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Prove that the number is the product of two successive positive integers Prove that the number $\underbrace{11\cdots 1}_{100\text{ digits}}\underbrace{22\cdots 2}_{100\text{ digits}}$ is the product of two successive positive integers.
What is the general method for this class of proof (big integers)?
|
Hint:
$$
\begin{align}
\underbrace{11\cdots 1}_{100\text{ digits}}\underbrace{22\cdots 2}_{100\text{ digits}} & = 1 \cdot 10^{199} + 1 \cdot 10^{198} + \cdots + 1 \cdot 10^{100} + 2 \cdot 10^{99} + 2 \cdot 10^{98} + \cdots + 2 \cdot 10^0 \\
& = 2 \sum_{k=0}^{99} 10^k + \sum_{k=100}^{199} 10^k \\
& = \frac{(10^{100}-1)(10^{100}+2)}{9}
\end{align}
$$
[ EDIT ] To conclude the proof, just note that $\frac{10^{100}-1}{3}$ is an integer, and $\frac{10^{100}+2}{3} = \frac{10^{100}-1}{3} + 1$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2027522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Basic geometry: find the angle Question: Find $\angle BDE$ when
$\angle A=20^\circ$, $\overline{AB}=\overline{AC}$, $\angle CBD=65^\circ$, and $\angle BCE=25^\circ$.
So far I figured out that $\angle BDE=\angle AED-15^\circ$ and $\overline{BD}\bot\overline{CE}$ that don't seem to be helping.
|
This is a variation of the notorious Langley's problem.
We follow a trigonometric approach. Let $\angle BDE=x$.
i) Use the Law of sines in $\triangle EBC$,
$$\frac{\sin 25^\circ}{EB}=\frac{\sin75^\circ}{AB}.$$
ii) Use the Law of sines in $\triangle DBC$,
$$\frac{\sin 80^\circ}{BD}=\frac{\sin 35^\circ}{AB}.$$
iii) Use the Law of sines in $\triangle EBD$,
$$\frac{\sin x}{EB}=\frac{\sin (165^\circ-x)}{BD}=\frac{\sin (15^\circ+x)}{BD}.$$
It follows that
$$\frac{\sin (15^\circ+x)}{\sin x}=\frac{\sin 80^\circ}{\sin 35^\circ}\cdot \frac{\sin 75^\circ}{\sin 25^\circ}$$
Since $\sin (15^\circ+x)=\sin 15^\circ\cos x+\cos 15^\circ\sin x$,
we obtain that
$$\cot x=\frac{1}{\sin 15^\circ}\cdot\frac{\sin 80^\circ}{\sin 35^\circ}\cdot \frac{\sin 75^\circ}{\sin 25^\circ}-\cot 15^\circ.$$
Therefore $x=5^\circ$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2029820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$? How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?
|
The factorization of $ab^2+a^2b+a^2c+ac^2+b^2c+bc^2+2abc$
is $(a + b) (a + c) (b + c)$.
First, take out the $(a+b)$ (because a and b are factors to exactly half of the polynomial):
$$((a^2b+a^2c+ac^2+abc)+(ab^2+b^2c+bc^2+abc))*(\frac{a+b}{a+b})$$Then we have:$$(a+b)(ab+ac+bc+c^2)$$Then take out $(a+c)$:$$(a+b)(a(b+c)+c(b+c))$$And voila! By the commutative property, we have: $$(a+b)(a+c)(b+c)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2030326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
finding local behaviour as x tends to $0^+$ of particular solution of ode we have
$$ y' + xy = \cos x $$
I am asked to find the first three terms in local behaviour as $x \to 0^+$.
thought:
As $ x \to 0^+$, then $\cos x \sim 1 $ so
$$ y' + xy \sim 1 $$
which can be solved by using integrating factor $e^{x^2/2}$, thus
$$ ( e^{x^2/2} y ) \sim \int e^{x^2/2} = \int \sum \frac{x^{2n} }{2^n n!} = \sum \frac{x^{2n+1} }{2^n (2n+1) n!} \sim x$$
thus,
$$ y(x) \sim cx e^{-x^2/2} $$
is this a correct?
|
\begin{equation}
y^\prime+xy=\cos x
\end{equation}
Let $y=\sum_{n=0}^{\infty}c_nx^n$. Then
\begin{eqnarray}
y^\prime+xy&=&\sum_{n=1}^{\infty}nc_nx^{n-1}+x\sum_{n=0}^{\infty}c_nx^n\\
&=&\sum_{n=1}^{\infty}nc_nx^{n-1}+\sum_{n=0}^{\infty}c_nx^{n+1}
\end{eqnarray}
Replacing each $n$ in the first sum with $n+1$ and each $n$ in the second with $n-1$ yields
\begin{eqnarray}
y^\prime+xy&=&\sum_{n=0}^{\infty}(n+1)c_{n+1}x^{n}+\sum_{n=1}^{\infty}c_{n-1}x^{n}\\
&=&c_1+\sum_{n=1}^{\infty}\left[(n+1)c_{n+1}+c_{n-1}\right]x^n
\end{eqnarray}
Therefore the DE can be written in summation form
\begin{equation}
c_1+\sum_{n=1}^{\infty}\left[(n+1)c_{n+1}+c_{n-1}\right]x^n=\sum_{k=0}^{\infty}\dfrac{(-1)^kx^{2k}}{(2k)!}
\end{equation}
Immediately we see that $c_1=1$, so we know that the first two terms of the solution are $y=c_0+x.$ Now, when $n=1$ the coefficient of $x$ in the cosine expansion must be $0$ so we know that $(n+1)c_{n+1}+c_{n-1}=2c_2+c_0=0$. Therefore, $c_2=-\frac{1}{2}c_0$. Thus we now have the first three terms as required
\begin{equation}
y=c_0+x-\frac{1}{2}c_0x^2
\end{equation}
Since we are on a roll, let's find the fourth term of $y$.
When $n=2$ we get $(n+1)c_{n+1}+c_{n-1}=3c_3+c_1=-\frac{1}{2}$, the coefficient of $x$ in the cosine expansion. We already know that $c_1=1$ so we have $c_3=-\frac{1}{2}$ in the expansion of $y$. Therefore we have the first four terms:
\begin{eqnarray}
y&=&c_0+x-\frac{1}{2}c_0x^2-\frac{1}{2}x^3\\
&=&(c_0+x)\left(1-\frac{x^2}{2}\right)
\end{eqnarray}
Thanks to @Winther for the hint.
|
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"timestamp": "2023-03-29T00:00:00",
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|
total number of positive and negative roots of the equation $ax^3+bx^2+cx+d=0$ Suppose $a,b,c,d$ are non zero real numbers and $ab>0,$ and
$\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx = \int^{2}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
Then total number of positive and negative roots of the equation $ax^3+bx^2+cx+d=0.$
what i have try $\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx = \int^{2}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
$\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=\int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx +\int^{2}_{1}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
Am I not able to go further. Could someone help me with this? Thanks
|
We know $a,b$ have the same sign, so the number of sign changes in the cubic is either $0, 1$ or $2$. Thus by Descartes rule of signs, at most it can have only $2$ positive roots.
From the integrals given, it is clear there must be sign changes in both the intervals $(0,1)$ and $(1,2)$. So there are two positive roots, then the remaining root has to be real and negative.
|
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"url": "https://math.stackexchange.com/questions/2035617",
"timestamp": "2023-03-29T00:00:00",
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|
find the value of given trigonometric equation If $\alpha$ and $\beta$ are two different values of $\theta$ which satisfy $$bc \cos{θ} \cos{\phi} + ac \sin{\theta} \sin{\phi} = ab,$$ then what is the value of $$(b^2 + c^2 - a^2) \cos{\alpha} \cos{\beta} + (c^2 + a^2 - b^2) \sin{\alpha} \sin{\beta}\,?$$
|
Considering an ellipse
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
The equations of tangents at $[\alpha]$ and $[\beta]$ are
$$\frac{x\cos \alpha}{a}+\frac{y\sin \alpha}{b}=1$$
$$\frac{x\cos \beta}{a}+\frac{y\sin \beta}{b}=1$$
If the tangents meet at $(c\cos \phi, \, c\sin \phi)$, then
$$(c\cos \phi , \, c\sin \phi)=
\left(
a\frac{\cos \frac{\alpha+\beta}{2}}{\cos \frac{\alpha-\beta}{2}} \, , \;
b\frac{\sin \frac{\alpha+\beta}{2}}{\cos \frac{\alpha-\beta}{2}}
\right)$$
Refer to this link
I
\begin{align*}
\frac{c^2\cos^2 \phi}{a^2}+\frac{c^2\sin^2 \phi}{b^2} &=
\sec^2 \frac{\alpha-\beta}{2} \\
\cos^2 \frac{\alpha-\beta}{2} &=
\frac{a^2b^2}{c^2(a^2\sin^2 \phi+b^2\cos^2 \phi)} \\
\cos (\alpha-\beta) &=
\frac{2a^2b^2}{c^2(a^2\sin^2 \phi+b^2\cos^2 \phi)}-1 \\
\end{align*}
II
\begin{align*}
\tan \frac{\alpha+\beta}{2} &=
\frac{a}{b} \tan \phi \\
\cos (\alpha+\beta) &=
\frac{1-\frac{a^2}{b^2} \tan^2 \phi}{1+\frac{a^2}{b^2} \tan^2 \phi} \\
&= \frac{b^2\cos^2 \phi-a^2\sin^2 \phi}{a^2\sin^2 \phi+b^2\cos^2 \phi}
\end{align*}
III
\begin{align*}
E &=
(b^2+c^2-a^2)\cos \alpha \cos \beta+
(c^2+a^2-b^2)\sin \alpha \sin \beta \\[5pt]
&= c^2\cos (\alpha-\beta)+(b^2-a^2)\cos (\alpha+\beta) \\[5pt]
&= \frac{a^4\sin^2 \phi+a^2b^2+b^4\cos^2 \phi}
{a^2\sin^2 \phi+b^2\cos^2 \phi}-c^2 \\[5pt]
&= \frac{a^4\sin^2 \phi+
a^2b^2\color{red}{(\sin^2 \phi+\cos^2 \phi)}+
b^4\cos^2 \phi}
{a^2\sin^2 \phi+b^2\cos^2 \phi}-c^2 \\[5pt]
&= \fbox{$a^2+b^2-c^2$}
\end{align*}
Note Briefly:
*
*For real distinct $\alpha$ and $\beta$, $$c > \frac{ab}{\sqrt{a^2\sin^2 \phi+b^2\cos^2 \phi}} \ge \min(a,b) > 0$$
*If $c^2=a^2+b^2$, then $x^2+y^2=c^2$ is the director circle of the ellipse so that $AC \perp BC$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2040565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How many four letter words can be formed using the letterss of the word 'INEFFECTIVE'?
How many four letter words can be formed using the letterss of the word 'INEFFECTIVE'?
Please explain with details. Also, how can we clearly differentiate between permutations and combinations?
|
There are $11$ letters in 'INEFFECTIVE'.
CASE $(1)$: There is only set of three same letters of the letter $E$. So $3$ same letters can be selected in one way. Out of the six remaining distinct letters we can select one in $\binom{6}{1}=6$ ways. Hence there are $6$ groups each of which contains the three same letters. These four letters can be arranged in $\binom{4}{3}=4$ ways. Hence the total number of words with $3$ same and one different letters is $6\times 4=24$ ways.
CASE $(2)$: There are $3$ sets of two same letters $E,F,I$. Out of these $3$ sets, two can be selected in $\binom{3}{2}=3$ ways. So there are $3$ groups each of which contains $4$ letters two of which are same of one type and the other two of a different same type. Now $4$ letters in each group can be arranged in $\binom{4}{2}=6$ ways. Hence the total number of words with two letters of one kind and the other two of the different kind is $3\times 6=18$ times.
CASE $(3)$: Out of $3$ sets of $2$ same letters one set can be chosen in $\binom{3}{1}=3$ ways. Now from the remaining $6$ letters, $2$ distinct letters can be chosen in $\binom{6}{2}=15$ ways. So there are $3\times 15=45$ groups of $4$ letters each. Now letters of each group can be arranged in $\frac{4!}{2!}=12$ ways. Hence, the total number of ways consisting of two same letters and $2$ different are $45\times 12=540$ ways.
CASE $(4)$: There are $\binom{7}{4}$ groups of words with $4$ different letters each. In each group each letter can be arranged in $4!=24$ ways. So the total number of $4$ letter words in which all letters are different is $\binom{7}{4}\times 24=840$ ways.
There are thus a total of $24+18+540+840=1422$ ways.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\sum_{t=1}^j \sum_{r=1}^{t} (-1)^{r+t} \binom{j-1}{t-1} \binom{t-1}{r-1} f(r) = f(j)$ I've run into the following identity and trying to prove it:
Let $j \in \mathbb{N}$ and $f:\mathbb{N} \to \mathbb{R}$, then
$$
\sum_{t=1}^j \sum_{r=1}^{t} (-1)^{r+t} \binom{j-1}{t-1} \binom{t-1}{r-1} f(r) = f(j)
$$
I've so far tried to find some connection with the multinomial theorem, since the product of the two binomials in the expression is just multinomial $\binom{j-1}{k1,k2,k3}$. So perhaps something like $(f(j)-1+1)^{j-1}$, but it does not quite fit. I am missing something, any ideas how to proceed?
Edit: Further attempt, trying to collect "coefficients" of $f(r)$ yields
$$
\sum_{t=r}^j(-1)^{r+t}\binom{j-1}{t-1}\binom{t-1}{r-1} = \sum_{t=r}^j (-1)^{r+t}\binom{j-1}{t-1}
$$
It should be enough to show that this is $1$ when $j=r$ and $0$ otherwise. First one is simple
$$
\sum_{t=r}^r (-1)^{r+t}\binom{r-1}{t-1} = (-1)^{2r} \binom{r-1}{r-1} = 1
$$ but how to show it is equal to $0$ in other cases ($j\neq r $) ...
|
Note that it is:
$$
\begin{gathered}
f(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)g(k)} \quad \Leftrightarrow \quad g(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,n - k} \left( \begin{gathered}
n \\
k \\
\end{gathered} \right)f(k)} \hfill \\
f(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
n \\
k \\
\end{gathered} \right)g(k)} \quad \Leftrightarrow \quad g(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
n \\
k \\
\end{gathered} \right)f(k)} \hfill \\
\end{gathered}
$$
because
$$
\begin{gathered}
f(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)g(k)} = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)\left( { - 1} \right)^{\,k - j} \left( \begin{gathered}
k \\
j \\
\end{gathered} \right)f(j)} } = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( {\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,k - j} \left( \begin{gathered}
n \\
k \\
\end{gathered} \right)\left( \begin{gathered}
k \\
j \\
\end{gathered} \right)} } \right)f(j)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
n \\
j \\
\end{gathered} \right)\left( {\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,k - j} \left( \begin{gathered}
n - j \\
k - j \\
\end{gathered} \right)} } \right)f(j)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
n \\
j \\
\end{gathered} \right)\left( {\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
n - j \\
k - j \\
\end{gathered} \right)1^{\,n - j - \left( {k - j} \right)} \left( { - 1} \right)^{\,k - j} } } \right)f(j)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
n \\
j \\
\end{gathered} \right)0^{\,n - j} f(j)} = f(n) \hfill \\
\end{gathered}
$$
where from 2nd to 3rd line we apply the known trinomial revision.
In your case we have
$$
\begin{gathered}
f(j) = \sum\limits_{\left( {1\, \leqslant } \right)\,t\,\left( { \leqslant \,j} \right)} {\sum\limits_{\left( {1\, \leqslant } \right)\,r\,\left( { \leqslant \,t} \right)} {\left( { - 1} \right)^{\,r + t} \left( \begin{gathered}
j - 1 \\
t - 1 \\
\end{gathered} \right)\left( \begin{gathered}
t - 1 \\
r - 1 \\
\end{gathered} \right)f(r)} } = \hfill \\
= \sum\limits_{\left( {1\, \leqslant } \right)\,t\,\left( { \leqslant \,j} \right)} {\sum\limits_{\left( {1\, \leqslant } \right)\,r\,\left( { \leqslant \,t} \right)} {\left( { - 1} \right)^{\,r - t} \left( \begin{gathered}
j - 1 \\
t - 1 \\
\end{gathered} \right)\left( \begin{gathered}
t - 1 \\
r - 1 \\
\end{gathered} \right)f(r)} } = \hfill \\
= \sum\limits_{\left( {1\, \leqslant } \right)\,t\,\left( { \leqslant \,j} \right)} {\sum\limits_{\left( {1\, \leqslant } \right)\,r\,\left( { \leqslant \,t} \right)} {\left( { - 1} \right)^{\,r - 1 - \left( {t - 1} \right)} \left( \begin{gathered}
j - 1 \\
t - 1 \\
\end{gathered} \right)\left( \begin{gathered}
t - 1 \\
r - 1 \\
\end{gathered} \right)f(r)} } = \hfill \\
= \sum\limits_{\left( {1\, \leqslant } \right)\,t\,\left( { \leqslant \,j} \right)} {\sum\limits_{\left( {1\, \leqslant } \right)\,r\,\left( { \leqslant \,t} \right)} {\left( { - 1} \right)^{\,r - 1 - \left( {t - 1} \right)} \left( \begin{gathered}
j - 1 \\
t - 1 \\
\end{gathered} \right)\left( \begin{gathered}
t - 1 \\
r - 1 \\
\end{gathered} \right)g(r - 1)} } = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,t - 1\,\left( { \leqslant \,j - 1} \right)} {\sum\limits_{\left( {0\, \leqslant } \right)\,r - 1\,\left( { \leqslant \,t - 1} \right)} {\left( { - 1} \right)^{\,r - 1 - \left( {t - 1} \right)} \left( \begin{gathered}
j - 1 \\
t - 1 \\
\end{gathered} \right)\left( \begin{gathered}
t - 1 \\
r - 1 \\
\end{gathered} \right)g(r - 1)} } = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,j - 1} \right)} {\sum\limits_{\left( {0\, \leqslant } \right)\,l\,\left( { \leqslant \,k} \right)} {\left( { - 1} \right)^{\,l - k} \left( \begin{gathered}
j - 1 \\
k \\
\end{gathered} \right)\left( \begin{gathered}
k \\
l \\
\end{gathered} \right)g(l)} } = \hfill \\
= g(j - 1) = f(j) \hfill \\
\end{gathered}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$
Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$.
So if we substitute $z=1-x-y$ into the equation $x^2+y^2=z$ we get $x^2+y^2=1-x-y$ which can be written in the form $\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=1$.
Is this the curve of intersection? Because it seems to me (by geometrical intuition) that the curve should be an ellipse. What is the logical flaw here. Help please!
|
There is an error in your algebra
$$\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac32$$
so parametrically continue from here.
$$ x = \sqrt{\frac32} \cos t -\frac12\,,y =\sqrt{\frac32} \sin t -\frac12 $$
Plug these into the second equation and find $z$ in terms of $t$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show $\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$? I tried to find the right handside of the equation by manipulating the series but I failed at getting the right handside of it.
$$\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$$
Closed form of the left handside in the parantheses would be
$$\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n$$
Any hint would be appreciated.
|
Well, you can carefully multiply and don't forget any terms:
$$\left( 1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots \right)^2 = \left( \color{blue}{1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots }\right)\left(\color{red}{ 1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots }\right)$$
Expanding and combining terms of the same degree:
*
*constant term: $$\color{blue}{1} \cdot \color{red}{1} = 1$$
*linear term: $$\color{blue}{1} \cdot \color{red}{\tfrac{1}{3}x}+\color{blue}{\tfrac{1}{3}x} \cdot \color{red}{1} = \tfrac{2}{3}x$$
*quadratic term: $$\color{blue}{1} \cdot \color{red}{\tfrac{1}{24}x^2}+\color{blue}{\tfrac{1}{3}x} \cdot \color{red}{\tfrac{1}{3}x} + \color{blue}{\tfrac{1}{24}x^2} \cdot \color{red}{1} = \tfrac{7}{36}x^2$$
*...
|
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"timestamp": "2023-03-29T00:00:00",
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|
Interesting limit involving gamma function $$\displaystyle \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{2} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{2} + 1\right)}}$$
This is from here.
If we add one more variable -
$$\displaystyle L(p,q,m,t) = \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{t} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{t} + 1\right)}}$$
Then, the results I got from W|A are really interesting. They seem to follow the pattern which says
$\displaystyle L(p,q,m,t) = \dfrac{1}{t^{|p-q|/m}}$
If it is indeed true, how to prove it? Factorial approximations did not lead me anywhere.
|
From Abramowitz and Stegun we know
$$\ln \Gamma(x) = x \ln(x) - \frac{1}{2} \ln(x) - x + \frac{\ln(2\pi)}{2} + O(\tfrac{1}{x}) \qquad (x \to \infty)$$
Applying this formula here yields
$$\begin{align*}
\ln \Gamma\left(\frac{p}{m} + \frac{n}{t}\right) &= \left(\frac{p}{m} + \frac{n}{t}\right) \ln\left(\frac{p}{m} + \frac{n}{t}\right) - \left(\frac{p}{m} + \frac{n}{t}\right) + \frac{\ln(2\pi)}{2} + O(\tfrac{1}{n}) \\
&= \left(\frac{p}{m} + \frac{n}{t}\right) \ln(n) - \frac{n}{t} - \frac{p}{m}
+ \left(\frac{p}{m} + \frac{n}{t}\right) \ln \left(\frac{1}{t} + \frac{p}{mn}\right) + \frac{\ln(2\pi)}{2} + o(1).
\end{align*}$$
After cancelling some terms, we get
$$\ln \frac{\Gamma\left(\frac{p}{m} + \frac{n}{t}\right)}{\Gamma\left(\frac{q}{m} + \frac{n}{t}\right)} = \frac{p-q}{m} \ln(n) - \frac{p - q}{n} + \frac{p - q}{m} \ln\left(\frac{1}{t}\right) + \frac{n}{t} \ln \left(\frac{\frac{1}{t} + \frac{p}{mn}}{\frac{1}{t} + \frac{q}{mn}}\right) + o(1).$$
Now note that
$$\frac{n}{t} \ln \left(\frac{\frac{1}{t} + \frac{p}{mn}}{\frac{1}{t} + \frac{q}{mn}}\right) = \frac{1}{t} \ln\left(\frac{\left(1 + \frac{pt}{mn}\right)^n}{\left(1 + \frac{qt}{mn}\right)^n}\right) = \frac{1}{t}\ln\left(\frac{\exp(pt/m)}{\exp(qt/m)}\right) + o(1) = \frac{p - q}{m} + o(1).$$
After cancelling some more terms we get
$$\ln \frac{\Gamma\left(\frac{p}{m} + n\right) \Gamma\left(\frac{q}{m} + \frac{n}{t}\right)}{\Gamma\left(\frac{q}{m} + n\right)\Gamma\left(\frac{p}{m} + \frac{n}{t}\right)} = \frac{q - p}{m} \ln(\tfrac{1}{t}) + o(1) = \frac{p - q}{m} \ln(t) + o(1).$$
Using the formula $\Gamma(x + 1) = x \Gamma(x)$ we see that the limit in the last equation is equal to $\ln L(p, q, m, t)$, which finally shows
$$L(p, q, m, t) = t^{(p-q)/m}.$$
|
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|
Prove the inequality $\sum\limits_{cyc}\sqrt{x+yz}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$
Let $x,y,z$ is positive real numbers, such that $\frac1x+\frac1y+\frac1z=1$. Prove the inequality
$$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$
My work so far:
$\frac1x+\frac1y+\frac1z=1\Rightarrow xyz=xy+yz+zx$
$\sqrt{x+yz}=\sqrt{x+xyz-xz-xy}=\sqrt x \cdot \sqrt{(y-1)(z-1)}$
|
An homogenization gives
$$\sum\limits_{cyc}\sqrt{x+yz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\geq\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$ or
$$\sum\limits_{cyc}\sqrt{yz(x+y)(x+z)}\geq xy+xz+yz+\sum\limits_{cyc}x\sqrt{yz},$$ which is C-S:
$$\sqrt{yz(x+y)(x+z)}=\sqrt{(xy+yz)(zx+yz)}\geq x\sqrt{yz}+yz$$
Done!
|
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|
What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $\ln(2)=0.693$.
Now, let's do some rearrangement:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......$$
$$
(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}.......$$
$$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}......$$
$$\frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......)$$
$$\frac{1}{2}\ln(2).$$
I know that mathematics can't be wrong, and I have done something wrong. But here is my question: where does the argument above go wrong?
|
In your new series
$$\left(\frac 11 - \frac 12\right) - \frac 14 +
\left(\frac 13 - \frac 16\right) - \frac 18 +
\left(\frac 15 - \frac 1{10}\right) - \frac 1{12} +
\left(\frac 17 - \frac 1{14}\right) - \frac 1{16} \dots$$
Terms of the form $\frac{1}{2z - 1}$ occur 1/3 of the time instead of 1/2 the time like they do in the original series, so they are underweighted.
Terms of the form $\frac{1}{2z}$ occur 2/3 of the time instead of 1/2 the time like they do in the original series, so they are overweighted.
By changing the density at which those terms appear, you are effectively converting the sum from
$$\lim_{n \to \infty} \sum_{k = 1}^n \frac{1}{2k - 1} - \sum_{k = 1}^n \frac{1}{2k}$$
into
$$\lim_{n \to \infty} \sum_{k = 1}^{n \cdot 2/3} \frac{1}{2k - 1} - \sum_{k = 1}^{n \cdot 4/3} \frac{1}{2k}$$
which naturally gives a smaller value since the subtracted terms have increased density. If instead you rearrange the series without changing the density of the subseries, then you'll get an equal result.
|
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|
Difficult coordinate geometry problem The base of a triangle passes through a fixed point $P(a,b)$ and its sides are respectively bisected at right angles by the lines $x+y=0$ and $x=9y$. If the locus of the third vertex is a circle, then find its equation.
Apart from the fact that the circumcentre of the given triangle is the origin $(0,0)$ , I havent been able to find anything else. Also since the geometry given in the problem is fixed, its a little difficult to plot. Any hints or clues??
Thanks in advance!!...
The correct answer given in the key is :
$4x^2 + 4y^2 + (5a+4b)x + (4a-5b)y=0$
|
Let $A(x,y)$ be the required locus.
First reflect $A(x,y)$ about $x+y=0$, we get
$$B=(-y,-x)$$
Now reflect $A(x,y)$ about $x-9y=0$, we get $C=(x',y')$
$$\frac{y-y'}{x-x'}=-9 \tag{1}$$
Also,
$$\frac{x-9y}{\sqrt{1+9^2}}=-\frac{x'-9y'}{\sqrt{1+9^2}}$$
$$x+x'=9(y+y') \tag{2}$$
Solving $(1)$ and $(2)$,
$$C(x',y')=\left( \frac{40x+9y}{41},\frac{9x-40y}{41} \right)$$
Now $BC$ contains $P$, then
\begin{align*}
\frac{-x-b}{-y-a} &= \frac{-x-\dfrac{9x-40y}{41}}{-y-\dfrac{40x+9y}{41}}
\\[5pt]
\frac{x+b}{y+a} &= \frac{50x-40y}{40x+50y} \\[5pt]
(x+b)(4x+5y) &= (y+a)(5x-4y)
\end{align*}
Hence, the locus is
$$\fbox{$4x^2+4y^2-(5a-4b)x+(4a+5b)y = 0$} \\$$
|
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|
Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$?
Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$ ? with $a,b\ge 0$
If this is true (I don't know whether it is true, just inserted some values in wolfram alpha) then I can show that $\frac{1}{a+b+1}<\int_{a}^{a+1}\int_{b}^{b+1}\frac{1}{x+y}dxdy$
Is it possible to see it from Taylor ?
|
Set $a+b+1=1/c$, so
$$
\frac{2a+2b+3}{2a+2b+1}=
\frac{\frac{2}{c}+1}{\frac{2}{c}-1}=\frac{2+c}{2-c}
$$
and the inequality is
$$
\log\frac{2+c}{2-c}>c
$$
for $0<c\le1$.
Consider the function
$$
f(x)=\log(2+x)-\log(2-x)-x
$$
defined over $[0,1]$. Then $f(0)=0$ and
$$
f'(x)=\frac{1}{2+x}+\frac{1}{2-x}-1=\frac{x^2}{4-x^2}>0
$$
over the interval $(0,1)$. So the function is increasing.
|
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|
How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$
How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$
I have tried this
$$f(x)=x^2\ln(x+1)$$
$$ f'(x) = 2x\ln(x+1) + \frac{x^2}{x+1}$$
$$ f''(x)=2\ln(x+1) + \frac{4x}{x+1} - \frac{x^2}{(x+1)^2}$$
However I do not see any pattern :(
|
Taking one more derivative results in
$$
f^{(3)}(x) = \frac2{x+1} + \frac{(x+1)4-4x(1)}{(x+1)^2} - \frac{(x+1)^22x-x^22(x+1)}{(x+1)^4} = \frac{2 \left(x^2+3 x+3\right)}{(x+1)^3}.
$$
However, that's not the most convenient form for taking further derivatives: using partial fractions gives us
$$
f^{(3)}(x) = \frac{2}{x+1}+\frac{2}{(x+1)^2}+\frac{2}{(x+1)^3}.
$$
And this should now be easy to take as many derivatives of as you want.
|
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|
Find minimal value for product of the sums Find minimal value for product of the sums
$$\left( x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)
\left( \frac{1}{x}+\frac{1}{2!x^2}+\frac{1}{3!x^3}+\dots \right)
$$
Here what we did about solution. We know these series equals Taylor expansion of $(e^x-1)(e^{\frac{1}{x}}-1)$. We tried to apply derivative test to find minimal value but it is hard to find extreme values from the equation we get.
Do you have any trick to find the minimal? Thank you for your help.
|
Hint. Set, for $x>0$,
$$
f(x)=(e^x-1)(e^{\frac{1}{x}}-1)=e^{x+\frac{1}{x}}-e^x-e^{\frac{1}{x}}+1
$$ giving
$$
f'(x)=\left(1-\frac1{x^2} \right)e^{x+\frac{1}{x}}-e^x+\frac1{x^2}e^{\frac{1}{x}}=\frac{e^{\frac{1}{x}}-e^{\frac{1}{x}+x}-e^x x^2+e^{\frac{1}{x}+x} x^2}{x^2}
$$ then observe that $f'(1)=0$, $f'(x)<0$ for $0<x<1$ and $f'(x)>0$ for $x>1$.
The minimum is thus
$$f(1)=(e-1)^2.$$
|
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|
Limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$
Find the limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$.
I would like to see a solving method without l'Hopital or Taylor expansion.
|
When $x$ is large, the leading behavior of
$\sqrt{x+1}$, $\sqrt{4x+1}$ and $\sqrt{9x+1}$ is $\sqrt{x}$, $2\sqrt{x}$ and $3\sqrt{x}$ respectively. One approach is isolate them from the sum and see
whether you can control the remainder. After you do this, you get
$$\begin{align}
&a\sqrt{x+1} + b \sqrt{4x+1} + c\sqrt{9x+1}\\
= & (a+2b+3c)\sqrt{x} + a(\sqrt{x+1}-\sqrt{x}) + b(\sqrt{4x+1} - \sqrt{4x}) + c(\sqrt{9x+1} -\sqrt{9x})\\
\end{align}
$$
You see something like $\sqrt{X + u} \pm \sqrt{X}$ where $u$ is a small number. This suggest us to apply following identities and see whether we can
simplify the expression.
$$\sqrt{X+u} \pm \sqrt{X} = \frac{u}{\sqrt{X+u} \mp \sqrt{X}}$$
Indeed we can!
$$\begin{align}
&a\sqrt{x+1} + b \sqrt{4x+1} + c\sqrt{9x+1}\\
= & (a+2b+3c)\sqrt{x} +
\underbrace{\frac{a}{\sqrt{x+1}+\sqrt{x}} + \frac{b}{\sqrt{4x+1}+\sqrt{4x}} + \frac{c}{\sqrt{9x+1}+\sqrt{9x}}}_{\to 0 \text{ as } x \to \infty}
\end{align}
$$
Since all but the first pieces goes to $0$ as $x \to \infty$, we have
$$\lim_{x\to\infty}
a\sqrt{x+1} + b \sqrt{4x+1} + c\sqrt{9x+1}
= \lim_{x\to\infty} (a+2b+3c)\sqrt{x} =
\begin{cases}
0, & a+2b+3c = 0\\
\infty, & \text{ otherwise }
\end{cases}
$$
About the question why the limit is $0$ when $a+2b+3c = 0$.
In that case, the expression we are taking limit
$(a+2b+3c)\sqrt{x}$ is $0$ for all finite $x$, so its limit is $0$.
In general, if $\lim\limits_{x\to\infty}f(x) = L \in \mathbb{R} \cup \{ \infty \}$ and $\alpha$ is a constant, we have
$$\lim_{x\to\infty}\alpha f(x) = \begin{cases}
\alpha L & L \ne \infty\\
\infty & L = \infty, \alpha \ne 0\\
0 & L = \infty, \alpha = 0
\end{cases}$$
The second case is not obtained from multiplying $\alpha$ by $\infty$ at all.
It is a result of taking this form of limit. The usual rule $\alpha \times \infty = \infty$ is not really a rule, it is simply a mnemonic of this particular result!
For the third case, it is not obtained by $0 \times \infty = 0$ again. In fact, the expression $0 \times \infty$ is indeterminate (should I say meaningless). To make it meaningful, the first thing one need to do is define $\infty$ rigorously. If you do that consistently to the point you can do arithmetics on them, you will find no matter what value you assigned to the expression $0\times \infty$, there are cases that will give you wrong result.
My advise is forget $0 \times \infty = $ anything, it will only cause confusion.
|
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|
Integral of $\int{\tan^{5}(x)\sec^4(x)}dx$? Here's my attempt at the problem:
$\int{\tan^{5}(x)\sec^4(x)}dx=
\int{\frac{\sin^5(x)}{\cos^9(x)}}\,dx=
\int{\frac{\sin(x)(\sin^2(x))^2}{\cos^9(x)}}\,dx=
\int{\frac{\sin(x)(1-\cos^2(x))^2}{\cos^9(x)}}\,dx=
\int{\frac{(1-u^2)^2}{u^9}}\,du=
\int{\frac{u^4-2u^2+1}{u^9}}\,du=
\int{\Big(\frac{1}{u^5}-\frac{2}{u^7}+\frac{1}{u^9}}\Big) \,du=
-\frac{1}{4u^4}+\frac{1}{3u^6}-\frac{1}{8u^8}+C=
-\frac{\sec^4(x)}{4}+\frac{\sec^6(x)}{3}-\frac{\sec^8(x)}{8}+C$
It seems, however, that the actual answer should be: $\frac{\sec^4(x)}{4}-\frac{\sec^6(x)}{3}+\frac{\sec^8(x)}{8}+C$
What am I doing wrong?
|
Substituting $u=\tan(x),\,du=\sec^2(x)\,dx$
\begin{eqnarray}
\int\tan^5(x)\sec^4(x)\,dx&=&\int\tan^5(x)\sec^2(x)\sec^2(x)\,dx\\
&=&\int\tan^5(x)\left[\tan^2(x)+1\right]\sec^2(x)\,dx\\
&=&\int u^5\left(u^2+1\right)\,du\\
&=&\int u^7+u^5\,du\\
&=&\frac{1}{8}u^8+\frac{1}{6}u^6+c\\
&=&\frac{1}{8}\tan^8(x)+\frac{1}{6}\tan^6(x)+c
\end{eqnarray}
|
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|
lagrange multiplier inequality question
Prove for every $x,y,z>0, that\
f(x,y,z)= \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\geq \frac{3}{2} $
So since $f$ is a homogeneous function we can prove this on the following constraint $g(x,y,z)=x+y+z-1=0$.
However, I seem to be stuck now with this equation set:
$\frac{1}{y+z}-\frac{y}{(x+z)^2}-\frac{z}{(y+x)^2}= \lambda$
$\frac{1}{x+z}-\frac{x}{(y+z)^2}-\frac{z}{(y+x)^2}= \lambda$
$\frac{1}{x+y}-\frac{x}{(y+z)^2}-\frac{y}{(z+x)^2}= \lambda$
any help?
|
$$\frac{x}{y+z}=\frac{x+y+z}{y+z}-1$$
So we get $$LHS=\frac{2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})}{2}-3 \ge \frac{9}{2}-3$$
Using AM-GM inequality,
$$LHS \ge \frac{9}{2}-3=\frac{3}{2}$$
|
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|
Prove that $\sum\limits_{cyc}\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}\geq5$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}+\sqrt{\frac{8ab+8bc+9ac}{(2a+c)(a+2c)}}+\sqrt{\frac{8ac+8bc+9ab}{(2a+b)(a+2b)}}\geq5$$
I tried Holder, but without success.
For example, Holder even with $(ka^2+b^2+c^2+nab+nac+mbc)^3$ does not help.
SOS or C-S seems very ugly here.
The equality occurs also for $(a,b,c)=(1,1,0)$, which adds a problems.
|
The Buffalo Way works. Due to symmetry, assume that $a\ge b\ge c$.
Let
$$X = \frac{9}{25}\frac{8ab + 9bc + 8ca}{(2b+c)(b+2c)}, \quad
Y = \frac{9}{25}\frac{8bc + 9ca + 8ab}{(2c+a)(c+2a)}, \quad
Z = \frac{9}{25}\frac{8ca + 9ab + 8bc}{(2a+b)(a+2b)}.$$
We need to prove that $\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge 3$.
We will use the following bounds:
$$\sqrt{x} \ge f(x) = \frac{22x(5x+6)}{25x^2 + 181x+36}, \quad \forall x \ge 0$$
and
$$\sqrt{x} \ge g(x) = \frac{16x(5x+3)}{25x^2+94x+9}, \quad \forall x\ge 0$$
since
$$x - \Big(\frac{22x(5x+6)}{25x^2 + 181x+36}\Big)^2 = \frac{x(x-1)^2(25x-36)^2}{(25x^2 + 181x+36)^2}$$
and
$$x - \Big(\frac{16x(5x+3)}{25x^2+94x+9}\Big)^2 = \frac{x(25x-9)^2(x-1)^2}{(25x^2 + 94x + 9)^2}.$$
It suffices to prove that $f(X) + f(Y) + g(Z) \ge 3$.
After clearing the denominators, it suffices to prove that $F(a,b,c)\ge 0$
where $F(a,b,c)$ is a homogeneous polynomial of degree $12$.
If $c = 0$, we have
\begin{align}
F(a,b,0) &= 64a^2b^2(a-b)^2\left(52500a^6+1729775a^5b+13521612a^4b^2\right.\\
&\quad \left. +27797474a^3b^3+13521612a^2b^4+1729775ab^5+52500b^6\right).
\end{align}
The inequality is true.
If $c > 0$, let $c = 1, \ b = 1+s, \ a = 1 + s + t$ for $s, t \ge 0$,
then $F(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. The inequality is true.
We are done.
|
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|
Another simple rule satisfied by the Fibonacci $n$-step constants? Given,
$$x^n(2-x)=1\tag1$$
for $n=2,3,4,\dots$
$$(x-1)(x^2-x-1)=0\\
(x-1)(x^3-x^2-x-1)=0\\
(x-1)(x^4-x^3-x^2-x-1)=0$$
the roots of which are the golden ratio, the tribonacci constant, the tetranacci constant, and so on.
Q: Is it true that for all integer $n>1$, if $y=x^{-1}$ then,
$$2y\,(1-y^{n-1})(1-y^{2n+2}) = (1-y^{n+1})^3\tag2$$
and
$$(1-y^{n-1})(1-y^{2n+2})^2 = (1-y^{n+1})^4\tag3$$
such that the RHS is a cube and fourth power, respectively?
P.S. See also this post for special relations for $n=2,3,5$.
|
For all integer $n$ define $a_n(x) := 1-2x^n+x^{n+1} = (x-1)(x^n - \dots -x^2 - x - 1).$
For $y=x^{-1}$ we have the two algebraic identities
$$ \frac{a_n(x)(1-x^{n+1})(1+2x+x^{n+1})}{x^{3n+3}} =
2y(1-y^{n-1})(1-y^{2n+2}) - (1-y^{n+1})^3,$$
$$ \frac{-a_n(x)(1 - x^{n + 1})^2(1 + 2 x^n + x^{n + 1})}{x^{5n+3}} =
(1-y^{n-1})(1-y^{2n+2})^2 - (1-y^{n+1})^4$$
which prove equations (2) and (3), respectively, if we assume $a_n(x)=0$.
|
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|
Given $ab+bc+ca=3abc$, prove $\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\leq 3$ $a, b, c$ are positive real numbers such that $ab+bc+ca=3abc$
Prove∶
$$\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\;\;\leq\; 3$$
|
We need to prove that $$\sum\limits_{cyc}\sqrt{\frac{a+b}{c(a^2+b^2)}}\leq\frac{ab+ac+bc}{abc}$$ or
$$\sum\limits_{cyc}\sqrt{\frac{ab(a+b)}{a^2+b^2}}\leq\frac{ab+ac+bc}{\sqrt{abc}}$$
By C-S $\left(\sum\limits_{cyc}\sqrt{\frac{ab(a+b)}{a^2+b^2}}\right)^2\leq(ab+ac+bc)\sum\limits_{cyc}\frac{a+b}{a^2+b^2}$.
Thus, it remains to prove that $\sum\limits_{cyc}\frac{a+b}{a^2+b^2}\leq\frac{ab+ac+bc}{abc}$ or $\sum\limits_{cyc}\frac{a+b}{a^2+b^2}\leq\frac{1}{2}\sum\limits_{cyc}\left(\frac{1}{a}+\frac{1}{b}\right)$,
which is just $\sum\limits_{cyc}\frac{(a+b)(a-b)^2}{2ab(a^2+b^2)}\geq0$.
Done!
|
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|
Using Divergence theorem to calculate flux
Let $W$ be the region bounded by the cylinder $x^2+y^2=4$, the plane $z=x+1$, and the $xy$-plane. Use the Divergence Theorem to compute the flux of $F = \langle z,x,y+z^2 \rangle$ through the boundary of $W$.
So far I've gotten to the point of computing div$(F)$ and integrating from $0$ to $x+1$ to obtain $$\iint_{D}(x+1)^2 dA.$$
My problem is finding the bounds of the domain which is the circle of radius $2$ centered at the origin. I understand I must use polar coordinates but since the circle is cut off by the line $x=-1$ I'm having trouble figuring out what the bounds for the radius should be. I think $\theta$ goes from $2\pi/3$ to $4\pi/3$ (somewhat guessing the bound for theta when the radius is cut off by the line $x = -1$)
|
If $x = r\cos\theta = -1$, then
\begin{align*}
r = -\frac{1}{\cos\theta} = -\sec\theta
\end{align*}
But since this means you will eventually have two compute two integrals, maybe polar coordinates is not the way to go.
\begin{align*}
\iint_D (x+1)^2\,dA &= \int_{-1}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(x+1)^2\,dy \,dx \\
&= 2 \int_{-1}^2 (x+1)^2\sqrt{4-x^2}\,dx \\
\end{align*}
Then substitute $x=2\sin\theta$.
|
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|
continued fraction for $\sqrt{n^2 − 1}$ Question:
Find the continued fraction for $\sqrt{n^2 − 1}$, where $n \ge 2$ is
an integer.
My attempt:
$n - 1 = \sqrt{n^2} - 1 \lt \sqrt{n^2 − 1} \lt \sqrt{n^2}$
So far, $[n-1; ]$
$\sqrt{n^2 − 1} = n - 1 + \frac{1}{x}$
$\to \frac{1}{x} = \sqrt{n^2 - 1} - (n-1) \to \frac{1}{x^2} = n^2-1-n^2-1+2n-2\times(n-1)\times\sqrt{n^2-1} = 2\times(n-1)\times(1-\sqrt{n^2-1})$
But, it doesn't really help to find continued fraction.
|
$$\begin{align*}\sqrt{n^2-1} &= (n-1) + \sqrt{n^2-1} - (n-1) \\& = (n-1) + \dfrac{1}{\dfrac{1}{\sqrt{n^2-1}-(n-1)}}\\ &=(n-1) + \dfrac{1}{\dfrac{\sqrt{n^2-1}+(n-1)}{2(n-1)}} \\ &= (n-1) + \dfrac{1}{1 + \dfrac{\sqrt{n^2-1}-(n-1)}{2(n-1)}} \\& = (n-1) + \dfrac{1}{1 + \dfrac{1}{\dfrac{2(n-1)}{\sqrt{n^2-1}-(n-1)}}} \\&=(n-1) + \dfrac{1}{1 + \dfrac{1}{\dfrac{2(n-1)(\sqrt{n^2-1} + (n-1))}{2(n-1)}}} \\&=(n-1) + \dfrac{1}{1 + \dfrac{1}{\sqrt{n^2-1} + (n-1)}}\end{align*}$$
So $\sqrt{n^2-1} = [n-1;\ \overline{1,\ 2(n-1)}]$
|
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|
If $a+b+c+d=0$ and $\{a,b,c,d\}\subset[-1,1]$ so $\sum\limits_{cyc}\sqrt{1+a+b^2}\geq4$ Let $\{a,b,c,d\}\subset[-1,1]$ such that $a+b+c+d=0$. Prove that:
$$\sqrt{1+a+b^2}+\sqrt{1+b+c^2}+\sqrt{1+c+d^2}+\sqrt{1+d+a^2}\geq4$$
I tried Holder and more, but without success.
|
Let $ {x,y,z}\in[-1, 1]$,and $x+y+z=0 $
Prove: $$\sqrt{1+x+y^2}+\sqrt{1+y+z^2}+\sqrt{1+z+x^2}\geq 3$$
Proof:
First we will show:
If $ab\geq 0, $ then:$\sqrt{1+a}+\sqrt{1+a}\geq 1+\sqrt{1+a+b}$
This is obvious！
Note that at least two of $ x+y^2, y+z^2 $ and $ z+x^2 $ have the same positive and negative values。Without loss of generality, we assume:$(x+y^2)(y+z^2)\geq 0$ then we have:$$\sqrt{1+x+y^2}+\sqrt{1+y+z^2}+\sqrt{1+z+x^2}$$ $$\geq 1+\sqrt{1+x+y^2+y+z^2}+\sqrt{1+z+x^2}$$ $$=1+\sqrt{(\sqrt{1-z+z^2})^2+y^2}+\sqrt{(\sqrt{1+z})^2+x^2}$$ $$\geq 1+\sqrt{(\sqrt{1-z+z^2}+\sqrt{1+z})^2+(x+y)^2}$$ $$ =1+\sqrt{(\sqrt{1-z+z^2}+\sqrt{1+z})^2+z^2}$$
So, just prove:$$(\sqrt{1-z+z^2}+\sqrt{1+z})^2+z^2\geq 4$$ $$\Longleftrightarrow 2z^2+2\sqrt{1+z^3}\geq 2$$ $$\Longleftrightarrow z^2(2-z)(z+1)\geq 0$$
This is clearly true because $|z|\leq 1$.Equal sign is established if and only if $x=y=z=0$
That's all,I hope it's works for you.
PS:Even if this method can be used, but the calculation is too complex that it is almost impossible to complete. The case of three variables is a good example, so I will not withdraw the answer.
|
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|
Prove by induction that $n^3 < 3^n$ The question is prove by induction that $n^3 < 3^n$ for all $n\ge4$.
The way I have been presented a solution is to consider:
$$\frac{(d+1)^3}{d^3} = (1 + \frac{1}{d})^3 \ge (1.25)^3 = (\frac{5}{4})^3 = \frac{125}{64} <2 < 3$$
Then using this $$(d+1)^3 = d^3 \times \frac{(d+1)^3}{d^3} < 3d^3 < 3 \times 3^d = 3^{d+1}$$ so we have shown the inductive step and hence skipping all the easy parts the above statement is true by induction.
However I don't find this method very intuitive or natural; is there another way to attack this problem?
The approach I wish to take involves starting from $$ 3^{d+1} = 3 \times3^d > 3d^3$$ but then I do not know how show further that $3d^3 > (d+1)^3 $ to complete the inductive step. I have looked around at the proofs related to showing that $2^n > n^2 $ inductively for $n \ge 5$ but cannot relate the proof for that case to this case.
Also, is there a more general method that could be used to solve, say $a^n > n^a $ for $ n \ge k $ for some $k\in \Bbb R$
|
The best solution that I can produce using some of the ideas from above is:
Proving the base case :
For $n=4$, we have $P(4): 3^4 > 4^3 \Leftrightarrow 81 > 64$ which is true.
Assume that the inequality is true for some d:
Assume $P(d): 3^d > d^3$.
Then show that $P(d+1)$ is true:
$P(d+1): 3^{d+1} > (d+1)^3 $
Starting from the RHS, $$(d+1)^3 = d^3 + 3d^2 + 3d +1 < 3^d + 3d^2 + 3d +1 $$ (using our inductive hypothesis)
Now if we can prove $3d^2 + 3d +1 < 3^d$ then we will be done.
So attempting to do this using induction again;
First if we prove that $6n+6 < 3^n$, we will be able to use this result later.
Proving the base case:
For $n=4$, $$6(4)+6 <3^4 \Leftrightarrow 30 < 81 $$ which is true.
Assume that the inequality is true for some j:
Assume $P(j): 6j+6 <3^j$
Then show that $P(j+1)$ is true:
$P(j+1): 6(j+1) +6 = 6j +6 +6 < 3^j+6 < 2(3^j) <3^{j+1}$
as $j \ge 4$ so 3^j > 6 for all j.
Thus, $6n+6 < 3^n$ is true for all $n \ge 4$.
Next, proving the base case for $3n^2 + 3n +1 < 3^n$:
For $n=4$, $$3(4^2) +3(4)+1 <3^4 \Leftrightarrow 61 < 81 $$ which is true.
Assume that the inequality is true for some k:
Assume $P(k): 3k^2 +3k +1 <3^k$
Then show that $P(k+1)$ is true:
$$P(k+1): 3(k+1)^2 +3(k+1) +1 = 3k^2 +3k +1 +(6k+6) < 3^k + 6k+6 < 2(3^k) <3^{k+1}$$
Therefore $3n^2 + 3n +1 < 3^n$ is true for all $n \ge 4$.
Going back to the question now,
$(d+1)^3 <3^d + 3d^2 +3d+1 <3^d +3^d <3^{d+1}$
So $3^n < n^3 $ for all $n \ge 4$ by mathematical induction!
Although the working out for this approach is very long winded I find that it is more logical to a beginner at induction.
|
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prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$ let $a,b,c\ge 0$ and such $a+b+c=3$ show that
$$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
My research:It seem use Holder inequality,so
$$(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3$$
Use AM-GM
$$abc\le\dfrac{(a+b+c)^3}{27}=1$$
what? then I think this method is wrong
|
BY Jensen inequality with $\phi\left(t\right)=\ln(t^3+1)$
$$\phi\left(\frac13\sum_{i=1}^{3}x_i\right)\le \frac13\sum_{i=1}^{3}\phi\left(x_i\right) $$
since $a+b+c= 3$ we get
$$\ln(2)=\phi\left(1\right)\le \frac13\left(\ln(a^3+1)+\ln(b^3+1)+\ln(c^3+1)\right) $$
hence $$\ln8= 3\ln(2)\le \ln \left((a^3+1)(b^3+1)(c^3+1)\right)$$
that is $$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
|
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What is the remainder left after dividing $1! + 2! + 3! + ... + 100!$ by $5$? I have this question as a homework.
What is the remainder left after dividing $1! + 2! + 3! + \cdots + 100!$ by $5$?
I tried this: I noticed that every $n! \equiv 0 \pmod{5}$ for every $n\geq 5$.
For $n < 5$:
$$\begin{align*}
1! &\equiv 1 \pmod{5} \\
2! &\equiv 2 \pmod{5} \\
3! &\equiv 1 \pmod{5} \\
4! &\equiv 4 \pmod{5} \\
\end{align*}$$
So $1! + 2! + \cdots + 100! \equiv 8 \equiv 3 \pmod{5}$.
Therefore the remainder is $3$.
Am I thinking properly?
Thanks a lot.
|
You are right.
Notice that the unit digit of the sum is $3$, because all the terms in the sum end up with a zero from $5!$ and hence to find the unit digit, we just need to find the unit digit of the sum $1!+2!+3!+4!=33$ which is $3$.
So, when the sum $1! + 2! + 3! + \cdots + 100!$ is divided by $5$, it leaves a remainder $3$.
|
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Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
Let $a$ and $b$ be integers. Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
I saw that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $(a^2+ab+b^2) = (a+b)^2-ab$. How can we use the fact that $10 \mid (a^2+ab+b^2)$ to solve this question?
|
Here's a more general result:
Let $\Psi_n(x,y)$ denote the $n$th homogenized cyclotomic polynomial. If $x,y\in\mathbb Z$, every prime divisor of $\Psi_n(x,y)$ is either:
*
*$1\pmod n$
*a prime divisor of $n$
*a prime divisor of both $x$ and $y$
This is a simple corollary of the same result for the usual cyclotomic polynomial $\Phi_n(x)=\Psi_n(x,1)$ (where the third case doesn't occur).
In our case $2\mid\Psi_3(a,b)=a^2+ab+b^2$. As $3\nmid2$ and $2\not\equiv1\pmod3$, we have $2\mid a,b$. Same argument for $5$.
Entirely self-contained proof of that result in the case of prime $n$ (here $n=3$):
We then have $\Psi_n(x,y)=x^{n-1}+xy^{n-2}+\cdots+y^{n-1}$. Let $p\mid\Psi_n(x,y)\mid x^n-y^n$. If $p\mid x$ or $y$, we are done. If not, $y$ is invertible mod $p$, and $(x/y)^n\equiv1$. Let $k$ be the order of $x/y$ mod $p$. $k\mid n$.
If $k=1$, $x\equiv y$ and $p\mid x^{n-1}+xy^{n-2}+\cdots+y^{n-1}\equiv nx^{n-1}$, so $p\mid n$, so $p=n$.
If $k=n$, $n\mid p-1$ by Fermat's Little Theorem.
|
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A fair coin is tossed repeatedly and independently until two consecutive heads or two consecutive tails appear? A fair coin is tossed repeatedly and independently until two consecutive heads or two consecutive tails appear . What is the probability of the number of tosses ?
I tried it as :
For success, either we end up getting
*
*HH,THH,HTHH,............
*TT,HTT,THTT,.............
Then, Add up both the successes. Am I right with my understanding ?
|
While other answers are correct, they don't explain the solution.
I am assuming your question is about the expected number of coin tosses. Let $X$ be the discreet random variable that is the number of throws until two of the same coin are observed. Your question is asking about the expected value of $X$ i.e. $E(X)$.
It is evident $P(X=1)= 0$.
So what is $P(X=2)$? In other words, what is the probability the game terminates after two tosses? Well whatever the first coin was, the probability of the second toss being the same is $1 \over 2$ So $P(X=2) = \frac{1}{2}$.
Similarly consider $P(X=3)$ which is the probability the game terminates after $3$ tosses. Whatever is tossed first, there is a $1 \over 2$ chance that the second throw is different and then a $1 \over 2$ chance the third throw is the same as the second. So the total probability is $P(X=3) = \frac{1}{ 2^2}$.
Continuing the pattern, we get that the probability of the game terminating after $n$ throws, for $n \geq 2$, is $P(X=n) = \frac{1}{2^{n-1}}$. Alternatively, the probability of the game terminating after $n+1$ throws, for $n \geq 1$, is $P(X=n+1) = \frac{1}{2^{n}}$.
Thus the expected value $E(X)$ is given by:
$$
\begin{align}
E(X) &= \sum_{n = 1}^{\infty}(n+1) \cdot P(X = n+1) \\
&= \sum_{n = 1}^{\infty}(n+1) \cdot \frac{1}{2^{n}} \\
&= \sum_{n = 1}^{\infty}\frac{n}{2^n} + \sum_{n = 1}^{\infty}\frac{1}{2^{n}}
\end{align}
$$
We know the right term: $\sum_{n = 1}^{\infty}\frac{1}{2^{n}}$ is a geometric progression and is given by:
$$
\begin{align}
\sum_{n = 1}^{\infty}\frac{1}{2^{n}} &= \frac{1}{2} \cdot \frac{1}{1+ \frac{1}{2}} \\
&= 1
\end{align}
$$
The other sum, $\sum_{n = 1}^{\infty}\frac{n}{2^n}$ is slightly harder but we can do a bit of trickery. Let us call the sum $S$ and consider $\frac{S}{2}$:
$$
\begin{align}
S &= \sum_{n = 1}^{\infty}\frac{n}{2^n} \\
\frac{S}{2} &= \sum_{n = 1}^{\infty}\frac{n}{2^{n+1}} = \sum_{n = 1}^{\infty}\frac{n-1}{2^n} \\
\end{align}
$$
Note how we shifted the indices in the second line for easier manipulation. Subtracting the two lines gives:
$$
\begin{align}
S - \frac{S}{2} = \frac{S}{2} &= \sum_{n = 1}^{\infty}\frac{n}{2^n} - \sum_{n = 1}^{\infty}\frac{n-1}{2^n} \\
&= \sum_{n = 1}^{\infty}\frac{n}{2^n} - \frac{n-1}{2^n} \\
&= \sum_{n = 1}^{\infty}\frac{1}{2^n} \\
&= 1
\end{align}
$$
and so $\frac{S}{2} = 1$ giving $S = 2$.
So in total:
$$
\begin{align}
E(X) &= \sum_{n = 1}^{\infty}\frac{n}{2^n} + \sum_{n = 1}^{\infty}\frac{1}{2^{n}} \\
&= 2 + 1 = 3\\
\end{align}
$$
Addendum (thanks to @bof)
An alternative, and excellent explanation offered by @bof is that:
$$
E(X)=\sum_{k=1}^\infty P(X\ge k)=1+1+\frac12+\frac14+\frac18+\cdots=3.
$$
To see why this is, consider the following (quoted from bof's comment):
Since $X$ takes only positive integer values,
$$
\begin{array}
\ P(X\ge1)= &P(X=1)&+ & P(X=2) & + & P(X=3)+\cdots \\
P(X\ge2)= & & & P(X=2) & + & P(X=3)+\cdots \\
P(X\ge3)= & & & & & P(X=3)+\cdots \\
\end{array}
$$
Adding by columns:
$$\sum_{k=1}^\infty P(X\ge k)=P(X=1)+2P(X=2)+3P(X=3)+\cdots=E(X).$$
In other words: Let the "indicator variable" $X_k=1$ if the kth toss is needed (no HH or TT in first $k-1$ tosses), $X_k=0$ otherwise; then $X=\sum_{k=1}^\infty X_k$ so
$$
E(X)=\sum_{k=1}^\infty E(X_k)=\sum_{k=1}^\infty P(X\ge k).
$$
|
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The sum of $\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}+\cdots \cdots $
The sum of $$\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}-\cdots \cdots \cdots $$
$\bf{My\; Try::}$ We can write above sum as $$\sum^{n}_{k=0}(-1)^k\binom{n}{k}\binom{3n-3k}{2n-3k} = \sum^{n}_{k=0}(-1)^k\binom{n}{k}\binom{3n-3k}{n}$$
So sum $$ = \sum^{n}_{k=0}(-1)^k\cdot \frac{n!}{k!\cdot (n-k)!}\times \frac{(3n-2k)!}{n!\cdot (2n-3k)!} = \sum^{n}_{k=0}(-1)^k \cdot \frac{(3n-2k)!}{k!\cdot (n-k)!\cdot (2n-3k)!}$$
Now i did not understand how can i solve it after that , help Required, Thanks
|
\begin{align*}
&\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}-\cdots \\
&=\binom{n}{0}\binom{3n}{n}-\binom{n}{1}\binom{3n-3}{n}+\binom{n}{2}\binom{3n-6}{n}-\cdots
\end{align*}
The right hand side is the coefficient of $x^n$ in
\begin{align*}
&\binom{n}{0}(1+x)^{3n} - \binom{n}{1}(1+x)^{3n-3} + \cdots \\
&= (1+x)^{3n}\left(1- \frac{1}{(1+x)^3}\right)^n \\
&= (1+x)^{3n-3}(3x+3x^2+x^3)^n \\
&= x^n(1+x)^{3n-3}(3+3x+x^2)^n
\end{align*}
and hence equals $3^n$.
|
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|
Choosing one type of ball without replacement. Suppose I have $9$ balls, among which $3$ are green and $6$ are red. What is the probability that a ball randomly chosen is green?
It is $\dfrac{3}{9}=\dfrac{1}{3}$.
If three balls are randomly chosen without replacement, then what is the probability that the three balls are green?
Is it $\dfrac{3}{9}\times\left\{\dfrac{2}{8}+\dfrac{3}{8}\right\}\times\left\{\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}\right\}=\dfrac{90}{504}$?
But in hypergeometric distribution formula, it is
$$f(X=3)=\dfrac{\binom{3}{3}\binom{9-3}{3-3}}{\binom{9}{3}}=\dfrac{1}{84}.$$
|
a) Your first answer is correct.
b) All 3 are green without replacement
= $\dfrac39 \cdot \dfrac28 \cdot \dfrac17$ = $\dfrac1{84}$
c) 2 balls are green out of 3 without replacement.
= $ \left(\dfrac{3}{9} \cdot \dfrac{2}{8} \cdot \dfrac{6}{7} \right ) + \left(\dfrac{3}{9} \cdot \dfrac{6}{8} \cdot \dfrac{2}{7} \right ) + \left(\dfrac{6}{9} \cdot \dfrac{3}{8} \cdot \dfrac{2}{7} \right )$
= $3 \cdot \left(\dfrac{3 \cdot 2 \cdot 6}{9 \cdot 8 \cdot 7}\right)$ = $\dfrac{3}{14}$
Or
= $ \binom{3}{2} \cdot \left(\frac{3}{9} \cdot \dfrac{2}{8} \cdot \dfrac{6}{7} \right )$ = $\frac{3}{14}$
Here $\binom{3}{2}$ because 2 greens can come to any place out of 3.
|
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|
Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?
|
$x^2 - 2^2 = 0$
$(x+2)(x-2)=0$
$x+2=0$ or $x-2=0$
$\sqrt4 = 2$
$x-\sqrt4=0$
$x=2$
Why can't the square root of $4$ be $-2$ instead of $2$, if $-2$ times $-2$ also equals $4$?
|
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|
System of $3$ nonlinear equations
Find positive solutions to the following:
\begin{align*}
x^2+y^2+xy=&1\\
y^2+z^2+yz=&3\\
z^2+x^2+xz=&4
\end{align*}
I simplified and got $x+y+z=\sqrt{7}$ and $x^2+y^2+xy=1$. How do I continue?
|
All positive real solutions are given by
$$
(x,y,z)=\frac{1}{\sqrt{7}}(2,1,4).
$$
I obtained the solutions by finding one linear equation (note that $x+y+z= 7$ does not hold), namely $2y + z - 3x=0$.
Then replacing $z=3x-2y$ yields two quadratic equations, which are easy to solve with the formula for quadratic equations. This gives all solutions, namely
$$
(x,y,z)=\pm\frac{1}{\sqrt{7}}(2,1,4), \pm (0,-1,2).
$$
|
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|
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.
Any help would be greatly appreciated.
|
Many questions with
sum or difference of square roots
can be solved with
conjugating.
So,
if $s = \sqrt{4 + 2\sqrt{3}} - \sqrt{3}$,
and $t = \sqrt{4 + 2\sqrt{3}} + \sqrt{3}$,
$\begin{array}\\
st
&=(\sqrt{4 + 2\sqrt{3}} - \sqrt{3})(\sqrt{4 + 2\sqrt{3}} + \sqrt{3})\\
&=4 + 2\sqrt{3}-3\\
&=1 + 2\sqrt{3}\\
\end{array}
$
Since
$t-s = 2\sqrt{3}$,
$st = 1+t-s$
or
$s(t+1) = 1+t$,
and, by magic,
we get
$s = 1$
(unless $t+1 = 0$
which it does not since
$t > 0$).
I am surprised that this worked so well.
I will suppress my need to generalize
and submit this as is.
|
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|
Unexpected Proofs Using Generating Functions I recently came across this beautiful proof by Erdős that uses generating functions in a unique way:
Let $S = \{a_1, \cdots, a_n \}$ be a finite set of positive integers such that no two subsets of $S$ have the same sum. Prove that $$\sum_{i=1}^n \frac{1}{a_i} < 2.$$
Question: Are there any more examples of surprising or unexpected proofs using generating functions that this community is aware of?
(Please refrain from posting answers that are widely known such as change making, closed form for Fibonacci, etc.)
The proof of the above theorem:
Proof:
Suppose $0< x < 1$. We have
$$\prod_{i=1}^n (1 + x^{a_i}) < \sum_{i = 0}^{\infty} x^i = \frac{1}{1-x}.$$ Thus,
$$\begin{align*}
\sum_{i=1}^n \log(1+x^{a_i}) &< - \log(1-x) \\
\sum_{i=1}^n \int_0^1 \frac{\log(1+x^{a_i})}{x} \ dx &< - \int_0^1 \frac{\log(1-x)}x \ dx .
\end{align*}$$
Putting $x^{a_i} = y$, we obtain
$$\begin{align*}
\sum_{i=1}^{n} \frac{1}{a_i} \int_0^1 \frac{\log(1+y)}{y} \ dy < - \int_0^1 \frac{\log(1-x)}{x} \ dx
\end{align*}$$ i.e.,
$$\sum_{i=1}^n \frac{1}{a_i} \left( \frac{\pi^2}{12} \right) < \frac{\pi^2}6.$$
Thus, $\sum_{i=1}^n \frac{1}{a_i} < 2$ and the theorem is proved.
|
Sicherman dice are the only pair of 6-sided dice that are not normal dice, bear only positive integers, and have the same probability distribution for the sum as normal dice.
The faces on the dice are numbered 1, 2, 2, 3, 3, 4 and 1, 3, 4, 5, 6, 8.
(Source: Wikipedia article on Sicherman dice)
We can prove this fact using generating functions.
Let the non-standard dice $A$ and $B$ have faces $(a_1,a_2,a_3,a_4,a_5,a_6)$ and $(b_1,b_2,b_3,b_4,b_5,b_6)$.
Let $a(x) = x^{a_1}+x^{a_2}+x^{a_3}+x^{a_4}+x^{a_5}+x^{a_6}$ and $b(x) = x^{b_1}+x^{b_2}+x^{b_3}+x^{b_4}+x^{b_5}+x^{b_6}$ be the generating functions for the number rolled on dice $A$ and dice $B$ respectively.
When the product $$a(x)b(x) = (x^{a_1}+x^{a_2}+x^{a_3}+x^{a_4}+x^{a_5}+x^{a_6})(x^{b_1}+x^{b_2}+x^{b_3}+x^{b_4}+x^{b_5}+x^{b_6})$$ is expanded as a polynomial, the $x^n$ coefficient is the number of ways these non-standard dice can have a sum of $n$. If the sum of the non-standard dice has the same distribution as two standard dice, then the above product must be equal to the product of the generating functions for two standard dice, i.e.
\begin{align}
a(x)b(x) &= (x+x^2+x^3+x^4+x^5+x^6)^2
\\
&= [x(1+x+x^2+x^3+x^4+x^5)]^2
\\
&= [x(1+x^3)(1+x+x^2)]^2
\\
&= [x(1+x)(1-x+x^2)(1+x+x^2)]^2.
\end{align}
We just need to figure out how to distribute the factors between the polynomials $a(x)$ and $b(x)$.
Since each dice must have only positive integer faces, $x$ divides both $a(x)$ and $b(x)$. Hence, $a(x)$ and $b(x)$ must each get one of the two $x$ factors.
Since dice $A$ and $B$ both have $6$ faces, $a(1) = b(1) = 6$. Notice that $(1+x)\left.\right|_{x = 1} = 2$, $(1-x+x^2)\left.\right|_{x = 1} = 1$, and $(1+x+x^2)\left.\right|_{x = 1} = 3$. Clearly, $a(x)$ and $b(x)$ must each get one of the two $1+x$ factors and one of the two $1+x+x^2$ factors.
This leaves only the two $1-x+x^2$ factors. If we distribute one to each of $a(x)$ and $b(x)$, then we will have $a(x) = b(x)$, and we'll end up with the standard dice. Thus, we have to give both $1-x+x^2$ factors to the same dice generating function (WLOG $b(x)$).
This gives us
\begin{align}
a(x) &= x(1+x)(1+x+x^2)
\\
&= x+2x^2+2x^3+x^4
\\
\left.\right.
\\
b(x) &= x(1+x)(1+x+x^2)(1-x+x^2)^2
\\
&= x+x^3+x^4+x^5+x^6+x^8
\end{align}
Therefore, the only pair of 6-sided dice that are not normal dice, bear only positive integers, and have the same probability distribution for the sum as normal dice have faces numbered $(1,2,2,3,3,4)$ and $(1,3,4,5,6,8)$.
|
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|
Simple limit. Used L'Hôpital's rule. Didn't work. $\lim_{x\rightarrow 0}f(x)$
$f(x)=\frac{\exp (\arcsin \left (x \right ))-\exp (\sin \left (x \right ))}{\exp (\arctan \left (x \right ))-\exp (\tan \left (x \right ))}$
I tried the L'Hôpital's rule as mentioned in the title and replaced ($\arcsin x$,$\arctan x$,$\sin x$,$\tan x$) with $\sim_{0}$ $x$ but still have an indeterminate form. Any hints please ?
|
We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{e^{\arcsin x} - e^{\sin x}}{e^{\arctan x} - e^{\tan x}}\notag\\
&= \lim_{x \to 0}e^{\sin x - \tan x}\cdot\frac{e^{\arcsin x - \sin x} - 1}{e^{\arctan x - \tan x} - 1}\notag\\
&= \lim_{x \to 0}\frac{e^{\arcsin x - \sin x} - 1}{\arcsin x - \sin x}\cdot\frac{\arcsin x - \sin x}{\arctan x - \tan x}\cdot\frac{\arctan x - \tan x}{e^{\arctan x - \tan x} - 1}\notag\\
&= \lim_{x \to 0}\frac{\arcsin x - \sin x}{\arctan x - \tan x}\notag\\
&= \lim_{x \to 0}\dfrac{\left(x + \dfrac{x^{3}}{6} + o(x^{3})\right) - \left(x - \dfrac{x^{3}}{6} + o(x^{3})\right)}{\left(x - \dfrac{x^{3}}{3} + o(x^{3})\right) - \left(x + \dfrac{x^{3}}{3} + o(x^{3})\right)}\notag\\
&= -\frac{1}{2}\notag
\end{align}
|
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|
Linear transformation matrix of vector space $\mathbb R^{2x2}$
Select a basis B of a vector space $\mathbb R^{2x2}$ and for linear transformation $f:\mathbb R^{2x2}→\mathbb R^{2x2}$ given by $f(X) = \begin{pmatrix}1 & 0 \\ 1 & 0 \end{pmatrix}X+\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}X^T$ compute the matrix relative to the base B.
I selected canonical basis $B=\{\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix},\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}\}$
Then I computed transformation of each vector in the basis according to the task.
Got $f(B)=\{\begin{pmatrix}2 & 0 \\ 2 & 0 \end{pmatrix},\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix}\}$, if I'm not wrong.
What should I do now?
|
Speedding: your calculations are correct. Another way. If $X=\begin{pmatrix}{x_1}&{x_2}\\{x_3}&{x_4}\end{pmatrix}\in\mathbb{R}^{2\times 2}$ then, $$f(X)=\begin{pmatrix}1 & 0 \\ 1 & 0 \end{pmatrix}X+\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}X^T=\ldots=\begin{pmatrix}{2x_1+x_2}&{x_2+x_3+x_4}\\{2x_1+x_2}&{x_2+x_3+x_4}\end{pmatrix}.$$ But for $M=\begin{pmatrix}{a}&{b}\\{c}&{d}\end{pmatrix}$, its coordinates on $B$ are $(a,b,c,d)^T$ so, we can write $$f\begin{pmatrix}{x_1}\\{x_2}\\{x_3}\\ x_4\end{pmatrix}=\begin{pmatrix}{2x_1+x_2}\\{x_2+x_3+x_4}\\{2x_1+x_2}\\ x_2+x_3+x_4\end{pmatrix}=\underbrace{\begin{pmatrix}{2}&{1}&{0}&0\\{0}&{1}&{1}&1\\{2}&{1}&{0}&0\\{0}&{1}&{1}&1\end{pmatrix}}_{A_B^B}\begin{pmatrix}{x_1}\\{x_2}\\{x_3}\\ x_4\end{pmatrix}.$$
|
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|
Want to show $e_1 (x) := \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu} = \frac{1}{x} - \sum_{m=1}^{\infty} \gamma_m x^{2m-1}$ I want to prove
\begin{align}
e_1 (x) := \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu}
\end{align}
for $\mu\neq 0$, $|x|<1$,
\begin{align}
e_1(x) = \frac{1}{x} - \sum_{m=1}^{\infty} \gamma_m x^{**2**m-1}
\end{align}
where $\gamma_m = 2 \sum_{\mu=1}^{\infty} \mu^{-2m}$.
This comes from A. Weil, "Elliptic functions according to Eisenstate and Kronecker"
The factor $2$ is missing in the original textbook. From its general expression for $e_n(x) $ i guess it is a typo.
|
My trial is as follows
\begin{align}
e_1 (x) &:= \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu} \\
&= \frac{1}{x} + \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \frac{2x}{x^2-\mu^2} \\
& = \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \frac{2x}{\mu^2-x^2} \\
&= \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \frac{2x}{\mu^2} \frac{1}{1-\frac{x^2}{\mu^2}} \\
&= \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \frac{2x}{\mu^2}
\sum_{m=1}^{\infty} \left( \frac{x^2}{\mu^2} \right)^{m-1} \\
& = \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \sum_{m=1}^{\infty} \frac{2}{\mu^{2m}} x^{2m-1} \\
& = \frac{1}{x} - \sum_{m=1}^{\infty} 2\zeta(2m) x^{2m-1} \\
\end{align}
Since $\mu$ is integer, thus $\mu\geq 1$, thus $\frac{x}{\mu} < 1$
|
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|
For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$? For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$?
1.$1$
2.$2$
3.$3$
4.$4$
5.$5$
My attempt:It's clear that it is true for $n=2$.Because:
$(a^3-1)(a^2-1) \ge 0$
Is true because $a^3-1$ and $a^2-1$ are both negative or both positive.But there is a problem:How can we make sure it doesn't hold for any other $n$?
|
Note that equality always holds for $a = 1$.
$
a^5 + 1 \geq a^3 + a^n \\
\implies a^5 - a^3 \geq a^n - 1 \\
\implies a^3(a + 1)(a - 1) \geq a^n - 1
$
Case 1: $a > 1$. Then,
$$
a^4 + a^3 \geq 1 + a + a^2 + \cdots + a^{n-1}
$$
Case 2: $0 < a < 1$. Then,
$$
a^4 + a^3 \leq 1 + a + a^2 + \cdots + a^{n-1}
$$
Now, both cases hold true for $n = 2$ and certainly for $a = 1$.
Case 2 doesn't hold for $n=1$:
$$
a^4+a^3 \leq 1.
$$
Let $a = 1/x$ where $x>1$. Then, the inequality becomes $f(x) = x^4 - x - 1 \geq 0$. Now, $f(x)$ is a curve opening upwards which is negative at $x = 1$. Hence, $f(x)$ has a positive real root, say $\alpha > 1$. This means that $f(x) < 0$ in the interval $(1, \alpha)$, i.e. the inequality doesn't hold.
Case 1 doesn't hold for $n=3$:
$
a^4+a^3 \geq 1 + a + a^2 \\
\implies a^4 + a^3 + a^2 + a + 1 \geq 2(1 + a + a^2) \\
\implies a^5 - 1 \geq 2(a^3 - 1) \\
\implies a^5 - 2a^3 + 1 \geq 0.
$
Now, $p(a) = a^5 - 2a^3 + 1$ has a minima at $a_0 = \sqrt{6/5}$, where $p(a_0) = -0.05162... < 0$.
Case 1 doesn't hold for $n = 4$:
$
a^4 + a^3 > 1 + a + a^2 + a^3 \\
\implies q(a) = a^3 - a^2 - 1 \geq 0.
$
Now, $q(a) \rightarrow \infty$ as $a \rightarrow \infty$, and $q(1) = -1$. Hence, if a root of $q(a)$ be $a_0 > 1$, then $q(a)$ is negative in the interval $(1, a_0)$.
Case 1 clearly doesn't hold for $n>4$. Thus the inequality holds for the entire $\mathbb{R}^+$ only for $n=2$.
|
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|
Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$. Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$.
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
$a=2^{x_{1}}3^{y_{1}}5^{z_{1}}7^{w_{1}}11^{t_{1}}$
$b=2^{x_{2}}3^{y_{2}}5^{z_{2}}7^{w_{2}}11^{t_{2}}$
$c=2^{x_{3}}3^{y_{3}}5^{z_{3}}7^{w_{3}}11^{t_{3}}$
where $x_{1}+x_{2}+x_{3}=1$; $x_{1},x_{2},x_{3}$ being non-negative integers.
Number of solutions of this equation clearly $3$ and similarly for other variables as well.
So,number of solutions $=3^5$.
But answer given is $40$ .
Am I misinterpreting the question
|
$5=3+1+1, 1+2+2.$ So making sets without integer $1,$ first possibility, $10$ ways. Second possibility, $15$ ways. If integer $1$ is taken once $, 5=3+2, 4+1.$
First possibility $10$ ways, second possibility $5$ ways. So total $40$ ways.
|
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|
Identifying the digits of $37 \cdot aaaa\ldots a$. With a calculator, I have noticed that the integer $37$ multiplied with some particular numbers yields numbers with some structures.
For instance, let $aaaa\ldots a$ be a natural number of $n$ identical digits. Then, $ 37 \cdot aaaa\ldots a$ is a number with $n+1$ or $n+2$ digits of the form
$$\underbrace{4\cdot a}_{1\text{ or }2}~ \underbrace{aaa\ldots a}_{n-2} ~\underbrace{7\cdot a}_{2^{*}}.$$
$*$ if $a=1$, then $7 \cdot a$ is the sequence of digits $07$.
I am wondering whether the result above can be proven using some number-theory tool.
Thanks in advance!
|
Observe the patterns in the written calculation of these products
$$a=1\to\begin{matrix}3&7\\&3&7\\&&3&7\\&&&3&7\end{matrix}$$
$$a=2\to\begin{matrix}7&4\\&7&4\\&&7&4\\&&&7&4\end{matrix}$$
$$a=3\to\begin{matrix}1&1&1\\&1&1&1\\&&1&1&1\\&&&1&1&1\end{matrix}$$
$$a=4\to\begin{matrix}1&4&8\\&1&4&8\\&&1&4&8\\&&&1&4&8\end{matrix}$$
$$a=5\to\begin{matrix}1&8&5\\&1&8&5\\&&1&8&5\\&&&1&8&5\end{matrix}$$
$$\cdots$$
$$a=9\to\begin{matrix}3&3&3\\&3&3&3\\&&3&3&3\\&&&3&3&3\end{matrix}$$
A digit of the product, in a complete column, is the sum of the digits of $37\cdot a$ plus a possible carry from the previous column, i.e. the sum of the digits of the sum of the digits of $37\cdot a$.
The initial and final digit pairs correspond to partial sums.
We have
$$
37\cdot1=37\to10\to1,\\
37\cdot2=74\to11\to2,\\
37\cdot3=111\to3\to3,\\
37\cdot4=148\to13\to4,\\
37\cdot5=185\to14\to5,\\
\cdots\\
37\cdot9=333\to9\to9.$$
The phenomenon is explained by the fact that $37=4\cdot9+1\to10\to1$, and this occurs for all factors of the form $k\cdot9+1$. For example,
$$73\cdot66666666666666=4866666666666618.$$
|
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|
When is the fraction $\frac{n+7}{2n+7}$ the square of a rational number?
Find all positive integers $n$ such that the fraction $\dfrac{n+7}{2n+7}$ is the square of a rational number.
The answer says $n = 9,57,477$, but how do we prove this? I wrote $\dfrac{n+7}{2n+7} = \dfrac{a^2}{b^2}$ for some $a,b \in \mathbb{Z^+}$ and $\gcd(a,b) = 1$. We know that $\gcd(n+7,2n+7) = \gcd(n+7,7) = 1,7$ and so $n+7$ and $2n+7$ must both be perfect squares or their greatest common divisor is $7$. How do we find all values for which this is the case?
|
$$\gcd(n+7,2n+7)=\gcd(n+7,n)=\gcd(7,n)\in\{1,7\}$$
If the $\gcd=1$, then $n+7,2n+7$ are both squares:
$$n+7=a^2,2n+7=b^2\implies 2a^2-b^2=7$$
If the $\gcd=7$, then $\frac{n+7}{7},\frac{2n+7}{7}$ are both integer squares. Write $n=7m$:
$$m+1=a^2,2m+1=b^2\implies2a^2-b^2=1$$
Finding exact solutions from here requires study of Pell's equation, as indicated in the comments.
*
*The first equation has solutions starting with
$\{(2,1),(8,11),\cdots\}$
*The second equation has solutions starting with
$\{(1,1),(5,7),(41,29),\cdots\}$.
These can be generated by looking at $(3+\sqrt2)(1+\sqrt2)^{2k},(1+\sqrt2)^{2k+1}$ respectively for integer $k$. Noting that $\frac{1}{1+\sqrt2}=-(1-\sqrt2)$ tells us that for the second of these sequences, we cann afford to simply look at $k\ge0$.
In the first case, given $2a^2-b^2=7$, note that taking $n=b^2-a^2$ gives $a^2=n+7,b^2=2n+7$ as desired, i.e. the fraction in question is a square. A slight adaptation of this resolves the second case.
Clarification: When I say "these can be generated by looking at $(a+b\sqrt2)^k$", I refer to the fact that by writing $(a+b\sqrt2)^k=A_k+B_k\sqrt2$ for integers $A_k,B_k$, we will have $2B_k^2-A_k^2=1$ or $7$, depending on this sequence. This method makes use of techniques from algebraic number theory, where we learn that the function $N:a+b\sqrt2\to a^2-2b^2$ is multiplicative.
|
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|
Integral of $2x^2 \sec^2{x} \tan{x}$ I've been trying for a while to find $\int{( 2x^2 \sec^2{x} \tan{x} )} dx$, using integration by parts.
I always end up getting a more complicated integral in the second part of the equation.
For example:
$$ \int{( 2x^2 \sec^2{x} \tan{x} )} dx =
\\ 2x^2 \tan^2x - \int{\tan{x} \cdot \frac{d}{dx}(2x^2 \tan{x})}
\\ \frac{d}{dx}(2x^2 \tan{x})=4x\tan{x} + 2x\sec^2{x} \rightarrow
\\ 2x^2 \tan^2x - \int{4x \tan^2{x}+2x\tan{x}\sec^2{x}}
$$
I've tried integrating with different value for $u$ and $v$, such as:
$$ 1:( 2x^2 \sec^2{x} \tan{x} ),
\\ \tan{x} : 2x^2 \sec^2{x},
\\ 2x^2: \sec^2{x} \tan{x},
\\ \sin{x}: 2x^2 \sec^3{x} $$ etc, however, haven't succeeded.
|
We have,
$\int \sec ^{2}x\tan {x}dx$
Put v = $\sec{x}$
Then dv = 2$\sec{x}\tan{x}$
So $\int (v) dv=\frac{1}{2}v^{2}=\frac{1}{2}\sec ^{2}x.$
Now use this result in question,
2$\int{(x^2 \sec^2{x} \tan{x} )} dx$
Integrating by parts, integrate $\sec^2x \tan{x}$ and differentiate $x^2$
We have,
2$\left[\frac{1}{2}x^{2}\sec ^{2}x-\int x\sec
^{2}x\,dx \right]$
Now again integration by parts on $ x\sec
^{2}x\,dx $ by integrating $\sec
^{2}x$ and differentiate x.
You got an answer.
|
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|
Integer solutions to nonlinear system of equations $(x+1)^2+y^2 = (x+2)^2+z^2$ and $(x+2)^2+z^2 = (x+3)^2+w^2$
Do there exist integers $x,y,z,w$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2?\end{align*}
I was thinking about trying to show by contradiction that no such integers exist. The first equation gives $y^2 = 2x+3+z^2$ while the second gives $z^2 = 2x+5+w^2$. How can we find a contradiction from here?
|
We need to find (or prove that it does not exist) an odd number $a$ that is the difference of two squares:
$$y^2-z^2=a$$
This is quite easy. We only need to write $a$ as the product of two different numbers: $a=mn$. If $m>n$, say, then the easy possbility is that $m=y+z$ and $n=y-z$. We only need to solve for $y$ and $z$ and we are done. The values for $y$ and $z$ are integers because $m$ and $n$ have the same parity (both are odd).
Now the hard step is that in addition to the former equation we have to find another square $w^2<z^2$ such that $z^2-w^2=a+2$.
In addition, the solution for $z$ in both cases has to be the same, of course.
Then:
*
*$y=(m+n)/2$
*$z=(m-n)/2=(u+v)/2$
*$w=(u-v)/2$
*$mn-uv=-2$
Since $u=m-n-v$ we have
$$mn-uv=mn-(m-n-v)v=v^2-(m-n)v+mn=-2$$
If we see this as a second degree equation on $v$, the discriminant (which must be a square) is
$$(m-n)^2-4mn-8=m^2-6mn+n^2-8=(m-3n)^2-8(n^2+1)$$
Let $s=m-3n$. There exists some $t$ such that
$$s^2-t^2=8(n^2+1)$$
Note that $s$ and $t$ are even.
We can say for example (there are more possibilities) that
$$s+t=2n^2+2$$
$$s-t=4$$
Then we have $s=n^2+3$, $t=n^2-1$. Therefore, $m=n^2+3n+3$. Also
$$v=\frac{n^2+2n+3\pm(n^2-1)}2=\begin{cases}n+2\implies u=n^2+n+1\\n^2+n+1\implies u=n+2\end{cases}$$
The correct solution is what makes $v\le u$, that is $u=n^2+n+1$, $v=n+2$.
Then, for any odd value for $n$ we have a solution (at least). The first of Will Jagy's solutions can be obtained for $n=3$.
Namely:
$$\begin{cases}y=\frac{(n+1)(n+3)}2\\z=\frac{n^2+2n+3}2\\w=\frac{(n+1)(n-1)}2\\x=\frac{y^2-z^2-3}2\end{cases}$$
Example: For $n=13$ we have $m=211$, $u=183$, $v=15$. So $y=112$, $z=99$, $w=84$. And from this,
$$1371^2+112^2=1372^2+99^2=1373^2+84^2=1892185$$
|
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|
Prove that there do not exist integers that satisfy the system
Prove that there do not exist integers $x,y,z,w,t,$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2\\(x+3)^2+w^2 &= (x+4)^2+t^2.\end{align*}
I thought about using a modular arithmetic argument. The given system is equivalent to \begin{align*}y^2 &= 2x+3+z^2\\z^2 &= 2x+5+w^2\\w^2 &= 2x+7+t^2.\end{align*} Thus $y^2 = 6x+15+t^2$. How can we find a contradiction from this?
|
My solution is similar, in spirit, to Fimpellizieri's, but organized a little differently ...
Suppose $x,y,z,w,t$ are integers such that
\begin{align}
2x + 3 &= y^2 - z^2 \tag{eq1}\\
2x + 5 &= z^2 - w^2 \tag{eq2}\\
2x + 7 &= w^2 - t^2 \tag{eq3}
\end{align}
\begin{align}
\text{Equation } (1) &\implies y,z \text{ have opposite parity.}\\
\text{Equation } (2) &\implies z,w \text{ have opposite parity.}\\
\text{Equation } (3) &\implies w,t \text{ have opposite parity.}
\end{align}
Thus one of the following cases must hold:
*
*$y,w$ are odd, and $z,t$ are even.
*$y,w$ are even, and $z,t$ are odd.
Consider each case separately.
The key is that an even square must be a multiple of 4, and an odd square must be congruent to 1 (mod 8).
First suppose $y,w$ are odd, and $z,t$ are even.
Then $(\text{eq}2) -(\text{eq}1)$ yields:
\begin{align}
&2z^2 - y^2 - w^2 = 2\\
\implies &2z^2 - y^2 - w^2 \equiv 2 \text{ }(\text{mod }8)\\
\implies &{-2} \equiv 2 \text{ }(\text{mod }8)
\end{align}
contradiction.
Next suppose $y,w$ are even, and $z,t$ are odd.
Then $(\text{eq}3) -(\text{eq}2)$ yields:
\begin{align}
&2w^2 - z^2 - t^2 = 2\\
\implies &2w^2 - z^2 - t^2 \equiv 2 \text{ }(\text{mod }8)\\
\implies &{-2} \equiv 2 \text{ }(\text{mod }8)
\end{align}
contradiction.
It follows that the given system of equations has no integer solutions.
|
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|
How to calculate $w = F(v)$ direct and also over the relation $c_{B}(w) = A\cdot c_{A}(v)$ We have linear transformation $F = \mathbb{P_{2}} \rightarrow \mathbb{R^{2}} $, where $ F(p(t)) = \begin{pmatrix} p(0) \\ P(1) \end{pmatrix},$
$ A = \{1,t,t^{2}\},$
$B=\left \{ \begin{pmatrix} 1\\ 0\end{pmatrix},\begin{pmatrix} 0\\ 1\end{pmatrix} \right \}$.
From this, transformation matrix is
$$M_B^A=\begin{pmatrix}{1}&{0}&{0}\\{1}&{1}&{1}\end{pmatrix}.$$
How should I calculate now $w = F(v)$ direct and also over the relation $c_{B}(w) = A\cdot c_{A}(v)$, when $v = a+bt+ct^{2}$?
|
By definition, we have
$$ F(v) = \begin{pmatrix} v(0) \\ v(1) \end{pmatrix} = \begin{pmatrix} a \\ a + b + c \end{pmatrix}. $$
Since $v = a \cdot 1 + b \cdot t + c \cdot t^2$, we also have
$$ c_A(v) = \begin{pmatrix} a \\ b \\ c \end{pmatrix} $$
and so
$$ c_B(F(v)) = M^A_B c_A(v) = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} a \\ a + b + c \end{pmatrix} $$
and indeed
$$ c_B(F(v)) = c_B \left( \begin{pmatrix} a \\ a + b + c \end{pmatrix} \right) = \begin{pmatrix} a \\ a + b + c \end{pmatrix} $$
as $B$ is the standard basis of $\mathbb{R}^2$.
|
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|
About an integration technique My professor wrote this :
$\displaystyle \int x^2\log(\sqrt{1-x^2}) \,\mathrm{d}x = \frac{1}{3}\int \log(\sqrt{1-x^2}) \, \mathrm dx^3$
Can someone explain me what is going on with the integration variable ?
|
$$\int x^2 \log\sqrt{1-x^2} dx = \int x^2 \log\sqrt{1-(x^3)^{2/3}} dx$$
Let $u = x^3$, so $du = 3x^2 dx$ or
$$dx = \frac{du}{3x^2}$$
so the integral becomes
$$\int \frac{x^2}{3x^2} \log\sqrt{1-u^{2/3}} du = \frac{1}{3} \int \log\sqrt{1-u^{2/3}} du \ .$$
Resubstituting $u = x^3$ yields the rather confusing expression of your professor.
|
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|
Show that if $2+2\sqrt{28n^2+1}$ is an integer then it must be perfect square. As written in title, I want to prove that
If $n$ is an integer, show that if $2+2\sqrt{28n^2+1}$ is an integer than it must be perfect square.
I m struggling in making a start . Please help.
|
With suggestions of @barak manos and @ Alqatrkapa I Noted that I can improve my post.
Notice that $2+2\sqrt{28n^2+1}$ is an even integer. Also, $28n^2+1$ is a perfect square of an odd integer say $m$ (Because $28n^2+1$ is odd itself).
Now, $$28n^2=m^2-1=(m+1)(m-1)\implies 7n^2=(\frac{m+1}{2})(\frac{m-1}{2})$$
Hence, $(\frac{m+1}{2})=7a^2$,$(\frac{m-1}{2})=b^2$ or $(\frac{m+1}{2})=b^2$, $(\frac{m-1}{2})=7a^2$. This is because $7n^2$ is $7$ times of a square and thus the right side is also $7$ times a perfect square. This is only possible when one of them is $7$ times of a square and other is simply a square as $(Square * Square=Square)$, you can say that there is possibility that they both are not squares but the product is (like $2*8=16=4^2$), but notice that $(\frac{m+1}{2})$ and $(\frac{m-1}{2})$ are consecutive integers and hence coprime
If $\frac{m+1}{2}=7a^2$ and $\frac{m-1}{2}=b^2$ then $b^2\equiv-1\mod (7)$, a contradiction.
Hence, $\frac{m-1}{2}=7a^2$ and $\frac{m+1}{2}=b^2$. Hence, $2+2m=4b^2$ a perfect square.
|
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|
Solving quadratic congruences mod powers of a prime. The problem statement is to prove that there are exactly two solutions in the 5-adic numbers to $x^2+1 =0$.
My investigation suggests that there should be solutions, one based on each solution to $x^2+1\equiv 0 \pmod 5$, which are 2 and 3 (the additive inverse of 2).
I found the solutions to the congruence for higher powers of 5 to be
7 and 18 $\pmod{25}$,
57 and 68 $\pmod{125}$ and
182 and 443 $\pmod{625}$.
The things that stands out to me are these: once I have one solution, its additive inverse is another (since $x^2 = (-x)^2$). Moreover, each of the solutions is congruent to only one of solutions to the previous equation modulo the previous power of 5. In light of this, my strategy is to prove this in three parts.
Firstly, prove (by induction on $k$) that there is a solution to $x^2+1 \equiv 0 \pmod{5^k}$ for all $k\geq 1$. Secondly, prove that I can choose my solutions to satisfy $x_{k+1} \equiv x_k \pmod{5^k}$, which may also require induction. Thirdly, prove that this property leads to two solutions, which are "generated by" the solutions $x_1 = 2$ and $x_1=3$.
Here is where I am struggling. The inductive step for part I assumes that I have a solution to $x^2 +1 \equiv 0 \pmod{5^k}$ for all $k\leq n$. How can I use this to construct a solution to $x^2+1 \equiv 0 \pmod{5^{k+1}}$?
The best attempt I have made thus far is to observe that $x_n^2+1 \equiv 0 \pmod{5^n}$ implies that $5(x_n^2 +1) \equiv 0 \pmod{5^{n+1}}$. Thus, this reduces to proving that $5x_n^2 +4$ is a square mod $5^{n+1}$. In the cases I have considered, this number turns out to be congruent to $-1$, so I need only prove that this always happens and that $-1$ is always a square.
Anyway, I am stuck, so some help would be appreciated.
|
Suppose $x^2\equiv -1 \bmod 5^k$, so that $x^2\equiv d\cdot5^k-1 \bmod 5^{k+1}$ for some $d \in \{0,1,2,3,4\}$. Now consider:
$x^2 \bmod 5^{k+1}$
$(x+5^k)^2 \bmod 5^{k+1}$
$(x+2\cdot5^k)^2 \bmod 5^{k+1}$
$(x+3\cdot5^k)^2 \bmod 5^{k+1}$
$(x+4\cdot5^k)^2 \bmod 5^{k+1}$
Multiplying these out, we get (mod $5^{k+1}$):
$d\cdot5^k-1$
$d\cdot5^k+2x\cdot5^k-1$
$d\cdot5^k+4x\cdot5^k-1$
$d\cdot5^k+x\cdot5^k-1$
$d\cdot5^k+3x\cdot5^k-1$
These are all different mod $5^{k+1}$, because $x$ and $5$ are coprime. But they are all equal to $-1$ mod $5^k$. So they must take all five values $-1, 5^k-1, 2\cdot5^k-1,3\cdot5^k-1,$ and $4\cdot5^k-1$ mod $5^{k+1}$. Hence one of them must equal $-1$ mod $5^{k+1}$.
|
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|
A simple proof by induction $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N}$ Verify by induction that $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N} \quad \forall n \ge 1 \in \mathbb{N}$
Basis: $P(1) \Rightarrow \frac{33}{11} \in \mathbb{N}$.
Induction: if the statement holds for some $n$ $\Rightarrow$ holds for $n+1$
I would appreciate some small hints because I'm stuck at $P(n+1) = \frac{6^{2(n+1)} - 3^{n+1}}{11}$
|
So the statement $P(n)$ could also be formulated that $11$ divides $6^{2n} -3^n$. So $P(n+1)$ would consider
\begin{align}
6^{2(n+1)} - 3^{n+1} & = \\
6^{2n} \times 36 - 3 \times 3^n & = \\
3(12 \times 6^{2n} - 3^n )&= \\
3(11\times 6^{2n} + 6^{2n} - 3^{n}) & = \ldots
\end{align} Can you use $P(n)$ in order to prove $P(n+1)$?
|
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|
How did he simplify this problem like this? $$\left(\frac{x}{12}+\frac{x}{18}\right)t=x$$
$$\left(\frac{1}{12}+\frac{1}{18}\right)t=1$$
I want a solid rule, how did he simplify the problem in the image like this!
|
If
$$\left(\frac{x}{12}+\frac{x}{18}\right)t=x\tag{1}$$
then in general it is not true that
$$\left(\frac{1}{12}+\frac{1}{18}\right)t=1\tag{2}$$
Counterexample: if $t=36$ and $x=0$, then you see that $(1)$ is equivalent to $0=0$ which is true, but $(2)$ is equivalent to $5=1$ which is not.
However, if you already know that $x\neq0$, then the implication is true since in that case
$$
\left(\frac{x}{12}+\frac{x}{18}\right)t=x
$$
is equivalent to (this is just a factorization)
$$
\left(\left(\frac{1}{12}+\frac{1}{18}\right)t\right)x=x
$$
and
you can divide both sides (which are equal, after all!) by $x$ (since $x\neq0$... otherwise division by $0$ doesn't make sense!) to maintain equality and infer
$$
\frac{\left(\left(\frac{1}{12}+\frac{1}{18}\right)t\right)x}{x}=\frac{x}{x}
$$
which is the same as
$$
\left(\left(\frac{1}{12}+\frac{1}{18}\right)t\right)\frac{x}{x}=\frac{x}{x}
$$
or, since $\frac{x}{x}=1$ always,
$$
\left(\left(\frac{1}{12}+\frac{1}{18}\right)t\right)\cdot1=1
$$
that is
$$
\left(\frac{1}{12}+\frac{1}{18}\right)t=1
$$
Of course, at a certain level of mathematical maturity, all of these intermediate steps are very rarely all written out since they should be obvious.
Note that if you have
$$
\left(\frac{1}{12}+\frac{1}{18}\right)t=1\tag{3}
$$
then it must always be true that
$$
\left(\frac{x}{12}+\frac{x}{18}\right)t=x
$$
(just multiply both sides of $(3)$ by $x$... this will always be a valid operation, unlike division by an a priori arbitrary real number $x$ which is not valid if $x=0$).
In fact, a well-known false proof that $2=1$ makes use of an invalid division by $0$.
Suppose that $x=y\neq0$. Then
\begin{align}
x^2&=xy\\
x^2-y^2&=xy-y^2\\
(x+y)(x-y)&=y(x-y)\\
x+y&=y\\
2y&=y\\
2&=1
\end{align}
right? No! From
$$
(x+y)(x-y)=y(x-y)\tag{4}
$$
you can't infer that
$$
x+y=y
$$
because dividing both sides of $(4)$ by $(x-y)$ and simplifying doesn't make sense since $x-y=0$.
|
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|
Find the bound of the following sequences Find the bound of the following sequences:
$$a)\ \frac{n^2}{n^2+1}$$
$$b)\ \frac{n^3}{n+1}$$
My attempt:
$a)$$$0<\frac{n^2}{n^2+1}<1, \forall n\in \Bbb N $$
$$\frac{n^2}{n^2+1}>0 \Leftrightarrow (n^2>0) \land (n^2+1)>0$$
which is true $\forall n\in \Bbb N$.
$$\frac{n^2}{n^2+1}<1 \Leftrightarrow \frac{n^2}{n^2+1}-1<0 \Leftrightarrow \frac{-1}{n^2+1}<0 \Leftrightarrow n^2>-1$$
which is also true $\forall n\in\Bbb N$.
Is this correct?
Similarly I managed to show that in $b)\ \frac{n^3}{n+1}>0, \forall n\in\Bbb N$ but I don't know what to do about the upper bound. I assume it doesn't exist, but how do I prove it?
|
$a)$ is correct, and so are the bounds.
For $b)$, it is true that $\forall n \in \mathbb N$, both $n^3 \geq 0$ and $n+1 \geq 0$ (infact, $n+1 \geq 1$) hold true. Therefore, their quotient is also positive i.e. $\frac {n^3}{n+1} \geq 0$ is also true.
However, to analyse the other direction, we can use the following trick:
$$
\frac{n^3}{n+1} = \frac{n^3+1}{n+1} - \frac 1{n+1} = n^2-n+1 - \frac{1}{n+1}
$$
Now, note that :$\frac 1{n+1} \leq 1$, so $-\frac 1{n+1} \geq -1$. Finally, we can say that:
$$
n^2-n+1 - \frac{1}{n+1} \geq n^2-n+1 -1 \geq n^2-n \geq n(n-1)\geq n-1
$$
Therefore, as $n$ becomes larger, $n-1$ also becomes infinitely large. But then, we have prove that $n-1 \leq \frac{n^3}{n+1} $, so it follows that $\frac{n^3}{n+1} $ is unbounded.
There are many tricks to show these kinds of limits, I have shown you one.
|
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|
For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the best method. I'm looking for a cleverer way to prove that inequality.
|
If $a,b\in \mathbb{R}$, then $(a+b)^2\leq 2(a^2+b^2)$ since $2ab\leq a^2+b^2$.
Applying this with $x$ and $y$ we get
$$ 1=(x+y)^2\leq 2(x^2+y^2)$$
so $x^2+y^2\geq \frac{1}{2}$, and then applying the inequality again with $x^2$ and $y^2$ we get
$$ \frac{1}{4}\leq (x^2+y^2)^2\leq 2(x^4+y^4)$$
which is the desired result.
|
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|
Write a product of two numbers in non decimal base If $A=(b-1) (b-1) (b-1)$ and $B=(b-1) (b-1)$ are written in base $b$, what is $A\times B$ in base $b$?
I've tried developing $A \times B$ in decimal base but couldn't get this product back in base $b$.
Beside this, is there a general procedure to compute product in non decimal base.
Thanks in advance for your help,
|
Your two numbers as the largest possible three digit and largest possible two digit number written in base $b$. I.e. they are $1000_b-1_b$ and $100_b-1_b$. Algebraically these are equal to $b^3-1$ and $b^2-1$ respectively. As such their product will be given by:
$$(b^3-1)(b^2-1)=b^5-b^3-b^2+1$$
This however has negatives in it so its hard to read it as a base $b$ number. Lets manipulate until it is all positive values multiplied by powers of $b$.
$$=(b-1+1)b^4-b^3-b^2+1$$
$$=(b-1)b^4+b^4-b^3-b^2+1$$
$$=(b-1)b^4+(b-2+2)b^3-b^3-b^2+1$$
$$=(b-1)b^4+(b-2)b^3+2b^3-b^3-b^2+1$$
$$=(b-1)b^4+(b-2)b^3+b^3-b^2+1$$
$$=(b-1)b^3+(b-2)b^3+(b-1+1)b^2-b^2+1$$
$$=(b-1)b^3+(b-2)b^3+(b-1)b^2+b^2-b^2+1$$
$$=(b-1)b^3+(b-2)b^3+(b-1)b^2+1$$
This is now a sum of positive numbers times a power of $b$ so we can read off the coefficients to give up the digits in base $b$. So the digits in your product are $b-1$, $b-2$, $b-1$, $0$, $1$. E.g. in base 10 it would be 98901, in base 4 it would be 32301, etc.
|
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|
Decomposition of this partial function I came across this
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book.
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way:
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
So is there a quicker or a more practical way that I can use?
|
One way to simplify this a bit, is to multiply numerator and denominator of your integrand by $x$:
$$\int \frac{x}{x^2(x^2+1)^2}dx$$
Now substituting $y = x^2$; $dy = 2x dx$, you get the integral
$$\frac{1}{2}\int \frac{1}{y(y+1)^2}dy$$
Splitting this in partial fractions is more straightforward than what you had before:
$$\frac{1}{y(y+1)^2} = -\frac{1}{(y+1)^2} -\frac{1}{y+1} + \frac{1}{y}$$
Solving the integral then gives
$$\frac{1}{2}\left(\frac{1}{y+1} + \log(y) -\log(y+1)\right)+c$$
Subsituting $y = x^2$ then gives
$$\frac{1}{2}\left(\frac{1}{x^2+1} + \log(x^2) -\log(x^2+1)\right)+c$$
which can be slightly simplified to
$$\frac{1}{2}\left(\frac{1}{x^2+1} -\log(x^2+1)\right)+ \log(x)+c$$
|
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|
Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$ Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2}{\sin(x+h)}-\frac{x^2}{\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2\sin x-x^2\sin(x+h)}{\sin(x+h)\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}$$
$$f'(x)=\lim_{h\to0}\frac{x^2}{\sin x}\frac{(1+\frac{h}{x})^2\sin x-\sin(x+h)}{h\sin(x+h)}$$
I am stuck here.
|
\begin{eqnarray}
\dfrac{f(x+h)-f(x)}{h}&=&\dfrac{1}{h}\left[\dfrac{(x+h)^2}{\sin(x+h)}-\dfrac{x^2}{\sin x}\right]\\
&=&\dfrac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}\\
&=&\dfrac{[(x+h)^2-x^2]\sin x+x^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}\\
&=&\dfrac{[(x+h)^2-x^2]\sin x-x^2[\sin(x+h)-\sin x]}{h\sin x\sin(x+h)}\\
&=&\dfrac{1}{\sin x\sin(x+h)}\left[\dfrac{(x+h)^2-x^2}{h}\cdot\sin x-x^2\cdot\dfrac{\sin(x+h)-\sin x}{h}\right]
\end{eqnarray}
Since
\begin{eqnarray}
\lim_{h\to0}\dfrac{1}{\sin x\sin(x+h)}&=&\dfrac{1}{\sin^2x}\\
\lim_{h\to0}\dfrac{(x+h)^2-x^2}{h}&=&\lim_{h\to0}\dfrac{x^2+2xh+h^2-x^2}{h}=\lim_{h\to0}\dfrac{2xh+h^2}{h}=\lim_{h\to0}(2x+h)=2x\\
\lim_{h\to0}\dfrac{\sin(x+h)-\sin x}{h}&=&\dfrac{d}{dx}(\sin x)=\cos x
\end{eqnarray}
it follows that
$$
\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{\sin^2x}[2x\cdot\sin x-x^2\cdot\cos x]=\dfrac{2x\cdot\sin x-x^2\cdot\cos x}{\sin^2x}
$$
Added:
Using the limits
$$
\lim_{h\to0}\dfrac{\cos h-1}{h^2}=-\frac12,\quad \lim_{h\to0}\dfrac{\sin h}{h}=1,
$$
we get:
\begin{eqnarray}
\lim_{h\to0}\dfrac{\sin(x+h)-\sin x}{h}
&=&\lim_{h\to0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\
&=&\lim_{h\to0}\left[\dfrac{\sin x\cos h-\sin x}{h}+\dfrac{\cos x\sin h}{h}\right]\\
&=&\sin x\cdot\lim_{h\to0}\dfrac{\cos h-1}{h}+\cos x\cdot\lim_{h\to0}\dfrac{\sin h}{h}\\
&=&\sin x\cdot 0+\cos x\cdot 1\\
&=&\cos x
\end{eqnarray}
|
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|
Compute $5^{10,000}\equiv \mod 52$ I'm unsure how to solve questions like these, we are given an example below but I'm still confused about how they got from $5^{100\times10+0}$ to $5^{0}$.
ie) Compute $5^{1000} \bmod 77$
$$
5^{1000}=5^{(166⋅6+4)}=5^4=25^2=4^2=16=2 \pmod 7\\
5^{1000}=5^{(100⋅10+0)}=5^0=1 \pmod{11}\\
5^{1000}=2⋅2⋅11+1⋅8⋅7=44+56=100=23 \pmod{77}
$$
|
One uses the effective Chinese remainder theorem:
Let $7u+11v=1$ a Bézout's relation between $7$ and $11$. Then $$(x\equiv a\mod 7 \;\text{ and }\; x\equiv b\mod 11)\iff (x\equiv 7bu+11av\mod 7\times11).$$
One ingredient is Lil' Fermat: as $7$ and $11$ are prime, $x^6\equiv1\mod 7$ for all $x\not \equiv 0\mod 7$ and $x^{10}\equiv1\mod 11$ for all $x\not \equiv 0\mod 11$, hence $5^{10,000}\equiv 5^{10,000\bmod 6}\mod 7$ and $5^{10,000}\equiv 5^{10,000\bmod 10}\mod 11$.
This is the general method. However, in the present case, it is faster to observe that $$5^4=625\equiv 1\mod 52, \;\text{ hence }\; 5^{10,000}=(5^4)^{2500}\equiv 1^{2500}\mod 52.$$
|
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|
What's the name of the following method for dividing polynomials? It's not long-division nor synthetic division I saw this method in some random PDF and am intrigued of the exact method used. I can't find any page of this method on the web because I'm not sure what you'd call this method.
Here is the method:
For solving
$$\int \frac{2x^4 + x^3}{x^2 + x - 2} \,\text{d}x$$
Observe that
$$\begin{align*}
2x^4 + x^3
&= 2x^2 (x^2+x-2) - x^3+4x^2 \\
&= 2x^2 (x^2+x-2) - x(x^2+x-2) + 5x^2-2x \\
&= 2x^2 (x^2+x-2) - x(x^2+x-2) + 5(x^2+x-2) - 7x+10 \\
&= (2x^2-x+5)(x^2+x-2) - 7x+10
\end{align*}$$
and then
$$
\int \frac{2x^4-x^3}{x^2+x-2} \,\text{d}x
= \int (2x^2-x+5)\,\text{d}x
+ \int \frac{-7x+10}{x^2+x-2}\,\text{d}x$$
|
What you have is literally polynomial long division written out without division signs. Indeed, what is written out here is the essence of polynomial long division, which is all about finding the coefficient of the factor that returns the highest degree term in the original polynomial.
|
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|
Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a convex function, so I can use jensen's inequality, but I can't come up with one which works.. Has anyone an idea?
|
WLOG $a\geq b\geq c.$ $$\text {Let } x=1/(a+3b+3c),\; y=1/(b+3c+3a),\; z=1/(c+3a+3b).$$ We have $a+3b+3c\leq b+3c+3a\leq c+3b+3a.$ Therefore $$(1).\quad x\geq y\geq z>0.$$ Let $f(a,b,c)=ax+by+cz.$
Differentiating $f$ by $c,$ keeping $a$ and $b$ constant, we have $$\partial f/\partial c=-3ax^2-3by^2+3(a+b)z^2=3(a(z^2-x^2)+b(z^2-y^2)).$$ This is not positive by (1). Therefore $$(2).\quad f(a,b,c)\geq f(a,b,b).$$ Differentiating $f$ by $a,$ keeping $b$ and $c$ constant, we have $$\partial f/\partial a=(3b+3c)x^2-3by^2-3cz^2=3(b(x^2-y^2)+c(x^2-y^2)).$$ This is not negative by (1). Therefore $$(3). \quad f(a,b,c)\geq f(b,b,c).$$ By(2) and (3) we have $$f(a,b,c)\geq f(b,b,c)\geq f(b,b,b)=3/7.$$
|
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|
Fibonacci numbers divisible by p for some prime p? This came up in a series of notes on number theory I'm reading. The question is:
Prove that the $(p-1)$th or the $(p+1)$th Fibonacci number is divisible by $p$ for some prime $p$
I'm very much lost on this question, so I thought I'd see if I could find a proof here.
|
Fix a prime $p\ge 7$. We have by Cassini's identity that
$$
F_{p-1}F_{p+1} = F_p^2-(-1)^{p-1} \,\,\,\,(=\, F_p^2-1).
$$
Hence, it is enough to prove that
$$
F_p^2 \equiv 1\pmod{p}.
$$
But we have
\begin{align*}
F_p^2&=\left(\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^p-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^p\right)^2 \\\
&=\frac{1}{5}\left(\frac{1+\sqrt{5}}{2}\right)^{2p}+\frac{1}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{2p}-\frac{2}{5}(-1)^p \\\
&=\frac{1}{5\cdot 2^{2p}}\left(1+\sqrt{5}\right)^{2p}+\frac{1}{5\cdot 2^{2p}}\left(1-\sqrt{5}\right)^{2p}+\frac{2}{5} \\\
&=\frac{1}{5\cdot 2^{p}}\left(3+\sqrt{5}\right)^{p}+\frac{1}{5\cdot 2^{2p}}\left(3-\sqrt{5}\right)^{2p}+\frac{2}{5} \\\
&=\frac{1}{5\cdot 2^{p}}\left(2\sum_{i=0, i\text{ even}}^p\binom{p}{i}3^{p-i}\sqrt{5}^i\right)+\frac{2}{5} \\\
&=\frac{1}{5\cdot 2^{p-1}}\left(\sum_{j=0}^{\frac{p-1}{2}}\binom{p}{2j}3^{p-2j}5^j\right)+\frac{2}{5}.
\end{align*}
At this point, considering that $p\mid \binom{p}{j}$ for $0<j<p$, we conclude that
$$
F_p^2 \equiv \frac{3^p}{5\cdot 2^{p-1}}+\frac{2}{5}\equiv \frac{3}{5}+\frac{2}{5}\equiv 1\pmod{p}.
$$
|
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|
Finding out the range of values of the argument of a complex number Let $z,z^2,z^3,z^4$ be the four complex numbers.If these taken in order form a cyclic quadrilateral then the question is to find out the range of values of $\theta$ where $\theta$=$arg(z)$ and $\theta €(0,2\pi)$.
Since it forms a cyclic quadrilateral i applied coni's theorem to get $$\frac{(z^3-z^2)(z-z^4)}{(z-z^2)(z^3-z^4)}$$ as a purely real number.I simplified to get $z+1/z$ as a real number and from this i guessed $ |z|=1$.But i could proceed after this..Please help me in this regard.Thanks.
|
You've already got $|z|=1$ correctly.
Note here that we have to have
$$z\not=z^2,z\not=z^3,z\not=z^4,z^2\not=z^3,z^2\not=z^4,z^3\not=z^4$$
giving
$$\theta\not=\frac{2}{3}\pi,\pi,\frac{4}{3}\pi$$
Also, we have to consider the expression "these taken in order".
For $0\lt\theta\le\frac{\pi}{2}$, since we have $0\lt\theta\lt 2\theta\lt 3\theta\lt 4\theta\le 2\pi$, these $\theta$ are sufficient.
For $\frac{\pi}{2}\lt\theta\lt\frac{2}{3}\pi$, since we have $\frac{\pi}{2}\lt\theta\lt 2\theta\lt 3\theta\lt 2\pi\lt 4\theta\lt 4\pi$ and $0\lt 4\theta-2\pi\lt \theta\lt 2\theta\lt 3\theta\lt 2\pi$, these $\theta$ are sufficient.
For $\frac{2}{3}\pi\lt\theta\lt\pi$, since we have $\frac{2}{3}\pi\lt \theta\lt 2\theta\lt 2\pi\lt 3\theta\lt 4\theta\lt 4\pi$ and $0\lt 3\theta-2\pi\lt\theta\lt 4\theta-2\pi\lt 2\theta\lt 2\pi$, these $\theta$ are not sufficient.
For $\pi\lt\theta\lt\frac{4}{3}\pi$, since we have $\pi\lt\theta\lt 2\pi\lt 2\theta\lt 3\theta\lt 4\pi\lt 4\theta\lt 6\pi$ and $0\lt 2\theta-2\pi\lt 4\theta-4\pi\lt \theta\lt 3\theta-2\pi\lt 2\pi$, these $\theta$ are not sufficient.
For $\frac{4}{3}\pi\lt\theta\lt\frac{3}{2}\pi$, since we have $\frac{4}{3}\pi\lt \theta\lt 2\pi\lt 2\theta\lt 4\pi\lt 3\theta\lt 4\theta\lt 6\pi$ and $0\lt 3\theta-4\pi\lt 2\theta-2\pi\lt \theta\lt 4\theta-4\pi\lt 2\pi$, these $\theta$ are sufficient.
For $\frac{3}{2}\pi\le\theta\lt 2\pi$, since we have $\frac{3}{2}\pi\le \theta\lt 2\pi\lt 2\theta\lt 4\pi\lt 3\theta\lt 6\pi\le 4\theta\lt 8\pi$ and $0\le 4\theta-6\pi\lt 3\theta-4\pi\lt 2\theta-2\pi\lt \theta\lt 2\pi$, these $\theta$ are sufficient.
It follows from these that the answer is
$$\color{red}{0\lt\theta\lt\frac{2}{3}\pi\quad\text{or}\quad \frac{4}{3}\pi\lt\theta\lt 2\pi}$$
|
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|
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? $F =$ The probability that the dice land on different numbers
$F = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)}$
$F = \frac{30}{36} = \frac{5}{6}$
$E =$ The event that at least one lands on 6
$E = {(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}$
$E = \frac{11}{36}$
What is the probability of E, when given that F has occurred?
$P = \frac{P(EF)}{P(F)}$
$P = \frac{1-P(EF^c)}{P(F)}$
$P = \frac{\frac{11}{36}}{\frac{5}{6}}$
$P = \frac{11}{30}$
This answer was incorrect it is $\frac{1}{3}$ I think there is an issue in identifying $P(EF)$, what should that be and why? Identifying F is easy because that is what is given to have occurred.
Added: Is $EF$ equivalent to $E \cap F$?
|
There are only 10 cases where the dice have different numbers: (1, 6), (2, 6), ..., (5, 6), (6, 5), (6, 4), ..., (6, 1).
So the answer is then $\frac{\frac{10}{36}}{\frac{5}{6}}=\frac{1}{3}$.
|
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|
Determine limit: $\lim_{x \to -\infty} (\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) $ Determine the limit of:
$$\lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
I've tried a few times, most notably the following two versions. I'm looking for a comment on both, since both amount to a wrong answer.
First attempt
$$L = \lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
I thought it'd be nice to get rid of as much ugliness as possible by moving $x$ out of the roots:
$$
\begin{split}
L &= \lim_{x \to -\infty} \left(x\left(\sqrt{1 + \frac{2}{x}}\right)
- x\left(\sqrt{1 - \frac{2}{x}}\right)\right)\\
&= \lim_{x \to -\infty} \left(x\left(\sqrt{1 + \frac{2}{x}} - \sqrt{1 - \frac{2}{x}}\right)\right)
\end{split}
$$
Seems simple enough, since $\frac{1}{x}$ should be an (infinitely) small number I thought it could be discarded with regard to $\sqrt1$, yielding:
$$L = \lim_{x \to -\infty} \left(x\left(\sqrt{1} - \sqrt{1}\right)\right) = 0$$
Wrong, Second attempt
Multply both sides with the conjugate:
$$
\begin{split}
L &= \lim_{x \to -\infty} \sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) \\
&= \lim_{x \to -\infty} \frac{4x}{(\sqrt{x^2 +2x} + \sqrt{x^2 - 2x})} \\
&= \lim_{x \to -\infty} \frac{4x}{x\sqrt{1 + \frac{2}{x}}
+ x\sqrt{1 - \frac{2}{x}}} \\
&= \lim_{x \to -\infty} \frac{4}{\sqrt{1 + \frac{2}{x}}
+ \sqrt{1 - \frac{2}{x}}}\\
&= \frac{4}{\sqrt{1} + \sqrt{1}} = \frac{4}{2} = 2
\end{split}
$$
Wrong as well... The answer should be $-2$, but I don't get how to 'get the negative' in...
|
The second is almost good, to mistake from the fact that if $x<0$, $\sqrt{x^2}=-x$.
|
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|
How to compute the sum $\sum_{n=0}^\infty \tfrac{n^2}{2^n}$?
How to find this sum :
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}$
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{9}{8}+\dfrac{16}{16}+\dfrac{25}{32}+\dfrac{36}{64}+\dfrac{49}{128}+\dots$
Now $\sum_{n=0}^\infty \dfrac{n}{2^n}\leqslant \sum_{n=0}^\infty \dfrac{n^2}{2^n}$
And I know that $\sum_{n=0}^\infty \dfrac{n}{2^n}=2$.
But how to find this sum ? I am confused.Please give some hints.
|
Hint: For $|x|<1$
\begin{align*}
\sum_{n=0}^{\infty }n^{2}x^{n}&=\sum_{n=0}^{\infty }n\left ( n+1 \right )x^{n}-\sum_{n=0}^{\infty }nx^{n} \\
&=x\sum_{n=0}^{\infty }n\left ( n+1 \right )x^{n-1}-x\sum_{n=0}^{\infty }nx^{n-1}\\
&=x\left ( \sum_{n=0}^{\infty }x^{n+1} \right )''-x\left ( \sum_{n=0}^{\infty }x^{n} \right )'\\
&=-\frac{x\left ( x+1 \right )}{\left ( x-1 \right )^{3}}
\end{align*}
then let $x=\dfrac{1}{2}$ you will get the answer.
|
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|
Calculate velocity vector along a curve (as a function of time) I have a curve defined by an equation: $y = f(x)$
In my case, the equation is a polynomial $y = ax^3 + bx^2 + cx + d$.
I also have $2$ boundary conditions. Point $A$ and Point $B$. I know the velocity and position of $A$ and $B$ and I know the time taken to travel from $A$ to $B$ along the curve. With some simple calculus I have been able to the determine the variables, $a$, $b$, $c$, $d$. This means I know $\dfrac{\dot{y}}{\dot{x}}= f'(x)$ and $y=f(x)$.
What I really want to know is position and velocity as a function on time:
\begin{align*}
\dot{x} &= v_x(t) \\
\dot{y} &= v_y(t) \\
x &= x(t) \\
y &= f(x(t))
\end{align*}
How would I determine this for either my polynomial or for an equation $y=f(x)$?
|
Using cubic spline:
\begin{align*}
p(x) &=
\frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b) \\
& \quad +
(x-a)(x-b)\left \{
\frac{x-b}{(a-b)^{2}} \left[ f'(a)-\frac{f(a)-f(b)}{a-b} \right]+
\frac{x-a}{(b-a)^{2}} \left[ f'(b)-\frac{f(b)-f(a)}{b-a} \right]
\right \} \\[5pt]
p'(x) &=
\frac{x-b}{a-b}f'(a)+\frac{x-a}{b-a}f'(b)+
3(x-a)(x-b)\left \{
\frac{f'(a)+f'(b)}{(b-a)^2}-\frac{2[f(b)-f(a)]}{(b-a)^3}
\right \}
\end{align*}
Note that $p(x) \equiv f(x)$ if $f(x)$ is a cubic (or lower) polynomial, else $p(x)$ is just an interpolating polynomial.
Boundary conditions:
*
*$A=\begin{pmatrix} a \\ f(a) \end{pmatrix}$
*$B=\begin{pmatrix} b \\ f(b) \end{pmatrix}$
*$\boldsymbol{v}_A=\dfrac{v_A}{\sqrt{1+f'(a)^2}}
\begin{pmatrix} 1 \\ f'(a) \end{pmatrix}$
*$\boldsymbol{v}_B=\dfrac{v_B}{\sqrt{1+f'(b)^2}}
\begin{pmatrix} 1 \\ f'(b) \end{pmatrix}$
where $y$ is a single-valued function of $x$.
Caution:
$a$, $b$ here are NOT the coefficients of the cubic polynomial but two distinct boundaries.
In general for space curve,
\begin{align*}
\boldsymbol{p}(t) &=
\frac{t-b}{a-b}\boldsymbol{x}(a)+\frac{t-a}{b-a}\boldsymbol{x}(b) \\
& \quad +
(t-a)(t-b)
\left \{
\frac{t-b}{(a-b)^{2}}
\left[
\boldsymbol{x}'(a)-\frac{\boldsymbol{x}(a)-\boldsymbol{x}(b)}{a-b}
\right]+
\frac{t-a}{(b-a)^{2}}
\left[
\boldsymbol{x}'(b)-\frac{\boldsymbol{x}(b)-\boldsymbol{x}(a)}{b-a}
\right]
\right \}
\end{align*}
|
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|
Finding two possible values of $z^2 + z + 1$ given that $z$ is one of the three cube roots of unity. Factorise $z^3 - 1 $.
If $z$ is one of the three cube roots of unity, find the two possible values of $z^2 + z + 1$.
Factorising gives you :
$(z - 1)(z^2 + z + 1) = 0$ since $z$ is one of the three cube roots of unity.
z is complex so $z \neq 1$ so $z-1 \neq 0$
Hence, $z^2 + z + 1 = 0$
Where do I go from here?
Any help is appreciated!!
|
$$
0 = z^2 + z + 1 = (z+1/2)^2 - (1/2)^2 + 1 = (z+1/2)^2 + 3/4 \iff \\
z + \frac{1}{2} = \pm i \frac{\sqrt{3}}{2} \iff \\
z = \frac{-1 \pm i \sqrt{3}}{2}
$$
By the way:
$$
1 + z + z^2 = \frac{z^3-1}{z-1}
$$
as a finite term geometric series.
|
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|
Probability of cards drawn from a deck
The cards are drawn from a well shuffled deck of 52 cards one after
the other without replacement. The probability of first card being a
spade and the second a black king is ?
Here, is my approach,
Upon first draw we got a black spade king
$$P(\text{first card is spade}) = \frac{13}{52}$$
$$ P(\text{second black king}) = \frac{1}{51} $$
Upon first draw we don't get a black spade king
$$P(\text{first card is spade}) = \frac{13}{52}$$
$$ P(\text{second black king}) = \frac{2}{51} $$
Now, the total probability,
$$P= \frac{13}{52} \frac{1}{51} + \frac{13}{52} \frac{2}{51} = \frac{39}{2652}$$
But the actual answer is $\frac{25}{2652}$
|
Upon first draw we got a black spade king
$$P(\text{first card is $K \spadesuit$}) = \frac{1}{52}$$
$$ P(\text{second black king i.e $K\clubsuit$}) = \frac{1}{51} $$
Upon first draw we don't get a black spade king
$$P(\text{first card is $\spadesuit$ but not $K\spadesuit$}) = \frac{12}{52}$$
$$ P(\text{second black $K$}) = \frac{2}{51} $$
Now, the total probability,
$$P= \frac{1}{52} \frac{1}{51} + \frac{12}{52} \frac{2}{51} = \frac{25}{2652}$$
|
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|
Evaluate the series $\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}$ I would like to show that the following sum converges $\forall x \in \mathbb{R}$ as well as calculate the sum:
$\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}$
First for the coefficient:
$\frac{n^2-1}{n!(n-1)}=\frac{n+1}{n!}$
Then, what I did was to try and formulate this series to a series which I know:
$\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}=\sum_{n=0}^{\infty} \left( \frac{n+1}{n!} \right)x^n=\cdots = \frac{1}{x} \sum_{n=0}^{\infty}\frac{(n+1)^2 x^{n+1}}{(n+1)!}$
I have ended up with this formula, which reminds me somehow the expansion of the exponential $e^x$
$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$
but I cannot see how the term $(n+1)^2$ affects the result.
Thanks.
|
I think it might be easier to not incur the $(n+1)^2$, but to simply break up the sum into
$$
\sum_{n=0}^\infty \frac{n+1}{n!} x^n \;\; =\;\; \sum_{n=0}^\infty \frac{n}{n!} x^n + \sum_{n=0}^\infty \frac{x^n}{n!}.
$$
Clearly the second sum goes to $e^x$, but we can notice that
$$
\sum_{n=0}^\infty \frac{nx^n}{n!} \;\; =\;\; 0 + x + \frac{2}{2!}x^2 + \frac{3}{3!}x^3 + \frac{4}{4!}x^4 + \ldots \;\; = \;\; x \left (1 + x + \frac{1}{2!}x^2 + \frac{1}{3!} x^3 + \ldots \right ) \;\; =\;\; xe^x.
$$
|
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|
Find a minimum of $x^2+y^2$ under the condition $x^3+3xy+y^3=1$ As in the title, I've tried to find a maximum and mininum of $x^2+y^2$ when $x^3+3xy+y^3=1$ holds. It is not too hard to show that $x^2+y^2$ has no maximum, but I can't find a minimum.
Lagrange multiplier gives a dirty calculation so I can't handle it. Is there any elegant way to find it? Thanks for any help.
p.s. Sorry. I make a typo in the $xy$-coefficient.
|
Noting that
$$x^3+y^3=(x+y)^3-3xy(x+y)$$
and
$$(x+y)^3-1=(x+y-1)\left[(x+y)^2+(x+y)+1\right]$$
gives
$$x^3+3xy+y^3=1 \\\\ \implies\ (x+y)^3-1-3xy(x+y)+3xy=0 \\\\ \implies\ (x+y-1)\left[(x+y)^2+(x+y)+1-3xy\right]=0 \\\\ \implies\ (x+y-1)(x^2+y^2-xy+x+y+1)=0$$
So either $x+y=1$ or $x^2+y^2-xy+x+y+1=0$.
Treating the latter as a quadratic equation in $x$ gives its discriminant as $(y-1)^2-4(y^2+y+1)=-3(y+1)^2$; so a real solution exists only when $y=-1$, at which value $x=-1$. This gives $x^2+y^2=(-1)^2+(-1)^2=2$.
As for $x+y=1$, we want the circle of $x^2+y^2=r$ $(r\ge0)$ to be tangent to that straight line. This occurs at $x=y=\dfrac12$, giving $r_{\min}=\left(\dfrac12\right)^2+\left(\dfrac12\right)^2=\dfrac12$.
Hence the minimum value is
$$\min\left(2,\,\frac12\right)\ =\ \frac12$$
|
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|
computation of minimum value of a function
How will we calculate the minimum value ilof the equation with the given condition.
|
Using Holder,s Inequality
$\displaystyle (x^2+8y^2+27z^2)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq \bigg[\bigg(x^2\cdot \frac{1}{x}\cdot \frac{1}{x}\bigg)^{\frac{1}{3}}+\bigg(8y^2\cdot \frac{1}{y}\cdot \frac{1}{y}\bigg)^{\frac{1}{3}}+\bigg(27z^2\cdot \frac{1}{z}\cdot \frac{1}{z}\bigg)^{\frac{1}{3}}\bigg]^{3}=(1+2+3)^3=216$
and equality hold when $\displaystyle x^3=8y^3=27z^3$
|
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|
How to find $\lim_{x\to1}\frac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$ without using L'Hospital Rule? Compute the following limit. I've tried using l'Hospital. And it work the result was $\dfrac{7}{3}$. But how can I do this without this rule? I am trying to help I friend who hasn't done derivatives yet.
$$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$$
|
Let $x=(u+1)^4$:
$$\frac{\sqrt[3]{(u+1)^4+7}+2\sqrt{3(u+1)^4+1}-6}u\\=\frac{\sqrt[3]{u^4+4u^3+6u^2+4u+8}+2\sqrt{3u^4+12u^3+18u^2+12u+4}-6}u$$
then some binomial expansions:
$$\sqrt[3]{8+(u^4+4u^3+6u^2+4u)}=2+\frac1{12}(u^4+4u^3+6u^2+4u)+\mathcal O(u^2)$$
$$\sqrt{4+(3u^4+12u^3+18u^2+12u)}=2+\frac14(3u^4+12u^3+18u^2+12u)+\mathcal O(u^2)$$
Thus, the limit reduces down to
$$\frac{2+\frac1{12}(u^4+4u^3+6u^2+4u)+4+\frac12(3u^4+12u^3+18u^2+12u)-6+\mathcal O(u^2)}u\\=\frac1{12}(u^3+4u^2+6u+4)+\frac12(3u^3+12u^2+18u+12)+\mathcal O(u)\\\to\frac1{12}(4)+\frac12(12)=\frac{19}3$$
which is the correct limit.
|
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|
How do I find the limit of the following sequence? $a_0$ to $a_5$ are set independently, all positive.
Then let us define the sequence as $a_n=\frac{\sum_{i=1}^6a_{n-i}}{6},\;\forall n\geq6$.
How do I find $\lim_{n\rightarrow\infty}a_n$?
Or, will it converge at all? Or will it just be exactly the limit value after some $n$?
|
Hint:
Note that if
$$x_n = \begin{pmatrix}
a_{n-5}\\
a_{n-4}\\
a_{n-3}\\
a_{n-2}\\
a_{n-1}\\
a_{n}
\end{pmatrix} ; \, \, \, \, \, A = \begin{pmatrix}
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1& 0 & 0 \\
0 & 0 & 0 & 0 & 1& 0 \\
0 & 0 & 0 & 0 & 0 & 1\\
1/6 & 1/6 &1/6 &1/6 &1/6 &1/6
\end{pmatrix},
$$
We have $x_n = Ax_{n-1}$.
You can check that $1$ is the single eigenvalue of $A$ with greatest absolute, and its associated eigenvector is
$$v = \begin{pmatrix}
1\\
1\\
1\\
1\\
1\\
1
\end{pmatrix}.$$
From here, if $a_0+\dots+a_5 \ne 0$, then $x_n$ converges to a multiple of $v$, say $x_n \rightarrow c v$ for some $c$. This proves that if $a_0+\dots+a_5 \ne 0$, the sequence converges.
To find $c$, keeping in mind that $w = \frac{1}{21} (1, 2, 3, 4, 5, 6)$ is the left eigenvector associated with eigenvalue $1$, try to show that $$c = w\, x_5,$$
which means $a_n \rightarrow \frac{1}{21} (a_0 + 2a_1+\dots+6a_5)$.
|
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|
Find the limit: $\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$ Find the limit:
$$\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$$
I really have no idea what to do here, since I obviously can't plug in $x=0$ nor divide by highest power...Any help is appreciated!
|
$$\lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} = \lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} \cdot \frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{6^x-1-(5^x-1)}\cdot\frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{x}\cdot\lim_{x\to0}\frac{x}{6^x-1-(5^x-1)} = \bigg(\lim_{x\to0}\frac{8^x-1}{x}-\frac{7^x-1}{x}\bigg)\cdot\lim_{x\to0} \frac{1}{\frac{6^x-1}{x}-\frac{5^x-1}{x}} = (\ln8-\ln7)\cdot\frac{1}{\ln6-\ln5} = \frac{\ln\frac{8}{7}}{\ln\frac{6}{5}}$$
|
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|
In right triangle $ABC$ ( $BC$ is hypotenuse),$D$ is a point on $BC$.Calculate $AB$ given that...
In right triangle $ABC$ ( $BC$ is hypotenuse),$D$ is a point on $BC$.Calculate $AB$ given that: $AD$=5,$BD=3$,$CD=6$.
|
You also can use Stewart's Theorem:
$$\frac{AC^2}{CD\cdot BC}-\frac{AD^2}{CD\cdot BD}+\frac{AB^2}{BD\cdot BC}=1\\ \frac{81-AB^2}{6\cdot 9}-\frac{5^2}{6\cdot 3}+\frac{AB^2}{3\cdot 9}=1$$
|
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|
Difficult trigonometry problem to find the minimum value Find the minimum value of $5\cos A + 12\sin A + 12$.
I don't know how to approach this problem. I need help.
I'll show you how much I got...
$$5\cos A +12\sin A + 12 = 13(5/13\cos A +12/13\sin A) + 12$$
|
$(p\cos A+q\sin A)^2+(p\sin A-q\cos A)^2=p^2+q^2$ See Brahmagupta-Fibonacci Identity
$\implies(p\cos A+q\sin A)^2\le p^2+q^2$ the equality occurs if $p\sin A-q\cos A=0$
$\iff-\sqrt{p^2+q^2}\le p\cos A+q\sin A\le \sqrt{p^2+q^2}$
|
{
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|
If $abc=1$ so $\sum\limits_{cyc}\frac{a^3c}{(a+c)(b+c)}\geq\frac{3}{4}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a^3c}{(a+c)(b+c)}+\frac{b^3a}{(b+a)(c+a)}+\frac{c^3b}{(c+b)(a+b)}\geq\frac{3}{4}$$
I tried C-S and BW. It does not help.
|
Let $a = \frac{x}{y}, \ b = \frac{y}{z}, \ c = \frac{z}{x}; \ x, y, z > 0$.
It suffices to prove that $f(x, y, z)\ge 0$ where
\begin{align*}
f(x,y, z) &= 4\, x^7\, z^5 - 3\, x^6\, y^3\, z^3 + 4\, x^6\, y^2\, z^4 + 4\, x^5\, y^7 - 3\, x^5\, y^5\, z^2 - 3\, x^5\, y^2\, z^5 + 4\, x^4\, y^6\, z^2 - 6\, x^4\, y^4\, z^4\\
&\quad - 3\, x^3\, y^6\, z^3 - 3\, x^3\, y^3\, z^6 - 3\, x^2\, y^5\, z^5 + 4\, x^2\, y^4\, z^6 + 4\, y^5\, z^7.
\end{align*}
WLOG, assume that $z = \min(x,y,z).$ There are two possible cases:
1) $z \le y \le x$: Let $z = 1, \ y = 1+s, \ x = 1+s + t; \ s,t \ge 0$. We have
\begin{align*}
&f(1+s+t, 1+s, 1)\\
=\ & 4\, t^7 + \left( - 3\, s^3 - 5\, s^2 + 27\, s + 29\right)\, t^6 + \left(4\, s^7 + 28\, s^6 + 81\, s^5 + 107\, s^4 + 62\, s^3 + 99\, s^2 + 175\, s + 88\right)\, t^5 \\
&\quad + g(s,t)\\
\ge \ & 0 ,
\end{align*}
since
\begin{align*}
&4\cdot 4 \cdot (4\, s^7 + 28\, s^6 + 81\, s^5 + 107\, s^4 + 62\, s^3 + 99\, s^2 + 175\, s + 88) -
( - 3\, s^3 - 5\, s^2 + 27\, s + 29)^2\\
=\ & {\left(s + 1\right)}^2\, \left(64\, s^5 + 311\, s^4 + 580\, s^3 + 378\, s^2 + 100\, s + 567\right)\\
\ge \ & 0,
\end{align*} where $g(s,t)$ is a polynomial with non-negative coefficients.
2) $z\le x \le y$: Let $z = 1, \ x = 1+s, \ y = 1+s+t; \ s,t\ge 0.$
$f(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients. The inequality is true. This completes the proof.
|
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|
How did Descartes come up with the spoof odd perfect number $198585576189$? We call $n$ a spoof odd perfect number if $n$ is odd and and $n=km$ for two integers $k, m > 1$ such that $\sigma(k)(m + 1) = 2n$, where $\sigma$ is the sum-of-divisors function.
In a letter to Mersenne dated November $15$, $1638$, Descartes showed that
$$d = {{3}^2}\cdot{{7}^2}\cdot{{11}^2}\cdot{{13}^2}\cdot{22021} = 198585576189$$
would be an odd perfect number if $22021$ were prime.
Here is my question:
How did Descartes come up with the spoof odd perfect number $198585576189$?
|
If $22021$ were prime, we would have $$\sigma(d)=\sigma(3^2\cdot 7^2\cdot 11^2\cdot 13^2)\cdot 22022=(1+3+3^2)\cdot(1+7+7^2)\cdot(1+11+11^2)\cdot(1+13+13^2)\cdot 22022=2d$$ which can be verified by multiplication
I guess Descartes calculated $\sigma(3^2\cdot 7^2\cdot 11^2\cdot 13^2)=3^2\cdot 7\cdot 13\cdot 19^2\cdot 61$ , tried to multiply with the $19^2\cdot 61$ not fitting to the $3^2\cdot 7^2\cdot 11^2\cdot 13^2$-part and was lucky.
|
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|
How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1) = 13 (x +1)^3\\
(x+1)^4 +36 = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!
|
Assuming that you don't see the direct substitution $t=x+1$ for whatever reason, note that $(x+1)^5 +36x+36 - 13(x+1)^3$ is a monic polynomial with the free term $1+36-13=24\,$. By the rational root theorem, try the divisors of $24$ for possible rational roots, and enjoy your luck.
|
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|
How does this simplify to 1? I am working on this differetiation problem:
$ \frac{d}{dx}x(1-\frac{2}{x})$
and I am currently stuck at this point:
$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x$
Symbolab tells me this simplifies to $1$ but I do not understand how. I am under the impression that;
$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x \equiv 1- 2x^{-x^2}-2^{-x}$
|
$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x $
= $1\cdot 1- 1 \cdot \frac{2}{x} + \frac{2}{x}$
= $1- \frac{2}{x} + \frac{2}{x}$
=$1$
|
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|
Find the polynomials which satisfy the condition $f(x)\mid f(x^2)$
I want find the polynomials which satisfy the condition
$$f(x)\mid f(x^2).$$
I want to find such polynomials with integer coefficients, real number coefficients and complex number coefficients.
For example, $x$ and $x-1$ are the linear polynomials which satisfy this condition.
Here is one way to find the $2$-degree polynomials with integer coefficients.
Let the quadratic be $p=ax^2+bx+c$, so its value at $x^2$ is $q=ax^4+bx^2+c$. If $p$ is to be a divisor of $q$ let the other factor be $dx^2+ex+f.$ Equating coefficients gives equations
[1] $ad=a,$
[2] $ae+bd=0,$
[3] $af+be+cd=b,$
[4] $bf+ce=0,$
[5] $cf=c.$
Now we know $a,c$ are nonzero (else $p$ is not quadratic, or is reducible). So from [1] and [5] we have $d=f=1.$ Then from [2] and [4] we obtain $ae=ce.$ Here $e=0$ leads to $b=0$ from either [2] or [4], and [3] then reads $a+c=0$, so that $p=a(x^2-1)$ which is reducible. So we may assume $e$ is nonzero, and also $a=c.$
At this point, [2] and [4] say the same thing, namely $ae+b=0.$ So we may replace $b=-ae$ in [3] (with its $c$ replaced by $a$) obtaining
$a+(-ae)e+a=-ae,$ which on factoring gives $a(2-e)(e+1)=0.$ The possibility $e=2$ then leads after some algebra to $2a+b=0$ and $p=a(x-1)^2$ which is reducible, while the possibility $e=-1$ leads to $a=b$ and then $p=ax^2+ax+a$ as claimed.
Should we list out all the irreducible degree polynomials and then check if these polynomials satisfy the condition
*
*$x$
*$x+1$
*$x^2 + x + 1$
*$x^3 + x^2 + 1$
*$x^3 + x + 1$
*$ x^4 + x^3 + x^2 + x + 1 $
*$ x^4 + x^3 + 1 $
*$ x^4 + x + 1 $
With the real number coefficients which can be factored into
$$(x-c_1)(x-c_2)\cdots(x^2-2a_1x-(a_1^2+b_1^2))(x^2-2a_2x-(a_2^2+b_2^2))\cdots$$ If all of these linear terms and quadratic terms satisfy $$f(x)\mid f(x^2),$$ this polynomial satisfy too? So what's pattern in the real number polynomials?
|
I don't have "the answer" per se, but I think I can provide a new perspective on it.
Let's work over the complex numbers $\mathbb{C}$. That way, we can write every polynomial as the product of its roots, i.e. in the form $f(x) = (x - r_1)^{e_1}...(x - r_n)^{e_n}$ for distinct $r_1, ... , r_n \in \mathbb{C}$.
Let's first deal with the case of $n=1$, where again, $n$ is the number of distinct roots of $f$. Then $f(x^2) = (x^2 - r_1)^{e_1} = (x - \sqrt{r_1})^{e_1}(x + \sqrt{r_1})^{e_1}$. Therefore since $r_1$ has to be a root of $f(x^2)$ if $f(x) | f(x^2)$, then $r_1 = \pm \sqrt{r_1}$ so $r_1 = 0$ or $r_1 = 1$.
How about two roots? Let's call them $r$ and $s$. Then similarly either $r = \pm \sqrt{r}$ or $r = \pm \sqrt{s}$, and either $s = \pm \sqrt{r}$ or $s = \pm \sqrt{s}$. The only interesting case is if $r = \pm \sqrt{s}$ and $s = \pm \sqrt{r}$. But then $r^2 = s = \pm \sqrt{r}$ so $r^4 - r = 0$, so $r \in \{0, 1, \omega, \omega^2\}$, where $\omega$ is a primitive third root of unity. Similarly $s \in \{0, 1, \omega, \omega^2\}$. By only going after the "new" case, we see that the only choice is to have one of the roots be $\omega$ and the other $\omega^2$. So an obvious candidate is $(x - \omega)(x - \omega^2) = x^2 + x + 1$, which holds since $\omega^2 = -\omega$.
I think it's possible to continue this way, and in particular I think it wouldn't be so hard to show any nonzero solution has complex norm 1.
|
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|
Prove $a_n = n 3^{n-1}$, recurrence relation. Let $a_0 = 0, a_1 = 1$, and let $a_{n+2} = 6a_{n+1} - 9a_n$ for $n\geq 0.$
Prove $a_n = n 3^{n-1}$ for $n\geq 0.$
attempt: Suppose $a_0 = 0, a_1 = 1$, and let $a_{n+2} = 6a_{n+1} - 9a_n$.
Then notice when $n = 0$, we have $a_2 = 6_{1} - 9a_0 = 6-0 = 6$.
And $a_2 = 2*(3^{2-1}) = 6.$
Suppose as inductive hypothesis this is true for all $n \geq 0.$
Then we need to show $a_{n+1} = (n+1) 3^{(n+1)-1}= (n+1)*3^n$.
Then $a_{n+1} = 6a_{n}-9a_{n-1}$ by the recurrence relation.
So $a_{n+1} = 6a_{n}-9a_{n-1} = 6(n 3^{n-1})- 9a_{n-1} = 3[2(n 3^{n-1})] - 3a_{n-1}]= $ I am stuck after this, I want to show $a_{n+1} = (n+1)*3^n$.
Can someone please help ? Thank you
|
Induction.
True for $n=0,n=1$.
We will assume true up to $n$, will show holds for $n+1$:
$$a_{n+1} = 6 a_n -9 a_{n-1} = 6 n 3^{n-1} - 9 (n-1) 3^{n-2}=(2n -(n-1))3^n=(n+1)3^{(n+1)-1}.$$
|
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|
Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$ Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.
i simplified and reach to expression as follows :
$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?
Thanks
|
HINT:
Let $y=\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$
Divide both sides by $\cos^2x$ and use $\sec^2x=1+\tan^2x$ to form a Quadratic Equation in $\tan x$
Now as $\tan x$ is real, the discriminant must be $\not<0$
Or divide both sides by $\sin^2x$ to form a Quadratic Equation in $\cot x$
|
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|
Find the GCD of ... Find the GCD of:
$y^2-10y+24+6x-9x^2$, $2y^4-18x^2y^2-48xy^2-32y^2$.
My Attempt:
$$
\begin{split}
1^{st} \text{expression} &= y^2-10y+24+6x-9x^2\\
&={(y)^2-2\cdot(y)\cdot5+(5)^2}-(5)^2+24+6x-9x^2\\
&=(y-5)^2-{25-24-6x+9x^2}\\
&=(y-5)^2-{(1)^2-2\cdot(1)\cdot(3x)+(3x)^2}\\
&=(y-5)^2-(1-3x)^2\\
&=(y-5+1-3x)(y-5-1+3x)\\
&=(y-3x-4)(y+3x-6).
\end{split}
$$
But, I could not factorize the $2^{nd}$ expression. Please help.
|
The problem is that sometimes you can't find a factorization for the algebric expressions. If you can is just to use Mark's approach.
If not, a more general way is use the same approach that we use to find the $\gcd$ between integer numbers. Which is based on Euclides's division algorithm.
Step $1)$: Divide $$a(y)=2y^4-18x^2y^2-48xy^2-32y^2$$ by $$b(x)=y^2-10y+24+6x-9x^2$$ w.r.t. the variable $y$ and find the quotient $q_1(x)$ and the remainder $r_1(x)$.
$$\text{So,}\quad a(x)=b(x)q_1(x)+r_1(x)$$
Step $2)$: Divide $b(x)$ by $r_1(x)$ and get quotient $q_2(x)$ and remainder $r_2(x)$
$$\text{So,}\quad b(x)=r_1(x)q_2(x)+r_2(x)$$
Step $3)$: Divide $r_1(x)$ by $r_2(x)$ and get quotient $q_3(x)$ and remainder $r_3(x)$
$$\text{So,}\quad r_1(x)=r_2(x)q_3(x)+r_3(x)$$
and go on untill find $r_n(x)=0$. Then $\gcd(a(x),b(x))=r_{n-1}(x)$.
|
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|
How to find this integral... $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\left(3x^2+2\sqrt2xy+3y^2\right)}dxdy$$
I have no idea how to integrate this function. If the middle $xy$ term would not have been present it would have been easy. But the $xy$ term is causing a problem.
|
$$I=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(3x^2+2\sqrt2xy+3y^2)}dxdy$$
Let $x=r\cos\theta$, $y=r\sin\theta$.
Converting to polar form,
$$\begin{align}I&=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-(3r^2\cos^2\theta+2\sqrt2r^2\sin\theta\cos\theta+3r^2\sin^2\theta)}rdrd\theta\\
&=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2(3+2\sqrt2\sin\theta\cos\theta)}rdrd\theta\\
&=\frac12\int_{0}^{2\pi}\int_{0}^{\infty}e^{-u(3+2\sqrt2\sin\theta\cos\theta)}dud\theta\\
&=\frac12\int_{0}^{2\pi}-\frac1{(3+2\sqrt2\sin\theta\cos\theta)}\left[e^{-u(3+2\sqrt2\sin\theta\cos\theta)}\right]^{\infty}_0d\theta\\
&=\frac12\int_0^{2\pi}\frac1{3+2\sqrt2\sin\theta\cos\theta}d\theta\\
&=\frac12\int_0^{2\pi}\frac1{3+\sqrt2\sin2\theta}d\theta\\
&=\frac14\int_0^{4\pi}\frac1{3+\sqrt2\sin v}dv\\
&=\frac12\int_0^{2\pi}\frac1{3+\sqrt2\sin v}dv\\
&=\frac12\frac{2\pi}{\sqrt{3^2-2}}\\
&=\frac{\pi}{\sqrt7}\end{align}$$
|
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|
Find all points where $f$ is continuous $f:\mathbb{R}^2 \rightarrow \mathbb{R}$
$f(x,y)=\begin{cases}
\frac{\sin{y}+2yx^2}{y}, & y \gt0 \\
\frac{y+3x}{1-y}, & y\le 0
\end{cases}$
I have to check
a) What points is $f$ continuous at?
b) Is $f$ class $C^1$ at $(1,0)$
My idea for a)
Function $f$ is continuous at all points $(x_0,y_0)$ where $y_0 \neq 0$ as sum, difference and quotient of continuous functions. Now for $y_0=0$, I have to check if $\lim_{x \to x_0} f(x,0) = f(x_0,0)$. I don't know what to plug in instead of $f(x,0)$ in limit.
b) I can find differential $Df(1,0)$ and check if it is continuous in $(1,0)$? I am not sure about b)
|
As you say, the function is continuous for all $y \neq 0$, and so we need to consider $y=0$.
$$\lim_{y \to 0^+} \left(\frac{\sin y +2x^2y}{y}\right) \ = \ \lim_{y\to 0^+}\left(\frac{\sin y}{y} + \frac{2x^2y}{y}\right) \ = \ 1+\lim_{y \to 0^+}\left(\frac{2x^2y}{y}\right) \ = \ 1+2x^2$$
$$\lim_{y\to 0^-}\left(\frac{3x+y}{1-y}\right) \ = \ 3x$$
For the function to be continuous, we need the $x$- and $y$-limits to agree.
We need (at least) $1+2x^2=3x$, i.e. $2x^2-3x+1=0$, i.e. $x=\frac{1}{2}$ or $x=1$.
The function is discontinuous at the the points $y=0$ and $x \neq \frac{1}{2},1$.
We need to consider $x=\frac{1}{2}$ and $x=1$ separately.
$$f(\tfrac{1}{2},y) = \left\{ \begin{array}{ccc} \frac{2\sin y + y}{2y} & : & y > 0 \\ \frac{2y+3}{2-2y} & : & y \le 0 \end{array}\right.$$
As $y \to 0^-$, $f \to \frac{3}{2}$. As $y \to 0^+$, $f \to \frac{3}{2}$ and so $f(\frac{1}{2},y)$ is continuous at $y=0$.
$$f(1,y) = \left\{ \begin{array}{ccc} \frac{\sin y + 2y}{y} & : & y > 0 \\ \frac{3+y}{1-y} & : & y \le 0 \end{array}\right.$$
As $y \to 0^-$, $f \to 3$. As $y \to 0^+$, $f \to 3$. It follows that $f$ is continuous at $(x,y)=(1,0)$.
Although the function is continuous as $(1,0)$, it needs to be continuous in a neighbourhood of $(1,0)$ for it to be differentiable. The function is not continuous at $(1+\varepsilon,0)$ for arbitrarily small $\varepsilon>0$.
|
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|
How to integrate ${dx}/{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}}$
If
$\displaystyle\frac{dx}{dt}=-\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}$
and $x(0)=R$, find $t$ when $x=0$.
I have no idea how to integrate $\displaystyle\frac{dx}{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}}$. Any suggestions?
|
Starting with
$$\int\frac{dx}{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}}$$
We will replace all the constant with simpler ones. We let $\frac{2K}{m} = a^2$ and we let $R=b^2$
$$\int\frac{dx}{a\sqrt{\left(\frac{1}{x}-\frac{1}{b^2}\right)}}$$
The obvious thing now is to let $x \to b^2 x$
$$\frac{b}{a}\int\frac{dx}{\sqrt{\left(\frac{1}{x}-1\right)}}$$
And now we let $u = 1/x \implies dx= -du/u^2 $
$$\frac{-b}{a}\int\frac{du}{u^2\sqrt{u-1}}$$
A standard trig substitution will suffice now
|
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|
interval of convergence of the power series $\sum_{n=1}^\infty(1+\frac{1}{n})^{n^2}x^n$ Find the radius of convergence and the interval of convergence of the power series
$\sum_{n=1}^\infty(1+\frac{1}{n})^{n^2}x^n$
My attempt: The root test brings us to
$\lim_{n\rightarrow\infty}|(1+\frac{1}{n})^nx|$,
but I'm not sure where this brings me. Is the ratio test any help?
Any help appreciated!
|
Note: Both tests, root test as well as ratio test can be used to analyze the convergence behavior. But, the first one is in this case considerably simpler to use.
We recall the validity of the limit
\begin{align*}
\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e
\end{align*}
Root test:
We obtain
\begin{align*}
\lim_{n\rightarrow\infty}\left|\sqrt[n]{\left(1+\frac{1}{n}\right)^{n^2}x^n}\right|
=|x|\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}
=|x|e
\end{align*}
and we conclude the series is convergent for
\begin{align*}
|x|<\frac{1}{e}
\end{align*}
and divergent for $|x|>\frac{1}{e}$.
$$ $$
Ratio test:
We obtain
\begin{align*}
\lim_{n\rightarrow\infty}\left|\frac{\left(1+\frac{1}{n+1}\right)^{(n+1)^2}x^{n+1}}
{\left(1+\frac{1}{n}\right)^{n^2}x^n}\right|
&=|x|
\lim_{n\rightarrow\infty}\frac{\left(1+\frac{1}{n+1}\right)^{n^2+2n+1}}
{\left(1+\frac{1}{n}\right)^{n^2}}\\
&=|x|\lim_{n\rightarrow\infty}\left(\frac{n(n+2)}{(n+1)(n+1)}\right)^{n^2}
\left(1+\frac{1}{n+1}\right)^{2n+1}\\
&=|x|\lim_{n\rightarrow\infty}
\underbrace{\frac{1}{\left(1+\frac{1}{n^2+2n}\right)^{n^2}}}_{\longrightarrow \ \ \frac{1}{e}}
\underbrace{\left(1+\frac{1}{n+1}\right)^{2n}}_{\longrightarrow\ \ e^2}
\underbrace{\left(1+\frac{1}{n+1}\right)^{1}}_{\longrightarrow \ \ 1}\\
&=|x|e
\end{align*}
and we conclude as above.
|
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|
Limit of function with natural logarithm I need help solving this problem. I tried L'hospital and rearranging but nothing worked.
$$
\lim_\limits{x→∞} x^2\left(\ln\left(1 + \frac{1}{x}\right)- \frac{1}{x+1}\right)
$$
|
Using L'Hospital rule we get $$\\ \lim _{ x\rightarrow \infty }{ \frac { \ln { \left( 1+\frac { 1 }{ x } \right) -\frac { 1 }{ x+1 } } }{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow \infty }{ \frac { \frac { x }{ x+1 } \cdot \left( -\frac { 1 }{ { x }^{ 2 } } \right) +\frac { 1 }{ { \left( x+1 \right) }^{ 2 } } }{ \frac { -2 }{ { x }^{ 3 } } } } =\\ =\lim _{ x\rightarrow \infty }{ \frac { -x-1+x }{ { x\left( x+1 \right) }^{ 2 } } \cdot \frac { { x }^{ 3 } }{ -2 } } =\lim _{ x\rightarrow \infty }{ \frac { 1 }{ { \left( x+1 \right) }^{ 2 } } \cdot \frac { { x }^{ 2 } }{ 2 } } =\frac { 1 }{ 2 } $$
|
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|
Is there a quicker way to evaluate $\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$? The integral is: $$\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$$
My procedure:
$$4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3)^2-5}\ dx=4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3-\sqrt5)(2x^2+3+\sqrt5)}\ dx$$
$$$$Using partial fractions to separate the integral, we obtain:
$$\frac1{\sqrt5}\int_{0}^{\infty} \left(\frac{5-\sqrt5}{2x^2+3-\sqrt5}-\frac{5+\sqrt5}{2x^2+3+\sqrt5}\right)\ dx$$
$$=\frac{5-\sqrt5}{\sqrt5}\int_{0}^{\infty} \frac1{2x^2+3-\sqrt5}\ dx-\frac{5+\sqrt5}{\sqrt5}\int_{0}^{\infty}\frac1{2x^2+3+\sqrt5}\ dx$$
$$$$We need write the fractions in the form $\frac1{u^2+1}$:
$$=\frac{5-\sqrt5}{\sqrt5}\int_{0}^{\infty} \frac1{(3-\sqrt5)\left(\frac{2x^2}{3-\sqrt5}+1\right)}\ dx-\frac{5+\sqrt5}{\sqrt5}\int_{0}^{\infty}\frac1{(3+\sqrt5)\left(\frac{2x^2}{3+\sqrt5}+1\right)}\ dx$$
$$=\frac{5-\sqrt5}{\sqrt5(3-\sqrt5)}\int_{0}^{\infty} \frac1{\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)^2+1}\ dx-\frac{5+\sqrt5}{\sqrt5(3+\sqrt5)}\int_{0}^{\infty}\frac1{\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)^2+1}\ dx$$
$$\frac{5-\sqrt5}{\sqrt{10(3-\sqrt5)}}\int_{0}^{\infty} \frac{\sqrt2}{\sqrt{3-\sqrt5}\left(\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)^2+1\right)}\ dx-\frac{5+\sqrt5}{\sqrt{10(3+\sqrt5)}}\int_{0}^{\infty}\frac{\sqrt2}{\sqrt{3+\sqrt5}\left(\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)^2+1\right)}\ dx$$
$$$$Evaluating the indefinite integrals, we obtain:
$$\arctan\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)+C$$
$$$$Evaluating the improper integral:
$$\lim_{z \to \infty}\left[\arctan\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)\right]_0^z$$
$$=\lim_{z \to \infty}\left(\arctan\left(\frac{\sqrt2z}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2z}{\sqrt{3+\sqrt5}}\right)\right)$$
$$=\lim_{z \to \infty}\left(\arctan\left(\infty\right)-\arctan\left(\infty\right)\right)=\frac{\pi}2-\frac{\pi}2=0$$
The way that I evalauted the integral is pretty long and has a lot of calculations. Is there an easier way?
|
We have
\begin{align*}
\int \frac{1-x^2}{x^4 + 3x^2 + 1} \, dx
&= - \int \frac{1 - x^{-2}}{(x + x^{-1})^2 + 1} \, dx \\
&= - \arctan\left(x + \frac{1}{x}\right) + C.
\end{align*}
Here, we utilized the substitution $u = x + x^{-1}$ when we move to the second line. For the definite integral, we have
$$ \int_{0}^{\infty} \frac{1-x^2}{x^4 + 3x^2 + 1} \, dx = \left[ - \arctan\left(x + \frac{1}{x}\right) \right]_{0^+}^{+\infty} = -\frac{\pi}{2} + \frac{\pi}{2} = 0. $$
|
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|
Solving and proving inequalities? If $a,b$ and $c$ are positive real numbers, how do I prove that:
$$\frac{a^3}{b^2-bc+c^2}+
\frac{b^3}{c^2-ca+a^2} + \frac{c^3}{a^2-ab+b^2} \geq 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$$
and when is equality?
Are there general techniques to solve these symmetric and or cyclic inequalities?
|
By C-S $$\sum_{cyc}\frac{a^3}{b^2-bc+c^2}=\sum_{cyc}\frac{a^4}{b^2a+c^2a-abc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b+a^2c-abc)}.$$
Thus, it remains to prove that
$$(a^2+b^2+c^2)^2(a+b+c)\geq3(ab+ac+bc)\sum\limits_{cyc}(a^2b+a^2c-abc)$$ or
$$\sum_{cyc}(a^5+a^4b+a^4c-a^3b^2-a^3c^2-6a^3bc+5a^2b^2c)\geq0.$$
Since by Schur $\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0$, it remains to prove that
$$\sum_{cyc}(2a^4b+2a^4c-a^3b^2-a^3c^2-7a^3bc+5a^2b^2c)\geq0$$ or
$$2\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)+\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)-5abc\sum_{cyc}(a^2-bc)\geq0$$ or
$$\sum_{cyc}(a-b)^2\left(2ab(a+b)+c^3-\frac{5}{2}abc\right)\geq0,$$
which is true by AM-GM:
$$2ab(a+b)+c^3-\frac{5}{2}abc\geq4\sqrt{a^3b^3}+c^3-\frac{5}{2}abc=2\cdot2\sqrt{a^3b^3}+c^3-\frac{5}{2}abc\geq$$
$$\geq3\sqrt[3]{4a^3b^3c^3}-\frac{5}{2}abc=\left(3\sqrt[3]4-\frac{5}{2}\right)abc>0.$$
Done!
|
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|
Evaluating $\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ Let $A(z_1),B(z_2),C(z_3)$ are complex numbers satisfying $|z-\sqrt{3}i|=1$ and $3z_1+\sqrt{3}i=2z_2+2z_3$. The question asks to find the value of $$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$$ I tried to shift the origin to $\sqrt{3}i$ and transformed the original condition in this system but it does not help me. Any ideas how to proceed? Thanks.
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We can rewrite the equation as $3(z_1 - \sqrt {3} i) = 2(z_2-\sqrt {3}i)+2(z_3-\sqrt {3}i)$
i.e. $3\vec{OZ_1} = 2\vec{OZ_2}+2 \vec{OZ_3}$
So if $\theta = \pm \arg \frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ is the angle between $\vec{OZ_2}$ and $\vec{OZ_3}$, we have from the above equation that
$9 = 2^2+2^2+8 \cos \theta \Rightarrow \cos \theta = \frac{1}{8} \Rightarrow \sin \theta = \pm \frac{3\sqrt 7}{8}$
Since $\left| \frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}\right|=1$, we have that
$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i} = \frac{1}{8} \pm \frac{3\sqrt 7}{8}i$
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|
Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know where to go next. Anyway, this is one of my many attempts to solve this question, and most of them ends with a complicated solution. I don't want to use modular arithmetic for this question. A hint or anything will help me.
|
The tool you need is modular arithmetic.
$3^{11}$ is congruent to $0$ mod $9$
$3^{11} - 1$ is congruent to $8$ mod $9$
The only hard part is the division. But, fortunately, you can uniquely divide by $2$ mod $9$:
$0*2 = 0,
1*2 = 2,
2*2 = 4,
3*2 = 6,
4*2 = 8,
5*2 = 1,
6*2 = 3,
7*2 = 5,$ and
$8*2 = 7$
so $8/2 = 4$ modulo $9$, thus the remainder is $4$.
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"timestamp": "2023-03-29T00:00:00",
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|
An interesting limit: $\lim_{n\rightarrow \infty}\frac{1}{n^{2}+1}+\frac{2}{n^{2}+2}+\cdots+\frac{n}{n^{2}+n}$ $$\lim_{n\rightarrow \infty}\frac{1}{n^{2}＋1}＋\frac{2}{n^{2}＋2}＋…＋\frac{n}{n^{2}＋n}$$
Thanks in advance
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HINT:
$$\frac{1}{n^2+n}\sum_{k=1}^n(k)\le \sum_{k=1}^n\frac{k}{n^2+k}\le \frac{1}{n^2+1}\sum_{k=1}^n (k)$$
SPOILER ALERT: Scroll over the highlighted area to reveal the solution
We can write the limit as$$\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k}$$Then, using $n^2+n\ge n^2+k\ge n^2+1$ we have $$\frac{1}{n^2+n}\sum_{k=1}^n(k)\le \sum_{k=1}^n\frac{k}{n^2+k}\le \frac{1}{n^2+1}\sum_{k=1}^n (k)\tag 1$$We can sum the arithmetic progression to obtain $$\sum_{k=1}^n (k)=\frac{n(n+1)}{2}\tag 2$$Using $(2)$ in $(1)$ reveals $$\frac{n(n+1)}{2(n^2+n)}\le \sum_{k=1}^n\frac{k}{n^2+k}\le \frac{n(n+1)}{2(n^2+1)}\tag 3$$whereupon applying the squeeze theorem to $(3)$ yields the coveted limit $$\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k}=\frac12$$
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|
Find $N$ given a sequence $a_n$
In the sequence {$a_n$}, $$a_n=\frac{a_1+a_2+...+a_{n-1}}{n-1} \, , \quad n \ge 3$$
If $a_1+a_2 \ne 0$ and the sum of the first $N$ terms is $12(a_1+a_2)$, find $N$.
Kind of lost on where to start with this one. My initial thought was, $$\sum^{N}_{n=3}\frac{a_1+a_2+...+a_{n-1}}{n-1}=12(a_1+a_2)$$ $$\sum^{N}_{n=3}\frac{a_1+a_2+...+a_{n-1}}{n-1}=\frac{1}{2}(a_1+a_2)+\frac{1}{3}(a_1+a_2+a_3)...$$but someting seems wrong here, or I do not know where to go from here.
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\begin{align*}
S_2 &= a_1+a_2 \\
S_n &= a_1+a_2+\ldots+a_n \\
a_{n+1} &= \frac{a_1+\ldots+a_{n}}{n} \\
&= \frac{S_n}{n} \\
S_{n+1} &= S_{n}+\frac{S_n}{n} \\
&= \frac{n+1}{n} S_{n} \\
&= \frac{n+1}{n} \cdot \frac{n}{n-1} \cdot \ldots \cdot \frac{4}{3} \cdot \frac{3}{2} S_{2} \\
&= \frac{n+1}{2} S_2 \\
S_{N} &= \frac{N}{2} S_2 \\
&= \frac{N(a_1+a_2)}{2} \\
\end{align*}
Also $\forall n\ge 3$, $$a_{n}=\frac{a_1+a_2}{2}$$
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|
Find the greatest integer not exceeding $A$ if $A = \sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}$ and $x=20062007$.
Find the greatest integer not exceeding $A$ if $$A = \sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}$$ and $x=20062007$.
The nested square roots are very tricky and I am not able to find any pattern or something except that every $x^2$ term is a perfect square. As far as I think, it is not plausible to denest the radical by calculating the square roots.
Is there any simpler approach ? I am interested in the general case when there is no particular given value of $x$. I have tried my best, but there is nothing I can come up with.
Thanks in Advance ! :-)
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A).
Show that $A>20062008$:
$$E:=100x^2+39x+\sqrt{3}>64x^2+16x+1 = (8x+1)^2;$$
$$D:=16x^2 + \sqrt{E} > 16x^2 + (8x+1) = (4x+1)^2;$$
$$C:=4x^2+ \sqrt{D} > 4x^2+(4x+1) = (2x+1)^2;$$
$$B:=x^2+\sqrt{C} > x^2+(2x+1) = (x+1)^2;$$
$$A=\sqrt{B}>x+1.$$
B). Show that $A<20062009$. More rough estimation:
$$E<(10x+2)^2;$$
$$D<16x^2+(10x+2)<(4x+2)^2;$$
$$C<4x^2+(4x+2)<(2x+2)^2;$$
$$B<x^2+(2x+2)<(x+2)^2;$$
$$A<x+2.$$
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|
Proving that $c$ is the mean proportional between $a$ and $b$.
If $a≠b$ and $a:b$ is the duplicate ratio of $(a+c):(b+c)$, prove that $c$ is the mean proportional between $a$ and $b$.
My attempt:
I have assumed that $c$ is the mean proportional between $a$ and $b$, expressed $a$ and $c$ in terms of $b$ and then substituted them in $(a+c)^2:(b+c)^2$ and $a:b$ to get equal expressions on L.H.S and R.H.S, i.e., i have worked in a reverse manner.
My question:
How to prove this starting from $(a+c)^2:(b+c)^2=a:b$?
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Hint:
$$
\require{cancel}
\begin{align}
\frac{a}{b}=\frac{(a+c)^2}{(b+c)^2} \quad & \implies \quad a(b+c)^2=b(a+c)^2 \\
& \iff \quad a(b^2+2bc+c^2)=b(a^2+2ac+c^2) \\
& \iff \quad ab^2 +\cancel{2abc}+ac^2-a^2b-\cancel{2abc}-bc^2=0 \\
& \iff \quad ab(b-a)+(a-b)c^2=0 \\
& \iff \quad (a-b)(c^2-ab)=0
\end{align}
$$
All that's left is to remember that $a \ne b\,$.
|
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|
How come rearrangement of a convergent series may not converge to the same value / or does not converge at all By looking at Riemann's Rearrangement theorem I wonder, How come a convergent series particular re-arrengement may not converge to the same value of the original series.
Isn't the below true?
Let $\phi : N \to N$ be a bijection. Where $N$ is the natural number set
A re-arrengement of a sequence $\sum a_n$ would be $\sum a_{\phi(n)}$
Then Give me an example of a series that have a rearrangement which does not converge or does not converge to the same value.
I mean... 4+1+2+2+9+6 = 1+2+2+6+4+9 = 24
??
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Here's a concrete example where rearrangement of a convergent series results in a series that converges to a different value.
Consider
$$
1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3}
+ \frac{1}{4} - \frac{1}{4} + \frac{1}{5} - \frac{1}{5} + \ldots
$$
The partial sums of this series alternate 0 between values of the form $1/n$, so it converges to zero.
Now consider
$$
1 + \frac{1}{2} - 1 + \frac{1}{3} + \frac{1}{4} - \frac{1}{2}
+ \frac{1}{5} + \frac{1}{6} - \frac{1}{3}
+ \frac{1}{7} + \frac{1}{8} - \frac{1}{4} + \ldots
$$
This is clearly a rearrangement of the first, and it's not too hard to see that you don't change the convergence behavior if you sum the terms in groups of 3:
$$
\left(1 + \frac{1}{2} - 1\right) + \left(\frac{1}{3} + \frac{1}{4} - \frac{1}{2}\right)
+ \left(\frac{1}{5} + \frac{1}{6} - \frac{1}{3}\right)
+ \left(\frac{1}{7} + \frac{1}{8} - \frac{1}{4}\right) + \ldots \\
= \frac{1}{2} + \frac{1}{12} +\frac{1}{30} + \frac{1}{56} + \ldots \\
$$
Now by standard methods you can check that this converges to some value greater than zero. However, expanding each term as a difference shows that it actually converges to a known value.
$$
\frac{1}{2} + \frac{1}{12} +\frac{1}{30} + \frac{1}{56} + \ldots \\
=\left(1 - \frac{1}{2} \right) + \left(\frac{1}{3} - \frac{1}{4} \right)
+ \left(\frac{1}{5} - \frac{1}{6} \right)
+ \left(\frac{1}{7} - \frac{1}{8} \right) + \ldots \\ = \ln(2)
$$
|
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|
Proving G.Bauer's Pi Formula
Question: How do you prove this $\pi$ formula found by G.Bauer in $1859$$$\dfrac 2\pi=1-5\left(\dfrac {1}{2}\right)^3+9\left(\dfrac {1\cdot3}{2\cdot4}\right)^3-13\left(\dfrac {1\cdot3\cdot5}{2\cdot4\cdot6}\right)^3+\&\text{c}.$$
I'm not sure what to do in this case. I found it in a wikipedia article and would like to know how to prove it.
And do you think it's possible to use the expansion$$(1-x)^{-5}=1+5x+\dfrac {5\times6}{1\times2}x^2+\dfrac {5\times6\times7}{1\times2\times3}x^3+\dfrac {5\times6\times7\times8}{1\times2\times3\times4}x^5+\&\text{c}.\tag{2}$$
Thanks in advance!
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This is what I have so far:
Start with the Hypergeometrical Series$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)}\tag1$$
And let $n=-x=-y=\frac 12$. Therefore, $(1)$ takes the form
$$\begin{align*}_4F_3\left[\begin{array}{c c}\frac 54,\frac 12,\frac 12,\frac 12\\\frac 14,1,1\end{array};-1\right] & =\dfrac {1}{\Gamma\left(\frac 32\right)\Gamma\left(\frac 12\right)}\\ & =\dfrac 2\pi\tag{2}\end{align*}$$
And since the general Hypergeometrical series obeys$$_pF_q\left[\begin{array}{c c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_p\end{array};x\right]=\sum\limits_{k=0}^\infty\dfrac {(\alpha_1)_k(\alpha_2)_k\cdots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\cdots(\beta_p)_k}\dfrac {x^k}{k!}\tag{3}$$
We have the LHS as$$\begin{align*}_4F_3\left[\begin{array}{c c}\frac 54,\frac 12,\frac 12,\frac 12\\\frac 14,1,1\end{array};-1\right] & =\sum\limits_{k=0}^{\infty}\dfrac {\left(\frac 54\right)_k\left(\frac 12\right)_k\left(\frac 12\right)_k\left(\frac 21\right)_k(-1)^k}{\left(\frac 14\right)_k\left(1\right)_k\left(1\right)_kk!}\\ & =1-5\left(\dfrac 12\right)^3+9\left(\dfrac {1\cdot3}{2\cdot4}\right)^3-13\left(\dfrac {1\cdot3\cdot5}{2\cdot4\cdot6}\right)^3+\&\text c.\tag{4}\end{align*}$$
Therefore, the identity is established. However, now the question simplifies into
Question: How do we prove$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)}\tag{5}$$
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.