Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 &= x^2 - 3x - 4 \\
0 &= (x-4)(x+1) \\
\end{align}
$$
Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.
But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:
$$
\begin{align}
\sqrt{4 + 5} &= 4 - 1 \\
\sqrt{9} &= 3 \\
3 &= 3
\end{align}
$$
So using 4 works as expected, but when using $-1$:
$$
\begin{align}
\sqrt{-1 + 5} &= -1 - 1 \\
\sqrt{4} &= -2 \\
2 &\ne -2
\end{align}
$$
At what stage am I going wrong?
And according the WolframAlpha, the solution is $x = 4$.
| Plugging $x= -1$ into each step of your work, you get :
$$\begin{align}2 = -2 \\ 4 = 4 \\ 4 = 4 \\ 4 = 4 \\ 0 = 0 \\ 0 = 0 \end{align}$$
Thus without any thinking, you know that the error is in the first step, because $2^2 = 4 = 4 = (-2)^2$, but $2 \neq -2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/41152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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} |
Linear Algebra: different determinant answers I'm having a problem verifying my answer to this question:
Solve for x:
$$\left| \begin{array}{cc}
x+3 & 2 \\
1 & x+2 \end{array} \right| = 0$$
I get:
$(x+3)(x+2)-2=0$
$(x+3)(x+2)=2$
thus:
$x+3=2$ and $x+2=2$
$x=-1$ and $x=0$
The book says that $x=-1$ and $x=-4$ is the correct answer.
I tried doing it a different way by expanding and got totally different answers:
$x^2+5x=-4$
$x(x+5)=-4$
$x=-4$ and $x=-9$
What is going on here?
| The problem here is that if both $(x+3)$ and $(x+2$ are equal to $2$, then $(x+3)(x+2) = (2)(2) = 4 \not = 2$. So you must first distribute, recollect the terms, and factor it again.
So $(x+2)(x+3) - 2= x^2 + 5x + 4 = x^2 + 4x + x + 4 = (x+1)(x+4) = 0$.
I don't know how you did the second way, so I can't comment on that. But I can say that I appreciate you showing your work. Were only everyone to do so...
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Quick way to find all roots of a polynomial? I'm following an example of finding the splitting field for $f(X)=X^3-2$ over $\mathbb{Q}$. It reads, if $\alpha$ is the positive cube root of $2$, then the roots of $f$ are $\alpha$, $\alpha(-\frac{1}{2}+i\frac{1}{2}\sqrt{3})$, and $\alpha(-\frac{1}{2}-i\frac{1}{2}\sqrt{3})$.
My question is, how does one know what those two latter roots are so quickly? Is there a general way to find all the other roots of a polynomial if you can find the obvious one, in this case $\sqrt[3]{2}$?
| The other two roots are $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$, where $\omega$ is a primitive cubic root of unity. It just so happens that the primitive cubic roots of unity are easy to express, because they are the "other two" roots of $x^3-1 = (x-1)(x^2+x+1)$, so they can be found using the quadratic formula.
In general, the $n$ complex roots of $x^n-a$, with $a\in\mathbb{R}$, are given by
$$\sqrt[n]{a},\quad \zeta_n\sqrt[n]{a},\quad \zeta_n^2\sqrt[n]{a},\quad\ldots,\quad \zeta_n^{n-1}\sqrt[n]{a},$$
where $\zeta_n$ is a primitive $n$th root of unity.
Also, if you happen to know one root of a cubic polynomial, then you can always divide and solve the resulting quadratic. Here, you have $x^3-2$, and you know that $x-\sqrt[3]{2}$ is a factor (because $\sqrt[3]{2}$ is a root). Factoring, you have
$$x^3 - 2 = (x-\sqrt[3]{2})(x^2 + \sqrt[3]{2}x + \sqrt[3]{4}).$$
So the other two roots are the roots of the quadratic, which can be found using the quadratic formula:
$$r_1 = \frac{-2^{1/3} + \sqrt{2^{2/3} - 4\cdot 2^{2/3}}}{2},\qquad r_2 = \frac{-2^{1/3} - \sqrt{2^{2/3} - 4\cdot 2^{2/3}}}{2}.$$
Simplify the square root, factor out $2^{1/3}$, and you get the expressions as well. E.g.,
$$\begin{align*}
r_1 &= \frac{-2^{1/3}+\sqrt{2^{2/3}-4\cdot 2^{2/3}}}{2}\\
&= \frac{-2^{1/3}+\sqrt{2^{2/3}(-3)}}{2}\\
&= \frac{-2^{1/3} + 2^{1/3}\sqrt{-3}}{2}\\
&= \left(\frac{-1+\sqrt{-3}}{2}\right)2^{1/3}\\
&= \left( - \frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\sqrt[3]{2}.
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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All functions $\frac{1}{f\left(y^2f(x)\right)} = \big(f(x)\big)^2\left(\frac{1}{f\left(x^2-y^2\right)} + \frac{2x^2}{f(y)}\right)$ How can I find all continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}^+$ such that
$$\frac{1}{f\left(y^2f(x)\right)} = \big(f(x)\big)^2\left(\frac{1}{f\left(x^2-y^2\right)} + \frac{2x^2}{f(y)}\right)$$
for all reals $x,y$?
| Let's first show that $f(x) = f(-x)$ for all $x \in \mathbb{R}$. Put $x=y=0$, and we get
$$\frac{1}{f(0)} = f(0),$$
which implies that $f(0) = 1$. Now letting $x=0$ we get
$$\frac{1}{f(y^2)} = \frac{1}{f(-y^2)},$$
which implies that $f(y^2) = f(-y^2)$, proving our first claim.
Now since $f(x) = f(-x)$ there exists a function $g\colon [0,\infty) \to \mathbb{R}^+$ such that $f(x) = g(x^2)$. The condition on $f$ becomes
$$\frac{1}{g(y^4g^2(x^2))} = (g(x^2))^2 \left( \frac{1}{g((x^2 - y^2)^2)} + \frac{2x^2}{g(y^2)} \right).$$
By changing $x^2 \mapsto x$ and $y^2 \mapsto y$, this turns to
$$\frac{1}{g(y^2 g^2(x))} = (g(x))^2 \left( \frac{1}{g((x-y)^2)} + \frac{2x}{g(y)}\right).$$
Now notice that we already know that $g(0) = f(0) = 1$.
Also, by setting $x=1$, $y=0$ we get
$$1 = (g(1))^2 \left( \frac{1}{g(1)} + 2 \right) \Leftrightarrow g(1) = (g(1))^2 + 2 (g(1))^3 \Leftrightarrow$$
$$1 = g(1) + 2 (g(1))^2 \Leftrightarrow 2(g(1) - \frac{1}{2})(g(1) + 1) = 0.$$
Hence $g(1) = \frac{1}{2}$.
We will show by induction that $g(n) = \frac{1}{1+n}$ for all $n \in \mathbb{N}$. Suppose that the claim holds for some $n \in \mathbb{N}$, then
$$\frac{1}{g((n+1)^2 g^2(n))} = (g(n))^2 \left( \frac{1}{g((n-(n+1))^2)} + \frac{2n}{g(n+1)}\right) \Leftrightarrow$$
$$\frac{1}{g(1)} = \frac{1}{(n+1)^2} \left( \frac{1}{g(1)} + \frac{2n}{g(n+1)}\right) \Leftrightarrow$$
$$2 (n+1)^2 = 2 + \frac{2n}{g(n+1)} \Leftrightarrow g(n+1) = \frac{1}{n+2},$$
and by induction we have that
$$g(n) = \frac{1}{n+1}$$
for all $n \in \mathbb{N}$.
Consider now the original condition on $g$ and let $x$ and $y$ be natural numbers. We get that
$$\frac{1}{g(\frac{y^2}{(x+1)^2})} = \frac{1}{(x+1)^2} \left( (x-y)^2 + 1 + 2x(y+1) \right) \Leftrightarrow$$
$$\frac{1}{g(\frac{y^2}{(x+1)^2})} = \frac{1}{(x+1)^2} \left( (x+1)^2 + y^2 \right) \Leftrightarrow$$
$$g(\frac{y^2}{(x+1)^2}) = \frac{(x+1)^2}{(x+1)^2 + y^2} = \frac{1}{\frac{y^2}{(x+1)^2} + 1},$$
and hence the formula
$$g(x) = \frac{1}{x+1}$$
holds for all squares of rational numbers. But they are dense in $[0,\infty)$ and since $g$ was continuous, we get that the only solution is
$$g(x) = \frac{1}{x+1}.$$
(Checking that this is indeed a solution is straightforward.)
Thus
$$f(x) = \frac{1}{x^2 + 1}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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question about division The question is from the following two problems:
Let $x$ and $y$ be positive integers such that $3x+7y$ is divisible by $11$. Which of the following must also be divisible by $11$?
A. $4x+6y$ B. $x+y+5$ C. $9x+4y$ D. $4x-9y$ E. $x+y-1$
If $x,y$ and $z$ are positive integers such that $4x-5y+2z$ is divisible by $13$, then which one of the following must also be divisible by $13$?
A. $x+13y-z$ B. $6x-10y-z$ C.$x-y-2z$ D.$-7x+12y+3z$ E. $-5x+3y-4z$
One way I know is that one can substitute some specific number to rule out the wrong answers. Or alternatively, one can try the properties of congruence(addition and subtraction), which I think is not easy to figure out the steps. Just as for the second problem, how on earth can one figure out one needs to add
$$-39x\equiv 0\pmod{13}$$
$$65y\equiv 0\pmod{13}$$
$$13z\equiv 0\pmod{13}$$
to
$$4x-5y+2z\equiv 0\pmod{13}$$
and then divides by $5$
in order to get the answer
$$-7x+12y+3z\equiv 0\pmod{13}?$$
So, here is my question:
Is there a general method (which might be also quick) to solve this kind of problems?
| This is what I would do:
If $3x+7y$ is divisible by $11$, then $3x + 7y\equiv 0\pmod{11}$, so $3x\equiv -7y = 4y\pmod{11}$. Multiplying through by $4$, we get $12x \equiv 16y\pmod{11}$, which simplifies to $x\equiv 5y\pmod{11}$.
Then one can plug into the congruences: $4x+6y \equiv 26y\equiv 3y\pmod{11}$; $x+y+5\equiv 6y+5\pmod{11}$; $9x+4y\equiv 49y\equiv 5y\pmod{11}$; $4x-9y\equiv 11y\equiv 0\pmod{11}$; and $x+y-1\equiv 6y-1\pmod{11}$. Clearly, the answer is D.
For the second one, from $4x-5y+2z\equiv 0\mod{13}$ we have $4x\equiv 5y-2z\pmod{13}$; multiplying through by $-3$ we get
$$-12x \equiv -15y+6z\pmod{13}$$
which is equivalent to $x \equiv -2y+6z\pmod{13}$.
Plugging into the different possibilities, we have:
$$\begin{align*}
\text{A. }&& x+13y - z &\equiv -2y+5z\pmod{13}\\
\text{B. }&& 6x-10y-z &\equiv 4y +9z\pmod{13}\\
\text{C. }&& x-y-2z &\equiv -3y+4z\pmod{13}\\
\text{D. }&&-7x+12y+3z &\equiv 26y -39z \equiv 0\pmod{13}\\
\text{E. }&& -5x+3y-4z &\equiv 13y-34z \equiv 5z\pmod{13},
\end{align*}$$
so again, the answer is D.
Alternatively.
Once you have $x-5y\equiv 0\pmod{11}$ in the first problem, multiply by a suitable $k$ to try to match the coefficient of $x$ in the alternatives, until you get a match:
*
*For $4x+6y$, multiply $x-5y$ by $4$ to get $4x-20y \equiv 4x+9y$, not equal.
*For $x+y+5$, not equal.
*For $9x+4y$, multiply $x-5y$ by $9$ to get $9x-45y \equiv 9x-y$; not equal.
*For $4x-9y$, multiply $x-5y$ by $4$ to get $4x - 20y \equiv 4x -9y$; equal. This is the answer.
For the second one, again, from $x+2y-6z\equiv 0$, we try to match the coefficient of $x$ in each:
*
*Multiply by $1$ to try to get $x+13y-z$; no luck.
*Multiply by $6$ to try to get $6x-10y-z$; we get $6x+12y-36z\equiv 6x+12y-3z$; no luck.
*Multipy by $1$ to try to get $x-y-2z$; no luck.
*Multiply by $-7$ to try to get $-7x+12y+3z$; we get $-7x-14y+42z \equiv -7x -y+3z \equiv -7x+12y+3z$. Equal, so this is the answer.
| {
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Will any value of a free variable satisfy a system of equation? Say I have a reduced row echelon form matrix like this:
$$A=\begin{bmatrix}
1 & 0 & \frac{1}{2} & 0\\
0 & 1 & -\frac{1}{3} & 0\\
0 & 0 & 0 & 1
\end{bmatrix}$$
The number of unknowns is more than the number of known equations. So I can expect an infinite number of solutions.
And $Ax=b$ is like this:
$$
\begin{bmatrix}
1 & 0 & \frac{1}{2} & 0\\
0 & 1 & -\frac{1}{3} & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}
=
\begin{bmatrix}
b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix}
$$
Then I can say that my $x$ is like this with $x_{3}$ being a free variable in the equation:
$$\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}
=
x_{3}\begin{bmatrix}
\frac{1}{2}\\
\frac{1}{3}\\
1\\
0
\end{bmatrix}+
\begin{bmatrix}
b_{1}\\
b_{2}\\
0\\
b_{3}
\end{bmatrix}
$$
Now, if I let $\; \begin{bmatrix}
b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix}=\begin{bmatrix}
5\\
2\\
7
\end{bmatrix}$, then...
$$\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}
=
x_{3}\begin{bmatrix}
\frac{1}{2}\\
\frac{1}{3}\\
1\\
0
\end{bmatrix}+
\begin{bmatrix}
5\\
2\\
0\\
7
\end{bmatrix}
$$
At this stage, I can say that for any value that I put into the variable $x_{3}$, I would get an answer that is equals to $b$, right?
So assume I just randomly throw a value into $x_{3}=2$, then...
$$\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}
=
\begin{bmatrix}
1\\
\frac{2}{3}\\
2\\
0
\end{bmatrix}+
\begin{bmatrix}
5\\
2\\
0\\
7
\end{bmatrix}=
\begin{bmatrix}
6\\
\frac{8}{3}\\
2\\
7
\end{bmatrix}
$$
From here, I need to tally if the equation really gets back my intended values of $b$, which is $\begin{bmatrix}
b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix}=\begin{bmatrix}
5\\
2\\
7
\end{bmatrix}$. So,
$x_{4} = b_{3} =7$
This is correct.
$b_{2}=x_{2}-\frac{1}{3}x_{3}$
$b_{2}=\frac{8}{3}-\frac{2}{3}=2$
This is also correct.
BUT NOW,
$b_{1}=x_{1}+\frac{1}{2}x_{3}$
$b_{1}=6+1=7$
$b_{1}$ suppose to be 6 but somehow, why am I getting 7 instead? This is weird. Where did I go wrong?
Thanks for any help!
| I think you've confused yourself somewhere.
You have a mapping from $\mathbb{R}^4 \to \mathbb{R}^3$. $x_3$ is free $x_4 = b_3$, the only stipulation is that:
$x_1/3 + x_2/2 = 8/3$.
The vector you've given: $[6, 8/3, 2, 7]^T$ does not satisfy the last condition.
Edit: Solve for $[x_1,x_2,x_3,x_4]^T$ given a vector $[b_1,b_2,b_3]^T$:
$x_4=b_3$
$x_1 + \dfrac{x_3}{2} = b_1$
$x_2 - \dfrac{x_3}{3} = b_2$
These all imply that
$\dfrac{x_1}{3} + \dfrac{x_2}{2} = \dfrac{b_1}{3} + \dfrac{b_2}{2}$
Now you want the solution to the vector $[5,2,7]^T$, so
$\dfrac{x_1}{3} + \dfrac{x_2}{2} = \dfrac{5}{3} + 1 = \dfrac{8}{3}$.
The vector you said mapped to $\vec{b}$ was $[6, 8/3, 2, 7]$.
Your proposed $x_1 = 6$ and $x_2 = \dfrac{8}{3}$ do not satisfy this.
| {
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$
Multiplying both sides by $(a + b + c)$:
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?
| The following uses what is perhaps the most practical formulation of the AM-GM inequalities, which can be found as Theorem 2.6a in Ivan Niven's excellent Maxima and Minima without Calculus.
Theorem 2.6a If $n$ positive functions have a fixed product, their sum is minimum if it can be arranged that the functions are equal. On the other hand, if $n$ positive functions have a fixed sum, their product is maximum if it can be arranged that the functions are equal.
By "positive" functions, Niven means functions that are positive on the domain we care about.
To see how this applies to the problem at hand, we see that it is enough to show that $(x+y+z)^3\geq 27k$ where $k$ is the product of $xyz$. Evidently, the theorem applies as the product is constant, so the minimum on the left-hand side is given by when $x=y=z=\sqrt[3]{k}$, and hence the minimum of the left-hand side is in fact $27k$ as desired.
The statement and (one) proof of the AM-GM inequalities can be found on page 21 of Niven's book, while the proof of Theorem 2.6a begins at the bottom of page 27.
| {
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Factoring $X^{16}+X$ over $\mathbb{F}_2$ I just asked wolframalpha to factor $X^{16}+X$ over $\mathbb{F}_2$. The normal factorization is
$$
X(X+1)(X^2-X+1)(X^4-X^3+X^2-X+1)(X^8+X^7-X^5-X^4-X^3+X+1)
$$
and over $GF(2)$ it is
$$
X(X+1)(X^2+X+1)(X^4+X+1)(X^4+X^3+1)(X^4+X^3+X^2+X+1).
$$
Does the second form follow from the first, or is there a different way to factor over $\mathbb{F}_2$? I noticed that simply replacing the $-$ signs with $+$ signs in the first factorization doesn't yield the second one.
| It shouldn't be a surprise that switching from integer (or rational) coefficients to modular coefficients allows for further factorization. The simplest example is probably the factorization
$$
x^2+1=x^2+2x+1=(x+1)^2
$$
over $F_2$.
Here the key difference between the two factorizations is that the polynomial of degree 8 splits into a product of two quartic polynomials over $F_2$: $$(x^4+x+1)(x^4+x^3+1)=x^8+x^7+x^5+x^4+x^3+x+1$$ that is equal to (up to sign changes) your last factor.
Once you start on finite fields in your studies you will immediately learn that all the elements of $GF(16)$ are roots of the polynomial $x^{16}+x$. As that field is a degree 4 extension of $F_2$, all those elements have minimal polynomials of degree a factor of 4. Furthermore, all such irreducible polynomials appear as factors of $x^{16}+x$. Now, we easily see that both $x^4+x+1$ and its reciprocal polynomial $x^4+x^3+1$ are both irreducible in the ring $F_2[x]$, so they must appear as factors.
| {
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Laplace transform and Differentiation help I need to know how to differentiate something like these
a) Differentiate with respect to s $$ b/(s^2 +b^2)$$
b) Solve the given equation
$$f(t) = s(1/s^2 + 2/s^3)$$
c) Reverse Laplace transform of $$(3s - 15)/(2s^2 -4s + 10)$$
| Note: This is an extensive edit to Shai Covo's answer.
For part ( C ) it is the following:
Pulling out a factor of $2$ from the denominator and the constant multiple of $3$ from the numerator leads to: $$ \frac{{3s - 15}}{{2s^2 - 4s + 10}} = \frac{3}{2}\frac{{s - 5}}{{s^2 - 2s + 5}} $$
Next we do Completing the Square for the polynomial $(s^2-2s+5)~$ in the denominator and re-writing the numerator constant $-5$ as $(-1-4)$ giving us:
$$ = \frac{3}{2} \cdot \frac{{s - 1 - 4}}{{(s - 1)^2 + 4}} $$
Let denominator equal $K$.
Separating the fraction into parts with them being $\dfrac{3}{2}\cdot\dfrac{(s-1)}{K}$ and $-3\cdot\dfrac{2}{K}$; after distributing the $\dfrac{3}{2}$ to the second term $-4$. This leads to the expression below now.
$$ =\frac{3}{2}\cdot\frac{{(s - 1)}}{{(s - 1)^2 + 2^2}} - 3\cdot\frac{2}{{(s - 1)^2 + 2^2 }}. $$ Now notice that $a = 1$ and $b = 2$. To see this, take a look at this table of Laplace Transforms for the form $e^{at} \cos(b t)$ and $e^{at} \sin (b t)$ located here $\longmapsto$ Table of $\mathcal Laplace~Transforms$ .
From the table of transforms, the inverse transform of $\dfrac{s-a }{(s-a )^2 + b ^2 }~$ and $~\dfrac{b}{{(s-a)^2 + b^2 }}$
are $e^{at} \cos(bt)$ and $e^{at} \sin (bt)$, respectively.
Hence the inverse laplace transform of our original problem is, $$ \mathcal{L^{-1}} \left\{ \dfrac{{3s - 15}}{{2s^2 - 4s + 10}} \right\} = \frac{3}{2}e^{t} \cos(2t) - 3e^{t} \sin(2t). $$
Hope this helps out.
Good~Luck
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I need explanation for this solution for the proof. (Perfect square ends with 0,1,4,5,6,9) Give a proof to the sentence: "The final decimal digit of a perfect square is 0, 1, 4, 5, 6 or 9."
Solution:
A integer $n$ can be expressed as $10a+b$, where $a$ and $b$ are positive integers and $b$ is 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.
Here $a$ is the integer obtained by subtracting the final decimal digit of $n$ and dividing by 10. (so $a=(n-b)/10$)
Next note that $(10a + b)^2 = 100a^2+20ab+b^2=10(10a^2+2b)+b^2$ so the final decimal digit of $n^2$ is same as the final decimal digit of $b^2$.
I understand until this point but not below:
Furthermore, note that the final decimal digit of $b^2$ is the same as final decimal digit of $(10-b)^2 = 100 - 20ab + b^2$. (how did you get this equation?) consequently we cab reduce our proof to the consideration of fix cases.
final digit of:
1) $n$ is 1 or 9 is 1
2) $n$ is 2 o 8 is 4
3) $n$ is 3 or 7 is 9
4) $n$ is 4 or 6 is 6
5) $n$ is 5 is 5
6) $n$ is 0 is 0
THANKS!
| The point of the final step is to prove that, upon squaring, $\rm\:b\:$ and $\rm\: 10-b\:$ leave the same remainder modulo $10\:,\:$ i.e. $\rm\:(\pm b)^2 = b^2\pmod{10}\:.\:$ Indeed $\rm\:(10-b)^2 =\: b^2 + 10\ (10-2\:b)\:.\:$ Therefore to find all of the squares modulo $10\:,\:$ it suffices to square only the integers between $0$ and $5\:,\:$ i.e.
$$\:\{0,1,2,3,4,5,6,7,8,9\}^2\equiv\{0,\pm1,\pm2,\pm3,\pm4,5\}^2 \equiv \{0,1,4,9,6,5\}\pmod{10}$$
since $\rm\:\ 9\equiv -1,\:\ 8\equiv -2,\:\ 7\equiv -3,\:\ 6\equiv -4\pmod{10}\:.$
| {
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Non-modular Power Residue Sums Given $a, b, k, n \in \mathbb{N}$, what techniques are available for computing sums of the form:
\begin{align}
\sum_{i = 1}^{n} (ai^{k} \ \text{mod} \ b ) ?
\end{align}
NB: Here, only the summands are reduced modulo $b$, not the overall summation.
I'm specifically interested in the $k = 1$ case. For example, if $a = 2$, $b = 3$ and $k = 1$, I can prove that
\begin{align}
\sum_{i = 1}^{n} (2i \ \text{mod} \ 3) = 3 + 2 \left \lfloor \frac{n-1}{3} \right \rfloor + \left \lfloor \frac{n-2}{3} \right \rfloor
\end{align}
by splitting the sum into sums over residue classes. In general, for prime $p$, one has
\begin{align}
\sum_{i =1}^{n} (ai \ \text{mod} \ p) = \binom{p}{2} + \sum_{k = 1}^{p-1} (k a \ \text{mod} \ p) \left \lfloor \frac{n-k}{p} \right \rfloor,
\end{align}
provided that $a$ and $p$ are coprime (otherwise the right side is identically $0$). Although this identity is nice, it is somewhat impractical if $p$ is large. Can the right-side be further simplified?
Edit 1: The formula above continues to hold in the case that $a$ and $b$ are coprime. What can be said about such sums if $a$ and $b$ are not coprime (and $b$ does not divide $a$)?
Edit 2: A slight modification of the formula above continues to hold in the case that $a$ and $b$ are not coprime. In this case, the constant is no longer $\binom{b}{2}$. What is the constant?
Edit 3: Using the comment below I can answer the question about the constant asked in Edit 2, although in this case $a$ and $b$ in the relation above must be replaced with $a/\gcd(a,b)$ and $b/\gcd(a,b)$, respectively, and both terms are multiplied by $\gcd(a,b)$. The constant is then $\gcd(a,b) \binom{b/\gcd(a,b)}{2}$.
Any help or hints are quite welcome!
| Applying the first transformation formula suggested in the comments with the above relation, we have the following identity.
Given $a, b \in \mathbb{N}$ with $d = \gcd(a,b)$, if $b \nmid a$, then for any $n \in \mathbb{N}$,
\begin{align}
S(n;a,b) = d \left( \binom{b / d}{2} + \sum_{k = 1}^{b / d-1} \left( \frac{k a}{d} \ \text{mod} \ \frac{b}{d} \right) \left \lfloor d \left( \frac{n - k}{b} \right) \right \rfloor \right).
\end{align}
The second transformation gives an equivalent, but slightly more involved, right side,
\begin{align}
d \left( \left \lfloor \frac{n}{b} \right \rfloor + 1 \right) \binom{b / d}{2} + d \sum_{k = 1}^{b / d-1} \left( \tfrac{k a}{d} \ \text{mod} \ \tfrac{b}{d} \right) \left( \left \lfloor \frac{n}{b} \right \rfloor \left \lfloor d \left( \frac{b - k}{b} \right) \right \rfloor + \left \lfloor d \left( \frac{n \ \text{mod} \ b - k}{b} \right) \right \rfloor \right).
\end{align}
NB: Applying the transformations repeatedly, I find
\begin{align}
S(n;a,b) = \left( d^{2} \left \lfloor \tfrac{n}{b} \right \rfloor + d \left \lfloor \tfrac{(n \ \text{mod} \ b)}{b/d} \right \rfloor \right) \binom{b/d}{2} + d S(n \ \text{mod} \ (b/d); a/b, b/d),
\end{align}
which thusfar gives me the best possible identity,
\begin{align}
S(n; a, b) & = d ( d \lfloor \tfrac{n}{b} \rfloor + \left \lfloor \tfrac{(n \ \text{mod} \ b)}{b/d} \right \rfloor + 1 ) \binom{b/d}{2} + d \sum_{k =1}^{b/d - 1} (k \tfrac{a}{d} \ \text{mod} \ \tfrac{b}{d}) \left \lfloor \tfrac{(n \ \text{mod} \ (b/d) ) - k}{b/d} \right \rfloor.
\end{align}
| {
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A multiple integral $$\iiint'_{\mathbb{R}^3} \frac{du \; dv \; dw}{|u^3+v^3+w^3|^{\frac{2}{3}}}$$ where the $'$ integration constraint is that only $u,v,w$ for which $|(u^3+v^3)(u^3+w^3)(v^3+w^3)| \leq 1 $ are taken into account. Is it possible to determine whether this integral converges or not?
Do you think that under the transformation $x:=u^3+v^3,y:=u^3+w^3$ etc.. one could even compute a value for the integral ?
| First of all, the integration region looks nasty:
As you suggest, let $x=u^3+v^3$, $y=u^3+w^3$ and $z=v^3+w^3$. Then
$$
\mathrm{d} u \,\, \mathrm{d} v \,\, \mathrm{d} w = \frac{4 \mathrm{d} x \,\, \mathrm{d} y \,\, \mathrm{d} z }{ ( (x+y-z)^2 (x+z-y)^2 (y+z-x)^2 )^{1/3}}
$$
The region becomes $\vert x y z \vert < 1$, and integral becomes
$$
\int_{( x \cdot y \cdot z )^2 < 1} \frac{4 \mathrm{d} x \,\, \mathrm{d} y \,\, \mathrm{d} z }{ ( 4(x+y+z)^2 (x+y-z)^2 (x+z-y)^2 (y+z-x)^2 )^{1/3}}
$$
From this form it seems like the integral would diverge at the origin, since the integrand scales as as $t^{-8/3}$ if all coordinates are rescaled by scale factor $t$.
Added: The integrand, as well as the region of integration, is symmetric with respect to each of $x\to-x$, $y \to -y$, $z \to -z$ as well as any permutations of $(x, y, z)$. Hence we can restrict the integration to $0 < x < y < z$ for the sake of determining the convergence. Consider
$$
\int_0^\infty \mathrm{d} x \int_x^\infty \mathrm{d} y \int_y^\infty \mathrm{d} z \frac{\chi_{x y z < 1}}{{ ( (x+y+z)^2 (x+y-z)^2 (x+z-y)^2 (y+z-x)^2 )^{1/3}} }
$$
In this region $x y z < 1$ implies $0<x<1 \land x < y < \frac{1}{\sqrt{x}} \land y < z < \frac{1}{x y}$:
$$
\int_0^1 \mathrm{d} x \int_x^{\frac{1}{\sqrt{x}}} \mathrm{d} y \int_y^{\frac{1}{x y}} \mathrm{d} z \frac{1}{{ ( (x+y+z)^2 (x+y-z)^2 (x+z-y)^2 (y+z-x)^2 )^{1/3}} }
$$
The only factor in the denominator which could vanish here is $(x+y-z)^2$. This occurs in
the following sub-region:
$$
0 < x < \frac{1}{\sqrt[3]{2}} \land x < y < -\frac{x}{2} + \sqrt{\frac{1}{x} + \frac{x^2}{4}}
$$
Numerically, the integral does seem to diverge, but the proof escape me.
| {
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Solution for a quadratic equation The following is the solution from my book, but without the calculation path.
$$
\frac{(18 \pm \sqrt{-180})}{6} = 3 \pm\sqrt{-5}
$$
Why is it $-5$?
| $$\begin{align}
\frac{1}{6}(18\pm\sqrt{-180})
&= \frac{1}{6} \times (18) \pm \frac{1}{6} \times \sqrt{-180}\\
&= \frac{1}{6} \times (18) \pm \frac{1}{6} \times \sqrt{-4 \times 9 \times 5}\\
&=\frac{1}{6} \times (18) \pm \frac{1}{6} \times (2 \times 3) \times\sqrt{-5}\\
&= 3 \pm\sqrt{-5}
\end{align}$$
| {
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Product of an even number of consecutive positive integers Let $z$ be a positive integer. How should one compute all $z$ such that $5^z-1$ can be written as the product of an even number of consecutive positive integers?
| Here is the case of four consecutive integers. Suppose that the integer $x$ solves
$x(x-1)(x-2)(x-3)=5^z-1$ with $z$ a positive integer. Then we have $x(x-1)(x-2)(x-3)+1=5^z$, where the
left hand side can be factored as $(x^2-3x+1)^2$. Thus $x^2-3x+1$ must also be
a positive power of $5$.
Working modulo $5$, we have
$$0\equiv x^2-3x+1\equiv x^2+2x+1\equiv (x+1)^2,$$
so that $x\equiv -1\pmod 5$.
Thus we can write $x=5q-1$ for some integer $q$.
Substituting this back into $x^2-3x+1$ gives
$25q(q-1)+5$ which can only be a power of $5$ if $q=0$ or $q=1$.
Thus $x$ must be either $-1$ or $4$, and this gives two solutions
$$(-1)(-2)(-3)(-4)=(4)(3)(2)(1)=5^2-1.$$
| {
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A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How could we find the sum of reciprocals of the roots of this equation?
A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How
could we find the sum of reciprocals of the roots of this equation?
I need some hints for solving this problem.
| Since the roots are equal, you must have $b^2 = 4ac = 8c^2$. Also, the roots satisfy
$$0=x^2 + \frac{b}{a}x + \frac{c}{a} = x^2 + \frac{b}{a} + \frac{1}{2},$$
so the product of the roots is $\frac{1}{2}$; that is, $4c^2 = \frac{1}{2}$. Therefore, $b^2=2(4c^2) = 1$.
The sum of the roots is $-\frac{b}{a}$; but it also is equal to $2a$. So $2a = -\frac{b}{a}$. This gives $2a^2 = -b$, and we know $|b|=1$, so we must have $b=-1$. So the sum of the reciprocals is $-4a/b = 4a$.
Now, $a^2 = \frac{1}{2}$, so either $a=\frac{\sqrt{2}}{2}$ or $a=-\frac{\sqrt{2}}{2}$. Thus, the equation is either
$$\frac{\sqrt{2}}{2}x^2 - x + \frac{\sqrt{2}}{4}\quad\text{or}\quad -\frac{\sqrt{2}}{2}x^2 - x -\frac{\sqrt{2}}{4}.$$
Both satisfy the desired conditions: they have a double root at $a=2c$. In one case, the sum of the reciprocals is $2\sqrt{2}$, in the other it is $-2\sqrt{2}$.
| {
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finding the coefficient of $x^{14}$ in the expression: $\frac{5x^2-x^4}{(1-x)^3}$ I have a homework question which requires me to find the coefficient of $x^{14}$ in the expression: $\dfrac{5x^2-x^4}{(1-x)^3}$
I have not figured out a way to do this (I believe this is because my algebra is weak).
This is a question in a combinatorics class and has to do with Generating Functions if that is any help.
Help is greatly appreciated. Thanks, Jason
| First pull a $5$ and an $x^2$ from the rational function. This gives you
$$
\frac{5x^2\left(1 - \left(\frac{1}{5}x^2\right) \right)}{(1-x)^3}.
$$
Then, use the expansions
$$
(1+x^m)^n = 1 - \binom{n}{1}x + \binom{n}{2}x^2 + \dots + (-1)^k\binom{n}{k}x^{km} + \dots + (-1)^n \binom{n}{n}x^{nm}
$$
and
$$
\frac{1}{(1-x)^n} = 1 + \binom{1 + n -1}{1}x + \binom{2 + n -1}{2}x^2 + \dots + \binom{r + n -1}{r}x^r + \dots.
$$
To find the coefficient of $x^{12}$ (Why $x^{12}$?)
| {
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How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
| If $\ c> a ,a^2+c^2 \gt 2ac $, because $\ (a-c)^2 \gt 0$
If $\ c>b>a ,c^2+b^2/2+a^2/2 \gt ac+bc $ and $\ b^2/2+a^2/2 \gt ab $, sum of them $\ a^2+b^2+c^2 \gt ab+bc+ac $
If $\ c=b \gt a $ or $\ a=b=c $, it can be solved with same logic.
| {
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Prove using mathematical induction: $1 \cdot 2 \cdot 3 + \ldots + n \cdot (n+1) \cdot (n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$ So I have the easy stuff done. However, I'm not sure how to go about doing the inductive step with such an awkward proof. :/
Statement:
$$ 1 \cdot 2 \cdot 3 + … + n \cdot (n+1) \cdot (n+ 2) = \frac{n \cdot (n + 1) \cdot (n + 2) \cdot (n +3)}{4} .$$
Base step: $n = 1$:
$$ 0 \cdot (0 + 1) \cdot (0 + 2) + 1 \cdot (1 + 1) \cdot (1 +2) = \frac{ 1 \cdot (1+ 1) \cdot (1 + 2) \cdot (1 + 3) }{4} .$$
$$ 0 \cdot (0 + 1) \cdot (0 + 2) +1 \cdot (1 + 1) \cdot (1 +2) = 0 + 6 = 6$$
$$ \frac{ 1 \cdot (1+ 1) \cdot (1 + 2) \cdot (1 + 3) }{4} = \frac{24}{ 4} = 6 .$$
$$6 = 6.$$
Inductive step:
$n = n + 1$
| $$1 \cdot 2 \cdot 3 + \cdots + n(n+1)(n+2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}.$$
Step 1: (Must prove that the given relation is true for $n=1$)
$$p(1): 1\cdot 2\cdot 3=\frac{1\cdot (1+1)\cdot (1+2)\cdot(1+3)}{4} \Rightarrow 1\cdot 2\cdot 3= 1\cdot 2\cdot 3.$$
The base case is true.
Step 2: Suppose thaat the given relation is valid for $n=k$.
$$p(k): 1\cdot 2 \cdot 3 + \cdots + k(k+1)(k+2) = \frac{k(k + 1)(k + 2)(k +3)}{4}.$$
Step 3: Now we must prove that the given relation is correct for $n=k+1$.
$$p(k+1): 1\cdot 2 \cdot 3 + \cdots + (k+1)(k+2)(k+3) = \frac{(k+1)(k+2)(k + 3)(k +4)}{4}.$$
The above relation can be written:
$$\underbrace{1\cdot 2 \cdot 3 + \cdots + k(k+1)(k+2)} + (k+1)(k+2)(k+3) = \frac{(k+1)(k + 2)(k + 3)(k +4)}{4}.$$
The underlined partion in fact is the step 2. So we have
\begin{align*}
& 1\cdot 2 \cdot 3 + \cdots + k(k+1)(k+2) + (k+1)(k+2)(k+3) \\
& = \frac{k(k + 1)(k + 2)(k +3)}{4} + (k+1)(k+2)(k+3) \\
& = \frac{k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)}{4} \\
& = \frac{(k+1)(k+2)(k+3)(k+4)}{4}.
\end{align*}
| {
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How to find the area of this square?
In a square $ABCD$, say there is a point $P$ which lies inside
it, the point $P$ is located at distances $x$, $y$ and $z$ meters from
$A,B$ and $C$ respectively.
Using this information how could we compute a form for the area of the square?Please explain your approach.
| Using coordinate geometry. Assume that the vertices (in order) are $(0,a), (0,0), (a,0), (a,a)$ and that $P$ is $(u,v)$. Then $$
\begin{align}
u^2+(v-a)^2 &= x^2,\tag{1}
\\ u^2+v^2 &= y^2, \tag{2}
\\ (u-a)^2+v^2 &= z^2 \tag{3}.
\end{align}
$$
We can "solve" for $v$ in terms of $a$ by subtracting $(2)$ from $(1)$; similarly for $u$. Since the steps are straightforward, I am just posting the answer here:
$$
(u, v) = \left( \frac{a^2 + y^2 - z^2}{2a}, \frac{a^2 + y^2 - x^2}{2a} \right).
$$
Plugging this in $(2)$, we get:
$$
(a^2+y^2 - z^2)^2 + (a^2 + y^2 - x^2)^2 = 4a^2y^2.
$$
$$
\implies a^4 + y^4 - (a^2+y^2)(x^2+z^2)+\frac{x^4+z^4}{2}=0.
$$
This is a quadratic equation in $a^2$(=area of the square) and can be solved.
Note. I do not yet understand whether both the roots of the quadratic are true solutions or not.
| {
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Evaluating $ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $ I would like to evaluate: $$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $$
$$ \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}=\frac{\sqrt{1-x}+\sqrt{1+x}-2}{2(\sqrt{1-x^2}-1)} $$
The substitution $ x \rightarrow \sin(x) $ or $ \cos(x) $ can only simplify the denominator, and $ x \rightarrow \sqrt{1+x}$ or $ \sqrt{1-x} $ is also useless...
Can you help me find a useful substitution?
$$ x=\cos(2t) $$
$$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx=-\int {\frac{\sqrt{2}\sin(t)\cos(t)}{\sqrt{2}+\sin(t)+\cos(t)}}\mathrm dt $$
$$ u=\tan(t/2) $$
$$ -4\sqrt{2}\int \frac{u(1-u^2)}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})}\mathrm du $$
But now it looks even more complicated... ?
| Let $w_+ = \sqrt{1+x}$ and $w_- = \sqrt{1-x}$. Then
$$
\begin{eqnarray}
\frac{1}{2+w_+ + w_-} &=& \frac{(2 - w_+ + w_- )(2 + w_+ - w_- )(2 - w_+ - w_- )}{(2 + w_+ + w_- )(2 - w_+ + w_- )(2 + w_+ - w_- )(2 - w_+ - w_- )} \\ &=& \frac{4 w_- w_+ - 2 x w_- - 4 w_- +2 x w_+ - 4 w_+ + 4}{4 x^2}
\end{eqnarray}
$$
This can now be integrate term-wise.
| {
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Find the limit as $x$ tends to $-\infty$ Find the limit as $x$ tends to $-\infty$ of
$$ f(x)=\frac{\sqrt{x^2+1}}{x+1} $$
I did
$$ f(x) = \frac{\sqrt{1+1/x^2}}{1+1/x}\to \frac{\sqrt{1+0}}{1-0} =1 $$
(as $x$ tends to -infinity, $1/x^2$ tends to $0$ and $1/x$ tends to $0, 0$ gets negative sign fro negative infinity)
But the correct answer is $-1$, where did I go wrong?
| If $x<0$, then $x=-\left\vert x\right\vert =-\sqrt{x^{2}}$. Therefore, for $x<0$,
$$f(x)=\frac{\sqrt{x^{2}+1}}{x+1}=\dfrac{\dfrac{\sqrt{x^{2}+1}}{x}}{\dfrac{x+1}{x}}=
\dfrac{\dfrac{\sqrt{x^{2}+1}}{-\sqrt{x^{2}}}}{\dfrac{x+1}{x}}=\dfrac{-\sqrt{
\dfrac{x^{2}+1}{x^{2}}}}{\dfrac{x+1}{x}}=-\dfrac{\sqrt{1+\dfrac{1}{x^{2}}}}{1+
\dfrac{1}{x}}\rightarrow -1,$$ as $x\to -\infty$.
Plot of $f(x)$
| {
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Integral solutions of $x^2+y^2+1=z^2$ I am interested in integral solutions of $$x^2+y^2+1=z^2.$$ Is there a complete theory comparable to the one for $x^2+y^2=z^2?$
| All of numbers can be any character.In Equation: $qX^2+Y^2=Z^2+a$
If the ratio is factored so: $a=(b-c)(b+c)$
Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$
where: $f=(q+1)k^2-2kt-(q-1)t^2$
Then the solutions are of the form:
$X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$
$Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$
$Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$
All of numbers can be any character.
| {
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Formula for completing the square? My math teacher said that this was the formula for completing the square.
Original function: $$ax^2 + bx + c$$
Completed square: $$a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$
However, using this formula I'm not getting the same answers that I would get just by determining the stuff myself. Is this correct?
| Note that
$$\begin{align*}a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c&=a\left(x^2+2\left(\frac{b}{2a}\right)x+\left(\frac{b}{2a}\right)^2\right)- \frac{b^2}{4a} + c\\ &=a\left(x^2+\left(\frac{b}{a}\right)x+\frac{b^2}{4a^2}\right)- \frac{b^2}{4a} + c\\
&=\left(ax^2+bx+\frac{b^2}{4a}\right)- \frac{b^2}{4a} + c\\\\
&= ax^2+bx+c \end{align*}$$
so the formula is correct. Try plugging in the numbers $a$, $b$, and $c$ you are using to each step here and seeing where they begin to differ; that will be where your error is.
| {
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"url": "https://math.stackexchange.com/questions/76772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Where is the mistake in this logarithmic equation? The problem is:
Given this equation find the possible values of $x$.
$$\log_{2}\left ( \frac{x+2}{x-1} \right )+\log_{2}\left [ (x+2)(x-1) \right ]=2$$
First I defined the domain of the function.
*
*The denominator $x-1$ must not be zero: $x-1\neq0$.
*The fraction $\frac{x+2}{x-1}$ must be bigger than zero: $\frac{x+2}{x-1}>0$.
*The product $(x+2)(x-1)$ must be bigger than zero: $(x+2)(x-1)>0$.
In the end I concluded that the domain should be $\left ]-\infty,-2\right [\cup \left ]1,+\infty\right[$.
Then I sarted to solve:
$$\begin{align*}
\log_{2}\left ( \frac{x+2}{x-1} \right )&+\log_{2}\left [ (x+2)(x-1) \right ]=2
\\
&\iff
\log_{2}(x+2)-\log_{2}(x-1)+\log_{2}(x+2)+\log_{2}(x-1)=2\\
&\iff
2 \cdot \log_{2}(x+2)=2\\
&\iff \log_{2}(x+2)=1 \\
&\iff x+2=2^1\\
&\iff x=2-2\\
&\iff x=0.
\end{align*}$$
This result is not in the domain!
Where I miss?
| If $x>1$, then $\log_2\left(\dfrac{x+2}{x-1}\right) = \log_2(x+2)-\log_2(x-1)$.
If $x<-2$ then $\log_2\left(\dfrac{x+2}{x-1}\right) = \log_2\left(\dfrac{-x-2}{1-x}\right) = \log_2(-x-2)-\log_2(1-x)$.
A similar thing applies to the logarithm of the product.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Irreducible polynomial which is reducible modulo every prime
How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$?
For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not solutions of $x^4+1=0$ then if it is reducible the factors are of degree $2$. This gives that $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and solving this system of equations $\bmod 3$ gives that $x^4+1=(x^2+x+2) (x^2+2x+2) \pmod 3$. But is there a simpler method to factor $x^4+1$ modulo a prime $p$?
| If $-1$ is a square in $\Bbb F_p$ (which includes the case $p=2$), say $a^2=-1$, then we have $$X^4+1=X^4-a^2=(X^2+a)(X^2-a).$$
If $p$ is odd and $2$ is a square in $\Bbb F_p$, say $2=b^2$, then we have
$$X^4+1=(X^2+1)^2-(bX)^2=(X^2+bX+1)(X^2-bX+1) $$
If $p$ is odd and neither $-1$ nor $2$ is a square, then their product $-2$ is a square, say $-2=c^2$. (Without using anything even remotely as deep as quadratic reciprocity, this follows immediately from the fact that $\Bbb F_p^\times$ is a cyclic group of even order). Then we have
$$ X^4+1=(X^2-1)^2-(cX)^2=(X^2-cX-1)(X^2+cX-1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the equation $\frac{2x+3}{x+1} = \frac{2x+2}{x-1}$ I'm a college student and I've got a problem solving this:
$$\frac{2x+3}{x+1} = \frac{2x+2}{x-1}.$$
Since neither the numerators nor the denominators are equal, I figured that $(2x+3)(x-1) = (x+1)(2x+2)$ would provide me with a suitable $x$.
However calculating this I found that $x = -\frac13$ while my answer sheet states that $x = -\frac23$. What am I doing wrong?
| Well you have by cross multiplying that,$$2(x+1)^{2} = 2x^{2} + x -3$$ $$ \Longrightarrow 2x^{2}+4x+2 = 2x^{2} + x -3$$ $$ \Longrightarrow 3x = -5 \Rightarrow x= -\frac{5}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Writing complete elliptic integral of first kind as a hypergeometric function I am trying to show $K=\int_{0}^{\frac{\pi}{2}}\frac{dz}{\sqrt{1-k^{2}\sin^{2}(z)}}$ can be written as $\frac{\pi}{2} \mathstrut_{2}F_{1}(\frac{1}{2},\frac{1}{2};1;k^{2})$.
First I used $(1+z)^{n}=\mathstrut_{2}F_{1}(-n,b;b;-z)$ to expand the integrand to get $\frac{dz}{\sqrt{1-k^{2}\sin^{2}(z)}} = 1+\frac{1}{2}k^{2}\sin^{2}(z) + \ldots +\frac{(2n-1)!!}{2^{n}n!}k^{2n}\sin^{2n}(z)$ but when I go to integrate with respect to $z$ I have to integrate $\sin^{2n}(z)$. I'm not sure where to go since even integrating the first few terms I get a $\cos(z)$ and plugging in $\frac{\pi}{2}$ kills the term which it shouldn't. Thanks for any help or tips you can give.
| As an alternative way of deriving the representation:
$$
K(k) = \int_0^\frac{\pi}{2} \frac{\mathrm{d} z}{\sqrt{1-k^2 \sin^2 (z)}}
\stackrel{u = \sin^2(z)}{=} \int_0^1 \frac{1}{2}\frac{\mathrm{d} u}{\sqrt{\left(1-k^2 u\right) \left( 1-u\right) u }}
$$
The latter is exactly the Euler's integral representation of the Gauss's hypergeometric function with $b=\frac{1}{2}$, $c-b=\frac{1}{2}$ and $a=\frac{1}{2}$, $z=k^2$, i.e. $b=\frac{1}{2}$, $c=1$, $a=\frac{1}{2}$:
$$
K(k) = \frac{1}{2} B\left(\frac{1}{2}, \frac{1}{2}\right) {}_2 F_1\left( \frac{1}{2}, \frac{1}{2}; 1; k^2 \right) = \frac{\pi}{2} {}_2 F_1\left( \frac{1}{2}, \frac{1}{2}; 1; k^2 \right)
$$
| {
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Need help finding smallest value of $x^2 + y^2$ I need to find the smallest value of $x^2 + y^2$ with the restriction $2x + 3y = 6$. This chapter focuses on the vertex formula.
| Using the Cauchy-Schwarz inequality, we have $6^2=(2x+3y)^2\leq (x^2+y^2)(2^2+3^2)$, which gives the minimum value of $x^2+y^2$ to be $\frac{36}{13}$.
Edit:Equality occurs for $\frac{x}{2}=\frac{y}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $r^4\cos^2\theta\sin\theta$ Over an Ellipse I was tutoring someone and one of the problems was to integrate $r^4\cos^2\theta\sin\theta$ over the portion of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ that lies in the first quadrant. So, $0\leq \theta\leq\frac{\pi}{2}$. But, given a $\theta$, $r$ runs from $0$ to
$$
\sqrt{\frac{36+36\tan^2\theta}{9+4\tan^2\theta}}.
$$
After integrating with respect to $r$ first, it looks very nasty.
Is there some clever method, such as a change of variables, that makes this easy to integrate?
| The integral yields to standard substitutions. Sometime when I am feeling in a clever mood I may come up with a quick to verify solution. At this point, I am interested in showing that the integral can be done by absolutely routine substitutions. By analyzing the substitutions, we could come up with shortcuts
We will not pay attention to multiplicative constants, since we would probably get them wrong anyway. After the integration with repsect to $r$, we end up with an $r^5$ term. But $r^2=36/(9\cos^2\theta+4\sin^2\theta)$. So we end up wanting something that (apart from multiplicative constants) looks like
$$\int_0^{\pi/2}\frac{\cos^2\theta\sin\theta}{(9\cos^2\theta+4\sin^2\theta)\sqrt{9\cos^2\theta +4\sin^2\theta}}d\theta.$$
Replace $9\cos^2\theta+4\sin^2\theta$ by $4+5\cos^2\theta$ and make the substitution $u=\cos\theta$. We end up wanting an integral of the shape
$$\int_0^1 \frac{u^2}{(4+5u^2)^2\sqrt{4+5u^2}}du.$$
In this situation, it is standard to make a substitution of the shape $\sqrt{5}u=2\tan\phi$, or something similar with hyperbolic sine. After a little work we end up with something that, apart from constants, looks like the absolutely familiar
$$\int \sin^2\phi\cos\phi\; d\phi.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int_0^\infty \frac{x^2}{1+x^5} \mathrm{d} x$ by contour integration Consider the following integral:
$$\int_{0}^{\infty} \frac{x^{2}}{1+x^{5}} \mathrm{d} x \>.$$
I did the following: Since $-1$ is a pole on the real axis, I took $z_{1}=e^{3\pi/5}$ then constructed an arc between $Rz_{1}$ and $R$ :
$f(e^{2\pi i /5}z)=f(z)$ it follows that : $(1-e^{2pi i/5})\int_{\alpha}f(z)dz + \int_{\beta}f(z)dz = 2\pi i \text{ Res } z_{2} f $ ,
where $\alpha$ is the way along the eral axis, $\beta$ is the way along the arc and $z_{2}=e^{i\pi/5}$
This gives: $\int_{0}^{\infty}...dx = 2\pi i (\frac{e^{2i\pi/5}}{-5})\cdot \frac{1}{(1-e^{2\pi i/5})}$
The real part of the right side in wolframalpha does not give the same result as the left side.
Does anybody see why? Please do tell.
| I ran out of patience, but the partial fraction approach can give an indefinite integral in closed form, as the tenth roots of 1 have real and imaginary parts that can be written with nothing worse than square root signs. I got
$$ \frac{x^2}{1 + x^5} \; = \; \; \frac{A}{x + 1} + \frac{B_1 x + B_0}{x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1} + \frac{C_1 x + C_0}{x^2 + \frac{1}{2}(\sqrt 5 - 1)x + 1}.$$
Then I got $$ A = \frac{1}{3}, \; B_0 = B_1 = \frac{1}{6}(\sqrt 5 - 1), \; \; C_0 = C_1 = -\frac{1}{6}(\sqrt 5 + 1).$$
The two quadratic denominators terms are positive over the reals. The rest of the task is also cookbook, you rewrite, for example $B_1 x + B_0$ as $\frac{B_1}{2}$ times the derivative of $x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1$ plus a constant term , call it $B_3.$ So, for this part, you get the indefinite integral being
$$ \frac{B_1}{2} \log(x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1) + B_3 \arctan(linear) $$
I imagine you can force your computer algebra system to show all the steps in this calculation while getting the specific numbers right. The beginning is to define
$$ \omega = e^{\pi i / 5} = \frac{1}{4}(1 + \sqrt 5) + \frac{i}{4} \sqrt{10 - 2 \sqrt 5} $$
with
$$ 1 + x^5 = (x + 1)(x - \omega)(x - \omega^3)(x - \omega^7)(x - \omega^9) $$
| {
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Finding the Derivative of |x| using the Limit Definition Please Help me derive the derivative of the absolute value of x using the following limit definition.
$$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}
$$
I have no idea as to how to get started.Please Help.
Thank You
| We can write $|x| = \sqrt{x^2}$. Using the chain rule we then get
$$|x|' = \frac{1}{2\sqrt{x^2}} \cdot 2x = \frac{x}{\sqrt{x^2}} = \frac{x}{|x|}$$
EDIT Using limit:
$$\begin{align}
\frac{\sqrt{(x+\Delta x)^2}-\sqrt{x^2}}{\Delta x}
& = \frac{(\sqrt{(x+\Delta x)^2}-\sqrt{x^2})(\sqrt{(x+\Delta x)^2}+\sqrt{x^2})}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{(x+\Delta x)^2-x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{2 x \Delta x + \Delta x^2}{\Delta x (\sqrt{(x+\Delta x)^2}+\sqrt{x^2})} \\
& = \frac{2x + \Delta x}{\sqrt{(x+\Delta x)^2}+\sqrt{x^2}} \\
& \to \frac{2x}{\sqrt{x^2}+\sqrt{x^2}} = \frac{2x}{2|x|} = \frac{x}{|x|}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Square Roots of Complex Number $3-4i$
What I did
$z^2=3-4i$
$(a+bi)^2 = 3-4i$
$a^2-b^2+2abi = 3-4i$
Then got 2 simultaneous equations
$a^2-b^2=3$ and $2ab=-4$
Solve for $a^2$ in 1st equation: $a^2=3+b^2$
Subbed into 2nd equation to power of 2
$(3+b^2)b^2=4$
$b^4+3b^2-4=0$
Is there a better way than below? Solving power 4 equation then cubic?
Then solved solved power 4 equation ...
$(b-1)(Ab^2+Bb^2+Cb+D)=b^4+3b^2-4$
$(b-1)(b^3+b^2+4b+4)=b^4+3b^2-4$
then solved for cubic equation, getting ...
$(b-1)(b+1)(b^2+4)=0$
So
$b=\pm 1 \text{ or } 2i$
What did the book do now?
I did: (subbing into $a^2=3+b^2$)
When $b=1, a^2=3+1^2 \Longrightarrow a = 2$
When $b=-1, a^2=3+(-1)^2 \Longrightarrow a=2$
When $b=2i, a^2=3+(2i)^2=3+4(-1)=-1, a=i$
So I will have 3 equations
*
*$2+i$ // why is this not in the book?
*$2-i$
*$i-2$
UPDATE (In response to @David Mitra)
Ok. I let $x=a^2$ and $b=b^2$ giving the quadratic equation: $x=3+y$ and $xy=4$. Then (after subbing) $(y-1)(y+4)=0$
Then when $y=1, x=4$. $y=-4, x=-1$
Then $b=\pm 1 or \pm 2i$ and $a=\pm 2i or \pm i$
Finally testing equations:
$(2+i)^2 = 3+4i$ (rej)
$(2-i)^2=5$ what to do?
$(i-2)^2=3$
$(-2-i)^2=3$
... 8 equations (but 4 unique)
| $b^4+3b^2-4=0$ is a quadratic equation in $b^2$, i.e. you can write $c=b^2$, and then the equation says $c^2+3c-4=0$. If you know how to solve quadratic equations, you've got it.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find x in $4^{\sin^2x}+4^{\cos^2x}=8$ $$4^{\sin^2x}+4^{\cos^2x}=8$$
I solved like this:
\begin{align*}4^{\sin^2x}+4^{\cos^2x}=8&\Rightarrow4^{\sin^2x}+4^{1-\sin^2x}=8\\
&\Rightarrow4^{\sin^2x}+\frac{4}{4^{\sin^2x}}=8 |\cdot4^{\sin^2x}\\
&\Rightarrow4^{2\sin^2x}-8\cdot4^{\sin^2x}+4=0\\
y=4^{\sin^2x}&\Rightarrow y^2-8y+4=0\\
&\Rightarrow\Delta=64-16=48\\
&\Rightarrow y_{1,2}=\frac{8\pm 4\sqrt{3}}{2}\\
&\Rightarrow y_{1,2}=4\pm 2\sqrt{3}\\
&\Rightarrow 4^{\sin^2x}=4 \pm 2\sqrt{3}
\end{align*}
But now I'm stuck.
| This equality cannot happen. Since $\sin^2 x , \cos^2 x \in [0,1]$ it follows that $$4^{\sin^2 x}+4^{\cos^2 x}\leq 4+4=8$$ with equality when $\sin^2x=\cos^2x=1$. The last equality is impossible.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to get from $2^{99} \cdot (\cos{(99\times \frac{5\pi}{6}) + i\cdot \sin{(99\times \frac{5\pi}{6})}})$ to $0+2^{99}i$?
How do I get from 2nd last to last step? How did they simplify cos & sin $99\times \frac{5\pi}{6}$?
| $\cos \left(99 \times \frac{5 \pi}{6} \right) = \cos \left(33 \times \frac{5 \pi}{2} \right) = \cos \left(\frac{165 \pi}{2} \right) = 0$ since $\cos(\frac{n \pi}{2}) = 0$ whenever $n$ is odd.
$\sin \left(99 \times \frac{5 \pi}{6} \right) = \sin \left(33 \times \frac{5 \pi}{2} \right) = \sin \left(\frac{165 \pi}{2} \right) = \sin \left(82 \pi + \frac{\pi}{2} \right) = 1$ since $\sin(2n \pi + \frac{\pi}{2}) = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{n^n}{3^n} < n! < \frac{n^n}{2^n}$ holds by induction I am Computer Science person, and I'm trying to beef up my mathematical skills a little bit. This isn't homework and I don't have a professor I can go to to ask for help. I hope this is an appropriate forum for this type of question.
To learn more about inductive proofs, I am reading the Wikipedia article on them. However, I am having trouble solving one of the examples on the page:
$$P(n) :\quad \frac{n^n}{3^n} < n! < \frac{n^n}{2^n} .$$
Prove $P(n) \; \forall n \in \mathbb{N}, n \ge 6$
Here is what I've done so far, with the "???" indicating where I don't know the next step.
Base case:
$$\begin{array}{rrcccl}
P(6)\colon& \frac{6^6}{3^6} &<& 6! &<& \frac{6^6}{2^6}\\
\iff& 2^6 &<& 6! &<& 3^6\\
\iff& 64 &<& 720 &<& 729
\end{array}$$
Inductive step:
$$\begin{array}{rrccclc}
P(n+1) \colon&\frac{(n+1)^{n+1}}{3^{n+1}} &<& (n+1)! &<& \frac{(n+1)^{n+1}}{2^{n+1}}\\
\iff& \frac{(n+1)^{n+1}}{3^{n+1}} &<& (n+1)n! &<& \frac{(n+1)^{n+1}}{2^{n+1}}
&\text{Defn factorial}\\
\iff& \frac{(n+1)(n+1)^n}{3 \cdot 3^n} &<& (n+1)n! &<& \frac{(n+1)(n+1)^n}{2 \cdot 2^n}
&\{a^{m+n} = a^m a^n\}\\
\iff& \frac{(n+1)^n}{3 \cdot 3^n} &<& n! &<& \frac{(n+1)^n}{2 \cdot 2^n}
&\text{Divide by n+1}\\
???
\end{array}$$
Perhaps this just looks hopelessly foolish from the point of view of mathematicians, but I'm trying to learn.
Specifically what I am looking for is either guidance to arrive at the answer on my own, or pointers to documentation I should read to further my own understanding.
| For the Inductive Step:
Suppose that we have
$$
\frac{n^n}{3^n} < n!
$$
and
$$
n! < \frac{n^n}{2^n}.
$$
We want to show that this implies that
$$
\frac{(n+1)^{n+1}}{3^{n+1}} < (n+1)!
$$
and
$$
(n+1)! < \frac{(n+1)^{n+1}}{2^{n+1}}.
$$
Under these assumptions, to see the first inequality holds, write:
$$\begin{align}
\frac{(n+1)^{n+1}}{3^{n+1}} &= \frac{(n+1)}{3} \frac{(n+1)^n}{3^n}
\\
&= \frac{n+1}{3} \left(\frac{n+1}{n}\right)^n \frac{n^n}{3^n}
\\
&< \frac{n+1}{3} \left(\frac{n+1}{n}\right)^n n! \qquad (\text{using inductive hypothesis})
\\
&= \left( 1 + \frac{1}{n} \right)^n \frac{1}{3} (n+1)!
\end{align}$$
and this reduces to showing that
$$
\left( 1 + \frac{1}{n} \right)^n \frac{1}{3} < 1.
$$
The other inequality can be handled similarly. You can probably save some paper by doing both inequalities at the same time, but if you are just starting out, I would suggest working with one inequality at a time.
| {
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Infinitely many primes of the form $6\cdot k+1$ , where $k$ is an odd number? How to prove that there are infinitely many primes of the form $6k+1$ , where $k$ is an odd number ?
Here is a proof that there are infinitely many primes of the form $6k+1$ :
We will assume that there are a finite number of primes of the form $6k+1$ .
Let $p_1 = 6k_1+1, p_2 = 6k_2+1$, ... Then,
$p_1p_2 = (6k_1+1)(6k_2+1) = 36k_1k_2 + 6k_1 + 6k_2 + 1$
$p_1p_2 = 6\cdot f(k_1,k_2) + 1$, where $f(k_1,k_2) = 6k_1k_2 + k_1 + k_2$
i.e. the product of two numbers of the form $6k+1$ is also a number of the form $6k+1$.
Hence $p_1p_2...p_n$ will be a number of the form $6k+1$
i.e. $ p_1\cdot p_2...p_n = 6f(k_1,k-2,...,k-n) + 1$
$(2p_1p_2...p_n)^2 = 4(36f^2 +12f + 1) = 6g + 4$, where $g = 24f^2 + 8f$,
$(2p_1p_2...p_n)^2 + 3 = 6g + 4 + 3$
$(2p_1p_2...p_n)^2 + 3 = 6g + 6 + 1$
$(2p_1p_2...p_n)^2 + 3 = 6g' + 1$
i.e. $N = 6g' + 1$, a number of the form $6k+1$
Now, by construction, $N = (2p_1p_2...p_n)^2 + 3$, and is not divisible by any of the $p_i$.
Hence it is either prime itself, or divisible by another prime greater than $p_n$, contradicting the assumption.
I.e. There are an infinite number of primes of the form $6k+1$
Can I use the method of this proof to prove that there are infinitely many primes of the form $6k+1$ , where $k$ is an odd number or to choose some other proof strategy ?
| edit, Monday, December 5: Pedja's argument can me made to work with the addition of a quadratic reciprocity step...
original: Pedja, you use Dirichlet on the arithmetic progression $12 n + 7,$ which is what you are describing by saying $6k+1$ with $k$ odd. Write $k = 2n+1,$ then $6k+1 = 6 (2n+1) + 1 = 12n + 6 + 1 = 12 n + 7.$
Meanwhile, if you are willing to accept the Cebotarev Density Theorem, asymptotically the two positive quadratic forms $x^2 + 12 y^2$ and $3 x^2 + 4 y^2$ represent the same proportion of primes. The first form represents primes $p \equiv 1 \pmod {12},$ while the second form represents 3 and the primes $q \equiv 7 \pmod {12}.$ I am quoting Theorem 9.12 on page 188 of Primes of the Form $x^2 + n y^2$ by David A. Cox.
EDIT: Actually, your attempt is almost correct, there is an elementary proof of this one. As you write above, take the collection of primes $q_j \equiv 7 \pmod{12}$ that is assumed to be complete, up to some largest we will call $q_r.$ Then, just as you wrote, define
$$ w = 3 + 4 \left( q_1 q_2 \ldots q_r \right)^2 $$
We know that $w \equiv 7 \pmod {12}.$ We also know that $w$ is primitively represented by the quadratic form $3 x^2 + 4 y^2.$ That means that $\gcd(x,y)=1,$ where here $x=1,$ and we are writing $w = 3 x^2 + 4 y^2.$
Now, $w$ is not divisible by 2 or 3. The fact that $\gcd(x,y)=1$ means precisely that $w$ is not divisible by any prime $s \equiv 2 \pmod 3.$ This fact comes under the general heading of quadratic reciprocity. All prime factors of $w$ are, in fact , $1 \pmod 3.$ That is, all prime factors of $w$ are either $1 \pmod {12}$ or $7 \pmod {12}.$ If all prime factors of $w$ were $1 \pmod {12},$ then $w$ itself would also be $1 \pmod {12},$ which it is not.
So, in fact, either $w$ is prime, or $w$ has at least one prime factor, call it $q_{r+1},$ such that $q_{r+1} \equiv 7 \pmod {12}.$ By construction, for any $j \leq r,$ we have $\gcd(q_j,q_{r+1} ) = 1.$ That is, either $w$ itself or $q_{r+1}$ is not in the assumed finite list of primes $7 \pmod {12},$ contradicting the assumption that there is a finite list. $\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$
EDIT TOO: Maybe not everyone will know this, so I will give the short proof. Suppose we have some prime $s \equiv 2 \pmod 3,$ with $s \neq 2,$ that divides some $ z = 3 u^2 + 4 v^2.$ Then suppose that $s | z,$ which is also written $ z \equiv 0 \pmod s.$ I claim that, in fact, both $s | u$ and $s | v,$ with the result that $\gcd(u,v)$ is larger than one. PROOF: We have $ 3 u^2 + 4 v^2 \equiv 0 \pmod s.$ Assume that $v \neq 0 \pmod s.$ Then $v$ has a multiplicative inverse $\pmod s.$ So we get
$$ 3 u^2 + 4 v^2 \equiv 0 \pmod s,$$
$$ 3 u^2 \equiv - 4 v^2 \pmod s,$$
$$ \frac{3 u^2}{v^2} \equiv - 4 \pmod s,$$
$$ \frac{ u^2}{v^2} \equiv \frac{- 4}{3} \pmod s,$$
$$ \left(\frac{ u}{v}\right)^2 \equiv \frac{- 4}{3} \pmod s.$$
This says that the right hand side is a quadratic residue $\pmod s.$ It also means that $-12$ is a quadratic residue $\pmod s.$ This is false, though, as we have Legendre symbol
$$( -12 | s) = ( -3 | s) = ( -1 | s ) \cdot ( 3 | s).$$
Now, if $s \equiv 1 \pmod 4,$ we can ignore the $-1$ and switch the other pair. If $s \equiv 3 \pmod 4,$ then $( -1 | s ) = -1$ and $(3 | s) = - (s | 3)$.
In either case, we get
$$ (-3 | s) = (s | 3).$$
As $s \equiv 2 \pmod 3,$ we have
$$ ( s | 3) = ( 2 | 3) = -1.$$
So, in the end we have
$$( -12 | s) = -1,$$
and this contradicts the assumption that $v$ is nonzero $\pmod s.$ Thus, $v \equiv 0 \pmod s.$ From $ 3 u^2 + 4 v^2 \equiv 0 \pmod s,$ we then have $u \equiv 0 \pmod s.$ So $ s | \gcd(u,v)$ and $\gcd(u,v) \neq 1.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve recurrence equations-homework extras Extras from my homework. The first one should be easier, but still hard enough.
1) $a_{n+3}-(3/2)a_{n+2}-a_{n+1}-(1/4)a_n=0$
2) $a_{n+3}-3a_{n+2}-3a_{n+1}+a_n=n^2+2^n$
| Use Wilf's "generatingfunctionology" techniques. Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$ to get for (1):
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
- \frac{3}{2} \frac{A(z) - a_0 - a_1 z}{z^2}
- \frac{A(z) - a_0}{z}
- \frac{1}{4} A(z) = 0
$$
Solve for $A(z)$, split into partial fractions. The expansions:
$$
(1 - u)^{-m} = \sum_{n \ge 0} \binom{n + m - 1}{m - 1} u^n
$$
will be all that is needed. It works with complex roots too.
For (2) you'll need:
$$
\sum_{n \ge 0} n^2 z^n
= z \frac{d}{dz} ( z \frac{d}{dz} \sum_{n \ge 0} z^2 )
= z \frac{d}{dz} ( z \frac{d}{dz} \frac{1}{1 - z} )
= \frac{z + z^2}{(1 - z)^3}
$$
and also:
$$
\sum_{n \ge 0} 2^n z^n
= \sum_{n \ge 0} (2 z)^n
= \frac{1}{1 - 2 z}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/91242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that if a prime $p$ divides a Fermat Number then $p=k\cdot 2^{n+2}+1$?
If a prime $p$ divides a Fermat Number then $p=k\cdot 2^{n+2}+1$?
Does anyone know a simple/elementary proof?
| Here's an explanation that avoids group theory and gives a pointer to the number theory results used.
To start with: we know that $2$ has a multiplicative order $\text{ord}_p(2)$ such that $2^m \equiv 1 \pmod p$ if and only if $m$ is a multiple of $\text{ord}_p(2)$. We are given that $p$ divides $2^{2^n} + 1$; therefore $2^{2^n} \equiv -1 \pmod p$ and (squaring that) $2^{2^{n+1}} \equiv 1 \pmod p$. Therefore $2^{n+1}$ is a multiple of $\text{ord}_p(2)$, but $2^n$ is not; this can only happen if $\text{ord}_p(2) = 2^{n+1}$.
By Fermat's little theorem, we know that for any prime $p$, $2^{p-1} \equiv 1 \pmod p$. Therefore $p-1$ is a multiple of $\text{ord}_p(2) = 2^{n+1}$. This tells us that $p = k \cdot 2^{n+1} + 1$ for some $k$, which is almost what we want; we just want to show that $k$ is even.
To do this, we turn to quadratic residues: the question of when $x^2 \equiv a \pmod p$ has a solution $x$. By Euler's criterion, a solution exists if and only if $a^{\frac{p-1}{2}} \equiv 1 \pmod p$. In our particular case, does $x^2 \equiv 2 \pmod p$ have a solution? Applying Euler's criterion, we test if $2^{\frac{p-1}{2}} = 2^{k \cdot 2^n} \equiv 1 \pmod p$. Well, $2^{2^n} \equiv -1 \pmod p$, so $2^{k \cdot 2^n} \equiv (-1)^k \pmod p$. So we conclude that $x^2 \equiv 2 \pmod p$ has a solution if and only if $k$ is even.
Finally, by the $\pm 2$ supplement to the law of quadratic reciprocity, we have a second way of testing if $x^2 \equiv 2 \pmod p$ has a solution: this happens if and only if $p \equiv \pm 1 \pmod 8$. In our case, $k \cdot 2^{n+1} +1 \equiv 1 \pmod 8$ as long as $n \ge 2$. Therefore $x^2 \equiv 2 \pmod p$ must have a solution, and from the previous paragraph, $k$ must be even, finishing the proof.
(The result does not hold for $2^{2^0} +1 = 3$ and $2^{2^1} = 5$, since these are prime themselves, but do not have the correct form.)
| {
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"url": "https://math.stackexchange.com/questions/91413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Sum of powers-of-integers bound I am looking for a nice proof of this inequality:
\begin{equation}
\sum_{k=1}^{n-1} k^n < n^n, \quad n > 0.
\end{equation}
Example: $1^4 + 2^4 + 3^4 = 1+16+81 = 98 < 4^4 = 256.$
| For all $0 \leqslant r \leqslant n-1$, we have
$$
\frac{(r+1)^n}{r^n} = \left( \frac{r+1}{r} \right)^n = \left( 1 + \frac{1}{r} \right)^n \geqslant 1 + \frac{n}{r} \geqslant 2,
$$
so $r^n \leqslant \frac{1}{2} (r+1)^n$. Now by a simple induction on $k$, we obtain
$$
(n-k)^n \leqslant \frac{n^n}{2^k}.
$$
Summing over $k = 1, 2, \ldots, n-1$, we get
$$
\sum_{k=1}^{n-1} (n-k)^n \leqslant n^n \sum_{k=1}^{n-1} 2^{-k} < n^n \sum_{k=1}^{\infty} 2^{-k} = n^n,
$$
which is what we wanted to show.
This approach can squeeze out a little more. In the very first step, assuming $\varepsilon > 0$ and $n$ sufficiently large, we have
$$
\frac{(r+1)^n}{r^n} = \left( 1 + \frac{1}{r} \right)^n \geqslant \left( 1 + \frac{1}{n} \right)^n \geqslant \mathrm e - \varepsilon.
$$
Plugging in this refined bound in the above proof, we get the upper bound
$$
\sum_{k=1}^{n-1} k^n \lt n^n \cdot \sum_{k=1}^{\infty} (\mathrm{e} - \varepsilon)^{-k} = \frac{n^n}{\mathrm e - \varepsilon - 1}.
$$
This gives
$$
\limsup_n \ n^{-n} \cdot \sum_{k=1}^{n-1} k^n \leqslant \frac{1}{\mathrm e - 1},
$$
that matches (one half of) the limit obtained by Mike Spivey.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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About one generating function Initially, I have the following problem: find $$\sum_{k=0}^{n+1}(−1)^{n−k}4^k{n+k+1 \choose 2k}.$$ I thought, if I found the function $g_n(x) = \sum_{k=0}^{n}{n+k \choose 2k}x^k$, the answer would be $(-1)^ng_{n+1}(-4)$. I tried to factorize ${n + k \choose 2k}$ to ${n+k \choose k}{n \choose k}/{2k \choose k}$. It is known that $$(1-x)^{-n-1} = \sum_{k=0}^{\infty}{n + k \choose k}x^k,$$ $$(1-4x)^{-1/2} = \sum_{k=0}^{\infty}{2k \choose k}x^k,$$ $$(1+x)^n = \sum_{k=0}^n{n \choose k}x^k.$$ I have tried to combine these formulae, but nothing interesting has been found yet.
| Suppose we seek to evaluate
$$\sum_{k=0}^{n+1} {n+1+k\choose 2k} (-1)^{n-k} 4^k
= (-1)^n \sum_{k=0}^{n+1} {n+1+k\choose n+1-k} (-4)^k.$$
Start from
$${n+1+k\choose n+1-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2-k}} (1+z)^{n+1+k} \; dz.$$
This yields the following expression for the sum
$$\frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k=0}^{n+1} \frac{1}{z^{n+2-k}} (1+z)^{n+1+k}
\times (-4)^k\; dz
\\ = \frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\sum_{k=0}^{n+1} (1+z)^k \times z^k
\times (-4)^k\; dz.$$
Observe that the defining integral of the binomial coefficient is zero
when $n+2 -k < 1.$ This condition is precisely $n-k < -1$ or $k >
n+1.$ Therefore we may extend the sum to infinity.
We thus have
$$\frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\sum_{k=0}^\infty (1+z)^k \times z^k
\times (-4)^k\; dz
\\ = \frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\frac{1}{1+4(1+z)z} \; dz.
\\ = \frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\frac{1}{(1+2z)^2} \; dz.$$
Therefore the value of the integral is given by
$$(-1)^n [z^{n+1}] \frac{(1+z)^{n+1}}{(1+2z)^2}
= (-1)^n \sum_{q=0}^{n+1} {n+1\choose n+1-q} (q+1) (-2)^q
\\= (-1)^n \sum_{q=0}^{n+1} {n+1\choose q} (q+1) (-2)^q.$$
Recall that $$(z (1+z)^n)'
= \sum_{q=0}^n {n \choose q} (q+1) z^q
= (1+z)^n + nz(1+z)^{n-1}$$
so that the end result for the sum is
$$(-1)^n \times ((-1)^{n+1} - 2(n+1)(-1)^n)
= -1 - 2(n+1) = -3 - 2n.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
The reader is invited to try this method on the binomial coefficient
$${n+1+k\choose 2k}$$ and examine carefully what happens.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/94536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find a sum of appropriate values of $\cos$ and $\sin$ to determine the value of a series The task is to find a sum of multiple values $\cos$ and $\sin$ to determine the value of
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}$$
Since I had no clue how to approach this I consulted Wolfram|Alpha which returned this result:
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \frac{\sin(1)}{2} + \cos(1) - \cos(0)$$
So I wrote down the partial sums of the given series and $\sin(1)$ and $\cos(1)$:
$$
\qquad\qquad\quad\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \quad - \frac{1}{4!} + \frac{2}{6!} - \frac{3}{8!} \cdots
$$
$$
\;\;\sin(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n+1}}{(2n+1)!} = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \cdots
$$
$$
\cos(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n}}{(2n \qquad)!} = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} \cdots
$$
Looking at the numbers I can see that Wolfram|Alpha's result is correct: $\frac{1}{2}1 - \frac{1}{2!} = 0$ and $\frac{1}{2}\frac{-1}{3!} + \frac{1}{4!} = \frac{1}{4!}$, so the $\cos(1)$-series is shifted by $1$ since there is no $1$ at the beginning of the given series, so it needs to be subtracted from $\cos(1)$: $-cos(0)=-1$. But how do I get here without Wolfram|Alpha?
| Start with the known Maclaurin series
$$\cos x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$$
and
$$\sin x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}\;.$$
The given series has factorials of even numbers in the denominator, so I began by shifting indices to make them $(2n)!$ to match the cosine series. After that it was mostly a matter of following my nose: at each step in the calculation below there’s really only one thing that suggests itself strongly.
$$\begin{align*}
\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}&=\sum_{n=2}^\infty(-1)^{n-1}\frac{n-1}{(2n)!}\\
&=\sum_{n=2}^\infty(-1)^{n-1}\frac{n}{(2n)!}-\sum_{n=2}^\infty(-1)^{n-1}\frac1{(2n)!}\\
&=\sum_{n=2}^\infty(-1)^{n-1}\frac{n}{2n(2n-1)!}+\sum_{n=2}^\infty\frac{(-1)^n}{(2n)!}\\
&=\frac12\sum_{n=2}^\infty\frac{(-1)^{n-1}}{(2n-1)!}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}-1+\frac12\\
&=\frac12\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)!}+\cos 1-\frac12\\
&=\frac12\left(\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}-1\right)+\cos 1-\frac12\\
&=\frac12\sin 1 + \cos 1 - 1\\
&=\frac12\sin 1+\cos 1-\cos 0
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/96571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Convergence of the series $\sum \frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ Could you help me to understand for which $x$ this series converge $\displaystyle\sum \frac{\sqrt{n+1}-\sqrt{n}}{n^x}$?
| Note that $$\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{\sqrt{n+1}-\sqrt{n}}{n^x}\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}.$$
If $x>1/2$, then
$$\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}\leq\frac{1}{n^x(2\sqrt{n})}=\frac{1}{2n^{x+\frac{1}{2}}}.$$
Since $\displaystyle\sum\frac{1}{2n^{x+\frac{1}{2}}}$ converges when $x>1/2$ (see p-series), by comparison test, $\displaystyle\sum\frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ converges.
On the other hand, if $x\le 1/2$,
$$\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}\geq\frac{1}{n^x(\sqrt{n})}=\frac{1}{n^{x+\frac{1}{2}}}\geq\frac{1}{n}.$$
Since the harmonic series diverges, by comparison test again, $\displaystyle\sum\frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Primes of the form : $\sqrt{x^2+7\cdot y^2}$ How to prove following statement :
Conjecture:
Odd prime $p$ is expressible as : $p=\sqrt{x^2+7\cdot y^2}$ , $x,y > 0$
if and only if : $p\equiv 1 \pmod {7}$ or $p\equiv 2 \pmod {7}$ or $p\equiv 4 \pmod {7}$ .
Note that $7$ is an Idoneal Number .
| First, check that $p=7$ is not expressible as such a square root.
Next, suppose $p$ is an odd prime other than $7$.
$p^2 = x^2 + 7y^2$ means that $p² = (x + y \sqrt{-7})(x - y \sqrt{-7})$.
So you are looking for non-trivial factorisations of $p^2$ in $\mathbb{Z}[\sqrt{-7}]$.
Note that $\mathbb{Q}(\sqrt{-7})$ has class number 1, so its ring of integers, $\mathbb{Z}[\frac{1+\sqrt{-7}}2]$, is a unique factorisation domain.
If $-7$ is a square mod $p$ then $p$ factors into $q \bar{q}$, and you have $p^2 = (q^2)(\bar q^2)$.
Now, if $q = \frac{a+b\sqrt{-7}}2$ with $a$ and $b$ odd, then $a^2,b^2 \equiv 1 \mod 8$, so $q\bar{q} = \frac{a^2+7b^2}4 \equiv 0 \mod 2$. Since $p$ is odd, this can't be the case, thus you get $p^2 = (q)^2(\bar q)^2$ with $q^2, \bar{q}^2 \in \mathbb{Z}[\sqrt{-7}]$
Furthermore, write $q^2 = x+y\sqrt{-7}$, so $p^2 = x^2+7y^2$.
If $x=0$ then $(p/y)^2 = 7$, which is impossible because $7$ is not a square in $\mathbb{Q}$.
If $y=0$ then $q^2 = x = p$ which is impossible because $p$ is not a square in $\mathbb{Q}[\sqrt{-7}]$. Thus, changing the signs of $x,y$ if needed, $p = \sqrt{x^2+7y^2}$ with $x,y>0$.
If $-7$ is not a square mod $p$, then $p$ is irreducible, so $p^2 = p*p$ is the only possible factorization (up to sign), so $p = \sqrt{x^2+7y^2}$ implies $x= \pm p, y=0$.
The quadratic reciprocity law does the rest : $-7$ is a square mod $p$ if and only if $p$ is a square mod $7$.
| {
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"url": "https://math.stackexchange.com/questions/97143",
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"source": "stackexchange",
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finding values of $a$ for which the limit exists I've been banging my head all day long over this question, would apprecate some help :)
$$f(x) = \frac{1}{x^2-1} - \frac{a}{x^3-1},\qquad a\in\mathbb{R}.$$
For which values of $a$ does $\lim\limits_{x \to 1} f( x)$ exist?
Also it was requested to use only the $\epsilon$-$\delta$ definition.
| Both denominators are $0$ when $x=1$; therefore $(x-1)$ is a factor of both. So factor them, then find the common denominator:
$$
\begin{align}
& \frac{1}{x^2-1} - \frac{a}{x^3-1} = \frac{1}{(x-1)(x+1)} - \frac{a}{(x-1)(x^2+x+1)} \\ \\
& = \frac{x^2+x+1}{(x-1)(x+1)(x^2+x+1)} - \frac{a(x+1)}{(x-1)(x+1)(x^2+x+1)}
\end{align}
$$
The limit can exist only if the numerator
$$
x^2+x+1 - a(x+1)
$$
is $0$ when $x=1$. And it is equal to $3-2a$ when $x=1$.
Can you take it from there?
| {
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Creating a zero matrix through matrix multiplication From an assignment:
Let $A = \left[ \begin{matrix} 3 & -6 \\ -2 & 4 \end{matrix}\right] $ Construct a $2 * 2$ matrix $B$ such that $AB$ is the zero matrix. Use two different nonzero columns for $B$.
The value of $AB$ would be:
$$
AB = \left[
\begin{matrix}
3b_{11} -6b_{12} & 3b_{21} -6b_{22} \\
-2b_{11} + 4b_{12} & -2b_{21} + 4b_{22}
\end{matrix}\right]
$$
I was thinking of using substitution, but the following equations just result in the variables equalling $0$:
$$\begin{align*}
3b_{11} -6b_{12} &= 0\\
-2b_{11} + 4b_{12} &= 0
\end{align*}$$
Any hints on how I can solve this?
| Consider the solutions to $A\mathbf{x}=\mathbf{0}$. These can be found easily by Gaussian elimination:
$$\left(\begin{array}{rr}
3 & -6\\
-2 & 4
\end{array}\right)\rightarrow \left(\begin{array}{rr}
1 & -2\\
1 & -2
\end{array}\right) \rightarrow \left(\begin{array}{rr}
1 & -2\\
0 & 0
\end{array}\right).$$
So $\mathbf{x}=\left(\begin{array}{c}r\\s\end{array}\right)$ satisfies $A\mathbf{x}=\mathbf{0}$ if and only if $r-2s = 0$, if and only if $r=2s$.
Now notice that if $B=[\mathbf{b}_1|\mathbf{b}_2]$, where $\mathbf{b}_1$ is the first column and $\mathbf{b}_2$ is the second column, then
$$AB = [A\mathbf{b}_1|A\mathbf{b}_2].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/102127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the minimum possible value of $(a+b+c)$? $a, b$ and $c$ are real positive numbers satisfying $ \frac 13 \le ab+bc+ca \le 1 $ and $abc \ge \frac 1{27}$ then what is the minimum possible value of $(a+b+c)$?
Applying AM $\ge$ GM gives $(a+b+c) \ge 1$ and if we apply AM $\ge$ HM gives $(a+b+c) \ge\frac 13$ but apparently $1$ is the answer, so my question is why are we not taking the second one as minimum, (since $ \frac13 \lt 1) $?
| As you observed, AM-GM says that, $a+b+c\ge 1$.
*In addition, AM-GM says that equality is attained precisely when $abc=1/27$ and $a=b=c=1/3$.
The inequalities $1/3 \le ab+bc+ca\le 1$ are, luckily, satisfied by this choice of $a,b,c$. So the minimum achievable value of $a+b+c$ is indeed $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Some double angle identity to solve $(2x^{2}+y^{2})\frac{dy}{dx}=2xy$? For some reason, I cannot see a clever way to solve this (I know the way of doing it like in Wolframalapha) but I am pretty sure there is a double angle identity to crack this puzzle. Could someone hint a bit to get this puzzle onwards?
Firstly, I thought to use some rules such as $(x+y)^{2}=x^{2}+2xy+y^{2}$ or $(x-y)^{2}=x^{2}-2xy+y^{2}$ but I think some trigonometric substitution could solve this problem.
| If we write the equation as,
$$\frac{{dy}}{{dx}} = \frac{{2xy}}{{2{x^2} + {y^2}}}$$
and then divide through $x^2$ we will get:
$$\frac{{dy}}{{dx}} = \frac{{2\dfrac{y}{x}}}{{2 + {{\left( {\dfrac{y}{x}} \right)}^2}}}$$
This suggests that we simplify the previous equation in terms of
$$f\left( v \right) = \frac{{2v}}{{2 + {v^2}}}$$
So putting
$$\eqalign{
& \frac{y}{x} = v \cr
& y = vx \cr
& y' = v'x + v \cr} $$
We get
$$\frac{{dv}}{{dx}}x + v = \frac{{2v}}{{2 + {v^2}}}$$
Then
$$\eqalign{
& \frac{{dv}}{{dx}}x = - \frac{{{v^3}}}{{2 + {v^2}}} \cr
& \frac{{dx}}{x} = - \frac{{2 + {v^2}}}{{{v^3}}}dv \cr
& \frac{{dx}}{x} = \left( { - \frac{2}{{{v^3}}} - \frac{1}{v}} \right)dv \cr} $$
Upon integration we have:
$$\log x + C = \frac{1}{{{v^2}}} - \log v$$
Let's substitute back
$$\eqalign{
& \log x + C = \frac{{{x^2}}}{{{y^2}}} - \log \frac{y}{x} \cr
& \log x + C = \frac{{{x^2}}}{{{y^2}}} - \log y + \log x \cr
& \log y = \frac{{{x^2}}}{{{y^2}}} - C \cr
& y = {C_1}\exp \left( {{x^2}{y^{ - 2}}} \right) \cr} $$
You can find $y$ in terms of $x$, but I don't think the inverse is possible, at least with everyday functions.
$$y\sqrt {\log y + C} = x$$
Ok, using the Lambert W we have
$${y^2}\left( {\log y + C} \right) = {x^2}$$
Use the exponential:
$${e^{{y^2}}}y{e^C} = {e^{{x^2}}}$$
Square and multiply by two
$$2{y^2}{e^{2{y^2}}}{e^{2C}} = 2{e^{2{x^2}}}$$
Use the Lambert W
$$2{y^2} = W\left( {\frac{{2{e^{2{x^2}}}}}{{{e^{2C}}}}} \right)$$
$$y = \sqrt {\frac{1}{2}W\left( {\frac{{2{e^{2{x^2}}}}}{{{e^{2C}}}}} \right)} $$
Another aproach would be
$$\eqalign{
& \log y + C = \frac{{{x^2}}}{{{y^2}}} \cr
& y{e^C} = {e^{\frac{{{x^2}}}{{{y^2}}}}} \cr
& {y^2}{e^{2C}} = {e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr
& 2\frac{{{x^2}}}{{{y^2}}}{y^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr
& 2{x^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr
& W\left( {2{x^2}{e^{2C}}} \right) = 2\frac{{{x^2}}}{{{y^2}}} \cr
& {y^2} = \frac{{2{x^2}}}{{W\left( {2{x^2}{e^{2C}}} \right)}} \cr
& y = \frac{{\sqrt 2 x}}{{\sqrt{W\left( {2{x^2}{e^{2C}}} \right)}}} \cr} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding root of a polynomial Let's say I've got $ax^3 + bx^2 + cx + d = 0$
I want to find x for a general-case.
Mathematica gave me a very long solution, and even longer for one which has $kx^4$.
Can you please help me reproduce the algorithm used by Mathematica to generate formula of:
$$P(x) = a_0x^0 + a_1x^1 + a_2x^2 + a_3x^3 + \cdots+ a_nx^n$$
i.e., general case of a polynomial.
| The trick is transforming the general cubic equation into a new cubic equation with missing $x^2$ term is due to
Nicolo Fontana Tartaglia.
Apply the substitution
$$ x = y - \frac{b}{3a} $$
then we get
$$ a\left(y-\frac{b}{3a} \right)^3 + b\left(y-\frac{b}{3a} \right)^2+c\left(y-\frac{b}{3a} \right)+d = 0 $$
which simplifies to
$$ ay^3 + \left( c-\frac{b^2}{3a}\right)y+ \left(d + \frac{2b^3}{27a^2} - \frac{bc}{3a} \right) = 0 $$
This is called a depressed cubic equation, because the square term is eliminated. It is much easier to use this and then
find the roots. (back substitute to get the roots in terms of $x$)
For example $$2x^3-18x^2+46x-30=0$$
Substitute $ x=y+3$ and simplify this cubic equation to
$2y^3-8y=0 \Rightarrow y=0,2,-2$ which then gives the roots as
$x=1,3, $ and $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving the inequality $\frac{x}{\sqrt{x+12}} - \frac{x-2}{\sqrt{x}} > 0$ I'm having troubles to solve the following inequality..
$$\frac{x}{\sqrt{x+12}} - \frac{x-2}{\sqrt{x}} > 0$$
I know that the result is $x>0$ and $x<4$ but I cannot find a way to the result..
Thanks in advance!
| We can combine the two fractions on the left-hand side to get
$$\frac{x\sqrt x-(x-2)\sqrt{x+12}}{\sqrt{x+12}\sqrt{x}}>0$$
and since we need for the denominator to be defined and not $0$, this gives us $x>0$. We can now multiply both sides by $\sqrt{x+12}\sqrt{x}$ to get
$$x\sqrt x-(x-2)\sqrt{x+12}>0$$
which we rewrite as
$x\sqrt x>(x-2)\sqrt{x+12}$ and square both sides giving us
$$x^3>\pm(x-2)^2(x+12)=\pm(x^3+8x^2-44x+48)$$
where the $\pm$ is determined by the sign of $x-2$, since if $x\geq 2$ then squaring does not change the sign of either side while if $0<x<2$ then it does.
When $x\geq 2$ this simplifies to $0>8x^2-44x+24$. Applying the quadratic equation, we get that $0=8x^2-44x+48$ at $x=\frac{3}{2},4$ and we can see that between these two values the inequality holds, so we know that our desired inequality holds for $x\geq 2$. When $0<x<2$ this simplifies to $2x^3+8x^2-44x+48>0$, which is true for all positive $x$, so the desired inequality holds for $0<x<2$. Thus it holds in the region $0<x<4$, but not outside.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}$ using Mclaurin polynomial How would I calculate $$\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}$$ using a Maclaurin polynomial? For the integral in the numerator, the polynomial doesn't seem to exist beyond degree 1.
Thanks!
| We have, for $x$ near zero,
$$
\sin t = t-t^3/6+o(t^5), \ \ \tan x =x+x^3/3+o(x^5),\ \ \cos x =1-x^2/2+o(x^4).
$$
Then
$$
\frac{\int_0^x\frac{\sin t}t dt-\tan x}{2x(1-\cos x)}
=\frac{\int_0^x(1-t^2/6+o(t^4))dt-(x+x^3/3+o(x^5))}{2x(x^2/2+o(x^4))}
=\frac{x-x^3/18+o(x^5)-(x+x^3/3+o(x^5))}{2x(x^2/2+o(x^4))}
=\frac{-7x^3/18+o(x^5)}{x^3+o(x^5))}
=\frac{-7/18+o(x^2)}{1+o(x^2))}
\to -\frac7{18}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\frac{d}{dx} \frac{x^2}{y}$ $$\frac{d}{dx} \frac{x^2}{y}$$
According to Wolframalpha
I "factor out constants"
$$\frac{\frac{d}{dx} x^2}{y}$$
Then I will get $\frac{2x}{y}$. Is that right? But $y$ is not a constant? What I did actually (quotient rule got me stuck)
The actual question is "Find $\frac{d^2y}{dx^2}$ of $2x^3 - 3y^2 = 8$"
I got
$$\frac{dy}{dx} = \frac{x^2}{y}$$
Then
$$\frac{d^2y}{dx^2} = \frac{y \cdot 2x - x^2 \cdot \frac{dy}{dx}}{y^2}$$
$$ = \frac{2xy - x^2 \cdot \frac{x^2}{y}}{y^2}$$
$$ = \frac{2xy^2 - x^4}{y^3}$$
Is this correct? It doesn't look like a "simple" answer (or whats in wolfram)?
| Your first attempt (second in order presented) is correct. There is some simplification that you can do though. Since you will only ever be evaluating this expression for $(x,y)$ on the originally defined curve, you can use that relation to simplify. Some "simplifications" would be
$$
\begin{align*}
\frac{d^2y}{dx^2} & = \frac{2xy^2-x^4}{y^3}&\frac{d^2y}{dx^2} & = \frac{2xy^2-x^4}{y^3}\\
& = \frac{2x(3y^2)-3x^4}{3y^2y}&& = \frac{x(4y^2-2x^3)}{2y^3}\\
& = \frac{2x(2x^3-8)-3x^4}{(2x^3-8)y}&& = \frac{x(4y^2-(8+3y^2))}{2y^3}\\
& = \frac{x^4-16x}{(2x^3-8)y}&& = \frac{x(y^2-8)}{2y^3}\\\\
& = \frac{x(x^3-16)}{2(x^3-4)y}\\
\end{align*}
$$
This is only "simpler" in that lower powers of $y$ (respectively $x$) are used.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^2 + b^2 + c^2 $ is not a prime number I am having difficulty solving this problem:
Let $a, b, c \in\mathbb{Z}$, $abc \neq 0$ and $a\neq c$ be such that
$$\frac{a}{c} = \frac{a^2+b^2}{c^2+b^2}.$$
Prove that $a^2 + b^2 + c^2$ is not a prime number.
Thanks in advance!
| NOTE: The "solution" below addressed the original quesion, which was "Prove that $a^2 +b^2 +ab$ is not a prime number", and was later changed to its present form. In fact, it is the case that every prime congruent to $1$ (mod $3$) has the form $a^2 +b^2 +ab$ for integers $a$ and $b,$ while no prime congruent to $2$ (mod $3$) has this form. The former (well-kown) statement can be proved in a fashion rather similar to Euler's proof that every prime congruent to $1$ (mod $4$) is a sum of two integer squares. In this case, however, one works with the ring of Eisenstein integers, $R = \mathbb{Z}[\omega],$ where $\omega$ is a primitive (complex) cube root of unity. This is a principal ideal domain. If $p \equiv 1$ (mod $3$) is a rational prime, then the multiplicative group of the field $\mathbb{Z}/p\mathbb{Z}$ contains an element of order $3$. Hence there is an integer $n$ such that $p$ divides $n^{3}- 1,$ but $p$ does not divide $n-1.$ Then $p$ divides $n^{2}+n+1,$ which factors as $(n- \omega)(n-\omega^{2})$ in $R.$
Since $p$ does not divide either of the two factors in $R,$ we must conclude that $p$ is not a prime in $R.$ Hence there are integers $a,b,c,d$ such that $p = (a - b \omega)(c- d\omega)$ in $R,$ where neither $a-b\omega$ nor $c-d\omega$ are units in $R.$ Then multiplying this expression by its complex coinjugate , we see that $p^2 = (a^2 +ab + b^2)(c^2 +cd +d^2).$ Now $a^2 +ab +b^2 \neq 1$ and $c^2 +cd +d^2 \neq 1$ as $a-b\omega$ and $c-d\omega$ are non-units in $R.$ Hence $a^2 +ab +b^2 = p$ (note that it is a positive quanitity). It is an easy exercise that if $q \equiv 2$ (mod $3$), then $q$ remains prime in $R,$ so $q$ can certainly not be written in the form $a^2 +ab+b^2$ for ratonal integers $a$ and $b.$
| {
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Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem:
Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$
I attempted to solve it by induction, but when proving for $n= x+1$ the algebra gets very messy very fast. I was wondering if this is the only way or if there is a quicker way to prove this. I guess I am a little unsure as to how to prove something is an integer.
I also noticed that letting $f(x) = \frac{x^5}5 + \frac{x^4}2 + \frac{x^3}3 - \frac x{30}$ and deriving $f(x)$ yields a fairly clean result, but I don't know if this helps me at all. Any help would be great.
| Induction is the way to go, but I'd rather show the following fact:
$\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ being an integer is the same as $a_n=6n^5+15n^4+10n^3-n$ being divisible by 30 for all $n\geq 0$. The base case is trivial. Assume that the statement is true for $n$, that is, $6n^5+15n^4+10n^3-n$ is divisible by $30$. Now, we can show that $a_{n+1}-a_n$ is also divisible by $30$ and we will be done.
$$a_{n+1}-a_n = 6((n+1)^5-n^5)+15((n+1)^4-n^4)+10((n+1)^3-n^3)-((n+1)-n)$$
$$=6(5n^4+10n^3+10n^2+5n+1)+15(4n^3+6n^2+4n+1)+10(3n^2+3n+1)-1$$
$$=30n^4+120n^3+180n^2+120n+30$$
| {
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Determine invertible elements of a monoid I have the following excercise.
In the set $\mathbb{Z}_6\times\mathbb{Z}_6$ consider the follow:
$$(\overline{a}, \overline{b})\cdot(\overline{c}, \overline{d}) = (\overline{a}+\overline{c}+\overline{3}, \overline{b}\overline{d})$$
Prove that $(\mathbb{Z}_6\times\mathbb{Z}_6, \cdot)$ is a monoid,
and determine the invertible elements.
Yes, it's a monoid.
Associative:
$$\big((\overline{a}, \overline{b})\cdot(\overline{c}, \overline{d})\big) \cdot (\overline{e}, \overline{f})=(\overline{a}, \overline{b})\cdot\big((\overline{c}, \overline{d}) \cdot (\overline{e}, \overline{f})\big)$$
through various steps (and considering $\overline{3}+\overline{3}=\overline{6}=\overline{0} $, is that correct???):
$$(\overline{a}+\overline{c}+\overline{e}, \overline{b}\overline{d}\overline{f})=(\overline{a}+\overline{c}+\overline{e}, \overline{b}\overline{d}\overline{f})$$
Identity:
$$(\overline{a}, \overline{b})\cdot(\overline{e_1}, \overline{e_2}) = (\overline{a}, \overline{b})\quad \forall (\overline{a}, \overline{b}) \in \mathbb{Z}_6\times\mathbb{Z}_6$$
follow:
$$(\overline{a}+\overline{e_1}+\overline{3}, \overline{b}\overline{e_2})=(\overline{a}, \overline{b})$$
thus:
$$\overline{a}+\overline{e_1}+\overline{3} = \overline{a}\Rightarrow \overline{e_1}=\overline{-3}$$
and
$$\overline{b}\overline{e_2}= \overline{b}\Rightarrow \overline{e_2}=\overline{1}$$
The structure is even commutative, than exist right and left identity.
Invertible elements:
$$(\overline{a}, \overline{b})\cdot(\overline{i_1}, \overline{i_2}) = (\overline{-3}, \overline{1})\quad \forall (\overline{a}, \overline{b}) \in \mathbb{Z}_6\times\mathbb{Z}_6$$
so:
$$(\overline{a}+\overline{i_1}+\overline{3}, \overline{b}\overline{i_2})=(\overline{-3}, \overline{1})$$
follow (and considering: $\overline{3}+\overline{3}=\overline{6}=\overline{0}$):
$$\overline{a}+\overline{i_1}+\overline{3} = \overline{-3}\Rightarrow \overline{a}+\overline{i_1}+\overline{3} +\overline{3} = 0 \Rightarrow \overline{i_1}=\overline{-a}$$
and
$$\overline{b}\overline{i_2}=\overline{1} \Rightarrow \overline{i_2}=\overline{b^{-1}}$$
Is (until now) all right? Now, how can I find $\overline{b^{-1}}$?
| Since you are in $\mathbb Z_6, \overline{b}^{-1}$ only exists for $\overline b= \overline 1$ or $\overline b= \overline5$.
| {
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Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
| The $x^2 y^{n-2}$ term from $x \cdot x y^{n-2}$ is cancelled by the term from $(-y) \cdot
x^2 y^{n-3}$. Similarly, the $(-y) \cdot x^{n-2} y$ is cancelled by the $x \cdot x^{n-3} y^2$.
| {
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convergence of $\sum \limits_{n=1}^{\infty }\bigl\{ \frac {1\cdot3 \cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \frac {4n+3} {2n+2}\bigr\} ^{2}$ I am investigating the convergence of
$$\begin{split}\sum _{n=1}^{\infty }\left\{ \dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \dfrac {4n+3} {2n+2}\right\} ^{2} &= \sum _{n=1}^{\infty }\left\{ \dfrac {\prod _{t=1}^n (2t-1)} {\prod _{t=1}^n (2t)}\cdot \dfrac {4n+3} {2n+2}\right\} ^{2} \\
&=\sum _{n=1}^{\infty }\left\{ \prod _{t=1}^n\left( 1-\dfrac {1} {2t}\right) \dfrac {4n+3} {2n+2}\right\} ^{2}
\end{split}$$
which after some manipulations I have reduced to
$$\sum _{n=1}^{\infty }e^ \left\{ 2\ln \left(2 -\dfrac {1} {2n+2}\right) +2\cdot \sum _{t=1}^{n}\ln \left( 1-\dfrac {1}{2t}\right) \right\} $$
and from an alternative approach I was able to reduce it to
$$\sum _{n=1}^{\infty } \dfrac{\left( 4n+3\right) ^{2}}{4\left(n+1\right)^{2}} \prod _{t=1}^n\left( 2+\dfrac{1}{2t^{2}}-\dfrac{2}{t}\right)$$
I am unsure how to proceed from here in either of the two cases. Any help would be much appreciated.
| We can prove by induction that
$$\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)} \ge \frac{1}{\sqrt{4n}}$$
and so your series diverges.
You can also notice that
$$\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)} = \dfrac{\binom{2n}{n}}{4^n}$$
and try using the approximation
$$ \dfrac{\binom{2n}{n}}{4^n} = \frac{1}{\sqrt{\pi n}} \left(1 + \mathcal{O}\left(\frac{1}{n}\right)\right)$$
| {
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Expressing an algebraic function as a single expression I have the following question in my text book:
Express as a single fraction
$$ \frac{3}{x-4} - \frac{2}{(x-4)^2} $$
The answer the book gives is this:
$$ \frac{3x-14}{(x-4)^2} $$
I understand how they get to this through these steps:
$$ \frac{3(x-4)-2}{(x-4)^2} = \frac{3x-12-2}{(x-4)^2} $$
My question is why can you not cancel the $$ (x-4) $$ instead like this:
$$ \frac{3(x-4)-2}{(x-4)^2} = \frac{3-2}{(x-4)} = \frac{1}{(x-4)} $$
| in your last line you can't cancel the $(x-4)$! You have
$$
\frac{3(x-4) - 2}{(x-4)^2} = \frac{3 - \frac 2{x-4}}{x-4} \ne \frac{1}{x-4}.
$$
Just as in, say $\frac{3+4}9$, you can't cancel the $3$ to obtain $\frac{1+4}{3} = \frac 53$.
| {
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Induction Proof: Proof of Strict Inequality involving Exponents of 3 in the Denominator. I have been trying to do this problem by using induction but I became stuck halfway through:
Use induction to show that
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{n^3}\right) \lt 3 - \frac{1}{n^2}$$
for $n\geq 2$.
Does the series $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$ converge? Justify your conclusions.
So far I have this:
Base Case of Induction, $n=2$
$$\begin{align*}
\frac{1}{n^3} &\lt 3- \frac{1}{n^2}\\
\frac{1}{8} &\lt 3 - \frac{1}{4}\\
\frac{1}{8} &\lt \frac{11}{4}
\end{align*}$$
Induction Step: Assume true for some $k \geq 2 :$
$$
2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}$$
Show true with $n= k+1$
$$ 2 + 4 + \frac{2}{27} + \cdots + \frac{2}{k^3} + \frac{2}{(k+1)^3} \lt 3-\frac{1}{(k+1)^2}.$$
From there, I have no idea what to do.
I was planning to have
$$3 - \frac{1}{k^2} + \frac{2}{(k+1)^3} \lt 3 - \frac{1}{(k+1)^2}$$
but I feel like it won't work since
$$2\left(1 + \frac{1}{8} + \frac{1}{27}+\cdots+\frac{1}{k^3}\right)$$
does not equal
$$3 - \frac{1}{k^2}.$$
| It doesn't matter that the sum does not equal $ 3- \frac{1}{k^3}$. What matters is that the sum is less than $3 - \frac{1}{k^3}$.
We have:
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}.$$
Adding $\frac{2}{(k+1)^3}$ to both sides does not change the inequality, so we have
$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) + \frac{2}{(k+1)^3} \lt 3 - \frac{1}{k^2} + \frac{2}{(k+1)^3}.$$
The left hand side is the quantity you want.
With the right hand side, we have
$$\frac{1}{k^2}\gt \frac{1}{(k+1)^2},$$
so
$$-\frac{1}{k^2} \lt -\frac{1}{(k+1)^2},$$
hence
$$\frac{2}{(k+1)^3} - \frac{1}{k^2} \lt \frac{2}{(k+1)^3} - \frac{1}{(k+1)^2} = \frac{1}{(k+1)^2},$$
so
$$3 - \frac{1}{k^2}+\frac{2}{(k+1)^3} \lt 3 - \frac{1}{(k+1)^2}.$$
And since $a\lt b$ and $b\lt c$ implies $a\lt c$, then...
For the last part of the question, note that the quantity on the left hand side is twice the $n$th partial sum of
$$\sum_{n=2}^{\infty} \frac{1}{n^3}.$$
So you should be able to conclude that the partial sums of
$$\sum_{n=1}^{\infty}\frac{1}{n^3}$$
are bounded above; since all terms are positive, that means that...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding $\int_0^1{\frac{x^4(1-x)^4}{1+x^2}}dx$ The question I am working on: Evaluate
$$\frac{1}{2} \int^1_0{x^4 (1-x)^4 } dx \le \int^1_0{\frac{x^4 (1-x)^4}{1+x^2}} dx \le \int^1_0{x^4 (1-x)^4 } dx$$
So using integration by parts to solve:
(letting $u=(1-x)^4$ and $dv=x^4$)
$$\int{x^4 (1-x)^ 4} dx = \frac{4}{5}x^5(x-1) - \frac{4}{5}\left(\frac{x^7}{7} - \frac{x^6}{6}\right)+c$$
Is it correct so far? If so ...
$$\int^1_0{ x^4 (1-x)^4 } dx = \frac{4}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac2{105}$$
$$\frac{1}{2} \int^1_0{ x^4 (1-x)^4 } dx = \frac{2}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac1{105}$$
But
$$\int\frac{{x^4(1-x)^4}}{1+x^2} dx = ??$$
Since I found $\int{{x^4(1-x)^4}} dx$. I thought of integration by parts, letting $u=\frac{1}{1+x^2}$, $dv = x^4(1-x)^4$. But I will get a very complicated $v$ to integrate later? Same if I did it the other way around?
| Using the binomial theorem and then long division,
$$
\frac{x^4(1-x)^4}{1+x^2}=
\frac{x^4(x^4-4x^3+6x^2-4x+1)}{1+x^2}=
x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}
$$
so that
$$
\int_0^1
\frac{x^4(1-x)^4}{1+x^2}dx=
\frac17-\frac46+\frac55-\frac43+\frac41-4\arctan\frac\pi4
=\frac{22}7-\pi\approx\frac1{790.~833~125~927~563}
$$
There's also a problem with your integration by parts since
$$
\eqalign{
u &=& (1-x)^4=(x-1)^4 &\qquad& v &=& x^4 \\\\
du &=& 4(x-1)^3 \, dx &\qquad& dv &=& \frac15 x^5 dx
}
$$
so
$$
\eqalign{
I_1
&= \int_0^1 x^4 (1-x)^4 dx \\
&= \int u\,dv = uv - \int v\,du \\
&= \left[ \frac45 x^5 (1-x)^4 \right]_0^1
+ \frac45 \int_0^1 x^5 (1-x)^3 dx \\
&= \frac45 \int_0^1 x^5 (1-x)^3 dx \\
&= \frac45\cdot\frac36 \int_0^1 x^6 (1-x)^2 dx \\
&= \frac45\cdot\frac36\cdot\frac27 \int_0^1 x^7 (1-x) dx \\
&= \frac45\cdot\frac36\cdot\frac27\cdot\frac18 \int_0^1 x^8 dx \\
&= \frac45\cdot\frac36\cdot\frac27\cdot\frac18\cdot\frac19 = \frac1{630}
}
$$
where we had to apply integration by parts repeatedly using
$$u=(1-x)^n,~dv=x^m\,dx$$
$$du=-n(1-x)^{n-1}\,dx,~v=\frac{x^{m+1}}{m+1}\implies$$
$$\int x^m(1-x)^ndx=\frac{x^n(1-x)^{m+1}}{m+1}
+\frac{n}{m+1}\int x^{m+1}(1-x)^{n-1}dx$$
until the powers of $(1-x)$ went away. In fact,
we just made a special case of the calculation
$$
\int_0^1 x^a (1-x)^b = B(a+1,b+1) = \frac{a!~b!}{(a+b+1)!}
$$
of the well-known Beta function for $a=b=4$
(our method works for $a,b\in\mathbb{N}$ and,
with an adaptation of the above formula
using the Gamma function, also for real $a,b\ge0$).
Now we can see that the reciprocal of the central integral
is certainly beween $630$ and $1260$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Indefinite Integral for $\cos x/(1+x^2)$ I have been working on the indefinite integral of $\cos x/(1+x^2)$.
$$ \int\frac{\cos x}{1+x^2}\;dx\text{ or } \int\frac{\sin x}{1+x^2}\;dx $$
are they unsolvable(impossible to solve) or is there a way to solve them even by approximation?
Thank you very much.
| $\int\dfrac{\sin x}{1+x^2}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^{n-k}(x^2+1)^k}{2(2n+1)!(x^2+1)}d(x^2+1)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(x^2+1)$
$=\int\left(\dfrac{1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$
$=\int\left(\dfrac{1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\dfrac{1}{2(2n+1)!(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$
$=\int\left(\sum\limits_{n=0}^\infty\dfrac{1}{2(2n+1)!(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$
$=\int\left(\dfrac{\sinh1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$
$=\dfrac{\sinh1\ln(x^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^k}{2(2n+1)!k!k(n-k)!}+C$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
| If $\displaystyle x=\frac\pi5,5x=\pi\implies3x=\pi-2x\implies\cos3x=\cos(\pi-2x)\ \ \ \ (1)$
But as $\displaystyle\cos(\pi-2x)=-\cos2x$, (1)$\displaystyle\implies\cos3x+\cos2x=0\ \ \ \ (2)$
If $\displaystyle\cos x=c(2)$ becomes $4c^3+2c^2-3c-1=0\ \ \ \ (3)$
Again from $(1)\implies3x=2n\pi\pm(\pi-2x)$
$\displaystyle'-'\implies x=(2n-1)\pi\implies\cos x=-1$
$\displaystyle'+'\implies x=\frac{(2n+1)\pi}5$ where $n=0,1,2,3,4$
Observe that $\displaystyle\cos\frac{(2\cdot2+1)\pi}5=\cos\pi=-1$(which converges with $'-'$ case )
and $\displaystyle\cos\frac{(2\cdot0+1)\pi}5=\cos\frac{(2\cdot4+1)\pi}5$ as $\displaystyle\cos(2\pi-y)=\cos y$
Similarly, $\displaystyle\cos\frac{(2\cdot1+1)\pi}5=\cos\frac{(2\cdot3+1)\pi}5$
So, the quadratic equation whose roots are $\displaystyle\cos\frac{\pi}5=\cos\frac{9\pi}5,\cos\frac{3\pi}5=\cos\frac{7\pi}5$ will be $\displaystyle\frac{4c^3+2c^2-3c-1}{c+1}=0\iff4c^2-2c-1=0\ \ \ \ (4)$
Now use the fact $\displaystyle c^2=\cos^2x=\frac{1+\cos x}2$
| {
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"question_score": "11",
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"answer_id": 11
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Integral (universal substitution) I need a little help, this integral looks like very simple, but i have problem with it
$\int_{0}^{2\pi }\frac{1}{2-\cos x}dx$ and i want to solve it by universal substitution, e.g. $t=\tan\frac{x}{2}$
but what about limits of integration?
What is a solution? Thanks.
| Yeah if you put $t = \tan\frac{x}{2}$, then you have $dt = \frac{1}{2}\cdot \sec^{2}\frac{x}{2} \ dx$. And from here note that $dx = \frac{2 \: dt}{1+t^{2}}$. And when $x = 0$ you have $t = \tan(0)=0$. And when $x=2\pi$ you have $t = \tan(\pi) = 0$.
I think the value of your integral will be $0$. But in case you want to evaluate something like
\begin{align*}
\int \frac{1}{2-\cos{x}} \ dx &= \int\frac{1}{2 - \frac{1-\tan^{2}x/2}{1+\tan^{2}x/2}} \ dx \\ &= \int \frac{1}{2 - \frac{1-t^{2}}{1+t^{2}}} \cdot \frac{2}{1+t^{2}} \ dt \\ &=\int\frac{2}{2+2t^{2}-1+t^{2}} \ dt \\ &= \int\frac{2}{3t^{2} +1} dt = \frac{2}{3} \int \frac{1}{t^{2}+\frac{1}{3}} \ dt
\end{align*}
I guess you can evaluate the integral now by putting $t =\frac{1}{\sqrt{3}}\:\tan{v}$. This is the way how one generally evaluates integrals of the form $$\int \frac{dx}{a+b\cos{x}} \qquad \text{and} \qquad \int \frac{dx}{a+b\sin{x}}$$ While evaluating integrals of the form $a+b\sin{x}$ one uses the formula $\sin{2x} = \frac{2\tan{t}}{1+\tan^{2}{t}}$
| {
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Proving $\sqrt 3$ is irrational. There is a very simple proof by means of divisibility that $\sqrt 2$ is irrational. I have to prove that $\sqrt 3$ is irrational too, as a homework. I have done it as follows, ad absurdum:
Suppose
$$\sqrt 3= \frac p q$$
with $p/q$ irreducible, then
$$\begin{align}
& 3q^2=p^2 \\
& 2q^2=p^2-q^2 \\
&2q^2=(p+q)(p-q) \\
\end{align}$$
Now I exploit the fact that $p$ and $q$ can't be both even, so it is the case that they are either both odd, or have different parity. Suppose then that $p=2n+1$ and $q=2m+1$
Then it is the case that
$$\begin{align}
&p-q=2(n-m) \\
&p+q=2(m+n+1) \\
\end{align}$$
Which means that
$$\begin{align}
&2q^2=4(m-n)(m+n+1) \\
&q^2=2(m-n)(m+n+1) \\
\end{align}$$
Then $q^2$ is even, and so is then $q$, which is absurd. Similarly, suppose
$q=2n$ and $p=2m+1$.
Then $p+q=2(m+n)+1$ and $p-q=2(m-n)+1$. So it is the case that
$$\begin{align}
&2q^2=(2(m-n)+1)(2(m+n)+1)\\
&2q^2=4(m^2+m-n^2)+1 \\
\end{align}$$
So $2q^2$ is odd, which is then absurd.
Is this valid?
| A proof of R. Dedekind, from 1901.
(He proves that if $n\in\mathbb{N}$ is not a square number, then $\sqrt{n}$ is irrational.)
$\sqrt{3}$ is irrational.
Proof: Suppose that $\sqrt{3}=\frac{x}{y}$, $x$ and $y$ are integers, when $x$ is the smallest integer which satisfies that equation. (If there exist $x$ and $y$ such that $\sqrt{3}=\frac{x}{y}$ then exist that minimal $x$, $\mathbb{N}$ is well-ordered set and $x$ is there). $\frac{x}{y}$ is not an integer ($3$ is not a perfect square), hence exist $k\in\mathbb{N}$, such that:
$$k-1<\frac{x}{y}<k$$
Now, let us look at
$$x'=(k-\frac{x}{y})x=kx-3y \ \ \ \ ; \ \ \ \ y'=(k-\frac{x}{y})y=ky-x $$
Furthermore, $x$ and $y$ are integers, and:
$$\frac{x'}{y'}=\frac{x}{y}=\sqrt{3}$$
But, $k-\frac{x}{y}<1$, therefor $x'<x$, and that is a contradiction to the fact that out $x$ chosen to be the minimal which satisfies $\sqrt{3}=\frac{x}{y}$.
$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
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How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
| It suffices to show that $k^5-k$ is always divisible by each of 2, 3, and 5. That is, we want $k^5-k\equiv 0 \mod d$ where $d$ is each of 2, 3, and 5.
We know that $k\equiv r\pmod 2$ where $r$ is 0 or 1. Then $k^5 - k \equiv r^5-r \equiv 0 \pmod2$. The last equivalence follows instantly from a consideration of the two cases: either $r=0$ in which case we have $0^5-0\equiv 0\pmod 2$, or $r=1$ in which case we have $1^5-1\equiv 0\pmod 2$.
Similarly $k\equiv r\pmod 3$ where $r$ is one of 0, 1 or 2. In each case consider $r^5-r\pmod 3$. The first two cases are completely trivial, and the same as in the previous paragraph. The $r=2$ case gives $32-2= 30\equiv 0\pmod 3$.
And again $k\equiv r\pmod 5$ where $r$ is one of $\{0,1,2,3,4\}$; now consider $r^r-r\pmod 5$. Again cases 0 and 1 are trivial; the other three give $30\equiv 0\pmod 5$, $3^5-3 = 243-3 = 240\equiv 0\pmod 5$, and $4^5-4 = 1024-4 = 1020 \equiv 0\pmod 5$, so we are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many numbers? How many numbers from 7, 13, 14, 21, 26, 39 and 91 divide $$(69^{12} − 38^{12})$$?
I am able to get the answer by calculating it.
But is there is any shorter way by which i can solve this?
Thanks in advance.
| You can calculate the expression in question modulo 7, 13, etc.
For example, $69^{12}-38^{12} \mod 7 = (-1)^{12}-3^{12} \mod 7 = 1-2^{6} \mod 7 = 0$.
Another way is to note that $x^{12}-y^{12} = (x^6)^2-(y^6)^2 = (x^6-y^6)(x^6+y^6) = (x^3-y^3)(x^3+y^3)(x^6+y^6) = (x-y)(x^2+xy+y^2)(x^3+y^3)(x^6+y^6)$, and $69^2+69\times38+38^2 = 4761+2622+1444=8827=7\times13\times97$.
Also note that $14=2\times7$, $21=3\times7$, $26=2\times13$, $39=3\times13$ and $91=7\times13$. Yet $69^{12}-38^{12} = 3^{12}\times23^{12}-2^{12}\times19^{12}$ obviously does not divides neither $2$ nor $3$, so you have to check only divisibility by $7$ and $13$ (and conclude the divisibility by $91$ from that).
| {
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Prove $\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$ How to prove: $$\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$$ Help please. Don't know where to start.
| Use the following facts:
*
*$\sin(A+B)= \sin{A} \cdot \cos{B} + \cos{A} \cdot \sin{B}$.
*$\displaystyle\frac{2 \tan\theta}{1+\tan^{2}\theta} = 2 \cdot \frac{\sin\theta}{\cos\theta} \cdot \cos^{2}\theta = 2 \cdot \sin\theta \cdot \cos\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133568",
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"source": "stackexchange",
"question_score": "5",
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How to $\int \sqrt{10x-x^2}dx$ Theres a hint to use $x=5+5\sin{t}$. Ok, but how do I know what substitution to use if a hint wasn't given? Is it "trivial" or perhaps, its very unlikely that that will appear?
Anyways, I did:
$\int \sqrt{10(5+5\sin{t}) - (5+2\sin{t})^2} dx \\
= \int \sqrt{50+50\sin{t} - (25+50\sin{t} + 25\sin^2{t})} dx\\
= \int \sqrt{ 25-25\sin^2{t} } dx \\
= 5 \int \sqrt{1-\sin^2{t}} dx \\
= 5\sin^{-1}{\sin{t}} \\
= 5t \\
= 5 \sin^{-1}{\frac{x-5}{5}}$
But the answer was:
$$\frac{25}{2}\sin^{-1}{\frac{x-5}{5}}+\frac{x-5}{2}\sqrt{10x-x^2}+c$$
What did I do wrong? Or is the answer wrong perhaps?
| Try to complete the square: $10x-x^2=10x-x^2+25-25 = -(x-5)^2+25$. After a change of variables $y=x-5$, your integral becomes $\int \sqrt{25-y^2}\, dy$. And yhis should be familiar to you.
| {
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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.
| Let $y = \cos\theta + i \cdot \sin\theta$, where $\theta$ has either of the values $$\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}, \frac{11\pi}{7} \ \text{and} \ \frac{13\pi}{7}$$ Then $y^{7} = \cos{7\theta} + i \:\sin{7\theta} =-1$. Then you have $$y^{7}+1=0 \Rightarrow (y+1)(y^{6}-y^{5}+y^{4}-y^{3}+y^{2}-y +1)=0$$ Now the root $y=-1$ corresponds to the value $\theta=\pi$. The roots of the equation $$y^{6}-y^{5}+y^{4}-y^{3}+y^{2}-y +1 =0 \qquad \cdots (1)$$ are therefore $\cos\theta + i\: \sin\theta$ where $\theta$ takes other values than $\pi$. Put $2x = y + \frac{1}{y} = 2 \cos\theta$ and note that:
*
*$\displaystyle y^{2}+\frac{1}{y^2} = 4x^{2}-2$
*$\displaystyle y^{3} + \frac{1}{y^3} = 8x^{3}-6x$.
Now divide $(1)$ by $y^{3}$ and get the answer.
Another method to look at this:
$7 \theta = \text{an odd multiple of} \: \pi$, therefore $\cos{4\theta} = -\cos{3\theta}$. Take $c=\cos{\theta}$. From this you have
\begin{align*}
2\cdot (2c^{2}-1)^{2} -1 &= -(4c^{2}-3c) \\\ 8c^{4} + 4c^{2}-8c^{2} -3c + 1 &=0 \\\ (c+1) \cdot (8c^{3} - 4c^{2} -4c+1) &=0
\end{align*}
The following page also may be helpful: http://www.isibang.ac.in/~sury/luckyoct10.pdf
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Root of Complex Number I just want to clarify something here. Using elementary computation we can verify that for $x,y\in\mathbb{R}$
$$\sqrt{x+iy}=\pm\left(\sqrt{\frac{r+x}{2}}+i \sqrt{\frac{r-x}{2}}\right)$$
where $r=\sqrt{x^2+y^2}$.
However, in wikipedia the algebraic formula for the root is given by
$$\sqrt{x+iy}=\sqrt{\frac{r+x}{2}}\pm i \sqrt{\frac{r-x}{2}}$$
By squaring $\sqrt{\frac{r+x}{2}}- i \sqrt{\frac{r-x}{2}}$ , We will get $x+iy=x-i\sqrt{y^2}$.
Wikipedia refers to the book Handbook of Mathematical Functions: With Formulas, Graphs, and Mathematical Table , the book also giving the same formula as wikipedia's.
Is this something related to "Principal Value" or just a double typographic error?
I ask this because it is rare to see two same mistakes from two different sources, so I have a doubt.
| Since $r\ge|x|$,
$$
\left(\sqrt{\dfrac{r+x}{2}}\right)^2=\dfrac{r+x}{2} \text{ and } \left(\sqrt{\dfrac{r-x}{2}}\right)^2=\dfrac{r-x}{2}
$$
However, since $y$ may be positive or negative,
$$
\sqrt{\dfrac{r+x}{2}}\sqrt{\dfrac{r-x}{2}}=\dfrac{\sqrt{r^2-x^2}}{2}=\dfrac{|y|}{2}
$$
Thus,
$$
\left(\sqrt{\dfrac{r+x}{2}}+i\sqrt{\dfrac{r-x}{2}}\right)^2=x+i|y|
$$
If $y>0$,
$$
\left(\sqrt{\dfrac{r+x}{2}}+i\sqrt{\dfrac{r-x}{2}}\right)^2=x+iy
$$
However, if $y<0$,
$$
\left(\sqrt{\dfrac{r+x}{2}}-i\sqrt{\dfrac{r-x}{2}}\right)^2=x+iy
$$
So to be correct, we should incorporate $\newcommand{\sgn}{\operatorname{sgn}}\sgn(y)$:
$$
\left(\sqrt{\dfrac{r+x}{2}}+i\sgn(y)\sqrt{\dfrac{r-x}{2}}\right)^2=x+iy
$$
Negate as necessary to get both solutions.
| {
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Solving a Diophantine Equation using factorisation of ideals I am stuck on the following question which is given as follows:
Prove that the only integer solutions to the equation
\begin{equation}
x^2 + 13 = y^3
\end{equation}
are $(70,17)$ and $(-70, 17)$.
(Hint: first show that $x$ is even and $y$ is odd)
I have seen a solution to this and first part to it is as follows.
Let $x$ and $y$ satisfy the given equation. A odd square is $1$ modulo $4$ and a cube cannot be $2$ modulo $4$ so $x$ is even, which immediately implies $y$ is odd.
So we have a factorisation of ideals $(x + \sqrt{-13}) \cdot (x - \sqrt{-13}) = (y)^3. \ $ The gcd of $(x + \sqrt{-13})$ and $(x - \sqrt{-13})$ divides $(2\sqrt{-13}) = (2, 1+\sqrt{-13})^2 \, (\sqrt{-13}). \ $ As $y$ is odd, $(y)$ cannot be divisible by $(2, 1+\sqrt{-13})$ hence
\begin{equation}
(x + \sqrt{-13}, \, x - \sqrt{-13}) = (1) \hspace{5pt} \text{or} \hspace{5pt} (\sqrt{-13}) \hspace{5pt} (*)
\end{equation}
In the latter case, $x$ would be divisible by $13$, hence $y$ would be divisible by $13$ but then $13 \ \big| \ y^3$ and $x^2 + 13 \equiv 13 \pmod{13^2}$ which is a contradiction so $(x + \sqrt{-13}, \, x - \sqrt{-13}) = (1)$. It follows that the principal ideal $(x + \sqrt{-13})$ is the cube of an ideal $I$.
So firstly why does $(*)$ hold given that $(y)$ cannot be divisible by $(2, 1 +\sqrt{-13})$ and in the latter case why does this mean $x$ is divisible by $13$?. Finally how do we know that $(x + \sqrt{-13})$ is the cube of an ideal $I$?
Any help would be very much appreciated!
| For your first question: Let $P$ denote the gcd of $(x + \sqrt{-13})$ and $(x - \sqrt{-13})$. You have shown that $P$ divides $(2,1 + \sqrt{-13})^2(\sqrt{-13})$. But also $$P\mid (x - \sqrt{-13})(x + \sqrt{-13}) = (y)^3.$$ Since $P$ divides both $(y)^3$ and $(2,1 + \sqrt{-13})^2(\sqrt{-13})$, it must divide their gcd. You know that $(2,1 + \sqrt{-13})\nmid (y)$, and hence $(2,1+\sqrt{-13})\nmid (y)^3$, so the gcd of $(2,1+\sqrt{-13})^2(\sqrt{-13})$ and $(y)^3$ divides $(\sqrt{-13})$. Therefore $P\mid (\sqrt{-13})$. I hope that clarifies the first point.
For the second question: assuming $P = (\sqrt{-13})$. Since $P$ divides both $(x - \sqrt{-13})$ and $(x + \sqrt{-13})$, you get $$(13) = P^2 \mid (x + \sqrt{-13})(x - \sqrt{-13}) = (x^2 + 13),$$ so $13\mid x^2 + 13$. It follows that $13\mid x$.
For the third question. We assume that $P = (1)$, and want to understand why $(x + \sqrt{-13})$ is the cube of an ideal. This will follow from the starting equation $$\tag{$**$}(x -\sqrt{-13})(x + \sqrt{-13}) = (y)^3.$$ Since $P = (1)$, the ideals on the left hand side are coprime. Thus any primes dividing $(y)$ must divide either $(x + \sqrt{-13})$ or $(x - \sqrt{-13})$, but not both. Let $(y) = P_1,\ldots, P_mQ_1,\ldots, Q_n$, where the $P_i\mid(x + \sqrt{-13})$ and the $Q_i\mid (x - \sqrt{-13})$. Then from ($**$) it follows that $P_1^3\cdots P_m^3 = (x + \sqrt{-13})$ and $Q_1^3\cdots Q_n^3 = (x - \sqrt{-13})$. In particular, both $(x + \sqrt{-13})$ and $(x - \sqrt{-13})$ are cubes of ideals.
| {
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Complex roots of $z^3 + \bar{z} = 0$ I'm trying to find the complex roots of $z^3 + \bar{z} = 0$ using De Moivre.
Some suggested multiplying both sides by z first, but that seems wrong to me as it would add a root ( and I wouldn't know which root was the extra ).
I noticed that $z=a+bi$ and there exists $\theta$ such that the trigonometric representation of $z$ is $\left ( \sqrt{a^2+b^2} \right )\left ( \cos \theta + i \sin \theta \right )$ .
It seems that $-\bar{z} = -\left ( \sqrt{a^2+b^2} \right )\left ( \cos (-\theta) + i \sin (-\theta) \right )$
However, my trig is pretty rusty and I'm not quite sure where to go from here.
| OK, I'm going to take a stab at this.
Given: $z^3 + \bar{z} = 0$
Therefore $z^3 = -\bar{z}$ and $|z^3| = |-\bar{z}|$ .
We have that $|-\bar{z}|=|\bar{z}|=|z|$ therefore $|z^3|=|z|$ and $|z||z||z|=|z|$.
Therefore $|z|=0$ or $|z|=1$.
Case 1: Assume $|z|=0$ then if
$z=a+bi\leftrightarrow|z|=\sqrt{a^2+b^2}=0\leftrightarrow a=0 \wedge b=0\leftrightarrow a+bi=0$
thus $z = 0$.
Case 2: Assume $|z|=1$
Let's multiply by $z$ and we get $z^4 = -z\bar{z}$. We see that $z\bar{z} = |z|^2$ so we get $z^4 = -( |z|^2 )$ so from the above either:
therefore $z^4 = -|z| = -1$
The trigonometric representation of $-1$ is $1*( \cos \pi + i \sin \pi )$ so according to De Moivre:
$z^4 = r^4(\cos 4\theta + i \sin 4\theta ) = 1*( \cos \pi + i \sin \pi )$
These are two complex numbers in trigonometric form so:
$r^4 = 1$ and $4\theta = \pi + 2\pi*k$ or
$r=1$ and $\theta = \frac{\pi + 2\pi*k}{4}$ and each solution has the form:
$z_k = \cos( \frac{\pi + 2\pi*k}{4} ) + i \sin (\frac{\pi + 2\pi*k}{4})$
for $0\leq k \leq 3$.
Which together with Case 1 gives the following values for $z$:
$0,\pm\frac{1+i}{\sqrt{2}},\pm\frac{1-i}{\sqrt{2}}$
| {
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How are the sum and product of root formulas derived? For a polynomial of degree n,say
$$F(x)=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+....+a_{n}x^n$$
The sum of roots is $-\frac{a_{n-1}}{a_n}$ and the product of roots is $(-1)^n \frac{a_{0}}{a_n}$ where the $a_i%$ are the coefficients of the polynomial. I've been using these formulas for some problems, but I don't get why they are true. How are they derived? Thanks.
| $P(x)=a_nx^n+a_{n-1}x^{n-1}+a_1x+a_0=a_n(x-x_1)(x-x_2)\cdot \cdot (x-x_n)$
To show the product of roots:
Need to calculate $P(0)$
$P(0)=a_n0^n+a_{n-1}0^{n-1}+a_10+a_0=a_0$
$P(0)=a_n(0-x_1)(0-x_2)\cdot \cdot (0-x_n)=a_n(-x_1)(-x_2)\cdot \cdot (-x_n)=a_n(-1)x_1(-1)x_2\cdot \cdot (-1)x_n=a_n(-1)^nx_1x_2\cdot \cdot x_n$
$P(0)=a_n(-1)^nx_1x_2\cdot \cdot x_n=a_0$
$x_1x_2\cdot \cdot x_n=\frac{a_0(-1)^n}{a_n}$
To show Sum of roots:
Need to focus the coefficient of $x^{n-1}$
$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$
$(x-x_1)(x-x_2)(x-x_3)=(x^2-(x_1+x_2)x+x_1x_2)(x-x_3)=x^3-(x_1+x_2)x^2-x_3x^2 \cdot \cdot=x^3-(x_1+x_2+x_3)x^2 \cdot \cdot$
$(x-x_1)(x-x_2)(x-x_3)(x-x_4)=(x^3-(x_1+x_2+x_3)x^2 \cdot \cdot)(x-x_4)=x^4-(x_1+x_2+x_3)x^3-x_4x^3 \cdot \cdot=x^4-(x_1+x_2+x_3+x_4)x^3 \cdot \cdot\cdot \cdot$
$a_n(x-x_1)(x-x_2)\cdot \cdot (x-x_n)=a_n(x^n-(x_1+x_2+x_3+x_4+\cdot \cdot+x_n)x^{n-1} \cdot \cdot \cdot \cdot )=a_nx^n-a_n(x_1+x_2+x_3+x_4+\cdot \cdot+x_n)x^{n-1} \cdot \cdot \cdot \cdot=a_nx^n+a_{n-1}x^{n-1}+a_1x+a_0$
If we equal the coefficient of $x^{n-1}$ the result
is: $-a_n(x_1+x_2+x_3+x_4+\cdot \cdot+x_n)=a_{n-1}$
$x_1+x_2+x_3+x_4+\cdot \cdot+x_n=\frac{-a_{n-1}}{a_n}$
| {
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Linear equation is equal to a cubic equations, question about number of solutions? $f(x) = -x^3 +3x^2 +9x - 11 $
$f(x) = 9x + b$
$ -x^3 +3x^2 +9x - 11 = 9x + b$
For what values of b does this have three different solutions...
From the looks of it the linear equation needs to cross the cubic one i three places. The rise of the linear equation is 9, is that random?
From the part task before this one I found two equations that I was suppose to use in the solving part of this equation. I just don't see how they help!
$Y = 9x -11$
$Y = 9x - 7$
These are just the equations for the two tangents on f(x) with rise equal to 9.
| I interpret the problem as asking the following. Let $f(x)=-x^3+3x^2+9x-11$ and let $g(x)=9x+b$. For what values of $b$ does the equation $f(x)=g(x)$ have three different solutions? Equivalently, for what values of $b$ does the curve $y=f(x)$ meet the line $y=g(x)$ at three different points?
We arrive at the equation $-x^3+3x^2+9x-11=9x+b$. This simplifies to
$$x^3-3x^2=-(11+b).$$
We now draw the graph of $y=x^3-3x^2$. Note that $\frac{dy}{dx}=3x^2-6x$. This is $0$ at $x=0$ and $x=2$.
So $x^3-3x^2$ is steadily increasing until $x=0$, where it reaches $0$. Then it steadily decreases until $x=2$, where it reaches the value $-4$. After that, the curve steadily increases.
By looking at the graph, we can see that the horizontal line $y=-(11+b)$ meets the curve $y=x^3-3x^2$ at $3$ distinct places precisely if $-(11+b)$ lies between $0$ and $-4$ (but not including these points). So the $b$'s that work are all $b$ in the open interval $(-11, -7)$.
There are in a sense three points of intersection when $b=-11$, but two happen to coincide. The same is true when $b=-7$. For values of $b$ not in $[-11,-7]$, there is only one point of intersection.
Remark: An equivalent solution that is not as evident geometrically is that we will have three meeting points between the values of $b$ for which the line $y=9x+b$ is tangent to the curve $y=-x^3+3x^2+9x-11$. For tangency, slopes must match, so we want
$$-3x^2+6x+9=9,$$
from which we get $x=0$ and $x=2$.
In order for $y=9x+b$ to be tangent to $y=f(x)$ at $x=0$, values must match, meaning that $-11=b$. For tangency at $x=2$, values must match, giving $-8+12+18-11=18+b$, meaning that $b=-7$.
| {
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Parametric question of the curve $x^2 + y^2 + 2x - 4y = 0$? What is the parametric form of the curve above? If I had to solve it, what I would say is that the first step is to complete the square. However, where would I go from there?
| $$x^2 + y^2 + 2x - 4y = 0$$
$$x^2 + 2x + y^2- 4y = 0$$
$$x^2 + 2x + 1 -1 +y^2- 4y +4 -4= 0$$
$$(x+1)^2 -1 + (y-2)^2-4= 0$$
$$(x+1)^2 + (y-2)^2= 5$$
For a circle with centre at (h,k) and radius r, the a parametric form is $x = h + r\cos(\theta)$ and $y = k + r\sin(\theta)$.
| {
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Solving $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$ My question is: Solve $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$
I deduced that:$LHS= x+1-(x-2)$
I am unable to solve this equation. I would like to get some hints to solve it.
| $$\sqrt {x^2 +2x + 1}-\sqrt { x^2-4x+4}= \sqrt{(x+1)^2} - \sqrt{(x-2)^2}=|x+1|-|x-2|$$
You have to consider three cases:
*
*$x \geq 2$
*$-1<x<2$
*$x \leq -1$
| {
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Circumcenter and incenter I need a hint on this problem:
Given a triangle ABC. CH is the altitude. CM and CN are the bisectors of $\angle ACH$ and $\angle BCH $. The circumcircle of $\triangle MNC$ and the incircle of $\triangle ABC$ have common center. Find the $\angle ACB$.
| Let $O$ be center of circles. Denote $\angle{CAB}=\alpha$, $\angle{CBA}=\beta$, then
$$
\angle{ACH}=\frac{\pi}{2}-\angle{CAB}=\frac{\pi}{2}-\alpha,\quad\angle{BCH}=\frac{\pi}{2}-\angle{CBA}=\frac{\pi}{2}-\beta
$$
$$
\angle{MCH}=\frac{1}{2}\angle{ACH}=\frac{\pi}{4}-\frac{\alpha}{2},\quad\angle{NCH}=\frac{1}{2}\angle{NCH}=\frac{\pi}{4}-\frac{\beta}{2}
$$
$$
\angle{CMH}=\frac{\pi}{2}-\angle{MCH}=\frac{\pi}{4}+\frac{\alpha}{2},\quad\angle{CNH}=\frac{\pi}{2}-\angle{NCH}=\frac{\pi}{4}+\frac{\beta}{2}
$$
$$
\angle{MCN}=\angle{MCH}+\angle{NCH}=\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\beta}{2}
$$
$$
\angle{ACB}=\pi-\angle{CAB}-\angle{CBA}=\pi-\alpha-\beta.
$$
Since $O$ is the center of incircle of $\triangle{ABC}$, then
$$
\angle{ACO}=\frac{1}{2}\angle{ACB}=\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\beta}{2},\quad
\angle{HCO}=\angle{ACO}-\angle{ACH}=\frac{\alpha}{2}-\frac{\beta}{2}.
$$
Denote $CH=x$, then
$$
AC=\frac{CH}{\sin\angle{CAB}}=\frac{x}{\sin\alpha},\quad
BC=\frac{CH}{\sin\angle{CBA}}=\frac{x}{\sin\beta}
$$
$$
AH=CH\cot\angle{CAB}=x\cot\alpha,\quad BH=CH\cot\angle{CBA}=x\cot\beta
$$
$$
MH=CH\cot\angle{CMH}=x\cot\left(\frac{\pi}{4}-\frac{\alpha}{2}\right),\quad NH=CH\cot\angle{CNH}=x\cot\left(\frac{\pi}{4}-\frac{\beta}{2}\right)
$$
$$
AB=AH+BH=x(\cot\alpha+\cot\beta)
$$
$$
MN=MH+NH=x\left(\cot\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+\cot\left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right).
$$
Since $CO$ is the radius of circumcircle of $\triangle{CMN}$, then
$$
CO=\frac{MN}{2\sin\angle{MCN}}=\frac{x\left(\cot\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+\cot\left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right)}{2\sin\left(\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{x}{\sin\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}.
$$
In order to compute radius of incircle of $\triangle{ABC}$ we need to know its area and perimeter
$$
S_{\triangle{ABC}}=\frac{1}{2}AB\; CH=\frac{1}{2}x^2(\cot\alpha+\cot\beta)
$$
$$
P_{\triangle{ABC}}=AB+BC+CA=x\left(\cot\alpha+\cot\beta+\frac{1}{\sin\alpha}+\frac{1}{\sin\beta}\right).
$$
Hence
$$
r=\frac{2S_{\triangle{ABC}}}{P_{\triangle{ABC}}}=\frac{x^2(\cot\alpha+\cot\beta)}{x\left(\cot\alpha+\cot\beta+\frac{1}{\sin\alpha}+\frac{1}{\sin\beta}\right)}=
\frac{x\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}
$$
Consider altitudes $OP$ and $OQ$ on sides $CH$ and $AB$, then
$$
CP=CO\cos\angle{HCO}=\frac{x\sin\left(\frac{\alpha-\beta}{2}\right)}{\sin\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}
$$
$$
PH=OQ=r=\frac{x\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}.
$$
Since $CH=CP+PH$ we get the following equation
$$
x=\frac{x\sin\left(\frac{\alpha-\beta}{2}\right)}{\sin\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}+\frac{x\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}
$$
After some laborious algebra we get
$$
\sin\left(\frac{\alpha+\beta}{2}\right)=\cos\left(\frac{\alpha+\beta}{2}\right)
$$
$$
\frac{\alpha+\beta}{2}=\frac{\pi}{2}-\frac{\alpha+\beta}{2}
$$
Hence
$$
\alpha+\beta=\frac{\pi}{2}
$$
$$\angle{ACB}=\pi-\alpha-\beta=\frac{\pi}{2}.
$$
| {
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Find the vertical and horizontal asymptotes of the function. I am asked to find the vertical and horizontal asymptotes of the equation:
$$f(x)=(a^{-1}+x^{-1})^{-1}$$
I simplify this to
$$f(x)=\frac{1}{a^{-1}+x^{-1}}$$
$$f(x)=a^1+x^1$$$$f(x)=a+x$$Which is some constant, graphed as horizontal line - that will not have a vertical or horizontal asymptote. Is my algebra terribly off?
| If $\displaystyle f(x) = \frac{1}{a^{-1} + x^{-1}}$, we can't distribute the inversion $x \mapsto x^{-1}$ to say that $f(x) = a + x$.
A concrete example:
$\displaystyle \frac{1}{\frac{1}{2} + \frac{1}{3}} = \frac{1}{\frac{5}{6}} = \frac{6}{5} \neq \frac{1}{\frac{1}{2}} + \frac{1}{\frac{1}{3}} = 2 + 3 = 5$.
(We can, however, distribute the divisor like this: $\displaystyle \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$.)
Note that the function is defined in $\mathbb{R} \setminus\{-a\}$ (the only "problem" we might run into in order to evaluate $f(x)$ at an arbitrary $x \in \mathbb{R}$ is dividing by zero; and that would only happen when $a^{-1} + x^{-1} = 0 \Leftrightarrow -a = x$), so to check if there are vertical asymptotes, we calculate $\lim_{x \to -a} f(x)$.
$\displaystyle \lim_{x \searrow -a} \frac{1}{a^{-1}+x^{-1}} = \lim_{x \nearrow -a^{-1}} \frac{1}{a^{-1} + x} = \lim_{x \nearrow 0} \frac{1}{x} = - \infty$, and likewise
$\displaystyle \lim_{x \nearrow -a} \frac{1}{a^{-1}+x^{-1}} = \lim_{x \searrow -a^{-1}} \frac{1}{a^{-1} + x} = \lim_{x \searrow 0} \frac{1}{x} = + \infty$.
(Note that $x \geq -a \Rightarrow x^{-1} \leq (-a)^{-1}$ and $x \leq -a \Rightarrow x^{-1} \geq (-a)^{-1}$ if $x$ and $-a$ have the same sign, which will happen when they are close enough, for example, in this case because we're taking a limit; that's why when we start approaching a limit from the left and make the substitution we end up approaching the limit from the right and vice versa.)
We deduce that there is a vertical asymptote with equation $x=-a$.
As for the horizontal asymptotes, we calculate $\lim_{x \to \pm \infty} f(x)$, and we'll get:
$\displaystyle \lim_{x \to \pm \infty} \frac{1}{a^{-1} + x^{-1}}= \lim_{x \to 0} \frac{1}{a^{-1}+x} = \frac{1}{a^{-1}} = a$, so there is a horizontal asymptote with equation $y=a$.
| {
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When does $q^b = (p^a - 1)/(p - 1)$? $\DeclareMathOperator{\Ord}{Ord}$When does $q^b = (p^a - 1)/(p - 1)$ for $p, q$ odd primes and $a, b$ odd integers $> 1$ ? If no examples are possible, please give a simple proof.
A proof of this for $q=3, p=5$ might be
Assume $3^b = (5^a - 1)/4$; then
\begin{equation}
5^a - 4 \cdot 3^b = 1. \tag{1}
\end{equation}
We show that the exponents $a,b$ in (1) are even.
We see that $3^b \mid 5^a - 1$ $\Rightarrow$ $5^a \equiv 1 \bmod 3$. Also $5 \mid 4\cdot3^b+1$ or $4 \cdot 3^b \equiv -1 \bmod 5$ so that $3^b \equiv 1 \bmod 5$.
In short:
\[
3^b\equiv1 \bmod 5, \quad 5^a\equiv1 \bmod 3.
\]
By a well known theorem: if $X^c \equiv 1 \bmod p$, then $\Ord(c,p) \mid c$.
By inspection, $\Ord(3,5) = 4$ so that $4 \mid b$.
Similarly, $\Ord(5,3) = 2$ so that $2 \mid a$.
This violates our assumption that $a,b$ are odd. Proof complete.
However, moving further:
As $a,b$ are both even we can write (1) as $1 = 5^{2A} - 4\cdot 3^{2B} = (5 - 2\cdot3^B)(5 + 2\cdot3^B)$. Thus, $5 + 2\cdot3^B = 1$ but there is no positive value of $B$ satisfying this. Therefore (1) is not true.
Note: I want to be clear that the above proof, if it is correct, is not mine but was given to me by someone I am not free to name. If it is incorrect, the fault is entirely mine.
| Well, there are examples, say $$\frac{5^3-1}{5-1}=31^1=31$$or $$\frac{7^5-1}{7-1}=2801^1=2801$$
| {
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Prove that $\tan^{-1}\left(\frac{x+1}{1-x}\right)=\frac{\pi}{4}+\tan^{-1}(x)$ The question is:
Prove that $\tan^{-1}\left(\frac{x+1}{1-x}\right)=\frac{\pi}{4}+\tan^{-1}(x)$.
It's from A-level further mathematics.
| The identity should read $$\tan^{-1} \left(\dfrac{x+1}{1-x} \right) = \tan^{-1}(x) + \pi/4$$
Let $\tan^{-1}(x) = \theta$ i.e. $x = \tan(\theta)$. Then we get that $$\dfrac{x+1}{1-x} = \dfrac{\tan(\theta) + 1}{1-\tan(\theta)}$$
Recall that $\tan(A+B) = \dfrac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$.
Taking $B = \pi/4$, we get that $\tan(A+ \pi/4) = \dfrac{\tan(A) + 1}{1 - \tan(A)}$.
Hence, we get that
$$\dfrac{x+1}{1-x} = \dfrac{\tan(\theta) + 1}{1-\tan(\theta)} = \tan(\theta + \pi/4)$$
Hence, $$\tan^{-1} \left(\dfrac{x+1}{1-x} \right) = \theta + \pi/4 = \tan^{-1}(x) + \pi/4$$
Proof of $\tan(A+B) = \dfrac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$
$$\tan(A+B) = \dfrac{\sin(A+B)}{\cos(A+B)} = \dfrac{\sin(A) \cos(B) + \cos(A) \sin(B)}{\cos(A) \cos(B) - \sin(A) \sin(B)}$$
Assuming $\cos(A) \cos(B) \neq 0$, divide numerator and denominator by $\cos(A) \cos(B)$, to get that
$$\tan(A+B) = \dfrac{\sin(A) \cos(B) + \cos(A) \sin(B)}{\cos(A) \cos(B) - \sin(A) \sin(B)} = \dfrac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$$
| {
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Find the limit of: $\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}$ Could be the following limit computed without using Stirling's approximation formula?
$$\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}$$
I know that the limit is $e$, but I'm looking for some alternative ways that doesn't require to resort
to the use of Stirling's approximation. I really appreciate any support at this limit. Thanks.
| This completes Jonas's answer, here is an idea. This is too long for a comment.
To prove that the limit exists, we can prove that $a_n$ is decreasing and positive:
$$\frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}} \geq 0 \Leftrightarrow $$
$$\sqrt[n+1]{(n+1)!} \geq \sqrt[n]{(n)!} \Leftrightarrow $$
$$(n+1)!^n\geq (n)!^{n+1} \Leftrightarrow $$
$$(n+1)^n\geq (n)! \Leftrightarrow $$
$$(n+1)\cdot(n+1)...\cdot(n+1)\geq 1\cdot 2..\cot n \checkmark $$
Now for decreasing
$$\frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}} \geq \frac{1}{\sqrt[n+2]{(n+2)!} - \sqrt[n+1]{(n+1)!}} \Leftrightarrow $$
$$\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!} \leq \sqrt[n+2]{(n+2)!} - \sqrt[n+1]{(n+1)!} \Leftrightarrow $$
$$2\sqrt[n+1]{(n+1)!} \leq \sqrt[n]{(n)!}+ \sqrt[n+2]{(n+2)!}$$
Now, by AM-GM inequality
$$\frac{\sqrt[n]{(n)!}+ \sqrt[n+2]{(n+2)!}}{2} \geq (n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+1}$$
So if we can prove that
$$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!}$$
we are done.
Now
$$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!} \Leftrightarrow $$
$$(n!)^\frac{1}{2n}[(n+2)]^\frac{1}{2n+4} \geq (n+1)!^{\frac{1}{n+1}-\frac{1}{2n+4}} \Leftrightarrow $$
$$(n!)^{\frac{1}{2n}+\frac{1}{2n+4}-\frac{1}{n+1}}[(n+2)]^\frac{1}{2n+4} \geq (n+1)^{\frac{1}{n+1}-\frac{1}{2n+4}}\,. $$
To keep it simple:
The power of $n!$ is
$$\frac{(2n^2+6n+4)+(2n^2+2n)-(4n^2+8n)}{(n+1)2n(2n+4)}=\frac{2}{n(n+1)(2n+4)}$$
The power of $n+1$ is
$$\frac{1}{n+1}-\frac{1}{2n+4}=\frac{n+3}{(n+1)(2n+4)}$$
Thus, after bringing the inequality to the $n(n+1)(2n+4)$, it becomes:
$$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!} \Leftrightarrow $$
$$(n!)^2(n+2)^{n(n+1)} \geq (n+1)^{n(n+3)} $$
Now, I ma not sure that this is true, but might work....
| {
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Which of these statements about $Q=\frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000}$ is correct? Pick the correct option regarding $Q$.
$$Q=\frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000}$$
Pick one option:
*
*$Q>1\qquad$ 2. $Q\leq \frac{1}{3}\qquad$ 3. $\frac{1}{3}<Q\leq \frac{2}{3}\qquad$ 4. $\frac{2}{3}<Q\leq 1$
My approach :
$$Q = \frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000} > \frac{1}{1000}+\frac{1}{1000}+\cdots+\frac{1}{1000} = \frac{901}{1000}$$
$$\implies \frac{1}{100}+\frac{1}{101}+\cdots+\frac{1}{1000} > \frac{9}{10}$$
So, $Q > 1$. Option 1 is correct.
Now, my question is: can this be proved with some other approach?
| Well, let's proceed with the same sort of approach you tried, but a bit finer.
There are $100$ terms greater than $1/200$, another $100$ terms greater than $1/300$, and so on, up to $100$ terms greater than $1/1000$. But this gives that the sum is greater than $1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10$, which is about $1.9$, but more importantly is very easily calculatable to be greater than $1$.
| {
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For prime $p>2: 1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$
Possible Duplicate:
Why is the square of all odds less than an odd prime $p$ congruent to $(-1)^{(p+1)/(2)}\pmod p$?
If p is an odd prime, prove that $1^2 \times 3^2 \times 5^2 \cdots \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}$
I'd love your help with proving the following claim:
For prime $p>2$:
$$1^23^25^2\cdot\cdot\cdot(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p.$$
I instantly thought of Wilson Theorem which says that $1\cdot2\cdot3\cdot\cdot\cdot\cdot(p-1) \equiv (-1) \pmod p$, but I can't see how to use it.
I also tried to divide it to two cases, for $p \equiv 1 \pmod4$, and $p \equiv 3 \pmod4$, but again I didn't reach the conclusion.
Thanks a lot!
| It isn't true. Take $p = 5$, then $1^2 3^2 \equiv -1 \not \equiv 1 \equiv (-1)^\frac{5-1}{2} \pmod 5$
| {
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congruent to mod p $1^{p-2}+2^{p-2}+\cdots+\left(\frac{p-1}{2}\right)^{p-2}\equiv\frac{2-2^p}{p}\pmod p.$ Let $p$ be an odd prime.How to prove that
$$1^{p-2}+2^{p-2}+\cdots+\left(\frac{p-1}{2}\right)^{p-2}\equiv\frac{2-2^p}{p}\pmod p.$$
| I find that this identity is stated in this paper:
Lehmer, E. "On Congruences Involving Bernoulli Numbers and the Quotients of Fermat and Wilson." Ann. Math. 39, 350-360, 1938 :
http://www.jstor.org/discover/10.2307/1968791?uid=2134&uid=2&uid=70&uid=4&sid=56284309973
Equation (30) in the paper answers your question.
EDIT:
In fact,we can use a simpler method to solve this problem.
Let $q(a)=\frac{a^{p-1}-1}{p}$ for all $a$ such that $\gcd(a,p)=1$.
It is not difficult to show that $q(ab) \equiv q(a)+q(b) \pmod{p}...(1)$.
Let $a$ be an arbitrary integer which is relatively prime to $p$.
For each $v \in \{1,2,...,p-1\}$,let $av=\lfloor{\frac{av}{p}} \rfloor p+r_v$.
Then we can see that $r_v$,($r=1,2,...,p-1$) also runs over ${1,2,...,p-1}$.
So $q(av)=((\lfloor{\frac{av}{p}} \rfloor p+r_v)^{p-1}-1)/p\equiv q(r_v)-\lfloor{\frac{av}{p}} \rfloor r_v^{p-2} \equiv q(r_v)-\frac{1}{av} \left \lfloor \frac{av}{p} \right \rfloor \pmod{p}$
So, $\sum_{v=1}^{p-1} q(av) \equiv \sum_{v=1}^{p-1} (q(r_v)-\frac{1}{av} \left \lfloor \frac{av}{p} \right \rfloor)\equiv \sum_{v=1}^{p-1} q(v)-\sum_{v=1}^{p-1}\frac{1}{av} \left \lfloor \frac{av}{p} \right \rfloor \pmod{p}...(2)$
Then by $(1),(2)$,we get $q(a) \equiv \sum_{v=1}^{p-1}\frac{1}{av} \left \lfloor \frac{av}{p} \right \rfloor \pmod{p}...(3) $
Take $a=2$, $\frac{2^p-2}{p} \equiv \sum_{v=1}^{p-1}\frac{1}{v} \left \lfloor \frac{2v}{p} \right \rfloor \equiv \sum_{p/2 <v} \frac{1}{v} \equiv -\sum_{v=1}^{(p-1)/2} \frac{1}{v} \equiv -\sum_{v=1}^{(p-1)/2} v^{p-2} \pmod{p}$
As a result,
$1^{p-2}+2^{p-2}+\cdots+\left(\frac{p-1}{2}\right)^{p-2}\equiv\frac{2-2^p}{p}\pmod p.$
| {
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Largest number that divides $n^2(n^2 - 1)(n^2 - n - 2)$ for all $n$ Obtain the greatest natural that divides $n^2(n^2 - 1)(n^2 - n - 2)$ for all natural numbers $n$.
What should be the approach in these type of questions?
Should I equate with prime factorization $2^a 3^b 5^c \cdots$ etc?
| Factor as $(n-2)(n-1)n^2(n+1)^2$. It is easy to see that no prime greater than $3$ must divide this product. (For if $p$ is a prime greater than $3$, let $n-3=p$.)
Our product is obviously divisible by $3$. By choosing $n-1=3$, we can make sure that $3^2$ does not divide our product.
To minimize the power of $2$ that divides our product, we make $n-2$ (respectively, $n-1$) divisible by $4$ but not by $8$. Then $n^2$ (respectively, $(n+1)^2$) is divisible by $4$ but not by $8$. Thus $2^4$ must divide our product, and a higher power of $2$ need not.
We conclude that the largest positive integer that always divides our product is $(3^1)(2^4)$.
| {
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$ | a^3 + ab^2 + a^2 b + b^3 | \leqslant |a + b|^3 $ holds?? Let $a , b \in \mathbb R$. Then $$ | a^3 + a^2 b + ab^2 + b^3 | \leqslant |a + b|^3 $$ holds?
| The left hand side is:
$$|a^3+a^2b+ab^2+b^3|=|(a+b)(a^2+b^2)|=|a+b|\cdot(a^2+b^2)$$
The right hand side is:
$$|a+b|^3=|a+b|\cdot(a+b)^2=|a+b|\cdot(a^2+2ab+b^2)$$
So the inequality is true if and only if $RHS-LHS\ge 0$:
$$|a+b|\cdot 2ab\ge 0$$
$$a=-b\quad\text{or}\quad ab\ge 0$$
| {
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Convergence of the sequence $\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots$ I recently came across this problem
Q1 Show that $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }} = 3$
After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists.
So my question is
Q2 Prove that the sequence$$\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}$$ converges.
Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .
If the solution from that book is required I can post it but it is not complete as I mentioned.
Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist.
| This is a known question of Putnam competition: I give a complete proof with details:
$f_n(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+n-1)\sqrt{1+(x+n)}}}}$
it is clear that for $m>n$, we have $f_m(x)>f_n(x)$. Moreover
$x+1=\sqrt{1+x(x+2)}=\sqrt{1+x\sqrt{1+(x+1)(x+2)}}=\cdots=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\cdots+(x+n-1)(x+n+1)}}}\geq f_n(x)$.
So this tells us that $\lim_{n\to \infty}f_n(x)$ exists. Now let $\lim_{n\to \infty}f_n(x)=f(x)$ hence $f(x)\leq x+1$ and $f(x)>\sqrt{1+x\sqrt{1+x\sqrt{1+x\sqrt{1+...}}}}=\frac{x+\sqrt{x^2+4}}{2}>x$ hence $x<f(x)<x+1$. Now, let $g(x)=x+1-f(x)$ then $0\leq g(x)<1$, it is easy to see that $(f(x))^2=1+xf(x+1)$ hence $(x+1+f(x))g(x)=xg(x+1)$, so
$$\frac{g(x)}{x}\leq \frac{g(x+1)}{x+1}\leq \cdots\leq \frac{g(x+n)}{x+n}<\frac{1}{x+n}\to 0$$ hence $f(x)=x+1$, so $f(2)=3$
| {
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If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
This seems clear, but I don't know how to prove this..
I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..
| The proof is by induction. Let us denote $\gcd(a,b) = d$.
For $n=1$, there is nothing to prove. Assume that it is true for all $n \leq k$ i.e. $\gcd(a^n,b^n) = d^n$ for all $n \leq k$. We now need to prove that $\gcd(a^{k+1},b^{k+1}) = d^{k+1}$. Since $\gcd(a^k,b^k) = d^k$, there exists $x_k,y_k \in \mathbb{Z}$ such that $a^k x_k + b^k y_k = d^k$. We also have $x,y \in \mathbb{Z}$ such that $ax+by = d$. Hence, we get that $$(ax+by) \left( a^k x_k + b^k y_k \right)^2 = d^{2k+1}.$$
Expanding this, we get that $$a^{k+1} \left( a^k x_k^2 x + a^{k-1} b x_k^2 y + 2 b^k x_k y_k x \right) + b^{k+1} \left( b^k y_k^2 y + ab^{k-1} y_k^2 x + 2 a^k x_k y_k y \right) = d^{2k+1}.$$ Since $d|a$ and $d|b$, we have that $a = d e$ and $b = df$. Hence, we get that $$a^{k+1} \left( d^k e^k x_k^2 x + d^k e^{k-1} f x_k^2 y + 2 d^k f^k x_k y_k x \right) + b^{k+1} \left( d^k f^k y_k^2 y + d^k ef^{k-1} y_k^2 x + 2 d^k e^k x_k y_k y \right) = d^{2k+1}.$$
This gives us that
$$a^{k+1} \left( e^k x_k^2 x + e^{k-1} f x_k^2 y + 2 f^k x_k y_k x \right) + b^{k+1} \left( f^k y_k^2 y + ef^{k-1} y_k^2 x + 2 e^k x_k y_k y \right) = d^{k+1}.$$
Hence, we have found integers $$x_{k+1} = \left( e^k x_k^2 x + e^{k-1} f x_k^2 y + 2 f^k x_k y_k x \right), \, y_{k+1} = \left( f^k y_k^2 y + ef^{k-1} y_k^2 x + 2 e^k x_k y_k y \right)$$ such that $$a^{k+1} x_{k+1} + b^{k+1} y_{k+1} = d^{k+1}.$$
Hence, we have that $\gcd(a^{k+1}, b^{k+1}) \vert d^{k+1}$. It is also true that $d^{k+1} \vert a^{k+1}$ and $d^{k+1} \vert b^{k+1}$, since $d \vert a$ and $d \vert b$. Hence, $d^{k+1} \vert \gcd(a^{k+1}, b^{k+1})$. Hence, we get that $$\gcd(a^{k+1}, b^{k+1}) = \gcd(a,b)^{k+1}.$$
Hence, by the principle of mathematical induction, we have that $$\gcd(a^n,b^n) = \gcd(a,b)^n, \,\, \forall n \in \mathbb{Z}^+.$$
| {
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Proving that an expression divides a number How do you prove that
$$n(n+1)(n+2)$$
is divisible by 6 by using the method of mathematical induction?
According to my book
$$\begin{aligned}
(n+1)(n+2)(n+3) &= n(n+1)(n+2)+3(n+1)(n+2)\\
&= 6k + 3*2k'\\
&= 6(k+k')\\
&=6k''
\end{aligned}$$
But I wonder, where does that k come from anyway?
| Since one number out of every $2$ consecutive integers is divisible by $2$ and one out of every $3$ consecutive integers is divisible by $3$, therefore, in the product of $3$ consecutive integers $n(n+1)(n+2)$, atleast one number is divisible by $2$ and one divisible by $3$ and $\gcd(2,3)=1 $ $ \implies n(n+1)(n+2)$ is divisible by $2\times3=6$. $$$$ You can also do it this way: Since every integer is one of the forms: $6k,6k+1,6k+2,6k+3,6k+4 $ or $6k+5$, therefore, possibilities the product of three consecutive numbers is : $$6k(6k+1)(6k+2)=6(k(6k+1)(6k+2))$$ $$(6k+1)(6k+2)(6k+3)=(6k+1)2(3k+1)3(2k+1)=6((6k+1)(3k+1)(2k+1))$$ $$(6k+2)(6k+3)(6k+4)=2(3k+1)3(2k+1)(6k+4)=6((3k+1)(2k+1)(6k+4))$$ $$(6k+3)(6k+4)(6k+5)=3(2k+1)2(3k+2)(6k+5)=6((2k+1)(3k+2)(6k+5)).$$ Thus product of every possible combination of $3$ consecutive integers is coming out to be divisible by $6$.
| {
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Find $\lim \ a_n$ if $a_n = \frac{1}{\sqrt[3]{n^3+1}} + \frac{1}{\sqrt[3]{n^3+2}}+\cdots+\frac{1}{\sqrt[3]{n^3+n}}.$ I would greatly appreciate some help in finding $\lim \ a_n$ if
$$a_n = \frac{1}{\sqrt[3]{n^3+1}} + \frac{1}{\sqrt[3]{n^3+2}}+\cdots+\frac{1}{\sqrt[3]{n^3+n}}.$$
| HINT:
Note that $$\dfrac1{\sqrt[3]{n^3+n}} \leq \dfrac1{\sqrt[3]{n^3+k}} \leq \dfrac1{\sqrt[3]{n^3+1}}$$
for all $k \in \{1,2,3,\ldots,n\}$.
Hence,
$$\dfrac1{\sqrt[3]{1+1/n^2}} = \dfrac{n}{\sqrt[3]{n^3+n}} \leq \sum_{k=1}^{n} \dfrac1{\sqrt[3]{n^3+k}} \leq \dfrac{n}{\sqrt[3]{n^3+1}} = \dfrac1{\sqrt[3]{1+1/n^3}}$$
Now apply the squeeze/sandwich theorem.
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Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.
I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$
I am trying to go ahead of this step.
| Wanted to add yet another method to the "catalogue"; I learned this by solving a problem from the book Numbers and Functions: Steps into Analysis by R P. Burn.
First, we have the following lemma:
Let $0 < a < b$. Then $$\frac{b^{n+1}-a^{n+1}}{b-a} < (n+1)b^n$$
This can be proved by noting that $$b^{n+1}-a^{n+1} = (b-a)(b^n + b^{n-1} a + b^{n-2} a^2 + \dots + a^n)$$ and since $0 < a < b \implies 0 < a^n < b^n$, we get $$b^n + b^{n-1} a + b^{n-2} a^2 + \dots + a^n < b^n + b^{n-1} b + b^{n-2} b^2 + \dots + b^n = (n+1)b^n$$.
Now plug in $$a = 1 + \frac{1}{n+1}, \hspace{10pt} b = 1 + \frac{1}{n}$$ into the lemma, and let $$t_n = \left(1 + \frac{1}{n}\right)^n$$
This gives
$$\frac{\overbrace{\left(1+\frac{1}{n}\right)^{n+1}}^{(1+\frac{1}{n})t_n} - \overbrace{\left(1 + \frac{1}{n+1}\right)^{n+1}}^{t_{n+1}}}{\frac{1}{n}-\frac{1}{n+1}} < (n+1)\overbrace{\left(1+\frac{1}{n}\right)^n}^{t_n}$$
which after some straightforward algebraic simplification yields $t_n < t_{n+1}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 14,
"answer_id": 7
} |
Find the closed form of the sequence Let $\{x_{n}\}_{n=1}^\infty$, with $x_{1}=a$ where $a>1$ be a sequence that satisfies the relation:
$$ x_{1}+x_{2}+...+x_{n+1}= x_{1}x_{2}\cdots x_{n+1}$$
For this problem, the requirement is to prove that $x_{n}$ is convergent, and then find its limit when $n$ goes to $\infty$. I think I can handle with these two requirements, but my curiosity is related to the way $x_{n}$ looks like and wonder if there is a nice closed form to it.
| For $n \ge 2$, $x_n = \dfrac{a^{2^{n-2}}}{P_n(a)}$ where $\displaystyle P_{n}(a) = a^{2^{n-2}} - \prod_{j=2}^{n-1} P_j(a)$ is a polynomial in $a$ of degree $2^{n-2}$.
$$\eqalign{P_2(a) &= a-1 \cr
P_3(a) &= a^2-a+1 \cr
P_4(a) &= a^4-a^3+2 a^2-2 a+1\cr
P_5(a) &= a^8-a^7+3a^6-6a^5+9a^4-10a^3+8a^2-4a+1\cr}$$
It looks like:
The coefficient of $a^0$ in $P_n(a)$ is $1$ for $n \ge 3$.
The coefficient of $a^1$ in $P_n(a)$ is $-2^{n-3}$ for $n \ge 3$.
The coefficient of $a^2$ in $P_n(a)$ is $2^{2n-7}$ for $n \ge 4$.
The coefficient of $a^3$ in $P_n(a)$ is $\dfrac{2^{n-3}-8^{n-3}}{6}$ for $n \ge 3$.
EDIT: The coefficient of $a^4$ in $P_n(a)$ is $\dfrac{2^n}{32} - \dfrac{4^n}{384} + \dfrac{16^n}{98304}$ for $n \ge 5$.
All this should be provable by induction. I got these by looking at the first few members of the sequence using Maple.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is $10^{8}!$ greater than $10^{10^9}$? My question is:
$10^8! > 10^{10^9}$ ?
I know that factorial is greater than exponential, but I am not sure about this specific case.
Thanks,
| By Stirling’s approximation,
$$\begin{align*}
10^8!&\approx\sqrt{2\cdot10^8\pi}\left(\frac{10^8}e\right)^{10^8}\\
&<10^5\cdot\left(10^8\right)^{10^8}\\
&<\left(10^8\right)^{10^8+1}\\
&=10^{8\cdot10^8+8}\\
&<10^{10^9}\;.
\end{align*}$$
Alternatively,
$$\begin{align*}
\ln 10^8!&=\sum_{k=1}^{10^8}\ln k\\
&<\int_1^{10^8}\ln x\,dx\\
&=\left[x\ln x-x\right]_1^{10^8}\\
&=8\cdot10^8\ln 10-10^8+1\\
&<10^9\ln 10\\
&=\ln 10^{10^9}\;,
\end{align*}$$
and $f(x)=\ln x$ is strictly increasing, so $10^8!<10^{10^9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\int^1_0 \frac{dx}{x^2+x+1}$ Here is an integral which I don't know the answer:
$$\int^1_0 \frac{dx}{x^2+x+1}$$
I tried to use complex number to solve i.e. the root of $x^2+x+1$ is $(-1/2 + \sqrt {3}i/2)$.
Let w=$(-1/2 + \sqrt {3}i/2)$ , then it becomes $ \int_{0}^{1} 1/(x-w)^2\,\mathrm{d} (x-w)$ , the answer is in terms of complex number. I wanna ask whether my method is correct and seek for another method. Thank you.
| Your idea of using complex functions can work, but you seem to believe the polynomial has only one root whereas it actually has two different (complex conjugate, of course, as the polynomial is real) roots:
$$x^2+x+1=\left(x+\frac{1+\sqrt 3i}{2}\right)\left(x+\frac{1-\sqrt 3i}{2}\right)=(x+w)(x+\overline w)\,\,,\,w:=\frac{1+\sqrt 3i}{2}$$
(note $\,w\,$ is a root of unit of order 6, though this is important here only to write $\,|w| = 1\,$) and from here we get
$$\frac{1}{x^2+x+1}=\frac{1}{(x+w)(x+\overline w)}=\frac{1}{\sqrt 3\,i}\left(\frac{1}{x+\overline w}-\frac{1}{x+w}\right)\Longrightarrow$$
$$\Longrightarrow \int_0^1\frac{dx}{x^2+x+1}=\frac{1}{\sqrt 3\,i}\left(\int_0^1\frac{dx}{x+\overline w}-\int_0^1\frac{dx}{x+w}\right)=\frac{1}{\sqrt 3\,i}\left.\left(Log(x+\overline w)-Log(x+w)\right)\right|_0^1$$
Now choose the principal branch of the complex logarithm, and we get:
$$\int_0^1\frac{dx}{x^2+x+1}=\frac{1}{\sqrt 3\,i}\left.Log\left(\frac{x+\overline w}{x+w}\right)\right|_0^1=\frac{1}{\sqrt 3\,i}\left.Log\left(\frac{(x+\overline w)^2}{|x+ w|^2}\right)\right|_0^1=$$
$$=\frac{1}{\sqrt 3\,i}\left(Log\frac{(1+\overline w)^2}{|1+w|^2}-Log\frac{\overline w^2}{|w|^2}\right)=\frac{1}{\sqrt 3\,i}\left[Log\left(\frac{1}{2}-\frac{\sqrt 3}{2}i\right)-Log\left(-\frac{1}{2}-\frac{\sqrt 3}{2}i\right)\right]=$$
$$=\frac{1}{\sqrt 3\,i}\,i\left[\frac{5\pi}{3}-\frac{4\pi}{3}=\right]=\frac{\pi}{3\sqrt 3}$$
You get, of course, the same result as the first, much simpler, answer by Joe, and there you don't have to mess with complex functions, their branches, conjugates and other beasts...yet it is doable!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Compute the trigonometric integrals How would you compute the following integrals?
$$ I_{n} = \int_0^\pi \frac{1-\cos nx}{1- \cos x} dx $$
$$ J_{n,m} = \int_0^\pi \frac{x^m(1-\cos nx)}{1- \cos x} dx $$
For instance, i noticed that the first integral is convergent for any value of $n$ since $\lim_{x\to0} \frac{1- \cos nx}{1 - \cos x}= n^2$. This fact may also be extended to the second integral, as well.
| The exponential generating function of $J_{m,n}$ for fixed $n$ is
$$G_n(t) = \sum_{m=0}^\infty J_{m,n} t^m/m! = \int_0^\pi e^{tx} \dfrac{\sin(nx)^2}{\sin(x)^2}\ dx$$
We then have $(\sin(nx)/\sin(x))^2 = n + 2 \sum_{j=1}^{n-1} (n-j) \cos(2 j x)$
and $$G_n(t) = (e^{\pi t} - 1) \left(\frac{n}{t} + 2 \sum_{j=1}^{n-1} (n-j) \dfrac{t}{t^2 + 4 j^2}\right)$$
EDIT:
For example, $G_3(t) = (e^{\pi t} - 1) \left(\dfrac{3}{t} + \dfrac{4t}{t^2+4} + \dfrac{2t}{t^2+16}\right)$. Now $$e^{\pi t} - 1 = \sum_{j=1}^\infty \pi^j \dfrac{t^j}{j!} = \pi t + \dfrac{\pi^2 t^2}{2} + \dfrac{\pi^3 t^3}{6} + \ldots$$
$$\dfrac{3}{t} + \dfrac{4t}{t^2+4} + \dfrac{2t}{t^2+16} = \dfrac{3}{t} + \sum_{j=0}^\infty (-1)^j(4^{-j}+ 16^{-j}/8) t^{2j+1} = \dfrac{3}{t} + \left(1+\dfrac{1}{8}\right) t - \left(\dfrac{1}{4} + \dfrac{1}{16 \times 8}\right) t^3 + \ldots$$
and so $J_{m,3}$ is $m!$ times the coefficient of $t^m$ in the product of these.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Finding the values of $\cos \frac{n\pi}{2}$ and $\sin \frac{n\pi}{2}$. i know that the values of $\cos n\pi=(-1)^{n}$ and $\sin n\pi=0$. Now i want to know that what is the general expressions of $\cos \frac{n\pi}{2}$ and $\sin \frac{n\pi}{2}$.
| $\displaystyle \cos\frac{n\pi}{2}=\{1,0,-1,0\}=2\left\lfloor\frac{n}{4}\right\rfloor+2\left\lfloor\frac{n+1}{4}\right\rfloor+1-n$
$\displaystyle \sin\frac{n\pi}{2}=\{0,1,0,-1\}=n-2\left\lfloor\frac{n+1}{4}\right\rfloor-2\left\lfloor\frac{n+2}{4}\right\rfloor$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
an infinite series involving odd zeta I ran across a cool series I have been trying to chip away at.
$$\sum_{k=1}^{\infty}\frac{\zeta(2k+1)-1}{k+2}=\frac{-\gamma}{2}-6\ln(A)+\ln(2)+\frac{7}{6}\approx 0.0786\ldots$$
where A = the Glaisher-Kinkelin constant. Numerically, it is approx. $1.282427\ldots$
I began by writing zeta as a sum and switching the summation order
$$\sum_{n=2}^\infty \sum_{k=1}^\infty \frac{1}{(k+2)n^{2k+1}}$$
The first sum is the series for $-n^3\ln(1-\frac{1}{n^2})-n-\frac{1}{2n}$
So, we have $-\sum_{n=2}^\infty \left[\ln(1-\frac{1}{n^2})+n+\frac{1}{2n}\right]$
This series numerically checks out, so I am onto something. At first glance the series looks like it should diverge, but it does converge.
Another idea I had was to write out the series of the series:
$$1/3(1/2)^{3}+1/4(1/2)^{5}+1/5(1/2)^{7}+\cdots
+1/3(1/3)^{3}+1/4(1/3)^{5}+1/5(1/3)^{7}+\cdots +1/3(1/4)^{3}+1/4(1/4)^{5}+1/5(1/4)^{7}+\cdots$$
and so on.
This can be written as
$$1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots +1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots + 1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots $$
where $x=1/2,1/3,1/4,\ldots$
This leads to the series representation for:
$$\frac{-\ln(1-x^2)}{x^3}-\frac{1}{x}-\frac{x}{2}$$
Since $x$ is of the form $1/n$, we end up with the same series as before.
Now, my quandary. How to finish?. Where in the world does that Glaisher-Kinkelin constant come in, and how can that nice closed from be obtained?. Whether from the series I have above or some other means. As usual, it is probably something I should be seeing but don't at the moment.
The GK constant has a closed form of $$e^{\frac{1}{12}-\zeta^{'}(-1)}$$.
Which means an equivalent closed form would be $\frac{-\gamma}{2}+\ln(2)+6\zeta^{'}(-1)+\frac{2}{3}$
Thanks all.
| We have
$$
\begin{eqnarray*}
&&\sum_{n=2}^\infty\left( -n^3\log(1-\frac{1}{n^2})-n-\frac{1}{2n}\right)\\
&=&\lim_{N\to\infty}\sum_{n=2}^N\left( -n^3\log(1-\frac{1}{n^2})-n-\frac{1}{2n}\right)\\
&=&\lim_{N\to\infty}\left(\sum_{n=2}^N -n^3\log(1-\frac{1}{n^2})-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}+O\left(\frac{1}{N}\right)\right)\right)\\
&=&\lim_{N\to\infty}[\sum_{n=2}^N (2n^3\log(n)-n^3\log(n+1)-n^3\log(n-1))-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]
\end{eqnarray*}
$$
In the sum on the last line, we may gather together the coefficients of each logarithm (terms at the boundary of the sum are a little funny), giving
$$
\begin{eqnarray*}
&&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)+(N^3+3N^2+3N+1)\log(N)-N^3\log(N+1)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\
&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)-N^3\log\left(1+\frac{1}{N}\right)+(3N^2+3N+1)\log(N)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\
&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)-N^3\left(\frac{1}{N}-\frac{1}{2N^2}+\frac{1}{3N^3}+O\left(\frac{1}{N^4}\right)\right)+(3N^2+3N+1)\log(N)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\
&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\left(3N^2+3N+\frac{1}{2}\right)\log(N)+\left(\frac{-3}{2}N^2+\frac{7}{6}-\frac{\gamma}{2}+\log(2)\right)]\\
&=&-6\log\left(\lim_{N\to\infty}\left(\frac{\prod_{n=1}^N n^n}{N^{N^2/2+N/2+1/12}e^{-N^2/4}}\right)\right)+\frac{7}{6}-\frac{\gamma}{2}+\log(2)\\
&=&-6\log(A)+\frac{7}{6}-\frac{\gamma}{2}+\log(2)
\end{eqnarray*}
$$
Here, I am taking
$$
A=\lim_{N\to\infty}\frac{\prod_{n=1}^N n^n}{N^{N^2/2+N/2+1/12}e^{-N^2/4}}
$$
as the definition of the Glaisher-Kinkelin constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ How would I verify the following double angle identity.
$$
\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B
$$
So far I have done this.
$$
(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)
$$But I am not sure how to proceed.
| \begin{eqnarray}
\sin(A+B)\sin(A-B) &=& (\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)\\\\
&=& \sin^2 A \cos^2 B -\sin^2 B \cos^2 A\\\\
&=& \sin^2 A \cos^2 B -\sin^2 B (1-\sin^2 A)\\\\
&=& \sin^2 A (\cos^2 B + \sin^2 B) - \sin^2 B\\\\
&=& \sin^2 A - \sin^2 B
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 2
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Finding shortest distance between a point and a surface Consider the surface $S$ (in $\mathbb R^3$) given by the equation $z=f(x,y)=\frac32(x^2+y^2)$. How can I find the shortest distance from a point $p=(a,b,c)$ on $S$ to the point $(0,0,1)$.
This is what I have done: Define $d(a,b,c)=a^2+b^2+(c-1)^2$, for all points $p=(a,b,c)\in S$. Then $\sqrt d$ is the distance from $S$ to $(0,0,1)$. I think that the method of Lagrange multipliers is the easiest way to solve my question, but how can I find the Lagrangian function? Or is there an easier way to find the shortest distance?
|
I think that the method of Lagrange multipliers is the easiest way to
solve my question, but how can I find the Lagrangian function?
As shown by other answers and in note 1 there are easier ways to find the shortest distance, but here is a detailed solution using the method of Lagrange multipliers. You need to find the minimum of the distance function
$$\begin{equation}
d(x,y,z)=\sqrt{x^{2}+y^{2}+(z-1)^{2}} \tag{1a}
\end{equation}$$
subject to the constraint given by the surface equation $z=f(x,y)=\frac32(x^2+y^2)$
$$\begin{equation}
g(x,y,z)=z-\frac{3}{2}\left( x^{2}+y^{2}\right) =0. \tag{2}
\end{equation}$$
Since $\sqrt{x^{2}+y^{2}+(z-1)^{2}}$ increases with $x^{2}+y^{2}+(z-1)^{2}$ you can simplify the
computations if you find the minimum of
$$\begin{equation}
[d(x,y,z)]^2=x^{2}+y^{2}+(z-1)^{2} \tag{1b}
\end{equation}$$
subject to the same constraint $(2)$. The Lagrangian function is then
defined by
$$\begin{eqnarray}
L\left( x,y,z,\lambda \right) &=&[d(x,y,z)]^2+\lambda g(x,y,z) \\
L\left( x,y,z,\lambda \right) &=&x^{2}+y^{2}+(z-1)^{2}+\lambda \left( z-
\frac{3}{2}\left( x^{2}+y^{2}\right) \right), \tag{3}
\end{eqnarray}$$
where $\lambda $ is the Lagrange multiplier. By this method you need to
solve the following system
$$\begin{equation}
\left\{ \frac{\partial L}{\partial x}=0,\frac{\partial L}{\partial y}=0,
\frac{\partial L}{\partial z}=0,\frac{\partial L}{\partial \lambda }
=0,\right. \tag{4}
\end{equation}$$
which results in
$$\begin{eqnarray}
\left\{
\begin{array}{c}
2x+3\lambda x=0 \\
2y+3\lambda y=0 \\
2z-2-\lambda =0 \\
-z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0
\end{array}
\right. &\Leftrightarrow &\left\{
\begin{array}{c}
x=0\vee 2+3\lambda =0 \\
y=0\vee 2+3\lambda =0 \\
2z-2-\lambda =0 \\
-z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0
\end{array}
\right. \\
&\Leftrightarrow &\left\{
\begin{array}{c}
x=0 \\
y=0 \\
\lambda =2 \\
z=0
\end{array}
\right. \vee \left\{
\begin{array}{c}
\lambda =-2/3 \\
z=2/3 \\
x^{2}+y^{2}=4/9
\end{array}
\right. \tag{5}
\end{eqnarray}$$
For $x=y=x=0$ we get $d(0,0,0)=1$. And for $x^2+y^2=4/9,z=2/3$ we get the minimum distance subject to the given conditions
$$\begin{equation}
\underset{g(x,y,z)=0}\min d(x,y,z)=\sqrt{\frac{4}{9}+(\frac{2}{3}-1)^{2}}=\frac{1}{3}\sqrt{5}. \tag{6}
\end{equation}$$
It is attained on the intersection of the surface $z=\frac{3}{2}\left(
x^{2}+y^{2}\right) $ with the vertical cylinder $x^{2}+y^{2}=\frac{4}{9}$ or equivalently with the horizontal plane $z=\frac{2}{3}$.
$$\text{Plane }z=\frac{2}{3} \text{(blue) and surface }z=\frac{3}{2}\left(
x^{2}+y^{2}\right) $$
Notes.
*
*As the solution depends only on the sum $r^{2}=x^{2}+y^{2}$ we could just find
$$\begin{equation}
\min [d(r)]^2=r^{2}+(\frac{3}{2}r^{2}-1)^{2} \tag{7}
\end{equation}$$
and then find $d(r)=\sqrt{[d(r)]^2}$ at the minimum.
*The surface $z=\frac{3}{2}\left(
x^{2}+y^{2}\right) $ is a surface of revolution around the $z$ axis.
| {
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"url": "https://math.stackexchange.com/questions/175624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 0
} |
From the given arc is known the start point the end point and a random point on the arc how we can find the center point coordinates? From the given arc is known the start point the end point and a random point
on the arc how we can find the center point coordinates?
| This point is called the circumcenter of the triangle formed by these three points.
Given $\triangle ABC$ with sides $a$, $b$, and $c$ opposite the corresponding vertices, the circumcenter of $\triangle ABC$ is
$$
\frac{a^2(b^2+c^2-a^2)A+b^2(c^2+a^2-b^2)B+c^2(a^2+b^2-c^2)C}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}\tag{1}
$$
Justification of $\mathbf{(1)}$:
The barycentric coordinates of $O$ are the proportions of the red, green, and blue triangles to the whole triangle. That is,
$$
O=\frac{\color{#C00000}{|\triangle BOC|}A+\color{#00A000}{|\triangle COA|}B+\color{#0000FF}{|\triangle AOB|}C}{|\triangle ABC|}\tag{2}
$$
$\hspace{4.5cm}$
Suppose we know the coordinates of $A$, $B$, and $C$, and that $|OA|=|OB|=|OC|=r$.
Since inscribed angles are half the corresponding central angles, $\left\{\small\begin{array}{ccc}\angle BAC=\angle BOC/2\\\angle CBA=\angle COA/2\\\angle ACB=\angle AOB/2\end{array}\right\}$ and
$$
\small\begin{array}{c|c|c}
a=2r\sin(\angle BAC)\quad&\quad\cos(\angle BAC)=\frac{b^2+c^2-a^2}{2bc}\quad&\quad|\triangle BOC|=\frac{r^2}{2}\sin(\angle BOC)\\
b=2r\sin(\angle CBA)\quad&\quad\cos(\angle CBA)=\frac{c^2+a^2-b^2}{2ca}\quad&\quad|\triangle COA|=\frac{r^2}{2}\sin(\angle COA)\\
c=2r\sin(\angle ACB)\quad&\quad\cos(\angle ACB)=\frac{a^2+b^2-c^2}{2ab}\quad&\quad|\triangle AOB|=\frac{r^2}{2}\sin(\angle AOB)\\
\end{array}
$$
Thus, we can compute quantities proportional to the areas of the red, green, and blue triangles
$$
\begin{align}
\frac{4abc}{r}|\triangle BOC|=2abcr\sin(\angle BOC)=a^2(b^2+c^2-a^2)\\
\frac{4abc}{r}|\triangle COA|=2abcr\sin(\angle COA)=b^2(c^2+a^2-b^2)\\
\frac{4abc}{r}|\triangle AOB|=2abcr\sin(\angle AOB)=c^2(a^2+b^2-c^2)
\end{align}\tag{3}
$$
Plugging $(3)$ into $(2)$ yields $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do I solve this $\lim \limits_{n\to\infty} (n/\pi) \cos (2\pi/3n) \sin (4\pi/3n) $? I have to find its value
$$\lim \limits_{n\to\infty} \frac{n}\pi \cos \left( \frac{2\pi}{3n}\right) \sin \left( \frac{4\pi}{3n}\right)$$
Can you please give just clues for solving it?
| \begin{equation*}
\begin{split}
\lim_{n\to\infty}\frac{n}{\pi} \cos \frac{2\pi}{3n}\sin \frac{4\pi}{3n}&= \lim_{n\to\infty} \frac{4}{3} \cos \frac{2\pi}{3n}\frac{\sin \frac{4\pi}{3n}}{\frac{4\pi}{3n}}\\
&= \frac{4}{3}.(\because \lim_{n\to\infty}\frac{\sin \frac{4\pi}{3n}}{\frac{4\pi}{3n}}=1\ \text{and} \lim_{n\to\infty}\cos \frac{2\pi}{3n}=\cos 0=1)
\end{split}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/176689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.