Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What will be the minimum value of $\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}$? What will be the minimum value of
$$\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}$$ if
$$p+q+r+s=5$$ where $p, q, r, s... | $(\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ) \left ( \frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ} \right ) \geq {(p+q+r+s)}^2=5^2$
From Cauchy–Schwarz
Hence
$$\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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How to see $\sin x + \cos x$ $$\sin x + \cos x = \sqrt{2} \sin(x + \pi/4)$$
Is there an easy way to visualize this identity or to convert the left-hand side to the right-hand side?
In general, can $p \sin x + q \cos x$ for some integers $p$ and $q$ be easily expressed as $\sin($something$)$ or $\cos($something$)$? If ... | Given $p \sin x + q \cos x$, divide the expression by $\sqrt{p^2+q^2}$ to get $a \sin x + b \cos x$ for $a^2 + b^2 = 1$. Now name $a = \cos \alpha$ and $b = \sin \alpha$. Notice that what you got is the expansion of $\sin(x+\alpha)$. In the given case, $a = b = \frac{\sqrt{2}}{2}$ so $\alpha = \frac{\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 3
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Calculating $x$ from $x^3 = 2x^2 - x$, getting wrong result I am working with a text book problem, trying to find the solutions to $x^3 = 2x^2 - x$, but I keep ending up with a different result than the book gives. I'm thinking the problem lie in getting the wrong square root towards the end, seeing as the text book gi... | You can not just divide by $x$, since $x$ could be $0$.
You have $x^3=2x^2-x$, which is equivalent to:
$$0=x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$$
Hence $x=0$ or $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluate $\sum_{i=1}^n \frac{i}{n^2}$ My book asks to simplify this problem:
$$\sum_{i=1}^n\frac{i}{n^2}$$
This equals:
$$\sum_{i=1}^n\frac{1}{n^2}i$$
Now, shouldn't
$$\sum_{i=1}^n\frac{1}{n^2}=\frac{1}{n}\text{ ?}$$
Because you are summing $1/n^2$, $n$ times, so that $n / n^2 = 1/n$.
So it should be
$$\frac{1}{n}\... | It is well known that $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$. Therefore by factoring out $n^2$.
$\sum_{i = 1}^n \frac{i}{n^2} = \frac{1}{n^2}\sum_{i = 1}^n i = \frac{1}{n^2}\frac{n(n + 1)}{2} = \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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limit of a sequence $(-1)^n$ instead $1$ I should calculate the limit of a sequence. A friend told me, that the solution is $1$. But I got $(-1)^n$.
The exercise is: $\lim\limits_{n \to \infty} \frac{1}{n^2} + (-1)^n \cdot \frac{n^2}{n^2+1}$
I did following:
$$\begin{align*}
&=\frac{n^2 ((-1)^n n^2 + 1 + \frac{1}{n^2})... | The limit of a sequence indexed by $n$ cannot contain $n$, as a limit is a number, not a sequence. So your answer $(-1)^n$ as outcome of the limit cannot be correct. However by writing
$$
\frac1{n^2} + (-1)^n \frac{n^2}{n^2+1}=\frac1{n^2} + (-1)^n -(-1)^n\frac1{n^2+1}
$$
you can see that you sequence is the sum of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/187368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Partial fraction: is my final answer and computation correct? I need some help at this exercise (partial fraction): $$\int{\frac{4x + 1}{x^3(x+2)}} dx$$
First of all I calculated the roots of the denominator.
$$
x_{1, 2, 3} = 0 \\
x_4 = -2
$$
After that I assigned the roots to the partial fraction.
$$
\frac{A_1}{x}, \f... | Reduce the RHS to a common denominator
$$\begin{eqnarray*}
\frac{4x+1}{x^{3}(x+2)} &=&\frac{A_{1}}{x}+\frac{A_{2}}{x^{2}}+\frac{A_{3}}{
x^{3}}+\frac{B}{x+2} \\
&=&\frac{A_{1}x^{2}\left( x+2\right) +A_{2}x\left( x+2\right) +A_{3}\left(
x+2\right) +Bx^{3}}{x^{3}(x+2)} \\
&=&\frac{\left( A_{1}+B\right) x^{3}+\left( 2A_{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/187437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods? I have seen the Fresnel integral
$$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$
evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral. I was wondering, howe... | Let $$F(t)=\int_0^{\infty} \sin(tx^2)\ dx$$ Taking the Laplace Transform of $F(t)$: $$\mathscr{L}\{F(t)\}=\int_0^{\infty}F(t)e^{-st}\ dt$$ By being slightly less than rigorous this can be rewritten as $$\int_0^{\infty}\mathscr{L}\{\sin (tx^2)\}\ dx=\int_0^{\infty}\frac{x^2}{s^2+x^4}\ dx$$ This can be put into partial f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/187729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "75",
"answer_count": 8,
"answer_id": 0
} |
Showing that $ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $ I would like to show that:
$$ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $$
Using AM-GM inequality:
$$ \left( \frac{a+2b}{a+2c}\ri... | I will prove that this is true for $a,b,c>0$. If they're allowed to be non-positive, there are counterexamples; e.g., $(a,b,c)=(1,0,-1)$.
Since this expression is homogeneous in $a,b,c$, we may assume without loss of generality that $a+b+c=1$. Then setting $x=c-b$, $y=a-c$, $z=b-a$ yields:
$$
\left(\frac{a+2b}{a+2c}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve $x^2+x^2t^2+2xt^2+t^2=1$ for $x$. Solving
$$x^2+x^2t^2+2xt^2+t^2=1$$
for $x$ yields
$$x=\frac{1-t^2}{t^2+1}.$$
I had to use software because I could not manipulate it algebraically, and sadly the program did not tell me how it reached that solution.
What is the method?
| Collect the coefficients in $x$ to get a polynomial in $x$:
$$ (t^2+1) x^2 +(2t^2)x+(t^2-1) = 0 $$
Now solve using quadratic formula:
$$ \frac{-2t^2 \pm \sqrt{(2t^2)^2 - 4(t^2+1)(t^2-1)}}{2(t^2+1)} = \\
\frac{-2t^2 \pm 2}{2(t^2+1)} = \frac{-t^2 \pm 1}{t^2 + 1}$$
So either $x = -1$ or $x = \frac{-t^2 + 1}{t^2 + 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Linear Algebra: Change of Basis Let $A[a,b,c]$ and $B[d,e,f]$ be two non-standard bases. I have to find the $3\times3$ matrix that will convert a vector defined in terms of $A$ to $B$.
My solution is:
Let's assume a standard basis $O[e_1,e_2,e_3]$.
Now, the rotation matrix $R(A\to O)$ to change basis from $A$ to $O$ is... | Let $\alpha=\{a,b,c\}$ and $\beta=\{d,e,f\}$ be bases for $\mathbb{R}^3$ (thinking of elements of $\mathbb{R}^3$ as column vectors). Then the matrix $A=[a|b|c]$ transitions from $\alpha$-coordinates to standard-coordinates. Likewise the basis $B=[d|e|f]$ transitions from $\beta$-coordinates to the standard-coordinates.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/190097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Little help with some algebra Note: This isn't homework, I'm skipping ahead of class. Please answer all these equations, I'm deathly stuck on them.
Use the substitution method only please. (Find $x$ and $y$.)
\begin{align}
ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2... | First Equation:
$$ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2 $$
$$ax\left({\frac {a+b-a+b}{a^2-b^2}}\right)+by\left({\frac {b+a-b+a}{b^2-a^2}}\right)=2$$
$$2abx-2aby=2(a^2-b^2)$$
$$x-y=\frac{a}{b}-\frac{b}{a}$$
From this the obvious solution is $x=a/b \text{ and } y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/190420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Limit (without series expansion and l'Hôpital's rule) $$\lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2}=\frac{1}{8}$$
Any suggestion to find this limit without series expansion and l'Hôpital's rule? Thanks and regards.
Note:
WolframAlpha confirms that the result... | I guess the only way to avoid using series expansion or l'Hôpital's rule is to reduce the limit to known limits. Let $y=1/x$. Then
$$
\lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2} = \lim_{y \to 0} \frac{\ln \frac{1+\sqrt{1+y^2}}{1+\sqrt{1-y^2}}}{\left(\ln \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Inequality. $ab^2+bc^2+ca^2 \geq a+b+c.$ Using rearrangement inequalities prove the following inequality:
Let $a,b,c$ be positive real numbers satisfying $abc=1$. Prove that
$$ab^2+bc^2+ca^2 \geq a+b+c.$$
Thanks :)
| Just by Hölder inequality;
$(ab^2+bc^2+ca^2)(\frac1a+\frac1b+\frac1c)\ge(a+b+c)^2\tag1$
If $(\frac1a+\frac1b+\frac1c)\le(a+b+c)$ we're done. Otherwise;
$(ab^2+bc^2+ca^2)(a+b+c)\ge(ab+bc+ca)^2=(\frac1a+\frac1b+\frac1c)^2\tag2$
since $(a+b+c)<(\frac1a+\frac1b+\frac1c)$ we're done again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the minimum values of x, y, and z If:
\begin{aligned}
\ {n} & = \ 2x ^2 \\
\ {n} & = \ 3y^3 \\
\ {n} & = \ 5z^5 \\
\end{aligned}
Find the minimum value for $x, y,$ and $z$.
Note: $n$ is the minimum value you can get by solving the set of equations.
| I take it that $n$ is a positive integer, as are $x$, $y$, and $z$. Then $n$ must be divisible by the primes $2$, $3$, and $5$. To minimize $n$, we allow no other primes.
Let $n=2^a3^b5^c$. Since $n=2x^2$, $a$ must be odd, and $b$ and $c$ even. This is because in the prime power factorization of a perfect square, eac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Factoring the expression $x^6 + 64$ Alright, so apparently I've factored this out wrong...
$x^6 + 64 =$
$x^6 + 2^6$
Then I continued, using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ to get ...
$$(x^2)^3 + 64 =
(x^2)^3 + 4^3 =
(x^2 + 4)(x^4 - 4x^2 + 16)$$
How is this incorrect?
| A different approach is to first factorize it over complex numbers:
$$x^6+2^6=\prod_{k=0}^5(x-2\cdot e^{\pi i(1/6+k/3)}) $$
And then pair conjugate roots, yielding:
$$\prod_{k=0}^2(x-2\cdot e^{\pi i(1/6+k/3)})(x-2\cdot e^{-\pi i(1/6+k/3)})=\prod_{k=0}^2(x^2-4\cos(\pi(1/6+k/3))+4)$$
Which equals
$$(x^2-2\sqrt 3x+4)(x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$
Let $a,b,c$ be positive numbers . Prove the following inequality:
$$\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3.$$
What I tried:
I used Cauchy-Schwarz in the following form
$\sqrt{Ax}+... | It's a bit ugly (actually more than a bit), but it works.
By AM-GM we get
$$
abc = \sqrt[3]{ab^2\cdot bc^2 \cdot ca^2} \leq \frac 1 3 (ab^2 + bc^2 + ca^2)
$$
and therefore
$$
9(5a + 6b)(5b + 6c)(5c + 6a) - 11^3(a + b + c)(ab + bc + ca) =\\ 289(ab^2 + bc^2 + ca^2) + 19(a^2b + b^2c + c^2a) - 924abc =\\ 289(ab^2 + bc^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
Expressing in the form $A \sin(x + c)$
Express in the form $A\sin(x+c)$
a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$
sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
b) $\sqrt 2\sin(x-\frac\pi4)$
Can someone please explain the method used in the provi... | Solution for item b of the question (this is the item on Problem Set 1 on MITOCW 18.01SC:
$$A\sin(x+c) = A\sin x \cos c + A \sin c \cos x$$
We can see that if $A \cos c = 1$ in the first term and $A \sin c = -1$ in the second term then we will end up with $\sin x - \cos x$, so we will have shown that $f(x)$ can be writ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Prove $ 1 + 2 + 4 + 8 + \dots = -1$
Possible Duplicate:
Infinity = -1 paradox
I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said:
Do I have to?
Okay so, Let $x = 1+2+4+8+\dots$
$2x-x=x$
$2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$
Therefore, ... | The infinite series $\displaystyle 1 + 2 + 4 + 8 + \ldots$ diverges. However, the sum
$f(z) = 1 + z + z^2 + z^3 + \ldots$, which converges to $1/(1-z)$ for $|z| < 1$, has an
analytic continuation to the complex plane with the point $1$ removed, and indeed $f(2) = -1$.
So in that sense you could regard $-1$ as the valu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\cos(2\pi/n)+\cos(4\pi/n)+\cdots+\cos(2(k-1)\pi/n)=-1$ How do you prove that
$$\cos(2\pi/n)+\cos(4\pi/n)+\cdots+\cos(2(n-1)\pi/n)=-1,$$ where $n \geq 2$?
| A proof using trigonometric identity proceeds by multiplying the identity with $2 \sin\left(\frac{\pi}{n}\right)$ and using $2 \cos(\alpha) \sin(\beta) = \sin(\alpha+\beta) - \sin(\alpha-\beta)$:
$$ \begin{eqnarray}
\sum_{m=1}^{n-1} 2 \sin\left(\frac{\pi}{n} \right) \cos\left(\frac{2 \pi}{n} m\right) &=& \sum_{m=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Limit of quotients with square roots: $\lim_{x\to2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ You can't use L'Hospital's rule :S
$$\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}$$
I've tried to multiply by conjugates but ended up with a so complex equation, please help, anyone? :S
| The key is to multiply by the conjugates of both radicals, in order to eliminate what makes the problem annoying: the two radicals whose limits both go to zero. And then you see what cancels out.
$$\begin{align*}
\lim_{x \to 2} \;\frac{\sqrt{6-x} - 2}{\sqrt{3-x} - 1}
\;&=\;
\lim_{x \to 2} \;\frac{\bigl((6-x) - 4\bigr)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Calculating Average Case Complexity I am trying to find the average case complexity of a sequential search. I know that the value is calculated as follows:
Probability of the last element is $\frac{1}{2}$
Probability of the next to last is $\frac{1}{4}$
Probability of any other elements is $\frac{1}{4n-8}$
I can assum... | Let $X \in \{1,2,\dots,n\}$ be the position of the element you're searching for. This random variable has the following probability mass function
$$p_X (k) = \left\{\begin{array}{cl} \frac{1}{4 n - 8} & \textrm{if} \quad{} k \in \{1,2,\dots,n-2\}\\ \frac{1}{4} & \textrm{if} \quad{} k = n-1\\ \frac{1}{2} & \textrm{if} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$? $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\fra... | Let $H_n=\sum_{k=1}^n \frac1k$ denote the $n$-th harmonic number.
It is well known that $H_n \sim \ln n + \gamma$, where $\gamma$ is Euler-Mascheroni constant. (To be more precise, $\lim\limits_{n\to\infty} (H_n-\ln n) = \gamma$, which is what we actually need below.)
You are interested in
$$\sum_{k=1}^n \left(\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/197595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
finding the sum of the absolute values for the roots How to find the sum of the absolute values for the roots of this equation:
$$x^4-4x^3-4x^2+16x-8=0$$
| Notice that
$$\begin{align}x^4-4x^3-4x^2+16x-8 &= (x-1)^4 - 10(x-1)^2 + 1 \\
&= ((x-1)^2-5)^2-24 \end{align}$$
so you can actually calculate the roots explicitly and sum their absolute values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/200298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Let $a,b,c>0$ and $a+b+c=3$. Find the minimum of $P$. Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$$P=\frac{{a}^{2}}{\sqrt{{b}^{3}+1}}+ \frac { { b }^{ 2 } }{ \sqrt{ { c }^{ 3 }+1 } }+\frac{{c}^{2}}{\sqrt{{a}^{3}+1}}\geq \frac{3\sqrt{2}}{2}$$
| By AM-GM, C-S and Rearrangement we obtain:
$$\sum_{cyc}\frac{a^2}{\sqrt{b^3+1}}=\sum_{cyc}\frac{2\sqrt2a^2}{2\sqrt{2(b^2-b+1)(b+1)}}\geq\sum_{cyc}\frac{2\sqrt2a^2}{2(b^2-b+1)+(b+1)}=$$
$$=\sum_{cyc}\frac{2\sqrt2a^2}{2b^2-b+3}=\sum_{cyc}\frac{2\sqrt2a^4}{2a^2b^2-a^2b+3a^2}\geq\frac{2\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/203309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric proof If $4 \tan(\alpha - \beta) = 3 \tan \alpha $, then prove that
$$\tan \beta = \frac{\sin(2 \alpha)}{7 + \cos(2 \alpha)}$$
This is not homework and I've tried everything so I would just like a straight answer thank you in advance.
| Expanding out the left side of the first equation,
$$ \frac{4 \tan \alpha - 4 \tan \beta}{1 + \tan \alpha \tan \beta} = 3 \tan \alpha$$
Thus
$$ \tan \beta = \frac{\tan \alpha}{4 + 3 \tan^2 \alpha}$$
Writing $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha}$, this becomes
$$ \tan \beta = \frac{\sin \alpha \cos \alpha}{4 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/206421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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which of the following is true, number theory problem
$a$ is correct as $n=2m+1 \text{(say)} \Rightarrow n^2=4m^2+4m+1\equiv1(\mod 8),\ m\ge1$
$b$ is true ingeneral, say $n=5, m=3$, for $c$ I dont know how to prove, please help
| (a)$(2a+1)^2-1=4a^2+4a+1=8\frac{a(a+1)}2$
(b)$(2x+1)^2+(2y+1)^2=4(x^2+y^2+x+y)+2\equiv2\pmod 4$
But $(2z)^2\equiv 0\pmod 4,(2z\pm1)^2\equiv1\pmod 4$
(c)$$\frac{n^5}5+\frac{n^3}5+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+\frac{7n}{15}+\frac n 5+\frac n 3$$
$$=\frac{n^5-n}5+\frac{n^3-n}3+n$$
Now using Femat's little theo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
$$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$
by Mathematical induction，n$\ge $1
| EDIT: This approach is unfortunately wrong, as pointed out in the comments.
Let $S(n) = \frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}.$
(1) Base step. Let $n=1$. Then $n+1 = 2n$, so the series only has one term. $S(1) = \frac{1}{2} < \frac{25}{36}$, hence the statement is true for $n=1$.
(2) Inductive step. Suppose... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 7
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When is $\left\lfloor \frac {7^n}{2^n} \right\rfloor \bmod {2^n} \ne 0\;$? Is
$$\left\lfloor \frac {7^n}{2^n} \right\rfloor \bmod{2^n} \ne 0\;$$
always true when $n \ge 3$.
Baker's theorem on transcendental numbers that provide bounds for diophantine equations may be useful, but I will leave that to the experts.
| The following may be a starting point:
Write $7^n \mod 4^n$:
$$ 7^n =a\cdot 4^n + b \quad b<4^n $$
$$ \left\lfloor \frac{7^n}{2^n} \right\rfloor \equiv \left\lfloor \frac b{2^n} \right\rfloor \pmod{2^n}$$
and this is $0$ iff $b<2^n$.
Well, $7^n=(2^3-1)^n=\displaystyle\sum_{0\le i\le n} \binom{n}{i} 2^{3i} \cdot (-1)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/214319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Finding the limit $ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2}$ I cannot find the following limit:
$$ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2} \, . $$
Please, help me.
| $$\cos(x) \cos(2x) = \left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots \right)\left(1 - \dfrac{(2x)^2}{2!} + \dfrac{(2x)^4}{4!} - \cdots \right)\\ = 1 - \dfrac{x^2 + 4x^2}{2!} + \mathcal{O}(x^4)$$
Hence, $$1 - \cos(x) \cos(2x) = \dfrac52 x^2 + \mathcal{O}(x^4)$$
Can you now find the limit?
EDIT
Alternately, you coul... | {
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"answer_id": 2
} |
Coefficent of an equation I am suppose to find the coefficent of x^2 in this equation
after doing the calculation I ended up with this
but that is the wrong answer what is that I am missing?
| It is actually a bit less confusing (you don't have to expand the brackets with three terms) if you write
$$
f(f(x))=(x^2+bx+c)^2+b(x^2+bx+c)+c=(x^2+bx+c)(x^2+bx+c+(b+c))
=x^4+bx^3+(c+b)x^2+bx^3+b^2 x^2+b(c+b)x+cx^2+bcx+c(b+c)+c
$$
The it is clear tha the coefficient at $x^2$ is $c+b+b^2+c=2 \cdot 15 -23+23^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\frac{a+b}{b+c} + \frac{c+d}{d+a} ≤ 4(\frac{a+c}{b+d}) ; a , b , c , d ∈ [1 , 2]$ Is it true that
if all the real numbers $a , b , c , d$ are from the closed interval $[1 , 2]$ then we always have the inequality
$$
\frac{a+b}{b+c} + \frac{c+d}{d+a} ≤ 4\Big(\frac{a+c}{b+d}\Big)
$$
| Since $a, c \geq 1$, we have
$$
a - \frac {a + b} {b + c} = \frac{ab + ac - a - b} {b + c} = \frac {b(a - 1) + a(c - 1)} {b + c} \geq 0
$$
A similar argument leads to
$$
c - \frac{c + d} {d + a} \geq 0
$$
Finally $b + d\leq 4$ implies
$$
\frac {a + b} {b + c} + \frac{c + d} {d + a} \leq a + c \leq \frac 4 {b + d} (a + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Where do I go wrong when solving this integral involving a quadratic? Simple question. I have an arithmetic mistake somewhere in here but can't find it.
$\int \frac{3x-1}{x^2+10x+28}dx$
$\frac{1}{3}\int \frac{3x-1}{\frac{1}{3}(x+5)^2+1}dx$
Set:
$u=x+5$
$x=u-5$
$dx = du$
$\frac{1}{3}\int \frac{3u-16}{\frac{u^2}{3}+1}du... | Note that $\ln(g(x)/c)=ln(g(x))-ln(c)$ for any function $g(x)$. Since you're finding an antiderivative, any extra added/subtracted constants are irrelevant.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sides of triangle and an altitude Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Let $h$ be the altitude drawn on the side of length $a$ Then is $a^2 + 4h^2 - (b+c)^2$ always negative ?
| I also have been able to come up with a proof that the expression indeed is negative ;
$-(b-c)^2 ≤ 0 $
$\implies a^2 - (b-c)^2 ≤ a^2 $
$\implies (a+b-c)(a-b+c) ≤ a^2 $
$\implies (a+b+c)(b+c-a)(a+b-c)(a-b+c) ≤ a^2 ( a+b+c)(b+c-a) $
{since $a+b+c >0$ and for a triangle $b+c > a$ , the multiplication in the above line ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/216804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the greatest powers of $2$ dividing $10!$, $20!$, $30!$, $40!$ I'm trying to find the greatest powers of $2$ dividing $10!$, $20!$, $30!$, $40!$, as part of a basic number systems course.
I'm rather lost with this question. For $10!$ I tried writing the terms out and just extracting powers of $2$ manually, getting... | For any prime $p$ and integer $n$, let
$$
n=d_0+d_1p+d_2p^2+d_3p^3+\dots+d_kp^k\tag{1}
$$
where $0\le d_j\lt p$. $(1)$ is the base-$p$ representation of $n$.
The number of multiples of $p$ not greater than $n$ would be
$$
\left\lfloor\frac np\right\rfloor=d_1p^0+d_2p^1+d_3p^2+d_4p^3+\dots+d_kp^{k-1}\tag{2}
$$
The numbe... | {
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$a+b=c \times d$ and $a\times b = c + d$ There is a 'nice' relationship between the integers (1,5) and (2,3) as
$$1+5=2 \times 3;$$
$$1\times 5 = 2 + 3.$$
So I tried to find all positive integers pairs $(a, b)$ and $(c, d)$ such that $$a+b=c \times d;$$
$$a\times b = c + d.$$
To find this, $a, b, c, d$ must satisfy $... | Define $r$, $s$ by $(x-a)(x-b)=x^2-rx+s$. Then $r=a+b=cd$ and $s=ab=c+d$, so $(x-c)(x-d)=x^2-sx+r$. Thus, the polynomials $x^2-rx+s$ and $x^2-sx+r$ both have only integer roots. Thus, $r^2-4s$ and $s^2-4r$ are both squares: $r^2-4s=u^2,s^2-4r=v^2$.
EDIT: Continuing, from $r^2-4s=u^2$ we get $4s=r^2-u^2\ge r^2-(r-2)^2=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating a partial fraction How do I compute the integral
$$\int_0^{\pi/2} \cfrac 1 {12\cos x + 9 \sin x} dx?$$
Please can you help.
| If all else fails, the Weierstrass substitution will do it.
However, let's try something that begins with what Marvin suggested in his answer. First notice that $12^2+9^2=15^2$, so $\left(\frac{12}{15}\right)^2+\left(\frac{9}{15}\right)^2=1$. Hence we can find $\theta$ such that $\cos\theta=12/15$ and $\sin\theta=9/1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Contour integration of $\int_{-\infty}^{\infty} \frac{1-b+x^2}{(1-b+x^2)^2 + 4bx^2}dx = \pi$ Given $0< b <1$, derive the equality:
$\displaystyle \int_{-\infty}^{\infty} \frac{1-b+x^2}{(1-b+x^2)^2 + 4bx^2}dx = \pi$
by integrating the function $(1+z^2)^{-1}$ around the rectangle with vertices $R, -R, R+i\sqrt{b}, -R+i\... | Note
$$ (1-b+x^2)^2+4bx^2=(1-b+x^2+2\sqrt{b}xi)(1-b+x^2-2\sqrt{b}xi)=[(x+\sqrt bi)^2+1][(x+\sqrt bi)^2+1] $$
and hence
$$ \frac{1-b+x^2}{(1-b+x^2)^2+4bx^2}=\frac12\left(\frac{1}{(x+\sqrt bi)^2+1}+\frac{1}{(x+\sqrt bi)^2+1}\right). $$
It is easy to use contour integral to check
$$ \int_{-\infty}^\infty \frac{1}{(x+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/226567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Partial Fractions continued... Hi asked the following question yesterday: Obtaining the sum of a series
Given the answers to that question by wj32, I am now trying to solve the following problem:
Consider the series
$$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$
Use partial fractions to write the general term
$$u_n=\frac{n... | First note that we have a nice factorization of $n^4 + n^2 + 1$.
$$n^4 + n^2 + 1 =(n^2+n+1)(n^2-n+1)$$
Hence, $n = \dfrac{(n^2+n+1) - (n^2-n+1)}2$. This gives us $$u_n = \dfrac{(n^2+n+1) - (n^2-n+1)}{2(n^2+n+1)(n^2-n+1)} = \dfrac12 \left(\dfrac1{n^2-n+1} - \dfrac1{n^2+n+1}\right) = \dfrac12 \left(\dfrac1{(n-1)n+1} - \d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\log_9 71$ or $\log_8 61$ I am trying to know which one is bigger :$$\log_9 71$$ or $$\log_8 61$$ how can i know without using a calculator ?
| If we are allowed to use calculus, we can get a somewhat formal estimate of both numbers: $$\log_9 (71) = \log_9 (81) + \log_9(71/81) = 2 + \log_9 \left(1 - \frac{10}{81}\right) \approx 2 - \frac{1}{2 \ln 3}\left(\frac{10}{81}\right), \\ \log_8 (61) = \log_8 (64) + \log_8 (61/64) = 2 + \log_8 \left(1 - \frac{3}{64}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/227642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 3
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Squaring inequality $a>b+c$ Given $a,b,c > 0$ and $a > b + c$ , is it true that $a^2 > b^2 + c^2$?
Im tried proving it. I followed the below steps and not sure whether im right or wrong?
Squaring on both sides
$\implies a^2 > (b+c)^2$
$\implies a^2 > b^2 + c^2 + 2*b*c$
since $b,c > 0 $, we have $2*b*c > 0$, then
$\impl... | Another way of proving is ; since $a$ , $b$ , $c$ $>$ $0$ then $a$ $>$ $b$ $+$ $c$ implies $a$ $>$ $b$ and $a$ $>$ $c$ ; so $ab$ $>$ $b^2$ and $ca$ $>$ $c^2$
, and $a$ $>$ $b$ $+$ $c$ $\implies$ $a^2$ $>$ $ab$ $+$ $ca$ $>$ $b^2$ $+$ $c^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to represent fraction $\frac{1}{a+b\cdot \sqrt[3]{2} + c\cdot (\sqrt[3]{2})^2}$ as $a_1 + b_1 \cdot \sqrt[3]{2} + c_1 \cdot (\sqrt[3]{2})^2$? What do I have to multiply both numerator and denominator with to get the representation is asked? $a, b, c, a_1, b_1, c_1$ are rational numbers.
| I think, the easiest if you solve this:
$$(a_1+b_1\cdot \sqrt[3]2 + c_1\sqrt[3]{2^2})(a+b\cdot \sqrt[3]2 + c\sqrt[3]{2^2}) =1 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorise $f(x)=x^5-x$ into a product of irreducibles in $\mathbb Z_{5}[x]$ So plugging in $1$ gives $f(1) = 0$ which means $1$ is a root and $f$ has a factor $(x-1)$ which is $\equiv (x+4)$ in $\mathbb Z_{5}[x]$ ?
I then divide $f(x)$ by $(x+4)$ using polynomial long division and I get
$x^4-4x^3+16x^2-64x+255$ with r... | $f(x) = x (x-1)(x^3+x^2 + x+1)=x(x-1)(x+1)(x^2+1)$
$(x+a)(x+b)=x^2 + (a+b)x + ab = x^2 +1$ in ${\bf Z}_5[x]$
So $a+b =0(5)$ and $ab=1(5)$ Hence $a=2$ and $b=3$
$ f(x) =x(x-1)(x+1)(x+2)(x+3)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}$
Determine
$$\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}.$$
Multiplying and dividing by $\sqrt{x}$ yields
$$\frac{(x+7)^2\sqrt{x^2+2x}}{7x^3-2x^2}$$
where I would like to approximate the squareroot for sufficientl... | Hint: A standard thing to do is to divide top and bottom by $x^2\sqrt{x}$.
The new top is
$$\left(1+\frac{7}{x}\right)^2 \sqrt{1+\frac{2}{x}},$$
and its behaviour for large $x$ is clear.
The new bottom is $7$ plus something tiny.
Remark: In the solution given in the OP, $\sqrt{x^2+2x}$ was replaced by $x+1$. True, th... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $1/x^n$ is differentiable Consider the function $g(x)=\frac{1}{x^n}$ where $n \in\Bbb N$, prove that g is differentiable.
I tried to use the definition,
Let $c \in\Bbb R$, then:
$$\frac{g(x)-g(c)}{x-c}=\frac{\frac{1}{x^n}-\frac{1}{c^n}}{x-c}=\frac{c^n-x^n}{(x-c)(x^n \cdot c^n)}=\frac{(c-x)^n}{(x-c)(x^n \cdot c^n)... | You should evidently assume $x\neq 0$.
Then, for $|h|<\delta$ so that $x+h\neq 0$
$$\begin{align}
\mathop {\lim }\limits_{h \to 0} \frac{{\dfrac{1}{{{{\left( {x + h} \right)}^n}}} - \dfrac{1}{{{x^n}}}}}{h} &= \mathop {\lim }\limits_{h \to 0} \frac{{\dfrac{{{x^n} - {{\left( {x + h} \right)}^n}}}{{{x^n}{{\left( {x + h... | {
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How to find the other vertices of an equilateral triangle given one vertex and centroid If I know the coordinates of the center and one vertex of an equilateral triangle, how do I find the coordinates of the other vertices?
I'm thinking I need to find (x,y) such that the distance to the known vertex is the square root ... | Suppose we have centroid $M = (x_0,\ y_0)$ and vertex $A=(x_1,\ y_1)$.
First let us center the triangle at the origin with shifted vertex $$A' = (x_1',\ y_1') = (x_1 - x_0,\ y_1 - y_0)$$
The other vertices will be reached from this one by a rotation about the origin $120^\circ$ clockwise and counter-clockwise. The cou... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\cot^2 (2x) + \cos^2 (2x) + \sin^2 (2x) = \csc^2 (2x)$ I'm having massive issues proving this identity:
$$\cot^2 (2x) + \cos^2 (2x) + \sin^2 (2x) = \csc^2 (2x)$$
How is this proven?
| Recall the following identities
$$\cos^2{\alpha} + \sin^2{\alpha} = 1$$
$$\cot^2{\beta} + 1 = \csc^2 \beta$$
Hence, $$\cot^2 (2x) + \cos^2 (2x) + \sin^2 (2x) = \underbrace{\cot^2 (2x) + \underbrace{\left(\cos^2 (2x) + \sin^2 (2x) \right)}_{=1}}_{\cot^2(2x)+1 = \csc^2(2x)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solving for unknown When solving for unknowns why does the symbols sometimes change from $a + $ to $a -$ or vice versa or stays the same?
For example (using solutions from text):
$2x+6 = x+16\quad$ and the solution is $2x-x = 16-6\quad \Rightarrow \quad x=10$.
Or
$8x-12 = -3x+21\quad $Solution: $8x+3x = 21+12\quad \Ri... |
For example: $2x+6 = x+16$, and the solution is $2x-x = 16-6\Rightarrow x=10$.
Here, let's start by adding $-x$ to both sides of the equation $2x +6 = x+16$: $$-x + 2x+6 = -x + x + 16$$ $$\leftrightarrow x+6 = 16.$$
Next: Adding $-6\,$ to each side of the equation gives us: $$x + 6 + (-6) = 16 + -6$$ $$\leftrightarr... | {
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"answer_id": 1
} |
Why ${ \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent , but ${ \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent? I don't understand why ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent, but ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent and its limit i... | We can see that the harmonic series, $\displaystyle\sum_{k=1}^\infty\frac1k$, diverges using the classical observation:
$$
\frac11+\frac12+\underbrace{\frac13+\frac14}_{\mbox{$2$ terms}}+\underbrace{\frac15+\frac16+\frac17+\frac18}_{\mbox{$4$ terms}}+\dots+\underbrace{\frac1{2^n+1}+\dots+\frac1{2^{n+1}}}_{\mbox{$2^n$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/243482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 4,
"answer_id": 2
} |
Prove that if $ |a_n-a_{n-1}| < \frac{1}{2^{n+1} }$ and $a_0=\frac12$, then $\{a_n\}$ converges to $0I try to solve this question but I don't know how.
given $ a_0 = \frac12 $ and for each $n\geq 1$:
$$ |a_n-a_{n-1}| < \frac{1}{2^{n+1}} $$
show that $\{a_n\}$ converges and the limit is $a$ such that $0<a<1$
Update (Edi... | You can also prove this generalization the same way:
if $|a_{n+1}-a_n| < c_n$
where $c_n$ is decreasing
and $\sum_{n=1}^{\infty} c_n$ converges,
then $\lim_{n \to \infty} a_n$ exists
and is less than
$a_1+\sum_{n=1}^{\infty} c_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/243761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the sum of a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ I was thinking about this problem:
Given a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ ,how can i show that its sum is a/an rational/irrational number,given that the series converges?
Could someone point me in the right direction? Thanks everyone in a... | $1/(n-1)-1/(n+1) = 2/(n^2-1)
$, so
$$\sum_{n=4}^{\infty} \frac{1}{n^2-1}
= (1/2)\sum_{n=4}^{\infty} \left(\frac{1}{n-1} - \frac{1}{n+1}\right)
= \frac{1}{3} + \frac{1}{4}
$$
since all the later terms are cancelled out.
Whoops - as pointed out by Limitless, this should be
$\frac{1}{2}\left(\frac{1}{3} + \frac{1}{4}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/249040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ has real solutions iff $7a^2 - 50a + 7\leq 0 $ How can we show that $(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ has real solutions if and only if $7a^2 - 50a + 7\leq 0$?
I know these are quadratics and can solve them, but I'm not entirely sure what the question is asking of me and how to lay... | From the quadratic formula you know that the solutions of
$$(a-1)x^2+3(a+1)x+4(a-1)=0$$
are given by
$$x=\frac{-3(a+1)\pm\sqrt{9(a+1)^2-16(a-1)^2}}{2(a-1)}\;.$$
These will be real if and only if
$$9(a+1)^2-16(a-1)^2\ge 0\;.$$
Expanding the lefthand side, we see that this inequality reduces to
$$-7a^2+50a-7\ge 0\;.$$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/250361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Taylor Polynomial of $\arctan$ of given Degree and Error
Replace the following function by its taylor polynomial of the given
grade, and approximate the error in the given interval:
$$f(x) = \arctan(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$$
My solution and thoughts
We only need the first three... | This should be considered a minor addition to the answer by Mike Spivey.
Note that the fourth derivative is $0$ at $0$. So the Taylor polynomial that we get by expanding until the third derivative is exactly the same as the Taylor polynomial we get by expanding until the fourth derivative!
Thus you can use the formu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/252360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\lim\limits_{x\rightarrow+\infty}\left[(x^3+x^2+1)^{1/3}- (x^2+x+1)^{0.5}\right] = -1/6$? Every method I use seem to get me to something to the extent of $0/0$, stuff I can't work with. Wolfram Alpha claims the answer to this is $-1/6$ but they offer no step by step solution.
Would appreciate any tips and help.
| Since $$\alpha-\beta=\frac{\alpha^6-\beta^6}{\alpha^5+...+\beta^5}$$
we have that
\begin{gather}\lim\limits_{x\rightarrow+\infty}({x^3+x^2+1})^{1/3}-{(x^2+x+1)}^{1/2}=\lim\limits_{x\rightarrow+\infty}x[{(1+1/x+1/{x^2})}^{1/3}-{(1+1/x+1/{x^2})^{1/2}}]=
\lim\limits_{x\rightarrow+\infty}x\frac{(1+1/x+1/{x^2})^2-(1+1/x+1/{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/254496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
5 linear equations in 5 unknowns I need an example of 5 linearly independent equations with 5 variables. How can I write such a equation set. As an example:
0 0 0 0 1 | -4
0 0 1 -1 0 | 3
1 0 0 0 1 | 2
... | Begin with a 5x5 random matrix, with positive diagonal. For example,
$$
A=\begin{pmatrix}
1&0&-3&2&4\\
5&6&7&8&-9\\
1&1&1&1&1\\
0&0&0&1&0\\
2&-3&2&-3&4
\end{pmatrix}
$$
Then compute the absolute row sums (i.e. sum up the absolute values of the entries in each row):
\begin{align}
1+0+3+2+4=10\\
5+6+7+8+9=35\\
1+1+1+1+1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the greatest value... The sum of the first three terms of a geometric progression is 21, and the sum of their squares is 1281. What is the greatest value can be the sum of the cubes?
| Let $a,ar,ar^2$ be the three terms
So, $a+ar+ar^2=21$
So, $(a)^2+(ar)^2+(ar^2)^2=1281\implies a(r^2+r+1)a(r^2-r+1)=1281$
On division, $a-ar+ar^2=\frac{1281}{21}=61$
So, $ar=-20,a+ar^2=41$
$a^3+(ar)^3+(ar^2)^3=a^3+(ar^2)^3+(-20)^3$
$=(a+ar^2)^3-3a\cdot ar^2(a+ar^2)+(-20)^3$
$=(41)^3-3(-20)^2(41)+(-20)^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/256119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Scale modified Bessel functions to then unscale later So I have some variables $\,x_{1},\, x_{2},\, \nu\, =\, 12.654,\, 13.487,\, 0\,$ and the following function:
$\dfrac{(x_{1}\cdot(-BesselK(\nu,x_{1}\cdot125))\cdot BesselI(\nu,x_{2}\cdot125))-(x_{2}\cdot BesselK(\nu+1,x_{1}\cdot 125)\cdot BesselI(\nu+1,x_{2}\cdot 125... | If $125 x_1$ and $125 x_2$ are fairly large compared to $\nu^2$, you might want to use the asymptotic forms of these Bessel functions:
$$ \eqalign{K_\nu(x) &\sim \sqrt{\frac{\pi}{2x}} e^{-x} \left(1+{\frac {4\,{\nu}^{2}-1^2}{1!\; 8x}}+{\frac {(4 \nu^2-1^2)(4 \nu^2-3^2)}{2!\; (8x)^{2}}}\right.\cr &\left.+{\frac {(4 \nu^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Bound of a function $f_n=\frac{x^2}{x^2+(1-nx)^2}$ Let $\displaystyle f_n=\frac{x^2}{x^2+(1-nx)^2}$ where ($0\le x\le 1,\; n=1,2,3,...$)
Then $|f_n(x)| \le M$.
Find this $M$.
The answer is $1$.
Without any restriction of n, how can we find that bound?
| Let $\frac{x^2}{x^2+(1-nx)^2}=y$
or $x^2(y+yn^2-1)-2nyx+y=0$
As $x$ is real, $(2ny)^2\ge 4y(y+yn^2-1)$
$\implies y^2-y\le 0\implies y(y-1)\le 0\implies 0\le y\le 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/257626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving square root of a square is the same as absolute value Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:
$\... | It is the definition of square root of a number. The square root is defined in the sense that $s(x^2) = \sqrt{ x^2 } = |x|$ for all real $x$. Thus, the domain is the real numbers and the codomain is the non-negative real numbers. The reason it is defined this way is to make sure that $s$ is a function. Assume for a min... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/258876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
} |
Finding the sum of all solutions $2x + 3y = n$ has exactly $2011$ non-negative integral solutions. Determine the SUM of the possible values of $n$.
| The integer solutions to $2x+3y=1$ are given by
$$\left\{\begin{align*}
&x=-1+3k\\
&y=1-2k
\end{align*}\right.$$
for $k\in\Bbb Z$, so the integer solutions to $2x+3y=n$ are given by
$$\left\{\begin{align*}
&x=-n+3k\\
&y=n-2k
\end{align*}\right.$$
for $k\in\Bbb Z$. Clearly $n>0$, so $x\ge 0$ iff $3k\ge n$ iff $k\ge\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/262452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to make a matrix a magic square? Suppose I have a matrix
$$\begin{pmatrix}
* & 3 & 6\\
5 & * & 5\\
4 & 7 & *
\end{pmatrix}$$
How can I find the three numbers on the main diagonal such that the sum of the numbers on every row and every column is equal (i.e., it's a magic square). Thank you.
| Let's say we put in the numbers $a$, $b$, and $c$:
$$\begin{pmatrix}
a & 3 & 6\\
5 & b & 5\\
4 & 7 & c
\end{pmatrix}$$
The sums of the rows are $a+9$, $b+10$, and $c+11$. The sums of the columns are the same (indeed, if that were not the case, it'd be impossible to make it into a magic square). Thus, we want the num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
upper and lower bounds of a complex expression How do I prove that $$\sqrt{\frac{7}{2}}\leq |1+z|+|1-z+z^2|\leq 3\sqrt{\frac{7}{6}}$$ for all complex numbers $|z|=1$?
I don't really know how to grapple with it.
P.. I am extremely sorry, the condition should actually be $|z|=1$ and it was previously incorrectly stated a... | Let $z=\cos 2t+i\sin 2t,$
$|1+z|=|1+\cos 2t+i\sin 2t|=|2\cos^2t+i2\sin t\cos t|=|2\cos t||\cos t+i\sin t|=|2\cos t|$
$|1-z+z^2|=|1-(\cos 2t+i\sin 2t)+(\cos 2t+i\sin 2t)^2$
$=|1-(\cos 2t+i\sin 2t)+(\cos 4t+i\sin 4t)|$
$=|2\cos^22t-\cos 2t+i(2\sin2t\cos2t-\sin2t)|$
$=|(2\cos2t-1)(\cos 2t+i\sin2t)|=|(2\cos2t-1)||(\cos 2t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
asymptotics of the expected number of edges of a random acyclic digraph with indegree and outdegree at most one A recent discussion, which may be found here, examined the problem of counting the number of acyclic digraphs on $n$ labelled nodes and having $k$ edges and indegree and outdegree at most one. It was establis... | We can expand this calculation to the variance in addition to the expected value. To do so we compute $$E[X(X-1)] = \frac{ [z^n] \left. \left(\frac{d}{du}\right)^2 G(z, u) \right|_{u=1}}{[z^n] G(z, 1)}$$ where $X$ is the RV representing the number of edges.
Now $$\left. \left(\frac{d}{du}\right)^2 G(z, u) \right|_{u=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
How can I find maximum and and minimum values of $ f(x,y)=xye^{-(x+y)}$? How can I find maximum and and minimum values of $ f(x,y)=xye^{-(x+y)}$ in the region $(x-1)^2+(y-1)^2 \leq4$ ?
| First look for critical points of $f$ in the interior $(x-1)^2 + (y-1)^2 < 4$. Do this by comparing the partial derivatives of $f$ to zero. Next search for critical values on the boundary $(x-1)^2 + (y-1)^2 = 4$ using Lagrange multipliers. Compute $f$ in these points and, by putting everything together, you will find t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find the square root of a permutation Observe that
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 1 & 5 & 3
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 1 & 5 & 3
\end{pmatrix} =
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
4 & 5 & 2 & 3 & 1
\end{pmatrix},$$
so $\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
4... | Any permutation can be written as a product of disjoint cycles $\sigma = c_1\cdot\ldots\cdot c_n$. Because disjoint cycles commute, $\sigma^2 = c_1^2\cdot\ldots\cdot c_n^2$.
So a square permutation is one that consists of a product of disjoint square cycles. Now when is a cycle square?
If $c=(i_1 i_2\ldots i_k)$, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 1
} |
Integral of $1/\sqrt{x^2 - a^2}$ where $a > 0$, why does $a$ has to be greater than $0$? I know how to solve the integral (set $x = a\sec(\theta)$ then $dx = a\sec(\theta)\tan(\theta)\,d\theta$ where $0 < \theta < \pi/2$ in order to have a one-to-one function), anyway, the problem specifies that $a > 0$ but I don't see... | If $x=a\sec \theta, dx=a\sec \theta\tan\theta \,d\theta $
$\sqrt{x^2-a^2}=\sqrt{a^2(\sec^2\theta-1)}=|a\tan \theta|=|a||\tan \theta|$
As $0< \theta< \frac\pi 2, |\tan \theta|=\tan \theta$
So, $\sqrt{x^2-a^2}=|a|\tan \theta$ if $0< \theta< \frac\pi 2$
$$\int \frac{dx}{\sqrt{x^2-a^2}}=\frac{a\sec\theta\tan\theta d\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Demonstrate: $ \sqrt{2}=\lim_{n\rightarrow\infty}{\sum_{i=0}^n{\frac{\left(-1\right)^i\left(-\frac{1}{2}\right)_i}{i!}}} $ Demonstrate that $\sqrt2$ can be expressed as:
$$ \sqrt{2}=\lim_{n\rightarrow\infty}{\sum_{i=0}^n{\frac{\left(-1\right)^i\left(-\frac{1}{2}\right)_i}{i!}}} $$ Where $\left(z\right)_i$ is the Pochha... | If we Taylor expand $f(x)=\sqrt{1+x}$, around $x=0$, we get
$$
\sqrt{1+x}=\sum_{k=0}^\infty \frac{(-1)^k\big(-\frac{1}{2}\big)_k}{k!}x^k=\sum_{k=0}^\infty
a_kx^k, \tag{1}
$$
and this series converges for $|x|<1$, as
$$
|a_k|=\frac{1}{k!}\cdot\frac{1}{2}\left(\frac{1}{2}\frac{3}{2}\cdots\frac{k-1-\frac{1}{2}}{2}\right)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Can the integral $\int \frac{6x-1}{3x^3-4} \, dx$ solved with a simple solution? I got the integral
$$\int \frac{6x-1}{3x^3-4} \, dx$$
as an exercise for exam preparation. I tried to get closer to the result with third binomial formula, but could not get anywhere. So I tried to find the solution with wolframalpha, but... | $$\int \frac{6x-1}{3x^3-4} \, dx$$
$$\frac{1}{4}\int \frac{6x-1}{\frac{3}{4}x^3-1} \, dx$$
$\frac{3}{4}x^3=u^3$
$\sqrt[3]{\frac{3}{4}}x=u$
$\sqrt[3]{\frac{4}{3}}u=x$
$\sqrt[3]{\frac{4}{3}}du=dx$
$$\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{6\sqrt[3]{\frac{4}{3}}u-1}{u^3-1} \, du=\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Help in manipulating Integrals I try to express : $\displaystyle 1+2\sum _{ k=1 }^n \cos(2k\theta ) $
as : $\dfrac { \sin\left( \theta +2\theta n \right) }{ \sin\left( \theta \right) } $
I tried to use the exponential function :
$$I=\sum _{ k=1 }^n \cos(2k\theta ) ;\quad J=\sum _{ k=1 }^n \sin(2k\theta) $$
$I+i\cdot... | There are many equivalent ways to proceed. Below are a couple of equivalent methods.
First method:
Let $$S = \sum_{k=1}^n \cos(2k \theta)$$
We then have
$$S \sin(\theta) = \sum_{k=1}^n \sin(\theta) \cos(2k \theta)$$
Now recall that
$$\sin(A) \cos(B) = \dfrac{\sin(A+B) - \sin(B-A)}2$$
Hence, we get that
$$S \sin(\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/271694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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For $n∈ N$, determine the real part of $(1 + i\sqrt{3})^{n}$ For $n∈ N$, determine the real part of $(1 + i\sqrt{3})^{n}$.
I just can't find the regularity within it. Thanks.
| Use De Moivre's formula
\begin{align}
\left[ 1+i\cdot \sqrt{3} \right]^n =
&
2^n \left[ \frac{1}{2}+i\cdot \frac{\sqrt{3}}{2} \right]^n
&
\\
=
&
2^n\cdot \left[\cos\left(\frac{\pi}{3}\right)+i\cdot \sin\left(\frac{\pi}{3}\right) \right]^n
&
\\
=
&
2^n\cdot \left[\cos\left(n\cdot\frac{\pi}{3}\right)+i\cdot \sin\left(n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/273099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solve this inequality $\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}$ Prove that
$ \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdot \frac{9}{10}\cdot \frac{11}{12}\cdot \frac{13}{14}...\cdot \frac{91}{92}\cdot \frac{93}{94}\cdot \frac{95}{96}\frac{97}{98}\cdot \frac{99}{100} <\frac{1}{10}$
| Multiply the numerators of the RHS and LHS by $49!2^{49}$. This converts the numerator on the LHS to $99!$. We then have $$\frac{99!}{2\cdot4\cdot8...\cdot100}<\frac{49!2^{49}}{10}$$ Next multiply the denominators on the RHS and LHS by $$\frac{100!}{50!2^{50}}$$ The formula to convert the above multiplications of even ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/274753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
calculate $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos2x}{\cos x-\sin x}$ $$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(2x)}{\cos(x)-\sin(x)}=\lim_{x\rightarrow\frac{\pi}{4}}\frac{2\cos^{2}(x)-1}{\cos(x)-\sqrt{1-\cos^{2}(x)}}$$
$$t=\cos(x)$$
$$\lim_{x\rightarrow\frac{\pi}{4}}\frac{2t^{2}-1}{t-\sqrt{1-t^{2}}}=\lim_{x\righ... | writing $\cos 2x = \cos^2 x - \sin^2x $ and expanding $a^2-b^2$ and cancelling $\cos x - \sin x$ you get $\cos x + \sin x$. Putting value $x \to \frac{\pi}{4}$ we get $ \frac{1}{\sqrt 2} + \frac{1}{\sqrt 2} = \sqrt 2$.
EDIT::
Looks like you've been writing $\sqrt{t^2-1}$ for $\sqrt{1 -t^2}$ and also there is substituti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Inequality. $\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$ Prove that :
$$\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.
I try to apply Cauchy-Buniakowski and I obtaine the followin:
$$\sum_{x,y,z}{\frac{1}{\sqrt{... | I try to apply Cauchy-Buniakowski and I obtaine the followin:
$$\sum_{x,y,z}{\frac{1}{\sqrt{x^2+yz+3}}}\cdot \sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}\geq 9$$
So I have to prove that : $$\displaystyle\frac{9}{\sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.
$$\left(\sum_{x,y,z}{\sqrt{x^2+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding all solutions of the equation $(x^2 + y^2 + z^2 − 1)^2 + (x + y + z − 3)^2 = 0$. Find all solutions of the equation
$(x^2 + y^2 + z^2 − 1)^2 + (x + y + z − 3)^2 = 0$.
not getting any clue.can somebody help me please.thanks.
| I assume we are looking for real solutions. Then our equation is equivalent to the system $x^2+y^2+z^2=1$, $x+y+z=3$.
Note that because $x^2+y^2+z^2=1$, the absolute values of $x$, $y$, and $z$ are $\le 1$, and they are not all $1$.
Thus $|x+y+z|\le |x|+|y|+|z| \lt 3$.
| {
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How do we solve the system of equations?18-1-2013 Giải hệ phương trình:
(Editor's note: This translates to "Solve a system of equations:")
$$ \begin{cases}\sqrt[4]{x}\left(\dfrac{1}{4}+\dfrac{2\sqrt{x}+\sqrt{y}}{x+y}\right)=2 \\[8pt] \sqrt[4]{y}\left(\dfrac{1}{4}-\dfrac{2\sqrt{x}+\sqrt{y}}{x+y}\right) =1\end{cases} $$... | Divide the first equation by $\sqrt[4]{x}$ and the second by $\sqrt[4]{y}$. Then adding and subtracting the two resulting equations gives us the new pair of simultaneous equations:
$$
\frac{2}{\sqrt[4]{x}}+\frac{1}{\sqrt[4]{y}}=\frac{1}{2} \\
\frac{2}{\sqrt[4]{x}}-\frac{1}{\sqrt[4]{y}}=2\frac{2\sqrt{x}+\sqrt{y}}{x+y} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Two different solutions to integral Given the very simple integral
\begin{equation}
\int -\frac{1}{2x} dx
\end{equation}
The obvious solution is
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln{|x|} + C
\end{equation}
However, by the following integration rule
\begin{equation... | They are both correct: they differ only in terms of their constants of integration.
$$\begin{align}
-\frac{1}{2}\ln|-2x| + \color{red}{C} &= -\frac{1}{2}(\ln 2 + \ln |x|) + \color{red}{ C} \\&= -\frac{1}{2}\ln |x| + \color{red}{(C - \frac{1}{2} \ln 2)}\\ &= -\frac{1}{2}\ln |x| + \color{red}{K}
\end{align}
$$
$C \neq K\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the limit of $\frac{e^{6x}-2e^{3x} + 1}{x^2}$, as $x \rightarrow 0$? I am to calculate $\frac{e^{6x}-2e^{3x} + 1}{x^2}$ when $x$ goes towards $0$.
I find that
$$\frac{e^{6x}-2e^{3x} + 1}{x^2} = \frac{(e^{3x}-1)^2}{x^2} = \left(\frac{e^{3x}-1}{x}\right)^2$$
$$\left(\frac{e^{3x}-1}{x}\right)^2 \rightarrow 1^2$$
... | Another approach to write $$\frac{e^{3x}-1}{x}=\frac{e^{3x}-e^{3\cdot 0}}{x}=\frac{f(x)-f(0)}{x}$$
Where $f(x)=e^{3x}$ then$$\lim_{x\to 0} \frac{f(x)-f(0)}{x} = f'(0)=3e^{0}=3$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A trigonometric integral inequality $$\displaystyle\frac{4\sin 1}{\pi }<\int_{0}^{1}{\frac{\cos x}{\sqrt{1-{{x}^{2}}}}}\text{d}x\le \frac{\pi }{2}\ln \left( \sec 1+\tan 1 \right)$$
I've got no ideas for this one.
| Numerically, the integral evaluates to $\approx 1.201969715$. Your RHS is $1.926096588$ which seems like a fairly bad estimate. Here's one way to do better:
Partial integration gives
\begin{align}
\int_0^1 \frac{\cos x}{\sqrt{1-x^2}}\,dx &= \left[ \arcsin x \cos x \right]_0^1 + \int_0^1 \sin x \arcsin x \,dx \\
&\le \f... | {
"language": "en",
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"source": "stackexchange",
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Find $a$ and $b$ in the given cubic polynomial Find $a$ and $b$ such that $x+1$ and $x+2$ are factors of the polynomials $x^3+ax^2-bx+10$.
Here I am not sure that how can I obtain the value of $a$ and $b$, I tried to multiply $x+1$ and $x+2$ to obtain a quadratic equation by which I divided the obtained quadratic polyn... | Let $x+c$ be the 3rd/last factor as the expression is of degree three.
So, $$(x+c)(x+1)(x+2)=x^3+ax^2-bx+10$$
But, $$(x+c)(x+1)(x+2)=(x+c)(x^2+3x+2)=x^3+x^2(c+3)+x(3c+2)+2c$$
So, $$x^3+x^2(c+3)+x(3c+2)+2c=x^3+ax^2-bx+10$$
Comparing the coefficients of the different powers of $x,$
$2c=10\implies c=5$
$a=c+3=5+3=8$
$-b=3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/284364",
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Product of sums of square is a sum of squares. Given $a,b,c,d \in \mathbb{Z}$, there is $x,y$ such that
$$(a^2 + b^2)(c^2 + d^2) = x^2 + y^2$$
One can show this by considering the complex number $a + bi$ and $c+ di$, using complex properties to deduce that $x = ac - bd, y = ad + bc$ is a solution. However, given that... | If $a=b=0$ or $c=d=0$, obviously we cannot do it.
If one of (say) $a$ or $b$ is $0$, and one of $c$ or $d$ is $0$, there will be situations in which we cannot do it, for instance when $a=c=3$, $b=d=0$.
If one of $a$ or $b$ is $0$, but the other is not, and neither $c$ nor $d=0$, there is no difficulty.
Now examine the... | {
"language": "en",
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Is a monotone differentiable function continuously differentiable? If $f:\mathbb{R}^+\to\mathbb{R}$ is monotonic and differentiable, does it follow that $f$ is continuously differentiable?
(This question arose from discussion here: problem on continuous and differentiable function)
| or every $x\in (-1,1).$ Define,
$$ f(x) =100000+ 100x+ x^2\sin\frac{1}{x} \qquad\text{with $f(0)=100000$}$$
which is differentiable, and
$$ f'(x) = 100+ 2x\sin\frac{1}{x}-\cos\frac{1}{x} \qquad\text{with $f'(0)=100$}$$
$f'$ is not continuous a $x=0$. Also $f'\ge 0$ indeed,
$$ f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/285944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2}... | You get that
$$I_n = \int_0^1 \dfrac{t^n}{1+t^2} dt = \int_0^1 t^n \left(\sum_{k=0}^{\infty} (-t^2)^k\right) dt = \sum_{k=0}^{\infty}(-1)^k \int_0^1 t^{n+2k} dt$$
Now $$\int_0^1 t^{n+2k} dt = \dfrac1{n+1+2k}$$
Hence,
$$I_n = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{n+1+2k}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Calculus integration problem: $\int \sin^5 (x) \cos^2 (x)\,dx$ What's the integration of $$\int \sin^5 (x) \cos^2 (x)\,dx?$$
| Using trig identities, you can show that:
$$\sin ^5(x) \cos ^2(x)=\frac{5 \sin (x)}{64}+\frac{1}{64} \sin (3
x)-\frac{3}{64} \sin (5 x)+\frac{1}{64} \sin (7 x)$$
To do this, first use the "Power-reduction formulas" to reduce to get:
$$\sin^5(x)=\frac{10 \sin x - 5 \sin 3 x+ \sin 5 x}{16}$$
$$\cos^2(x)=\frac{1 + \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/293892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to go by this, to solve for X $$\begin{bmatrix}3 & -2\\-1 & -3\end{bmatrix}X + \begin{bmatrix}1 & 5\\0 & 7\end{bmatrix} = \begin{bmatrix}-5 & 8\\-1 & 3\end{bmatrix}X$$
Not sure how I am supposed to go at this ? Do I move the + to the other side first ? Can someone let me know how to tackle this, thanks.
| Taking the $x$ to LHS, we get $$\begin{pmatrix}3+5&-2-8\\-1+1&-3-3\end{pmatrix}X=-\begin{pmatrix}1&5\\0&7\end{pmatrix}$$ $$\Rightarrow \begin{pmatrix}8&-10\\0&-6\end{pmatrix}X=-\begin{pmatrix}1&5\\0&7\end{pmatrix}$$
Now note that $\begin{pmatrix}8&-10\\0&-6\end{pmatrix}$ is invertible and let A be its inverse. (To find... | {
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How do I integrate $\int x\sqrt{A^2 - x^2} \mathrm dx$ $$ \int x\sqrt{A^2 - x^2} \mathrm dx $$
I never dealt with 2 variables on integration, what should I do here?
| Alternative way is using the trigonometric substitution $x=A\sin{t}.$
Then $$dx=A\cos{t}\;dt,\\
{A^2 - x^2}=A^2-A^2{\sin^2{t}}=A^2{\cos^2{t}},$$ therefore
$$\int x\sqrt{A^2 - x^2}\; dx = \int {A\sin{t}\,A{\cos{t}}}A\cos{t}\;dt =\\
= -A^3 \int {\cos^2{t}\;d(\cos{t})}=-\frac{A^3}{3}\cos^3{t}+C= -\frac{A^3}{3} \left(1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$? Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$ for $k \ge 1$?
Suppose $p,q,r$ are positive integers satisfying $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}<1$, prove that $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\le \dfrac{41}{42} $ ?
| $$(1)$$
$$3^{2^{k}}-1=(3^2-1)(3^2+1)(3^4+1)\cdots (3^{2^{k-1}}+1)=2^3 \cdot\displaystyle\prod_{m=1}^{k-1} (3^{2^m}+1)$$
Since each $3^{2^m}+1$ is even, $2$ divides each of these $k-1$ terms.
It follows that $2^3\cdot 2^{k-1}=2^{k+2}$ divides $\displaystyle 2^3\cdot \prod_{m=1}^{k-1} (3^{2^m}+1)=3^{2^k}-1$
$$(2)$$
We w... | {
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Find a basis for a solution set of a linear system I'm trying hard with this exercise, but it is breaking my back.
Find a basis for the solution set of the given homogeneous linear system
$3x_1+x_2+x_3=0$
$6x_1+2x_2+2x_3=0$
$-9x_1-3x_2-3x_3=0$
I do what I know I need to do. First I get the solution set of the sys... | Check your expression for $\vec x$. I think it should be $\vec x=\begin{bmatrix} x_1 \\x_2\\x_3\end{bmatrix}=\begin{bmatrix} -\frac{1}{3}r-\frac{1}{3}s \\r \\s \end{bmatrix}=r\begin{bmatrix} -\frac{1}{3} \\1 \\0 \end{bmatrix}+s\begin{bmatrix} -\frac{1}{3} \\0 \\1 \end{bmatrix}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove $p^n\ne a^3+b^3$ where $p\in\mathbb{P}\ge5$, $n,a,b\in\mathbb{N}$
Given prime $p\geq5$, prove that $p^n$ can't be represented as sum of two positive cubes for any $n\in\mathbb{N}$. What about $p=2$ and $p=3$?
| Let $p\ge 5$, and suppose that there is an $n\ge 1$ such that $p^n$ can be represented as $a^3+b^3$, where $a$ and $b$ are positive. Then there is a smallest such $n$.
From the factorization $a^3+b^3=(a+b)(a^2-ab+b^2)$, we conclude that $a+b$ and $a^2-ab+b^2$ are each powers of $p$. Neither can be the $0$-th power, ... | {
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Find max and min of $2\sqrt{x+y} - \sqrt x - \sqrt y$ Find $max$ and $min$ of $M=2\sqrt{x+y} - \sqrt{x} - \sqrt {y}$ with $x + y = constant$
I tried to call $c = \sqrt{x+y} \Rightarrow M= 2c - (\sqrt x+\sqrt {y})\ge (2-\sqrt{2})c$ ?
| Since $x+y$ is a constant, to maximize $2\sqrt{x+y}-\sqrt{x}-\sqrt{y}$, we need to minimize what we take away from $2\sqrt{x+y}$, so we need to minimize $\sqrt{x}+\sqrt{y}$.
Similarly, to minimize $2\sqrt{x+y}-\sqrt{x}-\sqrt{y}$, we need to maximize $\sqrt{x}+\sqrt{y}$.
So let us first minimize $\sqrt{x}+\sqrt{y}$. Thi... | {
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How many solutions to prime = $2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4$ Let $a,b,c$ be integers, no sign restriction.
Let $p$ be a given prime.
How to find the number of solutions to $p = 2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4$ ?
Note, from Heron's formula,
$$ 2 b^2 c^2 + 2 c^2 a... | It seems the only prime representable this way is $p=3$, which has eight such representations by using $a,b,c=\pm 1$ (independent sign choices). [Note that $-3$, which is prime, has no such representations.] I ran a check of all choices of $a,b,c$ between $-100$ and $100$ just to make sure of this.
It wouldn't be hard ... | {
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Proving that $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ How would we go about proving that $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots +\frac{1}{n(n+1)} = \frac{n}{n+1}$$
| Hint: $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$
| {
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Construct a convergent series of positive terms with $\displaystyle\limsup_{n\to\infty} {a_{n+1}\over{a_n}}=\infty$ Construct a convergent series of positive terms with $\displaystyle\limsup_{n\to\infty} {a_{n+1}\over{a_n}}=\infty$
My thoughts:
By the theorem: Suppose $a_n\ge0$ for all $n$, and let $l=\displaystyle\li... | Try the sequence $( \frac{1}{2})^3 + ( \frac{1}{2})^2+ ...+ ( \frac{1}{k})^3+ (\frac{1}{k})^2+ ...$.
Since $\frac{a_{2n+2}}{a_{2n+1}}=\frac{(\frac{1}{n+1})^2}{(\frac{1}{n+1})^3} = n+1$, we have $\limsup_n \frac{a_{n+1}}{a_n} = \infty$. The sequence is convergent since both $\sum \frac{1}{k^2}$ and $\sum \frac{1}{k^3}$ ... | {
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Two improper log integrals Evaluate
$$\int_0^{\frac{\pi}{2}}\ln ^2(\tan x)\text{d}x$$
$$\int_0^{\frac{\pi}{2}}\ln ^2(\sin x)\text{d}x$$
| Per @julien's comment: Let $u=\tan{x}$ and transform the integral into
$$\int_0^{\infty} du \frac{\log^2{u}}{1+u^2} $$
This can be evaluated via the residue theorem. Consider the integral
$$\oint_C dz \frac{\log^3{z}}{1+z^2}$$
where $C$ is a keyhole contour about the positive real axis. The integral about the contou... | {
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Using Parseval' s theorem to evaluate a sum.. On the function $f(x)$ = $x^3$ on $(-1,1)$ find Fourier coefficients for this function and then use Parseval's Theorem to evaluate:
$$\sum_{n=1}^{\infty}\frac{1}{n^6}$$
Current work:
I have used Mathematica to find the sum, which results in $\frac{\pi^6}{945}$
Since $a_0$ a... | When you integrate (using integration by parts or a table)
\begin{equation*}
b_n=\int_{-1}^{1}x^3\sin (n \pi x) \, dx = 2\int_{0}^{1}x^3 \sin (n \pi x) \, dx
\end{equation*}
you get
\begin{equation*}
-\frac{2(-1)^n}{n\pi}\left( 1-\frac{6}{n^2\pi^2} \right).
\end{equation*}
Squaring this, we obtain
\begin{equation*}... | {
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Simplification of $\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$ Simplify the expression $$\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$$
I have done this way : $(1-\cos\alpha \cos\beta)^2 = 1-2\cos\alpha \cos\beta +\cos^2\alpha \cos^2\beta$
Please guide further....
| $$\begin{align*}(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta&=1+\cos^2{\alpha}\cos^2{\beta}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\
&=1+(1-\sin^2{\alpha)\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\
&=1+\cos^2{\beta}-\sin^2{\alpha\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alp... | {
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Verifying Trigonometric Identities: $2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$ Verify that $$ 2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$$
| Putting $z=e^{ix}$ we have
$$ \cos x = \frac{1}{2} \left( z + \frac{1}{z} \right) =
\frac{1}{2z} (z^2+1)$$
and
$$ \tan x = \frac{1}{i} \frac{z - \frac{1}{z}}{ z + \frac{1}{z} } =
\frac{1}{i} \frac{z^2-1}{z^2+1}.$$
Now $$2\cos^2 x - 1 = \frac{1}{2z^2} (z^2+1)^2 -1 $$
and $$\frac{1-\tan^2 x}{1 + \tan^2 x} =
\frac{1+\left... | {
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The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ I was solving some problems and I came across this question:
Q: The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ is
(a) 0 (b) 2
(c) 4 (d) none of these
Solution: put $y = x + (3+5... | $$(y-1)^4+(y+1)^4$$
$$=y^4-\binom 41y^3+\binom 42y^2-\binom 43y+1+(y^4+\binom 41y^3+\binom 42y^2+\binom 43y+1)$$
$$=2(y^4+6y^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Trig substitution integral I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$
The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substit... | The first approach is easier in fact:
$$
I= \int \frac{(5x +1)dx}{x^2+4}=I_1 + I_2
$$
where
$$
I_1 = \int\frac{5x dx}{x^2+4}\\
I_2 = \int \frac{dx}{x^2+4}
$$
hence,
$$
I_1 = \frac{5}{2} \int \frac{2xdx}{x^2+4} = \frac{5}{2} \int \frac{d(x^2+4)} {x^2+4}=\frac{5}{2} \log( x^2 +4) +C_1
$$
In fact you can see that the nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Tangent line to a hyperbola Can anyone explain and prove the following statement?
Any tangent line to a hyperbola touches the hyperbola halfway between the points of intersection of the tangent and the asymptotes.
Thank you very much for the help!
| Derivating implicity the equation of the canonical hyperbola:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Longrightarrow \frac{2x}{a^2}dx-\frac{2y}{b^2}dy=0\Longrightarrow \frac{dy}{dx}=\frac{b^2x}{a^2y}$$
Thus, at any point $\,(x_1,y_1)\,$ on the hyperbola, the tangent line is
$$\text{I}\;\;\;y-y_1=\frac{b^2x_1}{a^2y_1}(x-x_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A two variable Inequality Show that $(a^2+b^2)(a^4+b^4) \geq (a^3+b^3)^2$
Can we go like this : $(a^2+b^2)=(a+b)^2-2ab \& (a^4+b^4) = (a^2+b^2)^2-2a^2b^2$ and right hand side of the inequality : $[(a+b)^3-3ab(a+b)]^2$
| Multiplying out and cancelling leaves $$a^2b^4+a^4b^2\ge2a^3b^3$$ which let's write as $${a^2b^4+a^4b^2\over2}\ge a^3b^3$$ which is true by the inequality of the arithmetic and geometric means.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/312544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $x^2+x+7 \equiv 0\pmod{81}$ Solve $x^2+x+7\equiv 0 \pmod{81}$
My work:
Prime factorization $81 = 9^2 = 3^4$
Test the value $x\equiv0,1,2$ for $x^2+x+7\equiv0\mod{3}$
we have $x\equiv1\mod{3}$ works.
Now life this to $\mod{3^2} = \mod{9}$
Let $x=1+3k$ for some integer$k$
$(1+3k)^2 + (1+3k) +7 \equiv0\mod{9}$
$1+6... | You can write the equation as $(x +41)^{2} \equiv -27$ (mod 81). But this means there is no solution, as $x+41$ would need to be divisible by $9.$ I'll put in more explanation, in view of comments and questions, including correcting a sign error pointed out in the comments.
We have $x^{2} + x + 7 \equiv x^{2}+ 82x + 7 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.