Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Bennett's Inequality to Bernstein's Inequality Bennett's Inequality is stated with a rather unintuitive function,
$$
h(u) = (1+u) \log(1+u) - u
$$
See here. I have seen in multiple places that Bernstein's Inequality, while slightly weaker, can be obtained by bounding $h(u)$ from below,
$$
h(u) \ge \frac{ u^2 }{ 2 + \f... | Here is how the approximation
could be derived.
$\begin{align}
h(u)
&=(1+u)\log(1+u)−u\\
&=(1+u)\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}−u\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+u\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}−u\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+\sum_{n=1}^{\infty} \dfrac{(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/114784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
How to find the maximum of this equation $x(1 - (1 - \frac{1}{x})^K)$ For x in reals and K some positive integer, can you find the maximum of this function analytically?
$f(x) = x(1 - (1 - \frac{1}{x})^K)$
| When $K = 1$, $f(x) = 1$. So assume $K > 1$ (not necessarily an integer).
The derivative of your function is
$$ 1 - \left(1 - \frac{1}{x}\right)^K - \frac{K}{x}\left(1 - \frac{1}{x}\right)^{K-1} =
1 -\left(1 + \frac{K-1}{x}\right)\left(1 - \frac{1}{x}\right)^{K-1} . $$
The derivative approaches $1$. In particular, whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
What is the least value of the function $y= (x-2) (x-4)^2 (x-6) + 6$? What is the least value of function:
$$y= (x-2) (x-4)^2 (x-6) + 6$$
For real values of $x$ ?
Does $\frac{dy}{dx} = 0$, give the value of $x$ which will give least value of $y$?
Thanks in advance.
| $$y= (x-2) (x-4)^2 (x-6) + 6$$
$let , t = x-4$
$$y= t^2(t-2)(t+2)+ 6$$
$$y= t^4-4t^2+6 $$
$$y=(t^2-2)^2+2$$
$$y(min)=2$$
attained when $t^2=2$
$(x-4)^2=2$ which gives, $ x=4-\sqrt2$ and $ x=4+\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Are there any $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square? Are there any positive $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?
I tried to simplify
\begin{align*}
n^4+n^3+n^2+n+1 &= n^2(n^2+1)+n(n^2+1)+1\\
&= (n^2+n)(n^2+1)+1 \\
&= n(n+1)(n^2+1)+1
\end{align*}
Then I assumed that the above expressi... |
Here is my answer!
H. Bensom, Germany
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/116064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Can someone please explain the cube to sphere mapping formula to me? I am wondering if anyone could explain how the following formula works, it is supposed to take the input as a point on a cube then map that to points on a sphere, please go gentle on me, I'm in 9th grade O_O
$$\begin{bmatrix}x' \\ y' \\ z'\end{bmatrix... | Thanks for the explanation by joriki. The above formula has been explained in 1. But the formula has some limitations. The cube length can be 2 (-1 to 1 in all directions). From the above concept, mentioned by joriki or 1 we can generalize for the cube of any length (i.e 2a (-a to a in all directions)). Here 2a is the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
how to prove that a solution of the equation is consider the equation
$ x^n - 4 x^{n-1} - 4 x^{n-2} - 4 x^{n-3} - \cdots-4x-4=0$
for $n = 1$ :: solution is : $x = 4$
for $n = 2$, ($x^2 - 4 x - 4 = 0$) :: solution is : $x = 4.8$
for $n = 3$, ($x^3 - 4x^2 - 4 x - 4 =0 $) :: solution is : x = $4.96$
how to prove that as... | $ x^n - 4 x^{n-1} - 4 x^{n-2} - 4 x^{n-3} - \cdots 4x-4=0 $
$ x^n$ - 4{ $x^{n-1} + x^{n-2} + x^{n-3} + \cdots+x+1$}=0
Using the property
$1+x+x^2......x^n-1 $= $\frac{1-x^n}{1-x}$
$ x^n$ - 4{$\frac{1-x^n}{1-x}$} =0
$x^{n+1} - 5x^n + 4 $= 0
$x - 5 + \frac{4}{x^n} $= 0
now if n tends to infinity x will tend to fiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$ I'm learning integral. Here is my homework:
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)
| Integrate by parts to reduce to table integral:
$$
\int_0^1 \sqrt{1+x^2} \mathrm{d} x = \left. x \sqrt{1+x^2} \right|_0^1 - \int_0^1 \frac{x^2 {\color\green{+1-1}}}{\sqrt{1+x^2}} \mathrm{d} x = \sqrt{2} - \int_0^1 \sqrt{1+x^2} \mathrm{d} x + \int_0^1 \frac{\mathrm{d} x}{\sqrt{1+x^2}}
$$
Now solving the equation fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
Finding $y$ from $y'=\sqrt{5x+2y-3}$. I'd love your help with finding the function $y$ from the following differential equation:
$y'=\sqrt{5x+2y-3}$.
I tried to use $z=5x+2-3$, so $z'=5+2y'$ , and $y'=\frac{z'}{2}-2.5$
and from the equation $\frac{z'}{2}-2.5=y'=\sqrt z$, and then $z'=2 \sqrt z+5$, so $\frac{dz}{2 \sqrt... | Integration here is simple
$$
\int\frac{dz}{2\sqrt{z}+5}=\{t=\sqrt{z}\}=\int\frac{2t dt}{2t+5}=\int\left(1-\frac{5}{2t+5}\right)dt=t-\frac{5}{2}\int\frac{d(2t+5)}{2t+5}=
$$
$$
t-\frac{5}{2}\ln(2t+5)+C=\sqrt{z}-\frac{5}{2}\ln(2\sqrt{z}+5)+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I find if $\frac{e^x}{x^3} = 2x + 1$ has an algebraic solution? Is there some way of solving $$\frac{e^x}{x^3} = 2x + 1 $$
non-numerically?
How would I go about proving if there exists a closed form solution? Similarly how would I go about proving if there exists an analytic solution?
| If you write your equation as, say, $f(x)=t$ where $f(x) = e^x - 2 x^4 - x^3$, you can use the Lagrange inversion formula to get power series solutions. A convenient starting point is $f(1)=e-3$. Taking $a=1$ and $b=f(a) = e-3$, the Lagrange inversion formula says that for $t$ near $b$, one solution of $f(x)=t$ is
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Finding Radius of convergence of $\sum_{n=2}^{\infty} \frac{n^{2n}}{4^n(2n+1)!} (3-2x)^n$ Radius of convergence of
$$\sum_{n=2}^{\infty} \frac{n^{2n}}{4^n(2n+1)!} (3-2x)^n$$
The answer was to let $$c_n = \frac{(n^{2}(-2))^n}{4^n(2n+1)!}$$
But how do I get the $-2$? Its probably from $(3-2x)^n$. But how do I remove the... | I've seen something similar around. Since you need the $x$ part to be of the form $(x-a)^n$, you have to extract the $2$ in the $x$:
$$(3-2x)^n=(-2)^n \cdot (x-3/2)^n$$
Now you have
$$c_n = \frac{(-2n^2)^n}{4^n(2n+1)!}$$
$$c_n = \frac{(-2)^nn^{2n}}{4^n(2n+1)!}$$
You now need to evaluate the limit:
$$\lim\limits_{n \to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is the permutation $\alpha : \mathbb Z_{11} \rightarrow \mathbb Z_{11}$ given by $\alpha(x)=4x^2-3x^7$ a cycle or not? The permutation $\alpha : \mathbb Z_{11} \rightarrow \mathbb Z_{11}$ given by $\alpha(x)=4x^2-3x^7$ is a cycle? If it´s not a cycle then write it as a product of disjoint cycles.
| Of course martini's comment is very true, as the easiest thing to do would just be computing $\alpha(x) \bmod 11$ for $x = 0, 1, \ldots, 10$, and seeing if we get a cycle. But let us try to be clever, and try to obtain the cycles without doing 'brute force' calculations. (Maybe it's too much, in which case you can just... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Riemann sum of $\sin(x)$ I would like to calculate the Riemann sum of $\sin(x)$. Fun starts here:
$$R = \frac{\pi}{n} \sum_{j=1}^n \sin\left(\frac{\pi}{n}\cdot j\right)$$
What would be the simplest way to calculate the sum of $\sin\left(\frac{\pi}{n}\cdot j\right)$, so that one could proceed to evaluating the limit and... | Use
$$
2 \sin\left(\frac{\pi}{2 n} \right) \sin\left(\frac{\pi}{n} \cdot j \right) = \cos\left( \frac{\pi}{2n} (2j-1) \right) - \cos\left( \frac{\pi}{2n} (2j+1) \right)
$$
Thus the sum telescopes $\sum_{j=1}^n \left(g(j) - g(j+1) \right) = g(1) - g(n+1) $:
$$
R_n =\frac{\pi}{n} \sum_{j=1}^n \sin\left(\frac{\pi}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/128186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Partial fraction with a constant as numerator I am trying to express this as partial fraction:
$$\frac{1}{(x+1)(x^2+2x+2)}$$
I have a similar exaple that has $5x$ as numerator, it is easy to understand. I do not know what to do with 1 in the numerator, how to solve it?!
| $$
\frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}.
$$
Then you need to find $A$, $B$, and $C$.
The polynomial $x^2+2x+2$ factors as $(x+1+i)(x+1-i)$, where $i$ is a square root of $-1$. You could go on to write
$$
\frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2} = \frac{A}{x+1} + \frac{D}{x+1+i} + \frac{E}{x+1-i},
$$
and the numbers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Counting the number of squares on an $n\times n$ Board Yesterday I was asked by a friend how many squares are in a chess board of $8\times 8$.
I thought about 64 immediately, but of course this is not the solution, there are much more.
So the number of squares is: $$8\times 8 + 7\times 7 + 6\times 6 + 5\times 5 + 4\tim... | The first step is to recognize that there are $8^2$ squares of size $1$ by $1$, $7^2$ squares of size $2$ by $2$ and so on. That justifies the total number being, as you say, $1^2+2^2+3^2+\ldots 8^2$. Sums of powers are calculated by Faulhaber's formula. There are several ways to derive them. One way is to know or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$ Evaluate $$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$$
So ...
$$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011} = (-\sqrt{2}+\sqrt{2}i)^{-2011}$$
$$\theta=\pi - \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{3\pi}{4}$$
$$-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$$
$$(-\sqrt{2}... | $$
\frac{1}{-\sqrt{2}+i\sqrt{2}}\frac{-\sqrt{2}-i\sqrt{2}}{-\sqrt{2}-i\sqrt{2}}=\frac{-\sqrt{2}-i\sqrt{2}}{4}=\frac12\frac{-1-i}{\sqrt{2}}
$$
which has absolute value $\dfrac12$ and argument $\dfrac{5\pi}{4}$, so
$\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{2011}$ has absolute value $1/2^{2011}$ and argument $\frac{5\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Problem about solving infinity limit with square root (I)
$$\lim_{x \to \infty } \, \left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=$$
$$\lim_{x \to \infty } \, \left(x\sqrt{1+1/x}-x\sqrt{1-1/x}\right)=$$
$$\lim_{x \to \infty } \, \left(x\sqrt{1}-x\sqrt{1}\right)=\lim_{x \to \infty } \, \left(x-x\right)=0$$
(II)
$$\lim_{x \to \... | By the same erroneous logic you'd conclude $\rm\:2\: =\: x(1+1/x)-x(1-1/x)\to\: 0$
In your example $\rm\:\displaystyle \sqrt{1+\frac{1}x}-\sqrt{1-\frac{1}x}\ =\ \frac{1}x + \frac{1}{8\: x^3} +\: \cdots\:$ which makes the error clear.
If the dominant terms in a sum of series cancel, then you need to look at subsequent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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How to prove $|F(x)|\leq\frac{M(b-a)^2}{8}$ $f(x)$ is derivable in $[a,b]$, $|f^{'}(x)|\leq M$.
$\int_a^b f(x)dx=0$.
Let $F(x)=\int_a^x f(t)dt$.
Try to prove $|F(x)|\leq\frac{M(b-a)^2}{8}$
I want to use Taylor expansion at $f(\xi)=0$, but I can't continue.
| I can prove that $\left | F( \frac{a+b}{2})\right | \le \frac{M(b-a)^2}{8}$.
By Taylor expansion at $ x_0 = \frac{a+b}{2}$,we have
$$F(x) = F(\frac{a+b}{2}) + F'(\frac{a+b}{2})(x-\frac{a+b}{2}) + \frac{F''(\xi)}{2}(x-\frac{a+b}{2})^2$$
Substituting $x=a$ and $x=b$ into the formula above,we get
$$F(a) = F(\frac{a+b}{2})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Solving an equation with nested ceilings How would I go about solving something like the following for $x$ in terms of $y$?
$$1+\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 2$$
I've never had to solve an equation with a ceiling before, never mind nested ceilings. Even ... | $$1+\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 2$$
$$\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 1$$
$$0 \lt \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \le 1$$
assuming $x \gt 2$:
$$1 \lt \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/136135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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A highschool factoring problem $x+y+z=0$
$x^3+y^3+z^3=9$
$x^5+y^5+z^5=30$
$xy+yz+zx=?$
I solved this problem by setting $xy+yz+zx=k$ and using the cubic equation with roots $x,y,z$. But is there any other methods?
| Here is one way of making progress, which uses the cubic as part of the solution. There are other routes which involve knowing some standard factorisations.
First note that $z=-(x+y)$ from the first equation and substitute in the second, obtaining:
$$-3x^2y-3xy^2 = 9$$
Divide by 3 to get:
$$-xy(x+y) = xyz = 3$$
Now $x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
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Finding area of a triangle with integrals I am suppose to find the area of a triangle using integrals with vertices 0,0 1,2 and 3,1
This gives me
$y= 2x$
$y=\frac{1}{3}x$
$y= \frac{-1}{2}x+\frac{5}{2}$
for my slopes
I know that I can calculate the area of the first part by finding
$$\int_{0}^{1}2x-\frac{1}{3}x$$
and th... | Working with what you have:
$$\int_{0}^{1}2x-\dfrac{1}{3}x dx$$
$$\int_{0}^{1}\dfrac{5x}{3}dx$$
$$\dfrac{5}{3}\int_0^{1}x dx$$
$$=\dfrac{5}{6}$$
The second integral before being evaluated from $1$ to $3$ is:
$$\dfrac{5x}{2} - \dfrac{5x^{2}}{12}$$
When evaluated at $x = 3$, you get $\dfrac{15}{4}$. When evaluated at $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$ How do I begin to evaluate this limit: $$\lim_{n\to \infty}\ \frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}\;?$$
Thanks a lot.
| Well known equality :
$$\frac{1}{2\sqrt{n}} \leqslant \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\leqslant \frac{1}{\sqrt{2n}}$$
Proof of right side. Having
$$\frac{1}{2}<\frac{2}{3},\frac{3}{4}<\frac{4}{5},\frac{5}{6}<\frac{6}{7},\cdots,\frac{2n-3}{2n-2}<\frac{2n-2}{2n-1},\frac{2n-1}{2n}<1$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/139494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \fra... | We want to show that our sum is equal to $1-\frac{1}{2^n}$. The base step is easy. Let's do the induction step.
Suppose that the result holds when $n$ is the particular number $k$. We want to show that the result holds for the "next" $n$, namely $k+1$.
Look at the sum
$$\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^k}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. Prove that if $x, y$ are rational numbers and
$$ x^5 +y^5 = 2x^2y^2$$
then $1-xy$ is a perfect square.
| Here is a proof I really like:
Using "polar coordinates", we have $x=r \cos\theta$ and $y=r\sin\theta$, where $r\geq 0$ and $\theta\in [0,2\pi)$. If $xy=0$, then $1-xy=1$ which is clearly a perfect square of a rational number. Therefore, we assume that $xy\neq 0$, which implies that $x^5 +y^5\neq 0$ thanks to $x^5 +y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 1
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List the $x$-axis intercepts for this trigonometric function
Sketch the graphs of each of the following for $x$ in [0,2$\pi$]. list the $x$-axis intercepts of each graph for this interval.
$$y=\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+1$$
I tried to solve the equation $y = 0$ by performing the following steps:
$$\be... | As mentioned in the comments, you can't add $\frac{\pi}{4}$ to both sides, you can think of it as being "trapped" inside the cosine. Your work is correct up to your second step:
\begin{align*} -1&=\sqrt{2} \cos\left(x-\frac{\pi}{4}\right) \\\\ -\frac{1}{\sqrt{2}}&=\cos\left(x-\frac{\pi}{4}\right) \end{align*}
Now, thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/142594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Puzzle: Cumulative Sum Divisible by 10 If we sum the first $4$ positive integers, we get $4 + 3 + 2 + 1 = 10$, which I think is pretty cool.
I'm interested in seeing solutions to the following puzzle: If we take the cumulative sum of the first $N$ positive integers, how many times along the way will the sum be divisib... | You need $20$ to divide $n(n+1)$. Note $n$ and $n+1$ are of opposite parity. Hence, $4$ divides either $n$ or $n+1$. $5$ divides either $n$ or $n+1$ since $5$ is a prime.
Hence, we have $4$ cases which gives us as discussed below.
$1$. $4|n$ and $5|n$. This case is trivial. Since $(4,5)=1$, we get that $20|n$ and hence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/142898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 3
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Need help with an integral need help with calculating this:
$$\int_{0}^{2\pi}\frac{1-x\cos\phi}{(1+x^2-2x\cos\phi)^{\frac{3}{2}}}d\phi$$
Thanks in advance!
| Let's assume $-1<x<1$, for simplicity.
Changing integration variable $\phi \to 2 \phi$, and using $\cos(2\phi) = 1- 2 \sin^2(\phi)$ we get:
$$\begin{eqnarray}
\int_0^{2\pi} \frac{1-x \cos(\phi)}{\left(1+x^2-2x \cos(\phi)\right)^{3/2}}\mathrm{d} \phi &=& 2 \int_0^{\pi} \frac{1-x \cos(2\phi)}{\left(1+x^2-2x \cos(2\ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Linear algebra exam questions Pick out the true statements:
*
*There exist $n\times n$ matrices $A$ and $B$ with real entries such that $(I-(AB-BA)^n) = 0$.
*If $A$ is symmetric and positive definite matrix then $tr(A)^n\geq n^n \det(A)$.
:(
I am stucked, unable to solve this problem.
| For the first statement, if $n=2$, let $A=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2}\end{pmatrix}$ and $B= \begin{pmatrix} 1 & 1 \\ -1 & -1\end{pmatrix}$
Then
$$AB-BA= \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2}& \frac{1}{2}\end{pmatrix} - \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the Cartesian equation corresponding to $r = \frac{5}{3-2\cos(\theta)}$
Find the Cartesian equation corresponding to $r =
\frac{5}{3-2\cos(\theta)}$
I got it into the form:
$(x^2 + y^2)(3-2x)^2 = 25$
and can see that maybe the equation of a circle will appear, but I can't seem to get much further without compli... | Rearranging we get that $3r - 2r\cos(\theta) = 5$. Note that $r = \sqrt{x^2 + y^2}$ and $r \cos(\theta) = x$.
This gives us $3 \sqrt{x^2+y^2} -2x = 5$.
Hence,
\begin{align}
9(x^2+y^2) & = (2x+5)^2\\
9x^2+9y^2 & = 4x^2 + 20x + 25\\
5x^2 -20x + 9y^2 & = 25\\
5(x-2)^2 + 9y^2 & = 45\\
\frac{(x-2)^2}{3^2} + \frac{y^2}{\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculating $\tan15\cdot \tan 30 \cdot \tan 45 \cdot \tan 60 \cdot \tan 75$ What is $\tan15\cdot \tan 30 \cdot \tan 45 \cdot \tan 60 \cdot \tan 75$ equal to (in degrees)?
Here is how I tried to solve it:
I assumed $x = \tan 15 (x= 0.27)$, so I rewrote it as:
$x\cdot 2x \cdot 3x \cdot 4x \cdot 5x = 120x^5$
$0.27^5 = 0... | Note that if $0\leq \alpha\leq \frac{\pi}{2}$ ($0^{\circ}$ to $90^{\circ}$), then
$$\cos\left(\frac{\pi}{2}-\alpha\right) = \cos\frac{\pi}{2}\cos\alpha + \sin\frac{\pi}{2}\sin\alpha = \sin\alpha.$$
In terms of degrees, $\cos(90-\alpha) = \sin(\alpha)$.
So if you look at $\tan(15^{\circ})$ and $\tan(75^{\circ}) = \tan(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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What equation intersects only once with $f(x)=\sqrt{1-(x-2)^2}$ Being $f(x)=\sqrt{1-(x-2)^2}$ I have to know what linear equation only touches the circle once(only one intersection), and passes by $P(0,0)$.
So the linear equation must be $y=mx$ because $n=0$.
I have a system of 2 equations:
\begin{align}
y&=\sqrt{1-(x-... | You were doing fine up through $mx=\sqrt{1-(x-2)^2}$. Starting there, square both sides to get $m^2x^2=1-(x-2)^2$, and rewrite this as $(m^2+1)x^2-4x+3=0$. Think of this as a quadratic in $x$; solving yields
$$x=\frac{4\pm\sqrt{16-12(m^2+1)}}{2(m^2+1)}=\frac{2\pm\sqrt{4-3(m^2+1)}}{m^2+1}\;,$$
which has two solutions or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/147874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Evaluating $\int\frac{x^{1/2}}{1+x^2}\,dx.$
Compute $$\int\frac{x^{1/2}}{1+x^2}\,dx.$$
All I can think of is some integration by substitution. But ran into something scary. Anyone have any tricks?
| $$I = \int \dfrac{x^{1/2}}{1+x^2} dx$$ Let $x = t^2$. We get $$I = \int \dfrac{2t^2 dt}{1+t^4}$$ Now factorize $(1+t^4)$ as $(t^2 + \sqrt{2}t+1)(t^2 - \sqrt{2}t+1)$ and use partial fractions.
$$I = \dfrac{1}{\sqrt{2}} \int \left( \dfrac{t}{t^2 - \sqrt{2} t+1} - \dfrac{t}{t^2 + \sqrt{2} t+1}\right)$$
Now $$\int \dfrac{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Is $k=3$ the only solution to $\sum_{r=1}^n r^k =\left( \sum_{r=1}^nr^1 \right)^2 $? $f(k) = \sum_{r=1}^{n} r^k$. Find an integer $x$ that solves the equation $f(x) = \bigl(f(1)\bigr)^2$.
Problem credit: http://cotpi.com/p/2/
I understand why $x = 3$ is a solution. $1^3 + 2^3 + \dots + n^3 = \left(\frac{n(n + 1)}{2}\ri... | More generally, one can look for expressions of the form
$f(k) = \sum_{j=1}^{\lfloor(n-1)/2\rfloor} a_j f(j) f(n+1-j)$. It seems to work for all odd $k$. For example:
$$\eqalign{f \left( 3 \right) &= \left( f \left( 1 \right) \right) ^{2}\cr
f \left( 5 \right) &=4\,f \left( 1 \right) f \left( 3 \right) -3\,
\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/149223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Perrin numbers in terms of the generalized hypergeometric function? Given the roots of $x^3=x^2+1$, we have sequence A001609,
$M(n) = x_1^n+x_2^n+x_3^n = \,_3F_2\left(\frac{-n}{3}, \frac{1-n}{3}, \frac{2-n}{3};\; \frac{1-n}{2}, \frac{2-n}{2};\; -\frac{3^3}{2^2}\right) = 1, 1, 4, 5, 6, 10, 15, 21,\dots$
for $n = {1,2,3,... | A solution for calculating P(n) when n is prime using the hypergeometric function can be found at Perrin088.org (Chapter 15)
This equation is derived from an incomplete beta function giving the nth term of the Perrin sequence when n is prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 1... | Using induction it is obvious that the statement is true for $n=1$.
Now suppose that it is true for $n=k$,
then we have $k(k^2+8)=3m$, where m is an integer.
Considering the case where $n=k+1$, we have;
$$(k+1)[(k+1)^2+8]=k(k^2+2k+9)+k^2+2k+9$$
$$=k(k^2+8+2k+1)+k^2+2k+9=k(k^2+8)+k(2k+1)+k^2+2k+9$$
$$=k(k^2+8)+3k^2+3k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/150425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 5
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Computing $ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$ I would like to compute the integral:
$$ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$$
where $$ a<b$$ and $$ p\in\mathbb{N}$$
$$ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx=\frac{2}{b-a} \int_a^b \frac{x^p \mathrm dx}{\sqrt{1-\left(\frac{2x-(a+b... | I would probably find it easiest to reduce the given integral to Euler's integral for Gauss's hypergeometric function:
$$
\int_a^b \frac{ x^p \mathrm{d} x}{\sqrt{(b-x)(x-a)}} \stackrel{x = a+(b-a) u}{=} \int_0^1 \frac{(a+(b-a)u)^p }{\sqrt{u(1-u)}} \mathrm{d} u = \\ a^p \int_0^1 u^{-\frac{1}{2}} \left(1-u\right)^{-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Ratio and number theory The question goes as follows:
Let $K$ be a three digit number such that the ratio of the number to the sum
of its digit is least. What is the difference between the hundreds and
the tens digits of $K$?
Now I was able to do this question by trial and error, assuming hundredth digit place t... | Let the number be $100a + 10 b +c$, where $1 \leq a \leq 9$ and $0 \leq b,c, \leq9$. Hence, we want to minimize
$$L=\dfrac{100a + 10b + c}{a+b+c} = 1 + \dfrac{99a+9b}{a+b+c}$$
This means that you should choose $c$ to be maximum as possible since $c$ appears only in the denominator and the term is positive. Hence, $c = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Limit of $\frac{x^{x^x}}{x}$ as $x\to 0^+$ I've encountered the following problem: Evaluate $$\lim_{x\to 0^+}\cfrac{x^{x^x}}{x}.$$
This is readily a "$\frac{0}{0}$" form, so I used L'Hopital's rule, but it got seriously messy, and fast. Can anyone recommend an alternative approach?
| All we need is a nice enough series expansion for $x^x$ about $0$, which can be obtained by rewriting $x^x$ as $\exp \left( x \log (x)\right)$ and looking at the Taylor series of $\exp \left( y\right)$. Now if we look at $$f(x) = \log \left( \dfrac{x^{x^x}}{x} \right) = \log \left( x^{x^x}\right) - \log x = \log (x) \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 3
} |
$3x^3 = 24$ quadratic equation Completing the square I know by factoring
$$x^3 - 8 = 0\\
x-2 = 0$$
that one of the solutions is 2.
but the other solutions is $1 ± i \sqrt 3$.
Can someone explain to me how to get that?
| May be you know this already but it's not clear to me from your post that you have the progression from the original problem statement to the final two factors exactly correct.
$3x^3 = 24 \\ \Rightarrow 3x^3 - 24 = 0 \\ \Rightarrow 3\left(x^3 - 8\right) = 0 \\ \Rightarrow x^3 - 8 = 0 \\ \Rightarrow \left(x-2\right)\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$... | I think your substitutions from the first line to the second aren't quite right. It looks like you used $x=a\cos\theta$ and $y=b\sin\theta$, but you probably wanted $$\begin{align}x&=r\cos\theta\\y&=r\sin\theta.\end{align}$$
Using those, $$\begin{align}
(x-a)^2 + (y - b)^2 &= a^2 + b^2 \\
(r\cos\theta-a)^2+(r\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
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Find the square root of the polynomial My question is:
Find the square root of the polynomial-
$$\frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left(\frac{x}y + \frac{y}x\right) + 3$$
| Let $t=\dfrac xy +\dfrac yx$.
Then $t^2=\dfrac {x^2}{y^2} +\dfrac{y^2}{x^2}+2$.
Your "polynomial" becomes finally an actual polynomial:
$$t^2-2-2t+3=t^2-2t+1=(t-1)^2=\left(\frac xy+\frac yx -1\right)^2.$$
So, the two square roots are $$\pm \left(\frac xy+\frac yx -1\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Integral of $\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx$ So, from here $$\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)} dx$$
I divided by cos(x) and I got
$$\int \frac{\tan(x)}{2\cos^2(x)+1} dx$$
But I'm stuck here.
I tried to substitute $t=\cos(x)$
$$\int \frac{-1}{t\cdot(2t^2+1)} dt$$
Any help w... | $$
\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx = \int \frac{1}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big)
$$
$$
= \int \frac{1}{3\cos^3(x)+(1-\cos^2 x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) = \int \frac{1}{3u^3 + (1-u^2)u}\,(-du)
$$
Then use partial fractions.
Later edit in response to comment... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Summation of a series. I encountered this problem in Physics before i knew about a thing called Taylor Polynomials My problem was that i had to sum this series :
$$\sum^\infty_{n=1}\frac{(-1)^{n+1}}{n}$$
basically $$1,-\frac{1}{2},\frac{1}{3},-\frac{1}{4},\frac{1}{5},-\frac{1}{6},\frac{1}{7}.....$$
So now i know that t... | $\left[ - \sum\limits_{k = 0}^\infty {\ln \left( {(1 - q)(x{q^k} - \alpha ) + 3q - 1} \right)} + \sum\limits_{k = 0}^\infty {\ln \left( {\frac{{(1 - q)}}{2}(x{q^k} - (\alpha + 1))} \right)} \right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac... | This can be solved by probability theory: Let your matrix be $\Sigma$ with diagonal elements all 1 and the off-diagonal elements all equal to $\rho$.
This is the variance-covariance matrix of a random vector $X$ where
all components have variance 1 and all pairs of distinct elements have correlation $\rho$, where $|\rh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$ I don't know how to find an explicit form for this sum, anyone can help me?
$$\sum_{k=-\infty}^{\infty}
{1 \over \left\vert\,x - k\,x_{\atop{ \small 0}}\,\right\vert}
$$
Here are the calculations I made, but don't bring me anywhere:
(original image)
$$\begi... | The sum is divergent.
Let's assume $x/x_0\notin\mathbb{Z}$ to avoid a trivial divergence.
Note that for $k\ne 0$
$$\begin{eqnarray*}
|x-k x_0| &=& |k| |x/k-x_0| \\
&\le& |k|(|x/k| + |x_0|) \\
&\le& |k|(|x|+|x_0|).
\end{eqnarray*}$$
We have used the triangle inequality and the fact that
$1/|k| \le 1$ for $|k|\ge 1$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Integration and Limits I suspect the following integration to be wrong. My answer is coming out to be $3/5$, but the solution says $1$.
$$\int_0^1\frac{2(x+2)}{5}\,dx=\left.\frac{(x+2)^2}{5}\;\right|_0^1=1.$$
Please help out. Thanks.
| $$\left. \dfrac{(x+2)^2}5 \right \vert_0^1 = \dfrac{(1+2)^2}5 - \dfrac{(0+2)^2}5 = \dfrac{3^2}5 - \dfrac{2^2}5 = \dfrac{9}5 - \dfrac45 = \dfrac{9-4}5 = \dfrac55 = 1$$
Note that we can integrate $\displaystyle \int_{x_1}^{x_2} (x+a) dx$ in seemingly two different ways.
The first method is to treat $x+a$ together as one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Need to find $\lim_{x\to4}\bigl((1/\sqrt x)-(1/2)\bigr)/(x-4)$ This one is frustrating $$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$$
| $\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$.
We substition $\sqrt {x}=t$, hance we:
If $ x\longrightarrow 4\Rightarrow t\longrightarrow 2. $
From here for the given limits have:
$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$=$\lim\limits_{t\to 2} \frac{\frac{1}{t} - \frac12}{t^2-4}$=$\lim\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$
Here is my attempt:
Letting
$$f(z)=\frac{1}{az^2+bz+c}$$
The poles of $f(z)$ are located at
$$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$
and
$$z_1 = \frac{-b-\sqrt{b^2-4ac}}... | Here is my answer in terms of the digamma function (cf. Abramowitz and Stegun). Using the decomposed version of the sum given by Eric Naslund♦, we find:
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}=
\frac{1}{\sqrt{b^{2}-4ac}}\sum_{n=0}^{\infty}\left(\frac{1}{n-z_{0}}-\frac{1}{n-z_{1}}\right)=
\frac{\psi (-z_0)-\psi(-z_1)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 2
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Evaluating $\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$ It's a question (not hw) I bumped into few years back. Couldn't make any real progress with. Maybe you can help?
$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}... | For your original question, it might be worth noting that
$$\frac{1}{\sqrt{k^2}+\sqrt{k(k+1)}+\sqrt{(k+1)^2}}=\frac{1}{2k+1+\sqrt{k(k+1)}}=\frac{2k+1-\sqrt{k(k+1)}}{4k^2+4k+1-k^2-k}=\frac{2k+1-\sqrt{k(k+1)}}{3k^2+3k+1}$$
Thus
$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$ Here is another interesting integral inequality :
$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$
According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference ... | Maybe this is easier:
$$\eqalign{
& \int\limits_0^1 {\frac{{{x^4}\log x}}{{{x^2} - 1}}dx} = \int\limits_0^1 {\frac{{\left( {{x^4} - 1} \right)\log x}}{{{x^2} - 1}}dx} + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx} \cr
& = \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx} + \int\limits_0^1 {\frac{{\log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
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Integral - using Euler Substitution I've been trying to solve one simple Integral with Euler substitution several times, but can't find where I'm going wrong. The integral is (+ the answer given here, too):
$$\int\frac{1}{x\sqrt{x^2+x+1}} dx=\log(x)-\log(2\sqrt{x^2+x+1}+x+2)+\text{ constant}$$
The problem is, I cannot ... | What you have done is correct. All you need to do is to rewrite it a different form.
$$\begin{align}
\ln \left( \sqrt{x^2+x+1} + x - 1\right) & = \ln \left( \dfrac{\left(\sqrt{x^2+x+1} + x - 1 \right) \left(\sqrt{x^2+x+1} - x + 1 \right)}{\left(\sqrt{x^2+x+1} - x + 1 \right)}\right)\\
& = \ln \left(\left(x^2 + x + 1 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
finding the rational number which the continued fraction $[1;1,2,1,1,2,\ldots]$ represents I'd really love your help with finding the rational number which the continued fraction $[1;1,2,1,1,2,\ldots]$ represents.
With the recursion for continued fraction $( p_0=a_0, q_0=1, p_{-1}=1, q_{-1}=o), q_s=a_sq_{s-1}+q_{s-2},p... | We have that
$$\Large x=1+\frac{1}{1+\frac{1}{2\,+\,\frac{1}{1\,+\,\frac{1}{1\,+\,\frac{1}{2\,+\,\cdots}}}}}=1+\frac{1}{1+\frac{1}{2+\frac{1}{x}}}$$
so that
$$\frac{1}{x-1}=1+\frac{1}{2+\frac{1}{x}}$$
hence
$$\frac{1}{\frac{1}{x-1}-1}=\frac{1}{\frac{1}{x-1}-\frac{x-1}{x-1}}=\frac{1}{\frac{2-x}{x-1}}=\frac{x-1}{2-x}=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/164379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Summing over a cyclic subgroup of a multiplicative group mod n Let $x$ be a unit in $\mathbb Z/ n \mathbb Z$ of multiplicative order $m$.
I am trying to determine when it is that
$$
\sum_{i=0}^{m-1} x^i \equiv 0 \mod n .
$$
Is this kind of situation something that has been studied? If so, I would be very interested in... | (Below I treat $x$ as an integer whose reduction $\bmod n$ is a unit of order $m$. This allows me to treat $\frac{x^m - 1}{x - 1}$ as an integer and consider its reduction $\bmod n$ without dividing by a zero divisor.)
As usual, the Chinese Remainder Theorem is your friend. We can reduce to the case that $n$ is a prime... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/164433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
proving there exists an integer $ k$ such that $ak^2+bk+c\equiv 0 \pmod{2^n}$ by induction $n$ is a given positive integer.
how to prove by using mathematical induction that there exists an integer $ k$ such that $ak^2+bk+c\equiv 0 \pmod{2^n}$, where $b$ is odd and at least one of $a, c$ is even
| First check directly that the result holds for $n=1$.
We do the induction step. Let $x$ be an integer such that
$$ax^2+bx+c\equiv 0\pmod{2^{n-1}}.$$
We would like to show that there is an integer $y$ such that $ay^2+by+c\equiv 0\pmod{2^n}$.
If $2^n$ already divides $ax^2+bx+c$, we are finished. So we need only wo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/165711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Want to test the series for convergence/divergence $\sum_n \frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Want to test the series $$\frac{\sqrt{2} - 1}{3^3 - 1} + \frac{\sqrt{3} - 1}{4^3 - 1} + \frac{\sqrt{4} - 1}{5^3 - 1} + \cdots$$ for convergence/divergence.
My attempt is
$U_n =\frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Now I wish ... | Here, $\frac{\sqrt{n+1}-1}{(n+2)^3-1} \leq \frac{1}{n^2}$. For n=1, it is true, and then the denominator of given series increases more rapidly than numerator and this rate is greater than the rate in $1/n^2$, so the given series is bounded by $1/n^2$ and hence $\sum_{1}^{\infty}\frac{\sqrt{n+1}-1}{(n+2)^3-1} \leq (\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/166094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Write down the sum of sum of sum of digits of $4444^{4444}$ Let $A = 4444^{4444}$;
Then sum of digits of $A = B$;
Then sum of digits of $B = C$;
Then sum of digits of $C = D$;
Find $D$.
What should be the approach here?
| It's well-known that, because $10 \equiv 1 \bmod 9$, and therefore $10^k = 1 \bmod 9$ for all $k>0$, we have that the sum of digits of $n$, $S(n) \equiv n \bmod 9$.
So what is $4444^{4444} \bmod 9$? The above equivalence gives us that $4444 \equiv 7 \bmod 9$.
Now we need the order of $7$ modulo $9$, the smallest $s$ s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
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Condensing logarithms
Simplify:
$2\log_{10}\sqrt{x}+3\log_{10}x^{\frac{1}{3}}$
I got to this: $2\log_{10}x^{\frac{1}{2}}+3\log_{10}x^{\frac{1}{3}}$.
Now, usually you bring the exponent the the front and that would yield:
$$\frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x=\log_{10}x+\log_{10}x=2\log_{10}x$$ ... | $$ \frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x \neq 2\log_{10}x+3\log_{10}x$$
$1/2 * 2 = 1$ so do $1/3 * 3.$
So the correct thing is
$$\frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x = \log_{10}x+\log_{10}x = 2\log_{10}x$$
Bring the exponent inside the $\log$ if you like.
The other way around: bring $1/2$ into ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all real solutions to $8x^3+27=0$
Find all real solutions to $8x^3+27=0$
$(a-b)^3=a^3-b^3=(a-b)(a^2+ab+b^2)$
$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
Now, to find solutions you must set each part $=0$. The first set of parenthesis is easy $$(2x+3)=0 ; x=-\left(\frac{3}{... | Using the discriminant of $ax^3+bx^2+cx+d=8x^3+27$:
$$\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2=0-0+0-0-27(8^2)(27^2) < 0$$
Thus there is only one real root, and that is, as Alex Becker found, $x=-3/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Computing a Laurent series Let $$f(z) = \frac{1}{(2z-1)(z-3)} $$. Compute the Laurent series about the point z = 1 in the annular domain $$ \frac{1}{2} < |z-1| < 2$$
My attempt:
I broke f(z) up into the partial fraction decomposition:
$$ -\frac{2}{5(2z-1)} + \frac{1}{5(z-3)} = -\frac{2}{5}*\frac{1}{(1-\frac{(z+\frac{1}... | $$\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\left(\frac{1}{z-3}-\frac{1}{z-\frac{1}{2}}\right)=-\frac{1}{10}\frac{1}{1-\frac{z-1}{2}}-\frac{1}{5(z-1)}\frac{1}{\left(1+\frac{1}{2(z-1)}\right)}$$
Now only check that
$$\left|\frac{z-1}{2}\right|<1\,\,,\,\,|2(z-1)|^{-1}<1$$
and you'll be able to use the developments
$$\frac{1}{1-z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\left \{ \frac{n}{n+1}\sin\frac{n\pi}{2} \right \}$ is divergent. I want to prove whether the sequence $\{a_n\} = \left \{ \dfrac{n}{n+1}\sin\dfrac{n\pi}{2} \right \}$ (defined for all positive integers $n$) is divergent or convergent.
I suspect that it diverges, because the $\sin\frac{n\pi}{2}$ factor osci... | Claim for you to prove: if a sequence converges then any subsequence converges and to the very same limit as the whole sequence.
Well, what about choosing two very special subsequences here and proving they don't have the same limit? In fact, one of the "obvious" subsequences doesn't even have a limit...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Minimum value of given expression What is the minimum value of the $$ \frac {x^2 + x + 1 } {x^2 - x + 1 } \ ?$$
I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can algebraic manipulation is possible
| Here is an 'algebra solution':
$\frac {x^2 + x + 1 } {x^2 - x + 1 } = \frac{(x+1)^2-(x+1)+1}{(x+1)^2-3(x+1)+3} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{x^2-x+1} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{(x-\frac{1}{2})^2+\frac{3}{4}}$.
Since the last term is greater than zero when $x\neq -1$, we see that the minimu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}$ How would I solve the following double angle identity.
$$
\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}
$$
So far my work has been.
$$
\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B+\sin A\sin B}
$$
But what would I do to contin... | What i get is, how to solve the problem?? Is that correct then here u are:
$$\dfrac{\dfrac{\sin x\cos y + \cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y + \sin x\sin y}{\cos x\cos y}}$$
$$\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y} +\dfrac{\cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y} {\cos x\cos y}+ \dfrac{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solving $5^n > 4,000,000$ without a calculator
If $n$ is an integer and $5^n > 4,000,000.$ What is the least possible value of $n$? (answer: $10$)
How could I find the value of $n$ without using a calculator ?
| Taking square roots of both sides we solve $5^r>2000=25\cdot80$. The right side is approximated from below by $5^2\cdot5^2\cdot3$ so we want $5^{r-4}>3$ or $r=5$, so $n=2r=10$. Check $n=9$ is too small: $5^9<2^9\cdot 3^9<2^9\cdot 3^5\cdot 3^4=512\cdot 243\cdot 81<125,000\cdot 100<4,000,000$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/176998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 11,
"answer_id": 10
} |
Prove $\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$ How would I prove the following?
$$\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$$
I do not know how to do do the problem I do know $\sin(3x)$ can be $\sin(2x+x)$ and such yet I am not sure how to commence.
| $2\sin x+\sin3x-\sin5x$
$=2\sin x-2\cos4x\sin x$ applying $\sin2C-\sin2D=2\sin(C-D)\cos(C+D)$ formula
$=2\sin x(1-\cos4x)$
$=4\sin x \cdot \sin^22x$, applying $\cos2A=1-2\sin^2A$
$=4\sin x (2\sin x\cos x)^2$
$=16\sin^3x\cos^2x$
Alternatively, we know $e^{iy}=\cos y+i\sin y$ => $e^{-iy}=\cos y - i\sin y$
So, $\cos y=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Proof by induction of Bernoulli's inequality $ (1+x)^n \ge 1+nx$ I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:
Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality:
$$ (1+x)^n \ge 1+nx$$
Proof: ... | What you have is perfectly acceptable. The calculations could be organized a little more neatly:
$$\begin{align*}
(1+x)^{k+1}&=(1+x)(1+x)^k\\
&\ge(1+x)(1+kx)\\
&=1+(k+1)x+kx^2\\
&\ge1+(k+1)x\;,
\end{align*}$$
since $kx^2\ge 0$. This completes the induction step.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
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$\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$. Find the positive integer solutions (x,y). $\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$.
Find the positive integer solutions (x,y).
| You can make it $(x-12)(y-48)=12\cdot 48$ so $y-48$ can be $1,3,9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/182882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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The probability of bricks arranged randomly. If someone have $4$ red bricks and $8$ blue bricks and arranges them randomly in a circle,
what is the probability that two red bricks are not side by side?
Is the sample space $\frac{12!}{4!\,8!} \,=\, 495$ ?
and the answer $P(\text{Reds non-adjacent}) \:=\:\frac{70}{495} ... | Suppose there $r$ red bricks, and $b$ blue bricks. Counting clock-wise, let
$$
(k_1, k_2, \ldots, k_r)
$$
be the tuples, representing the number of blue bricks between two red bricks. The number of favorable configurations is:
$$
N_F = \sum_{k_1=1}^{b-r+1} \sum_{k_2=1}^{b-r+1} \cdots \sum_{k_r=1}^{b-r+1} \delt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/183461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove Lagrange trigonometric identity I would to prove that
$$1+\cos \theta+\cos 2\theta+\ldots+\cos n\theta =\displaystyle\frac{1}{2}+
\frac{\sin\left[(2n+1)\frac{\theta}{2}\right]}{2\sin\left(\frac{\theta}{2}\right)}$$
given that
$$1+z+z^2+z^3+\ldots+z^n=\displaystyle\frac {1-z^{n+1}}{1-z}$$
where $z\neq 1$... | $2\sin\frac{\theta}{2}\cos r\theta=\sin\frac{(2r+1)\theta}{2} - \sin\frac{(2r-1)\theta}{2}$
Putting r=1,2,....,n,
$2\sin\frac{\theta}{2}\cos \theta=\sin\frac{3\theta}{2} - \sin\frac{\theta}{2}$
$2\sin\frac{\theta}{2}\cos 2\theta=\sin\frac{5\theta}{2} - \sin\frac{3\theta}{2}$
...
$2\sin\frac{\theta}{2}\cos n\theta=\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/183859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
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A little help with algebra (homework) I was brushing up on my algebra a little, and I've seen that I've become rusty with some of the concepts. Indications for solving these would perhaps be better suited rather than full answers, as the goal is to figure it out myself.
*
*Show that for any $a, b \in \mathbb{R}$ (wi... | The simplest way to verify an identity like $a^3-b^3=(a-b)(a^2+2ab+b^2)$ is to multiply out the righthand side and verify that you do indeed get the lefthand side. If you try it in this case, however, you’ll fail:
$$\begin{align*}
(a-b)(a^2+2ab+b^2)&=a(a^2+2ab+b^2)-b(a^2+2ab+b^2)\\
&=\left(a^3+2a^2b+ab^2\right)-\left(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
$1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$. $1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$.
NOTE: $1-\frac{1}{n} \lt 3 + \frac{1}{n}$
| when $n=1$,
$1-\frac{1}{n}=0$, $3+\frac{1}{n}=4$
when $n=1$, we have, $0 \lt x_{1} \le 4$
when $n=2$, we have, $\frac{1}{2} \lt x_{2} \le 3+\frac{1}{2}$
when $n \to \infty, \frac{1}{n} \to 0$, and then we have, $1 \lt x_{\infty} \le 3$
we can now conclude that, $0 \lt x \le 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplify $\tan^{-1}[(\cos x - \sin x)/(\cos x + \sin x)]$
Write the following functions in simplest form:
$$\tan^{-1}\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right), \quad 0<x<\pi$$
Please help me to solve this problem. I have been trying to solve this from last 3 hours. I can solve simple inverse trigonomet... | Let's multiply the numerator and denominator by $\ \cos\bigl(\frac {\pi}4\bigr)=\sin\bigl(\frac {\pi}4\bigr)\ $ :
$$\tan^{-1}\left(\frac{\sin\bigl(\frac {\pi}4\bigr)\cos(x)-\cos\bigl(\frac {\pi}4\bigr)\sin(x)}{\cos\bigl(\frac {\pi}4\bigr)\cos(x)+\sin\bigl(\frac {\pi}4\bigr)\sin(x)}\right),\quad 0<x<\pi$$
$$=\tan^{-1}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to approximate the value of $ \arctan(x)$ for $x> 1$ using Maclaurin's series? The expansion of $f(x) = \arctan(x)$ at $x=0$ seems to have interval of convergence $[-1, 1]$
$$\arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\mathcal{O}\left(x^{13}\right) $$
Does it mean that ... | Use, for $x>0$
$$
\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2}
$$
For $x<0$ use the parity.
Added: Use
$$
\tan\left(\frac{\pi}{2} - \phi\right) = \frac{\sin\left(\frac{\pi}{2}-\phi\right)}{\cos\left(\frac{\pi}{2}-\phi\right) } = \frac{\cos(\phi)}{\sin(\phi)} = \frac{1}{\tan(\phi)}
$$
Thus, for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, ... | Hint $\rm\quad S_n =\, \sum c_k n^k\ \Rightarrow\ S_{2n} + 4\, S_n\, =\: \sum\ (2^k\!+\!4)\ c_k\ =\ 4\, n^3+4\,n^2 + n,\:$ therefore
$$\rm S_n\, =\ \frac{4}{2^{\color{#C00}3}\!+\!4} n^\color{#C00}3 +\frac{4}{2^\color{#0A0}2\!+\!4}n^{\color{#0A0}2} + \frac{1}{2^\color{brown}1\!+\!4} n^{\color{brown}1}\ =\ \frac{n^3}3+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
prove by induction : $\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a^n}=n$ How to prove the following question by using induction?
$${\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+\frac{(1-a^n)(1-a^{n-1})(1-a^{n-2})}{1-a^3}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a... | I applied Andre Nicolas's method. Please see steps below.
$$H_a(n)=\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+\frac{(1-a^n)(1-a^{n-1})(1-a^{n-2})}{1-a^3}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a^n}$$
$$H_a(n+1)=\frac{(1-a^{n+1})}{1-a}+\frac{(1-a^{n+1})(1-a^{n})}{1-a^2}+\frac{(1-a^{n+1})(1-a^{n})(1-a^{n-1})}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Show $ \int_0^\infty\left(1-x\sin\frac 1 x\right)dx = \frac\pi 4 $ How to show that
$$
\int_0^\infty\left(1-x\sin\frac{1}{x}\right)dx=\frac{\pi}{4}
$$
?
| Use
$$
\int \left(1-x \sin\left(\frac{1}{x}\right)\right) \mathrm{d} x = x - \int \sin\left(\frac{1}{x}\right) \mathrm{d} \frac{x^2}{2} = x - \frac{x^2}{2}\sin\left(\frac{1}{x}\right) - \frac{1}{2} \int \cos\left(\frac{1}{x}\right) \mathrm{d}x
$$
Integrating by parts again $\int \cos\left(\frac{1}{x}\right) \mathrm{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/190730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 3
} |
Calculate $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$ Please help me solving $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})]$, in the region $|x|<1$.
| Here is one way. One can write the product as the telescoping product
$$
\frac{1-x^2}{1-x}\times \frac{1-x^4}{1-x^2}\times \frac{1-x^8}{1-x^4}\times \dotsb\frac{1-x^{2^{n+1}}}{1-x^{2^n}}\times \dotsb.
$$
so that the $n$th partial product (indexed from $0$) equals
$$
\frac{1-x^{2^{n+1}}}{1-x}=1+x+\dotsb+x^{2^{n+1}-1}
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to prove Cauchy-Schwarz Inequality in $R^3$? I am having trouble proving this inequality in $R^3$. It makes sense in $R^2$ for the most part. Can anyone at least give me a starting point to try. I am lost on this thanks in advance.
| You know that, for any $x,y$, we have that
$$(x-y)^2\geq 0$$
Thus
$$y^2+x^2\geq 2xy$$
Cauchy-Schwarz states that
$$x_1y_1+x_2y_2+x_3y_3\leq \sqrt{x_1^2+x_2^2+x_3^3}\sqrt{y_1^2+y_2^2+y_3^3}$$
Now, for each $i=1,2,3$, set
$$x=\frac{x_i}{\sqrt{x_1^2+x_2^2+x_3^2}}$$
$$y=\frac{y_i}{\sqrt{y_1^2+y_2^2+y_3^2}}$$
We get
$$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Inequality. $a^7b^2+b^7c^2+c^7a^2 \leq 3 $
Let $a,b,c$ be positive real numbers such that $a^6+b^6+c^6=3$. Prove that
$$a^7b^2+b^7c^2+c^7a^2 \leq 3 .$$
| Frank Science has showed that the inequality is equivalent to $$ 3(\sum a^7b^2)^2 \leqslant (\sum a^6)^3\tag{0}\\ $$
Use AM-GM for $a^{12}b^{6}$, $a^{12}b^{6}$ and $a^{18}$ we have the following:
$ 3a^{14}b^4 \leqslant 2a^{12}b^6+a^{18}\\ $
Combine with 2 other similar inequalities we have
$$3(\sum a^{14}b^4) \leqsla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Polynomial factors Why must $x^2 + x + 1$ be a factor of $x^5+x^4+x^3+x^2+x+1$?
I know that when we divide $x^5+x^4+x^3+x^2+x+1$ by $x^3+1$ we get $x^2 + x + 1$, but is there an argument/theorem or anything that could tell that $x^2+x+1$ must divide $x^2 + x + 1$?
| $(x^2+x+1)(x-1)=x^3-1$ and $(x^5+x^4+x^3+x^2+x+1)(x-1)=x^6-1$. Then note that $x^6-1=(x^3-1)(x^3+1)$. Thus in effect it is the third binmoial formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/198991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric identity and roots of a polynomial. Prove that
$$(\operatorname{cosec} A–\sin A) (\sec A–\cos A) = \frac {1}{\tan A + \cot A} $$
Also help me with this question please
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2–2x–15$ then find a quadratic polynomial whose series [roots?] are $2\alpha$ a... | Have:
$\csc A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Now:
$(\csc A-\sin a)(\sec A-\cos A)=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)=(\frac{1-\sin^2{A}}{\sin a})(\frac{1-\cos^2{a}}{\cos A})=\frac{\cos^2 A}{\sin A}\cdot\frac{\sin^2 A}{\cos A}=\cos A\sin A=\frac{\cos A\sin A}{1}=\frac{\cos A\sin A}{\cos^2 A+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$... | Note that the equation $\sin x + \sqrt 3 \cos x = 1$ is not equivalent to the equation $(\sin x)^2=(1−\sqrt{3} \cos x)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/201399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
| Substition
$x=4\sin y$ $\Rightarrow$ $dx=4\cos y dy$
$\sin y=\frac{x}{4}$$\Rightarrow$ $y=\arcsin\frac{x}{4}$
$\int\frac{1}{(16-x^2)^2}dx=\int\frac{1}{(4^2-x^2)^2}dx$=$\int\frac{4\cos y dy}{(16-(4\sin y)^2)^2}$=$\int\frac{4\cos y dy}{(16-16\sin^2y)^2}$ = $\int\frac{4\cos y dy}{16^2(1-\sin^2 y)^2} $=$\frac{1}{64} \int\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/204961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Show that the span of two vectors is T-invariant Show that the span of \begin{bmatrix}1\\1\\0\\1\end{bmatrix} and \begin{bmatrix}1\\0\\2\\0\end{bmatrix} is a $T$-invariant subspace of the linear map given by
\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}
I tried to take some general vector in... | As the map given by $x\mapsto Tx$ is linear, it suffices to prove that the image of the spanning vectors is again in the span of the two vectors. This is done by the following computation:
\begin{align*}\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}\begin{bmatrix}1\\1\\0\\1\end{bmatrix}&=\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/205359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$ $$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$$
Attempt
This is indeterminate of the form $\frac{0}{0}$. Applying L'Hopital's rule twice results in,
$$\lim_{x \to 0} \frac{e^{\sin^2 x}(\sin^2(2x) + \sin^2(4x)) + \cos x}{6x}.$$
Wolfram Alpha says the two sided limi... | $\lim_{x\to 0}\frac{e^x-1}{x}=1\implies e^x\approx 1+x $when $x\to 0$,thus, $e^{\sin^2x}\approx 1+\sin^2x$ as $x\to 0$
Hence, $$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}=\lim_{x \to 0} \frac{1+ \sin^2 x - \cos x}{x^3}$$
Now you can easily apply L'Hopital's rule to conclude that limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/205952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\lim_{x \rightarrow 0} \sqrt{9-x} + 3 = 6$ Prove that $\lim_{x \rightarrow 0} \sqrt{9-x} + 3 = 6$
So far my difficulty is trying to find a $\delta$ that will allow for this function to be less than $\epsilon$. I keep getting that $|\sqrt{9-x}+3-6|=|\sqrt{9-x}-3| < \epsilon$, but from here I do not know how... | Firstly, observe that for any $ x \in (- \infty,9] $, we have
\begin{equation}
|\sqrt{9 - x} - 3||\sqrt{9 - x} + 3| = |(9 - x) - 9| = |x|.
\end{equation}
Fix an $ \epsilon > 0 $. Then choosing any $ x $ that satisfies $ |x| < \min(9,3 \epsilon) $, we obtain
\begin{align}
|\sqrt{9 - x} - 3| &= \frac{|x|}{\sqrt{9 - x} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
general equation of a tangent line to a hyperbola Suppose that there is a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$.
I would like to figure out an equation that describes tangent line to this hyperbola.
How would I be able to do this using calculus? My calculus trials are bring me some gibberish answe... | (1a)
Let $y=mx+c$ be a tangent of $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
Let $(h,k)$ be the point of intersection.
So, $k=mh+c, b^2h^2-a^2k^2=a^2b^2\implies b^2h^2-a^2(mh+c)^2=a^2b^2$
or,$h^2(b^2-a^2m^2)- 2a^2mch -a^2b^2-a^2c^2=0 $
This is a quadratic equation in $h,$ for tangency, the roots need to be same, to make t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/214977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series?
thanks :)
| First, an identity. For $x\ne0$,
$$
\begin{align}
\frac{2\sin(x)-\sin(2x)}{x^3}
&=\frac{2\sin(x)-2\sin(x)\cos(x)}{x^3}\tag{1a}\\
&=2\frac{\sin(x)}x\frac{1-\cos(x)}{x^2}\tag{1b}\\
&=\frac2{1+\cos(x)}\frac{\sin^3(x)}{x^3}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\sin(2x)=2\sin(x)\cos(x)$
$\text{(1b)}$: algebra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/217081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
How to verify this trigonometric identity? Please help,
$9\cos^2 B − 9\sin^2 B = 18\cos^2 B − 9$
Thanks
| \begin{align*}
\text{LHS} &= 9\cos^2 B − 9\sin^2 B \\
&= 9(\cos^2 B-\sin^2 B) \\
&= 9(\cos^2 B - (1-\cos^2 B)) \\
&= 9(2\cos^2 B- 1) \\
&= 18\cos B - 9 \\
&= \text{RHS}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/218180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{n\to\infty} (1+\frac{1}{2}+...+\frac{1}{n})\frac{1}{n}$
Find the following limit:
$$\lim_{n\to\infty} \left(1+\frac{1}{2}+...+\frac{1}{n}\right)\frac{1}{n}$$
My intuition says that this goes to zero, because $1/n$ goes much faster to zero than the harmonic series go to infinity, but how can I prove this... | Here is a completely elementary
and self-contained proof
(subject to accepting that
$\frac{k}{2^k} \to 0$
as $k \to \infty$).
Let
$a_n
= \frac{1}{n}\sum_{k=1}^n \frac1{k}
$.
Then
$na_n
= \sum_{k=1}^n \frac1{k}
$.
Therefore
$\frac1{n+1}
=(n+1)a_{n+1}-na_n
=n(a_{n+1}-a_n)+a_{n+1}
$
or
$a_{n+1}-a_n
=\frac1{n}(\frac1{n+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/221114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Frobenius Method to solve $x(1 - x)y'' - 3xy' - y = 0$ So, Im trying to self-learn method of frobenius, and I would like to ask if someone can explain to me how can we solve the following DE about $ x = 0$ using this method.
$$ x(1 - x)y'' - 3xy' - y = 0 $$
| The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.
Let $y(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$. Then we get that $$y'(x) = \sum_{n=0}^{\infty} na_n x^{n-1}$$
$$3xy'(x) = \sum_{n=0}^{\infty} 3na_n x^{n}$$
$$y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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primitive residue classes modulo 32 $\mathbb Z_{32}^*$ is the primitive residue classes modulo 32. How is it possible to show that $\mathbb Z_{32}^*$ is generated by 5 and -1, without showing it for every element of $\mathbb Z_{32}^*$=$\{1,3,5,7,9,11,...,31\}$
| Let $n > 1$ be an integer.
We consider the group $G = (\mathbb{Z}/2^n\mathbb{Z})^*$.
Clearly $|G| = 2^{n-1}$.
Let $e \ge 2$ be an integer.
Let $k$ be an odd integer.
$(1 + 2^e k)^2 = 1 + 2^{e+1}k + 2^{2e} k^2 = 1 + 2^{e+1}(k + 2^{e-1}k^2)$.
Note that $k + 2^{e-1}k^2$ is odd.
Hence, by induction on $m \ge 1$, $(1 + 2^e ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/223925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help with Trigonometry homework - prove an identity I need to prove the following identity:
$\sin^2 2\alpha-\sin^2 \alpha = \sin 3\alpha \sin \alpha$
What I have tried, is to work on each side of the identity. I have started with the left side:
\begin{align}
\sin^2 2\alpha-\sin^2 \alpha &= (\sin 2\alpha-\sin \alpha)(\s... | There is a funny identity that goes as $$\sin(x+y) \sin(x-y) = \sin^2(x) - \sin^2(y)$$
Take $x = 2\alpha$ and $y = \alpha$ to get what you want.
Proof:
\begin{align}
\sin(x+y) \sin(x-y) & = \left( \sin(x) \cos(y) + \sin(y) \cos(x) \right) \left( \sin(x) \cos(y) - \sin(y) \cos(x) \right)\\
& = \sin^2(x) \cos^2(y) - \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/225499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Lost in second derivative simplification My book has the following problem:
Find the second derivative of y for $x^2 + xy + y^2 = 3$
I follow them this far:
$y'' = -\dfrac{ 6(x^2 + xy + y^2) }{ x + 2y }$
But then how do they get here?
$y'' = -\dfrac{ 18 }{ (x + 2y)^3}$
Updated
Assuming the $x^2 + xy + y^2$ is substitu... | I must have a different method than your book since I never reached the same step as you, but here is my derivation if it helps:
$x^2 + xy + y^2 = 3$
Find $\dfrac{dy}{dx}$:
$2x + [(x)(y') + (y)(1)] + 2y(y') = 0$
$2x + xy' + y + 2yy' = 0$
$(2x + y) + y'(x + 2y) = 0$
Solve for $y'$:
$y' = -\dfrac{ 2x + y }{ x + 2y }$
Fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/227269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Binomial sum of a sequence I have the following sequence:
$$
\sum\limits_{k=1}^n k\binom{n-1}{k-1}
$$
What is the sum of this sequence.
| $\sum_{k=1}^n \binom{n-1}{k-1}\cdot k$
$=\sum_{k=1}^n \binom{n-1}{k-1}+\sum_{k=1}^n \binom{n-1}{k-1}\cdot(k-1)$
Now, $ \binom{n-1}{k-1}\cdot(k-1)=\frac{(n-1)!}{\{n-1-(k-1)\}!(k-1)!}\cdot(k-1)$
$=(n-1)\frac{(n-2)!}{\{n-2-(k-2)\}!(k-2)!}=(n-1)\binom{n-2}{k-2}$
$\sum_{k=1}^n \binom{n-1}{k-1}\cdot k$
$=\sum_{k=1}^n \binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/228255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
limit of adjacent series Suppose $U_n = \prod_{k=2}^n \cos\left(\frac{\pi}{2^k}\right)$ and $V_n = U_n\cdot\cos\left(\frac{\pi}{2^n}\right)$.
How can I prove that $U_n$ and $V_n$ are adjacent?
What is the limit of $U_n$ and $v_n$?
Note: $W_n = U_n\cdot\sin\left(\frac{\pi}{2^n}\right)$ and $W_n$ is geometrical.
| By the double angle formula, we have that
$$\sin x=2\sin \frac x 2 \cos \frac x 2 $$
Using the double angle formula once more
$$\sin x=2\cdot 2 \sin \frac x 4\cos \frac x 4 \cos \frac x 2 $$
You should realize that by induction,
$$\sin x=2^n \sin \frac x {2^n} \prod_{k=1}^n \cos \frac x {2^k} $$
This means that, for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing the indefinite integral $\int x^n \sin x\,dx$ $\newcommand{\term}[3]{
\sum_{k=0}^{\lfloor #1/2 \rfloor} (-1)^{#2} x^{#3} \frac{n!}{(#3)!}
}$
I am trying to prove that for $n \in\mathbb N$,
$$
\int x^n \sin x \, dx
= \cos x \term{n}{k+1}{n-2k}
+ \sin x \term{(n-1)}{k}{n-2k-1}
$$
I started with differentiatio... | Integrate by parts twice, and use two inductions (odd and even case). You may be able to unite these if you are skillful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/231100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Number of consecutive zeros at the end of $11^{100} - 1$. How many consecutive zeros are there at the end of $11^{100} - 1$?
Attempt
Trial and error on Wolfram Alpha shows using modulus shows that there are 4 zeros (edit: 3 zeros, not 4). Otherwise, I have no idea even where to start.
| Find the highest power of 2 in its factorization and find the highest power of 5 in its factorization. If you take the minimum of these two numbers, you will get the highest power of 10 in the factorization.
Now, to find these highest powers, as you have already mentioned, we use modulus.
$11 \equiv 1 \mod 2$ so that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/231350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 2
} |
How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$ With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:
1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$
2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$
3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq... | For the first one:
Set $\left(\sqrt{\frac{bc}{a}},\sqrt{\frac{ca}{b}},\sqrt{\frac{ab}{c}}\right)\rightarrow (x,y,z)$
Thus $xy+yz+zx=3$ and we must prove $$\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\ge \frac{3}{2}$$
It's equivalent to $\sum x^2+3\ge 3x^2y^2z^2+\sum y^2z^2$
$$\Leftrightarrow (x+y+z-xyz)(x+y+z+3xyz)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find the area of largest rectangle that can be inscribed in an ellipse The actual problem reads:
Find the area of the largest rectangle that can be inscribed in the ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
I got as far as coming up with the equation for the area to be $A=4xy$ but then when trying to fin... | Suppose that the upper righthand corner of the rectangle is at the point $\langle x,y\rangle$. Then you know that the area of the rectangle is, as you say, $4xy$, and you know that $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\;.\tag{1}$$
Thinking of the area as a function of $x$, we have $$\frac{dA}{dx}=4x\frac{dy}{dx}+4y\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/240192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 4
} |
Evaluating $\lim_{n\to\infty}\left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right)$ Evaluate
$$\lim_{n\to\infty}\left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right)$$
| $$\sum_{k=0}^{n} \dfrac{(-2)^k}{k!} = e^{-2} - \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)$$
Hence, $$1 - e^2 \left(\sum_{k=0}^{n} \dfrac{(-2)^k}{k!} \right) = e^2 \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)$$
Hence,
$$f(n) = \left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/242201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Inequality with logarithm Hi can you help me with the following
Show that for every $u \ge 0$
$$(1 + u)\log(1 + u) - u \ge \cfrac {u^2} {2 + 2u/3}$$
Obviously i tried to see something clear from Taylor series of $\log(1+x)$ but didn't see something clear.
Thank you!
| Reading the book Concentration Inequalities, I found the exact inequality in Ex. 2.8. I solved it using series expansion (of the $\log$ function) and a carefully chosen geometric series thanks to an inequality proved by induction. Both are tricky and less straightforward than the direct approach. I'm posting this fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/242303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"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.