Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Probability of ordered dice Suppose you throw five dice, order them from lowest to highest number, then select the third die. What is the probability that the third die was number 2?
Or, in general, when throwing $n$ dice and selecting the $i$th die (sorted from low to high), what is the probability that the number of that die is $k_i$? So the above example was $n=5,i=3,k_i=2$. How do you determine a general formula for finding the probability?
| We can use the multinomial law to compute this probability. If we split the probability on the number of 2 you draw we have
$$p=\mathbb P(\textrm{exactly one 2 and two 1}) + \mathbb P(\textrm{exactly two 2 and (one or two) 1}) + \mathbb P(\textrm{exactly three 2}) + \mathbb P(\textrm{exactly four 2}) + \mathbb P(\textrm{exactly five 2})$$
$$p=\left(5 \cdot 6 \cdot \left( \dfrac{1}{6} \right)\left( \dfrac{1}{6} \right)^2\left( \dfrac{4}{6} \right)^2 \right) + \left(10 \cdot 3 \cdot \left( \dfrac{1}{6} \right)^2\left(\left( \dfrac{1}{6} \right)^2\left( \dfrac{4}{6} \right)+\left( \dfrac{1}{6} \right)\left( \dfrac{4}{6} \right)^2 \right) \right) + \left(10 \cdot \left( \dfrac{1}{6} \right)^3\left( \dfrac{5}{6} \right)^2 \right) + \left( 5\left( \dfrac{1}{6} \right)^4\left( \dfrac{5}{6} \right)\right) + \left( \left( \dfrac{1}{6} \right)^5\right)$$
$$p=\left( \dfrac{30 \cdot 16}{6^5}\right) + \left( \dfrac{30 \cdot 20}{6^5}\right) + \left( \dfrac{10 \cdot 25}{6^5}\right) + \left( \dfrac{5 \cdot 5}{6^5}\right)+\left( \dfrac{1}{6^5}\right)$$
$$p=\dfrac{1356}{6^5} \approx 0.17438$$
And we get the probability that @Henry found with R. We can also verify it with a simple Python program that gives the same result
from random import randint
ok = 0
nb_draws = 100000
for j in range(nb_draws):
l = [randint(1,6) for i in range(5)]
l.sort()
if l[2] == 2:
ok += 1
print(ok * 1.0 / nb_draws)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$
Find the closed formula for the sequence and hence find the limit.
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}
\implies a_{n+1} = \frac{4a_n} {2a_n - 3}
\implies a_{n+1} = \frac{\frac{4a_n} {a_n} } {\frac{2a_n}{a_n} - \frac{3} {a_n} }
\implies a_{n+1} = \frac{4} {2 - \frac{3} {a_n} }
\implies \frac{1 } {a_{n+1}} = \frac{2 - \frac{3} {a_n} } {4}
\implies \frac{1 } {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}$$
Then go to where????
| hint
If $a_n=0$ then $a_{n+1}=0$.
assume that for $n>0 \;\;a_n\ne 0$.
dividing by $2a_na_{n+1} $, we get
$$1=\frac {2}{a_{n+1}}+\frac {3}{2a_n} $$
Put $b_n=2/a_n $.
then
$$b_{n+1}=1-\frac {3}{4}b_n $$
Now look for $p$ such that
$$b_{n+1}-p=-3/4(b_n-p) $$
thus
$$p=4/7$$
$$b_n=4/7+(-3/4)^{n-1}(b_1-4/7) $$
$$=4/7+10/7 (-3/4)^{n-1} $$
$$a_n=\frac {7}{2+5 (-3/4)^{n-1}} $$
$$\lim_{n\to+\infty}=7/2$$
| {
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"url": "https://math.stackexchange.com/questions/2319174",
"timestamp": "2023-03-29T00:00:00",
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Solve the equation $x^4+(x-1)(x^2-2x+2)=0$ With $x \in \mathbb{R}$. Solve the equation $$x^4+(x-1)(x^2-2x+2)=0$$
Idea $1$ : $(x^2-x+1)(x^2+2x-2)=0$
Idea $2$ :$(2x^2+x-1)^2=(3x-3)^2$
Idea $3$ :$<=>x^4(x-1)x^2-2(x-1)^2=0$
Let $y=x-1 \dots$
I need another way because I am collecting methods of solving with quartic equations
| Let $t=x-1$. Then, $$x^4+(x-1)(x^2-2x+2)= x^4+(x-1)((x-1)^2+1)$$$$=(t+1)^4+t(t^2+1)=t^4+5t^3+6t^2+5t+1=0$$ Divide by $t^2$: $$\displaystyle \left(t^2+\frac{1}{t^2}\right)+5\left(t+\frac 1t\right)+6=0$$$$\implies \left(\left(t+\frac 1t\right)^2-2\right) +5\left(t+\frac 1t\right)+6=0$$ Let $u=\displaystyle t+\frac 1t$ so that $$u^2+5u+4=0.$$ This has solutions $u=-1,-4$ but will not admit the $-1$ solution because the range of $u=\displaystyle t+\frac 1t$ does not include $-1$. Thus, the roots are of $t^2-4t+1=0$. Add 1 to each root to get the values of x.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int\limits_0^1\frac{-2\ln(x)(1+x^2)^2-2(1-x^4)}{(1-x^2)^3}\,\mathrm{d}x$? $$\int\limits_0^1\frac{-2\ln(x)(1+x^2)^2-2(1-x^4)}{(1-x^2)^3}\,\mathrm{d}x$$
Are there any easy methods to solve this integral? I tried to solve this by series, but I don't know how to complete. I hope the answer does not contain th $\operatorname{Li}$ function.
The last answer from WolframAlpha is $\dfrac{\pi^2}{8}$.
| Let $f(x)$ be your integrand.
Maple gives the antiderivative as
$$ \frac{1}{2(x+1)}+\frac{1}{2(x-1)}-\frac{\text{dilog}(x+1)}{2} -\frac{\ln \left( x \right) \ln \left(
x+1 \right)}{2} -{\frac {\ln \left( x \right) x \left( x+2 \right)
}{2\; \left( x+1 \right) ^{2}}}-\frac{{\text {dilog}} \left( x \right)}{2} +{
\frac {\ln \left( x \right) x \left( x-2 \right) }{2\; \left( x-1
\right) ^{2}}}+{\frac {\ln \left( x \right) x}{2\;(x+1)}}-{
\frac {\ln \left( x \right) x}{2\;(x-1)}}
$$
where
$$ \text{dilog}(x) = \int_1^x \frac{\ln(t)}{1-t}\; dt $$
In fact, if $F_1(x)$ contains the terms without dilog,
$$ F_1(x) = \frac{1}{2(x+1)}+\frac{1}{2(x-1)}-\frac{\ln \left( x \right) \ln \left(
x+1 \right)}{2} -{\frac {\ln \left( x \right) x \left( x+2 \right)
}{2\; \left( x+1 \right) ^{2}}} +{
\frac {\ln \left( x \right) x \left( x-2 \right) }{2\; \left( x-1
\right) ^{2}}}+{\frac {\ln \left( x \right) x}{2\;(x+1)}}-{
\frac {\ln \left( x \right) x}{2\;(x-1)}}$$
you can verify that
$$F_1'(x) - f(x) = - \frac{\ln(x)}{2(x-1)} - \frac{\ln(x+1)}{2x}$$
and $$\lim_{x \to 0} F_1(x) = \lim_{x \to 1} F_1(x) = 0$$
So your integral becomes
$$ \int_0^1 \frac{\ln(x)\; dx}{2(x-1)} + \int_0^1 \frac{\ln(x+1)\; dx}{2x} $$
Using the change of variables $u=x+1$ in the second of these and combining the two, we find your integral is
$$ \frac{1}{2}\int_0^2 \frac{\ln(x)\; dx}{x-1} $$
which I think is fairly "well-known".
| {
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"url": "https://math.stackexchange.com/questions/2322847",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Ceiling inequality? I know $\lceil$$\frac{a}{3}$$\rceil$ + $\lceil$$\frac{b}{3}$$\rceil$ $\le$ $\lceil$$\frac{a + b}{3}$$\rceil$ + 1
what do we know about $\lceil$$\frac{a}{3}$$\rceil$ + $\lceil$$\frac{b}{3}$$\rceil$ + $\lceil$$\frac{c}{3}$$\rceil$ + $\lceil$$\frac{d}{3}$$\rceil$?
| For $x_i=3q_i+r_i$ we have $\lceil \frac {x_i}3\rceil=q_i+\lceil \frac {r_i}3\rceil$.
For a sum you'll get $\sum\limits_{i=1}^4 \lceil \frac{x_i}{3}\rceil=\sum\limits_{i=1}^4 q_i+\sum\limits_{i=1}^4 \lceil \frac{r_i}{3}\rceil$
While $\bigg\lceil\sum\limits_{i=1}^4 \frac{x_i}{3}\bigg\rceil=\sum\limits_{i=1}^4 q_i+\bigg\lceil\sum\limits_{i=1}^4 \frac{r_i}{3}\bigg\rceil$
So we have to compare $A=\sum\limits_{i=1}^4 \lceil \frac{r_i}{3}\rceil$ with $B=\bigg\lceil\sum\limits_{i=1}^4 \frac{r_i}{3}\bigg\rceil$ for all possible values of the $r_i\in\{0,1,2\}$.
$\begin{array}{|l|cc|c|}
\hline r_i & A & B & A-B\\ \hline
0000 & 0 & 0 & 0 \\
0001 & 1 & 1 & 0 \\
0002 & 1 & 1 & 0 \\
0011 & 2 & 1 & 1 \\
0012 & 2 & 1 & 1 \\
0022 & 2 & 2 & 0 \\
0111 & 3 & 1 & 2 \\
0112 & 3 & 2 & 1 \\
0122 & 3 & 2 & 1 \\
0222 & 3 & 2 & 1 \\
1111 & 4 & 2 & 2 \\
1112 & 4 & 2 & 2 \\
1122 & 4 & 2 & 2 \\
1222 & 4 & 3 & 1 \\
2222 & 4 & 3 & 1 \\ \hline
\end{array}$
Thus the maximum is $2$ and finally we get:
$\displaystyle \big\lceil\frac a3\big\rceil+\big\lceil\frac b3\big\rceil+\big\lceil\frac c3\big\rceil+\big\lceil\frac d3\big\rceil\le\big\lceil\frac {a+b+c+d}3\big\rceil+2$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to find the shortest distance from line to parabola? How to find the shortest distance from line to parabola?
parabola: $$2x^2-4xy+2y^2-x-y=0$$and the line is: $$9x-7y+16=0$$
Already tried use this formula for distance:
$$\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^2+b^2}}$$
| $$2x^2-4xy+2y^2-x-y=0$$
At a point $(x_0, y_0)$ on the parabola, the gradient is
$\nabla f(x_0,y_0) = (4x_0-4y_0-1, -4x_0+4y_0-1)$
The direction of a normal to the line $9x-7y+16=0$ is $(7,9)$. So we must have
\begin{align}
(4x_0-4y_0-1, -4x_0+4y_0-1) \circ (7,9) &= 0 \\
-8 x_0 + 8 y_0 - 16 &= 0 \\
y_0 = 2+x_0 \\
\hline
2 x_0^2 - 4 x_0 (2+x_0) + 2 (2+x_0)^2 - x_0 - (2+x_0) &= 0 \\
6 - 2x_0 &= 0 \\
(x_0, y_0) &= (3, 5)
\end{align}
The distance from the line $9x-7y+16=0$ to the point $(3,5)$ is
$$\frac{|9x_0-7y_0+16|}{\sqrt{9^2+7^2}}
=\frac{|27-35+16|}{\sqrt{130}} =\frac{8}{\sqrt{130}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can I break this limit into individual terms? $$\lim_{x\to \infty} {\frac{x}{x^2+1} +\frac{x}{x^2+2} + ... + \frac{x}{x^2+x} }$$
It seems obvious that the result is zero for each term but in order to break the limit into its individual parts we must know that every term's limit exists .
| Note that the series of interest can be written as
$$\frac{x}{x^2+1}+\frac{x}{x^2+2}+\cdots +\frac{x}{x^2+x}=\sum_{k=1}^x \frac{x}{x^2+k}$$
The summand clearly satisfies the bounds
$$\frac{x}{x^2+x}\le\frac{x}{x^2+k}\le \frac{x}{x^2+1}$$
Hence we can assert that
$$\frac{x^2}{x^2+x}\le \sum_{k=1}^x\frac{x}{x^2+k}\le \frac{x^2}{x^2+1}$$
whereby application of the squeeze theorem yields the coveted limit
$$\lim_{x\to \infty}\left(\frac{x}{x^2+1}+\frac{x}{x^2+2}+\cdots +\frac{x}{x^2+x}\right)=1$$
| {
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"url": "https://math.stackexchange.com/questions/2325093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Probability: 2 dice rolled, two players
Two students play a game based on the total roll of two standard dice.
Student A says that a 12 will be rolled first. Student B says that
two consecutive 7s will be rolled first. The students keep rolling
until one of them wins. What is the probability that A will win?
I am unable to formulate a repeating sequence or an appropriate formula. How can I solve this question?
| The probability of throwing a 12 equals $\frac{1}{36}$, the probability of throwing a 7 equals $\frac{6}{36}$ and the probability of throwing anything else equals $\frac{29}{36}$. In every turn, we must take into account these three possibilities. Let us denote $A$ the event in which a 12 is thrown first and $B$ the event in which two consecutive 7s are thrown first. We then get:
$$P[A] = \frac{1}{36} 1 + \frac{6}{36} \bigg(\frac{1}{36} 1 + \frac{6}{36} 0 + \frac{29}{36} P[A]\bigg) + \frac{29}{36} P[A]$$
$$\iff P[A] = \frac{6}{216} + \frac{1}{216} + \frac{29}{216} P[A] + \frac{174}{216} P[A] \iff \frac{13}{216} P[A] = \frac{7}{216} \iff P[A] = \frac{7}{13}$$
We can now either use $P[B] = 1 - P[A] = \frac{6}{13}$, or use a similar approach for the event B:
$$P[B] = \frac{1}{36} 0 + \frac{6}{36} \bigg(\frac{1}{36} 0 + \frac{6}{36} 1 + \frac{29}{36} P[B]\bigg) + \frac{29}{36} P[B]$$
$$\iff P[B] = \frac{6}{216} + \frac{29}{216} P[B] + \frac{174}{216} P[B] \iff \frac{13}{216} P[B] = \frac{6}{216} \iff P[B] = \frac{6}{13}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can you prove the asymptotic formula of the following function? In the following equation,
$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
)cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$
if $ 0<N_F(a) ≪ 1$ and $(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$
how do you arrive at
$$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2}
+\frac{1}{\frac{x^2}{η^2} − a^2} sin^2
(πN_F(a)\frac{x}{a}))$$
.
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.
.
.
.
.
.
.
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Please see what I have done so far and check if I have errors in it.
Relevant Equations
$\alpha (x; a) = \sqrt{N_F(a)\eta } (1 - \frac{x}{a\eta})$
where $\eta = 1 + L/D$, $N_F(a) = \frac{2a^2}{\lambda L}$ and (additional definition) $\gamma = \eta - 1$
Starting with this:
$P(x; a)=\frac{1}{2\lambda(L+D)} ([C(α(x; a)) + C(α(x; −a))]^2 + [S(α(x; a)) + S(α(x; −a))]^2)$
$C[α(x; +a)] + C[α(x; −a)] ≃ \frac{1}{πα(x; a)} sin (\frac{πα(x; a)^2}{2}
) + \frac{1}{πα(x; -a)} sin (\frac{πα(x; -a)^2}{2}
) $
$S[α(x; +a)] + S[α(x; −a)] ≃ \frac{-1}{πα(x; a)} cos (\frac{πα(x; a)^2}{2}
) - \frac{1}{πα(x; -a)} cos (\frac{πα(x; -a)^2}{2}
) $
Here is what I have done so far:
$(C[α(x; +a)] + C[α(x; −a)])^2 + (S[α(x; +a)] + S[α(x; −a)])^2 = \frac{1}{π^2α^2(x; a)} + \frac{1}{π^2α^2(x; -a)} + \frac{2}{π^2α(x; +a)α(x; −a)} [sin (\frac{πα(x; a)^2}{2})sin (\frac{πα(x; -a)^2}{2}) + cos (\frac{πα(x; a)^2}{2}
)cos (\frac{πα(x; -a)^2}{2})]$
$=\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
)cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]$
And then the outer factor:
$\frac{1}{2\lambda (L+D)} = \frac{\gamma}{2\lambda L \eta}$
So the new equation for P is:
$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
)cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$
But this is where I am not sure what to do anymore with $N_F (a) ≪ 1$ and if $(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$ to arrive at this equation.
$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2}
+\frac{1}{\frac{x^2}{η^2} − a^2} sin^2
(πN_F(a)\frac{x}{a}))$
For more info about these derivations, please visit Page 14 of https://arxiv.org/pdf/1110.2346.pdf
Picture of the final result in the paper
| $P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2}
- \frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$
$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2(1+\frac{x^2}{a^2\eta^2})}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [cos (2πN_F(a)\frac{x}{a})]]$
$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [\frac{1 + \frac{x^2}{a^2\eta^2}}{1 - \frac{x^2}{a^2\eta^2}}+cos (2πN_F(a)\frac{x}{a})]] $
$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{ \frac{x^2}{a^2\eta^2}}{ \frac{x^2}{a^2\eta^2}-1}+\frac{1 }{\frac{x^2}{a^2\eta^2}-1}-cos (2πN_F(a)\frac{x}{a})]] $
$P(x; a)≃ \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +1-cos (2πN_F(a)\frac{x}{a})]] $
$P(x; a)≃ \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +2 sin^2 (πN_F(a)\frac{x}{a})]] $
$P(x; a)≃ \frac{\gamma}{ \eta} \frac{1}{π^2a^2\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{2 }{\frac{x^2}{a^2\eta^2}-1} +2 sin^2 (πN_F(a)\frac{x}{a})] $
$P(x; a)≃ \frac{2\gamma}{π^2 \eta^2} (\frac{1}{a^2( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} + sin^2 (πN_F(a)\frac{x}{a})]) $
$P(x; a)≃ \frac{2\gamma}{π^2 \eta^2} (\frac{1}{( \frac{x^2}{\eta^2}-a^2)} [\frac{a^2 }{\frac{x^2}{\eta^2}-a^2} + sin^2 (πN_F(a)\frac{x}{a})]) $
$P(x; a)≃ \frac{2\gamma}{π^2 \eta^2} ( \frac{a^2 }{(\frac{x^2}{\eta^2}-a^2)^2} + \frac{1}{( \frac{x^2}{\eta^2}-a^2) }sin^2 (πN_F(a)\frac{x}{a})) $
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$
Solve for $x:0\leq x \leq \frac{\pi}{2}$
$$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$$
My attempt,
I changed it into $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$
| Let $\sin{x}+\cos{x}=t$ and we get a quadratic equation:
$$4(t^2-1)-\sqrt6t+1=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Polynomials factor Theorem If a cubic polynomial $p(x)$ with leading co-efficient 1 is divided by $x-1$, $x-2$ and $x-3$ respectively it leaves $r(x)$ $1$, $4$ and $9$. What will be the remainder when it is divided by $x-4$?
I've searched for this everywhere but I couldn't find an answer.
Help would be much appreciated.
| Let $f(x)=p(x)-x^2$. Then $f(x)$ is a cubic polynomial.
Note that $f(1)=p(1)-1^2=1-1=0$, $f(2)=p(2)-2^2=4-4=0$ and $f(3)=p(3)-3^2=9-9=0$
$f(x)$ has roots $1$, $2$ and $3$. So,
$$f(x)=A(x-1)(x-2)(x-3)$$
for some constant $A$.
$$p(x)=A(x-1)(x-2)(x-3)+x^2$$
When $p(x)$ is divided by $x-4$, then remainder is
$$p(4)=6A+16$$
Without other information on $p(x)$, we cannot proceed further.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $I=\int_{-\infty}^\infty \frac{u^2}{5u^2\left(u^2+1\right)+2}\,\mathrm{d}u$? This integral popped up when I was trying to solve this integral : $$\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{4(\cos^4x+2\sin^4x)+2\sin^2 2x}\,dx $$
I simplified it a little bit, substituted $\tan(x)=u$ and came up with,
$$I= \int_{-\infty}^\infty \frac{u^2}{5u^2\left(u^2+1\right)+2}\,\mathrm{d}u$$
any suggestions for how I can take it from here ?
| This is more a comment than an answer.
Considering
$$\frac{\sin^2(x)}{4(\cos^4(x)+2\sin^4(x))+2\sin^2 (2x)}$$ the denominator can be simplified as $2\cos^2(2x)-4\cos(2x)+10$ and the numerator as $\frac{1-\cos(2x)} 2$.
So, for the time being, let $X=\cos(2x)$ and the integrand is then $$\frac{1-X}{4 (X-a) (X-b)}=\frac 1{4(a-b)}\left(\frac{1-a}{X-a} -\frac{1-b}{X-b}\right)$$ where $a=1-2i$, $b=1+2i$ are the roots of $2X^2-4X+10=0$.
So, we are basically facing simpler integrals looking like $$I=\int \frac{dx}{\cos(2x)-c}=\frac 1{\sqrt{1-c^2}}{\tanh ^{-1}\left(\sqrt{\frac{1+c}{1-c}}\tan (x)\right)}$$ and the problem becomes easy (assuming that you enjoy complex numbers).
| {
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"url": "https://math.stackexchange.com/questions/2328886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find when $\frac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}$ is an integer
Find all natural numbers $n$ such that $$\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}$$ is an integer.
For $n = 1$ we have $\dfrac{18}{3} = 6$.
For $n = 2$ we have $\dfrac{126}{18} = 7$.
For $n = 3$ we have $\dfrac{972}{126} = \dfrac{54}{7}$
For $n = 4$ we have $\dfrac{7938}{972} = \dfrac{49}{6}$.
I don't seem to find any other integers for the fraction for $n > 2$. How can we find all of the $n$ such that it is an integer?
| $\begin{array}\\
\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}
&=\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}-9+9\\
&=\dfrac{3^n+6^n+9^n-9(3^{n-1}+6^{n-1}+9^{n-1})}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{3^n+6^n+9^n-9\cdot 3^{n-1}-9\cdot 6^{n-1}-9^{n}}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{3^n+6^n-9\cdot 3^{n-1}-9\cdot 6^{n-1}}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{3^n+6^n- 3^{n+1}-\frac23 6^{n}}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{\frac13 6^n-2\cdot 3^{n}}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{3\cdot (2/3)^{n-1}-9\cdot (1/3)^{n-1}}{(1/3)^{n-1}+(2/3)^{n-1}+1}+9\\
\end{array}
$
For
$n \ge 3$,
$0 \lt 3\cdot (2/3)^{n-1}-9\cdot (1/3)^{n-1}
\lt 1$,
so this is never an integer
for $n \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x^{2a-3} y^2 = x^{6-a} y^{5a}$ then $a\log(\frac{x}{y}) = ?$ If $x^{2a-3} y^2 = x^{6-a} y^{5a}$ then $a\log(\frac{x}{y}) = ?$
I have Tried this and stucked here.
| You are given
$x^{2a-3} y^2=x^{6-a} y^{5a}
$.
Multiplying by
$x^{a-6}y^{-2}
$,
this becomes
$x^{3a-9}=y^{5a-2}
$.
Looking at this,
I notice that,
if the
$y^2$
were
$y^{2a}$,
the right side would be
$y^{3a}$.
This indicates to me that
there might be
an error in
the statement of the problem.
However,
working with the problem
as written,
$(3a-9)\log x=(5a-2)\log y
$
so that
$\log y
=\frac{3a-9}{5a-2}\log x
$.
Therefore
$\begin{array}\\
a\log(x/y)
&=a(\log x-\log y)\\
&=a(\log x-\frac{3a-9}{5a-2}\log x)\\
&=\log x(a-\frac{3a-9}{5a-2})\\
&=a\log x(1-\frac{3a-9}{5a-2})\\
&=a\log x\frac{(5a-2)-(3a-9)}{5a-2}\\
&=a\log x\frac{2a+7}{5a-2}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Asymptotic expansion of the sum $ \sum\limits_{k=1}^{n} \frac{\binom{n+1}{k} B_k}{ 3^k-1 } $ The situation :
I am looking for an asymptotic expansion of the sum $\displaystyle a_n=\sum_{k=1}^{n} \frac{\binom{n+1}{k} B_k}{ 3^k-1 } $ when $n \to \infty$.
(The $ B_k $ are the Bernoulli numbers defined by $ \displaystyle \frac{z}{e^{z}-1}=\underset{n=0}{\overset{+\infty }{\sum }}\frac{B_{n}}{n!}z^{n}$).
Context :
The initial problem was that I need to calculate a radius of convergence of a power series $\displaystyle \sum_{k=1}^{} a_n z^n $. I have almost tried everything to calculate this asymptotic expansion of the $a_n$, but to no avail.
The numerical test (computing) shows that $\displaystyle \lim_{n\to +\infty} \frac{a_{n+1}}{a_n} = 1$, that is, the convergence radius of the series is equal to $1$. But I can not analytically prove it.
My attempts to solve it :
$\displaystyle
\begin{align*}
a_n=\sum_{k=1}^{n} \frac{\binom{n+1}{k} B_k}{3^k-1 } &= \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k3^{-k}}{ 1- 3^{-k} } \\
&= \sum_{k=1}^{n} \binom{n+1}{k}B_k3^{-k} \sum_{p=0}^{+\infty}3^{-pk} \\
&= \sum_{p=0}^{+\infty} \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k} {3^{(p+1)k}} \\
\end{align*}
$
Using the Faulhaber's formula : $\displaystyle \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k} {N^k} = \frac{n+1}{N^{n+1}} \sum_{k=1}^{N-1} k^n -1$
We replace $N$ by $3^{p+1}$
$\displaystyle \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k} {3^{(p+1)k}} = \frac{n+1}{3^{(p+1)(n+1)}} \sum_{k=1}^{3^{p+1}-1} k^n -1$
That is to say
$\displaystyle a_n = \sum_{p=0}^{+\infty} \left(\frac{n+1}{3^{(p+1)(n+1)}} \sum_{k=1}^{3^{p+1}-1} k^n -1\right)$
Or
$\displaystyle a_n = \sum_{p=0}^{+\infty} \left( \frac{n+1}{3^{p+1}} \sum_{k=1}^{3^{p+1}-1} \left( \frac{k}{3^{p+1}} \right)^n -1\right)$
If I come by your help, to answer this question, I will publish a new formula of Riemann zeta function that I find elegant.
Thank you in advance for your help.
| If we expand $\frac{1}{3^k-1}$ as a geometric series the problem boils down to estimating
$$ \sum_{k=1}^{n}\binom{n+1}{k}\frac{B_k}{3^{\eta k}} \tag{1}$$
and by Faulhaber's formula
$$ S_n(m)=\sum_{k=1}^{m}k^n = \frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k}B_k^+ m^{n+1-k} \tag{2} $$
so by choosing $m=3^{\eta}$ we get:
$$ (n+1) S_n(3^\eta) = (n+1)\left[1+2^n+\ldots+3^{\eta n}\right] = \sum_{k=0}^{n}\binom{n+1}{k}\frac{3^{\eta(n+1)}}{3^{\eta k}}B_k^+\tag{3} $$
and:
$$ \frac{n+1}{3^{\eta(n+1)}}\left[1+2^n+\ldots+3^{\eta n}\right] = 1+\frac{n+1}{3^\eta}+\sum_{k=1}^{n}\binom{n+1}{k}\frac{B_k}{3^{\eta k}} \tag{4}$$
from which:
$$ (1) = \sum_{k=1}^{n}\binom{n+1}{k}\frac{B_k}{3^{\eta k}}=-1+\frac{n+1}{3^{\eta(n+1)}}\left[1+2^n+3^n+\ldots+(3^\eta-1)^n\right]\tag{5} $$
and it is not difficult to finish from here. You may notice that
$$ \frac{1}{3^{p+1}}\sum_{k=1}^{3^{p+1}-1}\left(\frac{k}{3^{p+1}}\right)^n -\frac{1}{n+1}$$
is the error term of the rectangle method applied to the function $f(x)=x^n$ over the interval $(0,1)$, partitioned into $3^{p+1}$ equal intervals.
| {
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"url": "https://math.stackexchange.com/questions/2333563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Orthonormalbasis of a isometry $2 \times 2$ matrix I am a bit stuck in this easy question.
Let $A\in \mathbb R^{2 \times2}$ be a isometry so $A^{-1} = \bar{A}^T$. Show that $\left\{\begin{pmatrix} a \\ c \end{pmatrix},\begin{pmatrix} b \\ d \end{pmatrix}\right\}$ is a orthonormal basis.
$A = \begin{pmatrix} a&b \\ c&d \end{pmatrix} \Longrightarrow \bar{A}^T = \begin{pmatrix} a&c \\ b&d \end{pmatrix}$
$A \cdot \bar{A}^T = \begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd&c^2+d^2 \end{pmatrix} = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} $
I tried to use Gram-Schmidt but too many variables were left so I couldn't prove it.
Now i also know that is has to be
$\left\langle\begin{pmatrix} a \\ c \end{pmatrix}, \begin{pmatrix} b \\ d \end{pmatrix}\right\rangle =0 \Longrightarrow ab+cd = 0 $ and $\left\langle\begin{pmatrix} a \\ c \end{pmatrix}, \begin{pmatrix} a \\ c \end{pmatrix}\right\rangle =1 \Longrightarrow a^2+c^2 = 1$ but I also can't prove that because I only have $a^2+b^2 = 1 = c^2 + d^2$ and $ac+bd = 0$.
| Hint:
Since $A$ is an isometry therefore $\left\|A\mathbf{x}\right\|=\left\|\mathbf{x}\right\|$ for every $\mathbf{x} \in \mathbb{R}^2$. So
\begin{align*}
\left\| A \begin{bmatrix}1\\0\end{bmatrix}\right\|&=\left\|\begin{bmatrix}1\\0\end{bmatrix}\right\|\\
\left\| \begin{bmatrix}a\\c\end{bmatrix}\right\|&=\left\| \begin{bmatrix}1\\0\end{bmatrix}\right\|\\
a^2+c^2 & =1.
\end{align*}
Likewise you can get $b^2+d^2=1$.
Can you proceed from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How prove this inequality $\sum_{cyc}\sqrt{\frac{2b+2c}{a}-1}\ge 3\sqrt{3}$
Let $a,b,c>0 ,2b+2c-a\ge 0,2c+2a-b\ge 0,2a+2b-c\ge 0$ show that
$$\sqrt{\dfrac{2b+2c}{a}-1}+\sqrt{\dfrac{2c+2a}{b}-1}+\sqrt{\dfrac{2a+2b}{c}-1}\ge 3\sqrt{3}$$
I try use AM-GM and Cauchy-Schwarz inequality and from here I don't see what to do
| Since our inequality is homogeneous, we can assume that $a+b+c=3$.
Hence, $2b+2c-a=2(3-a)-a=3(2-a)\geq0$, which gives $\{a,b,c\}\subset(0,2]$.
Thus, we need to prove that
$$\sum_{cyc}\sqrt{\frac{2(b+c)}{a}-1}\geq3\sqrt3$$ or
$$\sum_{cyc}\sqrt{\frac{2(3-a)}{a}-1}\geq3\sqrt3$$ or
$$\sum_{cyc}\sqrt{\frac{2}{a}-1}\geq3$$ or
$$\sum_{cyc}\left(\sqrt{\frac{2}{a}-1}-1\right)\geq0$$ or
$$\sum_{cyc}\frac{1-a}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})}\geq0$$ or
$$\sum_{cyc}\left(\frac{1-a}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})}+\frac{a-1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)(\sqrt{a(2-a)}+a-2)}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})}\geq0$$ or
$$\sum_{cyc}\frac{(a-1)\sqrt{2-a}(\sqrt{a}-1)}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})}\geq0$$ or
$$\sum_{cyc}\frac{(a-1)^2\sqrt{2-a}}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})(\sqrt{a}+1)}\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A solution to a certain differential equation I was looking for the general solution of the differential equation $$nx \frac{dy}{dx} -2y=-kxy^{1-n}$$
This is my method:
I recognize that the differential equation above is a Bernoulli equation with $n = 1-n$ because it has the form $\frac{dy}{dx} + P(x)y = Q(x)y^n.$ Armed with that thought process, I divided the given differential equation by $nx$ to resemble the form $\frac{dy}{dx} + P(x)y = Q(x)y^n.$ It looks like this: $$\frac{dy}{dx} - \left(\frac{2}{nx}\right) y = \frac{ky^{1-n}}{n}$$
I see that $Q(x)= 1.$ Divide again the differential equation above by $y^{1-n}$, so it becomes: $$y^{-(1-n)} \frac{dy}{dx} - \left(\frac{2}{nx}\right) y^{-n} = \frac{k}{n}$$ or $$y^{n-1} \frac{dy}{dx} - \left(\frac{2}{nx}\right) y^{-n} = \frac{k}{n}$$
Let $y^{-n} = z, $ so it becomes $$\frac{1}{1-(n-1)} \frac{dz}{dx} - \left(\frac{2}{nx}\right)z = \frac{k}{n}$$ or $$-\frac{1}{n} \frac{dz}{dx} - \left(\frac{2}{nx}\right)z = \frac{k}{n}$$
The differential equation above must have the form $\frac{dz}{dx} + P(x)z = Q(x)$ to get its integrating factor $I(x), $ so the differential equation becomes $$ \frac{dz}{dx} + \left(\frac{2}{x}\right)z = -k$$
The integrating factor then is $I(x) = e^{\int P(x)dx} = e^{\int \frac{2n}{x}dx} = e^{\frac{2}{n} lnx} = x^{\frac{2}{n}} $. Multiplying this integrating factor to the differential equation above, it becomes: $$x^{\frac{2}{n}}\frac{dz}{dx} + x^{\frac{2}{n}}\left(\frac{2}{nx}\right)z = -kx^{\frac{2}{n}}$$ or $$x^{\frac{2}{n}}\frac{dz}{dx} + \frac{2}{n} x^{\frac{-n+2}{n}}z = -kx^{\frac{2}{n}}$$
Then, doing this: $$ d \left(x^{\frac{2}{n}} z \right) = -kx^{\frac{2}{n}}$$ Getting the integral of the terms in the equation, we get: $$x^{\frac{2}{n}} z = \frac{-kn}{2+n} x^{\frac{2+n}{2}} +c$$ or $$\frac{x^{\frac{2}{n}}}{y^n} = \frac{-kn}{2+n} x^{\frac{2+n}{2}} +c$$
Getting the $y^n,$ the solution to the given differential equation is $$y^n = \frac{\frac{-kn}{2+n} x^{\frac{2+n}{n}} + c}{x^{\frac{2}{n}}}$$
The correct general solution of the differential equation $nx \frac{dy}{dx} -2y=-kxy^{1-n}$ written in the book I used is $y^n = kx+cx^2$
Where did I go wrong?
| The integrating factor to solve he linear equation
$$
\frac{dz}{dx} + \frac{2}{x}\,z = -k
$$
is incorrect. It should be $x^2$ (there is no $n$ in the equation.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T$?
Find the basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T:M^{\mathbb R}_{2 \times 2} \to M^{\mathbb R}_{2 \times 2}$ defined as $T(A)=A-A^t$ for all $A \in M^{\mathbb R}_{2 \times 2}$.
Let $A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$. Then:
$$
T(A)=\begin{pmatrix} a&b\\c&d \end{pmatrix}-\begin{pmatrix} a&c\\b&d \end{pmatrix}=\begin{pmatrix} 0&b-c\\c-b&0 \end{pmatrix}\stackrel{R_2 \gets-1\cdot R_2}{=}\begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix}
$$
In order to find $\ker T$ we'll evalute $T(A)=0$:
$$
\begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix}=\begin{pmatrix} 0&0\\0&0 \end{pmatrix}\Rightarrow b=c \Rightarrow \ker T=span \Biggl\{ \begin{pmatrix} 1&0\\0&0 \end{pmatrix}, \begin{pmatrix} 0&1\\1&0 \end{pmatrix},\begin{pmatrix} 0&0\\0&1 \end{pmatrix} \Biggr\}
$$
Thus $\dim(\ker T)=3$.
Regarding $\mathrm{Im} T$:
$$
ImT=\Biggl\{ \begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix} \text{such that}\quad b,c, \in \mathbb R \Biggr\}=span\Biggl\{ \begin{pmatrix} 0&1\\1&0 \end{pmatrix} \Biggr\}
$$
Thus $\dim(\mathrm{Im} T)=1$.
Are my calculations correct?
| I think an answer will help clarify things. You computed that $$T(A) = \begin{pmatrix} 0 & b-c \\ c-b & 0 \end{pmatrix} = \begin{pmatrix} 0 & b-c \\ -(b-c) & 0 \end{pmatrix} = (b-c) \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}_.$$
From this we see that $\textrm{Im}(T) = \textrm{span}\left\{ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\right\}_.$
We also see that $T(A) = 0$ if and only if $b-c = 0$. In other words when $b=c$. From there you see that $$\ker(T) = \textrm{span}\left\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\}_.$$
Note you shouldn't actually be row reducing $T(A)$, you want that matrix to be zero. You could find the matrix of $T$ with respect to some basis, and row reduce that to find the kernel of $T$.
The reason that row reduction can be used to find the kernel of a matrix is because row reduction corresponds to multiplying on the left by an invertible matrix. This doesn't change the kernel.
An alternative approach to the problem. Let us compute the matrix of $T$, call it $B$, with respect to the standard basis of $M_{2\times 2}(\mathbb{R})$, $\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\}_.$
$$T\left(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}_. $$
$$T\left(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \right) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}_. $$
$$T\left(\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}_. $$
$$T\left(\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}_. $$
Using this, we calculate that $$ B = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & -1 & 0 \\ 0 & -1 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}_.$$
You could row reduce this or just see by inspection that $A \in \ker (T)$ precisely when $b = c$.
| {
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Show that $\{a + b\sqrt{5} \mid a,b \in \mathbf{Q} \}$ is a field. The question asked to prove that $F$ is a commutative field, I got everything except proving that there is a multiplicative inverse in $F$ for every $x\in F$. Any help is appreciated!
Here is specifically what I am asking:
Let $F = \{a + b\sqrt{5} \mid a,b \in \mathbf{Q} \}$.
Show that $\exists$ a multiplicative inverse $y\in F$ such that $xy = 1$ for all $x \in F$.
| We need that $(a+b\sqrt{5})(c+d\sqrt{5})=1$.
Consider $(a+b\sqrt{5})(a-b\sqrt{5})=a^2-5b^2$.
This is a rational number, since both $a$ and $b$ are.
We have that $\displaystyle (a+b\sqrt{5}) \times \boxed{\frac{1}{a^2-5b^2}(a-b\sqrt{5})}=1$.
Therefore, $\displaystyle c=\frac{a}{a^2-5b^2}$, $\displaystyle d=-\frac{b}{a^2-5b^2}$ forms the multiplicative inverse.
The inverse exists, because if $a^2-5b^2=0$, then $a^2=5b^2$, and $a=\sqrt{5}b$.
This cannot happen, because both $a$ and $b$ are rational.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the angle $\angle ACD=\theta$. Problem: Triangle $\triangle ABC$ has $AC=BC$ and $\angle ACB=96^{\text{o}}$. $D$ is a point in $\triangle ABC$ such that $\angle DAB=18^{\text{o}}$ and $\angle DBA=30^{\text{o}}$. Compute $\angle ACD.$
Drawing the problem first:
From the law of cosine we have that $b^2=2a^2-2a^2\cos{96}=2a^2(1-\cos{96})$ which leads to $b=|a|\sqrt{2(1-\cos{96})}.$
Since $\angle CAB=\angle ABC=(180-96)/2=42$, it follows that $\angle CAD=42-18=24$ and $\angle CBD = 42-30=12$. This information allows us to express $f$ in two ways by the law of cosine: \begin{array}{lcl}
f^2 & = & e^2+a^2+2ea\cos{12} \ \ \ \ \ \ \ \ \ \quad \ \quad (1)\\
f^2 & = & d^2+a^2+2da\cos{24} \ \ \ \ \ \ \ \ \ \quad \ \quad (2)\\
\end{array}
Subtracting (1)-(2) we get $e^2-d^2+2a(e\cos{12}-d\cos{24})=0$.
Here I'm stuck. Any suggestion on how to proceed?
| Using Law of Sines in triangle $\triangle{ABC}$ we can write:
$$
\frac{a}{\sin 42}=\frac{b}{\sin96} \Rightarrow b=\frac{a\sin 96}{\sin42}
$$
Similarly in triangle $\triangle{ABD}$ we have:
$$
\frac{b}{\sin 132}=\frac{d}{\sin30} \Rightarrow b=2d\sin 132=2d\cos42
$$
Equating $b$'s gives us:
$$a = \frac{2d\sin 42 \cos 42}{\sin 96}=\frac{d \sin84}{\cos6}=\frac{d\cos6}{\cos6} \Rightarrow a=d$$
So triangle $\triangle{ACD}$ is an isosceles triangle which yields $\measuredangle{ACD}=78^{\circ}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Correct answer of an indefinite integral Find the value of
$$ \int{\frac{dx}{x\sqrt{1-x^3}}} $$
I assumed $x^3 = \sin^2\theta$ and found the solution as
$$\frac{2}{3} \log\left|\frac{1}{x\sqrt{x}} - \frac{\sqrt{1-x^3}}{x\sqrt{x}} \right| + c$$
but the solution is given as
$$\frac{1}{3} \log{\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|} + c$$
Any help to reach to this provided solution will be appreciated.
| \begin{align}
u & = \sqrt{1-x^3} \\
u^3 & = 1-x^3 \\
3u^2\,du & = -3x^2\,dx \\
u^2\,du & = -x^2 \, dx \\[10pt]
\int \frac{dx}{x\sqrt{1-x^3}} & = \int \frac{x^2\,dx}{x^3\sqrt{1-x^3}} = -\int \frac{u^2\,du}{(1-u^3) u} = -\int \frac{u\,du}{1-u^3} \\[10pt]
& = - \int \frac{u\,du}{(1-u)(1+u+u^2)} = \int \left( \frac A {1-u} + \frac{Bu+C}{1+u+u^2} \right) du \quad \text{etc.}
\end{align}
$$ {} $$
\begin{align}
x^3 & = \sin^2\theta \\
3x^2\,dx & = 2\sin\theta\cos\theta\,d\theta \\[10pt]
\int\frac{dx}{x\sqrt{1-x^3}} & = \int \frac{x^2\,dx}{x^3\sqrt{1-x^3}} = \frac 2 3 \int \frac{\sin\theta\cos\theta\,d\theta}{\sin^2\theta \cos\theta} = \frac 2 3 \int \frac{d\theta}{\sin\theta} \\[10pt]
& = -\frac 2 3 \log|\csc\theta+\cot\theta| + C = - \frac 2 3 \log\left| \frac{1+\cos\theta}{\sin\theta} \right| + C \\
& = -\frac 2 3 \log\left| \frac{1 + \sqrt{1-x^3}}{x^{3/2}} \right| + C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify this expression using logarithms I want to find the value of this expression in as simple way as possible.
$$ \frac{1+ 2\log_3 2}{(1+ \log_3 2)^2}+ \log^2_6 2 $$
I simplified and I am stuck at
$$ \frac{1+2\log_3 2+2\log_6 2+4\log_3 2×2\log_6 2}{1+4\log_3 2} $$
| Knowing that $\log_ab = \frac{\log b}{\log a}$
Then the expression becomes:
$$\frac{1 + 2\frac{\log 2}{\log 3}}{(1+\frac{\log 2}{\log 3})^2} + (\frac{\log 2}{\log 6})^2$$
$$= \frac{1 + 2\frac{\log 2}{\log 3}}{1+2\frac{\log 2}{\log 3} + \frac{\log^2 2}{\log^2 3}} + (\frac{\log 2}{\log 3 + \log 2})^2$$
$$= \frac{\log^23 + 2\log 2\log 3}{\log^23+2\log2\log 3 + \log^2 2} + \frac{\log^2 2}{\log^23 + 2\log2\log3 + \log^22}$$
$$= \frac{\log^23 + 2\log 2\log 3 + \log^22}{\log^23+2\log2\log 3 + \log^2 2}$$
$$=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate : $1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots$ Evaluate :
$$1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots$$
thought: I have tried to manipulate the series so that the sum upto n terms , can be simplified. But I can't figure out any profitable result.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&1 + {2 \over 6} + {2 \cdot 5 \over 6 \cdot 12} +
{2 \cdot 5 \cdot 8 \over 6 \cdot 12 \cdot 18} +\cdots =
1 + \sum_{n = 0}^{\infty}{\prod_{k = 0}^{n}\pars{3k + 2} \over
\prod_{k = 0}^{n}\pars{6k + 6}}
\\[5mm] & =
1 + \sum_{n = 0}^{\infty}{1 \over 2^{n + 1}\pars{n + 1}!}\
\overbrace{\prod_{k = 0}^{n}\pars{k + {2 \over 3}}}
^{\substack{\ds{\pars{2/3}^{\overline{n + 1}} =}\\[1.5mm]
\ds{\Gamma\pars{5/3 + n}/\Gamma\pars{2/3}}}} =
1 +
\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}
{\pars{n + 2/3}! \over \pars{n + 1}!\pars{-1/3}!}
\\[5mm] & =
1 + \sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}{n + 2/3 \choose n + 1} =
1 +
\sum_{n = 0}^{\infty}{1 \over 2^{n + 1}}{-2/3 \choose n + 1}\pars{-1}^{n + 1} =
1 +
\sum_{n = 1}^{\infty}{-2/3 \choose n}\pars{-\,{1 \over 2}}^{n}
\\[5mm] & =
1 + \braces{\bracks{1 + \pars{-\,{1 \over 2}}}^{-2/3} - {-2/3 \choose 0}} =
\bbx{2^{2/3}} \approx 1.5874
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Get an approximation of $\int_0^1\int_0^1\frac{1+x+y^2+x^3+\ldots}{1+y+x^2+y^3+\ldots}dxdy$ I am interested in to know how get an approximation of the double integral defined as limit of these $$\int_0^1\int_0^1\frac{1+x}{1+y}dxdy\,,$$
$$\int_0^1\int_0^1\frac{1+x+y^2}{1+y+x^2}dxdy\,,$$
$$\int_0^1\int_0^1\frac{1+x+y^2+x^3}{1+y+x^2+y^3}dxdy\,,\ldots$$
Question. Provide me an approximation of the definite integral defined as limit of previous sequence. Thanks in advance.
Using the formula to get the sum of a geometric series I know how get in closed-form the integrand. Addtitionally I presume that a it's possible to get a closed-form of it, since Wolfram Alpha can deduce the antiderivatives. Any case even the evaluation of the integral limits seem to me now complicated.
| The numerator can be written as
$$\sum _{n=0}^{\infty } y^{2 n}=\frac{1}{1-y^2};\quad \sum _{n=1}^{\infty } x^{2 n-1}=-\frac{x}{x^2-1}$$
that is
$$\frac{1}{1-y^2}-\frac{x}{x^2-1}=\frac{-x^2-x y^2+x+1}{\left(x^2-1\right) \left(y^2-1\right)}$$
The denominator is pretty the same work swapping $x$ and $y$
$$\frac{1}{1-x^2}-\frac{y}{y^2-1}=\frac{-x^2y-y^2+y+1}{\left(x^2-1\right) \left(y^2-1\right)}$$
The fraction becomes simply
$$\frac{-x^2-x y^2+x+1}{-x^2y-y^2+y+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why do partial fractions sometimes switch the sign? According to my solution manual this a valid partial decomposition:
$$\frac1{(x^2-1)^2}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{x+1}+\frac D{(x+1)^2}$$
I'm not sure how I was supposed to know to flip the signs of the $1$ for $C$ and $D$? Is there some general rule I'm missing?
| You're not “flipping the signs”—you are using each of the factors of $x^2-1$. In this case, they happen to be $x-1$ and $x+1$.
The partial fraction decomposition can be stated very generally. But maybe it's best to illustrate with cases. If the denominator were just $(x^2-1) = (x-1)(x+1)$, you would have each of those factors in a denominator:
$$
\frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}
$$
This works for any product of distinct linear factors: just add one term for each factor, with a constant numerator. It doesn't work for repeated factors, though. In that case, you need to add one term to the PFD for each power of the repeated factor, up to the repeated power shown. So:
\begin{align*}
\frac{x}{(x-1)^2} &= \frac{A}{x-1} + \frac{B}{(x-1)^2} \\
\frac{x}{(x-1)^3} &= \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}\\
\end{align*}
and so on. The reason why is complicated, but basically you need enough independent constants to fit all possible numerators.
In your problem, $(x^2-1)^2$ factors as $(x-1)^2(x+1)^2$. So we have to do the repeated-factor thing twice, once for $x-1$ and one for $x+1$. Hence:
$$
\frac{1}{(x^2-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}+ \frac{D}{(x+1)^2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Relatively prime numbers are prime The problem is to find all numbers $n$ such that all numbers $k>1$ smaller than $n$ and coprime with $n$ are prime.
| Let $n$ have the property and let $p$ be the smallest prime not dividing $n$. Then $n<p^2$ as otherwise $\gcd(p^2,n)=1$ destroys the property.
On the other hand, this implies that $n$ is a multiple of the product of all primes $<p$.
For $p\ge13$, by Bertrand's postulate, the prime preceding $p$ is $>\frac p2$ and the one preceding that is $>\frac p4$ (and $>5$). Hence $p^2>2\cdot 3\cdot5\cdot \frac p4\cdot\frac p2=\frac{15}{4}p^2$, contradiction.
But also for $p=11$, we find $11^2=121>2\cdot3\cdot 5\cdot 7=210$, contradiction.
For $p=7$, we only get $n<49$ and $2\cdot3\cdot 5=30\mid n$,. Instead of a contradiction, this implies $n=30$. We verify that this does indeed have the property.
For $p=5$, we get $6\mid n<25$, so $n\in\{6,12,18,24\}$.
For $p=3$, we get $2\mid n<9$, so $n\in\{2,4,6,8\}$.
For $p=2$, we get $n<4$.
In summary:
The $n$ with this property are precisely
$$ n\in\{1,2,3,4,6,8,12,18,24,30\}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Quick doubt on converting a decimal to a fraction I'm trying to solve a problem that requires me to find the roots of the equation
$$-x^4+10x^2-x-20=0$$
By making use of Newton's method, symbolab's equation calculator has found one root to be
$x\approx -2.79129... : x=-\frac{\sqrt{21}+1}{2}$
My question is pretty basic, actually: how did they convert the decimal number on the left to the neat faction on the right? I couldn't get to it the way I've learned to convert decimals. Thanks very much in advance.
| Bit late; I would emphasize that we can factor the polynomial ourselves, and this one is not difficult. We check that there are no rational roots. After that, we need only try integer constants $a,b,c,d$ with
$$ (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + \cdots $$
Since we are trying to get $x^4 - 10 x^2 + x + 20,$ we see that $c+a = 0.$
$$ (x^2 + ax + b)(x^2 -ax + d) = x^4 + (b+d - a^2)x^2 + a(d-b)x + bd. $$
From $a(d-b)= 1$ we have $a = \pm 1.$ Both work, they just switch $b,d.$
First, $a=1$ gives
$$ (x^2 + x + b)(x^2 -x + d) = x^4 + (b+d - 1)x^2 + (d-b)x + bd, $$
so $b+d = -9, d - b = 1.$ Then $2d = -8,$ $d= -4,$ $b=-5,$ and we check
$$ (x^2 + x -5)(x^2 -x -4) = x^4 -10 x^2 + x + 20. $$
Repeat: YOU CAN DO THIS
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of $\log|x| \sqrt{4 - x^2}$ How can I prove that $$ \frac{1}{2\pi} \int_{-2}^2 \log|x| \sqrt{4 - x^2} dx = - \frac{1}{2}$$? I have tried the substitution $x = 2\sin \theta$, which leads to $$ \frac{2}{\pi} \int_{-\pi/2}^{\pi/2} \log|2\sin\theta| \cos^2\theta d\theta.$$ But I don't know how to proceed from here.
| Noting
$$ \int\sqrt{4 - x^2} dx=\frac12x\sqrt{4-x^2}+2\arcsin(\frac x2)+C$$
one has
\begin{eqnarray}
\int_{-2}^2 \log|x| \sqrt{4 - x^2} dx&=&2\int_0^2 \log x \sqrt{4 - x^2} dx\\
&=&2\int_0^2 \log x d\bigg[\frac12x\sqrt{4-x^2}+2\arcsin(\frac x2)\bigg]\\
&=&2\bigg\{\log x \bigg[\frac12x\sqrt{4-x^2}+2\arcsin(\frac x2)\bigg]\bigg|_0^2-\int_0^2 \frac1x \bigg[\frac12x\sqrt{4-x^2}+2\arcsin(\frac x2)\bigg]dx\bigg\}\\
&=&2\bigg\{\pi\log 2 -\int_0^2 \bigg[\frac12\sqrt{4-x^2}+\frac2x\arcsin(\frac x2)\bigg]dx\bigg\}\\
&=&2\bigg\{\pi\log 2 -\frac\pi2-2\int_0^2\frac1x\arcsin(\frac x2)dx\bigg\}\\
&=&2\bigg\{\pi\log 2 -\frac\pi2-2\int_0^1\frac1x\arcsin xdx\bigg\}.
\end{eqnarray}
Since
\begin{eqnarray}
\int_0^1\frac1x\arcsin xdx&=&\int_0^1\arcsin xd\ln x\\
&=&\arcsin x\ln x|_0^1-\int_0^1\frac{\ln x}{\sqrt{1-x^2}}dx\\
&=&-\int_0^{\pi/2}\ln\sin xdx\\
&=&\frac12\pi\ln 2
\end{eqnarray}
from here, it is easy to obtain $I=-\frac12$.
| {
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How to solve $n^2+1 \mid 2^n+1$ for positive integer $n$?
How to solve following divisibility relation: $$n^2+1 \mid 2^n+1$$ for positive integer $n$?
You might have seen the similar IMO problem. Find all positive integers $n$ such that $\frac{2^n+1}{n^2}$ is an integer.
But I am sure that the extra $1$ has made this problem difficult. $n$ must be even. Let $p$ be the smallest prime divisor of $n^2+1$, then
$$p \mid \gcd \left(2^{p-1}-1, 2^{2n}-1 \right)=4^{\gcd \left((p-1)/2, n \right)}-1$$
Can we say something about $\gcd \left((p-1)/2, n \right)$? Any ideas?
| Here is a partial solution, leaving out the case when the remainder of the division in question is 1 more than a multiple of 16 (Which I cannot solve!).
The only solutions for $n \leq 5$ are $n=2,4$, so let's consider the equation for $n \geq 6$. First rewrite as $2^n +1 = a(n^2+1) \ $with$\ a\geq1$. Now clearly n must be even so let $n=2m, \ m \geq 3$. Equation now becomes$ \ 4^m+1=a(4m^2+1) \ $so we must have $a \equiv 1 \mod4 \ $. Now begins a series of successive substitutions that eventually lead to something useful. $$ a=4b-3, \ b\geq 1 \\ \implies 4^{m-1} +1=(4b-3)m^2+b \\ \implies 3m^2+1 \equiv b \mod 4 \\ \implies b \equiv 0,1 \mod 4$$
This last line is because the quadratic residues modulo 4 are 0 and 1. Considering the case where b is 1 mod 4:
$$b=4c-3, \ c\geq 1 \\ \implies 4^{m-1}+1=16(c-1)m^2+4c-3 \\ \implies 4^{m-2}=(c-1)(4m^2+1)$$
There are no solutions for $c \geq 2$ ince the RHS has on odd factor ($4m^2+1$) but the LHS does not. Clearly c=1 yields no solutions.
So for $n \geq 6$ we must have $b \equiv 1 \mod 4$ (Or equivalently $a \equiv 1 \mod 16$).
| {
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Tricky trig problem If $\sin x \cos x + \sin y \cos y + \sin x \sin y + \cos x \cos y = 1$ and $\cos (x-y)$ is the smallest possible, what is the value of $2x-y$, expressed in degrees, that is closest 360º? (AMC 2012, Senior)
I tried using a number of trigonometric identities, going backwards from $\cos (x-y)$, but didn't really get anywhere. Other than that, I'm not too sure of how to approach this question.
| The equation $\sin (x) \sin (y)+\cos (x) \cos (y)+\sin (x) \cos (x)+\sin (y) \cos (y)=1$ can be written in this way look here
$\cos(x-y) (\sin (x+y)+1)=1$
As they say that $\cos(x-y)$ is the minimum, then $\sin (x+y)+1$ is the maximum, which means that $\sin (x+y)=1$ and $\cos(x-y)=\dfrac{1}{2}$
Then we solve the systems
$$\left\{x-y=\frac{\pi }{3}+2 \pi k,x+y=\pi h+\frac{\pi }{2}\right\}$$
and
$$\left\{x-y=-\frac{\pi }{3}+2 \pi k,x+y=\pi h+\frac{\pi }{2}\right\}$$
which give infinitely many solutions
$$x= \frac{1}{12} (6 \pi h+12 \pi k+5 \pi ),y= \frac{1}{12} (6 \pi h-12 \pi k+\pi )$$
and
$$x\to \frac{1}{12} (6 \pi h+12 \pi k+\pi ),y\to \frac{1}{12} (6 \pi h-12 \pi k+5 \pi );\;k,\;h\in\mathbb{Z}$$
The value of $2x-y$ closest to 360° is $\dfrac{7\pi}{4}=315^{\circ}$
| {
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Find the series of the function $f(x) = \cos(x)$ on the interval $[0,\pi]$
Find the sine series of the function $f(x) = \cos(x)$ on the interval $0\leq x \leq \pi$. For each $x$ in this interval, to what value does the sine series for $f(x)$ converge?
Attempted solution - The Fourier since series on $0\leq x \leq \pi$ is given by $$f(x) = \sum_{n=1}^\infty B_n \sin(nx)$$
where $$B_n = \frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)\,dx$$
Thus for $f(x) = \cos(x)$ on $[0,\pi]$ we have
\begin{align*}
B_n &= \frac{2}{\pi}\cos(x)\sin(nx) \,dx\\
&= \frac{1}{\pi}\int_{0}^{\pi}(\sin((n-1)x) + \sin((n+1)x))\,dx \ \ \ (\text{Using trig identity } \sin(\alpha)\cos(\beta) = \frac{1}{2}(\sin(\alpha - \beta) + \sin(\alpha + \beta))\\
&= \frac{1}{\pi}\int_0^\pi \sin((n+1)x)\,dx + \frac{1}{\pi}\int_0^\pi \sin((n-1)x)\,dx\\
&= \frac{1}{\pi(n+1)}\int_0^{\pi(n+1)}\sin(u)du + \frac{1}{\pi}\sin((n-1)x)\,dx \ \ (\text{substituting} \ u = (n+1)x \ du = (n+1)\,dx)\\
&= \frac{\cos(n\pi) + 1}{n\pi + \pi} + \frac{1}{\pi}\int_{0}^{\pi}\sin((n-1)x)dx\\
&= \frac{\cos(n\pi) + 1}{n\pi + \pi} + \frac{1}{\pi(n-1)}\int_0^{\pi(n-1)}\sin(s)ds \ \ (\text{substituting} \ s = (n-1)x \ ds = (n-1)\,dx)\\
&= \frac{\cos(n\pi) + 1}{n\pi + \pi} + \frac{\cos(n\pi) + 1}{\pi(n-1)}\\
&= \frac{2n(\cos(n\pi) + 1)}{\pi(n^2 - 1)}\\
&= \frac{2n((-1)^n + 1)}{\pi(n^2 - 1)}
\end{align*}
Therefore the Fourier sine series is $$\cos(x) = \sum_{n=1}^\infty \frac{2n((-1)^n + 1)}{\pi(n^2 - 1)} \sin(nx)$$
By the question's statement I believe we find the Fourier sine series for $f(x) = \cos(x)$. What I do not understand is the next question, "to what value does the sine series for $f(x)$ converge? Doesn't that depend on $n$ and $x$? I am not sure if what I did was correct any suggestions are greatly appreciated.
| You are in fact expanding the "odd-extension" of $\cos x$.
The series converges to $\frac{1}{2}(f(x^+)+f(x^-))$.
So it converges for all $x\in [0,\pi]$.
But since there is discontinuity at $0$, and $\pi$. So it converges to $\cos x$ only for $x\in (0,\pi)$.
At $0$, and $\pi$, it converges to the mean value of the discontinuity, viz., zero.
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What is the value of the expression $[1 + \cos(\frac{\pi}{8})][1 + \cos(\frac{3\pi}{8})][1 + \cos(\frac{5\pi}{8})][1 + \cos(\frac{7\pi}{8})]$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of the following expression?
$$\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 5\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 7\pi }{ 8 } \right) } \right) $$
Edit: Do I manually put the values of $\cos\frac{\pi}{8}$ and all other cosine terms? Is there a better and shorter way?
| Recall that
$$
1+\cos\alpha=2\cos^2\frac{\alpha}{2}
$$
If $\beta=\pi/16$ (so $8\beta=\pi/2$), you get
$$
16\cos^2\beta\cos^23\beta\cos^25\beta\cos^27\beta=
(2\cos\beta\cos7\beta)^2(2\cos3\beta\cos5\beta)^2
$$
Now, with the product-to-sum formula,
$$
2\cos\beta\cos7\beta=\cos8\beta+\cos6\beta=\cos6\beta
$$
and
$$
2\cos3\beta\cos5\beta=\cos8\beta+\cos2\beta=\cos2\beta
$$
Hence your expression is
$$
(\cos6\beta\cos2\beta)^2=
\frac{1}{4}(\cos8\beta+\cos4\beta)^2=
\frac{1}{4}\cos^2\frac{\pi}{4}=\frac{1}{8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $\alpha =2\pi/7$, find $\tan\alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha$.
If $\alpha =2\pi/7$, find $\tan\alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha$.
I tried by $7\alpha = 2\pi$
$4\alpha =2\pi-3\alpha$
$\sin 4\alpha = \sin 3\alpha$
But couldn't reach solution Please help
| $\displaystyle \tan \frac{k \pi}{7}; 1 \le k \le 6$ are roots of
$\tan 7 \theta = 0$ which gives the polynomial $$y^6 - \binom{7}{2} y^4 + \binom{7}{4} y^2- 7=0$$ with the substitution $ y = \tan \theta$
This also yields that $\displaystyle \tan^2 \frac{k \pi}{7}; k={2,4,8}$ are roots to the polynomial $$y^3 - \binom{7}{2} y^2 + \binom{7}{4} y - 7=0$$ so that from Vieta's formula $$\tan^2 \frac{2 \pi}{7}+\tan^2 \frac{4 \pi}{7}+\tan^2 \frac{8 \pi}{7} = 21$$
Further we have $$\tan \frac{2 \pi}{7}+\tan \frac{4 \pi}{7}+\tan \frac{8 \pi}{7} = \tan \frac{2 \pi}{7}\tan \frac{4 \pi}{7}\tan \frac{8 \pi}{7}$$
But from Vieta, $$tan^2 \frac{2 \pi}{7} \tan^2 \frac{4 \pi}{7} \tan^2 \frac{8 \pi}{7} = 7$$ and hence $$\tan \frac{2 \pi}{7}\tan \frac{4 \pi}{7}\tan \frac{8 \pi}{7} = -\sqrt 7$$
Using the identity 2$$\sum ab =(a+b+c)^2 -(a^2+b^2+c^2)$$ we obtain the required sum as $-7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
solve trigonometric lim with tan Hi guys I tried to solve the following problem lately and got stucked.
I'd love to get some help and guidance.
consider this:
$$\lim_{x \rightarrow\infty}\tan\left(\frac{\pi x}{2x+1}\right)^{\frac{1}{x}}
$$
I did the following:
1. break to sin and cos :
$$
\lim_{x \rightarrow\infty}\tan\left(\frac{\pi x}{2x+1}\right)^{\frac{1}{x}} = \lim_{x\rightarrow\infty}\frac{\left(\sin\left(\frac{\pi x}{2x+1}\right)\right)^\frac{1}{x}}{\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)^\frac{1}{x}}$$
*use log rules:
$$\lim_{x\rightarrow\infty}\frac{\left(\sin\left(\frac{\pi x}{2x+1}\right)\right)^\frac{1}{x}}{\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)^\frac{1}{x}} = \lim_{x\rightarrow\infty}\frac{e^{\frac{1}{x}\ln\left(\sin\left(\frac{\pi x}{2x+1}\right)\right)}}{e^{\frac{1}{x}\ln\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)}}$$
*Numerator converges to one so I am going to work with the denominator:
$$
\lim_{x \rightarrow \infty}e^{\frac{1}{x}\ln\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)} = e^\left(\lim_{x \rightarrow \infty}{\frac{1}{x}\ln\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)}\right)
$$
$$
\lim_{x \rightarrow \infty}{\frac{1}{x}\ln\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)} = \frac{-\infty}{\infty}$$
*at this point I tried to apply l'hopital but encountered a huge mess, and now I feel that maybe I chose the wrong way. Please help me guys.
| Set $1/x=h$ to find $$A=\lim_{h\to0^+}\left(\tan\dfrac{\pi}{2+h}\right)^h$$
$$\ln A=\lim_{h\to0^+}\dfrac{\ln\tan\dfrac{\pi}{2+h}}{1/h}=\lim_{h\to0^+}\dfrac{h^2\pi\cdot\sec^2\dfrac{\pi}{2+h}}{(2+h)^2\tan\dfrac{\pi}{2+h}}$$
$$=\pi\lim_{h\to0^+}\dfrac1{(2+h)^2\sin\dfrac\pi{2+h}}\lim_{h\to0^+}\dfrac{h^2}{\cos\dfrac\pi{2+h}}$$
As $\cos\dfrac\pi{2+h}=\sin\left(\dfrac\pi2-\dfrac\pi{2+h}\right)=\sin\dfrac{\pi h}{2(2+h)}$
$$\ln A=\dfrac\pi{2^2}\cdot\lim_{h\to0^+}\dfrac{h^2}{\sin\dfrac{\pi h}{2(2+h)}}$$
Apply L'Hospital on $\lim_{h\to0^+}\dfrac{h^2}{\sin\dfrac{\pi h}{2(2+h)}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that the largest root of the following equation is $Show that the largest root of the following equation is $<n;n>2$.
$x^2-(n-2)x-2(n-2)$.
On solving I got the largest root $a=\frac{(n-2)+\sqrt {(n-2)^2+8(n-2)}}{2}$ but unable to show how is it $<n$.
I tried for various values of $n$ where the result is holding .But to prove it.
Please help
| Assuming $n \ge 2$ . . .
One can just simplify the inequality $a < n$ until it admits to being true:
\begin{align*}
&a < n\\[5pt]
\iff\;&\frac{(n-2)+\sqrt {(n-2)^2+8(n-2)}}{2} < n\\[6pt]
\iff\;&(n-2) + \sqrt {(n-2)^2+8(n-2)} < 2n\\[6pt]
\iff\;&\sqrt {(n-2)^2+8(n-2)} < n+2\\[6pt]
\iff\;&(n-2)^2+8(n-2) < (n+2)^2\qquad\text{[since $n \ge 2$]}\\[5pt]
\iff\;&(n^2 - 4n + 4) + (8n-16) < n^2 + 4n + 4\\[5pt]
\iff\;&n^2 + 4n -12< n^2 + 4n + 4\\[5pt]
\iff\;&-12 < 4\\[5pt]
\end{align*}
which is clearly true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$?
Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$
MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtraction was written in numerator. so I did the same thing as we do in the method of partial fraction, and started putting $1,2,3$ and so on. my answer came didn't match with the right answer.
| We can write the sum as:
$$\sum _{r=1}^{\infty} \frac{8r}{4r^4+1}$$
$$= \sum _{r=1}^{\infty} \frac{8r}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}$$
$$= \sum _{r=1}^{\infty} \frac{2}{2r^2 - 2r + 1} - \frac{2}{2r^2 + 2r + 1}$$
$$= \sum _{r=1}^{\infty} \frac{2}{2r^2 - 2r + 1} - \frac{2}{2(r + 1)^2 - 2(r + 1) + 1}$$
$$=\frac{2}{2\cdot1^2 - 2\cdot1 + 1}$$
$$ = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Find the minimum of the value $3x^2-2xy$ if $\frac{x^2}{4}-y^2=1$ Let $x,y\in R$,such
$$\dfrac{x^2}{4}-y^2=1$$
find the minium of the
$$3x^2-2xy$$
I think $x=2\sec{t},y=\tan{t}$,then
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$
| Let $x=2(u+v)$ and $y = u-v$. Then $\frac{x^2}{4}-y^2 = 1 \Rightarrow uv = \frac{1}{4}$
Now $3x^2 - 2xy = 8u^2+16v^2+24uv = 8\left(u^2+\frac{1}{8u^2} \right)+6 \ge 4\sqrt 2+6$ by AM-GM Inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$
Let $\{a,b,c\}\subset\mathbb{R^+}$ such that $a+b+c =1$. Prove that :
$$a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$$
$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$
$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$
Now what ?
| Hint: by the generalized means inequality:
$$
\begin{cases}
\begin{align}
\sqrt[3]{\frac{a^3+b^3+c^3}{3}} &\ge \frac{a+b+c}{3} \\
\sqrt[3]{\frac{a^3+b^3+c^3}{3}} &\ge \sqrt{\frac{a^2+b^2+c^2}{3}}
\end{align}
\end{cases}
$$
Square the second inequality, multiply together, and use that $a+b+c=1\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2 -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2 -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
| Since
$$\frac{2x}{x^2+1}=\frac{(x+1)^2}{x^2+1}-1$$
it suffices to show that $\frac{(x+1)^2}{x^2+1}\le2$, as this expression is clearly positive. This is obvious, since
$$\frac{(x+1)^2}{x^2+1}-2=-\frac{(x-1)^2}{x^2+1}\le 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 8
} |
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}}$
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}$.
My Attempt:
\begin{align}
\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta }{\theta - \dfrac {\pi}{4}}
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \cos \dfrac {\pi}{4} + \sin \dfrac {\pi}{4} - \sin \theta}{\theta - \dfrac {\pi}{4}}
\\
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {2\sin \dfrac {\pi-4\theta }{8}\cos \dfrac {\pi+4\theta}{8} - 2\sin \dfrac {4\theta + \pi}{8}\sin \dfrac {4\theta -\pi}{8}}{\theta - \dfrac {\pi}{4}}.
\end{align}
How do I proceed?
| Both numerator and denominator go to zero, so use L'hopital's. Take derivative of numerator and denominator. Get $(-\sin\theta-\cos\theta)/1$. Take $\lim_\theta\rightarrow \frac {\pi}{4} $. Get $(-\sqrt {2}/2-\sqrt {2}/2)/1=-\sqrt 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Prove that $\frac{1}{15}<\frac{1}{2}*\frac{3}{4}* \dots *\frac{99}{100}<\frac{1}{10}$
Prove that $\frac{1}{15}<\frac{1}{2}*\frac{3}{4}* \dots *\frac{99}{100}<\frac{1}{10}$
My attempt:If we name the value $A$ we have:
$A^2<\left(\frac{1}{2}*\frac{3}{4}* \dots *\frac{99}{100}\right)\left(\frac{2}{3}*\frac{4}{5}* \dots \frac{100}{101}\right)=\frac{1}{101} \Rightarrow A<\frac{1}{10}$
But I don't know, how to prove the other side?
| $$A^2=\prod_{k=1}^{50}\frac{(2k-1)^2}{(2k)^2}=\frac{1}{200}\prod_{k=2}^{50}\frac{(2k-1)^2}{(2k-2)\cdot(2k)}>\frac{1}{200}\prod_{k=2}^{50}\frac{(2k-1)^2}{(2k-1)^2}=\frac{1}{200}>\frac{1}{225}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Rational integral $$\int\dfrac{x^2}{1+x^4}dx$$
I tried many standard approaches, but I didn't get too far! Here's the most promising of them:
$$\int\dfrac{dx}{\frac{1}{x^2}+x^2}$$
knowing that $\left(1/x+x\right)^2=\frac{1}{x^2}+x^2+2$ we can change variables $1/x+x=t$. Unluckily this doesn't work either.
| HINT:
Like $\int \frac{x^2}{x^4+x^2+1}\ dx$,
$$\dfrac{2x^2}{x^4+1}=\dfrac{1-1/x^2}{x^2+1/x^2}+\dfrac{1+1/x^2}{x^2+1/x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2367945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to transform $y = 2 \sin(x) - \cos(x)$ into the format $y = z \sin(x - b)$ How can I transform $y = 2 \sin(x) - \cos(x)$ into the format $y = z \sin(x - b)$ for some $z,b\in\Bbb R$? I have been playing with various trig identities and looking through various trigonometry references but cannot so much as find a starting point. Any hints or pushes in the right direction would be much appreciated. Helping a struggling high school study and find myself greatly out of practice.
| $z \sin (x-b)=z \cos b \sin x-z \sin b \cos x$
so we have to solve
$z \cos b =2;\;z \sin b =1$
from the second we get $b=\arcsin\frac{1}{z}$. We plug this in the first
$z\cos\left(\arcsin\frac{1}{z}\right)=2\to z\sqrt{1-\frac{1}{z^2}}=2 \to z=\sqrt{5}$
$b=\arcsin\frac{1}{\sqrt{5}}$
Therefore
$$y = 2 \sin x - \cos x \longrightarrow y=\sqrt{5} \sin\left(x-\arcsin\frac{1}{\sqrt{5}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Method of difference (Series)
Find $$\sum_{r=1}^{n}\frac{1}{r(r+3)}$$ by the method of differencing.
My attempt,
$$\sum_{r=1}^{n}\frac{1}{r(r+3)}=\sum_{r=1}^{n}\left(\frac{1}{3r}-\frac{1}{3(r+3)}\right)$$
$$=\left(\frac{1}{3}-\frac{1}{12}\right)+\left(\frac{1}6-\frac{1}{15}\right)+\dots$$
This question is different as I cant see the common fraction. Hope someone can point it out.
| $$\require{cancel}
\begin{align}
\sum_{r=1}^n\frac 1{r(r+3)}
&=\frac 13\sum_{r=1}^n\frac 1r-\frac 1{r+3}\\
&=\frac 13 \bigg[\bigg(\color{red}{\frac 11}-\cancel{\frac 14}\bigg)\\
&\;\;\quad+\bigg(\color{red}{\frac 12}-\cancel{\frac 15}\bigg)\\
&\;\;\quad+\bigg(\color{red}{\frac 13}-\cancel{\frac 16}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 14}-\cancel{\frac 17}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 15}-\cancel{\frac 18}\bigg)\\
&\qquad\qquad\vdots\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-3}}-\cancel{\frac 1n}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-2}}\color{blue}{-\frac 1{n+1}}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-1}}\color{blue}{-\frac 1{n+2}}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n}}\color{blue}{-\frac 1{n+3}}\bigg)\bigg]\\
&=\frac 13\bigg[\color{red}{1+\frac 12+\frac 13}\color{blue}{-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}}\bigg)\bigg]\\
&=\frac 13\bigg[\frac {11}6-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}\bigg)\bigg]
\end{align}$$
Alternatively, in less graphical fashion,
$$\begin{align}
\sum_{r=1}^n\frac 1{r(r+3)}
&=\frac 13\bigg[\sum_{r=1}^n\frac 1r-\sum_{r=1}^n\frac 1{r+3}\bigg]\\
&=\frac 13\bigg[\sum_{r=1}^n \frac 1r-\sum_{r=4}^{n+3}\frac 1r\bigg]\\
&=\frac 13\bigg[\bigg(\sum_{r=1}^3 \frac 1r+\sum_{r=4}^n \frac 1r\bigg)-\bigg(\sum_{r=4}^n \frac 1r+\sum_{r=n+1}^{n+3} \frac 1r\bigg)\bigg]\\
&=\frac 13\bigg[\sum_{r=1}^3 \frac 1r-\sum_{r=n+1}^{n+3}\frac 1r\bigg]\\
&=\frac 13\bigg[\frac {11}6-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}\bigg)\bigg]
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the value of $ \frac{ arg(z_1) +arg(z_2) + \cdots + arg(z_{25})}{25}$
If the equation $(1+x+x^2+x^3+\cdots x^{17})^2-x^{17}=0 $ has roots $z_1,z_2,\cdots ,z_{34}$ and $0 \lt arg(z_1) \lt arg(z_2) \lt \cdots \lt arg(z_{34}) $ then find
$$ \frac{ arg(z_1) +arg(z_2) + \cdots + arg(z_{25})}{25}$$
$$(x^{18}-1)^2-x^{17}(x-1)^2=0$$
$$(x^{17}-1)(x^{19}-1)=0$$
Neglecting $x=1$ as it came out to be extra
roots .
$$arg \rightarrow \frac{2\pi}{19},\frac{4\pi}{19},\frac{6\pi}{19},\cdots ,\frac{36\pi}{19}$$
$$arg \rightarrow \frac{2\pi}{17},\frac{4\pi}{17},\frac{6\pi}{17},\cdots ,\frac{32\pi}{17}$$
Now I am stuck at predicting first 25 $arg$ satisfying inequality
| There is a bug in this problem. Roots are 34 but only 16 have positive arguments: after all they are roots of the unity and are all around the unit circle.
Here we have the 34 arguments sorted ($\pi$ omitted to keep the list shorter)
$-\dfrac{18}{19},\;-\dfrac{16}{17},\;-\dfrac{16}{19},\;-\dfrac{14}{17},\;-\dfrac{14}{19},\;-\dfrac{12}{17},\;-\dfrac{12}{19},\;-\dfrac{10}{17},\;-\dfrac{10}{19},\;-\dfrac{8}{17},\;-\dfrac{8}{19},\;-\dfrac{6}{17},\;-\dfrac{6}{19},\;-\dfrac{4}{17},\;-\dfrac{4}{19},\;-\dfrac{2}{17},\;-\dfrac{2}{19},\;0,\;0,\;\dfrac{2}{19},\;\dfrac{2}{17},\;\dfrac{4}{19},\;\dfrac{4}{17},\;\dfrac{6}{19},\;\dfrac{6}{17},\;\dfrac{8}{19},\;\dfrac{8}{17},\;\dfrac{10}{19},\;\dfrac{10}{17},\;\dfrac{12}{19},\;\dfrac{12}{17},\;\dfrac{14}{19},\;\dfrac{14}{17},\;\dfrac{16}{19},\;\dfrac{16}{17},\;\dfrac{18}{19}$
If we consider the arithmetic mean of these positive result, we get $\dfrac{6058 \pi }{8075}\approx 2.357$
I don't know
I hope this helps
| {
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"timestamp": "2023-03-29T00:00:00",
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show that $xyz(x-y)(x-z)(y-z)\le 27$
Let $\{x,y,z\}\subset[0,+\infty)$,and $x+y+z=6$. Show that:
$$xyz(x-y)(x-z)(y-z)\le 27$$
I tried AM -GM but without success.
$$xyz\le\left(\dfrac{x+y+z}{3}\right)^3=8$$
maybe $$(x-y)(x-z)(y-z)\le \dfrac{27}{8}$$ it doesn't always true。
| We can assume that $(x-y)(x-z)(y-z)\geq0$.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, $xyz=w^3$ and $u=tw$.
Hence, we need to prove that
$$(x+y+z)^6\geq1728xyz(x-y)(x-z)(y-z)$$ or
$$27u^6\geq64w^3(x-y)(x-z)(y-z)$$ or
$$729u^{12}\geq4096w^6(x-y)^2(x-z)^2(y-z)^2$$ or
$$27u^{12}\geq4096w^6(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $f(v^2)\geq0$, where
$$f(v^2)=27u^{12}-4096w^6(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6).$$
But $f'(v^2)=24576w^6(2v^4-u^2v^2-uw^3)$, which says that $(v^2)_{min}=\frac{u^2+\sqrt{u^4+8uw^3}}{4}$.
Thus, it's enough to prove that $$f\left(\frac{u^2+\sqrt{u^4+8uw^3}}{4}\right)\geq0$$ or
$$27t^{12}\geq4096\left(3t^2\left(\tfrac{t^2+\sqrt{t^4+8t}}{4}\right)^2-4\left(\tfrac{t^2+\sqrt{t^4+8t}}{4}\right)^3-4t^3+6t\left(\tfrac{t^2+\sqrt{t^4+8t}}{4}\right)-1\right)$$ or
$$(t^3+8)\left(27t^9-216t^6+1216t^3+512-512\sqrt{t^3(t^3+8)}\right)\geq0.$$
Let $t^3=a$.
Hence, we need to prove that
$$27a^3-216a^2+1216a+512\geq512\sqrt{a(a+8)}$$ or
$$(3a-8)^2(81a^4-864a^3+7296a^2-10240a+4096)\geq0,$$
which is obvious.
Done!
| {
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Given 2 roots and a remainder of a cubic polynomial, find the remainder of $f(3)$
$f(x)$ is a cubic polynomial and the coefficient of $x^3$ is $k$. If the polynomial has exactly $2$ roots and these roots are $2$ and $-5$ and $f(1) = -48$, find the possible values of $f(3)$.
I've gotten part of the cubic equation to be $(kx+a)(x^2+3x-10)$, then I let $f(x) = (kx+a)(x^2+3x-10)$, substitute in $f(1)$ to end up with $-6k-a=-48$, and I am stuck from here onward. Any hint will be appreciated.
Thanks!
| You know that either
$$f(x)=k(x-2)(x+5)^2\tag{1}$$
or
$$f(x)=k(x-2)^2(x+5)\tag{2}$$
So either $f(1)=-36k=-48$ or $f(1)=6k=-48$ so $k=\frac{4}{3}$ in $(1)$ or $k=-8$ in $(2)$.
So there are two possibilities for $f(x)$
*
*$f(x)=\frac{4}{3}(x-2)(x+5)^2$
*$f(x)=-8(x-2)^2(x+5)$
Use these to find the two possible values of $f(3)$.
| {
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Proving $|\cos z\ |^2 + |\sin z\ |^2 \geq 1$ My attempt:
For $z :=x+iy$,
$$\cos z =\cos x \cos iy - \sin x \sin iy \\ = \cos x \cosh y - i \sin x \sinh y\\ \sin z = \sin x \cosh y + i \cos x \sinh y$$
So
$$|\cos z\ |^2 + |\sin z\ |^2 = \cos ^2 x \cosh ^2 y + \sin ^2 x \sinh ^2 y + \cosh^2 y \sin^2x + \cos ^2x \sinh ^2 y\\ = \cos ^2x ( \cosh ^2 y + \sinh ^2 y) + \sin ^2 x (\cosh^2 y + \sinh^2 y) \\ = \cos 2y \leq 1.$$
I'm not sure where I went wrong, and a numerical check on Wolfram Alpha shows that the inequality should be $\geq$ as suggested.
| From $\cosh^2 x\ge1$ and $\sinh^2 x\ge0$, the equations for $\cos z$ and $\sin z$ immediately yield $|\cos z|^2\ge\cos^2x$ and $|\sin z|^2\ge\sin^2x$. The result follows.
| {
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Ramanujan's Infinite Root While I was surfing on the Internet yesterday, I watched a video about Ramajuan's infinite root. After that I had tried on my own and I got the point.
$ 3=\sqrt9$
$3=\sqrt{1+8}$
$3=\sqrt{1+2 \cdot 4}$
$3=\sqrt{1+2\cdot \sqrt{16}}$
$3=\sqrt{1+2\cdot \sqrt{1+15}}$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot 5} }$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{25}} }$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+24}} }$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot 6}} }$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{36}}} }$
$\vdots$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$
and I also tried
$ 4=\sqrt{16}$
$4=\sqrt{1+15}$
$4=\sqrt{1+2 \cdot \frac{15}{2}}$
$4=\sqrt{1+2\cdot \sqrt{\frac{225}{4}}}$
$4=\sqrt{1+2\cdot \sqrt{1+\frac{221}{4}}}$
$4=\sqrt{1+2\cdot \sqrt{1+ 3\cdot \frac{221}{12}}}$
$4=\sqrt{1+2\cdot \sqrt{1+ 3\cdot \sqrt{\frac{48841}{144}}}}$
$\vdots$
$4=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$
for the value 4. Isn't there a contradiction? I also tried with starting $2=\sqrt4$ and it does not work for infinetely times . I sensed that I can do same actions for every number which is greater than and equals to 3. For every help and opinion, thanks in advance :)
| There's no contradiction, there's just ill-defined terms.
The row
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt36}} }$
is correct, but the following
$\vdots$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$
is meaningless.
There is no clear way to define what the three dots at the end of that expression mean.
To provide a counterexample, when a mathematitian writes $\dots$, he is usually capable of defining exactly what those dots mean. For example, writing
$$1+\frac12+\frac 14 + \cdots$$
is the same as
$$\lim_{n\to\infty} 1+\frac12+\frac14+\cdots + \frac{1}{2^n}.$$
This is a well defined expression, because
*
*The concept of limit is well defined using $\epsilon-\delta$ definitions
*The concept of a finite sum of $n$ elements is defined through induction
If you want to go down deeper, every concept can be broken down into small pieces until only axioms and definitions remain.
On the other hand, there is no clear way how to strictly mathematically (using more basic concepts) write down what
$$\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$$
represents, and until you write it down strictly, you shouldn't be surprised when you get "weird" results. But those weird results aren't strictly speaking "contradictions", since they are not real mathematical expressions.
Another example of why you can't just write down three dots and hope everything sticks in infinity:
$$0.9\neq 1
\\0.99\neq 1
\\0.999\neq 1
\\\vdots
\\0.999999\dots \neq 1$$
is also not true, because the conclusion is false.
Or, even more simply
*
*$\{1\}$ is finite
*$\{1,2\}$ is finite
*$\{1,2,3\}$ is finite
*$\vdots$
*$\{1,2,3,\dots\}$ is finite.
| {
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Given $\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$, what is the value of $a+b$?
Given that
$$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$
What is the value of $a+b$
My try
$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$
$\lim\limits_{x\to 0} (\frac {(1+a×cos(x)}{x^2}-\frac {b}{x^2})=1$
$\lim\limits_{x\to 0} \frac {1+a\cos x-bx}{x^2}=1$
Apply L'Hôpital's rule:
$\lim\limits_{x\to 0} \frac {-a\cos x-b}{2x}=1$
Apply L'Hôpital's rule again:
$\lim\limits_{x\to 0} \frac {-a\sin x}{2}=1$ $\to$ $a=-2$
Is my approach right?
| Series methods are always useful.
$$\cos(x)=1-\frac12x^2+\mathcal O(x^3)\\1+a\cos(x)=1+a-\frac a2x^2+\mathcal O(x^3)\\x(1+a\cos(x))=(1+a)x-\frac a2x^3+\mathcal O(x^4)\\\sin(x)=x-\frac16x^3+\mathcal O(x^4)\\x(1+a\cos(x))-b\sin(x)=(1+a-b)x+\frac{b-3a}6x^3+\mathcal O(x^4)\\\frac{x(1+a\cos(x))-b\sin(x)}{x^3}=\frac{1+a-b}{x^2}+\frac{b-3a}6+\mathcal O(x)$$
For this to go to $1$ as $x\to0$, we need $1+a-b=0$ and $\frac{b-3a}6=1$, which gives us
$$\begin{cases}b-3a=6\\1+a-b=0\end{cases}\implies a=-\frac52,b=-\frac32\\\implies a+b=-4$$
| {
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Given $a+b+c=0$ find the value of $\big(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\big)\big(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\big)$ I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.
First we just expand and find
$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big)+\frac{c-a}{b}\big(\frac{a}{b-c}+\frac{c}{a-b}\big)+\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big)$
Each of these "not yet a number" terms can be expanded to give
$\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big) = \frac{2(c-b)^2}{(c-a)(a-b)}$
$\frac{c-a}{b}\big(\frac{a}{b-c}\big)= \frac{2(a-c)^2}{(c-a)(a-b)}$
$\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big) = \frac{2(b-a)^2}{(b-c)(c-a)}$
Adding each of these terms together and factoring out the two we find that they equal
$2\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=6$
So our total is $9$.
This is problem 160 of The USSR olympiad problem book by Shklarsky, Chentzov and Yaglom, they give the answer and $9$ is correct. Obviously they don't give the method.
| plugging $$c=-a-b$$ in the given Expression we get two factors
$$\frac{b+a+b}{a}+\frac{-2a-b}{b}+\frac{a-b}{-a-b}=-\frac{(a-b) (2 a+b) (a+2 b)}{a b (a+b)}$$
and the other factor $$\frac{a}{b+a+b}+\frac{b}{-2a-b}+\frac{-a-b}{a-b}=-\frac{9 a b (a+b)}{(a-b) (2 a+b) (a+2 b)}$$
putting Things together we obtain $$9$$ if the denominators are $\ne 0$
| {
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Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ I have the following question.
Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ for all positive integer $n$.
My attempt:
Firstly, notice that $$n^{30}-n^{14}-n^{18}+n^2$$
$$=n^2(n-1)^2(n+1)^2(n^2+n+1)(n^2-n+1)(n^2+1)^2(n^4-n^2+1)(n^4+1)(n^8+1).$$
Also, $$46410=2\times3\times5\times7\times13\times17$$
My first thought was to use induction, maybe substitute $n+1$ into $n$, then find terms that was already divisible by $46410$, but then I don't think this is a smart way to do it.
Are there any better ways to do it?
| Let $$f(n):=n^{30}-n^{18}-n^{14}+n^2\,.$$
Thus,
$$f(n)=\left(n^{13}-n\right)\,\left(n^{17}-n\right)\,.$$
Set $r(n):=n^{13}-n$ and $s(n):=n^{17}-n$.
From here, you can see that
$$D(13,1)=\{3,5,7,13\}\,.$$
Thus,
$$g(13,1)=2\cdot 3\cdot 5\cdot 7\cdot 13\,.$$
Therefore, $2\cdot 3\cdot 5\cdot 7\cdot 13$ is a factor of $r(n)$, whence also a factor of $f(n)$, for every integer $n$.
On the other hand, the same link gives
$$D(17,1)=\{3,5,17\}\,.$$
Consequently,
$$g(17,1)=2\cdot 3\cdot 5\cdot 17\,.$$
Ergo,
$2\cdot 3\cdot 5\cdot 17$
is a factor of $s(n)$, whence also a factor of $f(n)$, for each $n\in\mathbb{Z}$.
As a result,
$$(2\cdot 3\cdot 5\cdot 7\cdot 13)\cdot (2\cdot 3\cdot 5\cdot 17)=2^2\cdot 3^2\cdot 5^2\cdot 7\cdot 13\cdot 17=1392300=30\cdot 46410$$
is a factor of $f(n)$ for every $n\in\mathbb{Z}$. In fact,
$$\underset{n\in\mathbb{Z}}{\gcd}\,f(n)=\gcd\big(f(2),f(3)\big)=2^2\cdot 3^2\cdot 5^2\cdot 7\cdot 13\cdot 17=1392300\,.$$
| {
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Indices/Exponents: Simplify the following expression I've been asked to simplify this expressing it with positive indices/exponents. Could someone please show me in steps how to do this? The answer is shown next to it. Thanks a lot.
$$\dfrac{{(\large a^\frac{-5}{2})}^2 b^2}{\large {a}^3(b^\frac{1}{4})^6} = \dfrac{b^\frac{1}{2}}{a^8}$$
| Power of $a$ in the numerator = $\dfrac{-5}{2} \cdot 2 = -5$
Power of $a$ in the denominator = $3$
Overall power of $a$ in the final equation = $-5 - 3 = -8$
Power of $b$ in the numerator = $2$
Power of $b$ in the denominator = $\dfrac{1}{4} \cdot 6 = \dfrac{3}{2}$
Overall power of $b$ in the final equation = $2 - \dfrac{3}{2} = \dfrac{1}{2}$
Hence the answer is $a^{-8} \cdot b^{\frac{1}{2}} = \boxed{\dfrac{\large{b}^{\frac{1}{2}}}{\large{a}^8}}$
| {
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Determine all pairs of $(x,y)$ such that $\frac{x^2y+x+y}{xy^2+y+11}$ is an integer Here is a problem that I got:
Determine all pairs of positive integer $(x,y)$ such that $\dfrac{x^2y+x+y}{xy^2+y+11}$ is an integer.
My attempt:
First notice that $(11, 11)$ is a solution. And if $\dfrac{x^2y+x+y}{xy^2+y+11}$ is an integer, then $xy^2+y+11|x^2y+x+y$. Hence $x^2y-xy^2+x-11|xy^2+y+11$.
And then I'm stuck from here. Can anyone help me? Thanks.
| From $\dfrac{x^2 y+x+y}{x y^2+y+11}=k$ we get
$x^2 y+x+y=k x y^2+k y+11 k\quad(*)$
from which we deduce that
$x^2 y=k x y^2$ and then $x=ky$
plug this in $(*)$
$k^2 y^3+k y+y=k^2 y^3+k y+11 k$
that is $y=11k$ and then $x=11k^2$
Thus $(11k^2;\;11k)$ for any $k\in\mathbb{N};\;k>0$ satisfy the given equation
There are three more solutions that do not enter in the previous formula
$(7;\;1)$ which gives $3$
$(10;\;2)$ which gives $4$
$(121;\;1)$ which gives $111$
BTW
plugging $(11k^2;\;11k)$ in the given equation gives $k$ as result
hope this helps
| {
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Find the number of positive divisors of $N^2$ such that the positive divisors are less than N and do not divide N for $N=2^{17}*3^9*5^3$ Find the number of positive divisors of $N^2$ such that the positive divisors are less than N and do not divide N for $N=2^{17}*3^9*5^3$
My doubts
What I did firstly:-
$N^2=2^{34}*3^{18}*5^6$
Total number of factors will be $(34+1)(9+1)(3+1)=4655$
As given in the solution
There will be $(4655-1)/2=2327$ pairs of factors.
Why are we subtracting $1$ and why are we dividing it by $2$?
How these $2337$ factors of $N^2$ will be less than N?
Total number of factors will be $N=18*10*4=720$
So $720-1=719$ factors of N which are less than N and all of these will be a factors of $$N^2.
Why subtracting $1$ again from $720$?
I will be so glad if anyone of you can please clarify my doubts.
| The reasonning step by step:
*
*You correctly found the total number of positive divisors of $N^2$: $4655$.
*One of these is $N$. Let's discard it. The other ones come in pairs whose products is $N^2$; there are $(4655-1)/2=2327$ such pairs.
*In each of these pairs one factor is above $N$ and one factor is below $N$. Hence the number is positive divisors of $N^2$ that are inferior to $N$ equals the number of pairs: $2327$
edit: for clarification about these two steps, let's see what happens with $M=6$ instead of $N$. The positive divisors of $M^2=36$ are $1,2,3,4,6,9,12,18,36$. $M=6$ stands alone, and all the other ones can be regrouped in four pairs whose product is $M^2=36$: $1*36=2*18=3*12=4*9=36$. That's how we find the number of divisors of $M^2$ strictly inferior to $M$: one per pair, hence $(9-1)/2=4$.
*
*Now we want to know how many of these do not divide $N$. The easiest way is to find the complement, which is exactly the number of positive divisors of $N$, since a divisor of $N$ is necessarily a divisor of $N^2$.
There are $720$ divisors of $N$, one of them is $N$ itself so there are $719$ divisors of $N$ inferior to $N$.
*Finally, the solution is $2327-719=1608$
There are 1608 positive divisors of $N^2$ which are (strictly) smaller than $N$ and do not divise $N$.
| {
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How many ways can 4 numbers be arranged using +,-, ÷, and ×? How many ways is it possible to arrange 4 of the same number using addition, subtraction, multiplication and division. Not all of the signs must be used in a combination, for example: x+x+x+x is acceptable. Reverse combinations count as one (x+x+x-x and x-x+x+x).
| I assume parentheses are not considered here, and the standard order of operations is used to evaluate the expression. The result is of the form: $$a_4x^4+a_3x^3+a_2x^2+a_1x+a_0+a_{-1}x^{-1}+a_{-2}x^{-2}$$
I also assume two of the expressions OP describes are equivalent if they are equal as rational expressions in $x$.
Think about the pair $(m,d)$ where $m$ counts how many times $\cdot$ was used, and $d$ counts how many times $\div$ was used. $(m,d)$ must be among the ten options: $$(0,0), (1,0), (2,0), (3,0), (0,1), (1,1), (2,1), (0,2), (1,2), (0,3)$$
$(m,d)=(3,0)$ implies the expression is $\color{blue}{x^4}$.
$(m,d)=(0,3)$ implies the expression is $\color{blue}{x^{-2}}$.
$(m,d)=(2,1)$ implies the expression is $\color{blue}{x^2}$.
$(m,d)=(1,2)$ implies the expression is $\color{blue}{1}$.
$(m,d)=(2,0)$ implies the expression is $\color{blue}{x^3+x}$, $\color{blue}{x^3-x}$, $\color{blue}{2x^2}$, $\color{blue}{0}$, or $\color{blue}{x-x^3}$.
$(m,d)=(0,2)$ implies the expression is $\color{blue}{x^{-1}+x}$, $\color{blue}{x^{-1}-x}$, $\color{blue}{2}$, $0$, or $\color{blue}{x-x^{-1}}$.
$(m,d)=(1,1)$ implies the expression is $\color{blue}{2x}$, $0$, $\color{blue}{x^2+1}$, $\color{blue}{x^2-1}$ or $\color{blue}{1-x^2}$.
$(m,d)=(0,0)$ implies the expression only uses $+$ and $-$, so it's of the form $nx$, and would be one of $\color{blue}{4x}$, $2x$, $0$, or $\color{blue}{-2x}$.
$(m,d)=(1,0)$ implies the expression must be one of $\color{blue}{x^2+2x}$, $x^2$, $\color{blue}{x^2-2x}$, $\color{blue}{2x-x^2}$, $\color{blue}{-x^2}$ or $\color{blue}{-2x-x^2}$.
$(m,d)=(0,1)$ implies the expression must be one of $\color{blue}{1+2x}$, $1$, $\color{blue}{1-2x}$, $\color{blue}{2x-1}$, $\color{blue}{-1}$ or $\color{blue}{-2x-1}$.
I count $29$ unique expressions (in blue). Of course this approach may come with human error on my part. Comments with corrections are welcome.
| {
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Chebyshev Inequality toughnut Let $a^2 + b^2 + c^2 + d^2 = 1$, where $(a,b,c,d \geq 0)$. Prove that:
$$ \frac{a^{2}}{b+c+d}+\frac{b^{2}}{a+c+d}+\frac{c^{2}}{b+a+d}+\frac{d^2}{b+c+a} \geq \frac{2}{3}$$
| I think C-S is better here:
$$\sum_{cyc}\frac{a^2}{b+c+d}=\sum_{cyc}\frac{a^4}{a^2(b+c+d)}\geq\frac{(a^2+b^2+c^2+d^2)^2}{\sum\limits_{cyc}a^2(b+c+d)}=\frac{1}{\sum\limits_{cyc}a(b^2+c^2+d^2)}.$$
Thus, it remains to prove that
$$3\geq2\sum_{cyc}a(1-a^2)$$ or
$$\sum_{cyc}\left(2a^3-2a+\frac{3}{4}\right)\geq0$$ or
$$\sum_{cyc}(2a-1)(4a^2+2a-3)\geq0$$ or
$$\sum_{cyc}\left((2a-1)(4a^2+2a-3)+\frac{1}{2}(4a^2-1)\right)\geq0$$ or
$$\sum_{cyc}(2a-1)^2(4a+5)\geq0.$$
Done!
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If $y^3 + 3a^2x + x^3 = 0,$ then prove that $y'' + \frac{2a^2x^2}{y^5} = 0$ I am comfortable with second derivatives but I am just unable to set all the variables up in such a format that I get the latter (the part which needs to be proven). A hint would be a lot of help.
| You want to use "implicit differentiation" to start, here:
$$ y^3 + 3a^2x + x^3 = 0 $$
$$ 3y^2y' + 3a^2 + 3x^2 = 0 \implies y^2y' + a^2 + x^2 = 0 $$
$$ 6yy'^2 + 3y^2y'' + 6x = 0 \implies 2yy'^2 + y^2y'' + 2x = 0 $$
Then a bit of algebra gives:
$$ y'' + \frac{2yy'^2 + 2x}{y^2} = 0 \quad \text{or} \quad y = 0 $$
$$ y'' + \frac{2y\left(\frac{-a^2 - x^2}{y^2}\right)^2 + 2x}{y^2} = 0 \quad \text{or} \quad y = 0 $$
$$ y'' + \frac{2\left(-a^2 - x^2\right)^2 + 2\left(-3ax^2 - x^3\right)x}{y^5} = 0 \quad \text{or} \quad y = 0 $$
If you then simplify further, you should find you get the desired result.
| {
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"url": "https://math.stackexchange.com/questions/2395747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x \to \infty}{\frac{1}{2^x-5^x+3^x}}$. Why is my solution wrong?
Evaluate $$\lim_{x \to\infty}{\frac{1}{2^x-5^x+3^x}}. $$
My attempt:
$$\lim\limits_{x \rightarrow \infty}{\frac{1}{2^x-5^x+3^x}}
= \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{1-\frac{5^x}{2^x}+\frac{3^x}{2^x}}} \\= \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{\frac{1}{5^x}-\frac{5^x}{10^x}+\frac{3^x}{10^x}}\cdot\frac{1}{5^x}} = 0
\cdot\infty \cdot 0 = 0.$$
Someone told me that it is a wrong way to show the limit. Can anyone explain why?
| $0\cdot \infty$ is $inderterminate$. To see this, calculate the following limits:
$$\lim_{x\to \infty} \frac{1}{x} \cdot x\qquad \text{and} \lim_{x\to \infty} 0\cdot x,$$
which are both of the form $0\cdot \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Asymptotic behavior of the coefficients of expansion of $\sqrt{1-4a(1-a) \cdot x+(2a-1)\sqrt{1-4a(1-a) \cdot x^2}}$ as a power series Consider the following generating function of some sequence $(f_n)_n$
$$
G(f_n;x) = \sqrt{1-4a(1-a) \cdot x+(2a-1)\sqrt{1-4a(1-a) \cdot x^2}}
$$
for $0\leq a\leq1/2$. How can I deduce the asymptotic formula of the coefficients corresponding to the above generating function?
The first five coefficients of the expansion around zero are:
\begin{align}
&x^0:\ \sqrt{2a}\\
&x^1:\ -\sqrt{2}(1-a)\sqrt{a}\\
&x^2:\ -\sqrt{2}(1-a)^2\sqrt{a}-\sqrt{2}(1-a)\sqrt{a}(2a-1)\\
&x^3:\ -3\sqrt{2}(1-a)^3\sqrt{a}-3\sqrt{2}(1-a)^2\sqrt{a}(2a-1)\\
&x^4:\ -15\sqrt{2}(1-a)^4\sqrt{a}-18\sqrt{2}(1-a)^3\sqrt{a}(2a-1)-12\sqrt{2}(1-a)^2a^{3/2}(2a-1)\\
&\ \ \ \ \ \ \ \;\ -3\sqrt{2}(1-a)^2\sqrt{a}(2a-1)^2\\
&x^5:\ -105\sqrt{2}(1 - a)^5\sqrt{a} -
150 \sqrt{2} (1 - a)^4 \sqrt{a} (2a-1) -
60 \sqrt{2} (1 - a)^3 a^{3/2} (2a-1) -
45 \sqrt{2} (1 - a)^3 \sqrt{a} (2a-1)^2
\end{align}
| Let $G_a(x):=\sqrt{1-4ab \cdot x+(a-b)\sqrt{1-4ab \cdot x^2}},\;$ where $b:=1-a$. Then we find also that
$G_a(x) = \sqrt{a/2}(f_a(x)+f_a(-x))-\sqrt{b/2}(f_a(x)-f_a(-x)),\;$
where $f_a(x):=\sqrt{1+x\sqrt{4ab}}.$
The coefficient of $x^n$ in $G_a(x)$ is asymptotic to
$\;-\sqrt{c_a(4ab)^n/(2n^3\pi)},\;$ where $c_a=a$ if $n$ is even, else $b$ if $n$ is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$
Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$
Why does this work? I am confused as to why this happens.
| If you want something step by step use the distributive property and associative property:
$$
\begin{array}{rll}
(a+b)^{3}=
&
\underbrace{(\color{red}{a}+\color{red}{b})}_{=x}
\cdot
\underbrace{(\color{green}{a}+\color{green}{b})}_{=y}
\cdot
(\color{blue}{a}+\color{blue}{b})
&\\
=&\underbrace{\color{red}{x}
\cdot
\color{green}{y}}_{=z}
\cdot
(\color{blue}{a}+\color{blue}{b})
&\\
=&
z\cdot (\color{blue}{a}+\color{blue}{b})
&\\
=&
z\cdot \color{blue}{a}+z\cdot\color{blue}{b}
&
\mbox{distributive property}
\\
=&
(x\cdot y)\cdot \color{blue}{a}+(x\cdot y)\cdot\color{blue}{b}
&
z=(x\cdot y)
\\
=&
x\cdot (y\cdot \color{blue}{a})+x\cdot (y\cdot\color{blue}{b})
&
\mbox{associative property}
\\
=&
x\cdot \big((\color{green}{a}+\color{green}{b})\cdot \color{blue}{a}\big)
+
x\cdot \big((\color{green}{a}+\color{green}{b})\cdot\color{blue}{b}\big)
&
y=\color{green}{a}+\color{green}{b}
\\
=&
\underbrace{x\cdot \big(\color{green}{a}\color{blue}{a}+\color{green}{b}\color{blue}{a}\big)}_{}
+
\underbrace{x\cdot \big(\color{green}{a}\color{blue}{b}+\color{green}{b}\color{blue}{b}\big)}_{}
&
\mbox{distributive property}
\\
=&
x\cdot(\color{green}{a}\color{blue}{a})+x\cdot(\color{green}{b}\color{blue}{a})
+
x\cdot(\color{green}{a}\color{blue}{b})+x\cdot(\color{green}{b}\color{blue}{b})
&
\mbox{distributive property}
\\
=& x\color{green}{a}\color{blue}{a}\;+
\\
&\hspace{6mm} + 2x\color{green}{a}\color{blue}{b}\;+
\\
&\hspace{18mm} + x\color{green}{b}\color{blue}{b}
\\
=& (\color{red}{a}+\color{red}{b})\color{green}{a}\color{blue}{a}\;+
\\
&\hspace{5mm} + 2(\color{red}{a}+\color{red}{b})\color{green}{a}\color{blue}{b}\;+
\\
&\hspace{15mm} + (\color{red}{a}+\color{red}{b})\color{green}{b}\color{blue}{b}
\\
=& \color{red}{a}\color{green}{a}\color{blue}{a}+\color{red}{b}\color{green}{a}\color{blue}{a}\;+
&\mbox{distributive property}\\
&\hspace{5mm} + 2(\color{red}{a}\color{green}{a}\color{blue}{b}+\color{red}{b}\color{green}{a}\color{blue}{b})\;+
& \mbox{distributive property}\\
&\hspace{15mm} + \color{red}{a}\color{green}{b}\color{blue}{b}+\color{red}{b}\color{green}{b}\color{blue}{b}
&\mbox{distributive property}
\\
=& \color{red}{a}\color{green}{a}\color{blue}{a}+\color{red}{b}\color{green}{a}\color{blue}{a}\;+
&\\
&\hspace{8mm} + 2\color{red}{a}\color{green}{a}\color{blue}{b}+2\color{red}{b}\color{green}{a}\color{blue}{b}\;+
& \mbox{distributive property}\\
&\hspace{21mm} + \color{red}{a}\color{green}{b}\color{blue}{b}+\color{red}{b}\color{green}{b}\color{blue}{b}
\\
=&
a^3+a^2b+
&\\
&\hspace{8mm} + 2{a}^2{b}+2{a}{b}^2\;+
& \\
&\hspace{21mm} + {a}{b}^2+{b}^3
\\
=& a^3+3a^2b+3ab^2+b^3
\end{array}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$ I have no idea how hard this conjecture is to prove:
Any even number $n\ge 36$ can be written as $n=a+b+c+d$ where $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb Z^+$.
Small exceptions are $n=2, 4, 6, 10, 12, 14, 20, 26, 34.\,$ Tested for $n\le 10,000$.
A counter-example would be as interesting as a proof.
$n$ as above must be even
$d-c,c+d|a^2+b^2$
| If you rewrite your sum of squares as
$$a_4^2-a_3^2=a_1^2+a_2^2$$
Then all you need to do is find a difference of two squares that equals a sum of two squares.
If the sum of two squares is an odd number $b$ with the form $2k+1$ that is
$$a_1^2+a_2^2=b=2k+1$$
Then since any odd number can be trivially written as the difference of two squares we have
$$\left( \frac{b+1}{2}\right)^2-\left( \frac{b-1}{2}\right)^2=b$$
which immediately gives
$$\left( \frac{b+1}{2}\right)^2-\left( \frac{b-1}{2}\right)^2=a_1^2+a_2^2$$
To fit your constraint above $\left( \frac{b+1}{2}\right)^2$ is even and $\left( \frac{b-1}{2}\right)^2$ is odd.
Most sums of two squares have the form $4k+1$ as you will see by adding these in turn
$$(4k_1+1)^2+(4k_2)^2$$
$$(4k_1+1)^2+(4k_2+1)^2$$
$$(4k_1+1)^2+(4k_2+2)^2$$
$$(4k_1+1)^2+(4k_2+3)^2$$
$$(4k_1+2)^2+(4k_2)^2$$
$$(4k_1+2)^2+(4k_2+1)^2$$
$$(4k_1+2)^2+(4k_2+2)^2$$
$$(4k_1+2)^2+(4k_2+3)^2$$
$$(4k_1+3)^2+(4k_2)^2$$
$$(4k_1+3)^2+(4k_2+3)^2$$
Find the results $(\mod 4)$ and see if you can find other sums of the form $a_4^2-a_3^2=a_1^2+a_2^2$
Note Added to help explain why this approach does not work:
I thought the above might help lead to a proof/disproof of the conjecture. Hopefully the comments below will help clarify why this approach does not work.
if $a_1=\left( \frac{u-v}{2}\right)$, $a_2=\left( \frac{u+v}{2}\right)$, $a_3=\left( \frac{w-x}{2}\right)$ and $a_4=\left( \frac{w+x}{2}\right)$ then
$$n=a_1+a_2+a_3+a_4$$
$$n=\left( \frac{u-v}{2}\right)+\left( \frac{u+v}{2}\right)+\left( \frac{w-x}{2}\right)+\left( \frac{w+x}{2}\right)$$
$$n=u+w \tag 1$$
and
$$a_4^2=a_1^2+a_2^2+a_3^2$$
$$\left( \frac{w+x}{2}\right)^2=\left( \frac{u+v}{2}\right)^2+\left( \frac{u-v}{2}\right)^2+\left( \frac{w-x}{2}\right)^2$$
$$wx= \left( \frac{u+v}{2}\right)^2+\left( \frac{u-v}{2}\right)^2$$
or
$$2wx=u^2+v^2$$
Substituting for $w$ using (1) gives
$$ n=u+\frac{u^2+v^2}{2x}$$
$2x$ being a factor of $u^2+v^2$.
This result I think shines a little light on why the problem of finding $n$ is not possible using this approach; that is the difficulty in finding a general solution to the factorization of $u^2+v^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solving an infinite product of consecutive square roots Given $a$ and $b$ calculate $ab$
$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
I simplified the terms and further obtained that $ab$ is equal to:
$$ab=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}\cdot7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...}$$
How can I get a finite value?
| This seems easier than the half page proofs people are providing
$$a = \sqrt{7 b}$$
$$b = \sqrt{2 a}$$
$$a^2 = 7 b$$
$$b^2 = 2 a$$
$$a^2 b^2 = 14 a b$$
$$a b = 14$$
Unless I am missing something, a > 0 and b >0 we already know
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
Maclurin series for $e^{\sqrt{x^2+1}}$ i am giving $e^{\sqrt{x^2+1}}$ and asked to find the Maclaurin series for this term.
Here is my solution:
let $u=\sqrt{x^2+1}$, and given that we know that Maclaurin series for $e^x= 1+x+\frac{x^2}{2!} ...$
then: $$e^u= 1+u+\frac{u^2}{2}+...$$
hence:
$$e^{\sqrt{x^2+1}}=1+\sqrt{x^2+1}+\frac{x^2+1}{2}+...$$
Am I doing it right? please help
| You have to compute the derivatives
$f(x)=e^{\sqrt{x^2+1}}$
$f^1(x)=\dfrac{e^{\sqrt{x^2+1}} x}{\sqrt{x^2+1}},\\f^2(x)=\dfrac{e^{\sqrt{x^2+1}} \left(\sqrt{x^2+1} x^2+1\right)}{\left(x^2+1\right)^{3/2}},\\f^3(x)=\dfrac{e^{\sqrt{x^2+1}} x \left(x^4+x^2+3 \sqrt{x^2+1}-3\right)}{\left(x^2+1\right)^{5/2}},\\f^4(x)=\dfrac{e^{\sqrt{x^2+1}} \left(6 \left(3-2 \sqrt{x^2+1}\right) x^2+3 \left(\sqrt{x^2+1}-1\right)+\sqrt{x^2+1} x^6+\left(\sqrt{x^2+1}+6\right) x^4\right)}{\left(x^2+1\right)^{7/2}},\ldots$
And then evaluate them at $x=0$
$f^0(0)=e,\;f^1(0)=0,\;f^2(0)=e,\;f^3(0)=0,\;f^4(0)=0,\;f^5(0)=0,\;f^6(0)=15 e,\;f^7(0)=0,\;f^8(0)=-525 e,\;f^9(0)=0,\;f^{10}(0)=34020 e$
And use the formula
$$f(x)=\sum _{n=0}^{\infty } \frac{f^n(0) x^n}{n!}$$
and get
$$f(x)=e\left(1+\frac{x^2}{2} +\frac{x^6}{48}-\frac{5 x^8}{384}+\frac{3 x^{10}}{320}+O(x^{11})\right)$$
Hope this helps
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many solutions does the equation $n_1 + n_2 + n_3 + n_4 + n_5 = 20$ have in the positive integers if $n_1 < n_2 < n_3 < n_4 < n_5$? Let $n_1 < n_2 < n_3 < n_4 < n_5$ be positive integers such that $n_1 + n_2 + n_3 + n_4 + n_5 = 20$. Then the number of such distinct arrangements $(n_1, n_2, n_3, n_4, n_5)$ is......
I have no idea how to proceed. Manually, I have done it
$$1+2+3+4+10$$
$$1+2+3+5+9$$
$$1+2+3+6+8$$
$$1+2+4+5+8$$
$$1+2+4+6+7$$
$$1+3+4+5+7$$
$$2+3+4+5+6$$
But is there any way I can do it by Permutation and Combination method?
| Let $m_1 = n_1, m_2 = n_2 -1, m_3 = n_3 -2, m_4 = n_4 -3, m_5 = n_5-4$; then $m_1 \leq m_2 \leq \cdots \leq m_5$ and $m_1+m_2+m_3+m_4+m_5 = 10$. Thus we need the number of 5 partitions of 10, $P(10,5)$. Clearly, $P(10, 5) = 7$, using the recurrence $P(n,p) = P(n-1, p-1) + P(n-p,p)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$ We have
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot{2n-1\over 2n+1}=4-\pi\tag1$$
I would like to know if there exist a closed form for
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2 =\,??\tag2$$
I was able to roughly estimate it as $\approx\sqrt{8+2\pi}$ but it is not the closed form.
How can we find the closd form for $(2)?$
| A somewhat easier way is the following:
Let's start with
$$\sum_{n=0}^{\infty}\frac{(2x)^{2n}}{\binom{2n}{n}}=\frac{1}{1-x^2}+\frac{x\arcsin x}{(1-x^2)^{\frac{3}{2}}}$$
Now, all we need is to integrate this expression twice and to differentiate it twice with respect to $x$.
I skip these simple procedures and write down only the end result.
$$\sum_{n=0}^{\infty}\frac{(2)^{2n}(x)^{2n-2}}{\binom{2n}{n}}\left (\frac{2n-1}{2n+1}\right )^2=$$
$$=\frac{1}{x^2(1-x^2)}-\frac{4-5x^2}{x^3(1-x^2)^{\frac{3}{2}}}\arcsin x+\frac{2\arcsin^2 x}{x^4}+\frac{2}{x^3}\int_{0}^{x}\left (\frac{\arcsin t}{t}\right )^2dt$$
To get the original sum, wich we denote as $S$, we evaluate this expression at $x=\frac{1}{\sqrt{2}}$
So the final result:
$$S=\frac{\pi^2}{2}-3\pi+4+4\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\left (\frac{\arcsin t}{t}\right )^2dt$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$? I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$
I'm trying to prove that such pair of numbers does not exist, but I'm not sure my proof and my reasoning are correct. Could anyone please check my proof attempt?
First of all, $\exists\frac{1}{a+b}\in\Bbb{R}\iff a\neq -b$
and $\exists(\frac{1}{a}+\frac{1}{b})\in\Bbb{R}\iff a\neq 0 \text{ and } b \neq 0$
$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b} \iff \frac{1}{a+b}=\frac{a+b}{ab}$
$\iff \frac{ab}{a+b}=a+b$
$\iff ab=(a+b)^2$
$\iff ab= a^2 + 2ab + b^2$
$\iff 0 = a^2+2ab+b^2-ab$
$\iff 0= a^2+b^2+ab$
Since $a \neq 0$ and $b \neq 0$; $\exists (ab)^{-1}\in\Bbb{R}$. This allows me to continue like this:
$ 0= a^2+b^2+ab \iff 0(ab)^{-1}=(a^2+b^2+ab)(ab)^{-1}$
$\iff 0 = \frac{a^2+b^2}{ab} + 1$
$\iff -1 = \frac{a^2+b^2}{ab}$
$a^2+b^2 > 0$ because $a \neq 0$ and $b \neq 0$
If a and b have the same sign, then $ab>0$.
So $\frac{a^2+b^2}{ab}$ could be negative only if:
*
*$a>0 \text{ and } b<0$ or
*$a<0 \text{ and } b>0$
Taking into account that $a \neq -b$, any of these two options would be possible only if either:
*
*$|a| > |b|$
*$|b| > |a|$
Assuming that $|a| > |b|$ we have the following:
$|a| > |b| \implies |a|*|a| > |b|*|a|$
$\implies |a|^2 > |ab|$
$\implies \frac{|a|^2}{|ab|}>1$
$\implies |\frac{a^2}{ab}|>1$
And $b \neq 0$ so we also have $|\frac{b^2}{ab}|>0$. Therefore $|\frac{a^2}{ab}|+|\frac{b^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.
This implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.
Assuming $|b|>|a|$ we arrive at the same conclusion:
$|b| > |a| \implies |b|*|b| > |a|*|b|$
$\implies |b|^2 > |ab|$
$\implies \frac{|b|^2}{|ab|}>1$
$\implies |\frac{b^2}{ab}|>1$
And $a \neq 0$ so we also have $|\frac{a^2}{ab}|>0$. Therefore $|\frac{b^2}{ab}|+|\frac{a^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.
So this also implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.
$\blacksquare$
Is this correct??
| It gives $$(a+b)^2=ab$$ or
$$a^2+ab+b^2=0$$ or
$$\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2=0,$$
which gives $b=0$ and $a+\frac{b}{2}=0$, which gives $a=b=0$, which is impossible.
Thus, we have no these numbers.
| {
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Convergence of the series $\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )$ Given $a, b \in \mathbb{R}$ with $b > 0$, is the series
\begin{equation}
\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )
\end{equation}
convergent or divergent?
If we drop out the $n=0$ term and fold the remaining sum, the question can be equivalently asked for the series
\begin{equation}
\sum_{n = 1}^{\infty} (\sqrt{(a+n)^2+b^2} + \sqrt{(a-n)^2+b^2} - 2n ).
\end{equation}
| My gut reaction is to say divergent, since the asymptotic behaviour (assume $n>0$, since the series is only twice that, plus a finite number of terms) is something like
$$
\begin{align}
\sqrt{n^2 + a^2 + 2an + b^2} - n &= n\left(\sqrt{1 + \frac{a^2}{n^2} + \frac{2a}{n} + \frac{b^2}{n^2}} - 1\right) \\
&= n\left(1 + \frac{a}{n} + o\left(\frac{1}{n}\right) - 1\right) \\
&= a + o(1)
\end{align}
$$
which seems to diverge for any $a \ne 0$.
(note that this assumes that $[\cdot]$ is just a normal set of brackets, if it's the floor function, then there are only a finite number of non-zero terms).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluating the integral $\int_{0}^{1}\frac{x^n}{1 + x^n}\,\mathrm dx$. For each positive integer $n$, let $f_{n}$ be the function defined on the interval [0,1] by $$f_{n}(x) = \frac{x^n}{1 + x^n},$$ I want to find $$\int_{0}^{1}f_{n}(x)\,\mathrm dx,$$ I thought that I could add and subtract $1$ to the numerator, but then how can I integrate $\frac{1}{1 + x^n}$?, any help would be appreciated.
| As noted, it suffices to evaluate the definite integral
$$I_n=\int_0^1 \frac1{x^n+1}dx$$
Note that $x^n+1= \prod_{k=1}^n(x-x_k)$ with $x_k= e^{i\frac{(2k-1)\pi}n}$ and $$\frac1{x^n+1}=-\frac1n \sum_{k=1}^n\frac{x_k}{x-x_k}$$
Then
\begin{align}
I_n & = -\frac1n \sum_{k=1}^n\int_0^1 \frac{x_k}{x-x_k} dx
= -\frac1n \sum_{k=1}^nx_k\ln(1-x_k^{-1})\\
&= -\frac1n \sum_{k=1}^n x_k \left[i \frac{n-2k+1}{2n}+ \ln \left(2 \sin\frac{(2k-1)\pi}{2n}\right) \right]
\end{align}
Apply $\sum_{k=1}^nx^k=0$ along with the symmetry of $x^k$ to obtain
$$\int_0^1 \frac1{x^n+1}dx = \frac2n \sum_{k=1}^{[\frac n2]} ( \theta_k \sin2\theta_k + \cos2\theta_k \ln \cos\theta_k)$$
where $\theta_k= \frac{n-2k+1}{2n} $. The close-form
holds for any $n\ge2$ and produces
\begin{align}
& \int_0^1 \frac1{x^2+1} dx= \frac\pi4\\
& \int_0^1 \frac1{x^3+1} dx= \frac\pi{3\sqrt3}+\frac13\ln2\\
& \int_0^1 \frac1{x^4+1} dx= \frac\pi{4\sqrt2}+\frac1{2\sqrt2}\ln(1+\sqrt2)\\
& \int_0^1 \frac1{x^5+1} dx= \frac\pi{5\sqrt2}\sqrt{1+\frac1{\sqrt5}}+\frac1{\sqrt5}\ln\frac{1+\sqrt5}2 +\frac15\ln2\\
& \int_0^1 \frac1{x^6+1} dx= \frac\pi6 +\frac1{2\sqrt3}\ln(2+\sqrt3) \\
& \hspace{1cm}...
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding the Paramertric Coordinates of the Maximum and Minimum points of a Certain Curve I was studying for some quizzes when a wild question appears. It goes like this:
Find the paramertric coordinates of the maximum and minimum points of
the following curves. The curve is $x = 3 \cos \theta$ ;$\space$ $ y = 4 \sin \theta$
My work:
I need to get the rectangular equation described by the parametric equation.
I remember the relation $(\sin \theta)^2 + (\cos \theta)^2 = 1$
Getting the $\cos \theta:$ $\cos \theta = \frac{x}{3}$
Getting the $\sin \theta:$ $\sin \theta = \frac{y}{4}$
Substituting the newly-found variables to $(\sin \theta)^2 + (\cos \theta)^2 = 1$,
it becomes $$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1$$
By using the implicit differentiation, we got:
$$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1$$
$$\left( \frac{1}{9} \right)\left( 2x \right) + \left( \frac{1}{16} \right)\left( 2 y y' \right) = 0 $$
$$\left( \frac{2}{9} \right)\left( 2x \right) + \left( \frac{1}{8} \right)\left(y y' \right) = 0$$
$$\left( \frac{1}{8} \right)\left(y y' \right) = \left( \frac{-2}{9} \right)\left( 2x \right) $$
$$y' = \frac{-16x}{9y}$$
We got to make the derivative expressed as a single variable, so we need to get the $y$ in terms of $x$ in the equation
$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1.$
getting the $y$ from $\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1,$ it is:
$$y = \frac{4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
Then ,substituting it to the equation $y' = \frac{-16x}{9y},$ it becomes:
$$y' = \frac{-16x}{9\left( \frac{4}{3}\left((9 - x^2)^{\frac{1}{2}}\right) \right)}$$
$$ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
To get the minimum and maximum points of the rectangular equation, we set $\frac{dy}{dx} = y'$ to zero.
$$ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
$$ 0 = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
$$ 0 = (9 - x^2)^{\frac{1}{2}}$$
$$ 0 = (9 - x^2)$$
$$x^2 = 9$$
We get $x = \pm 3$
To determine whether $x = -3$ is a minimum or maximum point, we use the first derivative test.
Letting $x = -3$ as the critical/base point, we let $x = -99$ and $x = +99.$ If $x = -99$ is plugged into
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary.
If $x = +99$ is plugged into
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary too.
To determine whether $x = +3$ is a minimum or maximum point, we use the first derivative test.
Letting $x = -3$ as the critical/base point, we let $x = -99$ and $x = +99.$ If $x = -99$ is plugged into
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary.
If $x = +99$ is plugged into
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary too.
At this point, I'm stuck, because I can't determine if $x = \pm 3$ is a minimum or maximum point. I used large numbers
just to make sure I see the change of signs as I pass through the derivative (+, 0, -) or (-, 0, +)
How do you get the minimum and maximum points of the curve shown above?
| First of all, a comment: since no interval is given for $\theta$, the convention then is to choose the maximal one, meaning $[0, 2\pi)$. Notice that this a closed curve (i.e. it gets back at its starting point), and as such it makes no rigorous sense to speak about its extrema. On the other hand, I believe that the problem is informally asking you about the vertices of the curve, which I am going to find next.
Almost surely the problem does not expect you to use differential calculus to solve it, because the computations can get a bit ugly (and time-consuming during an exam).
You have correctly noticed that the given curve can be presented in Cartesian coordinates as
$$\frac {x^2} 9 + \frac {y^2} {16} = 1$$
which I guess that you are supposed to recognize as an ellipse centered in $(0,0)$ and having the coordinate axes as its own axes of symmetry. You are next supposed to know how to sketch such an ellipse, from which it becomes apparent that the vertices are $(\pm 3, 0)$ and $(0, \pm 4)$.
Finally, you are required to find the parametric coordinates of thse points. Keep in mind that when $\theta = 0$ you are on the $x$ axis to the right, when $\theta = \frac \pi 2$ you are upward on the $y$ axis, when $\theta = \pi$ you are on the $x$ axis to the left, and when $\theta = \frac {3\pi} 2$ you are downward on the $y$ axis. This means that $(3,0) = (3 \cos 0, \sin 0)$, $(0, 4) = (\cos \frac \pi 2, 4 \sin \frac \pi 2)$, $(-3,0) = (3 \cos \pi, \sin \pi)$ and $(0, -4) = (\cos \frac {3\pi} 2, 4 \sin \frac {3\pi} 2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of two constants in a probability density function I have this exercise:
The continuous random variable X has p.d.f. f(x) where:
$f(x) = \begin{cases} \frac{x}{3} - \frac{2}{3}, & \mbox{if } 2 \le x \le 3 \\ \alpha, & \mbox{if }3 \le x \le 5 \\ 2-\beta x & \mbox{if } 5 \le x \le 6 \\ 0 & \mbox{otherwise} \end{cases}$
find $\alpha$ and $\beta$.
How can I find $\alpha$ and $\beta$ ?
I tried to do that with the following equation:
$\int_2^3(\frac{x}{3}- \frac{2}{3})dx + \int_3^5 \alpha \, dx + \int_5^6 (2-\beta x)dx = 1 \\ \frac{1}{3} \int_2^3x \, dx - \frac{2}{3} \int_2^3 dx + \alpha \int_3^5 dx + 2\int_5^6 dx - \beta \int_5^6 x \, dx = 1 \\ \frac{1}{3}[\frac{x^2}{2}]_2^3 - \frac{2}{3}[x]_2^3 + \alpha[x]_3^5 + 2[x]_5^6- \beta[\frac{x^2}{2}]_5^6 = 1 \\ \frac{5}{6} - \frac{2}{3} + 2\alpha + 2 - \frac{11}{2} \beta = 1$
arrived here I have tried also to maintain $\alpha$ and $\beta$ to the left side and trying to do something in addition, and I have multiplied by 6 both sides to obtain a integer number:
$2\alpha - \frac{11}{2} \beta = -\frac{7}{6} \\ 12 \alpha - \frac{11}{2} \cdot 6 \beta = -7 \\ 12 \alpha - 33 \beta = -7$
arrived here I really don't know how to continue.
Please, can you help me? Thanks!
| With the hint of Med you get two equations. The argument is that you have $\texttt{overlapping intervals}$. And the value of the function must be equal for equal $x$ values.
1. The value of the function must be equal for the upper limit of the first interval and the lower limit of the second interval (common point):
$$\frac33-\frac23=\alpha$$
2. The value of the function must be equal for the upper limit of the second interval and the lower limit of the third interval (common point):
$$\alpha=2-5\beta$$
With these two equations you can evaluate the values of $\alpha$ and $\beta$.
Finally you can check if these values satisfy your equation $12 \alpha - 33 \beta = -7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$. If $x$ satisfies $x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$.
This equation becomes $x^4+3x^3-26x^2+3x+1=0$ which has four solutions. One of the solutions is $2+\sqrt3$ which has the form $a+\sqrt b$. So, $a+b=2+3=5$. How does this look?
| Let $t = x + \frac{1}{x}$, then the equation is $$t^2 - 2 + 3t = 26,$$ or $$t^2 + 3t -28 = 0.$$
Then, we have $t = 4$ or $t = -7$.
If $t=4$, then $x + \frac{1}{x} = 4$, then $x = 2 \pm \sqrt{3}$.
If $t=-7$, then $x + \frac{1}{x} = -7$, then $x = \frac{1}{2}(-7\pm \sqrt{45})$.
Because $a$ and $b$ are positive integers, then $x = 2 + \sqrt{3}$. And yes, $a+b=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Line integral $\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy$ Solve $\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy$ =?
K: $x^2+y^2=1$ is oriented positively.
My attempt:
$\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy = |x=cost, y=sint,x'=- sint,y'=cost | $ = *
I put for $dx = -sint,dy=cost$
$= \int_0^{2\pi} \frac{sint}{cost^2+sint^2} -sint -\frac{cost}{cost^2+sint^2} cost dt= -\int_0^{2\pi} 1 dt = - 2 \pi $
We can't use Green's theorem because $P(x,y)$ and $Q(x,y)$ aren't continuous and their diveratives aren't continuous at point. $(0,0)$?
| Seems ok. You can't use Green-Stokes because of the reason you correctly stated. Another way to see this result is to note that if $z = x+iy$ and ${\rm d}z = {\rm d}x+i\,{\rm d}y$, then $$\frac{y}{x^2+y^2}\,{\rm d}x - \frac{x}{x^2+y^2}\,{\rm d}y = {\rm Im}\left(-\frac{{\rm d}z}{z}\right),$$and so $$ -{\rm Im} \int_{\Bbb S^1}\frac{{\rm d}z}{z} = -{\rm Im}(2\pi i) = -2\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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use **similarity transformation** to find the matrix $ \ B \ $ with respect to the basis $ \ \{(1,2) , \ (3,5) \} \ \ of \ \ \mathbb{R}^2 $ . Consider the linear operator $ \ T(x,y)=(5x,x-y) \ \ on \ \ \mathbb{R}^2 $ . Find the matrix $ \ A \ $ of $ \ T \ $ relative to the standard basis on $ \ \mathbb{R}^2 \ $ . Then use similarity transformation to find the matrix $ \ B \ $ with respect to the basis $ \ \{(1,2) , \ (3,5) \} \ \ of \ \ \mathbb{R}^2 $ .
Note: The answer is given as $ B =\begin{pmatrix}-29 & -56 \\ 17 & 33 \end{pmatrix} $
Answer:
I have found $ \ A= \begin{pmatrix} 5 & 0 \\ 1 & -1 \end{pmatrix} $ .
Now $ \ B=P^{-1} AP \ $ ,
where $ \ P \ $ is the transition matrix from the basis $ \ \{(1,2) , (3,5) \} \ $ to the basis $ \ \{(1,0) , (0,1) \} \ $.
Now calculating I got ,
$ \ P=\begin{pmatrix}1 & 3 \\ 2 & 5 \end{pmatrix} $
Thus,
$ \ P^{-1} =\begin{pmatrix}-5 & 2 \\ 3 & -1 \end{pmatrix} $
Hence ,
$ B=P^{-1} AP = \begin{pmatrix}-5 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix}-27 & -79 \\ 16 & 47 \end{pmatrix} $
$
But the answer does not match with the given answer .
Was my calculation and appoach right ?
If not then how to solve the problem ?
Is there any help ?
| Both you and your textbook are wrong. The correct answer seems to be
$$
B = \pmatrix{-28&-81 \\ 11 & 32}
$$
As computed here. Your mistake is that you incorrectly computed $P^{-1}$; the $2$ and $3$ should be switched.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int\frac{dx}{x^3+1}$ without using partial Fractions Evaluate $$I=\int\frac{dx}{x^3+1}$$ without using partial Fractions
I tried in this way:
$$I=\int \frac{\frac{xdx}{x^4}}{1+\frac{1}{x^3}}$$
Put $1+\frac{1}{x^3}=t$
$$I=\frac{-1}{3}\int \frac{dt}{t(t-1)^{\frac{1}{3}}}$$
Any way to help me?
| If you really want to do it, you can do the following. This is really just a sketch, there are lots of fine details omitted.
Write
$$\int \frac{1}{x^3+1} dx = \int 1 - \frac{x^3}{x^3+1} dx = x - \frac{1}{3} \int x \frac{3x^2}{x^3+1} dx.$$
This second term can be integrated by parts: take $u=x,dv=\frac{3x^2}{x^3+1}$. The boundary term gives a contribution to the integral and leaves you to evaluate $\int \ln(x^3+1) dx$. This can be split into $\ln(x+1) + \ln(x^2-x+1)$. The first term is easy to integrate by parts. The second term is $\ln((x-1/2)^2+3/4)$. Upon a suitable change of variable and factoring out constants, it becomes enough to compute $\int \ln(x^2+1) dx$. This can be done by integrating by parts the opposite of how we did before: $dv=dx,u=\ln(x^2+1)$ forces us to evaluate $\int x \frac{2x}{x^2+1} dx$, but this is now easy after a long division.
This is way harder than doing it by partial fractions, basically because you wound up doing partial fractions implicitly anyway, but obfuscated by two integration by parts steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$ Evaluate
$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$
My Try:
$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$
we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$ and
$$\cos x+\sin x=\frac{7}{25}(3 \sin x+4 \cos x)+\frac{-1}{25}(3 \cos x-4 \sin x)$$
So
$$\cos 2x=\frac{7}{625}(3 \sin x+4 \cos x)^2+\frac{48}{625}\left((3 \sin x+4 \cos x)(3\cos x-4 \sin x)\right)-\frac{7}{625}(3 \cos x-4 \sin x)^2$$
So
$$I=\frac{7}{625}\int (3 \sin x+4 \cos x)dx+\frac{48}{625}\int (3 \cos x-4 \sin x)dx-J$$
where
$$J=\frac{7}{625}\int \frac{(3 \cos x-4 \sin x)^2dx}{3 \sin x+4 \cos x}$$
I got stuck up to integrate $J$.
| $$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x} = \dfrac15\int \frac{\cos 2x \: dx}{\dfrac35 \sin x+\dfrac45 \cos x} = \dfrac15\int \frac{\cos 2x \: dx}{\cos(x - \alpha)}$$
Where $\alpha = \cos^{-1} \dfrac45$
$$\dfrac15\int \frac{\cos 2x \: dx}{\cos(x - \alpha)} = \dfrac15\int \frac{2\cos^2 x - 1 \: }{\cos(x - \alpha)}\ dx = \dfrac15\int \frac{2\cos^2 x \: }{\cos(x - \alpha)}\ dx - \dfrac15\int \sec(x - \alpha) dx$$
Now for the first integral, let $u = x - \alpha$
$$\dfrac25\int \frac{\cos^2 (u + \alpha) }{\cos(u)}\ du = \dfrac25\int \frac{(\cos u \cos \alpha - \sin u \sin \alpha)^2 }{\cos(u)}\ du \\= \dfrac25\int\frac{(\cos^2 u \cos^2 \alpha + \sin^2 u \sin^2 \alpha) + 2\sin \alpha \cos \alpha \sin u \cos u }{\cos(u)}\ du\\=\dfrac25\int{\cos u \cos^2 \alpha}\ du + \dfrac25\int \frac{\sin^2 u \sin^2 \alpha}{\cos u} du+ \dfrac25\int {2\sin \alpha \cos \alpha \sin u }{}\ du $$
The middle integral,
$$\dfrac25\int \frac{(1 -\cos^2 u) \sin^2 \alpha}{\cos u}\ du = \dfrac25\int{\sec u \sin^2 \alpha}\ du - \dfrac25\int \cos u \sin^2 \alpha \ du$$
So the answer is,
$$I = \dfrac25\int{\sec u \sin^2 \alpha}\ du - \dfrac25\int \cos u \sin^2 \alpha \ du + \dfrac25\int{\cos u \cos^2 \alpha}\ du + \dfrac25\int {2\sin \alpha \cos \alpha \sin u }{}\ du -\dfrac15\int \sec(u) du$$
$$I = \dfrac25\sin^2 \alpha\int{\sec u }\ du + \dfrac25\cos (2\alpha)\int{\cos u }\ du + \dfrac25\sin (2\alpha)\int { \sin u }{}\ du -\dfrac15\int \sec(u) du$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove the following by epsilon delta definition? I have to prove that $$\underset{(x,y) \to (0,0)}{\lim} (x^2+y^2) \log(\sqrt{x^2+y^2}) = 0$$ by $\epsilon- \delta$ definition. I can easily prove it by L-Hospital rule, but unable to do with definition.
| We need to prove that $\forall \epsilon>0,\exists \delta >0$ such that $\forall (x,y):\sqrt{x^2+y^2}<\delta \Rightarrow |(x^2+y^2) \log{\sqrt{x^2+y^2}}|< \epsilon$
Let $\epsilon>0$.
For $\delta \leq 1$
$$|(x^2+y^2) \log{\sqrt{x^2+y^2}}| \leq \frac{(x^2+y^2)^2}{2}+ \frac{(\log{\sqrt{x^2+y^2}})^2}{2} $$
Now $$\frac{(\log{\sqrt{x^2+y^2}})^2}{2}< \epsilon \Rightarrow \log{\sqrt{x^2+y^2}}<\sqrt{2 \epsilon} \Rightarrow \sqrt{x^2+y^2} <e^{\sqrt{2\epsilon}}$$
Take $\delta \leq\min\{1,2\sqrt[4]{\epsilon},e^{\sqrt{2 \epsilon}}\}$
| {
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"timestamp": "2023-03-29T00:00:00",
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product formula for the sum of cosine and sine
Above is an exercise from Conway. I have proved the product formula for sine and cosine functions and they are like below. However, I am stuck at how to calculate the exercise problem. Could anyone help me how to compute product formula in the exercise?
| I think there is typo in Conway's book. It should be
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=
\prod_{n=1}^{\infty}\left(1+\frac{(-1)^n z}{2n-1}\right).$$
Note that
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=\frac{\sin\left(\frac{\pi (z-1)}{4}\right)}{\sin\left(-\frac{\pi}{4}\right)}$$
Then by using
$$\sin(\pi z) = \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n},$$
we obtain
\begin{align*}
\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)
&=\frac{\frac{\pi (z-1)}{4} \prod_{n\neq 0} \left(1-\frac{(z-1)}{4n}\right)e^{(z-1)/(4n)}}{\frac{\pi (-1)}{4} \prod_{n\neq 0} \left(1-\frac{(-1)}{4n}\right)e^{(-1)/(4n)}}\\
&=(1-z) \prod_{n\neq 0} \left(\frac{4n-(z-1)}{4n+1}\right)e^{z/(4n)}\\
&=(1-z) \prod_{n\neq 0} \left(1-\frac{z}{4n+1}\right)e^{z/(4n)}\\
&=(1-z)\prod_{n\geq 1} \left(1-\frac{z}{4n+1}\right)e^{z/(4n)}\left(1-\frac{z}{-4n+1}\right)e^{-z/(4n)}\\
&=(1-z)\prod_{n\geq 1} \left(1-\frac{z}{4n+1}\right)\left(1+\frac{z}{4n-1}\right)
\end{align*}
which is equal to
\begin{align*}
\prod_{n=1}^{\infty}\left(1+\frac{(-1)^n z}{2n-1}\right)
&=\prod_{k\geq 1}\left(1+\frac{(-1)^{2k-1} z}{2(2k-1)-1}\right)\prod_{k\geq 1}\left(1+\frac{(-1)^{2k} z}{2(2k)-1}\right)
\\
&=\prod_{k\geq 1}\left(1-\frac{z}{4k-3}\right)\prod_{k\geq 1}\left(1+\frac{z}{4k-1}\right)\\
&=(1-z)\prod_{k\geq 2}\left(1-\frac{z}{4(k-1)+1}\right)\prod_{k\geq 1}\left(1+\frac{z}{4k-1}\right)\\
&=(1-z)\prod_{k\geq 1} \left(1-\frac{z}{4k+1}\right)\left(1+\frac{z}{4k-1}\right).
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of this series? Show that the series $\displaystyle \sum_{n=0}^{\infty}\frac{x}{(1+x^2)^n} $ converges $\forall x\in \mathbb{R}$ and find the sum $\forall x\in \mathbb{R}$
So, I've shown that the series converges $\forall x\in \mathbb{R}$ and, through testing, suspect that the sum is equal to $\frac{x}{1+x^2}$. Heres what I did:
Let $s =\displaystyle \sum_{n=0}^{k}\frac{x}{(1+x^2)^n}$ = $\frac{x}{(1+x^2)}+ \frac{x}{(1+x^2)^2} + ... +\frac{x}{(1+x^2)^k} \Rightarrow ... \Rightarrow\frac{(x^2+1)^k}{x}s = (1+x^2)^{k-1}+(1+x^2)^{k-2} + ...+1 $.
Not quite sure what to do from here. Any tips ?
| $(1-a)(1+a+...+a^k)=1-a^{k+1}.$ Examples:
$(1-a)(1+a)=(1+a)-(a+a^2)=1-a^2.$
$(1-a)(1+a+a^2)=(1+a+a^2)-(a+a^2+a^3)=1-a^3.$
$(1-a)(1+a+a^2+a^3)=(1+a+a^2+a^3)-(a+a^2+a^3+a^4)=1-a^4.$
So if $a\ne 1$ then $(1+a+...+a^k)=(1-a^{k+1})(1-a)^{-1}.$
When $x\ne 0$ apply this to $s(k,x)=x\sum_{n=0}^k (1+x^2)^{-n}$ with $a=(1+x^2)^{-1}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Deriving the number of all even length palindromic sequences with length at most $n$ Derive the number of all even length palindromic sequences with length $\leq n$.
If $n=4$, I have to count the number of non-palindromes of length 0, 2, and 4.
Also, $2p = n$, so a string of length $n = 4$ has $p = 2$. This comes in handy later.
$x$ represents the number of possible characters in the given alphabet- so $x = 26$ for English, 2 for binary, etc.
I chose to approach by counting, using this thought process.
$$\sum_{k=0}^{p} (x^{2k} - x^k)$$
In other words - a summation of all possible even length strings up to length $n$, subtracting cases that are palindromes.
I tested this with binary, up to length 4. (Binary non-palindromes of length 0, 2, and 4). This means $p$ goes from 0, to 1, to 2.
This evaluates to:
$$ 2^0 - 2^0 +$$
$$2^2 - 2^1 + $$
$$2^4 - 2^2$$
Which sums to a total of 14 non palindromes.
All possible combinations are
{},
11,
00,
10,
01,
1111,
1100,
1000,
0000,
0001,
0011,
0111,
1010,
0101,
0110,
1001
1110,
1101,
0010,
1011,
0100.
14 of these are not palindromes, confirming my approach.
The solution states the formula is
$$\frac{x^{2p+1}-2x^{p+1}+x}{x-1}$$
Following the same logic...
With $p = 2$, $x = 2$, $n = 4$, we have
$$\frac{2^{4+1}-2(2)^{3}+2}{2-1}$$
Which works out to 18, not 14.
I have worked this out several times, and I'm not exactly sure what/if I am misunderstanding. Any help would be greatly appreciated. Thanks!
| A palindrome is a sequence of characters that reads the same forwards as backwards.
If the sequence has even length, say $n = 2k$, selecting the first $k$ characters completely determines the palindrome since the remaining $k$ characters can be found by repeating the sequence in the reverse order. Hence, the number of palindromes of even length at most $n$ in an alphabet with $x$ characters is
$$x^0 + x^1 + x^2 + x^3 + \cdots + x^k = \frac{1 - x^{k + 1}}{1 - x}$$
where we have used the formula for the sum of a geometric series with initial term $1$ and common ratio $x$.
In your example of binary sequences of even length at most $4$, there are seven palindromes:
$$\emptyset, 00, 11, 0000, 0110, 1001, 1111$$
Notice that our formula yields
$$2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7$$
or
$$\frac{1 - 2^{2 + 1}}{1 - 2} = \frac{1 - 2^3}{-1} = \frac{-7}{-1} = 7$$
Since there are a total of
$$x^0 + x^2 + x^4 + \ldots + x^{2k} = \frac{1 - x^{2k + 2}}{1 - x^2}$$
sequences of even length at most $n = 2k$ in an alphabet with $x$ characters, the number of sequences of even length at most $n = 2k$ that are not palindromes is
\begin{align*}
\frac{1 - x^{2k + 2}}{1 - x^2} - \frac{1 - x^{k + 1}}{1 - x} & = \frac{1 - x^{2k + 2}}{1 - x^2} - \frac{1 - x^{k + 1}}{1 - x} \cdot \frac{1 + x}{1 + x}\\
& = \frac{1 - x^{2k + 2}}{1 - x^2} - \frac{1 + x - x^{k + 1} - x^{k + 2}}{1 - x^2}\\
& = \frac{-x + x^{k + 1} + x^{k + 2} - x^{2k + 2}}{1 - x^2}
\end{align*}
In your example of binary sequences of even length at most $4$, the total number of sequences is
$$2^0 + 2^2 + 2^4 = 1 + 4 + 16 = 21$$
of which seven are palindromes, so there are $14$ non-palindromes as you found. Notice that our formulas yield
$$\frac{1 - 2^{2 \cdot 2 + 2}}{1 - 2^2} = \frac{1 - 2^6}{1 - 2^2} = \frac{1 - 64}{1 - 4} = \frac{-63}{-3} = 21$$
total sequences and
$$\frac{-2 + 2^{2 + 1} + 2^{2 + 2} - 2^{2 \cdot 2 + 2}}{1 - 2^2} = \frac{-2 + 2^3 + 2^4 - 2^6}{1 - 2^2} = {-2 + 8 + 16 - 64}{1 - 4} = \frac{-42}{-3} = 14$$
non-palindromes.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The ellipse $x^2+2y^2=2.$ Question: The ellipse $x^2+2y^2=2$ is given by $x=a\cos{t}, \quad y=b\sin{t}, \quad t\in[0,2\pi],$ for what values of $a$ and $b$?
a) $(a,b)=(2,1)$
b) $(a,b)=(1,2)$
c) $(a,b)=(1,\sqrt{2})$
d) None of the above.
Attempt: Substitute the trig-values for $x$ & $y$ in the elliptic equation:
$$x^2+2y^2=2\Longleftrightarrow a^2\cos^2{t}+2b^2\sin^2{t}=2.$$
Trig identities gives
$$a^2(1-\sin^2{t})+2b^2\sin^2{t}=a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2+2b^2)=2.$$
Solving for $\sin{t}$ gives
$$\sin{t}=\pm\sqrt{\frac{a^2-2}{a^2+2b^2}}=\pm f(a,b).$$
Since $$-1\leq\sin{t}\leq1\Longleftrightarrow -1\leq\pm f(a,b)\leq1.$$
Answer a) gives $f(2,1)=\frac{\sqrt{3}}{3} \in[-1,1]$. OK!
Answer b) gives $f(1,2)=\frac{i}{3}\in \mathbb{C}$. Disregard.
Answer c) gives $f(1,\sqrt{2}) = \frac{i\sqrt{5}}{5}\in \mathbb{C}$. Disregard.
So according to me, the correct answer should be a). But correct answer is d). Where did I go wrong?
| Write it in the standard form:
$$\frac{x^2}{(\sqrt 2)^2}+\frac{y^2}{1^2}=1$$
Note that the equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
is an ellipse with parametric representation $x=a\cos t$ and $y=b\sin t$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate the integral $\int_0^2 \frac{\ln x}{\sqrt {x^2-2x+2}}dx$? Yesterday's integral may be too difficult, I think the following integral should not be difficult.
$$I=\int_0^2 \frac{\ln x}{\sqrt {x^2-2x+2}}dx=\int_{-1}^1\frac{\ln(x+1)}{\sqrt {x^2+1}}dx=\int_{-1}^1\ln(x+1)d(\ln (x+\sqrt {x^2+1}))$$
$$=\ln (x+1)\ln (x+\sqrt {x^2+1})|_{-1}^1-\int_{-1}^1\frac{\ln (x+\sqrt {x^2+1})}{(x+1)}dx=......$$
So far, I don't know what to do, any help is appreciated.
| Let $\sqrt {{x^2} - 2x + 2} = x + t$, then we have
$$I=\int_0^2 {\frac{{\ln x}}{{\sqrt {{x^2} - 2x + 2} }}} = \int_{\sqrt 2 - 2}^{\sqrt 2 } {\ln \left( {\frac{{2 - {t^2}}}{{2(t + 1)}}} \right)\frac{1}{{1 + t}}dt} $$
This integral can be broken down into pieces of the form $\int \frac{\ln x}{x+a} dx$, which can be evaluated by dilogarithm. The final result being
$$\DeclareMathOperator{\Li}{Li}
I=\frac{{{\pi ^2}}}{{12}} - \frac{9\ln ^22}{8} + \frac{3}{2}\ln (1 + \sqrt 2 )\ln 2 - \frac{1}{2}{\ln ^2}(1 + \sqrt 2 ) - \Li_2\left( {2\sqrt 2 - 2} \right) - \Li_2\left( {\frac{{2 - \sqrt 2 }}{4}} \right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many numbers smaller than one milion? How many numbers smaller than 1 000 000 :
a) have digits in a non-decreasing order?
b) contains exactly three digits 9 and have an odd sum of numbers?
c) have digits in non-increasing order?
For the a. solution
Is it like:
$\binom{6+10-1}{10-1}$
Am I right?
Solution to the b.
So the first solution is exactly 999,
for the 4-digits number $\binom{4}{1}$$\binom{4}{1}$ +1 ,
for the 5-digits number $\binom{5}{1}$$\binom{4}{1}$$\binom{4}{1}$$\binom{4}{1}$
| Your answer for the first question is correct if the numbers are nonnegative integers. As @skyking pointed out, we get the same result for the third question.
How many nonnegative integers less than $1~000~000$ contain exactly three $9$s and have an odd digit sum?
We treat the nonnegative integer as a six-digit decimal string, by appending leading zeros to numbers with fewer than six digits as necessary.
There are $\binom{6}{3}$ ways to choose three of the six positions for the $9$s.
Since $3 \cdot 9 = 27$, the sum of the remaining three digits must be even. To get an even sum, we must either use three even digits or one even digit and two odd digits when filling the remaining three positions.
Three even digits: We have five choices for each even digit, so this can be done in $5^3$ ways.
One even digit and two odd digits: We have three ways to choose the position of the even digit and five ways to choose the even digit we place in that position. Since we cannot use $9$, we have four ways to fill each of the remaining open positions with an odd digit. Hence, there are $3 \cdot 5 \cdot 4^2$ such choices.
Thus, the number of nonnegative integers less than $1~000~000$ that contain exactly three $9$s and have an odd digit sum is
$$\binom{6}{3}\left[5^3 + 3 \cdot 5 \cdot 4^2\right]$$
Finally, since the number $0$ does not contain three $9$s, the same argument applies to positive integers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help finding the Marginal PDF of Y given a Density Function of Two Variables Problem:
The join pdf of $(X,Y)$ is given by
$
f(x,y) = \begin{cases}
\Big( \frac{5}{32} \Big) x^2(4 - y) & \text{for } x < y < 2x, 0 < x < 2 \\
0 & \text{otherwise} \\
\end{cases}
$
where $k = \frac{5}{32}$. Find the marginal pdf of $Y$.
Answer:
\begin{eqnarray*}
f_y(y) &=& \int_{-\infty}^{\infty} f(x,y) \,\, dx \\
f_y(y) &=& \int_{0}^{2} \Big( \frac{5}{32} \Big) x^2(4 - y) \,\, dx \\
f_y(y) &=& \Big( \frac{5}{32} \Big) \frac{x^3}{3}(4 - y) \Big|_{x = 0}^{x = 2} \\
f_y(y) &=& \Big( \frac{5}{32} \Big) \frac{2^3}{3}(4 - y) \\
f_y(y) &=& \Big( \frac{5}{12} \Big) (4 - y) \\
\end{eqnarray*}
My answer is:
\begin{eqnarray*}
f_y(y) &=& \begin{cases}
\Big( \frac{5}{12} \Big)(4x^3 - \frac{3x^4}{2}) & \text{for } 0 < y < 4 \\
0 & \text{otherwise} \\
\end{cases} \\
\end{eqnarray*}
but the books answer is:
\begin{eqnarray*}
f_y(y) &=& \begin{cases}
\Big( \frac{5}{12} \Big) (4 - y) & \text{for } 0 < y < 2 \\
\Big( \frac{5}{32} \Big)\Big( \frac{1}{3} \Big)(4-y)(8 - \frac{y^3}{8}) & \text{for } 2 < y < 4 \\
0 & \text{otherwise} \\
\end{cases} \\
\end{eqnarray*}
I would like to know what I did wrong.
Thanks
Bob
| The support for the joint distribution is $\{(x,y): 0<x<2, x<y<2x\}$
*
*That is the triangle $\triangle(0,0)(2,2)(2,4)$
This is the same region as $\{(x,y): 0<y<4, y/2< x< \min\{2,y\}\}$
Which is the union: $\{(x,y): 0<y<2, y/2<x<y\}\cup\{(x,y): 2\leqslant y< 4, y/2<x<2\}$
*
*That is $\triangle(0,0)(1,2)(2,2)\cup\triangle(1,2)(2,2)(2,4)$
Plot the points and you see how the triangle is divided into two parts at the $y=2$ horizon. Draw horizonal likes through the upper and lower triangles and see why the bounds for the integral producing the $Y$-marginal pdf are different for each part.
Hence why:
$$f_Y(y) =\frac{5}{32} \begin{cases}\displaystyle \int_{y/2}^y x^2(4-y)\;\mathrm d\, x &:& 0<y<2 \\[1ex]\displaystyle \int_{y/2}^2 x^2(4-y)\;\mathrm d\, x &:& 2\leqslant y< 4 \\[1ex] 0 &:& \text{otherwise}\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral $\int \arctan^2 x dx$ in terms of the dilogarithm function I read about the integral
$$\int \arctan^2 x dx$$ in this old post: Evaluation of $\int (\arctan x)^2 dx$
By replacing
$$\arctan x = -\frac{i}{2}\left[\log(1+ix) - \log(i-ix)\right],$$
as suggested there, I ended up with this solution
$$\int\arctan^2 x dx = x\arctan^2x - \frac{1}{2}\log(1+x^2)\arctan x -\log 2 \arctan x + \mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+ix}{2}\right)\right\} + K, \tag{1}\label{uno}$$
where, as usual, $\mbox{Li}_2(z)$ is the dilogarithm function
$$\mbox{Li}_2(z) = -\int_0^z \frac{\log(1-u)}{u}du=\sum_{k=1}^{+\infty}\frac{z^k}{k^2}.$$
Is this a correct development?
In that case, if I determine, using \eqref{uno}, the definite integral $\int_0^1 \arctan^2xdx$ I get the result
$$\int_0^1 \arctan^2xdx=\frac{\pi^2}{16}-\frac{3\pi}{8}\log 2 + \mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+i}{2}\right)\right\}.$$
If I now compare this result with the one given in Definite Integral of $\arctan(x)^2$, i.e.
$$\int_0^1 \arctan^2xdx=\frac{\pi^2}{16}+\frac{\pi}{4}\log 2 - C,$$
where $C$ is the Catalan constant
$$C = \sum_{k=0}^{+\infty} \frac{(-1)^k}{(2k+1)^2},$$
I get the following expression:
$$\mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+i}{2}\right)\right\} = \frac{5\pi}{8}\log 2 - C.$$
Is that reasonable?
| When $x\in\mathbb{R}^+$ we can write:
$$\arctan\left(x\right)=\frac{i}{2}\cdot\ln\left(\frac{1-xi}{1+xi}\right)\tag1$$
So, when we square both sides we get:
$$\arctan^2\left(x\right)=\left(\frac{i}{2}\cdot\ln\left(\frac{1-xi}{1+xi}\right)\right)^2=-\frac{1}{4}\cdot\ln^2\left(\frac{1-xi}{1+xi}\right)\tag2$$
For the integral we get, then:
$$\mathscr{I}:=\int\arctan^2\left(x\right)\space\text{d}x=-\frac{1}{4}\cdot\int\ln^2\left(\frac{1-xi}{1+xi}\right)\space\text{d}x\tag3$$
Let $\text{u}:=\frac{1-xi}{1+xi}$:
$$\mathscr{I}=-\frac{2i}{4}\cdot\int\frac{\ln^2\left(\text{u}\right)}{\text{u}^2+2\text{u}+1}\space\text{d}\text{u}=-\frac{i}{2}\cdot\int\frac{\ln^2\left(\text{u}\right)}{\left(1+\text{u}\right)^2}\space\text{d}\text{u}\tag4$$
Let $\text{v}:=1+\text{u}$:
$$\mathscr{I}=-\frac{i}{2}\cdot\int\frac{\ln^2\left(\text{v}-1\right)}{\text{v}^2}\space\text{d}\text{v}\tag5$$
Using integration by parts:
$$\mathscr{I}=-\frac{i}{2}\cdot\left\{-\frac{\ln^2\left(\text{v}-1\right)}{\text{v}}+2\cdot\int\frac{\ln\left(\text{v}-1\right)}{\text{v}\cdot\left(\text{v}-1\right)}\space\text{d}\text{v}\right\}=$$
$$-\frac{i}{2}\cdot\left\{-\frac{\ln^2\left(\text{v}-1\right)}{\text{v}}+2\cdot\left\{\int\frac{\ln\left(\text{v}-1\right)}{\text{v}-1}\space\text{d}\text{v}-\int\frac{\ln\left(\text{v}-1\right)}{\text{v}}\space\text{d}\text{v}\right\}\right\}\tag6$$
| {
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Conditions to make conic section positve in the region $0 \leq x+y \leq 1$ Suppose $0 \leq x+y \leq 1$ and let $a,b,c \in [0,1]$ be fixed. I've been stuck with this inequality:
$$ -2\,{b}^{2}{x}^{2}+ \left( -4\,{b}^{2}+4 \right) xy+ \left( -4\,ab-2\,
c-2 \right) x-2\,{b}^{2}{y}^{2}+ \left( -4\,ab+2\,c-2 \right) y-2\,{a}
^{2}-{c}^{2}+1 \ge 0 $$
for a while now. Basically I am trying to find conditions on $a,b,c$ so that there exists some $x,y$ in the given region that makes this inequality true. Clearly the function on the left is some sort of conic in the $x,y$-plane, however, I am not sure what it would look like for a fixed $a,b,c$. Is there a straightforward way to find conditions on the $a,b,c$ so that this conic is positive in the given region?
| After a bunch of completing-the-square, we can write the polynomial as $$\begin{bmatrix}x-x_0 & y-y_0\end{bmatrix} \begin{bmatrix}-2b^2 & 2-2b^2 \\ 2-2b^2 & -2b^2\end{bmatrix}\begin{bmatrix}x-x_0 \\ y - y_0\end{bmatrix} + z_0$$
where
$$x_0 = \frac{1-c+2b(a+bc)}{2-4b^2}, \qquad y_0 = \frac{1+c+2b(a-bc)}{2-4b^2}, \qquad z_0 = \frac{2(a+b)^2}{2b^2-1}.$$
The matrix $$\begin{bmatrix}-2b^2 & 2-2b^2 \\ 2-2b^2 & -2b^2\end{bmatrix}$$ has one eigenvalue of $-2$ with eigenvector $(1,-1)$, and one eigenvalue of $2-4b^2$ with eigenvector $(1,1)$. So under the substitution arising from $$\begin{bmatrix}x-x_0 \\ y-y_0 \end{bmatrix} = u \begin{bmatrix} 1 \\ -1\end{bmatrix} + v \begin{bmatrix}1 \\ 1\end{bmatrix}$$ we can rewrite the polynomial as
$$-4u^2 + (4 - 8b^2) v^2 + z_0.$$
The constraint that $0 \le x + y \le 1$ turns into $0 \le 2v + x_0 + y_0 \le 1$, with no restriction on $u$. Since $u$ contributes a nonpositive term $-4u^2$, we should set $u=0$, which is equivalent to saying that $x - x_0 = y - y_0$, or that $y = x+(y_0 - x_0) = x+c$.
Now the polynomial is a quadratic in $x$. Relevantly, the coefficient of $x^2$ is $4-8b^2$. So to maximize the quadratic, we can do one of two things.
*
*If $b \le \frac1{\sqrt2}$, then the coefficient of $x^2$ is nonnegative, so the function is maximized at a boundary point. Either $x+y = 2x+c = 0$, or $x+y = 2x+c=1$. At the first point $(x,y) = (-c/2, c/2)$, the polynomial is $1 - 2a^2$, which is sometimes positive. At the second point $(x,y) = ((1-c)/2, (1+c)/2)$, the polynomial is $-2(a+b)^2$, which is always negative except I guess when $a=b=0$.
*If $ b > \frac1{\sqrt2}$, then the coefficient of $x^2$ is negative, so the function is maximized at the vertex of the parabola. Do some algebra and you find that the vertex of the parabola is at $x = \frac{-1-2ab+c-2b^2c}{4b^2-2}$, where $x+y = 2x + c = \frac{1+2ab}{1-2b^2}$ is always negative. That's outside our allowed range, so once again we want to take one of the boundary points above.
As a result, the best point to use is always $(x,y) = (-c/2, c/2)$. At this point, the polynomial is equal to $1 - 2a^2$, which is nonnegative if $a \le \frac1{\sqrt2}$.
| {
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Which Quadric Surface Is $-9x'^2+3y'^2+6z'^2=1$
Classify the quadric surface
$$3x^2-2y^2-z^2-4xy-12yz-8xz=1$$
I have found the eigenvalues which are $-9,6,3$
So it is $$-9x'^2+3y'^2+6z'^2=1$$
According to a page I have $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$ is an elliptic hyperboloid of one sheet and $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=-1$ is an elliptic hyperboloid of two sheets.
So which is $3x^2-2y^2-z^2-4xy-12yz-8xz=1$?
| Note that $-9x'^2+3y'^2+6z'^2=1$ is equivalent to
$$\frac{X^2}{(1/\sqrt{6})^2}+\frac{Y^2}{(1/\sqrt{3})^2}-\frac{Z^2}{(1/3)^2}=1$$
where $X=z'$, $Y=y'$ and $Z=x'$ (we swapped $x'$-axis and $z'$-axis).
So it is a one-sheet elliptic hyperboloid.
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Proving $1+ \frac 14 + \frac 19 + \ldots + \frac 1{n^2} \lt 2 - \frac 1n $ with induction Prove $$1+ \frac 14 + \frac 19 + \ldots + \frac 1{n^2} \lt 2 - \frac 1n $$
What I've noticed is that it only holds $\forall n \in \mathbb Z_0 - \{1\}$ because it doesn't say in my textbook to satisty such particular conditions.
$\underline{ n= 2:} \qquad $ $1 + \frac 14 \lt 2 - \frac 12$
$\underline{I.H.:} (n=k)\qquad $ $$\sum_{i=1}^k \frac 1{i^2} \lt 2- \frac 1k $$
Now it comes
$\underline{n= k+1:}$ $$\sum_{i=1}^k \frac 1{i^2} + \frac 1{(k+1)^2} \lt 2- \frac 1k + \frac1{(k+1)^2}$$
$$\sum_{i=1}^k \frac 1{(i+1)^2} \lt 2 - \frac {(k+1)^2 + k} {k(k+1)^2}= \frac{2k(k+1)^2 - (k+1)^2 + k }{k(k+1)^2} $$
Am I right working in the right direction and could someone give me a hint on how to proceed?
| \begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i^2}+ \frac{1}{(n+1)^2}< 2-\frac{1}{n} +\frac{1}{(n+1)^2}
= 2- \frac{n^2+n+\color{orange}{1}}{n(n+1)^2}
< 2 - \frac{n^2+n}{n(n+1)^2} \\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $a, b, c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $a^2 + b^2 - c^2 \gt 6(c - a)(c - b)$ .
Let $a$, $b$ and $c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $$a^2 + b^2 - c^2 \gt 6(c - a)(c - b).$$
This problem is from the Indian MO 2009, however, I do not know how to approach it, any help?
| Hint: Consider the quadratic equation $f(t) = 7t^2-6t(a+b)+6ab-a^2-b^2.$ You are to show that $f(\sqrt[3]{a^3+b^3})<0.$ But instead of plugging it in, it easier to show that the number $c = \sqrt[3]{a^3+b^3}$ lies strictly between the two roots of the quadratic $f(t).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine all the critical points of the given system and discuss their type and stability.
Determine all the critical points of the given system and discuss their type and stability.
$$\frac{\mathrm{d} x}{\mathrm{d} t}=x(1-x-y)$$$$\frac{\mathrm{d} y}{\mathrm{d} t}=y({\frac{3}{4}-y-\frac{1}{2}x})$$$$(1,0),(\frac{1}{2},\frac{1}{2})$$$$A=\begin{pmatrix}
1-2x-y, &-x \\-\frac{y}{2},
&\frac{3}{4}-2y-\frac{x}{2}
\end{pmatrix}$$$$\begin{pmatrix}
\frac{1}{2} &-\frac{1}{2} \\-\frac{1}{4}
&-\frac{1}{2}
\end{pmatrix}$$
$$(\lambda-\frac{1}{2})(\lambda+\frac{1}{2})=\frac{1}{8}$$
$$V_y=\frac{3}{4}-2y-\frac{x}{2}$$$$V_{xx}=1-2x-y$$$$V_{yy}=-\frac{5}{2}$$$$V_{xy}=-x$$$$R(x,y)=V_{xx}V_{yy}-2V_{xy}^2$$$$=-\frac{5}{2}+5x+\frac{5}{2}y-2x^2$$$$R(1,0)=\frac{1}{2}$$
But where to from here?
| There is an issue with your critical points. We want to simultaneously solve
$$\tag 1 x(1-x-y) = 0 \\ y\left(\dfrac{3}{4}-y-\dfrac{1}{2}x\right) = 0 $$
We see that for $x = 0$ in the first equation, we solve the second and find $y = 0, y = \dfrac{3}{4}$.
We see that for $y = 0$ in the second equation, we solve the first and find $x = 0, x = 1$.
Next, we assume that $x \ne 0, y \ne 0$, and have to solve the system
$$1-x-y = 0 \\ \dfrac{3}{4}-y-\dfrac{1}{2}x = 0 $$
Negating the first equation and adding gives $x = \dfrac{1}{2}, y = \dfrac{1}{2}$.
Summing up, we have four critical points as
$$(x, y) = (0, 0), \left(0, \dfrac{3}{4}\right), (1, 0), \left(\dfrac{1}{2}, \dfrac{1}{2} \right)$$
Next we find the Jacobian matrix of $(1)$
$$J(x, y) = \begin{pmatrix}
\dfrac{\partial u}{\partial x}& \dfrac{\partial u}{\partial y}\\
\dfrac{\partial v}{\partial x}& \dfrac{\partial v}{\partial y}
\end{pmatrix}
=\begin{pmatrix}
1-2x-y &-x \\-\dfrac{y}{2} &\dfrac{3}{4}-2y-\dfrac{x}{2}
\end{pmatrix}$$
Hints on remaining parts
*
*Find the eigenvalues of the Jacobian at each critical point.
*Classify the stability using the following
If the eigenvalues are real
*
*Eigenvalues both positive = An Unstable Node: All trajectories in the neighborhood of the fixed point will be directed outwards and away from the fixed point.
*Eigenvalues both negative = A Stable Node: All trajectories in the neighborhood of the fixed point will be directed towards the fixed point.
*Eigenvalues opposite sign = An Unstable Saddle Node: Trajectories in the general direction of the negative eigenvalue's eigenvector will initially approach the fixed point but will diverge as they approach a region dominated by the positive (unstable) eigenvalue.
If the eigenvalues are complex conjugates - their real parts are equal and their imaginary parts have equal magnitudes but opposite sign.
*
*Real parts positive = An Unstable Spiral: All trajectories in the neighborhood of the fixed point spiral away from the fixed point with ever increasing radius.
*Real parts negative = An Stable Spiral: All trajectories in the neighborhood of the fixed point spiral into the fixed point with ever decreasing radius.
You might also like to review this and this and this.
Look at a phase portrait to verify the results
| {
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"url": "https://math.stackexchange.com/questions/2434127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Asymptotics of $\sum_{0 \leq j \leq k \leq n-1} {\binom{2n}{j}}{\binom{2n-1}{k}}^{-1}$ For any integer $n\geq 1$, define $$f(n) = \sum_{0 \leq j \leq k \leq n-1} \frac{\binom{2n}{j}}{\binom{2n-1}{k}}$$ Our lecture says that $f(n) = n +n\log n +O(1)$. But I cannot prove it, the best result I've gotten is $f(n) \leq cn^{\frac{3}{2}}$, and now I'm kind of doubtful about the result. Is it possible to prove it or deny it?
|
Conjecture. As $n\to\infty$, we have
$$ f(n) = \frac{1}{2}n\log n + \left(\frac{\gamma}{2}+\log 2 \right) n + o(1).$$
A partial proof. Write
$$ f(n) = 2n \sum_{j=0}^{n-1} \binom{2n}{j} \sum_{k=j}^{n-1} \frac{(2n-1-k)!k!}{(2n)!}. \tag{1}$$
We first find an integral represtentation of the inner sum. Utilizing the gamma integral,
\begin{align*}
\sum_{k=j}^{n-1} \frac{(2n-1-k)!k!}{(2n)!}
&= \sum_{k=j}^{n-1} \frac{1}{(2n)!}\int_{0}^{\infty} \int_{0}^{\infty} x^{2n-1-k}y^k e^{-(x+y)} \, dxdy \\
&= \frac{1}{(2n)!} \int_{0}^{\infty} \int_{0}^{\infty} x^{2n-1} \left( \sum_{k=j}^{n-1} (y/x)^k \right) e^{-(x+y)} \, dxdy\\
&= \frac{1}{(2n)!} \int_{0}^{\infty} \int_{0}^{\infty} \frac{x^{2n-j}y^j - x^n y^n}{x - y} e^{-(x+y)} \, dxdy.
\end{align*}
Now make the substitution $(r, p) = (x+y, \frac{x}{x+y})$. Then we have $dxdy = r dr dp$ and
\begin{align*}
&= \frac{1}{(2n)!} \int_{0}^{1} \int_{0}^{\infty} r^{2n} e^{-r} \frac{p^{2n-j}(1-p)^j - p^n(1-p)^n}{2p - 1} \, drdp \\
&= \int_{0}^{1} \frac{p^{2n-j}(1-p)^j - p^n(1-p)^n}{2p - 1} \, dp \\
\small[\text{substitute }s = 1-2p] \quad &= \frac{1}{2^{2n+1}} \int_{-1}^{1} \frac{(1-s^2)^n - (1-s)^{2n-j}(1+s)^j}{s} \, ds
\end{align*}
Now using the symmetry, we know that
$$ \int_{-1}^{1} \frac{1 - (1-s^2)^n}{s} \, ds = 0. $$
Adding the above integral to our final integral, we obtain
$$ \sum_{k=j}^{n-1} \frac{(2n-1-k)!k!}{(2n)!}
= \frac{1}{2^{2n+1}} \int_{-1}^{1} \frac{1 - (1-s)^{2n-j}(1+s)^j}{s} \, ds. $$
Plugging this whole sum back to the first identity $\text{(1)}$ yields
\begin{align*}
f(n)
&= n \sum_{j=0}^{n-1} \binom{2n}{j} \frac{1}{2^{2n}} \int_{-1}^{1} \frac{1 - (1-s)^{2n-j}(1+s)^j}{s} \, ds \\
&= n \sum_{j=n+1}^{2n} \binom{2n}{j} \frac{1}{2^{2n}} \int_{-1}^{1} \frac{1 - (1-s)^{j}(1+s)^{2n-j}}{s} \, ds \\
&= n \mathbb{E} \left[ \int_{-1}^{1} \frac{1 - (1-s)^{N}(1+s)^{2n-N}}{s} \, ds \ ; \ N > n\right], \tag{2}
\end{align*}
where $N$ is a random variable having binomial distribution $\operatorname{Bin}(2n,\frac{1}{2})$. To compute the inner integral, we perform integration by parts:
\begin{align*}
&\int_{-1}^{1} \frac{1 - (1-s)^{N}(1+s)^{2n-N}}{s} \, ds \\
&\hspace{1.5em} = \left[ \vphantom{\int} \left( 1 - (1-s)^{N}(1+s)^{2n-N} \right) \log(\sqrt{n}|s|) \right]_{s=-1}^{s=1} \\
&\hspace{4em} + \int_{-1}^{1} \Big\{ (2n-N)(1-s)^{N}(1+s)^{2n-N-1} \\
&\hspace{6.5em} - N (1-s)^{N-1}(1+s)^{2n-N} \Big\} \log(\sqrt{n}|s|) \, ds \\
&\hspace{3em} = 2^{2n-1} (\log n) \mathbf{1}_{\{N = 2n\}} \\
&\hspace{5.5em} - \int_{-1}^{1} \frac{2(N-n+ns)}{1-s^2}(1-s)^{N}(1+s)^{2n-N} \log(\sqrt{n}|s|) \, ds.
\end{align*}
To compute the last line, we apply substitutions $N = n + \sqrt{\smash[b]{n/2}} \, Z$ and $s = t/\sqrt{n}$. Then $\text{(2)}$ simplifies to
\begin{align*}
f(n)
&= \frac{1}{2}n \log n \\
&\hspace{1em} - 2n \underbrace{ \mathbb{E} \Bigg[ \int_{-\sqrt{n}}^{\sqrt{n}} \left(\frac{Z}{\sqrt{2}}+t \right)\left(1 - \frac{t^2}{n}\right)^{n-1} \left( \frac{1 - \frac{t}{\sqrt{n}}}{1 + \frac{t}{\sqrt{n}}} \right)^{\sqrt{\smash[b]{n/2}} \, Z} \log|t| \, dt \, ; \, Z > 0 \Bigg] }_{=\text{(*)}}
\end{align*}
Now we make a bit of loose computation. Let $n \to \infty$ to the inner integral and notice that the integrand converges pointwise nicely. Also, the classical CLT tells that $Z$ converges in distribution to the standard normal distribution and this convergence is not wild. So it is tempting to believe that the whole expectation $\text{(*)}$ also converges to
$$ \text{(*)} \xrightarrow[n\to\infty]{\text{let's believe!}}
\mathbb{E} \Bigg[ \int_{-\infty}^{\infty} \left(\frac{Z}{\sqrt{2}}+t \right)e^{-t^2 - \sqrt{2}Zt} \log|t| \, dt \, ; \, Z > 0 \Bigg], \tag{3}
$$
where now $Z$ has standard normal distribution. (I am kind of sure that this can be justified with some hard analysis, though I do not want to spare much time on this.) Computing the outer expectation first,
$$ \text{[RHS of (3)]}
= \int_{-\infty}^{\infty} \frac{e^{-t^2} \log|t|}{2\sqrt{\pi}} \, dt
= -\frac{1}{4} (\gamma + 2\log 2). $$
Therefore, modulo the claim $\text{(3)}$ we have proved that
$$ f(n) = \frac{1}{2}n\log n + \left( \frac{\gamma}{2} + \log 2 \right)n + o(n). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Double integral of floor function and exponential function over rectangular region I am not getting any idea how to deal with this two sums of double integral where functions and rectangular domain is given as
*
*$\lfloor x+y\rfloor$ and $1\leq x\leq 3$, $2\leq y\leq 5$
*$e^{\max(x^2,y^2)}$ and $0\leq x\leq 1,0\leq y\leq 1$.
For first one I use inequality $x+y-1<\lfloor x+y\rfloor\leq x+y$, and I just get hint that my answer will in between $27$ and $33$. But how I will get exact answer.
| For a given value of $x$ and integer $k$, $$\lfloor x+y\rfloor = \left\{ \begin{matrix} \lfloor x \rfloor + k & \text{for $k \le y < k + 1 - x + \lfloor x \rfloor $,} \\ \lfloor x \rfloor + k+1 & \text{for $k + 1 - x + \lfloor x \rfloor \le y < k + 1$.}\end{matrix}\right.$$
Consequently, for an integer $k$,
$$\int_{k}^{k+1} \lfloor x+y\rfloor \,dy = \int_{k}^{k + 1 - x + \lfloor x \rfloor} (\lfloor x \rfloor+ k) \,dy + \int_{k + 1 - x + \lfloor x \rfloor}^{k+1} (\lfloor x \rfloor +k +1 )\,dy = k +x.$$
Therefore,
$$\int_{1}^{3}\int_{2}^{5} \lfloor x+y\rfloor \, dy \, dx = \int_{1}^{3}\left(\int_{2}^{3} \lfloor x+y\rfloor \, dy+\int_{3}^{4} \lfloor x+y\rfloor \, dy+\int_{4}^{5} \lfloor x+y\rfloor \, dy\right)\,dx$$ implies
$$\int_{1}^{3}\int_{2}^{5} \lfloor x+y\rfloor \, dy \, dx = \int_{1}^{3} [(2+x)+(3+x)+(4+x) ]\,dx = \color{red}{30}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
| Set $y = x + 1$. The equation rewrites as $(y + 1)^2 + (y - 1)^2 = 82$, that is $$2(y^4 + 6y^2 + 1) = 82 \iff y^4 + 6y^2 - 40 = 0$$ Now let $z = y^2$: the equation has become $z^2 + 6z - 40 = 0$, which is a simple quadratic equation with solutions $$z_{1, \: 2} = -3 \pm \sqrt{9 + 40} = -3 \pm 7 \implies z_1 = 4, \; z_2 = -10$$ Since $z = y^2 \ge 0$, $z_2$ is not acceptable, so $z = 4$. Finally, substituting backwards we get $$x = y - 1 = \pm\sqrt{z} - 1 = \pm 2 - 1$$ Thus there are two real solutions, namely $x_1 = 1$ and $x_2 = -3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
$\displaystyle\lim_{x\to-\infty}2x+\sqrt{4x^2+x}$ How do I calculate
$$\lim_{x\to-\infty}2x+\sqrt{4x^2+x}?$$
I've gotten "close" many times but I just can't figure it out, I always get stuck with zero division or something similar.
| $$\begin{align*} \lim_{x \rightarrow -\infty} 2x + \sqrt{4x^2 + x} & = \lim_{x \rightarrow \infty} \sqrt{4x^2 - x} - 2x = \\ & = \lim_{x \rightarrow \infty} \frac{(\sqrt{4x^2 - x} - 2x)(\sqrt{4x^2 - x} + 2x)}{\sqrt{4x^2 - x} + 2x} = \\ & = \lim_{x \rightarrow \infty} -\frac{x}{\sqrt{4x^2 - x} + 2x} = \boxed{-\frac{1}{4}} \end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Is it possible to find a McLaurin Series for the following function? Since the limit as $y$ approaches $0$ is $\infty$ for the function $g(y$) = $\frac{1}{\sqrt{1+y^2} - \sqrt{1-y^2}}$, can we say the MacLaurin Series in this case does not exist? And if this is case, what methods are there for approximating the function near $0$?
| Since,
as Mark Viola showed,
$f(y)
=\dfrac{1}{\sqrt{1+y^2}-\sqrt{1-y^2}}
=\dfrac{\sqrt{1+y^2}+\sqrt{1-y^2}}{2y^2}
$,
you can approximate the function
for small $y$
($y$ away from $0$ is no problem).
We can use the generalized binomial theorem
in the special case
$\sqrt{1+x}
=\sum_{n=0}^{\infty} \binom{1/2}{n}x^n
=\sum_{n=0}^{\infty} \binom{2n}{n}\dfrac{(-1)^{n+1}x^n}{4^n(2n-1)}
=1-\frac12 x-\frac18 x^2+...
$.
For small $y$,
$f(y)
\approx \dfrac{(1+\frac12 y^2-\frac18 y^4)+(1-\frac12 y^2-\frac18 y^4)}{2y^2}
= \dfrac{2-\frac14 y^4}{2y^2}
= \dfrac{1}{y^2}-\frac18 y^2
$.
You can take as
many terms as you want:
$\begin{array}\\
\sqrt{1+y^2}+\sqrt{1-y^2}
&=\sum_{n=0}^{\infty} \binom{1/2}{n}((y^{2n}+(-1)^ny^{2n})\\
&=\sum_{n=0}^{\infty} \binom{1/2}{2n}2y^{4n}\\
&=2(1-\frac18 y^4 - \frac{5}{128}y^6+...)\\
\end{array}
$
so
$f(y)
=\frac1{y^2}-\frac18 y^2 - \frac{5}{128}y^4+...
$.
| {
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"url": "https://math.stackexchange.com/questions/2449531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find the last $3$ digits of $3^{352}$? Find the last $3$ digits of $3^{352}$ ?
Apart from Carmichael Function, any other way of solving it ?
| As you imply, the Carmichael function $\lambda(1000) = \text{lcm}(\,\lambda(5^3), \lambda(2^3)\,) = \text{lcm}(100,2) = 100$ is the least universal exponent.
All$\bmod 1000$, this gives us that $3^{352}\equiv 3^{52}$ and then:
$\begin{align}3^{52} = 9^{26} = (10-1)^{26}
& \equiv (-1)^{26} + 26\cdot 10\cdot (-1)^{25} + \overbrace{\frac{26\cdot25}2}^{\text{odd mult of 5}}\cdot 10^2\cdot (-1)^{24}\\
& \equiv 1-260+500 \equiv 241\\
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Radical equation $\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt[4]{4x-3}}=\frac{2}{3}$ I am stuck in the following difficult radical equation:
$$
\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt[4]{4x-3}}=\frac{2}{3}
$$
Attemption. The left hand side of the equation is a decrease function. Therefore, the equation has a unique solution. An approximation for the solution is 8.74874 (using Matlab).
Thank you for all kind help hint.
| Hint: $u=2x-1$ and $v=4x-3$
Rewrite the expression:
$$
\frac{1}{\sqrt{u}}+\frac{1}{\sqrt[4]{v}}=\frac{2}{3} \implies \frac{\sqrt[4]{v}+\sqrt{u}}{\sqrt{u}\sqrt[4]{v}}=\frac{2}{3}
$$
$$
\implies \sqrt[4]{v}+\sqrt{u}=\frac 23\sqrt{u}\sqrt[4]{v}
$$
$$
\implies \sqrt{u}=\left(\frac 23\sqrt{u}-1\right)\sqrt[4]{v} \implies \frac{\sqrt{u}}{\frac 23\sqrt{u}-1}=\sqrt[4]{v}
$$
Now, raise the equation to the fourth power:
$$
\implies \frac{u^2}{(\frac 23\sqrt{u}-1)^4}=v
$$
Now isolate $\sqrt{u}$ on one side of the equation and raise everything to the second power. The resulting polynomial in $x$ after inverting the substitutions might not be solvable by an analytical expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a monic quadratic polynomial which is a factor of two other polynomials Problem statement:
Find a monic quadratic polynomial, $f(x)$, which divides both $$g(x) = 12x^3 − 30x^2 + 18x − 12$$ and $$h(x) = 6x^4 + 3x^3 + 6x^2 + 3$$
My take on it:
I divided $h(x)$ by $g(x)$ to get the quotient and remainder such that $$6x^4 + 3x^3 + 6x^2 + 3 = (12x^3 − 30x^2 + 18x − 12)(\frac{1}{2}x + \frac{3}{2}) + 3(14x^2 - 11x + 7)$$
It is also the case that any polynomial divisor of both $g(x)$ and $h(x)$ must also divide the remainder polynomial when $h(x)$ is divided by $g(x)$.
So following on from that, our common factor for $g(x)$ and $h(x)$ that we're trying to find, would also have to be a factor of our remainder, $3(14x^2 - 11x + 7)$. Yet the remainder cannot be factorised any further to turn it into a monic quadratic polynomial.
How should I go about solving this problem?
So far, it's only this particular problem that I find myself unable to solve, other ones like it are fine.
Important: Please don't provide the full solution but rather only the way forward as I don't want to be accused of plagiarism. This exact problem is part of an assignment.
EDIT: I actually had an error in my calculation, the resulting equation is now as follows
$$
6x^4 + 3x^3 + 6x^2 + 3 = (12x^3 − 30x^2 + 18x − 12)(\frac{1}{2}x + \frac{3}{2}) + 42(x^2 - \frac{1}{2}x + \frac{1}{2})
$$
| =-=-=-=-==-
$$ \left( 6 x^{4} + 3 x^{3} + 6 x^{2} + 3 \right) $$
$$ \left( 12 x^{3} - 30 x^{2} + 18 x - 12 \right) $$
$$ \left( 6 x^{4} + 3 x^{3} + 6 x^{2} + 3 \right) = \left( 12 x^{3} - 30 x^{2} + 18 x - 12 \right) \cdot \color{magenta}{ \left( \frac{ x + 3 }{ 2 } \right) } + \left( 42 x^{2} - 21 x + 21 \right) $$
$$ \left( 12 x^{3} - 30 x^{2} + 18 x - 12 \right) = \left( 42 x^{2} - 21 x + 21 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 4 }{ 7 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( \frac{ x + 3 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x + 3 }{ 2 } \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ 2 x - 4 }{ 7 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{2} + x + 1 }{ 7 } \right) }{ \left( \frac{ 2 x - 4 }{ 7 } \right) } $$
$$ \left( x^{2} + x + 1 \right) \left( \frac{ 1}{7 } \right) - \left( x - 2 \right) \left( \frac{ x + 3 }{ 7 } \right) = \left( 1 \right) $$
$$ \left( 6 x^{4} + 3 x^{3} + 6 x^{2} + 3 \right) = \left( x^{2} + x + 1 \right) \cdot \color{magenta}{ \left( 6 x^{2} - 3 x + 3 \right) } + \left( 0 \right) $$
$$ \left( 12 x^{3} - 30 x^{2} + 18 x - 12 \right) = \left( x - 2 \right) \cdot \color{magenta}{ \left( 6 x^{2} - 3 x + 3 \right) } + \left( 6 x^{3} - 15 x^{2} + 9 x - 6 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( 6 x^{2} - 3 x + 3 \right) } $$
$$ \left( 6 x^{4} + 3 x^{3} + 6 x^{2} + 3 \right) \left( \frac{ 1}{7 } \right) - \left( 6 x^{3} - 15 x^{2} + 9 x - 6 \right) \left( \frac{ x + 3 }{ 7 } \right) = \left( 6 x^{2} - 3 x + 3 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $S_1^2+S_2^2+S_3^2+...+S_{2n-1}^2$ Question: If $S_1,S_2,S_3,...,S_n$ be the sums of $n$ infinite G.P. series respectively whose first terms are respectively $1,2,3,...,n$ and common ratios $\frac{1}{2},\frac{1}{3},\frac{1}{4},...,\frac{1}{n+1}$ respectively, then find the value of $S_1^2+S_2^2+S_3^2+...+S_{2n-1}^2$.
My attempt:
$S_k=\frac{k}{1-\frac{1}{k+1}}$
$\implies S_k=k+1$
$\implies S_k^2=(k+1)^2$
$\implies S=\sum\limits_{k=1}^{2n-1}S_k^2$
$\implies S=\sum\limits_{k=1}^{2n-1}(k+1)^2$
$\implies S=\sum\limits_{k=1}^{2n-1}(k^2+1+2k)$
$\implies S=\sum\limits_{k=1}^{2n-1}(k^2)+\sum\limits_{k=1}^{2n-1}(1)+\sum\limits_{k=1}^{2n-1}(2k)$
$\implies S=\frac{(2n-1)(2n)[2(2n-1)+1]}{6}\ \ \ \ \ \ \ +(2n-1)+\frac{2(2n-1)(2n)}{2}$
$\implies S=\frac{(4n^2-2n)(4n-1)}{6}\;\;\;+2n-1+4n^2-2n$
$\implies S=\frac{16n^3-4n^2-8n^2+2n}{6}\;\;+2n-1+4n^2-2n$
$\implies S=\frac{16n^3-4n^2-8n^2+2n-6+24n^2}{6}$
$\implies S=\frac{16n^3+12n^2+2n-6}{6}$
$\implies S=\frac{8n^3+6n^2+n-3}{3}$
My problem: The correct answer is $\frac{n(2n+1)(4n+1)}{3}$ which on simplification gives $\frac{8n^3+6n^2+n}{3}$. Where have i gone wrong? Please help.
| Since
$$
\begin{align}
(k+1)^2
&=(k+1)k+(k+1)\\
&=2\binom{k+1}{2}+\binom{k+1}{1}
\end{align}
$$
we have
$$
\begin{align}
\sum_k\binom{2n-1-k}{2n-1-k}\left[2\binom{k+1}{k-1}+\binom{k+1}{k}\right]
&=2\binom{2n+1}{2n-2}+\binom{2n+1}{2n-1}\\
&=\frac{8n^3+6n^2+n}3
\end{align}
$$
Therefore, leaving out the $k=0$ term, we get
$$
\sum_{k=1}^{2n-1}(k+1)^2=\bbox[5px,border:2px solid #C0A000]{\frac{8n^3+6n^2+n-3}3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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.