Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\... | Do $a=\tan A, b= \tan B$ and $y = \tan Y$. Then,
\begin{eqnarray}
\tan (A+B+Y) &=& \dfrac{\tan(A+B) + \tan(Y)}{1-\tan(A+B)y}\\
&=& \dfrac{\dfrac{a+b}{1-ab}+y}{1 - \Big(\dfrac{a+b}{1-ab}\Bigr)y}\\
&=& \dfrac{\dfrac{a+b+(1-ab)y}{1-ab}}{\dfrac{1-ab - (a+b)y}{1-ab}}\\
&=& \dfrac{a+b+y -aby}{1-ab-ay-by}.\\
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/177640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Evaluating $\int_1^3\frac{\ln(x+2)}{x^2+2x+15} \ dx$ Could you please give me a hint on how to compute:
$$
\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx
$$
Thank you for your help
| First, let's make a change of variables, $x=3 + u$, which maps $(1,3)$ into $(-1,1)$:
$$
\int_1^3 \frac{\log(x+2)}{x^2+x+15}\mathrm{d} x = \int_{-1}^1 \frac{\log(4+u)}{u^2 + 6 u+ 23}\mathrm{d} u
$$
Now use $\log(4+u) = \log(4) + \log\left(1+\frac{u}{4}\right)$:
$$\begin{eqnarray}
\int_1^2 \frac{\log(1+2 u)}{2 u^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Taylor polynomial of $f(x) = 1/(1+\cos x)$ I'm trying to solve a problem from a previous exam. Unfortunately there is no solution for this problem.
So, the problem is:
Calculate the Taylor polynommial (degree $4$) in $x_0 = 0$ of the function:
$$f(x) = \frac{1}{1+\cos(x)}$$
What I tried so far:
*
*calculate all ... | $$\cos x=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}=1+\cos x= 2-\frac{x^2}{2}+\frac{x^4}{24}-...\Longrightarrow$$
$$\frac{1}{1+\cos x}=\frac{1}{2-\frac{x^2}{2}+\frac{x^4}{24}-...}=\frac{1}{2}\frac{1}{\left[1-\left(\frac{x}{2}\right)^2\right]\left(1-\frac{x^2}{24}+...\right)}=$$
$$=\frac{1}{2}\left(1+\frac{x^2}{4}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2}$ How would I prove the following trig identity?
$$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2} $$
My work thus far has been:
$$\dfrac{2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}}{-2\sin\dfrac{A+B}{2} \sin\... | A couple of hints: use the $\cot$ identity on the RHS to put in terms of $\sin$ and $\cos$, use a product to sum identity, and pay attention to odd and even functions. This should simplify it to an easier expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/178823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Legendre symbol, second supplementary law $$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8}$$
how did they get the exponent. May be from Gauss lemma, but how.
Suppose we have a = 2 and p = 11. Then n = 3 (6,8,10), but not
$$15 = (11^2-1)/8$$
n is a way to compute Legendre symbols from Gauss lemma:
$$\left(\frac{a}{p}\righ... | Here is a proof from Ireland-Rosen Ch-6 (The key idea is to work in $ \overline{\mathbb{Z}} $, modulo $ p \overline{\mathbb{Z}} $) :
Let $ p $ be an odd prime. We have $ \left( \frac{2}{p} \right) \equiv 2^{\frac{p-1}{2}} (\text{mod } p\mathbb{Z} ) $. Notice we can write $ 2^{\frac{1}{2}} $ as $ 2^{\frac{1}{2}} = 2 \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
When a is less than c in $ \int_a^b \frac {dx} {x^4 - c^4} $ $$ \int_a^x \frac {dx} {x^4 - c^4} = \frac {1} {4c^3} \ln \left(\frac {x-c} {x+c} \right)_a^x - \frac {1} {2c^3} \tan^{-1} \Bigl(\frac {x} {c} \Bigr)_a^x $$
$$ =\frac {1} {4c^3} \Bigl[ \ln \Bigl(\frac {x-c} {x+c} \Bigr) - \ln \Bigl(\frac {a-c} {a+c} \Bigr) \B... | Yes. Notice that
$$
\int \frac{dx}{x+c} = \ln \left| x+c \right| + \text{constant}
$$
(where "constant" actually means a piecewise constant function that is constant on $(-\infty,-c)$ and also on $(-c,\infty)$). Similar remarks apply with $x-c$ in place of $x-c$. The partial-fraction decomposition of $1/(x^4-c^4)$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\sqrt{x} +y = 4$, $x+ \sqrt{y}= 6$, find the solution $(x,y)$ $\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, find the solution (x,y). $NOTE$ : $\sqrt{4}+1= 4-1$, $\sqrt{1} +4 =1+4$
| This is basically the method which was suggested in the comments above - turning this into a quartic equation. We will see whether someone suggest a substantially more elegant solution.
$$\sqrt{x}+y=4\\
x+\sqrt{y}=6$$
Using the substitution
$\sqrt{x}=s$ and $\sqrt{y}=t$
we get:
$$s+t^2=4\\
t+s^2=6$$
Which gives
$$s=4-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use the given equalities to derive trigonometric functions.
(A) $\sin(-x)=-\sin x$
(B) $\cos(-x)=\cos x$
(C) $\cos(x+y)=\cos x\cos y-\sin x\sin y$
(D) $\sin(x+y)=\sin x\cos y+\cos x\sin y$
Use these equalities to derive the following important trigonometric functions:
f) $\left|\cos\dfrac{x}{2}\right|... | I know from an earlier question of yours that you are familiar with the identities
$$\cos 2w =2\cos^2 w-1=1-2\sin^2 w,\tag{$1$}$$
which can be derived fairly quickly from (C).
(Yes, I have changed the name of the variable. That is deliberate.)
Now let $w=\frac{x}{2}$. Then the identities $(1)$ can be rewritten as
$$\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Geometry/Triangles Problem Let $A,B,C$ be a triangle's vertices (FIG.1), and $D$ is a point on $\overline {BC}$. It is given that $\angle B= 2 \angle C $, $\overline{AB}=\overline{DC}$ and also that $\overline{AD}$ is the bisector of angle $A$.. We have to prove that angle $\angle BAC=72°$.
$\hspace{0.54 cm}$
$$\begi... | As $B=2C \Rightarrow A=\pi-B-C=\pi-3C$ then
$$ 0<B<\pi \Rightarrow 0<2C<\pi \Rightarrow 0<C<\frac{\pi}{2}$$
and $A=\pi-3C$ so, $0<A<\pi \Rightarrow0<\pi - 3C <\pi \Rightarrow0< C< \frac{\pi}{3} $
So, $0< C< \frac{\pi}{3} $
If $∠BAD=x , ∠DAC=x$ and if $∠ADB=y, ∠ADC=\pi-y$
If $AB=c$, $BC=a$, $CA=b \Rightarrow BD=a-c$
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/181559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| Let's solve it in an elementary way and start from the fact that:
$$a^2 \ge 2a -1 \tag1$$
$$b^2 \ge 2b-1 \tag2$$
$$c^2 \ge 2c-1 \tag3$$
Then add up $(1)$ $(2)$ $(3)$ and get:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \tag4$$
By AM-GM we have
$$\frac{a+b+c}{3} \ge (abc)^\frac{1}{3}=1 $$
$$ a+b+c \ge 3 \tag5 $$
Finally, from $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 5
} |
Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$ I want to prove the following inequality :
$$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab, $$
$a,b,c,d \in \mathbb{R}_{+} .$
In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.
Thanks :)
| As noted by Zarrax, it suffices to show that
$$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = 4 \sum_{\mathrm{sym}}a^3 + 36 \sum_{\mathrm{sym}} a^2 b - 40 \sum_{\mathrm{sym}}abc$$
is nonnegative for all $a, b, c, d \ge 0$. Here I use the notation $\sum_{\mathrm{sym}}f(a,b ,c, d) := (1/4!) \sum_{\pi \in S_4} f(\pi(a), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Summing digits of powers of 2 to get 1 2 4 8 7 5 pattern If you add the decimal digits of multiples of 9,possibly repeatedly, you get 9. For instance, with 18 we have $$18\mapsto 1+8=9$$
and with 909 we have $$ 909\mapsto 9+0+9=18\mapsto 1+8=9.$$ This is proven.
If you do the same with powers of two, you develop a repe... | We look at $2^n$, where $n$ ranges over the non-negative integers.
The key is the fact that $2^6$ has remainder $1$ on division by $9$. Using congruence notation, we have $2^6\equiv 1\pmod{9}$. Let $n$ be any non-negative integer. We can express $n$ as $6q+r$, where $0\le r\le 5$ ($q$ stands for quotient, $r$ for rema... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
So, LS=
$$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$
$$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\... | Or you could:
$$
\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta} = \frac{\sin \theta}{\sin \theta \cos\theta} - \frac{\cos\theta\sin\theta}{\sin\theta}.
$$
Then cancel terms. Then common denominator. Then...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the... | If $2|k=>16|n^4$ and $4|n^2=>16|(n^4+4n^2)=>n^4+4n^2+11≡11\pmod{16}$
Else
$n$ is odd$=2k+1$(say), $n^2=(2k+1)^2=8\cdot\frac{k(k+1)}{2}+1≡1\pmod{8}=>8|(n^2-1)$
(i)So, $n^4+4n^2+11=(n^2-1)^2+6(n^2-1)+16≡0\pmod{16}$ if $n$ is odd.
(ii)When $n$ is odd, $2|(n^2+1)$ and $8|(n^2-1)$(already proved) $=>2\cdot8|(n^2-1)\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/187033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$ let $ABC$ be an acute triangle with all angles greater than $45^o$
Prove that
$$\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$$
I let $\tan A=a$, $\tan B=b$, $\tan C=c$ with $a+b+c=abc$ then the inequali... | What about Calculus:
$f(x)=\frac{1}{1+\tan(x)}$ has a positive second derivative on $(\frac{\pi}{4}, \frac{\pi}{2})$. This solves the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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If $a$, $a+2$ and $a+4$ are prime numbers then, how can one prove that there is only one solution for $a$? If $a$, $a+2$ and $a+4$ are prime numbers then, how can one prove that there is only one solution for $a$?
when, $a=3$
we have, $a+2=5$ and $a+4=7$
| You have observed that $a=3$ is a possible solution. Now assume $a>3$. What forms can $a$ take? Since $a$ is prime it should be of the form $3k+2$ or $3k+1$ where $k$ is a positive integer. If it was of the form $3k+2$ then $a+4=3k+2+4=3(k+2)$ will not be prime since $k+2 >1 $. If it was of the form $3k+1$ then $a+2=3k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 7
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Generalization of $\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1}$ I thought about a generalization for the formula
$$\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1}$$
It can be written as
$$\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1} = \sum_{i + j ... | I think that this formula is what you are looking for. If $\mathbf x = (x_1,\dotsc,x_r)$, then
$$ \sum_{|I|=n} \mathbf{x}^I = \sum_i\frac{x_i^{n}}{\prod_{j\neq i}(1-\frac{x_j}{x_i})}. $$
With 1 variable, it gives
$$ x^n = x^n, $$
for two,
$$ \sum_{i+j = n} x^i y^j = \frac{x^{n+1}}{x-y} + \frac{y^{n+1}}{y-x}, $$
and for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
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When does $\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$? I am trying to find the conditions under which $\lfloor(n-1)x\rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$. The trivial case is whenever $x \in \mathbb{Z}$. If $n = 2$, then $x - \lfloor x \rfloor \lt \frac{1}{2}$.
As $n$ increases however... | Let $m=\lfloor x\rfloor$ and $\alpha=x-m$. Then
$$\lfloor (n-1)x \rfloor + \lfloor x \rfloor =\lfloor(n-1)m+(n-1)\alpha\rfloor+m=nm+\lfloor(n-1)\alpha\rfloor\;,$$
and $$\lfloor nx \rfloor=\lfloor nm+n\alpha\rfloor=nm+\lfloor n\alpha\rfloor\;,$$
so $\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$ iff $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/190605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Find a positive number $\delta$? Find a positive number $\delta$ such that when $|x-1|<\delta$, then $|x^2-1| < 0.45$
Part 2:
Find the LARGEST number $\lambda$ with the property that when $|x-1|< \lambda$, then $|x^2-1| < 0.45$
I don't understand this at all. Help?
| Suppose $0 < \epsilon < 1$. Consider the case $x^2 - 1 \ge 0$ first, i.e., $x \in (-\infty, -1] \cup [1, \infty)$. In this case, the inequality $|x^2 - 1| < \epsilon$ becomes
$$
x^2 - 1 - \epsilon < 0.
$$
Factor this as
$$
(x - \sqrt{1 + \epsilon})(x + \sqrt{1 + \epsilon}) < 0.
$$
This is true when $x \in (-\sqrt{1 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$
| By the Cauchy-Schwarz inequality,
$$6=1*(a+b)+1*(b+c)+1*(a+c)\leq\sqrt{1^2+1^2+1^2}\sqrt{(a+b)^2+(b+c)^2+(a+c)^2}$$
In other words
$$(a+b)^2+(b+c)^2+(a+c)^2\geq 12$$
which is the same as the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
A function $f(x)$ on the real domain and real constants $a$ and $b\neq 0$ for which $f(x)-f(x-\delta)+a+bx^2=0$ for some real $\delta\neq 0$ Is there a function $f(x)$ on the real domain and real constants $a$ and $b\neq 0$ for which the following is true:
$$f(x)-f(x-\delta)+a+bx^2=0$$
for some real $\delta\neq 0$?
EDI... | $f(x)-f(x-\delta)+a+bx^2=0$
$f(x)-f(x-\delta)=-bx^2-a$
In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe1108.pdf.
The general solution of this functional equation is $f(x)=\Theta(x)+f_p(x)$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$
Luc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Finding $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$ I try to find $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$.
Left side:
$$\sin(4 \alpha + \frac{\pi}{6}) =$$
$$= \sin4\alpha \times \cos \frac{\pi}{6} + \cos 4\alpha \times \sin\frac{\pi}{6} =$$
$$= \... | $\sin(4\alpha+\frac{\pi}{6})=\sin(2\alpha+\frac{\pi}{5})$
$\sin(4\alpha+\frac{\pi}{6})-\sin(2\alpha+\frac{\pi}{5}) = 0$
Consider the formula: $\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$,
and $A\cdot B=0\Leftrightarrow A=0\vee B=0$
$2\cos\frac{6\alpha+\frac{11\pi}{30}}{2}\sin\frac{2\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2?
Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$.
My attempt
I proved that the sequence is increasing and bounded by $2$, can anyone help me show that the seque... | We have
$$\sqrt{2}=2^{\frac{1}{2}}.$$
Exciting, no?
We also have
$$\sqrt{2\sqrt{2}}=2^{\frac{1}{2}+\frac{1}{4}},$$
and
$$\sqrt{2\sqrt{2\sqrt{2}}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}.$$
For the next term, we take the previous term, multiply by $2$, getting exponent of $2$ equal to $1+\frac{1}{2}+\frac{1}{4}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 7,
"answer_id": 4
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Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$
I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.
| Nice solution, just for clarification
$\left(5+2\sqrt{6}\right)^{\frac{x}{2}}\cdot \left(5-2\sqrt{6}\right)^{\frac{x}{2}}=\left[\left(5+2\sqrt{6}\right)\cdot \left(5-2\sqrt{6}\right)\right]^{\frac{x}{2}}=\left[5^2-\left(2\sqrt{6}\right)^2\right]^{\frac{x}{2}}=\left(25-24\right)^{\frac{x}{2}}=1$
and so for $t=\left(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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understanding definition of limit for x_n= [(n+1)/n]^0.5; where x_n is the n-th term in the sequence. $x_n= \sqrt{\frac{n+1}{n}}$where $x_n$ is the $n^{th}$ term in the sequence.
Now, $x_n \rightarrow 1$ as $n\rightarrow\infty$.
Then applying definition of limit: $$\left|\sqrt{\frac{n+1}{n}}- 1 \right|= \frac{\frac{n+1... | I see that since $n(n+1)>n$ then $\sqrt{n(n+1)}+n>2n$ and so $$\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|<\frac{1}{2n}$$ But, $$\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|=\bigg|\frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}\bigg|$$ Therefore $$\bigg|\sqrt{\frac{n+1}{n}}-1\bigg|=\bigg|\frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to derive the equation for x in a quadratic equation?
Possible Duplicate:
Why can ALL quadratic equations be solved by the quadratic formula?
How to derive this:
$x = \frac{-b + {\sqrt{b^2 + 4ac}}}{2a}$
From this:
$ax^2 + bx + c = 0$
I know this may be a little elementary :)
| The usual approach is to complete the square:
$$\begin{align*}
ax^2+bx+c&=a\left(x^2+\frac{b}ax+\frac{c}a\right)\\
&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}a\right)\\
&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c\;,
\end{align*}$$
which equals $0$ if and only if $$a\left(x+\frac{b}{2a}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/203195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Level curves on ellipsoid Let $a,b,c>0$ with $a\leq b\leq c$. Let $E$ be the ellipsoid determined by $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$$
Is there a function $f:E\rightarrow \mathbb{R}$ such that $f\in C^{\infty}(E)$ and all level curves of $f$ are circles?
Thanks
| Generalizing @Rahul's solution (but, as we'll see, not-generalizing that solution), consider the plane, through the origin, with normal $(p,q,r)$ such that $p^2+q^2+r^2=1$. Of course, obtaining circular level curves in the case $p^2+q^2 = 0$ requires $a=b$ and is easily dispatched; similarly, $q^2+r^2 = 0$ and $r^2 + p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Arithmetic Mean & Geometric Mean Question 1: if the arithmetic mean of two numbers is twice of their geometric mean, their ratio of sum of numbers to the difference of numbers equals?
Question 2: if the quadratic equation:
$(b^2+c^2)x2-2(a+b)cx+(c^2+a^2)=0$
has equal roots then?what is its AP & GP?
Question 3: If the... | Question 1:
Arithmetic mean: $\frac{a+b}{2}$
Geometric mean: $\sqrt{ab}$
So the condition becomes $\frac{a+b}{2}=2\sqrt{ab}$. Square both sides to get
$$
\frac{a^2+2ab+b^2}{4}=4ab
$$
which, assuming $b\ne0$, results in
$$
\left(\frac ab\right)^2-14\frac ab+1=0
$$
and
$$
\frac ab=7\pm4\sqrt{3}
$$
Then we can compute
$$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
sum of the series
could anyone tell me how to calculate these sums?I am not finding any usual way to calculate them.
| 5.7:
$$\begin{align*}
1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots
&=1+\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n+1)}{4\cdot8\cdots4n}\\
&=1+\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n+1)}{2\cdot4\cdots2n}(\frac{\sqrt{2}}{2})^{2n}
\end{align*}$$
Set $f(x)=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)}{2\cdot4\cdots2n}x^{2n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can any integer can be written as the sum of 8 integer cubes? Can anyone offer an elementary proof why: $$\forall n \in {\mathbb Z} \space \exists a, b, c, d, e, f, g, h \in {\mathbb Z}$$such that$$n=a^3+b^3+c^3+d^3+e^3+f^3+g^3+h^3{\rm ?}$$
In other words, why every integer is the sum of eight cubes integers.
My first ... | Allowing negative cubes, five suffice. It is suspected that four suffice but this is an open problem. I will see if I can find the argument for five, it is just one or two explicit formulas. By the way, these are called the "Easier" Waring problems.
Can't find them at this point, so here is section D5 form Richard K. ... | {
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"timestamp": "2023-03-29T00:00:00",
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$16^{18}+16^{18}+16^{18}+16^{18}+16^{18}=4^x$ without a calculator How to solve the following equation without using calculator
$$16^{18}+16^{18}+16^{18}+16^{18}+16^{18}=4^x$$
| $16^{18}+16^{18}+16^{18}+16^{18}+16^{18}$
$=5\times 16^{18}$
$= 5\times {4^2}^{18} $
$= 5\times 4^{2\times 18} = 5\times 4^{36} = 4^x$
$ 5\times 4^{36} = 4^{\log_4{5}} \times 4^{36} $
$= 4^{\log_4{5} +36}$
$\rightarrow x= \log_4{5} +36$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta$ I need solve this integral, and I tried various methods of solving and did not get it. The integral is:
$$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta,$$
where $t$ is a positive integer.
| $$\frac{1}{1-2t\cos \theta +t^2}=\frac 1{1+t^2-2t\frac{(1-\tan^2\frac{\theta}2)}{(1+\tan^2\frac{\theta}2)}}$$
$$=\frac{\sec^2\frac{\theta}2}{(1+t^2)(1+\tan^2\frac{\theta}2)-2t(1-\tan^2\frac{\theta}2)}=\frac{\sec^2\frac{\theta}2}{(1-t)^2+\tan^2\frac{\theta}2(1+t)^2}$$
$$=\frac1{(1+t)^2}\frac{\sec^2\frac{\theta}2}{(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/211058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
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Common tangent to two circles
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$
and
$$(x - 5)^2 + (y - 4)^2 = 4.$$
I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equation... | For the circles $$(x - 2)^2 + y^2 = 9\text{ and } (x - 5)^2 + (y - 4)^2 = 4,$$
using Article $180$ of The elements of coordinate geometry by Loney,
Case $1:$ If $T_2$ is the point dividing internally the line joining the centres
in the ratio $2:3$ then its coordinates are $$\left(\frac{2\cdot2+5\cdot3}{2+3},\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/211538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Least Square Solutions
Find the closest point or points to b $= (-1,2)^T$ that lie on
the line $x + y = 0$ and also find the line $2x + y = 0$.
The question is obviously elementary but I do not know how to set it up properly to find the points. I know for $x + y = 0$ we have a slope of -1 going through the origin ... | Note that $x + y = 0$ describes a slope in the direction of $\left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$. We want to find the point $c \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$ on this line such that $\left( \left( \begin{matrix} -1 \\ 2 \end{matrix} \right) - c \left( \begin{matrix} 1 \\ -1 \end{matr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $ord_m b^2$ if $ord_m a = 10$ and $ab\equiv 1\pmod m$ If $ab \equiv 1 \pmod {m}$ and if $ord_ma=10$, find $ord_mb^2$.
Could somebody give me a hint? What I know is that $ab \equiv 1 \pmod {m}$ can be used when finding the multiplicative inverse. Would this mean that $10$ is the multiplicative inverse, so $b=10$?... | $\newcommand{\ord}{\operatorname{ord}}$No, $\ord_ma=10$ says that $a^{10}\equiv 1\pmod m$, but if $1\le n < 10$, then $a^n\not\equiv 1\pmod m$. That is, $10$ is the smallest positive integer $n$ such that $a^n\equiv 1\pmod m$. Similarly, $\ord_mb^2$ is the smallest positive integer $n$ such that $\left(b^2\right)^n\equ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Frequency of a trigonometric function - Where is my mistake? I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$
The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1):
$$\cos(2x)=1-\sin^2(x)$$
$$\sin... | $$\sin^4\left(x+\frac{\pi}{2}\right)=\left(\sin x\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\cos x\right)^4=\cos^4x$$
$$\cos^4\left(x+\frac{\pi}{2}\right)=\left(\cos x\cos\left(\frac{\pi}{2}\right)-\sin x\sin\left(\frac{\pi}{2}\right)\right)^4=(-\sin x)^4=\sin^4 x$$
Thus, the period of $\,\sin^4 x+\co... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expected Number of Successes in a Sample
$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.
Solution:
Chance of broken calculator: $\dfrac{1}{10}$.
Do I need to calculate the odds of $0$ - $1... | (See note below for the correct answer.)
Let $X_k = 1$ if the $k$th calculator is broken, and $0$ otherwise.
Assuming independence of 'brokenness', we have the probability of the $k$th calculator being broken is $P(X_k = 1) = \frac{20}{200} = \frac{1}{10}$.
Then the expected number of broken calculators is
$$E (\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/218086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many distinct real root does the equation $x^{4}-x^{3}\cdot\sin(x)-x^{2}\cdot\cos(x)=0 $ have? How many distinct real root does the equation $x^{4}-x^{3} \cdot \sin(x)-x^{2} \cdot \cos(x)=0 $ have?
Is there any quick solution(less than 2 minutes)?
| You can factor out $x^2$ to get $x^2(x^2-x \sin x - \cos x ) = 0$. So, clearly, there is a root of multiplicity $2$ at $x=0$. Letting $f(x) = x^2-x \sin x - \cos x $, we note that $f$ is even, $f(0) <0$, $\lim_{x\to\infty} f(x) = \infty$ and $f'(x)> 0$ when $x>0$. Consequently $f$ has one real root in $(0,\infty)$.
So ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Iterated Integral requiring u-substitution This is an iterated integral. I have tried solving it several times using u-substition, but I am not getting the correct answer. My latest result is (4/15)(10^(5/2)-33). Obviously something is off, but what? Could you please show me some steps and your final answer so I can wo... | You could start with letting
$$u = \sqrt {{x^2} + {y^2}}$$
so that
$$4\int {xy\sqrt {{x^2} + {y^2}} dy = \frac{4}{3}x{{({x^2} + {y^2})}^{3/2}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/220394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Reason behind particular induction step I'm wondering why and how you get some of these steps in an Induction proof.
Okay, so the question is to prove the following statement by induction:
$1 - x + x^2 - x^3$ $+ ... +$ $x^{2n-2} = \frac{x^{2n-1}+1}{x+1}$
The first step that sort of confuses me is having a base case of ... | Let $P(n)$ be the statement
$$1-x+x^2-x^3+\ldots+x^{2n-2}=\frac{x^{2n-1}+1}{x+1}\;;\tag{1}$$
you want to prove that $P(n)$ is true for all $n\ge 2$. (I’ll come back to the choice of $2$ later.) For the induction step you assume $P(k)$ for some $k\ge 2$ and try to prove $P(k+1)$. Let’s see what those statements really ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sum k! = 1! +2! +3! + \cdots + n!$ ,is there a generic formula for this? I came across a question where I needed to find the sum of the factorials of the first $n$ numbers. So I was wondering if there is any generic formula for this?
Like there is a generic formula for the series:
$$ 1 + 2 + 3 + 4 + \cdots + n = \fra... | In addition to the special functions given by J.M., an asymptotic expansion can be computed
$$
\begin{align}
\sum_{k=0}^n k!
&=n!\left(\frac11+\frac1n+\frac1{n(n-1)}+\frac1{n(n-1)(n-2)}+\dots\right)\\
&=n!\left(1+\frac1n+\frac1{n^2}+\frac2{n^3}+\frac5{n^4}+\frac{15}{n^5}+O\left(\frac1{n^6}\right)\right)\\
&=\sqrt{2\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/227551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "72",
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How to calculate the maximum value of: $\frac{25x}{x^2+1600x+640000}$? Wolfram says it's 800, but how to calculate it?
$$
\frac{25x}{x^2+1600x+640000}
$$
| Let $$\frac{25x}{(x^2+1600x+640000)}=y$$
or $$x^2y+x(1600y-25)+640000y=0$$
This is a quadratic equation in $x$.
As $x$ is real, the discriminant $(1600y-25)^2-4\cdot y\cdot 640000y\ge 0$
On simplification, $-128y+1\ge 0\implies 128y\le 1\implies y\le \frac1 {128}$
So, the maximum value of $y=\frac{25x}{(x^2+1600x+640... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Area enclosed by the curve $x^2+y^2=|x|+|y|$.
Find the area enclosed by the curve $x^2+y^2=|x|+|y|$ on the coordinate plane.
I have no idea how this curve looks like, therefore can't find the area. Please help. Thank you.
| Wolfram|Alpha shows you how the curve looks like. To obtain the area, first note that by symmetry we can concentrate on the first quadrant. There the equation is equivalent to $x^2 + y^2 = x+y\iff y^2 - y + (x^2-x) = 0$, so
\[
y_{1,2} = \frac 12 \pm \sqrt{\frac 14 - (x^2-x)}
\]
The randicand is nonnegative for $x^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/232051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to find the roots of $x^4 +1$ I'm trying to find the roots of $x^4+1$. I've already found in this site solutions for polynomials like this $x^n+a$, where $a$ is a negative term. I don't remember how to solve an equation when $a$ is a positive term as the equation above.
Thanks
| We have:
$$\begin{align}x^4+1 =0 \\
x^4+1+2x^2-2x^2 =0 \\
(x^2+1)^2-(\sqrt{2}x)^2 =0 \\
(x^2+1+\sqrt{2}x)(x^2+1-\sqrt{2}x) =0 \end{align}$$
Then on solving these factors by quadratic formula, we will have:
$$x= \frac{-1}{\sqrt{2}}+\frac{i}{\sqrt{2}} , \frac{-1}{\sqrt{2}}-\frac{i}{\sqrt{2}}, \frac{1}{\sqrt{2}}+\frac{i}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/233999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
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Fibonacci conjecture: $(F_{n+5})^2 - (F_n)^2 = 3((F_{n+3})^2 - (F_{n+2})^2) + 8 F_{n+2} F_{n+3} $. So this is the question I have
The Fibonacci sequence is a recurrence system given by $$F_1 = 1, \ F_2 = 1, \ F_{n+2} = F_{n+1} + F_n \qquad (n = 1, 2, 3, \ldots).$$ This question concerns the following conjecture: $$(F_... | Use $F_{n+4}-F_{n+2}=F_{n+3}$ in the first factor and $F_{n+4}=F_{n+3}+F_{n+2}$ in the second factor to obtain the asterisked line from the preceding one.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$ $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$
I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake... | You should use that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Take $a=\sqrt[3]{n^3+n^2}$, $b=\sqrt[3]{n^3+1}$ and then multiply your expression by $(a^2+ab+b^2)/(a^2+ab+b^2)$. Then use the trick you are trying to use.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Geometry Prove - two perpendicular lines in a circle In a circle of radius r, two lines (AB and CD) are perpendicular to each other and meet at X.
Show that:
| a^2 + y^2 = r^2\\
B: b^2 + y^2 = r^2 \\
C: x^2 + c^2 = r^2 \\
D: x^2 + d^2 = r^2
$$
From $A$ and $B$ one can find that $a = -b$, and analogously $c = -d$. Then
$XA^2 + XB^2 + XC^2 + XD^2 = (a-x)^2 + (b-x)^2 + (c-y)^2 + (d-y)^2 = \\
a^2+b^2+c^2+d^2-2x(a+b)-2y(c+d)+2x^2+2y^2 = a^2+b^2+c^2+d^2+2x^2+2y^2$
which is sum of a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trying to find $\det (B)$ Let $A=\left(\begin{matrix}1&1&-1\\-1&1&1\\1&-1&1\end{matrix} \right)$,and ${A}^{T}B{\left( \cfrac{1}{2}{A}^{T}\right)}^{T}-8{A}^{-1}B=I$, How to compute $\left|B \right|$?。
| Consider the permutation matrix $$P = \pmatrix{0 & 1 & 0\cr 0 & 0 & 1\cr 1 & 0 & 0\cr}$$
Note that $P^3 = I$ and $A = I + P - P^2$. Since $A P^2 = P^2 + I - P = -A + 2 I$, we see
that $A^{-1} = (P^2 + I)/2$, while $A^T = I + P^2 - P$. It seems reasonable, then, that a solution $B$ might also be a linear combination o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the fixed points of the two systems What are the fixed points of the two non linear systems below?
\begin{align}
x(t)& = x(3-x-2y)\\
y(t)& = y(2-x-y)
\end{align}
I know that $(0,0)$, $(0,2)$, $(3,0)$, $(1,1)$ are the fixed points. However, I don't understand some of the steps to obtain this. Any help will be appre... | Unless I missed something in the question, the fixed points are given by solutions of $x = x(3-x-2y)$, $y=y(2-x-y)$.
If $x=0$, the first equation is trivially satisfied, and the second gives $y = y(2-y)$, or in other words $y(y-1) = 0$. Hence $(0,0)$ and $(0,1)$ are solutions.
If $y=0$, the second equation is trivially... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/242738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What are $2222^{5555}+5555^{2222} \pmod 7$ and $9^{2n+1}+8^{n+2} \pmod{73}$? Tell me hint for solve :
1) $ 2222^{5555}+5555^{2222} \equiv \mathord? \pmod 7$
2) $ 9^{2n+1}+8^{n+2} \equiv \mathord ?\pmod{73}$
thank you.
| (2)I was trying to figure out how the problem came to being.
Observe that $73=8\cdot9+1$
Replacing $8$ with $a$ we get $a(a+1)+1=a^2+a+1$
Now as the root of $a^2+a+1=0$ are the imaginary cube roots of $1,$
let $\omega$ be one of the imaginary cube roots of $1.$
So, $\omega^3=1\implies \omega^{3n}=1$
$\omega^{3n+2m+n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/247169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integer solutions to $\sqrt{a} + \sqrt{b} = \sqrt{c}$ An example I can see using basic algebra is $\sqrt{2} + \sqrt{8} = \sqrt{18}$, but is there a general method to find integer solutions to the problem? Another question: say you are given the value of $c$; can you find the values of $a$ and $b$?
| Let $c=d n^2$ with $n\ge 0$ and with $d \gt 0$ square-free, i.e. $d$ has no repeated prime factors. $d$ is the product of the individual prime factors of $c$ which occur an odd number of times.
Then $a=d k^2$ and $b=d (n-k)^2$ with $0 \le k \le n$ are the only solutions to $\sqrt a + \sqrt b = \sqrt c$, and there are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/248595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 1
} |
Factor $1+x+x^2+x^3+...+x^{14}$ In a previous task, I was asked to factor $1+x+x^2+x^3$ for $x \in \mathbb{R}$, which I accomplished by solving
$1+x+x^2+x^3 = 0 \to $
$1+x(1+x+x^2) = 0 \to $
$x(1+x+x^2) = -1$
which has a solution $x = -1$, and thus I knew $(x+1)$ was a factor. A bit of guesswork gave me $(x+1)(x^2+1)$... | Note that the polynomial has 15 terms, so try grouping it in 5 groups of 3:
$$ 1 + x + x^2 + \cdots + x^{14} =\\ =(1+x+x^2) + x^3(1+x+x^2)+ x^6(1+x+x^2) + x^9(1+x+x^2)+ x^{12}(1+x+x^2)$$
and then factor out $(1+x+x^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/249988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$
Possible Duplicate:
How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$?
How can I calculate precise value of that series: $\sum\limits_{i=0}^{n-1} i2^i$ ?
So far, I tried to differentiate the $ \sum\limits_{i=0}^{n} 2^i = 2^{i-1} - 1 $ series, b... | $$\begin{array}{rll}
S &=1\cdot2^1+&2\cdot2^2+3\cdot2^3+\cdots+(n-2)\cdot2^{n-2}+(n-1)\cdot2^{n-1} \\
2S &= &1\cdot2^2+2\cdot2^3+\cdots+(n-3)\cdot2^{n-2}+(n-2)\cdot2^{n-1}+(n-1)\cdot2^{n}
\end{array}$$
Subtracting,
$$S-2S=1\cdot2^1+(2-1)\cdot2^2+\cdots+\{(n-2)-(n-3)\}\cdot2^{n-2}+\{(n-1)-(n-2)\}\cdot2^{n-1}-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/250746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I wa... | Here's another approach that is less intensive. Show that
$$ (a^2 + bc) ( b^2 + ca) \geq ab(c+b)(c+a) $$
This is equivalent to
$$ c (a-b)^2 (a+b) \geq 0 $$
Multiply the 3 cyclic versions of the first inequality, and you get your conclusion. Equality holds if and only if $a=b=c$.
Note: The factorization of the second... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/253015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| $\gcd(a+b, a^2 - ab + b^2) = \gcd(a+b, (a+b)^2 - 3ab) = \gcd(a+b, 3ab)$
Since $\gcd(a,b) = 1$ hence $\gcd(a+b,a) = 1$ and $\gcd(a+b, b)$ = 1.
Thus, $\gcd(a+b, 3ab) = \gcd (a+b, 3) $ by applying the Lemma below.
Hence, $\gcd(a+b, a^2 -ab + b^2) = \gcd(a+b, 3)=1$ or 3. (This is very simple to evaluate for given values.)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 0
} |
simplify summation of factorial (random walk) I suspect that the expression
$$\sum_{n=0}^N \frac{(N-2n)^2}{n!(N-n)!}$$
simplifies to
$$\frac{2^N}{(N-1)!}$$
But I cannot find the intermediate steps. Can someone give me a hint how I can deduce this result?
(The expression comes up when calculating the average final posi... | $$
\begin{align}
\sum_{n=0}^N\frac{(N-2n)^2}{n!(N-n)!}
&=
\sum_{n=0}^N\frac{((N-n)-n)^2}{n!(N-n)!}
\\
&=
\sum_{n=0}^N\frac{(N-n)^2}{n!(N-n)!}+\sum_{n=0}^N\frac{n^2}{n!(N-n)!}-2\sum_{n=0}^N\frac{(N-n)n}{n!(N-n)!}
\\
&=
\sum_{n=0}^N\frac{n^2}{n!(N-n)!}+\sum_{n=0}^N\frac{n^2}{n!(N-n)!}-2\sum_{n=1}^{N-1}\frac1{(n-1)!(N-n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/258882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Find the value of : $\lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx]$. I am stuck on the following problem:
Find the value of : $$\lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx].$$
My attempts: Let $$I_{n}= \lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx]=\lim_{n\to\infty}[\ln(2)-\int_{0}^{1}\frac{x^{n+1... | You can also use power series. Here's how Euler probably would have done it. I say that because the proof as I present it here is a bit informal but it can be patched up easily. We do term by term integration and then later interchange a limit and an infinite sum. Start with
$\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots$
Repla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/259079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Can someone show me why this factorization is true? $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$
Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this
| Since you said above that you already believe that the result is true, that makes me think you really want to know how to come up with the result (if it had not been presented to you).
My answer would be: play with several examples!
That is, sharpen your pencil and get busy factoring out $x^n-y^n$ for $n=1,2,3,\dots$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 1
} |
Indefinite integral involving arctan How should I evaluate this definite integral? I am unable to figure out how to start.
$$\int \tan^{-1} \left(1 + x + x^{2}\right) dx $$
| Proceed by integration by parts, we get
\begin{align*}
\int \arctan \left( 1 + x + x^2 \right) dx & = x \arctan \left(1+x+x^2 \right) - \int \dfrac{x(2x+1)}{(x^2+x+1)^2+1} dx
\end{align*}
Now note that $$\left(x^2+x+1 \right)^2+1 = \left(x^2+1 \right) \left(x^2+2x+2 \right)$$
Hence, $$\dfrac{x(2x+1)}{(x^2+x+1)^2+1} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Do there exist any odd prime powers that can be represented as $n^4+4^n$? Well, I wrote up a solution on it, but according to the place I found the problem, it isn't quite correct. Ah, I'm simply hoping someone will point out where I got wrong.
Now, let, $n^4+4^n = p^k$, where $p$ is an odd prime and $k$ is a positive ... | For odd $n>1$ the expression is never prime, it can be factored into a product where each factor is the sum of two squares.
$$[((2^m)+n)^2 +(2^m)^2][(2^m)^2 +((2^m)^2-n)^2]$$
where $m = (n-1)/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/261925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
incircle and circumcircle Given a triangle $\Delta ABC$ with circumcenter $O$, and incenter $I$.
If $∠ABC = 45°$, $OI = d$, and $c − b = d√2$,
Find the value of $\sin A$.
I've thought of using the fact that $d^2 = R(R-2r)$ where $R$ is the circumradius and $r$ is the inradius. But I didn't know where to apply it.
| keep going:$\ (4+2\sqrt{2})x^3-(2+2\sqrt{2})x^2-(2+3\sqrt{2})x+(3+\sqrt{2})=4x^3-2\sqrt{2}x^2+2\sqrt{2}x^3-2x^2-3\sqrt{2}x+3-2x+\sqrt{2}=2\sqrt{2}x^2*(\sqrt{2}x-1)+2x^2(\sqrt{2}x-1)-3(\sqrt{2}x-1)-(\sqrt{2})(\sqrt{2}x-1)=(\sqrt{2}x-1)[(2\sqrt{2}+2)x^2-(3+\sqrt{2})]$=0
$\ x=\sqrt{2}/2$
or
$\ x^2=(3+\sqrt{2})/(2\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$ Let a;b;c>0. Prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{a+c}\right)^2+\left(\frac{c}{a+b}\right)^2+\frac{10abc}{(a+b)(b+c)(c+a)}\geq 2$$
I think
$$\frac{2a}{b+c}=x;\frac{2b}{c+a}=y;\frac{2c}{a+b}=z$$
We ha... | You are almost there. You have $x^{2}+y^{2}+z^{2}\ge xy+yz+xz$ by matching squares. So $$x^{2}+y^{2}+z^{2}-5(xy+yz+zx)+12\ge 12-4(xy+yz+zx)\ge 0$$ In other words you need to show $xy+yz+zx\le 3$. Now by Cauchy-Schwarz the left hand side only obtain maximum of $x=ky, y=kz,z=kx$. So we have either $x=y=z$ or $x=-y, y=-z,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$N \equiv 3 (\textrm{mod } 4)$ and Collatz conjecture Can the Collatz conjecture also be interpreted as behaviour, transformation of number of form $N\equiv 3(\textrm{mod }4)$ to the form of $N\equiv 1(\textrm{mod }4)$
Because integers of the form $N\equiv 3(\textrm{mod }4)$ can be later always only ones be divided... | To add to @coffemath approach: we can separate the class of odd numbers a,b in the following way such that $b=(3a+1)/2^A$ where $A$ is the value which makes b an odd integer:
$$ \begin{array} {r|c|l||r|c|l}
& k \ge 0& & && k\gt 0\\
A & a=& b= & & A & a= & b= \\
\hline \\
0 & 2^0 \cdot 2 k +0 & k & & 1 & 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Range of a function $f(x)$
As x varies over all real numbers, the range of the function $$f(x)=\frac{x^2-3x+4}{x^2+3x+4}$$ is (1) $[\frac{1}{7},7]$, (2) $[-\frac{1}{7},7]$, (3) $[-7,7]$ (4) $(-\infty,\frac{1}{7})\bigcup(7,\infty)$.
Trial:$$\begin{align} \frac{x^2-3x+4}{x^2+3x+4} &=\frac{(x-\frac{3}{2})^2+\frac{7}{4}... | Answer to modified question: Note that the line $y=1$ is an asymptote to the curve $y=f(x)$.
Calculate $f'(x)$. It turns out to be $\frac{6x^2-24}{(x^2+3x+4)^2}$. So $f(x)$ is increasing from $-\infty$ to $-2$, then decreasing until $2$, then increasing.
For negative $x$, the curve is above the asymptote, and for pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Continued fraction of a square root If I want to find the continued fraction of $\sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. ... | Here the easiest method to generate continued fraction for any square (or more) root. Lets take $\sqrt{5}$:
$$\sqrt{5} \approx 2,2360679775...$$
$$\sqrt{5} = 2 + 0,2360679775$$
So $2=\left \lfloor \sqrt{5} \right \rfloor$
$$\sqrt{5} =2 + x$$ with $x=0,2360679775$
$$\sqrt{5}^{2} = (x + 2)^{2} \Rightarrow 5= x^2 + 4x + 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 7,
"answer_id": 1
} |
Calculating the integral $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ using complex analysis I need to calculate $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ $,n\geq2$ using complex analysis. I probably need to use the Residue Theorem.
I use the function $f(z)={\frac{\ln z}{1+z^n}}$ in the normal branch.
I've tried to use this con... | This is NOT an answer to the original question. This is a justification of the final identity in Peter's answer. This was requested in chat, but too long to put there or in a comment box.
@Peter, all you need is recurrence and reflection for the trigamma function. Taking $z = 1/2n,$ first we find
$$ \psi^{(1)} \left( 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 7,
"answer_id": 4
} |
$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$
Find $$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$$
I know that $\lim_{x \to 0} \frac{\sin x}{x}=1$ But here $\sin \left(\frac{1}{3 \sqrt{x}}\right)$ is given when $x \to 0$. Need help.
| The limit of the denominator is $\sqrt7$ so we just need to tame the numerator . Observe
$$\left|\left|x\right|\sin\frac{1}{3\sqrt{x}}\right|=\left|x\right|\left|\sin\frac{1}{3\sqrt{x}}\right|\le \left|x\right|\cdot 1$$
What does this tell you?
Also note that $x\to 0^+$ as $\sqrt x$ must be defined
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/269143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\epsilon$-$\delta$ proof that $f(x) = x \sin(1/x)$, $x \ne 0$, is continuous I'm doing an exercise that asks me to prove that $f$ is continuous using a $\epsilon$-$\delta$ proof. I have that
$$
f(x) = \begin{cases}
x\cdot \sin \frac1x,&x\neq 0
\\
0,&x = 0
\end{cases}
$$
I've already managed to show this property for... | Let $a\neq 0$. By the triangle inequality,
$$\begin{array}{ccc}
\left|f(x)-f(a)\right| &=& \left|x\sin \frac1x-a\sin \frac1a\right| \\
&=& \left|x\sin \frac 1x-a\sin \frac 1x+a\sin \frac 1x-a\sin \frac1a\right| \\
&\le& \left|x-a\right|\left|\sin \frac 1x\right|+a\left|\sin \frac 1x-\sin \frac1a\right| \\
&<& \delta+a\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/274594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Is there any trick to evaluate this integral?
Possible Duplicate:
Please help me to evaluate $\int\frac{dx}{1+x^{2n}}$.
Is there any trick to evaluate
$$\int_{-\infty}^\infty \frac{{\rm d} x}{x^{2n}+1}?$$
| This is just a try ...
Since the function is even, we have $2 \int_{0}^\infty \frac{1}{x^{2n}+1}dx$
$$ \int_0^1 \frac{1}{x^{2n}+1} dx + \int_1^\infty\frac{1}{x^{2n}+1} dx \\
= \int_0^1 \sum_{k=0}^\infty (-1)^k (x)^{2kn} + \int_1^\infty \frac{1}{x^{2n}} \sum_{k=0}^\infty (-1)^kx^{-2nk} \\
= \sum_{k=0}^\infty (-1)^k \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An integration to first order I am having some trouble evaluating an integral -- involving taking an approximation. It would be great if someone could help me.
I wish to evaluate
$$\int_0^\pi {\cos\theta\cos \left[\omega t-{\omega \over c}(a^2+r^2-2rb\cos \theta)^{1\over 2}\right]\over (a^2+r^2-2rb\cos \theta)^{1\over ... | The supposed answer is incorrect. Up to first order, we have $$\int_0^\pi\frac{\cos\theta\cos\left(\omega t-\frac\omega c\sqrt{a^2+r^2-2rb\cos\theta}\right)}{a\sqrt{a^2+r^2-2rb\cos\theta}}\,d\theta\approx\frac{\pi rb\omega\sin(\omega a/c-\omega t)}{2ca^2}.$$
Proof: Write $\sqrt{a^2+r^2-2rb\cos\theta}=a\sqrt{1+(r^2-2rb\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $ How can I calculate the following integral :
$$\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $$
I try to write the integral like :
$$\int\frac{1+x-x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{(\sqrt{x^2+1})'}{... | This is how I would go, term $\sqrt{1+x^2}$ makes me go for $x=\sinh(u)$, which will convert to $\cosh(u)$ and cancel with the one brought by $du$.
Then I convert to exponential form ($t=e^u$) to get a rational fraction.
$\begin{align}\require{cancel}\int\frac {dx}{(x+1)\sqrt{1+x^2}}
&=\int\frac {\cancel{\cosh(u)}du}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Cyclotomic polynomial simplification I try to work out what $\Phi_{12}(z)$ is:
By the fundamental theorem of arithmetic:
$$\Phi_{12}(z)=(z-\xi^0)(z-\xi^1)(z-\xi^2)(z-\xi^3)(z-\xi^4)(z-\xi^5)(z-\xi^6)\\(z-\xi^7)(z-\xi^8)(z-\xi^9)(z-\xi^{10})(z-\xi^{11})$$
$$\Phi_{12}(z)=(z-\xi^0)(z-\xi^1)(z-\xi^2)(z-\xi^3)(z-\xi^4)(z-\x... | I suppose $\xi$ is a primitive $12^{th}$ root of unity.
The polynomial $\Phi_{12}(x)$ is the product $\prod(z-\xi^k)$ over the primitive $12^{th}$ roots of unity (i.e. for $k=1,5,7,11$).
What you are calculating is, $\displaystyle{\prod_{k=0}^{11}(z-\xi^k)}$, the product over all $12^{th}$ roots of unity which is j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to determine the sign of : I've got some trouble: I don't know how to determine the sign of :
$$\frac{4t^2- \sqrt t}{t}$$
for $t > 0$.
Thank you in advance
| Since $\sqrt{t}$ is only defined when $t \ge 0$, we can only consider inputs from that range.
Now we have to consider the sign of both the numerator and the denominator. The fraction will be positive whenever the sign of $4t^2 - \sqrt{t}$ matches that of $t$. Clearly though, $t$ is only positive when $t \ge 0$, so we ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| I have a new answer. We will apply the Cebysev's inequality.
$$(a^3+a+1)(b^3+b+1)\leq3(a^3b^3+ab+1)$$
So:
$$\left[(a^3+a+1)(b^3+b+1)(c^3+c+1)\right]^{2} \leq 27(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1).$$
Now, we will prove that:
$$27(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq 27^{2}.$$ Equivalent with:
$$(a^3b^3+ab... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 8,
"answer_id": 3
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Integral of $\frac{\sqrt{x^2+1}}{x}$ So I have to do the integration of $\frac{\sqrt{x^2+1}}{x}$. Give me a hint. What should I replace? Should I do it with integration by parts?
| You could use the substitution
$$
\begin{align}
&&u&=x+\sqrt{x^2+1},&x&=\frac{u^2-1}{2u} \\
&&\sqrt{x^2+1}&=\frac{u^2+1}{2u},&dx&=\frac{u^2+1}{2u^2}du
\end{align}
$$
to get (unless I did some mistake)
$$
\int\frac{\sqrt{x^2+1}}{x}dx=
\int\frac{\frac{u^2+1}{2u}}{\frac{u^2-1}{2u}}\frac{u^2+1}{2u^2}du=
\int\frac{(u^2+1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Determinant of an $n\times n$ matrix with 5's on the diagonal and 2's on the superdiagonal and subdiagonal
Possible Duplicate:
Special determinant formula for a specific matrix
How to find $\det A_n$ as a function of $n$?
$$A_n=\begin{pmatrix} 5&2 &0& 0 & \ldots & 0\\
2& 5& 2& 0 & \ldots & 0\\
0 &2& 5 &2 & \ld... | Example:
$$
\left|\begin{matrix}
5&2&0&0\\
2&5&2&0\\
0&2&5&2\\
0&0&2&5
\end{matrix}\right|
=5\left|\begin{matrix}
5&2&0\\
2&5&2\\
0&2&5
\end{matrix}\right|
-2\left|\begin{matrix}
2&2&0\\
0&5&2\\
0&2&5
\end{matrix}\right|
=5\left|\begin{matrix}
5&2&0\\
2&5&2\\
0&2&5
\end{matrix}\right|
-2^2\left|\begin{matrix}
5&2\\
2&5... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Power Series and Radius of Convergence
Determine for following Power Series in $\mathbb{C}$ the radius of
Convergence.
a) $\sum _{ n=0 }^{ \infty }{ (2+\sqrt { n } )^{ n }z^{ n } } $
b) $\sum _{ n=1 }^{ \infty }{ (1-\frac { 1 }{ n } )^{ n }z^{ n } } $
c) $\sum _{ n=0 }^{ \infty }{ n!n^{ -n }z^{ n } } $
For the ra... | For a), the series diverges $\forall \, z \ne 0$ because $\lim_{n \rightarrow \infty} (2 + \sqrt{n}) = \infty$.
For b)
$$\lim_{n \rightarrow \infty} \left ( 1-\frac{1}{n} \right )^n = \frac{1}{e}$$
so by the comparison test with a geometric series, the radius of convergence is $1$, i.e. the series converges only when |... | {
"language": "en",
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"source": "stackexchange",
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Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ for all $n \le x \le n+1$, using the mean value theorem applied to $f(x) = \frac{1}{x}$
Immediately, I can recognize some components of the mean value theorem. $\frac{1}{n^2}$ likely comes from the slope of $f$ ... | Assume $n>0$.
By mean value theorem, there exists $c\in(n,n+1)$ such that
$$\frac{1}{n}-\frac{1}{x}=\frac{1}{c^2}(x-n)$$
Since $c>n$, we get $c^2>n^2$. So
$$\left|\frac{1}{n}-\frac{1}{x}\right|<\frac{x-n}{n^2}\le\frac{1}{n^2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Finding $\int^1_0 \frac{\log(1+x)}{x}dx$ without series expansion I was trying to evaluate $$\int^1_0 \frac{\log(1+x)}{x}dx.$$
I expanded $\log(1+x) $ as
$x -\frac{x^2}{2}... $ and got the answer. I would like to know if there is any way to do it without series expanding.
| With $\int^1_0 \frac{\ln(1+x)}{x}dx
= 2\int^1_0 \frac{\ln(1+x^2)}{x}dx
= 3\int^1_0 \frac{\ln(1+x^3)}{x}dx $
\begin{align}
&\int^1_0 \frac{\ln(1+x)}{x}dx\\= &\ \frac67\int^1_0 {\frac{\ln\frac{(1+x^2)(1+x)} {(1+x^3)}}{x} } dx
=\frac67\int^1_0 \frac{\ln\frac{1+x^2}{1-x+x^2}}{x}dx \\= &\ \frac67\int_0^1\int_0^{\frac\pi6}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 1
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Find $\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ I have to find $\displaystyle\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ using Taylor polynomials.
Here's what I've done so far:
*
*$e^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + o(x^3)$
*$\sin x = x - \frac{1}{6} x^3 + o(x^4)$
*$e... | Let's rewrite the limit as
$$\displaystyle\lim_{x \rightarrow 0} e^x\times\lim_{x \rightarrow 0}\frac{e^{\sin x-x} - 1}{\sin x-x}\times\lim_{x \rightarrow 0}\left(\frac{2x}{\sin 2x}\right)^3\times \lim_{x \rightarrow 0}\frac{\sin x-x}{8x^3}=\lim_{x \rightarrow 0}\frac{x-x^3/6+O(x^5)-x}{8x^3}=-\frac{1}{48}$$
Q.E.D.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum of $\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$ I would like to know the minimum value of $$\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$$
for $x \in \mathbb{R}$.
| Note that $-1\le\sin x\le1$ for all real $x$. This allows you to rewrite each of the $4$ summands without the absolute value bars. For example: $\sin x-3\leq 1-3=-2<0$, so $|\sin x-3|=3-\sin x$; and $\sin x+1\geq -1+1=0$, so $|\sin x+1|=\sin x+1$. Simplify the resulting expression, and then use the fact that $\sin x$ r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/289185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
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Equation of a line in homogenous coordinates given 2 points in affine coordinates So if I have 2 points $A$ and $B$ such that $F(A) = (1; a, a^3)$, and $F(B) = (1; b, b^3)$. how do I find the equation of this line in homogeneous coordinates?
So I know how to get a line the "normal" way. If I take the points $a$ and $b... | In planar geometry the line passing through two points A and B in homogeneous coordinates is found from their vector cross product:
$$ \begin{pmatrix} C_0 \\ C_1 \\ C_2 \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix} \times \begin{pmatrix} B_0 \\ B_1 \\ B_2 \end{pmatrix} = \begin{pmatrix} A_1 B_2 - A_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find five positive integers whose reciprocals sum to $1$ Find a positive integer solution $(x,y,z,a,b)$ for which
$$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$
Is your answer the only solution? If so, show why.
I was surprised that a teacher would assign this kind of problem to a 5th gr... | There are many ways to obtain many solutions. Here is one systematic way to obtain solutions.
First look at a class of solutions such that $x \leq y \leq z \leq a \leq b \leq 10$ (other solution can be obtained as permutation of these).
This implies that $2 \leq x \leq 5$. So now start with $x=2$. This now means that $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "373",
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"answer_id": 8
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How to find real roots? $4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$ How to find real roots?
$$4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$$
I try LMVT but vary difficult
| The following argument is admittedly on the intuitive side, using concavity mixed with "flatter than" comparisons.
If $x<0$ then since $x^2>0$ we have
$$5^x-4^x<0<6^{x^2}-5^{x^2}.$$
If $0<x<1$ then in fact $0<x^2<x<1$, and the functions $u^x$ and $u^{x^2}$ are both concave down, with $u^{x^2}$ being flatter between 5 a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Combinatorics. Please explain me how to do it. In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively? This isn't a homework question. Please explain me the steps. Thank you!
| If $A$ gives $a$ coins, clearly, $0\le a\le 6$
and $B+C=10-a$
Now, $0\le B\le 7\implies 0\le 10-a-C\le 7\implies 3-a\le C\le 10-a $
Also, $0\le C\le 8\implies$ max $(3-a,0)\le C\le $ min$(10-a,8)$
If $a=0,$ max $(3,0)\le C\le $ min$(10,8)\implies 3\le C\le 8$ so $C$ can assume $8-3+1=6$ values.
Similarly, for $a=1,2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Square root of surds: $\sqrt{12+2\sqrt{6}}$? I got this question
Find the square root of $12+2\sqrt{6}$ expressing your answer in the form $\sqrt{m}+\sqrt{n}$.
I have no idea what this means and how to go about it.
| Let $x = \sqrt{12 + 2\sqrt{6}} = \sqrt{n} + \sqrt{m}$. Then $x^2 = 12 + 2\sqrt 6 = n + m + 2 \sqrt{nm}$.
Find $n$ and $m$ such that $n + m = 12$ and $nm = 6$.
| {
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"url": "https://math.stackexchange.com/questions/294032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determinant of block matrix Here is a determinant of a $(k+m) \times (k+m)$ Block matrix.
\begin{align}
D=\begin{vmatrix}
a_{11} &a_{12} & \cdots & a_{1k} &0 &\cdots &0 \\
a_{21}& a_{22}& \cdots & a_{2k} & 0 &\cdots &0 \\
\vdots& \vdots & & \vdots & \vdots & &\vdots\\
a_{k1} & a_{k2} & \cdots & a_{kk} ... | You can think of it in this way: take column $k+1$ (the first one containing $b$s) and swap it with column $k$. Then swap column $k$ with column $k-1$, etc, until you have "bubbled" the original column $k+1$ all the way to the left, after a total of $k$ swaps.
Repeating this process for all $m$ columns on the right req... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/297288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Overlapping spheres Say you have two spheres that are partially overlapping. How would I find the volume of the portion of one of the spheres that is not overlapping with the other based on how far apart the two spheres are and the spheres' individual radii?
| Assume that the 2 spheres have equal radii. The volume of the intersection is given by
$$V_I = 2 \pi \int_{-a}^{-d/2} dx \: (a^2-x^2)=\frac{4 \pi}{3} a^3-\pi d \left (a^2-\frac{d^2}{12}\right)$$
where $a$ is the radius of each sphere and $d$ is the separation between the centers of the spheres. So the volume in a sph... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that $(1+1/x)^x$ is monotonic increasing How does one prove that $(1+\frac{1}{x})^x$ is monotonic increasing for any $x \in [1,\infty)$?
Thanks a million!
| $$f(x) = \left ( 1+ \frac{1}{x} \right )^x$$
$$\log{f(x)} = x \log{\left( 1+ \frac{1}{x} \right )}$$
$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{x}{1+\frac{1}{x}}\left (\frac{-1}{x^2} \right ) = \log{\left( 1+ \frac{1}{x} \right )} - \frac{1/x}{1+ \frac{1}{x}}$$
or
$$\frac{d}{dx}\log{f(x)} =... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate $\int\frac{\cos x}{\sin x + \cos x}\,\text{d}x$. I'm trying to figure out how to take this indefinite integral:
$$ \int\frac{\cos x}{\sin x + \cos x}\,\text{d}x.$$
I tried simplifying and rearranging it, and this is the best I got:
$$\int\frac{1}{\tan x + 1 }\,\text{d}x.$$
But I still can't figure out how to i... | $\displaystyle \int \frac{1}{1+ \tan x} \ dx $
$ \displaystyle = \int \frac{1}{1+u} \frac{1}{1+u^{2}} \ du$ (let $u = \tan x$)
$ \displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} + \frac{1-u}{1+u^{2}} \right) \ du$
$ \displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} - \frac{u}{1+u^{2}} + \frac{1}{1+u^{2}} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/298461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Finding limit of a function ($0$ appears in denominator) This is the function:
$$f(x)= \frac{x^3 + 3x^2+2x}{x^2-2x-8} $$
So I tried to find this limit:
$$\lim_{x\to -2}\frac{x^3 + 3x^2+2x}{x^2-2x-8}=\left[\frac{-8-12-4}{4+4-8}=\frac{-24}{0}\right]= \lim_{x\to -2}\frac{x(x-1)(x-2)}{(x-4)(x+2)}$$
This isn't a case where ... | I assume you meant $$f(x)=\frac{x^3+3x^2+2x}{x^2-2x-8}$$
Then $\lim_{x\to2-}f(x)$ is a indeterminate form or type $\frac{0}{0}$, therefore we apply l'Hopital rule and receive
$$\lim_{x\to2-}f(x)=\lim_{x\to2-}\frac{3x^2+6x+2}{2x-2}=\frac{3\cdot4-2\cdot6+2}{-2\cdot2-2}=\frac{2}{-6}=-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find the exact value of $ \cos(36^\circ) $? The problem reads as follows:
Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of
$$\cos(36^\circ)$$
it says that you may find useful the following identities:
$$\cos^2 t+\sin^2 t = 1,\\
\sin 2t = 2\sin t\cos t,\\
\sin 3t = 3\sin t - 4\sin... | To find $\cos{\pi/5}$, note that
$$\sin{(3 \pi/5)} = \sin{(2 \pi/5)}$$
and
$$\sin{(3 \pi/5)} = 3 \sin{(\pi/5)} - 4 \sin^3{(\pi/5)} = 2 \sin{(\pi/5)} \cos{(\pi/5)}$$
Thus
$$2 \cos{(\pi/5)} = 3 - 4 \sin^2{(\pi/5)} = 4 \cos^2{(\pi/5)} - 1$$
Let $y=\cos{(\pi/5)}$. Then
$$4 y^2-2 y-1=0 \implies y = \frac{1 + \sqrt{5}}{4}$$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ Prove by Mathematical Induction . . .
$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:
$(1)(1!) = (1+1)!-1$
$1 = (2\cdot1)-1$
$1 = 1 \checkmark$
b. Inducti... | Your LHS may not look much like your RHS yet, but that's because you haven't finished getting it into the simplest possible form. You have $(k+1)k! - 1 + (k+1)^2 k!$. You're looking to get something minus $1$, so that's somewhat promising. Now what factors do the other two terms (the ones involving $k$) have in comm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$. If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$.
Students and I are unsure how to go about this problem. Also will this be a problem I can solve and ... | The resultant
$$h(z)=\text{Res}( \ \text{Res} ( \ z-xy, \ x^4+x^3-1, \ x), \ y^4+y^3-1, \ y)$$
is a $16^{th}$degree polynomial in $z$ with roots $\{a_ia_j:i,j=1,2,3,4\}$, where $a_1,a_2,a_3,a_4$ are the roots of $x^4+x^3-1$.
By expanding the above resultant we get that the polynomial $h(z)$ is $$h(z)= \left( {z}^{4}-... | {
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"timestamp": "2023-03-29T00:00:00",
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The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
| Of course, you could just sum the series:
$$\sum_{k=1}^n (2 k-1)^2 = \frac{4 n (n+1)(n-1)}{3} + n$$
In this case, $n=507$, and you want $507 (4 \cdot 169 \cdot 506 + 1) \mod{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
} |
How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$ If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them.
| We have $a^2 + b^2 - a(a+b) = b^2 - ab = -b (a-b)$ and $a^2 + b^2 - b(a+b) = a^2 - ba = a (a-b)$.
So if $d$ divides both $a+b$ and $a^2+b^2$, then $d$ divides $$\gcd(a (a-b), b (a-b)) = \gcd(a, b) (a-b) = a - b.$$
So $d$ divides $a+b + a - b = 2a$ and $a+b - (a - b) = 2b$.
So $d$ divides $2\gcd(a,b)=2$.
So the possib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
} |
Calculation of polynomial $g(x)$ satisfies $x\cdot g(x+1)=(x-3)\cdot g(x)$
If a polynomial $g(x)$ satisfies $x\cdot g(x+1)=(x-3)\cdot g(x)$ for all $x$, and $g(3)=6$, then $g(25)=$?
My try: $x\cdot g(x+1)=(x-3)\cdot g(x)$,
Put $x=3$, we get $g(4)=0$, means $(x-4)$ is a factor of $g(x)$.
Similarly put $x=0$. We get $g... | A repeated use of $g(x+1)=\dfrac{x-3}x g(x)$ gives
$$\begin{array} gg(25)&=&\dfrac{21}{24} g(24)\\ &=&\dfrac{22}{24}\dfrac{21}{23} g(23)\\&=&\cdots \\ &=&\dfrac{21}{24}\dfrac{20}{23}\cdots \frac{1}{4} g(3)\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Values of $a$ for which $(a+4)x^2-2ax+2a-6 <0$ for all $x \in R$ How can we find all values of $a$ for which the inequality $(a+4)x^2-2ax+2a-6 <0$ is satisfied for all $x \in R$?
For the given condition, $D >0$, therefore $ (-2a)^2-4(2a-6)(a+4) >0$. Solving for $a$, I get $(a+6)(a-4) <0$, but the answer is $(-\infty, -... | Let $f(x)=(a+4)x^2-2ax+2a-6$. Then $f(x)<0$ for all real $x$ precisely when $a<-4$ and $(2a)^2-4(a+4)(2a-6)<0$. The last inequality reduces to $(a+6)(a-4)>0$, so that $a<-6$ or $a>4$. Putting these together, we have the condition $a<-6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Geodesic on a surface of revolution using Christoffel Symbols. I have the following problem:
For a function $f:[a,b]\rightarrow \mathbb{R}_{>0}$ and for the open set $U=\{(u_1,u_2)\vert\: a<u_1<u_2, 0\leq u_2<2\pi\}$ consider the (local) surface of revolution $M$ obtained as the image of $\sigma:U\rightarrow U'\subset\... | Thank you THW, although I ended up doing it another way:
First of all, lets compute the first fundemental form and its inverse:
\begin{equation}
(g_{ij}) = \begin{pmatrix}
e & f \\
f & g
\end{pmatrix} = \begin{pmatrix}
f'(u_1)^2+1 & 0 \\
0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/309069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A few definite integrals a) $\displaystyle \int_{-1}^0 x \sqrt{2 - 5x}dx. $
I think I should use some change of variables here, but it didn't work the way I tried to use it.
b) $\displaystyle \int_{0}^{2} \frac{1}{(x + 2)(x + 3)}dx.$
| Another way is by a rationalizing substitution:
$$
\begin{align}
u & = \sqrt{2-5x} \\[8pt]
u^2 & = 2 - 5x \\[8pt]
2u\,du & = -5\,dx \\[8pt]
\frac{-2}{5} u \, du & = dx \\[8pt]
x & = \frac{2-u^2}{5}
\end{align}
$$
When $x=-1$ then $u=\sqrt{7}$ and when $x=0$ then $u=\sqrt{2}$. So
$$
\int_{-1}^0 x \sqrt{2-5x} \, dx = \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/309392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.