{12}(2+\frac52r)=2\cdot12+\frac52\cdot\frac{12(12+1)}{2}=24+195=219.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1314638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to expand $\sqrt{x^6+1}$ using Maclaurin's series The expansion would be
$\sum_{n=0}^\infty$$\frac{1}{2}\choose n $$x^{6n}$
How to evaluate binomial coefficient with rational numbers?
If $\frac{1}{2}\choose n $=$2n\choose n $$\times \frac{(-1)^{n+1}}{2^{2n}(2n-1)}$ what would be the expression for binomial coefficient if rational number is $\frac{3}{2}$ instead of $\frac{1}{2}$?
This is one example of Macclaurin expansion for the above expression:
$$1+\frac{1}{2x^6}+r(\frac{1}{x^{11}})$$
Can someone show the steps for expanding the expression, and how to generate remain?
| When $n$ and $m$ are positive integers, the binomial coefficient $\binom{n}{m}$ is defined as:
$$ \binom{n}{m}=\frac{n!}{m!(n-m)!}\tag{1}$$
hence by replacing $a!$ with $\Gamma(a+1)$ it is natural to consider the extended definition:
$$\binom{n}{m}=\frac{\Gamma(n+1)}{\Gamma(m+1)\,\Gamma(n-m+1)}\tag{2}$$
from which it follows that:
$$\binom{\frac{1}{2}}{n}=\frac{\Gamma(3/2)}{n!\Gamma(3/2-n)}=\frac{\frac{1}{2}\cdot\left(-\frac{1}{2}\right)\cdot\ldots\cdot\left(\frac{3}{2}-n\right)}{n!}=\frac{1}{4^n}\binom{2n}{n}\frac{(-1)^{n+1}}{2n-1}\tag{3}$$
as you stated. Since:
$$ \frac{\binom{n+1}{m}}{\binom{n}{m}}=\frac{n+1}{n+1-m}\tag{4}$$
it follows that:
$$ \binom{\frac{3}{2}}{n}=\frac{\frac{3}{2}}{\frac{3}{2}-n}\binom{\frac{1}{2}}{n}=\frac{3}{4^n}\binom{2n}{n}\frac{(-1)^n}{(2n-1)(2n-3)}.\tag{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to prove the inequality $(1+a+a^2)(1+b+b^2)(1+c+c^2) \leq (1+a+b^2)(1+b+c^2)(1+c+a^2)?$ For $a,b,c>0$ prove the inequality
$$
(1+a+a^2)(1+b+b^2)(1+c+c^2) \leq (1+a+b^2)(1+b+c^2)(1+c+a^2).
$$
Seems the rearrangement inequality must help but I can't do it. Any ideas?
| Hint: You can use Karamata's inequality and the concavity of $\log$ function, with the fact that $(1+a+a^2, 1+b+b^2, 1+c+c^2) \succ (1+a+b^2, 1+b+c^2, 1+c+a^2)$ to conclude that:
$$\sum_{cyc} \log(1+a+a^2) \le \sum_{cyc} \log(1+a+b^2)$$
P.S: The same approach provides a simple proof for the more general statement Darij made in his post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Compute the limit of this expression of norms: Compute the limit, as n goes to infinity, of the quotient:
$$\frac{||A^{n+2}(x)||}{||A^n(x)||} $$, given the matrix $$
\begin{bmatrix}
0 & 3 \\
-2 & 5 \\
\end{bmatrix}$$
and the vector x = (1,0).
I diagonalized A, found its eigenvalues and eigenvectors, and arrive at the expression
$A^{n+2}$ = $SD^{n+2}S^{-1}$ = $$\begin{bmatrix}
3 & 1 \\
2 & 1 \\
\end{bmatrix} \begin{bmatrix}
2^{n+2} & 0 \\
0 & 3^{n+2} \\
\end{bmatrix}\begin{bmatrix}
1 & -1 \\
-2 & 3 \\
\end{bmatrix}$$,
where the columns of S form a basis of eigenvectors of A for the 2-dimensional space.
Now, how do I actually compute: $${||A^{n+2}(x)||}?$$
Can I do this (from following my intuition): $$\begin{bmatrix}
3 & 1 \\
2 & 1 \\
\end{bmatrix} \begin{bmatrix}
2^{n+2} & 0 \\
0 & 3^{n+2} \\
\end{bmatrix}\begin{bmatrix}
1 & -1 \\
-2 & 3 \\
\end{bmatrix}* (1,0)^T$$, and then just multiply everything through and arrive at some 2x1 column vector, with entries that depend on n, and then take the usual p-norm for vectors, say, the 2-norm, then take the limit as n goes to infinity?
I tried this method of "brute force", which seems valid, but I am not getting the desired answer.
What's going wrong?
Thanks,
| Hint: note that
$$
\|A^n(x)\|
=
\left\|
\pmatrix{
3\cdot 2^n-2\cdot 3^n \\
2 \cdot 2^n-2\cdot 3^n
}
\right\| =
3^{n}\left\|
\pmatrix{
3\cdot (2/3)^n-2 \\
2\cdot (2/3)^n-2
}
\right\|
$$
and that as $n \to \infty$, we have
$$
\left\|
\pmatrix{
3\cdot (2/3)^n-2 \\
2\cdot (2/3)^n-2
}
\right\| \to
\left\|
\pmatrix{
-2 \\
-2
}
\right\|
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction $(\frac{n}{e})^n$n!>(\frac{n}{e})^n$
$$(n+1)!=n!(n+1)>(\frac{n}{e})^n(n+1)=(\frac{n+1}{e})^{n+1}\times \frac{(\frac{n}{e})^n(n+1)}{(\frac{n+1}{e})^{n+1}}>(\frac{n+1}{e})^{n+1}$$ This implies, but I think $$\frac{(\frac{n}{e})^n(n+1)}{(\frac{n+1}{e})^{n+1}}$$ should be considered further (more induction steps)
$n!<e(\frac{n}{2})^n$
$$n!(n+1)<e(\frac{n}{2})^n(n+1)=e(\frac{n}{2})^n(n+1)\times \frac{(\frac{n+1}{2})^{n+1}} {(\frac{n+1}{2})^{n+1}}<e(\frac{n+1}{2})^{n+1}$$ Also,
$$\frac{e(\frac{n}{2})^n(n+1)}{(\frac{n+1}{2})^{n+1}}$$ I think more steps are needed.
Can someone check this and tell if something is missing.
| I had originally written this up for another question but it seems fitting here as well. Maybe this can help someone.
Depending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ with
$$\begin{align}
a_n ~~~&:=~~~ \left ( 1 + \frac{1}{n} \right ) ^n \\ ~ \\
b_n ~~~&:=~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n}
\end{align}$$
and
$$\underset{n \rightarrow \infty}{\lim} a_n ~~~=~~~ \underset{n \rightarrow \infty}{\lim} b_n ~~~=~~~ e \\ ~ \\$$
While both sequences converge to the same limit, $a_n$ approaches from the bottom and $b_n$ approaches from the top:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import rcParams
rcParams.update({'figure.autolayout': True})
pts = np.arange(0, 20, 1)
a_n = lambda n: (1+1/n)**n
b_n = lambda n: (1-1/n)**(-n)
plt.errorbar(x = pts, xerr = None, y = a_n(pts), yerr = None, fmt = "bx", markersize = "5", markeredgewidth = "2", label = "$a_n$")
plt.errorbar(x = pts, xerr = None, y = b_n(pts), yerr = None, fmt = "rx", markersize = "5", markeredgewidth = "2", label = "$b_n$")
plt.plot(pts, [np.exp(1)]*len(pts), color = "black", linewidth = 2, label = "$e$")
plt.xlim(1.5, 14.5)
plt.ylim(2.0, 3.5)
plt.legend(loc = "best")
plt.setp(plt.gca().get_legend().get_texts(), fontsize = "22")
plt.show()
So we're going to use the following inequality:
$$\forall n \in \mathbb{N} ~ : ~~~~~ \left ( 1 + \frac{1}{n} \right ) ^n ~~~~<~~~~ e ~~~~<~~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n} \tag*{$\circledast$} \\ ~ \\$$
Thesis
$$\forall n \in \mathbb{N}, ~ n \geq 2 ~ : ~~~~~ e \cdot \left ( \frac{n}{e} \right )^n ~~~~<~~~~ n! ~~~~<~~~~ n \cdot e \cdot \left ( \frac{n}{e} \right )^n \\ ~ \\$$
Proof By Induction
Base Case
We begin with $n = 2$ and get
$$\begin{align}
& ~ && e \cdot \left ( \frac{2}{e} \right )^2 ~~~~&&<~~~~ 2! ~~~~&&<~~~~ 2 \cdot e \cdot \left ( \frac{2}{e} \right )^2 \\ ~ \\
& \Leftrightarrow && e \cdot \frac{4}{e^2} ~~~~&&<~~~~ 1 \cdot 2 ~~~~&&<~~~~ 2 \cdot e \cdot \frac{4}{e^2} \\ ~ \\
& \Leftrightarrow && \frac{4}{e} ~~~~&&<~~~~ 2 ~~~~&&<~~~~ \frac{8}{e} \\ ~ \\
&\Leftrightarrow && 2 ~~~~&&<~~~~ e ~~~~&&<~~~~ 4 ~~~~ \\
\end{align}
$$
Which is a true statement.
Inductive Hypothesis
Therefore the statement holds for some $n$. $\tag*{$\text{I.H.}$}$
Inductive Step
$$\begin{align}
& ~ && e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\
& = && (n+1) \cdot \frac{1}{e} \cdot e \cdot \left ( \frac{n+1}{e} \right )^n\\ ~ \\
& = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( \frac{n+1}{n} \right )^n\\ ~ \\
& = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( 1 + \frac{1}{n} \right )^n\\ ~ \\
& \overset{\circledast}{<} && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot e\\ ~ \\
& \overset{\text{I.H.}}{<} && (n+1) \cdot n!\\ ~ \\
& = && (n+1)!\\ ~ \\
& = && (n+1) \cdot n!\\ ~ \\
& \overset{\text{I.H.}}{<} && (n+1) \cdot n \cdot e \cdot \left ( \frac{n}{e} \right )^n\\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n}{e} \right )^{n+1} \cdot e \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( \frac{n}{n+1} \right )^{n+1} \cdot e \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot e \\ ~ \\
& \overset{\circledast}{<} && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{-(n+1)} \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\
\end{align}
$$
Conclusion
Therefore the statement holds $\forall n \in \mathbb{N}, ~ n \geq 2$.
$$\tag*{$\square$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplify $\large{\frac{7x}{2x-1} +\frac{5x-2}{2x+2}}$ Simplify the rational expression.
$$\frac{7x}{2x-1} + \frac{5x-2}{2x+2}$$
My work:
\begin{eqnarray*} \frac{7x}{2x-1} + \frac{5x-2}{2x+2}
&=& \frac{7x(2x+2)+(5x-2)(2x-1)}{(2x-1)(2x+2)} \\ &=& \frac{7x(2x+2)+(5x-2)(2x-1)}{(2x-1)(2x+2)}.\end{eqnarray*}
Is this correct?
| You would also have to check that nothing cancels from FOILing the quadratics.
Is $\frac{24x^2+5x+2}{4x^2+2x-2}$ simplifiable?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve this equations $x+y+z=1$
$ya+zc=0.5$
$yb+zd=0.5$
$zc=\frac{1}{3}$
$ya^2+zc^2=\frac{1}{3}$
$yab+zdc=\frac{1}{3}$
$yb^2+zd^2=\frac{1}{3}$
At the moment i got $a=b,c=d,a+2c=2,ya=\frac{1}{6},zc=\frac{1}{3}$
but still stuck
| Your solution is almost correct. If you substitute your expressions into your equations, you will see that also $x$ is determined. Finally you can choose, how to express the solution. One possibility is to express every veriable in terms of $d\neq 0,1$, i.e.,
\begin{align*}
x & = \frac{12d^2 - 15d + 4}{12d(d - 1)}, \\
y & = \frac{- 1}{12(d - 1)}, \\
z & = \frac{1}{3d},\\
a & = - 2(d - 1), \\
b & = - 2(d - 1),\\
c & = d.
\end{align*}
For $d=1$ and $d=0$ there is no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solving inequalities with fractions on both sides Solve this inequality: $\frac{x^2 -2}{2} < \frac{6x^2 -8x - 1}{x+5}$.
My solution:
Multiply $(x+5)^2$ on both sides:
$\frac{(x^2 -2)(x+5)^2}{2} = \frac{x^4 + 10x^3 + 23x^2 -20x -50}{2}$
$(6x^2 -8x - 1)(x+5) = 6x^3 + 22x^2 -41x - 5 $
$\frac{x^4 + 10x^3 + 23x^2 -20x -50}{2} < 6x^3 + 22x^2 -41x - 5$
$0.5x^4 - x^3 - 10.5x^2 + 31x - 20 < 0 $
I don't think doing this way will lead to solving the question....
| HINT: use that $$\frac{6x^2-8x-1}{x+5}-\frac{x^2-2}{2}=\frac{(4-x)(x-2)(x-1)}{2(5+x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
this integral $\int\frac{2-\sin{x}}{\sin^2{x}-\sin{x}+1}dx$ Find integral
$$\int\dfrac{2-\sin{x}}{\sin^2{x}-\sin{x}+1}dx$$
As suggested ,we take the Weierstrass Substitution is often useful in integrals such as the original.
$$
\begin{align}
\sin(x)&=\frac{2z}{1+z^2}\\
\cos(x)&=\frac{1-z^2}{1+z^2}\\
\mathrm{d}x&=\frac{2\,\mathrm{d}z}{1+z^2}
\end{align}
$$
I known is equal to
$$\int\dfrac{4(z^2-z+1)}{z^4-2z^3+6z^2-2z+1}dz$$
but I don't know how I should calculate the last integral.
| Hint
Note
$$\dfrac{2-\sin{x}}{\sin^2{x}-\sin{x}+1}=\dfrac{1}{w_{1}-w_{2}}\left(\dfrac{2-\sin{x}}{\sin{x}-w_{1}}-\dfrac{2-\sin{x}}{\sin{x}-w_{2}}\right)$$
where $w_{1},w_{2}$ are equation $x^2-x+1=0$complex roots
and
$$\dfrac{2-\sin{x}}{\sin{x}-w_{1}}=-1+\dfrac{2-w_{1}}{\sin{x}-w_{1}}$$
$$\dfrac{2-\sin{x}}{\sin{x}-w_{2}}=-1+\dfrac{2-w_{2}}{\sin{x}-w_{2}}$$
use this following
$$\int\dfrac{1}{\sin{x}-A}dx=\dfrac{2\arctan{\left(\dfrac{1-A\tan{\frac{x}{2}}}{\sqrt{A^2-1}}\right)}}{\sqrt{A^2-1}}$$
then you can solve it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\lim_{n\rightarrow \infty}n\ \int_0^n \sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right) dx$ Calculate
$$\lim_{n\rightarrow \infty}n\ \int_0^n \sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right) dx$$
Maybe this is too hard for me, Any suggestions please?
EDIT.
Since
$$\lim_{n\rightarrow \infty}\int_0^n \sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right) dx=0$$
I used L'Hospital's rule on
$$\lim_{t\rightarrow \infty}\frac{\int_0^t \sin^2(\pi x)\left(\frac{1}{(x-1)(x-2)}+\frac{1}{(x+1)(x+2)}\right) dx}{1/t}=$$
$$=\lim_{t\rightarrow \infty}\frac{\sin^2(\pi t)\left(\frac{1}{(t-1)(t-2)}+\frac{1}{(t+1)(t+2)}\right)}{-1/t^2}$$
i.e.
$$=\lim_{n\rightarrow \infty}\frac{\sin^2(\pi n)\left(\frac{1}{(n-1)(n-2)}+\frac{1}{(n+1)(n+2)}\right)}{-1/n^2}=0$$
by $\sin(\pi n)=0$. Is it right?
| Lemma: $n\int_n^{n+1}\frac{\sin^2 (\pi x)}{x}\,dx \to \int_0^1\sin^2 (\pi x)\,dx = 1/2.$
Proof: The integrand lies between $\sin^2 (\pi x)/(n+1)$ and $\sin^2 (\pi x)/n.$ Because $\sin^2 (\pi x)$ has period $1,$ the expression of interest lies between
$$ \frac{n}{n+1}\int_0^1\sin^2 (\pi x)\,dx, \int_0^1\sin^2 (\pi x)\,dx,$$
giving the lemma.
To find the limit in the problem, note
$$\frac{1}{(x-1)(x-2)}= \frac{1}{x-2}-\frac{1}{x-1},\,\,\,\,
\frac{1}{(x+1)(x+2)}= \frac{1}{x+1}-\frac{1}{x+2}.$$
Integrate $\sin^2 (\pi x)$ against each of these fractions and make simple changes of variables. The integral equals
$$\left(\int_1^{n+1} - \int_2^{n+2}+\int_{-2}^{n-2} - \int_{-1}^{n-1}\right) \frac{\sin^2 (\pi x)}{x}\,dx.$$
After cancellation we have
$$\left(\int_1^{2} - \int_{n+1}^{n+2}+\int_{-2}^{-1} - \int_{n-2}^{n-1}\right)\frac{\sin^2 (\pi x)}{x}\,dx.$$
Because the integrand is odd, the first and third integrals cancel. Now multiply by $n$ and take limits using the lemma. We get $-1$ for the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A calculator is broken so that the only keys that still work are the basic trigonometric and inverse trigonometric functions
A calculator is broken so that the only keys that still work are the $\sin$, $\cos$, $\tan$, $\cot$, $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ buttons. The display initially shows 0. (Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.)
(a) Find, with proof, a sequence of buttons that will transform $x$ into $\frac{1}{x}$.
(b) Find, with proof, a sequence of buttons that will transform $\sqrt x$ into $\sqrt{x+1}$.
(c) Prove that there is a sequence of buttons that will produce $\frac{3}{\sqrt{5}}$.
This is a continuation of a closed problem a while ago. I have solved parts, a and b but c is a challenge.
(a) We know that $\tan(\arctan(x))= x$ so inversing the equation, you get $\frac{1}{\tan(\arctan(x))}=\boxed{\cot(\arctan(x))}$
(b) We can solve this using right-triangle trigonometry. With legs, 1 and $\sqrt{x}$, the hypotenuse would be $\sqrt{x+1}$. To have $\frac{1}{\sqrt{x+1}}$, you would write it as $\cos(\arctan(\sqrt{x}))$. To transform $\sqrt{x}$ ot $\sqrt{x+1}$, you would have $\frac{1}{\cos(\arctan(\sqrt{x}))}$. From part (a), we can find the reciprocal of anything. All we need is the reciprocal which is $\boxed{\cot(\arctan(\cos(\arctan(\sqrt{x})))}$
How can we solve c when the initial display is 0?
| The inverses of the operations $x \mapsto \dfrac{1}{x}$ and $\sqrt{x} \mapsto \sqrt{x+1}$ are $x \mapsto \dfrac{1}{x}$ and $\sqrt{x} \mapsto \sqrt{x-1}$.
Let $A$ denote the operation $x \mapsto \dfrac{1}{x}$ and $B$ denote the operation $\sqrt{x} \mapsto \sqrt{x-1}$.
Let's work backwards from $\dfrac{3}{\sqrt{5}} = \sqrt{\dfrac{9}{5}}$ by using $B$ if the current number is greater than or equal to $1$, and $A$ if the current number is less than $1$:
$\sqrt{\dfrac{9}{5}} \overset{B}{\to} \sqrt{\dfrac{4}{5}} \overset{A}{\to} \sqrt{\dfrac{5}{4}} \overset{B}{\to} \sqrt{\dfrac{1}{4}} \overset{A}{\to} \sqrt{4} \overset{B}{\to} \sqrt{3} \overset{B}{\to} \sqrt{2} \overset{B}{\to} \sqrt{1} \overset{B}{\to} \sqrt{0} = 0$.
Now, start with $0$ and apply the operations $B^{-1}$, $B^{-1}$, $B^{-1}$, $B^{-1}$, $A^{-1}$, $B^{-1}$, $A^{-1}$, $B^{-1}$ in that order to get $\dfrac{3}{\sqrt{5}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Suppose $\lim _{ x\to 0 } \frac { a-\sqrt { { a }^{ 2 }-{ x }^{ 2 } }-{ x }^{ 2 }/4 }{ { x }^{ 4 } } $ is finite then how to find the value of a? Suppose $\lim _{ x\to 0 } \frac { a-\sqrt { { a }^{ 2 }-{ x }^{ 2 } }-{ x }^{ 2 }/4 }{ { x }^{ 4 } } $ is finite then how to find the value of a ?
I'm having trouble understand the fact that even though the highest power of x in the denominator and numerator is not same,how can the limit be finite?
Yeah I know,probably i'm having a conceptual doubt.Can someone help please!
| Answering the question about why these may vanish to the same order:
First, since $x$ is approaching $0$, you should be looking at the lowest order terms (the highest order terms are used when $x$ approaches infinity).
The numerator can be broken into
$$
\left(a-\frac{x^2}{4}\right)-\sqrt{a^2-x^2}
$$
The lowest order term of the first part is $a$ and the lowest order term of the second term is also $a=\sqrt{a^2}$. Therefore, these two lowest order terms cancel and the vanishing is of higher order.
For a more hands-on example, consider $x+x^2$ and $x-x^3$. At zero, both of these vanish to first order because of the $x$'s, but their difference vanishes to second order because the difference is $x^2+x^3$.
To actually solve this problem, you can take the original limit:
$$
\lim_{x\rightarrow 0}\frac{a-\frac{x^2}{4}-\sqrt{a^2-x^2}}{x^4}
$$
and multiply the numerator and denominator by $a-\frac{x^2}{4}+\sqrt{a^2-x^2}$, the conjugate of the numerator. This gives
$$
\lim_{x\rightarrow 0}\frac{\left(a-\frac{x^2}{4}\right)^2-(a^2-x^2)}{x^4\left(a-\frac{x^2}{4}+\sqrt{a^2-x^2}\right)}
$$
Expanding out (and simplifying the numerator) we get
$$
\lim_{x\rightarrow 0}\frac{-\frac{ax^2}{2}+\frac{x^4}{16}+x^2}{x^4\left(a-\frac{x^2}{4}+\sqrt{a^2-x^2}\right)}
$$
Since $x$ is approaching zero, the smallest terms in the numerator and denominator matter the most. If $(1-a/2)\not=0$, then the numerator has a $x^2$ term and the denominator has an $x^4$ term, which result in $\frac{1}{x^2}$, which diverges near $0$. Therefore, the $x^2$ term in the numerator must be eliminated, so $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Does the series: $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge? does $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge? I think yes, it does, because the $a_n$ in the series converges to zero. but I'm trying to prove this by the help of the fact that:
$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$
any suggestions?
| $$
\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}<\sum_{n=1}^\infty \frac{2n+1}{n^4}=\sum_{n=1}^\infty \frac{2}{n^3} + \sum_{n=1}^\infty \frac{1}{n^4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
What to do next with this indefinite integral? I'd like to evaluate the following indefinite integral:
$$\displaystyle\int \frac{dx}{(3+x^2)(\sqrt{1+x})}$$
I started by letting $y^2=1+x$ and, after simplifying, got here:
$$ 2\int\frac{dy}{y^4-2y^2+4}$$
I'm not sure what to do next. Does anyone have any advice?
| The factorization you need is
\begin{align}
y^4-2y^2+4&=y^4+4y^2+4-6y^2\\[4px]
&=(y^2+2)^2-(\sqrt{3}\,y)^2\\[4px]
&=(y^2-\sqrt{3}\,y+2)(y^2+\sqrt{3}\,y+2)
\end{align}
Then, with partial fractions,
$$
\frac{1}{(y^2-\sqrt{3}\,y+2)(y^2+\sqrt{3}\,y+2)}=
\frac{Ay+B}{y^2-\sqrt{3}\,y+2}+
\frac{Cy+D}{y^2+\sqrt{3}\,y+2}
$$
and the integral becomes (almost) elementary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $\sum_{1}^{\infty} \frac{1}{n^3} \le 1.5$ I have this sequence:
$$\sum_{1}^{\infty} \frac{1}{n^3}$$
and I need to prove that: $\sum_{1}^{\infty} \frac{1}{n^3} \le 1.5$
So basically I know that this sequence converges using the integral test, but I don't know how to prove the above statement.
Some help?
| $$1+ \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} + \frac{1}{216} + \frac{1}{343} + \frac{1}{512} + \frac{1}{729} + \cdots < 1+ \frac{1}{8} + \frac{1}{8} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{512}+\frac{1}{512}+\cdots$$
In fact, this gets you $$\mathrm{sum} < \color{red}{1 \frac{1}{3} \approx 1.333}$$
And one could easily improve it by changing the first $\frac{1}{8}$ to $\frac{1}{27}$, this is the largest overestimation. This actually gives:
$$\mathrm{sum} < \color{red}{1 \frac{53}{216} \approx 1.245}$$
By summing the first 31 terms and then using above technique, we can reach
$$\mathrm{sum} < \color{red}{1.202205}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric Limit without L'Hopital I am having problems solving this limit without L'Hopital or series.
$$
\lim_{ x\to 0 } \frac{x\cos(x) - \sin(x)}{2 x^3}
$$
I tried some trigonometric manipulations without success. I tried Trigonometric identities with no luck and separating $$ \frac{x\cos(x)}{2 x^3} and \frac{sin(x)}{2 x^3} $$ lead me nowhere, each of this limits are infinity. I kow the result is
$$
\lim_{ x\to 0 } \frac{x\cos(x) - \sin(x)}{2 x^3} = \frac{-1}{6}
$$
| since
$$\lim_{x\to 0}\dfrac{x\cos{x}-\sin{x}}{2x^3}=\lim_{x\to 0}\dfrac{-\cos{x}(\tan{x}-x)}{2x^3}=-\lim_{x\to 0}\dfrac{\tan{x}-x}{2x^3}$$
Use this two inequality
$$\tan{x}>x+\dfrac{1}{3}x^3\Longrightarrow \dfrac{\tan{x}-x}{2x^3}>\dfrac{1}{6},x\in [0,\dfrac{\pi}{2})$$
Other hand
$$\tan{x}<\dfrac{3x}{3-x^2}\Longrightarrow \dfrac{\tan{x}-x}{2x^3}<\dfrac{1}{2(3-x^2)},x\in [0,\dfrac{\pi}{2})$$
so
$$\lim_{x\to\infty}\dfrac{\tan{x}-x}{2x^3}=\dfrac{1}{6}$$
so
$$\lim_{x\to 0}\dfrac{x\cos{x}-\sin{x}}{2x^3}=-\dfrac{1}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
help with wrong result for $v(x) = \frac{\sqrt[3]{x-1}}{(x+2)^2}$ I need to differentiate this:
$$v(x) = \frac{\sqrt[3]{x-1}}{(x+2)^2}$$
I used this formula:
$$ \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}$$
Where:
$$ f'(x) = \frac{1}{3\sqrt[3]{(x-1)^2}}$$
and
$$ g'(x) = 2(x+2)$$
so my result is:
$$ \frac{ \frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2 - \sqrt[3]{(x-1)} \cdot2(x+2)}{(x+2)^4} = \frac{ \frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2 - 2\sqrt[3]{(x-1)}(x+2) }{(x+2)^4} = \frac{ (x+2)^2 - 2\sqrt[3]{(x-1)}(x+2) }{3\sqrt[3]{(x-1)^2}(x+2)^4} = \frac{ (x+2) \big[ (x+2) - 2\sqrt[3]{(x-1)} \big] }{3\sqrt[3]{(x-1)^2}(x+2)^4} = \frac{ (x+2) - 2\sqrt[3]{(x-1)} }{3\sqrt[3]{(x-1)^2}(x+2)^3}$$
but according to wolframalpha the result should be:
$$\frac{ 8 - 5x }{3\sqrt[3]{(x-1)^2}(x+2)^3}$$
| Multiplying top and bottom of
$$\frac{\frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2-2\sqrt[3]{x-1}\ (x+2)}{(x+2)^4}$$
by $3\sqrt[3]{(x-1)^2}$ gives you
$$\frac{(x+2)^2-2\sqrt[3]{x-1}\ (x+2)\cdot\color{red}{3\sqrt[3]{(x-1)^2}}}{3\sqrt[3]{(x-1)^2}(x+2)^4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Infinite sum $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}$ I am interested in the following result
\begin{equation*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}
\end{equation*}
I want to know the fundamentals.
Thankful in advance.
| Here are the definitions of the Riemann zeta function and Dirichlet eta function:
$$\zeta(s)=1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}\cdots$$
$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\frac{1}{5^s}-\frac{1}{6^s}+\cdots$$
The following manipulation relates these two functions:
$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\frac{1}{5^s}-\frac{1}{6^s}+\cdots$$
$$=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)-2\left(\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\cdots\right)$$
$$=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)-\frac{2}{2^s}\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots\right) $$
$$=\left(1-\frac{2}{2^s}\right)\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)=(1-2^{1-s})\zeta(s) $$
In fact, since $\eta(s)$ has abscissa of convergence ${\rm Re}(s)>0$ whereas $\zeta(s)$ has abscissa of convergence ${\rm Re}(s)>1$, this can be used to provide an analytic continuation of $\zeta(s)$ to the so-called critical strip $0<{\rm Re}(s)<1$. You seek $\eta(2)$, and computing $\zeta(2)=\frac{\pi^2}{6}$ is known famously as the Basel problem. See the link for many solutions to the problem.
Thus $\eta(2)=(1-2^{1-2})\zeta(2)=\frac{\pi^2}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the minimum value of $A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$ Let $a, b$ and $c$ three positive real numbers such that $a+b+c=3$. Find the minimum value of $$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}.$$
Here is my attempt. By symmetry we can assume that $a\leq b\leq c$. The function $f(x)=\frac{2-x^3}{x}$ is decreasing and convex on $]0,2^{1/3}]$. So if $c\leq 2^{1/3}$ then $3f(1)=3f(\frac{a+b+c}{3})\leq f(a)+f(b)+f(c)=A$ with equality if $a=b=c=1$.
If $c>2^{1/3}$ I don't see how to proceed.
| $\dfrac{2-a^3}{a}+\dfrac{2-b^3}{b}=\dfrac{2(a+b)}{ab}+2ab-(a+b)^2$
let $c$ is max $, a+b \le 2 ,x=ab \le \dfrac{(a+b)^2}{4} \le 1$, consider $f(x)=\dfrac{a+b}{x}+x, a+b \ge 2\sqrt{ab}\ge 2ab >ab=x\ge x^2 \implies f(x)$ will get min when $x$ get max.
so $A$ will get min when $ab=\dfrac{(a+b)^2}{4}=\dfrac{(3-c)^2}{4}$
$A \ge \dfrac{8}{3-c}-\dfrac{(3-c)^2}{2}+\dfrac{2}{c}-c^2=g(c) $
$g'(c)=(1-c)(1-\dfrac{2(c+3)}{(3-c)^2c^2}), (3-c)c \le \dfrac{(3-c+c)^2}{4}=\dfrac{9}{4}< \sqrt{2(c+3)}, g'(c)=0 $ will have only solution $c=1 $ ,it is easy to verify $g(1)=3$ is min
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{(b−c)}{a} + \frac{(a+c)}{b} + \frac{(a−b)}{c}=1$ and $a-b+c \neq 0 $, then prove that $\frac 1a = \frac 1b + \frac 1c$ The question given is
If $\dfrac{(b−c)}{a} + \dfrac{(a+c)}{b} + \dfrac{(a−b)}{c}=1$ and $a-b+c \neq 0 $ then prove that $\dfrac 1a = \dfrac 1b + \dfrac 1c$
I tried to take $abc$ on the right hand side after taking the LCM, but ended up with $b^2(c-a)+a^2(b+c)+c^2(a-b)=abc$. I could not simplify any further. Please provide only hints, not complete solution.
| $$\dfrac{(b−c)}{a} + \dfrac{(a+c)}{b} + \dfrac{(a−b)}{c}=1$$ Add $1$ to $\dfrac{(a−b)}{c}$ and subtract $1$ from $\dfrac{(a+c)}{b}$ We get,
$$\dfrac{(b−c)}{a} + \left[\dfrac{(a+c)}{b}-1\right] + \left[\dfrac{(a−b)}{c}+1\right]=1$$
$$\implies\dfrac{(b−c)}{a} + \dfrac{(a+c-b)}{b} + \dfrac{(a+c−b)}{c}=1$$
$$\implies + \dfrac{(a+c-b)}{b} + \dfrac{(a+c−b)}{c}=1-\dfrac{(b−c)}{a}$$
$$\implies \dfrac{(a+c-b)}{b} + \dfrac{(a+c−b)}{c}=\dfrac{(a+c-b)}{a}$$
$$\implies \dfrac{1}{b} + \dfrac{1}{c}=\dfrac{1}{a}$$
Yo!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
A polynomial that satisfies $x^pf(1-x) + (1-x)^pf(x) = 1$ The context of this question is the construction of the Daubechies wavelet.
$f$ is a polynomial of degree $p-1$ which satisfies the equation:
$$
x^pf(1-x) + (1-x)^pf(x) = 1 \tag{1}
$$
Since
$$
f(x) = \frac{1}{(1-x)^p} - \frac{x^p}{(1-x)^p} f(1-x)
$$
and
$$
\frac{1}{(1-x)^p} = \sum_{k=0}^{\infty} \begin{pmatrix} p-1+k \\ k \end{pmatrix} x^k
$$
it is argued that
$$
f(x) = \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix} x^k + O(x^p) = \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix} x^k
$$
where higher order terms are zero because $f$ is degree $p-1$.
My problem is that I can't seem to verify that $(1)$ holds with this $f$. I've tried expanding the $(1-x)^p$ and $(1-x)^k$ terms. This turns into a mess. I've tried doing induction on $p$. Again, I get nowhere. I've tried other approaches with no success. I feel like I'm missing something simple!
Here is an example of one of my attempts:
\begin{align*}
x^pf(1-x) + (1-x)^pf(x) &= x^p \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix} (1-x)^k + (1-x)^p \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix} x^k \\
&= \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix}\left( x^p(1-x)^k + x^k(1-x)^p \right) \\
&= \sum_{k=0}^{p-1} \begin{pmatrix} p-1+k \\ k \end{pmatrix}\left( x^{p-k}[x(1-x)]^k + [x(1-x)]^k(1-x)^{p-k} \right)
\end{align*}
I get to this point, and I can't help but thinking of using the Binomial Theorem, but I cannot seem to manipulate the factorial terms in nice way.
| define $g(x) =x^pf(1-x)$. so
$g(x)+g(1-x)=1$. make $x=1/2+y$, and we have
$g(x+1/2)+g(1/2-x)=1$
define $P(x)=g(x+1/2)$, and we have
$P(x)+P(-x)=1$, being $P(x)$ a polynomial, it must have the form $P(x)=1/2+\sum_{n=0}^{p-1} a_n*x^{2n+1}$
then $f(x) =P(1/2-x)/(1-x)^p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$
If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$
I don't know how to find all possible values.
I tried making the lower of the two roots as negative. That does not seem to work out. As $4>0$ the curve will be concave up. There can be numerous cases possible.
| The function will be positive when $x$ is very large positive or very large negative. This means that there are two solutions, to your equation, of which at least one is negative.
The smallest root is:
$$\frac{-a -\sqrt{a^2-16(a-3)}}{8}$$
So we want to know when $a^2-16(a-3) > 0$, or $a^2-16a+48 > 0$, or $(a-8)^2-16 > 0$, or $(a-8)^2 > 16$, or $$a-8 > 4 \vee a-8 < -4$$
$$a > 12 \vee a < 4$$
When $a$ is positive, both $-a$ and $-a -\sqrt{a^2-16(a-3)}$ are negative, so then the smallest root is negative.
When $a$ is negative, $$\sqrt{a^2-16a+48}=\sqrt{(a-6)^2+12}>\sqrt{(a-6)^2}=|a-6|=6-a$$
With the the second because of the increasing nature of $\sqrt{x}$, the fourth because $a$ and thus $a-6$ negative is. Therefore we have $$\frac{-a -\sqrt{a^2-16(a-3)}}{8}<\frac{-a - (6-a)}{8}=\frac{-6}{8}=-0.75$$
Thus the solution is
$$a > 12 \vee a < 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 3
} |
$\ln $ and Taylor Series Expansion (what went wrong) Edited
Problem
I'm trying to express $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})}$ in terms of $\ln N$, where $K$ is a constant and $1 \leq N \leq K$. This also implies $\frac{N}{K} \leq 1$.
Anyone knows if it is possible to represent $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})}$ in terms of $\ln N$ ?
Cuz as seen below, using the Taylor Expansion just returns the original term back to us and it didn't seem to help a lot.
Taylor Series Expansion Used
The Taylor series I'll be using is: (refer to this Website for Taylor Series Expansions)
$$\ln (1+x) = x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots $$
where $-1 \leq x \leq 1$.
Solving
Then, observe that $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} = \ln{(1-(\frac{N}{K})^{\frac{1}{8}}) (1+(\frac{N}{K})^{\frac{1}{8}})} = \ln(1-\frac{N}{K})^{\frac{1}{8}} + \ln (1+\frac{N}{K})^{\frac{1}{8}}$.
By the series expansion: we get:
$$\ln(1-\frac{N}{K})^{\frac{1}{8}} = -(\frac{N}{K})^{\frac{1}{8}}-\frac{(\frac{N}{K})^\frac{1}{4}}{2} - \frac{(\frac{N}{K})^\frac{3}{8}}{3}- \frac{(\frac{N}{K})^\frac{1}{2}}{4} - \ldots$$
and
$$\ln(1+\frac{N}{K})^{\frac{1}{8}} = (\frac{N}{K})^{\frac{1}{8}}-\frac{(\frac{N}{K})^\frac{1}{4}}{2} + \frac{(\frac{N}{K})^\frac{3}{8}}{3}- \frac{(\frac{N}{K})^\frac{1}{2}}{4} - \ldots$$
Thus taking the sum of the 2 logarithms above, we get:
$$\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} =- (\frac{N}{K})^\frac{1}{4} - \frac{(\frac{N}{K})^\frac{1}{2}}{2} - \frac{(\frac{N}{K})^\frac{3}{4}}{3} - \ldots $$
Notice that if we were to differentiate the series, we get:
$\frac{d}{dx} x + \frac{1}{2}(x^2) + \frac{1}{3}(x^3) + \ldots = 1+x+x^2+ \ldots = \frac{1}{1-x}$, for $|x| \leq 1$.
Hence, $x + \frac{1}{2}(x^2) + \frac{1}{3}(x^3) + \ldots = \int \frac{1}{1-x} dx = -\ln{(1-x)} + C$ for some arbitrary constant $C$.
Hence, $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} = -[(\frac{N}{K})^\frac{1}{4} + \frac{(\frac{N}{K})^\frac{1}{2}}{2} + \frac{(\frac{N}{K})^\frac{3}{4}}{3} - \ldots] = \int -\frac{1}{1-(\frac{N}{K})^\frac{1}{4}} d\frac{N}{K} = \ln {(1-(\frac{N}{K})^\frac{1}{4})} + C$.
We get back the same term!
| Note that instead of going through the steps of adding the
Taylor series for $\ln\left(1+\left(\frac NK\right)^{1/8}\right)$
and $\ln\left(1-\left(\frac NK\right)^{1/8}\right)$,
you could have derived the series
$$\ln\left(1-\left(\frac NK\right)^{\frac14}\right)
= -\left(\frac NK\right)^\frac14
- \frac{\left(\frac NK\right)^\frac12}{2}
- \frac{\left(\frac NK\right)^\frac34}{3} - \ldots $$
simply by applying the known Taylor series expansion of $\ln(1+x)$
with $x = -\left(\frac NK\right)^{1/4}.$
Your method of differentiating and then integrating the series
is a partial confirmation that the series is the
correct Taylor series of $\ln\left(1-\left(\frac NK\right)^{1/4}\right).$
(You would completely confirm the correctness of the series
if you showed that $C=0$.)
So it's hardly surprising that this manipulation of the series
gives you back what you started with (plus the pesky unknown constant $C$).
Of course you conceivably could have started with a much more complicated
expression that happened to give the same Taylor series and therefore
was equal to $\ln\left(1-\left(\frac NK\right)^{1/4}\right).$
Getting back the same expression is a clue that perhaps the expression
$\ln\left(1-\left(\frac NK\right)^{1/4}\right)$ is already about as
simple as it should be.
There's no reason to expect $\ln N$ to show up anywhere in any of these
calculations unless you force it to occur by substituting $N = e^{\ln N}$.
In fact,
$$\ln\left(1-\left(\frac NK\right)^{\frac14}\right)
= \ln\left(1-e^{(\ln N - \ln K)/4}\right),$$
so unless you have a nicer way of writing $\ln\left(1-e^{(x - \ln K)/4}\right)$
in terms of $x$ then I think this may be as good as it gets.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$f(x) =\lim_{n \to \infty} \frac{(1+ \sin \frac{\pi}x)^n - 1} { (1+ \sin \frac{\pi}x)^n +1}$, $x \in (0,1]$. To show that $f$ is integrable on $[0,1]$ A function defined on $[0,1]$ by $f(0) = 0$ and
$f(x) = \lim_{n \to \infty} \frac{(1+ \sin \frac{\pi}x)^n - 1} { (1+ \sin \frac{\pi}x)^n +1}$, $x \in (0,1]$. To show that $f$ is integrable on $[0,1]$ and $\int_0^1f = 1 - \log4$.
My Try: We see that $f(x) = 0 , \forall x = 1, \frac 1 2, \frac 1 3, \ldots$ ,
for, $\frac 1 2 < x <1 \implies \pi < \frac{\pi}{x} < 2 \pi \implies 0 < 1+ \sin \frac{\pi}x <1$
how to find $f(x)$ from here?
and $\frac 1 3 < x < \frac 1 2 \implies 2\pi < \frac{\pi}{x} < 3 \pi \implies 1 < 1+ \sin \frac{\pi}x <2$
In a same way other will repeat... how to proceed from here?
| Let $\displaystyle f_n:(0,1]\to \mathbb R,x\to\frac{(1+ \sin \frac{\pi}x)^n - 1} { (1+ \sin \frac{\pi}x)^n +1}$ .
Let $x\in (0,1]$ be a fixed real number.
If $x\in (\frac{1}{2p+1},\frac{1}{2p})$ for some $p\in \mathbb N$, then $\sin (\frac{\pi}x)>0$. Hence $f_n(x)\to 1$ as $n\to \infty$
If $x\in (\frac{1}{2p+2},\frac{1}{2p+1})$ for some $p\in \mathbb N$, then $\sin (\frac{\pi}x)< 0$. Hence $f_n(x)\to -1$ as $n\to \infty$
This yields a closed form for $f$:
*
*if $x\in (\frac{1}{2p+1},\frac{1}{2p})$ then $f(x)=1$
*if $x\in (\frac{1}{2p+2},\frac{1}{2p+1})$ then $f(x)=-1$
To prove $f$ is integrable, you can use Lebesgue DCT theorem.
Then $\displaystyle \int_{0}^1 f = \lim_{n\to \infty} \int_{1/(n+1)}^1 f(t) dt=\lim_{n\to \infty} \sum_{k=1}^n \int_{1/(k+1)}^{1/k} (-1)^k=\sum_{k=1}^\infty (-1)^k (\frac{1}{k}-\frac{1}{k+1})=1-\ln(4).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What happens to the following limit when $b\in [0, 1)$ and $b > 1$? Problem:
Find all $a, b$ which make the following statement true: $\lim_{x\to 0}\frac{\exp{\left(\sin ax\right)} - \cos x}{x^b} = \frac{1}{2}$
Attempted solution:
Firstly, let's notice that if $b < 0$ then the limit is $0$ no matter what the value of $a$ is.
Secondly, let's rewrite the fraction using the Maclaurin Series for $\exp$ and $\cos$:
$$
\frac{1 + \sin(ax) + \frac{1}{2!}(\sin^2 ax) + \frac{1}{3!}\sin^3(ax) + \dots+ (-1)(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots)}{x^b} = \\
\frac{\sin(ax) + \frac{1}{2!}(\sin^2(ax) + x^2) + \frac{1}{3!}\sin^3(ax) + \frac{1}{4!}(\sin^4(ax) - x^4) + \dots}{x^b}
$$
From this representation it is clear that if $b = 1$ and $a = 0.5$:
$$
\begin{split}
&\lim_{x\to 0}\frac{\sin(0.5 x) + \frac{1}{2!}(\sin^2(0.5 x) + x^2) + \frac{1}{3!}\sin^3(0.5x) + \frac{1}{4!}(\sin^4(0.5x) - x^4) + \dots}{x} &= \\
&\lim_{x\to 0}\frac{\sin(0.5 x)}{x} + 0 &= \frac{1}{2}
\end{split}
$$
At this point let's remember that $\lim_{x\to 0}\sin x = \lim_{x\to 0} x$ so the fraction may be rewritten like so:
$$
\lim_{x\to 0}\frac{ax + \frac{1}{2!}((ax)^2 + x^2) + \frac{1}{3!}(ax)^3 + \frac{1}{4!}((ax)^4 - x^4) + \dots}{x^b}
$$
which makes this limit $0$ if $b\in[0, 1)$ and $\infty$ if $b > 1$.
| Hint: By using limiting arguments about the behavior of functions $e^x$, $\sin x$, and $\cos x$ for small $x$ (e.g. using Taylor series), you should be able to show that your limit is equal to $\lim_{x \to 0} \frac{ax}{x^b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $f:\mathbb{R} \to \mathbb{R}$ be $C^3$. Show the equivalence: $f^{(k)}(0) = 0$ for $k=0,1,2 \iff \lim_{x\to 0} \frac{f(x)}{x^3}$ exists Let $f:\mathbb{R} \to \mathbb{R}$ be $C^3$. Show the equivalence:
$$f^{(k)}(0) = 0 \quad k=0,1,2 \iff \lim_{x\to 0} \frac{f(x)}{x^3} \text{, exists.}$$
Trying:
Since $f \in C^3$, implies $f, f', f''$ are differentiable (therefore continuous) and $f'''$ is continuous in all domain.
Therefore we can approach $f$ with a order 3 Taylor polynomial, such as:
$$f(a+x) = f(a) + f'(a)x + \frac{f''(a)}{2}x^2 + r(x)$$
Where $\lim_{x \to 0} \frac{r(x)}{x^3} = 0$. (can I assume that? isn't what I'm trying to prove? This is the definition of Infinitesimal Taylor Polynomial. ).
$(\Rightarrow)$ Let's solve for $r(x)$
$$ r(x) = f(a+x) - f(a) - f'(a)x - \frac{f''(a)}{2}x^2 $$
Dividing by $x^3$
$$ \frac{r(x)}{x^3} = \frac{f(a+x)}{x^3} - \frac{f(a)}{x^3} - \frac{f'(a)}{x^3}x - \frac{f''(a)}{2x^3}x^2 $$
Lets take $a=0$
$$ \frac{r(x)}{x^3} = \frac{f(x)}{x^3} - \frac{f(0)}{x^3} - \frac{f'(0)}{x^3}x - \frac{f''(0)}{2x^3}x^2 $$
By hypothesis $f^{(k)}(0) = 0 \quad k=0,1,2$. Let's substitute:
$$ \frac{r(x)}{x^3} = \frac{f(x)}{x^3} - \frac{0}{x^3} - \frac{0}{x^3}x - \frac{0}{2x^3}x^2 $$
Taking the limit
$$ \lim_{x \to 0} \frac{r(x)}{x^3} = \lim_{x \to 0} \frac{f(x)}{x^3} - \lim_{x \to 0} \frac{0}{x^3} - \lim_{x \to 0} \frac{0}{x^3}x - \lim_{x \to 0} \frac{0}{2x^3}x^2 $$
Since $f$ is continuous
$$ \lim_{x \to 0} \frac{r(x)}{x^3} = \lim_{x \to 0} \frac{f(x)}{x^3} - 0 - 0 - 0 $$
$$ \lim_{x \to 0} \frac{r(x)}{x^3} = \lim_{x \to 0} \frac{f(x)}{x^3} $$
Therefore $ r(x) = f(x) $ since the limit of $ \lim_{x \to 0} \frac{r(x)}{x^3} $ exists (is a Taylor Poly) we conclude that lim of $ \lim_{x \to 0} \frac{f(x)}{x^3} $ also exists.
Is that correct? Also I don't have any idea in how to do the other direction.
| Lemma: Suppose $g\in C^2(\mathbb {R})$ and $g'''(0)$ exists. If $0=g(0)=g'(0)=g''(0)=g'''(0),$ then $g(x) = o(x^3)$ as $x\to 0.$ Proof:
$$g(x) = g(x)-g(0) = g'(c_x)x = (g'(c_x)-g'(0))x= g''(d_x)c_xx= (g''(d_x)-g''(0))c_xx = [(g''(d_x)-g''(0))/d_x]d_xc_xx$$
where we've used the MVT twice. Since $c_x,d_x$ are in between $0$ and $x,$ and the term in brackets $\to g'''(0)=0,$ we see the above is $o(x^3)$ as desired.
Now consider an arbirary $f\in C^2$ such that $f'''(0)$ exists. Let $P(x) = f(0)+f'(0)x + f''(0)x^2/2 + f'''(0)x^3/3!.$ Then by the lemma,
$$f(x) = P(x) + o(x^3).$$
The conclusion in this problem for $f$ follows easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$ If $$x+y+z=7$$ and $$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{10}$$ Find the value of $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$$
I tried but I got nothing
| Rewrite the final expression you are trying to solve for as:
$\frac{(x+y+z) - (y+z)}{y+z} + \frac{(x+y+z) - (x+z)}{x+z} + \frac{(x+y+z) - (y+x)}{y+x}$
This is equal to:
$\frac{(x+y+z) }{y+z} + \frac{(x+y+z) }{x+z} + \frac{(x+y+z) }{y+x} - 3$
Which simplifies to:
$\frac{7}{y+z} + \frac{7 }{x+z} + \frac{7}{y+x} - 3 = 7 (\frac{1}{y+z} + \frac{1 }{x+z} + \frac{1}{y+x}) - 3$
This is equal to:
$7(\frac{7}{10}) - 3 = \frac{19}{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is it possible to express $\left(\begin{array}{cc}a & -a \\ a-1 & 1-a \\ \end{array} \right)$ as a certain product of two matrices? Is it possible to express
$$\left(
\begin{array}{cc}
a & -a \\
a-1 & 1-a \\
\end{array}
\right), \ \ \ \ a\in\mathbb R$$
as a certain product of two matrices? Namely,
$$\left(
\begin{array}{cc}
a & -a \\
a-1 & 1-a \\
\end{array}
\right)=PQ$$
where $P$, $Q$ are two matrices with entries not all $1$.
I tried to do some product but without success. Any suggestions please?
| Consider the product of two matrices $A$ and $B$ in the following way.
\begin{align}
A \cdot B = \left(\begin{array}{cc} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array} \right)\left(\begin{array}{cc} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array} \right) = \left(\begin{array}{cc} a & -a \\ a-1 & 1-a \end{array} \right).
\end{align}
Now, it can be seen that
\begin{align}
a_{1} a_{2} + b_{1} c_{2} &= a = -(a_{1} b_{2} + b_{1} d_{2}) \\
a_{2} c_{1} + d_{1} c_{2} &= a-1 = -(c_{1} b_{2} + d_{1} d_{2})
\end{align}
for which $a_{1} a_{2} + b_{1} c_{2} = -(a_{1} b_{2} + b_{1} d_{2})$ leads to
\begin{align}
a_{1} = - b_{1} \, \left( \frac{c_{2} + d_{2} }{ a_{2} + b_{2} } \right)
\end{align}
and from the second set
\begin{align}
c_{1} = - d_{1} \, \left( \frac{c_{2} + d_{2} }{ a_{2} + b_{2} } \right).
\end{align}
It should be mentioned that $a_{2} \neq - b_{2}$. By using $a_{1} a_{2} + b_{1} c_{2} = a$ then
\begin{align}
b_{1} = \frac{a \, (a_{2} + b_{2})}{b_{2} c_{2} - a_{2} d_{2}}
\end{align}
and similarly
\begin{align}
d_{1} = \frac{(a-1) \, (a_{2} + b_{2})}{b_{2} c_{2} - a_{2} d_{2}}
\end{align}
where $b_{2} c_{2} \neq a_{2} d_{2}$. Putting this all together leads to
\begin{align}\tag{1}
\frac{1}{\left| \begin{array}{cc} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array} \right|} \, \left(\begin{array}{cc} a \, (c_{2} + d_{2}) & -a \, ( a_{2} + b_{2}) \\ a \, ( c_{2} + d_{2}) & -(a-1) \, (a_{2} + d_{2}) \end{array} \right)\left(\begin{array}{cc} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array} \right) = \left(\begin{array}{cc} a & -a \\ a-1 & 1-a \end{array} \right).
\end{align}
where $a_{2} d_{2} - b_{2} c_{2} \neq 0$ and $a_{2} + b_{2} \neq 0$. From here many choices can be made for the values of $\{a_{2}, b_{2}, c_{2}, d_{2}\}$. One example is $(a_{2}, b_{2}, c_{2}, d_{2}) = (1,0,0,-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding $(a, b, c)$ with $ab-c$, $bc-a$, and $ca-b$ being powers of $2$ This is a 2015 IMO problem. It seems difficult to solve.
Find all triples of positive integers $(a, b, c)$ such that each of the numbers $ab-c$, $bc-a$, and $ca-b$ is a power of $2$.
Four such triples are $(a,b,c)=(2,2,2),(2,3,2),(3,5,7),(2,6,11)$.
| A Sketch of a Possible Solution
Wlog order the solutions so that $a \ge b \ge c \ge 1$ with the system of equations
$\begin{align}
ab - c &= 2^k \\
ac - b &= 2^l \\
bc - a &= 2^m
\end{align}$
with $k \ge l \ge m \ge 0$.
We distinguish four cases:
*
*$a=b=c$
*$a=b>c$
*$a>b=c$
*$a>b>c$
Case 1: $a=b=c$
The main system of equations reduce to the single equation
$a(a-1) = 2^k \tag{1A}$.
$a,a-1$ cannot both be powers of 2 unless $a=2$. Therefore, the only solution for this case is
$\boxed{a=b=c=2}$
Case 2: $a=b>c$
Main equations reduce to
$\begin{align}
a^2 - c &= 2^k \tag{2A}\\
a(c - 1) &= 2^l \tag{2B}
\end{align}$
with $a > c, k > l$. Equation (2B) implies that $a$ is even since $a>c$, and (2A) now implies that $c$ is even since $k>l\ge0$. Then $a=2^l$ and so (2A) becomes: $2^{2l} = 2^k+2$ which can only be satisfied by $k=l=1$. However, this violates $k>l$ (it is just the first case again). No solutions for this case.
Case 3: $a>b=c$
Main equations reduce to
$\begin{align}
b(a - 1) &= 2^k \tag{3A}\\
b^2 - a &= 2^m \tag{3B}
\end{align}$
with $a > b, k > m$. Since $a>1$, (3A) implies $a$ is odd. On the other hand, (3B) implies $b>1$, so by (3A) $b$ must be even and $m=0$. Furthermore, we can put $a=2^p+1,b=2^q$ with $p\ge1,q\ge1$. So by (3B)
$b^2-a = 2^{2q}-(2^p+1) = 1$ so $2^{2q}-2^p = 2$. Whence $p=q=1$ must hold. So $a=3,b=2$ and the only solutions for this case are
$\boxed{a=3,b=c=2}$
Case 4: $a>b>c$
We restate the system of equations
$\begin{align}
ab - c &= 2^k \\
ac - b &= 2^l \tag{4A}\\
bc - a &= 2^m
\end{align}$
where we now have $a>b>c\ge1,k>l>m\ge0$. If exactly one of $a,b,c$ is even, then $ab-c,ac-b,bc-a$ are all odd, which cannot satisfy the system of equations with $k>l>m$ (see Theo Bendit's argument).
So now distinguish three other cases:
*
*$a,b,c$ all even
*Two of $a,b,c$ are even
*$a,b,c$ all odd
Case 4.1: $a>b>c$ with $a,b,c$ all even
We write $a=2p,b=2q,c=2r$ with $p>q>r\ge1$. Then system (4A) can be re-expressed as
$\begin{align}
4pq - 2r &= 2^k \\
4pr - 2q &= 2^l \\
4qr - 2p &= 2^m
\end{align}$
Since the left-hand sides are all integers, divide through by 2 to get:
$\begin{align}
2pq - r &= 2^{k-1} \\
2pr - q &= 2^{l-1} \\
2qr - p &= 2^{m-1}
\end{align}$
Since $p>q>r$, only $r$ can be odd. But if this is so, $m=1$ and this is excluded since $2pq-r\ge2\cdot3\cdot2-1=11$. So $p,q,r$ are all even. Then write $p=2s,q=2t,r=2u$ with $s>t>u$. Then system (4A) becomes
$\begin{align}
4pq - r &= 2^{k-2} \\
4pr - q &= 2^{l-2} \\
4qr - p &= 2^{m-2}
\end{align}$
Using similar arguments we can show that $s,t,u$ must all be even. Since these arguments can be repeated ad infinitum, we conclude that there are no solutions for Case 4.1.
Case 4.2: $a>b>c$ with two of $a,b,c$ even
In this case, exactly one of $ab-c,ac-b,bc-a$ are odd, so to satisfy (4A) it must be the smallest of these: $bc-a$. Hence we must have $a$ odd, and $b,c$ even. Furthermore, $m=0$. So in (4A) we must have
$bc - a = 1 \tag{4.2A}$ with the restriction that $b>c\ge2$.
Now substitute the equivalent expression for $a$ into (4A) to get
$\begin{align}
b^2c - b - c &= 2^k \\
bc^2 - b -c &= 2^l \tag{4.2B}\\
\end{align}$
Hence by subtraction
$\begin{align}
b^2c - bc^2 &= 2^k - 2^l \\
bc(b-c) &= 2^l(2^{k-l}-1) \tag{4.2C}\\
\end{align}$
This is satisfied by $b=6,c=2$ or the solution $(a,b,c)=(11,6,2)$.
There may be other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 0
} |
Infinite derivative of nested radicals $$\cdots \frac{d}{dx}\frac{d}{dx}\frac{d}{dx} \sqrt{x+\sqrt[3]{x+\sqrt[4]{x\cdots}}}$$ Its not super hard to find a finite number of derivatives, but I can not understand how to pull off infinite here. Please help. Thank you.(Is it even logical?)
| Consider the function
\begin{align}
S_{n}(x) = \sqrt[2]{x + \sqrt[3]{x + \sqrt[4]{x + \cdots + \sqrt[n+2]{x}}}}
\end{align}
For the case of $S_{2}(x)$ then
\begin{align}
D S_{2}(x) &= D\left[ \sqrt[2]{x + \sqrt[3]{x + \sqrt[4]{x}}} \, \right] \\
&= \frac{1}{2 \, S_{2}(x) } \, \left[ 1 + \frac{1}{3 \, \left( \sqrt[3]{x + \sqrt[4]{x}} \right)^{2}} \, \left( 1 + \frac{1}{4 \, \sqrt[4]{x^{3}}} \right) \right]
\end{align}
Further differentiation can be applied. If this problem was found in a book the line "...an exercise left for the reader." is applied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve: $-\frac{1}{\sqrt{2}} < \sin \theta + \cos \theta < \frac{1}{\sqrt{2}}$ The question is:
Solve $$-\frac{1}{\sqrt{2}} \lt \sin\theta + \cos\theta < \frac{1}{\sqrt{2}}$$ for values of $\theta$ between $0^\circ$ and $180^\circ$.
I realized that:
$$\begin{align}
-\frac{1}{\sqrt{2}} < \sin\theta + \cosθ &< \frac{1}{\sqrt{2}} \\[4pt]
\left|\sin\theta + \cosθ\right| &< \frac{1}{\sqrt{2}} \\[4pt]
\left(\sin\theta + \cos\theta\right)^2 &< \frac12 \\[4pt]
\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta &< \frac12 \\[4pt]
1 + 2\sin\theta\cos\theta &< \frac12 \\[4pt]
1 + \sin 2\theta &< \frac12 \\[4pt]
\sin 2\theta &< -\frac12
\end{align}$$
But I don't know where to go from here. Can someone help me figure out how to get to the answer in the book: $105^\circ < θ < 165^\circ$. Thanks!
| Hint: $$\sin \theta + \cos \theta \equiv \sqrt{2} \sin \left(\theta + \frac{\pi}{4}\right).$$
Or alternatively, in degrees $$\sqrt{2} \sin \left(\theta + 45^{\circ}\right).$$
Using your method, however, we get that $$\sin 2\theta < -
\frac{1}{2} \implies 210^{\circ} < 2\theta < 330^{\circ}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\int \frac{\left(2x^3-4x^2-x-3\right)}{\left(x^2-2x-3\right)}dx$ confusion $$\int \frac{\left(2x^3-4x^2-x-3\right)}{\left(x^2-2x-3\right)}dx$$
=$$x^2+3\ln \left(x-3\right)+2\ln \left(x+1\right)+C$$
Where did the $x^2$ come from?
What I did:
Partial fraction then integrate, got the answer without $x^2$ term.
| Hint. It comes from the partial fraction decomposition, you have
$$
\frac{2x^3-4x^2-x-3}{x^2-2x-3}=2x+\frac2{x+1}-\frac3{x-3} \tag1
$$ giving
$$
\begin{align}
\int\frac{2x^3-4x^2-x-3}{x^2-2x-3}dx&=\int 2xdx+\int\frac2{x+1}dx-\int\frac3{x-3}dx\\\\
&=\color{red}{x^2}+\int\frac2{x+1}dx-\int\frac3{x-3}dx\\\\
&=\color{red}{x^2}+2 \ln|x+1|-3 \ln|x-3|+C.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplify the Difference Quotient when $f(x)= \frac{15}{x^2}$ Simplify the Difference Quotient when $f(x)= \frac{15}{x^2}$. The difference quotient formula is $\frac{f(x+h)- f(x)}{ h}$.
I understand how to simplify basic difference quotient equations. In this case, you start with the formula: $\displaystyle\frac{\frac{15}{{(x+h)}^2}-\frac{15}{x^2}}{h}$. The answer to the problem is $\displaystyle\frac{-15(h+2x)}{x^2(h+x)^2}$.
I have tried multiplying by the conjugate and the Least Common Denominator, but my algebra skills aren't leading me to the correct answer. Please help!
| $f(x)=\frac{15}{x^2}$
Now
$\frac{f(x+h)-f(x)}{h}=\frac{\frac{15}{(x+h)^2}-\frac{15}{x^2}}{h}$
$=\frac{15}{h}[\frac{1}{(x+h)^2}-\frac{1}{x^2}]$
$=\frac{15}{h}[\frac{x^2-(x+h)^2}{x^2(x+h)^2}]$
$=\frac{15}{h}[\frac{x^2-x^2-2xh-h^2}{x^2(x+h)^2}]$
$=\frac{-15}{h}[\frac{2xh+h^2}{x^2(x+h)^2}]$
$=-15[\frac{2x+h}{x^2(x+h)^2}]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.
options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0 \\
\implies x<3\\
x\in \emptyset $
Case $2$ ,for $\boxed{x< 0}\\
\dfrac{x^2+x-12}{x-3}\geq 2x\\
\implies \dfrac{(x-4)(x-3)}{x-3}\leq 0 \\
\implies x\leq 4\\
\implies x< 0\\
$
But the answer given is option $d.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.
| The answer is (d), which is the union of the solutions for the two cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's what I did: \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2} \\ = \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{x^2 (4-6/x+7/x^2)} \\ \\ = \lim_{x \to \infty} \frac{\sqrt{x^2(1/x+1/x^2)}(1-\sqrt{x^2(2/x+3/x^2)}}{x(4-6/x+7/x^2)} \\ \\ \lim_{x \to \infty} \frac{\sqrt{1/x+1/x^2}(1-x \sqrt{2/x+3/x^2})}{4-6/x+7/x^2}
\end{align*} If I now evaluate this limit, everything in the numerator goes to zero except $1$. And the denominator leaves me with $4$. So I thought the answer should be $1/4$?
| You can divide the numerator and denominator by $x^2$ and distribute in an appropriate manner to obtain:
$$\begin{align*}\frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}& = \frac{\frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{x^2}}{\frac{7-6x+4x^2}{x^2}}\\\\& =\frac{\frac{x}{x}\cdot \frac{\sqrt{x+1}}{\sqrt{x}}\cdot\left(\frac{1-\sqrt{2x+3}}{\sqrt{x}}\right)}{\frac{7-6x+4x^2}{x^2}}\\\\ & = \frac{\sqrt{\frac{x+1}{x}}\cdot\left(\frac{1}{\sqrt{x}}-\sqrt{\frac{2x+3}{x}}\right)}{\frac{7-6x+4x^2}{x^2}} =\frac{\sqrt{1+\frac{1}{x}}\cdot\left(\frac{1}{\sqrt{x}}-\sqrt{2+\frac{3}{x}}\right)}{\frac{7}{x^2}+\frac{6}{x}+4}\end{align*}$$
This technique is usually performed when both, the numerator and denominator are polynomials, however, it can be still performed here and it becomes clearer if you think the numerator as a $1+\frac{1}{2}+\frac{1}{2}=2$ degree 'polynomial', in some sense.
At this point you can take the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solving following quartic equation Solve in $\mathbb{R}$ :
$$(x^2+2)^2+8x^2=6x (x^2+2) $$
My try: I tried to make the graph by calculating values for $x=1, 2, 3, 4$ and I found that the function is positive at $x=0$ but negative at $x=1$, so the graph must have crossed the $x$ axis, thus there will be a root between $0$ and $1$ and similarly this was the case between $3$ and $4$, but I was unable to solve it precisely. I was also unable to find about the other two roots.
What is the way to solve it just by algebra or rough plotting with the help of pen and paper and not using any computational tools?
| Expanding the equation:
$$ x^4 + 4x^2 + 4+ 8x^2 = 6x^3 + 12x$$
$$x^4 -6x^3 +12x^2-12x+4=0$$
You can then factorise the above expression
Alternatively, you can replace $x^2 + 2$ with $y$ which gives you:
$$ y^2 + 8x^2 = 6xy$$
$$ (y-4x)(y-2x) = 0$$
$$ y = 4x$$ or $$ y = 2x$$
Now replace $y$ with $x^2 + 2$
$x^2 + 2 = 4x $ or $x^2 + 2 = 2x$
$$(x^2 -4x +2)(x^2-2x+2) = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Epsilon-delta proof of $\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2}$ I have been doing $\varepsilon$-$\delta$ proofs for fun and I challenged myself to prove $$\displaystyle\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)=\frac{a+b}{2},\quad a,b\in\mathbb{R}$$
The definition says: We say that $\displaystyle\lim_{x\to\infty}f(x)=l$ if for any positive number $\varepsilon$ we can find a positive number $N$ (depending on $\varepsilon$ in general) such that $|f(x)-l|<\varepsilon$ whenever $x>N$.
So I started with: $\left|\sqrt{(x+a)(x+b)}-x-\dfrac{a+b}{2}\right|<\varepsilon$ whenever $x>N$.
Manipulating the first inequatlity
\begin{gather*}
-\varepsilon<\sqrt{(x+a)(x+b)}-x-\dfrac{a+b}{2}<\varepsilon\\
-\varepsilon+\dfrac{a+b}{2}<\sqrt{(x+a)(x+b)}-x<\varepsilon+\frac{a+b}{2}
\end{gather*}
At this point I thought about adding $x$, squaring the expressions and then expanding them. I did it and I got: $$\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}+ax+bx+x^2-a\varepsilon-b\varepsilon-2x\varepsilon+\varepsilon^2<x^2+ax+bx+ab<\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}+ax+bx+x^2+a\varepsilon+b\varepsilon+2x\varepsilon+\varepsilon^2$$
And here I'm not sure how to proceed. Am I on the right track?
Thanks for any help / hints.
Note: There might be other ways to prove this, but I'd like to do it using just algebra if possible, even if it's not the best method.
| The "rationalization" trick are the same, but this may be the rigorous $\varepsilon$-$\delta$ proof you want.
Write
$$\sqrt{(x + a)(x + b)} - x - \frac{a + b}{2} = \frac{(x + a)(x + b) - \left(x + \frac{a + b}{2}\right)^2}{\sqrt{(x + a)(x + b)} + \left(x + \frac{a + b}{2}\right)} = \frac{-\frac{(a - b)^2}{4}}{\sqrt{(x + a)(x + b)} + \left(x + \frac{a + b}{2}\right)}.$$
Therefore,
$$
\left|\sqrt{(x + a)(x + b)} - x - \frac{a + b}{2}\right| = \frac{C}{\sqrt{(x + a)(x + b)} + \left(x + \frac{a + b}{2}\right)} < \frac{C}{x + \frac{a + b}{2}} \tag{1}
$$
where $C = \dfrac{(a - b)^2}{4} \geq 0$, note the denominator is positive when $x$ is sufficiently large (say, $x > -(a + b)/2$) so the absolute value symbol can be removed. Given $\varepsilon > 0$, take $\delta = \max\left(\dfrac{C}{\varepsilon} - \dfrac{a + b}{2}, 1\right) > 0$, it then follows by $(1)$ that for all $x > \delta$,
$$\left|\sqrt{(x + a)(x + b)} - x - \frac{a + b}{2}\right| < \varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove using $ \varepsilon-\delta $ that $ \lim_{(x,y)\to(1,1)} \frac {x^2+2xy-3y^2}{x^2-y^2} = 2 $
Prove limit using $ \varepsilon-\delta $ definition that: $$ \lim_{(x,y)\to(1,1)} \frac
{x^2+2xy-3y^2}{x^2-y^2} = 2 $$
I've been reading quite a lot about how to prove limits; so I want to show what I've done so far in this one, so you can tell me any suggestion (tricks) and even point out any mistake.
What I've tried:
I want to find $ \delta $ for every $ \varepsilon $ that verifies: $$ 0 < \left \| (x,y) - (1,1) \right \| < \delta \Rightarrow \left | \frac {x^2+2xy-3y^2}{x^2-y^2} - 2 \right | < \varepsilon $$
So here is my attempt:
$$ \begin{align*}
\left | \frac {x^2+2xy-3y^2}{x^2-y^2} - 2 \right | &\overset{(1)}{=} \left | \frac {y-x}{y+x} \right | \\
&\overset{(2)}{\leq} \frac {|x|+|y|}{|x+y|} \\
&\overset{(3)}{\leq} \frac {2\left \| (x,y)-(1,1) \right \|}{|x+y|} \\
&\overset{(4)}{\leq} 4\left \| (x,y)-(1,1) \right \| \\
& < \ 4 \delta
\end{align*} $$
So I can take $ \delta = \varepsilon / 4 $. Is this right?
Justifications:
(1) Basic operations.
(2) Triangle inequality.
(3) I used $ |x| \leq \left \| (x,y)-(1,1) \right \| $.
(4) I supposed $ |x| < 1/2 $ and also $ |y| < 1/2 $ then $ |x+y| < 1/2 $. (I don't understand why this step holds though).
(5) The metric I'm using is: $ \left \| (x,y) \right \| = \sqrt{x^2 + y^2} $
| $$ \lim_{(x,y)\to(1,1)} \frac {x^2+2xy-3y^2}{x^2-y^2}$$
factorizing numerator and denominator we have,
$$ \lim_{(x,y)\to(1,1)} \frac {(x+3y)(x-y)}{(x+y)(x-y)}$$
$$\implies \lim_{(x,y)\to(1,1)} \frac {x+3y}{x+y} $$
$$=\frac{4}{2}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
If $3(4^h)=4(2^k)$ and $9(8^h)=20(4^k)$,show that $2^h = \frac{4}{5}$ If $3(4^h)=4(2^k)$ and $9(8^h)=20(4^k)$,show that $2^h = \frac{4}{5}$.
I tried to substitute the equation 1 into equation 2 so that I can find the value of $k$ or $h$, but it did not work as the base is not the same (I cannot compare to find the answer)
How can I find the answer?
| ${\left( {3({4^h}) = 4({2^k})} \right)^2 \rightarrow 9({16^h}) = 16({4^k})
\choose 9({8^h}) = 20({4^k})}$$\Rightarrow$ $\frac{{9({{16}^h})}}{{9({8^h})}} = \frac{{16({4^k})}}{{20({4^k})}} \Rightarrow {2^h} = \frac{4}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about the 0?
EDIT:
ok, following some of your comments/answers this is what I have done:
$$z^3 = - \overline z$$
$$\rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
So
$$
\begin{Bmatrix}
\rho^3 = \rho\\
3\theta = -\theta + 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho^3 = \rho\\
2\theta = 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho = 0 or \rho = 1\\
\theta = k\frac{\pi}{2}
\end{Bmatrix}$$
is this the right way?
| $$z^3+\overline z=0$$
But $z=a+bi$, $\overline z = a-bi$ so:
$$(a+bi)^3+a-bi=a^3-3ab^2+3a^2bi-b^3i+a-bi=i(3a^2b-b^3-b)+(a^3-3ab^2+a)$$
We know that $a+bi=0 \iff a=0 \land b=0$:
$$\begin{cases}
3a^2b-b^3-b=0\\
a^3-3ab^2+a=0
\end{cases}$$
$$\begin{cases}
3a^2-b^2-1=0\\
a^2-3b^2+1=0
\end{cases}$$
$$a^2=3b^2-1 \implies 3a^2-b^2-1=9b^2-3-b^2-1=8b^2-4=0 \implies b=\pm\frac{\sqrt2}{2}$$
$$a^2=3b^2-1=1.5-1=0.5\implies a=b=\pm\frac{\sqrt2}{2}$$
So finally:
$$z=\frac{\sqrt2}{2}\pm\frac{\sqrt2}{2}i$$
or
$$z=-\frac{\sqrt2}{2}\pm\frac{\sqrt2}{2}i$$
To be clear there may be other solutions. I have divided my equations by $a$ and $b$, but each may be equal to $0$.
*
*$a=0$ then $b^3+b=0$
*
*$b=0$ OK
*$b\not=0$ then $b^2+1=0$ and $b\not\in \mathbb{R}$ (but it's a contradiction)
*$b=0$ then $a^3+a=0$, same as above, only $(a;b)=(0;0)$ is valid.
The other solution is then $$z=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 3
} |
Integrating an Iterated Integral Given the iterated integral:
$$\int^{\sqrt{2}}_{-\sqrt{2}}\int^{\sqrt{2-x^2}}_{-\sqrt{2-x^2}}\int^{\sqrt{4-x^2-y^2}}_{\sqrt{x^2+y^2}}{(x^2+y^2+z^2)^{3/2}}dzdydx$$
Now, my question is, what are the two quadric surfaces that bound the region from above and below, and what are their equations? Also, how do I integrate this?
| Ok, let's work out your quadrics. Given the order you must be integrating in (can you see why?), you have three constraints which characterise your volume: $$\sqrt{x^2+y^2}\le z\le\sqrt{4-x^2-y^2} \\-\sqrt{2-x^2}\le y \le \sqrt{2-x^2}\\-\sqrt{2}\le x \le \sqrt{2}$$
So the LHS of our first equation gives $z^{2}-x^{2}-y^{2}\ge 0$ whith $z \ge 0$, and the RHS gives $x^{2}+y^{2}+z^{2}\le 4$. Do you recognise these quadrics? You can get further constraints from the second equation. Now for the integral.
Let's change to spherical polar co-ordinates $(r,\theta,\phi)$ before we get upset. Recall that: $$x=r\sin \theta \cos \phi, \quad y=r\sin \theta \sin \phi, \quad z=r\cos \theta$$ $$r^{2}=x^{2}+y^{2}+z^{2}, \quad dxdydz=r^{2}\sin\theta drd\theta d\phi $$
Now we just need to work out our region of integration. It's not hard to see that $x^{2}+y^{2}=r^{2}\sin^{2}\theta$, so we get the two constraints $r\le 2$ and $r\cos\theta\ge r\sin \theta$, i.e. $0 \le \theta \le \pi/4$. We also have $x^{2}+y^{2}\le 2$ (which is automatically true already) and $-\sqrt{2}\le r \cos\theta \sin \phi \le \sqrt{2}$ (which allows $\phi$ to take any value). Hence, the integral is $$\int_{0}^{\pi/4}\int_{0}^{2\pi}\int_{0}^{2}r^{5}\sin\theta drd\phi d\theta=\frac{2^{6}}{6}2\pi\frac{\sqrt{2}}{2}=\frac{32(2-\sqrt{2})}{3}\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| The sequence $y=(1,2,3,4,\ldots)$ is an output of the linear system
$$
y_{k+1}=y_k+u_k,\qquad y_0=1
$$
for the input $u=(1,1,1,1,\ldots)$. Perform the $Z$-transform (multiply by $z^k$ and add up for all $k$)
$$
\sum_{k=0}^\infty y_{k+1}z^k=\underbrace{\sum_{k=0}^\infty y_kz^k}_{y(z)}+\underbrace{\sum_{k=0}^\infty u_kz^k}_{u(z)}\quad\Rightarrow\quad
\frac{1}{z}(y(z)-y_0)=y(z)+u(z)\quad\Rightarrow
$$
$$
\Rightarrow\quad y(z)=\frac{1+z u(z)}{1-z}=\frac{1+z\frac{1}{1-z}}{1-z}=\frac{1}{(1-z)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 10
} |
$\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
*given $$ \frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d} = 65, \frac {1} {a^2} + \frac {1} {b^2} + \frac {1} {c^2} + \frac {1} {d^2} = 209$$
find $$\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$$
(A) $2006$ (B) $2007$ (C) $2008$ (D) $2009$ (E) $2010$
I've use the direct way (make them become 1 fraction)
$$\frac {abc+abd+acd+bcd} {abcd} = 65$$
$$\frac {a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2} {a^2b^2c^2d^2} = 209$$
$$\frac {a^2b^2c^2d^2(ab+ac+ad+bc+bd+cd)} {a^3b^3c^3d^3}$$
$$\frac {(ab+ac+ad+bc+bd+cd)} {abcd}$$
but it doesn't make sense with these power $abcd$ thing
| Hint: Try expanding $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Trigonometry question If $$\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}.$$ If $\alpha$ is a natural number.Find $\alpha$.
My attempt is:
$$\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}$$
convert it into sin,cos
$$\frac{3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}{\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}$$
$$\frac{3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}{\cos\frac{2\pi}{7}}=\alpha \cos\frac{\pi}{7}$$
$$3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$
$$2\cos^2\frac{\pi}{7}+\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$
$$2\cos^2\frac{\pi}{7}+\cos\frac{2\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$
but i got stuck and could not further solve it....
I would appreciate the help,thanks in advance.
| Use $\tan^2A=\cdots=\dfrac{1-\cos^2A}{\cos^2A}$ and replace $\dfrac\pi7$ with $B$
to get $$2\alpha\cos^3B-4\cos^2B-\alpha\cos B+1=0\ \ \ \ (1)$$
Now if $7C=(2n+1)\pi$ where $n$ is any integer
$C=\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2,3\pmod7$
and $\cos4C=\cdots=-\cos3C\implies8\cos^4C-8\cos^2C+1=-(4\cos^3C-3\cos C)$
$\iff8\cos^4C+4\cos^3C-8\cos^2C-3\cos C+1=0\ \ \ \ (2)$
whose roots will be $\cos\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2,3\pmod7$
Now $n\equiv3\implies\cos\dfrac{(2n+1)\pi}7=\cdots=-1$
So, the roots of $0=\dfrac{8\cos^4C+4\cos^3C-8\cos^2C-3\cos C+1}{\cos C+1}=8\cos^3C-4\cos^2C-4\cos A+1\ \ \ \ (3)$
will be $\cos\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2\pmod7$
Now compare $(1),(3)$ to get $\dfrac{2\alpha}8=\dfrac44=\dfrac{\alpha}4=\dfrac11$
Observation: The given problem holds true for $\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2\pmod7$
$(3)$ can be derived like factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Nonnegative Integer solutions of $x+y-xy=0$ I would like to see other methods, besides the one I use here to find all the nonnegative integer solutions of an equation like $$x+y-xy=0$$.
This is the one I used:
First we note that for $x=1$ we get $1+y-y=0$, but $1+y-y=1$ so $x\neq 1$.
Then as $x\neq 1$ we have $y=\frac{x}{x-1}=1+\frac{1}{x-1}$. Furthermore $y(0)=0$ and $y(2)=2$ so $x=y=0$ and $x=y=2$ are solutions. Then if $x>2$, $x-1>1>0$ so as $x-1>1$ we have $\frac{1}{x-1}<1$ and as $x-1>0$ we have $\frac{1}{x-1}>0$ so $0<\frac{1}{x-1}<1$ for $x>2$, and so $1<1+\frac{1}{x-1}<2$ for $x>2$. As $y=1+\frac{1}{x-1}$, we get that for $x>2$, $1<y<2$. As there is no integer between $1$ and $2$, for $x>2$, $y$ can't be a integer. So the only nonnegative solutions of the equations are $x=y=0$ and $x=y=2$.
How else can I prove this, using less theory (For example, I think it can be proven using only Peano and a little more)?
| $$ xy - x - y = 0 $$
$$ xy - x - y + 1 = 1 $$
$$ (x-1)(y-1) = 1. $$
With integer variables, the choices are
$$ x-1 = 1, y-1 = 1, $$ or
$$ x-1 = -1, y-1 = -1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve $\sin A +\sin 2A +\sin 3A + \sin 4A = 0$, for $0 \leq A \leq 180$ I've tried using factor formula but still did not manage to get the answer, not sure if factor formula is the right method.
I rearrange to $\sin 4A + \sin 2A + \sin 3A + \sin A = 0$,
and after applying factor formula,
$2 \sin 3A \cos A + 2 \sin 2A \cos A = 0$
$2 \cos A ( \sin 3A + \sin 2A) = 0$
$2 \cos A ( \sin \frac{5}{2} A \cos \frac{1}{2} A) = 0$
Then I'm stuck..
| we have, $$\sin A+\sin 2A+\sin 3A+\sin 4A=0$$ $$(\sin A+\sin 3A)+(\sin 2A+\sin 4A)=0$$ $$2\sin\left(\frac{A+3A}{2}\right)\cos\left(\frac{A-3A}{2}\right)+2\sin\left(\frac{2A+4A}{2}\right)\cos \left(\frac{A-3A}{2}\right)=0$$ $$2\sin 2A\cos A+2\sin 3A\cos A=0$$ $$2\cos A(\sin 2A+\sin 3A)=0$$ $$\implies \cos A=0\implies \color{blue}{A=2n\pi+\frac{\pi}{2}}$$ $$\implies \sin 2A+\sin 3A=0\implies \sin 3A=-\sin 2A$$ $$\implies 3A=2n\pi-2A\implies \color{blue}{A=\frac{2n\pi}{5}}$$ Or $$3A=(2n+1)\pi-(-2A)$$ $$\implies \color{blue}{A=(2n+1)\pi}$$ Where, $n$ is any integer.
Edit: For $0\leq A\leq 180^\circ$ put $n=0$, $n=1$ & $n=2$ in the solutions, we get the following $$\color{blue}{A\in \left\{0, \frac{2\pi}{5}, \frac{\pi}{2}, \frac{4\pi}{5}, \pi\right\}}$$ or $$\color{blue}{A\in \left\{0^\circ, 72^\circ, 90^\circ, 144^\circ, 180^\circ\right\}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Square root equation I have the equation $\sqrt{(7-x)} - \sqrt {(x+13)} = 2 $ The square root should be expanded so it is square root of $7-x$ - square root of $x+13 = 2$.
When i square both sides i get:
$7-x - x-13 = 4 $
then i clean on the LHS so i get
$-2x-6 = 4$ which leads to $(-2x-6)^2 = 4^2$ and after working that out i get:
$4x^2 + 24 x + 36 = 16$.
Next step will then be $4x^2 +24x +20 = 0$
Using the pq formula i get $x^2 = -6/2$ +- $\sqrt {6 \over 2}^2 -5$
And after that i get $x^2 = -6/2 +- 2$ which leads to $x^2 = -1$ or $x^2= -5$.
The answer according to the textbook is $-9$. Do you know how to solve this ?
| You probably mean
$$
\sqrt{7-x} - \sqrt{x+13} = 2.
$$
The steps:
$$
\big( \sqrt{7-x} - \sqrt{x+13} \big)^2 = 2^2\\
\Downarrow\\
\sqrt{7-x}^2 + \sqrt{x+13}^2 - 2 \sqrt{7-x} \sqrt{x+13} = 4\\
\Downarrow\\
\big(7-x\big) + \big(x+13\big) - 2 \sqrt{7-x} \sqrt{x+13} = 4\\
\Downarrow\\
20 - 2 \sqrt{7-x} \sqrt{x+13} = 4\\
\Downarrow\\
\sqrt{7-x} \sqrt{x+13} = 8\\
\Downarrow\\
\big(7-x\big)\big(x+13\big) = 64\\
\Downarrow\\
\big(10-3-x\big)\big(10+3+x\big) = 64\\
\Downarrow\\
100-\big(3+x\big)^2 = 64\\
\Downarrow\\
\big(3+x\big)^2 = 36\\
\Downarrow\\
3+x = \pm 6\\
\Downarrow\\
\bbox[16px,border:2px solid #800000] {x = -9 \vee x = 3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\int\frac{8x+9}{(2x+1)^3}\,dx$. Do I split $\displaystyle\int\frac{8x+9}{(2x+1)^3}\,dx$ into partial fractions? Or do I use $(2x+1)^{-3}$ by itself? Not sure what to do. Please advice.
The answer given is $\dfrac{-16x+13}{4(2x+1)^2} +C$
| write the numerator of the integrand as $$8x+4+5=4(2x+1)+5$$ then we have
$$4\int\frac{dx}{(2x+1)^2}dx+5\int\frac{dx}{(2x+1)^3}dx$$ now you can set $$u=2x+1$$ and proceed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Roots of cubic equation If$\frac{1+\alpha}{1-\alpha},\frac{1+\beta}{1-\beta},\frac{1+\gamma}{1-\gamma}$ are the roots of the cubic equation $f(x)=0$ where $\alpha,\beta,\gamma$ are the real roots of the cubic equation $3x^3-2x+5=0$,then find the number of negative real roots of the equation $f(x)=0$.
My attempt:I tried finding out solution of $3x^3-2x+5=0$ to get $\alpha,\beta,\gamma$ by rational root method and hit and trial method but could not get them.Is my approach correct? Or Descretes rule is to be applied?
Can someone help me solve this question?
| Note that $\frac{1+\alpha}{1-\alpha}<0$ is equivalent to $1+\alpha$ and $1-\alpha$ having opposite signs, which happens when $\alpha>1$ or $\alpha <-1$. So all we have to do is count the number of roots of $3x^2-2x+5$ in $[-1,1]$.
Let $p(x) = 3x^3-2x+5$. Now $p(-1)=4$ and $p(1)=6$. Moreover, $p'(x)=9x^2-2$, which has roots at $x=\pm\sqrt{2}/3$. Also, $p''(x)=18x$, which has the sign the same as $x$, so $x=-\sqrt{2}/3$ is a local maximum of $p$ while $x=\sqrt{2}/3$ is a local minimum of $p$. In particular, the minimum of $p$ on $[-1,1]$ is the minimum of $p(-1)=4$, $p(1)=6$, and $p(\sqrt{2}/3)$. Now $p(\sqrt{2}/3)=3(\sqrt{2}/3)^3-2\sqrt{2}/3+5=-(4/9)\sqrt{2}+5>0$. Therefore, $p$ has no roots on $[-1,1]$, so the answer to your question is simply the number of real roots of $p$, which is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sine/cosine series $$\frac{\sin²(1°) + \sin²(2°) + \sin²(3°) + .. + \sin²(90°)}{\cos²(1°) + \cos²(2°) + \cos²(3°) + .. + \cos²(90°)} = ?$$
I tried to use multiple identities but I couldn't simplify the expression. Where should I start?
| HINT:
$$\frac{\sum_{n=a}^{b} \sin^2(n)}{\sum_{n=a}^{b} \cos^2(n)}=$$
$$\frac{2-2a+2b-\csc(1)\sin(1-2a)-\csc(1)\sin(1+2b)}{2-2a+2b+\csc(1)\sin(1-2a)+\csc(1)\sin(1+2b)}$$
Now we know that $a=1$ and $b=90$:
$$\frac{\sum_{n=1}^{90} \sin^2(n)}{\sum_{n=1}^{90} \cos^2(n)}=$$
$$\frac{2-2\cdot 1+2\cdot 90-\csc(1)\sin(1-2\cdot 1)-\csc(1)\sin(1+2\cdot 90)}{2-2\cdot 1+2\cdot 90+\csc(1)\sin(1-2\cdot 1)+\csc(1)\sin(1+2\cdot 90)}=$$
$$\frac{180-\csc(1)\sin(-1)-\csc(1)\sin(181)}{180+\csc(1)\sin(-1)+\csc(1)\sin(181)}=$$
$$\frac{182}{178}=\frac{91}{89}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Evaluating the indefinite integral $\int\sqrt{16-9x^2}\,dx$ I need to solve the integral below, but I just can't figure how.
$$\int \sqrt{16-9x^2}\,dx$$
I have tried to replace $9x^2$ with $16\sin^2\theta$. I get to a point where I have the function below. Please let me know whether I'm on the right track, and please explain to me how to finish it...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$
| Notice, this standard formula $$\int\sqrt{a^2-x^2}dx=\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\left(\frac{x}{a}\right)\right)$$ Hence, we have
$$\int\sqrt{16-9x^2}dx=\int\sqrt{(4)^2-(3x)^2}dx$$ let $3x=u\implies 3dx=du$$$=\frac{1}{3}\int\sqrt{(4)^2-(u)^2}du $$ $$=\frac{1}{3}\cdot\frac{1}{2}\left(u\sqrt{(4)^2-u^2}+(4)^2\sin^{-1}\left(\frac{u}{4}\right)\right)$$ $$=\frac{1}{6}\left(3x\sqrt{(4)^2-(3x)^2}+16\sin^{-1}\left(\frac{3x}{4}\right)\right)$$ $$=\color{blue}{\frac{1}{2}x\sqrt{16-9x^2}+\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 6
} |
Explain 'expressing a number using its digits' While studying divisibilty and prime numbers in my maths book (IB Mathematic Higher Level Option 10: Discrete Mathematics), I came across an explanation of a way to '[express] a number using its digits'.
It says:
If $N$ is a $k$-digit number with digits $a_k, a_{k-1}...a_2, a_1, a_0$, then
$$N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10^2 a_2 + 10 a_1 + a_0.$$
Shorthand $N = (a_k a_{k-1} \cdots a_2 a_1 a_0)$.
So, I decided to write out an example: $2^{16}$.
This is where is ran into some trouble. $65536$ is a five-digit number, but I can't write $10^5 \cdot 6$ as this would give me $600000$. Can someone please help me out and explain?
| As Aretino comments, your book should say
If $N$ is a $(k+1)$-digit number with digits $a_k, a_{k-1}...a_2, a_1, a_0$, then
$$N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10^2 a_2 + 10 a_1 + a_0.$$
You can think of this as a way of "exploding" the number by expanding its expression, writing it as a sum of numbers each of which has only one nonzero digit; this makes more sense after seeing a few examples:
$$25= 20+5=10^1\cdot2+10^0\cdot 5 \\
343 = 300+40+3=10^2\cdot 3 + 10^1\cdot 4+ 10^0\cdot 3 \\
2048 = 2000+000+40+8 = 10^3\cdot 2 + 10^2\cdot 0 + 10^1\cdot 4+10^0\cdot 8 $$
Now, for your example, $$65535 = 60000+5000+500+30+5 = 10^4\cdot 6+ 10^3\cdot 5+ 10^2\cdot 5+ 10^1\cdot 3+10^0\cdot 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $(1+p/n)^n$ is a Cauchy sequence for arbitrary $p$ It is a generalization of this question. I am looking for a similar derivation as in here.
Can we prove that $(1+p/n)^n$ is a Cauchy sequence for any $p \in [a, b]$ by showing that
$$ \Bigg| \left( 1 + \frac{p}{n}\right)^n - \left( 1 + \frac{p}{m}\right)^m \Bigg| \leq f(n)$$
where $f(n)$ is something that tends to zero as $n$ goes to infinity?
Here is an attempt.
Let $m=n+1$. Then,
$$ \Bigg| \left( 1 + \frac{p}{n}\right)^n - \left( 1 + \frac{p}{n+1}\right)^{n+1} \Bigg| = \Bigg| \sum_{k=0}^{n} \binom{n}{k}\left( \frac{p}{n} \right)^{k} - \sum_{k=0}^{n+1} \binom{n+1}{k}\left( \frac{p}{n+1} \right)^{k} \Bigg| = \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \left[ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \right] + \frac{p^{n+1}}{(n+1)^{n+1}} \Bigg| \leq
\sum_{k=0}^{n} \frac{|p|^k}{k!} \left[ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \right] + \Bigg| \frac{p^{n+1}}{(n+1)^{n+1}} \Bigg|
$$
Now intuitively, $\sum_{k=0}^{n} \frac{|p|^k}{k!}$ tends to the constant $e^{|p|}$ while the expression in square brackets tends to zero as $n$ goes to infinity. I was trying to show that this expression is less than a constant divided by $n^2$ (because we will sum up the consecutive terms for $m >n$, and the sum should converge which $\frac{C}{n^2}$ would provide).
Can we also find an estimate of the constant $C$?
Thanks to all for good answers. Meanwhile, I'm still quite interested in working it out algebraically. My next idea was to use the triangle inequality. Let us ignore the last term in the inequality above, it obviously tends to zero as $n$ goes to infinity. We want the sum to go to zero fast enough. So,
$$ \Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \Bigg| \leq \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \Bigg| + \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \Bigg|
$$
Each product is evidently positive and smaller than one. So we end up having kind of a usual exponential series, but weighted. Otherwise, we could try to plug something instead of the products, what would, when subtracted from the products, give something which decreases fast enough. I feel this should be possible since we have all the freedom to plug in whatever we want.
| This shows the estimate
needed in Winther's answer.
It is too complicated to be a comment,
though that is what it really it.
Define
$\ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
$.
Since,
for
$1 \le t \le 1+x$
$\frac{1}{1+x} \le \frac{1}{t} \le 1$,
$\frac{x}{1+x}
\le \ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
\le x
$,
so that
$\frac{x}{1+x}-x
\le \ln(1+x)-x
\le 0
$
or,
reversing,
$0
\le x-\ln(1+x)
\le \frac{x^2}{1+x}
$ .
Therefore
$0
\le \frac{a}{x}-\ln(1+a/x)
\le \frac{(a/x)^2}{1+a/x}
=\frac{a^2}{x(x+a)}
$.
Since
$\frac{a}{x}-\frac{a}{x+a}
=\frac{a^2}{x(x+a)}
$,
$0
\le \frac{a}{x+a}+\frac{a^2}{x(x+a)}-\ln(1+a/x)
\le \frac{a^2}{x(x+a)}
$
or
$0
\le \ln(1+a/x)-\frac{a}{x+a}
\le \frac{a^2}{x(x+a)}
\le \frac{a^2}{x^2}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
The value of the integral $\int_0^2\left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\:\right)dx$ The value of definite integral $$\int\limits_{0}^{2}\left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\:\right)dx$$ is $$(A)\,4 \quad(B)\,5 \quad (C)\,6 \quad(D)\,7$$
My attempt:
I tried using $\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx$ but not working. I tried putting $x^3+1=\tan^2\theta$, its also not working.
Can someone help me solve this problem?
| I have one more way to do this:
Let $$I_1=\int_{0}^{2} \sqrt{x^3+1} \:dx$$ By Integration by parts we get
$$I_1=x\sqrt{x^3+1} \bigg|_{0}^{2}-\int_{0}^{2}\frac{3x^3 dx}{2\sqrt{x^3+1}} \tag{1}$$
Now let $$I_2=\int_{0}^{2} (x^2+2x)^{\frac{1}{3}}dx$$ Use substitution $x^2+2x=t^3$ we get
$(2x+2)dx=3t^2dt$ that is
$dx=\frac{3t^2dt}{2(x+1)}$
But $(x+1)^2-1=t^3$ $\implies$
$x+1=\sqrt{t^3+1}$
Also the limits will be again $0$ and $2$
Thus
$$I_2=\int_{0}^{2}\frac{3t^3dt}{2(x+1)}=\int_{0}^{2}\frac{3t^3dt}{2\sqrt{t^3+1}}$$ since $t$ is Dummy variable
$$I_2=\int_{0}^{2}\frac{3x^3dx}{2\sqrt{x^3+1}} \tag{2}$$
Adding $(1)$ and $(2)$ we get
$$I_1+I_2=x\sqrt{x^3+1} \bigg|_{0}^{2}=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
How can I solve this recurrence relation: $a_n = 3a_{n-1} + \frac{4^n}{4}$? How can I solve the following recurrence relation?
$$a_n = 3a_{n-1} + \frac{4^n}{4}$$
I know that $a_n^{(h)} = 3a_{n-1}$ and that the characteristic equation is:
$$r-3 = 0$$ and thus:
$$a_n^{(h)} = \alpha_1(3)^n$$
I am struggling with the particular solution $a_n^{(p)}$.
| Consider the generating function for $a_{n}$. The process is the following.
\begin{align}
\sum_{n=0}^{\infty} a_{n+1} \, t^{n} &= 3 \, \sum_{n=0}^{\infty} a_{n} \, t^{n} + \sum_{n=0}^{\infty} (4t)^{n} \\
\frac{1}{t} \, \sum_{n=1}^{\infty} a_{n} \, t^{n} &= 3 \, A(t) + \frac{1}{1-4t} \\
\frac{1}{t} \left( A(t) - a_{0} \right) &= 3 \, A(t) + \frac{1}{1-4t}
\end{align}
where $A(t) = \sum_{n=0}^{\infty} a_{n} \, t^{n}$, which leads to
\begin{align}
A(t) &= \frac{a_{0}}{1-3t} + \frac{t}{(1-3t)(1-4t)} \\
&= \frac{a_{0}}{1-3t} + \left( \frac{1}{1-4t} - \frac{1}{1-3t}\right) \\
&= \frac{a_{0} - 1}{1-3t} + \frac{1}{1-4t} \\
&= \sum_{n=0}^{\infty} \left((a_{0} - 1) \, 3^{n} + 4^{n}\right) \, t^{n}
\end{align}
and finally
\begin{align}
a_{n} = 3^{n} \, (a_{0} - 1) + 4^{n}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Quadratic equation with several variables How does
$$y^{2} - 4y -t^{2} - C = 0$$
Become
$$y = 2 \pm \sqrt{t^{2} +2C + 4}$$
I know its the quadratic formula but I dont know how it got it that point
The original equation is $$\frac{dy}{dt} = \frac{t}{y-2}$$.
| The equation
$$\frac{dy}{dt} = \frac{t}{y-2}$$
can be seen in the form
\begin{align}
(y-2) \, dy = t \, dt
\end{align}
and upon integration of both sides becomes
\begin{align}
\frac{y^{2}}{2} - 2 y = \frac{t^{2}}{2} + \frac{c_{0}}{2}
\end{align}
or $y^{2} - 4 y -(t^{2} + c_{0}) = 0$. Solving this quadratic equation the value of $y$ is of the form
\begin{align}
y(t) = 2 \pm \sqrt{t^{2} + 4 + c_{0}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral problem Find $$ \int e^{x \sin x+\cos x} \left(\frac{x^4\cos^3 x-x \sin x+\cos x}{x^2\cos^2 x}\right) \, dx$$
My attempt:I tried putting $x \sin x+\cos x=t$ and cannot express it in the form of $\int e^t(f(t)+f'(t)) \, dt$
| The integral is of the form
\begin{equation*}
\int e^{x\sin x+\cos x}\left( \frac{x^{4}\cos ^{3}x-x\sin x+\cos x}{
x^{2}\cos ^{2}x}\right) dx=\int e^{g(x)}h(x)dx.
\end{equation*}
This form recalls the well-known formula
\begin{equation*}
\int e^{g(x)}\left( f^{\prime }(x)+g^{\prime }(x)f(x)\right)
dx=f(x)e^{g(x)}+C.
\end{equation*}
Its proof maybe found at
Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$
So we are done if we find a function $f(x)$ such that
\begin{equation*}
h(x)=\frac{x^{4}\cos ^{3}x-x\sin x+\cos x}{x^{2}\cos ^{2}x}=f^{\prime
}(x)+g^{\prime }(x)f(x)=f^{\prime }(x)+(x\cos x)f(x)
\end{equation*}
In what follows, I will show that $f(x)=x-\frac{1}{x\cos x}$ and therefore
\begin{eqnarray*}
\int e^{x\sin x+\cos x}\left( \frac{x^{4}\cos ^{3}x-x\sin x+\cos x}{
x^{2}\cos ^{2}x}\right) dx &=&\left( f(x)\right) e^{g(x)}+C \\
&=&\left( x-\frac{1}{x\cos x}\right) e^{x\sin x+\cos x}+C.
\end{eqnarray*}
Remark. If $f(x)=f_{1}(x)+f_{2}(x)$ then
\begin{eqnarray*}
f^{\prime }(x)+g^{\prime }(x)f(x) &=&(f_{1}+f_{2})^{\prime }+g^{\prime
}(x)(f_{1}+f_{2}) \\
&=&\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +\left(
f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right)
\end{eqnarray*}
conversely, if
\begin{equation*}
h(x)=\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +\left(
f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right)
\end{equation*}
then
\begin{equation*}
h(x)=f^{\prime }(x)+g^{\prime }(x)f(x),\ \ with\ \ f(x)=f_{1}(x)+f_{2}(x).
\end{equation*}
This remark suggests to look for $f$ by pieces! that is, if we can write
\begin{equation*}
h(x)=\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +something
\end{equation*}
we reduce our task to look for $f_{2}$ such that
\begin{equation*}
something=\left( f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right)
\end{equation*}
and therefore $h(x)$ would be $(f_{1}+f_{2})^{\prime }+g^{\prime
}(x)(f_{1}+f_{2}),$ that is we can take $f=f_{1}+f_{2}.$
The procedure can be described as follows. First look for $g^{\prime }(x)$
inside $h(x).$ If some expression like $g^{\prime }(x)f_{1}(x)$ is found
\begin{equation*}
h(x)=g^{\prime }(x)f_{1}(x)+something
\end{equation*}
then add and subtract $f_{1}^{\prime }(x)$ and write
\begin{equation*}
h(x)=\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right)
+something-f_{1}^{\prime }(x).
\end{equation*}
Next take $h_{1}(x)=something-f_{1}^{\prime }(x)$ and try to find $g^{\prime
}(x)f_{2}(x)$ inside $h_{1}(x)$. If found, add and subtract $f_{2}^{\prime
}(x).$ And so on. Since the integral to be evaluated is a reasonable
integral which come from an exercise textbook, then at some moment this
procedure should stop, for example when $f_{n}(x)$ is obtained, and
therefore
\begin{equation*}
h(x)=\sum_{k=1}^{n}\left( f_{k}^{\prime }(x)+g^{\prime }(x)f_{k}(x)\right)
\end{equation*}
and
\begin{equation*}
f(x)=\sum_{k=1}^{n}f_{k}(x).
\end{equation*}
According to my own experience, $n=1$ or $2.$ I never get $n=3.$
Let's go.!
The unique thing which is given in the statement is $g(x)=x\sin x+\cos x.$
So, the first thing we start with is to look for $g^{\prime }(x)=x\cos x$
inside what would be $h(x)=f^{\prime }(x)+g^{\prime }(x)f(x),$ that is,
inside
\begin{equation*}
\frac{x^{4}\cos ^{3}x-x\sin x+\cos x}{x^{2}\cos ^{2}x}.
\end{equation*}
If we separate the fraction into 3 fractions we get
\begin{equation*}
x^{2}\cos x-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}.
\end{equation*}
It is easy to see $g^{\prime }(x)=x\cos x$ in the first term
\begin{equation*}
xg^{\prime }(x)-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}.
\end{equation*}
This suggests to take $f_{1}(x)=x.$ So, we have to add and subtract $
f_{1}^{\prime }(x)$ to get the package $\left( f_{1}^{\prime }(x)+g^{\prime
}(x)f_{1}(x)\right) ,$
\begin{equation*}
\left( x^{\prime }+g^{\prime }(x)x\right) -\frac{\sin x}{x\cos ^{2}x}+\frac{1
}{x^{2}\cos x}-1.
\end{equation*}
Now we take
\begin{equation*}
h_{2}(x)=-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}-1
\end{equation*}
and try to find inside it $g^{\prime }(x)=x\cos x.$ This one is not present
in the first two fractions, but can be in the third as follows
\begin{eqnarray*}
h_{2}(x) &=&-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}+(x\cos
x)\left( \frac{-1}{x\cos x}\right) \\
&=&-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}+g^{\prime }(x)\left(
\frac{-1}{x\cos x}\right) .
\end{eqnarray*}
So we take $f_{2}(x)=\left( \frac{-1}{x\cos x}\right) ,$ and we have to add
and subtract $f_{2}^{\prime }(x)$
\begin{eqnarray*}
h_{2}(x) &=&\left( \left( \frac{-1}{x\cos x}\right) ^{\prime }+g^{\prime
}(x)\left( \frac{-1}{x\cos x}\right) \right) -\frac{\sin x}{x\cos ^{2}x}+
\frac{1}{x^{2}\cos x}-\left( \frac{-1}{x\cos x}\right) ^{\prime } \\
&=&\left( f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right) -\frac{\sin x}{
x\cos ^{2}x}+\frac{1}{x^{2}\cos x}-\left( \frac{-1}{x\cos x}\right) ^{\prime
}
\end{eqnarray*}
However, easy computation shows that
\begin{equation*}
-\frac{\sin x}{x\cos ^{2}x}+\frac{1}{x^{2}\cos x}-\left( \frac{-1}{x\cos x}
\right) ^{\prime }=0.
\end{equation*}
(This happens because the exercise is from a textbook) Therefore
\begin{equation*}
h(x)=f_{1}(x)+f_{2}(x)=x-\frac{1}{x\cos x}
\end{equation*}
and
\begin{equation*}
\int e^{x\sin x+\cos x}\left( \frac{x^{4}\cos ^{3}x-x\sin x+\cos x}{
x^{2}\cos ^{2}x}\right) dx=\left( x-\frac{1}{x\cos x}\right) e^{x\sin x+\cos
x}+C.
\ \ \
\color{red}
\blacksquare
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Why is $\frac{1}{4/3} - \frac{1}{3/2}$ not the same as $\bigl(\frac{4}{3} - \frac{3}{2}\bigr)^{-1}$ If you have the problem:$$\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}} =?$$
Why can't you change the problem to $(\frac{4}{3} - \frac{3}{2})^{-1}$ and get the same answer?
In the first scenario, the answer is $1/12$
In the second scenario, the answer is $-6$
Why doesn't this work?
| You are basically asking why $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}$. Take $ a = b = 2$, for example. You would expect that adding a half plus a half would get you back to one, i.e. $\frac{1}{2}+\frac{1}{2} = 1$, and certainly not a quarter, $\frac{1}{2+2} = \frac{1}{4}$.
Let's say you have some function $f(x)$ that takes some number $x$ and spits out a different number $y$. Let's take $f(x) = 3\cdot x$. If we put in a number that is a sum of two other numbers, for example, $x = a+b$, then we have that $f(x) = f(a+b) = f(a) + f(b)$. Same if $x$ is some multiple of a different number, $x = 4\cdot a$, namely $f(x) = f(4\cdot a) = 4f(a)$. These properties only hold because we choose our $f(x)$ to be very specific, i.e. multiplication by $3$. In general they do not have to hold however! For example if $f(x) = x^2$, or $f(x) = \frac{1}{x}$.
You are probably too used to functions behaving in the way $f(x) = 3\cdot x$ did, but if you look carefully almost everything that you can come up with will not have these nice properties.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Elegant solution for $\int {\frac{\cos(y)}{\sin^2(y)+\sin(y)-6}}dy$ I have the following integral: $\int {\frac{\cos(y)}{\sin^2(y)+\sin(y)-6}}dy$
I already know the solution, but it needs three substitutions. Is there a simpler, more elegant way to go about this?
| $$\int { \frac { \cos { \left( x \right) } dx }{ \sin ^{ 2 }{ \left( x \right) } +\sin { \left( x \right) } -6 } } =\int { \frac { d\sin { \left( x \right) } }{ \left( \sin { \left( x \right) +3 } \right) \left( \sin { \left( x \right) -2 }
\right) } } $$
Let $t=\sin { \left( x \right) } $ then
$$\frac { 1 }{ \left( t+3 \right) \left( t-2 \right) } =\frac { A }{ t+3 } +\frac { B }{ t-2 } $$
$$1=A\left( t-2 \right) +B\left( t+3 \right) \\ 1=\left( A+B \right) t-2A+3B\\ \begin{cases} A+B=0 \\ -2A+3B=1 \end{cases}\Rightarrow \begin{cases} A=-B \\ 2B+3B=1 \end{cases}\Rightarrow B=\frac { 1 }{ 5 } ,A=-\frac { 1 }{ 5 } $$ so
$\frac { 1 }{ \left( t+3 \right) \left( t-2 \right) } =\frac { 1 }{ 5 } \left( \frac { 1 }{ t-2 } -\frac { 1 }{ t+3 } \right) $
$$\\ \\ \\ \frac { 1 }{ 5 } \int { \left( \frac { 1 }{ t-2 } -\frac { 1 }{ t+3 } \right) dt=\frac { 1 }{ 5 } \int { \frac { dt }{ t-2 } } -\frac { 1 }{ 5 } \int { \frac { dt }{ t+3 } = } } $$ $$=\frac { 1 }{ 5 } \int { \frac { d\left( t-2 \right) }{ t-2 } } -\frac { 1 }{ 5 } \int { \frac { d\left( t+3 \right) }{ t+3 } =\frac { 1 }{ 5 } \left[ \ln { \left| t-2 \right| } -\ln { \left| t+3 \right| } \right] =\frac { 1 }{ 5 } \ln { \left| \frac { t-2 }{ t+3 } \right| +C } } $$
Hence $t=\sin { \left( x \right) } $,$$\int { \frac { \cos { \left( x \right) } dx }{ \sin ^{ 2 }{ \left( x \right) } +\sin { \left( x \right) } -6 } } =\frac { 1 }{ 5 } \ln { \left| \frac { \sin { \left( x \right) -2 } }{ \sin { \left( x \right) +3 } } \right| } +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Prove that every integer $n\geq 7$ can be expressed as a sum of distinct primes. My teacher said to use Bertrand's postulate and I have tried this for so long and I seem to go nowhere. Help would be appreciated.
EDIT: Here's what I've done in my proof so far (I need help finalizing case 2)
First note that for $p_0=13$, we can express all integers $7 \leq n \leq 2p_0=26$ as a sum of distinct primes less than or equal to $13$.
Now, we will prove that we can construct these sums indefinitely. Assume we know some prime $p'$ exists such that every integer $7\leq n \leq 2p'$ can be expressed as a sum of distinct primes less than or equal to $p'$. Then, by Bertrand's postulate, there exists a prime $p$ such that $p'<p<2p'$. We claim now that every integer $7\leq n \leq 2p$ can be written as a sum of distinct primes less than or equal to $p$. Consider the two following cases
Case 1: $2p'-p\geq 7$, hence $2p'\geq p+7$ so the terms $p,p+1,\dots,p+7$ are less than or equal to $2p'$ which means they can be written as a sum of distinct primes $\leq p'$ by hypothesis. It is left to check whether the terms $p+8,p+9,\dots, 2p$ satisfy our claim. Note if we subtract $p$ from every term in the arithmetic progression above, it becomes $8,9,\dots, p<2p'$ which shows that each term can be written as the sum of $p$ plus some other distinct primes less than or equal to $p'<p$ by hypothesis.
Case 2: $2p'-p\leq 6$, hence $2p'\leq p+6$. Here we can see all terms $p+7,\dots, 2p$ satisfy our claim along with $p+2,p+3$ and $p+5$ by a similar argument as in Case 1. I'm not sure how to deal with $p+1,p+4,$ and $p+6$ though.
EDIT: Oh, since any prime $p \geq 13$ is odd, then the only possible values for $2p'$ in Case 2 are $p+1,p+3$ and $p+5$, so I don't have to worry about the other cases. I think I'm done!
EDIT: Nope, I still need to deal with $p+4$ and $p+6$.
| We shall inductively prove a stronger form, namely that every positive integer $n \ge 7$ can be written as the sum of distinct primes such that the largest is at most $\max(11,n-7)$. It turns out that strengthening makes the induction work!
Take $n \ge 28$.
Let $m = \lceil \frac{n-6}{2} \rceil= \lfloor\frac{n-5}{2} \rfloor$.
Let $p$ be a prime such that $m+1 \le p \le 2m-1$ [by Bertrand's postulate].
Then $\frac{n-5}{2} \le p \le n-7$.
By the induction to be established $n-p$ can be written as a sum of distinct primes such that the largest is at most $\max(11,n-p-7)$, which is less than $p$ because $p \ge \frac{28-5}{2} > 11$ and $2p \ge n-5 > n-7$.
Thus $n$ can be written as a sum of distinct primes such that the largest is at most $\max(7,n-7)$, and the induction holds as long as the claim is true for every $n$ from $7$ to $27$.
7 = 5+2
8 = 5+3
9 = 7+2
10 = 5+3+2
11 = 11
12 = 7+5
13 = 11+2
14 = 7+5+2
15 = 7+5+3
16 = 11+5
17 = 7+5+3+2
18 = 11+7
19 = 11+5+3
20 = 11+7+2
21 = 11+7+3
22 = 13+7+2
23 = 13+7+3
24 = 13+11
25 = 13+7+5
26 = 13+11+2
27 = 13+11+3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Prof Gould combinatorial identity 3.27 and its "cousin" formula In the book on Combinatorial Identities of Prof Gould I found the identity 3.27
$$\sum_{k=0}^{\rho}\binom{2x+1}{2k+1}\binom{x-k}{\rho-k}=\frac{2x+1}{2\rho+1}\binom{x+\rho}{2\rho}2^{2\rho}$$
I now cant handle its "cousin" formula
$$\sum_{k=0}^{\rho}\binom{2x+1}{2k}\binom{x-k}{\rho-k}=?$$ and am looking for a similar identity.
Maybe also the identity 3.26 is relevant or helpful in this context $$\sum_{k=0}^{\rho}\binom{2x}{2k}\binom{x-k}{\rho-k}=\frac{x}{x+\rho}\binom{x+\rho}{2\rho}2^{2\rho}$$ (Remark: to me looks this formula strangely different to 3.27 when I look at the denominator $x+\rho$)
| We can also prove the companion identity from above.
Suppose we seek to evaluate
$$Q(x,\rho) = \sum_{k=0}^\rho {2x+1\choose 2k+1}
{x-k\choose \rho-k}$$
where $x\ge\rho.$
Introduce
$${x-k\choose \rho-k} = {x-k\choose x-\rho} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+z)^{x-k} \; dz.$$
Note that this controls the range being zero when $\rho\lt k \le x$ so
we can extend the sum to $x$ supposing that $x\gt \rho$. And when
$x=\rho$ we may also set the upper limit to $x.$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+z)^{x}
\sum_{k=0}^x {2x+1\choose 2k+1} \frac{1}{(1+z)^k}
\; dz.$$
This is
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{x}}{z^{x-\rho+1}} \sqrt{1+z}
\left(\left(1+\frac{1}{\sqrt{1+z}}\right)^{2x+1}
- \left(1-\frac{1}{\sqrt{1+z}}\right)^{2x+1}\right) \; dz
\\ =
\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
\left((1+\sqrt{1+z})^{2x+1}-(1-\sqrt{1+z})^{2x+1}\right)
\; dz.$$
Observe that the second term in the parenthesis (i.e. $1-\sqrt{1+z}$)
has no constant term and hence starts at $z^{2x+1}$ making for a zero
contribution. This leaves
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+\sqrt{1+z})^{2x+1}
\; dz.$$
Now put $1+z=w^2$ so that $dz = 2w\; dw$ to get
$$\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w^2-1)^{x-\rho+1}}
(1+w)^{2x+1}
\; w \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho+1}}
\frac{1}{(w+1)^{x-\rho+1}}
(1+w)^{2x+1}
\; w \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho+1}}
(1+w)^{x+\rho} \; w \; dw.$$
Writing $w=(w-1)+1$ this produces two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho}}
\sum_{q=0}^{x+\rho}
{x+\rho\choose q} 2^{x+\rho-q} (w-1)^q
\; dw.$$
This is
$$[(w-1)^{x-\rho-1}]
\sum_{q=0}^{x+\rho}
{x+\rho\choose q} 2^{x+\rho-q} (w-1)^q
\\= {x+\rho\choose x-\rho-1} 2^{x+\rho-(x-\rho-1)}
= {x+\rho\choose x-\rho-1} 2^{2\rho+1}
= {x+\rho\choose 2\rho+1} 2^{2\rho+1}.$$
The second piece is
$$[(w-1)^{x-\rho}]
\sum_{q=0}^{x+\rho}
{x+\rho\choose q} 2^{x+\rho-q} (w-1)^q
\\= {x+\rho\choose x-\rho} 2^{x+\rho-(x-\rho)}
= {x+\rho\choose x-\rho} 2^{2\rho}
= {x+\rho\choose 2\rho} 2^{2\rho}.$$
Joining the two pieces we finally obtain
$$\left(2\times \frac{x-\rho}{2\rho+1} + 1\right)
\times {x+\rho\choose 2\rho} 2^{2\rho}
\\ = \frac{2x+1}{2\rho +1}
{x+\rho\choose 2\rho} 2^{2\rho}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding all functions $f: [0, +\infty)\to \mathbb{R}$ with the properties $f^2(x)=k^2+x\cdot f(x+k)$ and $\frac{x+k}{2}\le f(x) \le 2\cdot (x+k)$
Let $f: [0, +\infty)\to \mathbb{R}$ be a function such that for one $k\in [0, +\infty)$:
$$f^2(x)=k^2+x\cdot f(x+k) \quad \forall x\in [k, +\infty) \tag 1 \label 1$$
and
$$\frac{x+k}{2}\le f(x) \le 2\cdot (x+k) \quad \forall x\in [k, +\infty) \tag 2 \label 2$$
Find all such functions $f$.
I have tried the following:
From \eqref{2}, with substitution $x=-k$ we get $0\le f(-k) \le 0$, so $f(-k)=0$. Then, from \eqref{1} with substitution $x=-k$ we get $0=k^2-k\cdot f(0)$, so $f(0)=k$, which is easily verifiable even when $k=0$.
I also see that $f(x)=x+k$ is a valid function. Any hint how to find ALL such functions? (I believe the above is the only function.)
Edit: As Hagen Von Eitzen noted, I cannot use the substitution $x=-k$, because both $x$ and $k$ are non-negative numbers...
| If we require that $(1)$ and $(2)$ hold for all $x\ge 0$, then $f(x)=x+k$ is the only solution. (It is a solution in the first place because clearly $(x+k)^2=k^2+x(x+2k)$ and $\frac{x+k}{2}\le x+k\le 2(x+k)$).
First consider the case $k=0$. Then $(1)$ translates to $f(x)^2=xf(x)$, so $f(x)=x$ or $f(x)=0$; and $(2)$ becomes $\frac x2\le f(x)\le 2x$ so that indeed $f(x)=x$ for all $x\ge 0$.
Assume $k>0$.
For $x\ge 0$ let $c(x)=\frac{f(x)-(x+k)}{k}$.
Then
$$\tag3-\frac{x+k}{2k}\le c(x)\le \frac{x+k}{k}$$
For $x>0$ we have $$\begin{align}f(x+k)&=\frac{f(x)^2-k^2}{x}\\&=\frac{(x+(c(x)+1)k)^2-k^2}{x}\\&=\frac{x^2+2(c(x)+1)kx_0+((c(x)+1)^2-1)k^2}{x}\\
&=x+2k\,+\,2c(x)k+\frac{c(x)(c(x)+2)k^2}{x}\end{align}$$
i.e.,
$$c(x+k)=2c(x)+\frac{c(x)(c(x)+2)k}{x} $$
If for some $x>0$ we have $c(x)>0$ this implies $c(x+k)>2c(x)$ and by induction $c(x+nk)>2^nc(x)\to\infty$. As exponentials grow faster than polynomials, we obtain a contradiction with $(3)$. We conclude $c(x)\le 0$ (that is: $f(x)\le x+k$) for $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.