Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding the root of a complex number I have a complex number $8-6 i$ . I have to find the square root . I did all the steps and I got $\pm(-3+i)$. I also got $\pm(3-i)$. On squaring I am getting the same $8-6 i$, but what is the right square root . Please someone point out my mistake.
| $$\sqrt{8-6i}=\sqrt{\left|8-6i\right|e^{\arg\left(8-6i\right)i}}=$$
$$\sqrt{\sqrt{8^2+6^2}e^{-\tan^{-1}\left(\frac{6}{8}\right)i}}=\sqrt{\sqrt{100}e^{-\tan^{-1}\left(\frac{3}{4}\right)i}}=$$
$$\sqrt{10e^{-\tan^{-1}\left(\frac{3}{4}\right)i}}=\left(10e^{-\tan^{-1}\left(\frac{3}{4}\right)i}\right)^{\frac{1}{2}}=$$
$$\sqrt{10}e^{-\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)i}=$$
$$\sqrt{10}\cos\left(-\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)\right)+\sqrt{10}\sin\left(-\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)\right)i=3-i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.
$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4-2a^2c^2x+(b^2x)^2+d^4-2b^2d^2x+2a^2c^2x+2b^2d^2x+4abcdx\right]=0$$
So $$(a^2x-c^2)^2+(b^2x-d^2)^2+2x(ac+bd)^2=0$$
Now I did not understand How can I proceed further.
Although I have a knowledge of $\bf{Discriminant\; Method.}$
So plz explain me above method which i am trying above.
Thanks.
| Hint: consider the term $$-4 \left(a^4 c^4+a^4 d^4-4 a^2 b^2 c^2 d^2+b^4 c^4+b^4 d^4\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving the irrationality of $\sqrt{5} = x \sqrt{7} + y$ I need to prove that there are no rational numbers $x, y$ that
$$\sqrt{5} = x \sqrt{7} + y$$
We know that square root of prime is irrational so $y = 5 - 7x$ so the only number for it to be rational is if $y = 0$ so $x = \frac{\sqrt{7}} {\sqrt {5}}$ but that is irrational or when $x = 0$, $y = \sqrt{5}$ which is irrational.
$x = \sqrt{5} - y \sqrt {7}$, so I'm stuck here.
Or I need to try to prove each at turn, like
1 case : let $x =$ blah blah and do it as irrational prime proof
2 case : let $y =$ ???
Help would be appreciated.
| If $x,y\in Q$, then $\sqrt 5= x\sqrt 7+y\implies \sqrt 5-y=x\sqrt 7\implies 5-2 y\sqrt 5 +y^2=7x^2\implies$ $$(-2 y)\sqrt 5=7x^2-5-y^2\in Q\implies y=0\implies$$ $$\sqrt 5=x\sqrt 7\implies x=\sqrt {5/7}\not \in Q$$contradicting $x\in Q$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove inequality $\left| \frac{x+yz}{x^2+y^2} \right| \leq 1$ for $x^2+y^2-z^2=1$ Prove this:
If $x,y,z \in \mathbb R$ and $x^2+y^2-z^2=1$, then $$\left| \frac{x+yz}{x^2+y^2} \right| \leq 1 $$ holds.
Own ideas:
If we eliminate $z$ the inequality is equivalent to $$\left| \frac{x \pm y\sqrt{x^2+y^2-1}}{x^2+y^2} \right| \leq 1.$$
| $(1+z^2)(x^2+y^2)\geq(x+yz)^2$ by Cauchy Inequality
Since $(1+z^2)=(x^2+y^2)$ the result follows.
| {
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"url": "https://math.stackexchange.com/questions/1509145",
"timestamp": "2023-03-29T00:00:00",
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Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
MyApproach
Since I cannot form pattern above
I simplified this as $21^3$ = $3^3$ $. $ $7^3$
Similarly I did $27^3$ = $9^3$ $.$ $3^3$
Taking both I get $3^3$($7^3$+$9^3$)=$9$ $.$ $1072$/ $2^5$.
From solving this I get Remainder as $1$
And Similarly on solving $25^3$+$23^3$ separately on dividing by $96$.I get remainder as $73$ and $71$
On adding and Finding remainder I get Remainder as $49$.
I am getting wrong Ans.
Please correct me how to approach towards the problem.
| By exploiting $a^3+b^3=(a+b)(a^2-ab+b^2)$ we have:
$$ 21^3+23^3+25^3+27^3 = 48\left(27^2-21\cdot 27+21^2+23^3-23\cdot 25+25^2\right)$$
hence $ 21^3+23^3+25^3+27^3$ is a multiple of $96$, since $\left(27^2-21\cdot 27+21^2+23^3-23\cdot 25+25^2\right)$ is an even number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number Theory: Prove that $x^{p-2}+\dots+x^2+x+1\equiv 0\pmod{p}$ has exactly $p-2$ solutions I just completed this homework problem, but I was wondering if my proof was correct:
If $p$ is an odd prime, then prove that the congruence $x^{p-2}+\dots+x^2+x+1\equiv 0\pmod{p}$ has exactly $p-2$ incongruent solutions, and they are the integers $2,3,\dots,p-1$.
Proof
Let $f(x)=x^{p-2}+\dots+x^2+x+1$ and $g(x)=(x-1)f(x)$.
Note that $f(1)=(p-2)+1=p-1\not\equiv 0\pmod{p}$.
So, $g(x)=(x-1)(x^{p-2}+\dots+x^2+x+1)=x^{p-1}-1$.
Now, $x^{p-1}-1\equiv 0\pmod{p}$ has exactly $p-1$ incongruent solutions modulo $p$ by Lagrange's Theorem.
Note that $g(1)=(1-1)f(1)=0\equiv 0\pmod{p}$, so $1$ is a root of $g(x)$ modulo $p$. Hence, the incongruent roots of $g(x)$ modulo $p$ are $1,2,3,\dots,p-1$.
But every root of $g(x)$ other than $1$ is also a root of $f(x)$ (This is the part I'm concerned about. Is it clear that this is the case?), hence $f(x)$ has exactly $p-2$ incongruent roots modulo $p$, which are $2,3,\dots,p-1$. $\blacksquare$
| If $a \in Z_p \setminus \{0,1\}$ and $p$ is prime, then:
$$1+a+a^2+\cdots +a^{p-2}=(a^{p-1} - 1)\cdot(a-1)^{-1} = 0,$$because $a^{p-1}=1$.
On the other hand, if $a=1$, then
$$1 + a + a^2 + \cdots + a^{p-2} = \underbrace{1+1+ \cdots +1}_{p-1} =(p-1) \ne 0.$$
Finally if $a = 0$,
$$1 + a + a^2 + \cdots + a^{p-2} = 1 \ne 0.
$$
So the solutions are all $a \ne 0, 1$, which are $p-2$ values.
| {
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Simple Division Problem I have the equation:
$$(1-\frac{1}{2^2})...(1-\frac{1}{n^2}) = \frac{n+1}{2n}$$ for n ≥ 2
Trying to prove by induction and I get the following equation.
$$\frac{k+1}{2k} + \frac{k(k+2)}{(k+1)^2} = \frac{k+2}{2(k+2)}$$
I can't to simplify it to the final answer.
I multiplied $$\frac{k+1}{2k}.{(k+1)^2}$$ and $$\frac{k(k+2)}{(k+1)^2}.{2k}$$ but I know something is wrong somewhere!
BTW, this is part of the proof by Induction step!
| The statement is true for $n=2$. Suppose it holds for $n$; then
$$
\left(1-\frac{1}{2^2}\right)\dots
\left(1-\frac{1}{n^2}\right)
\left(1-\frac{1}{(n+1)^2}\right)=
\frac{n+1}{2n}\left(1-\frac{1}{(n+1)^2}\right)
$$
The final term becomes
$$
\frac{n+1}{2n}\frac{n(n+2)}{(n+1)^2}=\frac{(n+1)+1}{2(n+1)}
$$
which is exactly what was to be proved.
(Note: I prefer avoiding the change from $n$ to $k$, do as you like better.)
You had a $+$ in
$$
\frac{k+1}{2k} + \frac{k(k+2)}{(k+1)^2}
$$
where multiplication should be used.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 (n^2 +1)}\right)^n
}
$$
I think I got the correct limit by using fast converging limits to $e$.
In particular I used truncated Taylor series for the sqrt and 4th root.
Or squares and bisquares.
Example
$(1+1/2n)^{2n}$
Becomes
$(1 + 1/n + 1/4n^2)^n.$
In combination with l'hopital it gives me the answer.
But I guess that is not a very good (fast) method.
| Hint: (for the first version, before multiplying by 2)
$$
\left(1+\frac1n\right)^n\leq\left(1+\frac1n+\frac1{n^2}\right)^n\leq\left(1+\frac{1+\epsilon}n\right)^n
$$
for large $n$. Hence all partial limits are $e$ and the final one — 0.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Limit of $\frac{(n+1)^{2n}}{(n^2+1)^n}$ as $n\to \infty$ So it is given to find $$\lim_{n\to \infty}\dfrac{(n+1)^{2n}}{(n^2+1)^n}$$
So what I did is
$$\lim_{n\to \infty}\dfrac{(n+1)^{2n}}{(n^2+1)^n}=\lim_{n\to \infty}\dfrac{(n^2+2n+1)^{n}}{(n^2+1)^n}=\lim_{n\to \infty}\left(1+\frac{2n}{n^2+1}\right)^n$$
Now the rightmost form, is it $e^2$? I mean I am unable convince myself that it is some form of exponential function. Help me out.
| Your guess is right.
We have
$$
1+\frac{2n}{n^2+1} \le 1+\frac{2n}{n^2} = 1+\frac{2}{n}
$$
and so
$$
\left(1+\frac{2n}{n^2+1}\right)^n \le \left(1+\frac{2}{n}\right)^n \to e^2
$$
On the other hand,
$$
1+\frac{2n}{n^2+1} \ge 1+\frac{2n}{n^2+n} = 1+\frac{2}{n+1}
$$
and so
$$
\left(1+\frac{2n}{n^2+1}\right)^n
\ge \left(1+\frac{2}{n+1}\right)^n
= \frac{\left(1+\frac{2}{n+1}\right)^{n+1}}{\left(1+\frac{2}{n+1}\right)^{\hphantom{n+1}}}
\to \frac{e^2}{1} = e^2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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LU decomposition on 5 by 3 matrix. This is a problem given in a quiz. Even after reading up a related question, I cannot figure it out.
Determine an LU-factorization of $$
C =
\begin{bmatrix}
3 & 1 & -4 \\
6 & -3 & 10 \\
-9 & 5 & -11 \\
-3 & 0 & -7 \\
6 & -4 & 2
\end{bmatrix}
$$
In the related question, I can acquire an upper triangular matrix with row operations and permute the rows. However, I can't turn this into an upper triangular matrix, only a lower one, but then it becomes UL-factorization.
Any insight is welcome.
| Because $C$ is not square, $U$ will not be in upper triangular form but rather in row echelon form.
In your case, proceed as you would for a square matrix, that is, reduce $C$ to row echelon form to get $U$, and keep track of the multipliers you used to get $L$:
Use the first row, $R_1$, to make the first entries of the other rows $0$:
$$\begin{bmatrix}
3 & 1 & -4\\
0 & -5 & 18\\
0 & 8 & -23\\
0 & 1 & -11\\
0 & -6 & 10\end{bmatrix}
\begin{array}
\\ \\
R_2 - 2R_1\\
R_3 + 3R_1\\
R_4 + 1R_1\\
R_5 - 2R_1.\end{array}$$
Next, use the second row to make the second entries of subsequent rows $0$:
$$\begin{bmatrix}
3 & 1 & -4 \\
0 & -5 & 18 \\
0 & 0 & 29/5\\
0 & 0 & -37/5\\
0 & 0 & -58/5\end{bmatrix}
\begin{array}
\\ \\ \\
R_3 + 8/5R_2\\
R_4 + 1/5R_2\\
R_5 - 6/5R_2.\end{array}$$
Finally, use the third row to make the third entries of subsequent rows $0$:
$$U = \begin{bmatrix}
3 & 1 & -4 \\
0 & -5 & 18 \\
0 & 0 & 29/5\\
0 & 0 & 0 \\
0 & 0 & 0 \end{bmatrix}
\begin{array}
\\ \\ \\ \\
R_4 + 37/29R_3\\
R_5 + 2R_3.\end{array}$$
The lower entries of $L$ are the negatives of the multipliers you used to reduce $A$ to $U$, hence,
$$L = \begin{bmatrix}
1 & 0 & 0 & 0 & 0\\
2 & 1 & 0 & 0 & 0\\
-3 & -8/5 & 1 & 0 & 0\\
-1 & -1/5 & -37/29 & 1 & 0\\
2 & 6/5 & -2 & 0 & 1\end{bmatrix}.$$
You can check that $LU = C.$
If you use a program that requires a square matrix for input, give $C$ dummy fourth and fifth columns.
| {
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"timestamp": "2023-03-29T00:00:00",
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Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\text{mod} ~2)$ means that one of $m$ or $n$ is odd or even
But now I'm not too sure on what to do.
Any help will be appreciated.
| If $m+n\equiv 1 \bmod 2$, then $m-n\equiv 1 \bmod 2$, so $m^2-n^2\equiv 1 \bmod 2$, and $m^2+n^2=m^2-n^2+2n^2\equiv 1\bmod 2$.
If $d\mid m^2-n^2$ and $d\mid m^2+n^2$, then $d\mid 2m^2$ and $d\mid 2n^2$.
Since both $m^2+n^2$ and $m^2-n^2$ are odd, $d$ is not even, so $d\mid m^2$ and $d\mid n^2$, which implies $d\mid m$ and $d\mid n$, so $d=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding maximum $b$ in $x^5-20x^4+bx^3+cx^2+dx+e=0$ Let $b, c, d, e$ be real numbers such that the following equation
$$x^5-20x^4+bx^3+cx^2+dx+e=0$$
has real roots only. Find the largest possibe value of $b$.
What I have done is:
Let $x_1, x_2, x_3, x_4, x_5$ be the 5 real roots of the equation. Then we have
$$x_1+x_2+x_3+x_4+x_5=20$$ and
$$b=\sum_{0<i\le j \le 5} x_ix_j=\frac{1}{2}[(x_1+x_2+x_3+x_4+x_5)^2-(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)]$$
To find maximum $b$, we can find minimum $x_1^2+x^2_2+x_3^2+x_4^2+x_5^2$. Cauchy-Schwartz Inequality yields,
$$(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)(1+1+1+1+1)\ge(x_1+x_2+x_3+x_4+x_5)=20$$
Thus,
$$x_1^2+x^2_2+x_3^2+x_4^2+x_5^2\ge\frac{20}{5}=4$$
So,
$$b_{max}=\frac{1}{2}[20^2-4]=198$$
However, the answer was 160, yet I am pretty sure I am correct. Where did I go wrong?
Thanks in advance!
| The condition for your answer is when $x_1=x_2=...=x_5=4$,so we can find that they are roots of $$(x-4)^5=0$$
Expand this equation of left side we have the coefficient of $x^4$ is $$C(5,4)*(-4)^1=-20$$
and coefficient of $x^3$ is $$C(5,3)*(-4)^2=160$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$ Let $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$ for all $n\geq0$.
How to find the value of $\displaystyle\prod_{n=1}^\infty A_n$ ?
I don't have any idea.
Thank you.
| If we set $A_n = \cos^2(\theta_n)$, we get:
$$ A_{n+1} = \frac{1+\cos(\theta_n)}{2} = \cos^2\left(\frac{\theta_n}{2}\right),$$
and since $\theta_0=\frac{\pi}{6}$, induction gives:
$$ A_n = \cos^2\left(\frac{\pi}{6\cdot 2^n}\right)=\left(\frac{\sin\left(\frac{\pi}{6\cdot 2^{n-1}}\right)}{2\sin\left(\frac{\pi}{6\cdot 2^n}\right)}\right)^2 $$
from which it follows that:
$$ \prod_{n=1}^{+\infty}A_n = \left(\frac{3}{\pi}\right)^2 = \color{red}{\frac{9}{\pi^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514901",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Show that $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 }$ So I'm trying to show that:
$\dfrac{ -1 }{ 2 } \le \dfrac{ x }{ 1+x^2 } \le \dfrac{ 1 }{ 2 }$
for every value of x.
I know I have to use mean value theorem so I tried to show it with cases. First I tried showing that $\dfrac{-1}{2} \le \dfrac{x}{1+x^2}$ and then $\dfrac{1}{2} \ge \dfrac{x}{1+x^2}$ using MVT. Is that correct? Can someone guide me?
Thanks!
| $$0\leq(1-x^2)^2\iff 4x^2\leq(1+x^2)^2\iff\frac{x^2}{(1+x^2)^2}\leq\frac14
\iff\left|\frac{x}{1+x^2}\right|\leq\frac12.$$
| {
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Evaluate $\sum_{n=0}^\infty\left(-1+(n+1)\ln\left(\frac{2n+3}{2n+1}\right)\right)$ I proved that the given series is convergent and to find the sum I tried to compute the partial sum and then I pass to the limit but I didn't find a closed form to this partial sum. How I should proceed?
| Set $a_n=n\ln(2n+1)$, and rewrite the general term of the series as \begin{align*}b_n&=-1+(n+1)\ln\left(\frac{2n+3}{2n+1}\right)\\&=-1+(n+1)\ln(2n+3)-n\ln(2n+1)-\ln(2n+1)=-1-\ln(2n+1)+a_{n+1}-a_n.\end{align*} Summing from $k=0$ to $n$, we obtain that $$\sum_{k=0}^nb_k=\sum_{k=0}^n\left(-1-\ln(2k+1)+a_{k+1}-a_k\right)=-(n+1)-\sum_{k=0}^n\ln(2k+1)+a_{n+1},$$ since $a_0=0$, therefore \begin{align*}\sum_{k=0}^nb_k&=-(n+1)-\sum_{k=0}^n\ln(2k+1)+(n+1)\ln(2n+3)=\ln\left(\frac{(2n+3)^{n+1}e^{-n-1}}{3\cdot 5\cdot 7\cdots\cdot (2n+1)}\right)\\
&=\ln\left(\frac{(2n+3)^{n+1}e^{-n-1}\cdot 2^nn!}{(2n+1)!}\right).\end{align*} For this last term, use Stirling's formula: asymptotically, it is equal to
$$\ln\left(\frac{(2n+3)^{n+1}e^{-n-1}\cdot 2^nn^ne^{-n}\sqrt{2\pi n}}{(2n+1)^{2n+1}e^{-2n-1}\sqrt{2\pi(2n+1)}}\right)\simeq\ln\left(\frac{(2n+3)^{n+1}(2n)^n}{(2n+1)^{2n+1}\sqrt{2}}\right),$$ and $$
\frac{(2n+3)^{n+1}(2n)^n}{(2n+1)^{2n+1}\sqrt{2}}=\left(1+\frac{2}{2n+1}\right)^{n+1}\left(1+\frac{1}{2n}\right)^{-n}\frac{1}{\sqrt{2}}\xrightarrow[n\to\infty]{}\frac{e\cdot e^{-1/2}}{\sqrt{2}}=\frac{e^{1/2}}{\sqrt{2}},$$ therefore the series is equal to $$\ln\left(\frac{e^{1/2}}{\sqrt{2}}\right)=\frac{1}{2}-\frac{1}{2}\ln 2.$$
| {
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} |
The identity $\tan({\pi\over4}-{a\over2}) = \sec(a)-\tan(a)$ Today I was solving an exercise and while checking the solution on WolframAlpha, the website used the following identity:
$$\tan \left({\pi\over4}-{\alpha\over2} \right) = \sec(\alpha)-\tan(\alpha)$$
Since I've never seen that formula, I tried to calculate it with the identities I know, especially using
$$\tan \left({\pi\over4}-{\alpha\over2} \right) = {(1-\tan({\alpha\over2}))\over(1+\tan({\alpha\over2}))},$$
but I couldn't make it work.
Is this a known property of trigonometry, like for example $\cos({\pi\over2}-x)=\sin(x)$?
| Using the identity $\sin^2(a)+cos^2(a)=1$ and dividing by $\cos^2(a)$ gives you the identity $\tan^2(a)+1=\sec^2(a)$. So $\tan(a)=\sqrt{sec^2(a)-1}$. Using this we have:
$$
\begin{split} \tan\left(\frac{\pi}{4}-\frac{a}{2}\right)&=\sqrt{\sec^2\left(\frac{\pi}{4}-\frac{a}{2}\right)-1} \\ &=\sqrt{\frac{1}{\cos^2\left(\frac{\pi}{4}-\frac{a}{2}\right)} -1} \\ \end{split}
$$
Then using $\cos^2(a)=\frac{2\cos(2a)+1}{2}$ we get:
$$
\begin{split} \sqrt{\frac{1}{\cos^2\left(\frac{\pi}{4}-\frac{a}{2}\right)} -1} &= \sqrt{\frac{2}{\cos\left(\frac{\pi}{2}-a\right)+1}-1 } \\ &=\sqrt{\frac{2-1-\sin(a)}{\sin(a)+1} }\\ &= \sqrt{\frac{1-\sin(a)}{1+\sin(a)}} \\ &= \sqrt{\frac{(1-\sin(a))^2}{1-\sin^2(a)}} \\ &= \frac{1-\sin(a)}{\cos(a)} \\ &=\sec(a)-\tan(a) \end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
} |
Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series
$$\sum\limits_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^2}$$
Second derivative is
$$\left(\frac{1}{1-x^2}\right)^{''}=\frac{1}{(1-x^2)^4}\cdot x(1+8x-2x^2-8x^3+x^4)$$
This gives
$$\frac{1}{(1-x^2)^4}=\sum\limits_{n=0}^{\infty}\frac{2n(2n-1)x^{2n-2}}{x(1+8x-2x^2-8x^3+x^4)}$$
I have a proof in combinatorics which involves this series.
How to represent this series using binomial coefficients?
| Perhaps it could be easier expand $(1-x^2)^{-4}$ using the binomial series.
$\begin{align}
(1-x^2)^{-4} &= 1+4x^2 + \frac{(-4)(-4-1)}{2}(-x^2)^2+\frac{(-4)(-4-1)(-4-2)}{3!}(-x^2)^3+ \dots\\
&=1+4x^2+10x^4+20x^6+35x^8+\dots\\
&=\sum_{n=1}^{\infty} \binom{n+2}{3}x^{2n-2}\\
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ find the ratio of $x$, $y$ and $ z$ I have this question from higher algebra by Hall and Knight:
if $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $\frac x4 =\frac y2 =\frac z3$ and the second is $\frac x1 =\frac y{-1} =\frac z0 $. Now I solve this question in the following manner:
Adding numerator and denominator gives $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y}= 2\frac{x+y}{x+y} =2$ (when x+y is not zero).This gives the first answer.
When (x+y ) is zero => $ y= -x $ that is $\frac xy = -1$ and now there are two things
(i)If I put these values in the original expression I get $ \frac{y}{x-z} =\frac 0z= -1$ this implies 0= 1 where am I making the mistake?
(ii)Also from the original expression I have$ \frac{x+y}{z} = \frac{x}{y}$ multiplying by z I get $ x+y =z\frac xy$. This implies that z=0 and x:y:z =x:-x:0 = 1:-1:0 and this gives the second answer.But the problem is how z can be zero when it appears in the denominator in the expression
I think these things are very basic but still I am stuck.Could anyone please help me in knowing where the mistake is?
| It's not clear what the OP means by "adding" the numerator and denominator but there is a way using componendo and dividendo. If $\frac{x+y}{z} = \frac{x}{y}$ then $\frac{(x + y) - x}{z - y} = \frac{x}{y} \implies \frac{y}{z - y} = \frac{x}{y}$.
This implies that $\frac{y}{z - y} = \frac{y}{x - z} \implies z - y = x - z \implies 2z = x + y$. Thus $\frac{2z}{z} = \frac{x}{y}$ and we have two cases: either $z = 0$ which implies $\frac{y}{x} = \frac{x}{y} \implies (y + x)(y - x) = 0$. This results in $y = -x$, or $y = x$ in which case $z = 2x$ and $x = y = z = 0$ which does not make any sense.
The other case is when $z = 2y$ and hence $\frac{3y}{z} = 2$ so $x = 2y = \frac{2}{3} \cdot 2z$, which is equivalent to the form above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Where did I mistake to integrate $I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx\; ?$ It was given to integrate $$I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx.$$
Attempt:
\begin{align}I&= \int\sqrt{\frac{\sin((x+\alpha)-2\alpha)}{\sin(x+\alpha)}}\; dx\\&= \sqrt{\sin 2\alpha}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\;dx\; .\end{align}
Then I took \begin{align}\cot 2\alpha - \cot (x+\alpha)= z^2 \\ \implies [1 + \cot^2(x+\alpha)]\;dx= 2z\;dz\\ \implies [1+\{\cot 2\alpha - z^2\}^2]\; dx= 2z\;dz\; .\end{align}
Now,
\begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int\frac{z\cdot 2z}{1+\{\cot 2\alpha - z^2\}^2}\;dz\\ &= \int\frac{2}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz \; .\end{align}
This makes me to break the integral to get \begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ \\ &= \int\frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z-\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right) +\text{const.}}\; dz +\int\frac{\left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z+\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right) +\text{const.}}\; dz \\ &=-\frac{1}{z- \dfrac{(1+ \cot^2 2\alpha)}{z}}-\frac{1}{z+ \dfrac{(1+ \cot^2 2\alpha)}{z}} .\end{align}
So, $$I= -\sqrt{\sin 2\alpha}\left(\frac{1}{z- \dfrac{(1+ \cot^2 2\alpha)}{z}}+\frac{1}{z+ \dfrac{(1+ \cot^2 2\alpha)}{z}}\right)\; .$$
Unfortunately, my book gave the answer as $$I= \cos\alpha \cos^{-1} (\cos x\sec\alpha) - \sin\alpha\log\left(\sin x+ \sqrt{\sin^2 x - \sin^2\alpha}\right)\; .$$
This is not near to my solution; I've made somewhere a big plunder.
So, could anyone please tell me where I did mistake? Is my attempted technique wrong?
| HINTS:
We have $$\frac{\sin(x-\alpha)}{\sin(x+\alpha)}=\frac{\sin x\cos\alpha-\sin\alpha\cos x}{\sin x\cos\alpha+\sin\alpha\cos x}=\frac{(\sin x\cos\alpha-\sin\alpha\cos x)^2}{\sin^2x\cos^2\alpha-\sin^2\alpha\cos^2x}=\frac{(\sin x\cos\alpha-\sin\alpha\cos x)^2}{\sin^2x-\sin^2\alpha}$$
and $$u=(\sin^2x-\sin^2\alpha)^{1/2}\rightarrow du=\frac{\sin x\cos x}{\sqrt{\sin^2x-\sin^2\alpha}}dx$$
Can you go further?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{1 + (-2n)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}
\end{align}
since $\left(\frac{n^2+1}{n^2} \right) \to 1$ when $n\to\infty$
$$\lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)} = \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}$$
Also I tried to make similar simplifications into the braсkets but nothing happens and no proof that limit $= 1$ or whatever.
And here is a rule for the task. This is a limit of a sequence so no usage of functional simplifications and derives are allowed. If you have really beautiful solution for the task then post it anyway.
| $\frac{n^2}{n^2+1}=1-\frac{1}{n^2+1}$
As $n \to \infty$,
$\because(1-\frac{1}{n^2+1})^\frac{n-1}{n+1} > 1^\frac{n-1}{n+1} = 1$
$(1-\frac{1}{n^2+1})^\frac{n-1}{n+1} < (1-\frac{1}{n^2+1})^\frac{n}{n}=1-\frac{1}{n^2+1}$
$\because \lim \limits_{n \to \infty} 1 = 1$
$\lim \limits_{n \to \infty} (1-\frac{1}{n^2+1}) = 1$
$\therefore$ By Squeeze Theorem, $\lim \limits_{n \to \infty} \frac{n^2}{n^2+1} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
The value of $m$ for which straight line $3x-2y+z+3=0=4x-3y+4z+1$ is parallel to the plane $2x-y+mz-2=0$ is The value of $m$ for which straight line $3x-2y+z+3=0=4x-3y+4z+1$ is parallel to the plane $2x-y+mz-2=0$ is
$(A)-2\hspace{1cm}(B)8\hspace{1cm}(C)-18\hspace{1cm}(D)11$
My Attempt:The plane $2x-y+mz-2=0$ is parallel to the $3x-2y+z+3=0$ and $4x-3y+4z+1=0$.
So the lines are perpendicular to the normal of the plane.
So,$(3)(2)+(-2)(-1)+(1)(m)=0$ gives $m=-8$
And $(4)(2)+(-3)(-1)+(4)(m)=0$ gives $m=\frac{-11}{4}$
But the answer given in the book is $m=-2$
There is a method given in the book to answer this question.
Vector $((3\hat{i}-2\hat{j}+\hat{k})\times(4\hat{i}-3\hat{j}+4\hat{k}))$ is perpendicular to $2\hat{i}-\hat{j}+m\hat{k}$
So $m=-2$
But i could not understand this method.Please help me.Thanks.
| Note that if you write only $3x-2y+z+3=0$, then it represents a plane.
Here, we have
$$3x-2y+z+3=0=4x-3y+4z+1,$$
which represents a line.
We can write the line with the equation
$$3x-2y+z+3=0=4x-3y+4z+1$$
as
$$\frac{x-0}{1}=\frac{y-\frac{11}{5}}{\frac 85}=\frac{z-\frac 75}{\frac 15}$$
(you can get this by expressing $y,z$ by $x$)
So,
$$2\cdot 1+(-1)\cdot \frac 85+m\cdot \frac 15=0\iff m=-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the formula for the sequence $a(n)$ Find the formula for the sequence $a(n)$ given by the following recurrence relations and prove that the formula is correct:
$a(0) = 1, ~a(1) = -2,~ a(n) = -2a(n-1)-a(n-2)$
How to solve it? I don't know how to start?
| Let $$f(x) = \sum_{n = 0}^\infty a_nx^n$$ be the generating function for the sequence $\left\{a_n\right\}_{0}^\infty$. Then
$$\begin{align}f(x) &= a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \dots \\
&= 1 -2x + (-2a_1 - a_0)x^2 + (-2a_2 - a_1)x^3 + (-2a_3 - a_2)x^4 + \dots\\
&= 1 - 2x + (-2x)(f(x) - a_0) + (-x^2)f(x)\\
&= 1 - 2x + 2x - 2xf(x) - x^2f(x)\\
&=1 - 2xf(x) - x^2f(x)\end{align}$$
so that with some rearrangement,
$$f(x) + 2xf(x) + x^2f(x) = 1$$
$$(1 + 2x + x^2)f(x) = 1$$
$$\begin{align}f(x) &= \frac{1}{1 + 2x + x^2}\\
&=\frac{1}{(1 + x)^2}\end{align}$$
But we know that
$$\frac{1}{1 + x} = 1 - x +x^2 - x^3 + x^4 - \dots$$
Differentiating both sides,
$$-\frac{1}{(1 + x)^2} = -1 + 2x - 3x^2 + 4x^3 - \dots$$
$$f(x) = \frac{1}{(1 + x)^2} = 1 - 2x + 3x^2 - 4x^3 + \dots$$
$$a_0 + a_1x + a_2x^2 + a_3x^3 + \dots = 1 - 2x + 3x^2 - 4x^3 + \dots$$
Hence, by comparing the coefficients directly,
$$a_n = (-1)^n (n + 1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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When is EF longer than AC? (a generalization) ABC is an isosceles right triangle,
M is on AC,
and EMF is a straight line.
When is EF longer than AC?
]1
Note:
This is a generalization
of the following problem,
which has M in the
center of AC:
Prove that EF is longer than AC
In that case,
I showed that
EF is always longer than AC.
I can show the following:
If M is closer to C than A,
then EF is longer.
If M is closer to A than C,
then EF can be shorter,
but it will be longer when
EA is long enough.
If $|EB| \ge \sqrt{2}$,
then EF is always longer
no matter where M is.
If there are no answers in two days,
I will post mine.
| In my solution to
the original problem,
I used algebra and
analytic geometry.
I tried to do the same thing
for this generalization,
but the final simplification
does not occur here.
Here is what I've got.
Suppose
$AB = BC = 1$.
Then
$M
=(a, 1-a)
$,
where
$0 < a < 1$.
Let
$E = (0, 1+v)
$,
where $v > 0
$.
Then
line $EM$
is
$\frac{y-(1+v)}{x}
=\frac{(1-a)-(1+v)}{a}
=\frac{-a-v}{a}
=-1-\frac{v}{a}
$
or
$y
=1+v-x(1+\frac{v}{a})
$.
Putting
$y=0$,
$x
=\frac{1+v}{1+\frac{v}{a}}
$,
so
$F
=(\frac{1+v}{1+\frac{v}{a}}, 0)
$.
$|AC|^2
=2
$
$\begin{array}\\
|EF|^2
&=(\frac{1+v}{1+v/a})^2+(1+v)^2\\
&=(1+v)^2(1+\frac{1}{(1+v/a)^2})\\
&=(1+v)^2(\frac{1+(1+v/a)^2}{(1+v/a)^2})\\
&=(1+v)^2(\frac{a^2+(a+v)^2}{(a+v)^2})\\
&=(1+v)^2(\frac{2a^2+2av+v^2}{(a+v)^2})\\
\end{array}
$
According to Wolfy,
$\begin{array}\\
|EF|^2-|AC|^2
&=(1+v)^2 (\frac{2a^2+2av+v^2}{(a+v)^2}) -2\\
&=v \frac{2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v}{(a+v)^2}\\
\end{array}
$
Therefore,
when
$0
\lt 2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v\\
= a^2(2v+4)+a(2v^2+4v-2)+v^3+2 v^2-v\\
$
for
$0 < a < 1$
and
$v > 0
$,
|EM| > |AC|.
Again
according to Wolfy,
if
$a = 1-c$,
then
$f(v, a)\\
=2 a^2 v+4 a^2+2 a v^2+4 a v-2 a+v^3+2 v^2-v\\
=2 c^2 (v+2)-2 c (v^2+3 v+1)+v^3+3 v^2+3 v/2\\
$
If $c \le 0$
(i.e., $a \ge \frac12$),
all the terms are positive,
so |EM| > |AC| for all $v > 0$.
However,
if $c > 0$
(i.e., $a < \frac12$),
for small $v$
this is about
$4c^2-2c
=2c(2c-1)
\lt 0
$,
so
there are values of $v$
for which the expression is negative
so that
|EM| < |AC|.
Again according to Wolfy,
this occurs at the following values of $v$:
$
a=1/3: v< 0.19795\\
a=1/4: v< 0.27832\\
a=1/10: v< 0.38509\\
a=1/100: v< 0.41382\\
a=1/10000: v<0.41421\\
$
Looking at these values,
it seems that
the $v$ bound is approaching
$\sqrt{2}-1$.
To show this is true,
Wolfy gets
$f(\sqrt{2}-1, a)
=2(1+\sqrt{2})a^2
\gt 0
$
for $a > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Every sixth polynomial shares a factor of $(a^2-6)$ I currently looking at the polynomials you get from the series expansion of
$$
\frac{1-x^2}{1-ax+2x^2}=1+a x
+(a^2-3) x^2
+(a^3-5a) x^3
+\underbrace{(a^4-7a^2+6)}_{(a-1)(a+1)(a^2-6)} x^4+\dots
$$
W|A helped here...
What I found is that from $x^{4}$ onwards, every sixth polynomial shares a factor of $(a^2-6)$. How to prove that?
| Consider $\displaystyle f(a,x)=\frac{1-x^2}{1-a x+2x^2}+{1\over2}=\frac{3-ax}{2(1-a x+2x^2)}$, which is just your function augmented by $1/2$. What you ask amounts at proving that the series expansions of both $f(\sqrt6, x)$ and $f(-\sqrt6, x)$ around $x=0$ do not contain powers of $x$ with exponent ${4+6k}$.
That is true because, by multiplying both numerator and denominator of $f(\pm\sqrt6, x)$ by $(1 \pm \sqrt6 x + 2 x^2)(1 + 2 x^2)$, one gets:
$$
f(\pm\sqrt6, x)
={3/2 \pm \sqrt6 x + 3 x^2 \pm \sqrt6 x^3 \mp 2 \sqrt6 x^5\over 1+8x^6}\\
=(3/2 \pm \sqrt6 x + 3 x^2 \pm \sqrt6 x^3 \mp 2 \sqrt6 x^5)
\sum_{k=0}^\infty(-8x^6)^k
$$
and the polynomial factor does not contain $x^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Find number of roots of the equation $e^x(x^4 + 4x^3 + 12x^2 + 24x + 24) + 1 = 0$ Find number of roots of the equation $e^x(x^4 + 4x^3 + 12x^2 + 24x + 24) + 1 = 0$
Using Descartes rule, number of positive roots is zero and there can be a maximum of 4 negative roots.
Also, for the function $P(x)=x^4 + 4x^3 + 12x^2 + 24x + 24$, the double derivative $P''(x)>0$. So the function can have either 2 negative roots or no root at all.
But I still lack information required to prove that the function has no real roots.
| Just observe that
\begin{align}
P(x) &=(x^4+4x^3+4x^2)+(8x^2+24x+18)+6\\
&=(x^2+2x)^2+2(2x+3)^2+6>0
\end{align}
Therefore
$$e^xP(x)+1>0$$ for all $x \in \mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$ The line $\dfrac{x+6}{5}=\dfrac{y+10}{3}=\dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.
My Attempt:
Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.
Let the equation of $AB$ be $\dfrac{x-7}{a_1}=\dfrac{y-2}{b_1}=\dfrac{z-4}{c_1}$
and the equation of $BC$ be $\dfrac{x-7}{a_2}=\dfrac{y-2}{b_2}=\dfrac{z-4}{c_2}$
As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$
Also as the triangle is a isosceles right triangle, so angle between $AC$ and
$BC$ will be same as the angle between $CA$ and $BA$. Let it be $\theta$.
So $\cos\theta=\dfrac{5a_1+3b_1+8c_1}{\sqrt{5^2+3^2+8^2}\sqrt{a_1^2+b_1^2+c_1^2}}=\dfrac{5a_2+3b_2+8c_2}{\sqrt{5^2+3^2+8^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Squaring both sides we get
$\dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=\dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$
But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.
| Efficient way is take angle b/w line 45 as given triangle is 90 with isosceles.Take parametric point on given line as (5¥-6,3¥-10,8¥-18) and make vector with (7,2,4) and then then take cos45 of this vector and (5,3,8).You will get a simple quadratic I.e. ¥^2-5¥+6=0. Imply ¥=2,3.means you get the value of ¥ due to other 2 vertex of triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Form of a real orthogonal matrix Q. Let $A$ be a real orthogonal matrix of order $2$ with $|A|=1$. Show that $\exists\theta$ s.t. $$A=\begin{pmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}\tag{4 marks}$$
My answer:
Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ where $a,b,c,d\in\Bbb R$. Now, since $A$ is orthogonal, we have,
$$AA^T=I_2\implies \begin{pmatrix}a^2+b^2&ac+bd\\ac+bd&c^2+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\implies\begin{cases}a^2+b^2=c^2+d^2=1\quad(1)\\ ac+bd=0\qquad\qquad\quad~(2)\end{cases}$$
Also, since $|A|=1$, we have $ad-bc=1\tag3$
Using $(1)$ and $(3)$, we get,
$$(a-d)^2+(b+c)^2=0$$
Since $a,b,c,d\in\Bbb R$ and squares of reals are always non-negative, in order to maintain equality, we must have,
$$a-d=0\text{ and }b+c=0\implies d=a~\land~c=-b$$
Putting this back in $(3)$. we see that it's the equation of the unit circle in the $ab$ plane, hence the parametric representation $a=\cos\theta$ and $b=\sin\theta$. The conclusion follows.
My question is how much marks this solution will get (in general) ? Also, it'd be nice to see simpler formal proofs. Thanks.
| The solution in the original post is correct.
Here we present another way forward, which seems fairly efficient. We note that
$$a^2+b^2=1\implies a=\cos \theta\,\,\text{and}\,\,b=\sin \theta$$
$$c^2+d^2=1\implies c=\cos \phi\,\,\text{and}\,\,d=\sin \phi$$
for some $\theta$ and some $\phi$.
Then, we have
$$ac+bd=0\implies \cos (\theta -\phi)=0\implies \theta -\phi =(2k+1)\pi/2$$
for any integer $k$.
Eliminating $\phi$, we find that $c=\pm \sin \theta$ and $d=\mp \cos \theta$.
We can choose the value of $k$ such that $c=-\sin \theta$ and $d=\cos \theta$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if $A\ge B$ then $\left[ {\begin{array}{*{20}{c}} A & B \\ B & A \\ \end{array}} \right]\ge0$. Let $A$ and $B$ be $n \times n$ matrices, i.e., $A, B \in M_n$. Also, $A \ge 0$, $B \ge 0$, and $A-B \ge 0$ which mean all these matrices are semi-positive-definite.
Why does
$\left[ {\begin{array}{*{20}{c}}
A & B \\
B & A \\
\end{array}} \right] \ge 0$?
| $\left[ {\begin{array}{*{20}{c}}
A & B \\
B & A \\
\end{array}} \right] =\left[ {\begin{array}{*{20}{c}}
A-B & 0 \\
0 & A-B \\
\end{array}} \right] +\left[ {\begin{array}{*{20}{c}}
B & B \\
B & B \\
\end{array}} \right] \ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Product of generating functions Let $f(x) = \sum_{i=0}^\infty a_ix^i$ and $g(x) = \sum_{i=0}^\infty b_ix^i$ where $a_n = 1$ and $b_n = 2^n$ for all natural numbers $n$
What are the first three terms of the sequence generated by $f(x)g(x)$
?
So I know that $f(x)$ will generate $\{1,1,1,1,1,.... \}$ and $g(x)$ will generate $\{1,2,4,8,16,32,64,....... \}$
The first term $c_0$ of the sequence $f(x)g(x)$will be $1 \times 1 = 1$ and the second term $c_1$ will be $a_0 \times b_1 + a_1 \times b_0 = 1 \times 2 + 1 \times 1 = 3$ and finally $c_2 = 1 \times 4 + 1 \times 2 + 1 \times 1 = 7$
Now what is the general formula for the sequence generated by
$f(x)g(x)$, assume that it generates $\{c_0,c_1,c_2,c_3,..........\}$
If I calculates $c_3$ it would be easier to see that pattern, $c_3 = 8 + 4 + 2 + 1 = 15$
and so the sequence is $\{1,3,7,15,31,.................... \}$
It is basically $\{1,2^{2}-1,2^3-1,2^4-1,.... \}$ and so it is $\{1,a^2-1,a^3-1,a^4-1,.... \}$ where $a=2$, Is this general formula correct ?
I mean what is the formula , I just know the pattern here. So is the formula just $$c_k = 2^k-1$$ where $k>1$
What function generates the sequence $\{c_0,c_1,c_2,..... \}$?
Well I know that $\frac{1}{1-2x}$ generates the sequence $\{1,2,2^2,2^3,... \}$ and that $\frac{-1}{1-x}$ generates $\{-1,-1,-1,.... \}$ but adding these two functions will gives us the sequence $\{0,1,2^2-1,2^3-1 \}$
And what should I do to eliminate that zero, multiply by $x$ will add more zero, what should do I do eliminate one zero ?
| $f(x)=\frac{1}{1-x}$, $g(x)=\frac{1}{1-2x}$, $f(x)g(x)=\frac{1}{(x-1)(2x-1)}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\sum{\frac{1}{m^2+n^2}}$ I want to prove that $\sum_{1\leq m^2+n^2\leq R^2}{\frac{1}{m^2+n^2}}=2\pi\log R+O(1)$ as $R\rightarrow\infty$.
For this, I'm trying to approximate the sum by using the integral $\int_{1\leq r\leq R}{\frac{1}{x^2+y^2}}dxdy=\int_{0}^{2\pi}\int_{1}^{R}{\frac{1}{r}}drd\theta=2\pi\log R$
How can I show that $\sum_{1\leq m^2+n^2\leq R^2}{\frac{1}{m^2+n^2}}=\int_{1\leq r\leq R}{\frac{1}{x^2+y^2}}dxdy$ as $R\rightarrow\infty$ rigorously?
Any help will be appreciated.
| $$
\begin{align}
\sum_{m,n=1}^R\frac1{m^2+n^2}
&=2\sum_{m=1}^R\sum_{n=1}^m\frac1{m^2+n^2}-\sum_{n=1}^R\frac1{2n^2}\\
&=2\sum_{m=1}^R\frac1m\sum_{n=1}^m\frac{\frac1m}{1+\left(\frac nm\right)^2}-\sum_{n=1}^R\frac1{2n^2}\\
\end{align}
$$
Comparing the Riemann sum $\sum\limits_{n=1}^m\frac{\frac1m}{1+\left(\frac nm\right)^2}$ and the integral $\int_0^1\frac{\mathrm{d}x}{1+x^2}$, we see that
$$
\sum_{n=1}^m\frac{\frac1m}{1+\left(\frac nm\right)^2}=\frac\pi4+O\left(\frac1m\right)
$$
Since $\sum\limits_{m=1}^n\frac1m=\log(n)+O(1)$, and $\sum\limits_{n=1}^\infty\frac1{2n^2}=\frac{\pi^2}{12}$, we get that
$$
\sum_{m,n=1}^R\frac1{m^2+n^2}=\frac\pi2\log(R)+O(1)\tag{1}
$$
Furthermore, since $\log(R/\sqrt2)=\log(R)+O(1)$,
$$
\sum_{m,n=1}^{R/\sqrt2}\frac1{m^2+n^2}=\frac\pi2\log(R)+O(1)\tag{2}
$$
The sum over the terms where $m=0$ or $n=0$ is handled by
$$
\sum_{m=1}^\infty\frac1{m^2}=\frac{\pi^2}6=O(1)\tag{3}
$$
Since
$$
\overbrace{\left\{1\le m,n\le R/\sqrt2\right\}}^{(2)}\subset\left\{1\le m^2+n^2\le R^2\text{ and }m,n\ge1\right\}\subset\overbrace{\left\{1\le m,n\le R\right\}}^{(1)}
$$
and accounting for the four quadrants and the four borders where $m=0$ or $n=0$, we get
$$
\sum_{1\le m^2+n^2\le R^2}\frac1{m^2+n^2}=2\pi\log(R)+O(1)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Define a bijection Okay so I know that I asked this already but I want to ask how much progress I have made. so the question is
Construct a bijection between $[1,2]$ and $[3,5)$
So I have:
\begin{equation}
f(x) = \left \{ \begin{array}{ll} 3 + 2^{1-n} & \textrm{ if } x = 1+2^{1-n} \textrm{ for } n \in \mathbb{N} \\
2x+1 & \textrm{ if } x \neq 1+2^{1-n} \end{array} \right.
\end{equation}
Sorry about the formatting, but is this correct?
Edit: the linked post is mine. I'm just asking since I didn't get a really satisfactory answer if the solution I have is going towards the right direction.
Edit2: changing the equation to what improvements I'm getting, still any feedback is helpful. Thanks!
| The function $f$ is onto:
Case 1
Let $y \in [3,5) $ be of the form: $3 + 2^{1-n}$ for some $n \in \mathbb{N}$. Then $x = 1 + 2^{1-n}$ is such that $f(x) = y$. In particular, if $n$ is such that $3 + 2^{1-n} \in [3,5)$ then $n \geq 0$. This in turn implies that $1 \leq 1 + 2^{1-n} \leq 2$.
Case 2
Let $y \in [3,5) $ not be of the form: $3 + 2^{1-n}$ for some $n \in \mathbb{N}$. Then $x = (y-1)/2 \in [1,2]$ is such that $f(x) = y$. In particular if $y \in [3,5) $ then $(y-1)/2 \in [1,2]$.
The function $f$ is inyective:
Let $a,b \in [1,2]$ be such that $f(a) = y = f(b)$
Case 1
If $y \neq 3 + 2^{1-n}$ then $f(a) = 2a + 1 = 2b + 1 = f(b)$. It follows that $a = b$.
Case 2
If $y = 3 + 2^{1-n}$ then $f(a) = 3 + 2^{1-n_a}$ and $f(b) = 3 + 2^{1-n_b}$ where $a = 1 + 2^{1-n_a}$ and $b = 1 + 2^{1-n_b}$ respectively. Then:
\begin{equation}
\begin{aligned}
3 + 2^{1-n_a} = 3 + 2^{1-n_a} & \Leftrightarrow 2^{1-n_a} = 2^{1-n_a} \\ & \Leftrightarrow log_2\left(2^{1-n_a}\right) = log_2\left(2^{1-n_b}\right) \\ & \Leftrightarrow n_a = n_b \\ & \Leftrightarrow a = b.
\end{aligned}
\end{equation}
| {
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Evaluation of $ \int_{-k}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx$ Evaluation of $$\displaystyle \int_{-k}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx\;,$$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int_{-k}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx = 2\int_{0}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx$$
Now Substiute $$\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\;,$$ So we get $$\displaystyle I = 2\int_{0}^{k}\frac{\sec \frac{x}{2}}{\sqrt{1-\tan^2 \frac{x}{2}-\cos k(1+\tan^2 \frac{x}{2})}}dx$$
So we get $$\displaystyle I = 2\int_{0}^{k}\frac{\sec \frac{x}{2}}{\sqrt{(1-\cos k)-(1+\cos k)\tan^2\frac{x}{2}}}dx = 2\int_{0}^{k}\frac{\sec \frac{x}{2}}{\sqrt{2\sin^2\frac{k}{2}-2\cos^2 \frac{k}{2}\tan^2 \frac{x}{2}}}dx$$
Now How can I solve after that, Help me
Thanks
| I am trying to continue from your method but before that let me post my method:
use substitution $$\sqrt{cosx-cosk}=t$$ so $$\frac{dx}{\sqrt{cosx-cosk}}=\frac{-2dt}{sinx}=\frac{-2dt}{\sqrt{1-(t^2+cosk)^2}}=\frac{-2dt}{\sqrt{1-(t^2+cosk)}\sqrt{1+(t^2+cosk)}}=\frac{-2dt}{\sqrt{2sin^2\frac{k}{2}-t^2}\sqrt{2cos^2\frac{k}{2}+t^2}}$$
if we let $a=\sqrt{2}sin\frac{k}{2}$ and $b=\sqrt{2}cos\frac{k}{2}$ we have
$$I=\int\frac{-2dt}{\sqrt{a^2-t^2}\sqrt{b^2+t^2}}$$ which wolfram gives as
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the number of all ordered triplets $(A,B,C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C.$ Let $X$ be as set containing $n$ elements.Find the number of all ordered triplets $(A,B,C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C.$
I have no idea how to solve this problem.Please help me.Thanks.
| First let's solve this simplier problem : What's the number $\mathcal{N}$ of sets of $X$ such that $|A| = a$, $|B| = b$ and $|C| = c$ with $n \geq a>b>c >0$?
There is $n\choose a$ way to choose $A$, then $a\choose b$ ways to choose $B$ inside $A$, then $b\choose c$ way to choose $C$ inside $B$. This way, we have all the possible set without double.
$$\mathcal{N}(a,b,c) = {n\choose a}{a \choose b}{b\choose c}$$
So now we can answer the original question (with $n \geq 3$):
$$\mathcal{N} = \sum_{a=3}^n \sum_{b=2}^a \sum_{c=1}^b {n\choose a}{a \choose b}{b\choose c}$$
$$=\sum_{a=3}^n \left( {n\choose a} \sum_{b=2}^a \left( {a \choose b} \sum_{c=1}^b {b\choose c} \right) \right)$$
But with classical binomial formula, this is also equal to
$$=\sum_{a=3}^n \left( {n\choose a} \sum_{b=2}^a \left( {a \choose b} \left( \sum_{c=0}^b {b\choose c} - 1 \right)\right) \right)$$
$$=\sum_{a=3}^n \left( {n\choose a} \sum_{b=2}^a \left( {a \choose b} ( 2^b -1 ) \right) \right)$$
$$=\sum_{a=3}^n \left( {n\choose a} \left( \sum_{b=2}^a {a \choose b} 2^b - \sum_{b=2}^a {a \choose b} \right) \right)$$
$$=\sum_{a=3}^n \left( {n\choose a} \left( \sum_{b=0}^a {a \choose b} 2^b -( 1 + 2a ) - \sum_{b=0}^a {a \choose b} + (1+a) \right) \right)$$
$$=\sum_{a=3}^n \left( {n\choose a} \left( 3^a-2^a-a \right) \right)$$
$$=\sum_{a=0}^n {n\choose a} 3^a - (1+3n+9\frac{n(n-1)}{2}) -\sum_{a=0}^n {n\choose a} 2^a + (1+2n+4\frac{n(n-1)}{2}) - \sum_{a=0}^n {n\choose a} a + (1+2n) $$
Finally we get
$$= 4^n -3^n - 2^{n-1}n -5\frac{n(n-1)}{2} +n+1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Integrate $\int \frac{dx}{3^x-8}$ I have a problem with solving this Integrate
$$\int \frac{dx}{3^x-8}$$
At first I use substitute $$3^x =t $$
$$\log_3{t} = x$$
$$\frac{1}{t\ln{3}}dt = dx$$
Next:
$$\int \frac{dt}{t^2\ln3} - \int\frac18dx$$
$$=\frac{1}{\ln3}\int \frac{dt}{t^2} - \int\frac18dx$$
$$=\frac{1}{\ln3} \left(\frac{-1}{t}\right) - \frac18$$
$$=\frac{1}{\ln3} \left(\frac{-1}{3^x}\right) - \frac18$$
But Wolfram alpha gives me different result. Where did I go wrong?
| Let $3^x-8=u\implies 3^x\ln(3)\ dx=du$ or $dx=\frac{1}{(u+8)\ln (3)}\ du$
$$\int \frac{1}{3^x-8}\ dx=\int \frac{1}{u(u+8)\ln 3}\ du $$
$$=\frac{1}{\ln 3}\int \frac{1}{u(u+8)}\ du $$
$$=\frac{1}{8\ln 3}\int\left( \frac{1}{u}-\frac{1}{u+8}\right)\ du $$
$$=\frac{1}{8\ln 3}\ln\left| \frac{u}{u+8}\right| +C$$
$$=\frac{1}{8\ln 3}\ln\left| \frac{3^x-8}{3^x}\right| +C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is my integration of $\sqrt{4z^2 - 4z + 2}$ correct? I'm trying to
$$
\int \sqrt{4z^2 - 4z + 2}\ dz
$$
the integrand I first rewrite to (completing the square)
$$
4\left(z^2 - z +\frac{1}{4} - \frac{1}{4}\right) + 2 = 4\left(z - \frac{1}{2}\right)^2 + 1
$$
I apply $\int \sqrt{a^2 + x^2} \ dx = \frac{x}{2} \sqrt{a^2+x^2} + \frac{a^2}{2} \ln \left(x + \sqrt{a^2+x^2}\right)$.
So, before simplification, I get
$$
(z-1/2)\sqrt{4z^2-4z+2} + 1/2 \ln \left[ 2(z-1/2) + \sqrt{4z^2 - 4z +2} \right]
$$
According to my book, the solution is
$$
(z-1/2)\sqrt{z^2-z+1/2} + 1/4 \ln \left[ z - 1/2 + \sqrt{z^2 - z +1/2} \right]
$$
Is my solution equivalent to the book? I dont see how the book gets the `1/4' before the log.
| Your integration methods are all right.
Few silly mistakes that I observed are:
*
*No need to keep it $4(z-\frac{1}{2})^2+1$. You could have made it $(2z-1)^2+1$ directly.
*You should change the $dz$ to $\frac{1}{2}d(2z-1)$,which you have not done. If you take this into account, your result will be according to the textbook. That is, the extra $\frac{1}{2}$ is missing from your result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant of rank-$1$ update of multiple of identity matrix I've got to calculate determinant for such matrix:
$$ \begin{bmatrix}
a_1+b & a_2 & \cdots & a_n\\
a_1 & a_2+b & \cdots & a_n\\
\vdots & \vdots & \ddots & \vdots\\
a_1 & a_2 & \cdots & a_n+b\\
\end{bmatrix} $$
Please give me some tips how to calculate this.
| Set $A=\sum\limits_{i=1}^na_i$. By multilinearity,
\begin{align*}
&\begin{vmatrix}
a_1+b &a_2&\dots&a_n\\
a_1&b+a_2 &\dots&a_n \\
\vdots&\vdots&&\vdots\\
a_1 a_2&\dots &a_n+b
\end{vmatrix}=
\begin{vmatrix}
A+b &a_2&\dots&a_n\\
A+b&b+a_2 &\dots&a_n \\
\vdots&\vdots&&\vdots\\
A+b & a_2&\dots &b+a_n
\end{vmatrix}\\[1ex]
&=(A+b)\begin{vmatrix}
1 &a_2&\dots&a_n\\
1&b+a_2 &\dots&a_n \\
\vdots&\vdots&&\vdots\\
1 & a_2&\dots &b+a_n
\end{vmatrix}=(A+b)\begin{vmatrix}
1 &a_2&\dots&a_n\\
0&b &\dots& 0 \\
\vdots&\vdots&&\vdots\\
0 & 0&\dots &b
\end{vmatrix}\\[1ex]
&=\color{red}{(A+b)b^{n-1}}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
| In order to prove anything whatsoever about $\frac{0}{0}$, we need a definition of $\frac{a}{b}$ where $a$ and $b$ are, say, integers. The correct definition is $ \frac{a}{b} = ab^{-1}$, where the definition of $b^{-1}$ is the number (which is unique, when it exists, according to some basic properties of numbers) such that $bb^{-1} = 1$. But then one can see (again, assuming some basic properties of numbers) that $0^{-1}$ does not exist, so $\frac{0}{0}$ isn't even defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 5
} |
Prove that $\overrightarrow{GH}.\overrightarrow{IJ}=-2x^2+8x-2$ in a regular hexagon We know that the hexagon is regular and:
$\overline{AB}=1$;
$\overline{AG}=\overline{CI}=\overline{DH}=\overline{FJ}=x$;
How would you prove that $\overrightarrow{GH}.\overrightarrow{IJ}=-2x^2+8x-2$?
| $\qquad\qquad\qquad$
Consider $\triangle{GAK}$ where $K$ is the intersection point between $GH$ and $IJ$. ($K$ is the center of the hexagon.) Since $GA=x,AK=1,\angle{GAK}=60^\circ$, by the law of cosines, we have
$$GK=\sqrt{x^2+1^2-2\cdot x\cdot 1\cos(60^\circ)}=\sqrt{x^2-x+1}$$
Here, note that $GK=HK=IK=JK$.
Also, consider an isosceles triangle $BIG$ where $BG=BI=1-x,\angle{GBI}=120^\circ$. So, we have $GI=\sqrt 3(1-x)$.
Hence, applying the law of cosines to $\triangle{GIK}$, we have
$$\cos\angle{GKI}=\frac{KI^2+KG^2-IG^2}{2\cdot KI\cdot KG}=\frac{2(x^2-x+1)-3(1-x)^2}{2(x^2-x+1)}$$
Thus, we have
$$\vec{GH}\cdot\vec{IJ}=|\vec{GH}|\times |\vec{IJ}|\times \cos\angle{GKI}$$$$=4(x^2-x+1)\cdot\frac{2(x^2-x+1)-3(1-x)^2}{2(x^2-x+1)}=-2x^2+8x-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
| Using
$$\log_{10}(x) = \frac{\ln(x)}{\ln(10)}$$
then
\begin{align}
D \left[ 10^{x} \, \log_{10}(x) \right] &= D \left[ e^{x \, \ln(10)} \, \frac{\ln(x)}{\ln(10)} \right] \\
&= \frac{1}{\ln(10)} \, \left[ \frac{10^{x}}{x} + \ln(10) \, 10^{x} \, \ln(x) \right] \\
&= 10^{x} \, \left[ \frac{1}{ x \, \ln(10)} + \ln(10) \, \log_{10}(x) \right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Find a recurrence relation and generating function of....
Model the amount of crab being caught per year based on the assumption that the # of crab caught in a year is the average of the # caught in the 3 preceding years.
a.) Find a recurrence relation for $C_n$, the # of crab caught in a year $n$.
For this I got:
$$C_n = {C_{n-1} + C_{n-2} + C_{n-3}\over 3}$$
b.) Find a generating function:
$$f(x)= \sum_{n≥1} C_nx^n $$
With initial conditions $C_1 = 1$, $C_2 = 2$ and $C_3 = 2$.
We had gone over the auxiliary equation and the generating function methods in class but I am not sure how to proceed after my recurrence relation, any help is appreciated.
| Caveat: This is maybe not the best method. But it gets the job done.
Let the generating function of $f$ be given by $f(x)=\sum_{n=1}^\infty C_n x^n$. Then,
$$
\begin{align}
3f(x)&=3(C_1x+C_2x^2+C_3x^3)+\sum_{n=4}^\infty 3C_n x^n
= 3(x+2x^2+2x^3)+\sum_{n=4}^\infty (C_{n-1}+C_{n-2}+C_{n-3}) x^n \\
&= 3(x+2x^2+2x^3)+x\sum_{n=4}^\infty C_{n-1}x^{n-1} +x^2\sum_{n=4}^\infty C_{n-2}x^{n-2}+x^3\sum_{n=4}^\infty C_{n-3}x^{n-3}\\
&= 3x+6x^2+6x^3+x\sum_{n=3}^\infty C_{n}x^{n} +x^2\sum_{n=2}^\infty C_{n}x^{n}+x^3\sum_{n=1}^\infty C_{n}x^{n} \\
&= 3x+6x^2+6x^3+x(f(x)-x-2x^2) +x^2(f(x)-x)+x^3f(x) \\
&= 3x+6x^2+6x^3-x^2-2x^3 -x^3+(x+x^2+x^3)f(x) \\
&= 3x+5x^2+3x^3+(x+x^2+x^3)f(x) \\
\end{align}
$$
so
$$
f(x) = \frac{3x+5x^2+3x^3}{3-x-x^2-x^3}
$$
(wherever it is well-defined, i.e. $x\neq 3$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find Matrix T of a Linear Transformation $T: M_{2\times 2} \to P_3$ Been staring at this practice problem and scribbling nonsense for quite a while now googled extensively and such but nothing triggered any spark of understanding.
All in all a helpful shove off the correct cliff would be appreciated.
Consider the linear transformation $T: M_{2\times 2} \to P_3$ given by
$$
T\left ( \begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \right ) = (a + b + c) +(a - b - c)x + (a + d)x^2 + (b + c - d)x^3
$$
(a) Find the matix of T with respect to the usual bases for $M_{2x2}$ and $P_3$
(b) Now consider the basis $B$ of $M_{2x2}$ given by
$$
B =\left\{
\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix},
\begin{bmatrix}0 & 1 \\1 & 0 \end{bmatrix},
\begin{bmatrix}0 & 1 \\-1 & 0 \end{bmatrix},
\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}
\right\}
$$
and the basis of $C$ of $P_3$ given by
$C =\left\{ 1+x+x^2, 1-x+x^3, x^2+x^3,x^2-x^3\right\}$
find the matrix $\left[T\right]_{C\leftarrow B}$
| For the $2\times 2$ matrices, the standard basis would be
$$
\begin{pmatrix} 1 & 0 \\ 0&0 \end{pmatrix}, \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}, \begin{pmatrix} 0&0\\1&0 \end{pmatrix}, \begin{pmatrix} 0&0\\0&1 \end{pmatrix}
$$
and for the polynomials the standard basis is $\{1, x, x^2, x^3\}$. Now you want to see what $T$ does to the standard basis of $M_{22}$ and write it in terms of the standard basis of $P_3$. Using the definition of $T$ we get that:
$$
\begin{split} T\begin{pmatrix} 1&0\\0&0 \end{pmatrix}&=1+x+x^2 \\ T\begin{pmatrix} 0&1\\0&0 \end{pmatrix} &=1-x+x^3 \\ T\begin{pmatrix} 0&0\\1&0 \end{pmatrix} &= 1-x+x^3 \\ T\begin{pmatrix} 0&0\\0&1 \end{pmatrix} &=x^2-x^3 \end{split}
$$
Now in the standard basis of $P_3$, $1+x+x^2$ corresponds to the column vector:
$$
\begin{pmatrix} 1\\1\\1\\0 \end{pmatrix}
$$
Combining the column vectors for each of the basis matrices gives the matrix of $T$ in the standard basis:
$$
T=\begin{pmatrix} 1&1&1&0 \\ 1&-1&-1&0 \\ 1&0&0&1 \\0&1&1&-1 \end{pmatrix}
$$
It would be a good exercise for you to do part b). The process will be the same but you will need to write the polynomials you get in terms of the polynomials in the new basis. For example, in the new basis $C$, the polynomial $1+x+x^2$ now corresponds to the vector:
$$
\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$ $$ \lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$$
How do I solve this without using derivatives or integrals.
| Take a look at the following steps
$$\eqalign{
& L = \mathop {\lim }\limits_{x \to 0} {{1 - {{(3x + 1)}^{0.5}}} \over {2 - {{(5x + 4)}^{0.5}}}} = \mathop {\lim }\limits_{x \to 0} {{1 - {{(3x + 1)}^{0.5}}} \over {2 - {{(5x + 4)}^{0.5}}}}{{2 + {{(5x + 4)}^{0.5}}} \over {2 + {{(5x + 4)}^{0.5}}}}{{1 + {{(3x + 1)}^{0.5}}} \over {1 + {{(3x + 1)}^{0.5}}}} \cr
& \,\,\,\, = \mathop {\lim }\limits_{x \to 0} {{1 - (3x + 1)} \over {4 - (5x + 4)}}{{2 + {{(5x + 4)}^{0.5}}} \over {1 + {{(3x + 1)}^{0.5}}}} \cr
& \,\,\,\, = \mathop {\lim }\limits_{x \to 0} {{ - 3x} \over { - 5x}}{{2 + {{(5x + 4)}^{0.5}}} \over {1 + {{(3x + 1)}^{0.5}}}} \cr
& \,\,\,\, = \mathop {\lim }\limits_{x \to 0} {{ - 3x} \over { - 5x}}\mathop {\lim }\limits_{x \to 0} {{2 + {{(5x + 4)}^{0.5}}} \over {1 + {{(3x + 1)}^{0.5}}}} \cr
& \,\,\,\, = \left( {{3 \over 5}} \right)\left( {{4 \over 2}} \right) = {6 \over 5} \cr} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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A nice double integral on the square $[0,1]$x$[0,1]$ It is recalled that the fractional part of a real $x$, noted {$x$}, is defined by
{$x$} $=x-\lfloor x\rfloor $ where $\lfloor x\rfloor $ is the floor function (largest integer not greater than $x$, many times noted $[x]$).
I don’t remember where I saw the double integral below but I have never forgot it gives us a very nice equality: it is equal to a linear expression $a\gamma+b$
of the important Euler’s Constant $\gamma$ with very simple rational coefficients $a$ and $b$.
It is well known that the constant $\gamma$ is defined by
$$\gamma=\lim_{n\rightarrow \infty}(\sum_{k=1}^n \frac{1}{k}-\ln n)$$ being perhaps the most celebrated limit of the indeterminate form $\infty-\infty$
So we know $$\int_0^1\int_0^1\left\{\frac{x}{y}\right\}dxdy=a\gamma +b$$
Determine the values of rational $a$ and $b$.
| $$\int_0^1\int_0^1\{\frac{x}{y}\}dxdy \\
= \lim_{N\to\infty} \sum_{n=1}^{N}\int_{\frac{1}{n+1}}^{\frac{1}{n}}((\sum_{k=0}^{[\frac{1}{y}]-1}\int_{ky}^{(k+1)y} (\frac{x}{y}-[\frac{x}{y}]) dx) + \int_{ny}^1 (\frac{x}{y}-[\frac{x}{y}])dx) dy \\
= \lim_{N\to\infty} \sum_{n=1}^{N} \frac{1}{2}\ln(\frac{n+1}{n})-\int_{\frac{1}{n+1}}^{\frac{1}{n}}(\sum_{k=0}^{n-1}\int_{ky}^{(k+1)y} [\frac{x}{y}] dx + \int_{ny}^1 [\frac{x}{y}]dx) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \int_{\frac{1}{n+1}}^{\frac{1}{n}}(\sum_{k=0}^{n-1}\int_{ky}^{(k+1)y} k dx + \int_{ny}^1 ndx) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \int_{\frac{1}{n+1}}^{\frac{1}{n}}(\sum_{k=0}^{n-1}ky + n(1-ny)) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \int_{\frac{1}{n+1}}^{\frac{1}{n}}(y\frac{n(n-1)}{2} + n(1-ny)) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \int_{\frac{1}{n+1}}^{\frac{1}{n}}(-y\frac{n(n+1)}{2} + n) dy\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \sum_{n=1}^{N} \frac{2n-1}{4n(n+1)}\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)- \frac{1}{4} \sum_{n=1}^{N} (-\frac{1}{n}+\frac{3}{n+1})\\
= \lim_{N\to\infty} \frac{1}{2}\ln(N+1)-\frac{1}{2}\sum_1^{N+1}\frac{1}{n} + \frac{3}{4}-\frac{1}{4(N+1)}\\
=-\frac{1}{2}\gamma+\frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
On proving $\lim \frac{n^2}{n^2+n+1} = 1$
Prove that $\lim \frac{n^2}{n^2+n+1} = 1$
Let $\varepsilon > 0$ and let $N = \frac{1}{\varepsilon}.$ Then $n > N$ implies $n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon.$
But $\displaystyle \frac{1}{n} = \frac{n+1}{n(n+1)} = \frac{n+1}{n^2+n} > \frac{n+1}{n^2+n+1} = \bigg|\frac{n^2}{n^2+n+1}-1\bigg|.$
Therefore $\bigg|\frac{n^2}{n^2+n+1}-1\bigg| < \varepsilon$, hence $\displaystyle \lim \frac{n^2}{n^2+n+1} = 1.$
Could someone please verify whether the above is correct.
| if $\forall n.a_n \gt 0$
$$
\lim_{n\to\infty} a_n = 1 \Leftrightarrow \lim_{n\to\infty} \frac1{a_n} = 1
$$
set
$$
a_n = \frac{n^2+n+1}{n^2} = 1+\frac1{n}+\frac1{n^2}
$$
so for $n \gt 1$
$$
0 \lt a_n-1 \lt \frac2n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Maximum remainder $(a-1)^n+(a+1)^n\mod a^2$ for $3\le a\le 1000$ Here's the problem:
Let $r$ be the remainder when $(a−1)^n + (a+1)^n$ is divided by $a^2$.
For example, if $a = 7$ and $n = 3$, then $r = 42$ since $63 + 83 = 728 \equiv 42 \pmod{49}$. And as $n$ varies, so too will $r$, but for $a = 7$ it turns out that $r_{max} = 42$.
For $3 \le a \le 1000$, find $\sum r_{max}$.
Source
What I've tried:
$$X=(a-1)^n+(a+1)^n=2\sum_{r=0,2|n-r}^n {}^n{\rm C}_ra^r$$
So:
$$X\mod a^2 = \begin{cases}2&2|n\\2an&2\not|n\end{cases}$$
Now I think $n_{(r_{max})}$therefore should be the largest odd number strictly less than $a/2$ and $r_{max}$ should be $2an_{(r_{max})}$.
So answer should be :
$$\sum_{r=3}^{1000}2an_{(r_{max})}$$
Am I correct?
Do you have any other better ideas?
| As you said the remainder is
$$r\mod a^2 = \begin{cases}2&2|n\\2an&2\not|n\end{cases}.$$
Now observe that the maximum value of $2an \mod a^2$ could be $a(a-1)$. IT holds if and only if the following system of congruence
$$\begin{cases}n\equiv1&\mod 2\\
2an\equiv a(a-1)&\mod a^2
\end{cases}$$
has solution for some $n$.
For $a$ odd this system has solution. In fact write $a=2b+1$, then
$$\begin{cases}n\equiv1&\mod 2\\
2(2b+1)n\equiv 2b(2b+1)&\mod (2b+1)^2
\end{cases}$$
becomes
$$\begin{cases}n\equiv1&\mod 2\\
n\equiv b&\mod (2b+1)
\end{cases}$$
that admits solution because $\gcd(2,2b+1)=1$.
With a similar argument you can show that this system does not have solutions for $a$ even. Write $a=2b$. Then
$$\begin{cases}n\equiv1&\mod 2\\
4bn\equiv 2b(2b-1)&\mod 4b^2
\end{cases}$$
becomes
$$\begin{cases}n\equiv1&\mod 2\\
2n\equiv -1&\mod (2b)
\end{cases}$$
that has not solution. So in this case the maximum has to be less than
$a(a-1)$ and you can show with the same argument that
$$\begin{cases}n\equiv1&\mod 2\\
2an\equiv a(a-2)&\mod a^2
\end{cases}$$
has solution.
Hence the value that you are looking for is
\begin{align}&\sum_{a \mbox{ odd}}a(a-1) +\sum_{a \mbox{ even}} a(a-2) \\
= &\sum_{a=3}^{1000}a^2-\sum_{a \mbox{ odd}}a -\sum_{a \mbox{ even}}2a \\
= &\sum_{a=3}^{1000}a^2-\sum_{k=2}^{500}(6k-1) \\
= &\frac{1000\cdot1001 \cdot 2001}{6}-5-6\cdot\frac{500\cdot501}{2}-6-499
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit involving exponentials of $\arcsin(x)$ and $\arctan(x)$ How can I calculate this limit without L'Hospital rule and Taylor series?
$$
\lim_{x \to 0} \frac{e^{\arctan(x)} - e^{\arcsin(x)}}{1 - \cos^3(x)}
$$
| Let $L=\lim_{x \to 0} \frac{e^{\arctan(x)} - e^{\arcsin(x)}}{1 - \cos^3(x)}$
$L=\lim_{x \to 0} \frac{(e^{\arctan(x)} -1)- (e^{\arcsin(x)}-1)}{1 - \cos^3(x)}$
$L=\lim_{x \to 0} \frac{\arctan (x)\frac{(e^{\arctan(x)} -1)}{\arctan(x)}- \arcsin (x)\frac{(e^{\arcsin(x)} -1)}{\arcsin(x)}}{1 - \cos^3(x)}$
$L=\lim_{x \to 0}\frac{\arctan (x)-\arcsin (x)}{1 - \cos^3(x)}$
$L=\lim_{x \to 0}\frac{\arctan (x)-\arcsin (x)}{(1 - \cos(x))(\cos^2x+\cos x+1)}$
$L=\lim_{x \to 0}\frac{\arctan (x)-\arcsin(x)}{x^2\frac{1-\cos x}{x^2}(\cos^2x+\cos x+1)}$
Now use $\lim_{x \to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$
$L=\lim_{x \to 0}\frac{\arctan (x)-\arcsin(x)}{x^2\times\frac{3}{2}}.....(1)$
Now let
$L_1=\lim_{x \to 0}\frac{\arctan (x)-\arcsin(x)}{x^2}$
$L_1=\lim_{x \to 0}\frac{\arctan (2x)-\arcsin(2x)}{(2x)^2}$
$4L_1=\lim_{x \to 0}\frac{\arctan (2x)-\arcsin(2x)}{x^2}$
$4L_1-L_1=\lim_{x \to 0}\frac{\arctan (2x)-\arcsin(2x)}{x^2}-\lim_{x \to 0}\frac{\arctan (x)-\arcsin(x)}{x^2}$
$3L_1=\lim_{x \to 0}\frac{(\arctan 2x-\arctan x)-(\arcsin 2x-\arcsin x)}{x^2}$
$3L_1=\lim_{x \to 0}\frac{(\arctan \frac{x}{1+x^2})-\arcsin(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}{x^2}$
$3L_1=\lim_{x \to 0}\frac{\frac{x}{1+x^2}\frac{\arctan \frac{x}{1+x^2}}{\frac{x}{1+x^2}}-(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})\frac{\arcsin(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}{2x\sqrt{1-x^2}-x\sqrt{1-4x^2}}}{x^2}$
$\lim_{x \to 0}\frac{\arctan \frac{x}{1+x^2}}{\frac{x}{1+x^2}}=1$ and
$\lim_{x \to 0}\frac{\arcsin(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}{(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}=1$
$3L_1=\lim_{x \to 0}\frac{\frac{x}{1+x^2}-(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}{x^2}$
$3L_1=\lim_{x \to 0}\frac{x}{(1+x^2)x^2}-\frac{(2\sqrt{1-x^2}-\sqrt{1-4x^2})}{x}$
$3L_1=\lim_{x \to 0}\frac{1}{(1+x^2)x}-\frac{({4-4x^2}-1+4x^2)}{x(2x\sqrt{1-x^2}+x\sqrt{1-4x^2})}$
$3L_1=\lim_{x \to 0}\frac{1}{(1+x^2)x}-\frac{1}{x}$
$3L_1=\lim_{x \to 0}\frac{1-1-x^2}{(1+x^2)x}$
$3L_1=\lim_{x \to 0}\frac{-x}{(1+x^2)}=0$
$L_1=0$
$L=\frac{2L_1}{3}=0$
| {
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"url": "https://math.stackexchange.com/questions/1581408",
"timestamp": "2023-03-29T00:00:00",
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The chart-problem; problem solving In how many ways can we construct a $6\times 6$ chart with only $1$ and $-1$ such that in every row and column, the product is always positive?
| There are $2^{25} $ combinations. Like Jack's answer fill the $5 \times 5$ subgrid. I will prove that there is a unique $6 \times 6$ grid with the desired property.
First we add the correct $1$'s or $0$'s in the first five places of the last row and column. We only have to determine the last digit in the lower left corner.
We can chose this number so that there is an even amount of 1's in the last column. I will prove that in this case the last row also has an even amount of 1's.
Consider the 6x6 grid as constructud above and call it $A$.
I do all algebraic calculations in $\mathbb{Z}_2$. So $1+1=0$.
Denote $e_i = \begin{pmatrix}
0 \\
\vdots \\
1 \\
\vdots \\
0
\end{pmatrix}$ where $1$ is placed in the $i$-th row.
Because in each column we have an even amount of $1$'s:
$$\begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix} A .e_i= 0 , \forall i $$
Because in the first 5 rows we have an even amount of $1$'s:
$$ e_j .A \begin{pmatrix}
1 \\
\vdots \\
1
\end{pmatrix} = 0 , \forall j \neq 6 $$
If we can prove that the last one also holdsfor $j=6$ we are done.
We have: $$e_6 = \begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix} - e_1 -e_2- \cdots - e_5$$
We have:
$$ \: \: \: \: e_6 .A \begin{pmatrix}
1 \\
\vdots \\
1
\end{pmatrix} =\left( \begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix} - e_1 -e_2- \cdots - e_5 \right)A \begin{pmatrix}
1 \\
\vdots \\
1
\end{pmatrix} \: \: \: \: \: \: \: \: (1)$$
We calculate
$$\begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix}A\begin{pmatrix}
1 \\
\vdots \\
1
\end{pmatrix} = \begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix}A(e_1+e_2+\cdots+e_6) = 0+ 0+\cdots+0 = 0$$
If we plug this in in $(1)$ and use the property of the columns the result follows. Thus starting with the $5 \times 5$ subgrid and constructing the $6 \times 6$ grid always gives you a correct grid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
I would like to prove convergence of the following series: $\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)}$ I would like to prove the following series:
$$\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)} $$
is convergent (absolutely?) or divergent. I think $\arctan\left(\frac{n}{1+n^2}\right)$ is divergent, but I don't know how it interacts with $(-1)^n$ and how to prove it.
Any ideas would be greatly appreciated.
| To show that the terms are decreasing:
$\arctan\left(\frac{n}{1+n^2}\right)- \arctan\left(\frac{n+1}{1+(n+1)^2}\right)
=\arctan\left(\dfrac{\frac{n}{1+n^2}-\frac{n+1}{1+(n+1)^2}}{1+\frac{n}{1+n^2}\frac{n+1}{1+(n+1)^2}}\right)
$
and
$\dfrac{n}{1+n^2}-\dfrac{n+1}{1+(n+1)^2}
=\dfrac{n(1+(n+1)^2)-(n+1)(1+n^2)}{(1+n^2)(1+(n+1)^2)}
$
and
$\begin{array}\\
n(1+(n+1)^2)-(n+1)(1+n^2)
&=n(n^2+2n+2)-(n^3+n^2+n+1)\\
&=n^3+2n^2+2n-(n^3+n^2+n+1)\\
&=n^2+n-1\\
&> 0\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n})$ I'm struggling to find the limit at infinity of : $$\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}), n\in\Bbb N$$
I know it is $1$ but I don't understand why this is wrong :
$\sin^2(\pi\sqrt{n^2 + n}) = \sin^2(\pi*n\sqrt{1 + \frac{1}{n}})$.
So $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}) = \lim \limits_{n\rightarrow\infty}\sin^2(\pi*n\sqrt{1 + \frac{1}{n}}) = \lim \limits_{n\rightarrow\infty} \sin^2(\pi*n) = 0$.
Thanks.
| $$\sqrt{n^2+n} = n \sqrt{1 + \frac{1}{n} } = n (1 + \frac{1}{2n} + O(1/n^2)) =$$
$$ = n + \frac{1}{2} + O(1/n) $$
| {
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"source": "stackexchange",
"question_score": "3",
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} |
Proving $\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$, where $a\geq 0,\ b\geq 0$. Who can help me to prove this equation?
$$\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$$ Where, $$a\geq 0,\ b\geq 0$$
| We have
$$\left(\sqrt{a-\sqrt{b}}+\sqrt{a+\sqrt{b}}\right)^2 = a-\sqrt{b}+a+\sqrt{b} +2\sqrt{a^2-b} = 2a+2\sqrt{a^2-b}$$
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
$\sec\theta+\tan\theta=p$ and $\sec\theta\tan\theta=q$. Eliminate $\theta$ to form a equation between $p$ and $q$. $\sec\theta+\tan\theta=p$ and $\sec\theta\tan\theta=q$. Eliminate $\theta$ to form a equation between $p$ and $q$.
$\sec\theta+\tan\theta=p$
$(\sec\theta+\tan\theta)^2=p^2$
$\sec^2\theta+\tan^2\theta+2\tan\theta\sec\theta=p^2$
$\sec^2\theta+\tan^2\theta+2q=p^2$
$1+2\tan^2\theta+2q=p^2$
I am stuck here. Please help me. Thanks.
| Now, $\sec^2\theta \tan^2\theta=q^2$.
So, $(1+\tan^2\theta)\tan^2\theta=q^2$.
You have got that $1+2\tan^2\theta+2q=p^2$.
So, $\tan^2\theta=\frac{p^2-1-2q}{2}$.
Now plug in this $\tan^2\theta$ in the equation got above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Calculating the value of $\lfloor(1+\sqrt{2})^{2n}\rfloor$ Problem: Calculate the value of $\lfloor(1+\sqrt{2})^{2n}\rfloor$ where $n$ is an arbitrary non-negative integer and $\lfloor x\rfloor$ indicates the largest integer not greater than $x$.
What I have tried:
Based on the Binomial Theorem:
$$
(1+\sqrt{2})^{2n} + (1-\sqrt{2})^{2n} = 2\sum_{k=0}^n {2n\choose 2k}2^k \Rightarrow (1+\sqrt{2})^{2k} = 2\sum_{k=0}^n {2n\choose 2k}2^k - (1-\sqrt{2})^{2n}
$$
Also
$$
(1-\sqrt{2})^{2n} \gt 0 \Rightarrow -(1-\sqrt{2})^{2n} \lt 0 \Rightarrow 2\sum_{k=0}^n {2n\choose 2k}2^k -(1-\sqrt{2})^{2n} \lt 2\sum_{k=0}^n {2n\choose 2k}2^k
$$
$$
\Rightarrow 2\sum_{k=0}^n {2n\choose 2k}2^k - 1 \le 2\sum_{k=0}^n {2n\choose 2k}2^k -(1-\sqrt{2})^{2n} \lt 2\sum_{k=0}^n {2n\choose 2k}2^k
$$
$$
\Rightarrow 2\sum_{k=0}^n {2n\choose 2k}2^k - 1 \le (1+\sqrt{2})^{2n} \lt 2\sum_{k=0}^n {2n\choose 2k}2^k
$$
By the definition of floor function:
$$
\Rightarrow \lfloor(1+\sqrt{2})^{2n}\rfloor\ = -1+2\sum_{k=0}^n {2n\choose 2k}2^k
$$
Calculating $\sum_{k=0}^n {2n\choose 2k}2^k$ seems to be too difficult. My question is: Can I change $\sum_{k=0}^n {2n\choose 2k}2^k$ into something simpler to calculate (maybe a closed form)?
| You can get recursions, which might be easier to compute.
Define $a_n=1+\left\lfloor (1+\sqrt{2})^{2n}\right\rfloor$, for $n>0$, and get a recursion:
$$a_{n+1} = 6a_n - a_{n-1}; a_1=6; a_2=34$$
So if you need a bunch of values $a_{n}$, recursion will be faster. It still might be faster to use this formula than your original, since it amounts to $n$ additions and $n$ multiplications by $6$ to compute $a_n$ and multiplication by $6$ can be computed very quickly by most good compilers by computing $6x=4x+2x$.
There is an $O(\log n)$ computation, computing the matrix product:
$$\begin{pmatrix}a_{n-1}\\a_n\end{pmatrix}=\begin{pmatrix}0&1\\6&-1\end{pmatrix}^{n-2}\begin{pmatrix}a_1\\a_2\end{pmatrix}$$
You can compute that matrix power in $O(\log n)$ time by using the method of repeated squaring.
Possibly much simpler application of repeated squarting: Compute $(1+\sqrt{2})^{2n}=(3+2\sqrt{2})^n$ as $b_n+c_n\sqrt{2}$ then $a_n=2b_n$. You can again compute $b_n,c_n$ by the repeated squaring technique in $O(\log n)$ time.
[These formula don't quite work when $n=0$ because $(1-\sqrt{2})^{2\cdot 0}=1$ is not less than $1$.]
| {
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"url": "https://math.stackexchange.com/questions/1587962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-...}}}}}}$ by trigonometric function ?
The signs : + - - + - - + - - ... .
Thanks for helping.
| Hint
First let us give a name to the nested radicals
$$x=\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}$$
Then you can observe that
$$((x^2-4)^2-4)^2=4-x$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Find the value of the integral $\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$ Find the value of the integral $\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$
$\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$
$=\int_{0}^{\frac{\pi}{12}}\frac{1}{3}\frac{\tan^2x-\frac{1}{3}+\frac{1}{3}-3}{\tan^2x-\frac{1}{3}}dx$
$=\frac{1}{3}-\frac{8}{9}\int_{0}^{\frac{\pi}{12}}\frac{dx}{\tan^2x-\frac{1}{3}}dx$
$=\frac{1}{3}-\frac{8}{9}\int_{0}^{\frac{\pi}{12}}\frac{\cot^2x dx}{1-\frac{1}{3}\cot^2x}dx$
$=\frac{1}{3}-\frac{8}{9}\int_{0}^{\frac{\pi}{12}}\frac{\cot^2x dx}{\frac{4}{3}-\frac{1}{3}\csc^2x}dx$
I am stuck here and could not solve further.Please help me.
| $$u=\tan(x)$$
$$du=(\tan^2x+1) dx$$
$$\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx=\int_{0}^{\tan\frac{\pi}{12}}\frac{u^2-3}{(3u^2-1)(u^2+1)}du$$
$$=\int_{0}^{\tan\frac{\pi}{12}}(\frac1{u^2+1}-\frac2{3u^2-1})du$$
$$=\int_{0}^{\tan\frac{\pi}{12}}(\frac1{u^2+1}+\frac{\sqrt3}{3u+\sqrt3}-\frac{\sqrt3}{3u-\sqrt3})du$$
$$=(\tan^{-1}u+\frac{\sqrt3}3\ln |\frac{1}{3u+\sqrt3}|-\frac{\sqrt3}3\ln |\frac{1}{3u-\sqrt3}|)~~|_0^{\tan\frac{\pi}{12}}$$
$$=(\tan^{-1}u+\frac{\sqrt3}3\ln |\frac{1}{3u+\sqrt3}|-\frac{\sqrt3}3\ln |\frac{1}{3u-\sqrt3}|)~~|_0^{\tan\frac{\pi}{12}}$$
$$=\tan^{-1}\tan\frac{\pi}{12}-\frac{\sqrt3}{3}\ln \frac{\sqrt3-3\tan\frac{\pi}{12}}{\sqrt3+3\tan\frac{\pi}{12}}$$
$$=\frac{\pi}{12}-\frac{\sqrt3}{3}\ln \frac{\sqrt3-3\tan\frac{\pi}{12}}{\sqrt3+3\tan\frac{\pi}{12}}$$
$$\approx 0.8420667415917798898282846513866$$
Notice that different methods of solving lead into the same result but maybe with different formats. They must be numerically equal. As an example $\frac1{12}(\pi - 8 \sqrt3 \tanh^{-1}(3 - 2 \sqrt3))$ seems to be different but in fact is equal to the answer obtained above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a closed form for $n^k$ in terms of $\Delta n^{k+1},\Delta n^k$, ...? Let $\Delta$ be a sort of difference operator on a function $f(n)$ such that
$$\Delta f(n)=f(n+1)-f(n)$$
Take the basic power function $f(n)=n^k$, $k\in\mathbb{N}\cup\{0\}$. Then we get
$$\begin{cases}\Delta 1=0\\[1ex]
\Delta n=1\\[1ex]
\Delta n^2=2n+1\\[1ex]
\Delta n^3=3n^2+3n+1
\end{cases}$$
and so on, with the general form
$$\Delta n^k=(n+1)^k-n^k=\sum_{i=0}^{k-1}\binom kin^k$$
I'm wondering if there's a parallel closed form for $n^k$ in terms of $\Delta n^{k+1},\Delta n^k,\ldots,\Delta n$? What I mean by this is that for $n$, we can write
$$n=\frac{\Delta n^2}{2}-\frac{\Delta n}{2}$$
since
$$\frac{\Delta n^2}{2}-\frac{\Delta n}{2}=\frac{2n+1}{2}-\frac{1}{2}=n$$
Similarly, one can show that
$$\begin{cases}n^2=\dfrac{\Delta n^3}{3}-\dfrac{\Delta n^2}{2}+\dfrac{\Delta n}{6}\\[1ex]
n^3=\dfrac{\Delta n^4}{4}-\dfrac{\Delta n^3}{2}+\dfrac{\Delta n^2}{4}\\[1ex]
n^4=\dfrac{\Delta n^5}{5}-\dfrac{\Delta n^4}{2}+\dfrac{\Delta n^3}{3}-\dfrac{\Delta n}{30}\end{cases}$$
and so on, with no immediate pattern as far as I can tell. Is there one?
| We derive a formula for $n^k$ in terms of $\Delta n^k$ by use of binomial inverse pairs and their generating functions. We show
the following is valid
\begin{align*}
n^k=\frac{1}{k+1}\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}\qquad\qquad k\geq 0
\end{align*}
with $B_k$ the Bernoulli numbers.
$$ $$
We define
\begin{align*}
A(x)=\sum_{k=0}^\infty\Delta n^k\frac{x^k}{k!}\qquad\qquad B(x)=\sum_{k=0}^{\infty}n^k\frac{x^k}{k!}
\end{align*}
According to the definition of the $\Delta$ operator we obtain
\begin{align*}
A(x)&=\sum_{k=0}^{\infty}\Delta n^k\frac{x^k}{k!}\\
&=\sum_{k=0}^{\infty}\left((n+1)^k-n^k\right)\frac{x^k}{k!}\\
&=\sum_{k=0}^{\infty}\sum_{i=0}^{k-1}\binom{k}{i}n^i\frac{x^k}{k!}\\
&=\sum_{k=0}^{\infty}\sum_{i=0}^{k}\binom{k}{i}n^i\frac{x^k}{k!}-\sum_{k=0}^{\infty}n^k\frac{x^k}{k!}\\
&=\left(\sum_{k=0}^{\infty}n^k\frac{x^k}{k!}\right)\left(e^x-1\right)\tag{1}\\
&=B(x)\left(e^x-1\right)\tag{2}
\end{align*}
Comment:
*
*In (1) we use the multiplication of exponential generating functions
\begin{align*}
\left(\sum_{k=0}^\infty a_k \frac{x^k}{k!}\right)\left(\sum_{k=0}^\infty b_k \frac{x^k}{k!}\right)
=\sum_{k=0}^\infty\left(\sum_{i=0}^k\binom{k}{i}a_i b_{k-i}\right)\frac{x^k}{k!}
\end{align*}
with $a_k=n^k$ and $b_k=1$ for all $k\geq 0$.
*In (2) we see the relationship between $\Delta n^k$ und $n^k$ in terms of generating functions.
From (2) we obtain the generating function $B(x)$ for $n^k$ by the generating function $A(x)$ as
\begin{align*}
B(x)&=A(x)\frac{1}{e^x-1}\\
&=\frac{1}{x}A(x)\frac{x}{e^x-1}\tag{3}
\end{align*}
In (3) we see the generating function of the Bernoulli numbers
\begin{align*}
\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k\frac{x^k}{k!}
\end{align*}
and we obtain
\begin{align*}
B(x)&=\frac{1}{x}A(x)\frac{x}{e^x-1}\\
&=\frac{1}{x}\left(\sum_{k=0}^{\infty}\Delta n^k\frac{x^k}{k!}\right)\left(\sum_{k=0}^\infty B_k\frac{x^k}{k!}\right)\\
&=\frac{1}{x}\sum_{k=0}^{\infty}\left(\sum_{i=0}^k\binom{k}{i}\Delta n^i B_{k-i}\right)\frac{x^k}{k!}\tag{4}
\end{align*}
$$ $$
Comparing coefficients of $B(x)$ and the RHS of (4) gives
\begin{align*}
\frac{1}{k!}n^k&=\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}\frac{1}{(k+1)!}\qquad\qquad k\geq 0\\
n^k&=\frac{1}{k+1}\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}
\end{align*}
and the claim follows.
Epilogue Binomial inverse pairs
Let's summarise the result about this binomial inverse pair. We observe the following relationship
\begin{align*}
A(x)&=\sum_{k=0}^\infty\Delta n^k\frac{x^k}{k!}&B(x)&=\sum_{k=0}^{\infty}n^k\frac{x^k}{k!}\\
&=B(x)\left(e^x-1\right)&&=\frac{1}{x}A(x)\frac{x}{e^x-1}\\
\Delta n^k&=\sum_{i=0}^{k-1}\binom{k}{i}n^k&n^k&=\frac{1}{k+1}\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/1591890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
| First, show that this is true for $n=1$:
$\left(\sum\limits_{k=1}^{1}k\right)^2=\sum\limits_{k=1}^{1}k^3$
Second, assume that this is true for $n$:
$\left(\sum\limits_{k=1}^{n}k\right)^2=\sum\limits_{k=1}^{n}k^3$
Third, prove that this is true for $n+1$:
$\left(\sum\limits_{k=1}^{n+1}k\right)^2=$
$\left(\color\green{\left(\sum\limits_{k=1}^{n}k\right)}+\color\purple{(n+1)}\right)^2=$
$\color\green{\left(\sum\limits_{k=1}^{n}k\right)}^2+2\color\green{\left(\sum\limits_{k=1}^{n}k\right)}\color\purple{(n+1)}+\color\purple{(n+1)}^2=$
$\color\red{\left(\sum\limits_{k=1}^{n}k\right)^2}+2\left(\sum\limits_{k=1}^{n}k\right)(n+1)+(n+1)^2=$
$\left(\color\red{\sum\limits_{k=1}^{n}k^3}\right)+2\left(\sum\limits_{k=1}^{n}k\right)(n+1)+(n+1)^2=$
$\left(\sum\limits_{k=1}^{n}k^3\right)+n(n+1)(n+1)+(n+1)^2=$
$\left(\sum\limits_{k=1}^{n}k^3\right)+(n+1)^3=$
$\sum\limits_{k=1}^{n+1}k^3$
Please note that the assumption is used only in the part marked red.
| {
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"url": "https://math.stackexchange.com/questions/1592512",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$ I came across this problem:
Prove there arent't any $a$, $b$ integers that satisfy
equation $3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$
Firstly, I've thought something like this:
$$(a^3 + b^3)-3b^3 + 3ab - 4a + 4b - 4 + b = 3$$
$$(a+b)(a^2-ab+b^2) - 3(b^3 - ab) -4(a-b+1)+b = 3$$
I think it's just a tricky exercise, but I can't find a way to solve it.
Some help would be apreciated.
Thanks!
| $3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$
Consider the equation mod $3$. Since $x^3\equiv x\pmod 3$ for all $x$, we have
$3ab+a-2b-4a+2b-7\equiv 0$ which is the same as $2\equiv 0\pmod 3$.
Contradiction.
| {
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Find a continuous function $f$ such $\int_{a}^{a^2+1}f(x)dx=0\quad\forall a\in \Bbb R$ Find example : exsits a nonzero continuous function $f:\Bbb R\to \Bbb R$ satisfying
$$\int_{a}^{a^2+1}f(x)dx=0\quad\forall a\in \Bbb R$$
the book answer this: $f(x)=x^2(x-1)(x-\frac{3}{5}),0\le x\le 1$,
$f(x)=-f(-x),-1\le x\le 0$.
$$\int_{-1}^{1}f(x)dx=0$$
But
$$\int_{1/2}^{5/4}f(x)dx\neq 0,a=\dfrac{1}{2}$$
| I think the answer you have is the seed of a full answer. You have a formula for $f$ on $[-1,1]$, but then you would need to use the defining relation to extend $f$ to larger and larger domains.
Take $$f(x)=\begin{cases}x^2(x-1)(x-\frac{3}{5})&0\le x\le 1
\end{cases}$$ where we know that with $a=0$, $$\int_0^1 x^2(x-1)(x-\frac{3}{5})=0$$
For $a\in(0,1]$,
$$\begin{align}
\int_a^{1} x^2(x-1)\left(x-\frac{3}{5}\right)\,dx+\int_1^{a^2+1}f(x)\,dx&\equiv0\quad\text{(apply $\frac{d}{da}$ to both sides of this functional relation)} \\-a^2(a-1)\left(a-\frac{3}{5}\right)+f(a^2+1)\cdot2a&=0\\
f(a^2+1)&=\frac{a}{2}(a-1)\left(a-\frac{3}{5}\right)
\end{align}$$
So for $x\in(1,2]$, using $x=a^2+1$, you have $$f(x)=\frac{\sqrt{x-1}}{2}\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}-\frac{3}{5}\right)$$ and now you have a formula
$$f(x)=\begin{cases}x^2(x-1)(x-\frac{3}{5})&0\le x\le 1\\
\frac{\sqrt{x-1}}{2}\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}-\frac{3}{5}\right)&1<x\leq2
\end{cases}$$ that extends the book's answer to a larger domain. You can continue like this, extending the domain to $(2,5]$, then to $(5,26]$, and so on. For instance, the next extension gives
$$f(x)=\begin{cases}x^2(x-1)(x-\frac{3}{5})&0\le x\le 1\\
\frac{\sqrt{x-1}}{2}\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}-\frac{3}{5}\right)&1<x\leq2\\
\frac{\sqrt{\sqrt{x-1}-1}}{4\sqrt{x-1}}\left(\sqrt{\sqrt{x-1}-1}-1\right)\left(\sqrt{\sqrt{x-1}-1}-\frac{3}{5}\right)&2<x\leq5
\end{cases}$$
Once you've formalized this, check that everything is in order with negative input as well.
This reconciles with your example:
$$\begin{align}
\int_{1/2}^{5/4}f(x)\,dx&=\int_{1/2}^{1}x^2(x-1)(x-\frac{3}{5})\,dx+\int_{1}^{5/4}\frac{\sqrt{x-1}}{2}\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}-\frac{3}{5}\right)\,dx\\
&=\left[\frac{x^3}{5} -\frac{2 x^4}{5} + \frac{x^5}{5}\right]_{1/2}^1+\left[\frac{1}{5}\left((x-1)^{5/2}-2(x-1)^2+(x-1)^{3/2}\right)\right]_{1}^{5/4}\\
&=-\frac{1}{160}+\frac{1}{160}\\
&=0
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry with Quadratic Equations If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
I tried the question but with $q$ other terms came associated.
| Since $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, you know that
$$
\tan A+\tan B=-p,
\qquad
\tan A\tan B=q
$$
so
$$
\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=
\frac{-p}{1-q}=\frac{p}{q-1}
$$
provided $q\ne1$.
The relation to be proved can be written
$$
\sin^2C+p\sin C\cos C+q\cos^2C=q
$$
where $C=A+B$, or
$$
\sin^2C+p\sin C\cos C-q\sin^2C=0
$$
recalling that $\cos^2C-1=-\sin^2C$. If $\cos C\ne0$ (I'll return to this later), this is equivalent to showing that
$$
(1-q)\tan^2C+p\tan C=0
$$
But, substituting the value found before, we have
$$
(1-q)\frac{p^2}{(q-1)^2}+p\frac{p}{q-1}=-\frac{p^2}{q-1}+\frac{p^2}{q-1}=0
$$
The case $\cos C=0$ corresponds to $\tan C$ not existing, that is, $q=1$. In this case the relation to be proved is
$$
\sin^2C=1
$$
which is true, if $\cos C=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is there a formula for computing the number of arrangements from a set of N elements groupped in K groups? Eg: N=3, K=2
There will be two groups in each solution. We need to calculate the number of such possible solutions. Consider the set S={1,2,3}. The possible solutions are:
{1} {2,3}
{1} {3,2}
{2} {1,3}
{2} {3,1}
{3} {1,2}
{3} {2,1}
{1,2} {3}
{2,1} {3}
{1,3} {2}
{3,1} {2}
{2,3} {1}
{3,2} {1}
The result for this example is: 12.
Eg: N=4, K=3
{1} {2} {3,4}
{1} {2} {4,3}
{1} {2,3} {4}
{1} {3,2} {4}
....
Can we generalize this formula?
| Consider an arrangement of a set $S = \{ a_1, a_2, a_3, \dots, a_n\}$.
We need to partition this set into $K$ groups such that each group contains more than one element such that the relative order of elements remain same. For example for $S = \{1,2,3\}$, the partition $\{1\},\{2,3\}$ will be valid but $\{2\},\{1,3\}$ will be invalid.
Let $x_i$ be the number of elements in $i^{th}$ partition, so we have $$x_1 + x_2 + x_3 + \dots + x_k = n$$
where each $x_i \ge 1$. This is a classical problem and we know the number of ways for the above problem is $$P = \binom{n-1}{k-1}$$
There are $n!$ sets, S are possible so number of arrangements will be equal to $n!*P$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Determinant $n\times n$ problem $$
D_n = \left|
\begin{matrix}
n & -1 & -3 & 0 & 0 & \cdots & 0 & 0 & 0 \\
n & 1 & 2 & -3 & 0 & \cdots & 0 & 0 & 0 \\
n & 0 & 1 & 2 & -3 & \cdots & 0 & 0 & 0 \\
n & 0 & 0 & 1 & 2 & \cdots & 0 & 0 & 0 \\
n & 0 & 0 & 0 & 1 & \cdots & 0 & 0 & 0 \\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots \\
n & 0 & 0 & 0 & 0 & \cdots & 1 & 2 & -3 \\
n & 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 2 \\
n & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\
\end{matrix}
\right|
$$
Can anyone help me with this determinant? It is recursive formula but I can't get it.
| Add the second up to the last rows to the first. Then the first row becomes $(n^2,0,\ldots,0)$. The new matrix is lower block-triangular and is in the form of $\pmatrix{n^2&0\\ n\mathbf1&A}$, where $A$ is by itself an upper triangular matrix with determinant 1. Hence $\det D_n=n^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595974",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The value of the polynomial at given point. Given that:
$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1$.
Find the value of $f(x)$ at $x=\sqrt{2}-1$
Answer is an integer. I tried factorization but couldn't proceed towards anything promising.
| $$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1\\=x(x((x-1)(x+1)(x(x+2)-1)x^4+3)+6)+1$$
$\Longrightarrow f(\sqrt 2 -1)=\boxed{\color{red}{4}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
| With limits going to infinity, usually factoring out the greatest term is the way to go, since then you'll have a lot of terms going to $0$. Here the term that grows faster is $x!$. So:
\begin{align}
\require{cancel}
\lim_{x \to +\infty} \frac{(x+2)(x+1)x!\left(1+\frac{4^x}{(x + 2)!}\right)}{((2x+1)^2+\ln x)x!} &= \lim_{x \to +\infty} \frac{(x^2+3x+2)\left(1+\frac{4^x}{(x + 2)!}\right)}{4x^2\left(1 + \frac1x + \frac1{4x^2}+\frac{\ln x}{4x^2}\right)} =\\
&= \lim_{x \to +\infty} \frac{\cancel{x^2}\left(1+\frac3x+\frac2{x^2}\right)\left(1+\frac{4^x}{(x + 2)!}\right)}{4\cancel{x^2}\left(1 + \frac1x + \frac1{4x^2}+\frac{\ln x}{4x^2}\right)} = \frac14.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597093",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Using Lagrange's diagonalization on degenerate linear forms
Let $A=\begin{pmatrix}1 & 2 & 3\\
2 & 3 & 4\\
3 & 4 & 5
\end{pmatrix}$ be a real matrix.
Find an invertible matrix $P\in M_{3}(\mathbb{R})$ such that $P^TAP$ is diagonal whose elements on the diagonal are all in the set $\{-1, 0, 1\}$
I've been trying to solve this with Lagrange's squares method, but I'm stuck at the end where I seem to get "less" squares then required. I assume that this is because $A$ is degenerate, but I'm unable to complete my form to a basis that would satisfy the requirements as a result, and would love a general method for this.
These are my calculations so far:
Let $\begin{pmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{pmatrix}\in\mathbb{R}^{3}$.
Then
$$\begin{aligned}\begin{pmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{pmatrix}^{T}\begin{pmatrix}1 & 2 & 3\\
2 & 3 & 4\\
3 & 4 & 5
\end{pmatrix}\begin{pmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{pmatrix}= & x_{1}^{2}+4x_{1}x_{2}+6x_{1}x_{3}+3x_{2}^{2}+5x_{3}^{2}+8x_{2}x_{3}\\
= & x_{1}^{2}+2x_{1}\left(2x_{2}+3x_{3}\right)+\left(2x_{2}+3x_{3}\right)^{2}\\
& -\left(2x_{2}+3x_{3}\right)^{2}+3x_{2}^{2}+5x_{3}^{2}+8x_{2}x_{3}\\
= & \left(x_{1}+\left(2x_{2}+3x_{3}\right)\right)^{2}-4x_{2}^{2}-12x_{2}x_{3}-9x_{3}^{2}+3x_{2}^{2}+5x_{3}^{2}+8x_{2}x_{3}\\
= & \left(x_{1}+2x_{2}+3x_{3}\right)^{2}-x_{2}^{2}-4x_{2}x_{3}-4x_{3}^{2}\\
= & \left(x_{1}+2x_{2}+3x_{3}\right)^{2}-\left(x_{2}+2x_{3}\right)^{2}
\end{aligned}
$$
and you can see I have only two "elements" for the basis, and I'm stuck..
| Denote the quadratic form by
$$ q \left( \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \right) = (x_1, x_2, x_3) \cdot A \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = (x_1 + 2x_2 + 3x_3)^2 - (x_2 + 2x_3)^2. $$
We see that if we make the change of variables
$$ u_1 = x_2 + 2x_2 + 3x_3, \,\, u_2 = x_2 + 2x_3, u_3 = r_{31} x_1 + r_{32} x_2 + r_{33} x_3 $$
the quadratic form will become diagonal. You can choose the scalars $r_{31}, r_{32}, r_{33}$ in any way you like as long as you get a legitimite change of variables - that is, the corresponding map is invertible.
More formally, consider the matrix $R$ given by
$$ R = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ r_{31} & r_{32} & r_{33} \end{pmatrix}, \,\, R \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}. $$
If we choose $r_{31}, r_{32}, r_{33}$ so that $R$ will be invertible, then
$$ q \left( R^{-1} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} \right) = (u_1, u_2, u_3) \left( R^{-1} \right)^T A R^{-1} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = u_1^2 - u_2^2 $$
and so the $P$ you are looking for can be taken to be $R^{-1}$. The simplest choice (that will also make computing $R^{-1}$ easier) is to take $r_{31} = r_{32} = 0$ and $r_{33} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx,\alpha\in(0,\pi)$ For the equation $x^2-2x\cos\alpha+1=0$ solutions are
$$x_1=\cos\alpha-\sqrt{\cos^2\alpha-1},x_2=\cos\alpha+\sqrt{\cos^2\alpha-1}\Rightarrow$$
$$\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx=\int_{-1}^1 \frac{1}{(x-\cos\alpha+\sqrt{\cos^2\alpha-1})(x-\cos\alpha-\sqrt{\cos^2\alpha-1})}\mathrm dx$$
$\frac{1}{(x-\cos\alpha+\sqrt{\cos^2\alpha-1})(x-\cos\alpha-\sqrt{\cos^2\alpha-1})}=\frac{A}{(x-\cos\alpha+\sqrt{\cos^2\alpha-1})}+\frac{B}{(x-\cos\alpha-\sqrt{\cos^2\alpha-1})}\Rightarrow$
$$A=\frac{-1}{2\sqrt{\cos^2\alpha-1}},B=\frac{1}{2\sqrt{\cos^2\alpha-1}}\Rightarrow$$
$$\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx=\frac{-1}{2\sqrt{\cos^2\alpha-1}}\left(\int_{-1}^1 \frac{1}{x-\cos\alpha+\sqrt{\cos^2\alpha-1}}\mathrm dx - \int_{-1}^1 \frac{1}{x-\cos\alpha-\sqrt{\cos^2\alpha-1}}\mathrm dx\right)$$
How to evaluate these partial integrals?
| Since the denominator has no real roots you should complete the square and write it as $$(x-\cos\alpha)^2+\sin^2\alpha$$
Applying the standard integral $$\int\frac{1}{a^2+x^2}dx=\frac 1a \arctan(\frac xa),$$ we get
$$\frac{1}{\sin\alpha}\left[\arctan\frac{1-\cos\alpha}{\sin\alpha}-\arctan\frac{-1-\cos\alpha}{\sin\alpha}\right]$$
$$=\frac{1}{\sin\alpha}\left[\arctan(\tan\frac{\alpha}{2})+\arctan(\cot\frac{\alpha}{2})\right]$$
$$=\frac{1}{\sin\alpha}\left[\frac{\alpha}{2}+\frac{\pi}{2}-\frac{\alpha}{2}\right]$$
$$=\frac{\pi}{2\sin\alpha}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:
$$A = \left(
\begin{matrix}
5&-7&2&2\\
0&3&0&-4\\
-5&-8&0&3\\
0&5&0&-6\\
\end{matrix} \right)
$$
I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining:
$$
\begin{pmatrix}
3 & 0 & -4 \\
-8 & 0 & 3 \\
5 & 0 & -6 \\
\end{pmatrix}
$$
M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:
3 times $$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $3(0-0)= 0$
then:
0 times $$
\begin{pmatrix}
-8 & 3\\
5 & -6\\
\end{pmatrix}
$$
giving 0(48-15)=0
Then:
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving $4(0-0)=0$
adding the determinants we get $0+0+0=0$
So det M1 $= 0(1) = 0$
M2--> M(1,2)---> $-1^1+2= -1^3 = -1$
$$
\begin{pmatrix}
0 & 0 & -4 \\
-5 & 0 & 3 \\
0 & 0 & -6 \\
\end{pmatrix}
$$
o*
$$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $0(0-0)=0$
obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving 4(0-0)= 0
So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3
M3 --> $-1^4 = 1$
$$
\begin{pmatrix}
0 & 3 & -4 \\
-5 & -8 & 3 \\
0 & 5 & -6 \\
\end{pmatrix}
$$
for the determinant:
0 times
$$
\begin{pmatrix}
-8 & 3 \\
5 & -6 \\
\end{pmatrix}
$$
which gives $0(48-15)=0$
-3 times
$$
\begin{pmatrix}
-5& 3 \\
0 & -6 \\
\end{pmatrix}
$$
which gives $-3(30-0)= -90$
it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4.
which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving
Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases?
| New Method
assume this Matrics
\begin{bmatrix} 1&2&2&-3\\4&2&5&0\\1&3&4&2\\-3&1&-1&1 \end{bmatrix}
rewrite the matrix after changing order of columns of original matrix from 1234 to 1342
\begin{bmatrix} 1&2&-3&2\\4&5&0&2\\1&4&2&3\\-3&-1&1&1 \end{bmatrix}
rewrite the matrix after changing order of columns of original matrix from 1234 to 1423
\begin{bmatrix}1&-3&2&2\\4&0&2&5\\1&2&3&4\\-3&1&1&-1 \end{bmatrix}
after that for each matrix divide it to 4 2x2 matrices and find the det of each normally to reach those in respective order
\begin{bmatrix} -6&15\\10&6\end{bmatrix}
\begin{bmatrix} -3&-6\\11&-1\end{bmatrix}\begin{bmatrix} 12&6\\7&-7\end{bmatrix}
Now convert each of those 3 to values according to the rule of 2x2 determinant but with adding not subtracting for example the first one
(-6)(6)+(15)(10)=114
second one
(-3)(-1)+(-6)(11)=-63
Third one
(12)(-7)+(7)(6)=-42
finally the det of the original matrix = 114-68-42=9
https://youtu.be/vzEJ9sFO0SU
check my method it is my pure original work can not be found in today's books
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 4
} |
Integral $\int\left(\frac{1}{x^4+x^2+1}\right)dx$ Someone can halp me to solve this integral:
$$\int\left(\frac{1}{x^4+x^2+1}\right)$$
solution$$\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{1}{2\sqrt3}\arctan\frac {x^2-1}{x\sqrt3}$$
I don't manage using partial fraction because $${x^4+x^2+1}$$ has $\Delta\lt 0$ and substituing $t=x^2,$ $2x$ appears.That's a problem.
| Notice, $$\int \frac{1}{x^4+x^2+1}\ dx$$ $$=\int \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}\ dx$$
$$=\frac{1}{2}\int \frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1}\ dx$$
$$=\frac{1}{2}\left(\int \frac{\left(1+\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1}\ dx-\int \frac{\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1}\ dx\right)$$
$$=\frac{1}{2}\left(\int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+(\sqrt{3})^2}-\int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-(1)^2}\right)$$
$$=\frac12\left(\frac{1}{\sqrt 3}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt 3}\right)-\frac{1}{2}\ln\left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|\right)+C$$
$$=\frac{1}{2\sqrt 3}\tan^{-1}\left(\frac{x^2-1}{x\sqrt 3}\right)-\frac{1}{4}\ln\left|\frac{x^2-x+1}{x^2+x+1}\right|+C$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove that $f(x,y)=g(x+y,xy)$. Let $f\in \mathbb R[x,y]$ such that $f(x,y)=f(y,x)$. Prove that $f(x,y)=g(x+y,xy)$.
My try.
$f\in \Bbb R[x,y]$ and $f(x,y)=f(y,x)$ .If I take for example $f(x,y)=a_{00}+a_{01}y+a_{11}xy+a_{10}x+a_{13}xy^3$ then only the coefficients which are symmetric w.r.t $x $ and $y$ will remain in the expression of $f$ ,all other terms will cancel out because $f(x,y)=f(y,x)$.
But How can I write down the equation of $f$ from here?
Please provide some hints.
| Let $f\in \mathbb R[x,y]$ such that $f(x,y)=f(y,x)$. To prove that $f(x,y)=g(x+y,xy)$.
Let $$f(x,y)=a_{0,0}x^0y^0+a_{0,1}x^0y^1+a_{0,2}x^0y^2+a_{1,0}x^1y^0+a_{1,1}x^1y^1+a_{1,2}x^1y^2+a_{2,0}x^2y^0+a_{2,1}x^2y^1+a_{2,2}x^2y^2$$
We then have
$$f(y,x)=a_{0,0}y^0x^0+a_{0,1}y^0x^1+a_{0,2}y^0x^2+a_{1,0}y^1x^0+a_{1,1}y^1x^1+a_{1,2}y^1x^2+a_{2,0}y^2x^0+a_{2,1}y^2x^1+a_{2,2}x^2y^2$$
As $f(x,y)=f(y,x)$ we have $a_{0,1}=a_{1,0}, a_{2,0}=a_{0,2},a_{1,2}=a_{2,1}$.. So, we have
$$f(x,y)=a_{0,0}x^0y^0+a_{0,1}x^0y^1+a_{0,2}x^0y^2+a_{0,1}x^1y^0+a_{1,1}x^1y^1+a_{1,2}x^1y^2+a_{0,2}x^2y^0+a_{1,2}x^2y^1+a_{2,2}x^2y^2=a_{0,0}x^0y^0+a_{0,1}(x^0y^1+x^1y^0)+a_{0,2}(x^0y^2+x^2y^0)+a_{1,1}x^1y^1+a_{1,2}(x^1y^2+x^2y^1)+a_{2,2}x^2y^2=a +bxy+c(xy)^2+d(x+y)+e((x+y)^2-2xy)+fxy(x+y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that the sequence $a_0 = 1$, $a_{n+1 }= \sqrt{2+a_n}$ is monotonically increasing Given the sequence $a_0 = 1$, $a_{n+1}= \sqrt{2+a_n}$, how can I show that it is monotonically increasing?
I need it to show, that the sequence converges. I already proved boundedness but I can't figure out this last part.
| Option 1: First, prove that $1 \le a_n < 2$ for all $n \in \mathbb{N}$. This can be done using induction.
After that, prove $\sqrt{2+x} > x$ for $1 \le x < 2$, and thus, $a_{n+1} = \sqrt{2+a_n} > a_n$ for all $n \in \mathbb{N}$.
Option 2: Define $\theta_n = \arccos\dfrac{a_n}{2}$. Then, $a_n = 2\cos\theta_n$ for each $n \in \mathbb{N}$.
Hence, $2\cos\theta_{n+1} = a_{n+1} = \sqrt{2+a_n} = \sqrt{2+2\cos\theta_n} = 2\cos\tfrac{\theta_n}{2}$, and thus, $\theta_{n+1} = \dfrac{\theta_n}{2}$.
Now it is pretty easy to prove that $\theta_n = \dfrac{\theta_0}{2^n}$ is monotonically decreasing to $0$, and thus, $a_n = 2\cos\theta_n$ monotonically increases to $2\cos 0 = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ Find all $n$ such that
$3^{2n+1}+2^{n+2}$ is divisible by $7$
Prove that your answer is correct
So I am not allowed to use mods, as is a calculus question, I have tried by induction but can't get to prove that it works for $k+1$, by multiplying the equation by powers of $2$ and $3$.
Thank you for your help
| Just another way:
Inductive hypothesis: $3^{2n+1} = 7k - 2^{n+2}$
Inductive step:
$$3^{2n+3} + 2^{n+3} = 3^2 * 3^{2n+1} + 2 * 2^{n+2} $$
$=3^2(7k - 2^{n+2})+2*2^{n+2}$
$=7k(3^2) - 3^2*2^{n+2} + 2* 2^{n+2}$
$=7k(3^2) + 2^{n+2}(-3^2+2)$
$=7k(3^2) -7*(2^{n+2})$
Both terms are divisible by 7
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
prove that $x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$ is cubic order of convergence near $x_0=\sqrt{5}$ To solve the equation
$$x^2-5=0$$
There exitsts a iteration method
$$x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$$
I know that it is cubic convergence but I don't know how to prove it.
I have tried the following
$$
\begin{align}
e_{k+1}&=x_{n+1}-\sqrt{5} \\ &=\frac{x_n(x_n^2+15)}{3x_n^2+5}-\sqrt{5} \\ &=\frac{\frac{1}{3}x_n(3x_n^2+5+40)}{3x_n^2+5}-\sqrt{5} \\ &=\frac{1}{3}x_n(1+\frac{40}{3x_n^2+5})-\sqrt{5} \\ &=\frac{\frac{40}{3}x_n}{3x_n^2+5}+\frac{1}{3}x_n-\sqrt{5}
\end{align}
$$
I'm trying to extract $e_n=x_n-\sqrt{5}$ from $e_{n+1}$ in an effort of proving $lim \frac{e_{n+1}}{e_n^3}\rightarrow C$ . But it seems that I'm not on the right way.
Do you have any idea? Any discussion is appreciated. Thanks in advance.
| Hint: Define $f(x):=\frac{x(x^2+15)}{3x^2+5}$, and Taylor expand it around $x=\sqrt{5}$.
If you want to continue with the method you're currently working with, then you'll want to replace then $x_n$ terms by $e_n+\sqrt{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1
$$
$$
\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{3} } = \frac{65}{8}; a \neq 0
$$
| 1) $\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1\implies\frac{(x^2-5x+4)(x^2-5x+6)}{(x^2+5x+4)(x^2+5x+6)}=1$
$\implies(x^2-5x+4)(x^2-5x+6)=(x^2+5x+4)(x^2+5x+6)$
We know that the even-indexed terms will cancel out because the LHS is just the RHS with $x\to-x$
$\implies-10x^3-50x=10x^3+50x\implies10x^3+50x=0\implies10x(x^2+5)=0\implies x=0$
2) Note that the denominators of these can be written as $u+2,u-4$ with $u=x^2+3x$. We see:
$\frac{6}{u+2}+\frac{8}{u-4}=1\implies\frac{6}{u+2}-3+\frac{8}{u-4}+2=1-3+2\implies\frac{-3u}{u+2}+\frac{2u}{u-4}=0$
Remove a factor of $u$ and accept $u=0$ as a solution. Then clear fractions:
$\frac{-3}{u+2}+\frac{2}{u-4}=0\implies-3(u-4)+2(u+2)=0\implies16-u=0\implies u=16$
So, $x^2+3x=0, 16$ - we extract $x=0,-3$ for $u=0$ and solve the $u=16$ case with the quadratic formula / your method of choice.
3) Let $(ax-b)^{3/7}=u$, then $u-(-u)=65/8\implies u=65/16$. Nowhere further to go as $a,b$ aren't given - can say $x=(65/16+b)/a$ if you'd like, I suppose.
Perhaps worth noting that $\frac{65}{8}=8+\frac{1}{8}$, as an aside.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
$\lim _{x\to 0}\left(\frac{\sqrt[2]{\cos \left(x\right)}-\sqrt[3]{\cos \left(x\right)}}{\sin ^2\left(x\right)}\right)$ without L'Hospitals rule? $$\lim _{x\to 0}\left(\frac{\sqrt[2]{\cos \left(x\right)}-\sqrt[3]{\cos \left(x\right)}}{\sin ^2\left(x\right)}\right)=?$$
How to solve it without using L'Hospitals rule?
| Use equivalents and Taylor's formula at order $2$: we know that
*
*$\cos x=1-\dfrac{x^2}2+o(x^2)$,
*$\sqrt{1+u}=1+\dfrac u2+o(u)$,
*$\sqrt[3]{1+u}=1+\dfrac u3+o(u)$
Thus the numerator gives, by composition:
$$\sqrt[2]{\cos x}-\sqrt[3]{\cos x}=\Bigl(1-\frac{x^2}4\Bigr)-\Bigl(1-\frac{x^2}6\Bigr)+o(x^2)=-\frac{x^2}{12}+o(x^2)$$
so that $\;\sqrt[2]{\cos x}-\sqrt[3]{\cos x}\sim_0 -\dfrac{x^2}{12}$, and
$$\frac{\sqrt[2]{\cos x}-\sqrt[3]{\cos x}}{\sin^2 x}\sim_0 -\dfrac{x^2}{12 x^2}=-\frac1{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$? I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used(without Hopital if is possible)? Thanks
$$\lim_{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$$
| $$
\begin{aligned}
\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2\left(1-x^2\right)}{\left(ln\left(x^2-3\sqrt{2}x+5\right)\right)^2}\right)
& = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(1-\left(t+\sqrt{2}\right)^2\right)}{\left(ln\left(\left(t+\sqrt{2}\right)^2-3\sqrt{2}\left(t+\sqrt{2}\right)+5\right)\right)^2}\right)
\\& = \lim _{t\to 0}\left(\frac{e^{\left(t+\sqrt{2}\right)^2}+e^2\left(-t^2-2\sqrt{2}t-1\right)}{\ln \:^2\left(t^2-\sqrt{2}t+1\right)}\right)
\\& = \lim _{t\to 0}\left(\frac{e^2+2\sqrt{2}e^2t+5e^2t^2+o\left(t^2\right)+e^2\left(-t^2-2\sqrt{2}t-1\right)}{2t^2+o\left(t^2\right)}\right)
\\& = \color{red}{2e^2}
\end{aligned}
$$
Solved with substitution $t = x-\sqrt2$ and Taylor expansion
| {
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"url": "https://math.stackexchange.com/questions/1605491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all pairs $(n,k)$ such that $n(n+1) \, \mid\,(k+1)! \,(1^k+2^k+3^k+\cdots+n^k)$ How can I solve (find all the solutions) the following problem?
Find all pairs of postive integers $(n,k)$ such that
$$n(n+1) \,\mid \,(k+1)!\, (1^k+2^k+3^k+\cdots+n^k).$$
I included here what I had done so far. If $k=1$ and $n=1$, then
$$(k+1)!\cdot (1^k+2^k+\cdots+n^k)=n(n+1)$$ is one solution
Thanks for any help.
| It is true for all $n,k \in \mathbb{N}, n,k \geq 1$.
Induction by $k$.
1) For $k=1$ we have $(k+1)! (1^k + \ldots + n^k = 2(1+\ldots+n) = 2 \frac{(n+1)n}{2}=(n+1)n $ , so this is obviously true.
2) For $k \geq 1$ let we assume that that this conditions it true for all previous $j \geq 1, j<k$.
Cosider following sum:
$\sum_{x=1}^{n}[(x+1)^{k+1}-x^{k+1}] = 2^{k+1}-1^{k+1} +3^{k+1}-2^{k+1}+\ldots + (n+1)^{k+1}-n^{k+1} = (n+1)^{k+1}-1 $.
In other hand $\sum_{x=1}^{n}[(x+1)^{k+1}-x^{k+1}] = \sum_{x=1}^{n} \sum_{j=0}^{k+1}[{ {k+1}\choose j } x^j -x^{k+1}] =
\sum_{x=1}^{n} \sum_{j=0}^{k}{ {k+1}\choose j } x^j =
\sum_{j=0}^{k}{ {k+1}\choose j } \sum_{x=1}^{n} x^j
$
Hence we have:
$ (n+1)^{k+1}-1=
\sum_{j=0}^{k}{ {k+1}\choose j } \sum_{x=1}^{n} x^j$
$ (n+1)^{k+1}-1= { {k+1}\choose k } \sum_{x=1}^{n} x^k +
\sum_{j=0}^{k-1}{ {k+1}\choose j } \sum_{x=1}^{n} x^j$
$ (n+1)^{k+1}-1= (k+1) (\sum_{x=1}^{n} x^k) +
\sum_{j=1}^{k-1}{ {k+1}\choose j } (\sum_{x=1}^{n} x^j) + n$
Next we multiple both sides of this equality by $k!$ and transforming left side (using formula for $x^n-y^n$):
$ k! ((n+1)-1)((n+1)^{k}+ (n+1)^{k-1}+\ldots+(n+1)^{1}+ 1)= (k+1)! (\sum_{x=1}^{n} x^k) + k!
\sum_{j=1}^{k-1}{ {k+1}\choose j } (\sum_{x=1}^{n} x^j) + k! n$
$ k! n((n+1)^{k}+ (n+1)^{k-1}+\ldots+(n+1)^{1}+ 1)= (k+1)! (\sum_{x=1}^{n} x^k) + k!
\sum_{j=1}^{k-1}{ {k+1}\choose j } (\sum_{x=1}^{n} x^j) + k! n$
we move $k!n$ to left side end we get:
$ k! n((n+1)^{k}+ (n+1)^{k-1}+\ldots+(n+1)^{1})
= (k+1)! (\sum_{x=1}^{n} x^k) + k! \sum_{j=1}^{k-1}{ {k+1}\choose j } (\sum_{x=1}^{n} x^j) + k! n$
We can see that left side of the equality is divisible by $n(n+1)$ and on the right side the sum $k!
\sum_{j=1}^{k-1}{ {k+1}\choose j } (\sum_{x=1}^{n} x^j)$ also is divisible by $n(n+1)$ , from assumtion of induction (for $j < k)$.
Hence $ n(n+1) | (k+1)! (\sum_{x=1}^{n} x^k) $, which proves the second step of induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Example of normal linear maps Let $V$ be a real inner product space and $g:V\rightarrow V$ a normal linear map.
Can you give me an example of:
1- A map $g$ that is not self-adjoint?
2- A map $g$ that is not an isometry?
3- A map $g$ that is not orthodiagonalizable?
| Consider the matrix
$$ A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \in M_2(\mathbb{R}). $$
We have
$$ AA^T = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{pmatrix}, \\
A^TA = \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{pmatrix}
$$
so $A^TA = AA^T$ and $A$ is a normal matrix. If $b \neq 0$, then $A$ is not self-adjoint and not orthogonally diagonalizable (as this is equivalent). If $a^2 + b^2 \neq 1$, then $A$ is not orthogonal.
Thus, if $b \neq 0$ and $a^2 + b^2 \neq 1$, the corresponding map $T_A \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T_A(x) = Ax$ (considering $\mathbb{R}^2$ with the standard inner product) is not self-adjoint, not an isometry and not orthogonally diagonalizable. In fact, it is not diagonalizable (as the roots of the characteristic polynomial $a \pm ib$ are complex).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$. The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.
I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.
Now how do I rationalize the numerator?
It is giving the $\frac{0}{0}$ form after plugging in $3$.
| Your first step is good. Now you want to compute
$$
\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}
$$
because the factor $x+3$ at the denominator poses no problem. This should remind you of the definition of derivative and indeed it's the derivative of $f(x)=\sqrt{x+6}$ at $3$. Since
$$
f'(x)=\frac{1}{2\sqrt{x+6}}
$$
for $x>-6$, you have $f'(3)=1/6$ and so
$$
\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}=
\lim_{x\to 3}\left(1+\frac{\sqrt{x+6}-3}{x-3}\frac{1}{x+3}\right)=
1+\frac{1}{6}\frac{1}{6}
$$
If you don't know about derivatives, just do
$$
\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}=
\lim_{x\to3}\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x-3)(\sqrt{x+6}+3)}
=\lim_{x\to3}\frac{x-3}{(x-3)(\sqrt{x+6}+3)}=\dots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculte indefinte integral of $\int \frac{dx}{\sqrt{(x+2)(3-x)}}$ I have to calculate $$\int \frac{dx}{\sqrt{(x+2)(3-x)}}$$.
I tried to use -
$\int u'v = uv - \int v'u$, but im pretty stuck.
Thanks.
| $$\int\frac{1}{\sqrt{(x+2)(3-x)}}\space\text{d}x=\int\frac{1}{\sqrt{-x^2+x+6}}\space\text{d}x=\int\frac{1}{\sqrt{\frac{25}{4}-\left(x-\frac{1}{2}\right)^2}}\space\text{d}x=$$
Substitute $u=x-\frac{1}{2}$ and $\text{d}u=\text{d}x$:
$$\int\frac{1}{\sqrt{\frac{25}{4}-u^2}}\space\text{d}u=\int\frac{2}{5\sqrt{1-\frac{4u^2}{25}}}\space\text{d}u=\frac{2}{5}\int\frac{1}{\sqrt{1-\frac{4u^2}{25}}}\space\text{d}u=$$
Substitute $s=\frac{2u}{5}$ and $\text{d}s=\frac{2}{5}\space\text{d}u$:
$$\int\frac{1}{\sqrt{1-s^2}}\space\text{d}s=\arcsin\left(s\right)+\text{C}=\arcsin\left(\frac{2u}{5}\right)+\text{C}=$$
$$\arcsin\left(\frac{2\left(x-\frac{1}{2}\right)}{5}\right)+\text{C}=\arcsin\left(\frac{2x-1}{5}\right)+\text{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find complex roots of $\frac{2i}{1+i}$
Find 6th roots of $$\frac{2i}{1+i}$$
$$\frac{2i}{1+i}=\frac{2e^{i\pi/2}}{\sqrt 2 e^{i \pi/4}}=\sqrt 2 e^{i \pi/4}$$
Now if I set $z^{1/6}=\sqrt 2 e^{i \pi/4}$ and knowing the fact that the roots are distibuted equality with an angle $k\pi/3$ for $k=1,2,3,4,5,6$ I get the answer to be:
$$2^{1/12}( \cos(\pi/24 + k\pi/3)+i\sin(\pi/24 + k\pi/3))$$
Is this correct?
| $$z^6=\frac{2i}{1+i}\Longleftrightarrow$$
$$z^6=\left|\frac{2i}{1+i}\right|e^{\arg\left(\frac{2i}{1+i}\right)i}\Longleftrightarrow$$
$$z^6=\frac{\left|2i\right|}{\left|1+i\right|}e^{\left(\arg(2i)-\arg(1+i)\right)i}\Longleftrightarrow$$
$$z^6=\frac{2}{\sqrt{2}}e^{\left(\frac{\pi}{2}-\arctan\left(\frac{1}{1}\right)\right)i}\Longleftrightarrow$$
$$z^6=\sqrt{2}e^{\left(\frac{\pi}{2}-\frac{\pi}{4}\right)i}\Longleftrightarrow$$
$$z^6=\sqrt{2}e^{\frac{\pi}{4}i}\Longleftrightarrow$$
$$z=\left(\sqrt{2}e^{\left(2\pi k+\frac{\pi}{4}\right)i}\right)^{\frac{1}{6}}\Longleftrightarrow$$
$$z=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi k+\frac{\pi}{4}\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-5$.
So the solutions are:
$$z_0=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot0+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{\frac{\pi}{24}i}$$
$$z_1=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot1+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{\frac{3\pi}{8}i}$$
$$z_2=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot2+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{\frac{17\pi}{24}i}$$
$$z_3=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot3+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{-\frac{23\pi}{24}i}$$
$$z_4=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot4+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{-\frac{5\pi}{8}i}$$
$$z_5=\sqrt[12]{2}e^{\frac{1}{6}\left(2\pi\cdot5+\frac{\pi}{4}\right)i}=\sqrt[12]{2}e^{-\frac{7\pi}{24}i}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \ \ \ \ \ (0\leq \sin^{2} x\leq 1)
$
But that is not in options.
I look for a short and simple way.
I have studied maths upto $12$th grade.
| $$\left(\frac{2}{\sin x}-3\sin x\right)^2\geq0$$
$$\iff\csc^2 x+9\sin^2 x\geq12$$ with equality achieved when $\sin^2x=2/3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is $$\frac{\sqrt{2}\sqrt{3}}{9}$$ which isn't equal to $$\frac{2\sqrt{2}-\sqrt{3}}{5}$$
What did I do wrong?
| As one of the comments stated, your first error lies in the first step where you multiply "everything" by $\sqrt{3}$. While you claim to do this, you did not multiply both terms in the denominator by $\sqrt{3}$. This was an error that let you to an incorrect expression.
That having been said, your approach is not correct. To show this equality, you have to simplify the expression on the left hand side. In this case, because you have radical expressions in the denominator, that means multiplying the top and bottom by the "right value" which will eliminate the radicals. In the case of a single term in the denominator like $\dfrac{1}{\sqrt{2}}$, you would simply multiply the top and bottom by $\sqrt{2}$ which would yield $$\dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$$
The case you have, however, contains a binomial in the denominator (meaning two terms). In this case, you should multiply by the conjugate of the denominator. So for your example, the conjugate of $2\sqrt{2} + \sqrt{3}$ would be $2\sqrt{2} - \sqrt{3}$. Doing so yields the following:
\begin{align}
\frac{1}{2\sqrt{2} + \sqrt{3}} \cdot \frac{2\sqrt{2}-\sqrt{3}}{2\sqrt{2} - \sqrt{3}} &= \frac{2\sqrt{2}-\sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\
&= \frac{2\sqrt{2}-\sqrt{3}}{8 - 3} \\
&= \frac{2\sqrt{2}-\sqrt{3}}{5}
\end{align}
Thus we've shown that $$ \frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2}-\sqrt{3}}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Complex number - wrong result at the end I need to solve this:
$$ \frac{i^4+3}{i-1}$$
On my book the result should be: $-2-2i$ but I get: $-1-2i$ and I do not understand where the error is.
My steps:
$$ \frac{i^4+3}{i-1} = \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i}$$
$$ \frac{(i^4+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$i^4 = (i^2)^2 = 1$$
$$ \frac{(1+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$ \frac{4(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$ \frac{4(-1-i) + (-i+1+1+i)}{(-i+1+1+i)}$$
$$ \frac{-4-4i + 2}{2}$$
$$ \frac{-2-4i}{2} = -1-2i$$
Where is the error?
| $$\frac{i^4+3}{i-1}=\frac{iiii+3}{i-1}=\frac{i^2i^2+3}{i-1}=\frac{-1-1+3}{i-1}=\frac{1+3}{i-1}=\frac{4}{i-1}=$$
$$\frac{4(i+1)}{(i-1)(i+1)}=\frac{4i+4}{-1^2-1^2}=\frac{4+4i}{-2}=\frac{4}{-2}+\frac{4i}{-2}=-2-2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I'm trying to prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. However, I came up with this:
Since $\sqrt{4}=2$ and $\sqrt{9}=3$ we have a following inequality: $2 < \sqrt{6} < 3$.
On the other hand we have $\sqrt{1}=1$ and $\sqrt{4}=2$ so we have $1 < \sqrt{2} < 2$.
Let $n = \sqrt{6}-\sqrt{2}$,
Therefore $ 1<n<3$, which implies that $n>1$.
Is this a correct proof?
Thank you.
| $\sqrt{6}-\sqrt{2}$ is positive because $\sqrt{x}$ is minimal at $0$ and always increasing at $x>0$.
Because $x^2$ is $1$ at $x=1$ and always increasing at $x>0$, numbers greater than 1 remain greater than $1$ when squared and $8-\sqrt{48}$ results from squaring $\sqrt{6}-\sqrt{2}\,$.
Since $8-\sqrt{49}=1$ and $\sqrt{49}$ exceeds $\sqrt{48}$ given properties of $\sqrt{x}$, it then follows that $\sqrt{6}-\sqrt{2} > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Eigenvectors of $\begin{pmatrix}6&2\\-10&-1\end{pmatrix}$ (linear equations with complex numbers) I want to compute the eigenvectors of the matrix
$$\begin{pmatrix}6&2\\-10&-1\end{pmatrix}$$
and thus far I got the eigenvalues $\lambda_{1,2}=\frac{1}{2}(5\pm\sqrt{31}i)$. However solving for example the system
$$\begin{pmatrix}6-\frac{1}{2}(5+\sqrt{31}i)&2\\-10&-1-\frac{1}{2}(5+\sqrt{31}i)\end{pmatrix}\cdot v=0$$
for the first eigenvector was pretty difficult and I couldn't come to a proper solution whatsoever. Do you have suggestions on how to simplify the system?
| We have:
$$\begin{pmatrix}6-\frac{1}{2}(5+\sqrt{31}i)&2\\-10&-1-\frac{1}{2}(5+\sqrt{31}i)\end{pmatrix}\cdot v=0$$
Lets simplify:
$$\begin{pmatrix}\frac{1}{2}(7-\sqrt{31}i)&2\\-10&-\frac{1}{2}(7+\sqrt{31}i)\end{pmatrix}\cdot v=0$$
Divide the first row by $\frac{1}{2}(7-\sqrt{31}i)$ and the second row by $-10$:
$$\begin{pmatrix}1 & \dfrac{4}{(7-\sqrt{31}i)}\\1&\dfrac{1}{20}(7+\sqrt{31}i)\end{pmatrix}\cdot v=0$$
Multiply $a_{12}$ by the conjugate:
$$\begin{pmatrix}1 & \dfrac{1}{20}(7+\sqrt{31}i)\\1&\dfrac{1}{20}(7+\sqrt{31}i)\end{pmatrix}\cdot v=0$$
$R_2 - R_1 \rightarrow R_2$:
$$\begin{pmatrix}1 & \dfrac{1}{20}(7+\sqrt{31}i)\\0&0\end{pmatrix}\cdot v=0$$
So:
$$v_1 = \left(-\dfrac{1}{20}(7+\sqrt{31}i), ~~1 \right)$$
Given that the eigenvalues are complex cojugates, so too are the eigenvectors, so you already know $v_2$.
However, you might want to give that one a go and make sure you can verify that.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find coordinate vector for an ordered none standart basis of polynomials? I know that it's trivial to know the coordinate vector with respect to the standart basis of polynomials.
The standart basis of $P_2[R]$ is:
$$\{1,t,t^2\}$$
A general vector spanned by the above set is of the form:
$$a+bt+ct^2$$
such that $a,b,c \in R.$
the coordinate vector with respect to the standart basis of $R^3$ is:
$$\left( \begin{array}{l}a\\b\\c\end{array} \right) = a\left( \begin{array}{l}1\\0\\0\end{array} \right) + b\left( \begin{array}{l}0\\1\\0\end{array} \right) + c\left( \begin{array}{l}0\\0\\1\end{array} \right)$$
But what if I have for example this ordered basis of $P_2[x]$ :
$$B=\{1+t,t^2,t\}$$
And I have a linear transorfmation
$$T:{P_2}[R] \to {M_{2x2}}[R]$$
and I need to find
$$[T]_B^E$$
This is the matrix that represent $T$ with respect to the basis $B$ and $E$ the standart basis for ${M_{2x2}}[R]$.
How can I find the general form of the coordinate vector with respect to the basis $B$ ?
I can only work with the coordinate verctor since I'm looking for the representing matrix for $T$ with respect to this basis.
so I need to find:
$$\begin{array}{l}T({b_1})\\T({b_2})\\T({b_3})\end{array}$$
And I dont know How to find them because Ineed the coordinate vector with respend to the basis $B$.
If it was for example to find $T[b_i] $ with respect to the standart basis of $P_2[R]$ ,
I would perform :
$$\begin{array}{l}T\left( \begin{array}{l}1\\0\\0\end{array} \right)\\T\left( \begin{array}{l}0\\1\\0\end{array} \right)\\T\left( \begin{array}{l}0\\0\\1\end{array} \right)\end{array}$$
But B is not the standart basis for $P_2[R]$ so I cant do $T$ on the three vectors above.
| method $1$:
Find the change of basis matrix from $E$ to $B$:
$1+t=1(1)+1(t)+1(t^2)$
$t^2\;\;\;\;\,=0(1)+0(t)+1(t^2)$
$t\;\;\;\;\;\;= 0(1)+1(t)+0(t^2)$
tranpose the coefficients and put as columns in a $3x3$ matrix, you get:
$$C = \left( {\begin{array}{*{20}{c}}1&0&0\\1&0&1\\0&1&0\end{array}} \right)$$
take $C$ and multiply it by a coordinate vector corosponding to each coefficient in the polynomial. for example:
$$a+bt+ct^2$$ has the coordinate vector
$$\left( \begin{array}{l}a\\b\\c\end{array} \right)$$
which is :
$$a\left( \begin{array}{l}1\\0\\0\end{array} \right) + b\left( \begin{array}{l}0\\1\\0\end{array} \right) + c\left( \begin{array}{l}0\\0\\1\end{array} \right)$$
so multiply $C$ times each vector and you will get:
$$\left( \begin{array}{l}1\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)$$
now as was said above me, find general solution and you will get:
$$\,x=a\\y=c\\\;\;\;\;\,\,z=b-a$$
so any coordinate vector with respect to the basis $B$
his representation as a vercor in $R^3$ is :
$$x\left( \begin{array}{l}1\\1\\0\end{array} \right) + y\left( \begin{array}{l}0\\0\\1\end{array} \right) + z\left( \begin{array}{l}0\\1\\0\end{array} \right) = a\left( \begin{array}{l}1\\1\\0\end{array} \right) + c\left( \begin{array}{l}0\\0\\1\end{array} \right) + b - a\left( \begin{array}{l}0\\1\\0\end{array} \right)$$
example :
$$\left( \begin{array}{l}2\\1\\3\end{array} \right) = 2\left( \begin{array}{l}1\\1\\0\end{array} \right) + 3\left( \begin{array}{l}0\\0\\1\end{array} \right) + (1 - 2)\left( \begin{array}{l}0\\1\\0\end{array} \right)$$
Method 2
find the coefficient as mentioned above.
and noticed that according to the basis $B = \{1+t,t^2,t\}$
the cooridnate vectors representantiom as a basis of $R^3$ is :
$$\left\{ {\left( \begin{array}{l}1\\0\\0\end{array} \right) + \left( \begin{array}{l}0\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)} \right\} = \left\{ {\left( \begin{array}{l}1\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)} \right\}$$
| {
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"url": "https://math.stackexchange.com/questions/1624850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+sinx}-3}$? $$\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin x}-3}$$
I have tried with Taylor:
$$\lim _{x\to \:0^+}\frac{\left(1+x+\frac{x^2}{2}+1+2x+2x^2\right)^2-4}{\sqrt{9+x-\frac{x^3}{6}+\frac{x^5}{120}}-3}=\lim _{x\to \:0^+}\frac{\frac{25x^4}{4}+15x^3+19x^2+12x}{\sqrt{9+x-\frac{x^3}{6}+\frac{x^5}{120}}-3}$$
But now I dont know how to move forward.
I tried with Hopital and it was found to be so:
$$\lim _{x\to \:0+}\left(\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin \left(x\right)}-3}\right)=\lim _{x\to \:0+}\left(\frac{2e^{2x}\left(e^x+1\right)\left(2e^x+1\right)}{\frac{\cos \left(x\right)}{2\sqrt{\sin \left(x\right)+9}}}\right)$$
$$=\lim _{x\to \:0+}\left(\frac{4e^{2x}\left(e^x+1\right)\left(2e^x+1\right)\sqrt{\sin \left(x\right)+9}}{\cos \left(x\right)}\right)=\frac{4e^{2\cdot \:0}\left(e^0+1\right)\left(2e^0+1\right)\sqrt{\sin \left(0\right)+9}}{\cos \left(0\right)}=\color{red}{72}$$
But I wanted to know if you could get the same result even with Taylor? Thanks
| $$\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin x}-3}$$
$$=\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sin x}\cdot\lim _{x\to \:0^+}(\sqrt{9+\sin x}+3)$$
$$=\left(\lim _{x\to0^+}\frac{e^{2x}-1}x+\lim _{x\to0^+}\frac{e^{4x}-1}x+2\lim _{x\to0^+}\cdot\frac{e^{3x}-1}x\right)\cdot\dfrac1{\lim _{x\to0^+}\dfrac{\sin x}x}\cdot\lim _{x\to0^+}(\sqrt{9+\sin x}+3)$$
Now use $\lim_{h\to0}\dfrac{e^{mh}-1}h=m$ to get $$(2+4+2\cdot3)\cdot(\sqrt9+3)=72$$
| {
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"url": "https://math.stackexchange.com/questions/1628919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for every natural $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$ Prove that for every natural number $n$, $(n^2 + n)(n^2 + 2)$ can be divided by $6$.
I've noticed that $(n^2 + n) = n(n+1)$ so these are two successive numbers hence one of them can be divided by two.
I suppose that I should prove that $(n^2 + n)(n^2 + 2)$ can be divided by $3$ but I don't know how to do that.
| HINT:
$$\begin{align} & (n^2+n)(n^2+2) \\
&=n(n+1)(n^2+2) \\
&=n(n+1)(n^2-1+3)\\
&=\color{red}{n(n+1)(n-1)} \cdot (n+1)+3\cdot\color{blue}{n(n+1)}\end{align}$$
Get the clue?
| {
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"url": "https://math.stackexchange.com/questions/1633277",
"timestamp": "2023-03-29T00:00:00",
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Solve limit with Lagrange theorem I tried to solve this limit:
$$ \lim_{x \to +\infty} x^2\left(e^{\frac{1}{x+1}}-e^{\frac{1}{x}}\right) $$
Instead of solving it with Taylor series (using $u = 1/x$), I noticed that the difference within the parenthesis is the $\Delta f$ of the function $e^{\frac{1}{x}}$.
$\lim\limits_{x \to +\infty} x^2 * \Delta\left(e^{1/x}\right) = $
The Lagrange theorem says that $\Delta f = D\left[f\right] * \Delta x$, so
$=\lim\limits_{x \to +\infty} x^2 * \left( D\left[e^{1/x}\right]_{x_0} * \Delta x\right), x_0 \in \left(x, x+1\right)$
$$ x_0=x+r(x)$$
$$ 0<r(x)<1 $$
$=\lim\limits_{x \to +\infty} x^2 * \left( e^{\frac{1}{x+r(x)}} * \left(-\frac{1}{\left(x+r(x)\right)^2}\right)\right) $
$=\lim\limits_{x \to +\infty} -\frac{1}{\left(x+r(x)\right)^2} * x^2 * e^{\frac{1}{x+r(x)}}$
$\approx \lim\limits_{x \to +\infty} -\frac{1}{x^2} * x^2 * e^{\frac{1}{x}} $
$=\lim\limits_{x \to +\infty} -1 * e^{\frac{1}{x}} = -1$
Is it correct to proceed in this way?
EDIT: Using approximations is not a very elegant solution. I might be more precise using the Squeeze theorem (I'll think the solution, then I'll post it)
Instead of using the approximation, we consider these three functions:
*
*$f(x) = x^2 * e^{\frac{1}{x+1}} * \frac{1}{(x+1)^2}$
*$h(x) = x^2 * e^{\frac{1}{x+r(x)}} * \frac{1}{(x+r(x))^2}$
*$g(x) = x^2 * e^{\frac{1}{x}} * \frac{1}{x^2}$
Obviously, $f(x) \le h(x) \le g(x)$ for $x > 0$ and $r(x) \in (0, 1)$.
$\lim\limits_{x \to +\infty} f(x) = \lim\limits_{x \to +\infty} x^2 * e^{\frac{1}{x+1}} * \frac{1}{(x+1)^2} = 1$
(It is 1 and not -1 just because I left the minus sign out of the limit)
$\lim\limits_{x \to +\infty} g(x) = \lim\limits_{x \to +\infty} x^2 * e^{\frac{1}{x}} * \frac{1}{x^2} = 1$
So $\lim\limits_{x \to +\infty} h(x) = 1 \rightarrow \lim\limits_{x \to +\infty} -h(x) = \lim\limits_{x \to +\infty} x^2\left(e^{\frac{1}{x+1}}-e^{\frac{1}{x}}\right) = -1$
Here's how the f (the red function) and g (the blue function) look like:
| \begin{align}
& \lim_{x \rightarrow \infty} x^2 \left( e^{\frac{1}{x+1}} -e^{\frac{1}{x}} \right)\\
= & \lim_{x \rightarrow \infty} \left( 1+ \frac{1}{x+1} + \frac{1}{2 (x+1)^2} + O(x^{-3}) - 1 -\frac{1}{x} - \frac{1}{2 x^2} + O(x^{-3})\right) \\
= & \lim_{x \rightarrow \infty} \frac{-x^2}{(x+1)x} + \frac{-2x-1}{2(x+1)^2 x^2} +O(x^{-1})
= -1.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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the minimum value of $a^2+b^2-a-\frac{2b}{3}$ using known standard inequalities If $a,b$ are real numbers satisfying $a+2b=3,$ then the minimum value of $a^2+b^2-a-\frac{2b}{3}$
Here $a+2b=3\implies a=3-2b$
$a^2+b^2-a-\frac{2b}{3}=(3-2b)^2+b^2-(3-2b)-\frac{2b}{3}$
$=9+4b^2-12b+b^2-3+2b-\frac{2b}{3}$
I diiferentiated it with respect to $b$ and put it equal to $0$ and then found $a$ by using the relation $a=3-2b$ to get the minimum value.
But this is a long method.Can we find its minimum value by using known inequalities like AM-GM etc.
| Hint : complete squares !can you see the answer?
| {
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Is this relationship already known? I like math because it's a puzzle to me, but am really not very good at it. But I figured out the relationship below myself. Just curious, is this already pretty common knowledge? Kind of proud of myself for figuring it out, but my son who's getting math minor had never heard of it. Apologies in advance for any poor explanation.
$b^2 = a^2 + a + b$ for positive integers where $b - a = 1$
e.g.
$2^2 = 4$
$3^2 = 9$
$4 + (2 + 3) = 9$
$4^2 = 16$
$9 + (3 + 4) = 16$
and so on.
Edit in response to @fleablood comment:
Where b = a + 2
Even numbers...
2^2 = 4 -> a^2 = a2
4^2 = 16 -> b^2 = b2
16 - 4 = 12 -> b2 - a2 = c
(a + 1) * b = 3 * 4 = c
Odd Numbers...
3^2 = 9 -> a^2 = a2
5^2 = 25 -> b^2 = b2
25 - 9 = 16 -> b2 - a2 = c
(a + 1) * b = 4 * 5 = 20
c != 20
| I would prove it this way:
\begin{align}
b^2&=a^2+a+b&\iff\\
b^2-b&=a^2+a&\iff\\
(b-1)(b)&=(a)(a+1)
\end{align}
which is clearly true if $b=a+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $2a^nb^nc^n+1\geq a^{2n}+b^{2n}+c^{2n}$ Let $a,b,c\in[-1,1]$ be such that $$2abc+1\geq a^2+b^2+c^2.$$ Prove that $$2a^nb^nc^n+1\geq a^{2n}+b^{2n}+c^{2n}$$ for any positive integer $n$.
The case $n=1$ is of course the same as the assumption. For $n=2$, squaring the assumption gives $$4a^2b^2c^2+4abc+1\geq a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$$ which is quite different than what is desired.
| Here is a partial solution : $a>0,b>0,c>0$ case
$b=\cos{B},c=\cos{C}, 0 \le B,C \le \dfrac{\pi}{2}$
$a^2-2abc\le 1-b^2-c^2 \iff bc-\sqrt{(1-b^2)(1-c^2)} \le a \le bc+\sqrt{(1-b^2)(1-c^2)} \iff \cos{(B+C)} \le a \le \cos{(B-C)}$
now we need to prove:
$b^{n}c^{n}-\sqrt{(1-b^{2n})(1-c^{2n})} \le a^{n} \le b^nc^n+\sqrt{(1-b^{2n})(1-c^{2n})} \\$
upbound $\implies \cos^n{(B-C)} \le \cos^n{B}\cos^n{C}+\sqrt{(1-\cos^{2n}{B})(1-\cos^{2n}{C})} \\ \iff \cos^n{(B-C)}-(\cos{B}\cos{C})^n\le \sqrt{(1-\cos^2{B})(1+\cos^2 B+(\cos^2B)^2+...+(\cos^2B)^{n-1})(1-\cos^2{C})(1+\cos^2 C+(\cos^2C)^2+...+(\cos^2C)^{n-1})} \\ \iff (\cos {(B-C)}-\cos B \cos C)(\cos^{n-1} (B-C) +\cos^{n-2}(B-C) (\cos B \cos C)+\cos^{n-3}(B-C) (\cos B \cos C)^2+...+\cos (B-C) (\cos B \cos C)^{n-2}+ (\cos B \cos C)^{n-1}\le \sin B \sin C\sqrt{(1+b^2+(b^2)^2+...(b^2)^{n-1})(1+c^2+(c^2)^2+...+((c^2)^{n-1})} $
$\cos (B-C) -\cos B \cos C= \sin B \sin C \ge 0$ , it remains
$\cos^{n-1} (B-C) + \cos^{n-2} (B-C) (bc)+ \cos^{n-3} (B-C) (bc)^2+...\cos (B-C) (bc)^{n-2}+(bc)^{n-1} \le \sqrt{(1+b^2+(b^2)^2+...(b^2)^{n-1})(1+c^2+(c^2)^2+...+((c^2)^{n-1})} $
LHS $\le 1+bc+(bc)^2+...(bc)^{n-1}$ because $0 \le (\cos (B-C))^k \le 1$
RHS $\ge 1+bc+(bc)^2+...(bc)^{n-1}$ it is Cauchy.
for low bound, it is similar.
for $abc \le 0$ , it is easy to prove as $2(abc)^n+1 \ge 2abc+1 \ge a^2+b^2+c^2 \ge a^{2n}+b^{2n}+c^{2n}$
| {
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} |
Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$ is a root of the cubic.
Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$
is a root of the cubic.
My effort
Working backwards I let $P(x)$ be a polynomial with roots $a,a^2+2a-14$ and $r$.
Thus, $$P(x)=(x-a)(x-r)(x-(a^2+2a-14))$$
Expanding, I get
$$P(x) =(x^2-(a+r)x+ar)(x-(a^2+2a-14)) $$
$$P(x) =x^3+x^2[-(a^2+2a-14)-(a+r)]+x[(a+r)(a^2+2a-14)+ar]-ar(a^2+2a-14)$$
Equating coefficients of $P(x)$ with the given cubic $x^3-21x+35=0$ I have the following system of equations :
\begin{array}
\space (a^2+2a-14)+(a+r)&=0 \\
(a+r)(a^2+2a-14)+ar&=-21 \\
-ar(a^2+2a-14)&=35 \\
\end{array}
From the first equation I have $(a^2+2a-14) =-(a+r) $ which, substituted in the other two equations ,it yields
\begin{array}
\space -(a+r)^2+ar &=-21 \\
ar(a+r) &=35 \\
\end{array}
Rearranging the second equation for $ar$ I have $ar=\cfrac{35}{(a+r)}$ which I now substitute into the first eq. to get:
\begin{array}
\space -(a+r)^2+\cfrac{35}{(a+r)}&=-21 \\
-(a+r)^3+35 +21(a+r) &=0 \\
\end{array}
My problem now is that the last equation looks pretty darn close to
$x^3-21x+35=0$ but some signs are not in the right place,which makes
me wonder if I have made some careless mistake(I have already checked
but I don't see it) or if I have left some algebraic manipulations to do.
| Since the term in $x^2$ is missing, the sum of the three roots is zero; so $a^2+2a-14$ is a root if and only if $-a-(a^2+2a-14)=-a^2-3a+14$ is also a root.
Since
$$
(x-a^2-2a+14)(x+a^2+3a-14)=x^2+ax-a^4 - 5a^3 + 22a^2 + 70a - 196
$$
and the remainder of $-t^4 - 5t^3 + 22t^2 + 70t - 196$ divided by $t^3-21t+35$ is $t^2-21$, we have
$$
(x-a^2-2a+14)(x+a^2+3a-14)=x^2+ax+a^2-21
$$
so
\begin{align}
(x-a)(x-a^2-2a+14)(x+a^2+3a-14)
&=(x-a)(x^2+ax+a^2-21)\\
&=x^3-21x-a^3+21a\\
&=x^3-21x+35
\end{align}
is the required factorization.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
how to prove this inequality $(ab+bc+ac)^2 ≥ 3abc(a+b+c)$
Prove that if $a,b,c$ are non-negative real numbers, then
$(ab + bc + ca)^2 \geq 3abc(a+b+c)$.
I tried to compute from $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$.
| $a,b,c$ can be any real numbers.
Let $x=ab$, $y=bc$, $z=ca$. Then your inequality is $$(x+y+z)^2\ge 3(xy+yz+zx),$$
which is true, because $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
and the inequality is equivalent to $$x^2+y^2+z^2\ge xy+yz+zx,$$
which is true, because, as you said, it's equivalent to $$\frac{1}{2}\left((x-y)^2+(y-z)^2+(z-x)^2\right)\ge 0,$$
which is true; or you can use Rearrangement Inequality.
Equality holds if and only if $x=y=z$, i.e. iff either at least two of $a,b,c$ are equal to $0$ or $abc\neq 0$ and $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove for all integers n such that n ≥ 3, $ 4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$ I am trying to prove this using mathematical induction, but I'm lost once I get to comparing the two sides of the equation.
Proposition:
For all integers n such that n ≥ 3,
$ 4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$
Proof:
Let the property P(n) be the equation
$P(n) = 4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$
Show that P(3) is true:
$4^3 = \frac{4(4^3 - 16)}{3}$
64 = 64, thus P(3) is true
Show that for all integers where n ≥ 3, if P(k) is true, then P(k + 1) is also true:
Suppose that P(k) is true for some particular but arbitrary integer where k ≥ 3. Suppose that k is any integer where k ≥ 3 such that:
$4^3 + 4^4 + 4^5 … 4^k = \frac{4(4^k - 16)}{3}$
We must show that P(k + 1) is true. That is, we must show that:
$4^3 + 4^4 + 4^5 … 4^{k + 1} = \frac{4(4^{k + 1} - 16)}{3}$
The left hand side is:
$4^3 + 4^4 + 4^5 … 4^{k + 1}$
$4^3 + 4^4 + 4^5 … 4k + 4^{k + 1}$
$4^3 + 4^4 + 4^5 … \frac{4(4^k - 16)}{3} + 4^{k + 1}$
| Using the induction hypothesis, the last line you wrote should be
$\frac{4(4^k - 16)}{3} + 4^{k + 1}$. Then:
\begin{align*}
\frac{4(4^k - 16)}{3} + 4^{k + 1} &= \frac{(4^{k+1} - 4\cdot16)}{3} + \frac {3\cdot4^{k + 1}}3
\\ &= \frac{4^{k+1} - 4\cdot16+3\cdot4^{k + 1}}3
\\ &= \frac{4\cdot 4^{k+1} - 4\cdot16}3
\\ &= \frac{4\cdot (4^{k+1} - 16)}3 \end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What am I counting wrong? EDIT: I made a mistake in the beginning, the second condition has changed. Sorry for this.
I'm asked to count the number of sets of 4 elements that satisfy the two following conditions:
1) Each element of the set is a two-digit number (from 10 to 99).
2) There are no repeated digits in the set.
What I've done is separate the sets in two disjoint cases. The ones that have an element with the digit 0 and the ones that don't.
The sets from the first case have three elements that don't have a $0$ and one that does. For the first element of the three mentioned, there are $C(7,2)$ ways of choosing the digits it will have and then two possible numbers they can form, so $2C(7,2)$ posible elements. Applying the same reasoning to the rest there are $2C(5,2)$ and $2C(3,2)$ possible elements. Order doesn't matter in a set, so I divide by $4!$.
This means that the number of sets from the first case is: $C(9,7)C(7,2)C(5,2)C(3,2)\cdot2^3\cdot1/4!$
Then I counted the number of sets from the second case using a similar argument and add them together:
$C(9,7)C(7,2)C(5,2)C(3,2)2^3\cdot1/4! + C(9,8)C(8,2)C(6,2)C(4,2)\cdot2^4\cdot1/4!=2^3\cdot3^4\cdot5\cdot7$
The answer is supposed to be twice that. What is it I'm I doing wrong?
| Case $1$ - digit $0$ appears:$$\frac{4\cdot\binom97\cdot7!}{4!}$$
Case $2$ - digit $0$ does not appear: $$\frac{\binom98\cdot8!}{4!}$$
Total count:$$\frac{4\cdot\binom97\cdot7!+\binom98\cdot8!}{4!}=\frac{3\cdot9!}{4!}=45360=2^4\cdot3^4\cdot5\cdot7$$
As for the error in your counting, you seem to have mixed your cases.
For case $1$ you posted: $$C(9,7)C(7,2)C(5,2)C(3,2)\cdot2^3\cdot1/4!$$
And for case $2$ you have $$C(9,8)C(8,2)C(6,2)C(4,2)\cdot2^4\cdot1/4!$$
But case $1$ should have been:
The first element contains the digit $0$ and there are $C(9,1)$ choices for the other digit. Then $2C(8,2)$ for the second element, $2C(6,2)$ for the third and $2C(4,2)$ for the fourth. The element with $0$ is fixed in the first postion and the order of the other three does not matter, so we divide by $3!$
The number of sets in the first case is: $$\frac{C(9,1)\cdot2C(8,2)\cdot2C(6,2)\cdot2C(4,2)}{3!}\tag{1}$$
Similarly, case 2 should be:
There are $2C(9,2)$ choices for the first element, $2C(7,2)$ for the second, $2C(5,2)$ for the third and $2C(3,2)$ for the fourth. The order does not matter, so we divide by $4!$
The number of sets in the second case is: $$\frac{2C(9,2)\cdot2C(7,2)\cdot2C(5,2)\cdot2C(3,2)}{4!}\tag{2}$$
Adding these gives the correct answer of $2^4\cdot3^4\cdot5\cdot7=45360$
I hope that clears it up for you.
Now, to relate $(1)$ and $(2)$ with the solution I posted, I'll rewrite them using only factorials:
$$\require{cancel}\frac{\frac{9!}{1!\cdot\cancel{8!}}\cdot\bcancel2\cdot\frac{\cancel{8!}}{\bcancel{2!}\cdot\cancel{6!}}\cdot\bcancel2\cdot\frac{\cancel{6!}}{\bcancel{2!}\cdot\cancel{4!}}\cdot\bcancel2\cdot\frac{\cancel{4!}}{\bcancel{2!}\cdot2!}}{3!}=\frac{\frac{9!}{2!}}{3!}=\frac{4\cdot\frac{9!}{2!}}{4!}=\frac{2\cdot9!}{4!}\tag{1}$$
$$\frac{\bcancel2\cdot\frac{9!}{\bcancel{2!}\cdot\cancel{7!}}\cdot\bcancel2\cdot\frac{\cancel{7!}}{\bcancel{2!}\cdot\cancel{5!}}\cdot\bcancel2\cdot\frac{\cancel{5!}}{\bcancel{2!}\cdot\cancel{3!}}\cdot\bcancel2\cdot\frac{\cancel{3!}}{\bcancel{2!}\cdot1!}}{3!}=\frac{9!}{4!}\tag{2}$$
$$\frac{2\cdot9!}{4!}+\frac{9!}{4!}=\frac{3\cdot9!}{4!}\tag{1+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatrix}
a & c \\
b & d
\end{bmatrix}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$ = $\begin{bmatrix}
a^2 + c^2 & ab + cd \\
ab + cd & b^2 + d^2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}$
. Then $a^2 + c^2 = 1$ implies $a = \cos\theta$, and $c = \sin\theta$ or $c = - \sin\theta$ using the unit circle .
Similarly $ d = \cos\theta$, and $b = \sin\theta$ or $b = -\sin\theta$.
So know I am stuck in showing how $b = -\sin\theta$ has to be chosen and $c = \sin\theta$.
Can someone please help? Thank you!
| Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
MY ATTEMPT:
Since it is an orthogonal matrix So modulus of It's eigenvalue is equal to 1
Also since $det(A)=1$ So Multiplication of eigenvalues=1
Let eigenvalues of $A$ is $a+ib$ and $c+id$. So
we have:
$a+ib=\frac{1}{c+id}$ with $|a+ib|=1$ and $|c+id|=1$
So we have $a+ib=\frac{c-id}{c^2+d^2}$ =$c+id$ [since ${c^2+d^2}=1$]
That is $a+ib=c+i\times(-d)$
$\Rightarrow a=c , b=-d$
for some $\theta$, $a+ib=(\cos\theta +i\sin\theta)$
[ r=1 since modulus of eigenvalue is 1]
So we have $a=c=\cos\theta$ and $b=-d=\sin\theta$
Therefore eigenvalues of this matrix are: $\cos\theta+i\sin\theta$ and $\cos\theta-i\sin\theta$.
That means \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
is similar to
$D=\begin{bmatrix}
\cos\theta+i\sin\theta & 0 \\
0 & \cos\theta-i\sin\theta
\end{bmatrix} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Require assistance proving $n≥2 \Longrightarrow \frac{n!}{n^n} ≤ \frac{1}{2}^{\lfloor \frac{n}{2}\rfloor}$ Theorem: $n≥2 \Longrightarrow \frac{n!}{n^n} ≤ \frac{1}{2}^{\lfloor \frac{n}{2}\rfloor}$
Attempted Solution:
We use induction. Additionally, we prove the stronger inequality omitting the floor function. That is, $$n≥2 \Longrightarrow \frac{n!}{n^n} ≤ \frac{1}{2^\frac{n}{2}}$$
The base case is clear. Suppose it holds for some $n$. Then, consider the case for $n+1$. On the right side, we get $\frac{1}{2^\frac{n+1}{2}} = \frac{1}{2^{\frac{n}{2} + \frac{1}{2}}} = \frac{1}{2^\frac{n}{2}} \cdot \frac{1}{\sqrt{2}}$. This implies the right side is multiplied by $\frac{1}{\sqrt{2}}$ for each successive increase in $n$; hence we must show the left side is multiplied at most by $\frac{1}{\sqrt{2}}$.
Consider now the left side for $n+1$. It is $\frac{(n+1)!}{(n+1)^{(n+1)}} = \frac{n!}{(n+1)^{n}}$. This implies that for each successive increase in $n$, the left side is being multiplied by $\frac{n^n}{(n+1)^n}$.
So we must show $\frac{n^n}{(n+1)^n} < \frac{1}{\sqrt{2}}$ for all $n≥2$. It is clear for $n=2$ and it suffices to show the left hand side is strictly decreasing as $n$ increases. I am unable to do this, though Wolfram seems to agree with me. Note that I don't want to differentiate, since the author of the book I am reading has not yet introduced derivatives.
| Another approach is to show that
$$\left(1+\frac1n\right)^{2n}=\left(\frac{n+1}{n}\right)^{2n}>2.$$
The inequality follows directly from the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
On a hypothetical computer with a word length of three digits and truncation, compute the solution of a system of equations
On a hypothetical computer with a word length of three digits and truncation, compute the solution of
$$
\begin{matrix}
-3x & + & y & = & -2 \\
10x & - & 3y & = & 7 \\
\end{matrix}
$$
a. Without partial pivoting
b. With partial pivoting
c. Exactly
We consider the matrix $$
\begin{bmatrix}
-3 & 1 & -2 \\
10 & -3 & 7 \\
\end{bmatrix}
$$
and the operation $R_2 → R_2 – (\frac{a_{kj}}{a_{ij}})R_1:$ where $R_i$ are rows and $k > i, a_{ij} \neq 0.$ Then the new matrix looks like
$$
\begin{bmatrix}
-3 & 1 & -2 \\
10 – (-\frac {10}{3})(-3) & -3 & 7 - (-\frac73)(-2) \\
\end{bmatrix}
$$
We'll have to truncate $\frac {10}{3} = 3.3333333333333335$ to $3.333 = 0.333 \times 10^1$. In the same way, $\frac 73 = 0. 233 \times 10^1$, so this matrix is
$$
\begin{bmatrix}
-3 & 1 & -2 \\
10 – (-0.333 \times 10^1)(-3) & -3 & 7 - (-0. 233 \times 10^1)(-2) \\
\end{bmatrix}
$$
which is
$$
\begin{bmatrix}
-3 & 1 & -2 \\
0.01 & -3 & 2.34 \\
\end{bmatrix}
$$
Because of the truncation $0$ becomes $0.01$, so that if we solve $0.01x - 3y = 2.34$ for $(x, y)$ we get distorted final solution.
Now consider partial pivoting which calls for row swapping. Then our matrix is
$$
\begin{bmatrix}
10 & -3 & 7 \\
-3 & 1 & -2 \\
\end{bmatrix}
$$
Applying the row operation $R_2 → R_2 – (\frac{a_{kj}}{a_{ij}})R_1$ yields
$$
\begin{bmatrix}
10 & -3 & 7 \\
-3 - (-\frac {3}{10})10 & 1 – (\frac {1}{10})(-3) & -2 - (-\frac{1}{5})7\\
\end{bmatrix}
$$
We don't have to truncate $\frac {3}{10}, \frac {1}{10}, \frac{1}{5}$ because they are $0.3, 0.1, 0.2$ respectively. So, the matrix is
$$
\begin{bmatrix}
10 & -3 & 7 \\
0 & 1.3 & -0.6\\
\end{bmatrix}
$$
Then $y = \frac {-0.6}{1.3} = -0.4615384615384615$. Since we didn't lose any digits by truncation, this $y$ is also exact.
Does that make sense?
| You have arrived at the correct conclusion. It is however possible to argue that you have made mistakes and reduce your score during a test. This depends on how strict an instructor you have. You specifically state that $\frac{10}{3}$ has a finite decimal expansion and that the last digit is the digit $5$. Of course, you know that this is wrong, but you should have written $\frac{10}{3} = 3.\overline{3} \approx 3.33 \cdot 10^{-1}$. Moreover, it is possible to argue that while you computer makes large errors, it can keep track of the entries which should be zero. In fact this is what is done in all library implementations of Gaussian elimination. They do not compute those entries which should be zero. In fact, those memory locations are used to store the multipliers obtained during the process, i.e. the matrix $A$ is overwritten by its $LU$ factorization.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the following congruence: $x(x+1)(x+2) \equiv 0 \pmod{221}$
Find the first five solutions for,
$$x(x+1)(x+2) \equiv 0 \pmod{221}$$
I am very confused. By CRT,
$x(x+1)(x+2) \equiv 0 \pmod{13}$ and $x(x+1)(x+2) \equiv 0 \pmod{17}$
But these two congruences are also reckless.
| $$x(x+1)(x+2)\equiv 0\pmod{221}\iff \begin{cases}x(x+1)(x+2)\equiv 0\pmod{13}\\x(x+1)(x+2)\equiv 0\pmod{17}\end{cases}$$
By Euclid's Lemma:
$$\iff \begin{cases}x\equiv \{0,-1,-2\}\pmod{13}\\x\equiv \{0,-1,-2\}\pmod{17}\end{cases}$$
By the Chinese Remainder Theorem (CRT) there are $9$ solutions mod $221$.
E.g., if $x\equiv 0\pmod{13}$ and $x\equiv -1\pmod{17}$, then $x\equiv 169 \pmod{221}$.
You should already know how to use CRT. You'll get:
$$x\equiv \{-2,-1,0,50,51,102,117,168,169\}\pmod{221}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Covariance of dice tosses that result in 1 or 2 (fake proof) Question:
Consider n independent tosses of a $k$-sided fair dice. Let $X_i$ be the number of tosses that result in $i$.
What is the covariance $\mathrm{cov}(X_1,X_2)$ of $X_1$ and $X_2$.
\begin{align}
\mathrm{cov}(X_1,X_2) = \mathbf{E}[X_1X_2] - \mathbf{E}[X_1]\mathbf{E}[X_2]
\end{align}
I get a different $\mathbf{E}[X_1X_2]$ than the given solution.
The solution given is
Let $A_t$ (respectively, $B_t$) be a Bernoulli random variabe that is equal to 1 if and only if the $t$th toss resulted in 1 (respectively, 2). We have E$[A_tB_t] = 0$ (since $A_t \neq 0$ implies $B_t \neq 0$)
$$ \mathbf{E}[A_tB_s] = \mathbf{E}[A_t]\mathbf{E}[B_t] = \frac{1}{k} \cdot \frac{1}{k} \mathrm{for}\ s \neq t.$$
Thus,
\begin{align}
\mathbf{E}[X_1X_2] &= \mathbf{E}[(A_1+\cdots + A_n)(B_1+\cdots B_n)]\\
&=n\mathbf{E}[A_1(B_1+\cdots+B_n)] = n(n-1)\cdot \frac{1}{k} \cdot \frac{1}{k} \\
&= \frac{n(n-1)}{k^2}
\end{align}
and
My solution that gives slightly off answer
My approach uses iterated expectations.
\begin{align}
\mathbf{E}[X_1X_2] = \mathbf{E}[\mathbf{E}[X_1X_2|X2]]
\end{align}
If I had $k$ instead of $k-1$ in the following equation, I would get an answer identical to given solution but if I already know $X_2=x_2$ then dice tosses should be identically distributed among k-1 remaining options, right?
\begin{align}
\mathbf{E}[X_1|X_2=x_2] = \frac{n-x_2}{k-1}
\end{align}
Then
\begin{align}
\mathbf{E}[X_1X_2|X_2] = \frac{n-X_2}{k-1} \cdot X_2
\end{align}
\begin{align}
\mathbf{E}[\mathbf{E}[X_1X_2|X_2]] = \mathbf{E}[\frac{nX_2-{X_2}^2}{k-1}]
\end{align}
given $\mathbf{E}[{X_2}^2] = \mathbf{E}[{X_2}] = \frac{n}{k}$
\begin{align}
\mathbf{E}[X_1X_2] = \frac{n(n-1)}{k(k-1)}
\end{align}
So my answer differs to the solution on the matter of $\mathbf{E}[X_1X_2]$
\begin{align}
\mathbf{E}[X_1X_2] = \frac{n(n-1)}{k^2} \neq \frac{n(n-1)}{k(k-1)}
\end{align}
Whats wrong with my logic? Or maybe MIT is wrong.
| Unfortunately it turns out MIT was right and I was wrong.
Thanks to the help from JeanMarie I was able to come up with the following derivation.
\begin{align}
\mathbf{E}[{X_2}^2] &= \mathrm{var}(\mathbf{E}[{X_2}^2]) + \mathbf{E}[X_2]^2 \\
\mathrm{var}(\mathbf{E}[{X_2}^2]) &= \frac{n(k-1)}{k^2} \\
\mathbf{E}[X_2]^2 &= \big(\frac{n}{k}\big)^2 \\
\mathbf{E}[{X_2}^2] &= \frac{n(k-1)}{k^2} + \frac{n^2}{k^2} \\
&= \frac{n^2+n(k-1)}{k^2}
\end{align}
Following from where my own solution left off, bunch of algebra.
\begin{align}
\mathbf{E}[X_1X_2] &= \mathbf{E}[\frac{nX_2-{X_2}^2}{k-1}] \\
&= \frac{1}{k-1}\big(n \cdot \frac{n}{k} - \frac{n^2+n(k-1)}{k^2}\big) \\
&= \frac{1}{k-1} \big( \frac{n^2k}{k^2} - \frac{n^2}{k^2}\big) - \frac{n}{k^2} \\
&= \frac{n^2}{k^2} - \frac{n}{k^2} \\
&= \frac{n(n-1)}{k^2}
\end{align}
Which agrees with the MIT solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.