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Problem in trigonometry integral $$\int \frac{1}{(9-x^2)^{\frac{3}{2}}}dx$$
Let $x=3\sin (u)$
$dx=3\cos (u)$
So, $(9-x^2)^{\frac{3}{2}}=(9-9\sin ^2(u))^{\frac{3}{2}}$
$=27\cos ^3(u)$
$u=\sin ^-1(\frac{x}{3})$
My problem is how to substitute u into the integral to become $3\int \frac{\sec ^2(u)}{27}$du?
| You are pretty much there.
$\int \frac{1}{(9-x^2)^{\frac{3}{2}}}dx = \int \frac{3\cos u}{27 cos^3 u}du = \frac{1}{9}\int \sec^2 u \; du$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Is this 5th root in the set of natural numbers?
Is $$\sqrt[5]{x(x+1)(x^4 + x^2 + 1)} \in \mathbb{N}$$ for some $x$?
I am not asking for all $x$, but just for some natural number $x$?
I don't believe so, but I may be wrong?
Suppose $x=1$, you get $\sqrt[5]{6}$ which isnt natural.
$x=2$, you get $\sqrt[5]{126}$
$x=3$, you get: $\sqrt[5]{1092}$
I might just try a proof by induction. Suppose for $x \in \mathbb{N}$:
$$\sqrt[5]{x(x+1)(x^4 + x^2 + 1)} \notin \mathbb{N}$$
But I dont see how that would work out...
| Note $(x+1)(x^4+x^2+1)=x^5+x^4+x^3+x^2+x+1$. But for $x\in\mathbb N, x>0$ we have
$$
x^5 < x^5+x^4+x^3+x^2+x+1 < x^5+5x^4+10x^3+10x^2+5x+1 = (x+1)^5
$$
so that $x^5+x^4+x^3+x^2+x+1$ is not a perfect fifth power, since it it strictly between two consecutive fifth powers.
Next, note that $x$ and $x^5+x^4+x^3+x^2+x+1$ are relatively prime. If their product is a perfect fifth power, then both of them are perfect fifth powers.
Thus, $x(x+1)(x^4+x^2+1)$ is not a perfect fifth power.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201196",
"timestamp": "2023-03-29T00:00:00",
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A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
| Both sides are positive so you may square the inequality ot obtain $$\begin{align*}\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^2\le\left(\sqrt{\frac{x+y}{2}}\right)^2 &\iff \frac{x+2\sqrt{xy}+y}{4}\le \frac{x+y}{2} \\&\iff \frac{2\sqrt{xy}}{4}\le\frac{x+y}{2}-\frac{x+y}{4} \\&\iff \sqrt{xy}\le \frac{x+y}{2}\\[0.2cm]& \iff (\sqrt{x}-\sqrt{y})^2\ge 0\end{align*}$$ which is true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^3+11(k+1)=k^3+3k^2+3k+1+11k+11=(k^3+11k)+(3k^2+3k+12)$
Since $P(k)$ is true, $(k^3+11k)$ is divisible by 6 but I can't show that $(3k^2+3k+12)$ is divisible by 6
| For any positive integer $n$, let $S(n)$ denote the statement
$$
S(n) : 6\mid (n^3+11n).
$$
Base step: For $n=1, S(1)$ gives $1^3+11(1) = 12 = 2\cdot 6$. Thus, $S(1)$ holds.
Inductive step: Let $k\geq 1$ be fixed, and suppose that $S(k)$ holds; in particular, let $\ell$ be an integer with $6\ell = k^3+11k$. Then
\begin{align}
(k+1)^3 + 11(k+1) &= (k^3+3k^2+3k+1) + (11k+11)\tag{expand}\\[0.5em]
&= \color{red}{k^3+11k}+3k^2+3k+12\tag{rearrange}\\[0.5em]
&= \color{red}{6\ell} + 3k(k+1)+2\cdot 6.\tag{by ind. hyp.}
\end{align}
Since one of $k$ and $k+1$ is even, the term $3k(k+1)$ is divisible by $6$, and so the last expression above is divisible by $6$. This proves $S(k+1)$ and concludes the inductive step $S(k)\to S(k+1)$.
By mathematical induction, for each $n\geq 1$, the statement $S(n)$ is true. $\blacksquare$
| {
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How to simplify recurrence relation? I'm having trouble seeing how
$$5(2^{n-1} + 5\cdot 3^{n-1}) - 6(2^{n-2} + 5\cdot3^{n-2})$$
simplifies to:
$$2^{n-2}\cdot (10 - 6) + 3^{n-2} \cdot (75 - 30)$$
How can I simplify the above expression to get to the second version?
| hint: $$5\cdot 2^{n-1} = 5\cdot 2\cdot 2^{n-2} = 10\cdot 2^{n-2}, 5\cdot 3^{n-1} = 5\cdot 3\cdot 3^{n-2} = 15\cdot 3^{n-2}$$, and combine like terms.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that real multiplication distributes over addition using Dedekind cuts? Proving $\forall x,y,z\in \mathbb R,\:x(y+z)=xy+xz$
There is a very concise proof of this in ProofWiki using Cauchy sequences, but I was wondering whether the same would be possible using Dedekind cuts.
| There is, but as with most of the checks of the field axioms using Dedekind Cuts, it isn't pretty.
First recall that multiplication is given by, for $A,B$ positive
\begin{equation}
A\cdot B =\{a\cdot b \mid a\in A \wedge 0<a \wedge b \in B \wedge 0<b \} \cup \{ q\in \mathbb{Q}\mid q\leq 0\}.
\end{equation}
More generally, we define
\begin{equation}
A\cdot B=
\begin{cases}
\overline{0} & \text{If $A=\overline{0}$ or $B=\overline{0}$,} \\
-(A\cdot (-B)) & \text{If $\overline{0}<A$ and $B<\overline{0}$,} \\
-((-A)\cdot B) & \text{If $A<\overline{0}$ and $\overline{0}<B$,} \\
(-A)\cdot (-B) & \text{If $A<\overline{0}$ and $B<\overline{0}$.}
\end{cases}
\end{equation}
We want to show that $A\cdot (B+C)=A\cdot B+A\cdot C$. Sadly, since this means that we now must take into account three variables, we have ourselves a case-by-case nightmare.
First let us suppose that $A$, $B$, and $C$ are all positive. Let $d\in A\cdot (B+C)$; if $d\leq 0$, then $d \in A\cdot B+A\cdot C$ since it is the sum of two positive Dedekind cuts. Otherwise we have $0<d$; $d$ can hence be written as $d=a\cdot (b+c)$ where $0<a \in A$, $b \in B$, and $c \in C$ where $0<b+c$. We can assume that $b$ and $c$ are positive because at least one must be positive; without loss of generality assume $b$ is positive. If $c$ is negative, then $b+c<b$, so that because $B$ is downwards closed it follows that $b+c\in B$. Then take a positive element $c'<b+c$ of $C$ and let $b'=b+c-c'$ which is necessarily positive by our choice of $c'$. Then $b'+c'=b+c$ with $0<b'\in B$ and $0<c'\in C$. Thus, we will simply assume that $b$ and $c$ are positive. Hence, $ab\in A\cdot B$ and $ac\in A\cdot C$ and thus $d=ab+ac\in A\cdot B+A\cdot C$, so that $A\cdot (B+C) \subset A\cdot B + A\cdot C$.
Similarly, if $d\in A\cdot B+A\cdot C$, then there exists $d_1\in A\cdot B$ and $d_2\in A\cdot C$ such that $d=d_1+d_2$. If $d_1$ and $d_2$ are positive, then there are $0<a,a'\in A$, $0<b\in B$, and $0<c\in C$ such that $ab=d_1$ and $a'c=d_2$. We may assume that $a=a'$, as otherwise we can let $a''=\frac{ab+a'c}{a+a'}$ which gives us $a''b+a''c=d$ and replace $a$ and $a'$ with $a''$. Then $ab+a'c=ab+ac=a(b+c)$ so that $d=a(b+c)\in A\cdot (B+C)$. If $d_1$ and $d_2$ are both non-positive, then $d_1+d_2$ is non-positive and thus lies in $A\cdot (B+C)$. Finally, if one of $d_1$ and $d_2$ is non-positive while the other is positive -- say $d_1$ is positive and $d_2$ non-positive -- then $d_1=ab$ for $0<a\in A$ and $0<b\in B$, so taking $c$ to be any positive rational number in $C$ we find that $d_1+d_2<ab+ac=a(b+c)\in A\cdot (B+C)$, so that because $A\cdot (B+C)$ is downwards closed, we see that $d=d_1+d_2\in A\cdot (B+C)$, thus establishing the fact that $A\cdot B+A\cdot C\subset A\cdot (B+C)$.
A tool we shall make use of in the remaining cases is that $-(A+B)=(-A)+(-B)$: $(A+B)+(-(A+B))=\overline{0}$ and $A+B+(-A)+(-B)=\overline{0}$, so that $(A+B)+(-(A+B))=(A+B)+(-A)+(-B)$, and cancellation of the term $A+B$ gives us the desired result.
*
*If $A=\overline{0}$, then $A\cdot (B+C)=A\cdot B=A\cdot C=\overline{0}$, and thus $A\cdot B+A\cdot C=\overline{0}=A\cdot (B+C)$.
*When $A$, $B$, and $C$ are all negative, $B+C$ is negative, so $A\cdot (B+C)=(-A)\cdot (-(B+C))=(-A)\cdot ((-B)+(-C))=(-A)\cdot (-B)+(-A)\cdot (-C)=A\cdot B+A\cdot C$.
\item When $A$ is negative and $B$ and $C$ are positive, $B+C$ is positive and $A\cdot (B+C)=-((-A)\cdot (B+C))=-((-A)\cdot B+(-A)\cdot C)=(-((-A)\cdot B))+(-((-A)\cdot C))=A\cdot B+A\cdot C$.
*If $B=\overline{0}$, then $A\cdot (B+C)=A\cdot (\overline{0}+C)=A\cdot C=A\cdot \overline{0}+A\cdot C=A\cdot B+A\cdot C$. The same argument shows the case for $C=\overline{0}$.
*If $A$ and $C$ are negative and $B$ is positive, then we have three cases: if $B+C$ is positive, negative, or zero. In the last case, $B=(-C)$ and $A\cdot (B+C)=A\cdot \overline{0}=\overline{0}=(A\cdot B)+(-(A\cdot B))=A\cdot B+A\cdot (-B)=A\cdot B+A\cdot C$. If $B+C$ is negative, then
\begin{align}
A\cdot C
& =A\cdot ((B+C)+(-B)) \\
& =(-A)\cdot (-((B+C)+(-B))) \\
& =(-A)\cdot ((-(B+C))+B) \\
& = (-A)\cdot (-(B+C))+(-A)\cdot B \\
& = A\cdot (B+C)+(-(A\cdot B))
\end{align}
Adding $A\cdot B$ to both sides gives us the desired result.
If $B+C$ is positive, then
\begin{align}
A\cdot B & =A\cdot ((B+C)+(-C)) \\
& =-((-A)\cdot ((B+C)+(-C))) \\
& = -((-A)\cdot (B+C)+(-A)\cdot (-C)) \\
& = -((-(A\cdot (B+C)))+A\cdot C) \\
& = A\cdot (B+C)+(-(A\cdot C)).
\end{align}
Adding $A\cdot C$ to both sides gives us the desired result.
The same arguments hold for $B$ negative and $C$ positive by interchanged the roles of $B$ and $C$ above.
*If $A$ and $B$ are positive and $C$ is negative, then we have three cases: if $B+C$ is positive, negative, or zero. In the last case, $B=(-C)$ and $A\cdot (B+C)=A\cdot \overline{0}=\overline{0}=(A\cdot B)+(-(A\cdot B))=A\cdot B+A\cdot (-B)=A\cdot B+A\cdot C$. If $B+C$ is negative, then
\begin{align}
A\cdot C
& = A\cdot ((B+C)+(-B)) \\
& = -(A\cdot (-((B+C)+(-B)))) \\
& = -(A\cdot ((-(B+C))+B)) \\
& = -(A\cdot (-(B+C))+A\cdot B) \\
& = (-(A\cdot(-(B+C))))+(-(A\cdot B)) \\
& = A\cdot (B+C)+(-(A\cdot B))
\end{align}
Adding $A\cdot B$ to both sides gives us the desired result.
If $B+C$ is positive, then
\begin{align}
A\cdot B & =A\cdot ((B+C)+(-C)) \\
& = A\cdot ((B+C)+(-C)) \\
& = A\cdot (B+C)+A\cdot (-C)) \\
& = A\cdot (B+C)+(-(A\cdot C))
\end{align}
Adding $A\cdot C$ to both sides gives us the desired result.
The same arguments hold for $B$ negative and $C$ positive by interchanging the roles of $B$ and $C$ above.
(If there's a prettier way to prove this with Dedekind Cuts, I certainly don't know it. At first glance the proof that Rudin gives in Principles of Mathematical Analysis is shorter, but that's because he doesn't actually do it all (rather, says "similar to our above proofs"))
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem with nonhomogeneous recurrence relations I studying Discrete maths during this semester and I need your help.
I have been trying to solve one non-homogeneous recurrence relation and read many-many guides how to do this, but I haven't found the way.
However, it's here:
$A_n = -3A_{n-1} +4A_{n-2}+ 3.2^n$
Anyways, the characteristic equation here gives $x_1=1$ and $x_2=-4$ as results, but I'm in need to find the special thingy somehow with this one $3.2^n$.
I hope you understand, what I exactly need.
| Let $B_n = A_n - A_{n-1}$. Then
$$
B_n = 3.2^n - 4B_{n-1}.
$$
Let $C_n = B_n/(-4)^n$. Then
$$
C_n = (-0.8)^n + C_{n-1}.
$$
Therefore
$$
C_n = \sum_{k=1}^n (-0.8)^k + C_0 = -\tfrac{4}{9} (1-(-0.8)^n) + C_0 = C_0-\tfrac{4}{9} + \tfrac{4}{9} (-0.8)^n.
$$
This implies that
$$
B_n = (B_0-\tfrac{4}{9})(-4)^n + \tfrac{4}{9} \cdot 3.2^n.
$$
Therefore
$$
\begin{align*}
A_n &= \sum_{k=1}^n B_k + A_0 \\ &= \sum_{k=1}^n \left[(B_0-\tfrac{4}{9})(-4)^k + \tfrac{4}{9} \cdot 3.2^k\right] + A_0 \\ &=
-0.8(B_0-\tfrac{4}{9}) [1-(-4)^n] + \tfrac{64}{99} (3.2^n-1) + A_0 \\ &=
0.8(B_0-\tfrac{4}{9})(-4)^n + \tfrac{64}{99} 3.2^n -0.8(B_0-\tfrac{4}{9}) - \tfrac{64}{99} + A_0.
\end{align*}
$$
Here $B_0 = A_0 - A_{-1}$.
Morale: There is an inhomogeneous solution of the form $C 3.2^n$. Using this Ansatz, we get
$$
C 3.2^n = -3 C 3.2^{n-1} + 4 C 3.2^{n-2} + 3.2^n.
$$
Dividing by $3.2^{n-2}$, we get
$$
C 3.2^2 = -3C 3.2 + 4C + 3.2^2.
$$
This simplifies to $15.84 C = 10.24$ and so $C = 10.24/15.84 = 64/99$.
At this point we can apply the usual methods to obtain the solution calculated explicitly above.
| {
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Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
| Hint: Write
$$(x + 3 - 2)^2 + (x+3 - 1)^3 + (x+3)^4 = 2$$
Let $y = x+3$ then
$$p(y) = y^2 - 4y + 4 + y^3 - 3y^2 + 3y -1 + y^4 = 2 \implies p(y) =y^4 + y^3 -2y^2 - y +1 = 0 $$
Notice that $1$ and $-1$ are roots of $p(y)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximum value of function given minimum value Suppose there is a function $f(x)=\frac{x^2-2x+b}{x^2+2x+b}$ (the problem doesn't specify, but I am assuming $b$ is a real) that has a minimum value of $\frac{1}{2}$. What is the maximum value of $f(x)$?
My first instinct was to divide out everything, getting that $f(x)=1-\frac{4x}{x^2+2x+b}$. From there, I'm not sure what to do.
I am looking for a solution that does not involve calculus.
| Consider $$y=\frac{x^2-2x+b}{x^2+2x+b}$$
Or $$(y-1)x^2+2(y+1)x+b(y-1)=0$$
In this quadratic equation in $x$, for those values of $y$ which exist in range, there should exist real value(s) of $x$.
Hence, Discriminant $\geq 0 $.
$$4(y+1)^2-4b(y-1)^2\geq0$$
$$(1-b)y^2+2y(1+b)+(1-b)\geq0$$
Now this is a quadratic in $y$ which will provide us with the values of $y$ in range. More specifically, those values of $y$ satisfying this inequality will be in the range.
We know that $y=\frac{1}{2}$ is the starting point of the range, so the inequality must be exactly $0$ at $y=\frac{1}{2}$.
$$\frac{1-b}{4}+(1+b)+(1-b)=0$$
That is $b=9$.
That provides us the other extreme of the inequality by solving it,
$$-8y^2+20y-8\geq0$$
$$2y^2-5y+2\geq0$$
So $$\frac{1}{2}\leq y \leq 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problem in Differential Equation (How to proceed?) $$y\frac{dy}{dx}=\sqrt{1-y^2}$$, y=0 when x=0
$$\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}=1$$
$$\int\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}dx=\int 1dx$$
$$-\sqrt{1-y^2}=x+c_1$$
$$1-y^2=(x+c_1)^2$$
$$y^2=1-(x+c_1)^2$$
$$y=\pm \sqrt{1-(x+c_1)^2}$$
When x=0, y=0
$$0=\pm \sqrt{1-c^2_1}$$
| $$\int_0^y \frac{y\, dy}{\sqrt{1-y^2}} = \int_0^x dx \to \sqrt{1-y^2} - 1=x$$ solving it gives you $$y = \pm \sqrt{2x-x^2}$$ is the circle or radius $1$ centered at $(1,0).$
$\bf note:$ use the definite integral whenever possible.
| {
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"url": "https://math.stackexchange.com/questions/1210780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Mathematical Induction Proof Question dealing with integers How would you use mathematical induction to prove that
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n \cdot (n + 1) \cdot (n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$
I tried proving the base case of $n = 1$ but the left half is much larger than the right half after computing $n = 1$. Can someone please solve this problem and explain how and why the solution is correct? I've been stuck on it for hours.
| You want to prove
$$\underbrace{1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + (n + 1)(n + 2)(n+3)}_a = \underbrace{\frac{(n + 1)(n + 2)(n + 3)(n+4)}{4}}_b$$
and you are allowed to assume
$$\underbrace{1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + n(n + 1)(n + 2)}_c = \underbrace{\frac{n(n + 1)(n + 2)(n + 3)}{4}}_d$$
So you want to prove $a=b$ given that $c=d$. First notice the relationship between $c$ and $a$:
$$c+(n + 1)(n + 2)(n+3) = a$$
Now that you have some information, you can prove $a=b$ starting with our assumption:
$$c=d$$
$$a - (n + 1)(n + 2)(n+3) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$$
$$a = \frac{n(n + 1)(n + 2)(n + 3)}{4} + \frac{4(n + 1)(n + 2)(n+3)}{4}$$
$$a = \frac{\bigg((n + 1)(n + 2)(n+3)\bigg)\bigg(n + 4\bigg)}{4}$$
$$a = b$$
| {
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"url": "https://math.stackexchange.com/questions/1212364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4 \sqrt{2014}$ How to prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4\sqrt{2014}$ without using calculator?
| Hint:
$$\sqrt{k-r}+\sqrt{k+r}<2\sqrt{k},\ \ \ \forall k\ge 0, k\ge |r|, r\neq 0$$
Since:
$$\iff 2k+2\sqrt{k^2-r^2}<4k\iff 2\sqrt{k^2-r^2}<2k\iff k^2-r^2<k^2\iff -r^2<0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$ I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and will attempt to prove it. Let $\varepsilon>0$. We have that: $$\left\lvert\frac{x^2y^2}{x^2+y^2}\right\rvert=\frac{x^2y^2}{x^2+y^2}\leq\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}=x^2+y^2$$ However, I must find $\delta>0$ such that $0<\sqrt{x^2+y^2}<\delta$, and I cannot see a way to obtain $\sqrt{x^2+y^2}$ in the above inequality to complete the proof. Am I mistaken in my process? Thank you.
| Given $\varepsilon>0$, take $\delta=\sqrt{\varepsilon}$. Then,
\begin{align*}
0<\sqrt{x^2+y^2}<\delta\quad&\Rightarrow \quad x^2<\delta^2\text{ and }y^2<\delta^2\\
&\Rightarrow \quad\frac{1}{x^2}+\frac{1}{y^2}>\frac{2}{\delta^2}\\
&\Rightarrow \quad \left|\frac{x^2y^2}{x^2+y^2}-0\right|=\frac{1}{\frac{1}{y^2}+\frac{1}{x^2}}<\frac{1}{\frac{2}{\delta^2}}=\frac{\varepsilon}{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Denesting a square root: $\sqrt{7 + \sqrt{14}}$ Write:
$$\sqrt{7 + \sqrt{14}} = a + b\sqrt{c}$$
Form. $$7 + \sqrt{14} = a^2 + 2ab\sqrt{c} + b^2c$$
$a^2 + b^2c = 7$ and $2ab = 1$, and $c = 14$
But that doesn't seem right as $a, b,$ wont be integers?
| Set $r=\sqrt{7+\sqrt{14}}$; then $r^2=7+\sqrt{14}$ and so
$$
14=r^4-14r^2+49
$$
or
$$
r^4-14r^2+35=0
$$
The polynomial $X^4-14X^2+35=0$ is irreducible over the rational numbers by Eisenstein's criterion (with $7$), so the degree of $r$ over the rationals is $4$. A number of the form $a+b\sqrt{c}$ with rational $a,b,c$ has degree $2$ over the rationals.
Therefore you can't find rational $a,b,c$ that satisfy your request.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$, is $a=b=c=0$? If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$ , then is it true that $a=b=c=0$ ? I was thinking of infinite descent but can't actually proceed , please help. Thanks in advance
| We may assume that $a$, $b$, and $c$ are relatively prime. If not, we can divide out by the cube of the common factor. Observe that $$c^3=2(3abc-2a^3-b^3),$$ so $c$ is even. Let $c=2n$.
Then, the equation becomes $4a^3+2b^3+8n^3=12abn$. Therefore,
$$
2b^3=4(3abn-a^3-2n^3).
$$
Hence, $b$ is even. Let $b=2m$.
Then, the equation becomes $4a^3+16m^3+8n^3=24amn$. Therefore,
$$
4a^3=8(3amn-2m^3-n^3).
$$
Hence $a$ is even, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that $\sin \tfrac{2\pi x_1}{3}+\sin \tfrac{2\pi x_3}{3}=2\sin \tfrac{2\pi x_2}{3}$. Find $a$. $x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that
$\sin \dfrac{2\pi x_1}{3}+\sin \dfrac{2\pi x_3}{3}=2\sin \dfrac{2\pi x_2}{3}$.
Find $a$
(Bulgari 1998)
| Observe
\begin{align*}x^3-3x^2+(a^2+2)x-a^2&=x(x^3-3x+2)+a^2(x-1)\\ &=[x(x-2)+a^2](x-1)\\ &=(x^2-2x+a^2)(x-1)\end{align*}
And the roots of $x^3-3x^2+(a^2+2)x-a^2$ are $1+\sqrt{1-a^2},\;1-\sqrt{1-a^2}\;\text{and}\; 1.$
If we set $a=1$ or $a=-1$, then the three roots are equal to $1$ and the equality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculation of real root values of $x$ in $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}.$
Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $
$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get
$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.
$\displaystyle (1-2x)^2=4(x^2-1)\Rightarrow 1+4x^2-4x=4x^2-4\Rightarrow x=\frac{5}{4}.$
But when we put $\displaystyle x = \frac{5}{4}\;,$ We get $\displaystyle \frac{3}{2}-\frac{1}{2}=2\Rightarrow 1=2.$(False.)
So we get no solution.
My Question is : Can we solve above question by using comparision of expressions?
Something like $\sqrt{x+1}<\sqrt{x-1}+\sqrt{4x-1}\; \forall x\geq 1?$
If that way possible, please help me solve it. Thanks.
| For $x\ge1$,
$$l(x):=\sqrt{x+1}-\sqrt{x-1}=\dfrac2{\sqrt{x+1}+\sqrt{x-1}}\le\sqrt2$$ and
$$r(x):=\sqrt{4x-1}\ge\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
What is $2^{7!}\bmod{2987}$ Find the remainder when $2^{7!}$ is divided by $2987$.
I tried to factorise $2987$ to make it simple but it was in vain.
| $2^{\large 7!}\!\pmod{\color{brown}{29}87}\,$ can be computed using only simple mental arithmetic.
Note that $\ \color{brown}{29}\!\mid\! 87,\,$ i.e. $\, 29\cdot 3 = 87\ $ so $\ 2987 = 29(100\!+\!3) = 29\cdot 103$
Since $\, p = 29,103\,$ are prime, we can use little Fermat to reduce exponents, i.e.
$(1)\qquad\qquad\ \ N \equiv n\pmod{p\!-\!1}\ \Rightarrow\ 2^{N}\!\equiv 2^{n}\pmod p$
${\rm mod}\,\ \ 28\!:\ 7! \equiv 0\ $ by $\ 7,4\mid 7!\,\Rightarrow\, 28=7\cdot 4\mid 7!$
${\rm mod}\ 102\!:\, 7! \equiv (7\cdot 5\cdot 3)(6\cdot 4\cdot 2)\equiv 105(14\!+\!34)\equiv 3(14) \equiv 42,\, $ so, $ $ applying $(1)$
${\rm mod}\ 29\!:\ \ \ x\equiv 2^{\large 7!}\equiv 2^{0}\equiv\color{#c00} 1$
${\rm mod}\ 103\!:\ x\equiv 2^{\large 7!}\equiv 2^{42}.\ $ Let's compute this power:
$\qquad\ 2^{10}=1024\equiv 1030\!-\!6\equiv -6,\,$ so $\,2^{11}\equiv -12\,$ so $\,2^{21}\equiv (-6)(-12) \equiv 72\equiv -31$
$\ \Rightarrow\ 2^{42} \equiv (2^{21})^2\equiv (-31)^2\equiv (1\!+\!30)^2 \equiv 1 + 2(30)+9(-3)\equiv \color{blue}{34}$
Finally we apply $ $ CRT = Chinese Remainder Theorem:
${\rm mod}\ 29\!:\ \color{#c00}1\equiv x\equiv \color{blue}{34}\!+\!103j\equiv 5\!+\!16j\!\iff\! 16j\equiv -4\!\iff\! 4j\equiv -1\equiv 28\!\iff\! j\equiv\color{#0a0} 7$
So we deduce $\ x = 34\! +\! 103(\color{#0a0}7\!+\!29k)\, =\, 755 + 2987k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to factorize $x^4+2x^2-x+2$? look at this:
$$x^4+2x^2-x+2$$
How to factorize it?
It should be changed to be in the form of standard factorization formulas.
| $$f(x)=x^4+2x^2-x+1=(x^4-x^3+x^2)+(x^3 -x^2+x)+(2x^2-2x+2)$$
$$f(x)=x^2(x^2-x+1)+x(x^2-x+1)+2(x^2-x+2)=(x^2-x+1)(x^2+x+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the greatest common factor of $m+n$ and $m^2+n^2$ is 1 or 2 if $m$ and $n$ are relatively prime. Prove that the greatest common factor of $m+n$ and $m^2+n^2$ is $1$ or $2$ if $m$ and $n$ are relatively prime natural numbers.
Can anyone give a step-by-step answer for this?
| Let $d\lvert m+n$ and $d\lvert m^2+n^2$.
Then $d\lvert(m^2+n^2)-(m-n)(m+n)$, so $d\lvert 2n^2$;
$\;$and $d\lvert(m^2+n^2)+(m-n)(m+n)$, so $d\lvert 2m^2$.
Since $\gcd(2m^2, 2n^2)=2\gcd(m^2,n^2)=2\cdot1=2$,
$\hspace{.3 in}d\lvert2$ so $d=1$ or $d=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of subsets of length 7 I have the following summation:
$$\sum\limits_{k=7}^{n} {k-1\choose 6} $$
and apparently it counts the number of subsets of {1, 2, . . . , n} having size 7.
Why is this?
| It is because of Pascal's recurrence. The recurrence is that $\binom{i}{j}+\binom{i}{j-1} = \binom{i+1}{j}$ provided that $0< j\leq i$.
\begin{align*}
\binom{6}{6}+\binom{7}{6} &= \binom{7}{7} + \binom{7}{6} \\
&= \binom{8}{7} \\
\binom{6}{6}+\binom{7}{6} + \binom{8}{6} &= \binom{8}{7} + \binom{8}{6} \\
&= \binom{9}{7} \\
\binom{6}{6} + \binom{7}{6} + \binom{8}{6} + \binom{9}{6} &= \binom{9}{7} + \binom{9}{6} \\
&= \binom{10}{7}
\end{align*}
This process repeats until you get to $\binom{n-1}{6}+\binom{n-1}{7}=\binom{n}{7}$. It helps to draw Pascal's Triangle and follow the additions down the tree...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The height of two triangles My textbook says the height of the following triangles ($BC$) is $37.5$
Because
$$
\widehat{B_1}=\widehat{B_2}=30 \Rightarrow BD=50 \Rightarrow DC=25 \Rightarrow AC=75
$$
and since in right triangles the side opposite the 30 degrees angle is half the length of the hypotenuse, the height is $37.5$.
But in my approach the answer is $25 \sqrt{3}$ because
$$
\tan{30}=\frac{1}{\sqrt{3}}=\frac{BC}{50+DC}, \quad \tan{60}=\sqrt{3}=\frac{BC}{DC}\to DC=\frac{BC}{\sqrt{3}}\\
\to \frac{1}{\sqrt{3}}=\frac{BC}{50+\frac{BC}{\sqrt{3}}} \to BC=25\sqrt{3} \approx 43.3
$$
Am I right?
| Using their approach:
$BC^2 + 75^2 = 4BC^2$
$3BC^2 = 3^2 \times 25^2$
$BC = \sqrt{3} \times 25$
SO you may conclude that your book had an error in calculating $\sqrt3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \frac{dx}{1+\sin x+\cos x}$ Evaluate $$\int \frac{1}{1+\sin x+\cos x}\:dx$$
I tried several ways but all of them didn't work
I tried to use Integration-By-Parts method but it's going to give me a more complicated integral
I also tried u-substitution but all of my choices of u didn't work
Any suggestions?
| It is easy to recall the half-angle identities $$\begin{align*} \sin \frac{x}{2} &= \sqrt{\frac{1-\cos x}{2}}, \\ \cos \frac{x}{2} &= \sqrt{\frac{1+\cos x}{2}}, \end{align*}$$ from which the tangent half-angle identity can be written $$\tan^2 \frac{x}{2} = \frac{1-\cos x}{1+\cos x} = \frac{\sin^2 x}{(1+\cos x)^2},$$ and it follows that $$\color{red}{\boxed{\displaystyle \tan \frac{x}{2} = \frac{\sin x}{1+\cos x}, \quad \sec^2 \frac{x}{2} = \frac{2}{1+\cos x}}}.$$ (Indeed, there are numerous methods of proving the boxed identities.) Hence $$\frac{1}{1 + \sin x + \cos x} = \frac{\frac{1}{1+\cos x}}{1 + \frac{\sin x}{1+\cos x}} = \frac{\frac{1}{2} \sec^2 \frac{x}{2}}{1 + \tan \frac{x}{2}}$$ and the substitution $$u = \tan \frac{x}{2}, \quad du = \frac{1}{2} \sec^2 \frac{x}{2} \, dx$$ immediately leads to $$\int \frac{dx}{1+\sin x + \cos x} = \int \frac{du}{1+u} = \log |1+u| + C = \log \left| 1 + \tan \frac{x}{2} \right| + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding Surface area of a curve rotated around the x axis I need to calculate the surface area obtained by rotating $\sin\pi x$, $0\le x \le 1$ about the x-axis. So the surface area equation i think i have to use is:
$$A = \int_0^1 2\pi y \sqrt{1+(dy/dx)^2} \, dx$$
so what I did so far is
$$A = \int_0^1 2\pi \sin{\pi x} \sqrt{1+(\pi \cos \pi x)^2} \, dx$$
and I let $\tan{\theta} = \pi \cos{\pi x}$ so
$$\begin{align*}
A &= 2\pi \int_0^1 \sqrt{1+\tan^2{\theta}} \frac{-sec^2{\theta}}{\pi^2} \, d\theta \\
&= -2\pi^3 \int_0^1 \sec^3{\theta} d\theta \\
&= -2\pi^3\left( \frac{1}{2}\sec{\theta}\tan{\theta} + \frac{1}{2} \ln|\sec\theta + \tan\theta| \right)
\end{align*}$$
From here I am kind of lost but I can tell it doesn't really compare to my prof's answer of
$$A = 2\sqrt{1+\pi^2} + \frac{2}{\pi} \ln{\left( \pi + \sqrt{1+\pi^2}\right)}$$
| You're fine as far as $\tan{\theta} = \pi \cos{\pi x}$, so $\sec^2{\theta} \, d\theta = -\pi^2 \sin{\pi x}$. The limits require some more care: when $x=0$, $\tan{\theta} = \pi$, so we have the rather weird $\arctan{\pi}=P$, say, as the lower limit. $x=1$ gives $\theta=-P$ as the upper limit. Hence, switching the limits round, we have
$$ A = 2\pi \int_{-P}^{P} \sqrt{1+\tan^2{\theta}} \frac{\sec^2{\theta}}{\pi^2} \, d\theta = \frac{2}{\pi} \int_{-P}^{P} \sec^3{\theta} \, d\theta $$
This is clearly even, so we can do it as twice the integral from $0$ to $P$ to save confusion.
You have the correct integral for $\sec^3$, so now it's a matter of sticking the limits into
$$ 2\pi \left( \frac{1}{2}\sec{\theta} \tan{\theta} + \frac{1}{2} \log{(\sec{\theta}+\tan{\theta})} \right) $$
The bottom limit obviously gives $0$ since $\sec{0}=1,\tan{0}=0$. Now, $\sec{\theta} = \sqrt{1+\tan^2{\theta}}$ for $0\leqslant \theta \leqslant \pi/2$, so we have
$$ \tan{P} = \pi, \quad \sec{P} = \sqrt{1+\pi^2}, \tag{1} $$
so we get the answer as
$$ A = \frac{2}{\pi} \left( \sec{P} \tan{P} + \log{(\sec{P}+\tan{P})} \right), $$
and sticking the values in (1) into this gives you the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there another way to solve $\int \frac{x}{\sqrt{2x-1}}dx$? $$\int \frac{x}{\sqrt{2x-1}}dx$$
Let $u=2x-1$
$du=2dx$
$$=\frac{1}{2}\int \frac{u+1}{2\sqrt{u}}du$$
$$=\frac{1}{2}\int (\frac{\sqrt{u}}{2}+\frac{1}{2\sqrt{u}})du$$
$$=\frac{1}{4}\int \sqrt{u}du+\frac{1}{4}\int\frac{1}{\sqrt{u}}du$$
$$=\frac{u^{\frac{3}{2}}}{6}+\frac{\sqrt{u}}{2}+c$$
$$=\frac{1}{6}(2x-1)^{\frac{3}{2}}+\frac{1}{2}\sqrt{2x-1}+c$$
$$=\frac{1}{3}(x+1)\sqrt{2x-1}+c$$
Is there another way except this?
| While the approaches already posted work well, here is an approach that provides a straight forward way of deriving the result.
$$\begin{align}
\int \frac{x}{\sqrt{2x-1}} dx &= \frac12 \int \frac{2x}{\sqrt{2x-1}} \,\,dx\\\\
&=\frac12 \int \frac{2x-1+1}{\sqrt{2x-1}} \,\,dx\\\\
&=\frac12 \int \sqrt{2x-1}\,\,\,dx+\frac12\int \frac{1}{\sqrt{2x-1}}\,\,\,dx
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove this inequality? $(a+b+c=1)$ Show that if $a,b,c$ are positive reals and $a+b+c=1$, then the following must hold:
$$\frac{2(a^3+b^3+c^3)}{abc}+3 \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
What I have tried is using $abc \leq \frac{1}{27}$ $(a+b+c \geq 3\sqrt[3]{abc}) $ and multiplying everything by $abc$, but I don'think that's a good idea because $abc$ can be positive and negative.
I have also tried substituting $a^3+b^3+c^3 \geq 3abc$, but that isn't strong enough.
Any help/hints please??
| if $a,b,c>0$ then we get
$$\frac{2(a^3+b^3+c^3)}{abc}+3\geq \frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}$$
this is equivalent to
$$\frac{2(a^3+b^3+c^3)}{abc}\geq \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}$$
and this
$$2(a^3+b^3+c^3)\geq bc(b+c)+ac(a+c)+ab(a+b)$$
and now note that
$$a^3+b^3=(a+b)(a^2-ab+b^2)\geq ab(a+b)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? I understand that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
But I am struggling to algebraically show that $\sqrt{2}$,$\sqrt[3]{5}\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$ to conclude that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$
| Let $x=\sqrt2+\sqrt[3]5$ for simplicity. By manually expanding we get
\begin{array}{rrrrrrr}
1 = & 1 \\
x = & & 1\cdot 2^{1/2} & + 1\cdot 5^{1/3} \\
x^2 = & 2 & & & + 2\cdot 2^{1/2}5^{1/3} & +1\cdot 5^{2/3} \\
x^3 = & 5 & + 2\cdot 2^{1/2} & + 6\cdot 5^{1/3} & & & +3\cdot2^{1/2}5^{2/3} \\
x^4 = & 4 & + 20\cdot 2^{1/2} & + 5\cdot 5^{1/3} & + 8\cdot2^{1/2}5^{1/3} & +12\cdot5^{2/3} \\
x^5 = & 100 & + 4\cdot 2^{1/2} & + 20\cdot 5^{1/3} & + 25\cdot2^{1/2}5^{1/3} & +5\cdot5^{2/3} & +20\cdot 2^{1/2}5^{1/3} \\
\end{array}
We want to solve $\sqrt2 = a + bx + cx^2 + dx^3 + ex^4 + fx^5$, and this can be rewritten as the matrix equation:
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 0 & 0\\
2 & 0 & 0 & 2 & 1 & 0 \\
5 & 2 & 6 & 0 & 0 & 3 \\
4 & 20 & 5 & 8 & 12 & 0 \\
100 & 4 & 20 & 25 & 5 & 20
\end{bmatrix}^{\mathrm T}
\begin{bmatrix}a \\ b \\ c \\ d \\ e \\ f\end{bmatrix}
=
\begin{bmatrix}0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix},
$$
from which we can find that
$$ -1820 + 735x - 780x^2 - 320x^3 + 45x^4 + 48x^5 = 1187 \sqrt2. $$
$\sqrt[3]5$ is then simply $x-\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the closed-form for $\sum_{i=0}^n(-1)^i(\frac{1}{2})^i$ I start with simplifying:
$$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$
then:
$$S = 1 + (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n$$
$$(-\frac{1}{2})S = (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n+(-\frac{1}{2})^{n+1}$$
$$(-\frac{1}{2}-1)S = (-\frac{1}{2})^{n+1} - 1$$
am I on the right track?
| Here's another approach using the formula for finite geometric series
\begin{align*}
\sum_{i=0}^nx^n=\frac{1-x^{n+1}}{1-x}\tag{1}
\end{align*}
We obtain
\begin{align*}
\sum_{i=0}^n(-1)^i\left(\frac{1}{2}\right)^i&=\sum_{i=0}^n\left(-\frac{1}{2}\right)^i\\
&=\frac{1-\left(-\frac{1}{2}\right)^{n+1}}{1-\left(-\frac{1}{2}\right)}\\
&=\frac{2}{3}\left(1-\left(-\frac{1}{2}\right)^{n+1}\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Algebraic Manipulation What is the best method to get the LHS equal to RHS?
$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$
Thank you.
| Your equation is :
$$ \frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3} \tag{$\star$}$$
Notice that the product $(n+1)(n+2)$ is common to all the terms in $(\star)$. If $n \in \mathbb{Z} \smallsetminus \left\{-2,-1 \right\}$, then $(n+1)(n+2) \neq 0$ and you can $\color{blue}{\text{simplify}}$ by $(n+1)(n+2)$ as follows :
$$ \frac{\require{cancel} n\color{blue}{\cancel{(n+1)(n+2)}}}{3} + \require{cancel} \color{blue}{\cancel{(n+1)(n+2)}} = \frac{\require{cancel} \color{blue}{\cancel{(n+1)(n+2)}}(n+3)}{3} \tag{$\star$}$$
It gives :
$$ \frac{n}{3} + 1 = \frac{n+3}{3} = \frac{n}{3} + 1. $$
which is true for all $n \in \mathbb{Z} \smallsetminus \left\{-2,-1 \right\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Determinant of a 5 × 5 matrix I have a little problem with a determinant.
Let $A = (a_{ij}) \in \mathbb{R}^{(n, n)}, n \ge 4$ with
$$a_{ij} =
\begin{cases}
x \quad \mbox{for } \,i = 2, \,\, j \ge 4,\\
d \quad \mbox{for } \,i \ge j, \\
0 \quad \mbox{else.}
\end{cases}$$
So for example, if we choose $n = 5$, the matrix would look like this:
$$ A =
\begin{pmatrix}
d &0 &0 &0 &0 \\
d &d &0 &x &x \\
d &d &d &0 &0 \\
d &d &d &d &0 \\
d &d &d &d &d \\
\end{pmatrix}$$
How can I find the determinant of this matrix?
My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A = T \cdot R$. Then I wanted to use $$\det(A) = \det(T \cdot R) = \det(T) \cdot \det(R).$$
to figure out the determinant. This would be something like
$$ d^n \cdot \det(R)$$
But is this approach even possible (I don't think so)? Is there any intelligent way of solving this? Thanks in advance.
|
My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A=T⋅R$.
That's very much a way to do it. The technique is called LU Decomposition. It produces a lower and upper triangular matrix, allowing trivial determinate calculations. For this reason, you actually only need to find the diagonal elements to get your determinant.
In this case,
$$ A =
\begin{pmatrix}
d &0 &0 &0 &0 \\
d &d &0 &x &x \\
d &d &d &0 &0 \\
d &d &d &d &0 \\
d &d &d &d &d \\
\end{pmatrix} = \begin{pmatrix}
1 &0 &0 &0 &0 \\
1 &1 &0 &0 &0 \\
1 &1 &1 &0 &0 \\
1 &1 &1 &1 &0 \\
1 &1 &1 &1 &1 \\
\end{pmatrix} \cdot \begin{pmatrix}
d &0 &0 &0 &0 \\
0 &d &0 &x &x \\
0 &0 &d &-x &-x \\
0 &0 &0 &d &0 \\
0 &0 &0 &0 &d \\
\end{pmatrix}$$
So
$$
\det A = \det L \cdot \det U = 1^5 \cdot d^5
$$
The first few steps of the method used here (takes longer to texify than to do) are:
$$
\begin{eqnarray}
\left[\begin{smallmatrix}
d &0 &0 &0 &0 \\
d &d &0 &x &x \\
d &d &d &0 &0 \\
d &d &d &d &0 \\
d &d &d &d &d \\
\end{smallmatrix}\right]
&&= \left[\begin{smallmatrix}
d &0 &0 &0 &0 \\
\end{smallmatrix}\right] \left[\begin{smallmatrix}
1 \\
1 \\
1 \\
1 \\
1 \\
\end{smallmatrix}\right]
+
\left[\begin{smallmatrix}
0 &0 &0 &0 &0 \\
0 &d &0 &x &x \\
0 &d &d &0 &0 \\
0 &d &d &d &0 \\
0 &d &d &d &d \\
\end{smallmatrix}\right]
\\
&&= \left[\begin{smallmatrix}
d &0 &0 &0 &0 \\
\end{smallmatrix}\right] \left[\begin{smallmatrix}
1 \\
1 \\
1 \\
1 \\
1 \\
\end{smallmatrix}\right]
+
\left[\begin{smallmatrix}
0 &d &0 &x &x \\
\end{smallmatrix}\right] \left[\begin{smallmatrix}
0 \\
1 \\
1 \\
1 \\
1 \\
\end{smallmatrix}\right]
+
\left[\begin{smallmatrix}
0 &0 &0 &0 &0 \\
0 &0 &0 &0 &0 \\
0 &0 &d &-x &-x \\
0 &0 &d &d-x &-x \\
0 &0 &d &d-x &d-x \\
\end{smallmatrix}\right]
\\
&&= \left[\begin{smallmatrix}
d &0 &0 &0 &0 \\
0 &d &0 &x &x \\
\end{smallmatrix}\right] \left[\begin{smallmatrix}
1 & 0\\
1 & 1\\
1 & 1\\
1 & 1\\
1 & 1\\
\end{smallmatrix}\right]
+
\cdots
\end{eqnarray}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove equality of two numbers written in complex polar form. Show that these two numbers are equal:
$$
z_1=\frac{e^{\tfrac{2\pi i}{9}}-e^{\tfrac{5\pi i}{9}}}{1-e^{\tfrac{7\pi i}{9}}}
$$
and
$$z_2=\frac{e^{\tfrac{\pi i}{9}}-e^{\tfrac{3\pi i}{9}}}{1-e^{\tfrac{4\pi i}{9}}}=\frac{1}{e^{\tfrac{\pi i}{9}}+e^{\tfrac{-\pi i}{9}}}.
$$
In case it helps, I do know that they are both real.
Thanks in advance for any suggestions!
| Both complex numbers are of the form:$$\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}=\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}\times\frac{1+e^{-i(a+b)}}{1+e^{-i(a+b)}}=\frac{(e^{ia}-e^{-ia})-(e^{ib}-e^{-ib})}{-e^{i(a+b)}+e^{-i(a+b)}}\tag{1}$$We know that:$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$$$\therefore2i\sin(x)=e^{ix}-e^{-ix}\tag{2}$$Using (2) in (1) we get:$$\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}=\frac{2i\sin(a)-2i\sin(b)}{-2i\sin(a+b)}=\frac{\sin(b)-\sin(a)}{\sin(a+b)}=\frac{\sin(b)-\sin(a)}{2\sin(\frac{a+b}{2})\cos(\frac{a+b}{2})}\tag{3}$$Then, using Prosthaphaeresis Formulas in (3) we get:$$\require{cancel}\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}=\frac{\cancel{2\cos(\frac{b+a}{2})}\sin(\frac{b-a}{2})}{\cancel{2}\sin(\frac{a+b}{2})\cancel{\cos(\frac{a+b}{2})}}=\frac{\sin(\frac{b-a}{2})}{\sin(\frac{a+b}{2})}\tag{4}$$If we now use (4) and the fact that $\sin(x)=\cos(\frac{\pi}{2}-x)$ with the values you have for $z_1$ and $z_2$ we get:$$z_1=\frac{\sin(\frac{3\pi}{18})}{\sin(\frac{7\pi}{18})}=\frac{\sin(\frac{3\pi}{18})}{\cos(\frac{2\pi}{18})}=\frac{1}{2\cos(\frac{2\pi}{18})}$$$$z_2=\frac{\sin(\frac{2\pi}{18})}{\sin(\frac{4\pi}{18})}=\frac{\cancel{\sin(\frac{2\pi}{18})}}{2\cancel{\sin(\frac{2\pi}{18})}\cos(\frac{2\pi}{18})}=\frac{1}{2\cos(\frac{2\pi}{18})}$$Hence:$$z_1=z_2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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If $ \cos(\theta) = - \frac{2}{3} $ and $ 450^{\circ} < \theta < 540^{\circ} $, find... If $ \cos(\theta) = - \frac{2}{3} $ and $ 450^{\circ} < \theta < 540^{\circ} $, find:
*
*The exact value of $ \cos \! \left( \frac{1}{2} \theta \right) $.
*The exact value of $ \tan(2 \theta) $.
What I’ve tried:
*
*I took the square root of $ \sqrt{\frac{1}{2} \left( 1 - \frac{2}{3} \right)} $, which equals the square root of $ \frac{1}{6} $.
*I considered the formula $ \sin(x) = \sqrt{1 - {\cos^{2}}(x)} $ and took the square root of $ 1 - \frac{4}{9} $, but I don’t know where to go from there.
Thanks in advance!
| You were given that $\cos\theta = -\dfrac{2}{3}$, with $450^\circ < \theta < 540^\circ$.
Recall that
$$\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos\theta}{2}}$$
where the sign is determined by the measure of angle $\frac{\theta}{2}$. Since $450^\circ < \theta < 540^\circ$, we may conclude that $225^\circ < \frac{\theta}{2} < 270^\circ$, so $\frac{\theta}{2}$ is a third-quadrant angle. Hence, its cosine is negative. Thus,
\begin{align*}
\cos\left(\frac{\theta}{2}\right) & = -\sqrt{\frac{1 + \cos\theta}{2}}\\
& = -\sqrt{\frac{1 - \frac{2}{3}}{2}}\\
& = -\sqrt{\frac{\frac{1}{3}}{2}}\\
& = -\sqrt{\frac{1}{6}}
\end{align*}
so you found the correct magnitude but did not take into account into the sign of $\cos(\frac{\theta}{2})$.
To determine $\tan(2\theta)$, we can use the formula
$$\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)}$$
together with the double angle formulas
\begin{align*}
\sin(2\theta) & = 2\sin\theta\cos\theta\\
\cos(2\theta) & = \cos^2\theta - \sin^2\theta\\
& = 2\cos^2\theta - 1\\
& = 1 - 2\sin^2\theta
\end{align*}
Since $450^\circ < \theta < 540^\circ$, $\theta$ is a second-quadrant angle, so $\sin\theta > 0$. Hence,
\begin{align*}
\sin\theta & = \sqrt{1 - \cos^2\theta}\\
& = \sqrt{1 - \left(-\frac{2}{3}\right)^2}\\
& = \sqrt{1 - \frac{4}{9}}\\
& = \sqrt{\frac{5}{9}}\\
& = \frac{\sqrt{5}}{3}
\end{align*}
Therefore,
\begin{align*}
\tan(2\theta) & = \frac{\sin(2\theta)}{\cos(2\theta)}\\
& = \frac{2\sin\theta\cos\theta}{\cos^2\theta - \sin^2\theta}\\
& = \frac{2\left(\frac{\sqrt{5}}{3}\right)\left(-\frac{2}{3}\right)}{\left(-\frac{2}{3}\right)^2 - \left(\frac{\sqrt{5}}{3}\right)^2}\\
& = \frac{\frac{-4\sqrt{5}}{9}}{\frac{4}{9} - \frac{5}{9}}\\
& = \frac{-4\sqrt{5}}{4 - 5}\\
& = 4\sqrt{5}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247745",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If there must be at least one person in each table, in how many ways can 6 people be seated in 3 tables?
If there must be at least one person in each table, in how many ways can 6 people be seated in 3 tables?
I know there are three possible ways to split the set of people P into three distinct groups ($1+1+4,1+2+3,2+2+2$). Consider only the case $2+2+2$. At first I thought that the number of ways to choose the first pair of people to sit on a table is ${6\choose2}$ but for some reason it is $\frac{1}{3}\times {6\choose2}$. Can anyone explain why this is the case?
| We consider the 3 cases mentioned above:
1) If they are split 4/1/1, there are $\dbinom{6}{4}$ ways to select the 4 people sitting together, and $3!$ ways to arrange them at their table, so this gives $\dbinom{6}{4}(3!)=15(6)=90$ possibilities.
2) If they are split 3/2/1, there are $\dbinom{6}{3}$ ways to select the 3 people sitting together, $2!$ ways to arrange them at their table, and $\dbinom{3}{2}$ ways to select the 2 people sitting together, so this gives $\hspace{.2 in}\dbinom{6}{3}\cdot2\cdot\dbinom{3}{2}=20(2)(3)=120$ possibilities.
3) If they are split 2/2/2, then there are $\displaystyle\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}$ or $\;5\cdot3\cdot1=15$ ways to pair up the 6 people.
Therefore there are a total of $90+120+15=$225 possible ways to do this.
This is s(6,3), a Stirling number of the first kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Finding $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$
Find $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$
I used a bit of trig identities to get: $\int \frac {2+\sin (2x)}{-4+\cos(2x)}dx$ and using the substitution: $t= \tan (2x)$ I got to a long partial fractions calculation which doesn't seem right.
Any hints on how to do it please?
| The denominator of your second integral should be $-3+5\cos 2x$, because
from the identities $\sin x\cos x=\frac{\sin 2x}{2}$ and $\sin ^{2}x=\frac{
1-\cos 2x}{2}$ we obtain
\begin{equation*}
\frac{1+\sin x\cos x}{1-5\sin ^{2}x}=\frac{2+\sin 2x}{-3+5\cos 2x}.
\end{equation*}
To evaluate
\begin{equation*}
\int \frac{2+\sin 2x}{-3+5\cos 2x}dx
\end{equation*}
we can use the standard half-angle substitution $t=\tan x$. Since $dt=\left(
t^{2}+1\right) dx$, $\sin 2x=\frac{2t}{t^{2}+1}$ and $\cos 2x=\frac{1-t^{2}}{
t^{2}+1}$, we thus have
\begin{eqnarray*}\int \frac{2+\sin 2x}{-3+5\cos 2x}dx
=\int \frac{t^{2}+t+1}{-\left( 2t-1\right) \left( 2t+1\right) \left(
t^{2}+1\right) }dt
\end{eqnarray*}
Now, using partial fractions decomposition, we obtain
\begin{equation*}
\frac{t^{2}+t+1}{-\left( 2t-1\right) \left( 2t+1\right) \left(
t^{2}+1\right) }=-\frac{7/20}{t-1/2}+\frac{3/20}{t+1/2}+\frac{t/5}{t^{2}+1}.
\end{equation*}
As such,
\begin{eqnarray*}
\int \frac{2+\sin 2x}{-3+5\cos 2x}dx
&=&-\frac{7}{20}\ln \left| \tan x-\frac{1}{2}\right| +\frac{3}{20}\ln \left|
\tan x+\frac{1}{2}\right| \\&&+\frac{1}{10}\ln \left| \tan ^{2}x+1\right| +C.
\end{eqnarray*}
| {
"language": "en",
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"source": "stackexchange",
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How to prove that $ \sin \angle{GAB}+\sin \angle{GBC}+\sin \angle{GCA} \le \frac{3}{2} $ for a triangle $ABC$ with centroid $G$? Let $ G $ be the centroid of $ \triangle ABC $ , such that $ \measuredangle{GAB}=x,\measuredangle{GBC}=y,\measuredangle{GCA}=z $.
How do I prove that :
$$ \sin x +\sin y +\sin z\le \frac{3}{2} $$
| Let $AA'$, $BB'$ and $CC'$ be medians of the triangle.
Thus, from $\Delta ABA'$ we obtain:
$$\frac{\frac{a}{2}}{\sin{x}}=\frac{AA'}{\sin\beta}$$ or
$$\sin{x}=\frac{a\sin\beta}{2AA'}$$ or
$$\sin{x}=\frac{a\cdot\frac{2S}{ac}}{\sqrt{2b^2+2c^2-a^2}}$$ or
$$\sin{x}=\frac{2S}{c\sqrt{2b^2+2c^2-a^2}}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{1}{c\sqrt{2b^2+2c^2-a^2}}\leq\frac{3}{4S}$$ or
$$\sum_{cyc}\frac{1}{c\sqrt{2b^2+2c^2-a^2}}\leq\frac{3}{\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}}.$$
Now, by C-S
$$\left(\sum_{cyc}\frac{1}{c\sqrt{2b^2+2c^2-a^2}}\right)^2\leq\sum_{cyc}\frac{1}{c^2}\sum_{cyc}\frac{1}{2b^2+2c^2-a^2}.$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{1}{a^2}\sum_{cyc}\frac{1}{2b^2+2c^2-a^2}\leq\frac{9}{\sum\limits_{cyc}(2a^2b^2-a^4)},$$
which is
$$(a^2-b^2)^2(a^2-c^2)^2(b^2-c^2)^2\geq0.$$
Done!
| {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
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Evaluate the limit $\displaystyle\lim_{x \to 0}\frac{(e-\left(1 + x\right)^{1/x})}{\tan x}$. How to evaluate the following limit
$$\displaystyle\lim_{x\to 0} \dfrac{e-\left(1 + x\right)^{1/x}}{\tan x}$$
I have tried to solve it using L-Hospital's Rule, but it creates utter mess.
Thanks for your generous help in advance.
| This is more a comment than an answer since previous answers show how to obtain the limit.
In the same spirit as abel's anwer, let me use a few more terms (keeping abel's notation) $$\ln y = \frac 1 x\ln(1 + x) =1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right) $$ $$y=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$ $$e-y=\frac{e x}{2}-\frac{11 e x^2}{24}+O\left(x^3\right)$$ $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$\dfrac{e-\left(1 + x\right)^{\frac{1}{x}}}{\tan x}=\frac{\frac{e x}{2}-\frac{11 e x^2}{24}+O\left(x^3\right) }{x+\frac{x^3}{3}+O\left(x^4\right) }\approx \frac{\frac{e }{2}-\frac{11 e x}{24} }{1+\frac{x^2}{3} }\approx (\frac{e }{2}-\frac{11 e x}{24})(1-\frac{x^2}{3})$$ This will give $$\dfrac{e-\left(1 + x\right)^{\frac{1}{x}}}{\tan x}=\frac{e}{2}-\frac{11 e x}{24}+O\left(x^2\right)$$ which shows the limit but also the manner it is approached.
Edit
Just as a side result, suppose that you need to solve for $x$ the equation $$\dfrac{e-\left(1 + x\right)^{\frac{1}{x}}}{\tan x}=1$$ the given first order expansion gives an extimate of the solution $$x\approx\frac{12 (e-2)}{11 e}\approx 0.288$$ while the exact solution is $\approx 0.339$.
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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A real $2 \times 2 $ matrix $M$ such that $M^2 = \tiny \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$ , then :
A real $2 \times 2 $ matrix $M$ such that $$M^2 = \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$$
(a) exists for all $\epsilon > 0$.
(b) does not exist for any $\epsilon > 0$.
(c) exists for some $\epsilon > 0$.
(d) None of the above is true.
Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix.
Could someone please give me a hint on how to go about this problem?
Thank you very much for your help in this regard.
| no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$
for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$
pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = \epsilon \tag 1$$
that means if $\epsilon \neq 0,$ then $a+d \neq0\implies b = c = 0.$ from $(1), a^2 = -1, d^2 = -1 - \epsilon$ there are no real solutions as $a^2 = -1.$
if $\epsilon = 0, a^2 - d^2 = 0$ it forces $a = -d, b = c = 0$ but then (1) forces $d^2 = -1$ which cannot be satisfied. the other case is $a = d.$ the constraints are $$a^2 + bc = -1, 2ab = 0 = 2ac, bc + d^2 = -1 $$ two choices: (i) $a = 0$ gives $bc = -1, d = 0$
(ii) $a \neq 0$ which gives $b = 0, c = 0, a^2 = -1$ not possible.
| {
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Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$
I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\frac{x+y}{1-xy}=\frac{\tan a+\tan b}{1-\tan a\tan b}=\tan(a+b)$
$\implies a+b=\arctan \frac{\tan a+\tan b}{1-\tan a\tan b} $
$\implies \arctan x+\arctan y = \arctan\frac{x+y}{1-xy}$
I know $\pi$ is the period of $\tan x$ so $xy\gt 1$ constraint must have something to do with this. But I am not able to figure out how exactly these period and $xy\gt 1$ are related. Any help is appreciated. Thanks!
| The reason is that $\frac{\pi}{2} < |a+b| < \pi$. $|a+b| < \pi$ is easy to see.
Notice that when $x>0, y>0$, $a+b$ reaches minimum when $xy=1$. Also, $\tan^{-1} \frac{1}{x} = \cot^{-1} x$, and $\tan^{-1} x + \tan^{-1} \frac{1}{x} = \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.
Similarly, when $x<0, y<0$, we should have $-\pi$ instead of $\pi$.
| {
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Compute $\lim_\limits{n\to\infty}a_n$ where $a_{n+2}=\sqrt{a_n.a_{n+1}}$ I managed to show that the limit exists, but I don't know how to compute it.
EDIT:
There are initial terms: $a_1=1$ and $a_2=2$.
| We have $a_2\gt a_1\gt0$ and $a_{n+2}^2=a_{n+1}a_n$.
So,$$\begin{align}
&a_3^2=a_2a_1\\
\implies&a_3^2\gt a_1^2\\
\implies &a_3\gt a_1
\end{align}$$
and $$\begin{align}
&a_3^2=a_2a_1\\
\implies &a_3^2\lt a_2^2\\
\implies &a_3\lt a_2
\end{align}$$
So, we have $$\color{red}{a_2\gt a_3\gt a_1\gt 0}$$
Similarly, $$\begin{align}
&a_4^2=a_3a_2\\
\implies &a_4^2\lt a_2^2\\
\implies &a_4\lt a_2
\end{align}$$
and $$\begin{align}
&a_4^2=a_3a_2\\
\implies &a_4^2\gt a_3^2\\
\implies &a_4\gt a_3
\end{align}$$
So, we have $$\color{red}{a_2\gt a_4\gt a_3\gt a_1\gt 0}$$
Now, following this pattern, one can easily see that this sequence is bounded below by $0$ and bounded above by $a_2$.
Now, consider two subsequences $\color{blue}{v_n=a_{2n-1}}$ and $\color{blue}{w_n=a_{2n}}$.
one can easily see that $v_n$ is increasing subsequence bounded above by $a_2$, since, $\color{blue}{0\lt a_1\lt a_3\lt \dots \lt a_2}$, and $w_n$ is decreasing subsequebce and bounded below by $0$, since, $\color{blue}{a_2\gt a_4\gt \dots \gt 0}$.
So, they must converge. Suppose $\color{blue}{\lim v_n=l_1}$ and $\color{blue}{\lim w_n=l_2}$.
Now, we have, $$\begin{align}
&a_{n+2}^2=a_{n+1}a_n\\
\implies &\lim a_{n+2}^2=\lim a_{n+1}\times\lim a_n\\
\implies &l_2^2=l_1\times l_2\\
\implies &\color{blue}{l_2=l_1=l(\text{say})}
\end{align}$$
here, without loss of generality, we assumed $n$ is even. So, the mother sequence $\{a_n\}$ converges to $l$ i.e. $\color{blue}{\lim a_n=l}$
Now write $$\begin{align}
\color{green}{\cancel{a_3^2}}&=a_2\times a_1\\
\color{purple}{\cancel{a_4^2}}&=\color{green}{\cancel{a_3}}\times a_2\\
\color{orange}{\cancel{a_5^2}}&=\color{purple}{\cancel{a_4}}\times \color{green}{\require{cancel}\cancel{a_3}}\\
&\vdots\\
\color{navy}{a_{n+1}^\cancel{2}}&=\color{maroon}{\cancel{a_{n}}}\times \color{teal}{\cancel{a_{n-1}}}\\
a_{n+2}^2&=\color{navy}{\cancel{a_{n+1}}}\times \color{maroon}{\cancel{a_n}}.
\end{align}$$
Now, multiplying all the terms, we have $$\begin{align}
&a_{n+2}^2 \times a_{n+1}=a_2^2 a_1\\
\implies &\lim a_{n+2}^2 \times \lim a_{n+1}=a_2^2 a_1\\
\implies &l^2\times l= a_2^2 a_1\\
\implies &l^3=a_2^2 a_1\\
\implies &\color{red}{l=\sqrt[3]{a_2^2 a_1}}.
\end{align}$$
And, here the particular case, $a_1=1$ and $a_2=2$, so,
$$\lim_{n\to\infty} a_n = 1^{1/3}2^{2/3} = \sqrt[3]{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Evaluating trigonometric limit: $\lim_{x \to 0} \frac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$
Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $
This is what I've tried yet:
$$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\
=&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\tan x\right\}}{4\sin^4 x}\\
=& \cfrac{2x\tan x \left(\frac{\tan^2 x}{1 - \tan^2 x}\right) }{4\sin^4 x} \\
=& \cfrac{x\tan^3 x}{2\sin^4 x (1-\tan^2 x)} \\
=& \cfrac{\tan^3 x}{2x^3\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} \\
=& \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)}\end{align}$$
Taking limit of the above expression, we've :
$$\lim_{x\to 0} \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} = \lim_{x\to 0} \cfrac{\cos^2x}{2\cos 2x} = \cfrac{1}{2} $$
Firstly, is my answer right or am I doing somewhere wrong?
Secondly, this seems a comparatively longer method than expected for objective type questions. I'm seeking for a shortcut method for such type of questions. Is there any method I should've preferred?
| Your solution looks good! Give yourself a pat on your back.
Another way is to expand the trigonometric polynomials using Taylor series, i.e.,
\begin{align}
\dfrac{x(\tan(2x)-2\tan(x))}{(1-\cos(2x))^2} & = x \cdot \dfrac{2x+ \dfrac{(2x)^3}3+ \mathcal{O}(x^5) - 2x - 2\cdot\dfrac{x^3}3 + \mathcal{O}(x^5)}{\left(1-\left(1-\dfrac{(2x)^2}{2!} + \mathcal{O}(x^4)\right)\right)^2}\\
& = x \cdot \dfrac{2x^3 + \mathcal{O}(x^5)}{\left(2x^2 + \mathcal{O}(x^4)\right)^2} = \dfrac{2x^4\left(1+\mathcal{O}(x^2)\right)}{4x^4\left(1+\mathcal{O}(x^2)\right)^2}
\end{align}
Taking the limit as $x \to 0$, you get the limit as $\dfrac12$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 4,
"answer_id": 3
} |
How to prove that $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$? Let $a,b,c>0: (a+b)(b+c)(c+a)=ab+bc+ca$. How to prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$$
| Let $x=\frac{a}{\sqrt{(a+b)(a+c)}}$ $y=\frac{b}{\sqrt{(b+a)(b+c)}}$ and $z=\frac{c}{\sqrt{(c+a)(c+b)}}$.
Then we have $x^{2}+y^{2}+z^{2}+2xyz=1$. Note that for angles $A$, $B$ and $C$ in some acute triangle:$\cos^{2}A+\cos^{2}B+\cos^{2}C+2\cos A\cos B\cos C=1$. Hence $x$,$y$ and $z$ can be replaced by $\cos A$, $\cos B$ and $\cos C$ respectively and we have following equations (Note $(a+b)(b+c)(c+a)=ab+bc+ca$):
$\cos A=x=\frac{a}{\sqrt{(a+b)(a+c)}}$, $\sin A=\sqrt{b+c}$; $\cos B=y=\frac{b}{\sqrt{(b+a)(b+c)}}$, $\sin B=\sqrt{a+c}$; $\cos C=z=\frac {c}{\sqrt{(c+a)(c+b)}}$, $\sin C=\sqrt{a+b}$.
Now $\sum_{cyc}\frac{a}{\sqrt{a+b}}\leq\frac{3\sqrt{3}}{4}\Longleftrightarrow\sum_{cyc}\sin A\cos B\leq\frac{3\sqrt{3}}{4}$.
Since the inequality is symmetric, we assume that $a\leq b\leq c$ without loss of generality. Then $\sin A\leq \sin B \leq \sin C$ and $\cos A\geq \cos B \geq \cos C$. By rearrangement inequality and AM–GM inequality:
$\sin A\cos B+\sin B\cos C+\sin C\cos A\leq \sin A\cos C+\sin B\cos B+\sin C\cos A=\sin B+\sin B\cos B=4\sin\frac{B}{2}\cos^{3}\frac{B}{2}\leq\frac{3\sqrt{3}}{4}$
Added by Alex Ravsky
It seems you can finish your solution as follows.
Let X be the smallest among angles $A,$ $B,$ and $C$ and $Z$ be the largest.
Then $\sin X\leq \sin Y \leq \sin Z$ and $\cos X\geq \cos Y \geq \cos Z$. By rearrangement inequality and AM–GM inequality:
$\sin A\cos B+\sin B\cos C+\sin C\cos A\leq \sin X\cos Z+\sin Y\cos Y+\sin Z\cos X=\sin Y+\sin Y\cos Y=4\sin\frac{Y}{2}\cos^{3}\frac{Y}{2}\leq\frac{3\sqrt{3}}{4}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
find coefficient of $x^{50}$ Let $f(x)=\frac{1}{(1+x)(1+x^2)(1+x^4)}$then find the coefficient of term $x^{50}$ in $(f(x))^3$.I think that we can set $$(f(x))^3=\frac{a}{(1+x)^3}+\frac{b}{(1+x^2)^3}+\frac{c}{(1+x^4)^3}$$ and find a,b and c then use Taylor seri .
| It won't work like that: $f(x)^3 = \dfrac{1}{(1+x)^3 (1+x^2)^3 (1+x^4)^3}$, and then the partial-fraction expansion will include terms in $1/(1+x)$, $1/(1+x)^2$, ...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Induction Clarification I had this problem: Is it always necessary to go from n to (n + 1) or from (n - 1) to n in the inductive hypothesis? Is the "direction" always important?
Here is my solution to one such proof, which turns out to be totally wrong. Is it really so?
$\underline{\textrm{BS}}.$ for n = 1
$\binom{2\cdot 1}{1} = 2 = \frac{4^1}{2 \sqrt{1}}$ It`s ok!
$\underline{\textrm{IH}}.$ for n+1
Assumed $$\binom{2(n+1)}{n+1} \ge \frac{4^{n+1}}{2 \sqrt{n+1}}$$.
$\underline{\textrm{IS}}.$ Prove for n
$$\binom{2n}{n} \ge \frac{4^{n}}{2 \sqrt{n}}$$
$\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{2!n!}{n!n!}= \frac{2}{n!}$\
First it applies:\
$$\frac{2}{n!} \ge \frac{2}{(n + 1)!} \ \ \ \ \ \ \ \ \textcircled{1}$$\
Clear, because $(n+1)! > n!$\
$$\frac{4^{n+1}}{2 \sqrt{n+1}} \ge \frac{4^{n}}{2 \sqrt{n}} \textcircled{2}$$\
$$\iff \frac{4^{n+1}}{4^n}\geq \frac{\sqrt{n+1}}{\sqrt{n}}$$\
$$\iff 4\geq\sqrt{\frac{n+1}{n}}=\sqrt{1+\frac{1}{n}}$$\
Since $\frac{1}{n} \to 0$ (monotonically decreasing)$ \implies \sqrt{1+\frac{1}{n}} = 2$ for n = 1 and $\sqrt{1+\frac{1}{n}} = 1$ for $n \to \infty$, that's why the inequality \textcircled{2} always holds.
Then through Induction Hypothesis(IH) I get:
$$\frac{2}{n!} \overset{\textcircled{1}}{\ge} \frac{2}{(n + 1)!} \overset{IH}{\ge} \frac{4^{n+1}}{2 \sqrt{n+1}}\overset{\textcircled{2}}{\ge} \frac{4^{n}}{2 \sqrt{n}}$$
$$\iff \binom{2n}{n}\ge \binom{2(n + 1)}{n+1} \ge \frac{4^{n+1}}{2 \sqrt{n+1}}\ge \frac{4^{n}}{2 \sqrt{n}}$$
From there applies
$$\binom{2n}{n}\ge \frac{4^{n}}{2 \sqrt{n}}$$
| Yes, this is wrong. The idea of induction is like a row of dominos. If the first falls, then the one behind must fall to. But what you prove is that if the second falls then the first falls. But you can't prove that the second falls.
There is a variant of induction that can be used:
*
*Prove $P(1)$.
*Prove that if $P(n)$, then $P(n-1)$.
*Prove that if $P(n)$, then $P(2n)$.
Adding the third step will complete this proof.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
According to Buckingham Theorem the rank of $A$ should be $2$ A physical system is described by a law of the form $f(E,P,A)=0$, where $E,P,A$ represent, respectively, energy, pressure and surface area. Find an equivalent physical law that relates suitable dimensionless quantities.
That' what I have tried so far:
1st step:
Choice of quantities
Mass: $M$
Time: $T$
Length: $L$
So:
$$[E]=M L^2 T^{-2}$$
$$[P]=ML^{-1}T^{-2}$$
$$[A]=L^2$$
2nd step:
Construction of dimonsionless quantities
The matrix of dimensions:
\begin{equation*}
A=\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
2 & -1 &2
\end{bmatrix}
\end{equation*}
I tried to find the rank, determining the smallest $n$ for which $A^n=I$.
\begin{equation*}
\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
2 & -1 &2
\end{bmatrix}\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
2 & -1 &2
\end{bmatrix}=\begin{bmatrix}
-1 & -1 & 0\\
2 & 2 & 0 \\
8 & 2 &4
\end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix}
-1 & -1 & 0\\
2 & 2 & 0 \\
8 & 2 &4
\end{bmatrix}\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
2 & -1 &2
\end{bmatrix}=\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
12 & 0 &8
\end{bmatrix}
\end{equation*}
But I saw the solution and there should be only one dimensionless quantity, so according to Buckingham Theorem the rank of $A$ should be $2$.
Where is my mistake?
| You don't need to do $A^n$.
You can see the 1st and 2nd row are multiple of each other, so one of them can be eliminated to $0$. That makes the rank $2$.
Alternatively, switch the 1st row and last row to make it
$$\pmatrix{2&-1&2\\-2&-2&0\\1&1&0}$$
Then divide the second row by $-2$, and subtract from the last row the resulting 2nd row. The result will be
$$\pmatrix{2&-1&2\\1&1&0\\0&0&0}$$
Or, you can do Gaussian elimination. First add $2$ times the 1st row to the second row, then subtract from 3rd row $2$ times the 1st row:
$$\pmatrix{1&1&0\\0&0&0\\0&-3&2}$$
Switching the last two rows:
$$\pmatrix{1&1&0\\0&-3&2\\0&0&0}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
compute improper integrals using integration by parts Compute
\begin{equation*}
\int_0^\infty \frac{\sin^4(x)}{x^2}~dx\text{ and }\int_0^\infty \frac{\sin (ax) \cos (bx)}{x}~dx.
\end{equation*}
For the first integral I tried letting $u = \sin ^4 x$ and $dv= \frac{1}{x^2}~dx$, which simplified to $\int_0 ^\infty \frac{4 \sin^3(x) \cos(x)}{x}~dx$ and applying integration by parts again with $u = \sin^3(x) \cos (x)$ and $dv = \frac{1}{x}$, I got
$$\left.\vphantom{\frac11}[\sin ^3 (x) \cos (x) \ln (x)] \right|_0^\infty - \int_0^\infty \ln (x) [3 \sin ^2 (2x) - \sin^4 (x)]~dx$$
The only problem is that it appears that the term $[\sin ^3 (x) \cos (x) \ln (x)] \big\vert_0^\infty$ evaluates to infinity and neither the $3\int_0^\infty \ln (x) \sin ^2 (2x)~dx$ term nor the $3\int_0^\infty \ln (x) \sin^4 (x)~dx$ converges.This looks like a dead end but the other choice of $u$ and $dv$ looks even worse, so I'm not sure how to proceed.
For the second integral I tried $u = \sin (ax) \cos (bx)$ and $dv = \frac{1}{x}$, which gives me $[\sin (ax) \cos (bx) \ln (x)] \big\vert_0^\infty + \int_0^\infty \frac{b-a}{2} \ln (x) \cos ((a-b)x)~dx - \int_0 ^ \infty \frac{b+a}{2} \ln (x) \cos ((a+b)x)~dx$, which also looks like a dead end because all three of the terms go to infinity.
Any help is appreciated.
| For the first one, we have
\begin{align}
\int_0^{\infty} \dfrac{\sin^4(x)}{x^2}dx & = \int_0^{\infty} \dfrac{\sin^2(x)(1-\cos^2(x))}{x^2}dx = \int_0^{\infty} \dfrac{\sin^2(x)}{x^2}dx - \int_0^{\infty} \dfrac{\sin^2(x)\cos^2(x)}{x^2}dx\\
& = \int_0^{\infty} \text{sinc}^2(x) dx - \int_0^{\infty} \dfrac{\sin^2(2x)}{(2x)^2}dx = \int_0^{\infty} \text{sinc}^2(x) dx - \dfrac12\int_0^{\infty} \text{sinc}^2(x) dx\\
& = \dfrac12\int_0^{\infty} \text{sinc}^2(x) dx = \dfrac{\pi}4
\end{align}
From here, we have the value of $\int_0^{\infty} \text{sinc}^m(x)dx$
For the second one, we have
$$\sin(ax)\cos(bx) = \dfrac{\sin((a+b)x) + \sin((a-b)x)}2$$
We hence have
$$\dfrac12\int_0^{\infty}\dfrac{\sin((a+b)x)}xdx + \dfrac12\int_0^{\infty}\dfrac{\sin((a-b)x)}xdx =
\begin{cases}
\dfrac{\pi}2 & \text{ if }a>b\\
\dfrac{\pi}4 & \text{ if }a=b\\
0 & \text{ if }a < b
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square. I successfully thought of bounding our expression examining consecutive squares that attain values close to it, and this led to the solution I'll post as an answer, which was the one reported. However, before that, I had briefly tried manipulating $$n^4+n^3+n^2+n+1=m^2.$$Since $n=1$ is not a solution, I rewrote this as $$\frac{n^5-1}{n-1}=m^2 \\ n^5-1=nm^2-m^2 \\ m^2-1=n(m^2-n^4) \\ (m-1)(m+1)=n(m-n^2)(m+n^2),$$ but in vain. Does manipulation lead somewhere? Is there a different approach from both of mine?
| For all positive $n$ we have $$\require\cancel \left(n^2+\frac{n}{2}\right)^2=\cancel{n^4+n^3}+\frac{n^2}{4}<\cancel{n^4+n^3}+n^2+n+1 \\ \frac{n^2}{4}<n^2+n+1 $$ and on the other hand $$\require\cancel \left(n^2+\frac{n+2}{2}\right)^2=n^4+n^3+2n^2+\frac{n^2+4n+4}{4}>n^4+n^3+n^2+n+1 \\ \cancel{n^4+n^3}+\frac{9}{4}n^2\cancel{+n+1}>\cancel{n^4+n^3}+n^2\cancel{+n+1} \\ \frac{9}{4}n^2>n^2.$$ As a consequence, $n$ cannot be even, and for some odd $n$ we must have $$\left(n^2+\frac{n+1}{2}\right)^2=\cancel{n^4+n^3+n^2}+\frac{n^2+2n+1}{4}=\cancel{n^4+n^3+n^2}+n+1 \\ n^2+2n+1=4n+4 \\ n^2-2n-3=(n-3)(n+1)=0,$$ whence $n=3$. In particular, $$3^4+3^3+3^2+3+1=11^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$
Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$
Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?
| You can rewrite the determinant as $(b-a)(c-a)(c-b)$, so $\det(A)$ is nonzero precisely when $a,b,c$ are all distinct (note that one of $a,b,c$ can be zero).
Your matrix is the $3\times 3$ case of the Vandermonde matrix, whose determinant is generally a similar product of differences.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
How do you add two series together How do you add the series
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
?
is this right?
$$\begin{aligned}
&\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}}{(z-3)^{n+1}} + \frac{(z-3)^{n}}{4^{n+1}}\right)\\
=\> & \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{n} (z-3)^{n+1}}{(z-3)^{n+1} 4^{n+1}} \right)\\
=\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{n+1+n}}{(z-3)^{n+1} 4^{n+1}}\right)\\
=\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{2n+1}}{(z-3)^{n+1} 4^{n+1}}\right)\end{aligned}$$
| Well, you're doing fine so far. You can do a bit better by noting that $$4^{n+1}=\left(2^2\right)^{n+1}=2^{2(n+1)}=2^{2n+2}$$ and that $$4^{n+1}\cdot(z-3)^{n+1}=\bigl(4(z-3)\bigr)^{n+1}=(4z-12)^{n+1}.$$
You could also distribute the $\frac12$ through, if you like.
I do have to wonder what you gain from combining these two series, though. In combined form, they don't give nearly as much information as readily, in exchange for a little saved space.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The value of ${\sum_{k=0}^{20}}(-1)^k\binom{30}{k}\binom{30}{k+10}$ $\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\c}[1]{{}^{30}{\mathbb C}_{#1}}
\newcommand{\r}[1]{\frac1{x^{#1}}}$
The value of $$\sum_{k=0}^{20}(-1)^k\binom{30}{k}\binom{30}{k+10}$$
It is also the coefficient of $x^{10}$ in:
$$\b{\c0\r0-\c1\r1+\c2\r2-....\c{20}\r{20}}\b{\c{10}x^{10}+\c{11}x^{11}+...\c{30}x^{30}}$$
Adding some terms maybe won't harm:
$$\b{\c0\r0-\c1\r1+\c2\r2-....\c{30}\r{30}}\b{\c{0}x^{0}+\c{1}x^{1}+...\c{30}x^{30}}$$
Contracting:
$$\b{1-\r1}^{30}\b{1+x}^{30}=\b{x-\r1}^{30}=\sum_{n=0}^{30}(-1)^{n}\c nx^{2n-30}$$
The $20$th term interests me, giving:
$$\c{20}=\c{10}$$
Well are there any other interesting ways?
| Vandermonde's Identity yields
$$
\begin{align}
\sum_{k=0}^{20}\binom{30}{k}\binom{30}{k+10}
&=\sum_{k=0}^{20}\binom{30}{k}\binom{30}{20-k}\\[6pt]
&=\binom{60}{20}
\end{align}
$$
In the product,
$$
\begin{align}
\left(x-\frac1x\right)^{30}
&=\sum_{k=0}^{30}(-1)^k\binom{30}{k}x^{30-k}x^{-k}\\
&=\sum_{k=0}^{30}(-1)^k\binom{30}{k}x^{2(15-k)}\\
\end{align}
$$
and in the product
$$
\begin{align}
\left(1-\frac1x\right)^{30}\left(1+x\right)^{30}
&=\sum_{j=0}^{30}(-1)^j\binom{30}{j}x^{-j}\sum_{k=0}^{30}\binom{30}{k}x^k\\
&=\sum_{n=-30}^{30}\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+n}x^n
\end{align}
$$
Comparing the coefficients of $x^n$, we have for $n$ odd
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+n}=0
$$
and for $n$ even
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+n}=(-1)^{15-\frac n2}\binom{30}{15-\frac n2}
$$
For $n=10$, we get
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j+10}=\binom{30}{10}
$$
For $n=-10$, we get
$$
\sum_{j=0}^{30}(-1)^j\binom{30}{j}\binom{30}{j-10}=\binom{30}{20}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Vandermonde determinant and linearly independent Let $a_1,a_2,a_3,b_1,b_2,b_3,b_4,b_5,b_6\in \mathbb{C}$ such that $a_i\not=a_j$ for all $i\not=j.$
If $$\begin{vmatrix}
a_1 & a_2& a_3 & b_1 \\
a_1^2 & a_2^{2} & a_3^{2} & b_2\\
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
\end{vmatrix} =0,$$
$$\begin{vmatrix}
a_1^2 & a_2^{2} & a_3^{2} & b_2\\
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
a_1^5 & a_2^{5} & a_3^{5} & b_5\\
\end{vmatrix} =0,$$
and
$$\begin{vmatrix}
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
a_1^5 & a_2^{5} & a_3^{5} & b_5\\
a_1^6 & a_2^{6} & a_3^{6} & b_6\\
\end{vmatrix} =0,$$
then all minors of order $4$ of the matrix
$$\begin{bmatrix}
a_1 & a_2& a_3 & b_1 \\
a_1^2 & a_2^{2} & a_3^{2} & b_2\\
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
a_1^5 & a_2^{5} & a_3^{5} & b_5\\
a_1^6 & a_2^{6} & a_3^{6} & b_6\\
\end{bmatrix}$$
are $0$. It is stated in a paper that this is true without proof. I believe that it is related with Vandermonde determinant but I do not know how to prove it. Could you please help me or give me an idea? Thank you so much for your help.
Masik
| Suppose that none of $a_i$ is zero. Otherwise the proof is obvoius.
Let $x = a_1, y = a_2, z = a_3$.
$$
\begin{vmatrix}
x^k & y^k & z^k & b_k\\
x^{k+1} & y^{k+1} & z^{k+1} & b_{k+1}\\
x^{k+2} & y^{k+2} & z^{k+2} & b_{k+2}\\
x^{k+3} & y^{k+3} & z^{k+3} & b_{k+3}
\end{vmatrix} = x^k y^k z^k
\begin{vmatrix}
1 & 1 & 1 & b_k\\
x & y & z & b_{k+1}\\
x^2 & y^2 & z^2 & b_{k+2}\\
x^3 & y^3 & z^3 & b_{k+3}
\end{vmatrix} = 0
$$
Since first three columns are linearly independent (they contain nonzero minor $\begin{vmatrix}
1 & 1 & 1\\
x & y & z\\
x^2 & y^2 & z^2\\
\end{vmatrix}$) the last must be their linear combination:
$$
\begin{pmatrix}
b_k \\ b_{k+1} \\ b_{k+2} \\ b_{k +3}
\end{pmatrix} =
A_k \begin{pmatrix}
1 \\ x \\ x^2 \\ x^3
\end{pmatrix}
+B_k \begin{pmatrix}
1 \\ y \\ y^2 \\ y^3
\end{pmatrix}
+C_k \begin{pmatrix}
1 \\ z \\ z^2 \\ z^3
\end{pmatrix}, \qquad k = 1, 2, 3.
$$
Now, eliminating $A_2, B_2, C_2, A_3, B_3, C_3$:
$$
\begin{pmatrix}
b_{k+1} \\ b_{k+2} \\ b_{k +3}
\end{pmatrix} =
A_k \begin{pmatrix}
x \\ x^2 \\ x^3
\end{pmatrix}
+B_k \begin{pmatrix}
y \\ y^2 \\ y^3
\end{pmatrix}
+C_k \begin{pmatrix}
z \\ z^2 \\ z^3
\end{pmatrix} = A_{k+1} \begin{pmatrix}
1 \\ x \\ x^2
\end{pmatrix}
+B_{k+1} \begin{pmatrix}
1 \\ y \\ y^2
\end{pmatrix}
+C_{k+1} \begin{pmatrix}
1 \\ z \\ z^2
\end{pmatrix}
\qquad k = 1, 2.
$$
Since the system for $(A_{k+1}, B_{k+1}, C_{k+1})$ is nonsingular, the solution exists and is unique. One solution is obvoius: $A_{k+1} = xA_k, B_{k+1} = yB_{k}, C_{k+1} = yC_k$. Finally
$$
\begin{pmatrix}
b_1 \\ b_2 \\ b_3 \\ b_4 \\ b_5 \\ b_6
\end{pmatrix} =
A_1 \begin{pmatrix}
1 \\ x \\ x^2 \\ x^3 \\ x^4 \\ x^5
\end{pmatrix}
+B_1 \begin{pmatrix}
1 \\ y \\ y^2 \\ y^3 \\ y^4 \\ y^5
\end{pmatrix}
+C_1 \begin{pmatrix}
1 \\ z \\ z^2 \\ z^3 \\ z^4 \\ z^5
\end{pmatrix}
$$
The last column is a linear combination of the first three. No matter which rows we select we always have a zero minor.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find numbers $a, b, c$ given that $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$
Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$
Suppose $a, b, c$ are roots of $P(x)$.
$$P(x) = k(x - a)(x - b)(x - c)$$
But then I get $(k = 1)$
$$P(x) = x^3 - 12x^2 + x(ab + ac + bc) - abc$$
Cant go further...
| \begin{eqnarray}ab+bc+ca=\frac{1}{2}((a+b+c)^2-(a^2+b^2+c^2))=47\end{eqnarray}
\begin{align*}a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\ abc&=44\end{align*}
So $a, b, c$ are roots of the polynomial $x^3-12x^2+47x-44=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int \frac{x^5 dx}{\sqrt{1+x^3}}$ I took $1+x^3$ as $t^2$ . I also split $x^5$ as $x^2 .x^3$ . Then I subsituted the differentiated value in in $x^2$ . I put $x^3$ as $1- t^2$ . I am getting the last step as $2/9[\sqrt{1+x^3}x^3 ]$ but this is the wrong answer , i should get $2/9[\sqrt{1+x^3}(x^3 +2)]$.
Please help me. Thanks
| It looks like you've only made a minor mistake, but since I couldn't tell exactly where it happened, I have provided a step-by-step solution starting with the substitution you chose.
If $t^2=1+x^3$, then $2t\,dt=3x^2\,dx$, so
\begin{align*}
\int \frac{x^5\,dx}{\sqrt{1+x^3}}&=\int \frac{x^3\cdot 3x^2\,dx}{3\sqrt{1+x^3}}\\
&=\int \frac{2t(t^2-1)\,dt}{3t}\\
&=\frac{2}{3}\int (t^2-1)\,dt\\
&=\frac{2}{3}\left(\frac{t^3}{3}-t\right)+C\\
&=\frac{2}{9}\left(t^3-3t\right)+C\\
&=\frac{2}{9}\left((1+x^3)\sqrt{1+x^3}-3\sqrt{1+x^3}\right)+C\\
&=\frac{2}{9}(x^3-2)\sqrt{1+x^3}+C.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Quadratic solutions puzzle The equation $x^2+ax+b=0$, where $a\neq b$, has solutions $x=a$ and $x=b$. How many such equations are there?
I'm getting $1$ equation as I can only find $a=b=0$ as an equation, which is not allowed.
$$x=\frac{±\sqrt{a^2-4 b}-a}2$$
$x=a$ or $b$ so these are the equations
$$a=\frac{\sqrt{a^2-4 b}-a}2$$
$$b=\frac{-\sqrt{a^2-4 b}-a}2$$
$$a=\frac{-\sqrt{a^2-4 b}-a}2$$
$$b=\frac{\sqrt{a^2-4 b}-a}2$$
The only solution for all of these is $a=b=0$, but is this right?
| Let's take $a=\frac{\sqrt{a^2-4 b}-a}2$ for example:
$$a=\frac{\sqrt{a^2-4 b}-a}2$$
$$2a=\sqrt{a^2-4 b}-a$$
$$3a=\sqrt{a^2-4 b}$$
$$9a^2=a^2-4 b$$
$$8a^2=-4 b$$
$$b=-2 a^2$$
Now you can take that $b$ and substitute in $b=\frac{\sqrt{a^2-4 b}-a}2$ to find a.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$
Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $$a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$$
The inequality can be written in the condensed form
$$\sum\limits_{Sym}a\ge\sum\limits_{Sym}a^3+\sum\limits_{Sym}ab$$
I was told that this is a pretty inequality to prove, but I have been unable to do so.
I've tried naive things like multiplying both sides by $a+b+c+d$, and rewriting $(a^2+b^2+c^2+d^2)^2$, but nothing panned out (and the computations were relatively time-consuming). I also tried looking for clever applications of Cauchy-Schwarz (which seems like the way to go) and AM-GM, but nothing sprung out at me.
| Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $abc+abd+acd+bcd=4w^3$.
Hence, $16u^2-12v^2=1$ and our inequality is equivalent to $3v^6-4uv^2w^3+w^6\geq0$.
By Roll's theorem there are $x>0$, $y>0$ and $z>0$, for which
$x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
After this substitution we need to prove that
$\sum\limits_{cyc}(x^3y^3-x^3y^2z-x^3z^2y+x^2y^2z^2)\geq0$, which is Schur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
find the solutions to the equation $4\sin^2\theta + 1 = 6\sin\theta$ in the interval $0^\circ \leq \theta < 360^\circ$ Find the solutions to the following equation for $0^\circ \leq θ < 360^\circ$:
$$4\sin^2 θ + 1 = 6\sin θ$$
My work:
$$4\sin^2\theta - 6\sin\theta + 1 = 0$$
Factor
$$\sin\theta= \frac{1}{4}(3+ \sqrt{5}) \qquad \text{and} \qquad \sin\theta = \frac{1}{4}(3- \sqrt{5})$$
$$\theta = \arcsin\left[\frac{1}{4}(3+ \sqrt{5})\right] \qquad \text{and} \qquad \theta = \arcsin\left[\frac{1}{4}(3- \sqrt{5})\right]$$
When I take the inverse sin they both come out >1 (no solution), what did I do wrong?
\begin{align*}
2\cos^2\theta - 2\cos\theta - 1 & = 0\\
\cos\theta & = \frac{2 \pm \sqrt{4+8}}{4} && \text{quadratic formula}\\
\cos\theta & = \frac{1 \pm \sqrt{3}}{2}\\
\theta & = \arccos\left[\frac{1-\sqrt{3}}{2}\right]\\
\theta & = 111.5^\circ
\end{align*}
Is this correct?
| The domain of arcsin is $[-1,1]$, the co-domain is $[-\pi/2, \pi/2]$
So you have only the solution $a = arcsin((3-\sqrt5)/4)$. That's because you cannot take arcsin from $(3+\sqrt5)/4$, as it's not in the domain.
Then ... given that your $sin$ value $(3-\sqrt5)/4$ is in $(0, 1)$, you conclude that $0 < a < \pi/2$. From that follows that $\pi - a$ is also a solution. And obviously you have no more solutions in $[0, 2\pi)$. So you have 2 solutions here: $a$ and $\pi-a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to evaluate the product $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$? Evaluating the infinite product of $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$. Please Help.
| The product is
$$\prod_{n=2}^{\infty} \frac1{1-\frac1{n^2}} = \prod_{n=2}^{\infty} \frac{n^2}{n^2-1} = \prod_{n=2}^{\infty} \frac{n^2}{(n-1)(n+1)} = \frac{2 \cdot 2}{1 \cdot 3}\cdot \frac{3 \cdot 3}{2 \cdot 4}\cdot \frac{4 \cdot 4}{3 \cdot 5} \cdots = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Closed forms for definite integrals involving error functions I have been working for a while with these kinds of integrals
$$\int_0^\infty dx\,\text{erfc}\left(c +i x\right)\exp \left(-\frac{1}{2}d^2x^2+i cx\right)$$
$$\int_\Lambda^\infty dx\,\frac{1}{x}\text{erfc}\left(c +i x\right) \exp \left(-\frac{1}{2}d^2x^2+i cx\right)$$
where $c$ and $d$ are just real constants and $\Lambda>0$.
I have also been working with other similar integrals that have a closed-form expression, but I can't figure out the form of these ones. Does anyone know if these integrals have a closed-form solution?
| If $u =\operatorname{erfc}(c+ix)$ then
$$du = \dfrac {-2} {\sqrt\pi} e^{-(c+ix)^2} i \, dx = \dfrac {-2} {\sqrt\pi} e^{-(c^2-2icx - x^2)^2} i \, dx. \tag 1$$
We need to work with
$$
\exp\left( \frac {-1}2 d^2x^2 + icx\right).
$$
The exponent is
$$
\frac{-d^2} 2 \left( x^2 - \frac{2icx}{d^2} - \frac{c^2}{d^4} \right) - \frac{c^2}{2d^2} = \frac{-d^2}2\left( x - \frac{ic}{d^2} \right)^2 - \frac{c^2}{2d^2} = \frac{-d^2}2 w^2 - \frac{c^2}{2d^2}.
$$
Where we find $x$ in $(1)$, replace it with
$$
x = w + \frac{ic}{d^2}.
$$
Then do routine algebra and go on from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the 2x2 matrix given 2 equations. Find the $2 \times 2$ matrix ${A}$ such that ${A}^2 = {A}$ and
${A} \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$
I have tried to express A as a matrix with variables a,b,c, and d, but it gets too messy with the equations. Any help is appreciated.
| Hint: we have
$$A\begin{pmatrix} 7\\-1\end{pmatrix}=\begin{pmatrix} 6\\2\end{pmatrix}$$
And therefore
$$\begin{align}A\begin{pmatrix} 6\\2\end{pmatrix}
&=A^2\begin{pmatrix} 7\\-1\end{pmatrix}\\
&=A\begin{pmatrix} 7\\-1\end{pmatrix}\\
&=\begin{pmatrix} 6\\2\end{pmatrix}
\end{align}$$
So if $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$ we have
$$\begin{cases} 7a-b=6\\7c-d=2\\6a+2b=6\\6c+2d=2\end{cases}$$
And this equivalent to two systems of two equations with two unknowns $(a,b)$ on one hand and $(c,d)$ on the other and I am sure you can solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285881",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Another combined limit I've tried to get rid of those logarithms, but still, no result has came.
$$\lim_{x\to 0 x \gt 0} \frac{\ln(x+ \sqrt{x^2+1})}{\ln{(\cos{x})}}$$
Please help
| Using $\lim\limits_{x\to0}\frac{\log(1+x)}x=1$ and $\lim\limits_{x\to0}\frac{\sin(x)}x=1$,
$$
\begin{align}
&\lim_{x\to0^+}\frac{\log\left(x+\sqrt{x^2+1}\right)}{\log(\cos(x))}\\
&=\lim_{x\to0^+}\frac{\log\left(1+x+\sqrt{x^2+1}-1\right)}{\log(1+\cos(x)-1)}\\
&=\lim_{x\to0^+}\frac{\log\left(1+x+\frac{x^2}{\sqrt{x^2+1}+1}\right)}{\log\left(1-\frac{\sin^2(x)}{\cos(x)+1}\right)}\\
&=\lim_{x\to0^+}\frac{x+\frac{x^2}{\sqrt{x^2+1}+1}}{\frac{\sin^2(x)}{\cos(x)+1}}
\cdot\lim_{x\to0^+}\frac{\log\left(1+x+\frac{x^2}{\sqrt{x^2+1}+1}\right)}{x+\frac{x^2}{\sqrt{x^2+1}+1}}
\cdot\lim_{x\to0^+}\frac{\frac{\sin^2(x)}{\cos(x)+1}}{\log\left(1-\frac{\sin^2(x)}{\cos(x)+1}\right)}\\
&=\lim_{x\to0^+}\frac1x
\cdot\lim_{x\to0^+}\frac{x^2}{\sin^2(x)}
\cdot\lim_{x\to0^+}\frac{1+\frac{x}{\sqrt{x^2+1}+1}}{\frac1{\cos(x)+1}}\cdot1\cdot-1\\[6pt]
&=\lim_{x\to0^+}\frac1x\cdot1\cdot2\cdot1\cdot-1\\[18pt]
&=-\infty
\end{align}
$$
| {
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"url": "https://math.stackexchange.com/questions/1286651",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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show that $\frac{1}{F_{1}}+\frac{2}{F_{2}}+\cdots+\frac{n}{F_{n}}<13$ Let $F_{n}$ is Fibonacci number,ie.($F_{n}=F_{n-1}+F_{n-2},F_{1}=F_{2}=1$)
show that
$$\dfrac{1}{F_{1}}+\dfrac{2}{F_{2}}+\cdots+\dfrac{n}{F_{n}}<13$$
if we use
Closed-form expression
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right)$$
$$\dfrac{n}{F_{n}}=\dfrac{\sqrt{5}n}{\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right)}$$
Well and now I'm stuck and don't know how to proceed
| Clearly: $F_n > F_{n-1}$, for $n \geq 3$
Thus, $F_n = F_{n-1} + F_{n-2} > 2*F_{n-2} > 2^2*F_{n-4}>...>2^{(n-3)/2}*F_3 = 2^{(n-1)/2}$
Now, your inequality can be written as: $\sum\limits_{i=3}^n \frac{i}{F_i} < 10$
The $LHS < \sum\limits_{i=3}^n \frac{i}{2^{(i-1)/2}} = 2^{1/2}*\sum\limits_{i=3}^n \frac{i}{2^{i/2}} < 2^{1/2}*\sum\limits_{i=3}^\infty \frac{i}{2^{i/2}} = 2^{1/2}*\sum\limits_{i=3}^\infty f(i)$
It suffices to prove that $\sum\limits_{i=3}^\infty f(i) < 5*2^{1/2}$
It's easy to verify that $f(i)$ is non-negative, monotone decreasing on $[3;\infty)$, thus integral test is applicable and we have:
$\sum\limits_{i=3}^\infty f(i) \leq f(3)+ \int_{3}^{\infty} f(x)dx = \frac{3}{2^{3/2}}+ \frac{1+3*\ln a}{(\ln a)^2}*a^{-3} = 7.06457 < 7.0710 = 5*2^{1/2}$, where $a=2^{1/2}$.
This upper approximation is far worse than that of Simon Rose.
| {
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"source": "stackexchange",
"question_score": "14",
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Similar matrix for two projections Let there be two matrices $A = \frac{1}{2} \begin{pmatrix}
1 & 1\\
1 & 1\\
\end{pmatrix}$ and $B = \frac{1}{4} \begin{pmatrix}
2 & 4\\
1 & 2\\ \end{pmatrix}$ over the field $\mathbb{F}=\mathbb{Q}$. Find a invertible matrix $S$, s.t. $S^{-1} A S = B$.
I tried to solve this via multiplication with simple matrices but I can't seem to find such $S$.
These two matrices have the property that $A^2 = A$ and $B^2 = B$. Is there any way to use this property to find the matrix $S$ (over image/kernel properties) ?
| An example is $S= \left( \begin{array}{cc} 1 & 0\\ 0&2 \end{array} \right)$.
EDIT: A way to find this matrix is: Take $S= \left( \begin{array}{cc} a & 0\\ b&c \end{array} \right)$. It's very important to take that 0 for simplify your work. (Generally, when you want to find a matrix with some properties, try to take as many zeros as you can, and if taking, let's say $n$ zeros, you find nothing, try to take $n-1$ zeros, and so on.) Then $S^{-1}= \frac{1}{ac} \left( \begin{array}{cc} c & 0\\ -b&a \end{array} \right)$. The ginven relation becomes:
$ \frac{1}{ac} \left( \begin{array}{cc} c & 0\\ -b&a \end{array} \right) \cdot \left( \begin{array}{cc} 1 & 1\\ 1&1 \end{array} \right) \cdot \left( \begin{array}{cc} a & 0\\b&c \end{array} \right)= \left( \begin{array}{cc} 1 & 2\\ \frac{1}{2}&1 \end{array} \right)$, or $ \left( \begin{array}{cc} ac+bc & c^2\\ (a-b)(a+b)&(a-b)c \end{array} \right)=\left( \begin{array}{cc} ac & 2ac\\ \frac{ac}{2}&ac \end{array} \right)$.
From $ac+bc=ac$ we get that $bc=0$, so $b=0$ or $c=0$. Clearly, $c \neq 0$ (because $S$ is invertible), so $b=0$. From $c^2=2ac$ we get that $c=2a$. In this case, we have $(a-b)(a+b)=\frac{ac}{2}$ and $(a-b)c=ac$. So $S=\left( \begin{array}{cc} a & 0\\ 0&2a \end{array} \right)$.
| {
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How many possible words of this type can be formed? We are making $10$ letter words using the letters $A,C,G,T$.
How many possible words are there of the form $A...AC...CG...GT...T$
This is where all of the $A's$ go before the all of the $C's$ and all of the $C's$ go before all of the $G's$ and all of the $G's$ go before all of the $T's$?
I know there are $4^{10}$ possible words comprised of the letters $A,C,G,T$ but I have no idea how to account for the overcounting in this bit.
| This problem is equivalent to finding the number of ways to partition $10$ into four parts, in which the order of the partitioning matters. For example, $10 = 2 + 3 + 3 + 2$ is not the same as $10 = 3 + 2 + 2 + 3$. The first number represents the number of A's, the second number represents the number of C's, and so on.
In order to then count the number of ways, think of $10$ dots and $3$ lines. The $3$ lines will divide the $10$ dots into four parts. So, we have a total of $10 + 3 = 13$ spaces to choose from, and we need to place 3 lines. Therefore, the answer will be ${13 \choose 3} = 286$.
Illustrations of $10$ dots and $3$ lines:
$$\cdot\cdot|\cdot\cdot\cdot|\cdot\cdot\cdot\cdot|\cdot$$
$$\cdot\cdot\cdot|\cdot|\cdot|\cdot\cdot\cdot\cdot\cdot$$
$$\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot|\cdot\cdot||$$
$$\cdot\cdot\cdot|\cdot\cdot\cdot|\cdot\cdot|\cdot\cdot$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate $7^{8^9}\mod 100$ I'm preparing myself for discrete math exam and here's one of the preparation problems:
Evaluate $$7^{8^9}\mod 100$$
Here's my solution:
$7^2\equiv49 \mod 100\implies (7^2)^2\equiv49^2=2401\equiv 1\mod 100$
So, as it turns out $7^4\equiv 1\mod 100$. It will be useful later. Now let's examine $8^9\mod 100$.
We have $512=8^3\equiv 12 \mod 100$, so $(8^3)^3\equiv 12^3 = 1728 \equiv 28 \mod 100$. So, as $8^9\equiv 28 \mod 100$, then $8^9=100x+28$ for some natural $x$. So:
$7^{8^9}=7^{100x+28}=7^{100x}7^{28}=(7^4)^{25x}(7^4)^7\equiv 1^{25x}1^7 \equiv 1 \mod 100$
Which means the final answer is that $7^{8^9}\mod 100=1$.
That's my solution, but I'm curious about other ones. Has anybody have any ideas?
| $7^2\equiv -1\mod 50$, hence $7^4\equiv 1\mod 100$.
Thus $\,7^{8^9}=\bigl(7^8\bigr)^{8^8}=\bigl((7^4)^2\bigr)^{8^8}\equiv 1\mod 100$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify the expression and find the minimum value I want to simplify the expression
\begin{equation}
A = \frac{\sqrt{1-\sqrt{1-x^2}}\Big[\sqrt{(1+x)^3} + \sqrt{(1-x)^3} \Big]}{2-\sqrt{1-\sqrt{1-x^2}}}
\end{equation}
and find the minimum value of $\displaystyle \sqrt{2}A+\frac{1}{x}$.
I can only simplify to this point
\begin{equation}
A = \frac{2x\Big( 2-\sqrt{1-x^2} \Big)}{2\sqrt{2}-\sqrt{1+x}+\sqrt{1-x}}.
\end{equation}
However, I am not sure if I can go any further. The next part of the question is to find the minimum value of $\displaystyle \sqrt{2}A+\frac{1}{x}$, which is obviously not so easy since I am not supposed to use differentiation to find optimal values.
| Try this: $u = \sqrt{1+x}, v = \sqrt{1-x} \to \sqrt{1-x^2} = uv, 2 = u^2+v^2$, and consider $f(u,v) = ...$
| {
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Let $X$ be a set of primes $p$ so that $5^{p^2}+1 \equiv 0 \pmod {p^2}$ Which of these sets is $X$ equal to? $5^{p^2}+1\equiv 0\pmod {p^2}$
$1.$ $\emptyset $
$2.$ {$3$}
$3.$ All primes of the form $4k+3$
$4.$ All primes except $2$ and $5$
$5.$ All primes
This one is pretty easy to get right through the process of elimination, which is how I solved it (it's $3$). What would be a more rigorous way of solving this?
| Clearly $p\neq 2$, because $5^{2^2}+1\equiv 1^{2^2}+1\equiv 2\not\equiv 0\pmod{\! 2^2}$.
$\ p\mid 5^{p^2}+1\,\Rightarrow\, p\mid 5^{2p^2}-1$ and by little Fermat $p\mid 5^{p-1}-1$.
See this lemma to get $\,p\mid 5^{(2p^2,\,p-1)}-1=5^2-1=2^3\cdot 3\,\Rightarrow\, p=3$
$5^{3^2}+1\equiv 125^3+1\equiv (-1)^3+1\equiv 0\pmod{3^2}$, so $\{3\}$ is the set of solutions.
Relevant lemma: $\,a^{p^{k}}\equiv a^{p^{k-1}}\pmod{p^k}$ (for all $k\ge 1$. see this question).
So another way of checking: $\,5^{3^2}+1\equiv 5^3+1\equiv 126\equiv 0\pmod{3^2}$.
Or like user225222 did it: $\,\color{#0b4}{5^{p^2}}\stackrel{\text{FLT}}\equiv 5^p\stackrel{\text{FLT}}\equiv 5\equiv \color{#0b4}{-1}\pmod{\! p}\,\Rightarrow\, p\mid 6$
| {
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Finding the area under $\frac{3}{2x+1}$ and $3x-2$ Find the area between these curves.
$$y=\dfrac{3}{2x+1},\qquad y=3x-2;\qquad x=2\quad \text{et} \quad y=0 $$
Indeed,
I calculate the integral of the blue function between $1$ and $2$. Then, I will calculate the area of the triangle between the yellow line and the x-axis on the interval $\dfrac{2}{3}$ and $1$, which gives me:
$$\int _{\frac{3}{2}}^1\left(3x-2\right)dx+\int _1^2\frac{3}{2x+1}dx=\frac{3}{2}\ln \left(\frac{5}{3}\right)-\frac{7}{8}$$
Am I right in the way how to find calculate the area and if that so that $\frac{3}{2}\ln \left(\frac{5}{3}\right)-\frac{7}{8}$ is it correct?
| As mentioned in the comments, you are almost correct.
But your first integral should start at $2/3$, not $3/2$. So you should have
$$\int _{\frac{2}{3}}^1\left(3x-2\right)dx+\int _1^2 \frac{3}{2x+1} dx = \frac{3}{2}\ln \left(\frac{5}{3}\right) + \frac{1}{6}.$$
As a sanity check, this makes much more sense for a few reasons. Firstly, $3x-2$ is positive in your picture and so it doesn't make sense for its integral contribution to be negative. Secondly, $3/2$ is contained in the interval $[1,2]$, which is where the other integral is accounting for the area already.
| {
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If $a_1,a_2,a_3$ are roots $x^3+7x^2-8x+3,$ find the polynomial with roots $a_1^2,a_2^2,a_3^2$ If $a_1,a_2,a_3$ are the roots of the cubic $x^3+7x^2-8x +3,$ find the cubic polynomial whose roots are:
$a_1^2,a_2^2,a_3^2$ and the polynomial whose roots are $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}.$ Not really sure where to go. Hints appreciated.
| Suppose $x^3+7x^2-8x+3=(x-a_1)(x-a_2)(x-a_3)$. Using difference of squares,
$$
\begin{align*}
(x^2-a_1^2)(x^2-a_2^2)(x^2-a_3^2)&=\left[(x-a_1)(x-a_2)(x-a_3)\right]\left[(x+a_1)(x+a_2)(x+a_3)\right]\\
&= \left[x^3+7x^2-8x+3\right]\left[x^3-7x^2-8x-3\right]\\
&=x^6-65 x^4+22 x^2-9.
\end{align*}
$$
Therefore $(x-a_1^2)(x-a_2^2)(x-a_3^2)=x^3-65x^2+22x-9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the non-linear system of equations For real $x,y,z>0$ solve the system of equation
\begin{cases}
\dfrac{1}{x}-3 y+4 z=5,\\
\dfrac{1}{y}-4 z+5 x=3,\\
\dfrac{1}{z}-5 x+3 y=4,
\end{cases}
It is easy to check out that
$$
x =\frac 1 5, y= \frac 1 3, z=\frac 1 4
$$
is a solution.
How I could prove that there is no any other real solutions?
Edit. By adding all equations we get
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=12.
$$
| Multiply the first equation by $5x$, the second by $3y$, the third by $4z$. Then add all three up. This gives you $25x + 9y + 16z = 12$. Now this is clearly a useful intermediate result. For example, you can substitute it into the original equations to eliminate one variable. This yields six new relations:
[1] $4/x -25x - 21y = 8$;
[2] $3/x + 25x + 28z = 27$
[3] $5/y - 9y - 36z = 3$
[4] $4/y + 9y + 45x = 24$
[5] $5/z + 16z + 24y = 32$
[6] $3/z - 16z - 40x = 0$
From eqs. $1$, $3$ and $6$ we obtain these bounds: $0 < x < \frac {2}{5}$; $0 < y < \frac {1}{3} \sqrt{5}$; $0 < z < \frac {1}{4} \sqrt{3}$.
Eliminating $y$ from eqs. $1$, $4$ [or alternatively $z$ from eqs. $2$, $6$] yields a fourth order equation for $x$. In the same way fourth order expressions for $y$ and $z$ can be derived.
| {
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"timestamp": "2023-03-29T00:00:00",
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Decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$, partial fraction braindead decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$
the way my teacher wants us to solve is by substitution values for x,
I set it up like this:
(after setting the variables to the common denominator and getting rid of the denominator in the original equation)
$x^2-2x+3= \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+4}$
1)Let $x=1$, after plugging in for $x$ I got $2=5B$, or $B=\frac{2}{5}$
2)let the $x=0$, I get $3=-4A+4B+D$, and if I substitute B and simplify I get $D=4A-\frac{7}{5}$. Furthermore, no matter what value I plug in I still end up with two variables and cant seem to find a way to eliminate one. It seems like I have to set it up as the triple system of equations but I just dont know how to apply it here. Would really appreciate if you guys could give me a hint on how to go about it.
| Hint : Start with $$\frac{x^2-2x+3}{(x-1)^2(x^2+4)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{CX+D}{x^2+4}$$
Multiply with $(x-1)^2(x^2+4)$ and then insert the special values.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometry Identity proving If $\sin(x-y) =\cos y$ prove that $\tan y = \frac{1+ \sin y}{\cos y} $.
Is there an error with the question? I don't seem to be able to get the answer. Should it be $\tan x$ instead of $\tan y = \frac{1+\sin y}{\cos y}$ ?
$$\tan x\times \cos y = 1 + \sin y $$
$$\frac{\sin x}{\cos x} \times \cos y = \frac{\sin x\times\cos y}{\cos x}$$
$$\frac{\sin x\times\cos y}{\cos x}= 1 + \sin y $$
$$\frac{\sin x\times\cos y}{\sin(x-y)}= 1 + \sin y $$
I am stuck at this step.
| HINT:
$$\cos y=\sin(x-y)=\sin x\cos y-\cos x\sin y$$
$$\iff\cos x\sin y=\cos y(\sin x-1)\implies\dfrac{\sin y}{\cos y}=\dfrac{\sin x-1}{\cos x}$$
Now $\cos^2x=(1-\sin x)(1+\sin x)\implies\dfrac{1-\sin x}{\cos x}=\dfrac{\cos x}{1+\sin x}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this entity with operators correct? Let define the operators $A = \frac{1}{\sqrt{2}}(x+\partial_x)$ and $B = \frac{1}{\sqrt{2}}(x-\partial_x)$. I am suppossed to check the identity $AB-BA=1$ but I cannot proof it. Is the identity correct?
My attempt:
$$AB = \frac{x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x-\partial_x\right)\right) + \frac{\partial_x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x-\partial_x\right)\right) = \frac{x^2}{2}-\frac{x}{2}\partial_x+\frac{1}{2}-\frac{1}{2}\partial_x^2,$$
$$BA = \frac{x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x+\partial_x\right)\right) - \frac{\partial_x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x+\partial_x\right)\right) = \frac{x^2}{2}+\frac{x}{2}\partial_x-\frac{1}{2}-\frac{1}{2}\partial_x^2.$$
So,
$$AB-BA = -x\partial_x+1.$$
Is this correct?
| The identity is correct. The problem is $\partial_x\cdot x$ is not equal to $1$. Think about it this way:
$$(\partial_x\cdot x)(f) =\partial_x(xf)=f+x\partial_x f$$
So the operator $\partial_x\cdot x=1+x\partial_x$. If you do this for both $AB$ and $BA$, you will get the required identity.
| {
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First order differential equation: did i solve this equation right So i'm trying to solve:
$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
$$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$
I think i've solved it and got
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i will show you how i get there but firstly...
This was part of another question which i've already solved
Show that if $y$ is the solution of the above Bernoulli differential
equation and $u = y^{1−n}$, then $u$ satisfies the linear differential
equation:
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
Applying the chain rule to $u = y^{1-n}$ we obtain that
\begin{align}
\frac{d u}{dx}(x)&= \frac{du}{dy}\cdot\frac{dy}{dx}\\
&= (1-n)y^{-n}\cdot\frac{dy}{dx}
\end{align}
Futhermore using the Bernoulli equation we have
$$
\frac{dy}{dx}=q(x)y^n-p(x)y
$$
and
\begin{align}
\frac{d u}{dx}&= (1-n)y^{-n}\cdot\frac{dy}{dx}\\
&=(1-n)y^{-n}\cdot q(x)y^n - (1-n)y^{-n}\cdot p(x) y\\
&=(1-n)q(x) -(1-n)p(x)y^{1-n}\\
&=(1-n)q(x) -(1-n)p(x)u
\end{align}
Hence U satisfies the equation
$$
\frac{du}{dx}+(1-n)p(x)u = (1-n)q(x)
$$
$$x^2\frac{dy}{dx} + 2xy = y^3$$
Divide both sides by $x^2$
$$\frac{dy}{dx} + \frac{2}{x}y = x^{-2} y^3$$
Consider
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
We know that
*
*n = 3
*1- n = 1-3 = -2
*p(x) = $ \frac{2}{x}$
*q(x) = $x^{-2}$
*u = $y^{1-3} = y^{-2}$
Subbing these in...
$$
\frac{du}{dx} + (-2)\frac{2}{x}u = (-2)x^{-2}
$$
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
So...
$$ \text{integrating factor} = e^{\int p(x) \, dx} $$
- p(x) dx = $-\frac{4}{x}$
$$ -4 \int \frac{1}{x} = -4log(x) = log (x^{-4}) $$
$$ \text{integrating factor} = e^{log (x^{-4})}= x^{-4} = \frac{1}{x^4}$$
So multiply this to the equation
$$\frac{1}{x^4}\frac{du}{dx} + \left(\frac{-4}{x^5} \right)u = \frac{-2}{x^6}$$
So we want to solve
$$ \frac{d}{dx}\frac{1}{x^4}u = \frac{-2}{x^6} $$
$$ \int \frac{d}{dx}\frac{1}{x^4}u = \int \frac{-2}{x^6} $$
$$ \frac{1}{x^4}u = -2\int \frac{1}{x^6} $$
$$ \frac{1}{x^4}u = -2\frac{1}{-5x^5} + c $$
$$ \frac{1}{x^4}u = \frac{2}{5x^5} + c $$
$$ \therefore u= \frac{2}{5x} + cx^{4} $$
is this fine? Or do i need to somehow equate this y or sub $u=y^{1-n}$
As $$u=y^{-2}$$
$$\frac{1}{y^2}= \frac{2}{5x} + cx^{4} $$
$$y^2= \frac{5x}{2} + \frac{1}{cx^{4}} $$
$$y= \sqrt{\frac{5x}{2} + \frac{1}{cx^{4}}} $$
| After correction, the solution to the initial differential equation is
$$y(x) = \pm \sqrt{ \frac{5x} {5Cx^5 + 2} }$$
Regarding the $u$ transformation: from the B.E. we know (1) $y' = q y^n - p y$. From the $u$ definition we know that (2) $u' = (1-n) y^{-n} y'$. Substitute $y'$ from (1) into (2) to recover the new ODE for $u$.
| {
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Here are two fractions, $\frac{2}{3}$,$\frac{7}{8}$, which of these fractions are closer to $\frac{3}{4}$? I've been throwing this question around my family. No one has a clue, therefore can someone help?
I'm pretty sure this will be easy to do
| The distance between real numbers $x$ and $y$ is $|x-y|$. So the distance from $\frac{2}{3}$ to $\frac{3}{4}$ is
\begin{align*}
\Big|\frac{2}{3}-\frac{3}{4}\Big|=\Big|\frac{8}{12}-\frac{9}{12}\Big|=\frac{1}{12}.
\end{align*}
And the distance from $\frac{7}{8}$ to $\frac{3}{4}$ is
\begin{align*}
\Big|\frac{7}{8}-\frac{3}{4}\Big|=\Big|\frac{7}{8}-\frac{6}{8}\Big|=\frac{1}{8}.
\end{align*}
Since $\frac{1}{12}<\frac{1}{8}$, then $\frac{2}{3}$ is closer to $\frac{3}{4}$ than $\frac{7}{8}$ is.
| {
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Solution to $y'=y^2-4$ I recognize this as a separable differential equation and receive the expression:
$\frac{dy}{y^2-4}=dx$
The issue comes about when evaluating the left hand side integral:
$\frac{dy}{y^2-4}$
I attempt to do this integral through partial fraction decomposition using the following logic:
$\frac{1}{(y+2)(y-2)} = \frac{A}{y+2}+\frac{B}{y-2}$
Therefore,
$1=Ay-2A+By+2B$.
Since the coefficients must be the same on both sides of the equation it follows that:
$0=A+B$ and $1=-2A+2B$.
Hence, $A=-B$, $B=\frac14$, $A=-\frac14$.
Thus the differential equation should be transformed into:
$-\frac{1}{4} \frac{dy}{y+2} + \frac14 \frac{dy}{y-2} = x+C$
Solving this should yield:
$-\frac14 \ln|y+2| + \frac14 \ln|y-2| = x+C$
which simplifies as:
$\ln(y-2)-\ln(y+2)=4(x+c)$
$\ln[(y-2)/(y+2)]=4(x+c)$
$(y-2)/(y+2)=\exp(4(x+c))$
$y-2=y*\exp(4(x+c)+2\exp(4(x+c))$
$y-y\exp(4(x+c))=2+2\exp(4(x+c))$
$y(1-\exp(4(x+c)))=2(1+\exp(4(x+c)))$
$y= 2(1+\exp(4(x+c)))/(1-\exp(4(x+c)))$
However, when done in Mathematica/Wolfram Alpha the result is given as (proof in the attached image)
$\frac14 \ln(2-y) -\frac14 \ln(2+y) = x + C$
and returns an answer of:
$y= 2(1-\exp(4(x+c)))/(1+\exp(4(x+c)))$.
Can anyone figure out where I have made an error? The only thing I can think of is something with evaluating the absolute values of the natural logarithms.
| $$
\begin{align}
x
&=\int\frac{\,\mathrm{d}y}{y^2-4}\\
&=\frac14\int\left(\frac1{y-2}-\frac1{y+2}\right)\,\mathrm{d}y\\
&=\frac14\log\left(\frac{y-2}{y+2}\right)+C
\end{align}
$$
Therefore,
$$
\frac{y-2}{y+2}=ke^{4x}
$$
or, solving for $y$,
$$
y=2\,\frac{1+ke^{4x}}{1-ke^{4x}}
$$
We get your form of the answer by letting $k\lt0$. It is hard to see this since you need to use a complex $c$ to get the equivalent answer. Check your answer in the original equation. You'll see that it works.
| {
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Rolling two dice, what is the probability of getting 6 on either of them, but not both? Rolling two dice, what is the probability of getting 6 on one of them, but not both?
| Total outcomes: $6^2 = 36$.
Allowed outcomes is the complementary of unallowed outcomes which can be split into "both dice < 6" ( case 1 ) "and" "not both dice = 6" ( case 2 ):
$6^2 - 5^2 - 1^2 = 36-25-1 = 10$. We can write this as a binary number matrix with the indices equal to the dice results:
$\left(\begin{array}{cccccc} 1&1&1&1&1&0\\1&1&1&1&1&0\\1&1&1&1&1&0\\1&1&1&1&1&0\\1&1&1&1&1&0\\0&0&0&0&0&2\end{array}\right)$ we see that the number of which are 0 is 10.
| {
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Prove bounds of a strictly increasing sequence using integrals to approximate For the strictly increasing sequence $ \ x_n = \frac1{1^2} + \frac1{2^2} + \frac1{3^2} +\cdots+\frac1{n^2},$ for $n\ge1$.
(a) Prove the sequence is bounded above by $2$; deduce that is has a limit $L\le2$.
(b) Prove that the "tail" sequence $\{x_n\}$, $n\ge N$, is bounded below by $3/2$ for some number $N$.
*Don't just calculate $N$ explicitly, use geometric reasoning and calculus.
What I've tried:
(a) I'm not sure how to set up the integral for this. I know I have to find an area greater then $x_n$, and show it converges to $2$, but none of the boundaries I've tried worked.
(b) Typing this out I realized it is completely wrong, but this is what I have. We can estimate a lower bound using $\int_2^n \frac1{x^2} \, dx$.
As $n\rightarrow \infty$, $\int_2^n \frac1{x^2}\,dx\rightarrow \frac12$. Since this estimation excludes n=1, where $x_n=1$, we can add $1$ to include that term and get a lower bound of $3/2$.
| For the upper bound, notice that
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \ldots < 1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \ldots = \\ 1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots = \frac{7}{4}$$
For the lower bound, notice that
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \ldots > 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{16} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \ldots = \\ 1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots = \frac{3}{2}$$
It was my intention to not write it down formally, I'm leaving that step to you.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Changing-sided dice probability problem. Suppose you roll a fair $6$-sided dice, and that the number you roll is $m$.
If $m=1$, stop. Otherwise, roll an $m$-sided dice. The number you roll is $n$. If $n=1$, stop. Otherwise roll an $n$-sided dice... etc.
What is the probability it will take exactly $x$ rolls to roll a 1?
So far I see a pattern, but I'm wondering if there's a better way of expressing these nasty sums, or if it's even possible?
$$P(1)=\frac{1}{6}$$
$$P(2)=\frac{1}{6}(\frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2})$$
$$P(x)=\underbrace{\frac{1}{6}\sum_{j_1=2}^{6}\left(\frac{1}{j_1}\sum_{j_2=2}^{j_1}\left[\frac{1}{j_2}\sum_{j_3=2}^{j_2}(\ldots)\right]\right)}_{x-1 \mbox{ sigmas}}$$
| This looks like a classic Markov Chain problem.
There are 6 states (6,5,4,3,2,1), and you start at 6 and end at 1. The transition matrix is:
$P=\left(\begin{array}{cccccc}
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6}\\
0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5}\\
0 & 0 & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\
0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\\
0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2}\\
0 & 0 & 0 & 0 & 0 & 1\\
\end{array} \right)$
So the probability that the state will be 1 after n terms can be read off using matrix multiplication (which is basically a neat way of organising your summations!)
$(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0\end{array}) \cdot P^n$
Actually, this will give you the probabilities that the state will be [6,5,4,3,2,1] after n turns.
Edit: to ensure that the final state is captured on exactly $n$ throws:
$P=\left(\begin{array}{ccccccc}
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0\\
0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0\\
0 & 0 & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & 0\\
0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0\\
0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & 1\\
\end{array} \right)$
Then figure out:
$(\begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0\end{array}) \cdot P^n$
The 6th number should be the probability that you're after (and the 7th number will be the cases where it reached 1 before $n$)
It can be confirmed that this spits out the 29/120 answer for n=2 at wolfram alpha here:
Wolfram Alpha Calculation (change the power if you want)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
An estimate for dyadic numbers I would like to prove that for some positive $\delta<1/2$ we have the following inequalities
$$ |\frac{1}{2^{n+1}} - \frac{1}{2^{m+1}}| \leqslant \delta\left( |\frac{1}{2^{n}} - \frac{1}{2^{m+1}}| + |\frac{1}{2^{m}} - \frac{1}{2^{n+1}}|\right)\qquad (m,n\in \mathbb{N})$$
$$ |\frac{1}{2^{n+1}}+i\frac{1}{2^{n+1}} - \frac{1}{2^{m+1}}| \leqslant \delta\left( |\frac{1}{2^{n}} + i\frac{1}{2^{n}}- \frac{1}{2^{m+1}}| + |\frac{1}{2^{m}} - \frac{1}{2^{n+1}}- i\frac{1}{2^{n+1}}|\right).$$
I tried to make $n$ a continuous variable and use calculus but failed with computations. Is there any way to see those inequalities?
I would be happy with an existential result for $\delta$.
| you can take $\delta=\frac{1}{3}$.
for any $a,b$ we have
$$\left|a-b\right|=\frac{1}{3}\left|3a-3b\right|=\frac{1}{3}\left|\left(2a-b\right)+\left(a-2b\right)\right|$$
now using triangle inequality and noting that $\left|a-2b\right|=\left|2b-a\right|$ we get
$$\le\frac{1}{3}\left(\left|2a-b\right|+\left|2b-a\right|\right)$$
now set $b=\frac{1}{2^{m+1}}$ whereas a in the first is $\frac{1}{2^{n+1}}$ and in the second $a=\frac{1+i}{2^{n+1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos x+\sin x-1=0$ How does one solve this equation?
$$\cos {x}+\sin {x}-1=0$$
I have no idea how to start it.
Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?
Thanks in advance!
| As $\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(\cos x+\sin x\right)$ by the angle addition formula we find that:
\begin{equation}
\begin{aligned}
\cos x+\sin x-1&=0\\
\implies\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)&=1\\
\implies\sin\left(x+\frac{\pi}{4}\right)&=\frac{\sqrt{2}}{2}\\
\implies x+\frac{\pi}{4}&=\frac{\pi}{4}+2\pi n,\frac{3\pi}{4}+2\pi n\\
\implies x&=2\pi n,\frac{\pi}{2}+2\pi n
\end{aligned}
\end{equation}
for $n\in\mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
$\sin(2x)-\cos(2x)-\sin(x)+\cos(x)=0$ I would like to share a trigonometry question here. Wonder is there another way to solve it or not.
$\sin(2x)-\cos(2x)-\sin(x)+\cos(x)=0$
$2\cos(x)\sin(x)-(1-2\sin^2(x))-\sin(x)+\cos(x)=0$
$2\cos(x)\sin(x)-1+2\sin^2(x)-\sin(x)+\cos(x)=0$
$2\sin^2(x)+2\cos(x)\sin(x)-\sin(x)+\cos(x)-1=0$
$(2\sin(x)+1)(\sin(x)+\cos(x)-1)=0$
So, we have $$2\sin(x)+1=0 \space\text{and}\sin(x)+\cos(x)-1=0$$
$2\sin(x)+1=0$
$\sin(x)=-\frac{1}{2}$
$x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n$
$\sin(x)+\cos(x)-1=0$ which answered in my yesterday post $\cos x+\sin x=1$
and get $x=2n\pi,2n\pi+\frac{\pi}{2}$
Combine all the solutions, $$x=2\pi n,x=\frac{7\pi}{6}+2\pi n,x=2\pi n+\frac{\pi}{2}, x=\frac{11\pi}{6}+2\pi n$$
| If $\sin2A-\sin2B=\cos2A-\cos2B$
using Prosthaphaeresis Formulas, $$2\sin(A-B)\cos(A+B)=-2\sin(A-B)\sin(A+B)$$
$$2\sin(A-B)[\cos(A+B)+\sin(A+B)]=0$$
$\sin(A-B)=0\implies A-B=n\pi$ where $n$ is any integer
$\cos(A+B)+\sin(A+B)=0\iff\tan(A+B)=-1=\tan\left(-\dfrac\pi4\right)$
$A+B=m\pi-\dfrac\pi4$ where $m$ is any integer
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Evalute big determinant Today in exam I tried to evaluate this determinant but failed, only somehow "guessed" the answer I got here. Now in home I've managed to find something intuitive, just want to know whether the approach is correct, and is there more faster way exist. Given determinant
$$\det\begin{vmatrix}
1 & 2 & 3 & ... & n-2 & n-1 & n\\
2 & 3 & 4 & ... & n-1 & n & n\\
3 & 4 & 5 & ... & n & n & n \\
. & . & . & . & . & . &. \\
n & n & n & ... & n & n & n
\end{vmatrix}$$
First thing I did, was rearranging rows. I remember from another problem, where I used to evaluate determinant of matrix of this kind $\det\begin{vmatrix}
0 & 0 ... & 0 & 1\\
0 & 0 ... & 1 & 0\\
0 & 0 ... & 0 & 0\\
. & . & . & . \\
1 & 0 ... & 0 & 0
\end{vmatrix}$ is $(-1)^{\frac{n(n-1)}{2}}*\det\begin{vmatrix}
1 & 0 ... & 0 & 0\\
0 & 1 ... & 0 & 0\\
0 & 0 ... & 0 & 0\\
. & . & . & . \\
0 & 0 ... & 0 & 1
\end{vmatrix}=(-1)^{\frac{n(n-1)}{2}}$.
So it becomes
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n & n & ... & n & n & n \\
. & . & . & . & . & . &. \\
3 & 4 & 5 & ... & n & n & n \\
2 & 3 & 4 & ... & n-1 & n & n\\
1 & 2 & 3 & ... & n-2 & n-1 & n\\
\end{vmatrix}$$
And then I transposed it
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n-1 & n-2 & ... & 3 & 2 & 1 \\
. & . & . & . & . & . &. \\
n & n & n & ... & n & n-1 & n-2 \\
n & n & n & ... & n & n & n-1\\
n & n & n & ... & n & n & n\\
\end{vmatrix}$$
and tried to subtract first row from all.
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n-1 & n-2 & ... & 3 & 2 & 1 \\
0 & 1 & 1 & ... & 1 & 1 & 1 \\
0 & 1 & 2 & ... & 2 & 2 & 2\\
. & . & . & . & . & . &. \\
0 & 1 & 2 & ... & n-3 & n-2 & n-1\\
\end{vmatrix}$$
next step is subtracting second row from others below.
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n-1 & n-2 & ... & 3 & 2 & 1 \\
0 & 1 & 1 & ... & 1 & 1 & 1 \\
0 & 0 & 1 & ... & 1 & 1 & 1\\
. & . & . & . & . & . &. \\
0 & 0 & 1 & ... & n-4 & n-3 & n-2\\
\end{vmatrix}$$
doing this for finite n we'll get
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n-1 & n-2 & ... & 3 & 2 & 1 \\
0 & 1 & 1 & ... & 1 & 1 & 1 \\
0 & 0 & 1 & ... & 1 & 1 & 1\\
. & . & . & . & . & . &. \\
0 & 0 & 0 & ... & 0 & 0 & 1\\
\end{vmatrix}$$
So my final solution is $n*(-1)^{\frac{n(n-1)}{2}}$. Do you see any mistakes? And maybe there's more easier approach? thanks.
| First we can subtract from each row the following row
$$D=
\begin{vmatrix}
1 & 2 & 3 & \ldots & n-2 & n-1 & n\\
2 & 3 & 4 & \ldots & n-1 & n & n\\
3 & 4 & 5 & \ldots & n & n & n \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
n & n & n & \ldots & n & n & n
\end{vmatrix}
=
\begin{vmatrix}
-1&-1 &-1 & \ldots &-1 &-1 & 0\\
-1&-1 &-1 & \ldots &-1 & 0 & 0\\
-1&-1 &-1 & \ldots & 0 & 0 & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
n & n & n & \ldots & n & n & n
\end{vmatrix}$$
We can "pull out" $n$ from the last row, and then add the last row to all rows above it (to cancel out all $-1$'s)
$$D=n
\begin{vmatrix}
-1&-1 &-1 & \ldots &-1 &-1 & 0\\
-1&-1 &-1 & \ldots &-1 & 0 & 0\\
-1&-1 &-1 & \ldots & 0 & 0 & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 1 & 1 & \ldots & 1 & 1 & 1
\end{vmatrix}=
n
\begin{vmatrix}
0& 0 & 0 & \ldots & 0 & 0 & 1\\
0& 0 & 0 & \ldots & 0 & 1 & 1\\
0& 0 & 0 & \ldots & 1 & 1 & 1\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 1 & 1 & \ldots & 1 & 1 & 1
\end{vmatrix}
$$
Now we only need to subtract each row from the row following it to get
$$
D=
n
\begin{vmatrix}
0& 0 & 0 & \ldots & 0 & 0 & 1\\
0& 0 & 0 & \ldots & 0 & 1 & 1\\
0& 0 & 0 & \ldots & 1 & 1 & 1\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 1 & 1 & \ldots & 1 & 1 & 1
\end{vmatrix}=
n
\begin{vmatrix}
0& 0 & 0 & \ldots & 0 & 0 & 1\\
0& 0 & 0 & \ldots & 0 & 1 & 0\\
0& 0 & 0 & \ldots & 1 & 0 & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 0 & 0 & \ldots & 0 & 0 & 0
\end{vmatrix}=n(-1)^{\frac{n(n-1)}2}$$
(We got the matrix for which you already know the determinant. It can be obtained, for example, using several swaps of neighbouring rows.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Double integral over trapezoid Compute $$\iint_D \frac{1}{(1+(x+2y)^2)^2} \,dx\,dy$$ where $D$ is given by $x \geq 0 , \, y \geq 0, \, 1 \leq x+2y \leq 2 \\$.
I am supposed to solve it with the help of contour lines. By drawing $D$ we get a trapezoid with the corners $(1,0), \, (2,0), \, (0, 1/2), \, (0,1)$. A contour line $\gamma$ to $f(x,y) = x+2y$ would be a line segment with end points $(\lambda, \frac{\lambda}{2})$. Consider the area of a trapezoid bounded by the line segment and the line segment with corners $(0, \frac{1}{2}), \, (1,0)$:
$$ A(\lambda) = \frac{1}{2}(\sqrt{(1/2)^2+1}+\sqrt{\lambda + (\lambda /2)^2)})\sqrt{(\lambda-1)^2+(\lambda/2-1/2)^2} = \frac{1}{2} \sqrt{5/4} (\lambda +1)(\lambda -1)\sqrt{5/4}=\frac{5}{8} (\lambda^2-1) $$
where, on the right side of the first equality, we are taking the mean of the sides of the trapezoid times the height.
The derivative is $A'(\lambda) = \frac{5}{4}\lambda$. The double integral can be computed as a single integral: $$\int_{\lambda=1}^{\lambda=2} \frac{1}{(1+\lambda^2)^2} \frac{5}{4} \lambda \, d \lambda =\cdots=3/16$$
This answer is wrong. The answer sheet says it is $3/40$. What have I done wrong?
| The height of the trapezoid is incorrect. Consider the line $y=2x$ which is perpendicular to the contour lines. This line intersects the contour $f(x,y) = \lambda$ at $\left(\frac{\lambda}{5} , \frac{2\lambda}{5}\right)$. So, the height is given by
$$\sqrt{\left(\frac{\lambda}{5} - \frac{1}{5}\right)^2 + \left( \frac{2\lambda}{5} - \frac{2}{5}\right)^2} = \frac{1}{\sqrt{5}} (\lambda - 1).$$
Using this instead gives $A(\lambda) = \frac{1}{4} \left(\lambda^2-1\right)$ and so $A'(\lambda) = \frac{\lambda}{2}$. Hence,
$$\int_{1}^{2} \frac{1}{\left(1+\lambda^2\right)^2} \frac{\lambda}{2} \: d \lambda = \frac{3}{40}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Is there a proof show that : $\cos(z)$ and $\sin(z)$ are images of unbounded functions? if we knew that :cos and sin are bounded function $\mathbb{R}$ for any real
number $x$ .
let $z $ be a complex variable , Is there a proof show that :
$\cos(z)$ and $\sin(z)$ are images of unbounded functions ?
Any kind of help is appreciated.
| In the spirit of giving an explicit computation accessible to good high-school students:
If one accepts that
$$
\left.
\begin{gathered}
\exp(iz) = \cos z + i\sin z, \\
\exp(z + w) = \exp(z) \exp(w),
\end{gathered}\right\}\quad\text{for all complex $z$ and $w$,}
\tag{1}
$$
then unboundedness of both $\cos$ and $\sin$ along the imaginary axis follows from Yves Daoust's comment.
More generally, the real and imaginary parts of the circular functions are given by:
$$
\left.
\begin{aligned}
\cos(x + iy) &= \cos x \cosh y - i\sin x \sinh y, \\
\sin(x + iy) &= \sin x \cosh y + i\cos x \sinh y.
\end{aligned}\right\}\quad\text{for all real $x$, $y$.}
$$
Since the hyperbolic functions $\cosh$ and $\sinh$ are unbounded, $\cos$ and $\sin$ are also unbounded.
Proof: By Euler's formula,
$$
\left.
\begin{aligned}
\exp(iz) &= \cos z + i\sin z, \\
\exp(-iz) &= \cos z - i\sin z,
\end{aligned}\right\}\quad\text{for all complex $z$.}
\tag{2}
$$
Adding and subtracting, we find
$$
\left.
\begin{aligned}
\cos z &= \frac{\exp(iz) + \exp(-iz)}{2}, \\
\sin z &= \frac{\exp(iz) - \exp(-iz)}{2i},
\end{aligned}\right\}\quad\quad\text{for all complex $z$.}
\tag{3}
$$
Setting $z = x + iy$ with $x$ and $y$ real in (3), so that $iz = ix - y$ and $-iz = -ix + y$, the law of exponents and Euler's formula give
\begin{align*}
\cos(x + iy) &= \tfrac{1}{2} \bigl[\exp(ix - y) + \exp(-ix + y)\bigr] \\
&= \tfrac{1}{2}\bigl[e^{-y}(\cos x + i\sin x) + e^{y}(\cos x - i\sin x)\bigr] \\
&= \cos x \bigl[\tfrac{1}{2}(e^{y} + e^{-y})\bigr] - i\sin x \bigl[\tfrac{1}{2}(e^{y} - e^{-y})\bigr] \\
&= \cos x \cosh y - i\sin x \sinh y,
\end{align*}
and
\begin{align*}
\sin(x + iy) &= \tfrac{1}{2i} \bigl[\exp(ix - y) - \exp(-ix + y)] \\
&= -\tfrac{i}{2} \bigl[e^{-y}(\cos x + i\sin x) - e^{y}(\cos x - i\sin x)\bigr] \\
&= \sin x \bigl[\tfrac{1}{2}(e^{y} + e^{-y})\bigr] + i\cos x \bigl[\tfrac{1}{2}(e^{y} - e^{-y})\bigr] \\
&= \sin x \cosh y + i\cos x \sinh y.
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\rightarrow \infty }\frac{2s}{s}=2$.
Is that ok?
Thanks.
| With the Euler-MacLaurin formula, we have
numerator = $\ln(n)+O(1)$ and denominator = $\frac12 \ln(2n+1) + O(1)$, so the limit is
$\lim_{n \to \infty}\frac{\ln(n)+O(1)}{\frac12 \ln(2n+1) + O(1)}= 2.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
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Prove that $\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c$ How to prove that
\begin{equation*}\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c,\ where \ a,b,c>0\end{equation*}
I tried the following:
\begin{equation*}abc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge a+b+c\end{equation*}
Using Chebyshev's inequality
\begin{equation*}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\le3(\frac{1}{a}\frac{1}{a}+\frac{1}{b}\frac{1}{b}+\frac{1}{c}\frac{1}{c})\end{equation*}
from first inequality follows
\begin{equation*}\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2abc\ge a+b+c\end{equation*}
equivalent to
\begin{equation*}\frac{abc}{3(a+b+c)}\ge (\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}
and by amplifying both members by 9
\begin{equation*}\frac{3abc}{a+b+c}\ge (\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}
now using mean inequality
\begin{equation*}\sqrt[3]{abc}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\end{equation*}
the inequality in question becomes
\begin{equation*}3abc\ge (a+b+c)(\sqrt[3]{abc})^2\end{equation*}
which yields
\begin{equation*}3\sqrt[3]{abc}\ge a+b+c\end{equation*}
not what I wanted.
| Multiplying both sides of your inequality with $abc>0$, you get equivalently that:
$ \displaystyle (ab)^2 + (bc)^2 + (ca)^2 \geq abc (a+b+c) $
This holds by the basic inequality $ \displaystyle x^2 + y^2 + z^2 \geq xy +yz+ xz $, which holds for all $x,y,z $ real.
edit: The basic inequality holds for all real, thank's to user26486, for pointing this out.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
| Why quadratic? Linear is sufficient.
If $x > 1$, then $(x-1)$ is positive so we can multiply by that to get: $2(x-1) \lt x \le 3(x-1)$. Left-hand gives us $x < 2$ and right-hand gives us $x \ge \frac32$. Thus: $\frac32 \le x < 2$.
If $x < 1$, then we have to flip the signs and get $2(x-1) > x \ge 3(x-1)$. But there, the left-hand gives us $x > 2$, so there's no solution here.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.
What are all integral solutions of this equation?
I can only find $x=y=z=1$.
| Using identity: $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$, we have:
$3^3 = 3 + 3(x+y)(y+z)(z+x) \Rightarrow (x+y)(y+z)(z+x) = 8$. From this you should be able to deduce the answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finer asymptotic estimate of an integral I'm studying the asymptotic behaviour for large $n\in \mathbb N$ of $\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}$
Using the substitution $u=(n+1)\ln(t)$, $$\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}=\frac{1}{n+1}\int_0^\infty\exp(\frac{u}{n+1})\frac{1}{1+\exp(u)}du$$
And, using dominated convergence, one has $\displaystyle \int_0^\infty\exp(\frac{u}{n+1})\frac{1}{1+\exp(u)}du=\ln(2)+o(1)$, so that $$\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}=\frac{\ln(2)}{n}+o(\frac{1}{n})$$
I'm looking to refine that estimate, that is to say get a sharper estimate of $\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}-\frac{\ln(2)}{n}$.
It amounts to finding an estimate of $$\int_0^\infty \frac{1}{1+e^u}(\exp(\frac{u}{n+1})-1)du$$ but that is harder.
| As Lucian remarked,
$$ \int_0^\infty \dfrac{dt}{1+t^{n+1}} = \dfrac{\pi}{n+1} \csc\left(\dfrac{\pi}{n+1}\right) $$
That is asymptotically
$$ 1+{\frac {{\pi }^{2}}{6{n}^{2}}}-{\frac {{\pi }^{2}}{3{n}^{3}}
}+{\frac {{\pi }^{2} \left( 7\,{\pi }^{2}+180 \right) }{360\,{n}^{4}}}
+ \ldots$$
Now we have to subtract the integral from $0$ to $1$. For $0 < t < 1$,
$$ \eqalign{\int_0^1 \dfrac{dt}{1+t^{n+1}} &= \int_0^1 \sum_{k=0}^\infty (-1)^k t^{k(n+1)} \; dt\cr
&= \sum_{k=0}^\infty \dfrac{(-1)^k}{1+k(n+1)}\cr
&= \dfrac{\Psi\left(\dfrac{n+2}{2(n+1)}\right) - \Psi\left(\dfrac{1}{2(n+1)}\right)}{2(n+1)}\cr
&= 1-{\frac {\ln \left( 2 \right) }{n}}+{\frac {{\pi }^{2}+12\ln
\left( 2 \right) }{12{n}^{2}}}+{\frac {-2\,{\pi }^{2}-9\,\zeta
\left( 3 \right) -12\;\ln \left( 2 \right) }{12{n}^{3}}}+\dfrac{7\,{\pi }^{4}+180\,{\pi }^{2}+720\,\ln \left( 2 \right) +1620\,\zeta
\left( 3 \right)
}{720\;{n}^{4}}+\ldots
}$$
So the final result is
$$ {\frac {\ln \left( 2 \right) }{n}}+{\frac {{\pi }^{2}-12\,\ln
\left( 2 \right) }{12{n}^{2}}}+{\frac {-2\,{\pi }^{2}+12\,\ln
\left( 2 \right) +9\,\zeta \left( 3 \right) }{12{n}^{3}}}+{\frac {7\,
{\pi }^{4}+180\,{\pi }^{2}-720\,\ln \left( 2 \right) -1620\,\zeta
\left( 3 \right) }{720\,{n}^{4}}}+\ldots
$$
| {
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For which values of $a$ does $||x^2-16|-7|-2a=0$ have 3 roots? Here is the equation:
$$||x^2-16|-7|-2a=0$$
Please help solve this.
For which $a$ does the equation have 3 roots?
| Api, first, use the definition of absolute value.
$||x^2-16|-7|-2a=0$ implies that two cases:
1) $|x^2-16|-7=2a$
This equation gives us $a=\frac{x^2-23}{2} $ and $a=\frac{x^2-9}{2}$.
2) $|x^2-16|-7=-2a$
Again, this equation gives us $a=\frac{23-x^2}{2} $ and $a=\frac{x^2+9}{2}$.
We should have the following this case.
$x=0$ and $x=\pm x_0$.
Hence, if $x=0$, then the only possible case is $a=\frac{9}{2}$.
Why?
if put $0$ in the first equation in the second case, then you get $a=\frac{23}{2}$ and it implies that $x\geq 4$ and a contradiction. The rest is similar.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{1}{{9\choose r}} -\frac{1}{{10\choose r}} = \frac{11}{6{11\choose r}}$. Is there a way to find $r$ without using algebra?
$$\frac{1}{\dbinom 9r} -\frac{1}{{\dbinom{10}r}} = \frac{11}{6\times \dbinom{11}r}$$
I guess directly applying algebra for this problem would be enough. But are there any simpler and prettier approaches to finding $r$ ?
| Well, you can use Pascal's rule, or whatever it's called:
$$ \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} $$
to rewrite the left-hand side as
$$ \frac{\binom{10}{r}-\binom{9}{r}}{\binom{10}{r}\binom{9}{r}} = \frac{\binom{9}{r-1}}{\binom{10}{r}\binom{9}{r}}. $$
Multiplying up gives
$$ \frac{11}{6}\binom{10}{r}\binom{9}{r} = \binom{9}{r-1}\binom{11}{r} $$
Our next trick is another well-known property of binomial coefficients, namely
$$ \binom{n}{k+1}=\frac{n-k}{k+1}\binom{n}{k} $$
Use this to rewrite
$$ \binom{9}{r} = \frac{10-r}{r}\binom{9}{r-1}, $$
so
$$ \frac{11}{6}\binom{10}{r}\binom{9}{r} = 11\binom{10}{r}\frac{10-r}{r}\binom{9}{r-1} = \binom{9}{r-1}\binom{11}{r} \\
\frac{11}{6}\binom{10}{r}\frac{10-r}{r} = \binom{11}{r} $$
Now we have another binomial identity,
$$ \binom{n+1}{k}=\frac{n+1}{n+1-k}\binom{n}{k}, $$
so
$$ \binom{11}{6r} = \frac{11}{11-r}\binom{10}{r}, $$
reducing the equation to
$$ \frac{1}{11-r} = \frac{10-r}{6r}, $$
which is the quadratic you get normally. This is really not inherently different, but does avoid getting factorials out.
| {
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"timestamp": "2023-03-29T00:00:00",
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Coefficient of operator and how to do it This question stems from this
$$
\frac{1}{x+z}- \frac{1}{x} = \sum_{k=0}^\infty \frac{z^k}{k!}\frac{d^k}{dx^k}[\frac{1}{x}]
$$
Now, i need to find the Bell Polynomial of $\frac{1}{x}$,
$$
[z^n]\left(\frac{(-z)^k}{x^k(x+z)^k}\right) = [z^n]\sum_{n \geq k} Y^{\Delta}(n,k,x) z^n = \frac{k!}{n!} B_{n,k}(\frac{d}{dx}[\frac{1}{x}], \cdots,\frac{d^{n-k+1}}{dx^{n-k+1}}[\frac{1}{x}])
$$
$$
\frac{(-1)^k}{x^k} [z^n] z^k (x+z)^k = \frac{(-1)^k}{x^{k}} [z^n] (\frac{x}{z}+1)^{-k}
$$
$$
\frac{(-1)^k}{x^{k}} [z^n] (\frac{x}{z}+1)^{-k} = \frac{(-1)^k}{x^{k}} [z^n] \sum_{j=0}^\infty {-k \choose j} (\frac{x}{z})^j
$$
How do i take the coefficient of this power series?
|
Here's the calculation of $[z^n]$:
\begin{align*}
[z^n]\frac{(-z)^k}{x^k(x+z)^k}&=[z^n]\frac{(-z)^k}{x^{2k}\left(1+\frac{z}{x}\right)^k}\\
&=\frac{(-1)^k}{x^{2k}}[z^n]\frac{z^k}{\left(1+\frac{z}{x}\right)^k}\\
&=\frac{(-1)^k}{x^{2k}}[z^{n-k}]\frac{1}{\left(1+\frac{z}{k}\right)^k}\tag{1}\\
&=\frac{(-1)^k}{x^{2k}}[z^{n-k}]\sum_{j=0}^{\infty}\binom{-k}{j}\left(\frac{z}{x}\right)^j\tag{2}\\
&=\frac{(-1)^k}{x^{2k}}[z^{n-k}]\sum_{j=0}^{\infty}\binom{k+j-1}{k-1}(-1)^j\left(\frac{z}{x}\right)^j\tag{3}\\
&=\frac{(-1)^k}{x^{2k}}\binom{n-1}{k-1}(-1)^{n-k}\left(\frac{1}{x}\right)^{n-k}\\
&=\frac{(-1)^n}{x^{n+k}}\binom{n-1}{k-1}
\end{align*}
Comment:
*
*In (1) we use the relationship $[z^n]z^k=[z^{n-k}]$ of the coefficient of operator
*In (2) we apply the binomial series expansion $(1+t)^\alpha=\sum_{j=0}^{\infty}\binom{\alpha}{j}t^j$ with $\alpha=-k$
*In (3) we apply the binomial coefficient identity $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $ \lim\limits_{x\to\infty}y_{n}=\sqrt{x}$ if $y_{n}=\frac{1}{2}\left(y_{n-1}+\frac{x}{y_{n-1}}\right),n\in \mathbb{N},x>0,y_{0}>0$ Can someone say how to solve this problem? In solution, it says that it stars with $$\frac{y_{n}-\sqrt{x}}{y_{n}+\sqrt{x}}=\left(\frac{y_{n-1}-\sqrt{x}}{y_{n-1}+\sqrt{x}}\right)^2,n\ge 1$$
How to get to this formula?
| The desired formula comes directly from the definitions.
\begin{align}
\frac{y_{n} - \sqrt{x}}{y_{n} + \sqrt{x}} &= \dfrac{\dfrac{1}{2}\left(y_{n - 1} + \dfrac{x}{y_{n - 1}}\right) - \sqrt{x}}{\dfrac{1}{2}\left(y_{n - 1} + \dfrac{x}{y_{n - 1}}\right) + \sqrt{x}}\notag\\
&= \dfrac{y_{n - 1}^{2} + x - 2y_{n - 1}\sqrt{x}}{y_{n - 1}^{2} + x + 2y_{n - 1}\sqrt{x}}\notag\\
&= \left(\frac{y_{n - 1} - \sqrt{x}}{y_{n - 1} + \sqrt{x}}\right)^{2}\tag{1}
\end{align}
Using this recursive formula you get $$\frac{y_{n} - \sqrt{x}}{y_{n} + \sqrt{x}} = \left(\frac{y_{0} - \sqrt{x}}{y_{0} + \sqrt{x}}\right)^{2^{n}} = k^{2^{n}}$$ where the constant $k = (y_{0} - \sqrt{x})/(y_{0} + \sqrt{x}) < 1$. And since $2^{n} > n$ it follows that $k^{2^{n}} < k^{n}$ therefore we have $$0 < \frac{y_{n} - \sqrt{x}}{y_{n} + \sqrt{x}} < k^{n}\tag{2}$$ for all $n$. Noting that $k^{n} \to 0$ as $n \to \infty$ we can see that $y_{n} \to \sqrt{x}$.
There are two fine points to note in the whole argument. We have assumed that $y_{0} > \sqrt{x}$ which makes $k$ positive. If this was not the case the relation $$\frac{y_{1} - \sqrt{x}}{y_{1} + \sqrt{x}} = \left(\frac{y_{0} - \sqrt{x}}{y_{0} + \sqrt{x}}\right)^{2}$$ ensures that $y_{1} > \sqrt{x}$ and hence we can apply the recursion $(1)$ only till $y_{1}$ and continue the argument.
Another point to note is that the part $y_{n} + \sqrt{x}$ is denominator is bounded. This is essential to get the conclusion $y_{n} \to \sqrt{x}$ from equation $(2)$. To establish this we need to know that $y_{n - 1} > \sqrt{x}$ for all $n \geq 2$ and hence $$y_{n} - y_{n - 1} = \frac{x - y_{n - 1}^{2}}{2y_{n - 1}} < 0$$ so that $y_{n}$ is decreasing and hence $0 < \sqrt{x} < y_{n} < y_{1}$ for all $n$. This shows that $y_{n} + \sqrt{x}$ lies between $\sqrt{x}$ and $y_{0} + \sqrt{x}$ and is hence bounded.
The second point above brings us to a much simpler proof. Sequence $y_{n}$ is decreasing and $y_{n} > \sqrt{x}$, it follows that $y_{n}$ tends to some limit $L \geq \sqrt{x} > 0$. And $y_{n - 1} \to L$ also. The equation $$y_{n} = \frac{1}{2}\left(y_{n - 1} + \frac{x}{y_{n - 1}}\right)$$ now tells us that $$L = \frac{1}{2}\left(L + \frac{x}{L}\right)$$ or $L^{2} = x$. Since $L > 0$ it follows that $L = \sqrt{x}$. This proof does not need the use of complicated formula $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Need a hint to evaluate $\lim_{x \to 0} {\sin(x)+\sin(3x)+\sin(5x) \over \tan(2x)+\tan(4x)+\tan(6x)}$ I know that $\sin A + \sin B + \sin C = 4\cos({A \over 2})\cos({B \over 2})\cos({C \over 2})$ when $A+B+C=\pi$. If ${x \to 0}$ then I have a half circle, right?
If it is right then I have
$\tan(2x) + \tan(4x) + \tan(6x)=\tan(2x)\tan(4x)\tan(6x)$.
I got stuck at
$${4\cos({x \over 2})\cos({3x \over 2})\cos({5x \over 2}) \over \tan(2x)\tan(4x)\tan(6x)}$$
Is this correct? Do you have a hint to help me?
| This is a much simpler take on this question and it uses the following result $$\lim_{x\to 0}\sin x = 0\tag{1}$$ from which we get $$\lim_{x \to 0}\cos x = 1\tag{2}$$ using the relation $\sin^{2}x + \cos^{2}x = 1$. Further dividing $(1)$ by $(2)$ we get $$\lim_{x \to 0}\tan x = 0\tag{3}$$ We can see that the numerator $f(x)$ of the given expression can be simplified as
\begin{align}
f(x) &= \sin x + \sin 3x + \sin 5x\notag\\
&= \sin x + \sin 5x + \sin 3x\notag\\
&= 2\sin 3x\cos 2x + \sin 3x\notag\\
&= \sin 3x(2\cos 2x + 1)\notag
\end{align}
and the denominator $g(x)$ can be simplified as
\begin{align}
g(x) &= \tan 2x + \tan 4x + \tan 6x\notag\\
&= \tan 6x(1 - \tan 2x\tan 4x) + \tan 6x\notag\\
&= \tan 6x(2 - \tan 2x\tan 4x)\notag\\
\end{align} In the above we have used the formula $$\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a\tan b}$$ which can be rewritten as $$\tan a + \tan b = \tan(a + b)(1 - \tan a\tan b)$$
Hence the desired limit $L$ can be calculated as follows
\begin{align}
L &= \lim_{x \to 0}\frac{f(x)}{g(x)}\notag\\
&= \lim_{x \to 0}\frac{\sin 3x(2\cos 2x + 1)}{\tan 6x(2 - \tan 2x\tan 4x)}\notag\\
&= \lim_{x \to 0}\frac{\sin 3x}{\tan 6x}\cdot\lim_{x \to 0}\frac{2\cos 2x + 1}{2 - \tan 2x\tan 4x}\notag\\
&= \lim_{x \to 0}\frac{\sin 3x (1 - \tan^{2}3x)}{2\tan 3x}\cdot\frac{2\cdot 1 + 1}{2 - 0\cdot 0}\notag\\
&= \frac{3}{4}\lim_{x \to 0}\frac{\sin 3x}{\tan 3x}\cdot\lim_{x \to 0}(1 - \tan^{2}3x)\notag\\
&= \frac{3}{4}\lim_{x \to 0}\cos 3x\cdot (1 - 0\cdot 0)\notag\\
&= \frac{3}{4}\cdot 1 = \frac{3}{4}\notag
\end{align}
The calculation above has been presented in somewhat unnecessarily detailed fashion to highlight the fact that no other limits other than $(1), (2), (3)$ are used in the process and as mentioned before out of these the result $(1)$ is the fundamental one. Thus there is no need to invoke the slightly more powerful result $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying sum equation. (Solving max integer encoded by n bits) Probably a lack of understanding of basic algebra. I can't get my head around why this sum to N equation simplifies to this much simpler form.
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2
$$
Background
To give you some background I am trying to derive $MaxInt(n) = 2^n-1$ which describes that the maximum integer which can be created using the two's complement where $n$ is the number of bits the integer is encoded by.
How two's complement encodes numbers with 4 bits is explained by the images below:
Therefore:
$$
MaxInt(n) = \sum_{i=0}^{n-2} 2^i = (2^0 + 2^1 + ... + 2^{n-3} + 2^{n-2} )
$$
Maybe there is a way of integrating this or simplifying it but I figured that this is a similar problem to sum to N where
$$
\frac{T(n) + T(n)}{2} = T(n) = \sum_{i=1}^{n} n-i+1 = \sum_{i=1}^{n} i
$$
So following this logic $MaxInt(n)$ is also equal to:
$$
MaxInt(n) = \frac{MaxInt(n) + MaxInt(n)}{2}
$$
Since
$$
(2^0 + 2^1 + ... + 2^{n-3} + 2^{n-2}) = (2^{n-2} + 2^{n-3} + ... 2^2 + 2^1 + 2^0)
$$
Then
$$
MaxInt(n) = \sum_{i=0}^{n-2} 2^{n-2-i} = (2^{n-2} + 2^{n-3} + ... 2^2 + 2^1 + 2^0)
$$
Putting it all together:
$$
MaxInt(n) = \frac{\sum_{i=0}^{n-2} 2^{n-2-i} + \sum_{i=0}^{n-2} 2^i}{2}
$$
$$
MaxInt(n) = \frac{\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i}{2}
$$
Which is when I got stuck, cheating with wolfrom alpha I found that
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2
$$
But I don't know why. If you see a better alternative way (i.e. not using sum to N method) of deriving $MaxInt(n) = 2^n-1$ please let me know.
Thanks for reading.
| Applying the proof of the geometric series to this particular case:
$$
MaxInt(n) = S(n) = \sum_{i=0}^{n-2} 2^i = 2^0 + 2^1 ... + 2^{n-3} + 2^{n-2}
$$
Adding up each term in the series:
$$
2S(n) = (2^0 + 2^0) + (2^1 + 2^1) ... + (2^{n-3} + 2^{n-3}) + (2^{n-2} + 2^{n-2})
$$
Simplifying:
$$
2S(n) = 2^1 + 2^2 ... + (2 \cdot \frac{2^{n}}{2^3}) + (2 \cdot \frac{2^{n}}{2^2})
$$
$$
2S(n) = 2^1 + 2^2 ... + \frac{2^{n}}{2^2} + \frac{2^{n}}{2^1}
$$
$$
2S(n) = {\color{Blue} {2^1 + 2^2 ... + 2^{n-2}} } + 2^{n-1}
$$
Notice what is in blue is actually $S(n)$ except it is missing the first term, lets call this part in blue $y$:
$$
2S(n) = {\color{Blue} {y} } + 2^{n-1}
$$
$$
{\color{Blue} {y} } = S(n) - 2^0
$$
Substituting in:
$$
2S(n) = {\color{Blue} {S(n) -2^0} } + 2^{n-1}
$$
$$
S(n) = 2^{n-1} - 2^0
$$
$$
= 2^{n-1} - 1
$$
$$
\therefore MaxInt(n) = \sum_{i=0}^{n-2} 2^i = 2^{n-1} - 1
$$
The above is the equation desired by the background question, but the original question asks for $2S(n)$
$$
2S(n) = 2 \sum_{i=0}^{n-2} 2^i
$$
$$
= 2^1 \cdot (2^{n-1} - 1 )
$$
$$
= 2^{n-1+1} - 2
$$
$$
= 2^{n} - 2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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a limit problem? $$\lim_{x\to 0}{e^x+ \ln{1-x\over e }\over \tan x-x} $$
i tried doing l'hospital. but couldn't do ! could anyone help ?
| This screams "Taylor" to me. First, $\log\frac{1-x}e=\log (1-x)-\log e=\log(1-x)-1$. We have
$$
e^x=1+x+\frac{x^2}2+\frac{x^3}6+o(x^4),\ \ \log(1-x)=-x+\frac{x^2}2-\frac{x^3}3+o(x^4),
$$
$$\tan x=x-\frac{x^3}3+o(x^5).
$$
So
\begin{align}
{e^x+ \ln{1-x\over e }\over \tan x-x}&=\frac{1+x+\frac{x^2}2+\frac{x^3}6+o(x^4)+(-x+\frac{x^2}2-\frac{x^3}3+o(x^4))-1}{x-\frac{x^3}3+o(x^5)-x}\\
&=\frac{\frac{x^3}6+o(x^4)}{-\frac{x^3}3+o(x^5)}
=\frac{\frac12+o(x)}{-1+o(x)},
\end{align}
so the limit as $x\to0$ is $-1/2$.
| {
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Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) = \lim_{x \to \infty} \frac{x^2 + 3x - (x^2 + 1)}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Simplifying gives:
$$\lim_{x \to \infty} \frac{3x - 1}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Breaking out $\sqrt{x^2}$ from the denominator and $x$ from the numerator gives:
$$\lim_{x \to \infty} \frac{x(3 - \frac{1}{x})}{x(\sqrt{1 + 3x} + \sqrt{1 + 1})} = \lim_{x \to \infty} \frac{3 - \frac{1}{x}}{\sqrt{1 + 3x} + \sqrt{2}}$$
The result turns out to be $\frac{3}{2}$, but unsure how to proceed from here.
| There's an error in the factorisation of the denominator. No need to factor $x$ in the numerator, as you have a theorem for the limit at $\infty$ of a rational function. You should obtain, if $x>0$:
$$\lim_{x \to+ \infty}\frac{3x-1}{x\Bigl(\sqrt{1 + \dfrac3x} + \sqrt{1 + \dfrac1{x^2}}\Bigr)} = \lim_{x \to +\infty} \frac{3x - 1}x\cdot\frac1{\sqrt{1 + \dfrac3x} + \sqrt{1 + \dfrac1{x^2}}}= 3\cdot\frac12.$$
Similarly the limit as $x\to-\infty\,$ is $\,-\dfrac32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prime number of the form $A^{B^C}+D^{E^F}$ My Brother asked me what is the smallest prime number of the form $A^{B^C}+D^{E^F}$ where A,B,C are three distinct prime numbers, and D,E,F are 3 distinct primes that is Permutations of those 3 primes.I realize that we must arrange the exponents to be as smallest as possible in order to get smallest prime as possible.So what is the smallest such prime? (and second smallest :-)
| $A = 2$ OR $D = 2$
$A \neq D$, $B \neq E$, $C \neq F$
$a^n$ generally increases faster as $n \rightarrow \infty$ than as $a$ does.
If we say that $A = 2$, then $A^{B^C}$ must be even. Therefore, $D^{E^F}$ must be odd.
Therefore we have $2^{B^C} + D^{2^F}$ OR $2^{B^C} + D^{F^2}$
Let's try $2^{5^3}+3^{2^5}$. It is prime! Are there any smaller primes of the specified form? Taking $2^{B^C} + D^{2^F}$, we have two cases: is $5^3 < 3^5$ and is $2^5 < 5^2$. The first cases is true, but the second case is not. Therefore, let's try varying the second case. Is $2^{5^3}+3^{5^2}$ prime? It is not! Therefore, $2^{5^3}+3^{2^5}$ is the answer!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$ How to prove the following conjectured identity?
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$
It holds numerically with precision of at least $1000$ decimal digits.
Are there any other integers under the radical except $7$ and $1$ that result in a nice closed form?
| I will follow @user15302's idea. In this answer, I showed that
$$ \int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv, $$
where $b = \frac{a-1}{a+1}$. Now let $I$ denote the Vladimir's integral and set $s = \frac{1}{4}$ and $a = 7$. Then we have $b = \frac{3}{4}$ and
$$ I = 2^{-3/4} \int_{0}^{1} \frac{1}{v^{3/4}\sqrt{(1-v)(1-\frac{3}{4}v)}} \, dv. $$
The reason why the case $b = \frac{3}{4}$ is special is that, if we plug $v = \operatorname{sech}^2 t$ then we can utilize the triple angle formula to get the following surprisingly neat integral
$$ I = 2^{5/4} \int_{0}^{\infty} \frac{\cosh t}{\sqrt{\cosh 3t}} \, dt. $$
Now using the substitution $x = e^{-6t}$, we easily find that
$$ I = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{1} \frac{x^{-11/12} + u^{-7/12}}{\sqrt{1+x}} \, dx = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{\infty} \frac{dx}{x^{11/12}\sqrt{1+x}}. $$
The last integral can be easily calculated by the following formula
$$ \int_{0}^{\infty} \frac{x^{a-1}}{(1+x)^{a+b}} \, dx = \beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$
Therefore we obtain the following closed form
$$ I = \frac{\Gamma(\frac{1}{12})\Gamma(\frac{5}{12})}{3 \sqrt[4]{2}\sqrt{\pi}}. $$
In order to verify that this is exactly the same as Vladimir's result, We utilize the Legendre multiplication formula and the reflection formula to find that
$$ \Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12})
= \frac{\Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12})\Gamma(\tfrac{9}{12})}{\Gamma(\tfrac{3}{4})}
= \frac{2 \pi \cdot 3^{1/4} \Gamma(\frac{1}{4})}{\Gamma(\tfrac{3}{4})}
= 2^{1/2} 3^{1/4} \Gamma(\tfrac{1}{4})^2. $$
This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 4,
"answer_id": 1
} |
Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”
In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$
Should:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$
not be:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$
as everything in the left-hand side is cubed?
| Consider the infinite "multiplication table" array $M$ defined by $M_{i,j}=ij$
$M =
\left[\begin{array}{r}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 6 & 8 & 10 \\
3 & 6 & 9 & 12 & 15 & \cdots\\
4 & 8 & 12 & 16 & 20 \\
5 & 10 & 15 & 20 & 25 \\
&& \vdots &&& \ddots
\end{array}\right]$
Let $L(n) =
\{M_{i,j} : 1 \le i \le n \wedge j=n \} \cup
\{M_{i,j} : i=n \wedge 1 \le j \le n \}$
be sum of the the _| shaped region consisting of all elements of $M$ whose first or second index is $n$ and whose other index is less than or equal to $n$. For example, the elements of $L(4)$ are shown below.
$M =
\left[\begin{array}{r}
1 & 2 & 3 & \color{red}4 & 5 \\
2 & 4 & 6 & \color{red}8 & 10 \\
3 & 6 & 9 & \color{red}{12} & 15 & \cdots\\
\color{red}4 & \color{red}8 & \color{red}{12} & \color{red}{16} & 20 \\
5 & 10 & 15 & 20 & 25 \\
&& \vdots &&& \ddots
\end{array}\right]$
It turns out that $L(1)=1,\; L(2)=8,\; L(3)=27,\; L(4)=64,$ and $L(5)=125$.
We show that $L(n) = n^3$.
\begin{align}
L(n)
&= \sum_{i=1}^n M_{i,n} + \sum_{j=1}^n M_{n,j} - M_{n,n} \\
&= n\sum_{i=1}^n i + n\sum_{j=1}^n j - n^2 \\
&= n^2(n+1) - n^2 \\
&= n^3
\end{align}
Let $S(n)$ be the sum of the $n^2$ elements in the $n \times n$ subarray of elements in the upper-left corner of $M$.
For example, $S(3) = (1+2+3) + (2 + 4 + 6) + (3 + 6 + 9) = 36$
We will compute $ S(n)$ two ways.
\begin{align}
\sum_{i=1}^n \sum_{j=1}^n M_{i,j}
&= \sum_{i=1}^n \sum_{j=1}^n ij \\
&= \left(\sum_{i=1}^n i \right) \left(\sum_{j=1}^n j \right) \\
&=\left[\frac{n(n+1)}{2}\right]^2
\end{align}
Also,
\begin{align}
S(n) &= L(1) + L(2) + \cdots + L(n) \\
S(n) &= \sum_{i=1}^n i^3
\end{align}
So $\displaystyle \sum_{i=1}^n i^3 = \left[\frac{n(n+1)}{2}\right]^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
List the elements of the field $K = \mathbb{Z}_2[x]/f(x)$ where $f(x)=x^5+x^4+1$ and is irreducible Since $\dim_{\mathbb{Z}_2} K = \deg f(x)=5$, $K$ has $2^5=32$ elements. So constructing the field $K$, I get:
\begin{array}{|c|c|c|}
\hline \text{polynomial} & \text{power of $x$} & \text{logarithm} \\\hline
0 & 0 & -\infty \\\hline
1 & 1 & 0 \\\hline
x & x & 1 \\\hline
x^2 & x^2 & 2 \\\hline
x^3 & x^3 & 3 \\\hline
x^4 & x^4 & 4 \\\hline
x^4+1 & x^5 & 5 \\\hline
x^4+x+1 & x^6 & 6 \\\hline
x^4+x^2+x+1 & x^7 & 7 \\\hline
x^4+x^3+x^2+x+1 & x^8 & 8 \\\hline
x^3+x^2+x+1 & x^9 & 9 \\\hline
x^4+x^3+x^2+x & x^{10} & 10 \\\hline
x^3+x^2+1 & x^{11} & 11 \\\hline
x^4+x^3+x & x^{12} & 12 \\\hline
x^2+1 & x^{13} & 13 \\\hline
x^3+x & x^{14} & 14 \\\hline
x^4+x^2 & x^{15} & 15 \\\hline
x^4+x^3+1 & x^{16} & 16 \\\hline
x+1 & x^{17} & 17 \\\hline
x^2+x & x^{18} & 18 \\\hline
x^3+x^2 & x^{19} & 19 \\\hline
x^4+x^3 & x^{20} & 20 \\\hline
\end{array}
Have I constructed the field correctly? I'm sure the claim that $K$ has $32$ elements is true, yet when I actually construct the field, I only get $22$ elements as $x^{21} = x^0 = 1$.
| One problem is, that you need to list all polynomials $f(x)\in \mathbb{F}_2[x]$ of degree $n\le 4$. This gives $2^{4+1}=32$ polynomials. For degree $5$, and hence higher degree, you can "reduce" the degree of the polynomial by the rule $x^5:=-x^4-1=x^4+1$. For example, the polynomial $x^3+1$ is missing in your list.
Edit: And yes, $x^5+x^4+1=(x^2+x+1)(x^3+x+1)$ here is not irreducible !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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