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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
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Since $x := \cos \frac{2 \pi}{5} = \frac{z + z^{-1}}{2}$ where $z:=e^{\frac{2 i \pi}{5}}$, and $1+z+z^2+z^3+z^4=0$ (for $z^5=1$ and $z \neq 1$), $x^2+\frac{x}{2}-\frac{1}{4}=0$, and voilà.
|
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|
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
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I'll post the one I know since it is Euler's, and is quite easy and stays in $\mathbb{R}$. (I'm guessing Euler didn't have tools like residues back then).
Let
$$s = {\sin ^{ - 1}}x$$
Then
$$\int\limits_0^{\frac{\pi }{2}} {sds} = \frac{{{\pi ^2}}}{8}$$
But then
$$\int\limits_0^1 {\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx} = \frac{{{\pi ^2}}}{8}$$
Since
$${\sin ^{ - 1}}x = \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = x + \frac{1}{2}\frac{{{x^3}}}{3} + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{{{x^5}}}{5} + \frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}\frac{{{x^7}}}{7} + \cdots $$
We have
$$\int\limits_0^1 {\left\{ {\frac{{dx}}{{\sqrt {1 - {x^2}} }}\int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} } \right\}} = \int\limits_0^1 {\left\{ {x + \frac{1}{2}\frac{{{x^3}}}{3}\frac{{dx}}{{\sqrt {1 - {x^2}} }} + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{{{x^5}}}{5}\frac{{dx}}{{\sqrt {1 - {x^2}} }} + \cdots } \right\}} $$
But
$$\int\limits_0^1 {\frac{{{x^{2n + 1}}}}{{\sqrt {1 - {x^2}} }}dx} = \frac{{2n}}{{2n + 1}}\int\limits_0^1 {\frac{{{x^{2n - 1}}}}{{\sqrt {1 - {x^2}} }}dx} $$
which yields
$$\int\limits_0^1 {\frac{{{x^{2n + 1}}}}{{\sqrt {1 - {x^2}} }}dx} = \frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}}$$
since all powers are odd.
This ultimately produces:
$$\frac{{{\pi ^2}}}{8} = 1 + \frac{1}{2}\frac{1}{3}\left( {\frac{2}{3}} \right) + \frac{{1 \cdot 3}}{{2 \cdot 4}}\frac{1}{5}\left( {\frac{{2 \cdot 4}}{{3 \cdot 5}}} \right) + \frac{{1 \cdot 3 \cdot 5}}{{2 \cdot 4 \cdot 6}}\frac{1}{7}\left( {\frac{{2 \cdot 4 \cdot 6}}{{3 \cdot 5 \cdot 7}}} \right) \cdots $$
$$\frac{{{\pi ^2}}}{8} = 1 + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}} + \cdots $$
Let
$$1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \cdots = \omega $$
Then
$$\frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{8^2}}} + \cdots = \frac{\omega }{4}$$
Which means
$$\frac{\omega }{4} + \frac{{{\pi ^2}}}{8} = \omega $$
or
$$\omega = \frac{{{\pi ^2}}}{6}$$
|
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Proving an identity involving terms in arithmetic progression. If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:
$ (1)\large \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$
$(2) \large\frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \biggr)$
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Both identities can be proved quite easily with inductions. Let $d$ be the common difference, i.e. $d=a_{n+1}-a_n$.
*
*Use induction. So we need to prove
$$ \frac{n-2}{\sqrt{a_1}+\sqrt{a_ {n-1}}}+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_ n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_ n}}.$$
Rationalize the denominators and substitute $$a_{n-1}-a_1=(n-2)d, a_n-a_{n-1}=d, a_n-a_1=(n-1)d.$$ Upon cancellations, we find
$$\sqrt{a_{n-1} }-\sqrt{a_ 1}+\sqrt{a_n }-\sqrt{a_{n-1} }=\sqrt{a_ n}-\sqrt{a_ 1}$$ on the left and the same on the right.
*Use induction, assuming true for $m<n$. Canceling the sum $$ \frac{1}{a_1a_n}+\frac{1}{a_na_1}$$ from the left with the term $$\frac{2}{a_1+a_n}\left(\frac{1}{a_1}+\frac{1}{a_n}\right)$$ on the right and using the induction hypothesis on the remaining terms $$\frac{1}{a_2a_{n-1}}+\cdots +\frac{1}{a_{n-1}a_2},$$ we find that we need to prove $$\frac{2}{a_2+a_{n-1}}$$ on the left is equal to $$\frac{2}{a_1+a_{n}}$$ on the right. And this is true because $a_2+a_{n-1}=a_1+a_n=2a_1+(n-1)d$.
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Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
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Like Trigonometry Simplification,
$$\frac{2\sin60^\circ\cdot\cos20^\circ-2(2\sin20^\circ\cos20^\circ)}{\sin20^\circ}$$
Using Werner Formula we get,
$$\frac{\sin80^\circ+\sin40^\circ-2\sin40^\circ}{\sin20^\circ}$$
Using Prosthaphaeresis Formula, $\sin80^\circ-\sin40^\circ=2\sin20^\circ\cos60^\circ$
|
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How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$? Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$
how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$
Thank you
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Perhaps a sixth way...
$$\displaystyle S = \sum_{r=1}^{d} r\cdot 2^r$$
$$\displaystyle 2S = \sum_{r=1}^{d} r\cdot 2^{r+1} = \sum_{r=2}^{d+1} (r-1)2^{r}$$
$$\displaystyle 2S -S = d\cdot 2^{d+1} - \sum_{r=1}^{d} 2^r = d\cdot 2^{d+1} - 2^{d+1} +2 = (d-1)2^{d+1} + 2$$
|
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Average length of the longest segment This post is related to a previous SE post If a 1 meter rope …. concerning average length of a smallest segment.
A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment.
My answer is 11/18. Here is how I do it:
Here we have two independent random variables $X,Y$, both uniform on $[0,1]$. Let
$A=\min (X,Y), B=\max (X,Y)$ and $C=\max (A, 1-B, B-A)$. First we want to find the probability
density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then
$$ F_C(a) = P(C\le a)=P(A\le a, 1-B\le a, B-A\le a).$$ By rewriting this probability as area in the unit square, I get
$$F_C(a)=\left\{\begin{array}{ll} (3a-1)^2 & \frac{1}{3}\le a\le \frac{1}{2}\\ 1-3(1-a)^2 & \frac{1}{2}\le a\le 1\end{array}\right.$$ from which it follows that
$$f_C(a)=\left\{\begin{array}{ll} 6(3a-1) & \frac{1}{3}\le a\le \frac{1}{2}\\ 6(1-a) & \frac{1}{2}\le a\le 1\end{array}\right.$$ Therefore
the expected value of $C$ is
$$\int_{1/3} ^{1/2}6a(3a-1) da+\int_{1/2} ^{1}6a(1-a) da= \frac{11}{18}.$$
My questions are:
(A) Is there a "clever" way to figure out this number 11/18?
(B) What is the answer if the rope is divided into $n>3$ segments?
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Neat as it is, I don't think Rahul's answer can be correct. If we have $3x+2y+z=1$ and $3x+2y+z=9n$, then $n=1/9$, which means $x \leq 2/9$, which can't be right, as one solution is all pieces being of length 1/3 (however unlikely this exact solution may be, $x$ can take values in $(2/9,1/3]$).
Stefan's answer is wrong because the probability of any given point on the stick being in the longest piece (when cut in 2) is not uniform. That can be seen by considering the point halfway along, which is always in the longest piece. You can find the probability density function for the likelihood of any given point being in the longer half, then integrate to get the answer below.
My preferred solution is to let the cuts be at $X, Y$, with $Y \gt X$:
Image of cut positions
Then each piece is equally likely to be the longest, and the expected length of the longest piece doesn't depend on which piece we choose. Then we can calculate $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} )$.
We have the three inequalities:
$$X \gt Y-X \implies Y < 2X$$
$$X \gt 1-Y \implies Y > 1-X$$
and, from our setup,
$$Y \gt X$$
These can be represented by the following diagram:
Diagram of inequalities
Then the area satisfying our inequalities is the two triangles A and B. So we wish to find the expected value of $X$ within this area.
The expected value of $X$ in A is $\bar{X}_A = \frac{1}{2}-\frac{1}{3}(\frac{1}{2}-\frac{1}{3}) = \frac{8}{18}$.
The expected value of $X$ in B is $\bar{X}_B = \frac{1}{2}+\frac{1}{3}(\frac{1}{2}) = \frac{4}{6} = \frac{12}{18}$
The area of A is $A_A = \frac{1}{2} \times \frac{1}{2}\times (\frac{1}{2}-\frac{1}{3}) = \frac{1}{24}$.
The area of B is $A_B = \frac{1}{2} \times \frac{1}{2}\times \frac{1}{2} = \frac{1}{8} = \frac{3}{24} = 3 A_A$.
So $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} ) = \frac{\tfrac{8}{18} + 3\left(\tfrac{12}{18}\right)}{4} = \frac{11}{18}$
(Apologies for the comments on other solutions and the links, but I haven't got a high enough reputation to comment or embed directly)
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$\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$ Show that if $a$, $b$ are positive integers, then we have:
$\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$.
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Another Dubuquesque attempt; for legibility, write $d=\gcd(a,b)$:
\begin{align*}
\gcd\Bigl(d(a+b), ab\Bigr) &= \gcd\Bigl(d(a+b), ab, ab\Bigr)\\
&=\gcd\Bigl(d(a+b),\ ab-a(a+b),\ ab-b(a+b)\Bigr)\\
&=\gcd\Bigl(d(a+b),\ a^2,\ b^2\Bigr)\\
&=\gcd\Bigl(d(a+b),\ \gcd(a^2,b^2)\Bigr)\\
&=\gcd\Bigl(d(a+b),\ \gcd(a,b)^2\Bigr)\\
&=\gcd\Bigl(d(a+b),\ d^2\Bigr)\\
&= d\gcd\Bigl(a+b,d\Bigr)\\
&= d\gcd\Bigl(a+b,\gcd(a,b)\Bigr)\\
&= d\gcd(a,b)\\
&= \gcd(a,b)\gcd(a,b).
\end{align*}
(Second line uses the fact that $a(a+b)$ and $b(a+b)$ are both multiples of $d(a+b)$).
Now divide through by $\gcd(a,b)$ to get the desired result.
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How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
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Just for completeness, here is induction (just for divisibility by 7) :
Claim : $n^7 - n$ is divisible by 7
Base Case: True for n = 1,2
Induction Step:
Assume true for n = k. To prove true for n = k + 1.
Now, $$(k+1)^7 - (k+1) = k^7 + 7k^6 + 21k^5 + 35k^4 + 35k^3 + 21k^2 + 7k + 1 - k - 1 \\= (k^7 - k) + 7(k^6 + 3k^5 + 5k^4 + 5k^3 + 3k^2 + k)$$
We know by our assumption that $k^7 - k$ is divisible by 7. Therefore, $(k + 1)^7 - (k + 1)$ is divisible by 7.
This shows that $n^7 - n$ is divisible by 7.
To show divisibility by 2 and 3, unfortunately, one has to fall back on some of the earlier tricks.
$$n^7 - n = n(n^6 - 1) = (n-1)n(n+1)(n^2 - n + 1)(n^2 + n + 1)$$
$(n-1)n(n+1)$ is a product of three consecutive integers. Product of two consecutive integers is divisible by 2 and product of three consecutive integers is divisible by 3. Since, $(2,3) = 1$, product of three consecutive integers is divisible by 6. Since $(6,7) = 1$, $n^7 - n$ is divisible by 42.
QED
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How to prove a formula for the sum of powers of $2$ by induction? How do I prove this by induction?
Prove that for every natural number n, $ 2^0 + 2^1 + ... + 2^n = 2^{n+1}-1$
Here is my attempt.
Base Case: let $ n = 0$ Then, $2^{0+1} - 1 = 1$ Which is true.
Inductive Step to prove is: $ 2^{n+1} = 2^{n+2} - 1$
Our hypothesis is: $2^n = 2^{n+1} -1$
Here is where I'm getting off track. Lets look at the right side of the last equation: $2^{n+1} -1$ I can rewrite this as the following.
$2^1(2^n) - 1$ But, from our hypothesis $2^n = 2^{n+1} - 1$ Thus:
$2^1(2^{n+1} -1) -1$ This is where I get lost. Because when I distribute through I get this.
$2^{n+2} -2 -1$ This is wrong is it not? Am I not applying the rules of exponents correctly here? I have the solution so I know what I'm doing is wrong. Here is the correct proof.
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I don't see the answer I like here, so I'm writing my own.
Basic proof:
We wish to prove $2^0 + 2^1 + ... + 2^{n-1} = 2^n - 1$ for all $n$. We can verify by inspection this is true for n=1. Next, assume that $2^0 + 2^1 + ... + 2^{n} = 2^{n+1} - 1$.
$(2^0 + 2^1 + ... + 2^n) + 2^{n+1} = (2^{n+1} - 1) + 2^{n+1} = 2 \cdot 2^{n+1} - 1 = 2^{n+2} - 1$, so we have shown $2^0 + 2^1 + ... + 2^{n-1} = 2^{n} - 1$ is true for all n.
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Show that $3^{4n+2} + 1$ is divisible by $10$ I'm am a little bit stuck on this question, any help is appreciated.
Show that for every $n\in\mathbb{N}$, $3^{4n+2} + 1$ is divisible by $10$.
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What is the formation law of the remainders of the division by $10$ of the
powers $3^{n}$?
(For the notation see modular arithmetic.)
$$\left\{
\begin{array}{c}
3\equiv 3\quad \pmod{10} \\
3^{2}\equiv 9\quad \pmod{10} \\
3^{3}\equiv 7\quad \pmod{10} \\
3^{4}\equiv 1\quad \pmod{10}
\end{array}\right. $$
$$\left\{
\begin{array}{c}
3^{5}\equiv 3\quad \pmod{10} \\
3^{6}\equiv 9\quad \pmod{10} \\
\cdots \\
\cdots
\end{array}\right. $$
So, with respect to the divisor $10$ the remainders of $3^{n}$ are periodic ($3,9,7,1,3,9,7,1,\dots$) with period $4$. This together with $4n+2\equiv 2\quad \pmod{4}$ yields $3^{4n+2}\equiv 3^{2}\quad \pmod{10}$. Also $1\equiv 1\quad \pmod{10}$. Hence, for all $n\ge 1$ $$3^{4n+2}+1\equiv 3^{2}+1\equiv 0\quad \pmod{10},$$
which means the remainders of the devision of $3^{4n+2}+1$ by $10$ are $0$.
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Implicitly find the second derivative Given the formula $x^{2}y^{2}-8x=3$, find the second derivative.
I calculated the first derivative as $$-\frac{xy^{2}+4}{x^{2}y}$$
Working from that, I calculated the second derivative starting with
$$\frac{([x^{2}y)\frac{d}{dx}(-xy^{2}+4)]-[(-xy^{2}+4)\frac{d}{dx}(x^{2}y)]}{(x^{2}y)^{2}}$$
The left $\frac{d}{dx}$ was calculated by
$$\frac{d}{dx}(-xy^{2}+4)= [\frac{d}{dx}(-xy^{2})]+[\frac{d}{dx}(4)]= -x\frac{d}{dx}(y^{2})+ y^{2}\frac{d}{dx}(-x)= -2xy\frac{dy}{dx}- y^{2}$$
The right $\frac{d}{dx}$ was calculated by
$$\frac{d}{dx}(x^{2}y)= x^{2}\frac{d}{dx}(y)+ y\frac{d}{dx}(x^{2})= x^{2}\frac{dy}{dx}+2xy$$
Plugging everything into the formula resulted in
$$\frac{[(x^{2}y)(-2xy\frac{d}{dx}-y^{2}] - [(-xy^{2}+4)(x^2\frac{dy}{dx}+2xy)]}{(x^{2}y)^2}$$
$$\frac{[(-2x^{2}y^{2}\frac{dy}{dx}- 2x^{2}y^{3}]- [(x^{3}y^{2}\frac{dy}{dx}+2x^{2}y^{3}- 4x^2\frac{dy}{dx}-8xy]}{x^{4}y{2}}$$
Combining like terms resulted in
$$\frac{-x^{3}y^{2}\frac{dy}{dx}-4x^{2}\frac{dy}{dx}-8xy}{x^{4}y^{3}}$$
This is where I stall out. All my previous examples haven't been in quotient form, and how do I produce a $\frac{dy}{dx}$ from a quotient?
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It's actually a bit simpler to work directly with the original (since then you don't have to worry about the quotient rule). Simply take derivatives twice, and then solve for $y''$ in terms of $x$, $y$, and $y'$; only then plug in $y'$.
Start with
$$x^2y^2 - 8x=3.$$
Taking derivatives once, we get
\begin{align*}
\frac{d}{dx}\Bigl(x^2y^2 - 8x\Bigr) &= \frac{d}{dx}3\\
x^2\frac{d}{dx}y^2 + y^2\frac{d}{dx}x^2 - 8 &= 0\\
x^2\bigl(2yy'\bigr) + y^2\bigl(2x\bigr) -8 &= 0\\
2x^2yy' + 2xy^2 - 8&= 0\\
x^2yy' + xy^2 - 4&= 0.
\end{align*}
Now take derivatives again, and solve for $y''$:
\begin{align*}
\frac{d}{dx}\Bigl(x^2yy' + xy^2-4\Bigr) &= 0\\
x^2y\left(\frac{d}{dx}y'\right) + x^2y'\left(\frac{d}{dx}y\right) + yy'\left(\frac{d}{dx}x^2\right)\\
\quad+x\left(\frac{d}{dx}y^2\right) + y^2\left(\frac{d}{dx}x\right) &=0\\
x^2yy'' + x^2(y')^2 + 2xyy' + 2xyy' + y^2 &=0\\
x^2yy'' +x^2(y')^2 + 4xyy' + y^2 &=0\\
x^2yy'' &= -\Bigl( x^2(y')^2 + 4xyy' + y^2\Bigr)\\
y''&= -\frac{x^2(y')^2 + 4xyy' + y^2}{x^2y}\\
y''&= -\frac{(y')^2}{y} - \frac{4y'}{x} - \frac{y}{x^2}.
\end{align*}
If you want to get the value entirely in terms of $x$ and $y$, you can go back to the formula we had with the first derivative,
$$x^2yy' + xy^2 - 4 = 0,$$
we can solve for $y'$ to get
\begin{align*}
x^2yy' &= 4 - xy^2\\
y' &= \frac{4-xy^2}{x^2y}\\
y' &= \frac{4}{x^2y} - \frac{y}{x}.
\end{align*}
So we can plug in this value of $y'$ into the formula for $y''$:
\begin{align*}
y'' &= -\frac{(y')^2}{y} - \frac{4y'}{x} - \frac{y}{x^2}\\
y'' &= -\frac{1}{y}\left(\frac{4}{x^2y} - \frac{y}{x}\right)^2 - \frac{4}{x}\left(\frac{4}{x^2y}-\frac{y}{x}\right) - \frac{y}{x^2}\\
y'' &= -\frac{1}{y}\left(\frac{16}{x^4y^2} - \frac{8y}{x^3y} + \frac{y^2}{x^2}\right)-\frac{16}{x^3y} + \frac{4y}{x^2} - \frac{y}{x^2}\\
y'' &= -\frac{16}{x^4y^3} + \frac{8}{x^3y} - \frac{y}{x^2} - \frac{16}{x^3y} + \frac{3y}{x^2}\\
y'' &= -\frac{16}{x^4y^3} - \frac{8}{x^3y}+ \frac{2y}{x^2}.
\end{align*}
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If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? Where $p$ is an odd prime.
I read in the book, they claimed:
$$p - 1 \equiv -1 \pmod{p}$$
$$p - 2 \equiv -2 \pmod{p}$$
$$p - 3 \equiv -3 \pmod{p}$$
$$ ... $$
$$\frac{p - 1}{2} \equiv -(\frac{p - 1}{2}) \pmod{p}$$
However, I realized that the last statement was wrong because if it's true then:
$$\frac{p - 1}{2} + \frac{p - 1}{2}\equiv 0 \pmod{p}$$
$$p - 1 \equiv 0 \pmod{p}$$
Any idea?
Thanks,
|
In the 5th edition, they have got the signs right, at least thats what the scanned version of the ebook says.
EDIT
We have $p-k \equiv -k \pmod{p}$ and hence $$ \prod_{k=1}^{\frac{p-1}{2}} (p-k) \equiv \prod_{k=1}^{\frac{p-1}{2}} (-k) \pmod{p}$$
Note that $$(p-1)! = \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (p-k) \right)$$
Hence, $$(p-1)! \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (p-k) \right) \pmod{p} \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (-k) \right) \pmod{p}$$
Hence, $$(p-1)! \equiv (-1)^{\frac{p-1}{2}} \left(\prod_{k=1}^{\frac{p-1}{2}} k \right)^2 \pmod{p}$$
$$\left(\prod_{k=1}^{\frac{p-1}{2}} k \right) = \left(\frac{p-1}{2}\right)!$$
From Wilson's theorem, $$(p-1)! \equiv -1 \pmod{p}$$
Hence, $$-1 \equiv (-1)^{\frac{p-1}{2}} \left[ \left(\frac{p-1}{2}\right)! \right]^2 \pmod{p}$$
Hence, $$\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv (-1)^{\frac{p+1}{2}} \pmod{p}$$
Hence, if $p = 4k+3$, then $$(-1)^{\frac{p+1}{2}} = 1$$
Hence, if $p = 4k+3$, $$\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv 1 \pmod{p}$$
|
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|
Find the value of 'x' in a product of exponentiated logarithms Find the value of '$x$' if, $$\large \left(\frac{1}{2^{\log_x 4}} \right) \cdot \left( \frac{1}{2^{\log_x 16}} \right) \cdot \left(\frac{1}{2^{\log_x 256}} \right) \cdots = 2 $$
I tried to make it simple by resulting series is not converging, the suggested answers in my module is either ($2,\frac{1}{2},4,\frac{1}{4}$)
Also I tried this and this but not the correct answer in either case.
|
$$
\begin{align}
\left(\frac{1}{2^{\log_x 4}}\right)\left(\frac{1}{2^{\log_x 16}}\right)\left(\frac{1}{2^{\log_x16}}\right)\cdots&=2\\
2^{-\log_x 4}\times2^{-\log_x 16}\times2^{-\log_x 256}\ldots&=2\\
2^{-\left(\log_x 4+\log_x 16+\log_x 256+\cdots\right)}&=2\\
\log_x(4\times16\times256\times\cdots)&=-1\\
4\times 16\times256\times\cdots&=\frac{1}{x}\\
2^2\times2^4\times2^8\times\cdots&=\frac{1}{x}\\
2^{2+4+8+\cdots}&=\frac{1}{x}
\end{align}
$$
Now say we have, $2+4+8+\cdots$; this can be written as $2+4+8+\cdots+2^n$ when $n\to\infty$. $2+4+8+\cdots+2^n$ is a geometric series and:
$$2+4+8+\cdots+2^n=2\left(2^n-1\right)$$
So now we have:
$$
\begin{align}
x&=\lim_{n\to\infty}\left(\frac{1}{2^{2\left(2^n-1\right)}}\right)
\end{align}$$
|
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Generating functions in combinatorics I am not very familiar with how generating functions are used but for something I was needing I ran into the following constructions,
*
*Let $c_{nk}$ be the number of solutions in $\{1,2,3,4...,\}$ for the equation $x_1+x_2+x_3+...+x_k = n$. Then one can easily show that $c_{nk} = ^{n-1}C_{k-1}$ But the following is not clear to me..one defines a "generating function" $c_k(x)$ as, $c_k(x) = \sum_{n=k}^{\infty} c_{nk} x^n$. Then one claims that the following is true,
$c_k(x) = (x+x^2+x^3+...)^k = x^k(1-x)^k$
I can't see from where the above came!
From the above it follows that, $\sum_{k=1} ^\infty c_k(x) = \frac{x}{1-2x}$
The idea seems to be that from the above one can read off that $\sum {k=1} ^n c_{nk} = 2^{n-1}$. (This was obvious from the initial formula given for $c_{nk}$) But this way of getting that result is not clear to me.
*
*Let $p(n)$ be the number of unordered partitions of $n$ (a positive integer). Then clearly $p(n)$ is the number of solutions to the equation, $x_1+x_2+x_3+...+x_n = n$ with $x_1\geq x_2 \geq x_3 ...x_n > 0$ After appropriate variable redefinition one can see that $p(n)$ is the number of solutions of $y_1 + 2y_2 + 3y_3 + ... + ny_n = n$ with $y_i \geq 0$ for all $i$. Now apparently $p(n)$ can be given through a generating function $P(x)$ defined as,
$P(x) = \sum _{n=0} ^\infty p(n)x^n$
Now I can't see how it follows that,
$P(x) = \prod _{k=1} ^\infty \frac{1}{(1-x^k)}$
(apparently the above can somehow be mapped to the familiar question of number of ways in which $n$ balls can be arranged in a line where $r_i$ of them are of the colour $i$ ($\sum r_i = n$). This is equal to the coefficient of $\prod _i x_{i}^{r_i}$ in $\prod _i (1+x_i + x_i^2 + x_i ^3 + ...)$. But the connection between this question and the generating function for unordered partitions of an integer is not cleat to me.)
|
In both cases, the key is to recognize the power series identity (which should be familiar if you know about geometric series)
$$ 1+y+y^2+y^3+ \ldots = \sum_{i=0}^\infty y^i = \frac{1}{1-y}. $$
In your formula for $c_k(x)$, the last exponent should certainly be negative, so we have $c_k(x) = x^k(1-x)^{-k}$. Then applying the power series identity shows that $\frac{x}{1-x} = x(1-x)^{-1} = x +x^2 +x^3 + \ldots$.
For the partitions generating function, we use the same identity many times, with $y = x^k$ for all integers $k\geq 1$. Think of how to find the coefficient of $x^n$ in
$$(1+x+x^2 + x^3 + \ldots) (1+x^2+x^4+x^6 +\ldots) (1+x^3+x^6+x^9+ \ldots)\cdots$$
and you'll see that it gives you precisely the number of partitions of $n$.
EDIT:
Okay, think about the case $k=2$ for $c_k(x)$. The claim is that
$$c_2(x) = (x+x^2+x^3+\ldots)^2.$$
Just look at what happens when we expand this, using the distributive property:
\begin{align*}
c_2(x) &= (x+x^2+x^3+\ldots)(x+x^2+x^3+\ldots) \\
&=x(x+x^2+x^3+\ldots)+x^2(x+x^2+x^3+\ldots)+x^3(x+x^2+x^3+\ldots)+ \ldots \\
&=(x^{1+1}+x^{1+2}+x^{1+3}+\ldots)+(x^{2+1}+x^{2+2}+x^{2+3}+\ldots)+(x^{3+1}+x^{3+2}+x^{3+3}+\ldots)+ \ldots \\
&=(x^{1+1})+(x^{1+2}+x^{2+1})+(x^{1+3}+x^{2+2}+x^{3+1})+\ldots \\
&=x^2 + 2x^3 + 3x^4 + \ldots
\end{align*}
You see that to get each $x^n$ term, we simply need a solution to $a_1 + a_2 = n$ with $a_i \geq 1$. In the same way, if you have $c_k(x) = (x+x^2+x^3+\ldots)^k$, to get each $x^n$ term, we simply need a solution to $a_1 + a_2 + \ldots + a_k = n$ with $a_i \geq 1$. Do you see how it works?
Try expanding the partition generating function in the same way-- it's basically the same idea.
|
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|
Proof that there are infinitely many primes of the form $4m+3$ I am reading a proof of there are infinitely many primes of the form $4m+3$, but have trouble understanding it. The proof goes like this:
Assume there are finitely many primes, and take $p_k$ to be the largest prime of the form $4m+3$.
Let $N_k = 2^2 \cdot 3 \cdot 5 \cdots p_k - 1$, where $p_1=2, p_2=3, p_3=5, \dots$ denotes the sequence of all primes.
We find that $N_k$ is congruent to $ 3 \pmod {4}$, so it must have a prime factor of the form $4m+3$, and this prime factor is larger than $p_k$ — contradiction.
My questions are:
*
*Why is $N_k$ congruent to $3 \pmod{4}$?
*Why must $N_k$ have a prime factor of the form $4m+3$ if it's congruent to $3 \pmod{4}$?
It seems that those should be obvious, but I don't see it. Any help would be appreciated!
|
For the first question, if $N_k=2^2\cdot 3\cdot 5\cdots p_k-1$, then
$$
N_k-3=2^2\cdot 3\cdot 5\cdots p_k-1-3=4(3\cdot 5\cdots p_k-1)
$$
which implies $4|N_k-3$, that is, $N_k\equiv 3\pmod{4}$.
For the second question, suppose instead that $N_k$ has no prime factors of form $4m+3$. Then all its prime factors must be of the form $4m+1$, as they can not be of the form $4m$ or $4m+2$, (except possibly for $2$). But notice $$(4m+1)(4k+1)=16mk+4m+4k+1=4(4mk+m+k)+1$$ so by an inductive argument, you see that a product of primes of the form $4m+1$ again has the form $4m+1$. In terms of congruences, if $p_i\equiv 1\pmod{4}$ and $p_j\equiv 1\pmod{4}$, then $p_ip_j\equiv 1^2\equiv 1\pmod{4}$. This would contradict the fact that $N_k$ has form $4m+3$, since $N_k\equiv 3\pmod{4}$.
|
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|
How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? How can I evaluate
$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$?
I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.
In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
|
Note that $\int \{1 + 2x + 3x^2 + \cdots\} \, dx = x + x^2 + x^3 + \cdots + \text{const}$, i.e., a geometric series, which converges to $x/(1 - x)$ if $|x| < 1$. Therefore,
$$\frac{d}{dx} \left(\frac{x}{1 - x}\right) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1}{(1 - x)^2},$$
that is,
$$1 + 2x + 3x^2 + \cdots = \frac{1}{(1 - x)^2}.$$
Another proof: Let $S = 1 + 2x + 3x^2 + \cdots$ with $|x| < 1$. Then
$$xS = x + 2x^2 + 3x^3 + \cdots$$
so
$$S - xS = (1- x)S = 1 + x + x^2 + \cdots = \frac{1}{1- x}.$$
Therefore: $S = (1 - x)^{-2}$.
|
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|
Prove that $\lfloor \sqrt{p} \rfloor + \lfloor \sqrt{2p} \rfloor +...+ \lfloor \sqrt{\frac{p-1}{4}p} \rfloor = \frac{p^2 - 1}{12}$ Problem
Prove that $\lfloor \sqrt{p} \rfloor + \lfloor \sqrt{2p} \rfloor +...+ \lfloor \sqrt{\frac{p-1}{4}p} \rfloor = \dfrac{p^2 - 1}{12}$ where $p$ prime such that $p \equiv 1 \pmod{4}$.
I really have no idea how to start :(! The square root part really messed me up. Can anyone give me a hint?
Thank you
|
The sum $S(p)$ counts the lattice points with positive coordinates under $y=\sqrt{px}$ from $x=1$ to $x=\frac{p-1}{4}$. Instead of counting the points below the parabola, we can count the lattice points on the parabola and above the parabola, and subtract these from the total number of lattice points in a box. Stop here if you only want a hint.
Since $p$ is prime, there are no lattice points on that parabola (with that range of $x$ values).
The total number of lattice points in the box $1 \le x \le \frac{p-1}4, 1\le y \le \frac {p-1}2$ is $\frac{(p-1)^2}8$.
The lattice points above the parabola are to the left of the parabola. These are counted by
$T(p)= \lfloor 1^2/p \rfloor + \lfloor 2^2/p \rfloor + ... + \lfloor (\frac{p-1}2)^2/p \rfloor$.
$T(p)+S(p) = \frac{(p-1)^2}8$, so $S = \frac {p^2-1}{12}$ is equivalent to $T(p) = \frac{(p-1)(p-5)}{24}$.
Consider $T(p)$ without the floor function. This sum is elementary:
$$\sum_{i=1}^{(p-1)/2} \frac{i^2}p = \frac 1p \sum_{i=1}^{(p-1)/2} i^2 = \frac 1p \frac 16 (\frac{p-1}2)(\frac {p-1}2 + 1)(2\frac{p-1}2 +1) = (p^2-1)/24.$$
What is the difference between these? Abusing the mod notation, $\frac{i^2}p - \lfloor \frac{i^2}p \rfloor = 1/p \times (i^2 \mod p)$. So,
$$(p^2-1)/24 - T(p) = \sum_{i=1}^{(p-1)/2} \frac{i^2}p - \lfloor \frac{i^2}p \rfloor = \sum_{i=1}^{(p-1)/2} \frac 1p \times (i^2 \mod p) = \frac 1p \sum_{i=1}^{(p-1)/2} (i^2 \mod p).$$
Since $i^2 = (-i)^2$, this last sum is over the nonzero quadratic residues. Since $p$ is $1 \mod 4$, $-1$ is a quadratic residue, so if $a$ is a nonzero quadratic residue, then so is $p-a$. Thus, the nonzero quadratic residues have average value $p/2$ and the sum is $\frac{(p-1)}2 \frac p2$.
$$(p^2-1)/24 - T(p) = \frac 1p \frac{(p-1)}2 \frac p2 = \frac{p-1}4$$
$$T(p) = \frac{(p-1)(p-5)}{24}.$$
That was what we needed to show.
|
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|
Proving $2 ( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} )$ is a root of$ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ How can one show that the number
$2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right)$
is a root of the equation
$\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$?
|
This solution is basically the stupidest possible way to solve the problem. I'm just posting it in case the direct calculation here sheds any light on the problem, it probably doesn't though. Especially frustrating is this does not seem to show any connection between the discriminant $19^2$ and the cyclotomic field involved.
Define $e(q) = e^{2 i \pi q}$ so that $2 \cos(2 \pi q) = e(q) + e(-q)$ and $x = 2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right) = e(2/19) + e(3/19) + e(5/19) + \text{mi}$. "mi" means mirror image, I'm going to omit these terms because there are a lot of terms involved. Since $x$ is real its square and cube will be too, that means (by algebraic independence of roots of unity) we must have the mirror image there too.. this save a bit of work.
Squaring it is a simple process it just means multiplying out the bracket and using the identity $e(q)e(r) = e(q+r)$. We get
x^2 = 6 + 2 e(1/19) + 2 e(2/19) + 2 e(3/19)
+ e(4/19) + 2 e(5/19) + e(6/19)
+ 2 e(7/19) + 2 e(8/19) + e(9/19) + mi
Cubing is slightly more work but in the end we get
x^3 = 12 + 9 e(1/19) + 15 e(2/19) + 15 e(3/19)
+ 10 e(4/19) + 15 e(5/19) + 10 e(6/19)
+ 9 e(7/19) + 9 e(8/19) + 10 e(9/19) + mi
Note that I have also reduced terms, e.g. $e(11/19) = e((11-19)/19) = e(-8)$. There is another reduction we can make too, since the terms contain the complete sum $e(1/19)+e(2/19)+...+e(9/19) + \text{mi} = -1$ (which can be seen consider the points evenly spaced around the unit circle) we apply this reduction to get:
x^2 = 4 - e(4/19) - e(6/19) - e(9/19) + mi
x^3 = 3 + 6 e(2/19) + 6 e(3/19) + e(4/19)
+ 6 e(5/19) + e(6/19) + e(9/19) + mi
It is now very easy to see that this satisfies the polynomial $x^3+x^2-6x-7$ and in fact you could have figured out that polynomial from this method.
What I would really like to do is start with the polynomial and produce an exponential series solving it..
|
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Question regarding the product of quadratic residue modulo $p$, $p$ is prime Problem
Prove that the product of the quadratic residue modulo $p$ is congruent to $1$ modulo p if $p \equiv -1 \pmod{4}$ and is congruent to $-1$ modulo $p$ if $p \equiv 1 \pmod{4}$.
First of all, my question is what do they mean by product? Is it $a \cdot a$ or it could be $a^{k}$, where $k$ is an integer?
On the other hand, when attempting to solve it, I realized it could lead to a contradiction.
Proof
By Euler's criteria, we have $$\bigg( \dfrac{a}{p} \bigg) \equiv a^{\frac{p-1}{2}} \pmod{p}$$
Since $a$ is quadratic residue modulo $p$ as suppose, hence,
$$\bigg( \dfrac{a}{p} \bigg) \equiv a^{\frac{p-1}{2}} \pmod{p} = 1$$
which implies $a^{\frac{p-1}{2}} \equiv 1 \pmod{p}$, then
If $p \equiv -1 \pmod{4} \Leftrightarrow p \equiv 3 \pmod{4} \Leftrightarrow p = 4k + 3$, for some integers $k$. And If $p \equiv 1 \pmod{4} \Leftrightarrow p = 4q + 1$, for some integers $q$. So,
If $p = 4k + 3, \Rightarrow \dfrac{p-1}{2} = \dfrac{4k + 3 - 1}{2} = 2k + 1$, then $a^{2k + 1} \equiv 1 \pmod{p}$.
If $p = 4q + 1, \Rightarrow \dfrac{p-1}{2} = \dfrac{4q + 1 - 1}{2} = 2q$, then $a^{2q} \equiv 1 \pmod{p}$.
From here, I saw that the product of $a$, $a$ raises to either an odd power or an even power is $1$. So how the product could be $-1$ as the in the proof requirement? And what can we deduce from here? Any idea?
Thanks,
|
They mean product as in multiplication, e.g. the product of the numbers $a$, $b$, and $c$ is $abc$.
Hint: If $x$ is a quadratic residue modulo $p$, then so is $x^{-1}$ (recall that $x^{-1}$ is the $a$ such that $ax=1\bmod p$). Pair up each quadratic residue with its inverse in the product, and cancel each pair; what's different about the case when $\frac{p-1}{2}$ (the number of quadratic residues mod $p$) is even, vs. when it is odd? In particular, think about $-1$.
|
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|
Area of a circle externally tangent to three mutually tangent circles Given three identical circles, with three points of intersection. The line between two of these intersecting points is $3$ feet. They are inside a $4$th circle. All circles are tangent to each other.
What is the area of the $4$th circle? I don't understand if any of the Descartes Theorem or Three Tangent Circles applies and how.
Or perhaps more simply, if the "outer Soddy circle" equation will yield the needed answer. works
|
Given that the radii of the inner circles are $3$ feet, Descartes' Theorem (aka Soddy's Theorem) says that
$$
\left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4}\right)^2=2\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r_4^2}\right)
$$
Plugging in $r_1=r_2=r_3=3$, we get
$$
\left(1+\frac{1}{r_4}\right)^2=2\left(\frac{1}{3}+\frac{1}{r_4^2}\right)
$$
which turns into
$$
r_4^2+6r_4-3=0
$$
which gives $r_4=-3\pm2\sqrt{3}$.
$r_4=-3+2\sqrt{3}$ says that the radius of the small inner circle tangent to the three given circles is $-3+2\sqrt{3}$.
$r_4=-3-2\sqrt{3}$ says that the radius of the outside circle is $3+2\sqrt{3}$.
Thus, the area of the outside circle is $\pi(21+12\sqrt{3})$.
|
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What kind of series is this, and how do I sum it? $\displaystyle\sum{\frac1{a_n}}$, e.g.
$$\frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \dots + \frac 1 n$$ or
$$\frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 8 + \dots + \frac 1 {2n}$$
|
A nice approximation for $n \to \infty$ can be derived from
$$\eqalign{
& \log \left( {1 + x} \right) = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{{{x^{k + 1}}}}{{k + 1}}} = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - + \cdots \cr
& \log \left( {1 + \frac{1}{n}} \right) = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{n^{k + 1}}}}\frac{1}{{k + 1}}} = \frac{1}{n} - \frac{1}{{2{n^2}}} + \frac{1}{{3{n^3}}} - + \cdots \cr} $$
$$\log \left( {n + 1} \right) - \log n = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{n^{k + 1}}}}\frac{1}{{k + 1}}} = \frac{1}{n} - \frac{1}{{2{n^2}}} + \frac{1}{{3{n^3}}} - + \cdots $$
If we sum for $n=1,2,3,\dots,m $ we have
$$\log \left( {m + 1} \right) = \sum\limits_{n = 1}^m {\frac{1}{n}} - \frac{1}{2}\sum\limits_{n = 1}^m {\frac{1}{{{n^2}}}} + \frac{1}{3}\sum\limits_{n = 1}^m {\frac{1}{{{n^3}}}} - + \cdots $$ or
$$\sum\limits_{n = 1}^m {\frac{1}{n}} - \log \left( {m + 1} \right) = \frac{1}{2}\sum\limits_{n = 1}^m {\frac{1}{{{n^2}}}} - \frac{1}{3}\sum\limits_{n = 1}^m {\frac{1}{{{n^3}}}} + - \cdots $$
If we let $m\to \infty$ , we have that
$$\mathop {\lim }\limits_{m \to \infty } \left[ {\sum\limits_{n = 1}^m {\frac{1}{n}} - \log \left( {m + 1} \right)} \right] = \frac{1}{2}\zeta \left( 2 \right) - \frac{1}{3}\zeta \left( 3 \right) + \frac{1}{4}\zeta \left( 4 \right) - + = \sum\limits_{n = 2}^\infty {{{\left( { - 1} \right)}^n}\frac{{\zeta \left( n \right)}}{n}} $$
where $$\zeta \left( n \right) = \sum\limits_{k = 1}^\infty {\frac{1}{{{k^n}}}} $$
The value of the right hand side is know as Euler's constant, $\gamma \approx 0.577$, and this result tells us that for large values of $m$, we can approximate the $m$th harmonic number by
$$\sum\limits_{n = 1}^m {\frac{1}{n}} \sim \log \left( {m + 1} \right) + \gamma $$
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Computing the integral of $\log(\sin x)$ How to compute the following integral?
$$\int\log(\sin x)\,dx$$
Motivation: Since $\log(\sin x)'=\cot x$, the antiderivative $\int\log(\sin x)\,dx$ has the nice property $F''(x)=\cot x$. Can we find $F$ explicitly? Failing that, can we find the definite integral over one of intervals where $\log (\sin x)$ is defined?
|
There was a duplicate posted a while ago. Since I think my answer might be of some interest, here it goes:
By substituting $\sin{x}=t$, we can write it as:
\begin{align*}
\int_{0}^{\pi/2} \, \log\sin{x}\, dx &= \int_{0}^{1} \, \frac{\log{t}}{\sqrt{1-t^2}}\, dt \tag{1}
\end{align*}
Now, consider:
\begin{align*}
I(a) &= \int_{0}^{1} \, \frac{t^a}{(1-t^2)^{1/2}}\, dt \\
&= \mathrm{B}\left(\frac{a+1}{2},\; \frac{1}{2}\right) \\
\frac{\partial }{\partial a}I(a) &= \frac{1}{4}\left(\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right)\mathrm{B}\left(\frac{a+1}{2},\; \frac{1}{2}\right) \\
\implies I'(0) &= \frac{1}{4}\left(\psi\left(\frac{1}{2}\right)-\psi\left(1\right)\right)\mathrm{B}\left(\frac{1}{2},\; \frac{1}{2}\right) \tag{2}
\end{align*}
Putting the values of digamma and beta functions.
\begin{align*}
\psi\left(\frac{1}{2}\right) &= -2\log{2}-\gamma \\
\psi\left(1\right) &= -\gamma \\
\mathrm{B}\left(\frac{1}{2}, \frac{1}{2}\right) &= \pi
\end{align*}
Hence, from $(1)$ and $(2)$,
\begin{align*}
\boxed{\displaystyle \int_{0}^{\pi/2} \, \log\sin{x}\, dx = -\frac{\pi}{2}\log{2}}
\end{align*}
Using a CAS, we can derive for higher powers of $\ln\sin{x}$, e.g.
\begin{align*}
\int_{0}^{\pi/2} \, \left(\log\sin{x}\right)^2\, dx &= \frac{1}{24} \, \pi^{3} + \frac{1}{2} \, \pi \log\left(2\right)^{2} \\
\int_{0}^{\pi/2} \, \left(\log\sin{x}\right)^3\, dx &= -\frac{1}{8} \, \pi^{3} \log\left(2\right) - \frac{1}{2} \, \pi \log\left(2\right)^{3} - \frac{3}{4} \, \pi \zeta(3)\\
\int_{0}^{\pi/2} \, \left(\log\sin{x}\right)^4\, dx &= \frac{19}{480} \, \pi^{5} + \frac{1}{4} \, \pi^{3} \log\left(2\right)^{2} + \frac{1}{2} \, \pi \log\left(2\right)^{4} + 3 \, \pi \log\left(2\right) \zeta(3)
\end{align*}
We can also observe another interesting thing, for small values of $n$
\begin{align*}
\displaystyle \int_{0}^{\pi/2} \, \left(\log\sin{x}\right)^n\, dx \approx \displaystyle (-1)^n\, n!
\end{align*}
|
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|
Solve recursion $a_{n}=ba_{n-1}+cd^{n-1}$ Let $b,c,d\in\mathbb{R}$ be constants with $b\neq d$. Let
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}
\end{eqnarray}$$
be a sequence for $n \geq 1$ with $a_{0}=0$. I want to find a closed formula for this recursion. (I only know the german term geschlossene Formel and translated it that way I felt it could be right. So if I got that wrong, please correct me)
First I wrote down some of the chains and I got
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}\\
&=& b\left(ba_{n-2}+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(ba_{n-3}+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(b\left(ba_{n-4}+cd^{n-4}\right)+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& \dots\\
&=& \sum_{k=0}^{n}b^{k}cd^{n-k-1}\\
&=& \sum_{k=0}^{n}b^{k}cd^{n-\left(k+1\right)}
\end{eqnarray}$$
So I catched the structure in a serie. Now I am asking myself how to proceed. I took the liberty to have a little peek at what WolframAlpha wood say to this serie. I hoped for inspiration and I got
$$\sum_{k=0}^{n-1}b^{k} c d^{n-(k+1)} = (c (b^n-d^n))/(b-d)$$
How did this came to be? And more important: Is my approach useful?
Thank you in advance for any advice!
Edit: My final Solution (recalculated)
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}\\
&=& b\left(ba_{n-2}+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(ba_{n-3}+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(b\left(ba_{n-4}+cd^{n-4}\right)+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b^{4}a_{n-4}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+bcd^{n-2}+cd^{n-1}\\
&=& b^{5}a_{n-5}+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+bcd^{n-2}+cd^{n-1}\\
&=& b^{n}a_{0}+b^{n-1}c+\dots+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+cbd^{n-2}+cd^{n-1}\\
&=& \dots\\
&=& 0+b^{n-1}c+\dots+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+cbd^{n-2}+cd^{n-1}\\
&=& \sum_{k=0}^{n-1}b^{k}cd^{n-1-k}\\
&=& cd^{n-1}\sum_{k=0}^{n-1}b^{k}d^{-k}\\
&=& cd^{n-1}\sum_{k=0}^{n-1}\left(\frac{b}{d}\right)^{k}\\
&=& cd^{n-1}\frac{1-\left(\frac{b}{d}\right)^{n}}{1-\left(\frac{b}{d}\right)}\\
&=& cd^{n-1}\frac{1-\frac{b^{n}}{d^{n}}}{1-\frac{b}{d}}\\
&=& cd^{n-1}\frac{\frac{d^{n}-b^{n}}{d^{n}}}{\frac{d-b}{d}}\\
&=& cd^{n-1}\frac{d^{n}-b^{n}}{d^{n}}\cdot\frac{d}{d-b}\\
&=& \frac{c\left(d^{n}-b^{n}\right)}{d-b}
\end{eqnarray}$$
|
Argh... use Wilf's techniques from "generatingfunctionology". Start defining:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
Also write:
$$
a_{n + 1} = b a_n + c d^n
$$
Multiply by $z^n$, add over $n \ge 0$:
$$
\frac{A(z) - a_0}{z} = b A(z) + c \frac{1}{1 - d z}
$$
Solve for $A(z)$, express as partial fractions. The resulting terms are of the forms:
$$
(1 - \alpha z)^{-m}
= \sum_{n \ge 0} \binom{-m}{n} (- \alpha)^n z^n
= \sum_{n \ge 0} \binom{n + m - 1}{m - 1} \alpha^n z^n
$$
Note that:
$$
\binom{n + m - 1}{m - 1}
= \frac{(n + m - 1) (n + m - 2) \ldots (n + 1)}{(m - 1)!}
$$
This is a polynomial in $n$ of degree $m - 1$.
$$
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Polynomial equations with finite field arithmetic there are given 3 equations (they are connected with cyclic codes):
$$s(x)=v(x)+q(x)g(x)$$
$$g(x)h(x)=x^7+1$$
$$s(x)=v(x)h(x)\bmod(x^7+1)$$
I have following data (for $GF(8)$ with generator polynomial $p(x)=x^3+x+1$):
$$g(x)=x^4+\alpha^3x^3+x^2+\alpha x+\alpha^3$$
$$q(x)=\alpha^6 x^2 + \alpha^2 x + \alpha$$
$$v(x)=\alpha^6 x^6 + \alpha^3 x^3 + \alpha^6 x^2 + x + \alpha^4$$
So I use first equation and:
$$s(x)=v(x)+q(x)g(x)$$$$=(\alpha^6 x^6 + \alpha^3 x^3 + \alpha^6 x^2 + x + \alpha^4)+(\alpha^6 x^2 + \alpha^2 x + \alpha)(x^4+\alpha^3x^3+x^2+\alpha x+\alpha^3)$$
$$=(\alpha^6 x^6 + \alpha^3 x^3 + \alpha^6 x^2 + x + \alpha^4)+(\alpha^6 x^6 + \alpha^3 x^3 + \alpha^6 x^2 + \alpha^3x + \alpha^4)=\alpha x$$
And it's all right until now - result is correct. Now let's see second equation:
$$h(x)=\frac{x^7+1}{g(x)}=x^3+\alpha^3x^2+\alpha^2x+\alpha^4$$
And using third equation:
$$s(x)=v(x)h(x)\bmod(x^7+1)$$
$$s(x)=(\alpha^6 x^6 + \alpha^3 x^3 + \alpha^6 x^2 + x + \alpha^4)(x^3+\alpha^3x^2+\alpha^2x+\alpha^4)\bmod(x^7+1)$$
$$s(x)=\alpha^6 x^9 + \alpha^2 x^8 + \alpha x^7 + \alpha x^4 + \alpha^4 x^3 + \alpha^4 x^2 + \alpha^3 x + \alpha \bmod (x^7+1)$$
$$\frac{\alpha^6 x^9 + \alpha^2 x^8 + \alpha x^7 + \alpha x^4 + \alpha^4 x^3 + \alpha^4 x^2 + \alpha^3 x + \alpha}{x^7+1}$$$$=(\alpha^6 x^2 + \alpha^2 x + \alpha) + (\alpha x^4 + \alpha^4 x^3 + \alpha^3 x^2 + \alpha^5 x)$$
so
$$s(x)=\alpha x^4 + \alpha^4 x^3 + \alpha^3 x^2 + \alpha^5 x$$
Problem is that the result of $s(x)$ calculated with first and third equations is not the same. It should be the same for $s(x)$ of degree less than degree of $g(x)$.
|
Your third equation should be $\rm\ s(x)\ h(x)\equiv v(x)\ h(x)\ \ (mod\ x^7 + 1)\:,\:$ which agrees with the calculations. It arises by multiplying the first equation by $\rm\:h(x)\:,\:$ then using the second equation.
|
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|
Understanding a simplification in a theorem I'm trying to understand a theorem in a paper on page 14/24. We are given that
$$Z = (nq-1) \log \left(\frac{M+nq-1}{nq-1} \right) + M \log \left(\frac{M+nq-1}{M} \right) + \frac{1}{2} \log \left( \frac{M+nq-1}{(nq-1)M} \right) + O(1) .$$
Since $e^a > \left(1+\frac{a}{b} \right)^b$ for all $a > 0$ and positive integers $b$, the second term equals
$$\log \left(1 + \frac{nq-1}{M} \right)^M < \log e^{nq-1}$$
for all positive M and $nq-1 > 0$. This much I'm following, but then it goes on saying:
Since $0 < q < n$ and $n \leq M \leq qn^2$,
$$\dfrac{1}{2(nq-1)} \log \left( \frac{M+nq-1}{(nq-1)M} \right)\to 0 \quad \text{for } n \to \infty .$$
Where does the factor $\frac{1}{2(nq-1)}$ come from? This might even be trivial, but somehow I'm totally unable to see it.
|
OK, thanks to the paper, I figured it out. What they do is rewrite
$$Z = (nq-1) \log \dfrac{M+nq-1}{nq-1} + M \log \dfrac{M+nq-1}{M} + \dfrac{1}{2} \log \dfrac{M+nq-1}{(nq-1)M} + O(1)$$
with their formula for the second term
$$M\log\left(1 + \dfrac{nq-1}{M}\right) < (nq-1)\log e$$
which gives
$$Z < (nq-1) \log \dfrac{M+nq-1}{nq-1} + (nq-1)\log e + \dfrac{1}{2} \log \dfrac{M+nq-1}{(nq-1)M} + O(1)$$
Of course, it has become an inequality, particularly an upper bound for $Z$. Now separate out the factor $(np-1)$ and you'll get
$$Z < (nq-1) \left( \log \dfrac{M+nq-1}{nq-1} + \log e + \dfrac{1}{2(nq-1)} \log \dfrac{M+nq-1}{(nq-1)M} + O\left(\frac{1}{nq-1}\right) \right)$$
Now, see the third term is exactly what they are computing the limit of. Since that goes to zero, and the $O\left(\frac{1}{nq-1}\right)$ term does as well, you get the end result
$$Z < (nq-1)\left( \log \dfrac{M+nq-1}{nq-1} + \log e \right) \; .$$
|
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|
Find all prime $p$ such that $x^2=-1$ has a solution in $\mathbb{Z}/p\mathbb{Z}$ I have found by a numerical experiment that first such primes are:
$2,5,13,17,29,37,41$. But I cannot work out the general formula for it.
Please share any your ideas on the subject.
|
The following argument depends on knowing (or separately proving) Wilson's Theorem.
Theorem Let $p$ be prime. Then $(p-1)! \equiv -1 \pmod{p}$.
Now we use Wilson's Theorem to prove the result. For the sake of concreteness, I will use $p=17$, but the same idea exactly works for all primes congruent to $1$ modulo $4$. All congruences will be modulo $17$, so $17$ will mostly not be mentioned explicitly.
Note that
$$16!= (1)(2)(3)(4)(5)(6)(7)(8)(16)(15)(14)(13)(12)(11)(10)(9).$$
(Didn't really do anything.)
Note that $16\equiv -1$, $15\equiv -2$, $14\equiv -3$, and so on until $9 \equiv -8$.
It follows that
$$(15)(14)(13)(12)(11)(10)(9)\equiv (-1)(-2)(-3)(-4)(-5)(-6)(-7)(-8).$$
But the right-hand side is congruent to $(-1)^8(1)(2)(3)(4)(5)(6)(7)(8)$, and of course $(-1)^8=1$.
It follows that $16! \equiv (8!)^2 \pmod{17}$. But by Wilson's Theorem, $16!\equiv -1 \pmod{17}$, and therefore
$$(8!)^2 \equiv -1 \pmod{17}.$$
We have found an explicit expression for an $x$ such that $x^2\equiv - \pmod{17}$.
If $p \equiv 3 \pmod{4}$, the argument breaks down, because we end up with an odd number of minus signs. (Anyway, the result does not hold when $p\equiv 3\pmod 4$.)
In general, if $p\equiv 1 \pmod 4$, and $q=(p-1)/2$, then
$$(q!)^2 \equiv -1 \pmod{p}.$$
Other primes: It remains to show that the congruence is not solvable if $p \equiv 3\pmod{4}$. So suppose to the contrary that $b^2\equiv -1\pmod{p}$. Look at the numbers $b, 2b, 3b, \dots, (p-1)b$. It is easy to verify they are pairwise incongruent modulo $p$, so their product is congruent to $(p-1)!$ modulo $p$. Thus
$$b^{p-1}(p-1)! \equiv (p-1)! \pmod p,$$
giving $b^{p-1}\equiv 1 \pmod p$. Let $q=(p-1)/2$. Then $b^{p-1} \equiv (-1)^q \pmod p$. But if $p \equiv 3 \pmod 4$, then $q$ is odd, and we obtain the absurd $-1 \equiv 1 \pmod{p}$.
|
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|
Find $\sin \theta$ and $\cos \theta$ given $\tan 2\theta$ Can you guys help with verifying my work for this problem. My answers don't match the given answers.
Given $\tan 2\theta = -\dfrac{-24}{7}$, where $\theta$ is an acute angle, find $\sin \theta$ and $\cos \theta$
I used the identity, $\tan 2\theta = \dfrac{2\tan \theta}{1 - tan^2 \theta}$ to try and get an equation in $\tan \theta$.
$$
\begin{align}
-\dfrac{24}{7} &= \dfrac{2\tan \theta}{1 - \tan^2 \theta} \\
-24 + 24\tan^2 \theta &= 14 \tan \theta \\
24tan^2 \theta - 14\tan \theta - 24 &= 0 \\
12tan^2 \theta - 7\tan \theta - 12 &= 0 \\
\end{align}
$$
Solving this quadratic I got,
$$ \tan \theta = \dfrac{3}{2} \text{ or } \tan \theta = -\dfrac{3}{4}$$
$$\therefore \sin \theta = \pm \dfrac{3}{\sqrt{13}} \text{ and } \cos \theta = \pm \dfrac{2}{\sqrt{13}}$$
Or,
$$\therefore \sin \theta = \pm \dfrac{3}{5} \text{ and } \cos \theta = \mp \dfrac{4}{5}$$
The given answer is,
$$\sin \theta = \dfrac{4}{5} \text{ and } \cos \theta = \dfrac{3}{5}$$
I thought I needed to discard the negative solution assuming $\theta$ is acute. But they haven't indicated a quadrant. Do I assume the quadrant is I only? What am i missing? Thanks again for your help.
|
At Chandru's request:
*
*The quadratic $12z^2-7z-12$ factors as $(3z-4)(4z+3)$ so we should get $\tan\,\theta=4/3$ and $\tan\,\theta=-3/4$.
*"Acute angle" means "angle between 0 and $\pi/2$" means 1st quadrant.
|
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|
Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end.
Prove that,
$$
\dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2}
$$
|
Now that OP has understood how to prove this, here is a geometric proof for certain angles, just for fun :-)
Consider the figure:
$\displaystyle \triangle ABC$ is a right angled triangle with the right angle being at $\displaystyle C$.
$\displaystyle \angle{CAB} = A$ and $\displaystyle AB = 1$ and thus $\displaystyle BC = \sin A$ and $\displaystyle AC = \cos A$.
Now $\displaystyle AO$ is the angular bisector of $\displaystyle \angle{CAB}$. We select $\displaystyle O$ so that $\displaystyle O$ is the in-center (point of intersection of angular bisectors of a triangle). Let the in-radius $\displaystyle OD$ be $\displaystyle r$.
Now $\displaystyle BE = BF$ and $\displaystyle AE = AD$ and adding gives us $\displaystyle BF + AD = AB = 1$
Now $\displaystyle BF = BC - FC =BC - OD$ (as $\displaystyle ODCF$ is a square).
Thus $\displaystyle BF = \sin A - r$. Similarly $\displaystyle AD = \cos A - r$.
Thus $\displaystyle \sin A + \cos A - 2r = 1$.
Using $\displaystyle \triangle ADO$, $\displaystyle \tan \frac{A}{2} = \frac{OD}{AD} = \frac{r}{\cos A - r}$
Since $\displaystyle 2r = \sin A + \cos A -1 $ we get
$$\tan \frac{A}{2} = \frac{2r}{2\cos A - 2r} = \frac{\sin A + \cos A - 1}{\cos A - \sin A + 1}$$
It is easy to verify that
$$ \frac{\sin A + \cos A - 1}{\cos A - \sin A + 1} = \frac{\sin A - \cos A + 1}{\cos A + \sin A + 1}$$
Incidently, the fact that in a right triangle (hypotenuse $c$), the in-radius is given by $ a + b -c = 2r$ and also by $r = \frac{\triangle}{s}$ ($\triangle$ is the area, $s$ is the semi-perimeter) can be used to prove the pythagoras theorem ($a^2 + b^2 = c^2$)!
|
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|
Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}
|
Looking at the equation, it looks like it contains terms of the form
$$ \sqrt{\frac{| |x| - 1 |}{|x| - 1}} $$
which evaluates to
$$\begin{cases} 1 & |x| > 1\\ i & |x| < 1\end{cases} $$
Since any non-zero real number $y$ cannot be equal to a purely imaginary non-zero number, the presence of that term is a way of writing a piece-wise defined function as a single expression. My guess is that if you try to plot this in $\mathbb{C}^2$ instead of $\mathbb{R}^2$ you will get all kinds of awful.
|
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|
Can $A^2+B^2+C^2-2AB-2AC-2BC$ be a perfect square? I have come upon the trivariate polynomial $A^2+B^2+C^2-2AB-2AC-2BC$ and want to factor it. Because of the symmetry, I am wondering if it can be a perfect square or if there is some other nice factorization.
|
A more symmetric factorization would be:
$$A^2 + B^2 + C^2 - 2AB -2AC - 2BC =$$
$$(A + B + C)^2 - 4AB - 4AC - 4BC =$$
$$(A + B + C - 2 \sqrt{AB + AC + BC}) (A + B + C + 2 \sqrt{AB + AC + BC})$$
if
$$AB + AC + BC \ge 0$$
|
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|
How many real roots are there to $2^x=x^2$? How many real roots are there to
$2^x=x^2$?
|
An obvious solution is $x=2$.
If $2^x = x^2$, then $x\neq 1$ and $x\neq 0$. I'll treat the positive and negative cases separately.
If $x\gt 0$, then we get $x\ln(2) = 2\ln (a)$, or $\frac{x}{\ln x} = \frac{2}{\ln 2}$.
The derivative of $g(x) = \frac{x}{\ln x}$ is $\frac{\ln x - 1}{(\ln x)^2}$.
On $(1,\infty)$, the derivative is positive on $(e,\infty)$ and negative on $(1,e)$, so there is an absolute minimum at $x=e$, where the value is $e$; $\lim\limits_{x\to 1^+} g(x) = \lim\limits_{x\to\infty}g(x) = \infty$; since $\frac{2}{\ln 2}\gt e$, there are two values of $x$ where $g(x) = \frac{2}{\ln 2}$; one is $x=2$, which we had already found, the other is a value greater than $e$ (which as it happens is $4$).
On $(0,1)$, $g(x)$ is always negative, so there are no values where $g(x)=\frac{2}{\ln 2}$.
So for $x\gt 0$, there are two solutions.
For $x\lt 0$, the equation $2^x = x^2$ is equivalent to the equation $\left(\frac{1}{2}\right)^a = a^2$, where $a=-x\gt 0$. This time, the equation is equivalent to $\frac{a}{\ln a} = -\frac{2}{\ln 2}$. There are no solutions for $a\gt 1$, since $g(x)$ is positive there. On $(0,1)$, $g'(x)\lt 0$, so the function is strictly decreasing; we have $\lim\limits_{a\to 0^+}\frac{a}{\ln a} = 0$ and $\lim\limits_{a\to 1^-}\frac{a}{\ln a} = -\infty$, so there is one and only one value of $a$ for which $\frac{a}{\ln a} = -\frac{2}{\ln 2}$. Thus, there is one value of $x\lt 0$ which solves the equation.
In summary, there are three real solutions: one lies in $(-1,0)$, the second is $2$, and the third is $4$.
|
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|
Why is numerator re-defining necessary for proper solution of some partial fraction integrations? In my homework, one problem was the following formula. Using standard partial fraction techniques where you'll see my work, I came up with an almost correct answer in the fact the book solution had a three term integral rather than the two term I used. The only difference between my work and the correct solution was I didn't rewrite the numerator. My two-part question is
*
*Why did excluding the numerator re-defining result in the wrong answer?
*Under what circumstances should I use numerator re-defining?
Given the following formula, integrate using partial fractions:
$$\int\frac{2x^3-4x^2-15x+5}{x^2-2x-8}$$
$$
\begin{align*}
2x^3-4x^2-15x+5=\frac{A}{x-4}+\frac{B}{x+2}\\
2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\
\end{align*}
$$
Let $Ax$=-2
$$
\begin{align*}
2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\
2(-2)^3-4(-2)^2-15(-2)+5=A(-2+2)+B(-2-4)\\
3=-6B\\
B=-\frac{1}{2}
\end{align*}
$$
Let $Bx= 4$
$$
\begin{align*}
2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\
2(4)^3-4(4)^2-15(4)+5=A(4+2)+B(4-4)\\
9=6A\\
A=\frac{3}{2}
\end{align*}
$$
Using $A \text{ and } B$ values, plug them into the integral as
$$
\begin{align*}
\frac{3}{2}\int\frac{1}{x-4}\text{dx}-\frac{1}{2}\int\frac{1}{x+2}\text{dx}\\
=\frac{3}{2}\text{ln}\left |x-4 \right |-\frac{1}{2}\text{ln}\left |x+2\right |+C
\end{align*}
$$
This work matches up with the solution process except the original integral was re-written as
$$\int2x+\frac{x+5}{(x-4)(x+2)}$$
which resulted in the final integral terms
$$\int2x\text{dx}+\frac{3}{2}\int\frac{1}{x-4}\text{dx}-\frac{1}{2}\int\frac{1}{x+2}\text{dx}$$
|
Normally one uses the division algorithm to decompose into integral and proper fractional parts
$$\rm \dfrac{2x^3-4x^2-15x+5}{(x+2)\:(x-4)}\ =\ 2\:x + \dfrac{x+5}{(x+2)\:(x-4)} $$
Then one performs the partial fraction decomposition only on the second fractional part. But you seem to desire to skip the initial division algorithm step and, instead, jump immediately to the partial fraction decomposition, proceeding essentially as follows.
$$\rm \dfrac{2x^3-4x^2-15x+5}{(x+2)\:(x-4)}\ = \ \dfrac{A}{x-4}\ +\ \dfrac{B}{x+2}\ +\ \: C(x) $$
$$\rm \Rightarrow\ \ \ 2\:x^3-4\:x^2-15\:x+5\ = \ A\ (x+2) + B\ (x-4)\ +\ C(x)\:(x+2)\:(x-4) $$
Using the Heaviside cover-up method will still work to find the correct values of $\rm\:A\:$ and $\rm\:B\:$ because the term $\rm\:C(x)\:(x+2)\:(x-4)\:$ plays no role - it vanishes when evaluating the RHS at $\rm\:x = 4\:$ and $\rm\:x = -2\:$ while solving for $\rm\:A\:$ and $\rm\:B\:.\:$ Similarly for the LHS which is $\rm\:x+5 + 2\:x\:(x+2)\:(x-4)\:.\:$ Therefore you'll get the same equations for $\rm\:A\:$ and $\rm\:B\:$ whether you use this method or the standard method. Thus, as you do, one can safely set $\rm\:C(x) = 0\:$ from the start. But that yields more work than the standard way, since then you have to evaluate a cubic (vs. $\rm\:x+5\:$) on the LHS. Finally, to obtain the complete answer, you still have to calculate the "integral" part $\rm\:C(x) = 2\:x\:$ and add it to the partial fraction decomposition. The failure to do so explains why your answer is erroneous. You could get the integral part by subtracting the fractional part from the original expression, but that too would be more work than using the division algorithm from the start.
|
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|
Whether this matrix is positive definite Let $A$ be a nonsingular real square matrix. Is it true that the matrix
$$\frac{1}{2}(A+A')-2(A^{-1}+(A^{-1})')^{-1}$$
is positive semidefinite?
Here, $A'$ denotes the transpose of $A$.
Edited
Let $A,B$ be positive definite matrices of the same size, is it true that $\frac{1}{2}(AB+BA)−2((AB)^{−1}+(BA)^{−1})^{−1}$ is positive semidefinite?
|
Added: This answer the updated question.
Your updated conjecture is also false. Consider:
$$
A = \left(
\begin{array}{cc}
10 & -5 \\
9 & 3 \\
\end{array}
\right) \qquad\qquad\qquad
B = \left(
\begin{array}{cc}
1 & -6 \\
8 & 8 \\
\end{array}
\right)
$$
Then
$$
\frac{1}{2}(A.B + B.A) - 2 ( (A.B)^{-1} + (B.A)^{-1} )^{-1} = \frac{1}{7474} \left(
\begin{array}{cc}
-98938 & -164451 \\
247345 & -61502 \\
\end{array}
\right)
$$
The matrix above is negative-definite.
Added: This portion answers the first variant of the problem:
No, it is not true. Counterexample:
$$
A = \left(
\begin{array}{ccc}
3 & 3 & -4 \\
1 & -2 & 0 \\
-1 & -2 & -3 \\
\end{array}
\right)
$$
Then
$$
\frac{1}{2}( A + A^t) - 2 ( A^{-1} + (A^{-1})^t)^{-1} = \left(
\begin{array}{ccc}
\frac{127}{198} & \frac{4}{33} & -\frac{163}{198} \\
\frac{4}{33} & -\frac{2}{11} & \frac{5}{33} \\
-\frac{163}{198} & \frac{5}{33} & \frac{59}{99} \\
\end{array}
\right)
$$
It eigenvalues are, approximately, $1.44$, $-0.387$ and zero.
Added: In order to create such a counterexample I have used Mathematica:
f[a_] := 1/2 (a + Transpose[a]) -
2 Inverse[# + Transpose[#] &[Inverse[a]]]
Now this generates random integer-valued matrices until a non-degenerate one is generated with the combination in question having eigenvalues of opposite signs:
While[True,
While[Det[a = RandomInteger[{-4, 4}, {2, 2}]] == 0, Null];
If[ Intersection[Sign[Eigenvalues[f[a]]], {-1, 1}] == {-1, 1},
Break[]]
];
Here is a screenshot:
|
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|
How to find the minimum value of $\frac{x}{2x+3y}+\frac{y}{y+z}+\frac{z}{z+x}$? Let $x,y,z\in [1,4]$ such that $x \geq y$ and $x \geq z$.
Find the minimum value of this expression:
$$
P=\frac{x}{2x+3y}+\frac{y}{y+z}+\frac{z}{z+x}
$$
|
As has been mentioned in comments, we want to find the minimum value of
$$
P=\frac{1}{2+3u}+\frac{u}{u+v}+\frac{v}{v+1}
$$
for $u,v\in[\frac{1}{4},1]$. Take partials of $P$ with respect to $u$ and $v$:
$$
\begin{align}
\frac{\partial P}{\partial u}&=-\frac{3}{(2+3u)^2}+\frac{v}{(u+v)^2}\\
\frac{\partial P}{\partial v}&=-\frac{u}{(u+v)^2}+\frac{1}{(v+1)^2}
\end{align}
$$
To find an interior extremum, we need $\frac{\partial P}{\partial u}=\frac{\partial P}{\partial v}=0$. In that case, we need
$$
0=u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}=-\frac{3u}{(2+3u)^2}+\frac{v}{(v+1)^2}
$$
However, for $u,v\in[\frac{1}{4},1]$, we have
$$
\begin{array}{c}
\frac{12}{121}\le\frac{3u}{(2+3u)^2}\le\frac{3}{25}\\
\frac{4}{25}\le\frac{v}{(v+1)^2}\le\frac{1}{4}
\end{array}
$$
Thus, $u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}\ge\frac{1}{25}$, so there can be no interior extremum.
Because $(u,v)\cdot\nabla P=u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}\ge\frac{1}{25}$ everywhere in $[\frac{1}{4},1]\times[\frac{1}{4},1]$, the minimum must be taken on the left or bottom edge of that square; i.e. $u=\frac{1}{4}$ or $v=\frac{1}{4}$.
$u=\frac{1}{4}$: $\frac{\partial P}{\partial v}=-\frac{4}{(1+4v)^2}+\frac{1}{(v+1)^2}=\frac{12v^2-3}{(1+4v)^2(v+1)^2}$ which vanishes at $v=\frac{1}{2}$.
Because $P(\frac{1}{4},\frac{1}{4})=\frac{117}{110}$ and $P(\frac{1}{4},\frac{1}{2})=\frac{34}{33}$ and $P(\frac{1}{4},1)=\frac{117}{110}$, $P(\frac{1}{4},\frac{1}{2})$ is a local minimum.
$v=\frac{1}{4}$: $\frac{\partial P}{\partial u}=-\frac{3}{(2+3u)^2}+\frac{4}{(4u+1)^2}=\frac{-12u^2+24u+13}{(2+3u)^2(4u+1)^2}$ which does not vanish on $[\frac{1}{4},1]$.
Because $P(\frac{1}{4},\frac{1}{4})=\frac{117}{110}$ and $P(1,\frac{1}{4})=\frac{33}{20}$, $P(\frac{1}{4},\frac{1}{4})$ is a local minimum.
Thus, $P(\frac{1}{4},\frac{1}{2})=\frac{34}{33}$ is the minimum of P for $(u,v)\in[\frac{1}{4},1]\times[\frac{1}{4},1]$.
|
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|
Restricted Integer Compositions Let $c_{k}(N;[a,b])$ denote the number of compositions of $N$ into $k$ parts, where each part is restricted to the interval $[a,b]$, i.e., $N = \sum_{i = 1}^{k} s_{i}$ with $a \leq s_{i} \leq b$. The generating function of $c_{k}(N;[a,b])$ is
\begin{align}
G(c_{k}(N; [a,b]);t) = t^{ka} \left( \frac{1-t^{b-a+1}}{1-t} \right)^{k},
\end{align}
Hence, $c_{k}(N;[a,b]) = [t^{N}] G(c_{k}(N; [a,b]);t)$.
For example, if $N = 8$, $a = 1$, $b = 3$ and $k = 4$, then
\begin{align}
c_{4}(8;[1,3]) = [t^{8}]t^{4} \left( \frac{1-t^{3}}{1-t} \right)^{4} = [t^{8}] (t^{4} + 4 t^{5} +10 t^{6} + 16 t^{7} + 19 t^{8}) = 19.
\end{align}
The nineteen positive compositions of $8$ into $4$ parts no greater than $3$ are
$2 + 2 + 2 + 2$, as well as, $1 + 1 + 3 + 3$, $1 + 3 + 1 + 3$, $1 + 3 + 3 + 1$, $1 + 2 + 2 + 3$, $1 + 2 + 3 + 2$, $1 + 3 + 2 + 2$, $2 + 1 + 2 + 3$, $2 + 1 + 3 + 2$, $2 + 2 + 1 + 3$, $2 + 2 + 3 + 1$, $2 + 3 + 1 + 2$, $2 + 3 + 2 + 1$, $3 + 1 + 2 + 2$, $3 + 2 + 1 + 2$, $3 + 2 + 2 + 1$, $3 + 1 + 1 + 3$, $3 + 3 + 1 + 1$ and $3 + 1 + 3 + 1$.
What about products of generating functions? For example, what combinatorial information is encoded in the following:
\begin{align}
[t^{N}]t^{4} \left( \frac{1-t^{3}}{1-t} \right)^{4} t^{5} \left( \frac{1-t^{2}}{1-t} \right)^{3},
\end{align}
where $N$ is again a positive integer?
More generally, is there a nice combinatorial interpretation of the following coefficient in terms of restricted compositions,
\begin{align}
[t^{N}]\prod_{i = 1}^{m} t^{\alpha_i} \left( \frac{1-t^{\beta_i}}{1-t^{\gamma_i}} \right)^{k_i},
\end{align}
where $\alpha_i, \beta_i, \gamma_i, m, N$ and $k_i$ are positive integers? (NB: In general, the coefficient will be rational, but one can add simple relations on the coefficients $\beta_i, \gamma_i$ and $k_i$ to ensure integrality.)
|
The $\alpha_i$ really contribute nothing and can simply be subtracted from $N$.
For the rest, if $\gamma_i$ divides $\beta_i$, you are looking for compositions of $N$ (minus the sum of the alphas) that start with $k_1$ terms at most $\beta_1$ and divisible by $\gamma_1$ then $k_2$ terms with the analogous properties and so on.
If you have other "simple relations" in mind that ensure integrality you could clarifiy this in the question.
|
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|
evaluating $ \int_0^{\sqrt3} \arcsin(\frac{2t}{1+t^2}) \,dt$ $$\begin{align*}
\int \arcsin\left(\frac{2t}{1+t^2}\right)\,dt&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int\frac{2t}{1+t^2}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right) + \ln(1+t^2)+C
\end{align*}$$
So
$$ \int\nolimits_0^{\sqrt3} \arcsin\left(\frac{2t}{1+t^2}\right)=\pi/\sqrt3+2\ln2.$$
However the result seems to be $ \pi/\sqrt3 $ only.
Why is there this $ 2\ln2 $?
Detail:
$$ \begin{align*}
t \arcsin\left(\frac{2t}{1+t^2}\right)&- \int t \left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\frac{1}{\sqrt{1-\frac{4t^2}{(1+t^2)^2}}}\,dt\\
&= t\arcsin\left(\frac{2t}{1+t^2}\right)- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt{(t^2-1)^2}}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int \frac{2t}{1+t^2}\,dt
\end{align*} $$
|
I think you made a simplification error. We have
$$\begin{align*}
\frac{d}{dt}\arcsin\left(\frac{2t}{1+t^2}\right) &= \frac{1}{\sqrt{1 - \frac{4t^2}{(1+t^2)^2}}}\left(\frac{2t}{1+t^2}\right)'\\
&= \frac{(1+t^2)}{\sqrt{(1+t^2)^2-4t^2}}\left(\frac{2(1+t^2)-4t^2}{(1+t^2)^2}\right)\\
&= \frac{(1+t^2)}{\sqrt{(1-t^2)^2}}\left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\\
&= \frac{2(1-t^2)}{(1+t^2)\sqrt{(t^2-1)^2}}
=\frac{2(1-t^2)}{(1+t^2)|t^2-1|}.
\end{align*}$$
You then cancelled to get
$$-\frac{2}{1+t^2}.$$
However, that cancellation is only valid if $t^2-1\geq 0$, i.e., if $|t|\geq 1$. Yet your integral covers a region that includes places where you get $t^2\lt 1$, so that the cancellation is not valid over the entire interval. Try doing it by splitting the integral as an integral over $[0,1]$ and over $[1,\sqrt{3}]$, being careful with the signs.
|
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|
How would I solve $\frac{(n - 10)(n - 9)(n - 8)\times\ldots\times(n - 2)(n - 1)n}{11!} = 12376$ for some $n$ without brute forcing it? Given this equation:
$$
\frac{(n - 10)(n - 9)(n - 8)\times\ldots\times(n - 2)(n - 1)n}{11!} = 12376
$$
How would I find $n$?
I already know the answer to this, all thanks toWolfram|Alpha, but just knowing the answer isn't good enough for me. I want to know on how I would go about figuring out the answer without having to multiply each term, then using algebra to figure out the answer.
I was hoping that there might be a more clever way of doing this.
|
Using @pharmine's factorization, we have
$$\begin{eqnarray*}
n(n-1)(n-2)\cdots(n-10) &=& 17\cdot 13 \cdot 7 \cdot 2^3 \cdot 11! \\
&=& 17\cdot 13 \cdot 7 \cdot 2^3 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2
\end{eqnarray*}$$
Now, $17$ divides the LHS, so LHS consists of eleven consecutive integers including a multiple of $17$; given the size of the product, that multiple is likely "$17$" itself. The prime $19$ isn't a factor, so the consecutive integers are either "$18$" through "$8$" (that is, $n=18$), or "$17$" through "$7$" ($n=17$). In either case, the list of integers contains "$17$" through "$8$"; constructing these uses up the RHS factors:
$$(17)\cdot 13 \cdot 7 \cdot 2^3 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2$$
$$(17\cdot 16)\cdot 13 \cdot 7 \cdot 2^3 \cdot 11 \cdot 10 \cdot 9 \cdot 7\cdot 6\cdot 5\cdot 4\cdot 3$$
$$(17\cdot 16 \cdot 15)\cdot 13 \cdot 7 \cdot 2^3 \cdot 11 \cdot 10 \cdot 9 \cdot 7\cdot 6\cdot 4$$
$$\cdots$$
$$(17\cdot 16 \cdot 15 \cdot 14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8) \cdot 7$$
... and we conclude that $n=17$.
|
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|
Distributive Property on Fractions: Swapping Denominators I'm learning Algebra and am curious about some methodological fundamentals here. One, in particular is why the following equation:
6(2x + 1 / 3) = 6(x + 4 / 2)
results in:
2(2x + 1) = 3(x + 4)
It's obvious that the distributive property swaps the numerators of the fractions and chooses to use another distributive property to complete the equation. Is there a specific formula for this, and why does it work that way specifically?
|
May be the following steps help you see how you get the result from the given expression - Swapping is fine as long as you understand the meaning of it so that you don't make mistakes.
Given:
$6 \left ( \frac{2x+1}{3} \right )= 6 \left ( \frac{x+4}{2} \right )$
a-multiply both sides by 1/6 to get:
$ \left ( \frac{2x+1}{3} \right )= \left ( \frac{x+4}{2} \right )$
b-multiply both sides by 3 to get:
$ 3\left ( \frac{2x+1}{3} \right )= 3 \left ( \frac{x+4}{2} \right )$
c-This is equal to:
$ \left ( \frac{2x+1}{1} \right )= 3 \left ( \frac{x+4}{2} \right )$
d-Multiply both sides by 2
$2 \left ( \frac{2x+1}{1} \right )= 2 * 3 \left ( \frac{x+4}{2} \right )$
e-Simplifying the right hand side you get
$2 \left ( \frac{2x+1}{1} \right )= 3 \left ( \frac{x+4}{1} \right )$
f-Which is:
$2(2x+1)=3(x+4)$
|
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Why do definitions of distinct conic sections produce a single equation? I understand how to get from the definitions of a hyperbola — as the set of all points on a plane such that the absolute value of the difference between the distances to two foci at $(-c,0)$ and $(c,0)$ is constant, $2a$ — and an ellipse — as the set of all points for which the sum of these distances is constant — to the equation $$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1,\qquad\text{(A)}$$ and I also understand if $a>c$ we can define $b^2=a^2-c^2$, yielding $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad\text{(E)}$$ and if $a<c$ we can define $b^2=c^2-a^2$, yielding $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.\qquad\text{(H)}$$ I also understand (by simply "looking" at the graphs of these two cases) that the former, (E), corresponds to an ellipse, and the latter, (H), to a hyperbola.
However, it seems that the general equation, (A), obscures a distinction that was specified in the definitions: if two different definitions produce the same equation, hasn't something been lost in the process? At some point the derivations must have taken a step that wiped out the some feature of the equations — (B) and (C) below — that distinguishes the definitions. I see that one can "restore" a distinction by considering the relationship between $a$ and $b$, as above, but how that distinction maps back to the distinction between the definitions is obscure to me.
What steps in the derivations of (A) from the respective definitions, (B) and (C), is obscuring information that distinguishes those definitions? Is something going on here that can be generalized?
The derivations I'm referring to are pretty standard, they appear in many texts and also in several places on this site, but are repeated here for reference.
From Spivak's Calculus (p. 66): a point $(x,y)$ is on an ellipse if and only if
$$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}= 2a\qquad\text{(B)}$$
$$\sqrt{(x+c)^2+y^2}= 2a-\sqrt{(x-c)^2+y^2}$$
$$x^2+2cx+c^2+y^2=4a^2-4a\sqrt{(x-c)^2+y^2}+x^2-2cx+c^2+y^2$$
$$4(cx-a^2)=-4a\sqrt{(x-c)^2+y^2}$$
$$c^2x^2-2cxa^2+a^4=a^2(x^2-2cx+c^2+y^2)$$
$$(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)$$
$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\qquad\text{(A)}$$
From a related post: a point $(x,y)$ is on a hyperbola if and only if
$$\sqrt{(x+c)^2+y^2} -\sqrt{(x-c)^2-y^2}=\pm 2a\qquad\text{(C)}$$
$$\frac{4xc}{\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2-y^2}}=\pm 2a$$
$$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2-y^2}=\pm \frac{2cx}{a}$$
$$2\sqrt{(x+c)^2+y^2}=\pm 2\left(a+ \frac{xc}{a}\right)$$
$$x^2+2cx+c^2+y^2=a^2+ 2cx+ \frac{c^2x^2}{a^2}$$
$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\qquad\text{(A)}$$
|
This is the difference between real numbers, some of which are not squares of other real numbers, and complex numbers, all of which are squares of other complex numbers. Is the $a^2 - c^2$ in the denominator a negative number (i.e. one that is not a square) or a positive number (one that is a square) (the case where it's $0$, which also is a square, is not what we're examining here).
If you work with complex numbers rather than real numbers, then there's no difference between ellipses and hyperbolas.
(If you go a step further and work with projective spaces rather than affine spaces, then there's no difference between those and parabolas.)
|
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|
Problem about sum of arithemtic progression and geometric progression Question: An arithmetic sequence has a common difference of $1$ and a geometric sequence has a common ratio of $3$. A new sequence is formed by adding corresponding terms of these two progressions. It is given that the second and fourth term of the sequence are $12$ and $86$ respectively. Find, in terms of $n$,
*
*the $n^{th}$ term,
*the sum of the first $n$ terms
of the new sequence.
Answers:
*
*$n+1+3^n$.
*$\frac 1 2 n(n+3) + \frac{3}{2} 3^n - \frac{3}{2}$.
Working so far
I am stuck getting different values for $a$.
|
(i)
the first sequence is: $u_n=u_0+n$
the second sequence is $v_n=v_0 \cdot 3^n$
so the new sequence is : $w_n=u_0+n+v_0 \cdot 3^n$
since $w_2=12$ and $w_4=86$
then $u_0+2+v_0\cdot9=12$ and $u_0+4+v_0\cdot81=86$
then $u_0=1$ and $v_0=1$
so $w_n=1+n+3^n$
(ii)
$\sum\limits_{k=0}^{n-1}w_k = \sum\limits_{k=0}^{n-1}1+\sum\limits_{k=0}^{n-1} k+\sum\limits_{k=0}^{n-1}3^k = n-1 + n\cdot(n-1)/2 + (3^n-1)/2$
|
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|
Proof of dividing fractions $\frac{a/b}{c/d}=\frac{ad}{bc}$ For dividing two fractional expressions, how does the division sign turns into multiplication? Is there a step by step proof which proves
$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}=\frac{ad}{bc}?$$
|
Write $\frac{a}{b} \div \frac{c}{d}$ as
$$
\frac{\ \frac{a}{b}\ }{\frac{c}{d}}.
$$
Suppose you wanted to clear the denominator of this compound fraction. You could try multiplication by $\frac{d}{c}$, but you'll have to multiply the top and the bottom of the fraction to avoid changing it. So, you end up with
$$
\frac{\ \frac{a}{b}\ }{\frac{c}{d}} = \frac{\ \frac{a}{b}\ }{\frac{c}{d}} \cdot \frac{\ \frac{d}{c}\ }{\frac{d}{c}} = \frac{\ \frac{a}{b} \cdot \frac{d}{c}\ }{\frac{c}{d} \cdot \frac{d}{c}} = \frac{\ \frac{ad}{bc}\ }{1} = \frac{ad}{bc}.
$$
|
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|
How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$? How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$?
Should I use some geometrical approach or apagoge?
|
If $\cos \frac{2 \pi}{7}$ is rational, then $i\sin \frac{2 \pi}{7}=\sqrt{\cos^2 \frac{2 \pi}{7}-1}$ is a quadratic irrational, and hence so is $\cos \frac{2 \pi}{7}+i\sin \frac{2 \pi}{7}$. But $\cos \frac{2 \pi}{7}+i\sin \frac{2 \pi}{7}$ is a primitive $7$th root of $1$, and so has minimal polynomial $x^6+x^5+x^4+x^3+x^2+x+1$, which is not a quadratic.
|
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|
$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = n(n+1)(2n+7)/6$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.
$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.
Sol:
$P(n):\ 1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.
$P(1):\ \frac{1}{6}(2)(9) = \frac{1}{2}(2)(3)$.
$P(1): 3$.
Hence it is true for $n=n_0 = 1$.
Let it be true for $n=k$.
$P(k):\ 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$.
We have to prove
$P(k+1):\ 1\cdot 3 + 2\cdot 4 + \cdots + (k+1)(k+3)= \frac{1}{6}(k+1)(k+2)(2k+9)$.
Taking LHS:
$$\begin{align*}
1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right].
\end{align*}$$
|
You used the wrong formula in your induction hypothesis. You are assuming that
$$1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$$
but in your inductive argument, you wrote
$$1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3) = \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3).$$
That is, you substituted $1\cdot 3 + 2\cdot 4 + \cdots + k(k+2)$ with the formula for $k+1$, not the formula for $k$.
|
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|
Find $x$ if $(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$
Known that:
$$(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$$
What is the value of $x$
|
Let $a=(7+4\sqrt{3})^{x^2-8}$ then $\dfrac{1}{a}=(7-4\sqrt{3})^{x^2-8}$
So the given equation will be $a+\dfrac{1}{a}=14\implies a^2-14a+1=0$
Solving quadratic for $a=7+4\sqrt{3}$ or $a=7-4\sqrt{3}\implies x^2-8=1$ or $-1$
therefore possible values of $x=3,-3,\sqrt{7},-\sqrt{7}$
|
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|
What is the formula for $n^{th}$ derivative of $ \sin^{-1} x, \quad \tan^{-1} x,\quad \sec x \quad \text{and}\quad \tan x$? Are there formulae for the nth derivatives of the following functions?
$1)\quad$ $sin^{-1} x$
$2)\quad$ $tan^{-1} x$
$3)\quad$ $sec x$
$4)\quad$ $tan x$
Thanks.
|
A related problem. See Chapter 6 in this book for formulas for the nth derivative, $n$ is a non-negative integer, of $\tan(x)$ and $\sec(x)$ in terms of the $\psi$ function
\begin{equation}
{\tan}^{(n)} (z) = \frac{1}{{\pi}^{n+1}} \left({\psi}^{(n)} \left(
\frac{1}{2}+\frac{z}{\pi}\right) + (-1)^{n+1} {\psi}^{(n)} \left(
\frac{1}{2}-\frac{z}{\pi}\right) \right)\,,
\end{equation}
\begin{align}\nonumber
{\sec}^{(n)}(z) = \frac{1}{ (2 \pi)^{ n + 1 } } &\left(
(-1)^{n}{\psi}^{(n)}\left( -\frac{2z-3\pi}{4\pi}\right) -
(-1)^{n}{\psi}^{(n)}\left( -\frac{2z-\pi}{4\pi}\right) \right.&
\\ \nonumber
& \left. - {\psi}^{(n)}\left( \frac{2z+\pi}{4\pi}\right) +
{\psi}^{(n)}\left( \frac{2z+3\pi}{4\pi}\right) \right) \,. &
\end{align}
|
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|
If $x^2+y^2=z^2$, why can't $x$ and $y$ both be odd? What does the following mean: If $x^2 + y^2 = z^2$ some integers $z$, then $x$ and $y$ can't be both odd (otherwise, the sum of their squares would be $2$ modulo $4$, which can't be a square). So, one of them must be even?
I see that if $x$ and $y$ are both odd, then $z^2 = 4k+2 =2(2k+1)$. So $z^2$ is even. But why does it say above that... can't be square?
|
If $n = 2k$, then $n^2 = 4k^2$ is a multiple of $4$.
Likewise, if $n = 2k+1$, then $n^2 = 4(k^2 + k) + 1$.
Therefore, a square is congruent to either $0$ or $1 \pmod{4}$. In other words, a square is never of the form $4k + 2$, for some $k$.
More specifically, since you've seen that $x^2 + y^2 \equiv 2 \pmod{4}$ when $x$ and $y$ are odd, and since squares are never congruent to $2 \pmod{4}$, this shows that $x^2 + y^2$ is never a square when $x$ and $y$ are both odd.
|
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|
Pythagorean quadruples Another Project Euler problem has me checking the internet again. Among other conditions, four of my variables satisfy:
$$a^2+b^2+c^2=d^2 .$$
According to Wikipedia, this is known as a Pythagorean Quadruple. It goes on to say all quadruples can be generated from an odd value of $a$ and an even value of $b$ as:
$$c=\frac{a^2+b^2-p^2}{2}, \quad d=\frac{a^2+b^2+p^2}{2} ,$$
where $p$ is any factor of $a^2+b^2$ that satisfies $p^2<a^2+b^2$.
However, I can't see how or why this works. I also can't seem to see how this works for $\lbrace 4,4,7,9 \rbrace$. Am I missing something here?
|
I think $c$ and $d$ should have been
$$
\begin{split}
c &= \frac{a^2+b^2-p^2}{2 p}\qquad\qquad
d &= \frac{a^2+b^2+p^2}{2 p}
\end{split}
$$
|
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|
Finding $\lim\limits_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}$ $$\lim_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}.$$
With a first look this must give $1$ as a result but have a problem to explain it.
How can I do it?
Edit
I noticed that it is $\frac{\infty}{\infty}$.
$$\lim_{n \to \infty}{n^{n}\frac{(\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1)}{n^n}}= \lim_{n \to \infty}{\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1}=1$$
Is this correct?
|
Let $f(n) = (1^1 + 2^2 + 3^3 + \cdots + n^n)/n^n$. You want to show $\lim_{n \to \infty} f(n) = 1$.
It's obvious that $f(n) > 1$ for all $n$.
For an upper bound,
$$ f(n) \le {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} + {n^{n-1} \over n^n} + {n^n \over n^n} = {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} + {1 \over n} + 1.$$
Now I leave it to you to find some bound $g(n)$, with $\lim_{n \to \infty} g(n) = 0$ and
$$ {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} \le g(n). $$
So you have
$$ 1 < f(n) < 1 + {1 \over n} + g(n) $$
and apply the squeeze theorem to finish the proof.
|
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|
Find all integers $m$ such that $\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} $ How would you determine all integers $m$ such that the following is true?
$$\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} .$$
Note that $\lfloor \cdot \rfloor$ means the greatest integer function. Also, $x$ must be a positive real number.
|
Consider the reciprocal equation
$$ m = \dfrac{\lfloor 2x \rfloor \cdot \lfloor 5x \rfloor}
{\lfloor 2x \rfloor + \lfloor 5x \rfloor}$$
Letting $x=n+\delta$ where $0 \le \delta < 1$, we get
$$ m = \dfrac{(2n + \lfloor 2\delta \rfloor)(5n + \lfloor 5\delta \rfloor)}
{7n + \lfloor 2\delta \rfloor + \lfloor 5\delta \rfloor}$$
We list the possible values of m with respect to $\delta$.
\begin{array}{|c|c|c|c|c|c|}
\delta \in & \lfloor 2\delta \rfloor & \lfloor 5\delta \rfloor
& \lfloor 2\delta \rfloor + \lfloor 5\delta \rfloor
& \lfloor 2\delta \rfloor \cdot \lfloor 5\delta \rfloor
& m \\
\hline
[ 0, 0.2) & 0 & 0 & 0 & 0 & \dfrac{10n}{7}\\
[0.2, 0.4) & 0 & 1 & 1 & 0 & \dfrac{10n^2+2n}{7n+1}\\
[0.4, 0.5) & 0 & 2 & 2 & 0 & \dfrac{10n^2+4n}{7n+2}\\
[0.5, 0.6) & 1 & 2 & 3 & 2 & \dfrac{10n^2+9n+2}{7n+3}\\
[0.6, 0.8) & 1 & 3 & 4 & 3 & \dfrac{10n^2+11n+3}{7n+4}\\
[0.8, 1.0) & 1 & 4 & 5 & 4 & \dfrac{10n^2+13n+4}{7n+5}\\
\hline
\end{array}
Now its just a matter of finding when those rational polynomials have integer solutions.
|
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How can I prove the inequality $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$? For $x > 0$, $y > 0$, $z > 0$, prove:
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z} .$$
I can see that this is true, I also checked it with a few numbers. But I guess that is not enough to prove it. So how can I prove it?
(You don't need to show the whole proof, I think just a hint will be enough. I simply don't know how to start.)
|
You can also brute-force it along with some careful re-grouping of terms:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$
$\Leftrightarrow \frac{xy+xz+yz}{xyz}\geq \frac{9}{x+y+z}$
$\Leftrightarrow x^2 y + xyz + x^2 z + x y^2 + xyz + y^2 z + xyz + xz^2 + y z^2 \geq 9xyz$
$\Leftrightarrow x^2y-2xyz + yz^2 + x^2z-2xyz + yz^2 + xz^2-2xyz + xy^2 \geq 0$
$\Leftrightarrow y(x-z)^2 + z(x-y)^2 + x(y-z)^2 \geq 0$.
The last inequality holds (since $x,y,z>0$), therefore the first inequality holds.
|
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|
The limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0? Is there any elementary proof that the limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0?
|
The following requires no tools other than induction. In particular, machinery from the calculus is not used.
Let $f(n)$ be the product up to the term $1-\dfrac{1}{2^n}$.
We show by induction that
$$f(n)\ge\frac{1}{4} +\frac{1}{2^{n+1}}.$$
The result is true at $n=1$. For the induction step, note that
$$f(m+1)=f(m)\left(1-\frac{1}{2^{m+1}}\right)\ge\left(\frac{1}{4}+\frac{1}{2^{m+1}}\right)\left(1-\frac{1}{2^{m+1}}\right).$$
Expand the product on the right. We get
$$\frac{1}{4}+\frac{1}{2^{m+1}}-\frac{1}{4}\cdot\frac{1}{2^{m+1}}-\frac{1}{2^{2m+2}}.\qquad(\ast)$$
Rewrite $\dfrac{1}{4}\cdot\dfrac{1}{2^{m+1}}$ as $\dfrac{1}{2^{m+2}}-\dfrac{1}{2^{m+3}}$.
Then $(\ast)$ becomes
$$\frac{1}{4}+\frac{1}{2^{m+2}}+\frac{1}{2^{m+3}}-\frac{1}{2^{2m+2}}.$$
The term $\dfrac{1}{2^{m+3}}-\dfrac{1}{2^{2m+2}}$ is non-negative. Thus $f(m+1)\ge\dfrac{1}{4}+\dfrac{1}{2^{m+2}}$, which completes the induction step.
|
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|
Any trick on finding the inverse of this matrix? Supposing I have a matrix, $\pmatrix{0&0&\lambda\\0&\lambda&-1\\ \lambda&-1&0}$.
Without question you can work out the inverse if this matrix. But since it is highly structured, I suppose there should be some quick way to find out the inverse of it?
Can anyone show me a quick way of thinking about this problem?
|
You can partition the matrix into
$$ \begin{bmatrix} 0 & v^\top \\ v & A \end{bmatrix} $$
where $A$ is a symmetric $2\times2$ matrix and $v$ a $2\times1$ vector, yielding the solution of
$$ \begin{bmatrix} 0 & v^\top \\ v & A \end{bmatrix}^{-1} = \begin{bmatrix} -(v^\top A^{-1} v)^{-1} & T^\top \\ T & A^{-1}-T v^\top A^{-1} \end{bmatrix} $$
where $T = A^{-1} v (v^\top A^{-1} v)^{-1} $ is the weighted pseudo inverse of $v$, since $T^\top v=1$ and $v^\top T=1 $
Proof:
$$ \begin{bmatrix} -(v^\top A^{-1} v)^{-1} & T^\top \\ T & A^{-1}-T v^\top A^{-1} \end{bmatrix} \begin{bmatrix} 0 & v^\top \\ v & A \end{bmatrix} = \begin{bmatrix} T^\top v & -(v^\top A^{-1} v)^{-1} v^\top+T^\top A \\ A^{-1}v-T v^\top A^{-1} v & T v^\top + A^{-1} A - T v^\top \end{bmatrix} $$
with the simplification $T^\top A = (v^\top A^{-1} v)^{-1} v^\top$ the above becomes
$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $
Example:
$A = \begin{pmatrix} \lambda & -1 \\ -1 & 0 \end{pmatrix} $, $v = \begin{pmatrix} 0 \\ \lambda \end{pmatrix}$, $T = \begin{pmatrix} \lambda & -1 \\ -1 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 0 \\ \lambda \end{pmatrix} \left(\begin{pmatrix} 0 & \lambda \end{pmatrix} \begin{pmatrix} \lambda & -1 \\ -1 & 0 \end{pmatrix}^{-1} \begin{pmatrix} 0 \\ \lambda \end{pmatrix} \right)^{-1} = \begin{pmatrix} \frac{1}{\lambda^2} \\ \frac{1}{\lambda} \end{pmatrix}$
$$ \begin{bmatrix} -(v^\top A^{-1} v)^{-1} & T^\top \\ T & A^{-1}-T v^\top A^{-1} \end{bmatrix} = \begin{bmatrix}
\begin{pmatrix} \frac{1}{\lambda^3} \end{pmatrix} &
\begin{pmatrix} \frac{1}{\lambda^2} & \frac{1}{\lambda} \end{pmatrix} \\
\begin{pmatrix} \frac{1}{\lambda^2} \\ \frac{1}{\lambda} \end{pmatrix} &
\begin{pmatrix} \frac{1}{\lambda} & 0 \\ 0 & 0 \end{pmatrix}
\end{bmatrix} $$
|
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|
Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4} \frac{1}{x-4} \cdot \left (\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{4+3 \cdot 4}} \right )
\\ \\ & = \lim_{x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{16}} \right ) \\ \\ & = \lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-2\right)
\end{align*}$
But now I can't figure it out, how to end this limit.
I know that the derivative formula for this function is $-\frac{12}{(4+3x)\sqrt{4+3x}}$.
Thanks for the help.
|
All these answers are simple fact (except using L' Hospital, which is the best approach though) that you have to multiply by conjugate surds, which is greatly illustrated in @Michael Hardy's answer. So to sum up, you must make a factor $(x-4)$ in numerator, in order to cancel the same at denominator. But, if that factor was not there, how it comes after multiplication by conjugate ? The actual answer is We don't need $(x-4)$, but we need to cancel $(\sqrt{x}-2)$ (note that the other factor $(\sqrt{x}+2)$ gives no trouble) and surely, the numerator should have such a factor, though not explicitly. So, here is an alternate way (without multiplying by conjugate and this approach hopefully applicable to similar problems):
First set $y=4+3x$. Then as $x\to4,y\to16$ and yor limit becomes$$\lim_{y\to16}\frac{3}{y-16}.2\left(\frac{4}{\sqrt{y}}-1\right)$$ $$=\lim_{y\to16}\frac{6}{(\sqrt{y}-4)(\sqrt{y}+4)}\left(\frac{4-\sqrt{y}}{\sqrt{y}}\right)$$ $$=-\dfrac{3}{16}$$
|
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|
Asymptotic formula for $\sum_{n \le x} \frac{\varphi(n)}{n^2}$ Here is yet another problem I can't seem to do by myself... I am supposed to prove that
$$\sum_{n \le x} \frac{\varphi(n)}{n^2}=\frac{\log x}{\zeta(2)}+\frac{\gamma}{\zeta(2)}-A+O \left(\frac{\log x}{x} \right),$$
where $\gamma$ is the Euler-Mascheroni constant and $A= \sum_{n=1}^\infty \frac{\mu(n) \log n}{n^2}$. I might be close to solving it, but what I end up with doesn't seem quite right. So far I've got:
$$
\begin{align*}
\sum_{n \le x} \frac{\varphi(n)}{n^2} &= \sum_{n \le x} \frac{1}{n^2} \sum_{d \mid n} \mu(d) \frac{n}{d} \\
&= \sum_{n \le x} \frac{1}{n} \sum_{d \le x/n} \frac{\mu(d)}{d^2} \\
&= \sum_{n \le x} \frac{1}{n} \left( \sum_{d=1}^\infty \frac{\mu(d)}{d^2}- \sum_{d>x/n} \frac{\mu(d)}{d^2} \right) \\
&= \sum_{n \le x} \frac{1}{n} \left( \frac{1}{\zeta(2)}- \sum_{d>x/n} \frac{\mu(d)}{d^2} \right) \\
&= \frac{1}{\zeta(2)} \left(\log x + \gamma +O(x^{-1}) \right) - \sum_{n \le x} \frac{1}{n}\sum_{d>x/n} \frac{\mu(d)}{d^2}.
\end{align*}
$$
So. I suppose my main problem is the rightmost sum, I have no idea what to do with it! I'm not sure where $A$ comes into the picture either. I tried getting something useful out of
$$
\begin{align*}
& \sum_{n \le x} \frac{1}{n} \left( \frac{1}{\zeta(2)}- \sum_{d>x/n} \frac{\mu(d)}{d^2} \right) +A-A \\
&= \left( \frac{1}{\zeta(2)} \left(\log x + \gamma +O(x^{-1}) \right) - A \right) - \sum_{n \le x} \frac{1}{n}\sum_{d>x/n} \frac{\mu(d)}{d^2} + A,
\end{align*}
$$
but I quickly realized that I had no clue what I was doing. Any help would be much appreciated.
|
Consider Dirichlet series, related to the problem at hand:
$$
g(s) = \sum_{n=1}^\infty \frac{\varphi(n)}{n^{2+s}} = \frac{\zeta(s+1)}{\zeta(s+2)}
$$
We can now recover behavior of $A(x) = \sum_{n \le x} \frac{\varphi(n)}{n^s}$ by employing Perron's formula, using $c > 0$:
$$
A(x) = \frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} \frac{\zeta(z+1)}{\zeta(z+2)} \frac{x^z}{z} \mathrm{d} z = \mathcal{L}^{-1}_z\left( \frac{\zeta(z+1)}{\zeta(z+2)} \frac{1}{z} \right)(\log x)
$$
For large $x$, the main contribution comes from the pole of ratio of zeta functions at $z = 0$. Using
$$
\frac{\zeta(z+1)}{\zeta(z+2)} \sim \frac{1}{\zeta(2)} \left( \frac{1}{z} + \gamma - \zeta^\prime(2)\right) + O(z)
$$
Since $\mathcal{L}^{-1}_z\left( \frac{1}{z^{n+1}} \right)(s) = \frac{s^n}{n!}$ we have
$$
A(x) = \frac{\log x}{\zeta(2)} + \left( \frac{\gamma}{\zeta(2)} - \frac{\zeta^\prime(2)}{\zeta(2)} \right) + \mathcal{L}^{-1}_z\left( \frac{\zeta(z+1)}{\zeta(z+2)} \frac{1}{z} - \frac{1}{\zeta(2) z^2} - \frac{\gamma - \zeta^\prime(2)}{\zeta(2) z} \right)(\log x)
$$
Notice that $\frac{\zeta^\prime(s)}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n) \log(n)}{n^s}$, hence $\frac{\zeta^\prime(2)}{\zeta(2)} = A$.
It remains to be shown that the remainder term is small.
|
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|
System of first-order ordinary differential equations solve for $x,y,z$:
$$\frac{dx}{x^{2}+a^{2}}=\frac{dy}{xy-az}=\frac{dz}{xz+ay}$$
please give a hint. I am not able to formulate the steps required to proceed solving this one.
|
We have
$$
\begin{align}
\frac{dy}{xy-az} & = \frac{dz}{xz+ay}\\
\frac{dy/y}{x-a (z/y)} & = \frac{dz/z}{x+a (y/z)}
\end{align}
$$
This gives a motivation to let $z = ky$ where $k$ is a constant.
$$
\begin{align}
\frac{dy/y}{x-a k} & = \frac{dy/y}{x+a/k}
\end{align}
$$
This gives us that $k = \pm i$. Let $k=i$.
This gives us that
$$\begin{align}
\frac{dx}{x^2 + a^2} & = \frac{dy/y}{x - ia}\\
\frac{dx}{x + ia} & = \frac{dy}{y}\\
y & = c(x + ia)
\end{align}
$$
Hence, we get
$$
\begin{align}
z & = ic(x+ia)\\
y & = c(x+ia)
\end{align}
$$
and
$$
\begin{align}
z & = -ic(x-ia)\\
y & = c(x-ia)
\end{align}
$$
I don't know to justify my motivation why I chose $z = ky$ instead of $z=k(y)y$.
|
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|
Finding pairs of integers such that $x^2+3y$ and $y^2+3x$ are both perfect squares Can we find pairs $(x,y)$ of positive integers such that $x^2+3y$ and $y^2+3x$ are simultaneously perfect squares? Thanks a lot in advance. My progress is minimal.
|
Not a complete solution but an approach that seems like it will work.
Assume $y \gt x$
Then we have that
$(y+2)^2 \gt y^2 + 3y \gt y^2+3x \gt y^2$
If $y^2 + 3x$ was a perfect square, then we have that $y^2 + 3x = (y+1)^2$.
This gives us $3x = 2y+1$.
Substitute in the other expression, and form similar inequalities. This will narrow down to few small cases to consider.
To elaborate, given $3x = 2y+1$
$x^2 + 3y = x^2 + \frac{9x-3}{2} \lt (x+3)^2$
Thus $x^2 + 3y$ is either $(x+1)^2$ or $(x+2)^2$.
Thus we have that
$3y = 2x+1$ or $3y = 4x + 4$.
Substitute $y = \frac{3x -1}{2}$ and solve the linear equation in $x$ and compute $y$. In the end, don't forget to verify that both the expressions are indeed perfect squares.
The other case $y=x$ can be treated similarly.
|
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|
How to find the roots of $f(x)= \ln( \frac{x+1 }{x-2})$? I can't solve this equation:
$$\ln\left(\frac{x+1}{x-2}\right) = 0.$$
I do:
$$\begin{align*}
\ln \left( \frac{x+1}{x-2} \right)&=0\\
\frac{x+1}{x-2} &= 1 \\
x+1&=x-2 \\
x+1-x+2&=0 \\
x-x+3&=0 \\
3&=0
\end{align*}$$
Then $x$ is?
|
What you've shown is that $\frac{x+1}{x-2}$ is never equal to 1. Since 1 is the only value where natural log equals zero, the equation $\log \frac{x+1}{x-2} = 0 $ has no solutions.
|
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|
How to prove $(1+1/x)^x$ is increasing when $x>0$? Let $F(x)=(1+\frac{1}{x})^x$.
How do we prove $F(x)$ is increasing when $x>0$?
|
This proof is from theorem 140 from Hardy's Inequalities.
Let $f(x) = \ln\left[\left(1 + \frac{1}{x}\right)^x\right] = x(\ln(x+1) - \ln(x)).$ We refer to the mean value theorem: for each differentiable $g$,
$$
g(x + h) - g(x) = hg'(x + \theta h)
$$
for $\theta \in (0,1)$.
Applying the MVT to $g(x) = \ln(x)$, we get $\ln(x+1) - \ln(x) = \frac{1}{x + \theta}$. Since, $\frac{1}{x+\theta} > \frac{1}{x+1} $, we have that $\ln(x+1) - \ln(x) > \frac{1}{x+1}$, so
$$
f'(x) = \ln(x+1) - \ln(x) - \frac{1}{x+1} > 0.
$$
Hence, $f(x)$ and $e^{f(x)} = \left(1+\frac{1}{x}\right)^x$ are increasing with $x$.
|
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|
Proving $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$ Could anyone help me prove that $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$?
As $6=2*3=(1+\sqrt{-5})(1-\sqrt{-5})$ so $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. Therefore is not a PID or euclidean domain
|
The standard method is:
Define a function $N\colon \mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}$ by $N(a+b\sqrt{-5}) = (a+b\sqrt{-5})(a-b\sqrt{-5}) = a^2+5b^2$.
*
*Prove that $N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha,\beta\in\mathbb{Z}[\sqrt{-5}]$.
*Conclude that if $\alpha|\beta$ in $\mathbb{Z}[\sqrt{-5}]$, then $N(\alpha)|N(\beta)$ in $\mathbb{Z}$.
*Prove that $\alpha\in\mathbb{Z}[\sqrt{-5}]$ is a unit if and only if $N(\alpha)=1$.
*Show that there are no elements in $\mathbb{Z}[\sqrt{-5}]$ with $N(\alpha)=2$ or $N(\alpha)=3$.
*Conclude that $2$, $3$, $1+\sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible.
This is a common technique for dealing with rings of the form $\mathbb{Z}[\theta]$, where $\theta$ is an algebraic integer.
It can also be done directly, though it is a bit more laborious. Here's what I came up with on the fly:
Assume that $(a+b\sqrt{-5})(c+d\sqrt{-5}) = 2$. Then $ac+5bd = 2$ and $ad+bc=0$. If $a=0$, then we must have $c=0$ (since $bc=0$ but we cannot have $a=b=0$), but then $5bd=2$ is impossible. Thus, $a\neq0$ and $c\neq 0$. If $b=0$, then $d=0$, so the factorization occurs in $\mathbb{Z}$ and is trivial; symmetrically if $d=0$. So we may assume that all of $a,b,c,d$ are nonzero.
Then $ad=-bc$, so $2d=acd + 5bd^2 = -bc^2 + 5bd^2 = b(5d^2 - c^2)$. If $b$ is odd, then $b|d$, so writing $d=bk$ we get $abk + bc=0$, so $ak+c=0$, thus $c=-ak$. Hence $c+d\sqrt{-5} = -ak+bk\sqrt{-5} = k(-a+b\sqrt{-5})$. But this gives
$$2 = (a+b\sqrt{-5})(c+d\sqrt{-5}) = k(a+b\sqrt{-5})(-a+b\sqrt{-5}) = -k(a^2+5b^2).$$
Since $a$ and $b$ are both nonzero, $a^2+5b^2$ is at least 6, which is impossible. So $b$ is even, $b=2b'$.
Then $b'(5d^2-c^2) = d$, so setting $5d^2-c^2 = k$ we have
$$a+b\sqrt{-5} = a+2b'\sqrt{-5},\qquad c+d\sqrt{-5} = c+kb'\sqrt{-5}.$$
From $ad=-bc$ we get $ak=-2c$. If $a$ is even, then we have $a=2a'$, so
$$2 = (2a'+2b'\sqrt{-5})(c+d\sqrt{-5}) = 2(a'+b'\sqrt{-5})(c+d\sqrt{-5}),$$
which yields that $c+d\sqrt{-5}$ is a unit (in fact, this is impossible with $c$ and $d$ both nonzero, but that doesn't matter). If $a$ is odd, then $k$ is even and $c=ak'$, with $2k'=k$. So now we have
$$\begin{align*}
2 &= (a+b\sqrt{-5})(c+d\sqrt{-5})\\
&= (a + 2b'\sqrt{-5})(ak' + 2b'k'\sqrt{-5})\\
&= k'(a+2b'\sqrt{-5})(a+2b'\sqrt{-5})\\
&= k'(a^2 - 20b'^2) + 4ab'\sqrt{-5}
\end{align*}$$
which implies $a=0$ or $b'=0$ (hence $b=0$), contradicting our hypotheses.
Thus, the only factorizations of $2$ in $\mathbb{Z}[\sqrt{-5}]$ are trivial.
(And you can probably see why the "standard method" is so much better....)
|
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|
Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$? Let $a$ be a non-zero real number. Is it true that the value of $$\int\limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$$ is independent on $a$?
|
Let $\mathcal{I}(a)$ denote the integral. Then
$$ \begin{eqnarray}
\mathcal{I}(a) &=& \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + \int_1^\infty \frac{\mathrm{d} y}{(1+y^2)(1+y^a)} \\
&\stackrel{y=1/x}{=}&
\int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} +
\int_0^1 \frac{x^a \mathrm{d} x}{(1+x^2)(1+x^a)} \\
&=& \int_0^1 \frac{1+x^a}{(1+x^2)(1+x^a)} \mathrm{d} x =
\int_0^1 \frac{1}{1+x^2} \mathrm{d} x = \frac{\pi}{4}
\end{eqnarray}
$$
Thus $\mathcal{I}(a) = \frac{\pi}{4}$ for all $a$. I do not see a need to require $a$ to be non-zero.
|
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|
How to prove the Fibonacci sum $\sum \limits_{n=0}^{\infty}\frac{F_n}{p^n} = \frac{p}{p^2-p-1}$ We are familiar with the nifty fact that given the Fibonacci series $F_n = 0, 1, 1, 2, 3, 5, 8,\dots$ then $0.0112358\dots\approx 1/89$. In fact,
$$\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$$
How do we prove that, more generally, for $p > 1$ then,
$$\sum_{n=0}^{\infty}\frac{F_n}{p^n} = \frac{p}{p^2-p-1}$$
(The above simply was the case $p = 10$.)
|
I'll copy here an answer, which I posted before at AoPS. However the solution from wikipedia article (see the link in lhf's comment to J.M.'s answer) seems to be much more elegant. Basically these matrix proofs often can be rewritten to proofs using generating functions.
We denote by $F_i$ the i-th Fibonacci number.
Using the formula $$\sum_{i=0}^n F_i k^i = \frac{(k-1)k^{n+1}F_{n+1}-k^{n+2}F_{n+2}+k}{1-k-k^2} \qquad (*)$$
you get for $k<\frac{1-\sqrt{5}}{2}$ using Binet formula mentioned by mavropnevma that
$$\sum_{i=0}^\infty F_i k^i = \lim_{n\to\infty} \frac{(k-1)k^{n+1}F_{n+1}-k^{n+2}F_{n+2}+k}{1-k-k^2} = \frac k{1-k-k^2}.$$
In your case $k=1/10$ and this limit is equal to 10/89.
The above formula (*) can be derived by induction.
Inductive step:
$\frac{(k-1)k^{n+1}F_{n+1}-k^{n+2}F_{n+2}+k}{1-k-k^2}+k^{n+1}F_{n+1}=
\frac{-k^{n+3}F_{n+1}-k^{n+2}F_{n+2}+k}{1-k-k^2}=$
$\frac{-k^{n+3}(F_{n+3}-F_{n+2})-k^{n+2}F_{n+2}+k}{1-k-k^2}=
\frac{(k-1)k^{n+2}F_{n+2}-k^{n+3}F_{n+3}+k}{1-k-k^2}$
However, my method (which lead me to "discovering" the formula, by which I mean that I did not know the value of the sum in advance) was using matrix method.
For $A=\begin{pmatrix}
0 & 1\\
1 & 1
\end{pmatrix}$ we have $A^n=
\begin{pmatrix}
F_{n-1} & F_n\\
F_n & F_{n+1}
\end{pmatrix}
$.
From
$$\sum_{k=0}^n (Ak)^i = (A^{n+1}k^{n+1}-I)(Ak-I)^{-1}$$
using $(Ak-I)^{-1}=\begin{pmatrix}-1&k\\k&k-1\end{pmatrix}^{-1}=
\frac1{1-k-k^2}\begin{pmatrix} k-1& -k \\ -k & -1 \end{pmatrix}$
and $A^{n+1}k^{n+1}-I=\begin{pmatrix}
k^{n+1}F_n-1 & k^{n+1}F_{n+1}\\
k^{n+1}F_{n+1} & k^{n+1}F_{n+2}-1
\end{pmatrix}$ by matrix multiplication we get the desired result.
EDIT:
I have noticed that this sum is mentioned in wikipedia's article on Fibonacci numbers
They refer to this page - a proof (using matrices) is provided there.
|
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|
$\int \cos^{-1} x \; dx$; trying to salvage an unsuccessful attempt $$
\begin{align}
\int \cos^{-1} x \; dx &= \int \cos^{-1} x \times 1 \; dx
\end{align}
$$
Then, setting
$$\begin{array}{l l}
u=\cos^{-1} x & v=x \\
u' = -\frac{1}{\sqrt{1-x^2}} & v'=1\\
\end{array}$$
Then by the IBP technique, we have:
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &=\cos^{-1} (x) \cdot x - \int x \cdot -\frac{1}{\sqrt{1-x^2}} \; dx\\
&= x \cos^{-1} (x) - \int -\frac{x}{\sqrt{1-x^2}} \; dx\\
\end{array}$$
Now at this point suppose I have overlooked the possibility of using integration by substitution (setting $u=1-x^2$) to simplify the second integral. Instead, I attempt to reapply IBP to the second integral $\int -\frac{x}{\sqrt{1-x^2}} \; dx$.
I let
$$\begin{array}{l l}
u= -\frac{1}{\sqrt{1-x^2}} = -(1-x^2)^{-\frac{1}{2}} \qquad & v= \frac{x^2}{2} \\
u' = - \left( -\frac{1}{2} \right) (1-x^2)^{-\frac{3}{2}} \times -2x = -x(1-x^2)^{-\frac{3}{2}} \qquad & v'=x\\
\end{array}$$
Then by IBP again,
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &= x \cos^{-1} (x) - \left( -\frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot -x(1-x^2)^{-\frac{3}{2}} \; dx \right) \\
\end{array}$$
At this stage, I can see no way to proceed. Can anyone see a reasonable way to salvage this solution, continuing along this line of reasoning? Or was approaching the second integral by IBP doomed to fail?
|
The integral
$$
\int \frac{1}{\sqrt{1-x^2}} x\;dx
$$
is BEGGING for a simple substitution. Don't integrate by parts here. Instead, do this:
$$
\begin{align}
u & = 1 - x^2 \\ \\ du & = -2x\;dx \\ \\ \frac{-du}{2} & = x\;dx
\end{align}
$$
You get
$$
-\int \frac{1}{2\sqrt{u}} \;du = \sqrt{u}+C = \sqrt{1-x^2}+ C.
$$
The reason why some people are calling this substitution "obvious" is that you have $\big((\text{1st-degree polynomial})\cdot dx\big)$ and elsewhere in the expression you have a 2nd-degree polynomial, and---lo and behold---the derivative of the 2nd-degree polyonomial is the 1st-degree polyonomial, except for the factor of $1/2$, which is a constant. When you see something multiplied by $dx$ that is the derivative of another expression that's there, then that's what you do. You should be looking for that.
Later note: Suppose I am asked about
$$
\int \frac{1}{\sqrt{1-x^2}} x\;dx.
$$
If this were in class or the if the questioner were otherwise physically present, I might just say "Here's a hint", and then write
$$
\int \frac{1}{\sqrt{1-x^2}} {\huge(}x\;dx{\huge)}.
$$
When you understand why that's a hint, then you understand finding integrals by means of substitutions.
|
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|
Showing two matrices are similar I have to show that each of the following matrices
$$ \frac{1}{\sqrt{2}}
\begin{pmatrix}
0&1&0\\
1&0&1\\
0&1&0
\end{pmatrix}\quad ,
\frac{1}{\sqrt{2}}
\begin{pmatrix}
0&-i&0\\
i&0&-i\\
0&i&0
\end{pmatrix}
,
\begin{pmatrix}
1&0&0\\
0&0&0\\
0&0&-1
\end{pmatrix}$$
are equivalent to one of the following
$$\begin{pmatrix}0&0&0\\0&0&1\\0&-1&0\end{pmatrix},\begin{pmatrix}0&0&-1\\0&0&0\\1&0&0\end{pmatrix},\begin{pmatrix}0&1&0\\-1&0&\\0&0&0\end{pmatrix}$$
It is a relatively simple but time consuming problem to construct a similarity transformation. I have nine real independent variables and 9 equations to check for each matrix. I want to know if there is any process or software that could help me do this calculation?
|
Two matrices are similar if their traces are equal.
|
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|
Is this a proper use of induction? ($(n^2+5)n$ is divisible by 6) Just want to get input on my use of induction in this problem:
Question. Use mathematical induction to prove that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
Proof by mathematical induction.
(1) show base case ($n=1$) is true:
$$
((1)^2 + 5) (1) = 6
$$
$6$ is indeed divisible by $6$, so base case ($n=1$) is true
(2a) Assume case $n$ is true: $(n^2+5)n$ is divisible by $6$.
(2b) Show that case $n$ $\implies$ case $(n+1)$
$$
\begin{align*}
((n+1)^2+5)(n+1)
&\rightarrow
((n^2+2n+1)+5)(n+1)
\\ &\rightarrow
[(n^2+5)+(2n+1)](n+1)
\\ &\rightarrow
(n^2+5)n + (n^2+5)+(2n+1)n+ (2n+1)
\\ &\rightarrow
(n^2+5)n + [(n^2+5)+(2n^2+n)+ (2n+1)]
\\ &\rightarrow
(n^2+5)n + [(3n^2+3n)+6]
\end{align*}
$$
Now we can see case $(n+1)$ $= (n^2+5)n + (3n^2+3n)+6$.
We know $6$ is divisible by $6$ and we are assuming $(n^2+5)n$ is divisible by $6$ already, so all we need to do is show $(3n^2+3n)$ is divisible by $6$:
Letting $n=1$ for $(3n^2+3n)$ gives: $(3(1)^2+3(1)) = 6$
Thus, it has been demonstrated that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
I'm not sure if letting $n=1$ for that last expression is enough to prove it is divisible by $6$
|
It is definitively not sufficient, since "letting $n=1$" means "you don't know what to do in every other cases of values of $n$". Since saying that $6$ divides $3n^2 + 3n$ is equivalent to saying that $2$ divides $n^2 + n$, you only need to show the latter. Now why should $2$ divide $n^2+n = n(n+1)$, two consecutive integers? (You could also prove this by induction, but that would be a little useless, would it.)
Hope that helps,
|
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|
Drawing $z^4 +16 = 0$ I need to draw $z^4 +16 = 0$ on the complex numbers plane.
By solving $z^4 +16 = 0$ I get:
$z = 2 (-1)^{3/4}$ or $z = -2 (-1)^{3/4}$ or $z = -2 (-1)^{1/4}$ or $z = 2 (-1)^{1/4}$
However, the suggested solution by my teacher is:
Where you can see that the solutions that he found are:
$z = \sqrt{2} + i\sqrt{2}$ or $z = \sqrt{2} - i\sqrt{2}$ or $z = - \sqrt{2} + i\sqrt{2}$ or $z = - \sqrt{2} - i\sqrt{2}$
How did my teacher find these solutions?
Am I missing something here? Maybe I don't need to solve the equation in order to draw it on the plane of complex numbers?
|
The solutions that your teacher found are the $4$ solutions to $z^4=16$. Here we will find these roots algebraically.
As you point out, the solutions should be of the form $z=2\cdot (-1)^{1/4}$, where "$(-1)^{1/4}$" stands for any fourth root of $-1$. Thus, we need to find the solutions to
$$z^4=-1,$$
or equivalently, the solutions to $z^4+1=0$. Over the complex numbers, $-1$ is a square, namely $i^2=-1$, so
$$z^4+1 = z^4-(-1) = z^4-i^2$$
and therefore $z^4+1$ factors as $(z^2-i)(z^2+i)$. In order to factor this expression completely, you need to find the square roots of $i$ and the square roots of $-i$.
Let us find the square roots of $i$, i.e., we are looking for a complex number $a+bi$ such that $(a+bi)^2=i$. Since $(a+bi)^2=a^2-b^2+2abi$, we conclude that $a^2-b^2=0$ (so $a=\pm b$) and $1=2ab=\pm 2b^2$. Thus, $a=b=\pm \frac{\sqrt{2}}{2}$. In other words, the square roots of $i$ are
$$z_1=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i,\quad \text{and} \quad z_2=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i.$$
You can similarly find that the square roots of $-i$ are given by
$$z_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i,\quad \text{and} \quad z_4=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i.$$
Therefore, it follows from our previous discussion that the roots of $z^4+16=0$ are $2z_1$, $2z_2$, $2z_3$ and $2z_4$, which are the solutions your teacher found.
|
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|
Formula for the sum of the squares of numbers We have the well-known formula
$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$
If the difference between the closest numbers is smaller, we obtain, for example
$$\frac{n \times (n + 0.1) (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots + n^2 .$$
It is easy to check. Now if the difference between the closest numbers becomes smallest possible, we will obtain
$$ \frac{n \cdot (n + 0.0..1) \cdot (2 n + 0.0..1)}{6 \cdot 0.0..1} = 0.0..1^2 + 0.0..2^2 + \cdots + n^2$$
So can conclude that
$$\frac{2n ^ 3}{6} = \frac{n ^ 3}{3} = \frac{0.0..1 ^ 2 + 0.0..2 ^ 2 + \cdots + n ^ 2}{0.0..1}.$$
Is this conclusion correct?
|
Suppose we want to calculate the sum of squares with successive differences $\epsilon$ from $0$ to some fixed $n$ (we require $\frac{n}{\epsilon}\in\mathbb{N}$ for this particular calculation, however for the general formulation of integrals and Riemann sums, this is not required), that is
$$S_\epsilon = \sum_{i=0}^{\frac{n}{\epsilon}}(i\epsilon)^2$$
letting $m = \frac{n}{\epsilon}$ this sum is equivalent to
$$\sum_{i=0}^{m}i^{2}\left(\frac{n}{m}\right)^2$$
which we can write as
$$=\left(\frac{n^2}{m^2}\right)\sum_{i=0}^{m}i^2=\left(\frac{n^2}{m^2}\right)\frac{m(m+1)(2m+1)}{6} $$
$$= \frac{\left(\frac{n}{m}\right)m\left(\frac{n}{m}\right)(m+1)\left(\frac{n}{m}\right)(2m+1)}{6\left(\frac{n}{m}\right)}=\frac{n(n+\epsilon)(2n+\epsilon)}{6\epsilon}$$
taking the limit $\epsilon\rightarrow 0$ this is equivalent to $m\rightarrow\infty$ and $S_{\epsilon\rightarrow 0}$ is easily seem to be divergent to $+\infty$. However, $S_{\epsilon\rightarrow 0}\cdot \epsilon$ is convergent (which easily evaluated by simply substituting $\epsilon = 0$, which we can do by the continuity of the expression) and is of certain interest. In particular, we can write
$$S = S_{\epsilon\rightarrow 0}\cdot\epsilon=\lim_{m\rightarrow\infty}\ \sum_{i=0}^{m}\left(i\frac{n}{m}\right)^2\left(\frac{n}{m}\right)$$
we recognize this as the Riemann Sum which defines the integral
$$\lim_{m\rightarrow\infty}\ \sum_{i=0}^{m}\left[f\left(x_0 + i\frac{n}{m}\right)\frac{n}{m}\right]=\int_{x_0}^{x_0 + n}f(x)\ dx$$
for $f(x) = x^2$ and $x_0 = 0$. (In particular this is the left Riemann sum). By the Fundamental Theorem of Calculus,
$$\int_{0}^{n}x^2\ dx = \frac{x^3}{3}\bigg|_{0}^{n} = \frac{n^3}{3}$$
which is exactly the quantity you cite.
|
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|
Normal and Lower triangular matrix implies diagonal matrix A lower triangular complex matrix $A$ satisfies $AA^*=A^*A$.
I would like to show that $A$ is diagonal. I know there exists a unitary matrix $P$ such that $PAP^*$ is diagonal. But I don't know how to show $A$ itself is diagonal.
|
We can show it by induction on the dimension. For $n=2$, let $A=\begin{pmatrix}a&0\\\ b&c\end{pmatrix}$ such a matrix. Then
\begin{align*}A^* A-AA^* &=\begin{pmatrix}\bar a&\bar b\\ 0&\bar c\end{pmatrix}\begin{pmatrix}a&0\\ b&c\end{pmatrix}-\begin{pmatrix}a&0\\ b&c\end{pmatrix}\begin{pmatrix}\bar a&\bar b\\ 0&\bar c\end{pmatrix}\\
&=\begin{pmatrix}|a|^2+|b|^2-|a|^2&\bar bc-a\bar b\\ \bar cb-b\bar a&|c|^2-(|b|^2+|c|^2)\end{pmatrix}\\
&=\begin{pmatrix}|b|^2&\bar bc-a\bar b\\ \bar cb-b\bar a&-|b|^2\end{pmatrix},
\end{align*}
so $b=0$ and $A$ is diagonal. If the result is true for $n\geq 2$, let $A=\begin{pmatrix}T&0\\\ v&a\end{pmatrix}$, where $T$ is a $n\times n$ triangular matrix, $v$ a $1\times n$ matrix and $a$ a complex number. Since $A^*A=AA^*$, we have
\begin{align*}
\begin{pmatrix}0&0\\ 0&0\end{pmatrix}&=\begin{pmatrix}T^* &v^* \\ 0&\bar a\end{pmatrix}\begin{pmatrix}T&0\\ v&a\end{pmatrix}-\begin{pmatrix}T&0\\ v&a\end{pmatrix}\begin{pmatrix}T^* &v^* \\ 0&\bar a\end{pmatrix}\\
&=\begin{pmatrix}T^* T+v^* v &v^* a \\ \bar a v&|a|^2\end{pmatrix}-
\begin{pmatrix}TT^* &Tv^* \\ vT^* &|v|^2+|a|^2\end{pmatrix}\\
&=\begin{pmatrix}T^* T-TT^* +v^* v &v^* a-Tv^* \\ \bar av-v T^* &-|v|^2,\end{pmatrix}
\end{align*}
hence $v=0$ and $T$ is normal. Since $T$ is lower triangular, $T$ is normal and by induction hypothesis $T$ is diagonal. We conclude that $A$ is diagonal.
|
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|
How can I evaluate $\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx$? How can I solve this integral: $$\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx.$$ Can I solve this problem using the Laplace transform? How can I do this?
|
Writing $2x^2-1$ as $2x^2 + 2 - 3$, the integral simplifies to
$$\begin{align*}
\int_{-\infty}^{\infty}\exp(-x^2)\frac{2x^2-1}{1+x^2}\mathrm dx
&= 2\int_{-\infty}^{\infty}\exp(-x^2)\mathrm dx
-3 \int_{-\infty}^{\infty}\exp(-x^2)\frac{1}{1+x^2}\mathrm dx\\
&= 2\sqrt{\pi}
- \frac{3}{2\pi}\int_{-\infty}^{\infty}\sqrt{\pi}\exp(-\omega^2/4)
\pi\exp(-|\omega|)\mathrm d\omega\\
&= 2\sqrt{\pi}
- 3\sqrt{\pi}\int_{0}^{\infty}\exp(-\omega^2/4)
\exp(-\omega)\mathrm d\omega\\
&= 2\sqrt{\pi}
- 6\pi e \int_{0}^{\infty}\frac{1}{\sqrt{2}\sqrt{2\pi}}
\exp\left(-\frac{(\omega + 2)^2}{2\cdot(\sqrt{2})^2}\right)
\mathrm d\omega\\
\end{align*}$$
where in the second step, we have used a well-known result
(see e.g. the answer by bgins and the links therein) on the first
integral, and applied the inner-product form of Parseval's theorem
to convert the second integral to the integral
of the product of the Fourier transforms $\sqrt{\pi}\exp(-\omega^2/4)$
and $\pi\exp(-|\omega|)$ of $\exp(-x^2)$ and
$(1+x^2)^{-1}$ respectively, while the last step follows upon
completing the square in the exponent and writing the
integrand as the probability density function of a
normal random variable with mean $-2$ and variance $2$.
Hence we have that
$$\begin{align*}
\int_{-\infty}^{\infty}\exp(-x^2)\frac{2x^2-1}{1+x^2}\mathrm dx
&= 2\sqrt{\pi} -6\pi e \Phi(-2/\sqrt{2})\\
&= 2\sqrt{\pi} -6\pi e Q(\sqrt{2})
\end{align*}$$
where $\Phi(\cdot)$ is the cumulative standard normal
distribution function and $Q(x) = 1-\Phi(x)$ is its complement.
Since $\text{erfc}(x) = 2Q(x\sqrt{2})$, the value can also be
expressed
as $2\sqrt{\pi} - 3e\pi \text{erfc}(1)$ as in the answers by Bruno and bgins.
|
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|
Taylor polynomial of $\int_{0}^{x}\sin(t^2)dt$ I just learned about Taylor polynomials, and I am trying to estimate $\int_{0}^{1/2}\sin(x^2)dx$ using the 3rd degree Taylor polynomial of $F(x)=\int_{0}^{x}\sin(t^2)dt$ at $0$. I get the following:
$F'(x)=\sin(x^2)$,
$F''(x)=2x\cos(x^2)$,
$F^{(3)}(x)=2\cos(x^2)-4x^2\sin(x^2)$.
And so the estimate is:
$$\frac{F(0)}{0!}\cdot 1+\frac{F'(0)}{1!}\cdot \frac{1}{2}+\frac{F''(0)}{2!}\cdot \frac{1}{4}+\frac{F^{(3)}(0)}{3!}\cdot \frac{1}{8} = 0.$$
However this seems very weird (shouldn't it be closer to the actual value than $0$?), is my estimation correct?
|
We have
$$f(x)=f(0)+xf'(0)+\frac{x^2}2f''(0)+\frac{x^3}{3!}f^{(3)}(0)+\int_0^x\frac{(x-t)^3}{3!}f^{(4)}(t)dt,$$
so
\begin{align*}
\sin x^2&=x^2+\int_0^x\frac{(x-t)^3}{3!}(8\cos(t^2)(1-5t)+16t^4\sin(t^2))dt\\
&=x^2+\frac 43\int_0^x(x-t)^3(\cos(t^2)(1-5t)+2t^4\sin(t^2))dt,
\end{align*}
and for $x\geq 0$
\begin{align*}
\left|\int_0^x\sin t^2dt-\frac{x^3}3\right|&=\left|\int_0^x\sin t^2dt-\int_0^xt^2dt\right|\\
&\leq \frac 43\left|\int_0^x\int_0^t(t-s)^3(\cos(s^2)(1-5s)+2s^4\sin(s^2))dsdt\right|\\
&\leq \frac 43\int_0^x\int_0^t(t-s)^3(1+5s+2s^4)dsdt\\
&\leq \frac 43\int_0^x\int_0^tt^3(1+5s+2s^4)dsdt\\
&=\frac 13\left(\left[t^4\int_0^t(1+5s+2s^4)ds\right]_0^x-\int_0^xt^4(1+5t+2t^4)dt\right)\\
&=\frac 13x^4\left(x+\frac 52x^2+\frac 25x^5\right)-\frac 13\left(\frac{x^5}5+\frac 56x^6+\frac 29x^9\right)\\
&=\frac{x^5}3\left(\frac45+\frac 53x+\frac 25x^4-\frac 29x^4\right).
\end{align*}
|
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|
Trigonometric equality $\cos x + \cos 3x - 1 - \cos 2x = 0$ In my text book I have this equation:
\begin{equation}
\cos x + \cos 3x - 1 - \cos 2x = 0
\end{equation}
I tried to solve it for $x$, but I didn't succeed.
This is what I tried:
\begin{align}
\cos x + \cos 3x - 1 - \cos 2x &= 0 \\
2\cos 2x \cdot \cos x - 1 - \cos 2x &= 0 \\
\cos 2x \cdot (2\cos x - 1) &= 1
\end{align}
So I clearly didn't choose the right path, since this will only be useful if I become something like $a \cdot b = 0$.
All tips will be greatly appreciated.
Solution (Addition to the accepted answer):
The problem was in writing $\cos 3x$ in therms of $cos x$. Anon pointed out that it was equal to $4\cos^3 x - 3\cos x$ but I had to work may way thru it to actually prove that. So I write it down here, maybe it's of use to anybody else.
\begin{align}
\cos 3x &= \cos(2x + x)\\
&= \cos(2x)\cdot \cos x - \sin(2x)\sin x\\
&= (\cos^2 x - \sin^2 x) \cdot \cos x - 2\sin^2 x \cdot \cos^2 x\\
&= \cos^3 x - \sin^2x\cdot \cos x - 2\sin^2x\cdot \cos^2 x\\
&= \cos^3 x - (1 - \cos^2 x)\cos x - 2(1-\cos^2 x)\cos x\\
&= cos^3 x - \cos x + \cos^3 x - 2\cos x + 2\cos^3 x\\
&= 4\cos^3 x - 3\cos x
\end{align}
EDIT: Apparently you can write all of the formulas $\cos(n\cdot x)$ with $n \in \{1, 2, 3, …\}$ in terms of $\cos x$. Why didn't my teacher tell me that! I don't have time to proof it myself now, but I'll definitely adapt my answer tomorrow (or any time soon)!
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To solve an "equals zero" equation involving polynomials or trigonometric functions, it's generally prudent to keep the $0$ on one side and only work with the other side. Here you could use the double and triple angle formulas (which can be deduced from the addition formulas if need be) given by
$$\cos2x=2\cos^2x-1,\qquad \cos3x=4\cos^3x-3\cos x.$$
Substitute these expressions in and your equation will read $P(\cos x)=0$ for a polynomial $P$. Solve for the roots of the polynomial and then take inverse cosines where allowed. Alternatively, if you're at all familiar with complex analysis you could forego the use of angle formulas and go right to a polynomial by using $\cos\theta=(e^{i\theta}+e^{-i\theta})/2$ and then solving for $e^{i\theta}$ and then $\theta$.
|
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|
$(\cos \alpha, \sin \alpha)$ - possible value pairs We introduced the complex numbers as elements of $ \mathrm{Mat}(2\times 2, \mathbb{R})$ with
$$
\mathbb{C} \ni x =
\left(\begin{array}{cc}
a & -b \\
b & a \\
\end{array}\right) =
\frac{1}{\sqrt{a^2+b^2}}
\left(\begin{array}{cc}
\frac{a}{\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}} \\
\frac{b}{\sqrt{a^2+b^2}} & \frac{a}{\sqrt{a^2+b^2}} \\
\end{array}\right)
$$
Then we concluded that $0 \leq \frac{a}{\sqrt{a^2+b^2}} \leq 1$ and $0 \leq \frac{b}{\sqrt{a^2+b^2}} \leq 1$ and therefore we could find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$. Then we can write the matrix with $\cos$ and $\sin$ and can write it as the Euler form as well. So far so good.
My question ist about the following:
Why is it that we cand find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$ for every possible value combination of $a$ and $b$?
Can't be there a combination of $a$ and $b$ where we can't find one and the same angle $\alpha$ so that the identites are true?
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I suppose you could simply plot the point $(a,b)$ in the $x$-$y$ plane and let $\alpha$ be the angle formed by the positive $x$-axis and the ray joining the origin with $(a,b)$ (measured counterclockwise starting from the positive $x$-axis to the ray).
Then the distance from $(a,b)$ to the origin is $\sqrt{a^2+b^2}$ and $\cos\alpha ={a\over\sqrt{a^2+b^2}}$ and $\sin\alpha ={b\over\sqrt{a^2+b^2}}$, essentially by definition. (Use similar triangles and the unit circle definition of the trigonometric functions if you like.)
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|
Prove that $x_1^2+x_2^2+x_3^2=1$ yields $ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $ Prove this inequality, if $x_1^2+x_2^2+x_3^2=1$:
$$ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $$
So far I got to $x_1^4+x_2^4+x_3^4\ge\frac{1}3$ by using QM-AM for $(2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x_1^2)$, but to be honest I'm not sure if that's helpful at all.
(You might want to "rename" those to $x,y,z$ to make writing easier):
$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt3}4$$
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We assume that $x_i\geq 0$ and let $\theta_i\in \left(0,\frac{\pi}2\right)$ sucht that $x_i=\tan\frac{\theta_i}2$. We have $\sin(\theta_i)=\frac{2x_i}{1+x_i^2}$ and since $\sin$ in concave on $\left(0,\frac{\pi}2\right)$, we have
$$\sum_{i=1}^3\frac{x_i}{1+x_i^2}=\frac 32\sum_{i=1}^3\frac 13\sin(\theta_i)\leq \frac 32\sin\frac{\theta_1+\theta_2+\theta_3}3.$$
We have, using the convextiy of $x\mapsto \tan^2 x$:
$$\frac 13=\frac 13\sum_{i=1}^3\tan^2\frac{\theta_i}2\geq \tan^2\frac{\theta_1+\theta_2+\theta_3}6,$$
so $\tan\frac{\theta_1+\theta_2+\theta_3}6\leq \frac 1{\sqrt 3}$ and $\frac{\theta_1+\theta_2+\theta_3}6\leq \frac{\pi}6$. Finally
$$\sum_{i=1}^3\frac{x_i}{1+x_i^2}\leq \frac 32\sin \frac{\pi}3=\frac{3\sqrt 3}4,$$
with equality if and only if $(x_1,x_2,x_3)=\left(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3}\right)$.
|
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|
System $a+b+c=4$, $a^2+b^2+c^2=8$. find all possible values for $c$. $$a+b+c=4$$$$a^2+b^2+c^2=8$$
I'm not sure if my solution is good, since I don't have answers for this problem. Any directions, comments and/or corrections would be appreciated.
It's obvious that $\{a,b,c\}\in[-\sqrt8,\sqrt8]$. Since two irrational numbers always give irrational number when added, if we assume that one of $a,b $ or $c$ is irrational, for example $a$, then one more has to be irrational, for example $b$, such that $a=-b$. That leaves $c=4$ in order to satisfy the first equation, but that makes the second one incorrect. That's why all of $a,b,c$ have to be rational numbers. $(*)$
I squared the first equation and got $ab+bc+ca=4$. Then, since $a=4-(b+c)$, I got quadratic equation $b^2+(c4)b+(c-2)^2=0$. Its' discriminant has to be positive and perfect square to satisfy $(*)$.
$$ D= -c(3c-8) $$
From this, we see that $c$ has to be between $0$ and $8/3$ in order to satisfy definition of square root. Specially, for $c=0$ we get $D=0$ and solution for equation $b=\frac{-c+4}{2}=2$. Since the system is symmetric, we also get $c=2$. Similar, for $c=8/3$ we get $b=2$.
In order for $D$ to be perfect square, one of the following has to be true:
$$ -c=3c-8 $$
$$ c=n^2 \land 3c-8=1 $$
$$ 3c-8=n^2 \land c=1 $$
However, only the first one is possible, so $c=2$, and for that we get $b=\frac{(-2+4\pm2)}{2} \Rightarrow b=2 \lor b=0$.
So, all possible values for $c$ are $c\in\{0, 2, \frac{8}3\}$.
EDIT: For $a=b$, there is one more solution: $c=2/3$. Why couldn't I find it with method described above?
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We will show that any $c$ in the closed interval $[0,8/3]$ is achievable, and nothing else is. Let $c$ be any real number, and suppose $a$ and $b$ are real numbers such that $(a,b,c)$ satisfies our two equations. Note that in general
$$2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0. \qquad\qquad(\ast)$$
Put $a^2+b^2=8-c^2$ and $a+b=4-c$. Thus from $(\ast)$,
$$2(8-c^2)-(4-c)^2 =(a-b)^2\ge 0. \qquad\qquad(\ast\ast)$$
So we must have $8c-3c^2\ge 0$. This is only true for $0\le c \le 8/3$.
Now we show that any $c$ in the interval $[0,8/3]$ is achievable.
We want to make $(a-b)^2=8c-3c^2$. If $c$ is in $[0,8/3]$, then $8c-3c^2\ge 0$, so we can take the square root, and obtain
$$a-b=\pm\sqrt{8c-3c^2}.$$
We can now solve the system $a-b=\pm\sqrt{8c-3c^2}$, $a+b=4$ to find the values of $a$ and $b$. We might as well give the solutions $(a,b,c)$ explicitly. They are
$$a=2\pm \frac{1}{2}\sqrt{8t-t^2}, \qquad b=2\mp \frac{1}{2}\sqrt{8t-t^2},\qquad c=t,$$
where $t$ is a parameter that ranges over the interval $0\le t\le 8/3$.
There is something mildly ugly in the above general solution, since it breaks symmetry. That can be fixed.
Remark: It is maybe interesting to ask what are the possible values of $c$ under the conditions $c\ge a$, $c\ge b$. Of course we still must have $c\le 8/3$.
For given $a$, $b$, and $c$, look at the monic cubic polynomial $P(x)$ whose roots are $a$, $b$, and $c$. Then
$$P(x)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc.$$
You found that $ab+bc+ca=4$. So our polynomial has shape
$$P(x)=x^3-4x^2+4x-p,$$
where $p$ is the product of the solutions. From our previous work we know the solutions are $\ge 0$, so $p\ge 0$.
The derivative of $P(x)$ is $3x^2-8x+4$, which vanishes at $x=2/3$ and $x=2$. So $P(x)=0$ has a solution in the interval $[0, 2/3]$, a solution in the interval $[2/3,2]$, and a solution in the interval $[2,8/3)$ (there can be a double root at $2/3$ or at $2$). Thus at least one of $a$, $b$, $c$ is $\ge 2$.
Let $c$ be a maximal root. We can have $c=2$ (let $a=0$, $b=2$), and we can have $c=8/3$ ($a=b=2/3$). By continuity, or calculation, one can then see that for any $c$ in the interval $[2,8/3]$, there are $a$ and $b$ such that $(a,b,c)$ is a solution of our system, and $c\ge a$, $c\ge b$.
|
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"url": "https://math.stackexchange.com/questions/106530",
"timestamp": "2023-03-29T00:00:00",
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|
Count the number of divisions of a set recursively I'm trying to understand the reasoning behind the answer to the following question from an old exam.
Given a set $S_n = \{ 1, 2, 3,\dots n\}$ find how many divisions, $K_n$, there are of the set using only subsets of 1 or 2 members. For example, $\{ \{1,4\}, 6 , 2, \{5,3\} \}$ is one possible division of $S_6$.
Find a recursive definition of $K_n$.
So the first thing I did was to calculate a few terms combinatorially:
$K_1 = \binom{1}{1} = 1$
$K_2 = \binom{2}{2} + \binom{2}{2} = 2$
$K_3 = \binom{3}{3} + \binom{3}{2} = 4$
$K_4 = \binom{4}{4} + \binom{4}{2} + \frac{\binom{4}{2}}{2!} = 10$
$K_5 = \binom{5}{5} + \binom{5}{2} + \frac{\binom{5}{2}\binom{3}{2}}{2!} = 26$
$K_6 = \binom{6}{6} + \binom{6}{2} + \frac{\binom{6}{2}\binom{4}{2}}{2!} + \frac{\binom{6}{2}\binom{4}{2}}{3!} = 76$
And so on. The recursive definition given was $K_n = K_{n-1} + (n-1)\cdot K_{n-2}$
I understand the first half of the definition. You take the number $n$ and add it as a set of 1 to each of the sets in $K_{n-1}$ giving you $K_{n-1}$ new sets in $K_n$. However I'm completely at a loss as to the reasoning behind the second half of the definition.
|
Let's take a stab at the recurrence... define the exponential generating function $\hat{K}(z) = \sum_{n \ge 0} K_n \frac{z^n}{n!}$ (to compensate for the $n - 1$ factor), and write:
$$
K_{n + 1} = K_n + n K_{n - 1} \qquad K_0 = K_1 = 1
$$
Using properties of exponential generating functions:
$$
\hat{K}'(z) = \hat{K}(z) + z \frac{d}{dz} \int \hat{K}(z) dz
$$
Thus:
$$
\int_1^\hat{K} \frac{d \hat{K}'}{\hat{K}} = \int_0^z (1 + z) dz
$$
and so:
$$
\hat{K}(z) = e^{z + z^2/2}
$$
so that $K_n = 1, 1, 2, 4, 5, \ldots$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/106972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Writing a function $f$ when $x$ and $f(x)$ are known I'm trying to write a function. For each possible input, I know what I want for output. The domain of possible inputs is small:
$$\begin{vmatrix}
x &f(x)\\
0 & 2\\
1 & 0\\
2 & 0\\
3 &0\\
4 &0\\
5 &0\\
6 &1\\
\end{vmatrix}$$
My thought is to start with a function $g(x)$ that transforms $x$ to a value from which I can subtract a function $h(g(x))$ and receive my desired $f(x)$:
$$\begin{vmatrix}
x &g(x)& h(g(x)) &f(x)\\
0 & 7& 5 & 2\\
6 & 6 & 5 & 1\\
5 & 5 & 5& 0\\
4 & 4 & 4 & 0\\
3 & 3 & 3 & 0\\
2 & 2 & 2 & 0\\
1 & 1 & 1& 0
\end{vmatrix}$$
But I'm not sure where to go from there, or even whether I'm heading in the right direction. How should I approach creating this function? Is there a methodical way to go about it, or is it trial and error (and knowledge)? Feel free to suggest modifications to how I'm stating my problem, too. Thanks.
|
You can generalise the problem: suppose you know the value of $f(x)$ for a particular finite set of values of $x$. (Here, you know the value of $f(x)$ when $x=0,1,2,3,4,5,6$.) Then you can find a possible polynomial function $f$ which takes the given values using the following method.
Suppose you know the value of $f(x)$ when $x=x_1, x_2, \dots, x_n$, and that $f(x_j) = a_j$ for $1 \le j \le n$.
Let
$$P_i(x) = \lambda (x-x_1)(x-x_2) \cdots (x-x_{i-1})(x-x_{i+1}) \cdots (x-x_n)$$
Where $\lambda$ is some constant. That is, it's $\lambda$ times the product of all the $(x-x_k)$ terms with $x-x_i$ left out. Then $P_i(x_k) = 0$ whenever $k \ne i$.
We'd like $P_i(x_i) = 1$: then if we let
$$f(x) = a_1 P_1(x) + a_2 P_2(x) + \cdots + a_n P_n(x)$$
then setting $x=x_j$ sends all the $P_i(x)$ terms to zero except $P_j(x)$, leaving you with $f(x_j) = a_jP_j(x_j) = a_j$, which is exactly what we wanted.
Well we can set $\lambda$ to be equal to $1$ divided by what we get by setting $x=x_j$ in the product: this is never zero, so we can definitely divide by it. So we get
$$P_i(x) = \dfrac{(x-x_1)(x-x_2) \dots (x-x_{i-1})(x-x_{i+1}) \dots (x-x_n)}{(x_i-x_1)(x_i-x_2) \dots (x_i-x_{i-1})(x_i-x_{i+1}) \dots (x_i-x_n)}$$
Then $P_i(x_k) = 0$ if $k \ne i$ and $1$ if $k=i$, which is just dandy.
More concisely, if $f$ is to satisfy $f(x_j)=a_j$ for $1 \le j \le n$ then
$$f(x) = \sum_{j=1}^n \left[ a_j \prod_{i=1}_{i \ne j}^n \frac{x-x_i}{x_j-x_i} \right]$$
This method is called Lagrange interpolation.
So in this case, your $x_1, x_2, \dots, x_7$ are the numbers $0, 1, \dots, 6$ and your $a_1, a_2, \dots, a_7$ are, respectively, $2,0,0,0,0,0,1$. Substituting these into the above formula, we get:
$$\begin{align} f(x) &= 2 \times \dfrac{(x-1)(x-2) \dots (x-6)}{(0-1) (0-2) \dots (0-6)} + 0 \times (\text{stuff}) + 1 \times \dfrac{x(x-1) \dots (x-5)}{6(6-1)(6-2) \dots (6-5)}\\
&= 2 \dfrac{(x-1) (x-2) \dots (x-6)}{720} + \dfrac{x(x-1) \dots (x-5)}{720} \\
&= \dfrac{(x-1)(x-2)(x-3)(x-4)(x-5)}{720} \left[ 2(x-6) + x \right] \\
&= \boxed{\dfrac{(x-1)(x-2)(x-3)(x-4)^2(x-5)}{240}} \end{align}$$
You can check easily that this polynomial satisfies the values in your table.
In fact, in this particular case, all the above machinery wasn't necessary. It's plain that $f(x)=0$ when $x=1,2,3,4,5$, and so $x-j$ must divide $f(x)$ for $j=1,2,3,4,5$, and so
$$f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)g(x)$$
for some polynomial $g(x)$. Since we're only worried about $x=0,6$ beyond this, i.e. $2$ values of $x$, it suggests we have $2$ free parameters in $g(x)$ and hence $g(x)=ax+b$ is linear. That is, we have
$$f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)(ax+b)$$
Substituting $x=0$ and $x=6$, respectively, gives
$$\begin{align}2 &= -5! \cdot b \\ 1 &= 5! \cdot (6a+b) \end{align}$$
and solving simultaneously gives $b=-\dfrac{2}{120}$ and $a = \dfrac{3}{720}$, which (after simplification) yields the desired result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $ Is there any way to show that
$$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{1}{a} + \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\left( {\frac{1}{{a - k}} + \frac{1}{{a + k}}} \right)}=\frac{\pi }{{\sin \pi a}}} $$
Where $0 < a = \dfrac{n+1}{m} < 1$
The infinite series is equal to
$$\int\limits_{ - \infty }^\infty {\frac{{{e^{at}}}}{{{e^t} + 1}}dt} $$
To get to the result, I split the integral at $x=0$ and use the convergent series in $(0,\infty)$ and $(-\infty,0)$ respectively:
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{ - \left( {k + 1} \right)t}}} $$
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{kt}}} $$
Since $0 < a < 1$
$$\eqalign{
& \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to - \infty } \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} = \frac{1}{{k + a}} \cr
& \mathop {\lim }\limits_{t \to \infty } \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} = - \frac{1}{{a - \left( {k + 1} \right)}} \cr} $$
A change in the indices will give the desired series.
Although I don't mind direct solutions from tables and other sources, I prefer an elaborated answer.
Here's the solution in terms of $\psi(x)$. By separating even and odd indices we can get
$$\eqalign{
& \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \sum\limits_{k = 0}^\infty {\frac{1}{{a + 2k}}} - \sum\limits_{k = 0}^\infty {\frac{1}{{a + 2k + 1}}} \cr
& \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \sum\limits_{k = 0}^\infty {\frac{1}{{a - 2k}}} - \sum\limits_{k = 0}^\infty {\frac{1}{{a - 2k - 1}}} \cr} $$
which gives
$$\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)$$
$$\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) + \frac{1}{a}$$
Then
$$\eqalign{
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} + \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} - \frac{1}{a} = \cr
& = \left\{ {\frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)} \right\} - \left\{ {\frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right)} \right\} \cr} $$
But using the reflection formula one has
$$\eqalign{
& \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi a}}{2} \cr
& \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi \left( {a + 1} \right)}}{2} = - \frac{\pi }{2}\tan \frac{{\pi a}}{2} \cr} $$
So the series become
$$\eqalign{
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \frac{\pi }{2}\left\{ {\cot \frac{{\pi a}}{2} + \tan \frac{{\pi a}}{2}} \right\} \cr
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \pi \csc \pi a \cr} $$
The last being an application of a trigonometric identity.
|
Set $x=0$ in the Fourier series of $\cos(ax)$:
$$\cos(ax)=\frac{2a\sin(\pi a)}{\pi}\left[\frac1{2a^2}+\sum_{k=1}^\infty\frac{(-1)^k\cos(kx)}{a^2-k^2}\right],\quad a\notin\mathbb{Z}$$
we get
\begin{align}
\frac{\pi}{\sin(\pi a)}&=\frac1a+\sum_{k=1}^\infty\frac{2a(-1)^k}{a^2-k^2}\\
&=\frac1a+\sum_{k=1}^\infty\frac{(-1)^k}{a-k}+\sum_{k=1}^\infty\frac{(-1)^k}{a+k}\\
&=\frac1a+\sum_{k=-\infty}^{-1}\frac{(-1)^k}{a+k}+\sum_{k=1}^\infty\frac{(-1)^k}{a+k}\\
&=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{a+k}
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find smallest number when divided by $2,3,4,5,6,7,8,9,10$ leaves $1,2,3,4,5,6,7,8,9$ remainder Find smallest number when divided by $2,3,4,5,6,7,8,9,10$ leaves $1,2,3,4,5,6,7,8,9$ remainder.How to go about solving this problem??
|
The smallest such number is 2519. The next number can obtain by the equation 2519x+x-1. Where x can replaced by any whole number
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/111595",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
An integral evaluation I tried my luck with Wolfram Alpha, with $p \in \mathbb{R}$
$$\int_{-\infty}^{\infty} \frac{x^p}{1+x^2} dx = \frac{1}{2} \pi ((-1)^p+1) \sec(\frac{\pi p}{2})$$ for $-1<p<1$, and doesn't exist for other $p$.
I wonder how to integrate it myself? Especially given that $(-1)^p$ may be a non-real complex number. Thanks in advance!
PS: Does Mathematica or some other (free) CAS give the process of deriving the result?
|
Split integration over $\mathbb{R}$ into integration over $\mathbb{R}_{\geqslant 0}$ and $\mathbb{R}_{<0}$ and perform a change of variables $x \mapsto -x$ in the latter one:
$$
\int_{-\infty}^\infty \frac{x^p}{1+x^2} \mathrm{d} x = \left(1 + (-1)^p \right) \int_0^\infty \frac{x^p}{1+x^2} \mathrm{d} x
$$
Now the idea is to reduce the integral to the Euler's beta integral. To this end perform substitution $x^2 = \frac{t}{1-t}$ so that $t$ ranges from 0 to 1.
$$
x \mathrm{d} x = \frac{1}{2} \cdot \frac{\mathrm{d} t}{(1-t)^2}
$$
Thus
$$
\begin{eqnarray}
2 \int_0^\infty \frac{x^p}{1+x^2} \mathrm{d} x &=& \int_0^1 \left( \frac{t}{1-t} \right)^{(p-1)/2} \frac{1}{1+ \frac{t}{1-t}} \frac{\mathrm{d} t}{(1-t)^2} \\
&=& \int_0^1 t^{(p+1)/2 - 1} \left( 1-t \right)^{-1-(p-1)/2} \mathrm{d} t \\
&=& \operatorname{Beta}\left( \frac{p+1}{2}, \frac{1-p}{2} \right) = \frac{ \Gamma\left( \frac{1-p}{2} \right) \Gamma\left( \frac{p+1}{2}\right)}{\Gamma\left( \frac{p+1}{2} + \frac{1-p}{2} \right)} \\
&=& \frac{\pi}{ \sin\left( \pi \frac{p+1}{2} \right)} = \pi \sec\left( \frac{\pi p}{2} \right)
\end{eqnarray}
$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate using Trigonometric Substitutions Evaluate the integral using trigonometric substitutions.
$$\int{ x\over \sqrt{3-2x-x^2}} \,dx$$
I am familiar with using the right triangle diagram and theta, but I do not know which terms would go on the hypotenuse and sides in this case. If you can determine which numbers or $x$-values go on the hypotenuse, adjacent, and opposite sides, I can figure out the rest, although your final answer would help me check mine. Thanks!
|
$\int \frac{x}{\sqrt{4-(x+1)^2}}dx = \int \frac{2\sin\theta-1}{\sqrt{4-4\sin^2\theta}}(2\cos\theta)d\theta$
(using the substitution $x+1=2\sin\theta$)
$=\int\frac{2\sin\theta-1}{2\cos\theta}2\cos\theta d\theta$
$= \int (2\sin\theta-1) d\theta$
$=-2\cos\theta-\theta +C$
$=-2\left(\frac{\sqrt{3-2x-x^2}}{2}\right) - \sin^{-1}\left(\frac{x+1}{2}\right)+C$
$=-\sqrt{3-2x-x^2}- \sin^{-1}\left(\frac{x+1}{2}\right)+C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/111899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Puzzle: The number of quadratic equations which are unchanged by squaring their roots is My friend asked me this puzzle:
The number of quadratic equations which are unchanged by squaring their roots is
My answer is: 3
$x^2-(\alpha+\beta)x +\alpha\beta = 0$
where $\alpha$ and $\beta$ be the roots.
case 1:
$\alpha$ = 0
$\beta$ = 0
case 2: $\alpha$ = 1
$\beta$ =1
case 3:$\alpha$ = 0
$\beta$ = 1
But my friend says answer is 4. How come?
|
How about just writing: $(x-\alpha)(x-\beta)=(x-\alpha^2)(x-\beta^2)$, so either $\alpha = \alpha^2$ and $\beta=\beta^2$ or $\alpha=\beta^2$ and $\beta=\alpha^2$.
If $\alpha=\alpha^2$, then $\alpha=0\text{ or }1$. Similarly, $\beta=0\text{ or } 1$. So there are three such equations (because $(\alpha,\beta)=(0,1)$ and $(\alpha,\beta)=(1,0)$ yield the same quadratic.)
On the other hand, if $\alpha = \beta^2$ and $\beta=\alpha^2$, then $\alpha^4=\alpha$. If $\alpha=0\text{ or } 1$, then $\beta=\alpha$ and we already covered those quadratics above. So assume $\alpha\neq 0,1$. Then $0=\alpha^2+\alpha +1 = \frac{\alpha^4-\alpha}{\alpha^2-\alpha}$. But then, $\beta=\alpha^2$ is also a root of $x^2+x+1$, so that's one last quadratic. ($\beta = \alpha^2 = -(\alpha+1)$ so $\beta^2 + \beta + 1 = \alpha^2+2\alpha+1 - (\alpha+1) + 1 = \alpha^2 + \alpha +1 = 0$.)
So you get four quadratics, $x^2$, $x^2-x$, $x^2-2x+1$, and $x^2+x+1$.
So the total is $4$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the upper bound for this series I came across this series in a book and the author tells the upper bound and it checks out under the conditions but i could n't find the connection or the reason if u may, that had i not been presented with this info, that i could have used to find the upper bound myself.
For $s>1$ the sum of $2^{p}-1$ terms of the series is less than
$\dfrac {1} {1^{s-1}}+\dfrac {1} {2^{s-1}}+\dfrac {1} {4^{s-1}}+\dfrac {1} {8^{s-1}}+\ldots +\dfrac {1} {2^{\left( p-1\right) \left( s-1\right) }} < \dfrac {1} {1-2^{1-s}}$
Any hints or clues would be much appreciated.
|
Use the fact that if $|x|<1$, then $\frac{1}{1-x}$ has the power series expansion
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots +x^n+\cdots .$$
Or, in less fancy language, if $|x|<1$, then the infinite geometric series $1+x+x^2+\cdots+x^n+\cdots$ has sum $\frac{1}{1-x}$. In your particular example, $x=\frac{1}{2^{s-1}}$. Since $s>1$, we have $|x|<1$.
More explicitly, for this value of $x$, the left-hand side is just
$$1+x+x^2+\cdots+x^{p-1}.$$
This finite geometric series has sum
$$1+x+x^2+\cdots+x^{p-1}=\frac{1-x^p}{1-x}=\frac{1}{1-x}-\frac{x^p}{1-x}.$$
This says that the sum on the left-hand side is less than $\frac{1}{1-x}$. It even says by how much it is less, namely $\frac{x^p}{1-x}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What is the least value of the function $y= (x-2) (x-4)^2 (x-6) + 6$? What is the least value of function:
$$y= (x-2) (x-4)^2 (x-6) + 6$$
For real values of $x$ ?
Does $\frac{dy}{dx} = 0$, give the value of $x$ which will give least value of $y$?
Thanks in advance.
|
Note that $(x-2)(x-6)=x^2-8x+12$ and $(x-4)^2=x^2-8x+16$. This suggests the symmetrizing substitution $w=x^2-8x+14$. Thus
$$y=(w-2)(w+2)+6=w^2+2.$$
We want to minimize the absolute value of $w$. But $w=(x-4)^2-2$. So $w$ has minimum absolute value $0$, reached when $(x-4)^2=2$, that is, when $x=4\pm\sqrt{2}$. The minimum value of $y$ is $2$.
Remark: The curve $y=(x-2)(x-4)^2(x-6)+6$ is very special, with a beautiful double symmetry. It seems likely that the method used above was the intended one. The same idea works for $(x-a)(x-b)(x-c)(x-d)+k$, where $a+d=b+c$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Are there any $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square? Are there any positive $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?
I tried to simplify
\begin{align*}
n^4+n^3+n^2+n+1 &= n^2(n^2+1)+n(n^2+1)+1\\
&= (n^2+n)(n^2+1)+1 \\
&= n(n+1)(n^2+1)+1
\end{align*}
Then I assumed that the above expression is a square; then
$$ n(n+1)(n^2+1)+1 = k^2$$
$$
\begin{align*}
n(n+1)(n^2+1) &= (k^2-1) \\
&= (k+1)(k-1)
\end{align*}
$$
Then trying to reason with prime factors, but cannot find a concrete proof yet.
|
Assuming you want positive integers $n$,
I believe we can show that
$$(2n^2 + n)^2 \lt 4(n^4 + n^3 + n^2 + n + 1) \lt (2n^2 + n + 1)^2$$
for $n \gt 3$.
Note: A similar inequality can be given for negative $n$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
For complex $z$, $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0$
If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$
I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:
$$\frac{1 - x}{1 + x}$$
I tried a number of manipulations of the equation but couldn't seem to arrive at any point where I could link the two to show that the real part was = 0.
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Not as elegant as many other approaches, but one I thought I'd try out:
Taking the ratio $ \ \frac{1 - z}{1 + z } \ = \ w \ = \ \zeta + i·\eta \ $ to be equal to some complex number, we may arrange the equation into $ \ [ \ 1 - (\cos \theta + i·\sin \theta) \ ] \ = \ [ \ 1 + (\cos \theta + i·\sin \theta) \ ] · (\zeta + i·\eta) \ \ , \ \ z \neq -1 \ \ , $ writing $ \ z \ $ in "polar form" for $ \ |z| = 1 \ \ . $ Multiplying out the right side and equating real and imaginary parts produces the system of equations
$$ 1 - \cos \theta \ \ = \ \ \zeta·\cos \theta \ + \ \zeta \ - \ \eta·\sin \theta \ \ \ , \ \ \ -\sin \theta \ \ = \ \ \eta·\cos \theta \ + \ \eta \ + \ \zeta·\sin \theta $$
$$ \Rightarrow \ \ (1 + \zeta)·\cos \theta \ - \ \eta·\sin \theta \ \ = \ \ 1 - \zeta \ \ \ , \ \ \ \eta·\cos \theta \ + \ (1 + \zeta)·\sin \theta \ \ = \ \ -\eta \ \ . $$
We solve these to obtain
$$ \cos \theta \ \ = \ \ \frac{ (1 + \zeta)·(1 - \zeta) \ - \ \eta^2}{(1 + \zeta)^2 \ + \ \eta^2} \ \ = \ \ \frac{ 1 \ - \ \zeta^2 \ - \ \eta^2}{(1 + \zeta)^2 \ + \ \eta^2} \ \ , $$
$$ \sin \theta \ \ = \ \ \frac{ -\eta·(1 + \zeta) \ - \ \eta·(1 - \zeta)}{(1 + \zeta)^2 \ + \ \eta^2} \ \ = \ \ \frac{ -2·\eta }{(1 + \zeta)^2 \ + \ \eta^2} \ \ . $$
The sum of the squares of these expressions is
$$ \cos^2 \theta \ + \ \sin^2 \theta \ \ = \ \ \frac{ (1 \ - \ \zeta^2 \ - \ \eta^2)^2 \ + \ ( -2·\eta)^2 }{[ \ (1 + \zeta)^2 \ + \ \eta^2 \ ]^2} $$ $$ = \ \ \frac{ \zeta^4 \ + \ 2·\zeta^2·\eta^2 \ + \ \eta^4 \ - \ 2· \zeta^2 \ - \ 2·\eta^2 \ + \ 1 }{[ \ ( \zeta + 1)^2 \ + \ \eta^2 \ ]^2} $$
$$ = \ \ \frac{ [ \ ( \zeta + 1)^2 \ + \ \eta^2 \ ]·[ \ ( \zeta - 1)^2 \ + \ \eta^2 \ ]}{[ \ (\zeta + 1)^2 \ + \ \eta^2 \ ]^2} \ \ = \ \ \frac{ ( \zeta - 1)^2 \ + \ \eta^2 }{ (\zeta + 1)^2 \ + \ \eta^2 } \ \ , $$
which is only equal to $ \ 1 \ $ for all values of $ \ \theta \ $ when $ \ \zeta = 0 \ \ . $ (We can verify that this becomes a constant function by calculating
$$ \frac{\partial}{\partial \eta} \ \left[ \ \frac{ ( \zeta - 1)^2 \ + \ \eta^2 }{ (\zeta + 1)^2 \ + \ \eta^2 } \ \right]_{\zeta = 0} \ \ = \ \ \frac{ 8 \ · \zeta \ · \ \eta }{[ \ (\zeta + 1)^2 \ + \ \eta^2 \ ]^2} | _{\zeta = 0} \ \ = \ \ 0 \ \ . \ \ ) $$
Hence, $ \ w \ $ is purely imaginary.
[I believe that the converse of the proposition is shown to be true by the relation for $ \ \cos^2 \theta \ + \ \sin^2 \theta \ \ . \ ] $
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|
Solve $ x^2+4=y^d$ in integers with $d\ge 3$
Find all triples of integers $(x,y,d)$ with $d\ge 3$ such that $x^2+4=y^d$.
I did some advance in the problem with Gaussian integers but still can't finish it. The problem is similar to Catalan's conjecture.
NOTE: You can suppose that $d$ is a prime.
Source: My head
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What follows only takes care of $d=3$. We use the Gaussian integer approach that you mentioned. The method can be used with other $d$, but some details of the calculation depend on $d$. Thus one can only deal with one $d$ at a time.
We first examine the case $x$ odd. Factor $x^2+4$ in the Gaussian integers as $(x+2i)(x-2i)$. Since $x$ is odd, $x+2i$ and $x-2i$ are relatively prime. If their product is a perfect cube, each must be a (Gaussian) perfect cube. Here we are using the Unique Factorization Theorem for $\mathbb{Z}[i]$. Some care needs to be taken, because of units that may appear in front of the prime factorization. But since all four units are cubes, they give no difficulty.
Let $x+2i=(a+bi)^3$. Expand, and compare real and imaginary parts. We get $a^3-3ab^2=x$ and $3a^2b-b^3=2$. Since $3a^2b-b^3=b(3a^2-b^2)$, there are only the possibilities $b=\pm 1$, $b=\pm 2$. The case $b=1$ gives $a=\pm 1$. The case $b=-1$ gives no solution. The case $b=2$ also gives no solution, while $b=-2$ gives $a=\pm 1$. The case $b=1$ gives the solution $x=2$ which we were not looking for, and $b=-2$, $a=-1$ gives the solution $x=11$.
Next we deal somewhat more briefly with the case $x$ even. Let $x=2s$ and $z=2t$. Our equation becomes $(s+i)(s-i)=2t^3$. Note that $s$ and $t$ must be odd. By considering factorizations again, we get that $s+i$ must have shape $(1\pm i)(a+bi)^3$, and an analysis similar to the one above works. We get (for $(1+i)(a+bi)^3$) that $(a-b)(a^2+4ab+b^2)=1$, which forces $b=0$, $a=1$ or $b=-1$, $a=0$. Thus the only positive solution of $s^2+1=2t^3$ has $s=1$, and therefore the only positive even $x$ such that $x^2+4$ is a perfect cube is given by $x=2$.
Remark: Let $d>3$ be odd. We could, for any specific $d$, use exactly the same strategy. Think for example of $d=5$, and the case $x$ odd. Then from the fact that $x+2i$ must be a perfect (Gaussian) fifth power, we get $x=a^5-10a^3b^2+5ab^4$ and $2=5a^4b-10a^2b^3-b^5$. The second equation forces $b=\pm 1$ or $b=\pm 2$. For these four possibilities we can quickly find any integer solution $a$ by using the "Factor Theorem." If we get an integer solution $a$, that gives a solution of the original equation, and if we don't, we know the original equation has no solution.
For any fixed odd $d$, the kind of calculation described above can be carried out mechanically and quickly. And it is clear from the analysis that for any fixed $d$, there can be at most finitely many solutions. But unfortunately the strategy can only take care of $d$'s one at a time.
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|
Are there exact expressions for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$? I was just wondering if there is any way to get an exact expression (with radicals) for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$. In case it's relevant, I want to express $z = \sqrt[4]{8} e^{\frac{5\pi}{8}i}$ in binomial form, and I know that
$$
\begin{aligned}
\cos \frac{5\pi}{8} &= - \cos \frac{3\pi}{8} \\
\sin \frac{5\pi}{8} &= \sin \frac{3\pi}{8}
\end{aligned}
$$
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Yes, we can use the double angle formula
$$ 2\cos^2 x - 1 = \cos 2x$$
Pick $x = \frac{3\pi}{8}$ and you can solve it.
Once you have the cos value, you can easily get the sin value.
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|
Probability for roots of quadratic equation to be real, with coefficients being dice rolls. I really need help with this question.
The coefficients $a,b,c$ of the quadratic equation $ax^2+bx+c=0$ are determined by throwing $3$ dice and reading off the value shown on the uppermost face of each die, so that the first die gives $a$, the second $b$ and and third $c$. Find the probabilities that the roots the equations are real, complex and equal.
I was thinking about using the fundamental formula but i'm not sure how to go about doing it. Help would be greatly appreciated.
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For roots to be real,
$ b^2 - 4ac >= 0 $
the following values of {b,a,c} are possible,
{2,1,1}
{3,1,1} {3,1,2} {3,2,1}
{4,1,1} {4,1,2} {4,2,1} {4,2,2} {4,3,1} {4,13,}
{5,1,1} {5,1,2} {5,2,1} {5,2,2} {5,3,2} {5,2,3} {5,3,1} {5,1,3}
{6,1,1} {6,1,2} {6,2,1} {6,2,2} {6,3,1} {6,1,3} {6,3,3} {6,3,2} {6,2,3}
These are 27 cases.
While the total cases possible for the values of a, b, c are 6*6*6
hence probability equals $\frac{27}{216}$ = $0.125$
..........................................................................................................................................
For the roots to be complex, 60 - 27 = 33 cases are possible,
probability equals $\frac{3}{216}$ = $0.152$
..........................................................................................................................................
For roots to be equal, discriminant
$ b^2 - 4ac = 0 $
which is possible for the following values of {b,a,c} only {6,3,3}, {4,2,2}, {2,1,1},
probability equals $\frac{3}{216}$ = $0.0138$
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|
Determine the average value of the following function Please bear with me because I have only little experience in using codes to construct the symbols for the equations.
The question is:
Determine the average value of $f(x,y) = x^2 y^2$, in the region $$R: a\le x\le b, c\le y\le d,$$
where $a+b=5, ab=13, c+d=4, cd=7.$
The formula for average value is
$$\dfrac{\displaystyle\iint f(x,y)dA}{(b-a)(d-c)}.$$
Please use only elementary calculus, and no complex analysis. Much appreciated!
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I guess this example could be a weird way to show possible transformation formalities during (double-)integration. Let's just do it formally:
$$\dfrac{\displaystyle\iint f(x,y)dA}{(b-a)(d-c)} = \frac{\int\limits_a^b\int\limits_c^d x^2y^2 dxdy}{(b-a)(d-c)} = \frac{\int\limits_a^b x^2 dx}{b-a}\times\frac{\int\limits_c^d y^2 dy}{d-c} = \frac{\frac{x^3}{3}\big|_a^b}{b-a}\times\frac{\frac{y^3}{3}\big|_c^d}{d-c} = \frac{1}{9}\frac{b^3-a^3}{b-a}\frac{d^3-c^3}{d-c} = \frac{((a+b)^2-ab)((c+d)^2-cd)}{9} = \frac{12\cdot 9}{9} = 12$$
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|
How to expand $\cos nx$ with $\cos x$? Multiple Angle Identities:
How to expand $\cos nx$ with $\cos x$, such as
$$\cos10x=512(\cos x)^{10}-1280(\cos x)^8+1120(\cos x)^6-400(\cos x)^4+50(\cos x)^2-1$$
See a list of trigonometric identities in english/ chinese
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There were two corrections in the posting ($1102$ should be $1120$ and $\cos x$ term should be $\cos^2 x$ term:
$$\cos 10x = 512 (\cos x)^{10} -1280 (\cos x)^8 +1120 (\cos x)^6 -400(\cos x)^4+50(\cos x)^2-1$$
Follow Arturo's answer you should get the following
$$
\begin{align*}
\cos 2x &= 2 \cos^2 x -1 \\
\cos 3x &= 4 \cos^3 x - 3 \cos x\\
\cos 4x &= 8 \cos^4 x - 8 \cos^2 x +1 \\
\cos 5x &= 16 \cos^5 x - 20 \cos^3 x + 5 \cos x\\
\cos 6x &= 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x -1\\
\cos 8x &= 128 \cos^8 x - 256 \cos^6 x + 160 \cos^4 x -32 \cos^2 x +1\\
\cos 10x &= 512 \cos^{10} x -1280 \cos^8 x +1120 \cos^6 x -400 \cos^4 x +50 \cos^2 x-1\\
\end{align*}
$$
(Corrected $-1120 $ to $1120$ )
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|
Partial fraction with a constant as numerator I am trying to express this as partial fraction:
$$\frac{1}{(x+1)(x^2+2x+2)}$$
I have a similar exaple that has $5x$ as numerator, it is easy to understand. I do not know what to do with 1 in the numerator, how to solve it?!
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You want to find $A, B,$ and $C$ such that
$$\frac{1}{(x+1)(x^2 + 2x + 2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 2x + 2}
$$
That is such that
$$\begin{align}0x^2 + 0x + 1 &= A(x^2 + 2x + 2) + (x+1)(Bx+c)\\
&= (A+B)x^2 + (2A+B+C)x + 2A + C.
\end{align}$$
So you get three equations
$$\begin{align} 0 &= A + B \\
0 &= 2A + B + C \\
1 &= 2A + C.
\end{align}$$
Solving this I get $A =1, B = -1, C= -1$.
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Why does $(\frac{p-1}{2}!)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$? Assume $p$ is prime and $p\ge 3$.
Through experimentation, I can see that it's probably true. Using Wilson's theorem and Fermat's little theorem, it's equivalent to saying $2^2 4^2 6^2 \cdots (p-1)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$, but I can't figure out any more than that.
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$$(p-1)!=1\cdot2\cdots\frac{p-1}{2}\cdot\frac{p+1}{2}\cdots(p-2)(p-1)$$ We have the congruences
$$\begin{align*}
p-1&\equiv -1\pmod p\\
p-2&\equiv -2\pmod p\\
&\vdots\\
\frac{p+1}{2}&\equiv -\frac{p-1}{2}\pmod{p}\end{align*}$$
Rearranging the factors produces
$$(p-1)!\equiv 1\cdot(-1)\cdot2\cdot(-2)\cdots\frac{p-1}{2}\cdot\left(-\frac{p-1}{2}\right)\pmod p.$$
$\therefore (p-1)!\equiv (-1)^{\frac{p-1}{2}}\left(1\cdot2\cdots\frac{p-1}{2}\right)^{2}\pmod p.$
From Wilson's theorem, $(p-1)!\equiv -1\pmod{p}$
Thus $$-1\equiv (-1)^{\frac{p-1}{2}}\left[\left(\frac{p-1}{2}\right)!\right]^{2}\pmod{p}$$
It follows that $\left[\left(\frac{p-1}{2}\right)!\right]^{2}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}$.
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|
Meaning of this 4x4 determinant Let $p,q,r$ and $s$ be four points on the plane. Moreover, $p,q,r$ are given in clockwise order. My book said that the following determinant is positive if and only if $s$ lies inside the circle passing through $p,q,r$. Why?
$$\det
\begin{bmatrix}
p_x & p_y & p_x^2+p_y^2 & 1 \\
q_x & q_y & q_x^2+q_y^2 & 1 \\
r_x & r_y & r_x^2+r_y^2 & 1 \\
s_x & s_y & s_x^2+s_y^2 & 1 \\
\end{bmatrix}
$$
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Following the hints from J.M., I was able to get the answer.
$$\det
\begin{bmatrix}
p_x & p_y & p_x^2+p_y^2 & 1 \\
q_x & q_y & q_x^2+q_y^2 & 1 \\
r_x & r_y & r_x^2+r_y^2 & 1 \\
s_x & s_y & s_x^2+s_y^2 & 1 \\
\end{bmatrix}
$$
$$
=-a(s_x^2+s_y^2)-bs_x+cs_y+d\\
=-a(s_x^2+s_y^2+\frac{b}{a}s_x-\frac{c}{a}s_y+\frac{d}{a})\\
=-a((s_x+\frac{b}{2a})^2+(s_y-\frac{c}{2a})^2-\frac{b^2+a^2}{4a^2}+\frac{d}{a} )\\
=-a((s_x+\frac{b}{2a})^2+(s_y-\frac{c}{2a})^2-r^2 )\\
$$
where
$a=\det
\begin{bmatrix}
p_x & p_y & 1 \\
q_x & q_y & 1 \\
r_x & r_y & 1 \\
\end{bmatrix}$ and $r=\frac{\sqrt{b^2+c^2-4ad}}{2a}$. Since $p,q,r$ are in clockwise order, $a>0$ and $d<0$. Therefore, the determinant in question is positiver if and only if $s$ lies inside the circumcircle.
|
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|
How to find perpendicular vector to another vector? How do I find a vector perpendicular to a vector like this: $$3\mathbf{i}+4\mathbf{j}-2\mathbf{k}?$$
Could anyone explain this to me, please?
I have a solution to this when I have $3\mathbf{i}+4\mathbf{j}$, but could not solve if I have $3$ components...
When I googled, I saw the direct solution but did not find a process or method to follow. Kindly let me know the way to do it. Thanks.
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One way to do this is to express the vector in terms of a spherical coordinate system. For example
$$ \boldsymbol{e}= \pmatrix{a \\ b \\ c} = r \pmatrix{ \cos\varphi \cos\psi \\ \sin\varphi \cos\psi \\ \sin\psi} $$
where $r=\sqrt{a^2+b^2+c^2}$, $\tan(\varphi) = \frac{b}{a}$ and $\tan{\psi} = \frac{c}{\sqrt{a^2+b^2}}.$
Provided that $a \neq 0$ or $b \neq 0$ then
A choice of two orthogonal vectors can be found with $$ \begin{aligned} \boldsymbol{n}_1 & = \frac{{\rm d} \boldsymbol{e}}{{\rm d} \varphi}
= r\pmatrix{-\sin \varphi \cos\psi \\ \cos\varphi \cos\psi \\ 0}&
\boldsymbol{n}_2 & = \frac{{\rm d} \boldsymbol{e}}{{\rm d} \psi}
= r\pmatrix{-\cos\varphi \sin\psi \\ -\sin\varphi \sin\psi \\ \cos\psi}
\end{aligned}$$
Of course, any non-zero linear combination of these two vectors is also orthogonal
$$ \boldsymbol{n} = \cos(t) \boldsymbol{n}_1 + \sin(t) \boldsymbol{n}_2 $$
where $t$ is an rotation angle about the vector $\boldsymbol{e}$.
Put it all together to make a family of orthogonal vectors in terms of $t$ as
$$ \boldsymbol{n} = \pmatrix{-b \cos(t) - \frac{a c}{\sqrt{a^2+b^2}} \sin(t) \\ a \cos(t) - \frac{b c}{\sqrt{a^2+b^2}} \sin(t) \\ \sqrt{a^2+b^2} \sin(t)} $$
For $\boldsymbol{e} = \pmatrix{3 & 4 & -2}$ the aboves gives
$$ \boldsymbol{n} = \pmatrix{
\frac{6}{4} \sin(t)-4 \cos(t) \\ 3 \cos(t) - \frac{8}{5} \sin(t) \\ 5 \sin(t) } \longrightarrow \begin{cases} \boldsymbol{n} = \pmatrix{-4 & 3 & 0} & t =0 \\ \boldsymbol{n} =\pmatrix{\frac{6}{5} & \frac{8}{5} & 5} & t = \frac{\pi}{2} \end{cases} $$
For the case when $a=0$ and $b=0$ then you know can assign the perpendicular somewhat arbitrarily with
$$ \boldsymbol{n}_1 = \pmatrix{1 \\ 0 \\ 0} $$
and
$$ \boldsymbol{n}_2 = \pmatrix{0 \\ 1 \\ 0} $$
for the general solution
$$\boldsymbol{n} = \cos(t) \boldsymbol{n}_1 + \sin(t) \boldsymbol{n}_2$$
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|
How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$ Reference: http://xkcd.com/1047/
We tried various different trigonometric identities. Still no luck.
Geometric interpretation would be also welcome.
EDIT: Very good answers, I'm clearly impressed. I followed all the answers and they work! I can only accept one answer, the others got my upvote.
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To elaborate on Mathlover's comment, the three numbers $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$, and $\cos\frac{5\pi}{7}$ are the three roots of the monic Chebyshev polynomial of the third kind
$$\hat{V}_n(x)=\frac{\cos\left(\left(n+\frac12\right)\arccos\,x\right)}{2^n\cos\frac{\arccos\,x}{2}}=\frac1{2^n}\left(U_n(x)-U_{n-1}(x)\right)$$
where $U_n(x)=\frac{\sin((n+1)\arccos\,x)}{\sqrt{1-x^2}}$ is the usual Chebyshev polynomial of the second kind, and the last relationship is derived through the trigonometric identity
$$2\sin\frac{\theta}{2}\cos\left(\left(n+\frac12\right)\theta\right)=\sin((n+1)\theta)-\sin\,n\theta$$
The question then is essentially asking for the negative of the coefficient of the $x^2$ term of $\frac18(U_3(x)-U_2(x))$ (Vieta); we can generate the two polynomials using the definition given above, or through an appropriate recursion relation. We then have
$$\begin{align*}U_2(x)&=4x^2-1\\U_3(x)&=8x^3-4x\end{align*}$$
and we thus have
$$\hat{V}_3(x)=\frac18((8x^3-4x)-(4x^2-1))=x^3-\frac{x^2}{2}-\frac{x}{2}+\frac18$$
which yields the identities
$$\begin{align*}
\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}&=\frac12\\
\cos\frac{\pi}{7}\cos\frac{3\pi}{7}+\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}+\cos\frac{\pi}{7}\cos\frac{5\pi}{7}&=-\frac12\\
\cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}&=-\frac18
\end{align*}$$
|
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How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \frac{1}{2^{n+1}} = \frac{2^{n+1}-1}{2^{n+1}}$.
What should the approach be here?
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Just multiply the numerator and denominator in the first fraction by 2 and add the fractions.
|
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Simplifying Logarithmic Expression
Compute:
$$\frac{1-\log_a^{3}{b} }{(\log_a b+\log_b a+1)\log_a\frac{a}{b}}$$
I tried to expand it :
$$\frac{1-\log_a^{3}{b} }{(\log_a b+\log_b a+1)\log_a\frac{a}{b}}$$
$$=\frac{(1-\log_a{b})(\log_a^{2}b+\log_a b+1)}{(\log_a b+\log_b a+1)(1-\log_a{b})}$$
$$=\frac{(\log_a^{2}b+\log_a b+1)}{(\log_a b+\log_b a+1)}$$
But I got nothing.
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$$ \large{x = \log_a b}$$
Then $$ 1- \log_a^{3}{b} = (1-x^3) $$
Also
$$(\log_a b+\log_b a+1)(\log_a\frac{a}{b}) = (x+\frac{1}{x}+1)(1-x)$$
because
$$ \log_b a = \frac{1}{\log_a b} \hspace{8pt} \textit{and} \hspace{8pt} \log_a \frac{a}{b} = (1-\log_a b)$$
The whole thing gets simplified to
$$ \frac{(1-x^3)}{(x+\frac{1}{x}+1)(1-x)} = \frac{x(1-x^3)}{1-x^3}= x$$
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"url": "https://math.stackexchange.com/questions/142197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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|
How to find $\int{\frac{\sqrt{x^2+1}}{x+2}dx}$ I have got such integral $$\int{\frac{\sqrt{x^2+1}}{x+2}dx}$$ and with Maple I got something like this:
$$\int\frac{1}{2} + \frac{1+3u^2+4u^3}{-2u^2+2u^4-8u^3}du$$ And I want to know how to achive this changes.
I tried to use WolframAlpha, but there is scarier solution. This integral was for Gaussian quadrature method, so it's analytic solution is horrible.
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\begin{align}
&\int{\frac{\sqrt{x^2+1}}{x+2}dx}\\
= &\int \frac{{(x^2-4)+5}}{(x+2)\sqrt{x^2+1}}dx
= \int \frac{x-2}{\sqrt{x^2+1}} +\frac{5}{(x+2)\sqrt{x^2+1}}\ dx\\
=&\ \sqrt{x^2+1}-2\sinh^{-1}x+\sqrt5\tanh^{-1}\frac{2x-1}{\sqrt{5(x^2+1)}}+C
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/145066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding the integral of $x/\sqrt{4-x^2}$ Find the integral:
$$\int \frac{x}{\sqrt{4-x^2}} dx = \int \frac{x}{\sqrt{2^2-x^2}} dx$$
using $$\int \frac1{\sqrt{a^2-x^2}} dx = \arcsin(x/a) + C$$
I get $\displaystyle \frac{x^2}{2} \arcsin \left(\frac{x}{2} \right) + C$.
I'm not sure if the $\dfrac{x^2}{2}$ is right. Any suggestions and help would be great.
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Another way: $$\int\frac{x}{\sqrt{4-x^2}}dx=-\frac{1}{2}\int\frac{d(4-x^2)}{(4-x^2)^{1/2}}dx=-\frac{1}{2}\frac{\sqrt{4-x^2}}{1/2}+C=-\sqrt{4-x^2}+C$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/146543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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|
Does $\int_0^\infty\frac{\cos^2x}{x^2+5x+11}dx$ converge or diverge? When I'm learning convergence, my teacher just show me about condition to convergence or not. But I haven't meet a function that contain both trigonometric and normal polynomial. When I asked one of my friends, he tell me that using Taylor to developed $\cos x$, but I'm afraid that is not the good solution.
$$\int_0^\infty\frac{\cos^2(x)}{x^2+5x+11}dx$$
Thanks :)
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Let $$I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx$$ The integrand is non-negative and since $\cos^2(x) \in [0,1]$, $\forall x \in \mathbb{R}$, we get that $$0 \leq I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac1{x^2 + 5x+11} dx = \int_0^{\infty} \frac{dx}{(x+5/2)^2 + 19/4}$$
Now make use of the identity $$\int \frac{dx}{x^2 + a^2} = \frac{\arctan (x/a)}a$$ and bound the integral $I$. Hence, we get that $$0 \leq I \leq \left. \frac{2}{\sqrt{19}} \arctan \left(\frac{2x+5}{\sqrt{19}} \right) \right \rvert_{0}^{\infty} = \frac{2}{\sqrt{19}} \left( \frac{\pi}{2} - \arctan \left(\frac{5}{\sqrt{19}}\right) \right) = \frac{\pi - 2 \arctan \left( \dfrac{5}{\sqrt{19}} \right)}{\sqrt{19}} \approx 0.328984$$
EDIT
Another quick argument is along the lines of what mixedmath has done but there is no need to split the integral over two regions $$I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac1{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac{dx}{(x+5/2)^2} = \left. \left(-\frac1{x+5/2} \right) \right \rvert_{0}^{\infty} = \frac25$$
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"timestamp": "2023-03-29T00:00:00",
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|
What equation intersects only once with $f(x)=\sqrt{1-(x-2)^2}$ Being $f(x)=\sqrt{1-(x-2)^2}$ I have to know what linear equation only touches the circle once(only one intersection), and passes by $P(0,0)$.
So the linear equation must be $y=mx$ because $n=0$.
I have a system of 2 equations:
\begin{align}
y&=\sqrt{1-(x-2)^2}\\
y&=mx
\end{align}
So I equal both equations and I get \begin{align}
mx&=\sqrt{1-(x-2)^2}\\
m&=\frac{\sqrt{1-(x-2)^2}}{x}
\end{align}
$m$ can be put in the $y=mx$ equation, which equals to:
\begin{align}
y&=\left(\frac{\sqrt{1-(x-2)^2}}{x}\right)x\\
&=\sqrt{1-(x-2)^2}
\end{align}
But that equation has $\infty$ intersections, and I want only the equation who has $1$ interception.
What is the good way to know this? And how can it be calculated?
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As you have in your post, we have $y = mx$ as the straight line. For this line to touch the semi-circle, we need that $y = mx$ and $y = \sqrt{1 - (x-2)^2}$ must have only one solution. This means that the equation $$mx = \sqrt{1 - (x-2)^2}$$ must have only one solution.
Hence, we need to find $m$ such that $m^2x^2 = 1 - (x-2)^2$ has only one solution.
$(m^2 + 1)x^2 -4x +4 = 1 \implies (m^2+1) x^2 - 4x + 3 = 0$.
Any quadratic equation always has two solution. The two solutions collapse to a single solution when the discriminant of the quadratic equation is $0$. This is seen from the following reasoning.
For instance, if we have $ax^2 + bx+c = 0$, then we get that $$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
i.e. $\displaystyle x_1 = \frac{-b + \sqrt{b^2 -4ac}}{2a}$ and $\displaystyle x_2 = \frac{-b - \sqrt{b^2 -4ac}}{2a}$ are the two solutions. If the two solutions to collapse into a single solution i.e. if $x_1 = x_2$, we get that $$ \frac{-b + \sqrt{b^2 -4ac}}{2a} = \frac{-b - \sqrt{b^2 -4ac}}{2a}$$ This gives us that $\displaystyle \sqrt{b^2 -4ac} = 0$. $D = b^2 - 4ac$ is called the discriminant of the quadratic.
The discriminant of the quadratic equation, $(m^2+1) x^2 - 4x + 3 = 0$ is $D = (-4)^2 - 4 \times 3 \times (m^2+1)$.
Setting the discriminant to zero gives us that $(-4)^2 - 4 \times 3 \times (m^2 + 1) =0$ which gives us $ \displaystyle m^2 + 1 = \frac43 \implies m = \pm \frac1{\sqrt{3}}$.
Hence, the two lines from origin that touch the circle are $y = \pm \dfrac{x}{\sqrt{3}}$.
Since you have a semi-circle, the only line that touches the circle is $\displaystyle y = \frac{x}{\sqrt{3}}$. (Thanks to @Joe Johnson 126 for pointing this out).
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"timestamp": "2023-03-29T00:00:00",
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|
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