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Find all integer $n$ so that both $n+2008$ divides $n+2008^2$ and $n+2009$ divides $n+2009^2$ Find all integer $n$ so that both $n+2008$ divides $n+2008^2$ and $n+2009$ divides $n+2009^2$
Obviously $n$=$0$ is a solution, but are there any others? Thank you
|
$$a|b \iff \gcd(a, b) = a$$
$$\gcd(a, b) = \gcd(a, b-a)$$
So $\gcd(n + 2008, n+2008^2) = \gcd(n+2008, 2008^2 - 2008) = n + 2008$, similarly for 2009.
$$\begin{cases} n + 2008 ~|~ 2008 \cdot 2007 \\
n + 2009 ~|~ 2009 \cdot 2008\end{cases}$$
So $n=0$ is the only nonnegative solution. For negative solutions, brute force works as the range is pretty small. Otherwise find the values of $k$ where $k ~|~ 2008\cdot 2007$ and $k+1 ~|~ 2009\cdot 2008$ which are
$$k \in \{-9, -8, -3, -2, 1, 3, 6, 2007, 2008\}$$
which corresponds by $k = n + 2008$ to:
$$n \in \{-2017,-2016,-2015,-2010,-2007,-2005,-2002,-1,0\}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the inverse of a polynomial in a quotient ring Consider $\mathbb Z_5[x]/I$ with $I$ as ideal generated by $b=x^3+3x+2$. If $(x+2) + I$ is element of $\mathbb Z_5[x]/I$ that has an inverse. Find the inverse of $(x+2) + I$.
I stuck to get the inverse because the gcd of that is not $1$.
|
Hint: $$(x+2)(ax^2+bx+c) + I =ax^3+(2a+b)x^2+(2b+c)x+2c+I\\ = (2a+b)x^2 +(2b+c-3a)x+2(c-a) + I$$
Now solve
\begin{align}
2a+b &=0\\
2b+c-3a &= 0\\
2(c-a) &= 1
\end{align}
in $\Bbb Z_5$
|
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|
$1!+2!+3!+4!+5!+...+n!$ is a square of an integer At what n is the number $1!+2!+3!+4!+...+n!$ a square of another integer number?
All in all I have found $2$ values: $1$ and $3$. I think they are the only ones, but the only thing left is to prove it. How?
|
Let $1!+2!+3!+\cdots \cdots +n! = y^2$
$\star$ if $n=1\;,$ then $1=y^2$ (True)
$\star$ if $n=2\;,$ then $1+2=3=y^2$ (False)
$\star$ if $n=3,$ then $1+2+6=9=y^2$ (True)
$\star$ if $n=4,$ then $1+2+6+24=33=y^2$(False)
for $n\geq 5,$ then L .H .S end with $3$ and we now that square of any integer does not have last digit $3$
so we have only $n=1,n=3$ for which $y^2$ is a perfect square quantity
|
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|
Using the binomial theorem I'd like some help with proving the next equation:
$$\sqrt{1+x}=\sum_{0}^{\infty }\frac{(-1)^{n-1}}{2^{2n-1}\cdot n}\binom{2n-2}{n-1}\cdot x^{n}$$
|
Let's rewrite the expression in terms of the Gamma function
$$
\sqrt {1 + x} = \sum\limits_{0 \leqslant n} {\frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\left( \begin{gathered}
2(n - 1) \\
n - 1 \\
\end{gathered} \right)x^{\,n} } = \sum\limits_{0 \leqslant n} {\frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\frac{{\Gamma (2n - 1)}}
{{\Gamma (n)^{\,2} }}x^{\,n} }
$$
and then consider the Duplication formula
$$
\Gamma \left( {2z} \right) = \frac{{2^{\,2\,z - 1} }}
{{\sqrt \pi }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right) = \frac{{2^{\,2\,z - 1} }}
{{\Gamma \left( {1/2} \right)}}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)
$$
so that the second term in the summand will become
$$
\frac{{\Gamma (2n - 1)}}
{{\Gamma (n)^{\,2} }} = \frac{{\Gamma \left( {2\left( {n - 1/2} \right)} \right)}}
{{\Gamma (n)^{\,2} }} = \frac{{2^{\,2\,n - 2} }}
{{\Gamma \left( {1/2} \right)}}\frac{{\Gamma \left( {n - 1/2} \right)}}
{{\Gamma \left( n \right)}}
$$
Therefore the whole coefficient of $x^n$ reduces to:
$$
\begin{gathered}
\frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\left( \begin{gathered}
2(n - 1) \\
n - 1 \\
\end{gathered} \right) = \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\frac{{\Gamma (2n - 1)}}
{{\Gamma (n)^{\,2} }} = \hfill \\
= \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\frac{{2^{\,2\,n - 2} }}
{{\Gamma \left( {1/2} \right)}}\frac{{\Gamma \left( {n - 1/2} \right)}}
{{\Gamma \left( n \right)}} = \hfill \\
= \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2n}}\frac{{\Gamma \left( {n - 1/2} \right)}}
{{\Gamma \left( {1/2} \right)\Gamma \left( n \right)}} = \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2n}}\left( \begin{gathered}
n - 1 - 1/2 \\
n - 1 \\
\end{gathered} \right) = \hfill \\
= \frac{1}
{{2n}}\left( \begin{gathered}
- 1/2 \\
n - 1 \\
\end{gathered} \right) = \frac{{\frac{1}
{2}\Gamma (1/2)}}
{{n\;\Gamma (n)\;\Gamma (1 + 1/2 - n)}} = \frac{{\Gamma (1 + 1/2)}}
{{\Gamma (1 + n)\;\Gamma (1 + 1/2 - n)}} = \hfill \\
= \left( \begin{gathered}
1/2 \\
n \\
\end{gathered} \right) \hfill \\
\end{gathered}
$$
Note that the Duplication formula stems from splitting the product into even and odd components
as follows
$$
\begin{gathered}
\prod\limits_{0\, \leqslant \,k\; \leqslant \,n} {\left( {x + k} \right)} = \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x + 2j} \right)} \;\prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x + 2j + 1} \right)} = \hfill \\
= 2^{\,\left\lfloor {n/2} \right\rfloor + 1} \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x/2 + j} \right)} \;\;2^{\,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor + 1} \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x/2 + 1/2 + j} \right)} = \hfill \\
= \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x + 2j} \right)\prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x - 1 + 2\left( {j + 1} \right)} \right)} } = \hfill \\
= \quad \; \cdots \hfill \\
\end{gathered}
$$
with all the various "manipulations" you can do on the terms and on the multiplication bounds.
|
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Difficult definite integration $\int_{-2}^{0}\frac{x^2 + x - 5}{(x-1)^2}e^x\,\mathrm dx$ I found an integral in a contest that seems very difficult to compute. The answer is $-2$, however, I do not know how to arrive at this answer.
$$\int_{-2}^{0}\frac{x^2 + x - 5}{(x-1)^2}e^x\,\mathrm dx$$
At first I tried to make the substitution $u = x - 1$, but I did not get anywhere. I also tried to expand the denominator and perform synthetic division, which did not help so much either.
Also, I don't think it is possible to do partial fraction decomposition since the degree of the denominator is equal to the degree of the numerator.
Attempt with substitution (Moo's help) -
Let $u = x - 1$. Then, $du = 1$
$$\begin{align}
I &= \int_{-2}^{0} \frac{x^2 + x - 5}{(x - 1)^2} e^x\,\mathrm dx =\\
&= \int_{-2}^{0} \frac{x^2 + x - 5}{u^2} e^{u+1}\,\mathrm du =\\
&= \int_{-2}^{0} \frac{e^{u+1}(x^2 + x - 5)}{u^2}\,\mathrm du
\end{align}$$
Now, $x = u + 1$, $x^2 = u^2 + 2u + 2$ and therefore
$$I = \int_{-3}^{-1} \frac{e^{u+1}(u^2 + 3u - 2)}{u^2}\mathrm du = e \int_{-3}^{-1} e^{u}\frac{u^2 + 3u - 3}{u^2}\mathrm du$$
|
Integration by part we get
\begin{align*}
\int\frac{x^2 + x - 5}{(x-1)^2}e^x \, \mathrm{d}x&=-\frac{x^2 + x - 5}{x-1}e^x +\int e^{x}\left ( x+4 \right )\, \mathrm{d}x \\
&=-\frac{x^2 + x - 5}{x-1}e^x+e^{x}\left ( x+3 \right )\\
&=\frac{x+2}{x-1}e^{x}
\end{align*}
Then you can take it from here.
|
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|
Number of pairs of prime numbers $p$ and $q$ such that $p$ divides $q-1$ and $q$ divides $p^3-1$
Determine the number of pairs of prime numbers $p$ and $q$, both of which are less than $2017,$ such that $p$ divides $q-1$ and $q$ divides $p^3-1$.
If $p = 2$, then $q = 7$ and if $q = 2$, there are no solutions. Now suppose that $p,q > 2$. We have from the given that $q \equiv 1 \pmod{p}$ and $p^3 \equiv 1 \pmod{q}$. Thus, $q = 1+pk$ where $k$ is even and $p^3 = 1+qm$, where $m$ is even. Therefore, $p^3 = 1+m(1+pk)$ and so $m = \dfrac{p^3-1}{pk+1}$.
I didn't see how to continue from here.
|
If $p\mid q-1$ then certainly $p<q$. Then from $q\mid p^3-1=(p-1)(p^2+p+1)$, we conclude $q\mid p^2+p+1$.
As you wrote, $p=2$ leads to $q$ odd (wow!) and $q\mid p^-1=7$, so $q=7$.
Next, $p=3$ leads to $q\mid 9+3+1=13$, so $q=13$, which gives a solution.
Then, $p=5$ leads to $q\mid 25+5+1=31$, so $q=31$, which gives a solution.
On the other hand, $p=7$ leads to $q\mid 49+7+1=3\cdot 19$, but $q=19$ does not work, so our luck has come to a halt for the moment.
For a systematic approach:
Write $q=kp+1$ with $k\ge2$ (because we already treated the case $p=2$, $q=3$ above), and $p^2+p+1=mq=mkp+m$ with $m\ge1$. Then $p\mid p^2+p-mkp=m-1$, so $m=rp+1$ with $r\ge 0$.
If $r=0$, we are led to $m=1$, $p^2+p+1=q=kp+1$, $k=p+1$. As we want $q<2017$, we must have $p<\sqrt{2017}< 45$. By inspection, we find all primes in this range and check if $p^2+p+1$ is prime: $p=2,3,5,17,41$ (and accordingly $q=7,13,31,307,1723$)
If $r>0$, then $p^2+p+1=(rp+1)(kp+1)\ge krp^2\ge 2p^2>p^2+p+1$, and there is no solution.
|
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|
Generating function for counting compositions of $n$ I was given the next question: I was asked to find the generating function for the number of divisions of $n$ (a given natural number) with exactly 3 elements.
For example: if $n$ equals to 5 then 3,1,1 are a set of one of the options (3 + 1 + 1 = 5)
I tried thinking about using an exponential generating function but I'm failing to see whether it's right or wrong. Any help will be appreciated!
Repetitions are not allowed
|
We derive the wanted generating function for the number of partitions of $n$ with exactly three parts by starting with a seemingly different generating function.
The generating function for the number of partitions which consist of zero or more of $1,2,$ and $3$ is
\begin{align*}
&(1+x+x^2+x^3+\cdots)(1+x^2+x^4+x^6+\cdots)(1+x^3+x^6+x^9+\cdots)\\
&\qquad=\frac{1}{(1-x)(1-x^2)(1-x^3)}
\end{align*}
Note: The number of partitions consisting of numbers $\leq k$ is the same as the number of partitions with number of parts $\leq k$.
If we use Ferrer diagrams to visualise the situation we see that each partition containing numbers $\leq k$ which is reflected at the main diagonal corresponds with a partition containing $\leq k$ summands.
Since this correspondence is bijective the generating function is the same in both cases.
We conclude a generating function for the number of partitions with exactly three parts is
\begin{align*}
&\frac{1}{(1-x)(1-x^2)(1-x^3)}-\frac{1}{(1-x)(1-x^2)}\\
&\qquad=\frac{1}{(1-x)(1-x^2)}\left(\frac{1}{1-x^3}-1\right)\\
&\qquad=\frac{x^3}{(1-x)(1-x^2)(1-x^3)}\\
&\qquad=x^3+x^4+2x^5+3x^6+4x^7+\color{blue}{5}x^8+7x^9\cdots
\end{align*}
The last line was done with some help of Wolfram Alpha.
Example: There are $\color{blue}{5}$ partitions of $8$ with three summands
\begin{align*}
8&=1+1+6\\
&=1+2+5\\
&=1+3+4\\
&=2+2+4\\
&=2+3+3
\end{align*}
|
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if $x$ is rational and $x^2$ is natural, prove that $x$ is integer If $x$ is rational: There exists $\frac{a}{b}$ such that $a$, $b$ are integers.
If $x^2$ is natural:
$\left(\frac{a}{b}\right)^2$ is natural => $\frac{a^2}{b^2}$ is natural
Then $a^2$ divides $b^2$ => $a$ divides $b$
If $a$ divides $b$ and $a$, $b$ are integers, $\frac{a}{b}$ is integer
so $x$ is integer.
|
Let $x^2 = n \in \mathbb{N}\,$, then $x$ is a root of $x^2-n=0$. Since $x^2-n$ is a monic polynomial, any rational root must be an integer by the rational root theorem, so $x$ is an integer.
|
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Show the recursive sequence is increasing
How do I show that the recursive sequence
$$a_n = a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1, \quad
n\geq 2, \phantom{x} a_1 = 3$$
is an increasing sequence?
1. attempt:
If I can show that $a_{n+1}-a_n>0$, I would be able to show it is increasing.
\begin{align*}
a_{n+1} - a_n & = a_{\left\lfloor \frac{n+1}{2} \right\rfloor} + a_{\left\lceil \frac{n+1}{2} \right\rceil} +3(n+1)+1 - ( a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1) \\
& = a_{\left\lfloor \frac{n+1}{2} \right\rfloor} + a_{\left\lceil \frac{n+1}{2} \right\rceil}+3 - a_{\lfloor n/2 \rfloor} - a_{\lceil n/2 \rceil}
\end{align*}
I can't put $a$ together because the indexes are different (because of the ceils and floors).
2. attempt:
Proof by induction
Base case: $n=2$ then $a_2 = a_1 + a_1 + 3\cdot 2 + 1 = 3+3+7=13$ so $a_1<a_2$.
Testing with more gives: $a_1 < a_2 < a_3 =26 < a_4 = 39<... $
Assume $a_n<a_{n+1}$.
Now I want to show that $a_{n+1} < a_{n+2}$
$$ a_{n+2} = a_{\lfloor (n+2)/2 \rfloor} + a_{\lceil (n+2)/2 \rceil} +3(n+2)+1$$
Again I get stuck since I don't know how to handle the ceils and floors.
$\phantom{x}$
How do I go about showing the sequence is increasing? Maybe I'm making it harder than it actually is - is there by any chance an easier way?
|
For an even index we have
$$a_{2n}=2a_n+6n+1$$
and for an odd one,
$$a_{2n+1}=a_n+a_{n+1}+6n+4$$
so if we assume (using induction) that $a_{n+1}\ge a_n$ then
$$a_{2n+2}=2a_{n+1}+3(2n+2)+1=2a_{n+1}+6n+7>a_n+a_{n+1}+6n+4=a_{2n+1}$$
and
$$a_{2n+1}=a_n+a_{n+1}+6n+4>2a_n+6n+1=a_{2n}$$
Remark: Perhaps one of two more base cases are needed.
|
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Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\leq2$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\leq2$$
I tried C-S, BW, uvw and more, but without any success.
|
Here is my proof. Dedicated to dear Dr. Sonnhard Graubner.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$,$abc=w^3$ and $u=xw$.
Hence, $x\geq1$ and we need to prove that $$\sum_{cyc}\frac{a}{a+b}+\frac{w}{2u}\leq2$$ or
$$\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)\leq\frac{1}{2}-\frac{w}{2u}$$ or
$$(u-w)(9uv^2-w^3)\geq u\sum_{cyc}(a-b)(a+c)(b+c)$$ or
$$(u-w)(9uv^2-w^3)\geq u\sum_{cyc}(a-b)c^2$$ or
$$(u-w)(9uv^2-w^3)\geq u(a-b)(a-c)(b-c),$$
which says that it remains to prove our inequality for $a\geq b\geq c$.
Id est, it remains to prove that $$(u-w)^2(9uv^2-w^3)^2\geq u^2(a-b)^2(a-c)^2(b-c)^2$$ or
$$(u-w)^2(9uv^2-w^3)^2\geq 27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or
$f(v^2)\geq0$, where $f(v^2)=108u^2v^6-81u^2w(2u-w)v^4-$
$-18uw^3(10u^2-2uw+w^2)v^2+w^3(108u^5+28u^2w^3-2uw^4+w^5)$.
But
$$f'(v^2)=324u^2v^4-162u^2w(2u-w)v^2-18uw^3(10u^2-2uw+w^2),$$
which says that $v^2_{min}=\frac{6x^2-3x+\sqrt{36x^4+44x^3-7x^2+8x}}{12x}w^2$
and it remains to prove that $f\left(v^2_{min}\right)\geq0$ or
$$648x^5-396x^4+342x^3+107x^2+20x+8\geq\sqrt{x(36x^3+44x^2-7x+8)^3}$$ or
$$(x-1)^2(5832x^8+972x^7+2646x^6+1477x^5+581x^4+105x^3+51x^2-x+1)\geq0.$$
Done!
|
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|
Finding a closed form expression for $\sum_{i=1}^{n-1}\csc{\frac{i\pi}{n}}$ Consider
$\displaystyle{S =
\sum_{k = 1}^{n - 1}\csc\left(k\,{\pi \over n}\right) =
\frac{1}{\sin\left(\pi/n\right)} +
\frac{1}{\sin\left(2\pi/n\right)} +
\frac{1}{\sin\left(3\pi/n\right)} + \cdots +
\frac{1}{\sin\left(\left[n - 1\right]\pi/n\right)}}$
How can we find a general formula for $S$ using trigonometry identities or complex numbers ?.
|
Let we assume for first that $n$ is an odd number, $n=2N+1$. In such a case the given sum is
$$ \sum_{k=1}^{2N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)} = 2\sum_{k=1}^{N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)}=\frac{4N+2}{\pi}H_N+2\sum_{k=1}^{N}\left[\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)}-\frac{1}{\frac{\pi k}{2N+1}}\right] $$
and $\frac{1}{\sin(x)}-\frac{1}{x}$ is an integrable function on the interval $\left(0,\frac{\pi}{2}\right)$, whose integral equals $\log\frac{4}{\pi}$.$^{(*)}$
By Riemann sums is follows that:
$$ \sum_{k=1}^{2N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)} = \frac{4N}{\pi}\left[H_N+\log\frac{4}{\pi}\right]+O(\log N).$$
In the general case we get that the given sum behaves like $Cn\log n$.
$^{(*)}$ Since $\text{Res}\left(\frac{1}{\sin x},x=k\pi\right)=(-1)^k$, by Herglotz' trick we have
$$\frac{1}{\sin x}-\frac{1}{x}=\sum_{k\geq 1}\left(\frac{1}{x-k\pi}+\frac{1}{x+k\pi}\right)(-1)^k $$
and by termwise integration
$$ \int_{0}^{\pi/2}\left(\frac{1}{\sin x}-\frac{1}{x}\right)\,dx = \sum_{k\geq 1}(-1)^k \log\left(1-\frac{1}{4k^2}\right) $$
so $I=\log\frac{4}{\pi}$ by simplifying the partial sums of the last series and recalling Wallis product.
|
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How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ How do I find this particular sum?
$$\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$$
where $H_n = \sum_{k=1}^{n}\frac1k$.
This was given to me by a friend and I have absolutely no idea how to proceed as I have never done these questions before. If possible, please give a way out without using polylogarithmic functions or other non-elementary functions.
|
I thought it would be instructive to present a way forward that relies on elementary analysis, including knowledge of the sum $\sum_{k=1}^\infty \frac1{k^2}=\pi^2/6$, partial fraction expansion, and telescoping series. It is to that end we proceed.
The Harmonic number, $H_{n+2}$, can be written as
$$H_{n+2}=\sum_{k=1}^{n+2}\frac1k=\frac{1}{n+2}+\frac{1}{n+1}+\sum_{k=1}^n \frac1k$$
Therefore, we have
$$\begin{align}
\sum_{n=1}^\infty\frac{1}{n(n+2)}\sum_{k=1}^{n+2}\frac1k&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty \frac1k \sum_{n=k}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+2)}\right)\\\\
&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty\frac1{2k^2}+\frac12\\\\
&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\frac{\pi^2}{12}+\frac12\\\\
&=\color{red}{\sum_{n=1}^\infty\left(\frac{1}{4n}-\frac{1}{4(n+1)}-\frac{1}{2(n+2)^2}\right)}\\\\
&+\color{blue}{\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{2(n+1)}\right)}\\\\
&+\frac{\pi^2}{12}+\frac12\\\\
&=\color{blue}{1-\frac{\pi^2}{12}}+\color{red}{\frac14}+\frac{\pi^2}{12}+\frac12\\\\
&=\frac74
\end{align}$$
|
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"url": "https://math.stackexchange.com/questions/2104031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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|
Double summation equation involving a parameter While trying to solve a statistics probability density function question, I have reached a point where I unable to proceed with solving for the value of $c$ in the equation:
$\sum_{i=1}^\infty \sum_{j=1}^\infty \frac{c}{(i+j-1)(i+j)(i+j+1)} = 1$, where $c>0$
I went through the list of series and sequences, including Taylor series, but I could not find any which will help simplify the summations and solve for the value of $c$. How can I solve such an equation? I would really appreciate some help with this.
|
The starting point is partial fraction decomposition :
$$\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2(i+j-1)}-\frac{1}{i+j}+\frac{1}{2(i+j+1)}$$
Now, let's compute a partial sum (indexed by $j$) :
$$\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\sum_{j=1}^q\left(\frac{1}{i+j-1}-\frac{1}{i+j}\right)-\frac{1}{2}\sum_{j=1}^q\left(\frac{1}{i+j}-\frac{1}{i+j+1}\right)$$
After re-indexation :
$$\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\sum_{k=i+1}^{i+q}\left(\frac{1}{k-1}-\frac{1}{k}\right)-\frac{1}{2}\sum_{k=i+1}^{i+q}\left(\frac{1}{k}-\frac{1}{k+1}\right)$$
that is :
$$\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\left(\frac{1}{i}-\frac{1}{i+q}\right)-\frac{1}{2}\left(\frac{1}{i+1}-\frac{1}{i+q+1}\right)=\frac{1}{2}\left(\frac{1}{i}-\frac{1}{i+1}\right)-\frac{1}{2}\left(\frac{1}{i+q}-\frac{1}{i+1+q}\right)$$
Now the second summation :
$$\sum_{i=1}^p\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\sum_{i=1}^p\left(\frac{1}{i}-\frac{1}{i+1}\right)-\frac{1}{2}\sum_{i=1}^p\left(\frac{1}{i+q}-\frac{1}{i+1+q}\right)$$
that is :
$$\sum_{i=1}^p\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\left(1-\frac{1}{p+1}\right)-\frac{1}{2}\left(\frac{1}{q+1}-\frac{1}{p+q+1}\right)$$
Taking the limit as $p,q\to\infty$, we get :
$$\sum_{i=1}^\infty\sum_{j=1}^\infty\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}$$
Therefore, you should choose $c=2$
|
{
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"url": "https://math.stackexchange.com/questions/2104185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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|
If $f(x)=(x-2)q(x)-8$ for polynomial $q$, and $x+2$ is a factor of $f(x)$, find the remainder when $f(x)$ is divided by $x^2-4$
Given:
$f(x)=(x-2)q(x) -8$, where $q(x)$ is a polynomial
$(x+2)$ is a factor of $f(x)$
Find the remainder when $f(x)$ is divided by $(x^2-4)$.
I know the answer is $-2x-4$, but I do not know the working behind it?
|
We know that when $f(x)$ is divided by $x-2$ the remainder is $-8$, so
$$f(x) = (x-2)q(x) - 8$$
for some polynomial $q(x)$.
We also know that $x+2$ is a factor of of $f(x)$. This means that when we plug in $x=-2$ to $f(x)$ we should get $0$. So
$$0 = (-2 -2)q(-2) - 8$$
$$8 = -4q(-2)$$
$$-2 = q(-2)$$
This last result tells us that when $q(x)$ is divided by $x+2$, the remainder is $-2$, so for some polynomial $u(x)$ we have
$$q(x)=(x+2)u(x) - 2$$
Combining these results, we now know that
$$f(x) = (x-2)((x+2)u(x) - 2) - 8$$
$$f(x) = (x^2-4)u(x) -2(x-2) - 8$$
$$f(x) = (x^2 - 4)u(x) -2x - 4$$
and now you can see the remainder when $f(x)$ is divided by $x^2-4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2104458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding the matrix associated with a linear map
Find the matrix associated with the linear map $f:R^2 \rightarrow R^2$ defined by $f(x,y)=(3x-y,y-x)$ with respect to the ordered basis ${(1,0),(1,1)}$
Let the matrix be $A$ and let $f(x)=AX$ where
$$X=\begin{bmatrix}
x \\
y
\end{bmatrix}$$
and
$$A=\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$$
I tried solving for $a,b,c,d$ by using the basis vectors as $(x,y)$. That is, I took $(x,y)=(1,0)$ and $(x,y)=(1,1)$ to find $a,b,c,d$. But I am not getting the given answer.
|
Regarding to the standard basis $e_i$ the map has the matrix
$$
A =
\begin{pmatrix}
3 & -1 \\
-1 & 1
\end{pmatrix}
$$
because
$$
f(x)
= x A
= (x_1 e_1 + x_2 e_2) A
= x_1 e_1 A + x_2 e_2 A
= (x_1, x_2)
\begin{pmatrix}
e_1 A \\
e_2 A
\end{pmatrix}
\\
e_1 = (1,0):\quad f(1,0) = (3,-1) = e_1 A = (1,0) A = (a_{11} a_{12})\\
e_2 = (0,1):\quad f(0,1) = (-1,1) = e_2 A = (0,1) A = (a_{21} a_{22})
$$
What you are looking for is
$$
B = M A M^{-1}
$$
$M$ first transforms a vector regarding the basis $b_1 = (1,0)$ and $b_2 = (1,1)$ into standard coordinates (basis $e_i$). Then it applies the function $f$ via the matrix $A$ (representing $f$ if you use it with vectors regarding standard coordinates) and finally transforms the result into the same vector regarding the basis $b_i$ via $M^{-1}$.
The combined matrices have the same effect as $f$ for vectors regarding the basis $b_i$.
As the $b_i$ regarding themselves have the coordinates $(1,0)$ and $(0,1)$ these vectors must be mapped by $M$ to $(1,0)$ and $(1,1)$, thus
$$
M =
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}
\quad
M^{-1} =
\begin{pmatrix}
1 & 0 \\
-1 & 1
\end{pmatrix}
$$
and we calculate
\begin{align}
B
&= M A M^{-1} \\
&=
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}
\begin{pmatrix}
3 & -1 \\
-1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
-1 & 1
\end{pmatrix}
\\
&=
\begin{pmatrix}
3 & -1 \\
2 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
-1 & 1
\end{pmatrix}
\\
&=
\begin{pmatrix}
4 & -1 \\
2 & 0
\end{pmatrix}
\end{align}
Test:
$e_1 = (1,0)$ regarding basis $b_i$ still has coordinates $(1,0)$,
as $(1,0) = 1 \cdot b_1 + 0 \cdot b_2$.
Then
$$
e_1 B =
(1,0)
\begin{pmatrix}
4 & -1 \\
2 & 0
\end{pmatrix}
= (4,-1)
$$
And indeed
$$
4 b_1 - b_2 = 4 (1,0) - (1,1) = (4,0)-(1,1) = (3,-1)
$$
Further $e_2 = (0,1)$ regarding standard coordinates has coordinates $(-1,1)$ regarding the $b_i$:
$$
-1 \cdot b_1 + 1 \cdot b_2 = (-1,0) + (1,1) = (0,1)
$$
This is mapped as:
$$
(-1,1)
\begin{pmatrix}
4 & -1 \\
2 & 0
\end{pmatrix}
= (-2, 1)
$$
Calculating back to standard coordinates:
$$
-2 b_1 + b_2 = -2 (1,0) + (1,1) = (-2,0) + (1,1) = (-1,1)
$$
which is where $e_2$ should mapped to by $f$.
Note:
The above assumed from your question that you have to use row vectors. If you use column vectors you
have to transpose the matrices and vectors:
$$
f(x) = A x
$$
as $A = A^T$ is symmetric. Further
$$
B = (MAM^{-1})^T
= (M^{-1})^T A^T M^T
= (M^T)^{-1} A M^T
=
\begin{pmatrix}
4 & 2 \\
-1 & 0
\end{pmatrix}
$$
|
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"url": "https://math.stackexchange.com/questions/2105075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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|
Evaluating $\iint_V x^4+y^4+z^4$ with the divergence theorem
Use Gauss divergence theorem to evaluate
$$\iint_S \left(x^{4} + y^{4} + z^{4}\right)$$ over sphere S of radius $a$.
So I wrote this as
$$
\begin{align}
&a\iint_{\partial V} \Big(x^3 \hat{i}+y^3\hat{j}+y^3\hat{k}\Big)\cdot\Big(\frac{x\hat{i}+y\hat{j}+z\hat{k}}{a}\Big)\\
=\ &a\iiint_V \operatorname{div}(x^3,y^3,z^3)\\
=\ &3a\iiint_Va^2=3a^3\frac{4\pi}{3}a^3=4\pi a^6
\end{align}$$
But my answer is not matching. The answer key says it should be $\frac{12 \pi a^6}{5}$
|
The normal vector $\mathrm{n}$ at at point $(x,y,z)$ on a sphere is given by $\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$, hence it you want to write $x^4+y^4+z^4$ as the dot product $\mathrm{F}\cdot\mathrm{n}$ you have to take $\mathrm{F}$ as $a(x^3+y^3+z^3)$, then $\text{div }\mathrm{F}=3a(x^2+y^2+z^2)$ and the original integral becomes
$$ 3a\iiint_{x^2+y^2+z^2\leq a^2}(x^2+y^2+z^2)\,d\mu=3a\int_{0}^{a}4\pi\rho^2\cdot\rho^2\,d\rho=\frac{12\pi}{5}a^6$$
by integrating along shells.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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|
What is the minimum value of $x^2+y^2$ subject to $x^3+y^3+xy=1$?
What is the minimum value of $x^2+y^2$ under the constraint $x^3+y^3+xy=1$? Please do not use partial differentials (multivariable calculus) or Lagrange multipliers. You can use elementary algebra or single variable calculus.
I plotted the graph of $x^3+y^3+xy=1$. It seems the minima occurs when $x=y$ but I don't know why that will be true. Any ideas?
|
Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Hence, $x^3+y^3+xy=1$ gives $2u(4u^2-3v^2)+v^2=1$ or $v^2=\frac{8u^3-1}{6u-1}$.
But $u^2\geq v^2$.
Hence, $u^2\geq\frac{8u^3-1}{6u-1}$ or $\frac{2u^3+u^2-1}{6u-1}\leq0,$ which gives $\frac{1}{6}<u\leq u_1$,
where $u_1$ is a real root of the equation $2u^3+u^2-1=0$.
Hence, $x^2+y^2=4u^2-2v^2=4u^2-\frac{2(8u^3-1)}{6u-1}=\frac{2(4u^3-2u^2+1)}{6u-1}$, which is decreasing on $\left(\frac{1}{6},u_1\right]$,
which says that the answer is $\frac{2(4u_1^3-2u_1^2+1)}{6u_1-1}.$
We can get $u_1$ by the Cardano's formula.
|
{
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"url": "https://math.stackexchange.com/questions/2107117",
"timestamp": "2023-03-29T00:00:00",
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|
Prove $\sum_{k=1}^\infty\left \lfloor{\frac{n}{p^k}}\right\rfloor \leq 2\sum_{k=1}^\infty\left \lfloor{\frac{n-1}{p^k}}\right\rfloor$ I'm trying to prove that:
$$\sum_{k=1}^\infty\left \lfloor{\frac{n}{p^k}}\right\rfloor \leq 2\sum_{k=1}^\infty\left \lfloor{\frac{n-1}{p^k}}\right\rfloor $$ for composite $n > 4$ where $p$ is any prime less than $n$. The inequality is related to proving that $n|(n-1)!$ for composite $n >4$ and I know the standard proof for that but was wondering if this was an approach one could make.
I also know you can use the fact that $n|(n-1)!$ to prove this but I wanted an analytic solution. Any thoughts? I think it's true but I may be wrong.
Thanks!
|
We want to prove that
$$\sum_{k=1}^{\infty}\left(2\left\lfloor\frac{n-1}{p^k}\right\rfloor-\left\lfloor\frac{n}{p^k}\right\rfloor\right)\ge 0$$
Let $n=p^kM_k+R_k$ and $n-1=p^km_k+r_k$ where $M_k,R_k,m_k,r_k$ are integers such that $M_k\ge 0,m_k\ge 0,0\le R_k\lt p^k$ and $0\le r_k\lt p^k$.
Let us separate it into two cases :
*
*Case 1 : $n=p^aq$ where $a,q\ge 1\in\mathbb Z,\gcd(p,q)=1$
Since $R_k=0$ for $1\le k\le a$ and $R_k\ge 1$ for $k\gt a$, we have $m_k=M_k-1$ for $1\le k\le a$ and $m_k=M_k$ for $k\gt a$.
If $a=1$, then
$$\begin{align}\sum_{k=1}^{\infty}\left(2\left\lfloor\frac{n-1}{p^k}\right\rfloor-\left\lfloor\frac{n}{p^k}\right\rfloor\right)&=\sum_{k=1}^{1}(2m_k-M_k)+\sum_{k=2}^{\infty}(2m_k-M_k)\\\\&=\sum_{k=1}^{1}(M_k-2)+\sum_{k=2}^{\infty}m_k\\\\&\ge (q-2)+0\\\\&\ge 0\end{align}$$
since $q\ge 2$.
If $a=2$, then
$$\begin{align}\sum_{k=1}^{\infty}\left(2\left\lfloor\frac{n-1}{p^k}\right\rfloor-\left\lfloor\frac{n}{p^k}\right\rfloor\right)&=\sum_{k=1}^{2}(2m_k-M_k)+\sum_{k=3}^{\infty}(2m_k-M_k)\\\\&=\sum_{k=1}^{2}(M_k-2)+\sum_{k=3}^{\infty}m_k\\\\&\ge (pq+q-4)+0\\\\&\ge 0\end{align}$$
since $(p,q)\not=(2,1)$.
For $a\ge 3$,
$$\begin{align}\sum_{k=1}^{\infty}\left(2\left\lfloor\frac{n-1}{p^k}\right\rfloor-\left\lfloor\frac{n}{p^k}\right\rfloor\right)&=\sum_{k=1}^{a}(2m_k-M_k)+\sum_{k=a+1}^{\infty}(2m_k-M_k)\\\\&=\sum_{k=1}^{a}(M_k-2)+\sum_{k=a+1}^{\infty}m_k\\\\&\ge \sum_{k=1}^{a}(p^{a-k}q-2)+0\\\\&\ge \sum_{k=1}^{a}(p^{a-k}-2)\\\\&\ge (2^{a-1}+2^{a-2}+\cdots +2+1)-2a\\\\&=2^a-1-2a\\\\&\ge 0\end{align}$$
*
*Case 2 : $\gcd(n,p)=1$
Since $R_k\ge 1$ for every $k$, we have $m_k=M_k$ for every $k$.
Therefore,
$$\sum_{k=1}^{\infty}\left(2\left\lfloor\frac{n-1}{p^k}\right\rfloor-\left\lfloor\frac{n}{p^k}\right\rfloor\right)=\sum_{k=1}^{\infty}(2m_k-M_k)=\sum_{k=1}^{\infty}m_k\ge 0$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
If cosB=3/4, how do I find cos 2B and cos(B/2)? I got $\cos 2B$;
$\cos 2B=2 \cos^2 B - 1
=9/8 -1
=1/8$
but when I tried cos1/2B, I got: COS1/2B=cos^2*1/4B-1, then I solved it, and didn't get the answer which was in the book, why?
|
$\cos 2B = 2 \cos^2 B - 1
= \frac9{8} - 1 = \frac{1}{8}$
Second answer -
$\cos B = 2 \cos^2(\frac B2) - 1$
$\frac34 = 2 \cos^2(\frac B2) - 1$
$\frac 74 = 2 \cos^2(\frac B2)$
$\frac 78 = \cos^2(\frac B2)$
$\sqrt{\frac78} = \cos (\frac B2)$
$\frac12 \sqrt{\frac72} = \cos (\frac B2)$
|
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"url": "https://math.stackexchange.com/questions/2108822",
"timestamp": "2023-03-29T00:00:00",
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|
For all triangle prove that $\sum\limits_{cyc}\frac{a}{(a+b)^2}\geq\frac{9}{4(a+b+c)}$
Let $a$, $b$ and $c$ be a sides-lengths of the triangle. Prove that:
$$\frac{a}{(a+b)^2}+\frac{b}{(b+c)^2}+\frac{c}{(c+a)^2}\geq\frac{9}{4(a+b+c)}$$
The Buffalo way kills it, but I am looking for a nice proof for this nice inequality.
SOS (sums of squares) gives $\sum\limits_{cyc}\frac{(a-b)^2(ab+b^2-ac)}{(a+b)^2}\geq0$ and I don't see what comes next.
|
Have: $\Leftrightarrow \left(\sum _{cyc}a\right)\left(\sum _{cyc}\frac{a}{\left(b+c\right)^2}\right)\ge \frac{9}{4}$
By C-S and Nesbitt : $LHS\ge \left(\frac{a}{b+c}+\frac{b}{b+c}+\frac{c}{c+a}\right)^2=\frac{9}{4}$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of $\frac{z^{k+1}}{(1-z)^2}$ over the roots of $x^n+1=0$ equals $\frac{n}{2}(k-\frac{n}{2})$ The problem is to show that for every $k \in \{1,2,...,n\}$ if we sum $\frac{z^{k+1}}{(1-z)^2}$ over the roots $z$ of $x^n+1=0$, we obtain the result $\frac{n}{2}(k-\frac{n}{2})$.
One idea I had is to consider the polynomial $P(t)=(1-t)^n+1$. We can easily calculate the sum of roots of $P$ and the sum of their squares. By considering the polynomial $Q(t)=t^nP(1/t)$ we can easily compute the the sum of the reciprocals and sum of the squares of the reciprocals. Hence for $k=0,1,2$ we can compute the sum in question quite easily and we get the intended result. However this method doesn't really generalise because the formulas for the symmetric sums lead to really messy computations.
|
We have with
$$f(z) = \frac{z^{k+1}}{(1-z)^2} \frac{nz^{n-1}}{z^n+1}
= \frac{1}{(z-1)^2} \frac{nz^{n+k}}{z^n+1} $$
and $\zeta_{n,q} = \exp(\pi i/n+ 2\pi i q/n)$
$$\sum_{x^n+1=0} \frac{x^{k+1}}{(1-x)^2}
= \sum_{q=0}^{n-1}
\mathrm{Res}_{\large z=\zeta_{n,q}} f(z).$$
This is also given by (residues sum to zero)
$$-\mathrm{Res}_{z=1} f(z)
-\mathrm{Res}_{z=\infty} f(z).$$
For the first of these we differentiate to obtain
$$\frac{n(n+k) z^{n+k-1}}{z^n+1}
- \frac{nz^{n+k}}{(z^n+1)^2} n z^{n-1}$$
Evaluate at $z=1$ to get
$$\frac{1}{2} n(n+k) - \frac{1}{4} n^2
= \frac{1}{2} nk + \frac{1}{4} n^2.$$
For the residue at infinity we get
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{1}{(1/z-1)^2} \frac{n/z^{n+k}}{1/z^n+1}
\\ = -\mathrm{Res}_{z=0}
\frac{1}{(1-z)^2} \frac{n}{z^{n+k}+z^k}
\\ = -\mathrm{Res}_{z=0} \frac{1}{z^k}
\frac{1}{(1-z)^2} \frac{n}{z^{n}+1}$$
This is
$$-[z^{k-1}] \frac{1}{(1-z)^2} \frac{n}{z^{n}+1}.$$
With $k\le n$ only the constant term from $\frac{n}{z^{n}+1}$
contributes and we get
$$-[z^{k-1}] \frac{n}{(1-z)^2} = -nk.$$
Adding the contributions from the residues at one and at infinity
and flipping the sign we thus obtain
$$-\left(\frac{1}{2} nk + \frac{1}{4} n^2
- nk\right)
= -\left(-\frac{1}{2} nk + \frac{1}{4} n^2\right)
= -\frac{1}{2} n \left(\frac{1}{2}n - k\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2} n \left(k - \frac{1}{2} n\right)}$$
as claimed.
|
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|
Prove the inequality ${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $ given that $abc = 1$ Prove the inequality $${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $$ So, I know a proof for this, but I basically memorized it without understanding. It's
$$ {(\frac 1a + \frac 1b + \frac 1c)}^2 + {(a(b+c) + b(a+c) + c(a+b))}^-1 = {ab + bc + ac \over 2} \ge {\frac 32} $$
I don't know what inequalities were applied here or how. Another proof would also be appreciated.
|
By Holder and AM-GM
$$\sum\limits_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^3c^3}{b+c}\geq\frac{(ab+ac+bc)^3}{3\sum\limits_{cyc}(b+c)}\geq\frac{3abc(a+b+c)(ab+ac+bc)}{3\sum\limits_{cyc}(b+c)}\geq\frac{3}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the equation $2\arcsin x=\arcsin(\frac{3}{4}x)$ $$2\arcsin x=\arcsin(\frac{3}{4}x)$$ so $x\in[-1,1]$
so we have:
$2\arcsin x=y\Rightarrow\sin\frac{y}{2}=x$ and $\arcsin x=y \Rightarrow \sin y=\frac{3}{4}x\Rightarrow\frac{4}{3}\sin y=x$ ,
$y\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$$\sin\frac{y}{2}-\frac{4}{3}\sin y=0$$$$\sin\frac{y}{2}-\frac{4}{3}\cdot 2\sin\frac{y}{2}\cos\frac{y}{2}=0$$ $$\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$$
Is it done properly at this point?
1.$\sin \frac{y}{2}=0\Rightarrow x=0$
And what else?
|
HINT:
Take the sine of both sides, apply the double angle formula for the sine function, and use $\cos(\arcsin(x))=\sqrt{1-x^2}$.
|
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|
If $P=log_{x}xy $ and $Q=log_{y}xy$, then how is $P+Q=PQ$? If $P$ and $Q$ have different bases for the log, how do we prove $P+Q=PQ$?
|
Consider $P + Q$ , it is $log_{x}{xy}+log_{y}{xy}$ which is equal to
$ log_{x}{x} + log_{x}{y} + log_{y}{x} + log_{y}{y}= 2 + log_{x}{y} + log_{y}{x} $
now consider rightside $P.Q$ which is $log_{x}{xy} .log_{y}{xy} = (1 + log_{x}{y}).(1+log_{y}{x}) $
which is equal to $1 + log_{x}{y} + log_{y}{x} + log_{x}{y}.log_{y}{x}$ and this last term will be $1$ , equal to $2+log_{x}{y}+log_{y}{x}$,
Comparing bothsides you see they are equal.Hence, proved.
|
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|
Prove that if the sides $a,b,c$ of a triangle satisfy $a^2+b^2=kc^2$, then $k>\frac {1}{2}$.
Prove that if the sides $a,b,c$ of a triangle satisfy $a^2+b^2=kc^2$, then
$k>\frac {1}{2}$.
Source : CRUX (Page Number $1$;Question Number $74$)
Obviously, for $k=1$, a right angled triangle exists. I tried assuming $k<\frac {1}{2}$ and finding some contradiction, but all I got was $a \leq b, $ which seems pretty alright to me. I strongly suspect that there has to be some elegant proof by contradiction for this problem.
Can anyone provide a guideline as to what should be done ?
Any help would me gratefully acknowledged :) .
|
By triangle inequality we can write:
$$c<a+b \to c^2<a^2+b^2+2ab$$
Multiply both sides by $k$ and use $kc^2=a^2+b^2$:
$$kc^2<k(a^2+b^2)+(2ab)k \to (1-k)(a^2+b^2)<(2ab)k$$
If $k \le 1/2$ then $1-k \ge 1/2$, so we have:
$$\frac{a^2+b^2}{2}\le (1-k)(a^2+b^2)<(2ab)k \le ab$$
what give us
$$\frac{a^2+b^2}{2}< ab \to (a-b)^2<0$$
what is a contradiction.
|
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|
Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.
So, for $k=1$:
$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$
it is valid.
For $k+1$ here is the proof he does:
$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2} &(1)\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2} &(2)\\
&=\frac{k^2+2k+k+2}{2}&(3)\\
&=\frac{k^2+3k+2}{2}&(4)\\
&=\frac{(k+1)(k+2)}{2}&\text{factoring (4)}
\end{align}$$
Therefore this formula is valid for $k+1$.
But is this true? I think not. He is just undoing what he have just done. To prove it I think I need to do this:
$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)+2(k+1)}{2}&=\frac{(k+1)(k+2)}{2}\\
\frac{(k+1)(k+2)}{2}&=\frac{(k+1)(k+2)}{2}\\
(k+1)(k+2)&=(k+1)(k+2)\\
k+1&=k+1\\
k&=k\\
0&=0\text{ or }1=1
\end{align}$$
Therefore, $k+1$ is valid in this formula.
Is this right or am I making a mistake somewhere?
|
You don't to begin using equality!! You must to arrive in this...
$1+2+⋯+k+k+1= $(here you use the induction hipoteses) Than...We start correct:
$$1+2+⋯+k+k+1=**induction hipoteses(ih)**\frac{k(k+1)}{2}**finish(ih)**+**hereyourepeatlikeabove**k+1$$
I hope I've helped.
|
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|
System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt{5})/2, (1-\sqrt{5})/2);
\end{align*}
|
Notice that $x=0 \implies y=0$, hence $(0,0)$ is a trivial solution. We will consider the case where $x \neq 0$.
$$x=y^2(y-1)$$
$$y=x^2(x-1)$$
Suppose on the contrary that $x=1$, then from the second equation $y=0$ which contradicts the first equation $x=0$. Hence $x \neq 1$.
Suppose on the contrary that $x \in (0,1)$,
$$y=x^2(x-1) \in (-1,0)$$
and hence $$x=y^2(y-1)<0$$
which is a contradiction. Hence $x \notin (0,1).$ By symmetry, $y \notin (0,1)$.
Also, notice if $x>1$, then $y=x^2(x-1)>0$, and hence $y>1$.
and if $x<0$ then $y=x^2(x-1)<0$.
$$x=y^2(y-1)$$
$$x^2(x-1)=y$$
Multiply the two equations together, we have
$$x^3(x-1)=y^3(y-1)$$
Consider the function, $$f: (1,\infty) \rightarrow (0,\infty),\text{ where }f(t)= t^3(t-1).$$
We can easily see that this function is increasing and it is an injective function.
Hence, if $x>1$, we have $f(x)=f(y)$ and hence $x=y$.
Similarly, we can consider the function, $$g: (-\infty,0) \rightarrow (0,\infty),\text{ where }g(t)= t^3(t-1).$$
We can easily see that this function is decreasing and it is an injective function.
Hence, if $x<0$, we have $g(x)=g(y)$ and hence $x=y$.
Hence, we always have $x=y$.
$$x+x^2=x^3$$
$$x(1-x-x^2)=0$$
$x=0$ or $1-x-x^2=0$
|
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|
Find the summation of the following sequence Please give me an idea on how to proceed as I am really stuck with this, I have not encountered this type of question before yet my friend gave this summation to me and I am stuck.
$${4\choose 1} + \frac{5\choose2}{2}+\frac{6\choose3}{4}+\dots$$
|
Let $\displaystyle\binom{r+3}r\left(\dfrac12\right)^{r-1}=\binom{r+3}3\left(\dfrac12\right)^{r-1}=f(r+1)-f(r)$
where $f(m)=\left(\dfrac12\right)^m\sum_{r=0}^na_rm^r$ where $a_r$ are arbitrary constants
$\dfrac16(r+3)(r+2)(r+1)\left(\dfrac12\right)^{r-1}$
$=\left(\dfrac12\right)^{r+1}\left(a_0+a_1(r+1)+a_2(r+1)^2+a_3(r+1)^3+\cdots\right)-\left(\dfrac12\right)^r\left(a_0+a_1(r)+a_2(r)^2+a_3(r)^3+\cdots\right)$
$\dfrac{r^3+6r^2+11r+6}6$ $=\left(\dfrac12\right)^2\left(a_0+a_1(r+1)+a_2(r+1)^2+a_3(r+1)^3+\cdots\right)-\left(\dfrac12\right)\left(a_0+a_1(r)+a_2(r)^2+a_3(r)^3+\cdots\right)$
Clearly, $a_r=0$ for $r\ge4$
Comparing the coefficients of $r^3,$ $$\dfrac16=\dfrac{a_3}4-\dfrac{a_3}2\iff a_3=?$$
Comparing the coefficients of $r^2,$ $$1=\dfrac{a_2+3a_3}4-\dfrac{a_2}2\iff a_2=?$$
Similarly, comparing the coefficients of $r$ and the constants, we can find $a_1,a_0$
Using Telescoping Series, $$\sum_{r=1}^\infty\binom{r+3}3\left(\dfrac12\right)^{r-1}=-f(1)$$
|
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|
$a,b,c,d>0$ and $abcd=1$ prove $\sum \frac {1}{(1+a)(1+a^2)} \ge 1$ If $a,b,c,d>0$ and $abcd=1$ prove:
$$\sum \frac{1}{(1+a)(1+a^2)}\ge 1$$
Here's my solution:
I try to prove it by reverse:
$$3\ge\sum\frac{a^3+a^2+a}{(a+1)(a^2+1)}$$Then by AM-GM:
$$3\ge\sum\frac{a(1+a+a^2)}{2a(a+1)}$$$$$6\ge\sum\frac{1+a+a^2}{a+1}$$
Then We need to prove:
$$2\ge\sum\frac{a^2}{a+1}$$
This inequality but I still can't prove it.
I don't know the last inequality is right or wrong.
There's another solution from another user @Michael Rozenberg by Vasc's RCF theorem.
|
The inequality $\sum\limits_{cyc}\frac{a^2}{a+1}\leq2$ is wrong.
Try $b=c=d\rightarrow0^+$ and $a\rightarrow+\infty$.
The starting inequality we can prove also by the following two inequalities.
$$\frac{1}{(a^2+1)(a+1)}+\frac{1}{(b^2+1)(b+1)}\geq\frac{1}{1+\sqrt{a^3b^3}}$$ and
$$\frac{1}{(c^2+1)(c+1)}+\frac{1}{(d^2+1)(d+1)}\geq\frac{1}{1+\sqrt{c^3d^3}}$$
and the rest is smooth.
|
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|
How can I solve $x^2+2xy+y^2+3x-3y-18=0$ $$x^2+2xy+y^2+3x-3y-18=0$$
I don't know how to solve it.
|
When a general conic equation
$$ Ax^2+Bxy+Cy^2+Dx +Ey +F=0$$
has $B\ne0$ and $A=C$ then rotating it $45^\circ$ will eliminate the $xy$ term.
This will be accomplished by the substitution
\begin{eqnarray}
x&=&\frac{X-Y}{\sqrt{2}}\\
y&=&\frac{(X+Y)}{\sqrt{2}}
\end{eqnarray}
\begin{equation}
x^2+2xy+y^2+3x-3y-18=0
\end{equation}
\begin{equation}
\frac{(X-Y)^2}{2}+(X-Y)(X+Y)+\frac{(X+Y)^2}{2}+\frac{3}{\sqrt{2}}(X-Y)-\frac{3}{\sqrt{2}}(X+Y)-18=0
\end{equation}
$$X^2=\frac{3}{\sqrt{2}}Y+18$$
So it is the equation of a parabola rotated $45^\circ$ about the origin.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $\sin(a+b)=1$ and $\sin(a-b)=\frac{1}{2}$ wher $a \geq 0$ and $b \leq \frac{\pi}{2}$ , find value of $\tan(a+2b)$ and $\tan(2a+b)$ If $\sin(a+b)=1$ and $\sin(a-b)=\frac{1}{2}$ where $a \geq 0$ and $b \leq \frac{\pi}{2}$ , find the value of $\tan(a+2b)$ and $\tan(2a+b)$.
I have tried to apply the formula for $\tan(a+(a+b))$ but couldn't reach anywhere.
THANKS
|
Hint:
$$\sin(a + b) = 1$$
$$\sin(a + b) = \sin \frac π2$$
$$a + b = \frac π2 = 90^\circ$$
Also,
$$\sin(a - b) = \frac 12$$
$$\sin(a - b) = \sin \frac π6$$
$$a - b = \frac π6 = 30^\circ$$
Solve equations to find $a$, $b$.
$$a = \frac π3 = 60^\circ$$
$$b = \frac π6 = 30^\circ$$
Put values of $a$ and $b$ to solve.
|
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|
Evaluate: $\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^2}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$ The problem is to evaluate the following sum:
$$\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
My approach was to find the common denominator ($2^{100}$), then the series becomes:
$$ \frac{1\cdot 2^{99}-2\cdot 2^{98}+3\cdot 2^{97}-4\cdot 2^{96}+\ldots -100\cdot 2^0}{2^{100}}$$
And split it this way:
$$ \frac{(1\cdot 2^{99}+3\cdot 2^{97}+\ldots+99\cdot 2^{1})-(2\cdot 2^{98}+4\cdot 2^{96}+\ldots+100\cdot 2^{0})}{2^{100}}$$
But it seems that cacelling out the terms is not a promising approach. Any hints?
|
Notice that
$$\sum_{n=1}^{100}x^n=\frac{x-x^{101}}{1-x}$$
which is just a geometric series. Differentiate once to get
$$\sum_{n=1}^{100}nx^{n-1}=\frac d{dx}\frac{x-x^{101}}{1-x}$$
Multiply both sides by $x$ and set $x=-\frac12$ to get
$$\frac1{2^1}-\frac2{2^2}+\dots-\frac{100}{2^{100}}=-\frac12\left(\frac d{dx}\bigg|_{x=-\frac12}\frac{x-x^{101}}{1-x}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to solve the following recurrence? $t(n)=[4-t(n-1)]^{-1}$ I am trying to solve the following recurrence :
$T_n=\frac{1}{4-T_{n-1}}$
I tried various methods using range transformation but still can't figure it out.
|
If it has a limit $L$,
$T_n=\frac{1}{4-T_{n-1}}
$
becomes
$L=\frac{1}{4-L}
$
or
$L^2-4L+1 = 0$
so that
$L
=\dfrac{4\pm\sqrt{16-4}}{2}
=2\pm\sqrt{3}
$.
Let
$L_1
=2+\sqrt{3}
$
and
$L_2
=2-\sqrt{3}
$.
If
$0 < T_{n-1} < 1$,
then
$\frac14 < T_n
< \frac13
$.
After this,
$\frac1{4-1/3}
< T_{n+1}
< \frac1{4-1/4}
$
or
$\frac{3}{13}
< T_{n+1}
< \frac{4}{15}
$.
If $0 < x < \frac14$,
then
$1+x
<\frac1{1-x}
< 1+2x
$,
so that,
if $0 < T_{n-1}
< 1$,
then
$T_n
=\frac{1}{4-T_{n-1}}
=\frac14\frac{1}{1-T_{n-1}/4}
\gt\frac14(1+T_{n-1}/4)
$
and
$T_n
=\frac14\frac{1}{1-T_{n-1}/4}
\lt\frac14(1+2T_{n-1}/4)
=\frac14+\frac{T_{n-1}}{8}
\lt \frac38
$.
Let
$u_n
= T_n-L_2
$.
Then
$u_n+L_2
=\dfrac1{4-(u_{n-1}+L_2)}
$
or
$\begin{array}\\
1
&=(u_n+L_2)(4-L_2-u_{n-1})\\
&=u_n(4-L_2)-L_2u_{n-1}+L_2(4-L_2)\\
&=u_n(4-L_2)-L_2u_{n-1}+1
\qquad\text{since } L_2(4-L_2) = 1\\
\text{so}\\
u_n(4-L_2)
&=L_2u_{n-1}\\
\text{or}\\
u_n
&=\frac{L_2}{4-L_2}u_{n-1}\\
&=L_2^2u_{n-1}\\
\end{array}
$
Therefore,
if
$0 < u_n < 1$,
$u_{n+k}
=L_2^{2k}u_n
$.
Since
$L_2^2
=\frac{7-4\sqrt{3}}{4}
\approx 0.07179676972449
$,
this converges
quickly.
We then have
$T_{n+k}-L_2
=L_2^{2k}(T_n-L_2)
$
or
$T_{n+k}
=L_2^{2k}(T_n-L_2)+L_2
$.
Therefore,
if $0 \le T_0 < 1$,
$T_{k}
=L_2^{2k}(T_0-L_2)+L_2
$.
In particular,
if $T_0 = 0$,
$T_{k}
=L_2-L_2^{2k+1}
$.
|
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|
How many triples $(x,y,n)$ are there such that $x^n - y^n = 2^{100}$
How many triples $(x,y) \in \mathbb{N^+}^2$ and $n \gt 1$ are there such that $x^n - y^n = 2^{100}$
I dont know how to start. Any hint will be helpful.
I know the identity $x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2}+y^{n-1})$.
I think from here we need some combinatorics to get the rest of answer.
|
Since the question is being asked rather often and the solution in the comments seems to have a flaw I propose the following answer.
We claim: the equation
$$
a^n-b^n=2^k,\tag1
$$
where $a,b,n,k$ are positive integers ($n>1$), has exactly $\left\lfloor\frac{k-1}2\right\rfloor$ solutions $(a,b)$ for $n=2$ and no solutions for $n>2$.
First of all we note that from
$$
a^n-b^n=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^i=2^k,\tag2
$$
it immediately follows that the solution can exist for $n>1$ only if $k>0$.
Now let consider the problem case by case.
For $n=2$ from
$$
a-b=2^r,\quad a+b=2^{k-r}
$$
the only solutions are:
$$a=2^{k-r-1}+2^{r-1};\quad b=2^{k-r-1}-2^{r-1},\quad 0<r<\frac{k}2.\tag3$$
For $n=4$ we would have the equation:
$$
(a^2-b^2)(a^2+b^2)=2^k.
$$
This implies
$$a^2-b^2=2^m\implies a=2^{m-r-1}+2^{r-1};\quad b=2^{m-r-1}-2^{r-1}$$
for some $m$ and $r$.
But the equation
$$
2^{2(m-r-1)}+2^{2(r-1)}=\frac{a^2+b^2}2=2^{k-m-1}
$$
cannot hold in view of $r\ne\frac m2$ (cf. (3)).
For odd $n>1$ we note that equation (2) can hold only if both $a$ and $b$ are even: $a=2A$, $b=2B$. Assume now that equation (1) has an integer solution $(a,b)$ for some $k$. Then there must exist some minimal positive $K$ for which the equation holds. However from
$$
(2A)^n-(2B)^n=2^K\implies A^n-B^n=2^{K-n}
$$
it follows that such minimal $k$ does not exist (since $n<K$).
In general case an arbitrary $n>2$ can be represented as $2^p q$, where $q$ is an odd number. If $p>0$ repeating the process:
$$
2^k=a^n-b^n=(a^{n/2}-b^{n/2})(a^{n/2}+b^{n/2})\implies a^{n/2}-b^{n/2}=2^m;\ (
m>0)$$
one reduces the problem either to $n=4$ (if $q=1$) or $n$ odd (if $q\ne1$). The obtained contradiction finalizes the proof.
|
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|
Is there easy way to factor polynomials for calculus? factor of $$4y^9-4y$$
comes out to be $$4y(y^4+1)(y^2+1)(y+1)(y-1)$$
How would you approach to factor?
PS: My math is very rusty.
|
\begin{align*}
4y^9-4y&=4y(y^8-1)\\
&=4y((y^4)^2-1)\\
&=4y(y^4-1)(y^4+1)\\
&=4y((y^2)^2-1)(y^4+1)\\
&=4y(y^2-1)(y^2+1)(y^4+1)\\
&=4y(y-1)(y+1)(y^2+1)(y^4+1).
\end{align*}
I am simply using $a^2-b^2=(a-b)(a+b)$. For example, at the second equation we took $a=y^4,\ b=1$.
|
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|
Combinatorics and limit problem For every natural number $n$ let us consider $a_n$ the greatest natural non-zero number such that:
$$\binom {a_n}{n-1} \gt \binom {a_n-1}{n}.$$
Compute $$\lim_{n \to \infty} \frac {a_n}{n}.$$
I started by using the formula for the binomial coefficient, I obtained a second degree inequation in $a_n$, but I can't find the greatest $a_n$. That's where I got stuck. The equation I got is $a_n^2+a_n(1-3n)+n^2-n<0$.
|
$$
\begin{align}
&\, \binom{a_{\small n}-1}{n}\lt\binom{a_{\small n}}{n-1} \,\Rightarrow \frac{(a_{\small n}-1)!}{(n)!\,(a_{\small n}-n-1)!}\lt\frac{(a_{\small n})!}{(n-1)!\,(a_{\small n}-n+1)!} \\[4mm]
&\, \Rightarrow 1\lt\frac{a_{\small n}\,n}{(a_{\small n}-n)\,(a_{\small n}-n+1)} \,\Rightarrow\, a_{\small n}^2-(3n-1)a_{\small n}+(n^2-n)\lt0 \\[4mm]
&\, \qquad \left\{{\small\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{(3n-1)\pm\sqrt{(3n-1)^2-4(n^2-n)}}{2}=\frac{(3n-1)\pm\sqrt{5n^2-2n+1}}{2}}\right\} \\[4mm]
&\, \Rightarrow \left(a_{\small n}-\frac{(3n-1)\color{red}{-}\sqrt{5n^2-2n+1}}{2}\right)\left(a_{\small n}-\frac{(3n-1)\color{red}{+}\sqrt{5n^2-2n+1}}{2}\right)\lt0 \\[4mm]
&\, \Rightarrow \frac{(3n-1)\color{red}{-}\sqrt{5n^2-2n+1}}{2}\lt a_{\small n} \lt\frac{(3n-1)\color{red}{+}\sqrt{5n^2-2n+1}}{2} \\[4mm]
&\, \Rightarrow a_{\small n}=\color{red}{\left\lfloor\,{\small \frac{(3n-1)+\sqrt{5n^2-2n+1}}{2}}\,\right\rfloor} \\[4mm]
&\, \Rightarrow \lim_{n\to\infty}\frac{a_{\small n}}{n}=\lim_{n\to\infty}\frac{(3n-1)+\sqrt{5n^2-2n+1}}{2n} \\[2mm]
&\, \quad\qquad\qquad =\lim_{n\to\infty}\frac{3-(1/n)+\sqrt{5-(2/n)+(1/n^2)}}{2}=\color{red}{\frac{3+\sqrt{5}}{2}}
\end{align}
$$
|
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|
Evaluating $\int \frac{1-7\cos^2x}{\sin^7x\cos^2x}dx$ How do i evaluate $$\int \frac{1-7\cos^2x}{\sin^7x\cos^2x}dx$$. I tried using integration by parts and here is my approach
$\int \frac{ sinx}{(1-cos^2x)^4\cos^2x} dx$ and then put $cos x=t$ and then tried to use partial fractions.I applied similar logic for the other part.But that made it lengthy to solve as decomposition into partial fractions is very time consuming.This question came in an objective examination in which time was limited.Can anyone help me with a shorter way to solve this problem.Thanks.
|
Well, we know that:
$$\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}=\csc^7\left(x\right)\left(\sec^2\left(x\right)-7\right)\tag1$$
So, for the integral we get:
$$\int\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}\space\text{d}x=\int\csc^7\left(x\right)\sec^2\left(x\right)\space\text{d}x-7\int\csc^7\left(x\right)\space\text{d}x\tag2$$
Now, for the right integral you can use the reduction formula.
$\color{red}{\text{But}}$ using integration by parts:
$$\int\csc^7\left(x\right)\sec^2\left(x\right)\space\text{d}x=\csc^6\left(x\right)\sec\left(x\right)+7\int\csc^7\left(x\right)\space\text{d}x\tag3$$
So, we get that:
$$\int\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}\space\text{d}x=\csc^6\left(x\right)\sec\left(x\right)+\color{red}{7\int\csc^7\left(x\right)\space\text{d}x-7\int\csc^7\left(x\right)\space\text{d}x}\tag4$$
Which gives that:
$$\int\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}\space\text{d}x=\csc^6\left(x\right)\sec\left(x\right)+\text{C}\tag{5}$$
|
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|
Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$
*$\frac{3(1)^2}{2}-\frac{(1)^3}{3}$ = $\frac{3}{2}-\frac{1}{3}$
*$\frac{3(2)^2}{2}-\frac{(2)^3}{3}$ = $\frac{12}{2}-\frac{8}{3}$
*($\frac{3}{2}-\frac{1}{3}$)-($\frac{12}{2}-\frac{8}{3}$) = $-\frac{9}{2}-\frac{9}{3}$
*-$\frac{15}{2}$
|
Step 5. onwards is not ok
$$\int_a^bf(x) dx = F(b)-F(a) $$
$$\int_a^bf(x) dx = F(b-a) $$
The first result is correct but second one is in general wrong.
EDIT1:
The constant term in $ (x^2-3 x +2) $ is missing for integration. Factorization is not beneficial,
|
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|
Triple Integral $\iiint x^{2n}+y^{2n}+z^{2n}dV$ Evaluate:
$$\iiint_{x^2+y^2+z^2 \leqslant 1} x^{2n}+y^{2n}+z^{2n} dV $$
I have tried to convert to spherical polars and then compute the integral, but it gets really messy because of the 2n power. Any tips?
|
First observation: it is symmetric in $x,y,z$, so by linearity we have
$$\iiint_{x^2+y^2+z^2 \leqslant 1} x^{2n}+y^{2n}+z^{2n} dV =3\iiint_{x^2+y^2+z^2 \leqslant 1} z^{2n} dV.$$
Choosing spherical coordinates it becomes
$$3\iiint_{x^2+y^2+z^2 \leqslant 1} (r\cos \theta)^{2n} dV$$
where $dV= r^2 \sin \theta \ \text{d}r \ \text{d}\theta \ \text{d}\phi$.
Thus the integral simplifies to
$$3 \int_0^{2\pi}\int_0^{\pi}\int_0^1 r^{2(n+1)} (\cos \theta)^{2n} \sin \theta \ \text{d}r \ \text{d}\theta \ \text{d}\phi = \frac{3}{2n+3}2 \pi \int_0^{\pi}(\cos \theta)^{2n} \sin \theta \ \text{d}\theta. $$
Using that
$$\int_0^{\pi}(\cos \theta)^{2n} \sin \theta \ \text{d}\theta = \frac{2}{2n+1} $$
we have
$$\iiint_{x^2+y^2+z^2 \leqslant 1} x^{2n}+y^{2n}+z^{2n} dV= \frac{3}{2n+3}2 \pi \frac{2}{2n+1}.$$
|
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|
Euler's Method and induction Euler's Method for series associates with a given series $\sum\limits_{j=0}^\infty(-1)^ja_j$ the transformed series $\sum\limits_{n=0}^\infty\frac{\Delta^n a_0}{2^{n+1}}$ where $\Delta^0a_j=a_j$, $\Delta^na_j=\Delta^{n-1}a_j-\Delta^{n-1}a_{j+1}$, $j=1,2,\cdots.$
I am given that $\ln 2 = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+-\cdots = \frac{1}{1\cdot2}+ \frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\frac{1}{4\cdot 2^4}\cdots$.
I need to prove that $ \sum\limits_{n=0}^\infty\frac{\Delta^n a_0}{2^{n+1}} = \frac{1}{2}+ \frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\frac{1}{4\cdot 2^4}\cdots$, but I can not figure out how. I attempt to by induction and the base case is simple enough. However, when trying to use the induction hypothesis I get stuck.
If I assume that $\sum\limits_{n=0}^k\frac{\Delta^n a_0}{2^{n+1}} = \frac{1}{2}+ \frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\frac{1}{4\cdot 2^4}\cdots\frac{1}{(k+1)\cdot 2^{k+1}}$ and I add $\frac{1}{(k+2)\cdot 2^{k+2}}$ to both sides. Then I just need to show that $\Delta^{k+1}a_0 = \frac{1}{k+2}$. However I don't see how I can.
So I know that $\Delta^{k+1}a_0 = \Delta^{k}a_0-\Delta^{k}a_1$. From the induction hypothesis I know that $\Delta^{k}a_0 = \frac{1}{k+1}$. But I would have to work with $\Delta^{k}a_1$ recursively. So I am not sure how to prove this.
Any input would be very welcome!
|
A formula for $ \Delta ^n a_j$ is given by
\begin{eqnarray*}
\Delta ^n a_j = \frac{n! \ j!}{(n+j+1)!}
\end{eqnarray*}
We now need to use double induction ... it suffice to show that the formula $\Delta^na_j=\Delta^{n-1}a_j-\Delta^{n-1}a_{j+1}$ holds, which is easy
\begin{eqnarray*}
\Delta ^n a_j = \Delta^{n-1}a_j-\Delta^{n-1}a_{j+1} =\frac{(n-1)! \ j!}{(n+j)!}-\frac{(n-1)! \ (j+1)!}{(n+j+1)!}= \frac{n! \ j!}{(n+j+1)!}
\end{eqnarray*}
|
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|
Which integration techniques I should use for $\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$
$$\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$$
I can simplify it to:
$$\int{\frac{dx}{\sqrt{x^2+3}}} - 3\int{\frac{dx}{\sqrt{x^2-3}}}$$
but I can't go from here.
|
Hint :
$$\int{\frac{dx}{\sqrt{x^2+3}}} - 3\int{\frac{dx}{\sqrt{x^2-3}}}\\=\int{\frac{dx}{\sqrt{3}\sqrt{(\frac{x}{\sqrt3})^2+1}}} - 3\int{\frac{dx}{\sqrt{3}\sqrt{(\frac{x}{\sqrt3})^2-1}}}$$
|
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|
What are the solutions to $a^2+ab+b^2$ $=$ $3^n$? What are all known integer solutions ($a, b, n$) to $a^2+ab+b^2$ $=$ $3^n$ besides ($1, 1, 1$) and ($-2, 1, 1$)? Do any others even exist?
This question comes from the identity that ($a^3±b^3$)/($a±b$) $=$ $0, 1$ $\pmod 3$. If ($a^3±b^3$)/($a±b$) $=$ $0$ $\pmod 3$, then let $3^n$ be highest power of $3$ dividing ($a^3±b^3$)/($a±b$). ($a^3±b^3$)/(($a±b$)$*3^n$) $=$ $1$ $\pmod 3$. When does ($a^3±b^3$)/(($a±b$)$*3^n$) $=$ $1$?
|
Let $a=u+v$ and $b=u-v$. It follows:
$$a^2+ab+b^2=(u+v)^2+(u^2-v^2)+(u-v)^2$$
Hence
$$3u^2+v^2=3^n$$
If $n=0$, then, $u=0$ and $v=\pm 1$
If $n=1$, then $v^2=3(1-u)(1+u)$ $\implies$ $(u,v)=(\pm 1,0),(\pm 1, \pm 1 )$
If $n>1$, then $u,v$ are both divisible by 3.
|
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|
Find the Derivative of $f(x)=\frac{7}{\sqrt {x}}$ using the definition. I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
If someone or anyone could go step by step and do the problem, I would be eternally grateful.
|
\begin{align*}
\dfrac{\frac{7}{\sqrt{x+h}} - \frac{7}{\sqrt{x}}}{h} &= 7\left(\dfrac{\sqrt x - \sqrt{x+h}}{h\sqrt{x(x+h)}}\right) \\
&= 7\left(\dfrac{\sqrt x - \sqrt{x+h}}{h\sqrt{x(x+h)}}\right)\left(\dfrac{\sqrt x + \sqrt{x+h}}{\sqrt x + \sqrt{x+h}}\right) \\
&= 7\left(\dfrac{x-(x+h)}{h\sqrt{x(x+h)}}\right)\left(\dfrac{1}{\sqrt x + \sqrt{x+h}}\right) \\
&= 7\left(\dfrac{-h}{h\sqrt{x(x+h)}}\right)\left(\dfrac{1}{\sqrt x + \sqrt{x+h}}\right) \\
&= 7\left(\dfrac{-1}{\sqrt{x(x+h)}}\right)\left(\dfrac{1}{\sqrt x + \sqrt{x+h}}\right) \\
\end{align*}
Therefore,
$$\lim_{h \to 0}\dfrac{\frac{7}{\sqrt{x+h}} - \frac{7}{\sqrt{x}}}{h} = \lim_{h \to 0} 7\left(\dfrac{-1}{\sqrt{x(x+h)}}\right)\left(\dfrac{1}{\sqrt x + \sqrt{x+h}}\right) = \frac{-7}{2|x|\sqrt x} = \frac{-7}{2x^{3/2}}$$
|
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|
Prove for all real numbers $a, \ b,$ and $c$, $bc + ac + ab \leq a^2 + b^2 + c^2$. I've been having some trouble with this particular problem. Could someone provide a hint of some sort? My original attempt that turned out to be useless is to start off with
$$(b+c)^2 + (a+c)^2 + (a+b)^2 = (b+c)^2 + (a+c)^2 + (a+b)^2$$
and subtract from both sides until I get the above inequality, but I found out I'd end up subtracting that very same inequality and would be right back where I started.
|
$
a^2 + b^2 + c^2 \ge ab + bc + ca\\
\Longleftrightarrow \frac12(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca) \ge 0\\
\Longleftrightarrow (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) \ge 0 \\
\Longleftrightarrow (a-b)^2 +(b-c)^2 +(c-a)^2 \ge 0
$
which is obviously true. Giving a hint for this problem is as good as solving it entirely. :-)
|
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|
Homogeneous D.E show that $ x\frac{dy}{dx} = y + \frac{x}{lny-lnx} $ is a homogeneous D.E.
Find the general solution in the form $ye^{y/x} = f(x) $
The answer given is $ye^{y/x} = Ax^2 $ but I am getting $ (\frac{y}{x})^{y/x} = Axe^{y/x} $
Is the given answer correct?
|
Consider the given answer $\quad ye^{y/x} = Ax^2 \quad\to\quad \ln(y)+\frac{y}{x}=\ln(A)+2\ln(x) \quad$ and differentiate it :
$$\frac{dy}{y}+\frac{dy}{x}-\frac{ydx}{x^2}=2\frac{dx}{x}$$
$$\left( \frac{1}{y}+\frac{1}{x}\right)dy=\left(\frac{2}{x}+\frac{y}{x^2} \right)dx$$
$$\frac{dy}{dx}=\frac{\frac{2}{x}+\frac{y}{x^2}}{\frac{1}{y}+\frac{1}{x}} = \frac{y}{x}\left(\frac{2x+y}{x+y}\right)$$
$$x\frac{dy}{dx}= y\left(\frac{2x+y}{x+y}\right)=y\left(1+\frac{x}{x+y}\right)$$
$$x\frac{dy}{dx}= y+\frac{x}{\frac{x}{y}+1}\tag 2$$
Compared to the given ODE
$$x\frac{dy}{dx}= y+\frac{x}{\ln(y)-\ln(x)}\tag 1$$
we see that $\quad ye^{y/x} = Ax^2\quad$ should be the solution of $(1)$ if $\quad \frac{x}{y}+1=\ln(y)-\ln(x)\quad$
Except for particular values of $x$ and $y$, this isn't the case in general.
So, in general $\quad ye^{y/x} = Ax^2\quad$ isn't solution of $(1)$.
The general solution of $(1)$ is $ (\frac{y}{x})^{y/x} = Axe^{y/x} $
Probably there is a typo or a mistake in the wording of the problem. May be the question isn't to find the general solution of $(1)$, but only to find a particular solution on a specified form.
|
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|
If $47^x = 8$ and $376^y = 128$ , find $\frac{3}{x}-\frac{7}{y}$ What I know:
$x={\log_{47}8}$ and $y=\log_{376}128$
How do I do this without using a calculator?
|
No logs, no calculator. :)
$$47^x=8=2^3\Longrightarrow 47=2^{3/x}\\
376^y=128=2^7\Longrightarrow 376=2^{7/y}$$
Dividing,
$$\frac {2^{3/x}} {2^{7/y}}=\frac {47}{376}=\frac 18\\
2^{3/x-7/y}=2^{-3}\\
\frac 3x-\frac 7y=\color{red}{-3}$$
|
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|
Evaluate the given limit Given a function $f : R → R$ for which $|f(x) − 3| ≤ x^2$. Find
$$\lim_{ x\to0}\frac{f(x) - \sqrt{x^2 + 9}}{x}$$
Can the function $f(x)$ be considered as $x^2 + 3$ and go about evaluating the limit using the Limit laws?
|
We have $|f(x)-\sqrt{x^2 +9}| = |f(x)-3 + 3 - \sqrt{x^2 + 9}| \le |f(x)-3| + |3 - \sqrt{x^2+9}| \le x^2 + |3 - \sqrt{x^2 + 9}|$. So:
$$\left|\frac{f(x)-\sqrt{x^2+9}}{x}\right| \le |x| + \left| \frac{\sqrt{x^2+9}-3}{x}\right|$$
Now squeeze.
|
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|
What values for a prime $p$ are possible if $3p+1$ is a perfect cube? There is a similar case like this already on the site, but it deals with perfect squares and is relatively easy to solve.
But what about perfect cubes? Thus $3p+1= n^3$ ?
any help?
Thanks!
|
This is not the solution, I tried all four cases and didn't get anywhere.... sigh. Can someone help?
Case 1: assume p=n-1 and 3=(n^2)+n+1
Then n=p+1, so
3=(p+1)^2 + (p+1) +1
3= p^2 +2p +1 +p +1 +1
3= p^2 +3p +3
but then what?
0=p^2 +3p
0= p(p +3) so p is 0 or -3? would 3 be the answer? i thought it would have to be positive...
Case 2:
assume 3=n-1 and p=(n^2)+n+1
Then, n= 4
p=16 +4 +1 = 21, which is not prime.
Case 3:
assume 3p=n-1 and 1=(n^2)+n+1
Then, n= 3p+1
1= (3p+1)^2 +(3p+1) +1
1= 9p^2 +6p+ 1 +3p+1 +1
1= 9p^2 +9p +3
-2= 9(p^2-p)
-2/9 = p^2-p
which ... not sure about a solution
Case 4:
assume 1=n-1 and 3p=(n^2)+n+1
Then, n=2
3p = 4+2+1
3p = 7
p=7/3 which is not prime nor an integer....
Any help????
|
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|
Solve $\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1}$ without using L'Hôpital's I tried:
$$\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1} = \\
\frac{3\ln(x) + (1-x^2)}{-1(1-x)} = \\
\frac{3\ln(x)+ (1-x)(1+x)}{-(1-x)} = \\
\frac{3\ln(x)}{x-1} + \frac{1+x}{-1} = \\
\frac{\ln{x^3}}{x-1} - 1-x = \\
???$$
What do I do next? Remember, I can't use L'Hôpital.
|
Recall that
$$\ln(x^3)=3\ln(x)$$
and
$$\lim_{x\to1}\frac{\ln(x)}{x-1}=1$$
which should give you
$$\lim_{x\to1}\frac{3\ln(x)}{x-1}-1-x=3-1-1=\boxed1$$
|
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|
Probability that $x^2+y^2$ is divisible by $5$ and $7$ Two natural numbers $x$ and $y$ are chosen at random with a uniform distribution.Then the question is to find out the probability that $x^2+y^2$ is divisible by $5$ and $7$.
I understand that divisible by $5$ and $7$ is the same thing as divisible by $35$.But I am a bit confused on how to calculate probability on an infinite set.Any help shall be highly appreciated.Thanks.
|
$1^2≡1, 2^2≡4, 3^2≡2, 4^2≡2, 5^2≡4, 6^2≡1 \mod 7$.
Then $x,y$ are multiple of $7$. Since about one's place $1+7^2≡1+9,4^2+7^2≡6+9, 4^2+8^2≡6+4,1^2+8^2≡1+4≡0\pmod5$, these are multiple of 5. Therefore they are 9 patterns of $(0,0)(7,14)(7,21)(14,7)(14,28)(21,7)(21,28)(28,14)(28,21) \pmod{35}$$$\dfrac9{35^2}=0.735\%$$
|
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|
Epsilon-Delta: Prove $\frac{1}{x} \rightarrow 7$ as $x \rightarrow \frac{1}{7}$
Prove that $\displaystyle\frac{1}{x} \rightarrow 7$ as $\displaystyle x \rightarrow \frac{1}{7}$.
I need to show this with an $\epsilon-\delta$ argument. Still figuring these types of proofs out though, so I could use some tips/critiques of my proof, if it is correct at all.
It might not be so clear, but I use the fact that $\displaystyle\left|x - \frac{1}{7}\right| < \delta$ several times in the proof.
For $\varepsilon > 0$, let $\displaystyle\delta = \min\left\{\frac{1}{14}, \frac{\varepsilon}{98}\right\}$. Then $\displaystyle \left|x - \frac{1}{7}\right| < \delta$ implies:
$$\left|\frac{1}{7}\right| = \left|\left(-x + \frac{1}{7}\right) + x\right| \leq \left|x - \frac{1}{7}\right| + \left|x\right| < \frac{1}{14} + |x|,$$ and so $\displaystyle |x| > \frac{1}{14}$.
Also, $\displaystyle \left|x - \frac{1}{7}\right| < \delta$ implies:
$$\left|\frac{1}{x} - 7\right| = \left|\frac{1-7x}{x}\right| = 7\frac{\left|x - \frac{1}{7}\right|}{|x|} < 98\left|x - \frac{1}{7}\right| < \frac{98\varepsilon}{98} = \varepsilon.$$
Thus for $\varepsilon > 0$, $\displaystyle\left|\frac{1}{x} - 7\right| < \varepsilon$ if $\displaystyle\left|x - \frac{1}{7}\right| < \delta$, for $\displaystyle \delta = \min\left\{\frac{1}{14}, \frac{\varepsilon}{98}\right\}$.
|
The only logical error I could find is related to the definition of $\delta$. If you want to use strict inequalities, you should have $\delta<\min\left\{\frac{1}{14},\frac{\varepsilon}{98}\right\}$. Otherwise, at least one of the strict inequalities should be changed (depending on the value of $\varepsilon$).
Regardless, a good proof if you're still gaining familiarity with $\varepsilon-\delta$ proofs.
|
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|
Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
...?
|
You can even prove more: if $a,b,c$ are the sides and $\alpha,\beta,\gamma$ the respectively opposite angles, summing up the three identities of the law of cosine, we get
$$
a^2+b^2+c^2=2ab\cos\gamma+2bc\cos\alpha+2ca\cos\beta
$$
which implies the statement, as all the cosines are less than $1$.
|
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|
Keep getting wrong answer for showing $(a_n)=(\frac{2n^2-1}{n^2},\frac{1}{n})$ is $d^{(2)}$-convergent? A sequence in $\mathbb{R}^2$ is given by:
$a_n=(\frac{2n^2-1}{n^2},\frac{1}{n})$, for each $n∈\mathbb{N}$.
I must show that $(a_n)$ is $d^{(2)}$-convergent.
My textbook says that I can show $a_n$ is convergent by showing $d^{(2)}(a_n, 0)$ converges to $0$.
So I have tried:
$d^{(2)}((\frac{2n^2-1}{n^2},\frac{1}{n}),(0, 0))=\sqrt{(0-\frac{2n^2-1}{n^2})^2+(0-\frac{1}{n})^2}$
which when expanded ends up as:
$d^{(2)}((\frac{2n^2-1}{n^2},\frac{1}{n}),(0, 0))=\sqrt{\frac{4n^4-3n^2+1}{n^4}}$
and so:
$$\lim_{n\to\infty} \sqrt{\frac{4n^4-3n^2+1}{n^4}}=\sqrt{\lim_{n\to\infty} \frac{4n^4-3n^2+1}{n^4}}=\sqrt{\lim_{n\to\infty} {4+\frac{3}{n^2}+\frac{1}{n^4}}}=\sqrt{4}=2$$
This converges to $2$ rather than the expected $0$, so therefore it is not a real null sequence and hence $(a_n)$ is not $d^{(2)}$-convergent, which the question tells me is wrong. Where have I gone wrong?
|
The sequence does converge, but not to $(0,0)$. Observe that
\begin{align*}
d^{(2)}(a_n,(2,0))&=\sqrt{\left(\frac{2n^2-1}{n^2}-2\right)^2+\left(\frac{1}{n}-0\right)^2} \\
&=\sqrt{\left(\frac{1}{n^2}\right)^2+\left(\frac{1}{n}\right)^2} \\
&=\sqrt{\frac{1+n^2}{n^4}}.
\end{align*}
Can you take it from here?
|
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|
find $g^{-1}(x)$ in terms of $f^{-1}$ Consider $g(2x-3) = \frac{2f(x-2) + 3}{5 - f(x-2)}$ . Also $f$ and $g$ are invertible . Now find $g^{-1}(x)$ in terms of $f^{-1}$.
My try : Because in the parentheses we have $x-2$ and $2x-3$ instead of $x$ , I don't know how we can solve this problem.
|
HInt:
first $$x \to x+2 \\\to
g(2(x+2)-3)=\dfrac{2f(x+2-2) + 3}{5 - f(x+2-2)}\to \\
g(2x+1)=\dfrac{2f(x) + 3}{5 - f(x)}$$
now find $f(x)$ then apply $f^{-1}$
$$y=\dfrac{2f(x) + 3}{5 - f(x)} \to \\
5y-yf(x)=2f(x)+3 \to \\f(x)(y+2)=5y-3 \\ \to \\
f(x)=\dfrac{5y-3}{y+2}$$
apply $f^{-1}$to both sides
$$f^{-1}f(x)=f^{-1}(\dfrac{5y-3}{y+2})\\
x=f^{-1}(\dfrac{5y-3}{y+2})\\
x=f^{-1}(\dfrac{5g(2x+1)-3}{g(2x+1)+2})$$
|
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|
Producing circles with $r^2=x^2+y^2$ $r^2=x^2+y^2\ +\pi $ produces graph with two circles :
$r = 2$
This graph was produced using Desmos, is this correct or a quirk of the graph software ?
Ive noticed $\forall n \epsilon N [r^2=x^2+y^2+n\pi]$ appears to produce same circle, so $r^2=x^2+y^2\ +3\pi $ produces same circle as $r^2=x^2+y^2\ +4\pi $ :
Why is this the case, should each circle not be differing sizes ?
|
Your first picture is correct if you are graphing
\begin{align*}
2^2 &= x^2 + y^2\\
- \pi + 2^2 &= x^2 + y^2
\end{align*}
for red and yellow respectively. If we have the equation
$$
r^2 = x^2 + y^2 + c
$$
where $c$ is a constant, that is the same as
$$
r^2 - c = x^2 + y^2
$$
and so our new radius is $r' = \sqrt{r^2 - c}$, if $c > r^2$, the circle has no points as squares are positive. As $3\pi$ and $4\pi$ are larger than $2^2$, there are no solutions for
\begin{align*}
-3\pi + 2^2 = x^2 + y^2\\
-4\pi + 2^2 = x^2 + y^2.
\end{align*}
Desmos will therefore not draw anything.
|
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|
Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divisible by $4$ so $8|(n^3 - 1)(n^3 + 1)$. I'm stuck at proving divisibility by $9$ and $7$
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For $7$: $(n^3-1)(n^3+1)=n^6-1$ then apply Fermat's little theorem.
For $3^2$ : First, $(k-1)k(k+1)$ has to be divisible by $3$.
Second, $(n^3-1)\cdot(n^3+1)=(n-1)(n^2+n+1)\cdot (n+1)(n^2-n+1)$ which has the factor $(n^2-1)$ which is divisible by 3 by Fermat's LT.
|
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|
What operations can be performed on three variables $x$, $y$, and $z$ and produce one unique sum? I'm having a programming problem, but I feel this would be more relevant here. I have three unique variables, $x$, $y$, and $z$. What operations can I perform on these variables to produce a unique value? Are there any approaches I could take?
Thanks!
EDIT: To be more specific, I want some function $f(x, y, z)$ that produces a unique output for each set of inputs (assuming uniqueness in inputs, $1$, $2$, $3$ does not need to output something different than $3$, $2$, $1$). I realize that a generalized case of this is probably not possible, but I would definitely be able to co-opt similar solutions
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$f(x,y,z) = x\sqrt{2} + y\sqrt{5} + z\sqrt{7}$
This function would increase by a similar order of magnitude to its inputs. Moreover it's possible to prove that for $x,y,z \in \mathbb{Z}, f(x,y,z)$ must be unique $\forall x,y,z$.
Edit: Proof:
Suppose for contradiction that $\exists x_{1},y_{1},z_{1},x_{2},y_{2},z_{2} \in \mathbb{Z}$:
$f(x_{1},y_{1},z_{1}) = f(x_{2},y_{2},z_{2})$ with at least one of the following being true: $x_{1} \neq x_{2}, y_{1} \neq y_{2} , z_{1} \neq z_{2}$
Then we have $x_{1}\sqrt{2} + y_{1}\sqrt{5} + z_{1}\sqrt{7} = x_{2}\sqrt{2} + y_{2}\sqrt{5} + z_{2}\sqrt{7}$
$\Rightarrow (x_{1} - x_{2})\sqrt{2} + (y_{1} - y_{2})\sqrt{5} + (z_{1} - z_{2})\sqrt{7} = 0$
In the case that $x_{1} = x_{2}$ and $y_{1} \neq y_{2}$ and $z_{1} \neq z_{2}$:
$\frac{\sqrt{5}}{\sqrt{7}} = \frac{z_{2} - z_{1}}{y_{1}-y_{2}}$
This would imply that $\frac{\sqrt{5}}{\sqrt{7}} \in \mathbb{Q}$ which leads to a contradiction (distinct irrational divided by distinct irrational must also be irrational - simple to prove)
A similar proof applies to the case where $y_{1} = y_{2}, x_{1} \neq x_{2}$ and $z_{1} \neq z_{2}$ and the case when $z_{1} = z_{2}, x_{1} \neq x_{2}$ and $y_{1} \neq y_{2}$
If we have $x_{1} = x_{2}$ and $y_{1} = y_{2}$, then we get $z_{1} = z_{2}$ which is a contradiction. Therefore we have proved the following conditions:
$x_{1} \neq x_{2}, y_{1} \neq y_{2} , z_{1} \neq z_{2}$
let $a = x_{1} - x_{2}, b = -(y_{1}-y_{2}), c = -(z_{1} - z_{2})$
Then we have $a\sqrt{2} = b\sqrt{5} + c\sqrt{7}$
$\Rightarrow \sqrt{2} = \frac{b\sqrt{5} + c\sqrt{7}}{a}$
let $b_{1} = \frac{b}{a}, c_{1} = \frac{c}{a}$
$\Rightarrow b_{1}, c_{1} \in \mathbb{R}$
$2 = (b_{1}\sqrt{5} + c_{1}\sqrt{7})^{2}$
$\sqrt{35} = 2 - 5b_{1}^{2} - 7c_{1}^{2}$
$\Rightarrow \sqrt{35} \in \mathbb{Q}$ which is a contradiction.
This implies that $\forall x,y,z \in \mathbb{Z}, f(x,y,z)$ is unique.
|
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|
find extremum of $y = x(x-1)^{\frac{1}{3}}$ given function:
$$y = x(x-1)^{\frac{1}{3}}$$
steps:
$$y'=\frac{(x^2-x)^{\frac{-2}{3}}2x-1}{3}$$
After simplifying:
$$ y' = \frac{2x-1}{(3x^2-3x)^{\frac{2}{3}}}$$
therefore
$$ x = \frac{1}{2}$$ $$x \ne 0$$
Am I right with calculus?
|
The function is
$$
f(x)=x\sqrt[3]{1-x}
$$
not
$$
\sqrt[3]{x(1-x)}
$$
at least according to common notation. Thus the derivative is
$$
f'(x)=(1-x)^{1/3}+x\cdot\frac{1}{3}(1-x)^{-2/3}\cdot(-1)=
\frac{1}{3(1-x)^{2/3}}(3-3x-x)=\frac{3-4x}{3(1-x)^{2/3}}
$$
which vanishes at $x=3/4$ and is undefined at $x=1$.
Note that the function is increasing over $(-\infty,3/4]$ and decreasing over $[3/4,\infty)$ (the point where the function is not differentiable doesn't harm).
Alternatively, use the fact that $x\mapsto x^3$ is a continuous bijection $\mathbb{R}\to\mathbb{R}$, so the function $f$ has the same extremal points as
$$
g(x)=(f(x))^3=x^3(1-x)=x^3-x^4
$$
Since $g'(x)=3x^2-4x^3=x^2(3-4x)$, we get the same conclusions as before.
|
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|
Proof that $CE=2AD$ I already have a proof, but if you can please give another:
Let $ABC$ be an isosceles triangle with $AB=AC$ and point $D$ be on segment $AB$. The line perpendicular to $AB$ which passes $D$ intersects $BC$ (extended) and $AC$ at $E$ and $F$ respectively. $C$ is on segment $BE$, between $B$ and $E$. If the area of $CEF$ is twice that of $ADF$, prove that $CE=2AD$.
My proof involves the Pythagorean theorem and the similar triangles property.
|
Let $\measuredangle A=\alpha$, $AB=a$ and $AD=xa$.
Hence, $$FC=a-\frac{ax}{\cos\alpha}=\frac{a(\cos\alpha-x)}{\cos\alpha}$$ and
$$CE=BE-BC=\frac{a(1-x)}{\sin\frac{\alpha}{2}}-2a\sin\frac{\alpha}{2}=\frac{a(\cos\alpha-x)}{\sin\frac{\alpha}{2}}.$$
Thus, $$S_{\Delta{FCE}}=\frac{FC\cdot CE\sin\left(90^{\circ}+\frac{\alpha}{2}\right)}{2}=\frac{a^2(\cos\alpha-x)^2\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}\cos\alpha}$$ and
$$S_{\Delta ADF}=\frac{ax\cdot\frac{ax}{\cos\alpha}\sin\alpha}{2}=\frac{a^2x^2\sin\alpha}{2\cos\alpha}.$$
Hence,
$$\frac{a^2(\cos\alpha-x)^2\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}\cos\alpha}=\frac{a^2x^2\sin\alpha}{\cos\alpha}$$ or
$$(\cos\alpha-x)^2=4x^2\sin^2\frac{\alpha}{2},$$
which says that $$x=\frac{\cos\alpha}{1+2\sin\frac{\alpha}{2}}.$$
Thus, $AD=\frac{a\cos\alpha}{1+2\sin\frac{\alpha}{2}}$,
$$CE=\frac{a\left(\cos\alpha-\frac{\cos\alpha}{1+2\sin\frac{\alpha}{2}}\right)}{\sin\frac{\alpha}{2}}=\frac{2a\cos\alpha}{1+2\sin\frac{\alpha}{2}}$$
and we are done!
|
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|
$5x^2−10x+7$ in completed square form? I was studying quadratic equations and practicing to solve them using the technique of completing the squares.
My answer was as follows:
$$5x^2−10x=−7$$
$$5x^2−10x+25=18$$
$$x(x-5)^2−5(x-5)=18$$
$$\boldsymbol{(x-5)^2-18=0}$$
The answer in the book is:
$$\boldsymbol{5(x-1)^2+2=0}$$
I'm confused. I just started with these sums today. Could you please tell me where I've gone wrong?
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Usually, if it's an equation, then we can divide both sides to get a quadratic term of the form $x^2$ instead of $ax^2$:
\begin{eqnarray*}
5x^2 -10x &=& -7 \\
x^2 - 2x &=& - \frac{7}{5} \\ \\
\left(x-1\right)^2-1 &=& -\frac{7}{5} \\ \\
\left(x-1\right)^2 &=& - \frac{2}{5}
\end{eqnarray*}
This has no real roots. You are welcome - because it's an equation - to multiply both sides by $5$:
$$5(x-1)^2 = -2 \implies 5(x-1)^2 + 2 = 0$$
This is a rather odd mixture of methods. If you're solving the equation $5x^2-10x=-7$ then you can solve from $(x-1)^2 = -\frac{2}{5}$ (allowing complex numbers). There's no need to re-introduce the factor of $5$. If you're completing the square on the expression $5x^2-10x+7$ then you can't just divide by $5$, and you need to take out common factors. Then you get
$$5x^2-10x+7 \equiv 5(x-1)^2+2$$
|
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|
Beta function-like integral $$\int\limits_0^1 \frac{x^{1-\alpha} (1-x)^\alpha}{(1+x)^3} \, dx $$
After the substitution $z=\frac{1}{x} - 1$,
I've got this:
$$\int\limits_0^\infty \left(\frac{1}{z}-1\right)^\alpha\left(\frac{1}{z}+1\right)^{-3} dz $$
But it is still far from the Beta function.
|
The substitution $u = \frac{1}{x}-1$ is actually useful because it transforms it into an integral that can be evaluated fairly easily using contour integration.
Specifically, for $-1 <a <2$,
$$\int_{0}^{1} \frac{x^{1-a}(1-x)^{a}}{(1+x)^{3}} \, dx = -\int^{0}_{\infty} \frac{\left(\frac{1}{1+u} \right)^{1-a} \left(1- \frac{1}{1+u} \right)^{a}}{\left(1+ \frac{1}{1+u} \right)^{3}} \frac{du}{(1+u)^{2}} = \int_{0}^{\infty} \frac{u^{a}}{(2+u)^{3}} \, du.$$
Integrating the function $$f(z) = \frac{z^{a}}{(2+z)^{3}}$$ around a keyhole contour deformed around a branch cut along the positive real axis, we get
$$ \begin{align} \int_{0}^{\infty} \frac{u^{a}}{(2+u)^3} \, du + \int_{\infty}^{0} \frac{(ue^{2 \pi i})^{a}}{(2+u)^{3}} \, du &= 2 \pi i \, \operatorname{Res}[f(z), -2] \\ &= 2 \pi i \, \frac{1}{2!} \lim_{z \to -2} \frac{\mathrm{d}^{2}}{\mathrm{d}z^{2}} \, z^{a} \\ &=\pi i a(a-1)(-2)^{a-2} \\ &= \pi i a(a-1)e^{i \pi a}2^{a-2}. \end{align} $$
Therefore, $$\begin{align} \int_{0}^{\infty} \frac{u^{a}}{(2+u)^{3}} \, du &= \pi 2^{a-2} a(a-1) \, \frac{i e^{\pi ia}}{1-e^{2 \pi ia}}\\ &= \pi 2^{a-3} a(1-a) \csc(\pi a) . \end{align} $$
This is the result you'll get if you apply the functional equation and the reflection formula of the gamma function to David H's answer.
When $a=0$ or $a=1$, the result should be interpreted as a limit.
|
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|
Prove that if LCMs are equal, then the numbers are equal too. For $a,b \in \mathbb{N}$, how do I prove:
$$lcm(a,a+5)=lcm(b,b+5) \implies a=b$$
|
I assumed $a$ and $b$ were positive integers.
We have $\mathrm{lcm}(a,a+5) = \frac {a(a+5)}{\gcd(a,a+5)}$, so that the given equality implies
$$\frac {a(a+5)}{\gcd(a,a+5)} = \frac {b(b+5)}{\gcd(b,b+5)}.$$
But $\gcd(x,y) = \gcd(x,y+kx)$ for all $k \in \mathbb{Z}$ so that $\gcd(a,a+5) = \gcd(a,5)$ and $\gcd(b,b+5) = \gcd(b,5)$.
Now if $a$ and $b$ are not multiples of $5$ we have $\gcd(a,5) = 1$ and $\gcd(b,5) = 1$ then
$$\frac {a(a+5)}{\gcd(a,a+5)} = \frac {b(b+5)}{\gcd(b,b+5)} \implies \frac{a*(a+5)}{1} = \frac{b*(b+5)}{1} \implies a = b.$$
If both $a$ and $b$ are multiple of $5$ then $\gcd(a,5) = \gcd(b,5) = 5$ so that
$$\frac {a(a+5)}{\gcd(a,a+5)} = \frac {b(b+5)}{\gcd(b,b+5)} \implies a(a+5) = b(b+5).$$From here we can conclude that $a=b$ by completing the square both sides and using the fact that a,b are positive.
Edit :
If $a$ is a multiple of $5$ and not $b$ we have $$a(a+5)=5b(b+5).$$ But $a=5a′$, then the last equality becomes $$5a′(a′+1)=b(b+5).$$ The left side is a multiple of $5$ but not the right side so this case never happens (same reasoning for the case $b$ multiple of $5$ and not $a$).
|
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Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\\=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+4\\=(a+b)^2-2ab+\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)^2-\dfrac{2}{ab}+4\\=4-2ab-\dfrac{2}{ab}+1+\dfrac{1}{a^2b^2}\\=4-2\bigg(\dfrac{a^2b^2+1}{ab}\bigg)+\dfrac{a^2b^2+1}{a^2b^2}\\=4-\bigg(\dfrac{a^2b^2+1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)\\=4-\bigg(ab+\dfrac{1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)$
Seems to using Cauchy-Schwartz. Please help.
|
Using Cauchy-Schwarz Inequality we have that:
$$\left(\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\right)(1 + 1) \ge \left( a + b + \frac 1a + \frac 1b \right)^2 \ge 5^2 = 25$$
The last inequality follows as:
$$\frac 1a + \frac 1b = \frac{b + a}{ab} = \frac 1{ab} \ge \frac{4}{(a+b)^2} = 4$$
Another way to solve the inequality is to expand the squares and use the fact that:
$$a^2 + b^2 \ge \frac{(a+b)^2}{2} = \frac 12 \quad \text{ and } \quad \frac 1{a^2} + \frac 1{b^2} \ge \frac 1{2a^2b^2} \ge 8$$
|
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"url": "https://math.stackexchange.com/questions/2172634",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluating surface integral. The question is:
Let $S$ be that part of the surface of the paraboloid $z=x^2+y^2$ between the planes $z=1$ and $z=4$. Given $\vec{V}=x^3j+z^3k$, I want to evaluate the surface intergal
$$\iint_s\nabla\times V.\hat n dS$$
would the paramatisation be something like $x=r\cos\theta,y=r\sin\theta,z=r^2 $ or
$ z=r$
$z=x^2+y^2$ = constant when z=4 or z=2 does this mean we can calculate $\hat ndS$
|
alternative
Green's / Stokes theorem:
$\iint_s\nabla\times V.\hat n dS = \oint_c V\cdot dr_2 - \oint_c V \cdot dr_1$
$r_1 = (\cos\theta, \sin\theta, 0), r_2 = (2\cos\theta, 2\sin\theta, 0)$
$\int_0^{2pi} 2^4\cos^4\theta\ d\theta - \int_0^{2pi} \cos^4\theta\ d\theta\\
15\int cos^4 \theta\ d\theta\\
\frac {15}{2}\int (\frac 12 (1+cos2\theta))^2 d\theta\\
\frac {15}{4}\int 1+ 2cos2\theta + \frac 12 + \frac 12 cos 4\theta d\theta\\
\frac {45}{4}\pi$
Alternative number 2 Divergence theorem:
$\iint_s\nabla\times V.\hat n dS + \iint_{D_4}\nabla\times V \cdot (0,0,1) dA + \iint_{D_1}\nabla\times V \cdot (0,0,-1) dA = \iiint \nabla \cdot (\nabla \times V) dV = 0$
$\int_0^{2\pi} \int_0^2 (3r^3 cos^2\theta)\ dr\ d\theta - \int_0^{2\pi} \int_0^1 (3r^3 cos^2\theta)\ dr\ d\theta$
$\int_0^{2\pi} \int_1^2 3r^3 cos^2\theta\ dr\ d\theta\\
\int_0^{2\pi} \frac {3}{4}(15) cos^2\theta\ d\theta\\
\frac {45}{4} \pi$
|
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"timestamp": "2023-03-29T00:00:00",
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|
On the Diophantine equation $2(a^3+b^3+c^3)=abc(abc+6)$
I am sure that $(3, 2, 1)$ is the only natural triple satisfying the following Diophantine equation $$2(a^3+b^3+c^3)=abc(abc+6)\quad\text{for}\quad a\ge b\ge c.$$
But I can not prove or refute that. Help me, please.
What I've tried so far. Rearrange the equation to get
$$
2(a^3+b^3+c^3-3abc)=a^2 b^2 c^2
$$
and
$$
(a+b+c)((a-b)^2+(a-c)^2+(b-c)^2)=a^2 b^2 c^2
$$
Let $a-b=n$ and $a-c=m$. Hence $m \ge n$ and
$$
(3a-m-n)(m^2+n^2+(m-n)^2)=[a(a-m)(a-n)]^2
$$
or
$$
2(3a-(m+n))((m+n)^2-3mn)=[a(a^2-(m+n)a+mn)]^2
$$
Now let $m+n=p$ and $mn=q$, then we have
$$
2(3a-p)(p^2-3q)=[a(a^2-pa+q)]^2
$$
Now we have a quadratic in $q$ and I tried to use $\Delta_{q}=k^2$ (where $k$ is an integer) for bounding some of variables, but I was unsuccessful.
|
HINT
Using the identity
$$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca),$$
one can get the original equation in the form of
$$2(a+b+c)((a+b+c)^2-3(ab+bc+ac)) = (abc)^2,$$
or
$$r=\frac{2s^3-p^2}{6s},$$
where
$$s=a+b+c,\quad r=ab+bc+ac,\quad p=abc,$$
$$a>0,\quad b>0,\quad c>0,\quad s^3-27p\ge 0.$$
That allows to write the cubic equation for $x\in\{a, b, c\}$ in the form of
$$x^3 - sx^2 + \frac{2s^3-p^2}{6s}x - p = 0,$$
or
$$6sx^3 - 6s^2x^2 + (2s^3-p^2)x - 6sp =0.$$
For natural solutions
$$p=qx,$$
$$(6s-q^2)x^2 - 6s^2x + 2s(s^2-3q) = 0.\qquad(1)$$
The "linear" case in $x$ is
$$\begin{cases}
q^2 = 6s\\
x_1=\dfrac{s^2-3q}{3s}\\
x_{2,3}^2 - (s-x_1)x_{2,3} + q = 0\\
x>0,\quad s>0,\quad q>0.
\end{cases}\qquad(2)$$
System $(2)$ gives the only solution:
$$x_1=\frac{q^2}{18} - \frac 6q,\quad q=s=6,\quad x\in\{1,2,3\},$$
$$\boxed{(a,b,c) = (3,2,1)}.$$
The "quadratic" case in $x$ gives:
$$\begin{cases}
6s\not= q^2\\
x_1 = \dfrac{3s^2\pm d}{6s-q^2}\\
d^2 = 9s^4 - 2s(s^2-3q)(6s-q^2)\\
x_{2,3}^2 - (s-x_1)x_{2,3} + q = 0\\
x>0,\quad q>0,\quad s^3\ge 27qx_1.
\end{cases}\qquad(3)$$
UPD
Using AM-GM inequality:
$$x_1 = s-(x_2+x_3)\ge s-2\sqrt{x_2x_3},$$
$$x_1\ge s-2\sqrt q,$$
so
$$\dfrac{2s(s^2-3q)}{3s^2\mp d}\in\mathcal N,\quad \dfrac{2s(s^2-3q)}{3s^2\mp d}\ge s-2\sqrt q\ge0,\quad s^2\ge4q.\qquad(4).$$
Attempts to obtain other solutions either of $(1-4)$ in the case of $6s\not= q^2\\$ do not lead to a positive result. One can do substution
$$s=w\sqrt q,$$
obtaining
$$6w\not=q\sqrt q,\quad\dfrac{2w(w^2-3)}{3w^2\mp\sqrt{9w^4 - 2w(w^2-3)(6w-q\sqrt q)}}\ge w-2\ge0$$
and closed inequality in form $q(w)\ge0$, but even in the case
$$6w<q\sqrt q,\quad \text{"}\mp\text{"}=\text{"}+\text{"}$$
one get infinity set of inequality solutions, so this don't change the situation cardinally.
This gives grounds for considering that solution $(3,2,1)$ is an easy case of the OP task and reduces the chances of success in the general case.
|
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|
Probability of product isn't greater than a specific number I have a joint distribution of 2 random variables $X$ and $Y$:
$$ f(x, y) = \frac{3}{2}x + \frac{y}{2} $$ if $x,y \in [0;1]$ and $0$ otherwise.
What's the following probability:$$ P\left(XY < \dfrac{1}{2}\right) $$
I thought about sum of probabilities: $P(X < \frac{1}{2Y}) + P(Y < \frac{1}{2X})$, but appropriate integrals don't converge:
$$ \int^{1}_{0}\int^{\frac{1}{2y}}_{0}\left(\frac{3}{2}x + \frac{y}{2}\right) dxdy $$
$$ \int^{1}_{0}\int^{\frac{1}{2x}}_{0}\left(\frac{3}{2}x + \frac{y}{2}\right) dydx $$
Then I imagined space of $X, Y$ as a square on a plane with center in $(0.5, 0.5)$. Then the area when $X > 0.5$ and $Y > 0.5$ violate $ XY < \frac{1}{2}$, therefore I calculated
$$ 1 - P(0.5 < x < 1, 0.5 < y < 1)$$ as
$$ 1 - \int_{\frac{1}{2}}^{1}\int_{\frac{1}{2}}^{1}\left(\frac{3}{2}x + \frac{y}{2}\right) dxdy = \frac{5}{8}$$
But I'm not sure that it's right and therefore I'm here.
I will be grateful for any help you can provide.
|
Actually, if $X>\frac{1}{2}$ and $Y>\frac{1}{2}$, you may still have $XY<\frac{1}{2}$. For instance, if $X=Y=\frac{5}{8}$, $XY=\frac{25}{64}<\frac{1}{2}$.
What you really need is $Y<\frac{1}{2X}$. So you can integrate the joint density of $X$ and $Y$ over the region $R$ where $0 \leq x \leq 1$, $0 \leq y \leq 1$, and $y < \frac{1}{2x}$. The reason your integrals did not converge is because you did not account for the fact that $X$ and $Y$ are always $ \leq 1$. You will have to break $R$ up into two regions to write down the integral, for instance
$$P\left(Y< \frac{1}{2X}\right)=\int_R f(x,y) \ dy \ dx = \int_0^\frac{1}{2}\int_0^1 f(x,y) \ dy \ dx + \int_\frac{1}{2}^1 \int_0^\frac{1}{2x} f(x,y) \ dy \ dx.$$
Finally, you do NOT want $P\left(Y<\frac{1}{2X}\right)+P\left(X< \frac{1}{2Y}\right)$ because these are the same event; if $Y<\frac{1}{2X}$ then $X<\frac{1}{2Y}$ and vice versa. So just calculate one of them.
|
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|
Prove the inequality, power. $\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality:
$x^4 + y^4 \ge 2$
How I started
*
*$(x+y)^2 = 4$
*$x^2 + y^2 = 4 - 2xy$
*$(x^2+y^2)^2 - 2(xy)^2 \ge 2$
*$(4-2xy)^2 - 2(xy)^2 \ge 2$
*$16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$
*$2(xy)^2 - 16xy + 14 \ge 0$
*for $t=xy$
*$2t^2 - 16t + 14 \ge 0$
It isn't always true, I think that I should have a assumption for $t$, but I don't know how should I do this.
|
We need to prove that $\frac{x^4+y^4}{2}\geq\left(\frac{x+y}{2}\right)^4,$
which enough to prove for non-negatives $x$ and $y$.
Let $x^2+y^2=2uxy$.
Hence, $u\geq1$ and we need to prove that $2u^2-1\geq\left(\frac{u+1}{2}\right)^2,$
which is $(u-1)(7u+5)\geq0$.
Done!
|
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|
Rank of the $n \times n$ matrix with ones on the main diagonal and $a$ off the main diagonal I want to find the rank of this $n\times n$ matrix
\begin{pmatrix}
1 & a & a & \cdots & \cdots & a \\
a & 1 & a & \cdots & \cdots & a \\
a & a & 1 & a & \cdots & a \\
\vdots & \vdots & a& \ddots & & \vdots\\
\vdots & \vdots & \vdots & & \ddots & \vdots \\
a & a & a & \cdots &\cdots & 1
\end{pmatrix}
that is, the matrix whose diagonals are $1's$ and $a$ otherwise, where $a$ is any real number.
My first observation is when $a=0$ the rank is $n$ and when $a=1$ the rank is $1.$ Then I can assume $a\neq 0, 1$ and proceed row reduction to find its pivot rows. I obtain
\begin{pmatrix}
1 & a & a & \cdots & a \\
0 & 1+a & a & \cdots & a \\
0 & a & 1+a & \cdots & a \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & a & a & \cdots & 1+a
\end{pmatrix}
by subtracting the first row multiplied $a$ for each row below the first, and then divides the factor $(1-a)$,
and stuck there. Any hints/helps?
|
Let
$$\mathrm M_n (a) := \begin{bmatrix} 1 & a & a & \dots & a & a\\ a & 1 & a & \dots & a & a\\ a & a & 1 & \dots & a & a\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a & a & a & \dots & 1 & a\\ a & a & a & \dots & a & 1\end{bmatrix} = (1-a) \mathrm I_n + a 1_n 1_n^{\top}$$
The eigenvalues of rank-$1$ matrix $a 1_n 1_n^{\top}$ are
*
*$\color{blue}{0}$ with multiplicity $n-1$.
*$\color{blue}{n a}$ with multiplicity $1$.
Thus, the eigenvalues of $\mathrm M_n (a) = (1-a) \mathrm I_n + a 1_n 1_n^{\top}$ are
*
*$\color{blue}{1-a}$ with multiplicity $n-1$.
*$(1-a) + na = \color{blue}{(n-1) \, a + 1}$ with multiplicity $1$.
We could also have arrived at this conclusion computing the characteristic polynomial of $\mathrm M_n (a)$
$$\begin{array}{rl} \det ( s \mathrm I_n - \mathrm M_n (a) ) &= \det \left( (s-(1-a)) \mathrm I_n - a 1_n 1_n^{\top} \right)\\ &= \det \left( (s-(1-a)) \left( \mathrm I_n - \frac{a}{s-(1-a)} 1_n 1_n^{\top} \right) \right)\\ &= (s-(1-a))^n \cdot \det \left( \mathrm I_n - \frac{a}{s-(1-a)} 1_n 1_n^{\top} \right)\\ &= (s-(1-a))^n \cdot \left(1 - \frac{n a}{s-(1-a)}\right)\\ &= (s-(1-a))^{n-1} \cdot \left(s-(1-a) - n a\right)\\ &= (s-(1-a))^{n-1} \cdot \left( s - ((n-1) \, a + 1) \right)\end{array}$$
where the matrix determinant lemma was used. If
$$a \in \left\{ -\frac{1}{n-1}, 1 \right\}$$
then $\mathrm M_n (a)$ is singular. Thus, using the rank-nullity theorem, we conclude that
$$\boxed{\mbox{rank} (\mathrm M_n (a)) = \begin{cases} 1 & \text{if } a = 1\\ n-1 & \text{if } a = -\frac{1}{n-1}\\ n & \text{otherwise}\end{cases}}$$
|
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|
Find real $P$ s.t. $B=P^{-1}AP$ TASK:
$\frac{dx}{dt}=Ax$
Given $A=\begin{bmatrix}
1 & 0 & 0 \\
0 & 2 & -3 \\
1 & 3 &2
\end{bmatrix}$
Find real matrix $P$ s.t change of coordinates $x=Py$ transforms the system to
and
$\frac{dy}{dt}=By$
$B=\begin{bmatrix}
1 & 0 & 0 \\
0 & 2 & -3 \\
0 & 3 &2
\end{bmatrix}$
Solve explicitly for $y$ hence evaluate solution in terms of $x$
APPROACH:
Firstly $PB=AP$
I tried to solve the 9 simultanouesequations and ended up with matrix where there are 3 free variables of the form:
$P=\begin{bmatrix}
-10x & 0 & 0 \\
3x & y & -z \\
x & z &y
\end{bmatrix}$
Then I put this into $P^{-1}AP$ in software "SYMBOLAB" and the asnwer he gave me was $B$. So it turns out $x,y,z$may are free as long as $det(P)\neq0$ And since $det(P)=-10x(y^2+z^2)$, all I know is that $x\neq 0$ and at least one of the $y,z$ is not 0.
Is this the correct answer?
|
Besides the hint given in my comment to a related post, consider that the similarity relation is not unique, because in fact:
$$
\mathbf{A = R}\;\mathbf{C}\;\mathbf{R}^{\,\mathbf{ - 1}} \quad \quad \Leftrightarrow \quad \left\{ {\begin{array}{*{20}c}
{\mathbf{A} = \mathbf{A}\;\mathbf{A}\;\mathbf{A}^{\,\mathbf{ - 1}} } & \Leftrightarrow & {\mathbf{A = }\left( {\mathbf{A}\;\mathbf{R}} \right)\;\mathbf{C}\;\left( {\mathbf{A}\;\mathbf{R}} \right)^{\,\mathbf{ - 1}} } \\
{\mathbf{C} = \mathbf{C}\;\mathbf{C}\;\mathbf{C}^{\,\mathbf{ - 1}} } & \Leftrightarrow & {\mathbf{A = }\left( {\mathbf{R}\;\mathbf{C}} \right)\;\mathbf{C}\;\left( {\mathbf{R}\;\mathbf{C}} \right)^{\,\mathbf{ - 1}} } \\
\begin{gathered}
\mathbf{A} = \lambda \;\mathbf{A}\;\lambda ^{\, - 1} \hfill \\
\mathbf{C} = \mu \;\mathbf{C}\;\mu ^{\, - 1} \hfill \\
\end{gathered} & \Leftrightarrow & {\mathbf{A = }\left( {\lambda \;\mathbf{R}\;\mu } \right)\;\mathbf{C}\;\left( {\lambda \;\mathbf{R}\;\mu } \right)^{\,\mathbf{ - 1}} } \\
\end{array} } \right.
$$
|
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|
Help with a combinatoric formula Anyone can help solve th following? $\left( {\begin{array}{*{20}{c}}
{2n} \\
n
\end{array}} \right)$ means $2n$ chooses $n$. Thanks!
$\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}
{2k} \\
k
\end{array}} \right){2^{ - 2k}}} = \sum\limits_{k = 0}^n {\frac{{(2k)!}}{{k!k!}}{2^{ - 2k}}} = \frac{{(2n)!}}{{n!n!}}(2n + 1){2^{ - 2n}} = \left( {\begin{array}{*{20}{c}}
{2n} \\
n
\end{array}} \right){2^{ - 2n}}(2n + 1)$
The report from wolframalpha shows they indeed equal.
|
How about induction? The base case $n = 0$ is easily verified. Then
\begin{align}
\binom{2n}{n}2^{-2n}(2n+1) + \binom{2n+2}{n+1}2^{-2n-2} &= \binom{2n+2}{n+1}2^{-2n-2}\biggl[\frac{4(n+1)^2}{2n+2} + 1\biggr] \\
&= \binom{2n+2}{n+1}2^{-2n-2}\bigl(2n+2+1\bigr)
\end{align}
is the induction step.
|
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|
Proof: $n^2 - 2$ is not divisible by 4 I tried to prove that $n^2 - 2$ is not divisible by 4 via proof by contradiction. Does this look right?
Suppose $n^2 - 2$ is divisible by $4$. Then:
$n^2 - 2 = 4g$, $g \in \mathbb{Z}$.
$n^2 = 4g + 2$.
Consider the case where $n$ is even.
$(2x)^2 = 4g + 2$, $x \in \mathbb{Z}$.
$4x^2 = 4g + 2$.
$4s = 4g + 2$, $s = x^2, s \in \mathbb{Z}$ as integers are closed under multiplication.
$2s = 2g + 1$
$2s$ is even, and $2g + 1$ is odd (by definition of even/odd numbers). An even number cannot equal an odd number, so we have a contradiction.
Consider the case where $n$ is odd.
$(2x + 1)^2 = 4g + 2$, $x \in \mathbb{Z}$
$4x^2 + 4x + 1 = 4g + 2$
$4x^2 + 4x = 4g + 1$
$4(x^2 + x) = 4g + 1$
$4j = 4g + 1$, $j = x^2 + x, j \in \mathbb{Z}$ as integers are closed under addition
$2d = 2e + 1$, $d = 2j, e = 2g; d, e \in \mathbb{Z}$ as integers are closed under multiplication
$2d$ is even, and $2e + 1$ is odd (by definition of even/odd numbers). An even number cannot equal an odd number, so we have a contradiction.
As both cases have a contradiction, the original supposition is false, and $n^2 - 2$ is not divisible by $4$.
|
For odd $n$, $n^2-2$ is odd.
For even $n$, $n^2$ is divisible by $4$, so that $n^2-2$ is not.
By contradiction:
Let $n$ be odd. Then $n^2-2$ is both odd and a multiple of four.
Let $n$ be even. Then $n^2-2$ and $n^2$ are both multiples of $4$, so that $2$ is a multiple of $4$.
|
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|
$\int_{-\infty}^{\infty }\frac{x^{2} \, dx}{x^{4}+a^{4}}$ I'm considering the following integral:
$$\int_{-\infty}^{\infty }\frac{x^{2} \, dx}{x^{4}+a^{4}}$$
I do not know how to factor the denominator to calculate the residues.
I'm stuck here:
$$(x^{2}-|a|^{2}i)(x^{2}+|a|^{2}i)$$
Thank you so much.
|
Find the indefinite integral.
First, determine constants $p, q, r, s$ such that
$$
\frac{px+q}{x^2+\sqrt{2}ax+a^2}+\frac{rx+s}{x^2-\sqrt{2}ax+a^2} = \frac{x^2}{x^4+a^4}
$$
By comparing the coefficients of $x^3, x^2, x, 1$, we have the following relational expression:
$$
p+r=0, \\
-\sqrt{2}a(p-r)+q+s=1, \\
a^2(p+r)-\sqrt{2}a(q-s)=0, \\
(q+s)a^2=0
$$
Thus, $p=-\frac{1}{2\sqrt{2}a}, \ q=0, \ r=\frac{1}{2\sqrt{2}a}, \ s=0$.
Based on those,
$$
\int \frac{x^2}{x^4+a^4}dx \\
= \int \left\{ -\frac{\frac{1}{2\sqrt{2}a}(x+\frac{\sqrt{2}}{2}a)}{x^2+\sqrt{2}ax+a^2} + \frac{\frac{1}{2\sqrt{2}a}(x-\frac{\sqrt{2}}{2}a)}{x^2-\sqrt{2}ax+a^2} + \frac{1}{4}\frac{1}{x^2+\sqrt{2}ax+a^2} + \frac{1}{4}\frac{1}{x^2-\sqrt{2}ax+a^2} \right\} dx \\
= -\frac{1}{4\sqrt{2}a}\log{(x^2+\sqrt{2}ax+a^2)} + \frac{1}{4\sqrt{2}a}\log{(x^2-\sqrt{2}ax+a^2)} \\
+ \frac{1}{4}\int \left\{ \frac{1}{(x+\frac{1}{\sqrt{2}}a)^2+\frac{a^2}{2}} + \frac{1}{(x-\frac{1}{\sqrt{2}}a)^2+\frac{a^2}{2}} \right\} dx
$$
The remaining integral is,
$$
\frac{1}{4} \cdot \frac{2}{a^2} \int \left\{ \frac{1}{(\frac{\sqrt{2}}{a}x+1)^2+1} + \frac{1}{(\frac{\sqrt{2}}{a}x-1)^2+1} \right\} dx \\
= \frac{1}{2\sqrt{2}a} \left\{ \tan^{-1}(\frac{\sqrt{2}}{a}x+1) + \tan^{-1}(\frac{\sqrt{2}}{a}x-1) \right\} + Const.
$$
To sum up,
$$
\int \frac{x^2}{x^4+a^4}dx \\
= \frac{1}{4\sqrt{2}a}\log {\frac{x^2-\sqrt{2}ax+a^2}{x^2+\sqrt{2}ax+a^2}} + \frac{1}{2\sqrt{2}a} \left\{ \tan^{-1}(\frac{\sqrt{2}}{a}x+1) + \tan^{-1}(\frac{\sqrt{2}}{a}x-1) \right\} + Const.
$$
After that, check each item of the result when $x \rightarrow \infty$ or $-\infty$ and calculate.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a >0$. Then...
$x_2 = 2 + \frac{1}{2} = 2.5$
$x_3 = 2.5 + \frac{1}{2.5} = 2.9$
$x_4 = 2.9 + \frac{1}{2.9} = 3.2448$
$x_5 = 3.2448 + \frac{1}{3.2448} = 3.5530$
$x_6 = 3.5530 + \frac{1}{3.5530} = 3.8344$
$x_7 = 3.8344 + \frac{1}{3.8344} = 4.0952$
$x_8 = 4.0952 + \frac{1}{4.0952} = 4.3394$
$x_9 = 4.3394 + \frac{1}{4.3394} = 4.5698$
$x_{10} = 4.5698 + \frac{1}{4.5698} = 4.7887$
but others have said it converges so I'm confused on whether it converges or diverges? Can someone please explain.
|
We have $$x_{N+1}-x_1=\sum_{n=1}^N x_{n+1}-x_n =\sum_{n=1}^N \frac1{x_n}$$ so $x_n$ converges if and only if the rightmost series converges.
But the series converges only if $x_n$ diverges to $\pm\infty $, impossible: both the sequence and the series must diverge.
|
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|
Prove that $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
Prove that
$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
What I've tried doing :
If $\theta=\frac{\pi}{7}:$
$$
3\theta+4\theta=\pi
$$
This allowed me to prove that :
$$
\tan^2\frac{\pi}{7}+\tan^2\frac{2\pi}{7}+\tan^2\frac{3\pi}{7}=21
\\
\cot^2\frac{\pi}{7}+\cot^2\frac{2\pi}{7}+\cot^2\frac{3\pi}{7}=5
$$
Is my reasoning wrong or is this entirely the wrong way to approach this question ?
|
Since $\sin\frac{k\pi}{7}=\sin\frac{(7-k)\pi}{7}$ for k=1,2,3 we get:
$$(\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7})^2=
\prod_{k=1}^6 {\sin\frac{k\pi}{7}}=\frac{7}{2^6}$$
where the last equality is the case n=7 in this question:
Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$
So we get the desired result: $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $ I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x\log(1+x^3)=x\int(1-x^3+x^6-x^9+-...)\\\\
&=x\left[x-\frac{x^4}{4}+\frac{x^7}{7}-\frac{x^{10}}{10}+-...+ C\right]
\end{align}$$
$$
f(0)=0; C=0
$$
$$
f(x)=x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...
$$
$$\begin{align}
\int_0^{1/2}x\log(1+x^3)dx&=\int_0^{1/2}(x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...)\,dx\\\\
&=\left.\left[\frac{x^3}{3}-\frac{x^6}{6*4}+\frac{x^9}{7*9}-\frac{x^{12}}{10*12}+-...\right]\right|_0^{1/2}\\\\
&=\frac{1}{2^3(3)}-\frac{1}{2^6(6)(4)}+\frac{1}{2^9(7)(9)}-\frac{1}{2^{12}(10)(12)}+-...
\end{align}$$
Is my method correct?
|
Here is an efficient way forward. We write $\log(1+x^3)=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^{3n}}{n}$, which is valid for $|x|<1$.
The series converges uniformly on $[0,1/2]$ and we may exchange the series and integration to obtain
$$\begin{align}
\int_0^{1/2}x\log(1+x^3)\,dx&=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_0^{1/2}x^{3n+1}\,dx\\\\
&=\frac14\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n(3n+2)2^{3n}}
\end{align}$$
It might be of interest to note that the integrand has an antiderivative in terms of elementary functions, which can be obtained by integration by parts and partial fraction expansion (SEE HERE). Hence the integral can be evaluated over any interval in closed form. Here, we find that
$$\int_0^{1/2}x\log(1+x^3)\,dx=\frac{\sqrt 3 \pi}{12}-\frac{3\log(4e)}{16}$$
|
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|
Maximum value of $x^2+y^2$ Question
Find the maximum value of $x^2+y^2$ if
$4x^4+9y^4=64$
Now I really don't understand how to proceed or whether I should change my approach all together. Any help is appreciated.Thanks :)
|
$$4x^4+9y^4=64$$
$$\frac{(x^2)^2}{4^2}+\frac{(y^2)^2}{(8/3)^2}=1$$
Let $x=\pm \sqrt{4\cos(\theta)}$ and $y=\pm \sqrt{\frac{8}{3} \sin (\theta)}$. With $\theta \in [0,\frac{\pi}{2}]$.
Then,
$x^2+y^2=4\cos (\theta)+\frac{8}{3} \sin (\theta)$
$$=\langle 4, \frac{8}{3} \rangle \cdot \langle \cos (\theta), \sin (\theta) \rangle$$
$$=\sqrt{4^2+(\frac{8}{3})^2} \cos \left( \theta-\arctan(\frac{\frac{8}{3}}{4}) \right)$$
Because $\arctan(\frac{8}{12}) \in [0, \frac{\pi}{2}]$. A maximum of,
$$\sqrt{4^2+(\frac{8}{3})^2}$$
Is achievable.
|
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|
If $a>b>0$ and $a^3+b^3+27ab=729$ then $ax^2+bx-9=0$ has roots $\alpha,\beta,(\alpha<\beta)$. Find the value of $4\beta -a\alpha$.
If $a>b>0$ and $a^3+b^3+27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $\alpha,\beta,(\alpha<\beta)$. Find the value of $4\beta -a\alpha$.
By looking at the equation I figured out $a+b=9$. Hence one root of of the equation is 1. But I don't know how to proceed further. It would be great if I could get some help with this question.
|
Hint ( I haven't completely verified)
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
If this equal zero then
$$(a+b+c)=0$$ or
$$(a^2+b^2+c^2-ab-bc-ca)=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
$32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a gmat exam question.
|
Repeatedly squaring mod $19$ and noting that $3\cdot13\equiv1\pmod{19}$:
$$
\begin{align}
10^2&\equiv5&\pmod{19}\\
10^4&\equiv6&\pmod{19}\\
10^8&\equiv17&\pmod{19}\\
10^{16}&\equiv4&\pmod{19}\\
10^{32}&\equiv16&\pmod{19}\\
10^{33}&\equiv8&\pmod{19}\\
10^{33}-1&\equiv7&\pmod{19}\\
\frac{10^{33}-1}3&\equiv7\cdot13&\pmod{19}\\
&\equiv15&\pmod{19}
\end{align}
$$
Edit After a Clarification to the Question
Above, it is shown that $10^{33}\equiv8\mod{19}$. Squaring twice and multiplying by $10^{33}\equiv8\mod{19}$ gives
$$
\begin{align}
10^{66}&\equiv7&\pmod{19}\\
10^{132}&\equiv11&\pmod{19}\\
10^{165}&\equiv12&\pmod{19}\\
10^{165}-1&\equiv11&\pmod{19}\\
\frac{10^{165}-1}3&\equiv11\cdot13&\pmod{19}\\
&\equiv10&\pmod{19}\\
\end{align}
$$
Computing $\boldsymbol{\frac13\pmod{n}}$
If $n\ne0\pmod3$, then either $n$ or $2n$ is $\equiv-1\pmod3$. This means that either $\frac{n+1}3$ or $\frac{2n+1}3$ is an integer. That integer is $\frac13\pmod{n}$ (that is, multiplying it by $3$ gives $1\bmod{n}$).
|
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|
Suppose a is a real number for which all the roots of the equation $ x^4 - 2ax^2 + x +a^2 -a =0$ are real.Find the range of values of a? I have tried the question with $Rolle's$ theorem...
Let the equation have $2$ unequal real roots
$f'(x)= 4ax^3-4ax+1=0$
But cannot proceed after this.
|
we can write above equation as $x^4-2ax^2+x+a^2-a=0$ as $a^2-(2x^2+1)a+x^4+x=0$
So $$\displaystyle a= \frac{(2x^2+1)\pm\sqrt{(2x^2+1)^2-4(x^4+x)}}{2}=\frac{(2x^2+1)\pm (2x-1)}{2}$$
So $x^2+x-a=0$ or $x^2-x+1-a=0$
So for real roots , $1+4a\geq 0$ and $1-4(1-a)\geq 0 $
|
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|
proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
& = 5
\end{align}
Assumption: $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n = k \in \mathbb{N}$.
$5 \mid (3^{2n-1} + 2^{2n-1}) \implies 3^{2n-1} + 2^{2n-1} = 5m$, $m \in \mathbb{Z}$
Proof: Let $n = k + 1$
\begin{align}
3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\
& = 3^{2k+1} + 2^{2k+1}\\
& = 3^{2k} \cdot 3^1 + 2^{2k} \cdot 2^1\\
& = 3^{2k} \cdot 3 + 2^{2k} \cdot 2
\end{align}
And here I got stuck. I don't know how to get from the last line to the Assumption. Either I am overlooking a remodeling rule or I have used a wrong approach.
Anyway, I am stuck and would be thankful for any help.
|
From where you got to:
\begin{align}
3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\
& = 9\cdot3^{2k-1} + 4\cdot 2^{2k-1}\\
& = 5\cdot3^{2k-1} + 4(3^{2k-1}+ 2^{2k-1})\\
& =5\cdot3^{2k-1} + 4\cdot 5m \tag {from hypothesis}\\
& =5(3^{2k-1} + 4m)\\
\end{align}
... so divisible by $5$ as required
|
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|
Find the singular values of $T: p(x)\mapsto xp'(x)+2x^2p''(x)$ I would appreciate help with this problem I am self-studying this problem:
Find the singular values of the operator $T\in \mathcal P_2(\mathbb{C}): p(x)\mapsto xp'(x)+2x^2p''(x)$. The inner-product on $\mathcal P_2(\mathbb{C})$ is defined as $\langle p,q \rangle:=\int_{-1}^{1} p(x)\overline{q(x)}dx$.
I know the singular values are the eigenvalues of $\sqrt{T^*T}$, where $T=S \sqrt{T^*T}$ and $S$ is an isometry.
Thanks
EDIT I am a bit skeptical of the hint to construct a matrix, for one, because I don't see how that will utilize the defined inner-product. I think for a start, I should write:
$\langle Tp,q\rangle= \int_{-1}^{1} Tp(x)\overline{q(x)}dx= \int_{-1}^{1} p(x)\overline{T^*q(x)}dx=\langle p,T^*q\rangle$, where $Tp(x)=xp'(x)+2x^2p''(x)$
But, assuming this is on the right tract, I don't know how to proceed. Thanks again
|
Start with the basis $\{1, x, x^2\}$ for $\mathcal{P}_2(\mathbb{C})$. Performing Gram-Schmidt yields the following orthonormal basis for $\mathcal{P}_2(\mathbb{C})$: $$\mathcal{B} = \left\{ \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}x, \sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right\}_.$$
Let $A$ denote the matrix of $T$ with respect to $\mathcal{B}$. We then compute that:
$$A = \begin{pmatrix}
0 & 0 & \sqrt{45}\\
0 & 1 & 0\\
0 & 0 & 6
\end{pmatrix} = \begin{pmatrix}
0 & 0 & 3\sqrt{5}\\
0 & 1 & 0\\
0 & 0 & 6
\end{pmatrix}$$
Since $\mathcal{B}$ is an orthonormal basis, the matrix of $T^*$ with respect to $\mathcal{B}$ is the conjugate transpose of $A$ (Relevant link if you feel shaky about this). Therefore the matrix of $T^*T$ with respect to $\mathcal{B}$ is:
$$\overline{A^T}A = \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
3\sqrt{5} & 0 & 6
\end{pmatrix}\begin{pmatrix}
0 & 0 & 3\sqrt{5}\\
0 & 1 & 0\\
0 & 0 & 6
\end{pmatrix} = \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 81 \\
\end{pmatrix}_.$$
Finally, the matrix of $\sqrt{T^*T}$ with respect to $\mathcal{B}$ is then:
$$\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 9
\end{pmatrix}_.$$
Thus the singular values of $T$ are $0,1, \text{ and } 9$.
Per request: How to find the matrix $A$.
To do so, we find the image of each element of $\mathcal{B}$ under $T$, then express the image as a linear combination of the elements of $\mathcal{B}$. The coefficients of this linear combination become entries of the matrix.
$$T\left(\frac{1}{\sqrt{2}}\right) = 0 = 0\left(\frac{1}{\sqrt{2}}\right) + 0 \left(\sqrt{\frac{3}{2}}x \right) + 0 \left(\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right)_.$$
This corresponds to the first column of $A$ being $\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}_.$
$$T\left(\sqrt{\frac{3}{2}}x\right) = x\sqrt{\frac{3}{2}} + 2x^2(0) = 0 \left(\frac{1}{\sqrt{2}}\right) + 1 \left(\sqrt{\frac{3}{2}}x\right) + 0 \left(\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right)_.$$
This corresponds to the second column of $A$ being $\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}_.$
\begin{align*}
T\left(\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right) &= x\left(2\sqrt{\frac{45}{8}}x\right) + 2x^2\left(2\sqrt{\frac{45}{8}}\right)\\
&= 6\sqrt{\frac{45}{8}}x^2 \\
&= \sqrt{45}\left(\frac{1}{\sqrt{2}}\right) + 0 \left(\sqrt{\frac{3}{2}}x\right) + 6\left(\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right)_.
\end{align*}
This corresponds to the third column of $A$ being $\begin{pmatrix} \sqrt{45} \\ 0 \\ 6 \end{pmatrix}_.$
|
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|
Evaluating $\int \frac{x^2-1}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx$
The question is to evaluate $$\int \frac{x^2-1}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx$$
I tried to rewrite the integral as $$\frac{ab}{b-a}\int \frac{1/a(x^2+ax+1)-1/b(x^2+bx+1)}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx-2\int \frac{1}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx$$ For the second integral I rewrite it as $$\int \frac{1}{x^3\sqrt{\frac1{x^2}+\frac{a}{x}+1}\sqrt{\frac1{x^2}+\frac{b}{x}+1}} dx$$ Now I used the substitution $x\to 1/x$ to yield $$-\int \frac{1}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx$$ which makes the second integral to vanish.However i could not proceed with the first integral.Any ideas.Thanks.
|
Suppose $x>0$. Letting $x+\frac{1}{x}\to u$ and then $du=(1-\frac{1}{x^2})dx$, one has
\begin{eqnarray}
&&\int \frac{x^2-1}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx\\
&=&\int\frac{1-\frac1{x^2}}{\sqrt{x+a+\frac{1}{x}}\sqrt{x+b+\frac{1}{x}}} dx\\
&=&\int\frac{1}{\sqrt{u+a}\sqrt{u+b}}du\\
&=&\int\frac{1}{\sqrt{(u+\frac{a+b}2)^2-(\frac{a-b}{2})^2}}du\\
\end{eqnarray}
It is not hard to get the integral. I omit the detail.
|
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|
Approaching Tricky Combinatorial Proofs - Tough Example I'm struggling with finding effective ways to approach questions (especially proofs) that use summations and Taylor Series. I've worked through several simpler examples, but always get stuck once a non-trivial question arises. In particular, I'm hoping to get some help in proving the following statement
$$\sum_{i=0}^{n-1} {2i \choose i} \frac{1}{i+1}\frac{1}{2^{2i}}=2(1-\frac{1}{2^{2i}}{2i \choose i})$$
As a hint, it's stated that $ \frac{1-\sqrt{1-x}}{\frac{1}{2}x}=\sum_{i=0}^{\infty}{2i \choose i}\frac{1}{i+1}\frac{x^{i}}{2^{2i}} $ and that it may potentially be necessary to derive the Taylor seires centered at $x=0$ for $\frac{1-\sqrt{1-x}}{x} $ and $\frac{1}{\sqrt{1-x}} $
With some help, I've done the following work, but am not sure if I'm either headed in the right direction, or even correct:
$$\sum_{i=0}^{n-1} {2i \choose i} \frac{1}{i+1}\frac{1}{2^{2i}}=\sum_{i=0}^{\infty}{2i \choose i} \frac{1}{i+1}\frac{1}{2^{2i}}-\sum_{i=n}^{\infty}{2i \choose i} \frac{1}{i+1}\frac{1}{2^{2i}}=2-\sum_{i=n}^{\infty}{2i \choose i} \frac{1}{i+1}\frac{1}{2^{2i}}$$
From here, I've tried to get to the point of proving the following:
$$2-\sum_{i=n}^{\infty}{2i \choose i} \frac{1}{i+1}\frac{1}{2^{2i}}=2-\frac{2}{2^n}{2n \choose n}$$
$$\sum_{i=n}^{\infty}{2i \choose i} \frac{1}{i+1}\frac{1}{2^{2i}}=\frac{2}{2^n}{2n \choose n} $$
I really don't know where to proceed next, or which direction the proof will continue in. Any additions/corrections would be greatly appreciated. Thank you!
Edit: to add on another potential solution, would it hold any water to try to write down a recurrence relation, assuming the left side of the first equation to be $ a_n $?
|
Here is an approach based on the observation that
\begin{align*}
\sum_{i=0}^{n-1}\binom{2i}{i}\frac{1}{i+1}\cdot\frac{1}{2^{2i}}
\end{align*}
is a Cauchy-product which can be interpreted as the coefficient of the product of two (ordinary) generating functions. We recall the generating function of the Catalan numbers $C_i=\binom{2i}{i}\frac{1}{i+1}$
\begin{align*}
C(x)=\sum_{i=0}^\infty C_i x^i=\frac{1-\sqrt{1-4x}}{2x}\tag{1}
\end{align*}
It is also convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
\sum_{i=0}^{n-1}\binom{2i}{i}\frac{1}{i+1}\frac{1}{2^{2i}}
&=\frac{1}{4^{n-1}}\sum_{i=0}^{n-1}\binom{2i}{i}\frac{1}{i+1}4^{n-1-i}\tag{2}\\
&=\frac{1}{4^{n-1}}[x^{n-1}]\frac{C(x)}{1-4x}\tag{3}\\
&=\frac{1}{4^{n-1}}[x^{n-1}]\frac{1-\sqrt{1-4x}}{2x(1-4x)}\tag{4}\\
&=\frac{1}{2\cdot4^{n-1}}[x^{n}]\left(\frac{1}{1-4x}-\frac{1}{\sqrt{1-4x}}\right)\tag{5}\\
&=\frac{1}{2\cdot4^{n-1}}\left(4^n-\binom{-\frac{1}{2}}{n}(-4)^n\right)\tag{6}\\
&=2\left(1-\frac{1}{2^{2n}}\binom{2n}{n}\right)\tag{7}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we write the sum as Cauchy-product in the form $\sum_{i=0}^{n-1}a_ib_{n-1-i}$.
*In (3) we use the fact that the Cauchy-product is the coefficient of $x^{n-1}$ of $C(x)$ and the geometric series $\frac{1}{1-4x}$.
*In (4) we use the representation (1).
*In (5) we do a small rearrangement to easily derive the coefficients of the series and apply the rule $[x^{p+q}]A(x)=[x^p]x^{-q}A(x)$.
*In (6) we select the coefficient of $x^n$ of the geometric series and the binomial series with $\alpha=-\frac{1}{2}$.
*In (7) we apply the binomial identity
\begin{align*}
\binom{-\frac{1}{2}}{i}=\frac{1}{2^{2i}}\binom{2i}{i}(-1)^i
\end{align*}
See e.g. (1.9) in H.W. Goulds Binomial Identities, vol. 1.
|
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|
Solve the inequality $(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$
Solve the inequality $$(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$$
My work so far:
1) $a>0$
2) Let $f(x)=(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)$.
$f(x)=0$ if and only if $x^2-2\sqrt{a}\cdot x+1=0$ or $2^x+\lg a=0$
$$x^2-2\sqrt{a}\cdot x+1=0$$
$$x_{1,2}=\sqrt a \pm \sqrt{a-1}$$
and $$2^x+\lg a=0$$
$$x=\log_2\lg\frac1a$$
|
$1)$ If $0<a<1$ then we have $x^2-2\sqrt{a}\cdot x+1>0$ for all real number $x$.
Then we must have
$$2^x+\log a<0\to 2^x<\log a^{-1}\to x<\log_2(\log a^{-1})$$
$2)$ If $a>1$ then we have $2^x+\log a>0$ for all real number $x$.
Then we must have
$$x^2-2\sqrt{a}\cdot x+1<0\to \sqrt{a}-\sqrt{a-1}<x<\sqrt{a}+\sqrt{a-1}$$
$3)$ If $a=1$ then
$$(x^2-2x+1)(2)=2(x-1)^2\ge0$$
and then we have no solution.
|
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|
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+\tan^3(x)} dx$$
Then to evaluate $\int \frac{1}{1+\tan^3(x)}dx$, I let $\tan(x)=t$ and hence $dt = \sec^2(x) dx$, which implies $dx=\frac{1}{\sec^2(x)}dt$.
Then I don't know how to continue.
|
Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$ ......(1)
Using property -
$\int_a^{b} f(x) dx = \int_a^{b} f(a + b - x) dx$
We have,
$I = \int_0^{\frac{\pi}{2}} \frac{\sin^3(\frac{\pi}{2}-x)}{\cos^3(\frac{\pi}{2}-x)+\sin^3(\frac{\pi}{2}-x)} dx$
$I = \int_0^{\frac{\pi}{2}} \frac{\cos^3(x)}{\sin^3(x)+\cos^3(x)} dx$ .....(2)
Adding equation (1) and (2),
$2I = \int_0^{\frac{\pi}{2}} dx$
$2I = \left[x\right]_0^{\frac{\pi}{2}}$
$2I = \frac{\pi}{2}$
$I = \frac{\pi}{4}$
|
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|
If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How to prove further?
|
from your condition we get $$a^2+c^2-ac=b^2$$ apllying the theorem of cosines we get
$$a^2+c^2-ac=a^2+c^2-2ac\cos(\beta)$$
thus we get $$\cos(\beta)=\frac{1}{2}$$
|
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|
System of Equation. (Complicated) Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt,
I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone would provide some different solutions. Thanks in advance.
|
$$5x(1+\frac{1}{x^2+y^2})=12\cdots \cdots (1)$$
$$5y(1-\frac{1}{x^2+y^2})=4\cdots \cdots \cdots \cdots (2) \times i$$
$$5(x+iy)+5\cdot \frac{x-iy}{(x+iy)(x+iy)}=12+4i$$
Now adding these two equations
put $z=x+iy$ and $\bar{z} = x-iy$ and $|z|^2= z \cdot \bar{z} = x^2+y^2$
So $$5z+\frac{5\bar{z}}{z \cdot \bar{z}}=12+4i$$
So $$5z+\frac{5}{z} = 12+4i$$
So $$5z^2-(12+4i)z+5=0$$
after solving $$z=\frac{12+4i\pm \sqrt{(12+4i)^2-100}}{10} = \frac{12+4i\pm (8+6i)}{10} = 2+i\;,\frac{2-i}{5}$$
So $$z= x+iy = 2+i\;,\frac{2-i}{5}$$
So $$(x,y) = \left\{(2,1)\;\;, \left(\frac{2}{5},-\frac{1}{5}\right)\right\}$$
|
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|
If $c$ is equal to $\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$ then find $\log_2{(-c)}$. If $c$ is equal to $$\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$$then find $\log_2{(-c)}$.
I think the binomial theorem might help but the bottom numbers are skipping by 2's. How would I apply it now?
|
Since
$$
\begin{gathered}
\left( {1 + i\,} \right)^{\,n} = \sqrt 2 ^{\,n} \left( {\cos \left( {\frac{{n\,\pi }}
{4}} \right) + i\sin \left( {\frac{{n\,\pi }}
{4}} \right)} \right) = \sum\limits_{0\, \leqslant \,k\left( { \leqslant \;n} \right)} {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)i^{\,k} } = \hfill \\
= \sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( \begin{gathered}
n \\
2j \\
\end{gathered} \right)i^{\,2j} } + \sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( \begin{gathered}
n \\
2j + 1 \\
\end{gathered} \right)i^{\,2j + 1} } = \hfill \\
= \sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( { - 1} \right)^{\,j} \left( \begin{gathered}
n \\
2j \\
\end{gathered} \right)} + i\sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( { - 1} \right)^{\,j} \left( \begin{gathered}
n \\
2j + 1 \\
\end{gathered} \right)} \hfill \\
\end{gathered}
$$
Then
$$
\sum\limits_{0\, \leqslant \,j\left( { \leqslant \;n/2} \right)} {\left( { - 1} \right)^{\,j} \left( \begin{gathered}
n \\
2j \\
\end{gathered} \right)} = \sqrt 2 ^{\,n} \cos \left( {\frac{{n\,\pi }}
{4}} \right)
$$
i.e.:
$$
\begin{gathered}
\log _2 \left( { - c} \right) = \log _2 \left( { - 2^{\,99/2} \cos \left( {\frac{{99\,\pi }}
{4}} \right)} \right) = \log _2 \left( { - 2^{\,99/2} \cos \left( {\frac{{3\,\pi }}
{4}} \right)} \right) = \hfill \\
= \log _2 \left( {2^{\,99/2} \frac{{\sqrt 2 }}
{2}} \right) = \log _2 \left( {2^{\,98/2} } \right) = 49 \hfill \\
\end{gathered}
$$
|
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|
Partial sum back into series. If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n−1}{n+1}$
then find $a_n$ and find
the sum of this series. Fully justify your answer.
Using the definition.
$a_n = S_{n} - S_{n-1} $
$$a_n = \frac{n-1}{n+1} - \frac{n-2}{n}$$
$$= \frac{n(n-1) - (n-2)(n+1)}{n(n+1)}$$
$$= \frac{n^2-n - (n^2-n-2)}{n(n+1)}$$
$$= \frac{2}{n(n+1)}$$
Hmm, I can use PFD on this to make it.
$$= \frac{2}{n} - \frac{2}{n+1}$$
Is this right? What about n = 0
The sum is 1 right? cus lim at $\infty$ ?
|
Obviously, the sequence can be defined as
$$\begin{cases}a_0=S_0=-1,\\a_n=S_n-S_{n-1}=\dfrac{n-1}{n+1}-\dfrac{n-2}{n}=\dfrac2{n(n+1)},n>0.\end{cases}$$
(In particular, $a_1=1$, ensuring $S_1=a_0+a_1=0$.)
The sum is indeed $\lim_{n\to\infty}\dfrac{n+1-2}{n+1}=1$.
|
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|
Solution for continued fraction To solve the finite continued fraction $[2,....,2]$, can I use this solution.
Let $y$ be the value of $[2,....,2]$.
Then $y=[2;y]=2+\frac{1}{y}$.
So $y-2=\frac{1}{y}$.
Thus $y^2-2y-1=0$
My question is is my method correct. Is this approach suitable to solve $[2,....,2]$
|
Is this approach suitable to solve $[2,....,2]$
No, because your continued fraction is finite. Moreover, it is rational and the solutions of $y^2-2y-1=0$ are irrational.
Notation.
\begin{equation*}
x_{n}=a_0+\cfrac{1}{a_1+\cfrac{1}{\begin{array}{ccc}a_{2}+ & & \\& \ddots & \\& & +\cfrac{1}{a_{n-1}+\cfrac{1}{a_{n}}}\end{array}}}=
\left[ a_{0};a_{1},a_{2},\ldots ,a_{n}\right] =[2;\underset{n\text{
elements}}{\underbrace{2,\ldots ,2}}]=\frac{p_{n}}{q_{n}}.
\end{equation*}
We then have $x_{0}=[a_{0}]=\frac{p_{0}}{q_{0}}=\frac{a_{0}}{1}=\frac{2}{1}$
with $(p_{0},q_{0})=(2,1)$; and $x_{1}=\left[ a_{0};a_{1}\right] =\frac{p_{1}}{q_{1}
}=a_{0}+\frac{1}{a_{1}}=\frac{a_{0}a_{1}+1}{a_{1}}=\frac{5}{2}$; with $
(p_{1},q_{1})=(5,2)$.
For $n\geq 2$ the integers $p_{n},q_{n}$ satisfy the
fundamental recurrent relation
\begin{eqnarray*}
p_{n} &=&a_{n}p_{n-1}+p_{n-2}=2p_{n-1}+p_{n-2}, \\
q_{n} &=&a_{n}q_{n-1}+q_{n-2}=2q_{n-1}+q_{n-2}.
\end{eqnarray*}
The next two terms of the finite sequence $(x_n)$ are
\begin{eqnarray*}
x_{2} &=&\left[ a_{0};a_{1},a_{2}\right] =a_{0}+\frac{1}{a_{1}+1/a_{2}}=
\frac{p_{2}}{q_{2}}=\frac{2p_{1}+p_{0}}{2q_{1}+q_{0}}=\frac{2(5)+2}{2(2)+1}=
\frac{12}{5}, \\
x_{3} &=&\left[ a_{0};a_{1},a_{2},a_{3}\right] =a_{0}+\frac{1}{
a_{1}+1/(a_{2}+1/a_{3})}=\frac{p_{3}}{q_{3}}=\frac{2p_{2}+p_{1}}{2q_{2}+q_{1}
}=\frac{2(12)+5}{2(5)+2}=\frac{29}{12}, \\
&&\cdots
\end{eqnarray*}
and
\begin{equation*}
(p_{2},q_{2})=(12,5),(p_{3},q_{3})=(29,12),\ldots .
\end{equation*}
|
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|
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\beta=\frac{c}{a}$$
Therefore the roots of $9x^2-2x+7=0$ are $(\alpha+2)$ and $(\beta+2)$. So, $$\alpha+\beta+4=\frac{2}{9}\implies4-\frac{b}{a}=\frac{2}{9}\implies\frac{b}{a}=\frac{34}{9}$$ and
\begin{align*}
(\alpha+2)(\beta+2)=\frac{7}{9}\\
\Rightarrow\alpha\beta+2(\alpha+\beta)+4=\frac{7}{9}\\
\Rightarrow\frac{c}{a}-2\frac{b}{a}+4=\frac{7}{9}\\
\Rightarrow\frac{4a-2b+c}{a}=\frac{7}{9}\\
\end{align*}
So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.
There is another similar question: If the roots of $px^2+qx+r=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then what will be the expression of $r$ in terms of $a$, $b$, and $c$?
|
Hint: Find the roots of equation $9x^2-2x+7=0$ by solving this let say the roots be $\alpha$ and $\beta$ (You know the value of $\alpha$ and $\beta$.) then the roots of other equation must be $\alpha-2$ and $\beta-2$ form a equation using $$(x-(\alpha-2))(x-(\beta-2))$$ compare the coefficient. then put the value in $4a-2b+c.$
|
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|
Method to prove limit in $\mathbb{R}^2$ Given is the following limit $\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}$.
Now, it holds that
$0\leq\left\vert\frac{x^3y}{x^4+y^2}\right\vert\leq \frac{x^3y}{y^2}=\left\vert\frac{x^3}{y}\right\vert$.
It is clear that $(x,y)\mapsto\frac{x^3}{y}$ is homogeneous on $\mathbb{R^2}\setminus\{\vec0\}$. This implies that $\lim_{(x,y)\to(0,0)}\frac{x^3}{y}=0$.
The squeeze lemma would now imply that
$\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}=0$.
Where does it go wrong?
|
Let $S=\frac {|x^3y|}{x^4+y^2}.$ We have $x^2|y|\leq \frac {(x^2+|y|)^2}{4}$ so $$S=\frac {|x|\cdot x^2|y|}{x^4+y^2} \leq \frac {|x|(x^2+|y|)^2}{4(x^4+y^2)}=$$ $$=\frac {|x|(x^4+y^2)+2|x^3y|}{4(x^4+y^2)}=$$ $$ =\frac {|x|}{4}+\frac {S}{2}.$$ $$ \text {Therefore }\quad S\leq \frac {|x|}{2}.$$
Not as slick as the answer from Dr.MV but it works.
|
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|
Eigen value and eigen vectors I have given a matrix $$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{pmatrix}. $$
We have to find 3 linearly independent eigen vector. I have calculated the eigen values that is 1 with multiplicity 3 . I have found the eigen vector $ \begin{pmatrix}1 & 1 & 1 \end{pmatrix}^T$. But I can't find the other two linearly independent eigen vectors.
Any help is appreaciating.
|
The characteristic polynomial is
\begin{align}
\det\begin{pmatrix}
0-X & 1 & 0 \\
0 & 0-X & 1 \\
1 & -3 & 3-X
\end{pmatrix}
&=
-X\det\begin{pmatrix}-X & 1 \\ -3 & 3-X\end{pmatrix}
-\det\begin{pmatrix}0 & 1 \\ 1 & 3-X\end{pmatrix}
\\
&=-X(-3X+X^2+3)+1
\\[6px]
&=-X^3+3X^2-3X+1=(1-X)^3
\end{align}
There cannot exist three linearly independent vectors: the matrix would be diagonalizable and similar to
$$
\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}
$$
but the identity matrix is only similar to itself.
Actually, if we compute the rank of $A-I$ (where $A$ is the given matrix), we get, with Gaussian elimination,
$$
A-I=\begin{pmatrix}
-1 & 1 & 0 \\
0 & -1 & 1 \\
1 & -3 & 2
\end{pmatrix}
\to
\begin{pmatrix}
-1 & 1 & 0 \\
0 & -1 & 1 \\
0 & -2 & 2
\end{pmatrix}
\to
\begin{pmatrix}
-1 & 1 & 0 \\
0 & -1 & 1 \\
0 & 0 & 0
\end{pmatrix}
$$
so the rank is $2$ and the eigenspace relative to $1$ has dimension $3-2=1$. No set of two eigenvalues can be linearly independent.
|
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|
If $(a,b)=1$, how to find the suitable $u$ and $v$ such that $(a+b)u+(a^2-ab+b^2)v=3$? Suppose $(a,b)=1$. I want to prove that $(a+b,a^2-ab+b^2)=1$ or $3$.
Since $(a,b)=1$, so $(a^2,b^2)=1$ and therefore there are $x,y\in\mathbb Z$ such that $a^2x+b^2y=1$ or $3a^2x+3b^2y=3$ and by adding and subtracting $3abx+3aby+3a^2y+3b^2x$, we have
$$\begin{align*}3a^2x+3b^2y&+3abx+3aby+3a^2y+3b^2x\\
&-3abx-3aby-3a^2y-3b^2x\\
&=(a-b)(3bx-3ay)+(a^2-ab+b^2)(3x+3y)=3
\end{align*}$$
which implies that $(a-b,a^2-ab+b^2)|3$ and so $(a-b,a^2-ab+b^2)=1$ or $3$.
But all of my try to find the suitable combination for $a+b$ and $a^2-ab+b^2$ fail. Can anyone help me please?
|
\begin{align}a^{2} - a b + b^{2} &= (a + b)(a - 2 b) + 3 b^{2} \\&= (a + b) (b - 2 a) + 3 a^{2}.\end{align}
This you can see by first regarding $a^{2} - a b + b^{2}$ and $a + b$ as polynomials in $a$, and dividing $a^{2} - a b + b^{2}$ by $a + b$. And then doing the same regarding them as polynomials in $b$, or simply by symmetry.
So if $d \mid \gcd (a+b,a^2-ab+b^2)$, then $d \mid \gcd(3 a^{2}, 3 b^{2}) = 3 \gcd(a^{2}, b^{2})= 3$.
Clearly both cases occur, take first $a = b = 1$, and then $a = 1$, $b = -1$.
If $a^{2} x + b^{2} y = 1$, then
\begin{align}
3
&= 3 a^{2} x + 3 b^{2} y
\\&= ((\color{red}{a^{2} - a b + b^{2}}) - (\color{blue}{a + b}) (b - 2 a)) x + ((\color{red}{a^{2} - a b + b^{2}}) - (\color{blue}{a + b})(a - 2 b)) y
\\&= (\color{red}{a^{2} - a b + b^{2}})(x + y) + (\color{blue}{a + b}) (-(b-2a)x - (a - 2 b) y)
\end{align}
|
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|
Matrix and a Linear Function I'm a student of electrical engineering. At our math class we were given following assignment: given a matrix
$A = \begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}$ and a function $f(x) = x^5 - x^3 + 5x^2 + 8$, compute $f(A)$.
I've checked out this PDF document, but I don't think I understand it properly. I have tried by assuming that first element (let me denote it as $a_{11}$) should equal $$a_{11}=1^5-1^3+5\cdot 1+8=13$$ and second element $a_{12}$ should consequently equal $$a_{12} = 0 + 0+ 0 + 8=8$$
As it turns out (according to professor's solutions), $a_{11}$ really is $13$, but $a_{12}$ is zero. Final solution should be $$f(A) = \begin{pmatrix}
13 & 0 & 24 \\ 0 & 13 & 0 \\ 0 & 0 & 13
\end{pmatrix}$$
Now, I know that this site is not meant for students to ask questions and anticipate full solution to their problems, so instead I'm just asking for a guideline. Can somebody please give me any hint on how to solve this task?
|
I claim that
$$A^n = \begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}^n=\begin{pmatrix}
1 & 0 & 2n\\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}.\tag 1$$
By the principle of mathematical induction, first, for $n=2$
$$A^2 = \begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}^2=\begin{pmatrix}
1 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}$$
then assuming that
$$A^{n-1}= \begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}^{n-1}=\begin{pmatrix}
1 & 0 & 2(n-1) \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}$$
we have
$$A^n=A^{n-1}A =$$
$$= \begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}^{n-1}\begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}=$$
$$=\begin{pmatrix}
1 & 0 & 2(n-1) \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}\begin{pmatrix}
1 & 0 & 2\\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}=\begin{pmatrix}
1 & 0 & 2n\\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}.$$
So, $(1)$ is true. Then
$$A^5 =\begin{pmatrix}
1 & 0 & 10\\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix},A^3 =\begin{pmatrix}
1 & 0 & 6\\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}, 5A^2 =\begin{pmatrix}
5 & 0 & 20\\ 0 & 5 & 0 \\ 0 & 0 & 5\end{pmatrix}. $$
Finally,
$$f(A) = A^5 - A^3 + 5A^2 + 8I=$$
$$=\begin{pmatrix}
13 & 0 & 24\\ 0 & 13 & 0 \\ 0 & 0 & 13\end{pmatrix}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Find how many roots of $z^5+2z^2+7z+1=0$ lie inside $|z|=2$? How many are real? Find how many roots of $z^5+2z^2+7z+1=0$ lie inside $|z|=2$? How many are real?
Let $f(z)=z^5+2z^2+7z+1$, and $g(z)=z^5$. For $|z|=2,$ we note that
$$
|f(z)-g(z)|\leq2|z|^2+7|z|+1=23<32=2^5=|g(z)|.
$$
Since $g$ has $5$ zeros in $|z|\leq 2$, Rouche's theorem implies that $f$ also has $5$ zeros lie inside $|z|=2$.
However, I have no idea to find how many of them are real. Can anyone tell me how to do it? Thanks.
|
I'd never seen Sturm's theorem before, but it was interesting enough that I'll add the computation:
\begin{align}
p_0(x) &= x^5+2x^2+7x+1\\
p_0'(x) = p_1(x) &= 5x^4+4x+7\\
\text{remainder of $-p_0/p_1$:}\quad p_2(x) &= -\tfrac{6}{5}x^2 - \tfrac{28}{5}x - 1\\
\text{remainder of $-p_1/p_2$:}\quad p_3(x) &= \tfrac{12562}{27}x + \tfrac{8669}{108}\\
\text{remainder of $-p_2/p_3$:}\quad p_4(x) &= -\tfrac{87841935}{1262430752}\\
\end{align}
Computing at the end points,
$$p_0(-2)=-37,\;p_1(-2)=79,\;p_2(-2)=\tfrac{27}{5},\;p_3(-2)=-\tfrac{3401}{4},\;p_4(-2)=-\tfrac{87841935}{1262430752}$$
$$p_0(2)=55,\;p_1(2)=95,\;p_2(2)=-17,\;p_3(2)=\tfrac{109165}{108},\;p_4(2)=-\tfrac{87841935}{1262430752}$$
Hence at $-2$ we have the sign sequence $-++--$, which has $2$ sign changes and at $2$ we have $++-+-$, which has $3$ sign changes and so there are $3-2=1$ real roots in $[-2,2]$.
Alternatively, we may note that the second derivative $4(5x^3+1)$ is zero at $x = -\tfrac{1}{\sqrt[3]{5}}$, which is a global minimum of the derivative, which is positive there, hence $x^5+2x^2+7x+1$ is monotone increasing everywhere. As $p_0(-2) < 0 < p_0(2)$, by continuity of polynomials and intermediate value theorem there is precisely one root in $[-2,2]$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2218292",
"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int {t+ 1\over t^2 + t - 1}dt$
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$
Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {1- w_2\over w - w_2}$$
$$\int {t+ 1\over t^2 + t - 1}dt = A\int {1\over t+ w} dt + B\int {1 \over t+w_2}dt \\= {w - 1\over w - w_2} \ln|t + w| + {1- w_2\over w - w_2}\ln|t + w_2| + C $$
After finding the value of $w, w_2$ final answer I got is
$${\sqrt{5} + 1\over 2\sqrt{5}}\ln|t + 1/2 - \sqrt{5}/2| + {\sqrt{5} - 1 \over 2\sqrt{5}}\ln|t + 1/2 + \sqrt{5}/2| + C$$
But the given answer is :
$$\bbox[7px,Border:2px solid black]{ \frac{\ln\left(\left|t^2+t-1\right|\right)}{2}+\frac{\ln\left(\left|2t-\sqrt{5}+1\right|\right)-\ln\left(\left|2t+\sqrt{5}+1\right|\right)}{2\cdot\sqrt{5}}+C}$$
Where did I go wrong ? especially that first term of the answer is a mystery to me.
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To get to their answer more directly, you can write the integral as$$\frac 12\int \frac{2t+1}{t^2+t-1}dt+\frac 12\int\frac{1}{(t+\frac 12)^2-\frac 54}dt$$
$$=\frac 12 \ln|t^2+t-1|+\frac{1}{2\sqrt{5}}\ln\left|\frac{t+\frac 12-\frac{\sqrt{5}}{2}}{t+\frac 12+\frac{\sqrt{5}}{2}}\right|+c$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2222679",
"timestamp": "2023-03-29T00:00:00",
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|
Vector magnitude subtraction Please help! This was in a textbook and I cannot seem to make sense of it. Would it not be 2?
The two vectors a and b are perpendicular. If a has magnitude 8 and b has magnitude 3, what is |a−2b|??
I
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Since your vectors are perpendicular, they don't directly add and subtract like scalars would.
Denoting them in R$^2$:
$\textbf{a} - 2\textbf{b} = \begin{pmatrix}
8 \\
0
\end{pmatrix} -
2 \begin{pmatrix}
0 \\
3
\end{pmatrix} =
\begin{pmatrix}
8 \\
-6
\end{pmatrix}.$
Thus we need to consider what the value of $\begin{pmatrix}
8 \\
-6
\end{pmatrix}$.
Often, this is calculated with the Euclidean Distance, as $\left| \begin{pmatrix}
8 \\
-6
\end{pmatrix} \right| = \sqrt{8^2 + (-6)^2} = 10. $
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Knowing that $\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3}$ and $\tan \alpha > 0$, find the exact value of $\tan \alpha$ First I tried to solve for $\alpha$:
$$\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3} \Leftrightarrow \alpha = \arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{2} \lor \alpha = -\arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{2}$$
Then I put $\tan(\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ and $\tan(-\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ in the calculator and $\tan(\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ was positive while the other was negative, so I assumed the positive one was the correct expression.
But I don't know how to simplify it to get the exact value. My book says the solution is $\frac{\sqrt{2}}{4}$. How do I solve this?
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Hint: your first equation gives $\sin (\alpha)\,=\,\frac{1}{3}$ which can be easily solved to get $\tan(\alpha)\,=\,\frac{\sqrt2}{4}$
$$\cos(\frac{3\pi}{2}+\alpha)\,=\,-\sin(\alpha)$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$ We are given a function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that $$ f ( x ) + f \left( \frac 1 x \right) = 1 $$
and
$$ f ( 2 x ) = 2 f \big( f ( x ) \big) \text . $$
Find, with proof, an explicit expression for $f(x)$ for all positive rational numbers $x$.
Every number I have evaluated is of the form $ f ( x ) = \frac x { x + 1 } $ and this clearly fits the functional equations, but I can't prove that it's the only solution. Can anyone help me? I have put down the start of my workings which led me to the conjecture of $ f ( x ) = \frac x { x + 1 } $.
Plugging in $ x = 1 $ clearly gives $ f ( 1 ) = \frac 1 2 $ and $ f ( 2 ) = 2 f \big( f ( 1 ) \big) = 2 f \left( \frac 1 2 \right) $ which we can plug back into the first equation to get that $ f ( 2 ) = \frac 2 3 $. Working in this vein I have been able to show that $ f ( x ) = \frac x { x + 1 } $ for particular values of $ x $, but not in general.
The most difficult part appears to be proving it for the even integers. To prove $ x = 8 $, we have
$$ f ( 12 ) = 2 f \left( \frac 6 7 \right) = 4 f \left( \frac 3 { 10 } \right) = 4 - 4 f \left( \frac { 10 } 3 \right) \\
= 4 - 8 f \left( \frac 5 8 \right) = 8 f \left( \frac 8 5 \right) - 4 = 16 f \left( \frac 4 9 \right) - 4 \\
= 32 f \left( \frac 2 { 11 } \right) - 4 = 64 f \left( \frac 1 { 12 } \right) - 4 = 60 - 64 f ( 12 ) \text , $$
giving us $ f ( 12 ) = \frac { 12 } { 13 } $. This will probably be the main area of difficulty in the proof.
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A bit too long for a comment, but here's a start:
You have shown $f(1) = 1/2$, $f(2) = 2/3$ and hence $f(1/2) = 1/3$.
Plug $x = 1/2$ into your second identity to get $f(1) = 2f(f(1/2))$, i.e $1/2 = 2 f(1/3)$, i.e $\color{blue}{f(1/3) = 1/4}$. Thus, $\color{blue}{f(3) = 3/4}$ as well.
To get $f(4)$ and $f(1/4)$, we must use the second identity twice as well as the first identity. First $f(4) = 2f(f(2)) = 2f(2/3)$. In addition, $f(2/3) = 2f(f(1/3)) = 2f(1/4)$. Thus $f(4) = 4f(1/4)$ and since $f(4) + f(1/4) = 1$ by the first identity, $\color{blue}{f(4) = 4/5}$ and $\color{blue}{f(1/4) = 1/5}$. We also get $\color{blue}{f(2/3) = 2/5}$ and $\color{blue}{f(3/2) = 3/5}$.
Then $f(1/2) = 2f(f(1/4))$ gives $1/3 = 2f(1/5)$ to get $\color{blue}{f(1/5) = 1/6}$ and $\color{blue}{f(5) = 5/6}$.
To get $f(6)$ and $f(1/6)$, we try to repeat the arguments used for $f(4)$ and $f(1/4)$. First, $f(6) = 2f(f(3)) = 2f(3/4)$. In addition $f(2/5) = 2f(f(1/5)) = 2f(1/6)$. Alas, this is where my ad hoc calculations break down.
Hopefully, these simple calculations are useful to help others find the underlying structure necessary to continue.
|
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"timestamp": "2023-03-29T00:00:00",
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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.